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ADVANCED AND EXTENSION 1

Year 11 New South Wales

© 2025 Mathspace Group Holdings Ltd Copyright Notice This Work is copyright. All rights are reserved. Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. For permission to use material from this text or product, please email hello@mathspace.com.au For our full digital offering, visit mathspace.co Title: Mathspace New South Wales Curriculum - Year 11 Advanced and Extension 1: 2026 ISBN: 978-1-968666-57-6 Editors and lead authors: Erin Gallagher, Nathan Peter, Jaya Lal, Jonathan Weslake, Will Berry Writing and development team: Adam Humphreys, Neil Lopez, Shaira Llanita, Jelly Candido, Mikaela Nicolas, Valerie Baja, Rosalita Campilla, Julie Manzano, Mae Lucid, Christine Bibon, Dhave Fernandez, Grace Manzano, Dharent Fernandez, Paul Platero, Adrian Doctolero, Marielle Belen, Crystel Lontoc, Mary Tacud, Kyra Manzano, Adriane Abunda, Glady Mejias Images and design team: Scott Nolan, Chastine Marquez, Keith Gimeno, Jasper Jumawan, Mary Matheu, Jemark Orevillo, Jessa Ortega

Contents 0.A

Algebraic techniques 0.A1 0.A2 0.A3 0.A4 0.A5 0.A6 0.A7 0.A8 0.A9 0.A10 0.A11

0.B

Solve equations 0.B1 0.B2 0.B3

Index laws Negative and fractional indices Expand binomial products Factorise algebraic expressions Factorise quadratics Complete the square Algebraic fractions Further algebraic fractions Simplify surds Operations with surds Further operations with surds Chapter 0.A review

Systems of linear equations Solve quadratic equations Investigation: Derive the quadratic formula The discriminant Chapter 0.B review

Functions, relations and polynomials 1.01 1.02 1.03 1.04E 1.05E 1.06E

Functions and relations Variables and substitution Characteristics of functions Language of polynomials Behaviour and graphs of polynomials Sketch polynomials Chapter 1 review

4 15 22 28 37 47 56

Contents mathspace.co

iii

0.C

Linear functions 0.C1 0.C2 0.C3 0.C4 0.C5

Quadratic and cubic functions 2.A 2.B 2.C 2.01 2.02 2.03 2.04 2.05

62 69 75 81 88

Division of polynomials The remainder and factor theorems Further applications Chapter 3 review

92 100 109 118

Function properties and further polynomials 4.01 4.02 4.A 4.B 4.03E 4.04E

Characteristics of quadratics Completed square form Graph parabolas Equations of parabolas Solve quadratic systems Quadratic inequalities Quadratic models Cubic functions Chapter 2 review

Remainder and factor theorems 3.01E 3.02E 3.03E

Characteristics of linear graphs Equations of lines Linear inequalities Linear models Simultaneous equations Chapter O.C review

Further domain and range Even and odd functions Composite functions Piecewise functions Sums and products of zeroes Applications of sums and products of zeroes Chapter 4 review

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

120 122 130

138 151 158

Functions and relations 5.A 5.B 5.01 5.02 5.03 5.04

Direct variation models Inverse variation models Graphs of reciprocal functions Introduction to absolute value functions Absolute value functions Circles and semicircles Chapter 5 review

Inequalities 6.01E 6.02E 6.03E

160 162 169 174 182 190

194

Solve cubic inequalities Solve inequalities with variables in the denominator Solve absolute value inequalities Chapter 6 review

196 206 213 224

Trigonometry 226 7.A 7.B 7.01 7.02 7.03 7.04 7.05 7.06 7.07E

Exact trigonometric values Angles of elevation, depression and bearings Unit circle Related angles and identities Sine and cosine rules Radians Arc length and sector area Graphs of trigonometric functions Three-dimensional trigonometry Chapter 7 review

228 241 248 259 268 276 296

Contents mathspace.co

Trigonometric identities and equations

300

Secant, cosecant and cotangent Unit circle with secant, cosecant and cotangent Reciprocal and quotient identities Complementary angle identities Evaluate expressions with identities Simplify and prove identities Trigonometric equations Sum and difference expansions for trigonometric functions Double angle formulas Trigonometric equations The auxiliary angle method Apply the auxiliary angle method Applications of trigonometric equations Solve trigonometric equations graphically Chapter 8 review

302 311 321 326 334 340 349 356 366 375 382 394 405

8.01 8.02 8.03 8.04 8.05 8.06 8.07 8.08E 8.09E 8.10E 8.11E 8.12E 8.13E 8.14E

Parametric form of a function 9.01E 9.02E 9.03E

422 424 436 443 454

Probability 456 10.A 10.B 10.C 10.01 10.02 10.03 10.04 10.05

Parametric forms of equations Convert from parametric form to Cartesian form Graph linear functions, quadratic functions and circles in parametric form Chapter 9 review

416

Sets and notation Set operations and complements Venn diagrams Probability and events Mutually exclusive events Multistage events and conditional probability Conditional probability formulas Independent events Chapter 10 review

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

458 465 473 482 489 496

Permutations and combinations 11.01E 11.02E 11.03E 11.04E 11.05E 11.06E 11.07E 11.08E 11.09E 11.10E

Factorial notation Multiplication and addition principles Permutations Applications of permutations Circular arrangements Combinations Proofs involving combinations Applications of combinations Applications of combinations and permutations Probability applications Chapter 11 review

Statistics and the binomial theorem 12.01 12.02 12.03 12.04E 12.05E 12.06E 12.07E 12.08E 12.09E 12.10E 12.11E 12.12E

500 505 516 523 535 540 548 553 563 571 578

582

Random variables Organise and graph datasets Analyse data Binomial expansions Coefficients in a binomial expansion Properties of Pascal’s triangle The binomial theorem Apply the binomial theorem Specific terms in a binomial expression Simplify expressions with binomial coefficients Proofs with binomial expansions Prove further identities with binomial coefficients Chapter 12 review

Introduction to rates of change 13.01 13.02 13.03 13.04 13.05

498

584 588 597 602 608 613 619 624 628 633 640 646 653

658

Average rate of change Speed as a rate of change Instantaneous vs. average speed Instantaneous speed and tangents Linear and quadratic rates of change Chapter 13 review

660 665 671 676 683 692

Contents mathspace.co

vii

The derivative 14.01 14.02 14.03 14.04 14.05 14.06 14.07 14.08 14.09 14.10 14.11

Exponential and logarithmic functions 15.01 15.02 15.03 15.04 15.05 15.06 15.07 15.08 15.09

viii

Gradient of a curve Derivatives of basic functions First principles for derivatives Derivative notation and basic rules Tangents and normals Chain rule Extension: Proof of chain rule Product rule Extension: Proof of product rule Quotient rule Extension: Proof of quotient rule Apply differentiation rules Graphical behaviour of functions Derivatives as rates of change Chapter 14 review

Exponential graphs Tangent gradient at y-intercept Euler’s number and derivatives Investigation: Differentiation of exponential functions Applications Logarithms Exponential-logarithmic equivalence Logarithm laws and properties Logarithm expressions and equations Logarithmic graphs Investigation: Logarithms Chapter 15 review

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

696 698 710 714 721 731 744 751 756 763 770 783 792

796 798 810 814 819 825 834 841 847 854 862

Inverse functions 16.AE 16.01E 16.02E 16.03E 16.04E 16.05E 16.06E 16.07E

One-to-one functions Inverse functions Formal definition of an inverse function Determine the equation of an inverse function Graphs of functions and their inverse functions The horizontal line test Domain restrictions Solve problems involving a function and its inverse function Chapter 16 review

870 879 884 889 894 899 905 912

Transformations 916 17.01 17.02 17.03 17.A 17.04 17.05 17.06 17.07 17.08

868

Reflections in axes Horizontal and vertical translations Dilations Linear, quadratic and cubic functions Exponential and logarithmic functions Reciprocal and absolute value functions Circles and translations Order of transformations Multiple transformations Chapter 17 review

Graphical relationships 18.01E 18.02E 18.03E 18.04E 18.05E

918 924 930 939 952 960 967 978 986

992

Reciprocal functions Reciprocal trigonometric functions Absolute value functions Sum and difference of functions Graphical relationships Chapter 18 review

994 1000 1011 1025 1031 1037

Answers

1042

Contents mathspace.co

1 Functions, relations and polynomials Chapter outline 1.01 1.02 1.03 1.04E 1.05E 1.06E

Functions and relations Variables and substitution Characteristics of functions Language of polynomials Behaviour and graphs of polynomials Sketch polynomials Chapter 1 review

4 15 22 28 37 47 56

Every time you press a button on a vending machine, you’re using a function – input a code, get the same snack every time.

1.01 Functions and relations After this lesson, you will be able to… • describe a relation as an association between two sets. • represent relations using algebraic formulas, tables of values, ordered pairs, and graphs. • define a function as a special type of relation where each input has a unique output. • use the vertical line test to determine if a graph represents a function.

Relations and functions Relation An association between the elements of one set A and the elements of another set B. It may be represented as a set of ordered pairs (a, b), where a is in A, b is in B, and b is related to a. Set A collection of objects or elements, usually speciﬁed by listing its elements, e.g. {1, 2, 3, 4}; by describing it in words, e.g. ‘the set of primes’; or by using a rule such as y = 2x + 1. A collection of distinct, unordered objects, referred to as members or elements of the set. Element A member of a set. For example, 3 is a member of the set of natural numbers N = {0, 1, 2, 3, 4, …}. This relation can be written more concisely as 3 ∈ N (‘3 is an element of the set N’). Function A function f is a relation which assigns to each element of one set S precisely one element of a second set T.

A relation describes an association between two sets, typically the set of x-values (inputs) and the set of y-values (outputs). Relations can be represented by an algebraic formula, a table of values, a set of ordered pairs, or a graph. For example, consider the relation deﬁned by y = x + 1 for x ∈ {−2, −1, 0, 1, 2}.

−2

−1

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

This relation can also be written as ordered pairs: {(−2, −1), (−1, 0), (0, 1), (1, 2), (2, 3)} or graphed on a coordinate plane. Relations are classified by how inputs map to outputs: one-to-one, many-to-one, one-to-many, or many-to-many.

1 2 3

2 4 6

one-to-one

0 4 9

−3 −2 0 2 3

−1 0 1 2

1 0 4

Two or more inputs relate to one output. For example, y = x2.

one-to-many

many-to-many

One input relates to two or more outputs. An example might be the parabola in y expressed by

many-to-one

One input relates to one output. For example, the relationship y = 2x. x

Two or more inputs are related to two or more outputs. For example, 4 = x2 + y2 (circle).

A function is a relation where each x-value is associated with exactly one y-value. Only one-to-one and many-to-one relations are functions. For example, a vending machine sells juice bottles at $3 each. Let x represent the number of bottles and y the cost in dollars. x

The rule for this function is y = 3x, where each x-value corresponds to exactly one y-value, indicating a one-to-one function.

1.01 Functions and relations mathspace.co

Exploration Consider a vending machine that sells juice bottles, where the cost depends on the number of bottles purchased. Let x represent the number of bottles, and f (x) represent the cost in dollars. This table shows the cost for different numbers of bottles: x

f (x)

Explore these questions: 1. What is the rule for the function f (x) based on the table? Express it as an algebraic formula. 2. Why x must be a non-negative integer in this context? Discuss the implications if the vending machine allowed fractional bottle purchases. 3. If the vending machine introduced a discount for purchasing more than 3 bottles (e.g. each additional bottle costs $2), how would the function f (x) change? Describe the new rule.

Example 1 This table represents a relation between x and y: x

−8

−7

−6

−3

−18

−16

−15

−2

−4

−9

Determine if it represents a function.

Create a strategy A function requires each x-value to have exactly one y-value. Check the table for repeated x-values with different y-values.

Apply the idea The x-value 9 corresponds to two y-values: − 4 and 11. This one-to-many relationship indicates the relation is not a function.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary A relation associates elements of two sets and can be represented by a formula, table, ordered pairs, or graph. A function is a relation where each input has exactly one output, applicable to one-to-one and many-to-one relations.

Vertical line test Vertical line test Determines whether a relation or graph is also a function. If a vertical line intersects or touches a graph at more than one point, then the graph is not a function.

Passes the vertical line test and represents a function and a relation

Fails the vertical line test and represents a relation but not a function

1.01 Functions and relations mathspace.co

Example 2 Determine whether these graphs represent functions: a

y 4 3 2 1

−4 −3 −2 −1 −1

−2 −3 −4

Create a strategy Apply the vertical line test: a graph represents a function if no vertical line intersects it more than once.

Apply the idea A vertical line intersects the graph of y = one y-value.

at most once for all x-values, as each x-value produces y 4 3 2 1

−4 −3 −2 −1

−1

x 1

−2 −3 −4

The graph represents both a relation and a function.

Reﬂect and check All functions are relations, but not all relations are functions, as functions require exactly one output per input.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

y 4 3 2 1 −4 −3 −2 −1

x 1

−1

−2 −3 −4

Create a strategy Apply the vertical line test to check for multiple intersections.

Apply the idea A vertical line at x = 1 intersects the graph at (1, 1) and (1, −1), indicating multiple y-values for one x-value. y 4 3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

The graph represents a relation but not a function.

Reﬂect and check Restricting to y ≥ 0 would make the graph a function, as it would include only the upper branch.

1.01 Functions and relations mathspace.co

Example 3 Determine whether these equations describe relations and/or functions: a y = 9x

Create a strategy Graph the equation and apply the vertical line test to check if a vertical line intersects the graph at most once.

Apply the idea A vertical line intersects the graph of y = 9x at most once for all x-values, as each x-value produces one y-value. y 12 9 6 3 −1

−3

x 1

−6 −9

The equation describes both a relation and a function.

Reﬂect and check The linear equation y = 9x represents a one-to-one function, as each input has a unique output.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b y = x2 + 2

Create a strategy Graph the equation and apply the vertical line test to check if a vertical line intersects the graph at most once.

Apply the idea A vertical line intersects the graph of y = x2 + 2 at one point for all x-values, as each x-value produces one y-value. 11 10 9 8 7 6 5 4 3 2 1 −3 −2

−1 −1 −2

x 1

The equation describes both a relation and a function.

Reﬂect and check The quadratic equation y = x2 + 2 represents a many-to-one function, as multiple inputs may produce the same output.

Idea summary The vertical line test helps determine whether a relation is a function. If no vertical line intersects the graph at more than one point, then it passes the test, meaning each x-value corresponds to only one y-value.

1.01 Functions and relations mathspace.co

1.01 Practice squestions What do you remember? 1

Name the mapping where a single input value is mapped to multiple values of the output. Provide an example.

Determine whether these statements are true or false:

Mapping a value of x into a function gives only one value of y.

A vertical line can intersect the graph of a function at more than one point.

A relation always passes the vertical line test.

All functions are relations.

Determine whether each statement is correct: a

Some relations are functions.

All functions are relations.

No functions are relations.

The graph of every non-linear curve represents a function.

Practice Ex 1

These pairs of values in the table represent a relation between x and y: x

−9

−5

−4

−2

−9

−3

−5

−12

−14

Do they represent a function? Ex 2

Determine whether each graph represents a function, a relation, or both: a

4 2

x −6 −4 −2

−6 −4 −2

−2

−4

−6

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 3

Determine whether the following represent a function or just a relation: a

x2 + y2 = 9

y = x2 − 4

xy = 1

−4

−3

−2

−1

Plot the points on a Cartesian plane.

Do they represent a function?

A relation is deﬁned: y = 1 if x is positive, and y = −1 if x is zero or negative. a

b 9

Consider the points in the table:

a 8

y = 3x + 2

Complete the table for this relation: x

−4

−3

−2

−1

⬚

Do these values represent a function?

Consider these ordered pairs: {(−9, −5), (−5, −10), (−5, − 4), (−3, 7), (−2, − 4), (−1, 1)}

Plot the ordered pairs on a Cartesian plane.

Which ordered pair(s) would need to be removed to make the set represent a function?

CheapCalls Mobile charges $1.10 per minute plus a connection fee of 70 cents for an international call: a

Complete the table.

Is this relation a function?

Call length (minutes)

International call cost (dollars)

⬚ ⬚

⬚

5 11

A store offers one free t-shirt for every two t-shirts purchased, where each t-shirt costs $19: a

Complete the table.

Is this relation a function?

Number of t-shirts

Total cost (dollars)

⬚

2 3 4

⬚ ⬚ ⬚

1.01 Functions and relations mathspace.co

An internet provider charges $40 per month for a 20 GB plan and $10 for each additional 10 GB used: a

Complete the table.

Is this relation a function?

Total GB used

Total charge (dollars)

⬚ ⬚

⬚

50 13

Use the vertical line test to determine whether the graph represents a function: a

8 7

1 −4 −3 −2 −1 −1

x 1

4 3

−2

2 1

−3 −4

−4 −3 −2 −1

x 1

Extend your thinking 14

A vending machine uses a tiered pricing system. The total cost, y, for x bottles is deﬁned as: y = 3x,

if x ≤ 3

y = 9 + 2(x − 3),

if x > 3

A second vending machine is malfunctioning. It charges $3 per bottle for the ﬁrst three bottles, but every additional bottle after that is free:

For both machines, complete a table showing the total cost y for x = 1 to x = 6 bottles.

Use the vertical line test to determine whether the union of both machines’ graphs represents a function. Justify your reasoning.

A delivery service charges based on distance x (in km) as follows: $5 for up to 10 km, $5 + 0.5(x − 10) for 10 < x ≤ 20, and $10 + 0.75(x − 20) for x > 20. Let y be the total cost in dollars: a

Express the rule for this relation as a piecewise function.

Complete the table for this relation:

⬚

Is this relation a function? Explain using the vertical line test.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the set of ordered pairs: {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)} a

Is this set a function? Explain using the deﬁnition of a function.

Modify the set by adding or changing exactly one ordered pair to create a relation that is not a function. Write the new set and explain why it is not a function.

Modify the original set by changing exactly two ordered pairs to create a one-to-one function. Write the new set and explain why it is one-to-one.

1.02 Variables and substitution After this lesson, you will be able to… • use function notation, such as f (x), to represent a function. • identify the independent and dependent variables in a function. • substitute numerical values or algebraic expressions into a function to find and simplify the corresponding output. • interpret function notation and variables in real-world contexts.

Function notation Functions are often expressed using function notation, where f (x) denotes the y-value associated with the input x.

y = f (x) f (x)

is the output value for the input x

Functions are often expressed using function notation. Instead of writing y = 2x + 3, write f (x) = 2x + 3. Here, f (x) (read as ‘f of x’) denotes the unique output y-value associated with the input x, and is referred to as the ‘value of f at x’. For example, f (4) represents the value of the function when x = 4.

Interactive exploration Discover this concept in action online

mathspace.co

1.02 Variables and substitution mathspace.co

Example 1 An equation is given by y = 5x − 3: a Rewrite the equation using function notation, where f is the name of the function.

Create a strategy Replace y with the function notation f (x).

Apply the idea The function is f (x) = 5x − 3.

b Use function notation to represent ‘the value of y at x = 4’.

Create a strategy To ﬁnd the value of the function f at a speciﬁc point, replace the x inside the brackets with the given value.

Apply the idea The value of y at x = 4 is written as f (4).

Idea summary Function notation f (x) represents the output for an input x, used to evaluate functions at speciﬁc value of the input.

Variables in function notation Independent variable A variable used to represent values in the domain (input values) of a function. Dependent variable The variable used to represent the output values of a function. Variable (algebra) Things that are measurable or observable that are expected to either change over time or between individual observations.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A function f (x) describes a relationship where each input, the independent variable x, produces exactly one output, the dependent variable f (x). The independent variable is chosen freely, while the dependent variable’s value depends on the input and the function’s rule.

f (x) = y x

is the independent variable (input)

is the dependent variable (output)

For example, in the function f (x) = 2x + 1, x is the independent variable, and f (x) (or y) is the dependent variable.

Exploration Consider the function f (x) = x2. Discuss how changing the independent variable x affects the dependent variable f (x). For instance, what happens when x doubles or becomes negative?

Example 2 The cost, C, in dollars, to produce n t-shirts is given by the function C(n) = 12n + 50: a Identify the independent variable and determine what it represents in this context.

Create a strategy

Apply the idea

Identify the input variable of the function, which is the value that can be changed freely.

The independent variable is n. It represents the number of t-shirts produced.

b Identify the dependent variable and determine what it represents in this context.

Create a strategy

Apply the idea

Identify the output variable of the function, which value depends on the input.

The dependent variable is C (or C(n)). It represents the total cost of production in dollars.

c Explain why the cost is the dependent variable.

Create a strategy

Apply the idea

Describe the relationship between the two variables based on the context of the problem.

The cost is the dependent variable because its value depends on the number of t-shirts that are produced.

1.02 Variables and substitution mathspace.co

Idea summary In a function f (x) = x, x is the independent variable (input), and y or f (x) is the dependent variable (output), determined by the function’s rule.

Substitution into functions Substitution The process of replacing a variable in an algebraic expression, formula, equation or function consistently by a particular value, another variable, expression or function. Function notation allows eﬃcient substitution of numerical or algebraic expressions into a function’s rule to determine the corresponding output. To evaluate f (a) for a function f (x), replace x with a and simplify.

f (a) = f (x) a

with x replaced by a

is the value or expression substituted for x

Example 3 For the function f (x) = x2 − 4x + 7, evaluate: a f (3)

Create a strategy Substitute x = 3 into f (x) = x2 − 4x + 7 and simplify the expression.

Apply the idea f (x) = x2 − 4x + 7

Write the function

Substitute x = 3

f (3) = 3 − 4 (3) + 7 = 9 − 12 + 7

Evaluate the square and multiplication

Evaluate

Reﬂect and check Substituting a numerical value like 3 gives a numerical output.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b f (a + 2)

Create a strategy Substitute x = a + 2 into f (x) = x2 − 4x + 7 and simplify the expression.

Apply the idea f (x) = x2 − 4x + 7 2

f (a + 2) = (a + 2) − 4 (a + 2) + 7

Write the function Substitute x = a + 2

Expand the square and distribute

Combine like terms

= a + 4a + 4 − 4a − 8 + 7 =a +3

Reﬂect and check Substituting an expression like a + 2 results in an algebraic expression, showing the versatility of function notation.

Idea summary To evaluate f (a), substitute a (a number or expression) for x in the function’s rule and simplify. This applies to both numerical and algebraic substitutions.

1.02 Practice questions What do you remember? 1

In a function y = f (x), what is meant by the independent variable and the dependent variable?

Identify the independent and dependent variables in the function f (t) = 3t + 5.

A car travels at a constant speed. The distance d (in km) it covers is a function of time t (in hours), given by d(t) = 80t. Identify the independent and dependent variables in this context.

Practice Ex 1

The relationship between two variables is given by the equation y = x2 − 10: a

Rewrite this equation using function notation, naming the function g.

Use function notation to represent ‘the value of y at x = −3’.

1.02 Variables and substitution mathspace.co

Ex 2

Ex 3

The monthly cost, C, in dollars, of a phone plan is given by the function C(d) = 2d + 35, where d is the amount of data used in gigabytes (GB): a

Identify the independent variable and determine what it represents.

Identify the dependent variable and determine what it represents.

Explain why the total monthly cost is the dependent variable.

If f (x) = 9x2 + 7x − 4, evaluate: a

If g(x) = a

g(− 4)

g(0)

g(−1)

f (4)

f (−2)

f (m)

f (−b)

f (2)

f (−1)

f (3)

f (4)

, calculate the exact value of:

f (−3)

f (− 4)

f (a)

f (−x)

f (3)

If f (x) = 2x3 + 3x2 − 4, evaluate: f (0)

f (a + 3)

If j(x) = 3x − 3 − x, calculate each, rounded to two decimal places: j(0)

If m(x) = m(4)

j(1)

j(4)

j(7)

m(−x)

m(x + h)

R(t) = −3

g(x) = 0

, calculate: b

m(0)

If R(t) = 2t2 − 8, determine the value(s) of t for which: a

g(5)

For the function f (x) = x2 + 8x, determine an expression for:

a 15

f (2)

, evaluate:

a 14

If f (t) =

a 13

f (0.1)

a 12

f (0)

a 11

f (10)

, evaluate:

If f (x) = a

If f (x) = 4 + x3, evaluate: a

f (− 4)

R(t) = 10

R(t) = 0

R(t) = 5

If g(x) = x2 + x − 5, determine the value(s) of x for which: a

g(x) = 1

g(x) = 7

g(x) = −3

If h(t) = t2 − 8t − 7, evaluate h(5) + h(−2).

For the function f (x) = x2 − 49, evaluate f (7) − f (−7) + f (0).

A function is deﬁned as f (x) = 2 + 3x − x3. Evaluate f (−2) + f (4).

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Let the function f (x) be deﬁned by the equation x + 3y = 6, and the function g(x) be deﬁned by the equation y + 6x2 = 3 − x: a

Determine the rules for f (x) and g(x).

Calculate the value of f (3) + g(−1).

If A(x) = x2 + 1 and Q(x) = x2 + 9x, evaluate A(3) + Q(2).

A vending machine sells juice bottles at $3 each. The function C(x) represents the cost in dollars for x bottles: a

Evaluate C(5) + C(2) − C(1).

What does the value calculated in part (a) represent in this context?

Extend your thinking 23

The point (7, −6) satisﬁes the function f (x). Express this point using function notation.

A piecewise function is deﬁned by the rule:

Calculate the value of f (−2) + f (3) − f (6). 25

If f (x) = x2 + 5x, determine an expression for

If f (x) = x2 − 2x, determine

A company calculates the proﬁt, P (x), from producing x units of a product. The revenue is R(x) =

in simplest form.

+ 40x, and the cost is C(x) = 2.5x + 206.375. Proﬁt is calculated as

P (x) = R(x) − C(x). Determine the number of units such that the proﬁt is equal to the cost of producing a product. 28

The cost of a trip to two cities is calculated based on the number of days until travel, x. The cost for City A is S(x) = x2 − 200x + 10 227, and the cost for City B is U (x) = 18x + 15 423. A new total-cost function is deﬁned as: T (x) = S(2x − 3) + U (x + 5) Determine an explicit, simpliﬁed formula for T (x).

1.02 Variables and substitution mathspace.co

1.03 Characteristics of functions After this lesson, you will be able to… • define the domain and range of a function. • determine the domain of a function by identifying restrictions from denominators or even roots. • determine the range of a function from its algebraic rule or graph. • define a zero of a function and identify zeroes as the x-intercepts. • find the y-intercept of a function by evaluating f (0).

Domain and range Domain The set of allowable values of x in a function or relation. For a function or relation, it is the set of real numbers on which the function or relation is defined. Range (function) The set of values of the dependent variable for which a function is defined. The domain of a function is the set of all possible input values (x-values), and the range is the set of all possible output values ( y-values). To determine the domain, check: • Denominators must not be zero. For example, in f (x) = , x ≠ 0, so the domain is all real numbers except for 0. • Even roots require a non-negative radicand. For example, in f (x) = , x ≥ 0, so the domain is x ≥ 0. The range is determined by evaluating the function over its domain to find all possible outputs.

Example 1 Determine the domain and range for: x

Create a strategy

Apply the idea

List the unique x- and y-values.

Domain: {1, 2, 3} Range: {2, 3, 7}

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Determine the domain and range of the function f (x) = .

Create a strategy For the domain, identify values of x where the function is deﬁned, checking the denominator. For the range, evaluate the function over its domain to ﬁnd all possible outputs.

Apply the idea Determine the domain: Write the function

The denominator must not be zero

The domain is all real numbers where x ≠ 0. Determine the range: Consider the behaviour of the function as x approaches positive and negative values.

As x approaches 0 from the positive side, f (x) → + ∞

As x approaches 0 from the negative side, f (x) → − ∞

The function can take any real value except zero as x approaches zero from either side. Thus, the range is all real numbers except zero.

Reﬂect and check The function f (x) =

is undeﬁned at x = 0 due to division by zero. The range includes all real

numbers except zero because as x approaches zero, the function approaches positive or negative inﬁnity.

Idea summary The domain is the set of allowable values of x in a function or relation. The range is the set of all corresponding output values, y.

1.03 Characteristics of functions mathspace.co

Zeroes and intercepts Zero (of a function) A point in the domain of a function where the value of the function is zero. A solution of the equation f (x) = 0. Intercept The point at which a curve or function crosses an axis or other curve in a plane. The point at which a curve crosses the x-axis ( y = 0) is called the x-intercept and the point at which a curve crosses the y-axis (x = 0) is called the y-intercept. The x- and y-intercepts are sometimes taken to mean the signed distance from the point at which the curve crosses the axis to the origin, e.g. for the line y = mx + c, the y-intercept is c.

Example 3 For the function f (x) = x2 − 4, identify: a The zeroes

Create a strategy Solve x2 − 4 = 0.

Apply the idea x2 − 4 = 0

Set the function equal to zero

x =4

Add 4 to both sides

x=±2

Take the square root of both sides

The zeroes are x = −2 and x = 2.

b The y-intercept

Create a strategy Evaluate f (0).

Apply the idea f (x) = x2 – 4 2

f (0) = 0 − 4 = −4

Write the function Substitute x = 0 Evaluate

The y-intercept is − 4.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary A zero is a value where f (x) = 0, corresponding to the x-intercepts. The y-intercept is f (0).

1.03 Practice questions What do you remember? 1

Determine whether these statements are true or false: a

Every function is a relation.

The dependent variable is the output of a function.

A function’s graph always crosses both axes.

An intercept is where a function’s graph crosses an axis.

Deﬁne these terms for a function f : a

Domain

Range

Zero

y-intercept

For the relation {(1, 2), (2, 3), (3, 2), (4, 5)}, determine: a

The domain

The range

Whether it is a function

Practice Ex 1

For each relation, determine the domain, range, and identify whether it is a function or not: a

x 1 2 3 y

Ex 2

f (x) = −3x + 1

f (x) = x2 + 2

f (x) = (x − 1)2 − 2

The y-intercept

f (x) = x2 + 1

f (x) = 3

f (x) = 3x + 12

For the function f (x) = x2 − 9, identify: a

Determine the domain and range of each function: a

Ex 3

The zeroes

Find the zeroes of these functions: a

f (x) = 2x − 6

f (x) = x2 − 9

1.03 Characteristics of functions mathspace.co

For the graph of y = 2x, determine:

The y-intercept

The x-intercept

The domain

The range

4 3 2 1 −5 −4 −3 −2 −1

x 1 2 3 4 5

For the graph of y = −(x2 + 8x), determine: a

The maximum value of the range

The range

The domain

16 14 12 10 8 6 4 2 −9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4

For each graph: i

Determine whether each graph represents a function or not.

Identify its domain and range

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

For the function f (x) = a

The domain

10 8 6 4 2

x 2 4 6 8 10

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

, determine: b

The range

The x-intercept

The range

For the function y = x2 + 5, determine: a

10 8 6 4 2

The y-intercept

The domain

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x 1

Complete the table of values and determine the range for f (x) = x3 − 1: x

−2

−1

⬚

For f (x) = x + 3 with domain {−5, − 4, 0, 1}, complete the table and determine the range: x

−5

−4

f (x)

⬚

For f (x) = −2x with domain {−1, 0, 1, 2}, complete the table and determine the range: x

−1

f (x)

⬚

Describe the domain of each function in words: a

f (x) = x2 + 3x

g(x) = x2 − 4x + 5

The range

Extend your thinking 17

For the function f (x) = a

The domain

For the function a

, determine:

The domain

For the function f (x) = a

, determine:

The domain

, determine: b

The range

The zeroes

1.03 Characteristics of functions mathspace.co

1.04E Language of polynomials After this lesson, you will be able to… • define a polynomial function P ( x) and identify its key components: degree, leading term, leading coefficient, and constant term • distinguish between monic and non-monic polynomials, and recognise the definition and properties of the zero polynomial • evaluate polynomial functions for specific values • perform addition, subtraction, and multiplication operations with polynomials

Define polynomials An expression containing only one term is called a “monomial”. In the monomial Axn: • A is the coeﬃcient • x is the variable • n is the index or power Polynomial An expression made up of two or more terms with non-negative integer powers of the same variable and coeﬃcients combined using addition, subtraction and multiplication. Example: The expression x3 + 2x − 3 is a polynomial. But powers contain fractions and/or negative numbers.

is not, as its

Monic A polynomial in which the coeﬃcient of the leading term is 1. Example: P ( x) = x3 + 4x + 3 is a monic polynomial. A non-monic polynomial has a leading coeﬃcient that is not 1. For example, P ( x) = 5x4 + 4x + 3 is a non-monic polynomial. Polynomials are often expressed in function notation, such as: leading term

coefficient

quadratic term

constant term

p(x) = an xn + an − 1 xn − 1 + ... + a2 x2 + a1 x + a0 leading coefficient

degree

term

linear term

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Key polynomial terms include: Degree Highest index of the polynomial. Example: P ( x) = x3 + 4x + 7 is a polynomial of degree 3. Leading term The term that contains the degree of the polynomial. Example: P ( x) = 5x4 + 4x + 3 is a polynomial with a leading term of 5x4. Coefficient A numerical quantity which multiplies a variable in an algebraic expression. Example: For example, 5 is the coefficient of 5x. Variables with no specified coefficient have a coefficient of 1. Leading coefficient The coefficient of the term that contains the degree of the polynomial. Example: P ( x) = 5x4 + 4x + 3 is a polynomial with a leading coefficient of 5. Constant Coefficient of the term with index 0. Example: P ( x) = x3 + 4x − 9 is a polynomial with a constant of −9.

Polynomials are named by degree:

Polynomials are also named by number of terms:

Degree

Name

Example

Terms

Name

Example

Constant

Monomial

4x3

Linear

2x + 3

Binomial

x2 + 5

Quadratic

x2 − 4x + 7

Trinomial

x2 − 3x + 2

Cubic

x3 + 2x2 − x + 1

Quartic

3x4 − x3 + 5x − 2

To evaluate the polynomial P ( x) at x = a, we can use the notation P (a). For example, to evaluate the polynomial P ( x) = 3x4 + 5x2 + 7 at x = 2, we would write P (2) = 3(2)4 + 5(2)2 + 7 and then evaluate.

1.04E Language of polynomials mathspace.co

Example 1 For the polynomial

a Determine the degree.

Apply the idea The highest power of x is 7, so the degree of the polynomial is 7. b Identify the leading coeﬃcient.

Apply the idea The leading term of the polynomial is

, so the leading coeﬃcient is .

c Identify the constant term.

Apply the idea The term independent of x is 5, so the constant term is 5.

Reﬂect and check The constant term can also be viewed as the coeﬃcient of the term with index 0. The constant term could also be written as 5x0 since x0 = 1.

Example 2 Given P ( x) = 4x5 + 3x6 − 8: a Calculate P (0).

Create a strategy Substitute x = 0 into the polynomial.

Apply the idea P ( x) = 4x5 + 3x6 – 8 5

Write the polynomial 6

P (0) = 4 × 0 + 3 × 0 – 8 = −8

Substitute x = 0 Evaluate

Reﬂect and check P (0) is equal to the constant term, since the constant term is independent of x. This would also be the y-intercept of the graph of the polynomial.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Calculate P (−4).

Create a strategy Substitute x = −4 into the polynomial.

Apply the idea P ( x) = 4x5 + 3x6 – 8 5

Write the polynomial 6

P (−4) = 4 × (−4) + 3 × (−4) − 8

Substitute x = −4

= 8184

Evaluate

Idea summary Polynomial is an expression made up of non-negative integer powers of the same variable and coefficients combined using addition, subtraction and multiplication. A monic polynomial is one in which the coefficient of the leading term is 1. Polynomials are often expressed in function notation, such as: leading term

coefficient

quadratic term

constant term

p(x) = an xn + an − 1 xn − 1 + ... + a2 x2 + a1 x + a0 leading coefficient

degree

term

linear term

Key polynomial terms include: • Degree: Highest power of the polynomial. • Leading term: The term that contains the degree of the polynomial. • Coefficient: A numerical quantity which multiplies a variable in an algebraic expression. • Leading coefficient: The coefficient of the term that contains the degree of the polynomial. • Constant: Coefficient of the term with index 0.

1.04E Language of polynomials mathspace.co

Operations on polynomials Operations on polynomials include: • Addition: Combine like terms. Example: For P ( x) = x5 + 3x2 + 7 and Q( x) = 3x4 + 2x2 + 9: P ( x) + Q( x) = ( x5 + 3x2 + 7) + (3x4 + 2x2 + 9) 5

Write the sum

= x + 3x + 5x + 16

Combine like terms

• Subtraction: Subtract like terms. Example: For P ( x) = x5 + 3x2 + 7 and Q( x) = 3x4 + 2x2 + 9: P ( x) − Q( x) = ( x5 + 3x2 + 7) − (3x4 + 2x2 + 9) 5

Write the difference

= x + 3x + 7 − 3x − 2x – 9

Expand the brackets

= x − 3x + x – 2

Combine like terms

• Multiplication: Apply the distributive law. Example: For P ( x) = x5 + 3x2 + 7 and Q( x) = 3x4 + 2x2: P ( x) × Q( x) = ( x5 + 3x2 + 7) (3x4 + 2x2) 5

Write the product 2

= x (3x + 2x ) + 3x (3x + 2x ) + 7(3x + 2x ) 9

= 3x + 2x + 9x + 6x + 21x + 14x 4

= 3x + 2x + 9x + 27x + 14x

Apply the distributive law Expand the brackets Combine like terms

Adding, subtracting, or multiplying polynomials always results in a polynomial.

Example 3 Given P ( x) = −5x2 − 6x − 6 and Q( x) = −7x + 7, simplify P ( x) − Q( x).

Create a strategy Substitute the expressions for P ( x) and Q( x), and then subtract the like terms.

Apply the idea P ( x) − Q( x) = (−5x2 − 6x − 6) − (−7x + 7) 2

= −5x − 6x − 6 + 7x – 7

Substitute expressions Expand the brackets

= −5x + x – 13 Collect like terms in decreasing order of the power

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 4 Simplify: a P ( x) = ( x + 3) ( x2 − 5x + 3)

Create a strategy Use the distributive law to expand the brackets by multiplying each term in x + 3 by each term in x2 − 5x + 3, then combine like terms to simplify the resulting polynomial.

Apply the idea P ( x) = ( x + 3) ( x2 − 5x + 3) 2

Write the polynomial 2

Apply the distributive law

Expand the brackets

= x( x − 5x + 3) + 3( x − 5x + 3) 3

= ( x − 5x + 3x) + (3x − 15x + 9)

Collect like terms in decreasing order of the = x − 2x − 12x + 9 power b P ( x) = (2x − 5)2 + ( x − 4) ( x2 + 3x − 1)

Create a strategy Expand the two sets of brackets, then simplify by collecting like terms.

Apply the idea P ( x) = (2x − 5)2 + ( x − 4) ( x2 + 3x − 1) 2

Write the polynomial 2

Apply the distributive law

Expand

= (4x − 20x + 25) + ( x( x + 3x − 1) − 4( x + 3x − 1)) 2

= (4x − 20x + 25) + ( x + 3x − x) − (4x + 12x − 4) 2

= (4x − 20x + 25) + ( x − x − 13x + 4) 3

Simplify

Collect like terms in decreasing = x + 3x − 33x + 29 order of the power

Idea summary Operations on polynomials include: • Addition and subtraction which combine like terms. • Multiplication which uses the distributive law.

1.04E Language of polynomials mathspace.co

1.04E Practice questions What do you remember? 1

For the term 6x7: a

What is the coeﬃcient?

Are these polynomials? If not, explain why not. a c e

b 2x3 + 9x − 4 4

5x − 9x + 2x + 11x

d f

Give an example of a monic quadratic trinomial.

Identify the term used for:

What is the index?

A polynomial with degree 1

A polynomial with degree 3

A polynomial with 1 term

A polynomial with 2 terms

For a polynomial P ( x), what does P (3) represent?

Practice 6

Consider the polynomial P ( x) = 5: a

Ex 1

Identify the degree.

Identify the leading term.

iii

Identify the leading coeﬃcient.

Identify the constant term.

Identify whether it is a monic or non-monic polynomial.

P ( x) = 9x7 9

P ( x) = −8x7 − x

P ( x) = x + 4x − x

P ( x) = x3 − 6x2 + 8x − 5

P ( x) = 5x4 + 7x2 + 5x + 3

P ( x) = x5 − 3x4 + 6x3 − x2 + 20x − 11

Consider the polynomial P ( x) = 4x4 + 2x5 − 5, find: a

Identify the constant term.

For the polynomials:

g Ex 2

P (0)

P (1)

P (−4)

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the polynomial P ( x) = 4 − x2, find: a

P (0)

Consider the polynomial a

Consider the polynomial P ( x) = ( x4 + 3) (2 − 4x5), find: a

P (−1)

P (2)

, ﬁnd:

Simplify: a

(−3x2 − 8x − 2) + (5x2 − 5x + 5)

(−8x2 − 6) + (3x2 − 4x)

(−4x3 − 4x2 + 1) + (−x2 − 9x − 5)

6x3 − x2 + 5x − ( x3 + x + 2)

(−8x2 + 2x − 8) − (2x2 − 6x − 6)

(7x3 + 5x2 + 8x) − (6x3 + 7x2 + 7x + 9)

( x3 + 4x2) − (−2x + 9)

( x − 8) + (−3x3 + 6x2 + 3x − 1) − (−6x2 + 5x − 3)

3 (−2x + 7) − (6x2 + 8x − 5) + 7x

Let P ( x) = x2 − 6x + 6 and Q ( x) = x + 7. Find the simpliﬁed expression of P ( x) + Q ( x).

Ex 3

Let P ( x) = 2x2 + 5x − 7 and Q ( x) = x + 2. Find the simpliﬁed expression of P ( x) − Q ( x).

Ex 4a

Simplify: a

P (m) = (m − 5) (4m − 3 + m2)

P (u) = (5u + 2) (−2u − 2 + 4u)

P (c) = (5c2 − 2) (3 − 2c + 2c2)

P (v) = (−2v − 4 + 2v2) (5v + 4)

P (u) = (3u + 4) (−2u4 − 2u2 + 3u3 + 2u)

P ( x) = (2x − 3) ( x2 + 4x − 2)

g Ex 4b

P (v) = (v + 5) (5v2 − 3v − 5)

P (u) = (4u − 4u + 5) (5u − 4u − 1)

Simplify: a

P (b) = (3b − 7)2

P ( y) = (4y + 2) (y + 2)2

P ( x) = ( x − 4) ( x − 2) ( x − 2)

P ( x) = (2x3 + 3x2 − 9x + 6) ( x − 3) − ( x2 − 3x − 1) ( x − 3)

Let P ( x) = 2x3 + 4x2 − 8x + 5, Q ( x) = 9 − 8x3 and R ( x) = −2x2 + 9x − x3 + 8. Find the simpliﬁed expression: a

P ( x) − R ( x)

P ( x) − (R ( x) − Q ( x))

1.04E Language of polynomials mathspace.co

Find the polynomial that represents the perimeter of the shapes: a

b x+7

3x2 + 6x + 5 c

4x2 + x

3x3 + 4x2 + 5 3x + 4

A landscape design company makes large raised boxes for plantings. The height of the boxes they build is (5u + 3) m, and the area of the rectangular base is (u2 + 2u + 4) m2. If the box is ﬁlled with soil, what is the volume of soil needed?

Extend your thinking

Consider a rectangular cardboard measuring 9 cm by 5 cm. It is to be converted into a box with no lid by cutting out square corners measuring x cm in length and folding up the sides: Find an expression for the length of the cardboard box in terms of x.

Find an expression for the width of the cardboard box in terms of x.

Form an expression in terms of x for the volume of the cardboard box. Give your answer in expanded form.

9 cm

Fold Fold

Fold

x x

5 cm

From 2006 to 2016, the population of Australia, measured in millions, could be estimated by the polynomial 0.02x2 + 0.16x + 20.6, where x is the number of years since 2006: a

Estimate the population of Australia (in millions) in 2006.

Use the polynomial to estimate the population of Australia (in millions) in 2009.

According to the estimation given by the polynomial, how much (in millions) did the population of Australia increase by from 2006 to 2009? Round your answer to two decimal places.

Jenny’s tool manufacturing business sells its tools exclusively through two retailers. The profit generated through selling at Bargains Bonanza is modelled by the polynomial A (v) = 2.6v2 + 31.3v + 880 and the profit generated through selling at Just Stuff is modelled by B (v) = 4.4v2 − 40.3v + 720, where v is the number of tools sold. Form an expression for the polynomial, C (v), that models Jenny’s total profit.

Students were asked to identify the leading term for the polynomial P ( x) = 3x − 4x2 − x5 + 1. Holly’s answer is 3x while Nadine’s answer is −x5. Who is correct? Justify your answer.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Jessica was asked to evaluate P (−2) for P ( x) = x4 + 2x − 3. Her working out is shown. P (−2) = −24 + 2 × −2 – 3 = −16 − 4 − 3 = −23 In which step did she make an error? Explain what her mistake was and ﬁnd the correct answer for P (−2).

Consider the polynomials P ( x) = 2x3 − 4x2 + 5x − 7 and Q( x) = 3x2 − 2x + 6. Justify why P ( x) + Q( x) is a polynomial.

Explain how two trinomials can be added together to produce a binomial.

What polynomial would need to be added to x3 − 2x2 + 3x to give 5x3 − 8x2 − 5?

1.05E Behaviour and graphs of polynomials After this lesson, you will be able to… • determine the degree of the sum of two non-zero polynomials P ( x) and Q( x). • explain how the leading coefficient and the degree of a polynomial determine its end behaviour as x → ±∞. • define the zeroes of a polynomial P ( x) and the roots of the equation P ( x) = 0, and find these for factorable polynomials. • describe common polynomials (linear, quadratic, cubic, quartic) by their degree and general form.

Describe polynomials Operations like addition or subtraction yield polynomials with distinct features based on their degree. Understanding these forms aids in describing the results of those operations: • A linear polynomial is the equation of a straight line, which can be written in the form: f ( x) = mx + b • A quadratic polynomial is the equation of a parabola, which can be written in the form: f ( x) = ax2 + bx + c • A cubic polynomial, which can be written in the form: f ( x) = ax3 + bx2 + cx + d • A quartic polynomial, which can be written in the form: f ( x) = ax4 + bx3 + cx2 + dx + e

1.05E Behaviour and graphs of polynomials mathspace.co

Example 1 Describe 3x2 − 7x + 4 by its degree and number of terms.

Create a strategy

Apply the idea

Identify the degree and count the terms.

The degree is 2 (quadratic), and it has 3 terms (3x2, −7x, 4), making it a trinomial.

Example 2 Compare the polynomial P (x) = x3 + 2x − 5 with its sum when added to 4x2 − x + 3 based on degree.

Create a strategy Add the polynomials and compare the degrees of the original and the sum.

Apply the idea P ( x) = ( x3 + 2x − 5) + (4x2 − x + 3) 3

= x + 4x + x – 2

Write the sum Combine like terms

The original degree is 3 (cubic) and the degree after sum is 3 (cubic). The degree remains unchanged.

Idea summary Polynomial operations combine like terms, with absent terms as 0. Resulting polynomials are classiﬁed by degree: linear 1, quadratic 2, cubic 3, or quartic 4, each with unique shapes.

Zeroes of a polynomial Zeroes of polynomial The values of x which make P ( x) = 0. For a non-zero, factorable polynomial, zeroes can be found by factorising. For example, if P ( x) = ( x − 1) ( x − 3), the zeroes are x = 1 and x = 3. Verify that P (1) = 0 and P (3) = 0. Quadratic polynomials (ax2 + bx + c) can be solved using factorising, completing the square, or the quadratic formula. Higher-degree polynomials follow the same principle, solving P ( x) = 0.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 3 Find the zero of P ( x) = 5x − 10.

Create a strategy Set the polynomial equal to 0 and solve for x.

Apply the idea When P ( x) = 0: 5x − 10 = 0

Write the equation

5x = 10

Add 10 to both sides

x=2

Divide both sides by 5

The zero is x = 2, where the graph crosses the x-axis at (2, 0).

Example 4 Find the zeroes of Q( x) = x2 − 5x + 6.

Create a strategy

Apply the idea

Factorise the quadratic polynomial into two binomials and solve for x.

Factorise the polynomial x2 − 5x + 6 = ( x − 2) ( x − 3). When Q( x) = 0: For x − 2 = 0: x−2=0

Write the equation

x=2

Add 2 to both sides

For x − 3 = 0: x−3=0

Write the equation

x=3

Add 3 to both sides

The zeroes are x = 2 and x = 3, where the graph crosses the x-axis at (2, 0) and (3, 0).

Example 5 Find all the zeroes of P ( x) = x3 − 5x2 + 6x.

Create a strategy Factorise the polynomial by factoring out the greatest common factor, then substitute P ( x) = 0 to use null factor law.

1.05E Behaviour and graphs of polynomials mathspace.co

Apply the idea Factorise the polynomial by taking out the common factor x: P ( x) = x3 − 5x2 + 6x

Write the polynomial

= x( x − 5x + 6)

Factor out x

To factor x − 5x + 6, ﬁnd factors of 6 that sum to −5. The numbers are −2 and −3. x2 − 5x + 6 = ( x − 2) ( x − 3)

Factorise

So, P ( x) = x( x − 2) ( x − 3). Set P ( x) = 0:

x( x − 2) ( x − 3) = P ( x)

Write the polynomial

x( x − 2) ( x − 3) = 0

Substitute P ( x) = 0

0, x − 2

0, x − 3 = 0

Use null factor law

x−2=0

Write the equation

x=2

Add 2 to both sides

For x − 2 = 0:

For x − 3 = 0: x−3=0

Write the equation

x=3

Add 3 to both sides

The zeroes are x = 0, x = 2, and x = 3, where the graph crosses the x-axis at (0, 0), (2, 0), and (3, 0).

Reﬂect and check To verify these values are the correct zeroes, substitute each one into the polynomial P ( x). The result should be 0 in each case. • Check for x = 0: P ( x) = x3 − 5x2 + 6x 3

P (0) = 0 − 5(0) + 6(0) =0 • Check for x = 2: 2

P (2) = 2 − 5(2) + 6(2) = 8 − 20 + 12 =0 • Check for x = 3:

Write the polynomial Substitute x = 2 Evaluate each term Evaluate

P ( x) = x3 − 5x2 + 6x 3

Substitute x = 0 Evaluate

P ( x) = x3 − 5x2 + 6x 3

Write the polynomial

P (3) = 3 − 5(3) + 6(3)

Write the polynomial Substitute x = 3

= 27 − 45 + 18

Evaluate each term

Evaluate

All three values satisfy the condition P ( x) = 0. Since P ( x) is a cubic polynomial (degree 3), it can have at most 3 real zeroes. Therefore, we have found all the zeroes.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary • For non-zero polynomials, zeroes are found by solving P ( x) = 0 (like factorising quadratics). • The zero polynomial P ( x) = 0 has every real number as a zero and no deﬁned degree.

Behaviour of polynomials The end behaviour of a non-zero polynomial, as x → ±∞, depends on two features: • Degree (d): the highest exponent of x. • Leading coeﬃcient (ad): the coeﬃcient of xd. End behaviour rules: Degree

Leading coefficient

Even

Odd

y x

Positive

y Negative

y x

For the sum of two polynomials P ( x) and Q( x) with degrees m and n, respectively: • The degree of P ( x) + Q( x) is typically the maximum of n and m. • If leading terms cancel, the degree may be lower. This concept is useful when combining polynomials to see how the highest power might stay the same or drop if the highest terms happen to oppose each other.

Interactive exploration Discover this concept in action online

mathspace.co

Example 6 Given P ( x) = 3x4 + 2x2 − 1 and Q( x) = x − 3x4 + 5: a Find the degrees of P ( x) and Q( x).

Create a strategy Rewrite Q( x) so the powers are in descending order.

1.05E Behaviour and graphs of polynomials mathspace.co

Apply the idea Rewriting Q( x) so the powers are in descending order, results to Q( x) = −3x4 + x + 5. Both P ( x) and Q( x) have highest term x4, so their degrees are 4. b Find the degree of P ( x) + Q( x).

Create a strategy Add the polynomials by combining like terms, then identify the highest remaining exponent of x to ﬁnd the degree.

Apply the idea P ( x) + Q( x) = (3x4 + 2x2 − 1) + (−3x4 + x + 5) 2

= 2x + x + 4

Write the equation Combine like terms

The degree of 2x + x + 4 is 2.

Reﬂect and check Even though both polynomials were degree 4, the x4 terms cancelled out, giving a lower degree for P ( x) + Q( x).

Example 7 Describe the end behaviour of R( x) = −2x5 + 7x3 − x + 10 as x → ±∞.

Create a strategy Determine the degree and leading coeﬃcient, then apply the rules for end behaviour based on whether the degree is odd or even and the sign of the coeﬃcient.

Apply the idea The degree is 5 (odd) and the leading coeﬃcient is −2 (negative). For odd degree and negative ad: • As x → +∞, R( x) → −∞. • As x → −∞, R( x) → +∞.

Example 8 Use the zeroes and end behaviour of F ( x) = x4 − 1 to sketch its graph.

Create a strategy Find the real zeroes by solving F ( x) = 0, determine the end behaviour using the degree and leading coeﬃcient, then sketch the graph based on these features.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea F ( x) = x4 – 1 4

0=x –1 4

x −1=0

Factorise using the difference of squares

Factorise x2 − 1 further as ( x − 1) ( x + 1)

0, x + 1 = 0

Use null factor law

x−1=0

Write the equation

x=1

Add 1 to both sides

x+1=0

Write the equation

( x − 1) ( x + 1) ( x + 1) = 0 0, x + 1

Substitute F ( x) = 0 Swap sides

( x2 − 1) ( x2 + 1) = 0 x−1

Write the polynomial

For x − 1 = 0:

For x + 1 = 0: x = −1

Subtract 1 from both sides

For x2 + 1 = 0: x2 + 1 = 0

Write the equation

x2 = −1

Subtract 1 from both sides

No real solutions for x2 + 1 = 0. This means the only real zeroes are x = 1, −1. y

Determining the end behaviour, the degree is 4 (even) and the leading coeﬃcient is 1 (positive), so as x → −∞, F ( x) → ∞ and as x → ∞, F ( x) → ∞.

For the y-intercept, substitute x = 0 into F ( x): F ( x) = x4 – 1 4

F (0) = 0 – 1 = −1

Write the polynomial Substitute x = 0 Evaluate

The graph: • Has a y-intercept at (0, −1). • Crosses the x-axis at (−1, 0) and (1, 0). • Rises on both ends, forming a wider U-shape than a normal x2 parabola.

4 3 2 1 −2

−1

x 1

−1 −2

2 4

F(x) = x − 1

1.05E Behaviour and graphs of polynomials mathspace.co

Example 9 A cubic polynomial has zeroes at x = −2, x = 1, and x = 3. Find the equation of the polynomial if it passes through (4, −36).

Create a strategy Form the polynomial using the zeroes as factors, determine the leading coeﬃcient with the given point, and write the ﬁnal equation.

Apply the idea Since the polynomial has zeroes at x = −2, x = 1, and x = 3, its factors are x + 2, x − 1, and x − 3. P ( x) = a( x + 2) ( x − 1) ( x − 3)

Write the polynomial with factors from zeroes

P (4) = a(4 + 2) (4 − 1) (4 − 3)

Substitute x = 4

P (4) = a × 6 × 3 × 1

Evaluate each expression inside the brackets

P (4) = 18a

Simplify

−36 = 18a

Substitute P (4) = −36

−2 = a

Divide both sides by 18

a = −2

Make a the subject

Substitute a = −2 into the polynomial: P ( x) = a( x + 2) ( x − 1) ( x − 3)

Write the polynomial with factors from zeroes

P ( x) = −2( x + 2) ( x − 1) ( x − 3)

Substitute a = −2

Idea summary The degree and leading coeﬃcient determine a polynomial’s end behaviour as x → ±∞: • Even degree (d = 2, 4, 6, …): • If ad > 0, the polynomial rises to +∞ as x → ±∞. • If ad < 0, the polynomial falls to −∞ as x → ±∞. • Odd degree (d = 1, 3, 5, …): • If ad > 0, the polynomial falls to −∞ as x → −∞ and rises to +∞ as x → +∞. • If ad < 0, the polynomial rises to +∞ as x → −∞ and falls to −∞ as x → +∞. For P ( x) + Q( x), the degree is usually the maximum of their degrees, unless leading terms cancel.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

1.05E Practice questions What do you remember? 1

Identify the zeroes of each function: a

P ( x) = 3( x−3)

Q( x) = ( x − 1) ( x + 1)

For each polynomial description, decide whether y → +∞ or y → −∞ as x → +∞, and as x → −∞: a

A polynomial of degree 4 with negative leading coeﬃcient.

A polynomial of degree 4 with positive leading coeﬃcient.

A polynomial of degree 3 with negative leading coeﬃcient.

A polynomial of degree 5 with positive leading coeﬃcient.

Answer these questions about polynomial characteristics: a

What is the degree of 2x3 − 5x + 1?

How many terms does x2 + 3x − 4 have?

What type of polynomial is 4x − 7?

What is the leading coeﬃcient of −3x4 + x2 − 2?

Practice Ex 1

Describe each polynomial by its degree and number of terms: a c

Ex 2

Ex 3

Ex 4

5x3 + 2x2 − 3x + 1 4

−2x + x − 7

4x2 − 9

x+5

Compare the polynomial Q( x) = 2x4 − 3x2 + 5 with its sum when added to the given polynomial based on degree: a

3x3 + x2 − 2x + 1

−2x4 + 4x3 − x

x2 + 2x − 3

5x4 − 3x2 + 2

Determine the zero of each linear polynomial: a

P ( x) = 3x − 6

Q( x) = 2x + 8

R( x) = −4x + 12

S( x) = 5x − 15

Determine the zeroes of each factorised polynomial: a

T1 ( x) = ( x−2) ( x + 5) ( x−1)

T2 ( x) = 2( x + 3) ( x−4) ( x−7)

T3 ( x) = (3x + 1) ( x−5) ( x + 2) ( x−6)

T4 ( x) = ( x−1) ( x−1) ( x + 4)

T5 ( x) = (2x−3) ( x + 5) ( x−2) ( x−2) ( x + 1)

Determine the zeroes of these polynomials: a

P ( x) = 2x2 + 4x − 6

Q( x) = x2 − 5x + 6

R( x) = 3x2 − 12

S( x) = x2 − 2x − 8

1.05E Behaviour and graphs of polynomials mathspace.co

Ex 5

Ex 6

Determine the zeroes of these cubic polynomials: a

P ( x) = x3 + 2x2 − 8x

Q( x) = x3 − 7x2 + 12x

S( x) = x3 − 3x2 − 4x

R( x) = 2x − 6x − 8x

Consider these polynomials: • A1 ( x) = x3 + 2 • B1 ( x) = −x3 + 1 a

Ex 8

Ex 9

A3( x)

iii

B1( x)

B2( x)

A1( x) + A2( x)

A1( x) − B1( x)

iii

A2( x) − A3( x)

B2( x) + B3( x)

p1( x) = 4x3 − 2x + 1

p3( x) = x5 − x4

p2( x) = −3x2 + 10x

p4( x) = −2x4 + x2 − 1

For each polynomial: i

Determine the zero(es).

Determine the end behaviour.

iii

Sketch the graph of the polynomial.

m( x) = ( x + 4) ( x − 3)

m( x) = ( x − 3)2

m( x) = ( x + 2)3

m( x) = 2( x + 1)4

For the polynomial f ( x) = x4 − 5x2 + 4: a

Determine the y-intercept.

Determine the zero(es).

Identify the end behaviour.

Sketch the graph of the polynomial.

Let K( x) = ( x + 3) ( x + 1) (2x−1): a

How many real zeroes does K( x) have?

What is its degree?

What is the behaviour as x → ±∞?

Suppose a polynomial Q( x) has the form Q( x) = a ( x−2)2( x + 4) and Q(0) = −32: a

For each polynomial, describe the end behaviour as x → ±∞: a

A1( x)

Calculate each polynomial to determine the degrees: i

Ex 7

• A3 ( x) = 2x2 − x + 1 • B3 ( x) = 2x2 + x − 1

Determine the degrees of: i

• A2 ( x) = 5x4 − x2 • B2 ( x) = 5x4 + 3

Determine the value of a.

List the real zeroes of Q( x).

A polynomial f ( x) is deﬁned by f ( x) = (1 − x) ( x−4) (2x + 1): a

How many x-intercepts does the graph have?

Does the graph open “up” or “down” as x → ±∞?

Calculate: a

A quadratic has zeroes at x = 1, x = 4, and passes through (2, 6). Determine its equation in factored form.

A quadratic has zeroes at x = −2, x = 5, and passes through (0, 10). Determine its equation.

A cubic has zeroes at x = −1, x = 2, x = 3 and passes through (4, 10). Determine its equation.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking 18

Explain why every real number is a zero of R( x) = 0.

Let A( x) and B( x) be two degree-5 polynomials. If the degree of (A + B) = 3, determine the degree of (A − B)?

Analyse these polynomial scenarios: a

A polynomial P ( x) = 2x3 − x2 + 3x − 5 is added to another polynomial Q( x) of degree 2. What is the degree of the resulting polynomial?

If P ( x) = x4 − 2x2 + 1 and Q( x) = −x4 + 3x3 − x are added, explain why the resulting degree is not 4.

Can a polynomial with degree 3 be a binomial?

A polynomial Q( x) of even degree is negative as x → +∞ but satisﬁes Q(0) = 5:

Describe the leading coeﬃcient.

Must it have at least one real zero? Explain your thinking.

A polynomial H( x) is deﬁned by H( x) = 2( x − 1)2( x + 3): a

Determine the degree and classify the polynomial.

Determine the real zeroes.

Describe the end behaviour as x → ±∞.

1.06E Sketch polynomials After this lesson, you will be able to… • determine the degree of the sum of two non-zero polynomials P ( x) and Q( x). • explain how the leading coefficient and the degree of a polynomial determine its end behaviour as x → ±∞. • define the zeroes of a polynomial P ( x) and the roots of the equation P ( x) = 0, and find these for factorable polynomials. • describe common polynomials (linear, quadratic, cubic, quartic) by their degree and general form.

Sketch polynomials When sketching polynomials, consider these features: • The leading term, which determines end behaviour. • The zeroes ( x-intercepts). • The multiplicity of each zero, affecting the graph’s shape. • The y-intercept.

1.06E Sketch polynomials mathspace.co

Interactive exploration Discover this concept in action online

Leading coefficient

Degree

mathspace.co

Even

Odd

y x

Positive

y x

Negative

Multiplicity (of a root/zero) The number of times a particular number is a zero for a given polynomial or a root for a polynomial equation.

Zeroes of multiplicity 1: The graph passes through the x-axis.

Zeroes of multiplicity 2: The graph has a turning point at the x-axis.

Zeroes of multiplicity 3: The graph has a point of inﬂection at the x-axis.

Horizontal point of inﬂection

A point at which the concavity of the graph changes (similar to y = x3) x

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

To sketch a polynomial function, follow these steps: 1. Identify the leading term to determine the end behaviour as x → ∞ or x → −∞. 2. Determine the y-intercept by evaluating the polynomial at x = 0. 3. Determine the x-intercepts by setting P ( x) = 0 and solving for the zeroes. 4. Determine the multiplicity of each zero to understand the graph’s behaviour at each intercept. 5. Draw a smooth curve that matches the end behaviour, intercepts, and multiplicity effects. For the polynomial P ( x) = ( x + 3) ( x + 1)2( x − 2)3: • The degree is 1 + 2 + 3 = 6 (even) and the leading coefficient is positive. Therefore, P ( x) → ∞ as > x → ∞ and P ( x) → ∞ as → −∞. • The y-intercept is at y = −24. • Zero at −3 (multiplicity 1). • Zero at −1 (multiplicity 2). • Zero at 2 (multiplicity 3). The polynomial can then be sketched by connecting all these key features to form the curve: Zero at −1 (multiplicity 2)

y −3

−1 2

P(x) → ∞ as x → ∞

Zero at −3 (multiplicity 1) −24

P(x) → −∞ as x → ∞

Zero at 2 (multiplicity 3) y-intercept is at y = −24

Example 1 For the function y = ( x − 3) ( x − 2) ( x + 2): a Determine the y-intercept.

Create a strategy Substitute x = 0 into the function.

Apply the idea y = ( x − 3) ( x − 2) ( x + 2)

Write the equation

= (0 − 3) (0 − 2) (0 + 2)

Substitute x = 0

= (−3) × (−2) × (2)

Evaluate each expression inside the brackets

= 12

Evaluate

1.06E Sketch polynomials mathspace.co

b Determine the x-intercepts.

Create a strategy Substitute y = 0 into the function, then use null factor law to solve.

Apply the idea y = ( x − 3) ( x − 2) ( x + 2)

Write the equation

0 = ( x − 3) ( x − 2) ( x + 2)

Substitute y = 0

For x − 3 = 0: x−3=0

Use null factor law

x=3

Add 3 to both sides

0=0

Use null factor law

x=2

Add 2 to both sides

0=0

Use null factor law

x = −2

Subtract 2 from both sides

For x − 2 = 0: x−2 For x + 2 = 0: x+2

Reﬂect and check Each factor has a power of 1, so each zero has multiplicity 1 and passes through the x-axis. c Plot the graph of the curve.

Create a strategy The leading term has a positive coeﬃcient and an odd degree, so the graph rises as x → ∞ and falls as x → −∞. The y-intercept is y = 12. The x-intercepts are x = 3, 2, −2, each with multiplicity 1, so the graph passes through the x-axis at these points.

Apply the idea 12

10 8 6 4 2 −2

−1

−2 −4

x 1

y = (x − 3) (x−2) (x+2)

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Consider y = 2(1 − x) ( x + 2)2. a Determine the behaviour as x approaches inﬁnity.

Create a strategy Identify the leading term by multiplying the coeﬃcients and highest powers of x in each factor.

Apply the idea Leading term = 2 × (−x) × x2 3

= −2x

Multiply the coeﬃcients and highest powers Evaluate

The leading term has a negative coeﬃcient and an odd degree. b Determine the zeroes and identify their multiplicity.

Create a strategy The multiplicity of each zero is the exponent of its factor in the factored form.

Apply the idea Zero at x = 1 with multiplicity 1 and x = −2 with multiplicity 2. c Determine the y-intercept.

Create a strategy Evaluate the polynomial at x = 0.

Apply the idea y = 2(1 − x) ( x + 2)2 2

Write the equation

= 2(1 − 0) (0 + 2)

Substitute x = 0

=2×1×4

Evaluate each expression inside the brackets

Evaluate

d Sketch the graph.

Create a strategy The leading term has a negative coeﬃcient and an odd degree, so the graph falls as x → ∞ and rises as x → −∞. Zeroes at x = 1 (multiplicity 1, passes through) and x = −2 (multiplicity 2, turning point). The y-intercept is 8.

1.06E Sketch polynomials mathspace.co

Apply the idea y 10 (0, 8) 8 6 4 2

(−2, 0) −3

−2

(1, 0) x

−1

−2 −4 −6 −8 y = 2(1 − x) (x + 2)2 −10

Idea summary A polynomial P ( x) = ( x − α)Q( x) has a single zero at α if ( x − α) is not a factor of Q( x), and a repeated zero if it is. A zero α has multiplicity m if P ( x) = ( x − α)mQ( x) and Q( α) ≠ 0. To sketch a polynomial function: 1. Identify the leading coeﬃcient to determine end behaviour. 2. Determine the y-intercept by evaluating at x = 0. 3. Determine the x-intercepts by setting P ( x) = 0 and solving for zeroes. 4. Determine the multiplicity of each zero, which is the exponent of its factor. 5. Draw a smooth curve matching the end behaviour, intercepts, and multiplicity effects. The multiplicity of a zero affects the graph: • Multiplicity 1: The graph passes through the x-axis. • Multiplicity 2: The graph has a turning point at the x-axis. • Multiplicity 3: The graph has a point of inﬂection at the x-axis.

1.06E Practice questions What do you remember? 1

For each polynomial, identify whether the given zero is single or repeated. a

Polynomial P ( x) = ( x − 2) ( x + 1)2, zero x = 2.

Polynomial P ( x) = ( x − 3)2( x + 4), zero x = 3.

Identify the multiplicity of each root for the given polynomials. a

P ( x) = ( x + 5) ( x − 1) ( x − 3)

P ( x) = 4( x − 2)2( x + 1)3

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For each polynomial: i

Identify the degree of the polynomial.

Identify the leading coeﬃcient of the polynomial.

iii

Determine the end behaviour of f ( x) as x → ∞.

Determine the end behaviour of f ( x) as x → −∞.

f ( x) = 2x3 − 6x2 + 3x + 10

f ( x) = −x2 + 5x − 6

f ( x) = ( x + 3) (3x + 4) (2x − 5)

f ( x) = ( x + 3) ( x − 2) (5 − 3x)

f ( x) = x4 − 4x2

f ( x) = −x5 + x3

Practice 4

Match these functions to their respective graphs: i

f ( x) = 4x3 + 7x − 10

f ( x) = −5x8 − 2x7 + 4x3 + 7

iii

f ( x) = −3x5 + 6

f ( x) = 6x2 − 4x − 7

B 8

−8 −6 −4 −2 −2

2 4 6 8

−4

−6

−8

D 8

x 2 4 6 8

2 4 6 8

−8 −6 −4 −2 −2

−4

−6

−8

x 2 4 6 8

Determine the end behaviour as x → ∞ and as x → −∞ of a polynomial function with: a

A positive leading coeﬃcient and an odd degree.

A leading coeﬃcient of 2 and all real zeroes with even multiplicity.

A leading coeﬃcient of −1 and exactly one real zero with odd multiplicity.

All real zeroes with even multiplicity and a negative y-intercept.

A negative leading coeﬃcient of 2 and an odd number of real zeroes with odd multiplicity.

A positive leading coeﬃcient of 2 and no real zeroes. 1.06E Sketch polynomials mathspace.co

Ex 1

Ex 2

For the polynomial y = ( x + 4) ( x − 1) ( x − 5): a

Determine the y-intercept.

Determine the x-intercepts and their multiplicities.

Sketch the graph with labelled intercepts.

For the polynomial y = −3( x + 2)2( x − 4): a

Determine the end behaviour as x approaches positive and negative inﬁnity.

Determine the zeroes and their multiplicities.

Determine the y-intercept.

Sketch the graph with labelled intercepts.

For the polynomial y = 2( x − 3)3( x + 1): a

Determine the zeroes and their multiplicities.

Sketch the graph with labelled intercepts.

For the polynomial y = x2( x − 4): a

Determine the zeroes and their multiplicities.

Sketch the graph with labelled intercepts.

For each polynomial: i

Express in factored form.

Identify the zeroes and their multiplicities.

iii

Sketch the graph with labelled intercepts.

y = x3 − 3x2 + 2x

Determine whether each polynomial could describe the graph, assuming k is a non-zero real number. i

y = x4 − 4x2

A polynomial graph has intercepts shown: a

y = k( x − 1)2( x + 3) 2

y = k( x + 1) ( x − 3)

iii

y = k( x − 1)3( x + 3)

(−3, 0)

4 3 2 1

−4 −3 −2

−1 −1

Calculate the value of k.

Express the polynomial in factored form.

Given the y-intercept is −20, sketch the graph with labelled intercepts.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x 2

−2 (0, −1.5) −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 −14

A quartic polynomial P ( x) has a zero at x = 2 with multiplicity 2, a zero at x = −1 with multiplicity 1, and a zero at x = 5 with multiplicity 1. a

(1, 0)

Extend your thinking 13

Explain the conditions under which a polynomial P ( x) = ( x − α)Q( x) has a single or repeated zero at x = α, and deﬁne the multiplicity of a zero.

A polynomial has a graph with intercepts shown:

Determine whether the statement is true:

The polynomial can be described by y = k( x + 6)2( x + 2) ( x − 4) for some non-zero real number k. Justify the response using the graph’s intercepts and multiplicity behaviour.

56 48 40 32 24

(0, 24)

16 8

(−6, 0) −6

(4, 0) x

(−2, 0) −4

−2

−8

Explain why a polynomial of odd degree with real coeﬃcients always has at least one real zero.

Did you know?

Polynomial functions are used in bike navigation systems to help calculate smooth and efficient routes! Whether you’re cycling through city streets or winding trails, the curved graphs help ensure safe turns, gradual climbs, and steady descents. By interpreting the shape and direction of these graphs, navigation apps can avoid steep hills, sharp corners, and help cyclists optimise their journey. Polynomials shape the paths we ride every day—making travel smarter, safer, and more enjoyable!

1.06E Sketch polynomials mathspace.co

1 Chapter review 1

Consider the graph of a curve shown. Which statement best describes the graph? A

It represents a function only.

It represents a relation only.

It represents both a function and a relation.

It represents neither a function nor a relation.

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

−1

x=2

For the function f (x) = x2 − 16, which is a zero of the function? A

If f (x) = 3x − 5, select the value of f (4): A

5 4 3 2 1

x=0

x = 16

x=4

These pairs of values in the table represent a relation between x and y: x

−8

−6

−3

−1

−7

−1

−3

−10

−12

y2 = x + 1

Do they represent a function? 5

Determine whether these represent a function or just a relation: a

y = 5x − 3

x2 + y2 = 25

y = x2 + 7

Consider these ordered pairs: {(−8, −6), (−6, −11), (−6, −5), (− 4, 8), (−3, −5), (−2, 2)}

Plot the ordered pairs on a Cartesian plane.

Which ordered pair(s) would need to be removed to make the set represent a function?

Consider the set of ordered pairs: {(−1, 3), (0, 4), (0, 6), (1, 5), (2, 5)}

Is this set a function? Explain.

Modify the set by removing or changing exactly two ordered pairs to make it a one-to-one function. Write the new set.

If h(x) = 2x2 + 5x − 7, evaluate: a

h(−3)

h(5)

h(0.2)

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

h(x + a)

If f (x) =

, evaluate:

f (3)

a 10

If f (x) = a

f (−2)

f (7) − f (−1)

f (0) × f (−1)

⬚

Is this relation a function? Explain.

If f (x) = x2 − 4x, determine

Determine the domain and range for each function: a

g(x) = x2 − 4

Complete the table. Viewing hours (H)

Loyalty points (P )

⬚ ⬚

⬚

20 b

Is this relation a function? Explain.

For each relation, determine the domain, range, and whether it is a function: a

x 2 4 6 y

For the function f (x) = x2 − 25, identify: a

The zeroes

The y-intercept

f (x) = x2 + 4

Find the zeroes of these functions: a

k(x) = x3

An online streaming service converts viewing hours (H) to loyalty points (P ) using the formula P = 2.5H + 10: a

f (x − h) + f (h)

Complete the table for this relation:

If f (x) = x2 + 3x, determine f (b + k).

, calculate the exact value of:

f (0) + f (7)

f (0)

A coffee shop sells muffins at $4 each for up to 2 muffins, with additional muffins costing $3 each. Let x be the number of muffins and y the total cost in dollars. a

f (x) = 4x − 12

f (x) = x2 − 49

f (x) = 2x + 10

A rectangular garden bed has a ﬁxed width of w metres, where w > 0. The length is l metres, l > 0. The perimeter is given by P (l) = 2l + 2w. If the width w = 3, determine: a

Whether l = −1 is in the domain

The range of the function if l > 0 Chapter 1 review mathspace.co

20E Consider the polynomial P ( x) = 7 − 2x3 + 5x4 − x. Which of these statements is correct? A

The degree of P ( x) is 3.

The leading coeﬃcient of P ( x) is −2.

The constant term of P ( x) is 7.

P ( x) is a monic polynomial.

21E Let P ( x) = 3x4 − 2x + 1 and Q( x) = −3x4 + 5x3 − 7. What is the degree of the polynomial R( x) = P ( x) + Q( x)? A

22E The polynomial P ( x) = x( x − 2)3( x + 5)2 has a zero at x = 2. What is the behaviour of the graph of P ( x) at this zero? A

The graph crosses the x-axis like a straight line.

The graph touches the x-axis and turns around.

The graph crosses the x-axis with a point of horizontal inﬂection.

The graph has a vertical asymptote at x = 2.

23E Given the polynomial P ( x) = 4x − 9x3 + x5 − 11: a

Write the polynomial in descending order of powers of x.

State the degree of P ( x).

Identify the leading term of P ( x).

What is the constant term of P ( x)?

Is P ( x) a monic polynomial? Justify your answer.

24E If P ( x) = 2x3 − 5x2 + x − 6, determine: a

P (2)

P (−1)

P (0)

25E Let A( x) = 5x3 − 2x2 + 7x − 4 and B( x) = 3x3 + 6x2 − x + 9. Calculate, in the simplest form: a

A( x) + B( x)

A( x) − B( x)

2A( x) − 3B( x)

26E For each polynomial, describe the end behaviour as x → ∞ and as x → −∞: a

P ( x) = −3x5 + 2x2 − 10

Q( x) = 7x4 − x3 + 5x + 2

R( x) = (2 − x) ( x + 3) ( x − 1)

27E For the polynomial P ( x) = −2x( x + 3)2( x − 1)3: a

Identify the degree of the polynomial.

List each zero and its multiplicity.

Determine the y-intercept.

Sketch the graph of the polynomial.

28E A rectangular box has dimensions that are polynomials in x: length L( x) = x + 3, width W ( x) = x − 1, and height H( x) = 2x + 1.

Determine an expression for the volume V ( x) = L( x) × W ( x) × H( x) in expanded form.

What is the degree of V ( x)?

If the area of the base of the box is A( x) = L( x) × W ( x), determine A( x) in expanded form.

A smaller box has volume S( x) = x3 − 2x + 5. If the larger box V ( x) from part (a) is placed inside a much larger container, and the smaller box S( x) is also placed in the same container, what is the polynomial representing their combined volume?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

29E The graph of a polynomial P ( x) has x-intercepts at x = −2, x = 1, x = 3 and a y-intercept at y = 6: a

What is the minimum possible degree of P ( x)? Justify your answer.

Determine the equation of P ( x).

Is the leading coeﬃcient of P ( x) positive or negative? Explain.

Sketch the graph of y = P ( x), showing all intercepts.

Based on the graph, write down the real zeroes of P ( x).

30E A polynomial P ( x) has degree 4. It has a zero of multiplicity 2 at x = 1, and a zero of multiplicity 2 at x = −2. The graph of y = P ( x) passes through the point (0, −4).

31E

Write down the general form for P ( x) involving a constant a.

Determine the value of a.

Write the equation for P ( x).

Sketch the graph of y = P ( x), showing all intercepts and the correct behaviour at each zero.

Let P ( x) be a polynomial of degree m and Q( x) be a polynomial of degree n: a

What is the degree of the product P ( x)Q( x)?

If m = n = 5, and the leading coeﬃcient of P ( x) is a5 and the leading coeﬃcient of Q( x) is b5: i

Under what condition(s) would the degree of P ( x) + Q( x) be less than 5?

If a5 = 2 and b5 = −2, what is the maximum possible degree of P ( x) + Q( x)?

iii

If a5 = 2 and b5 = 3, what is the degree of P ( x) + Q( x)?

32E The graph of a polynomial y = P ( x) is shown. Determine the equation of the polynomial. 5

4 3 2 1 −2

−1

(2, 0) 1

−2 −3

Chapter 1 review mathspace.co

2 Quadratic and cubic functions Chapter outline 2.A 2.B 2.C 2.01 2.02 2.03 2.04 2.05

Characteristics of quadratics Completed square form Graph parabolas Equations of parabolas Solve quadratic systems Quadratic inequalities Quadratic models Cubic functions Chapter 2 review

62 69 75 81 88

Quadratic models help optimise sports — from golf swings to soccer kicks, the curve predicts the perfect shot.

2.01 Equations of parabolas After this lesson, you will be able to… • find the equation of a parabola in vertex form given its vertex and another point. • find the equation of a parabola in factored form given its x-intercepts and another point. • find the equation of a parabola in standard form given three points. • convert between vertex, factored, and standard forms of a quadratic equation. • solve problems by equating the coefficients of two equal quadratic functions.

Solve parabola equations from graphical features A quadratic function can be expressed as f (x) = ax2 + bx + c (standard form), f (x) = a(x − h)2 + k (vertex form), or f (x) = a(x − x1)(x − x2) (factorised form). Graphical features like the vertex, x-intercept, y-intercept, or other points help determine the equation. • Vertex form: Use vertex (h, k) and another point to ﬁnd a. • Factorised form: Use x-intercepts x1, x2 and another point to ﬁnd a. • Standard form: Use three points to solve for a, b, c.

Interactive exploration Discover this concept in action online

Example 1 A parabola has vertex (2, −1) and passes through (0, 3): a Find the equation in vertex form.

Create a strategy Use f (x) = a(x − h)2 + k with vertex (h, k) and substitute (0, 3) to ﬁnd a.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

mathspace.co

Apply the idea Vertex (2, −1) gives h = 2, k = −1. Substitute the coordinates of the point (0, 3) into the equation by setting x = 0 and f (x) = 3. f (x) = a(x − h)2 + k 2

Write the vertex form

3 = a(0 − 2) + (−1)

Substitute f (x) = 3, x = 0, h = 2 and k = −1

3 = 4a − 1

Evaluate each term

4 = 4a

Add 1 to both sides

a=1

Divide both sides by 4

Substituting the values, the equation in vertex form is: f (x) = a(x − h)2 + k

Write the vertex form

Substitute a = 1, h = 2 and k = −1

Simplify

f (x) = 1(x − 2) – 1 f (x) = (x − 2) − 1

b Determine the domain and range.

Create a strategy Use the fact that the domain is . For the range, use the vertex and concavity.

Apply the idea The domain is . Since a = 1 > 0, the parabola is concave up, with a minimum at y = −1. Thus, the range is [−1, ∞).

c Find the x-intercepts.

Create a strategy Set the function f (x) = 0 and solve for the values of x using the vertex form.

Apply the idea Set the equation to zero

Add 1 to both sides

Take the square root of both sides

Evaluate the square root

Solve for the two possible values of x: x

2+1=3

First solution

2−1=1

Second solution

The x-intercepts are at (1, 0) and (3, 0).

2.01 Equations of parabolas mathspace.co

d Sketch the graph, showing key features.

Create a strategy Plot the vertex, y-intercept, x-intercept, and axis of symmetry. Draw a concave up parabola.

Apply the idea The axis of symmetry is x = 2. The parabola is concave up (a = 1 > 0). y 3 (0, 3)

x=2

2 1

x 1 −1

Plot the vertex (2, −1) and y-intercept. Plot the x-intercepts at (1,0) and (3,0). Draw the axis of symmetry at x = 2. Sketch a concave up parabola.

(2, −1)

Idea summary Use the vertex, x-intercepts, or other points to ﬁnd a parabola’s equation in vertex, factorised, or standard form. The domain is , and the range depends on concavity and vertex.

Equate quadratic coefficients Two quadratic functions are equal for all x if and only if their corresponding coeﬃcients are equal. For a1 x2 + b1 x + c1 = a2 x2 + b2 x + c2, set a1 = a2, b1 = b2, c1 = c2. This method is useful when a parabola’s equation is given in one form (e.g., vertex) and needs to be expressed in another (e.g., standard), or when solving for parameters.

Interactive exploration Discover this concept in action online

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

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Example 2 A parabola has x-intercepts (−1, 0) and (3, 0), and vertex (1, −4): a Find the equation in standard form.

Create a strategy Substitute the x-intercepts and vertex into the factor form f (x) = a(x − x1)(x − x2) to ﬁnd a, then expand to standard form. (Alternatively, vertex form could be used, since the vertex is provided).

Apply the idea Using the x-intercepts, set x1 = −1 and x2 = 3. Then the vertex as x = 1 and y = −4, and since y = f (x), substitute f (x) = −4. f (x) = a(x − x1)(x − x2)

Write the factor formula

−4 = a(1 − (−1))(1 − 3)

Substitute f (x) = −4, x = 1, x1 = −1 and x2 = 3

−4 = (2a)(−2)

Evaluate each term

−4 = −4a

Evaluate the multiplication

a=1

Divide both sides by −4

Substituting the values, the equation in standard form is: f (x) = a(x − x1)(x − x2)

Write the factor formula

f (x) = 1(x − (−1))(x − 3)

Substitute a = 1, x1 = −1 and x2 = 3

f (x) = (x + 1)(x − 3)

Evaluate each term

Expand

Combine like terms to write in f (x) = ax2 + bx + c

f (x) = x − 3x + x − 3 f (x) = x − 2x − 3

b Determine the domain and range.

Create a strategy Use the fact that the domain is . For the range, use the vertex and concavity.

Apply the idea The domain is . Since a = 1 > 0, the parabola is concave up, with a minimum at y = −4. Thus, the range is [−4, ∞).

c Find the values of a, b, and c such that f (x) = ax(x + b) + c(x + b) is equivalent to the parabola.

Create a strategy Expand the expression ax(x + b) + c(x + b), collect like terms, and equate the coeﬃcients with the standard form f (x) = x2 − 2x − 3 to solve for a, b, and c.

2.01 Equations of parabolas mathspace.co

Apply the idea First, expand the given expression and write it in standard form: f (x) = ax(x + b) + c(x + b)

Write the given function

Expand the brackets

Group like terms

= ax + abx + cx + bc = ax + (ab + c)x + bc

From part (a), the standard form of the parabola is f (x) = x2 − 2x − 3. Equate the coeﬃcients of the two forms. ax2 + (ab + c)x + bc = 1x2 − 2x − 3 Comparing coeﬃcients for each power of x: • x2 term: a = 1 • x term: ab + c = −2 • Constant term: bc = −3 Substitute a = 1 into the second equation: ab + c = −2

Write the equation

(1)b + c = −2

Substitute a = 1

b + c = −2

Simplify

Now solve the system of simultaneous equations: 1

b + c = −2

bc = −3

From 1 , express c in terms of b: c = −2 − b. Substitute this into 2 : bc = −3 b(−2 − b) = −3 2

−2b − b = −3 2

Write equation 2 Substitute c = −2 − b Expand

b + 2b − 3 = 0

Rearrange into a quadratic equation

(b + 3)(b − 1) = 0

Factorise

This gives two possible values for b: b = −3 or b = 1. Find the corresponding value of c for each. Case 1: If b = −3, then c = −2 − (−3) = 1. Case 2: If b = 1, then c = −2 − 1 = −3. So, there are two possible sets of solutions: • a = 1, b = −3, c = 1 • a = 1, b = 1, c = −3

Idea summary Two quadratics are equal if their coefficients match in standard form. To solve for unknown parameters, equate the corresponding coefficients of each expression.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2.01 Practice questions What do you remember? 1

State whether each statement about quadratic functions is true or false: a

The vertex form y = a(x − h)2 + k gives the vertex at (h, k).

The factorised form y = a(x − x1)(x − x2) gives x-intercepts at x1, x2.

Two quadratics must be equal if their x-intercepts are the same.

The range of y = ax2 + bx + c is  if a > 0.

Match each quadratic equation with its form name: Equation

Form Name

y = a(x − x1)(x − x2)

Vertex form

y = ax + bx + c

Standard form

iii

Factorised form

y = a(x − h) + k

How to determine if two quadratic expressions ax2 + bx + c and px2 + qx + r are equivalent?

Practice Ex 1

Ex 2

A parabola has vertex (3, −2) and passes through (1, 2): a

Find the equation in vertex form.

Determine the domain and range.

Sketch the graph, labelling the vertex, point (1, 2), and axis of symmetry.

A parabola has x-intercepts (−2, 0) and (4, 0), and vertex (1, −9): a

Find the equation in standard form.

Determine the domain and range.

Find the values of a, b, and c such that y = ax(x + b) + c(x + b) is equivalent to the parabola.

A parabola has x-intercepts (2, 0) and (4, 0), and passes through (0, 8): a

Find the equation in factorised form.

Convert to standard form.

A parabola passes through (0, 2), (1, 3), and (2, 6): a

Find the equation in standard form.

Determine the range.

A parabola has vertex (−1, 5) and passes through (1, −3): a

Find the equation in vertex form.

Convert to standard form.

2.01 Equations of parabolas mathspace.co

A parabola has x-intercepts (−2, 0) and (4, 0), and y-intercept (0, −8): a

Find the equation in factorised form.

Find the vertex.

A parabola has vertex (0, 3) and passes through (2, −1): a

Find the equation in vertex form.

Determine the range.

A parabola passes through (−1, 0), (0, −2), and (1, 0): a

Find the equation in standard form.

Sketch the graph, labelling the given points and axis of symmetry.

Extend your thinking 12

A parabola has x-intercepts (−2, 0) and (6, 0), and a y-intercept at (0, −9/4): a

Find the equation in factorised form.

Find the axis of symmetry.

A parabola has vertex (2, 4) and passes through the origin: a

Find the equation in vertex form.

Find the other x-intercept.

The graph of a quadratic function is shown: y

(2, 0) 1

−1

−2

Find the equation in standard form.

Determine the domain and range.

(1, −2)

Find the values of a, b, and c if the quadratic functions f (x) = x2 + 4x and f (x) = a(x − 2)2 + bx + c + 5 have the same graph.

Find the values of a, b, and c if the quadratic functions f (x) = 2x2 − 3x and f (x) = a(x + 1)2 + bx + c − 4 have the same graph.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2.02 Solve quadratic systems After this lesson, you will be able to… • solve systems of equations involving one linear and one quadratic equation, or two quadratic equations, using both algebraic and graphical methods. • recognise that solving f (x) = k corresponds to finding the intersection of y = f (x) and y = k. • use the discriminant to determine the number of intersection points in a system. • interpret the solutions of a system as points of intersection on a graph.

Intersections of quadratics and linear graphs The intersection points of a quadratic function f (x) = ax2 + bx + c and a linear function g(x) = mx + d occur where f (x) = g(x). These points can be found algebraically or graphically. To ﬁnd the points algebraically, set f (x) = g(x), rearrange to form a quadratic equation, and solve. The solutions are the x-values of the intersection points. After graphing, the points can be found by observation if the intersections are clear and can be read accurately from the graph. • Algebraically: Solve ax2 + bx + c = mx + d, rearranging to ax2 + (b − m)x + (c − d) = 0. Once the x-values are found, substitute them into either equation to obtain the corresponding y-values. Using the linear equation is usually quicker. • Graphically: Plot both functions and identify intersection points (0, 1, or 2 possible). The discriminant quickly determines the number of intersections: • Δ > 0: Two intersections • Δ = 0: One intersection (tangent) • Δ < 0: No intersections

Interactive exploration Discover this concept in action online

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Example 1 Find the intersection points of f (x) = x2 − 4x + 4 and g(x) = 2x − 1: a Algebraically

Create a strategy Set f (x) = g(x), form a quadratic equation, and solve by factorising.

2.02 Solve quadratic systems mathspace.co

Apply the idea f (x) = g(x)

Set f (x) = g(x)

x − 4x + 4 = 2x – 1

Substitute the values of each function

x − 4x − 2x + 4 + 1 = 0

Subtract 2x and add 1 to both sides

x − 6x + 5 = 0

Collect like terms

(x − 5)(x − 1) = 0

Factorise

5, x = 1

Use null factor law

Substitute the x-values into g(x) = 2x − 1 to ﬁnd their corresponding y-coordinates. If x = 5: g(x) = 2x − 1

Write the function

g(5) = 2 × 5 − 1

Substitute x = 5

Evaluate

If x = 1: g(1) = 2 × 1 − 1 =1

Substitute x = 1 Evaluate

The intersections are at (5, 9) and (1, 1).

Reﬂect and check This means the solutions are x = 5 and x = 1.

b Graphically

Create a strategy Plot both functions on a Cartesian plane and identify intersection points.

Apply the idea y 9 8 7 6 5 4 f (x) g(x) 3 2 (1, 1) 1 −1

(5, 9)

The parabola f (x) = x2 − 4x + 4 and line g(x) = 2x − 1 intersect at (5, 9) and (1, 1). x 4

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary Intersections of a quadratic f (x) and a linear function g(x) occur where f (x) = g(x). Graphically, these are the points where the two graphs meet. Algebraically, set f (x) = g(x), solve the resulting quadratic for the x-values, then substitute into either equation to ﬁnd the corresponding y-values. The discriminant provides a quick way to determine the number of intersection points.

Intersections of two quadratics The intersection points of two quadratic functions f (x) = a1 x2 + b1 x + c1 and g(x) = a2 x2 + b2 x + c2 occur where f (x) = g(x). These can be found graphically or algebraically. Algebraically, form the equation f (x) = g(x), rearrange to set it equal to zero, and solve. Graphically, the solutions are the points of intersection, although these are not always exact when read from a graph. • Algebraically: Solve a1 x2 + b1 x + c1 = a2 x2 + b2 x + c2 by rearranging to (a1 − a2)x2 + (b1 − b2)x + (c1 − c2) = 0. • Graphically: Plot both parabolas and identify intersection points (0, 1, or 2 possible). The discriminant quickly determines the number of intersections: • Δ > 0: Two intersections • Δ = 0: One intersection (tangent) • Δ < 0: No intersections

Exploration Consider the quadratic functions f (x) = x2 − 2x − 3 and g(x) = −x2 + 3x + 1. 1. Sketch their graphs on paper and estimate intersection points. 2. Set f (x) = g(x) and form the quadratic equation to solve algebraically. Compare the graphical estimate with the algebraic solution.

Example 2 Find the intersection points of f (x) = x2 − 2x − 3 and g(x) = −x2 + 3x + 1 graphically and algebraically: a Algebraically

Create a strategy Set f (x) = g(x), set equal to zero and solve.

2.02 Solve quadratic systems mathspace.co

Apply the idea f (x) = g(x) 2

Set f (x) = g(x)

x − 2x − 3 = −x + 3x + 1 2

Substitute the values of each function Add x2 and subtract 3x + 1

x − 2x − 3 + x − 3x − 1 = 0 2x2 − 5x − 4 = 0

Collect like terms

Use the quadratic formula to ﬁnd the x-coordinates: Write the quadratic formula

Substitute a = 2, b = −5 and c = −4

Evaluate each term

Evaluate the expression inside the square root Substitute the x-values into f (x) = x2 − 2x − 3 to ﬁnd their corresponding y-coordinates. Substitute

: Write the function

Substitute

Simplify

Substitute

: Write the function

Substitute

Simplify and

The intersections are approximately

Reﬂect and check This means the solutions are

and

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Graphically

Create a strategy Plot both parabolas and identify intersection points.

Apply the idea 4

g(x)

3 2 1 −1

−1

(3.1, 0.6) 1

x 4

The parabolas f (x) = x2 − 2x − 3 and g(x) = −x2 + 3x + 1 intersect at approximately (3.1, 0.6) and (−0.6, −1.3).

(−0.6, −1.3)

−2

−3 −4

f (x)

Idea summary Intersections of two quadratics occur where f (x) = g(x). Graphically, these are the points where the curves meet. Algebraically, set f (x) = g(x), rearrange to form a quadratic equation equal to zero, and solve. The discriminant quickly determines how many intersections exist.

2.02 Practice questions What do you remember? 1

2 3

State whether each statement about intersections of functions is true or false: a

Intersection points of f (x) = ax2 + bx + c and g(x) = mx + d occur where f (x) = g(x).

The discriminant of the original quadratic function determines how many times it intersects a given straight line function.

Two quadratic functions always have two intersection points.

If Δ = 0, the functions intersect at one point.

What equation must be solved to ﬁnd the intersections of f (x) = x2 + 3x − 1 and g(x) = 2x + 1? If the equation f (x) = k is solved for x, what does each solution represent? A

The x-intercepts of the function f (x).

The x-values of the intersection points of the graphs of y = f (x) and y = k.

The maximum or minimum values of the function f (x).

The x-intercept and y-intercept of the function f (x). 2.02 Solve quadratic systems mathspace.co

Practice Ex 1

Find the intersection points of f (x) = x2 − 6x + 5 and g(x) = x − 1: a

Ex 2

Algebraically

Graphically

Solve algebraically.

Determine the discriminant and interpret its meaning.

Find the intersection points of f (x) = 2x − 3 and g(x) = x2 − 4x + 5: a

Solve algebraically.

Sketch the graphs, labelling the intersection points.

For f (x) = −2x2 + 4x + 4 and g(x) = x2 + x − 2: Solve algebraically.

Verify graphically.

For f (x) = x2 − 4 and g(x) = 2x − 4: a

Graphically

Find the intersection points of f (x) = x2 − 3x + 2 and g(x) = 1:

a 9

Find the intersection points of f (x) = x2 − 4x + 3 and g(x) = −x2 + 2x + 3: a

Algebraically

Solve algebraically.

For f (x) = 2x2 − 3x + 1 and g(x) = x2 − 2x + 1: a

Solve algebraically.

Determine the discriminant and interpret its meaning.

Use technology to ﬁnd the intersection points of f (x) = x2 − 4x + 5 and g(x) = 2x − 3.

Extend your thinking 12

Find the value of m such that f (x) = x2 − 2x + 5 and g(x) = mx + 1 intersect at exactly one point.

Find the values of k such that f (x) = x2 − 4x + 4 and g(x) = kx − 2 have two points of intersection.

Two quadratic functions, f (x) and g(x), intersect at exactly one point, on the line y = x. The function f (x) passes through the point (1.5, 2.5) and has its vertex at (2, 2.25), while the function g(x) satisﬁes the condition g(2) = 3: a

Find the equation of f (x).

Determine the coordinates of the unique point of intersection of f (x) and g(x).

Find the equation of g(x).

A quadratic f (x) = x2 − 4x + 3 intersects a linear function g(x) = mx + d at (1, 0) and is tangent at that point. Find m and d.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2.03 Quadratic inequalities After this lesson, you will be able to… • solve quadratic inequalities algebraically by finding boundary points and testing intervals. • solve quadratic inequalities graphically by identifying regions where one function is above or below another. • represent solutions to inequalities using interval notation and on a number line. • solve inequalities that involve a quadratic function and a linear or another quadratic function.

Quadratic inequalities with constants A quadratic inequality compares a quadratic function f (x) = ax2 + bx + c to a constant, e.g., f (x) > k. Solutions are found algebraically or graphically. • Algebraically: Solve f (x) = k to ﬁnd boundary points, then test intervals to determine where the inequality holds. • Graphically: Plot y = f (x) and y = k, and identify where the graph is above (> k), below (< k), or on (= k). The discriminant Δ = b2 − 4ac of f (x) = k determines the number of boundary points: • Δ > 0: Two boundary points • Δ = 0: One boundary point • Δ < 0: No boundary points (parabola entirely above or below y = k)

Interactive exploration Discover this concept in action online

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Example 1 Solve the inequality x2 − 4x − 5 ≤ 0: a Algebraically

Create a strategy Rewrite the inequality as an equation to solve for the boundary points. Test intervals to determine where x2 − 4x − 5 ≤ 0.

2.03 Quadratic inequalities mathspace.co

Apply the idea x2 − 4x − 5 ≤ 0

Write the inequality

x − 4x − 5 = 0

Write as an equation

(x − 5)(x + 1) = 0

Factorise

5, x = −1

Use null factor law

So the boundary points are at x = −1, 5. Select an easy test value from within each interval to check if the inequality is satisﬁed: (−∞, −1), (−1, 5), (5, ∞) Test point (−∞, −1)

−2

(−1, 5)

(5, ∞)

(x − 5)(x + 1) (−2 − 5)(−2 + 1) = (−7) × (−1) =7 (0 − 5)(0 + 1) = −5 × 1 = −5 (6 − 5)(6 + 1) = 1 × 7 =7

≤ 0? No Yes No

The inequality holds for −1 ≤ x ≤ 5.

b Graphically

Create a strategy Plot y = x2 − 4x − 5 and identify where the graph is on or below y = 0.

Apply the idea y 4

(−1, 0)

−3 −2 −1 −2 −4

(5, 0) 1

2 3 4 5 6

The parabola y = x2 − 4x − 5 is on or below the line y = 0 (the x-axis) between x = −1 and x = 5, so the solution is −1 ≤ x ≤ 5.

−6 −8

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary Quadratic inequalities with constants are solved by ﬁnding boundary points algebraically (solving f (x) = k and testing intervals) or graphically (where y = f (x) crosses y = k). The solution depends on the parabola’s concavity and the inequality sign.

Quadratic inequalities with linear or quadratic functions Quadratic inequalities comparing a quadratic function f (x) = ax2 + bx + c to a linear function g(x) = mx + d or another quadratic function h(x) = a2 x2 + b2 x + c2 are solved similarly. • Algebraically: Solve f (x) = g(x) or f (x) = h(x) to ﬁnd intersection points, then test intervals. • Graphically: Plot both functions and identify where f (x) is above, below, or equal to the other function.

Exploration Consider f (x) = x2 − 2x and g(x) = −x + 2: 1. Sketch their graphs on the same plane and estimate where f (x) ≤ g(x). 2. Solve x2 − 2x ≤ −x + 2 algebraically and compare results.

Example 2 Solve the inequality 3x2 + x ≥ 2x2 + 2: a Algebraically

Create a strategy Rewrite the inequality as an equation to solve for the boundary points. Test intervals to determine where 3x2 + x ≥ 2x2 + 2.

2.03 Quadratic inequalities mathspace.co

Apply the idea 3x2 + x ≥ 2x2 + 2 2

Write the inequality

3x + x = 2x + 2

Write as an equation

Subtract 2x2 + 2 from both sides

x +x−2=0 (x + 2)(x − 1) = 0 x

Factorise

−2, x = 1

Use null factor law

So the boundary points are at x = −2, 1. Test intervals: (−∞, −2), (−2, 1), (1, ∞) (x + 2)(x − 1)

Test point (−∞, −2)

−3

(−2, 1)

(1, ∞)

≥ 0?

(−3 + 2)(−3 − 1) = (−1) × (−4) =4 (0 + 2)(0 − 1) = 2 × (−1) = −2 (2 + 2)(2 − 1) = 4 × 1

Yes No Yes

The inequality holds for x ≤ −2 or x ≥ 1.

b Graphically

Create a strategy Plot y = 3x2 + x and y = 2x2 + 2 and identify where the ﬁrst parabola is above or on the second.

Apply the idea (−2, 10)

y 10 8 6

y = 2x2 + 2 4

y = 3x2 + x −2

(1, 4)

−1

x 1

−2

The parabola y = 3x2 + x is above or on y = 2x2 + 2 for x ≤ −2 and x ≥ 1.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary Quadratic inequalities with linear or quadratic functions are solved by ﬁnding intersection points algebraically or graphically and testing intervals. The solution regions depend on where one function is above, below, or equal to the other, depending on the sign of the inequality.

2.03 Practice questions What do you remember? 1

State whether each statement about quadratic inequalities is true or false: a

The solution to ax2 + bx + c > 0 is where the parabola y = ax2 + bx + c is above the x-axis.

If Δ < 0 and a > 0, then ax2 + bx + c > 0 for all x.

To solve f (x) ≤ g(x), ﬁnd where f (x) = g(x) and test intervals.

A quadratic inequality always has two boundary points.

What is the ﬁrst step to solve x2 − 6x + 8 ≥ 0 algebraically?

What information does the discriminant provide about the solutions to the inequality f (x) < k, where f (x) is a quadratic function?

Practice Ex 1

Solve the inequality x2 − 6x + 5 ≤ 0: a

Ex 2

Graphically

Algebraically

Graphically

Solve the inequality x2 − 2x − 8 > 0: a

Solve the inequality x2 + 3x ≥ x2 + x + 2: a

Algebraically

For the inequality x2 + 4x + 4 ≤ 0: a

Solve algebraically.

Determine the discriminant and interpret its meaning.

Solve the inequality x2 − 4x ≥ −3 algebraically and represent the solution on a number line.

Solve the inequality 2x2 − 3x + 1 ≤ x2 − x + 1 algebraically.

2.03 Quadratic inequalities mathspace.co

Solve the inequality x2 − 5x + 6 > −x + 3 algebraically and represent the solution on a number line.

Solve the inequality x2 + 2x − 3 ≥ 0 using technology to sketch the graph and verify the solutions algebraically.

Extend your thinking 12

For which values of k is the inequality x2 − 4x + k > 0 true for all real x?

Let f (x) = x2 + x − px − 3 and g(x) = x2 − x − 2, where p is a real constant: a

Solve the inequality f (x) > g(x) algebraically in terms of p.

Describe how the solution set changes as p varies.

Verify the results graphically by sketching f (x), for p = −2 and p = 1, and g(x) on the same set of axes.

A projectile’s height is modelled by h(t) = −5t2 + 20t + 1 metres. For what times t ≥ 0 is the height at least 10 metres?

Find the possible values of m such that the inequality x2 − 4x + 3 ≥ mx − 1 holds for all real values of x.

Did you know?

Quadratic inequalities help farmers optimise crop yields by showing the safe range of conditions for growth! For example, they can model how much water or fertiliser plants need so they thrive without being damaged by excess. By identifying these ranges, farmers can make smarter decisions that boost harvests while protecting the soil and environment.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2.04 Quadratic models After this lesson, you will be able to… • construct a quadratic function to model a real-world situation. • interpret the vertex of a quadratic model as a maximum or minimum value. • interpret the intercepts of a quadratic model in the context of the problem. • solve practical optimisation problems using quadratic functions. • determine and justify a realistic domain and range for a quadratic model.

Model with quadratics Quadratic functions f (x) = ax2 + bx + c model scenarios with parabolic behaviour, such as projectile motion, area, or profit. The vertex, intercepts, and concavity often make these situations easier to interpret. For example, the maximum or minimum of a parabola might represent the highest profit margin, the largest possible area, or the greatest height reached by a projectile. In real-world contexts, these are often the critical points of interest. • Standard form f (x) = ax2 + bx + c: Useful for finding y-intercept (0, c) and solving for roots. • Vertex form f (x) = a(x − h)2 + k: The vertex (h, k) gives maximum or minimum value. • Factorised form f (x) = a(x − x1)(x − x2): Roots x1, x2 indicate x-intercepts. Domain is often  or restricted by context (e.g., non-negative for time). Range is [k, ∞) for a > 0 or (−∞, k] for a < 0, where k is the vertex’s y-coordinate.

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 A ball is thrown upward from a height of 2 metres with velocity 10 m/s. Its height is modelled by f (x) = −4.9x2 + 10x + 2, where x is time in seconds and f (x) is height in metres: a Find when the ball hits the ground, rounded to three decimal places.

Create a strategy Solve f (x) = 0 using the quadratic formula to ﬁnd when height is zero.

2.04 Quadratic models mathspace.co

Apply the idea f (x) = −4.9x2 + 10x + 2

Write the function

0 = −4.9x + 10x + 2

Set f (x) = 0

Use the quadratic formula to ﬁnd the time: Write the formula

Substitute a = −4.9, b = 10 and c = 2

Evaluate each term

Evaluate the expression inside the square root

Separate into two solutions

Evaluate and round

The ball hits the ground at x = 2.224 seconds (discard negative time).

b Find the maximum height, rounded to two decimal places.

Create a strategy and evaluate f (x).

Find the vertex using

Apply the idea For the vertex: Write the formula Substitute a = −4.9 and b = 10

Evaluate and round

For the maximum height: f (x) = −4.9(x)2 + 10x + 2 2

f (1.02) = −4.9(1.02) + 10 × 1.02 + 2 = 7.10 At x = 1.02 seconds, the maximum height is 7.10 metres.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Write the function Substitute x = 1.02 Evaluate and round

c State the domain and range.

Create a strategy Determine the domain from context and the range using concavity and vertex.

Apply the idea The domain is [0, 2.224], as time is non-negative until the ball hits the ground. The range is [0, 7.10] , since a = −4.9 < 0, parabola opens downward, with minimum at f (0) = 2 and maximum at 7.10.

Reﬂect and check y

(1.02, 7.10)

7 6 5 4

The parabola shows height from x = 0 to x = 2.224, with maximum at (1.02, 7.10).

3 2 (0, 2) 1

(2.224, 0) 0.5

1.5

Idea summary Quadratic functions model scenarios like projectile motion. Use standard, vertex, or factorised forms to ﬁnd key features. The maximum or minimum is at the y-value of the vertex, that is

. Domain is context-dependent while the range depends

on concavity and vertex.

2.04 Quadratic models mathspace.co

Optimise with quadratics Quadratic functions model optimisation problems, such as maximising area or proﬁt. The vertex provides the optimal value, and roots indicate break-even points. Break-even point The point at which income from production and cost of production are equal.

Exploration A farmer has 100 metres of fencing to enclose a rectangular ﬁeld. 1. Let x be the width in metres, and model the area f (x). 2. Sketch the graph, estimate the maximum area, and consider domain constraints.

Example 2 A company’s proﬁt is modelled by f (x) = −2x2 + 40x − 150, where x is units sold (in hundreds) and f (x) is proﬁt in thousands of dollars: a Find the break-even points.

Create a strategy Solve f (x) = 0 to ﬁnd where proﬁt is zero, using factorisation.

Apply the idea f (x) = −2x2 + 40x – 150 2

Write the function

−2x + 40x − 150 = 0

Set f (x) = 0

−2(x2 − 20x + 75) = 0

Factorise −2

x − 20x + 75 = 0

Divide both sides by −2

(x − 15)(x − 5) = 0

Factorise

5, x = 15

Use null factor law

Break-even points are at x = 5 and x = 15 (500 and 1500 units).

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Find the maximum proﬁt.

Create a strategy Find the vertex using

and substitute into f (x).

Apply the idea For the vertex: Write the formula

Substitute a = −2, b = 40

Evaluate

For the maximum proﬁt: f (x) = −2(x)2 + 40x – 150 2

Write the function

f (10) = −2 × (10) + 40 × 10 − 150

Substitute x = 10

= −200 + 400 − 150

Evaluate each term

= 50

Evaluate

At x = 10 (1000 units), the maximum proﬁt is 50 thousand dollars ($50 000).

c State the domain and range.

Create a strategy Determine the domain from context and range using concavity and vertex.

Apply the idea The domain is [5, 15], as proﬁt is non-negative between break-even points. The range is [0, 50], since a = −2 < 0, parabola opens downward, with minimum at f (5) = f (15) = 0 and maximum at 50.

Reﬂect and check y 50

(10, 50)

The parabola shows proﬁt when 5 < x < 15, with maximum at (10, 50).

30 20 10 x (5, 0) (15, 0) 0 2 4 6 8 10 12 14 16 18

2.04 Quadratic models mathspace.co

Idea summary Optimisation problems use the vertex for maximum/minimum values and roots for break-even points. Domain and range are constrained by context, reﬂecting real-world limits.

2.04 Practice questions What do you remember? 1

State whether each statement about quadratic modelling is true or false: a

The vertex of f (x) = a(x − h)2 + k represents the maximum or minimum value of the function.

The roots of f (x) = a(x − x1)(x − x2) indicate break-even points in proﬁt models.

The domain of a quadratic function is always  regardless of context.

If a < 0, the range is [k, ∞) where k is the vertex’s y-coordinate.

In a quadratic model f (x) = ax2 + bx + c, what does the vertex represent in the context of projectile motion?

How can the domain be determined in a quadratic model for an area optimisation problem?

Practice Ex 1

Ex 2

A ball is thrown upward from a height of 5 metres with an initial velocity of 13 m/s. Its height is modelled by f (x) = −4.8x2 + 13x + 5, where x is time in seconds and f (x) is height in metres: a

Find the time when the ball hits the ground.

Find the maximum height and when it occurs.

State the domain and range in context.

A company’s proﬁt is modelled by f (x) = −x2 + 30x − 200, where x is units sold (in hundreds) and f (x) is proﬁt in thousands of dollars: a

Find the break-even points.

Find the maximum proﬁt and the number of units sold.

State the domain and range in context.

A company’s proﬁt is f (x) = −x2 + 50x − 600 dollars for x units sold. Find the maximum proﬁt earned by the company.

A projectile’s height is given by f (x) = −5x2 + 20x + 3 metres, where x is time in seconds:

Find the time when the projectile reaches the ground.

Find the maximum height.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A shop’s revenue is modelled by f (x) = −3x2 + 60x, where x is the price in dollars per item, and f (x) is revenue in dollars: a

Find the price that maximises revenue.

Find the maximum revenue.

A ball’s height is modelled by f (x) = −4x2 + 16x metres, where x is time in seconds: a

Find the times when the ball is at ground level.

State the domain and range in context.

A company’s cost is f (x) = x2 − 20x + 150 dollars, where x is units produced (in hundreds): a

Find the minimum cost and the number of units.

Find the domain in this context.

A diver’s height above water is f (x) = −5x2 + 10x + 3 metres, where x is time in seconds: a

Find the maximum height.

Find when the diver enters the water.

A farmer has 120 metres of fencing to enclose a rectangular garden. Let x be the width in metres: a

Write an expression for the length of the garden in terms of x.

Express the area f (x) as a quadratic function.

Find the maximum area and the dimensions of the garden.

State the domain in this context.

Extend your thinking 13

A rectangular ﬁeld has a perimeter of 80 metres, and its area is modelled by f (x) = 40x − x2, where x is the width in metres: a

Find the dimensions for maximum area.

Sketch the graph of f (x), labelling the vertex.

A rocket’s height is f (x) = −5x2 + 40x + 10 metres, where x is time in seconds. Find the time interval when the rocket is above 80 metres.

A farmer wants to enclose a rectangular ﬁeld with 200 metres of fencing, with one side along a river (no fencing needed). Let x be the width perpendicular to the river:

Model the area f (x) and ﬁnd the maximum area.

State the domain and range.

A bridge’s cable forms a parabola with equation f (x) = 0.01x2 − 0.4x + 10 metres, where x is horizontal distance in metres from one ends: a

Find the minimum height of the cable.

Find the domain where the cable is higher than 6 metres.

2.04 Quadratic models mathspace.co

2 Chapter review 1

A rectangular plot of land has the following dimensions: 3x + 2

x−2

Write the quadratic expression for the area in factored and expanded forms.

Find the area when x = 4 m.

Explain why x must be greater than 2.

For the equation x2 + 10x + k = 0, ﬁnd the values of k for which it has: a

One real solution

No real solutions

A parabola y = x2 − 6x + 10 intersects a line y = mx + 1: a

Form the quadratic equation for the points of intersection.

Find the values of m for which there are two distinct intersection points.

A rectangular paddock has a perimeter of 40 metres and area A = x(20 − x) m2, where x is the width in metres: a

Express the area in completed square form and ﬁnd the vertex.

Determine the maximum area and corresponding dimensions.

For the quadratic function y = x2 − 4x − 5: a

Find the points of x-intercepts.

Find the point of y-intercept.

Find the axis of symmetry and vertex.

Determine the domain and range.

Sketch the graph, labelling key features.

For the quadratic y = (x + 1)(x − 5): a

Find the x-intercepts and y-intercept.

Find the vertex and axis of symmetry.

Sketch the graph, labelling key features.

A quadratic function has vertex (1, −9) and passes through (4, 0). Find its equation in standard form.

A parabola has x-intercepts (−1, 0) and (3, 0), and passes through (0, −6): a

Find the equation in factored form.

Convert to standard form.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A parabola passes through (0, −1), (1, 1), and (2, 7): a

Find the equation in standard form.

Determine the range.

Find the integer values of a and b such that the quadratic identity x2 − 5x − 4 = (x − a)2 + (x − b) is true for all values of x.

Find the intersection points of f (x) = x2 − 5x + 7 and g(x) = x + 2: a

Algebraically

Graphically

Find the intersection points of f (x) = x2 − 2x − 3 and g(x) = −x2 + 4x − 3: a

Algebraically

Graphically

Find the value of m such that f (x) = x2 − 4x + 8 and g(x) = mx − 1 intersect at exactly one point.

Find the values of k such that f (x) = x2 + 2x + 3 and g(x) = kx − 1 have two points of intersection.

Solve the inequality x2 − x − 12 ≤ 0: a

Algebraically

Graphically

Solve the inequality x2 + 3x − 10 > 0: a

Algebraically

Solve the inequality x2 − 2x ≥ 8 algebraically and represent the solution on a number line.

For which values of k is the inequality x2 + 8x + k > 0 true for all real x?

A company’s proﬁt is modelled by P (x) = −x2 + 40x − 300, where x is units sold (in hundreds) and P (x) is proﬁt in thousands of dollars:

Find the break-even points.

Find the maximum proﬁt and the number of units sold.

State the domain and range for which the company is proﬁtable.

A farmer has 160 metres of fencing to enclose a rectangular garden. Let x be the width in metres: a

Express the area A(x) as a quadratic function.

Find the maximum area and the dimensions of the garden.

State the practical domain for this context.

A rocket’s height is h(t) = −5t2 + 50t + 5 metres, where t is time in seconds. Find the time interval when the rocket is above 125 metres.

Chapter 2 review mathspace.co

3 Remainder and factor theorems Chapter outline 3.01E Division of polynomials 3.02E The remainder and factor theorems 3.03E Further applications Chapter 3 review

92 100 109 118

Factoring polynomials is like cracking a secret code — if you find the right combination, everything unlocks.

3.01E Division of polynomials After this lesson, you will be able to… • divide a polynomial by a monomial, applying distributive property and exponent laws • perform polynomial long division, understanding and using the terms dividend, divisor, quotient, and remainder • express the result of polynomial division in the form P ( x) = A( x)Q( x) + R( x), ensuring deg(R( x)) < deg(A( x)) or R( x) = 0 • express the result of polynomial division in the quotient form

Divide by a monomial A monomial is a single term with non-negative integer exponents. Division of a polynomial by a monomial applies the distributive property of division over addition. For real numbers a, b, and c ≠ 0, the property is expressed as:

a is the ﬁrst term in the numerator b is the second term in the numerator c is the denominator, a non-zero value Each term of the polynomial (dividend) is divided by the monomial (divisor). Dividend

Quotient

The number that is divided or distributed in a division problem.

The result of dividing one number or algebraic expression by another.

Example:

9 is the dividend.

1 is the quotient.

Divisor

Remainder

A number which is divided into another number.

What is left over after a division.

Example:

4 is the remainder.

Example:

5 is the divisor.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

If the monomial includes a variable, the exponent law is applied:

x m, n

is the variable are the exponents (powers)

Exploration Compare the two divisions: • 24 + 16 by 8 numerically and • 6x2 + 4x by 2x algebraically 1. How are they similar? 2. What role does the exponent law play in simplifying the polynomial division? 3. Why must the divisor be non-zero in both cases?

Example 1 Determine the quotient of

Create a strategy Apply distributive property by dividing each term of the numerator by the monomial 3y, then apply exponent laws.

Apply the idea Divide each term by 3y

Apply

Simplify each term

Reﬂect and check Verify by multiplying the quotient by the divisor. (2y2 − 5y + 8) × 3y = 6y3 − 15y2 + 24y Evaluate This matches 6y3 − 15y2 + 24y, confirming the solution.

3.01E Division of polynomials mathspace.co

Idea summary Division of a polynomial by a monomial involves dividing each term using the distributive property and simplifying with exponent laws, resulting in a quotient polynomial.

Polynomial long division Polynomial division expresses a polynomial in the form:

P ( x) = A( x)Q( x) + R( x) P ( x) is the dividend polynomial A( x) is the divisor polynomial, not the zero polynomial Q( x) is the quotient polynomial R( x) is the remainder polynomial, where either R( x) = 0 or deg(R( x)) < deg(A( x)) The degree of the remainder is less than the divisor because division continues until the remaining polynomial’s leading term cannot be divided by the divisor’s leading term. For example, a linear divisor (deg(A( x)) = 1) yields a constant remainder, while a quadratic divisor (deg(A( x)) = 2) yields a linear remainder.

Exploration Consider why the remainder’s degree must be less than the divisor’s: 1. What happens if the remainder’s degree equals or exceeds the divisor’s? 2. What operation must be performed if the remainder’s degree is greater than or equal to the divisor’s? 3. Why does a linear divisor produce a constant remainder?

For non-monomial divisors (like linear or quadratic polynomials), polynomial long division is used, similar to numerical long division: 1. Divide the leading term of the dividend by the leading term of the divisor. 2. Multiply the result by the divisor. 3. Subtract the product from the dividend. 4. Bring down the next term. 5. Repeat until the remainder’s degree is less than the divisor’s or no terms remain. Terms are arranged in descending order of degree, with 0 coeﬃcients for missing terms.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Divide x3 + 7x2 + 14x + 3 by x + 2 and express in the form P ( x) = A( x)Q( x) + R( x).

Create a strategy Conﬁrm that the terms are in descending order with no missing degrees to apply polynomial long division.

Apply the idea The dividend x3 + 7x2 + 14x + 3 has degrees 3, 2, 1, 0 which is complete. The divisor x + 2 has degrees 1, 0 which is complete. Divide: x3 ÷ x = x2 Multiply: x2 × ( x + 2) = x3 + 2x2 Subtract: ( x3 + 7x2) − ( x3 + 2x2) = 5x2 Bring down: 14x, forming 5x2 + 14x Divide: 5x2 ÷ x = 5x Multiply: 5x × ( x + 2) = 5x2 + 10x Subtract: (5x2 + 14x) − (5x2 + 10x) = 4x Bring down: 3, forming 4x + 3

Divide: 4x ÷ x = 4 Multiply: 4 × ( x + 2) = 4x + 8 Subtract: (4x + 3) − (4x + 8) = −5 Remainder: −5 (degree 0, less than 1)

The quotient is x2 + 5x + 4, while the remainder is −5. Expressing in the polynomial form: P ( x) = A( x)Q( x) + R( x) 3

Write the formula

x + 7x + 14x + 3 = ( x + 2) ( x + 5x + 4) − 5

Substitute the values

Reﬂect and check Verify P ( x) = A( x)Q( x) + R( x). A( x)Q( x) + R( x) = ( x + 2) ( x2 + 5x + 4) − 5

Substitute the values

Expand the brackets

= x + 5x + 4x + 2x + 10x + 8 − 5 = x + 7x + 14x + 3

Collect like terms

= P ( x)

The result matches the polynomial

3.01E Division of polynomials mathspace.co

Idea summary Polynomial long division expresses P ( x) as A( x)Q( x) + R( x), where either R( x) = 0 or deg(R( x)) < deg(A( x)). The process involves dividing leading terms, multiplying, subtracting, and repeating until the remainder’s degree is less than the divisor’s.

Polynomial division in quotient form Polynomial division can also be expressed in quotient form:

P ( x) is the dividend polynomial A( x) is the divisor polynomial, not the zero polynomial Q( x)

is the quotient polynomial

R( x)

is the remainder polynomial, where either R( x) = 0 or deg(R( x)) < deg(A( x))

This form is derived from P ( x) = A( x)Q( x) + R( x) by dividing both sides by A( x). It represents the division result as a polynomial plus a fractional term involving the remainder.

Exploration 1. How does quotient form differ from the standard form of P ( x) = A( x)Q( x) + R( x)? 2. What happens to the fractional term if R( x) = 0? 3. How does the degree of R( x) affect the fractional term?

Example 3 Divide 2x3 + 7x2 + 10x + 14 by x − 1 and express the result in quotient form

Create a strategy Use long division to find the quotient and remainder, then write the result in the required quotient form.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea Performing the long division: Divide: 2x3 ÷ x = 2x2 Multiply: 2x2 × ( x − 1) = 2x3 − 2x2 Subtract: (2x3 + 7x2) − (2x3 − 2x2) = 9x2 Bring down: 10x, then do it the same step Remainder: 33 (degree 0 to 1)

So the quotient is Q( x) = 2x2 + 9x + 19 and the remainder is R( x) = 33. Expressing in quotient form: Write the formula

Substitute the values

Reﬂect and check Verify

. ubstitute the quotient, remainder and S divisor

Find a common denominator to combine the terms Expand the brackets in the numerator Collect like terms in the numerator The result matches the original expression

Idea summary Polynomial division can be expressed as P ( x) = A( x)Q( x) + R( x) by dividing by A( x).

, derived from

3.01E Division of polynomials mathspace.co

3.01E Practice questions What do you remember? 1

Deﬁne these terms: a

Dividend

Polynomial = 3x2 + 2x − 1, identify each:

Dividend

Divisor

Quotient

= 2x2 − x + 1 −

For the division a

Given the polynomial division a

Monomial

Divisor

Remainder

, identify each: Quotient

Divide 15 by 4 then identify the dividend, divisor, quotient and remainder.

Practice Ex 1

Determine the quotient: a

Fill in the boxes to make each monomial division statement true, for x ≠ 0: a c

Ex 2

Evaluate each expressing the answers in the form P ( x) = A( x)Q( x) + R( x): a

c e

b d

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 3

Determine the remainder when P ( x) is divided by A( x): a

P ( x) = x3 − 2x2 + 3x − 4, A( x) = x − 3

P ( x) = 2x4 + x3 − 5x2 + 2x − 1, A( x) = x + 2

P ( x) = x4 − x3 + 2x2 − x + 3, A( x) = x2 − 2

P ( x) = 3x3 − 4x2 + x − 5, A( x) = x2 + x − 1

Convert each result to the form

x3 − 5x2 + 3x − 2 = ( x − 1)( x2 − 4x − 1) − 3

2x3 + x2 − x + 3 = ( x + 2)(2x2 − 3x + 5) − 7

x4 − 2x3 + 3x2 − x + 1 = ( x2 − x + 1)( x2 − x + 1) + x

3x3 − x2 + 2x − 4 = ( x2 + x − 2)(3x − 4) + (12x − 12)

x3 + 4x2 − x + 2 = ( x + 3)( x2 + x − 4) + 14

Evaluate each expressing the answer in the form

Extend your thinking 11

List two similarities and one difference of the division of 27 by 4, which gives a quotient of 6 = 3x2 + 2x − 1.

and a remainder of 3 and 12

Given

= x2 + 2x − 3 +

, determine P ( x) and verify your solution by converting back

to quotient form. 13

Review these student responses and correct any mistakes: = 4x2 − 2x + 3x

Student A (Monomial division):

Student B (Long division): Dividing x3 − 2x2 + x − 3 by x − 1 gives quotient x2 − x and remainder x − 3.

A renovation project costs 2x4 − 3x3 + 5x2 − 2x + 1 dollars and x2 − x + 2 dollars per square metre. Determine the expression for the area of the renovated space.

Determine if the division

results in R( x) = 0. Justify your answer using the

divide by a monomial and polynomial long division methods. 16

A manufacturing process costs x4 − 2x3 + 4x2 − x + 3 dollars and x2 − x + 1 dollars per hour. Express the time taken in hours in the form

3.01E Division of polynomials mathspace.co

3.02E The remainder and factor theorems After this lesson, you will be able to… • explain why division of P ( x) by ( x − α) yields P ( x) = ( x − α )Q( x) + r, where r is a constant • prove and apply the remainder theorem to find the remainder when P ( x) is divided by ( x − α ) or (ax + b) • prove and apply the factor theorem, understanding that ( x − α ) is a factor of P ( x) if and only if P (α ) = 0 • use the remainder and factor theorems to solve related polynomial problems (e.g., find unknown coefficients, test for factors)

Remainder theorem Remainder theorem A rule to determine the remainder when one polynomial P ( x) is divided by a linear polynomial ( x − α). The remainder is equal to P (α ).

P ( x) = ( x − α)Q( x) + r P ( x) x−α Q( x) r

is the polynomial being divided is the linear divisor is the quotient polynomial is the remainder, a constant equal to P (α )

Exploration Given the two polynomials: • P ( x) = x3 + 2x2 − 5x − 6 • Q( x) = x3 − 7x2 − 7x + 20 1. Use long division to ﬁnd the remainder of P ( x) ÷ ( x − 1). 2. Use long division to ﬁnd the remainder of Q( x) ÷ ( x + 4). 3. Evaluate P (1). 4. Evaluate Q(−4). 5. Compare the remainder of P ( x) ÷ ( x − 1) with P (1). Identify any relationship. 6. Compare the remainder of Q( x) ÷ ( x + 4) with Q(−4). What does this suggest about the remainder theorem?

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To prove this, substitute x = α: P ( x) = ( x − α)Q( x) + r

Write the formula

P (α ) = (α − α)Q(α ) + r

Substitute x = α

= 0 × Q(α ) + r

Simplify

Evaluate

So, the remainder is r = P (α ). For a divisor ax + b, solve ax + b = 0 to determine

, then the remainder is given by

Example 1 Determine the remainder when P ( x) = 3x4 − 2x3 − 4x − 5 is divided by A( x) = x + 2.

Create a strategy Determine x to substitute into P ( x).

Apply the idea Since the divisor is x + 2, use

: Write the formula

Substitute b = 2 and a = 1

Evaluate

Determining P ( x): P ( x) = 3x4 − 2x3 − 4x − 5

Write the polynomial

P (−2) = 3(−2)4 − 2(−2)3 − 4(−2) − 5

Substitute x = −2

= 3 × 16 − 2 × (−8) − 4 × (−2) − 5

Evaluate the powers

= 48 + 16 + 8 − 5

Evaluate the multiplication

= 67

Evaluate

The remainder is 67.

Reﬂect and check Verify using polynomial long division to confirm that the remainder is 67.

Divide: 3x3 ÷ x = 3x3 Multiply: 3x3 × ( x + 2) = 3x4 + 6x3 Subtract: (3x4 − 2x3) − (3x4 − 6x3) = −8x3 Bring down: 0x2, then do the same step Remainder: 67

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Example 2 Determine k if the remainder when P ( x) = 2x3 − 2x2 − 3x + k is divided by x − 2 is 15.

Create a strategy Because the divisor is x − 2, substitute x = 2 into P ( x) and solve for k.

Apply the idea P ( x) = 2x3 − 2x2 − 3x + k 3

Write the polynomial

P (2) = 2(2) − 2(2) − 3(2) + k

Substitute x = 2

P (2) = 16 − 8 − 6 + k

Evaluate each term

P (2) = 2 + k

Collect like terms

15 = 2 + k

Substitute P (2) = 15

k = 13

Subtract 2 from both sides and make k the subject

Reﬂect and check Substitute k = 13 and evaluate P (2) to conﬁrm the remainder is 15. P ( x) = 2x3 − 2x2 − 3x + k 3

Write the polynomial

P (2) = 2(2) − 2(2) − 3(2) + 13

Substitute k = 13

= 15

Evaluate

The remainder is 15, conﬁrming the solution.

Idea summary The remainder theorem is a rule to determine the remainder when one polynomial P ( x) is divided by a linear polynomial ( x − α). The remainder is equal to P (α ). For ax + b, the remainder is

Factor theorem Factor theorem If P ( x) is a polynomial and P (a) = 0 for some number a, then P ( x) is divisible by x − a. The factor theorem can be used to obtain factors of a polynomial.

P ( x) = ( x − α)Q( x)

102

P ( x)

is the polynomial

x−α

is a linear factor

Q( x)

is the quotient polynomial

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Interactive exploration Discover this concept in action online

mathspace.co

To prove this, apply the remainder theorem: P ( x) = ( x − α)Q( x) + r

Write the formula

P (α ) = (α − α)Q(α ) + r

Substitute x = α

= 0 × Q(α ) + r

Simplify

Evaluate

So if P (α ) = 0, then r = 0. Therefore, P ( x) = ( x − α)Q( x) and x − α is a factor of P ( x). This follows from the remainder theorem: if P (α ) = 0, the remainder is zero, so x − α is a factor. Conversely, if x − α is a factor, then P (α ) = 0. Monic polynomial

Non-monic polynomial

A polynomial where the leading coeﬃcient (the coeﬃcient of the term with the highest degree) is equal to 1.

A polynomial where the leading coeﬃcient is not equal to 1.

To find factors of a monic polynomial, test integer factors of the constant term. For x3 − 6x2 + 11x − 6, test ±1, ±2, ±3, ±6. Note: A polynomial of degree n has at most n zeroes. To find factors of a non-monic polynomial, test factors of the constant term divided by the leading coefficient to find zeroes. For example, to factorise 3x3 − 4x2 + 6x − 8, test factors of

, such as ±1, ±2, ±4, ±8, ± , ± , ± ,

± . Start with integer factors, and once a zero is found, use polynomial division to ﬁnd remaining factors. The remainder and factor theorems can be combined to determine polynomial coeﬃcients or verify factors.

Example 3 For P ( x) = x3 − 6x2 + 11x − 6: a Show that x − 1 is a linear factor.

Create a strategy Substitute x = 1 into P ( x). If P (1) = 0, then x − 1 is a factor.

3.02E The remainder and factor theorems mathspace.co

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Apply the idea P ( x) = x3 − 6x2 + 11x − 6 3

Write the polynomial

P (1) = 1 − 6 × 1 + 11 × 1 − 6

Substitute x = 1

= 1 − 6 + 11 − 6

Evaluate each term

Evaluate

Since P (1) = 0, x − 1 is a factor. b Determine the quadratic factor using polynomial long division.

Create a strategy Divide P ( x) by x − 1 to determine the quotient.

Apply the idea Divide: x3 ÷ x = x2 Multiply: x2 × ( x − 1) = x3 − x2 Subtract: ( x3 − 6x2) − ( x3 − x2) = −5x2 Bring down: 11x, then do the same step Remainder: 0 (degree 0 to 1) The quotient is x2 − 5x + 6 with remainder 0. c Factorise P ( x) fully as a product of linear factors.

Create a strategy Factorise the quadratic x2 − 5x + 6 by determining two numbers that add to −5 and multiply to 6.

Apply the idea The numbers −2 and −3 satisfy −2 + (−3) = −5 and −2 × (−3) = 6. x2 − 5x + 6 = ( x − 2) ( x − 3)

Factorise the quadratic

Factorising P ( x): P ( x) = x3 − 6x2 + 11x − 6

Write the polynomial

= ( x − 1) ( x − 5x + 6)

Write the obtained factors

= ( x − 1) ( x − 2) ( x − 3)

Substitute x2 − 5x + 6 = ( x − 2) ( x − 3)

Reﬂect and check Expand to verify: ( x − 1) ( x2 − 5x + 6) = x3 − 5x2 + 6x − x2 + 5x − 6 3

= x − 6x + 11x − 6 This matches P ( x), conﬁrming the factorisation.

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Expand Collect like terms

Example 4 Given P ( x) = x3 + 4x2 + ax + b, where x − 2 is a factor and the remainder when divided by x + 3 is −5, determine a and b.

Create a strategy Apply the factor theorem to solve for P (2) = 0 and remainder theorem to solve for P (−3) = −5 to form two equations and use elimination and substitution methods to solve for a and b.

Apply the idea Applying the factor theorem: P ( x) = 0 3

x + 4x + ax + b = 0 23 + 4 × 22 + a × 2 + b = 0 8 + 16 + 2a + b = 0 24 + 2a + b = 0 2a + b = −24

Apply the factor theorem Substitute P ( x) = x3 + 4x2 + ax + b Substitute x = 2 Evaluate each term Collect like terms Subtract 24 from both sides

Applying the remainder theorem: P (α ) = −5 3

(−3) + 4 × (−3) − 3a + b = −5 −27 + 36 − 3a + b = −5 9 − 3a + b = −5 −3a + b = −14

Apply the remainder theorem Substitute α = −3 Evaluate each term Collect like terms Subtract 9 from both sides

Using elimination method to solve for a:

5a = −10 Write the equation a = −2 Divide both sides by 5 Using the ﬁrst equation to solve for b: 2a + b = −24 Write the ﬁrst equation 2 × (−2) + b = −24 Substitute a = −2 −4 + b = −24

Evaluate the multiplication

b = −20 Add 4 to both sides So, a = −2 and b = −20.

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Reﬂect and check Verify by substituting a = −2 and b = −20 into P ( x) and checking both conditions. Verifying through factor theorem: P ( x) = x3 + 4x2 + ax + b 3

P (2) = 2 + 4 × 2 + (−2) × 2 + (−20)

Write the polynomial Substitute x = 2, a = −2 and b = −20

= 8 + 16 − 4 − 20

Evaluate each term

Evaluate

Verifying through remainder theorem: P ( x) = x3 + 4x2 + ax + b 3

P (−3) = (−3) + 4 × (−3) + (−2) × (−3) + (−20)

Write the polynomial Substitute x = −3, a = −2 and b = −20

= −27 + 36 + 6 − 20

Evaluate each term

= −5

Evaluate

Both conditions are satisﬁed, conﬁrming the solution.

Idea summary The factor theorem states that if P ( x) is a polynomial and P (a) = 0 for some number a, then P ( x) is divisible by x − a. The factor theorem can be used to obtain factors of a polynomial. Non-monic polynomials are factorised by testing factors of the constant term divided by the leading coeﬃcient. The remainder and factor theorems can determine coeﬃcients or verify factors.

3.02E Practice questions What do you remember? 1

State the remainder theorem for a polynomial P ( x) divided by x − α.

According to the factor theorem, under what condition is x − α a factor of P ( x)?

If a polynomial is divided by (2x−5), what value of x should be substituted into P ( x) to ﬁnd the remainder?

Identify whether this statement is true or false: If P (2) = 0, then ( x − 2) is a factor of P ( x).

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Practice Ex 1

Determine the remainder when each polynomial P ( x) is divided by the given divisor A( x): a

P ( x) = x3 − 4x2 + 2x − 7

P ( x) = 2x4 + x2 − 3x + 5

A( x) = x − 4 A( x) = x + 2 c

P ( x) = 3x3 + 5x2 − x − 2

P ( x) = 5x3 − x2 + 3x − 9

A( x) = x + 1 A( x) = 3x + 1 e

P ( x) = 4x3 − 2x2 + x − 7

P ( x) = x4 + 2x3 − 5x + 1

A( x) = x − 3 A( x) = x + 4 Ex 2

Determine k such that the remainder is as given when each polynomial is divided by the divisor: a

x3 + 2x2 − kx + 4

Divisor: x − 1, remainder 6 c

3x + x − 2x + k

Divisor: x − 2, remainder 10 e

2x + kx − x + 7

Divisor: x − 3, remainder 13 Ex 3

2x4 − kx3 + x − 3 Divisor: x + 3, remainder −6 x3 − 3x2 + kx − 5 Divisor: x + 1, remainder 4 4x3 − x2 + kx + 1 Divisor: 2x − 1, remainder

For each dividend and divisor: i

Use the factor theorem to determine if the divisor is a factor.

If it is a factor, find the corresponding quadratic factor. Otherwise, find the quotient.

iii

Fully factorise P ( x) into linear factors.

P ( x) = x3 − 7x2 + 14x − 8

P ( x) = 2x3 + 5x2 − 4x − 3

A( x) = x − 4 A( x) = x + 3 c

P ( x) = 3x3 − 2x2 − 5x + 2

P ( x) = x3 + 2x2 − 5x − 6

A( x) = x − 2 A( x) = x + 3 Ex 4

For each polynomial P ( x), determine the coeﬃcients a and b given the factor and remainder when divided by the divisor: a

P ( x) = x3 + 3x2 + ax + b, factor x − 1, remainder 8 when divided by x + 2

P ( x) = x3 − 2x2 + ax + b, factor x + 2, remainder −3 when divided by x − 3

For each polynomial P ( x), determine the coeﬃcients a and b given the remainders when divided by the divisors: a

P ( x) = x2 + ax + b, remainder 5 when divided by x − 2, remainder 1 when divided by x + 1

P ( x) = x2 + ax + b, remainder 0 when divided by x − 1, remainder 6 when divided by x − 3

3.02E The remainder and factor theorems mathspace.co

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Extend your thinking 10

A monic cubic polynomial P ( x) has a remainder of 5 when divided by x − 2 and a remainder of −4 when divided by x + 1. When P ( x) is divided by x − 1, the remainder is zero. Determine P ( x).

Given P ( x) = 2x3 + ax2 + bx + c has a factor x − 1 and remainders 10 and 20 when divided by x + 2 and x − 3, respectively, determine a, b, and c.

A cubic polynomial P ( x) is such that P (1) = 0, P (2) = 5, and P (−1) = −4. The remainder when P ( x) is divided by x + 2 is 8. Determine the coeﬃcient of x2 in P ( x).

Identify the errors and correct the work of each student for factorising P ( x) = x3 − 5x2 + 7x − 3 into linear factors: a

Student A:

Write the polynomial

Tests x = 1 Evaluate Divides P ( x) Factorises Final answer: ( x − 1)( x − 5)( x − 1). b

Student B:

Write the polynomial

Tests x = 1 Evaluate Divides P ( x) Factorises Final answer: ( x − 1)( x − 4)( x + 1). 14

The cost of producing x units of a product is modelled by a cubic polynomial C( x). The cost is 10 dollars when x = 1, 20 dollars when x = 2, and has a factor x − 3. The remainder when divided by x + 1 is 4. Calculate the cost when no units are produced.

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3.03E Further applications After this lesson, you will be able to… • use the factor theorem to find integer and rational linear factors of higher-degree polynomials (monic and non-monic) • apply polynomial long division once after identifying sufficient factors using the factor theorem to obtain a quadratic quotient that can be easily factorised or solved • fully factorise higher-degree polynomials into a product of linear and/or irreducible quadratic factors • connect the full factorisation of a polynomial to the key features of its graph ( x-intercepts, y-intercept, end behaviour)

Factorise higher-degree polynomials The factor theorem states that x − α is a factor of a polynomial P ( x) if P (α ) = 0. For monic higherdegree polynomials, such as quartics or quintics, integer factors of the constant term are tested to ﬁnd zeroes. For non-monic polynomials, rational factors of the constant term divided by the leading coeﬃcient are tested. After identifying linear factors, polynomial long division yields a quotient, which is factorised to complete the factorisation.

P ( x) = ( x − α)Q( x) P ( x) x−α Q( x)

the polynomial to be factorised a linear factor, where α is a zero the quotient polynomial

Exploration Consider the non-monic quartic polynomial P ( x) = 3x4 − 7x3 + 2x2 + 7x − 4. Possible integer factors of the constant term are ±1, ±2, ±4, and rational factors (constant or leading coeﬃcient) include

1. Why is it efficient to test smaller integer factors first before rational factors in monic and non-monic polynomials? 2. How does finding two linear factors simplify the division process compared to finding one? 3. If a non-monic polynomial has a rational zero like , how does the leading coefficient affect the form of the linear factor, and what impact does this have on the subsequent division process?

3.03E Further applications mathspace.co

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Example 1 Factorise P ( x) = x4 − 4x3 − x2 + 16x − 12 and then state the solutions to P ( x) = 0.

Create a strategy Test integer factors of the constant term −12 (±1, ±2, ±3, ±4, ±6, ±12) to find zeroes using the factor theorem. Identify two linear factors, expand them to form a quadratic factor, and use long division to find the remaining quadratic factor. Factorise the resulting quadratic to obtain all linear factors. Finally, solve each factor set to zero to find the solutions to P ( x) = 0.

Apply the idea Testing x = 1 as an integer factor of −12: P ( x) = x4 − 4x3 − x2 + 16x − 12 4

Write the polynomial

P (1) = 1 − 4 × 1 − 1 + 16 × 1 − 12

Substitute x = 1

= 1 − 4 − 1 + 16 − 12

Evaluate each term

Evaluate

So, x − 1 is a factor. Testing x = 2 as an integer factor of −12: P ( x) = x4 − 4x3 − x2 + 16x − 12 4

Write the polynomial

P (2) = 2 − 4 × 2 − 2 + 16 × 2 − 12

Substitute x = 2

= 16 − 32 − 4 + 32 − 12

Evaluate each term

Evaluate

So, x − 2 is a factor. Expanding ( x − 1) ( x − 2) to form the quadratic divisor: ( x − 1) ( x − 2) = x2 − 3x + 2 Expand the linear factors Dividing P ( x) by x2 − 3x + 2: Divide: x4 ÷ x2 = x2 Multiply: x2 × ( x2 − 3x + 2) = x4 − 3x3 + 2x2 Subtract: ( x4 − 4x3 − x2) − ( x4 − 3x3 + 2x2) = −x3 − 3x2 Bring down: 16x, forming −x3 − 3x2 + 16x, then do it the same step Remainder: 0 (degree 0 to 2) The quotient is x2 − x − 6 with a remainder of 0. Factorising the quotient x2 − x − 6: Factorise by ﬁnding numbers that multiply to x2 − x − 6 = ( x − 3) ( x + 2) −6 and add to −1

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For the full factorisation: P ( x) = x4 − 4x3 − x2 + 16x − 12 = ( x − 1) ( x − 2) ( x − 3) ( x + 2)

Write the polynomial Factorise as linear factors

Solving P ( x) = 0: P ( x) = ( x − 1) ( x − 2) ( x − 3) ( x + 2)

Write the factored polynomial

0 = ( x − 1) ( x − 2) ( x − 3) ( x + 2) Set P ( x) = 0 x = 1, 2, 3, −2

Solve each factor for x

Reﬂect and check Verify by expanding ( x − 1) ( x − 2) ( x − 3) ( x + 2): ( x − 1) ( x − 2) ( x − 3) ( x + 2) = ( x2 − 3x + 2) ( x2 − x − 6) 4

Group and expand pairs of brackets

= x − 4x − x + 16x − 12

Expand fully and simplify

= P ( x)

The result matches the polynomial 2

The remainder is 0, which confirms that ( x − 3x + 2) is a factor of P ( x) and that the division is correct.

Idea summary Factor and remainder theorem have many applications extending to quartic polynomials and sketching.

Graph polynomials The factor theorem finds x-intercepts of a polynomial P ( x) at x = α where P (α ) = 0. For non-monic polynomials, rational roots (constant term divided by leading coefficient) are tested. The y-intercept is P (0). End behaviour depends on the degree and leading coefficient: for odd degree with positive leading coefficient, the graph rises as x → ∞ and falls as x → −∞. The multiplicity of a zero affects the graph’s behaviour at the intercept: multiplicity 1 means the graph crosses the x-axis, while higher multiplicities may indicate a turning point or inflection.

P ( x) = a( x − α1 ) ( x − α2 ) ⋯ ( x − αn) P ( x) a x − αi

the polynomial in factored form the leading coefficient linear factors giving x-intercepts at x = αi

Interactive exploration Discover this concept in action online

mathspace.co

3.03E Further applications mathspace.co

111

Example 2 Consider P ( x) = x4 − 2x3 − 12x2 + 8x + 32. a Determine the x-intercepts and the y-intercept.

Create a strategy Test integer factors of the constant term 32 to determine the roots. Divide P ( x) by each linear factor x − α where P (α ) = 0 using polynomial long division to ﬁnd all factors.

Apply the idea Testing x = 2: P ( x) = x4 − 2x3 − 12x2 + 8x + 32 4

Write the polynomial

P (2) = 2 − 2 × 2 − 12 × 2 + 8 × 2 + 32

Substitute x = 2

= 16 − 2 × 8 − 12 × 4 + 16 + 32

Evaluate each term

Simplify

Since P (2) = 0, x − 2 is a factor. Testing x = −2: P ( x) = x4 − 2x3 − 12x2 + 8x + 32 4

Write the polynomial 2

P (−2) = (−2) − 2 × (−2) − 12 × (−2) + 8 × (−2) + 32

Substitute x = −2

= 16 − 2 × (−8) − 12 × 4 − 16 + 32

Evaluate each term

Simplify

Since P (−2) = 0, x + 2 is a factor. Use polynomial long division to divide the polynomial by the two factors ( x − 2)( x + 2) = x2 − 4: Divide: x4 ÷ x2 = x2 Multiply: x2 × ( x2 − 4 ) = x4 − 4x2 Subtract: ( x4 − 2x3 − 12x2) − ( x4 − 4x2) = −2x3 − 8x2 Bring down: +8x, forming −2x3 − 8x2 + 8x, then repeat the process Remainder: 0 (degree 0 to 1) The quotient is x2 − 2x − 8. So, P ( x) = ( x2 − 4)( x2 − 2x − 8). To factorise x2 − 2x − 8, determine two numbers that multiply to −8 and add to −2. Those numbers are −4 and 2. x2 − 2x − 8 = ( x − 4) ( x + 2) Factorise For the full factorisation: P ( x) = x4 − 2x3 − 12x2 + 8x + 32 = ( x − 2) ( x + 2) ( x + 2) ( x − 4) 2

= ( x − 2) ( x + 2) ( x − 4)

112

Write the polynomial Fully factorise Rewrite with repeated factors

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

To determine the x-intercepts, use null factor law. P ( x) = ( x − 2) ( x + 2)2( x − 4)

Substitute P ( x) = 0

0, x − 4 = 0

Use null factor law

( x − 2) ( x + 2) ( x − 4) = 0 x−2

0, x + 2

Write the factorised polynomial

x = 2, −2, 4

Solve each factor for x

So the x-intercepts are at (2, 0), (−2, 0), (4, 0). For the y-intercept, evaluate P (0): P ( x) = x4 − 2x3 − 12x2 + 8x + 32 4

Write the polynomial

P (0) = 0 − 2 × 0 − 12 × 0 + 8 × 0 + 32

Substitute x = 0

= 32

Evaluate

So the y-intercept is at (0, 32). b Determine the end behaviour.

Create a strategy Consider if the degree is odd or even and if the leading coeﬃcient is positive or negative.

Apply the idea The degree is 4 (even) and the leading coeﬃcient is 1 (positive). So, as x → ∞, P ( x) → ∞ and as x → −∞, P ( x) → ∞. c Sketch the graph of P ( x).

Create a strategy Plot the points from part (a) and apply the end behaviour from part (b). Consider the multiplicity of roots for intercept behaviour (cut for power 1, bounce for power 2).

Apply the idea y 40 (0, 32) 30 20 10

(−2, 0) −6

−4

−2

−10

(2, 0) 2

x 4(4, 0)

−20 −30

P(x) = x4 − 2x3 − 12x2 + 8x + 32

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Idea summary The factor theorem identiﬁes x-intercepts by ﬁnding zeroes (rational for non-monic polynomials). The y-intercept, end behaviour, and multiplicity of zeroes determine the polynomial’s graph.

3.03E Practice questions What do you remember? 1

State the factor theorem for a polynomial P ( x) and a linear factor x − α.

What are the possible integer roots to test for a monic polynomial with a constant term of 8?

How is the y-intercept of a polynomial P ( x) determined?

Describe the end behaviour of a cubic polynomial with a positive leading coeﬃcient.

Practice Ex 1

Factorise each polynomial completely: a

x4 + 2x3 − 7x2 − 8x + 12

x4 − 2x3 − 3x2 + 8x − 4

4x4 − 12x3 + 13x2 − 12x + 4

2x5 − 10x4 + 12x3 + 4x2 − 8x

3x5 − 3x4 − 21x3 + 3x2 − 18x

x4 − 6x3 + 11x2 − 6x

x5 − 2x4 − 3x3 + 6x2

g 6

Ex 2

114

2x − 10x + 8

Solve each equation for x. a

x4 − x3 − 7x2 + x + 6 = 0

x4 − 2x3 − 13x2 + 14x + 24 = 0

x − 6x − 19x + 144x − 180 = 0

x4 − 4x3 + 16x − 16 = 0

x5 − x4 − 5x3 + x2 + 8x + 4 = 0

x4 − x3 − 37x2 − 47x + 84 = 0

x5 − 7x3 + 6x = 0

x5 + x4 − 9x3 − x2 + 8x = 0

For each polynomial: i

Determine the x-intercepts and the y-intercept.

Determine the end behaviour.

iii

Sketch the graph.

x3 − 4x2 + x + 6

x3 + 2x2 − 5x − 6

x − 5x + 4

x3 − x2 − 8x + 12

x4 − 8x2 + 12

x3 + x2 − 2x − 2

x3 − 3x2 − 4x + 12

x − 2x − 3x + 4x + 4

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Determine the value of k: a

P ( x) = x3 − 2x2 + kx − 6 has a factor x − 3.

P ( x) = 2x3 + x2 − 5x + k has a remainder of 6 when divided by x − 2.

P ( x) = x4 + kx3 − 2x2 + 3x − 4 has a root at x = 1.

P ( x) = x3 + 2x2 − kx + 12 has a remainder of 8 when divided by x + 2.

P ( x) = 2x4 − x3 + kx2 − 5x + 6 has a factor x − 3.

For each polynomial, determine the multiplicity of the root given and its effect on the graph. a

P ( x) = x3 − 4x2 + x + 6, root x = 2

P ( x) = x4 − 5x2 + 4, root x = −1

P ( x) = x4 − 8x2 + 12, root x =

P ( x) = x3 + x2 − 2x − 2, root x = −1

P ( x) = x4 − 2x3 − 3x2 + 4x + 4, root x = −1

For each polynomial, determine the required value: a

The volume of a rectangular box is modelled by V ( x) = x3 + 3x2 − 6x − 8 cubic metres, where x is the height. If one dimension is x − 2, ﬁnd the height when the volume is zero.

The area of a rectangular garden is modelled by A( x) = x3 − 4x2 − x + 4 square metres, where x is the width. If one dimension is x − 4, ﬁnd the width when the area is zero.

The profit of a company is modelled by P ( x) = x3 − 3x2 − 4x + 12 thousand dollars, where x is the number of units sold (in thousands). If one factor of the profit function related to sales volume is x − 3, find the number of units sold when the profit is zero.

Extend your thinking 11

The motion of a roller coaster’s height above the ground, h( x) in metres, at time x seconds, is modelled by the polynomial h( x) = x4 − 10x3 + 35x2 − 50x + 24. Factorise the polynomial to determine the times when the roller coaster is at ground level, and explain the signiﬁcance of these times in the context of the ride.

The population of a species in a habitat over time x years is modelled by the polynomial P ( x) = x5 − 2x4 − 13x3 + 14x2 + 24x thousand individuals. Factorise the polynomial to determine the years when the population is zero, and discuss the possible implications for the species’ survival.

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Identify and correct the errors in the following student work for factorising P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2 into linear and irreducible quadratic factors over the real numbers: a

Student A: P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2 4

Write the polynomial

P (1) = 2(1) − 5(1) + 4(1) − 5(1) + 2

Substitute x = 1

=2−5+4−5+2

Evaluate each term

= −2

Evaluate

P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2 4

Write the polynomial

P (−1) = 2(−1) − 5(−1) + 4(−1) − 5(−1) + 2

Substitute x = −1

=2+5+4+5+2

Evaluate each term

= 18

Evaluate

P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2 4

Write the polynomial

P (2) = 2(2) − 5(2) + 4(2) − 5(2) + 2

Substitute x = 2

= 32 − 40 + 16 − 10 + 2

Evaluate each term

Evaluate

P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2

Write the polynomial

Substitute x =

Evaluate each term Evaluate Divide by x − 2:

Factor the polynomial:

Final answer:

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(2x3 − x2 + 2x − 1).

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Student B: P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2 4

P (1) = 2 × 1 − 5 × 1 + 4 × 1 − 5 × 1 + 2

Write the polynomial Substitute x = 1

=2−5+4−5+2

Evaluate each term

= −2

Evaluate

P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2 4

P (2) = 2 × 2 − 5 × 2 + 4 × 2 − 5 × 2 + 2

Write the polynomial Substitute x = 2

= 32 − 40 + 16 − 10 + 2

Evaluate each term

Evaluate

Divide by x − 2:

Divide 2x3 − x2 + 2x − 1 by x − 1: Remainder 2 Final answer: ( x − 2)( x − 1)(2x2 + x + 3). c

Student C: P ( x) = 2x4 − 5x3 + 4x2 − 5x + 2 4

P (2) = 2 × 2 − 5 × 2 + 4 × 2 − 5 × 2 + 2

Write the polynomial Substitute x = 2

= 32 − 40 + 16 − 10 + 2

Evaluate each term

Evaluate

Divide by x − 2:

Factor the quotient: 2x3 − x2 + 2x − 1 = ( x − 1)( x + 1)(2x − 1) Final answer: ( x − 2)( x − 1)( x + 1)(2x − 1). 14

Find the remainder when P ( x) = 2x4 − 3x3 + 6x2 + x − 5 is divided by x2 − 7x + 12 without using long division.

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3 Chapter review 1

When the polynomial P ( x) = x3 − 2x2 + 5x − 3 is divided by x − 1, what is the remainder? A

−1

If P ( x) is a polynomial and it is found that P (−3) = 0, which of the following statements must be true? A

( x − 3) is a factor of P ( x).

When P ( x) is divided by ( x − 3), the remainder is 0.

( x + 3) is a factor of P ( x).

When P ( x) is divided by ( x + 3), the remainder is −3.

When P ( x) = x3 − x2 − 5x + 2 is divided by A( x) = x + 2, the result can be written as P ( x) = A( x)Q( x) + R( x). Which of the following correctly identiﬁes Q( x) and R( x)? A

Q( x) = x2 + x − 7, R( x) = 16

Q( x) = x2 − 3x + 1, R( x) = 0

Q( x) = x2 − 3x + 1, R( x) = 4

Q( x) = x2 + x − 3, R( x) = 8

Expressing your answer in the form P ( x) = A( x)Q( x) + R( x), divide: a

P ( x) = x3 − 7x2 + 10x − 6 by A( x) = x − 2

P ( x) = 3x3 + 2x2 − 5x + 4 by A( x) = x + 1

P ( x) = x4 − 2x3 + 3x2 − x + 7 by A( x) = x2 + x − 3

P ( x) = 2x3 − 9x2 + 12x − 7 by A( x) = x − 3

P ( x) = x3 + 5x2 − 10 by A( x) = x + 2

P ( x) = 2x4 − x3 − 2x2 + 6x − 5 by A( x) = x2 − x + 2

Find the remainder when P ( x) = 2x3 − 4x2 + 3x − 9 is divided by x − 2.

Given P ( x) = x3 + ax2 − 5x + 7, ﬁnd a if the remainder is 10 when P ( x) is divided by x − 1.

The polynomial P ( x) = 2x3 + kx2 − 7x + 10 has ( x − 2) as a factor. Find the value of k.

Given P ( x) = x3 − x2 − 8x + 12 and that ( x − 2) is a factor, ﬁnd the other linear factors.

The polynomial P ( x) = x3 + ax2 + bx + 10 has ( x + 2) as a factor. When P ( x) is divided by ( x − 1), the remainder is 18. Find the values of a and b.

The polynomial P ( x) = x4 + 3x3 − 3x2 + ax + b is divisible by ( x − 1) and ( x + 3): a

Determine the values of a and b.

Factorise P ( x) completely.

Solve the polynomial P ( x) = 2x4 − 9x3 + 6x2 + 11x − 6.

For the polynomial P ( x) = 2x4 + 3x3 − 6x2 − 5x + 6:

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Factorise P ( x) completely.

Sketch the graph of y = P ( x), showing all intercepts.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A polynomial P ( x) leaves a remainder of 7 when divided by x − 2 and a remainder of −5 when divided by x + 3. Find the remainder when P ( x) is divided by ( x − 2)( x + 3).

The proﬁt, in thousands of dollars, from selling x hundred items is modelled by P ( x) = x3 − 6x2 + 3x + 10 for x ≥ 0. a

Determine the values of x for which there is zero proﬁt.

What is the proﬁt or loss if no items are sold?

Determine the values of x for which a positive proﬁt is made.

Did you know?

Using the remainder and factor theorems is like finding the first book in a messy bookshelf series! Once you identify where the sequence starts, it becomes easier to organise and solve the rest. Each correct factor puts the next piece of the puzzle in place—just like sliding books into order. Just like recognising the first book helps you arrange the whole set, finding a single factor helps you break down a complex polynomial step by step. With each new factor, the equation becomes simpler, leading you closer to the full solution with logic and clarity.

Chapter 3 review mathspace.co

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4 Function properties and further polynomials Chapter outline 4.01 4.02 4.A 4.B 4.03E 4.04E

Further domain and range Even and odd functions Composite functions Piecewise functions Sums and products of zeroes Applications of sums and products of zeroes Chapter 4 review

122 130

138 151 158

Shoemakers use polynomial curves to shape the perfect sole — and zeroes help keep you grounded.

4.01 Further domain and range After this lesson, you will be able to… • extend the definitions of domain and range to relations. • recognise and use interval notation, inequalities and worded descriptions for domains and ranges. • determine the domain and range of functions and relations from their algebraic or graphical representations.

Domain and range notations Domain The set of allowable values of x in a function or relation. Range (function) The set of values of the dependent variable for which a function is deﬁned.

The domain and range of relations and functions can be expressed using interval notation, inequalities, or worded descriptions to precisely describe the set of possible inputs and outputs. Interval notation Notation for representing an interval on the real number line by its endpoints. Parentheses and/or square brackets are used respectively to show whether the endpoints are excluded or included. For a < b, (a, b) is an open interval, and [a, b] is a closed interval. For example, [1, 4] represents all real numbers from 1 to 4, inclusive, while (1, 4) excludes 1 and 4. Inequality notation employs symbols such as ≤ or <. For instance, [1, 4] is equivalent to 1 ≤ x ≤ 4, and (1, 4) corresponds to 1 < x < 4. Worded descriptions provide a verbal explanation, such as “all real numbers greater than 1 and less than or equal to 4.”

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Example 1 Express the domain [−2, 5) in: a Inequality notation

Create a strategy The x-value beside the square bracket is included, so use ≤ or ≥ , while the x-value beside the parenthesis is excluded, so use < or >.

Apply the idea The x-value beside the square bracket is − 2, so the lower boundary is − 2 ≤ x. The x-value beside the parenthesis is 5, so the upper boundary is x < 5. So, the inequality notation is −2 ≤ x < 5.

b Worded description

Create a strategy Use the fact that the domain of all real numbers is represented by x. Then use the inequality symbols used from part (a).

Apply the idea The domain of all real numbers greater than or equal to −2 and less than 5.

Example 2 The range of a relation is described as “All real numbers greater than or equal to −3 but less than 8.” Write this in: a Inequality notation

Create a strategy Use the fact that range of all real numbers is represented by y. Then, use the proper inequality symbols.

Apply the idea The ﬁrst part can be translated as y ≥ −3, so the lower boundary is −3 ≤ y. The second part can be translated as y < 8, which is the upper boundary. So, the inequality notation is −3 ≤ y < 8.

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b Interval notation

Create a strategy For ≤ or ≥ , use square brackets while parentheses for < or >.

Apply the idea [−3, 8)

Idea summary The domain of a function or relation is the set of all possible input values (x-values) for which the function or relation is deﬁned. The range is the set of all possible output values ( y-values) that result from the domain. Domain and range may be expressed using interval notation, inequalities, or worded descriptions. Each method conveys the same information in a different form.

Domain and range of complex relations Relations, unlike functions, may not pass the vertical line test, allowing multiple outputs for a single input. Determining the domain and range of complex relations involves analysing graphs, equations, or descriptions. For equations, the domain includes all x-values where the relation is deﬁned, accounting for restrictions such as division by zero, or negative square roots. The range includes all possible y-values produced by the relation. Similarly, the range can often be determined by rearranging the equation to make x the subject and applying the same restrictions, while taking into account the given domain. When the domain or range includes two or more intervals, the symbol ∪ (read as ‘union’) is often used to combine them into a single set when using interval notation. For example, the relation y =

is undeﬁned when x = 2, so the domain is x ≠ 2 or

(− ∞, 2) ∪ (2, ∞). The range is all real numbers except y = 0, as the function never equals zero, giving (− ∞, 0) ∪ (0, ∞).

Interactive exploration Discover this concept in action online

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mathspace.co

Example 3 For the relation y =

, determine in interval notation:

a The domain

Create a strategy Identify restrictions on x for the domain by determining where the denominator is zero.

Apply the idea The denominator x2 − 4 = (x − 2)(x + 2) is zero when x = 2 or x = −2, so the relation is undeﬁned at these points. Domain = (− ∞, −2) ∪ (−2, 2) ∪ (2, ∞) Express in interval notation

b The range

Create a strategy Solve for x in terms of y to determine the possible y-values for the range.

Apply the idea Write the equation

Multiply both sides by

Add 4 to both sides For x to be real,

+ 4 ≥ 0. Since

Take the square root of both sides ≥ − 4, then y ≤

or y > 0 when y ≠ 0.

Express in interval notation

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Reﬂect and check Graphing y =

shows vertical asymptotes at x = ± 2 and a horizontal asymptote at y = 0,

conﬁrming the domain and range. y 4 3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

Example 4 For the relation

, determine in interval notation:

a The domain

Create a strategy Solve for y in terms of x to determine the possible x-values for the domain.

Apply the idea Write the equation

Subtract x2 from both sides

Multiply both sides by 9

Take the square root of both sides

For y to be real, 1 − x2 ≥ 0, so x2 ≤ 1, meaning −1 ≤ x ≤ 1. Domain = [−1, 1] Express in interval notation

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b The range

Create a strategy Solve for x in terms of y to determine the possible y-values for the range.

Apply the idea Write the equation

Subtract

Take the square root of both sides

For x to be real, 1 −

from both sides

≤ 1, meaning y2 ≤ 9 or −3 ≤ y ≤ 3.

≥ 0, so

Range = [−3, 3] Express in interval notation

Idea summary The domain and range of complex relations are determined by analysing equations or graphs, identifying restrictions on x (domain) and possible y-values (range), and expressing them in interval notation, inequalities, or worded descriptions.

4.01 Practice questions What do you remember? 1

Determine whether each equation represents a function or just a relation: a

y = 2x + 1

x2 + y2 = 25

y = x2 − 2

What do square brackets [ ] and parentheses ( ) mean in interval notation?

Deﬁne the term “domain” in the context of a relation.

Determine whether these statements are true or false:

xy = 6

Interval notation uses only parentheses ( ) to describe sets of real numbers.

Inequality notation can represent the same set as interval notation, such as x > 3 being equivalent to (3, ∞).

A worded description is not a valid way to express the range of a relation.

The range of a relation includes all possible y-values produced by the relation for valid x-values in the domain.

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Practice Ex 1

Express the given interval notation of the domain in: i

Ex 2

Ex 3

Ex 4

128

Worded description

[0, 4)

(−3, 1]

[−2, 5]

(1, 7)

(− ∞, 0]

[3, ∞)

Interval notation

Convert the worded descriptions to: i

Inequality notation

The domain of real numbers greater than −2 but less than or equal to 8.

The range of all real numbers greater than or equal to 5 but less than 12.

The range of all real numbers less than 4.

The domain of all real numbers greater than or equal to −1.

Determine the domain of each relation based on the equation: a

Inequality notation

Convert the inequality notation to interval notation: a

−1 ≤ x < 7

x>−5

0 < x ≤ 4

x≤0

−3 < x ≤ 2

x≥6

Convert the interval notation to inequality notation: Range of (− 4, ∞)

Domain of [2, 8)

Domain of (− ∞, 1]

Range of (−2, 3]

Domain of [0, 5]

Range of (1, ∞)

For each relation, determine in interval notation: i

The domain

The range

y = x2 + 3

y = − x2 + 2

y = 2 − x2

y = x3 + 2

y = −(x − 1)3

y = 2x3 + 1

Determine in interval notation: i

The domain

The range

x2 + y2 = 25

x2 + y2 = 16

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Determine the domain and range of each relation based on the given description: a

A semicircle with radius 3, centred at the origin, above the x-axis.

A parabola opening upwards with vertex at (0, −2).

A semicircle with radius 2, centred at the origin, below the x-axis.

A parabola opening downwards with vertex at (0, 4).

A line segment from (−1, 2) to (3, 4).

A circle with radius 1, centred at the origin.

For each set of ordered pairs: i

Determine the domain.

Determine the range.

iii

Identify whether it represents a function or relation.

A = {(1, 2), (1, 3), (3, 4), (3, 5), (5, 5)}

B = {(−2, 0), (0, 1), (2, 2), (4, 3)}

C = {(0, 1), (1, 1), (2, 1), (3, 1)}

D = {(−1, −1), (−1, 1), (0, 0), (1, −1), (1, 1)}

Determine the domain and range in interval notation for the relation deﬁned by y2 = −2x2 + 4.

Determine the domain and range in interval notation for the relation deﬁned by

A region is deﬁned as the space bounded between the intersection points of y = x2 and y = 3 − x2. Determine in interval notation, the domain and range of this region.

Extend your thinking 17

A relation is deﬁned by

Determine the domain in interval notation.

Determine the range in interval notation.

Explain why x = 2 is excluded from the domain.

Explain why y = 0 is excluded from the range.

Consider the relation y2 = x: a

Determine the domain in interval notation.

Determine the range in interval notation.

Explain why this relation is not a function.

Describe how the graph of this relation differs from y =

A relation is described as all points on a circle with radius 2, centred at (1, −1): a

Write the equation of the circle.

Determine the domain in interval notation.

Determine the range in interval notation.

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A student states that the range of y = x2 − 4 is (− 4, ∞): a

Is the student’s answer correct? Why or why not? If not, state the correct range.

Describe how the graph conﬁrms this range.

Consider the relation y =

Find the domain in interval notation.

Find the range in interval notation.

Explain why the range does not include y = 0.

Determine the maximum value of y and when it occurs.

4.02 Even and odd functions After this lesson, you will be able to… • define an even function by its reflectional symmetry in the y-axis. • define an odd function by its rotational symmetry of 180° about the origin. • use the algebraic tests f (− x) = f (x) and f (− x) = − f (x) to classify functions. • solve problems involving even and odd functions.

Even and odd functions Functions are classified as even, odd, or neither based on their symmetry properties. These properties are defined algebraically and can be observed graphically. Even function A function is even if its graph is unchanged under reflection in the y-axis. An even function f (x) has the property f (− x) = f (x), for all values of x in the domain. Odd function A function is odd if its graph is unchanged under rotation of 180° about the origin. An odd function f (x) has the property f (− x) = −f (x), for all values of x in the domain.

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mathspace.co

y 8

f (x) = x2

6 4 2

−2

−1

x 1

−2

−4

An example of an even function is f (x) = x2. This function is symmetric about the y-axis, meaning that for every point (x, y) on the graph, the point (− x, y) also lies on the graph.

−6 −8 y 8 6 4 3 2

f (x) = x −2

−1

−2

x 1

−4

An example of an odd function is f (x) = x3. Its graph is unchanged after a 180° rotation about the origin, meaning that for every point (x, y) on the graph, the point (− x, − y) also lies on the graph.

−6 −8

Functions that do not satisfy either condition are classiﬁed as neither even nor odd.

Example 1 Determine whether the function f (x) = x4 − 2x2 + 5 is even, odd, or neither.

Create a strategy Substitute − x into f (x), then use the fact that if f (− x) = f (x), the function is even. If f (− x) = −f (x), the function is odd. Otherwise, it is neither.

Apply the idea f (x) = x4 − 2x2 + 5

Write the function

f (− x) = (− x)4 − 2(− x)2 + 5

Substitute x = − x

= x − 2x + 5

Evaluate each term

= f (x)

Substitute x4 − 2x2 + 5 = f (x)

Since f (− x) = f (x), the function is even.

Reﬂect and check The function has only even powers of x and a constant term, which typically indicates an even function, as conﬁrmed algebraically.

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Example 2 Determine whether the function f (x) =

is even, odd, or neither.

Create a strategy Determine f (− x) and compare it with f (x) and −f (x) to classify the function.

Apply the idea Write the function Substitute x = − x

Evaluate the power

Substitute

= f (x)

Since f (− x) = −f (x), the function is odd.

Example 3 Given that f (x) = x2 − 3 is an even function, determine the coordinates of the point on the graph of f (x) that corresponds to x = 2 under the symmetry property.

Create a strategy Since f (x) is even, use the property f (− x) = f (x) to determine the point symmetric to x = 2 then substitute into f (x).

Apply the idea Since the function is even, the symmetric point for x = 2 is at x = −2. Evaluate f (−2): f (x) = x2 − 3

Write the function

f (−2) = (−2) − 3

Substitute x = −2

=4−3

Evaluate the power

Evaluate

The corresponding point is (−2, 1).

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Idea summary The function f (x) is an even function if f (− x) = f (x), exhibiting reﬂective symmetry across the y-axis. It is an odd function if f (− x) = −f (x), exhibiting 180° rotational symmetry about the origin. Functions that satisfy neither condition are neither even nor odd.

Did you know?

Functions are used in car design to model how different parts move and respond under real-world conditions! Engineers use mathematical functions to simulate aerodynamics, suspension, and steering performance before building a physical prototype.

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4.02 Practice questions What do you remember? 1

Identify the algebraic condition for a function to be: a

Even

Odd

Identify the symmetry associated with each type of function: a

What type of symmetry does an even function exhibit?

What type of symmetry does an odd function exhibit?

Determine whether these statements are true or false: a

All even functions pass through the origin.

All odd functions pass through the origin.

A function can be both even and odd.

Practice 4

Determine whether each function is even, odd, or neither: a

−2

x 1

−2

−1

−4

−6

−8

−2

x 1

−2

−1

−2

−4

−6

−8

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−2

−4

−2

134

−1

−2

−1

x 1

−2

−1

−4 −6

−8

−2

x −2

−1

−2

−6

−4

−6

−8

−2

−4

Determine whether these functions are even, odd, or neither: a

Ex 2

−2

−4

−5 −4 −3 −2 −1

Ex 1

f (x) = 3x4 4

f (x) = −2x5

f (x) = 2x + 5

f (x) = x3 − x

f (x) = x2 + 2x

f (x) = 3x5 + x3

f (x) = 4x4 − 2x2 + 5

f (x) = 3x3 + 6x − 2

Determine whether these functions are even, odd, or neither: a

For each function: i

Simplify

Determine if even, odd, or neither

f (x) = x2 + x2

f (x) = x3 − x3

f (x) = x4 − x3 + x3

f (x) = x5 + x3 − x4

Determine whether these functions are even, odd, or neither using algebraic methods: a

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Ex 3

For each function, determine the coordinates of the point on the graph that corresponds to the given x-value under the symmetry property: a

Given that f (x) = x4 + 1 is an even function, point at x = 3

Given that f (x) = x3 − x is an odd function, point at x = 2

Match each graph to its function based on its symmetry properties: i

f (x) = x4 − 2x2 3

−1

x 1

−2

−1

−4

−6

−8

d 8

−2

−1

−2

−4

−6

−8

For each function, determine: i

f (− x)

Hence, determine if even, odd, or neither

f (x) = x8 − 4x4 + 2

f (x) = x5 − 3x

f (x) = x3 + x2

f (x) = 7x9 + x5

For each function: i

Simplify into one fraction.

Determine if even, odd, or neither.

x 1

−2

−4

−2

136

f (x) = x2 + x + 1

f (x) = x + x

−2

f (x) = −2

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x 1

Extend your thinking 13

A function f (x) is known to be odd. Prove that it must pass through the origin.

Consider the function f (x) = x2 + k, where k is a constant: a

Determine the values of k for which f (x) is even.

Can f (x) ever be odd? Explain.

A student claims that f (x) = x3 + 2 is an odd function because of the degree 3. Identify and correct the error in the reasoning.

Suppose f (x) is an even function and g(x) is an odd function. Determine whether these functions are even, odd, or neither: a

h(x) = f (x) + g(x)

h(x) = f (x) × g(x)

Explore whether a function can be both even and odd: a

Determine whether there exists a non-trivial function (i.e., not identically zero) that is both even and odd. Explain your reasoning.

Provide an example of a function that is both even and odd, and verify its properties algebraically.

Did you know?

Some flowers, like daisies, show mirror symmetry just like even functions, while pinwheels show rotational symmetry like odd functions. Mathematicians use these function properties to classify patterns in nature.

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4.03E Sums and products of zeroes After this lesson, you will be able to… • prove that if a quadratic P ( x) = ax2 + bx + c has zeroes α and β then

α+β=

, the sum of the zeroes, and αβ =

, the product of the zeroes

• use the formulas for the sum and product of zeroes of a quadratic equation to solve problems • state and use the formulas for the sum of zeroes (singly, pairwise, and three at a time for quartics) and the product of all zeroes for cubic and quartic equations • apply the formulas for the sums and products of zeroes to solve problems involving coefficients and zeroes of cubic and quartic polynomials

Quadratic roots Quadratic expression An expression of the form ax2 + bx + c, where a ≠ 0, b and c are constants. Polynomial An expression made up of non-negative integer powers of the same variable and coeﬃcients combined using addition, subtraction and multiplication. Degree (of a polynomial) The highest power of x that appears in a polynomial P ( x). Root The solution to an equation. For example, ( x − 1) ( x + 3) = 0 has roots x = 1 and x = −3. Zero polynomial A polynomial which has all coeﬃcients equal to zero.

A quadratic equation ax2 + bx + c = 0 has roots α and β, the values of x where the function y = ax2 + bx + c crosses the x-axis. The sum and product of zeroes are given by:

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These arise from the factored form a( x − α ) ( x − β ) = ax2 + bx + c. Expanding ( x − α ) ( x − β ) gives x2 − ( α + β )x + αβ, and equating coeﬃcients with

yields the formulas.

Exploration For x2 − 5x + 6 = 0, calculate the sum and product of zeroes using the coefficients. 1. How does the sum change if the coefficient of x becomes 10? 2. If the constant term is changed to 8 instead of 6, how does the product of the roots change, and what does this imply about zeroes’ relationship? 3. How would doubling the coefficient of x2 (from 1 to 2) affect the sum and product of zeroes, and why might this matter when comparing the original and new quadratics?

Example 1 For x2 − 3x + 7 = 0 with roots α and β, determine: a α+β

Create a strategy Substitute b = −3 and a = 1 into α + β =

Apply the idea Write the sum of zeroes formula

Substitute b = −3 and a = 1

Evaluate

b αβ

Create a strategy Substitute c = 7 and a = 1 into αβ = .

Apply the idea

Write the product of zeroes formula

Substitute c = 7 and a = 1

Evaluate

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c α2 + β2

Create a strategy Substitute α + β = 3 from part (a) and αβ = 7 from part (b) into α 2 + β 2 = ( α + β )2 − 2αβ.

Apply the idea α 2 + β 2 = ( α + β )2 − 2αβ

Write the identity

= 3 − 2 × 7 Substitute α + β = 3 and αβ = 7 = −5 Evaluate

Reﬂect and check The discriminant ∆ is: Write the discriminant formula ∆ = b2 − 4ac = (−3)2 − 4 × 1 × 7 Substitute a = 1, b = −3 and c = 7 = −19 Evaluate Since the discriminant, −19, is negative, zeroes are not real, so α 2 + β 2 can be negative.

Example 2 For x2 + mx + 18 = 0, determine: a The sum of zeroes in terms of m.

Create a strategy Substitute b = m and a = 1 into α + β =

Apply the idea Write sum of zeroes formula

Substitute b = m and a = 1

Simplify

b The product of zeroes.

Create a strategy Substitute c = 18 and a = 1 into αβ = .

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Apply the idea

Write product of zeroes formula

Substitute c = 18 and a = 1

Evaluate

c The value of m when the product is 6 times the sum.

Create a strategy Set αβ = 6 × ( α + β ) and solve for m using the sum and product from parts (a) and (b).

Apply the idea From part (a), α + β = −m. From part (b), αβ = 18.

αβ = 6 × ( α + β )

Write the condition

18 = 6 × (−m)

Substitute αβ = 18 and α + β = −m

18 = −6m

Evaluate the multiplication

−3 = m

Divide both sides by −6

m = −3

Make m the subject

Idea summary A quadratic equation ax2 + bx + c = 0 has roots α and β, the values of x where the function y = ax2 + bx + c crosses the x-axis. The sum and product of zeroes are given by:

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Cubic and quartic roots A cubic equation ax3 + bx2 + cx + d = 0 has roots α, β, γ. The relationships are:

A quartic equation ax4 + bx3 + cx2 + dx + e = 0 has roots α, β, γ, d :

Notice that the signs alternate: • sum of zeroes is

• triple products are

• pairwise products are

• and the product is

Exploration For a cubic x3 − 3x2 + 4x − 2 = 0, calculate α + β + γ and αβγ using the coefficients. 1. How do the values change if the constant term becomes 2? 2. If the coefficient of x2 is changed to 0 instead of −3, how does the sum of zeroes change, and what does this imply about the nature of zeroes? 3. How would halving the coefficient of x (from 4 to 2) affect the sum of the pairwise products αβ + βγ + γα, and why might this be significant for the polynomial’s behaviour?

Example 3 For P ( x) = x3 − 2x2 + 3x − 5 with roots α, β, γ, determine: a ( α − 2) ( β − 2) ( γ − 2)

Create a strategy Substitute a = 1, b = −2, c = 3, d = −5 into α + β + γ = ( α − 2) ( β − 2) ( γ − 2).

142

, αβ + βγ + γα = , αβγ =

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to expand

Apply the idea For α + β + γ : Write the formula

Substitute b = −2 and a = 1

Evaluate

For αβ + βγ + γα: Write the formula

Substitute c = 3 and a = 1

Evaluate

For αβγ : Write the formula

Substitute d = −5 and a = 1

Evaluate

For ( α − 2) ( β − 2) ( γ − 2): ( α − 2) ( β − 2) ( γ − 2) = αβγ − 2( αβ + βγ + γα ) + 4( α + β + γ ) − 8

Expand and group terms

=5−2×3+4×2−8

Substitute the values

=5−6+8−8

Evaluate the multiplication

= −1

Evaluate

Create a strategy Rewrite

using the cubic root formulas.

Apply the idea From part (a), αβ + βγ + γα = 3 and αβγ = 5.

Combine fractions Substitute αβ + βγ + γα = 3 and αβγ = 5

Reﬂect and check The result

is consistent with the ratio of the pairwise product sum to the product of roots.

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Example 4 For P ( x) = x4 − 4x3 − kx2 + 9x − 9 with roots α, β, γ, d, determine k if d = −α.

Create a strategy Substitute a = 1, b = −4, c = −k, d = 9, and e = −9 into the sum and product formulas for quartic roots. Then substitute the condition δ = −α into the formulas and solve the resulting system of equations to find the value of k.

Apply the idea For α + β + γ + d : Write the formula

Substitute b = −4, a = 1 and d = −α

Evaluate

For αβγ + αβd + αγd + βγd : Write the formula

Substitute d = 9, a = 1 and d = −α

Evaluate both sides

Evaluate αβγ − αβγ = 0

Factor out the common term

Substitute β + γ = 4

Divide both sides by −4

For αβγd : Write the formula Substitute e = −9, a = 1 and d = −α

Evaluate both sides

Substitute α 2 =

Multiply both sides by 4

Divide both sides by −9

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For αβ + αγ + αd + βγ + βd + γd : Write the formula

Substitute c = −k, a = 1 and d = −α

Evaluate

Remove the common terms

Substitute a2 =

Multiply both sides by −1

Evaluate

and βγ = 4

Idea summary A cubic equation ax3 + bx2 + cx + d = 0 has roots α, β, γ. The relationships are:

A quartic equation ax4 + bx3 + cx2 + dx + e = 0 has roots α, β, γ, d. The relationships are:

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Sigma notation The Greek letter (∑) (sigma) means “sum of “. Sigma notation is a faster way of writing out the sum of the roots one at a time, two at a time, etc. A quadratic equation ax2 + bx + c = 0 has two roots, α and β. ∑α

Sum of the roots one at a time

∑ αβ

Sum of the roots two at a time (product of roots)

A cubic equation ax3 + bx2 + cx + d = 0 has three roots, α, β, and γ. ∑α

Sum of the roots one at a time

∑ αβ

Sum of the roots two at a time

∑ αβγ

Sum of the roots three at a time (product of roots)

A quartic equation ax4 + bx3 + cx2 + dx + e = 0 has four roots, α, β, γ, and d. ∑α

Sum of the roots one at a time

∑ αβ

Sum of the roots two at a time

∑ αβγ

Sum of the roots three at a time

∑ αβγd

Sum of the roots four at a time (product of roots)

Example 5 For P ( x) = 3x4 − 2x3 + 5x − 7 with roots α, β, γ, and d, determine: a ∑α

Create a strategy

Apply the idea

Substitute the values a = 3 and b = −2 into the

Write the formula

formula: ∑ α =

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b ∑ αβ

Create a strategy

Apply the idea

Substitute the values a = 3 and c = 0 into the

Write the formula

formula: ∑ αβ =

Substitute the values

Evaluate

c ∑ αβγ

Create a strategy

Apply the idea

Substitute the values a = 3 and d = 5 into the

Write the formula

formula: ∑ αβγ =

Substitute the values

d ∑ αβγd

Create a strategy

Apply the idea

Substitute the values a = 3 and e = −7 into the

Write the formula

formula: ∑ αβγd =

Substitute the values

e ∑ α2

Create a strategy Derive a formula for ∑ α 2 using ∑ α and ∑ αβ. Substitute values from parts (a) and (b).

Apply the idea Derive the sum using the identity and substitute known values. Write the formula Apply the identity for the sum of squares

Simplify using sum notation

Using the answers from parts (a) and (b): Write the formula Substitute the values Evaluate

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Create a strategy Derive a formula for

using ∑ αβγ and ∑ αβγd. Substitute values from parts (c) and (d).

Apply the idea Derive the sum using the identity and substitute known values. Write the formula

Express the sum as a single fraction

Simplify using sum notation

Using the answers from parts (c) and (d): Write the formula

Substitute the values

Evaluate

Idea summary Sigma notation is a faster way of writing out the sum of the roots one at a time, two at a time, etc.

4.03E Practice questions What do you remember? 1

Deﬁne the following terms related to polynomials and their roots: a

148

Root

Write the corresponding formula for a quadratic polynomial ax2 + bx + c = 0 with roots α and β : a

Polynomial

Sum of roots

Product of roots

Determine the key fact about the relationship between the coeﬃcients of a quadratic polynomial ax2 + bx + c = 0 and its roots α and β in terms of the factored form.

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Practice Ex 1

Ex 2

For P ( x) = 2x2 + x − 4 with roots α and β, calculate the value of: a

α + β

(α − 2) (β − 2)

α2 + β2

For P ( x) = x2 − 2x + 7 with roots α and β, calculate the value of:

a Ex 3

(α + 4) (β + 4) (γ + 4)

α 2 + β 2 + γ 2 d Hint: Start with (α + β + γ )2. α2 β2 + β2 γ2 + γ2 α2

c 8

α3 + β3 + γ3 Hint: Start with (α + β + γ )3.

α + β + γ + d The value of k if d = −α.

αβγd

For P ( x) = x4 − 2x3 + 5x2 + 9x − 2 with roots α, β, γ, and d, calculate the value of: a

α2 + β2

For P ( x) = x4 − 3x3 + kx2 + x − 4 with roots α, β, γ, and d, determine: a

Ex 5

For P ( x) = x3 − 4x2 + 5x + 9 with roots α, β, and γ, calculate the value of:

e Ex 4

αβ

∑α ∑(1 − α )

∑ αβ

∑α

∑ αβγ

∑ αβγd

For α 3 + β 3 = ( α + β ) ( α 2 − αβ + β 2): a

Prove α 3 + β 3 = (α + β ) (α 2 − αβ + β 2).

Calculate α 3 + β 3 for x2 + 4x − 6 = 0 with roots α and β.

For ( α − β )2 = ( α + β )2 − 4αβ : a

Prove (α − β )2 = (α + β )2 − 4αβ.

Calculate α − β for x2 − 4x + 2 = 0 with roots α and β.

Determine the roots of these cubic polynomials given the speciﬁed conditions: a

2x3 − 11x2 + 17x − 6 = 0 has roots α, β, γ, with the condition that α = 2.

x3 − 6x2 + 11x − 6 = 0 has roots α, β, γ, with the condition that one root equals the sum of the other two.

The cubic 3x3 − 11x2 + kx − 4 = 0 has roots α, β, γ, with the condition that Determine k.

For x4 − 8x3 + 23x2 − 28x + 12 = 0 with roots α, β, γ, d, such that α + d = β + γ. Determine the roots given they are all integers.

= 3.

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For the cubic polynomial 2x3 − mx2 + x − 6 = 0 with roots α, β, and γ, determine: a

The sum of the zeroes in terms of m.

The sum of the pairwise products.

The value of m if ∑ α = ∑ αβγ.

For the quadratic polynomial 3x2 + 2x − 5 = 0 with roots α and β, determine: a

(α + 1)(β + 1)(γ + 1)

α2 + β2 + γ2

∑ α

∑ αβ

For x3 + 4x2 − 3x + 1 = 0 with roots α, β, and γ, calculate the value of: a c

α2 β + β2 α

For x2 − 2x + 7 = 0 with roots α and β, calculate the value of: a

For the cubic polynomial x3 − 6x2 + 11x − 6 = 0 with roots α, β, and γ, determine: a

α 3 + β 3

α + β + γ αβγ

αβ + βγ + γα

For x4 − 3x3 + 2x2 − 6x − 8 = 0 with roots α, β, γ, and d, calculate the value of: a

∑α

∑ αβ

∑ αβγ

∑ αβγd

Extend your thinking 20

Prove the formulas for the sum and product of zeroes for these polynomials: a

A quadratic P ( x) = ax2 + bx + c with roots α and β.

A cubic P ( x) = ax3 + bx2 + cx + d with roots α, β and γ.

A quartic P ( x) = ax4 + bx3 + cx2 + dx + e with roots α, β, γ and d.

Determine the monic polynomial P ( x) with roots α, β and γ, given the roots satisfy: • • ∑( α − 1) = 4 • αβγ = 2

For a cubic P ( x) = x3 − 2x2 + 7x − 3 with roots α, β and γ, determine: a

α 3 + β 3 + γ 3

α4 + β4 + γ4

Construct a quartic polynomial x4 + ax3 + bx2 + cx + d = 0 which roots α, β, γ, and d satisfy α + β + γ + d = 4 and αβγd = 1. Explain your reasoning.

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4.04E Applications of sums and products of zeroes After this lesson, you will be able to… • recall and use the formulas for the sum and product of roots for quadratic, cubic, and quartic polynomials. • apply the factor theorem to identify roots of polynomial equations. • use known root(s) and relationships between roots and coefficients to find the sum and product of the remaining roots. • form and solve a simpler polynomial (typically quadratic) to determine the remaining roots. • solve cubic and quartic polynomial equations and factorise them into linear factors using these combined techniques.

Applications of sums and products Factor theorem If p( x) is a polynomial and p(a) = 0 for some number a, then p( x) is divisible by x − a. The factor theorem can be used to obtain factors of a polynomial. Long division for polynomials can be tedious and error-prone. Using the sum and product of zeroes, we can factorise and solve polynomials without long division, unless explicitly required.

Interactive exploration Discover this concept in action online

mathspace.co

The relationships between the roots and coeﬃcients of polynomial equations are essential tools. Here is a summary of these formulas: For a quadratic equation ax2 + bx + c = 0 with roots α and β : Relationship

Formula

Sum of the roots: α + β Product of the roots: αβ

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For a cubic equation ax3 + bx2 + cx + d = 0 with roots α, β, and γ : Relationship

Formula

Sum of the roots: α + β + γ Sum of the products of the roots taken two at a time: αβ + βγ + γα Product of the roots: αβγ For a quartic equation ax4 + bx3 + cx2 + dx + e = 0 with roots α, β, γ, and d : Relationship

Formula

Sum of the roots: α + β + γ + d Sum of the products of the roots taken two at a time: αβ + αγ + αd + βγ + βd + γd Sum of the products of the roots taken three at a time: αβγ + βγd + γdα + dαβ Product of the roots: αβγd If a sufficient number of zeroes are known (e.g., one for a cubic, two for a quartic), we can use these relationships to find the remaining zeroes by forming a lower-degree polynomial.

Example 1 Factorise P ( x) = x3 − 6x2 + 11x − 6 as a product of linear factors.

Create a strategy Use the factor theorem to conﬁrm a zero, then apply sum and product of zeroes to ﬁnd the other zeroes and form a quadratic to factorise.

Apply the idea Using the factor theorem, test factors of the constant term −6: ±1, ±2, ±3, ±6. For this case, substitute x = 2 into P ( x) = x3 − 6x2 + 11x − 6: P ( x) = x3 − 6x2 + 11x − 6 3

P (2) = 2 − 6 × 2 + 11 × 2 − 6

Write the polynomial Substitute x = 2

= 8 − 24 + 22 − 6

Evaluate each term

Evaluate

Since P (2) = 0, use it to substitute γ = 2.

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For the sum of zeroes: Write the formula Substitute γ = 2, b = −6 and a = 1

Evaluate the substitution

Subtract 2 from both sides

For the product of zeroes: Write the formula

Substitute γ = 2, d = −6 and a = 1 Evaluate

Divide both sides by 2

Substitute the obtained values into the form x2 − ( α + β )x + αβ = 0: x2 − ( α + β )x + αβ = 0

Write the quadratic equation in sum and product form

x − 4x + 3 = 0

Substitute α + β = 4 and αβ = 3

( x − 1) ( x − 3) = 0

Factorise

Since x = 2 is also a zero, another factor is x − 2. P ( x) = x3 − 6x2 + 11x − 6

Write the polynomial

= ( x − 1) ( x − 2) ( x − 3)

Rewrite as linear factors

The factorised form of the polynomial is ( x − 1) ( x − 2) ( x − 3).

Reﬂect and check ( x − 1) ( x − 2) ( x − 3) = ( x − 1)( x2 − 5x + 6) 2

Expand ( x − 2) ( x − 3) 2

= x( x − 5x + 6) − 1( x − 5x + 6) 3

= x − 5x + 6x − x + 5x − 6 = x − 6x + 11x − 6 P ( x) = x3 − 6x2 + 11x − 6

Distribute ( x − 1) over the trinomial Expand Collect like terms The result matches the polynomial

Example 2 Solve x4 − 4x3 − x2 + 16x − 12 = 0.

Create a strategy Use the factor theorem twice to ﬁnd two zeroes, then use sum and product of zeroes to form a quadratic for the remaining zeroes.

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Apply the idea Using the factor theorem, test factors of the constant term −12: ±1, ±2, ±3, ±4, ±6, ±12. Substitute x = 1 into P ( x) = x4 − 4x3 − x2 + 16x − 12: P ( x) = x4 − 4x3 − x2 + 16x − 12 4

Write the polynomial

P (1) = 1 − 4 × 1 − 1 + 16 × 1 − 12

Substitute x = 1 into the polynomial

= 1 − 4 − 1 + 16 − 12

Simplify each term

Evaluate

Since P (1) = 0, x = 1 is a zero. Now test x = 2: P ( x) = x4 − 4x3 − x2 + 16x − 12 4

Write the polynomial

P (2) = 2 − 4 × 2 − 2 + 16 × 2 − 12

Substitute x = 2

= 16 − 32 − 4 + 32 − 12

Simplify each term

Evaluate

Since P (2) = 0, x = 2 is also a zero. Use these zeroes ( γ = 1, d = 2) to ﬁnd the remaining zeroes. For the sum of zeroes: Write the formula

Substitute γ = 1, d = 2, b = −4 and a = 1

Evaluate the substitution

Subtract 3 from both sides

For the product of zeroes: Write the formula Substitute γ = 1, d = 2, e = −12 and a = 1

Evaluate the substitution

Divide both sides by 2

Substitute the obtained values into the form x2 − (sum of zeroes)x + product of zeroes = 0: x2 − (sum of zeroes)x + product of zeroes = 0 Write the quadratic equation for the remaining zeroes x2 − ( α + β )x + αβ = 0 Substitute the sum and product of the remaining zeroes x2 − 1x + (−6) = 0 2

Substitute α + β = 1 and αβ = −6

x −x−6=0

Simplify

( x − 3) ( x + 2) = 0

Factorise

The zeroes of the resulting quadratic are 3 and −2. Thus, the solutions to the original equation are −2, 1, 2 and 3.

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Idea summary The sum and product of zeroes allows us to solve polynomial equations eﬃciently, bypassing long division by forming a quadratic for the remaining zeroes.

4.04E Practice questions What do you remember? 1

Consider a general quadratic polynomial P ( x) = ax2 + bx + c with roots α and β. Express the following in terms of the coeﬃcients a, b and c: a

The sum of the roots (α + β )

The product of the roots (αβ )

Let a general cubic polynomial be P ( x) = ax3 + bx2 + cx + d, with roots α, β, and γ. Express each relationship in terms of the coeﬃcients a, b, c and d: a

The sum of the roots (α + β + γ )

The sum of the products of the roots taken two at a time (αβ + αγ + βγ )

The product of the roots (αβγ )

For a general quartic polynomial P ( x) = ax4 + bx3 + cx2 + dx + e, let the roots be α, β, γ, and d. Express the following expressions in terms of the coeﬃcients a, b, c, d and e: a

The sum of the roots (∑ α )

The sum of the products of the roots taken two at a time (∑ αβ )

The sum of the products of the roots taken three at a time (∑ αβγ )

The product of the roots (αβγd)

Practice 4

If P ( x) = x2 − 4x + 3 has roots 1 and β, determine β using the sum of roots.

If P ( x) = x3 − 4x2 − x + 4 has roots 1, 4, and γ, determine γ using the product of roots.

If P ( x) = x4 − 9x2 − 4x + 12 has roots 1, −2, 3, and d, determine d using the sum of roots.

A polynomial has the given sum and product of roots. Find the monic polynomial in the form speciﬁed. a

Quadratic, sum of roots 4, product of roots 3

Quadratic, sum of roots −2, product of roots −5

Cubic, sum of roots 3, sum of pairwise products 2, product of roots −1

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Ex 1

Factorise the following polynomials as products of linear factors without using long division. b

P ( x) = x3 − 6x2 − x + 30

P ( x) = x − 9x + 23x − 15

P ( x) = x3 + 8x2 + 17x + 10

P ( x) = 4x3 − x2 − 29x + 30

P ( x) = x3 − 7x2 + 14x − 8

Ex 2

P ( x) = x3 + 2x2 − 13x + 10 3

Factorise each monic quartic polynomial completely into linear factors without using long division. a

x4 − 6x3 + 11x2 − 6x

x4 + 2x3 − 7x2 − 8x + 12

x + 2x − 21x − 22x + 40

x4 − 5x3 − 7x2 + 29x + 30

x4 − 22x2 + 24x + 45

x4 + x3 − 7x2 − x + 6

A cubic polynomial has roots −1, p, and q, where p and q are real numbers such that p + q = 3 and pq = 2. If the polynomial is monic, determine its coeﬃcients.

For the parabola y = x2 − 2x − 5 that has roots α, β where α > β, ﬁnd: a

The roots

The vertex

Determine the value of the unknown coeﬃcient(s) for each polynomial, and then solve the equation. a

The equation x3 + 6x2 + kx − 10 = 0 has roots −1, α and β. Find k and the other roots.

The equation 2x3 + ax2 + bx + 4 = 0 has roots 2 and −1. Find a, b, and the third root.

Solve x3 − 10x2 + 31x − 30 = 0, given that one root is the sum of the other two.

Solve 3x3 − 13x2 + 13x − 3 = 0, given that one root is the reciprocal of another.

The quadratic equation 2x2 − 4x + 1 = 0 has roots α and β. Without solving the equation, calculate the value of: a

α2 + β2

(α − β )2

The cubic equation x3 − 3x2 − 10x + 24 = 0 has roots α, β and γ. Form a monic cubic equation with roots: a

2α, 2β, 2γ

α + 1, β + 1, γ + 1

Extend your thinking 15

For the parabola y = x2 − 2x − 5 that has roots α, β where α > β : a

Show that the area of the triangle formed by the x-intercepts and vertex in terms of the roots is A = 3(α − β ).

Calculate the exact area.

The cubic polynomial y = x3 − 6x2 + 11x + b intersects the line y = 2x + 1 such that the resulting polynomial has two roots repeated. Determine the possible value(s) of b.

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Consider the parabola y = x2 + x − 4 and the line y = 3x − 2. a

Determine the quadratic equation whose roots are the x-coordinates of the points of intersection.

Using the sum and product of roots, determine the exact distance between the points of intersection.

The circle x2 + (y − k)2 = 25 is tangent to the parabola y = x2 at exactly two points. Determine the value of k using the sum and product of roots.

Did you know?

The sum and product of zeroes in a polynomial are like the total number of apples and how they’re grouped in a fruit basket! Without knowing the exact number in each pile, you can still work out the total and how they interact—just by knowing how they were arranged. This clever shortcut helps you understand complex equations by focussing on the relationship between the roots and their coefficients—no solving required!

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4 Chapter review 1

What is the range of the function f (x) = x2 − 4? A

f (x) = 2x4 + 3

(− 4, ∞)

(− ∞, − 4]

(− ∞, ∞)

f (x) = x3 − 5x

f (x) = x2 + x

f (x) = 5

Determine the domain of each relation based on the equation: a

Which of the following functions are odd? A

[− 4, ∞)

For each relation, determine in interval notation: i

The domain

y = − x2 + 5

ii x2 + y2 = 36

The range

A region is deﬁned as the space enclosed between the intersection points of y = x2 and y = 5 − x2. Determine in interval notation, the domain and range of this region.

A relation is deﬁned by y =

Determine the domain in interval notation.

Determine the range in interval notation.

Explain why x = 3 is excluded from the domain.

Explain why y = 0 is excluded from the range.

A relation is described as all points on a circle with radius 3, centred at (2, −1): a

Write the equation of the circle.

Determine the domain in interval notation.

Determine the range in interval notation.

Determine whether these functions are even, odd, or neither: a

f (x) = 5x4 − 2x2

f (x) = − 4x3 + 2x

f (x) = x4 + 3x

f (x) =

+ 2x

Given that f (x) = x5 − 2x is an odd function, ﬁnd the coordinates of the point on the graph that is symmetric to the point where x = 2.

A function f (x) is known to be odd. Prove that if its domain includes x = 0, it must pass through the origin.

A student claims that f (x) = x5 + 3 is an odd function because the highest power of x is the odd number 5. Identify and correct the error in the student’s reasoning.

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12E For the polynomial P ( x) = x3 − 5x2 + 2x − 8, which has roots α, β, and γ, what is the value of α + β + γ? A 13E

−5

−2

If the roots of the quadratic equation 2x2 − 6x + 3 = 0 are α and β, what is the value of αβ? A

−3

−6

α2 + β2

14E For P ( x) = 3x2 − x + 5 with roots α and β, calculate the value of: a

α+β

αβ

15E For P ( x) = x3 + 2x2 − 3x − 7 with roots α, β, and γ, calculate the value of: a

α + β + γ

αβ + βγ + γα

αβγ

(α + 1) (β + 1) (γ + 1)

e 16E For P ( x) = x4 + x3 − 2x2 + 3x − 5 with roots α, β, γ, and d, calculate the value of:

17E

α + β + γ + d

αβγd

α 2 + β 2 + γ 2 + d2

A monic quadratic polynomial P ( x) = 0 has roots whose sum is −5 and product is 4. Write down the polynomial P ( x).

18E A monic cubic polynomial P ( x) has roots 2, a, and b. It is known that a + b = 5 and ab = 7. Determine P ( x) in expanded form. 19E Consider the parabola y = x2 − 3x + 1 and the line y = x − 1: a

Determine the quadratic equation whose roots are the x-coordinates of the points of intersection.

Determine the exact distance between the points of intersection using the sum and product of the roots.

20E The roots of x2 − 5x + 3 = 0 are α and β : a

Show that α 3 − β 3 = (α − β )3 + 3αβ (α − β ).

Hence, ﬁnd all possible values of α 3 − β 3.

21E Given the sum of two of the roots is equal to the sum of the other two roots for P ( x) = x4 + 2x3 − 13x2 − 14x + 24, ﬁnd all roots using sum and product formulas.

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5 Functions and relations Chapter outline 5.A 5.B 5.01 5.02 5.03 5.04

Direct variation models Inverse variation models Graphs of reciprocal functions Introduction to absolute value functions Absolute value functions Circles and semicircles Chapter 5 review

162 169 174 182 190

Reciprocal graphs never touch zero — kinda like how cats always avoid water!

5.01 Graphs of reciprocal functions After this lesson, you will be able to… • graph functions of the form asymptotes

, and identify their hyperbolic shape and

• describe the behaviour of as x approaches positive or negative infinity • understand how the constant k affects the location and scale of the graph • determine the equation of a reciprocal function from its graph

Features of reciprocal functions A reciprocal function of the form known as a rectangular hyperbola.

, where k is a constant and k ≠ 0, produces a graph

k is the constant that scales the hyperbola, with k ≠ 0. x The simplest form, f (x) =

is the independent variable, where x ≠ 0.

(where k = 1), consists of two smooth curves: one in the ﬁrst quadrant

(x > 0, f (x) > 0) and one in the third quadrant (x < 0, f (x) < 0). Asymptote A straight line (or another curve) that a curve approaches as x tends to ±∞, or to some particular value. For example, the curve f (x) = 2x has an asymptote f (x) = 0 as x tends to −∞, and the curve f (x) =

has asymptotes x = 0 and f (x) = 0 as x tends to 0 and ±∞ respectively.

The function f (x) =

has two important characteristics based on how fractions work:

1. The denominator cannot be zero, because division by zero is undeﬁned. In this case, the denominator is x, so the function is undeﬁned at x = 0. This creates a vertical asymptote at x = 0. As x approaches 0 from the positive side (x → 0+), f (x) → ∞ if k > 0 or f (x) → −∞ if k < 0. As x approaches 0 from the negative side (x → 0−), f (x) → −∞ if k > 0 or f (x) → ∞ if k < 0.

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2. A fraction equals zero only if the numerator is zero. Here, the numerator is k, and we’re told that k ≠ 0, so f (x) can never be zero. This creates a horizontal asymptote at f (x) = 0. 3. The asymptotes intersect at right angles, giving the graph its “rectangular” hyperbola name.

4 3 2

y=0

k=3 k=1

−4 −3 −2 −1 1 2 −1 x = 0 −2 −3

The graph illustrates f (x) =

for k = 1, k = 3, and

k = −2, showing the hyperbolic shape and asymptotes.

k = −2

−4

The sign of k determines the quadrants of the curves: • If k > 0, curves lie in quadrants 1 and 3. • If k < 0, curves lie in quadrants 2 and 4. The magnitude of k scales the hyperbola: larger ∣k∣ values stretch the curves away from the origin.

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Example 1 Consider the function f (x) = . a Complete the table of values for f (x): x

−2

−1

f (x)

−2

−4

⬚

−2

Create a strategy Substitute each x value into f (x) =

to calculate f (x).

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Apply the idea If x =

If x = 1:

Write the function

Substitute x =

Evaluate

The completed table is:

−2

−1

f (x)

−2

−4

Substitute x = 1

Evaluate

−8

b Plot the points from the table and identify the asymptotes.

Create a strategy Plot the points on a Cartesian plane and determine the vertical and horizontal asymptotes by ﬁnding where the function is undeﬁned.

Apply the idea Determining the vertical asymptote: Write the function

Set the denominator equal to zero

A fraction with a zero denominator is undefined. So the vertical asymptote is x = 0. Determining the horizontal asymptote: A fraction is only equal to zero when its numerator is zero. Since 4 ≠ 0, f (x) ≠ 0. Therefore, a horizontal asymptote exists at f (x) = 0. Plotting the points and asymptotes: 8

6 4

y=0 −6 −4 −2

2 −2

x 2

−4 x = 0 −6 −8

Reflect and check The points lie in quadrants 1 and 3, consistent with k = 4 > 0.

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Idea summary The function f (x) =

forms a rectangular hyperbola with two curves, vertical

asymptote at x = 0, and horizontal asymptote at f (x) = 0. If k > 0, curves are in quadrants 1 and 3; if k < 0, in quadrants 2 and 4.

End behaviour The behaviour of f (x) = as x approaches infinity or negative infinity describes how f (x) approaches the horizontal asymptote. As x → ∞, the denominator becomes very large, making • If k > 0, f (x) → 0 from the positive side.

approach 0:

• If k < 0, f (x) → 0 from the negative side.

As x → −∞, the denominator becomes a large negative number: • If k > 0, f (x) → 0 from the negative side.

• If k < 0, f (x) → 0 from the positive side.

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Example 2 Describe the behaviour of f (x) =

as x → ∞ and as x → −∞.

Create a strategy Evaluate f (x) =

for large positive and negative x values, considering the sign of k = −3.

Apply the idea To consider x → ∞, test with x = 1000: Write the function Substitute x = 1000 Evaluate As x → ∞, f (x) → 0 from the negative side, since k = −3 < 0. To consider x → −∞, test with x = −1000: Write the function Substitute x = −1000 Evaluate As x → −∞, f (x) → 0 from the positive side, since k = −3 < 0.

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Reﬂect and check The negative k causes f (x) to approach 0 from opposite sides compared to positive k.

Idea summary For f (x) = : • As x → ∞, f (x) → 0 (positive side if k > 0, negative side if k < 0). • As x → −∞, f (x) → 0 (negative side if k > 0, positive side if k < 0).

5.01 Practice questions What do you remember? 1

A function has the form f (x) = , where k ≠ 0: a

What is the shape of the graph called?

What are the equations of the asymptotes?

True or false for f (x) = ? a

The graph intersects the axes.

The graph has asymptotes.

If k < 0, the graph lies in the second and fourth quadrants.

As x approaches 0 from the positive side, that is, x → 0+, then f (x) → ∞ if k > 0.

In a rectangular hyperbola centred at the origin, what happens to f (x) as x gets very large in the negative direction, given k > 0?

Practice Ex 1

For f (x) = : a

166

Complete the table of values: x

−2

−1

f (x)

⬚

Sketch the graph, clearly showing the asymptotes and key points from the table.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For f (x) = a

b Ex 2

Complete the table of values: −2

−1

f (x)

⬚

with x > 0:

As x increases, what happens to f (x)?

For f (x) =

As x → 0+, what happens to f (x)?

As x → −∞, what does f (x) approach?

As x → 0−, what does f (x) approach?

In which quadrants does the graph lie?

Determine f (x) when x = −1.

Express x in terms of f (x).

Identify the equations of the asymptotes.

For f (x) = : a c

As x → ∞, what does f (x) approach? +

As x → 0 , what does f (x) approach?

For f (x) =

For f (x) = :

In which quadrants does the graph lie?

For f (x) = a

, sketch the graph, clearly showing the asymptotes and a point on the curve.

Can x or f ( x) be 0?

Verify that x × f ( x) = 8 is equivalent to the given equation.

Compare the behaviour of f (x) =

and f (x) =

as x → ±∞ and as x → 0 from both sides.

Explain how the sign of k affects the graph.

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Extend your thinking 12

For each rectangular hyperbola, identify a point on the graph and substitute the point into f (x) = a

to ﬁnd k. Thus, write the equation of the graph:

1 −4 −3 −2 −1 −1

x 1

−4 −3 −2 −1 −1

−2

−3

−4

x 1

The graphs A, B, and C represent the functions f (x) = , g(x) = , and h(x) = , where a, b and c are positive integers. Given that c − b = 2a, identify the equations of the hyperbolas shown: y 3

2 1 −3 −2

−1

B A 1

x 2

−1 −2 −3

168

The time t (in hours) to complete a task is inversely proportional to the number of workers n. If 4 workers complete the task in 6 hours: a

Write a formula relating t and n.

How long will it take with 8 workers?

How many workers are needed to complete the task in 3 hours?

The cost c(in dollars) per person for a group event is inversely proportional to the number of attendees n. If 10 attendees pay $25 each: a

Write a formula relating c and n.

Sketch the graph for n > 0 and c > 0.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

5.02 Introduction to absolute value functions After this lesson, you will be able to… • define the absolute value of a number as its distance from the origin • use the piecewise definition of the absolute value function to evaluate expressions • graph the function y = ∣x∣ and identify its domain, range, and symmetry • use the identity to simplify expressions • write absolute value equations to model simple contextual problems

Piecewise form of absolute value functions Absolute value The magnitude or size of a real number, i.e. the distance of the number from the origin on a number line. An absolute value function is deﬁned as f (x) = ∣x∣, mapping each real number x to a unique non-negative output y. This establishes a relation between the set of all real numbers (inputs) and non-negative real numbers (outputs).

f (x) = ∣x∣ x

is a real number input

f (x) is the absolute value of x, always non-negative The absolute value function is deﬁned piecewise as:

is a real number input

The domain is all real numbers (−∞, ∞), and the range is [0, ∞). y 5 4

The graph of y = ∣x∣ forms a V-shape with vertex at (0, 0).

The graph is symmetric about the y-axis, hence has even symmetry.

2 1 x −4 −3 −2 −1

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Example 1 Write an absolute equation to represent “All real numbers x that are 2 units from −3.”

Create a strategy Represent the distance from x to −3 as an absolute value equation.

Apply the idea The distance from x to −3 is 2 units, so the absolute equation is: ∣x − (−3)∣ = 2 ∣x + 3∣ = 2

Express the distance using absolute value Evaluate the adjacent signs

Example 2 Evaluate f (x) = ∣x∣ for x = −3 and x = 2 using the piecewise deﬁnition.

Create a strategy Use the piecewise deﬁnition

to determine the output for each input.

Apply the idea For x = −3: f (x) = ∣x∣

Write the function

f (−3) = ∣−3∣

Substitute x = −3

= −(−3)

Since −3 < 0, use −x

Evaluate

For x = 2: f (x) = ∣x∣

Write the function

f (2) = ∣2∣

Substitute x = 2

Since 2 ≥ 0, use x

Reflect and check Verify: ∣−3∣ = 3 and ∣2∣ = 2 match the distance from 0 on a number line.

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Example 3 A weather station measures temperatures with a target of 20°C and a range of acceptable error of ±3°C. Write an absolute value equation for the acceptable temperatures t.

Create a strategy Represent the distance from the temperature t to 20 as an absolute value equation.

Apply the idea The distance from t to 20 is 3 degrees Celsius, so the absolute equation is: ∣t − 20∣ = 3 Express the distance using absolute value

Idea summary The absolute value function f (x) = ∣x∣ maps real numbers to non-negative outputs, deﬁned piecewise as x if x ≥ 0, or −x if x < 0.

Square root and absolute value relationship equals ∣x∣, as it always yields a non-negative result, matching the absolute

The expression value’s output.

is a real number input

Numerical substitutions demonstrate this: • For x = 3: • For x = −3: • For x = 0: This relationship holds because the square root function returns the non-negative root, aligning with the absolute value’s deﬁnition.

Example 4 Verify

for x = −5 and use this result to evaluate

Create a strategy Substitute x = −5 into both sides of the equation to conﬁrm they are equal. Then, apply this identity to simplify and evaluate the given expression

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Apply the idea Verify for x = −5: Write the equation Substitute x = −5

Evaluate each side

Evaluate the square root

This shows that the equality holds. :

Evaluate

Apply the identity

Evaluate the absolute value

Evaluate

Idea summary The identity holds as both yield non-negative equal outputs. Use numerical substitutions to verify, and apply to simplify expressions involving square roots of squared terms.

5.02 Practice questions What do you remember? 1

Deﬁne the absolute value of a number.

Explain the representation of ∣−5∣ on a number line.

Determine the possible values of the pronumeral for each equation and describe it as the deﬁnition of the absolute value function: a

∣a∣ = 6

∣b∣ =

∣c∣ = −4

Evaluate f (x) = ∣x∣ using the piecewise deﬁnition for the given values of x: a

x=4

x = −2

x=0

State the deﬁnition of the absolute value function as a piecewise function.

Are there any values of a for which

172

does not equal ∣a∣?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

∣d∣ = 0

Practice Ex 1

Write an absolute value equation to represent each situation: a

All real numbers x that are 7 units from 0.

All real numbers x that are 3 units from −1.

All real numbers x that are 2.5 units from 4.

Ex 2

Evaluate f (x) = ∣x∣ for x = −6, 0, 3 using the piecewise deﬁnition.

Ex 3

A factory produces rods with a target length of 50 cm, with a range of acceptable error of ±0.5 cm. Write an absolute value equation for the acceptable lengths l.

A thermometer has a target temperature of 20°C, with an acceptable deviation of ±1.5°C. Write an absolute value equation for the acceptable temperatures t.

Ex 4

Consider

= ∣x∣:

Verify for x = −7 and x = 4.

Use the result from part (a) to evaluate

Write an absolute value equation for each relation: a

The distance between a number and 2 is 4.

The distance between a number and −3 is 2.

Write an absolute value equation to represent each situation: a

All real numbers x that are 5 units from 3.

All real numbers x that are 1.2 units from −2.

All real numbers x that are

units from 0.

Extend your thinking 13

True or false: If ∣x∣ = a, then x = ± a. Explain your reasoning

A machine dispenses liquid with a target volume of 500 mL, with a range of acceptable error of ±10 mL:

Write an absolute value function f (v) for the deviation from the target volume v.

Represent the function as a piecewise function and state its vertex.

A quality control system monitors the diameter of ball bearings, targeting 10 mm with a range of acceptable error of ±0.2 mm: a

Write an absolute value equation for the acceptable diameters d.

Solve the equation to determine the minimum and maximum diameters.

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Consider the absolute value function f (x) = a∣x − h∣ + k: a

Explain how the parameters a, h, and k affect the graph’s shape and position.

For f (x) = 2∣x − 3∣ + 1, state the vertex and compare this graph with f (x) = ∣x∣, in terms of steepness, concavity, and shift of the vertex.

If the graph is reﬂected over the x-axis and shifted left by 2 units, write the equation of the new function.

5.03 Absolute value functions After this lesson, you will be able to… • graph absolute value functions of the form y = ∣ax + b∣ • identify the vertex, axis of symmetry, domain, and range of y = ∣ax + b∣ • solve absolute value equations of the form ∣ax + b∣ = k algebraically and graphically

Absolute value functions An absolute value function is deﬁned as f (x) = ∣ax + b∣, where a and b are constants, and a ≠ 0. The absolute value ensures that the output f (x) is always non-negative, resulting in a V-shaped graph. The graph of f (x) = ∣ax + b∣ has a vertex at the point where ax + b = 0. Solving for x gives x = and the corresponding y-value is 0. The vertex is thus

f (x) = ∣ax + b∣ a determines the slope of the arms of the V-shape; if ∣a < 0∣ the graph is identical to ∣a > 0∣ but reﬂected across the y-axis; if −a∣x∣ then the graph is reﬂected across the x-axis b shifts the vertex

units horizontally along the x-axis

The graph is symmetric about the vertical line through the vertex, x = numbers, , and the range is y ≥ 0.

. The domain is all real

x = −2

For f (x) = ∣2x + 4∣, the vertex is at 2x + 4 = 0, so x = −2, y = 0. The graph is symmetric about x = −2.

4 (0, 4)

For the y-intercept:

y = 2x + 4

=2×0+4 =4

1 x

(−2, 0) −4 −3 −2 −1

174

So, the y-intercept is y = 4.

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Example 1 Graph the function f (x) = ∣3x − 6∣ and state its symmetry, domain, and range.

Create a strategy Find the vertex by solving 3x − 6 = 0. Plot key points around the vertex, connect them to form the V-shape, and determine symmetry, domain, and range.

Apply the idea Find the vertex: 3x − 6 = 0

Set the expression inside the absolute value to zero

3x = 6

Add 6 to both sides

x=2

Divide both sides by 3

The vertex is at (2, 0). Choose points around x = 2 to plot: x

f (x) = ∣3x − 6∣

Plot points (0, 6), (1, 3), (2, 0), (3, 3), (4, 6) and connect to form a V-shape. y 6 5 4

The graph of f (x) = ∣3x − 6∣ is shown.

3 2 1

(2, 0) 1

x 4

Symmetry: The graph is symmetric about x = 2. Domain: All real numbers, . Range: y ≥ 0

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Reflect and check Verify using a graphing calculator or software to confirm the vertex at (2, 0) and the V-shape. Check symmetry by noting that points like (1, 3) and (3, 3) are equidistant from x = 2.

Example 2 Solve the equation ∣2x − 5∣ = 9 graphically.

Create a strategy Find the intersections of f (x) = ∣2x − 5∣ and y = 9.

Apply the idea y 10

(−2, 9)

(7, 9)

8 6

The intersections at (−2, 9) and (7, 9) conﬁrm the solutions x = −2 and x = 7.

4 2 x −4 −2

Example 3 Use a graphing application to graph the function f (x) = ∣2x − 4∣ and identify its vertex, symmetry, and intercepts.

Create a strategy Input f (x) = ∣2x − 4∣ into the graphing application. Use the graph to locate the vertex, determine the line of symmetry, and ﬁnd x- and y-intercepts.

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Apply the idea Enter f (x) = ∣2x − 4∣ into the graphing application. The graph displays a V-shaped function. For the vertex, zoom in or use the trace feature to ﬁnd the lowest point, which is at (2, 0) (where 2x − 4 = 0). For the symmetry, observe that the graph is symmetric about the vertical line x = 2, as points equidistant from x = 2 have the same y-value. For the intercepts, the y-intercept is at x = 0, so f (x) = ∣2(0) − 4∣ = 4, giving (0, 4). The x-intercept is at the vertex, (2, 0), where y = 0. y 5 4 (0, 4) 3

Graph of f (x) = ∣2x − 4∣ showing vertex at (2, 0) and y-intercept at (0, 4).

2 1 x

(2, 0) −1

Reﬂect and check Verify by calculating: • Vertex is at x =

= 2, y = 0.

• The y-intercept is at x = 0, f (x) = ∣0 − 4∣ = 4. Symmetry conﬁrmed by equal y-values at points like (1, 2) and (3, 2).

Idea summary The function f (x) = ∣ax + b∣ has its vertex at horizontally by

meaning the vertex is shifted

from the origin.

Its domain is all real x, and its range is y ≥ 0. To solve ∣ax + b∣ = k graphically, ﬁnd the x-values where the graph of f (x) = ∣ax + b∣ intersects the line y = k.

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Solutions of absolute value equations An absolute value equation involves an absolute value expression, for example, ∣ax + b∣ = k. Solutions are found by considering the expression inside the absolute value equalling k or −k.

∣ax + b∣ = k a, b are the constants defining the linear expression inside the absolute value k is the non-negative constant representing the equation’s right-hand side The number of solutions depends on k: • If k > 0, there are two solutions. • If k = 0, there is one solution. • If k < 0, there are no solutions, as absolute values are non-negative. Solutions can be verified graphically by finding the x-values where f (x) = ∣ax + b∣ intersects y = k.

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Example 4 Solve the equation ∣2x + 1∣ = 9.

Create a strategy Isolate the absolute value expression, then solve the two resulting equations: 2x + 1 = 9 and 2x + 1 = −9.

Apply the idea Solve the equation at k = 9: ∣2x + 1∣ = 9

Write the equation

2x + 1 = 9

Evaluate the absolute value

2x = 8

Subtract 1 from both sides

x=4

Divide both sides by 2

Solve the equation at k = −9: ∣2x + 1∣ = −9

Write the equation

2x + 1 = −9

Evaluate the absolute value

2x = −10

Subtract 1 from both sides

x = −5

Divide both sides by 2

The solutions are x = 4 and x = −5.

Reﬂect and check Verify by substituting: ∣2(4) + 1∣ = ∣8 + 1∣ = 9 and ∣2(−5) + 1∣ = ∣−10 + 1∣ = 9.

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Example 5 Solve the equation 2∣x + 1∣ + 3 = 21.

Create a strategy Isolate the absolute value expression, then solve the two resulting equations: x + 1 = k and x + 1 = −k.

Apply the idea 2∣x + 1∣ + 3 = 21

Write the equation

2∣x + 1∣ = 18

Subtract 3 from both sides

∣x + 1∣ = 9

Divide both sides by 2

x+1=9

Write the equation at k = 9

x=8

Subtract 1 from both sides

x + 1 = −9

Write the equation at k = −9

Solve the equation at k = 9:

Solve the equation at k = −9: x = −10

Subtract 1 from both sides

The solutions are x = 8 and x = −10.

Example 6 A quality control process requires a product’s weight to deviate by exactly 8 grams from the target weight of 120 grams. Write an absolute value function to model the deviation and ﬁnd the acceptable weights.

Create a strategy Model the deviation using f (x) = ∣x − c∣, where c is the target weight. Solve the equation ∣x − c∣ = 8 to ﬁnd the acceptable weights.

Apply the idea f (x) = ∣x − 120∣ Write the model for deviation, where x is the actual weight The acceptable weights satisfy ∣x − 120∣ = 8. Solve at k = 8: x − 120 = 8 x = 128

Write the equation for k = 8 Add 120 to both sides

Solve at k = −8: x − 120 = −8 x = 112

Write the equation for k = −8 Add 120 to both sides

The acceptable weights are x = 112 or x = 128 grams.

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Idea summary Absolute value equations ∣ax + b∣ = k are solved by isolating the absolute value and solving ax + b = k and ax + b = −k. Solutions depend on k: two if k > 0, one if k = 0, and none if k < 0. Inequalities like ∣ax + b∣ ≤ k yield a range of solutions.

5.03 Practice questions What do you remember? 1

Describe what the absolute value of a number represents on a number line.

State the formula for the vertex of the absolute value function f (x) = ∣ax + b∣.

What is the range of any absolute value function of the form f (x) = ∣ax + b∣ where a ≠ 0?

Describe the effect of the parameter a on the graph of f (x) = ∣ax + b∣ compared to f (x) = ∣x∣: a

When ∣a∣ > 1

When ∣a∣ < 1

When ∣a∣ = 0

How many solutions does the equation ∣ax + b∣ = k have for each case of k? a

When k > 0

When k = 0

When k < 0

Practice 6

Write an absolute value equation to represent the set of all real numbers x that are: a

Ex 1

f (x) = ∣2x + 6∣

f (x) = ∣x − 4∣

f (x) = ∣−2x + 2∣

∣2x − 4∣ = 2

∣−x + 3∣ = 9

∣x + 1∣ = 14

Use a graphing application to graph each function and identify the vertex, line of symmetry, x-intercept(s), and y-intercept: a

180

4 units away from −2

Solve these equations graphically: a

Ex 3

For each function, sketch the graph, and state its line of symmetry, domain, and range: a

Ex 2

5 units away from 0

f (x) = ∣1.5x + 2∣

f (x) = ∣3x − 9∣

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 4

Solve each equation algebraically: a

Ex 5

Ex 6

∣3x − 6∣ = 9

∣2x + 4∣ = 10

c g

∣2x − 3∣ = 8

∣x + 2∣ = 6

∣u + 2∣ = 5

∣v − 1∣ = 4

∣3x − 1∣ = 15

∣−4x + 8∣ = 12

∣3m∣ = 12

Write an absolute value equation to model each situation and solve for the unknown: a

A quality control machine accepts rods with a target length of 50 mm, with deviation for length of 48 mm.

A car’s target speed is 60 km/h, with deviation for speed of 55 km/h.

The distance between a number and 2 is 4.

Solve each equation algebraically: a

2∣x − 3∣ + 1 = 9

3∣2x + 4∣ − 2 = 16

4∣x + 1∣ + 3 = 15

5∣3x − 6∣ − 4 = 21

A manufacturer produces bolts with a target diameter of 25 mm. A quality control process requires the diameter to deviate by exactly 3 mm from the target. Write an absolute value function to model the deviation and ﬁnd the acceptable diameters.

Graph: a

f (x) = 2 and f (x) = ∣x − 3∣ on the number plane and solve ∣x − 3∣ = 2.

f (x) = 1 and f (x) = ∣2x + 1∣ on the number plane and solve ∣2x + 1∣ = 1.

Extend your thinking 15

For the equation ∣2x − 1∣ = x + 2: a

Solve algebraically and identify possible solutions.

Verify graphically by sketching f (x) = ∣2x − 1∣ and f (x) = x + 2.

A temperature sensor records a target temperature of 25°C. Write an absolute value function to model the deviation and evaluate the deviation for temperatures of 22, 25, and 28°C.

Explain why the equation ∣ax + b∣ = k has:

Two solutions when k > 0

No solutions when k < 0

One solution when k = 0

A delivery truck travels at an average speed of 50 km/h to a destination. The target time taken is 1 hour. Using distance = speed × time, write an absolute value function to model the deviation in time and evaluate the deviation for times of , 1, and corresponding distances.

hours. Then ﬁnd the

Solve each equation algebraically and identify possible solutions: a

∣m + 3∣ = m + 1

∣n − 2∣ = n − 2

∣p + 1∣ = p + 3

3∣q − 4∣ = q

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5.04 Circles and semicircles After this lesson, you will be able to… • derive the equation of a circle with radius r and centre at the origin using Pythagoras’ theorem • graph circles of the form x2 + y2 = r2 • determine the equation of a circle from its graph • identify and graph the four forms of semicircles derived from the circle equation

Derive circle equation A circle equation is derived using Pythagoras’ theorem for a set of points equidistant from a ﬁxed point. Any point (x, y) on a circle with its ﬁxed point at (0, 0) is a certain distance from that point. y

(0, 0) r

A right-angled triangle is formed with vertices at (0, 0), (x, 0), and (x, y). The hypotenuse is the distance r, with legs x and y.

(x, y)

Pythagoras’ theorem gives the equation:

x2 + y2 = r2 x

is the x-coordinate of a point on the circle

is the y-coordinate of a point on the circle

r is the distance from the ﬁxed point to the circumference 2

The equation x + y = r ensures all points are equidistant from the origin. Changing r scales the circle’s size, with larger r producing a larger circle.

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Example 1 Derive the equation of a circle with a distance of 7 from its ﬁxed point at the origin.

Create a strategy Use Pythagoras’ theorem to ﬁnd the equation.

Apply the idea x2 + y2 = r2

Write the Pythagoras’ theorem

x +y =7

Substitute r = 7

Evaluate

x + y = 49 2

The equation is x + y = 49.

Example 2 Verify that the point (3, 4) lies on a circle with a distance of 5 from its ﬁxed point at the origin.

Create a strategy Substitute the point into x2 + y2 = r2.

Apply the idea x2 + y2 = r2

Write the equation

3 +4 =5

Substitute x = 3, y = 4, and r = 5

9 + 16 = 25

Evaluate each term

25 = 25 2

Evaluate

The point satisﬁes x + y = 25.

Idea summary The equation of a circle with radius r and centre at the origin is x2 + y2 = r2, derived using Pythagoras’ theorem.

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Graph circles A set of points equidistant from a ﬁxed point is graphed using its equation x2 + y2 = r2. The ﬁxed point is at (0, 0), and the distance to the circumference is r. Key points include intercepts at (±r, 0) and (0, ±r).

Example 3 Graph the circle x2 + y2 = 16.

Create a strategy Plot the ﬁxed point and intercepts using the radius

Apply the idea The equation x2 + y2 = 16 has a ﬁxed point at (0, 0) and radius r =

= 4.

Key points are (4, 0), (−4, 0), (0, 4), and (0, −4). y 4

−4

(0, 0)

4 x

−4

Reﬂect and check Substituting x = 0 gives y = ±4, and y = 0 gives x = ±4, conﬁrming the intercepts.

Idea summary To graph a circle with equation x2 + y2 = r2, plot the centre at (0, 0) and points at distance radius r along the x- and y-axes.

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Find equation from graph The equation x2 + y2 = r2 of a set of points equidistant from a fixed point is found from its graph by identifying the distance from the fixed point to the circumference. The fixed point is at (0, 0), and the distance r is found from a point on the circumference, often using intercepts.

Example 4 A circle has its ﬁxed point at (0, 0) and passes through (0, 6). Find its equation.

Create a strategy Substitute the point (0, 6) into x2 + y2 = r2 to ﬁnd the radius.

Apply the idea x2 + y2 = r2

Write the equation

Substitute x = 0 and y = 6

36 = r

Evaluate

0 +6 =r

r=6 2

Take the positive square root

The equation is x + y = 36.

Reflect and check The point (6, 0) satisfies 62 + 02 = 36, confirming the equation.

Idea summary For a circle with centre at (0, 0), ﬁnd the radius r from a point on the graph to form the equation x2 + y2 = r2.

Graph semicircles A semicircle is half of a set of points equidistant from a ﬁxed point, deﬁned by equations derived from x2 + y2 = r2. Solving for y gives:

y is the y-coordinate of a point on the semicircle x is the x-coordinate of a point on the semicircle r is the distance from the ﬁxed point to the circumference

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represents the upper semicircle (above the x-axis), and the negative

The positive root

represents the lower semicircle below the x-axis.

root

Solving for x gives:

x is the x-coordinate of a point on the semicircle y is the y-coordinate of a point on the semicircle r is the distance from the ﬁxed point to the circumference represents the right semicircle (right of the y-axis) and the negative

The positive root

represents the left semicircle left of the y-axis.

root

with radius r = 5 has a domain of [−5, 5] and forms the upper semicircle,

For example,

with r = 4 has a range of [−4, 4] and forms the left semicircle.

while

Interactive exploration Discover this concept in action online

mathspace.co

Example 5 Graph the semicircle

Create a strategy Plot key points using the radius and domain.

Apply the idea The equation comes from x2 + y2 = 25, with radius r = 5. The positive root indicates the upper semicircle. 2

Domain: 25 − x ≥ 0, so x ≤ 25, or [−5, 5].

(−3, 4)

Key points: (0, 5), (5, 0), (−5, 0).

5 4

(0, 5) (3, 4)

3 2

(−5, 0)

−5 −4 −3 −2 −1 −1 −2 −3

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(5, 0) x 1 2 3 4 5

Example 6 Graph the semicircle

Create a strategy Plot key points using the radius and range.

Apply the idea The equation the left semicircle.

comes from x2 + y2 = 16, with radius r = 4. The negative root indicates

Range: 16 − y2 ≥ 0, so y2 ≤ 16, or [−4, 4]. Key points: (−4, 0), (0, 4), (0, −4). y 5 4 (4, 0) 3 2 1 x

(−3, 2.645) −4 −5 −4 −3 −2

−1

(−3, −2.645)

1 −1 −2 −3 −4 (−4, 0) −5

Reflect and check The point (−3, 2.645) (approximate) satisfies (−3)2 + (2.645)2 ≈ 16, confirming it lies on the semicircle.

Idea summary Semicircles are graphed using (right/left), with domain or range [−r, r].

(upper/lower) or

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5.04 Practice questions What do you remember? 1

Answer the following questions about circles and semicircles: a

State the general equation of a circle with radius r and centre at the origin.

Explain how Pythagoras’ theorem is used to derive the equation of a circle with centre at the origin.

State the equation of the upper semicircle for a circle with radius r centred at the origin.

What shape is formed by all points exactly 5 units from the origin?

State the radius of the circle x2 + y2 = 100.

Answer the following about graphing circles: a

Identify the centre and radius of the circle x2 + y2 = 81.

State the x-intercepts of the circle x2 + y2 = 16.

Practice Ex 1

Derive the equation of a circle with centre at the origin and radius 6 using Pythagoras’ theorem.

Ex 2

Verify that the point (8, 6) lies on a circle with a distance of 10 from its ﬁxed point at the origin.

Ex 3

Graph the circle x2 + y2 = 25 on the Cartesian plane, stating the radius and centre.

Ex 4

Determine the equation of the circle with centre at the origin given the following conditions:

Passes through the point (3, 0).

Has a radius of 7.

Identify whether each equation represents a circle or semicircle, and if a semicircle, specify which part (upper, lower, right, or left): a

Ex 5

For the circle x2 + y2 = 64, ﬁnd the equations of the: a

188

Graph the following semicircles: a

Ex 6

Upper semicircle

Lower semicircle

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Consider the circle with equation x2 + y2 = 81: a

Determine the y-values when x = 1.

Sketch the graph of the circle.

A circular garden bed has radius 5 metres: a

Write the equation of the circle representing the garden bed, with centre at the origin.

Find the equation of the upper semicircle to represent a raised section of the bed.

Extend your thinking 14

Derive the equations of the right and left semicircles for a circle with equation x2 + y2 = r2.

A semicircular arch has the equation and the maximum height above the ground.

A semicircular window has the equation base and the maximum height.

A circular pond with equation x2 + y2 = 36 has a semicircular path along its left edge:

metres. Calculate the width at ground level

metres. Calculate the width at the

Write the equation of the left semicircle representing the path.

Determine the range of the semicircle.

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5 Chapter review 1

What is the radius of the circle with the equation x2 + y2 = 81? A

40.5

For f (x) =

In which quadrants does the graph lie?

Determine f (x) when x = −5.

Express x in terms of f (x).

Identify the equations of the asymptotes.

For f (x) =

As x → ∞, what does f (x) approach?

As x → −∞, what does f (x) approach?

As x → 0+, what does f (x) approach?

As x → 0−, what does f (x) approach?

For each rectangular hyperbola, identify a point on the graph and substitute the point into f (x) = a

to ﬁnd k. Thus, write the equation of the graph:

1 −4 −3 −2 −1 −1

−2

−3

−4

2 x 1

−4 −3 −2 −1 −1

−2

−4 −3 −2 −1 −1

1 4

−2

190

1 −4 −3 −2 −1 −1 −2

−3

−4

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x 1

−4 −3 −2 −1 −1

−2

−3

−4

Consider the identity

Verify that this identity is true for a = −9 and a = 5.

Evaluate

A factory produces pistons with a target diameter of 75 mm. Write an absolute value function to model the deviation and evaluate the deviation for diameters of 74.7, 75, and 75.3 mm: a

Write an absolute value function f (d) for the deviation from the target diameter d.

Evaluate the deviation for the given diameters.

A machine dispenses liquid with a target volume of 450 mL: a

Write an absolute value function f (v) for the deviation from the target volume v.

Represent the function as a piecewise function and state its vertex.

Consider the absolute value function g(x) = a∣x − h∣ + k: a

Explain how the parameters a, h, and k affect the graph’s shape and position.

For g(x) = 2∣x − 5∣ + 3, state the vertex and describe the graph’s orientation.

If the graph is reﬂected over the x-axis and shifted right by 1 unit, write the equation of the new function.

Solve each equation algebraically: a

−4 −3 −2 −1 −1

∣2x − 1∣ = 5

∣x + 3∣ = 9

∣x − 6∣ = 2

∣x + 4∣ = 1

Graph each function and identify the vertex, line of symmetry, x-intercept, and y-intercept: a

f (x) = ∣2x + 5∣

f (x) = ∣3x − 12∣

Graph f (x) = 5 and f (x) = ∣x − 3∣ on the number plane and solve ∣x − 3∣ = 5.

For the function f (x) = ∣x − 3∣ + 2, determine the values of x where f (x) = 7. Represent the solutions as ordered pairs.

Verify that the point (8, 15) lies on a circle with a radius of 17 centred at the origin.

Graph the circle x2 + y2 = 9, stating its radius and centre.

For the circle x2 + y2 = 100, ﬁnd the equations of the: a

Upper semicircle

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A semicircular window has the equation and the maximum height above the ground.

A circular garden bed has the equation x2 + y2 = 16. A path runs along its left edge:

metres. Calculate the width at the base

Write the equation of the left semicircle representing the path.

Determine the range of y-values for the semicircle.

Did you know?

Functions can be like gardens, where every input grows into a unique flower! Injective functions ensure no two seeds produce the same bloom. Surjective functions make sure every patch of soil blossoms with at least one flower. And bijective functions create a perfect garden, where every seed has a unique bloom and every bloom comes from a unique seed.

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“Mathematics is the art of giving the same name to different things.” Henri Poincaré

6 Inequalities Chapter outline 6.01E Solve cubic inequalities 6.02E Solve inequalities with variables in the denominator 6.03E Solve absolute value inequalities Chapter 6 review

196 206 213 224

Ski patrols use inequalities to predict avalanches: if the snow load passes a certain threshold, the risk skyrockets.

6.01E Solve cubic inequalities After this lesson, you will be able to… • solve cubic inequalities where the cubic is expressed as a product of linear factors • determine the roots of a cubic polynomial from its linear factors and use a sign table to analyse intervals for solving the inequality • sketch the graph of a cubic polynomial from its factored form and interpret the graph to identify solution intervals for the inequality • differentiate solution sets based on strict (<, >) or non-strict (≤, ≥) inequalities, and analyse the effect of repeated roots on the solution

Solve cubic inequalities Inequality A statement that one number or algebraic expression is less than (or greater than) another. There are 4 types of inequalities: • a is less than b is written a < b • a is greater than b is written a > b • a is less than or equal to b is written a ≤ b • a is greater than or equal to b is written a ≥ b An order relation between one number or algebraic expression and another.

A cubic inequality compares a cubic polynomial (degree 3) to zero or another expression using <, ≤, >, or ≥. When expressed as a product of linear factors, ( x − a) ( x − b) ( x − c), it can be solved by: • Algebraic method: Determine roots where ( x − a) ( x − b) ( x − c) = 0 ( x = a, b, c). Use a sign table to test the intervals between roots, checking each factor’s sign to determine the sign of the product of factors. • Graphical method: Sketch y = ( x − a) ( x − b) ( x − c). Roots are x-intercepts. Identify where the graph is above ( y > 0), below ( y < 0), or on the x-axis ( y = 0). • Cubic behaviour: For a positive leading coeﬃcient, y → +∞ as x → +∞, and y → −∞ as x → −∞. For a negative leading coeﬃcient, y → −∞ as x → +∞, and y → +∞ as x → −∞. The graph crosses the x-axis at each root.

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• Roots in solutions: Include roots for ≤ or ≥; exclude for < or >. A cubic inequality can also be expressed as ( x − a)2 ( x − b). This inequality can be solved by: • Algebraic method: The roots of the inequality are x = a and x = b. Use a sign table to determine the sign of each factor around the roots and hence determine the sign of the product of the factors. • Graphical method: Sketch the graph. The graph touches the x-axis at x = a and bounces back in the same direction, whereas at x = b the graph crosses the x-axis. For solving y > 0, the solution includes the part of the graph that is above the x-axis and for y < 0, the solution includes the part of the graph that is below the x-axis. • Cubic behaviour: For a positive leading coeﬃcient, y → +∞ as x → +∞, and y → −∞ as x → −∞. For a negative leading coeﬃcient, y → −∞ as x → +∞, and y → +∞ as x → −∞. The graph touches the x-axis at the repeated root ( x = a) and crosses it at the single root ( x = b). • Roots in solutions: Include roots for ≤ or ≥; exclude for < or >.

Exploration Consider the cubic polynomial y = ( x − 2) ( x + 1) ( x − 4). Sketch the graph by identifying the roots, determining the end behaviour based on the leading coeﬃcient, and noting the y-intercept. Use the sketch to answer these questions: 1. Identify the intervals where y > 0. Explain how the graph’s position relative to the x-axis supports the response. 2. Identify the intervals where y < 0. Describe how the multiplicity of the roots affects the graph’s behaviour at the x-intercepts. 3. If the inequality changes to y ≥ 0, determine how the solution intervals differ from those for y > 0. Justify the inclusion of speciﬁc points in the solution.

Example 1 Solve ( x − 1) ( x + 1) ( x − 3) ≤ 0 using: a Algebraic method

Create a strategy Use null-factor law to determine the roots at ( x − 1) ( x + 1) ( x − 3) = 0. Test the intervals using a sign table to determine where the product is ≤ 0.

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Apply the idea ( x − 1) ( x + 1) ( x − 3) = 0

Write the equation

0, x + 1

0, x − 3 = 0

Use null factor law

x−1=0

Write the equation

x=1

Add 1 to both sides

x+1=0

Write the equation

x−1 For x − 1 = 0:

For x + 1 = 0: x = −1

Subtract 1 from both sides

For x − 3 = 0: x−3=0

Write the equation

x=3

Add 3 to both sides

x < −1

−1 < x < 1

1<x<3

x>3

x−1

−

x+1

−

x−3

−

Product

−

So the roots are x = −1, 1, 3. Testing the intervals:

The product is ≤ 0 for x ≤ −1 or 1 ≤ x ≤ 3.

Reﬂect and check Verify the solution by testing the boundary points. For x ≤ −1, substitute x = −1: ( x − 1) ( x + 1) ( x − 3) ≤ 0

Write the inequality

(−1 − 1) (−1 + 1) (−1 − 3) ≤ 0

Substitute x = −1

−2 × 0 × (−4) ≤ 0 0≤0

Evaluate each expression inside the brackets Evaluate

For 1 ≤ x ≤ 3, substitute x = 2: ( x − 1) ( x + 1) ( x − 3) ≤ 0 (2 − 1) (2 + 1) (2 − 3) ≤ 0 1 × 3 × (−1) ≤ 0 −3 ≤ 0

Write the inequality Substitute x = 2 Evaluate each expression inside the brackets Evaluate

Both satisfy ≤ 0, conﬁrming the inclusion of roots in the solution.

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b Graphical method

Create a strategy Sketch y = ( x − 1) ( x + 1) ( x − 3) and determine where y ≤ 0.

Apply the idea 8

6 4

(−1, 0)

−4 −3 −2 −1 −2

(1, 0) (3, 0) x 1

−4 −6 −8

Graph is below or on the x-axis for x ≤ −1 and 1 ≤ x ≤ 3.

Example 2 Solve −(2x + 1) ( x − 2) ( x + 3) > 0 using: a Algebraic method

Create a strategy Use null-factor law to determine the roots at −(2x + 1) ( x − 2) ( x + 3) = 0. Test the intervals using a sign table to determine where the product is > 0.

Apply the idea −(2x + 1) ( x − 2) ( x + 3) = 0 2x + 1

0, x − 2

0, x + 3 = 0

Write the equation Use null factor law

For 2x + 1 = 0: Write the equation

Subtract 1 from both sides

Divide both sides by 2

For x − 2 = 0: x−2=0

Write the equation

x=2

Add 2 to both sides

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For x + 3 = 0:

So the roots are x = −3,

x+3=0

Write the equation

x = −3

Subtract 3 from both sides

, 2.

The intervals are x < −3, −3 < x <

< x < 2, x > 2.

The negative sign affects the product’s sign, making it negative when the number of negative factors is odd. Testing the intervals: x < −3

x>2

−(2x + 1)

−

x−2

−

x+3

−

Product

−

The product is > 0 for x < −3 or

< x < 2.

Reﬂect and check Verify the solution by testing boundary points. For x < −3, substitute x = −4: −(2x + 1) ( x − 2) ( x + 3) > 0 −(2 × (−4) + 1) ((−4) − 2) ((−4) + 3) > 0 −(−7) × (−6) × (−1) > 0 7 × (−6) × (−1) > 0 42 > 0 For

Write the inequality Substitute x = −4 Evaluate each expression inside the brackets Evaluate adjacent signs Evaluate

< x < 2, substitute x = 1: −(2x + 1) ( x − 2) ( x + 3) > 0

Write the inequality

−(2 × 1 + 1) (1 − 2) (1 + 3) > 0

Substitute x = 1

−3 × (−1) × 4 > 0 12 > 0

Evaluate each expression inside the brackets Evaluate

Both satisfy > 0, conﬁrming the inclusion of roots in the solution.

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b Graphical method

Create a strategy Sketch y = −(2x + 1) ( x − 2) ( x + 3) and determine where y > 0.

Apply the idea The negative leading coeﬃcient inverts the cubic, so y → −∞ as x → +∞ and y → +∞ as x → −∞. 25

20 15 10

(−3, 0)

−4 −3 −2 −1 −5

(2, 0) 1

−10 −15

The graph is above the x-axis for x < −3 and

< x < 2.

Example 3 Solve ( x − 1)2( x + 2) ≥ 0 using: a Algebraic method

Create a strategy Use null-factor law to determine the roots at ( x − 1)2( x + 2) = 0. Test the intervals using a sign table to determine where the product is ≥ 0.

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Apply the idea ( x − 1)2 ( x + 2) = 0 ( x − 1)

Write the equation

0, x + 2 = 0

Use null factor law

( x − 1)2 = 0

Write the equation

For ( x − 1) = 0: x−1=0

Take the square root of both sides

x=1

Add 1 to both sides

x+2=0

Write the equation

For x + 2 = 0: x = −2

Subtract 2 from both sides

So the roots are x = −2, 1. The intervals are x < −2, −2 < x < 1, x > 1. Since ( x − 1)2 is always non-negative, the sign of the product depends on the sign of x + 2. Testing the intervals: x < −2

−2 < x < 1

x>1

( x − 1)2

x+2

−

Product

−

The product is ≥ 0 for x ≥ −2.

Reﬂect and check Verify the solution by testing boundary points. For x = −2: ( x − 1)2 ( x + 2) ≥ 0 2

(−2 − 1) (−2 + 2) ≥ 0

Write the inequality Substitute x = −2

Evaluate each expression inside the brackets

9×0≥0

Evaluate the index

(−3) × 0 ≥ 0 0≥0

Evaluate

For x = −1: ( x − 1)2 ( x + 2) ≥ 0 2

(−1 − 1) (−1 + 2) ≥ 0

Write the inequality Substitute x = −1

Evaluate each expression inside the brackets

4×1≥0

Evaluate the index

(−2) × 1 ≥ 0 4≥0

Evaluate

Both satisfy ≥ 0, conﬁrming the inclusion of roots in the solution.

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b Graphical method

Create a strategy Sketch y = ( x − 1)2 ( x + 2) and determine where y ≥ 0.

Apply the idea The leading coefficient is positive, so y → +∞ as x → +∞, and y → −∞ as x → −∞. The repeated root at x = 1 suggests a turning point. 8

6 4

(−2, 0)

−4 −3 −2 −1 −2

(1, 0) 1

−4 −6 −8

The graph is above or on the x-axis for x ≥ −2.

Idea summary For cubic inequalities expressed as a product of linear factors, ( x − a) ( x − b) ( x − c) ≤ 0: • Use a sign table to check the product’s sign across intervals deﬁned by roots. • Graph the cubic and identify regions above, below, or on the x-axis. • Include roots for ≤ or ≥; exclude for < or >.

6.01E Practice questions What do you remember? 1

Complete this table for the function f ( x) = ( x − 2) ( x + 1) ( x − 4): x

−2

−1

f ( x) = ( x − 2) ( x + 1) ( x − 4)

⬚

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203

Determine whether each statement about the cubic function y = f ( x) is true or false: a

The roots of f ( x) = 0 divide the number line into intervals for solving inequalities.

If f ( x) has a positive leading coeﬃcient, y → −∞ as x → +∞.

For f ( x) ≥ 0, the roots are included in the solution.

Determine the roots of ( x − 5) ( x + 2) ( x − 7) = 0.

Practice 4

Identify the intervals where the function f ( x) = ( x + 3) ( x − 1) ( x − 5) changes sign by completing the sign table:

x+3 x−1 x−5 Product

Ex 1

Ex 2

204

x < −3

−3 < x < 1

1<x<5

x>5

⬚

Match each cubic inequality to its solution interval: a

( x − 4) ( x + 2) ( x − 8) > 0

−2 ≤ x ≤ 4 or x ≥ 8

−( x − 4) ( x + 2) ( x − 8) ≤ 0

−2 < x < 4 or x > 8

iii

x ≤ −2 or 4 ≤ x ≤ 8

x < −2 or 4 < x < 8

Solve each inequality using: i

Algebraic method

Graphical method

( x + 1) ( x − 6) ( x − 8) > 0

( x − 7) ( x + 3) ( x − 10) ≤ 0

( x + 6) ( x − 9) ( x − 1) < 0

−( x − 2) ( x + 8) ( x − 10) ≥ 0

( x − 6) ( x + 2) ( x − 9) ≤ 0

−( x + 9) ( x − 1) ( x − 7) ≤ 0 ( x − 8) ( x + 4) ( x − 10) ≥ 0

( x + 7) ( x − 3) ( x − 10) > 0 h

For each inequality: i

Algebraic method

Graphical method

−(2x + 1) (3x − 7) (4x + 5) ≥ 0

(3x − 2) (4x + 7) (5x − 1) ≤ 0

−(4x + 3) (3x − 8) (5x + 1) > 0

(5x − 7) (2x + 9) (3x − 10) ≥ 0

−(2x + 3) (4x − 7) (3x + 10) < 0

(3x − 8) (5x + 1) (4x − 9) > 0

−(5x + 7) (2x − 9) (4x + 3) ≤ 0

(4x − 7) (3x + 8) (5x − 10) > 0

Determine the domain where y = ( x − 2) ( x + 1) ( x − 4) is non-negative.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 3

Determine the solution to the cubic inequalities ( x − 3) ( x + 1) ( x − 6) ≥ 0 and −( x − 3) ( x + 1) ( x − 6) ≥ 0. Comment on your observations.

Solve each inequality using:

Algebraic method

Graphical method

( x − 3)2( x + 1) > 0

( x + 2)2( x − 5) ≤ 0

−( x − 1)2( x + 4) < 0

( x + 5)2( x − 2) ≥ 0

( x − 4) ( x + 3) < 0

−( x + 1)2( x − 7) ≥ 0

( x − 2)2( x + 6) ≤ 0

−( x + 3)2( x − 8) > 0

A rectangular prism has a length of x metres, a width 2 metres less than the length, and a height 4 metres more than the length. The volume of the prism must be at least 21 cubic metres. a

Construct an inequality to represent this condition.

Solve the inequality to determine the possible values for the length, x.

Extend your thinking 12

The function y = ( x − 1) ( x − 3) ( x − 5) has roots at x = 1, 3, 5. Solve y ≥ 0.

The graph represents a function y = f ( x):

Solve f ( x) ≤ 0.

15 10 5 x −4 −3 −2 −1 −5

1 2 3 4 5 6

−10 −15

A box has a square base with side length x metres and a height 2 metres more than the side length. The volume of the box must be at least 45 cubic metres but no more than 96 cubic metres. a

Construct an inequality representing the possible values for the side length, x.

By solving the inequality, find the range of possible values for x.

Determine the values of k such that ( x − 2) ( x + 1) ( x − k) ≥ 0 includes the interval x ≥ 2.

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6.02E Solve inequalities with variables in the denominator After this lesson, you will be able to… • solve inequalities involving expressions with variables in the denominator by identifying critical values and using appropriate algebraic techniques (for example, case analysis, multiplying by the square of the denominator) or graphical analysis • identify critical values (roots of numerators and denominators, or points of undefinedness) and use them to establish intervals for testing inequalities • interpret solutions in the context of the original inequality, ensuring exclusion of values that make denominators zero

Solve inequalities with variables in the denominator When solving an inequality that has a variable in the denominator, like

> 0, we cannot simply

multiply both sides by the denominator. This is because we do not know if the denominator is positive or negative, and multiplying by a negative number would require us to reverse the inequality sign. To solve these types of inequalities correctly, use one of these reliable methods: • Use a number line and test intervals: Find the “critical values” where the numerator is zero or the denominator is zero. These values divide the number line into intervals. Test a point from each interval to see if it satisfies the inequality. • Multiply by the square of the denominator: Since the square of any non-zero number is always positive, multiply both sides by it without changing the direction of the inequality. • Use a sketch: Sketch the graph of the expression and identify the regions where the graph is above or below the required value (for example, above or below the x-axis).

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Exploration Consider the expression y = . The variable in the denominator creates a challenge because its sign changes at x = 0, which would affect the direction of an inequality. Sketch the graph and use one of the methods above to answer these questions: 1. Identify the intervals where

< 1. Explain how the graph’s position relative to the

line y = 1 supports the answer. 2. Identify the intervals where used to verify the solution.

> 1. Show how multiplying both sides by x2 can be

3. If the inequality changes to

≤ 1, how do the solution intervals differ from those

for

< 1? Remember that the original expression is undeﬁned when x = 0.

Example 1 Solve

Create a strategy The critical value where the denominator is zero is x = 1. Solve this while being careful about the sign of the denominator x − 1. Solve by: • Considering two separate cases: when x − 1 > 0 and when x − 1 < 0. • Multiplying both sides by ( x − 1)2 to eliminate the denominator and solving the resulting quadratic inequality. • Sketching the graph of y =

and determining where it lies below the line y = 1.

Apply the idea Method 1: Considering cases Note that x ≠ 1, as this would make the denominator zero. Case 1: The denominator is positive (when x > 1) Write the inequality

Multiply both sides by x − 1

Add 1 to both sides

We need to satisfy both x > 1 and x > 3. The intersection of these is x > 3.

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Case 2: The denominator is negative (when x < 1) Write the inequality

Multiply both sides by x − 1 and reverse the inequality sign

Add 1 to both sides

We need to satisfy both x < 1 and x < 3. The intersection of these is x < 1. Combining the results from both cases, the ﬁnal solution is x < 1 or x > 3. Method 2: Multiplying by the square of the denominator Write the inequality Multiply both sides by ( x − 1)2

Subtract 2( x − 1) from both sides

Factor out the common term ( x − 1)

Simplify the expression inside the brackets

This is a quadratic inequality. The graph of y = ( x − 1) ( x − 3) is an upward-opening parabola with roots at x = 1 and x = 3. The expression is greater than zero (the parabola is above the x-axis) when x < 1 or x > 3. Method 3: Graphical solution and the line y = 1 and determining the values

Solve the inequality by sketching the curve y = of x for which the curve is below the line. 4

y=1

(3, 1)

−4 −3 −2 −1 −1

x 1

−2 −3 −4

The graphs intersect where

= 1 ⟹ 2 = x – 1 ⟹ x = 3. The point of intersection is at (3, 1).

From the sketch, the graph of y =

208

is below the line y = 1 when x < 1 or when x > 3.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 ≤0

Solve

Create a strategy Solve this by finding the critical values for the numerator and the denominator. These values will divide the number line into intervals, which we can then test.

Apply the idea First, factor the denominator: x2 − 8x + 12 = ( x − 2) ( x − 6). The inequality is now

≤ 0.

The critical values are the numbers that make the numerator or denominator equal to zero: • Numerator zero: x − 4 = 0 ⟹ x = 4 • Denominator zero: x − 2 = 0 ⟹ x = 2 and x − 6 = 0 ⟹ x = 6 These values 2, 4, 6 divide the number line into four intervals. Test a value from each interval to see if the expression is negative. Remember that x ≠ 2 and x ≠ 6. Interval

Test value

Sign of expression

Result

x<2

x=0

Valid

2<x<4

x=3

Not Valid

4<x<6

x=5

Valid

x>6

x=7

Not Valid

From the table, the expression is negative on the intervals x < 2 and 4 < x < 6. Now check where the expression is equal to zero. This only happens when the numerator is zero (and the denominator is not). This occurs at x = 4. Since the inequality is ≤, include this point. Combining these results, the ﬁnal solution is x < 2 or 4 ≤ x < 6.

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Reﬂect and check The graphical method conﬁrms this. The graph of y =

is shown. The sections where

the curve is on or below the x-axis satisfy the inequality. The vertical asymptotes at x = 2 and x = 6 (where the denominator is zero) must be excluded. This matches the solution. 4

3 2 1 −1 −1

x 1

−2 −3 −4

Idea summary When solving inequalities with variables in the denominator, you cannot simply multiply by the denominator. Use one of these reliable methods: • Interval testing: Find critical values (where numerator or denominator is zero). Mark them on a number line and test a point in each interval to find where the inequality is true. • Multiply by a positive: Multiply both sides by the square of the denominator. This creates a new inequality that can be solved, as the inequality sign does not change. • Graphing: Sketch the function and identify the x-values for which the graph is in the correct region (e.g., below the x-axis or below a certain line). Always remember to exclude any values of x that make the denominator zero, as the expression is undeﬁned at these points.

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6.02E Practice questions What do you remember? 1

Explain why multiplying both sides of a rational inequality by the denominator directly can lead to errors, and describe a safer method to solve it.

How do you ﬁnd critical points for a rational inequality like expressions?

What is the role of the number line in solving rational inequalities?

Why is the graphical method useful for rational inequalities, and what key features should be identiﬁed on the graph?

< K, where A and B are linear

Practice Ex 1

Solve each inequality: a

For each inequality, sketch the graph of the function and use it to identify the solution intervals: a

For each inequality, sketch the graph of the function noting vertical asymptotes and use it to solve: a

Sketch the graph of y =

A production rate is modelled by

and use it to solve

≥ 0.

≤ 1. Solve using the critical point method.

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Ex 2

Solve each inequality: a

Sketch the graphs of the given functions on the same axes and use them to compare the solution intervals for each pair of inequalities: and

a 12

and

For each production model, sketch the graph of the function and solve the inequality, interpreting the solution in the context of production time (in hours):

a 13

A time estimate model gives

. Solve using all three methods (critical point,

denominator squaring, and graphical), and verify that the solutions are consistent.

Extend your thinking 14

Solve

using both the critical point method and the graphical method, and verify

that the solutions are consistent. 15

A cost model gives

. Solve using the method of multiplying by the square of the

denominator, and interpret the solution in the context of production time (in hours). 16

An eﬃciency model gives

. Solve using both the critical point method and the

graphical method, and interpret the solution in the context of production eﬃciency (in units). 17

Solve

using the method of multiplying by the square of the denominator, and

verify the solution using the graphical method.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

6.03E Solve absolute value inequalities After this lesson, you will be able to… • solve absolute value inequalities of the form ∣ax + b∣ ≥ k, ∣ax + b∣ ≤ k, ∣ax + b∣ < k, and ∣ax + b∣ > k using algebraic methods. • solve absolute value inequalities of the form ∣ax + b∣ ≥ k, ∣ax + b∣ ≤ k, ∣ax + b∣ < k, and ∣ax + b∣ > k using graphical methods. • interpret the absolute value ∣x∣ as the distance of x from zero, and ∣x − c∣ as the distance between x and c. • graph functions of the form y = ∣ax + b∣ and relate them to y = ax + b. • recognise and correctly handle special cases in absolute value inequalities, such as when the constant k is negative.

Solve absolute value inequalities algebraically Absolute value The magnitude or size of a real number, i.e. the distance of the number from the origin on a number line. Formally, it has the piecewise deﬁnition ∣x∣ =

The absolute value of a number, denoted ∣x∣, is its distance from zero on the number line, always non- negative. For an expression ax + b, ∣ax + b∣ is the absolute value of the linear expression ax + b. The expression ax + b itself equals zero when x =

Interactive exploration Discover this concept in action online

mathspace.co

Absolute value inequalities of the form ∣ax + b∣ ≥ k, ∣ax + b∣ > k, ∣ax + b∣ ≤ k, or ∣ax + b∣ < k, where a, b, and k are constants, can be solved algebraically: • For ∣ax + b∣ ≥ k, solve ax + b ≥ k or ax + b ≤ −k. For ∣ax + b∣ > k, solve ax + b > k or ax + b < −k. This inequality represents the value of the expression ax + b at least k units from 0. If a ≠ 0, this corresponds to x-values whose distance on either side of

is at least

units.

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• For ∣ax + b∣ ≤ k or ∣ax + b∣ < k, solve −k ≤ ax + b ≤ k. For ∣ax + b∣ < k, solve −k < ax + b < k. This means the value of the expression ax + b is within k units of 0. If a ≠ 0, this corresponds to x-values whose distance from • If k < 0:

is within

units.

Then ∣ax + b∣ ≥ k or ∣ax + b∣ > k holds for all real x, and ∣ax + b∣ ≤ k or ∣ax + b∣ < k has no solutions.

Example 1 Solve the inequality ∣x + 3∣ < 1.

Create a strategy Solve −1 < x + 3 < 1, representing points within 1 unit of x = −3.

Apply the idea −1 < x + 3 < 1

Write the compound inequality

−4 < x < −2

Subtract 3 from all parts

The solution is −4 < x < −2.

Example 2 Solve the inequality ∣x − 5∣ ≥ 2.

Create a strategy Solve x − 5 ≥ 2 or x − 5 ≤ −2, representing points at least 2 units from x = 5.

Apply the idea x−5≥2

Write ﬁrst inequality

x≥7

Add 5 to both sides

x − 5 ≤ −2

Write second inequality

x≤3

Add 5 to both sides

The solution is x ≤ 3 or x ≥ 7.

Idea summary Absolute value inequalities are solved algebraically using the distance from x = • For ∣ax + b∣ ≥ k or > k: Solve ax + b ≥ k or ax + b ≤ −k. • For ∣ax + b∣ ≤ k or < k: Solve −k ≤ ax + b ≤ k. • If k < 0, ∣ax + b∣ ≥ k holds for all x; ∣ax + b∣ ≤ k has no solutions.

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Solve absolute value inequalities graphically Interactive exploration Discover this concept in action online

mathspace.co

Absolute value inequalities of the form • ∣ax + b∣ ≥ k • ∣ax + b∣ > k • ∣ax + b∣ ≤ k • ∣ax + b∣ < k can be solved graphically by plotting y = ∣ax + b∣ and y = k. The graph of y = ∣ax + b∣ is V-shaped, with a vertex at x =

, representing the point where ax + b = 0.

To solve: • Determine the points of intersection of y = ∣ax + b∣ and y = k • Identify x-values where y = ∣ax + b∣ is above, below, or equal to y = k, based on the inequality. For example, ∣ax + b∣ ≥ k includes x-values where the graph is at or above y = k, corresponding to distances ≥ k from x = . • If k < 0, ∣ax + b∣ ≤ k has no solutions, as the graph is always non-negative.

Example 3 Solve ∣2x + 1∣ > 3 graphically.

Create a strategy Graph y = ∣2x + 1∣ and y = 3, determine the points of intersection, and identify x-values where the graph is above y = 3.

Apply the idea y = ∣2x + 1∣

Determine the points of intersection by solving the two equations. Solving 2x + 1 = 3: 2x + 1 = 3

Write the equation

2x = 2

Subtract 1 from both sides

x=1

Divide both sides by 2

2x + 1 = −3

5 4

y=3

(−2, 3)

Subtract 1 from both sides

x = −2

Divide both sides by 2

(1, 3)

2 1

Write the equation

2x = −4

−5 −4 −3 −2 −1

Solving 2x + 1 = −3:

−1

x 1

−2

The graph is above y = 3 for x < −2 or x > 1, representing distances > 3 from x =

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Example 4 Solve ∣3x − 2∣ ≤ −1 graphically.

Create a strategy Graph y = ∣3x − 2∣ and y = −1, check for regions below or at y = −1.

Apply the idea y = ∣3x − 2∣

4 3 2

(0, 2)

1 −4 −3 −2 −1 −1

y = −1

x 1

−2 −3 −4

The graph y = ∣3x − 2∣ has a range y ≥ 0, never reaching y = −1. There is no real x that satisﬁes ∣3x − 2∣ ≤ −1.

Example 5 Solve ∣x − 3∣ + ∣x + 1∣ ≤ 8: a Algebraically

Create a strategy Identify the critical points to determine the intervals, then solve for each interval.

Apply the idea • The critical points can be determined when each term inside it is equal to 0: For x − 3: x−3=0 x=3

Equate the ﬁrst term to 0 Add 3 to both sides

For x + 1: x+1=0 x = −1

216

Equate the second term to 0 Subtract 1 from both sides

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

If the critical points are x = 3 and x = −1, then the intervals are: • x < −1 • −1 ≤ x < 3 • x≥3 • For x < −1, this tells that both terms must be negative: ∣x − 3∣ + ∣x + 1∣ ≤ 8

Write the inequality

−( x − 3) + (−( x + 1)) ≤ 8

Rewrite the absolute values

−x + 3 − x − 1 ≤ 8

Evaluate the adjacent signs

−2x + 2 ≤ 8 −2x ≤ 6 x ≥ −3

Collect like terms Subtract 2 from both sides Divide both sides by −2

But this must also satisfy x < −1, so the ﬁrst interval is: −3 ≤ x < −1 • For −1 ≤ x < 3, analyse ﬁrst its lower and upper boundaries. For −1 ≤ x or x ≥ −1, this is the same as x + 1 ≥ 0, so ∣x + 1∣ = x + 1. For x < 3, this is the same as x − 3 < 0, so ∣x − 3∣ = −( x − 3). ∣x − 3∣ + ∣x + 1∣ ≤ 8

Write the inequality

−( x − 3) + ( x + 1) ≤ 8

Rewrite the absolute values

−x + 3 + x + 1 ≤ 8

Evaluate the adjacent signs

4≤8

Collect like terms

The statement is true, so the entire interval −1 ≤ x < 3 satisﬁes the inequality. • For x ≥ 3, this tells that both terms must be positive. ∣x − 3∣ + ∣x + 1∣ ≤ 8

Write the inequality

( x − 3) + ( x + 1) ≤ 8

Rewrite the absolute values

x−3+x+1≤8

Evaluate the adjacent signs

2x − 2 ≤ 8

Collect like terms

2x ≤ 10

Add 2 to both sides

x≤5

Divide both sides by 2

But this must also satisfy x ≥ 3, so the third interval is: 3≤x≤5 Combining these intervals results in −3 ≤ x ≤ 5.

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217

b Graphically

Create a strategy Graph y = ∣x − 3∣ + ∣x + 1∣ and y = 8, check for regions below or at y = 8.

Apply the idea From part (a), with the critical points x = −1 and x = 3, these are intervals: • x < −1 • −1 ≤ x < 3 • x≥3 For x < −1, this tells that both terms must be negative: y = ∣x − 3∣ + ∣x + 1∣

Write the equation

= −( x − 3) + (−( x + 1))

Rewrite the absolute values

= −x + 3 − x – 1

Evaluate the adjacent signs

= −2x + 2

Collect like terms

For −1 ≤ x < 3, this tells that ∣x + 1∣ is positive while ∣x − 3∣ is negative: y = ∣x − 3∣ + ∣x + 1∣

Write the equation

= −( x − 3) + ( x + 1)

Rewrite the absolute values

= −x + 3 + x + 1

Evaluate the adjacent signs

Collect like terms

For x ≥ 3, this tells that both terms must be positive: y = ∣x − 3∣ + ∣x + 1∣

Write the equation

= ( x − 3) + ( x + 1)

Rewrite the absolute values

=x−3+x+1

Evaluate the adjacent signs

= 2x − 2

Collect like terms

The inequality can be represented graphically as shown: 10

x = −3

x=5

8 7 6 5 4 3 2 1

−4

−3

−2

−1

x 1

• The green curve is y = ∣x − 3∣ + ∣x + 1∣. • The blue dashed is y = 8. • The shaded region shows where the inequality is ∣x − 3∣ + ∣x + 1∣ ≤ 8.

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Idea summary Graphically solve absolute value inequalities by comparing y = ∣ax + b∣ with y = k: • Plot y = ∣ax + b∣, V-shaped at x =

• Determine the points of intersection through ax + b = k, ax + b = −k. • Select x-values based on inequality (for example, ∣ax + b∣ ≥ k for above y = k). • If k < 0, ∣ax + b∣ ≤ k has no solutions.

6.03E Practice questions What do you remember? 1

Write an absolute value inequality that represents each statement: a

All real numbers x that are less than 8 units away from 0.

All real numbers x that are more than 8 units away from 0.

All real numbers x that are at least 2 units away from 8.

All real numbers x that are at most 5 units away from −2.

Rewrite these absolute value inequalities as compound inequalities or a union of intervals: a

∣x∣ < 2

∣x∣ ≥ 5

∣x∣ > 13

∣x∣ ≤ k, for a positive value of k

x < −8 or x > 8

Rewrite these as absolute value inequalities: a

−7 ≤ x ≤ 7

−0.6 ≤ 2x + 3 ≤ 0.6

For the absolute value inequalities below, test if each x-value is in the solution set: i

x=0

x=5

iii

x = −3

∣4x − 6∣ ≤ 12

∣1.2x + 4∣ ≤ 6.5

∣8 − 5x∣ ≥ 11

x = 2.5

For each absolute value inequality: i

Identify the centre c and distance k using the deﬁnition of the absolute value, which can be interpreted as the distance from a centre, often in the form ∣x − c∣ ≤ k.

Represent on a number line.

∣x − 3∣ ≤ 5

∣x + 4∣ ≤ 2

∣x − 1∣ ≤ 7

∣x + 6∣ ≤ 3

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Practice Ex 1

Ex 2

Ex 3

Ex 4

Solve these inequalities: a

4∣x∣ − 5 < 19

2∣3x + 1∣ < 8

0.3∣x − 4∣ > 0.6

3∣x∣ + 1 < 10

0.5∣2x + 5∣ > 1.5

∣2x + 3∣ ≤ 5

2 + ∣x∣ ≥ 9

0.4∣x − 2∣ ≥ 0.8

Solve these inequalities: a

∣x − 7∣ ≥ 2

3∣x − 7∣ − 12 ≥ 0

∣x − 4∣ ≤ 9

Solve these inequalities using graphical method: a

∣x − 5∣ > 11

∣x + 2∣ < 6

∣3x + 5∣ < 11

∣9 − 2x∣ > 7

∣8 + 6x∣ > 4

0.5∣x − 3∣ < 4

∣x + 3∣ ≤ 7

∣x∣ − 6 > 9

Solve these inequalities using graphical method: a

∣x − 6∣ ≥ 4

∣x − 8∣ ≤ 9

−3 + ∣x∣ ≥ 1

∣x − 7∣ − 6 ≥ −4

0.4∣2x − 3∣ ≥ 0.8

Solve each absolute value inequality algebraically, considering that k < 0: a

∣x − 2∣ < −3

∣x + 5∣ ≥ −1

∣2x − 4∣ ≤ −2

∣3x + 1∣ > −5

∣x − 7∣ < −4

∣x + 3∣ ≥ −6

∣4x − 2∣ ≤ −1

∣2x + 9∣ > −2

For each absolute value inequality, represent the solution on a number line and state the solution interval: a

∣11 − 2x∣ > 5

∣x − 4∣ > 3

∣x + 2∣ ≤ 5

∣2x − 1∣ ≥ 4

∣3x + 6∣ < 9

In a certain company, the measured thickness, m, of a helicopter blade must not differ from the standard, s, by more than 0.17 millimetres. The manufacturing engineer expresses this as the inequality ∣m − s∣ ≤ 0.17. Determine the range of values that m can take if s is 17.92 millimetres.

Mobile phone cases have dimension requirements to ensure the phone will ﬁt properly in the case. The manufacturing engineer has written a speciﬁcation that the new length, n, of the case can differ from the previous length, p, by at most 0.04 centimetres. Determine the range of values for the new length of a mobile phone case if the previous length was 18.9 centimetres.

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If a coin is tossed 100 times, we would expect approximately 50 of the outcomes to be heads. A coin is deemed to be unfair if h, the number of outcomes that result in heads, satisﬁes

≥ 1.645.

Solve the inequality.

Determine the largest and smallest integer values for the number of heads outcomes that could occur without the coin being labelled as unfair.

A freediver holds his breath and dives below the surface of the water. His depth, d in metres, after t seconds, is given by:

Where d = 0 means the diver is at the surface and d = −20 means the diver is 20 metres below the surface.

When the diver is 40 metres or less below the surface he is accompanied by a safety diver. Write an absolute value inequality to model the time when the diver is without a safety diver.

Solve for the range of times during the dive when the diver is without a safety diver.

Determine if t = 180 is a valid solution for the absolute value inequality. Explain what this means in the given context.

For each inequality, write two distinct absolute value inequalities that have the given solution: a

−2 < x < 2

x ≤ −9 or x ≥ 9

1 ≤ x ≤ 7

x < −4 or x > 0

Use the graph of f ( x) = ∣2x − 4∣ and g ( x) = 8 to solve the equation and inequalities: a

∣2x − 4∣ = 8

∣2x − 4∣ ≤ 8

∣2x − 4∣ > 8

12 10 8 6 4 2 −8 −6 −4 −2 −2

x 2 4 6 8 10

−4

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Floyd measures the distance between two cities on a map, which has a scale factor of 1 : 50 000. He determines that the cities are approximately 16 centimetres apart, with a possible error of measurement of up to half centimetre. a

If x represents the actual distance between the two cities, write an absolute value inequality to represent the situation.

The actual distance between the two cities is 12.9 miles (to one decimal place). Determine if Floyd’s measurement was accurate.

Ex 5

Solve the inequality ∣x − 1∣ + ∣x − 2∣ > 2: a

Algebraically

Graphically

Extend your thinking 20

Solve these inequalities: a

2∣4.5 − x∣ > x

∣3x + 4∣ ≥ x + 8

∣3.6x − 7∣ ≥ 1.2x + 5

e 21

Solve the inequality ∣3x + 4∣ ≥ x + 8 using graphical method.

Determine the solution(s) to the inequality ∣2x − 3∣ ≤ 1 − x. Explain what you conclude from the result.

The speedometer in Najah’s car is displaying 56.33 km/h. They know that the speedometer is accurate to within 10% of the actual speed. a

Write an absolute value inequality to represent the situation.

The road Najah is driving down has a speed limit of 64.37 km/h. Is it possible for Najah to be driving over the speed limit? Explain your thinking.

If x represents temperature in Nashville, Tennessee, explain using a suitable example or otherwise a possible interpretation of the inequality ∣x − 84∣ ≤ 7.

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“Every inequality speaks of a tension — and in mathematics, as in life, tension reveals structure.” André Weil

6 Chapter review 1

Which of the following represents the solution to the inequality ( x + 2) ( x − 4) ( x − 7) > 0? A

x < −2 or 4 < x < 7

−2 < x < 4 or x > 7

x < −2 or x > 7

−2 < x < 7 and x ≠ 4

What is the solution to the inequality A

x < 0

x > −3

−3 < x < 0

x < −3 or x > 0

Which inequality represents the solution set for 3∣x∣ − 2 < 10? A

> 5?

∣x∣ < 4

−4 < x < 4

x < −4 or x > 4

Solve the inequality ( x + 1) ( x − 5) ( x − 9) > 0 using: a

Algebraic method

Graphical method

Determine the values of k such that the solution to ( x − 4) ( x + 2) ( x − k) ≥ 0 includes the interval x ≥ 4.

A box has a square base with side length x metres and a height 4 metres more than the side length of the base. The volume of the box must be at least 70 m3 but no more than 120 m3.

Construct a compound inequality representing the possible volumes of the box in terms of x.

Estimate the solution set for the inequality using a graph.

Solve each inequality: b

a 8

Sketch the graph of y =

Solve

d ≤ 1.

and use it to solve

> 0 using both the critical point method and the graphical method, and verify

that the solutions are consistent. 10

A process quality index, Q, is modelled by Q ( x) =

, where x is a process parameter

( x ≠ 8). The process is considered stable if Q( x) ≤ 0. Solve this inequality using both the critical point method and a graphical sketch, and interpret the solution in the context of the process parameter x. 11

Solve these inequalities: a

224

∣x − 8∣ ≥ 3

∣x − 5∣ ≤ 7

∣2x + 5∣ ≤ 7

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3 + ∣x∣ ≥ 10

Solve using the graphical method: a

∣x + 3∣ < 7

For each absolute value inequality, represent the solution on a number line and state the solution interval: a

∣x − 6∣ > 10

∣x − 5∣ > 4

∣x + 3∣ ≤ 6

For a standardised test, scores s are analysed. A score is considered an outlier ≥ 1.96.

if a

Solve the inequality for s.

Determine the smallest integer score above the mean (70) and the largest integer score below the mean that would be considered outliers.

Solve the inequality ∣2x + 5∣ ≥ x + 7 using the graphical method.

A clinical thermometer displays a reading Tr. It is known to be accurate to within 4% of the actual temperature Ta.

Write an absolute value inequality to represent the possible range of the actual temperature Ta in terms of the reading Tr.

If the thermometer reading Tr is 37.8°C, determine the range of possible actual temperatures (to two decimal places). If a fever is deﬁned as an actual temperature greater than 37.5°C, is it possible the person has a fever based on this reading? Explain.

Solve

≥ ∣x − b∣, 0 < b < 2 using the graphical method.

Did you know?

Inequalities are essential in creating balanced and engaging video games! For example, developers use inequalities to set limits on player health, control the speed of characters, and define the boundaries of game worlds. By applying mathematical conditions, they can regulate enemy difficulty, manage scoring systems, and ensure players progress through levels at an appropriate pace. Inequalities make it possible to design challenging, fair, and immersive gaming experiences that keep players coming back for more!

Chapter 6 review mathspace.co

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7 Trigonometry Chapter outline 7.A 7.B 7.01 7.02 7.03 7.04 7.05 7.06 7.07E

228 241 248 259 268 276 296

You can measure a tree’s height without climbing it — just use its shadow and trigonometry.

7.01 Unit circle After this lesson, you will be able to… • define sine, cosine and tangent using the coordinates of a point on the unit circle. • determine the sign of a trigonometric ratio for an angle in any of the four quadrants. • evaluate trigonometric ratios for quadrant boundary angles (0°, 90°, 180°, 270°). • represent angles of any magnitude on the Cartesian plane. • evaluate trigonometric ratios for angles greater than 360° and for negative angles.

The unit circle If a right-angled triangle is placed on the Cartesian plane at the origin and the hypotenuse is made to have a length of 1, trigonometric ratios can be written in terms of x and y. Trigonometric ratios The relationship between the angles and sides of right-angled triangles. The 3 basic trigonometric ratios are sine, cosine and tangent.

P (x, y) 1

−1

θ x

The trigonometric ratios become: y 1

−1 If any right-angled triangle with a hypotenuse of 1 is looked at, the endpoint of the hypotenuse will lie on a circle with radius 1, known as the unit circle. Angles in the unit circle are measured from the positive x-axis to the line segment joining the origin to a point on the circle.

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1 P (cos θ , sin θ )

Unit circle

Radius

θ x

−1

1 x

θ is the angle formed by rotating anticlockwise from the positive x-axis. Since cos θ = x and sin θ = y, the coordinates of the endpoint can be represented by: P (cos θ, sin θ ) The ratio , which is the gradient of the hypotenuse, can be represented by:

Therefore, m = tan θ.

−1 This will be true in all quadrants.

Example 1 The graph shows an angle a with its terminal side intersecting the circle at

1 −1

5 12 , 13 13

1 a Determine the exact value of sin a.

Create a strategy The value of sin a is the y-coordinate of point P.

Apply the idea

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Reﬂect and check If a vertical line is drawn from point P to the x-axis, it gets the triangle shown. y

P 1

5 12 , 13 13

Write the sine ratio

12 13

5 13

Substitute O =

Simplify

b Determine the exact value of cos a.

Create a strategy The value of cos a is the x-coordinate of point P.

Apply the idea

c Determine the exact value of tan a.

Create a strategy Use tan a = , the gradient of the line from the origin to point P.

Apply the idea Write the gradient formula Substitute x =

230

and y =

Evaluate the division

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

and H = 1

Example 2 Determine the coordinates of P (x, y), rounded to two decimal places.

P 72°

−1

Create a strategy When the angle in the ﬁrst quadrant is not a known angle (0°, 30°, 45°, 60°, 90°), use technology to evaluate. Recall cos θ = x and sin θ = y.

Apply the idea

Reﬂect and check

Using technology, the values are:

When using technology, make sure it is set to degree mode when angles are given in degrees. Using the other mode (radian mode) will give different values because it is a different unit.

cos (72°) ≈ 0.31 sin (72°) ≈ 0.95 The coordinates of the point are (0.31, 0.95).

Idea summary The unit circle is a circle with a radius of 1. A point on the unit circle, after having rotated by an angle of measure θ in the anticlockwise direction, can be represented by P (cos θ, sin θ ). 1

P (x, y) 1

−1

θ x

The trigonometric ratios are: • sin θ = y, or the height of the triangle.

y 1

• cos θ = x, or the length of the base of the triangle. • tan θ = , or the gradient of the hypotenuse.

−1

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Signs of trigonometric ratios The unit circle shows that sin θ, cos θ, and tan θ are deﬁned for angles larger than what can be contained in a right-angled triangle.

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Second

First

Third

Fourth

Recall that sin θ is equal to the y-coordinate (height), cos θ is equal to the x-coordinate (length), and tan θ is the gradient of the line from the origin to the point.

(cos θ , sin θ )

cos θ sin θ

−1

Second quadrant: • y is positive: sin θ is positive • x is negative: cos θ is negative • Gradient is negative: tan θ is negative

−1

θ −1

1 sin θ

Third quadrant: • y is negative: sin θ is negative • x is negative: cos θ is negative • Gradient is positive: tan θ is positive

cos θ (cos θ , sin θ ) −1

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−1 sin θ

Fourth quadrant: • y is negative: sin θ is negative • x is positive, cos θ is positive • Gradient is negative: tan θ is negative

cos θ (cos θ , sin θ )

−1

S sin positive

A all positive

θ −1

T tan positive

−1

C cos positive

Notice that: • In Quadrant 1, all ratios are positive. (A) • In Quadrant 2, only sin θ is positive. (S) • In Quadrant 3, only tan θ is positive. (T) • In Quadrant 4, only cos θ is positive. (C ) A helpful mnemonic for positive trigonometric function values is “All Stations To Central” (ASTC ). This can help remember the signs of the trigonometric ratios in each quadrant.

Example 3 Will tan 195° have a positive or a negative answer?

Create a strategy Think about which quadrant the angle is in and what the sign of tangent will be in that quadrant.

Apply the idea The angle 195° is between 180° and 270°, so it lies in Quadrant 3. In Quadrant 3, both x and y are negative, so

So, tan 195° will have a positive answer.

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Reﬂect and check y 1

195°

When the angle is drawn, the ray from the origin slopes upwards from left to right, showing a positive gradient. This conﬁrms that tan θ is positive.

−1

1 −1

Idea summary y

The sign of trigonometric functions can be determined by relating the point P (x, y) on the unit circle to the trigonometric functions P (cos θ, sin θ ).

(−, +)

(+, +)

−x

In the quadrants where x is positive or negative, cos θ will be positive or negative.

(−, −) 3

(+, −)

In the quadrants where y is positive or negative, sin θ is positive or negative, while tan θ is the gradient of the line,

−y

. S sin positive

A all positive

A helpful mnemonic for positive trigonometric function values is “All Stations To Central” (ASTC ).

θ −1

• In Quadrant 1, all ratios are positive. • In Quadrant 2, only sin θ is positive. • In Quadrant 3, only tan θ is positive.

T tan positive

234

−1

C cos positive

• In Quadrant 4, only cos θ is positive.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Trigonometric ratios for any angle magnitude The unit circle extends the deﬁnitions of trigonometric ratios cos θ = , sin θ =

and tan

to angles of any magnitude, including positive, negative and angles beyond 360°. 1

P (cos θ , sin θ )

−1 x

−1

y 1 x

For any point P (x, y) on the unit circle (radius r = 1), rotated by angle θ anticlockwise from the positive x-axis: • cos θ = x • sin θ = y • tan θ =

(undeﬁned when x = 0)

These deﬁnitions apply to all angles, including negative angles (clockwise rotation) and angles greater than 360°.

Angles of any magnitude can be represented by adding or subtracting multiples of 360°:

θ ± 360° × n, where n is an integer For negative angles, rotate clockwise from the positive x-axis. For example, θ = −30° is equivalent to 360° − 30° = 330°. A special case occurs when the angle is 360°, it lies in the same position as 0°, so its related angle is 0°.

Example 4 Find the exact values of sin (−450°), cos (−450°), and tan (−450°).

Create a strategy Reduce the angle to an equivalent angle between 0° and 360° by adding or subtracting multiples of 360°. Identify the quadrant, determine the signs of the ratios, and evaluate using the unit circle coordinates.

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Apply the idea Reduce −450°: −450° = −450° + 360° + 360°

Add 360° twice

= 270°

Evaluate

At θ = 270°, point P (0, −1) on the unit circle (negative y-axis): • sin (270°) = y = −1 • cos (270°) = x = 0 • tan (270°) =

(undefined)

Thus: • sin (270°) = −1 • cos (270°) = 0 • tan (270°) = Undefined

Reﬂect and check Verify by sketching −450°, so rotate 450° clockwise (1.25 full rotations), landing at 270°. 1

−1

−450° 1 x

270° −1 The coordinates (0, −1) conﬁrm the results.

Example 5 Determine the coordinates of point P on the unit circle for θ = 780°, rounded to two decimal places.

Create a strategy Reduce the angle to fall within the interval 0° ≤ θ < 360°, ﬁnd the quadrant, and calculate cos θ and sin θ using technology in degree mode.

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Apply the idea Reduce 780°: 780° = 780° − 360° − 360°

Subtract 360° twice

= 60°

Evaluate

At θ = 60° (Quadrant 1): • cos (60°) =

= 0.50

• sin (60°) =

= 0.87

So the coordinates are P (0.50, 0.87).

Reflect and check Since 780° = 720° + 60°, two full rotations plus 60° lands in Quadrant 1, matching the positive coordinates.

Idea summary The unit circle extends trigonometric ratios to any angle magnitude: cos θ = x, sin θ = y, tan θ =

(undeﬁned at x = 0).

For angles θ : • Reduce to lie within 0° ≤ θ < 360° by adding or subtracting multiples of 360°, that is, θ ± 360 × n, where n is an integer. • If negative, rotate clockwise.

7.01 Practice questions What do you remember? 1

Identify the quadrants in which these angles are located: a

278°

25°

142°

Identify the angles that correspond to the quadrant boundaries.

For each trigonometric ratio, identify the quadrants where it is positive: a

Sine

Cosine

Tangent

tan θ > 0, sin θ < 0

208°

Identify the quadrant where angle θ is located: a

sin θ > 0, cos θ < 0

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Practice 5

Calculate the following, rounded to two decimal places: a

tan 172°

sin w and cos x are positive

sin w and cos x are negative

sin 410°

tan y is negative

w, x are in Quadrant 3

For each diagram, determine the value of: i

sin θ

−1

cos θ

iii

P −

8 15 , 17 17

1 1

−1

tan θ

4 9 P  ,   9 10 

238

Consider the angles w = 238° and x = 282°. Without using technology, determine whether the statements are true or false: a

Ex 1

cos 118°

Consider the angles w = 68°, x = 22° and y = −42°. Without using technology, determine whether the statements are true or false: a

sin 138°

3 4 ,− 5 5

−1

4 3 P − ,− 5 5

−1

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

Determine the coordinates of P (x, y), rounded to two decimal places: a

1 P

−1

35°

sin 305°

tan θ

in interval of 180° < θ < 270°, calculate the exact value of:

cos θ

sin (−270°)

θ = −510°

Given sin 45° = a

tan θ

sin θ

cos (540°)

tan (−90°)

sin (450°)

Determine the coordinates of point P on the unit circle for each angle, rounded to two decimal places: a

tan 245°

Find the exact values of these trigonometric ratios using the unit circle: a

Ex 5

Given tan θ = −1.8, 270° < θ < 360°, calculate the exact value of: a

Ex 4

cos 152°

Given sin θ = 0.7, tan θ < 0, calculate the exact value of: a

sin θ

Given sin θ = a

sin 28°

Given cos θ = , 270° < θ < 360°, calculate the exact value of: a

Without using technology, determine whether the values are positive or negative: a

−1

Ex 3

155°

−1

cos 135°

θ = 930°

θ = 420°

θ = −210°

tan 315°

sin (−45°)

, calculate the exact value: b

sin 225°

Determine the value of each trigonometric ratio for quadrant boundary angles, if deﬁned: a

sin 0°

cos 90°

tan 180°

sin 270°

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Extend your thinking 19

For a unit circle with point P (a, b) at 58°, express each in terms of a, b: a

sin 58°

cos 238°

tan 302°

Points P, Q, R, S represent angles 42°, 138°, 222°, 318°, respectively. If P (a, b), find the coordinates of Q, R, S in terms of a, b.

Simplify: a

sin (180° + x) − sin x

cos (360° − x) + cos x

Given point C(0.4067, 0.9135) on the unit circle and ∠COD = 48°, find the coordinates of D to three decimal places.

Given cos 15° ≈ 0.97, sin 15° ≈ 0.26, determine each value rounded to two decimal places: a

sin 165°

cos 195°

sin 345°

cos −165°

A point P (a, b) lies on the unit circle at θ = −420°. Show that sin2(−420°) + cos2(−420°) = 1 using the coordinates (a, b).

Did you know?

Radio and Wi-Fi signals travel as waves! Engineers use the unit circle to map their peaks and troughs, helping them strengthen coverage and reduce interference. By calculating precise angles, they can measure wavelength and frequency, then design antennas that match the wave patterns for a stronger signal. Every time your phone connects to Wi-Fi, it’s relying on tiny electromagnetic waves — and engineers use unit circle maths to keep those waves in sync, strong, and stable. Without these calculations, signals would be weak and unreliable. This precise maths makes modern communication fast, efficient, and seamless.

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7.02 Related angles and identities After this lesson, you will be able to… • define and find the related angle for an angle in any quadrant. • use the related angle and quadrant sign (ASTC) to evaluate trigonometric ratios. • apply the identities for angles of the form 180° ± A and 360° − A. • apply the identities for negative angles to simplify expressions. • evaluate trigonometric ratios for angles of any magnitude without a calculator.

Related angles and trigonometric ratios The related angle (or reference angle) for an angle θ (not a multiple of 90°) is the acute angle between the ray and the x-axis on the unit circle. This allows trigonometric ratios for any angle to be expressed using an acute angle. Second

First Angles are classiﬁed by quadrant: • Quadrant 1: 0° < θ < 90° • Quadrant 2: 90° < θ < 180° • Quadrant 3: 180° < θ < 270° • Quadrant 4: 270° < θ < 360°

Third

Fourth

Related angles are found as: • Quadrant 1: θ • Quadrant 2: 180° − θ • Quadrant 3: θ − 180° • Quadrant 4: 360° − θ For angles outside 0° to 360°, add or subtract multiples of 360° to bring them into this range, then ﬁnd the related angle.

7.02 Related angles and identities mathspace.co

241

Signs of trigonometric ratios depend on the quadrant (ASTC ): S sin positive

A all positive (cos θ , sin θ )

θ −1

T tan positive

−1

C cos positive

• Quadrant 1: All positive • Quadrant 2: sin θ positive • Quadrant 3: tan θ positive • Quadrant 4: cos θ positive

Angles can be expressed as θ = 180° ± A or θ = 360° − A, where A is an acute angle. The identities are: sin θ

cos θ

tan θ

180° − A

sin A

−cos A

−tan A

180° + A

−sin A

−cos A

tan A

360° − A

−sin A

cos A

−tan A

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Example 1 Find the related angle for 750° and express cos 750° using an acute angle.

Create a strategy Reduce 750° to an angle between 0° and 360°, identify the quadrant, and apply the cosine formula for the related angle.

Apply the idea Reduce the angle: 750° = 750° − 360° − 360° = 30°

Subtract 360° twice Evaluate

The angle 30° is in Quadrant 1, where cos θ = cos A and the related angle is 30°: cos 750° = cos 30°

242

Apply the quadrant rule

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Evaluate sin 300° without technology.

Create a strategy Identify the quadrant, ﬁnd the related angle, determine the sign, and use the exact value.

Apply the idea The angle 300° is in Quadrant 4 (270° to 360°). To find the related angle, apply the formula θ = 360° − A:

θ = 360° − A

Write the formula

= 360° − 300°

Substitute A = 300°

= 60°

Evaluate

In Quadrant 4, sin θ = −sin A and sin 60° =

: Apply the quadrant rule

Substitute the exact value

Example 3 Express tan 195° using an acute angle and determine its value, rounded to two decimal places.

Create a strategy Identify the quadrant, find the related angle, then apply the tangent formula.

Apply the idea The angle 195° is in Quadrant 3 (180° to 270°). To find the related angle, use the formula θ = 180° + A:

θ = 180° + A

Write the formula

195° = 180° + A

Substitute θ = 195°

A = 15°

Subtract 180° from both sides

For θ = 180° + A, tan θ = tan A. Thus: tan 195° = tan 15° = −0.86

Apply the formula Evaluate and round

Reﬂect and check In Quadrant 3, tangent is positive (ASTC ). The exact value of tan 15° requires further computation (e.g., using tan (45° − 30°)), but the expression is tan 15°.

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Idea summary Related angles are acute angles between the ray and the x-axis. Trigonometric ratios for any angle are found using the related angle and quadrant signs (ASTC ). Key identities include sin (180° ± A), cos (180° ± A), tan (180° ± A), and sin (360° − A), cos (360° − A), tan (360° − A), used to solve problems involving angles in any quadrant.

Negative angle identities Negative angles are formed by clockwise rotation on the unit circle. For an angle −θ, the coordinates of the point on the unit circle are (cos (−θ ), sin (−θ )). y

1 (cos 60°, sin 60°) 1 −1

60°

−60°

For θ = 60°, the point is (cos 60°, sin 60°). For −θ = −60°, the point is (cos 60°, −sin 60°), showing the x-coordinate is unchanged but the y-coordinate is negative.

−1 (cos 60°, −sin 60°) The x-coordinate remains the same, but the y-coordinate changes sign. This principle underpins the derivation of the following fundamental identities: This line of reasoning leads to these identities:

cos (−θ ) = cos θ sin (−θ ) = − sin θ

θ is acute angle between the ray and the x-axis

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Example 4 Find sin (−60°) and cos (−60°) using identities.

Create a strategy Apply sin (−θ ) = − sin θ and cos (−θ ) = cos θ with known values for 60°.

Apply the idea Use the fact that sin 60° =

, cos 60° = . Apply the identity Substitute sin 60° =

Apply the identity Substitute cos 60° =

Example 5 Express tan (−150°) using a positive acute angle and determine its sign.

Create a strategy Convert −150° to a positive angle, ﬁnd the related angle, apply tan (−θ ) = −tan θ, and check the quadrant’s sign.

Apply the idea Convert −150° to a positive angle: −150° + 360° = 210°

Add 360°

The angle 210° is in Quadrant 3 (180° to 270°). To find the related angle, use the formula θ = 180° + A:

θ = 180° + A

Write the formula

210° = 180° + A

Substitute θ = 210°

A = 30°

Subtract 180° from both sides

For Quadrant 3, tan θ = tan A (positive). Thus: tan (−150°) = tan 30°

Apply the identity tan θ = tan A

So, tan (−150°) = tan 30° is positive.

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Idea summary Negative angle identities are cos(−θ ) = cosθ, sin(−θ ) = −sinθ, and tan(−θ ) = − tanθ, derived from the unit circle’s coordinates. These allow trigonometric ratios for negative angles to be computed using positive angle.

7.02 Practice questions What do you remember? 1

Write the related angle formula for each quadrant: a

Quadrant 2

Quadrant 3

Quadrant 4

Identify which trigonometric ratio(s) are positive in each quadrant (ASTC ): a

Quadrant 1

Quadrant 2

Quadrant 3

tan (−θ )

Quadrant 4

Write the negative angle identities for: a

sin (−θ )

cos (−θ )

Practice Ex 1

For each angle, find the related angle and express it with the corresponding trigonometric ratio: a

Ex 2

sin 120°

cos 225°

−210°, cos(−210°)

tan 300°

tan 165°

sin 240°

cos 330°

sin (−30°)

cos (−45°)

tan (−60°)

sin 135°

cos 210°

tan 315°

480°

−420°

Express as a ratio of an angle between 0° and 90°: a

246

Reduce each angle to fall between 0° to 360° and ﬁnd its related angle: a

510°, tan 510°

Determine the quadrant and sign of each ratio: a

Find the exact value of each ratio using negative angle identities: a

225°, sin 225°

Express each ratio using an acute angle and determine its value, rounded to two decimal places: a

Ex 4

Evaluate each trigonometric ratio, without using technology: a

Ex 3

120°, cos 120°

sin 150°

cos 270°

tan 210°

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Identify the related angle for 135° and evaluate sin 135°.

135° x

Complete the table for θ = 150°: sin θ

cos θ

tan θ

⬚

150°

Evaluate each expression using negative angle identities: a

−300°

−480°

sin (180° + θ )

cos (360° − θ )

tan (180° + θ )

sin 200°

cos 150°

Find the exact value of tan 330°.

Express each ratio using a positive acute angle and determine its sign: a

sin 135°

cos 210°

tan 300°

tan 135° + cos 180°

Evaluate each expression: a

sin (−θ )

tan 240°

Determine if evaluating these ratios will give positive or negative answer: a

Ex 5

cos (−135°)

Rewrite each ratio as an equivalent trigonometric ratio of a positive reference angle θ : a

Express each as a positive angle and ﬁnd the related angle: a

sin (−120°)

sin 150° cos 30°

Complete the table for negative angles:

−120° −225°

sin θ

cos θ

tan θ

⬚

Find the related angle and evaluate cos 510°, rounded two decimal places.

Find the exact value of sin (−210°). 7.02 Related angles and identities mathspace.co

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Extend your thinking 23

Prove that sin (180° − θ ) = sin θ using the unit circle.

Show that tan (180° − θ ) = −tan θ using tan θ =

Find all angles θ between −360° and 360° where sin θ = .

Explain why tan 270° is undeﬁned using the unit circle.

7.03 Sine and cosine rules After this lesson, you will be able to… • apply the sine and cosine rules to find unknown sides and angles in nonright-angled triangles. • use the formula A =

ab sin C to calculate the area of a triangle.

• identify the ambiguous case of the sine rule and determine all possible solutions. • choose the most appropriate rule to solve problems involving non-rightangled triangles.

Sine rule and area of triangle formula Sine rule Relates the lengths of the sides of a triangle to the sines of its angles:

The sine rule applies to non-right-angled triangles to ﬁnd unknown sides or angles when two angles and a side, or two sides and a non-included angle, are known. Proof of the sine rule: Consider this triangle with sides a, b, and c opposite angles A, B, and C, respectively. To establish the sine rule, construct the altitude from vertex C to the opposite side AB, intersecting at point D. This altitude, labelled h, divides the triangle into two right-angled triangles: C b

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Using the deﬁnition of the sine function in right-angled triangles: • In △ACD, sin A =

, so h = b sin A.

• In △BCD, sin B =

, so h = a sin B.

Since both expressions equal h, they can be equated: b sin A = a sin B Since sin A ≠ 0 and sin B ≠ 0 in a non-degenerate triangle, divide both sides by sin A sin B to obtain:

Note: This result holds for obtuse angles since sin (180° − θ ) = sin θ. A similar argument applies by constructing the altitude from another vertex (for example, from A to BC ), leading to:

a Thus, combining both results leads to the conclusion that: c

Area of triangle Calculated using A =

bh, where b is the base length and h is the perpendicular height. For a

triangle ABC with sides a and b about the angle C, the area of the triangle is A =

ab sin C.

The area of the triangle formula calculates the area of a triangle using its two sides and the included angle. The area of triangle formula is:

is the area of triangle

a, b

are the sides adjacent to angle C

is the included angle

Proof of the area of non-right-angled triangles: Consider triangle ABC with base a (side BC ) and height h from A: A The area of a triangle with base and height is: c B

h a

A= C

base × height

But in this triangle, base = a and height h = b sin C.

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Write the formula Substitute base = a and h = b sin C

Similarly, the area can be expressed as A =

bc sin A, depending on the base and height used.

Summary of formulas: Formula Sine rule Area of triangle formula To apply the sine rule: • Know one angle-side pair and either another angle (for a side) or another side (for an angle). • Verify: smallest angle faces shortest side. To compute area: • Use two sides and the included angle.

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Example 1 Find side a in triangle ABC, with ∠A = 35°30′, ∠B = 72° and b = 20, rounded to two decimal places.

Create a strategy Given an angle-side pair and another angle, convert ∠A to decimals and use the sine rule to find a.

Apply the idea Convert A = 35°30′ to decimal degrees: Divide 30′ by 60′ to convert minutes to degrees

Evaluate the division

Evaluate

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Calculating side a: Write the sine rule

Substitute A = 35.5°, B = 72° and b = 20

Multiply both sides by sin 35.5°

Evaluate and round

Reﬂect and check Since 35.5° < 72°, side a should be shorter than b = 20, which 12.22 satisﬁes.

Example 2 Calculate the area of triangle ABC with a = 12, b = 18 and ∠C = 50°, rounded to two decimal places.

Create a strategy Use the area of triangle formula A =

ab sin C with given sides and included angle.

Apply the idea Write the formula

Substitute a = 12, b = 18 and C = 50°

Evaluate and round

Idea summary The sine rule ﬁnds sides or angles in non-right-angled triangles using:

The area of triangle formula A =

ab sin C uses two sides and the included angle.

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Ambiguous case of the sine rule Ambiguous case (of the sine rule) In trigonometry, refers to using the sine rule to calculate the size of an angle in a triangle where there are two possibilities for the angle, one obtuse and one acute, leading to two possible triangles. This occurs because, for an acute angle θ, sin (180° − θ ) = sin θ. The ambiguous case arises in SSA triangles (two sides and a non-included angle) when using the sine rule, as sin x = sin (180° − x) may yield two possible angles for a solution.

B1, B2 are the possible angles opposite side b in triangle ABC a, b

are the known side lengths in SSA conﬁguration

is the known angle in SSA conﬁguration

Always verify the triangle’s angle sum (A + B + C = 180°) to conﬁrm valid solutions.

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For triangle ABC with known a, b, and ∠A (acute), the ambiguous case occurs if: • No solution: a < b sin A (no possible triangle) • One solution: a ≥ b or a = b sin A • Two solutions: b sin A ≤ a < b (two triangles) Geometric construction: C2

C1 b

Possibility 1

a B2 = 180° − B1

Possibility 2

Fix ∠A and side b (AC1 ). Draw side a (B1C1 ). Point B can lie at two positions (B1, B2), forming angles x and 180° − x, if b sin A ≤ a < b.

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Example 3 In triangle ABC, ﬁnd all possible ∠B given a = 10, b = 12 and ∠A = 30°, rounded to one decimal place.

Create a strategy Use the sine rule to ﬁnd ∠B, then check for a second solution using B2 = 180° − B1 and verify triangle validity with the angle sum.

Apply the idea Apply the sine rule to ﬁnd B1: Write the inverted sine rule

Substitute A = 30°, a = 10 and b = 12

Multiply both sides by 12

Evaluate

Take the inverse sine of both sides

Evaluate and round

For the second solution, use the formula B2 = 180° − B1 : B2 = 180° − B1

Write the formula

= 180° − 36.9°

Substitute B1 = 36.9°

= 143.1°

Evaluate

Verify for B1 = 36.9°: ∠A + ∠B + ∠C = 180° 30° + 36.9° + ∠C = 180° 66.9° + ∠C = 180° ∠C = 113.1°

Use the triangle sum rule Substitute ∠A = and ∠B = 36.9° Combine like terms Subtract 66.9° from both sides

Verify for B2 = 141.1°: ∠A + ∠B + ∠C = 180°

Use the triangle sum rule

30° + 141.1° + ∠C = 180°

Substitute ∠A = and ∠B = 141.1°

173.1° + ∠C = 180° ∠C = 6.9°

Combine like terms Subtract 173.1° from both sides

Since the values are valid, the possible values are ∠B = 36.9° or 143.1°.

Reﬂect and check Check conditions: a = 10 < b = 12 and 10 ≥ 12 sin 30° = 6 conﬁrm the ambiguous case, supporting two valid triangles.

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Idea summary The ambiguous case (of the sine rule) occurs in SSA triangles, where the sine rule may yield two angles:

Verify triangle validity by ensuring A + B + C = 180°. The ambiguous case occurs when b sin A ≤ a < b, producing two triangles.

Cosine rule Cosine rule A formula c2 = a2 + b2 − 2ab cos C that relates the lengths of the sides of a triangle to the cosine of one of its angles.

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The cosine rule applies to non-right-angled triangles when two sides and the included angle, or three sides, are known. It ﬁnds a side or angle without an ambiguous case. To ﬁnd a side:

c2 = a2 + b2 — 2ab cos C c

is the side opposite to angle C

a, b

are the sides adjacent to C

is the side opposite to angle C

a, b

are the other two sides

To ﬁnd an angle:

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Proof of the cosine rule: Consider the triangle with sides a, b, and c opposite angles A, B, and C, respectively. To establish the cosine rule, construct the altitude from vertex B to the opposite side AC, intersecting at point D. This altitude h divides the triangle into right-angled triangles, △ABD and △BCD. Let CD = x, so that AD = b − x: B c

b−x

h D

Using Pythagoras’ theorem: • In △BCD, h2 + x2 = a2 • In △ABD, h2 + (b − x)2 = c2 h2 + b2 − 2bx + x2 = c2 a2 − x2 + b2 − 2bx + x2 = c2 a2 + b2 − 2bx = c2

Expand (b − x)2 Substitute h2 = a2 − x2 from the ﬁrst equation Simplify

In △BCD, which has a right angle at point D, cos C = , since x is the adjacent side and a is the hypotenuse. Thus, x = a cos C: a2 + b2 − 2ab cos C = c2 Substitute x = a cos C Thus, the cosine rule is given by: c2 = a2 + b2 − 2ab cos C Applications: • Two sides and included angle to ﬁnd the opposite side. • Three sides to ﬁnd an angle.

Example 4 In triangle ABC, ﬁnd side c given a = 14, b = 25 and ∠C = 38°, rounded to two decimal places.

Create a strategy Use the cosine rule to ﬁnd side c opposite to ∠C.

Apply the idea Write the cosine rule

Substitute a = 14, b = 25 and C = 38°

Evaluate each term

Take the square root of both sides

Evaluate and round

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Example 5 In triangle ABC, ﬁnd ∠A given a = 17, b = 20 and c = 22, rounded to one decimal place.

Create a strategy Use the cosine rule to ﬁnd ∠A opposite to side a.

Apply the idea Write the cosine rule

Substitute a = 17, b = 20 and c = 22

Evaluate the numerator and denominator

Take the inverse cosine of both sides

Evaluate and round

Reﬂect and check Since cos A is positive, ∠A is acute, consistent with 47.3°.

Idea summary Use cosine rule in non-right-angled triangles: • To finds side using: c2 = a2 + b2 − 2 ab cos C • To finds angles using:

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7.03 Practice questions What do you remember? 1

Write each rule or formula for triangle ABC: a

Sine rule

Cosine rule

Area formula

Identify which rule (sine or cosine) applies to ﬁnd the unknown in each case: a

Given two sides and the included angle, ﬁnd opposite side

Given two angles and one side, ﬁnd another side

Given three sides, ﬁnd an angle

Given two sides and a non-included angle, ﬁnd angle

Write the condition for the ambiguous case in triangle ABC with known a, b, ∠A (acute).

Write the area formula for a triangle given sides p, q and included angle R.

Practice Ex 1

Find side a in triangle ABC with ∠A = 42°15′, ∠B = 68°, and b = 15, rounded to two decimal places.

Ex 2

Calculate the area of triangle PQR with p = 10, q = 14, and ∠R = 55°, rounded to two decimal places.

Ex 3

Find all possible ∠C in triangle ABC with a = 14, b = 18, and ∠A = 35°, rounded to one decimal place.

Ex 4

Find side c in triangle ABC with a = 16, b = 20, and ∠C = 45°, rounded to two decimal places.

Calculate the area of triangle DEF with d = 8, e = 12, and ∠F = 60°, rounded to two decimal places.

Find side b in triangle ABC with a = 10, c = 14, and ∠B = 80°, rounded to two decimal places.

Find ∠A in triangle ABC with a = 9, b = 11, and c = 13, rounded to one decimal place.

Determine if triangle ABC with a = 12, b = 15, and ∠A = 50° has one, two, or no solutions for ∠B.

Find side q in triangle PQR with p = 7, r = 9, and ∠Q = 70°, rounded to two decimal places.

Calculate the area of triangle XY Z with x = 20, y = 25, and ∠Z = 45°30′, rounded to two decimal places.

Find ∠C in triangle ABC with a = 30, b = 40, and c = 50, rounded to one decimal place.

Ex 5

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Find all possible ∠B in triangle ABC with a = 8, b = 10, and ∠A = 30°, rounded to one decimal place.

Determine the number of possible triangles for ABC with a = 6, b = 8, and ∠A = 40°. Explain using the ambiguous case condition.

Extend your thinking 18

Prove the sine rule for triangle ABC using a perpendicular from A to BC.

Prove the cosine rule for triangle ABC using a perpendicular from B to AC.

Two hikers, A and B, start at the same point P. Hiker A walks 12 km on a bearing of 65°. Hiker B walks 15 km on a bearing of 160°. Calculate the bearing of Hiker A from Hiker B, rounded to the nearest degree.

Two ships leave a port. Ship A travels 30 km at 20° north of east, and Ship B travels 40 km at 50° north of east. Calculate the distance between them after 1 hour, rounded to two decimal places.

In a quadrilateral ABCD, diagonal AC divides it into triangles ABC and ADC. In triangle ABC, a = 20, ∠B = 65°, ∠C = 45°. In triangle ADC, d = 25, ∠D = 70°. Find the length of diagonal AC, rounded to two decimal places.

Prove the triangle area formula A = to BC.

A rhombus has side length 15 cm and one diagonal of 24 cm. Find the length of the other diagonal using the cosine rule, rounded to two decimal places.

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ab sin C for triangle ABC using a perpendicular from A

7.04 Radians After this lesson, you will be able to… • define a radian and explain the relationship 2π radians = 360°. • convert angle measures between degrees and radians. • recall and find the exact trigonometric ratios for key angles expressed in radians. • find the related angle in radians for an angle in any quadrant. • evaluate trigonometric expressions involving radian measure.

Degrees and radians Degrees A unit for measuring an angle. Angles are measured as a proportion of a full turn, which is equivalent to 360 degrees, so that one degree, written as 1° is equal to

of a full turn.

Radian A unit of angular measure frequently used in mathematics. 1 radian is the angle between two radii of a circle which cut off on the circumference an arc equal to the radius. Radian measure The size of an angle subtended by an arc of a circle in radian (or circular) measure is given by the ratio

One radian is the angle formed when the length of the arc created by the angle is equal to the radius of the circle. r

1 radian r O

To understand why a full circle corresponds to 2π radians, consider a unit circle where the radius is 1. The circumference of this circle is 2π × 1 = 2π. Since the entire circumference represents one complete revolution (or 360°), the angle subtended by this circumference is 2π radians. 7.04 Radians mathspace.co

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Thus, 360° is equivalent to 2π radians. Additionally, half a circle (or 180°) corresponds to half the circumference, which is π radians. Since the circumference of a circle is 2π r, a full circle (360°) contains 2π radians. Thus, 360° = 2π radians, and 180° = π radians. To convert between degrees and radians: • Degrees to radians: Multiply by • Radians to degrees: Multiply by ×

π 180°

Radians

Degrees

180°

Common radian measures and their degree equivalents are: Radian measure

Degree value

180°

90°

60°

45°

30°

Example 1 Convert −300° to radians in exact form.

Create a strategy Multiply −300° by

Apply the idea Apply conversion formula

Multiply

Simplify

Reﬂect and check Since 300° > 180°, the radian measure exceeds π, conﬁrming the result.

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Example 2 Convert

radians to degrees.

Create a strategy Multiply

Apply the idea Apply conversion formula

Simplify

Evaluate

Idea summary Radians measure angles based on arc length equalling the radius. Convert degrees to radians using Radian measure

Degree value

180°

and radians to degrees using

90°

60°

45°

30°

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Trigonometric ratios with radians Trigonometric ratios (sin θ, cos θ and tan θ ) apply to angles in radians, using the unit circle where the radius is 1.

π 2

Quadrant 2

Quadrant 1

0, 2π

π Quadrant 3

Quadrant 4 3π 2

The sign of trigonometric ratios depends on the quadrant: • Quadrant 1

: All positive

• Quadrant 2

: Sine positive, cosine and tangent negative

• Quadrant 3

: Tangent positive, sine and cosine negative

• Quadrant 4

: Cosine positive, sine and tangent negative

Exact values for multiples of

and

can be derived from special triangles.

π 2

π −θ

π +θ

0 or 2π

2π − θ 3π 2

Ensure calculators are in radian mode when evaluating trigonometric ratios in radians.

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Example 3 Write an equivalent ratio using a reference angle for sin

Create a strategy Locate

on the unit circle to determine its quadrant and reference angle, then apply the

appropriate sign.

π 2

π −θ

π +θ

0 or 2π

2π − θ 3π 2

Apply the idea Angle

is in Quadrant 3. So the reference angle is

π 2

7π 6

Sine is negative in Quadrant 3 so:

0 or 2π

3π 2

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Example 4 Evaluate the exact values of sin

, cos

, and tan

Create a strategy Use special triangles (30° − 60° − 90° and 45° − 45° − 90°) to ﬁnd exact values, noting that = 45° and

= 30°.

Apply the idea For sin

, use 30° − 60° − 90° triangle Convert radians to degrees

For cos

Evaluate the exact value

, use 45° − 45° − 90° triangle Convert radians to degrees

For tan

Evaluate the exact value

, use 30° − 60° − 90° triangle Convert radians to degrees

Apply tan θ =

Evaluate the exact values

Evaluate the division

Rationalise

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= 60°,

Idea summary Trigonometric ratios in radians use the unit circle. Exact values for multiples of and are derived from special triangles. Signs depend on the quadrant.

π 2

π −θ

π +θ

0 or 2π

2π − θ 3π 2

7.04 Practice questions What do you remember? 1

Determine if each statement is true or false: a

To convert degrees to radians, multiply by

An angle of 1 radian in a circle of radius r has an arc length of r.

sin (θ ) is positive in all quadrants.

Match each radian measure to its degree equivalent: a

60°

135°

iii

150°

180°

Determine the exact value of each trigonometric ratio: a

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Practice Ex 1

Convert these angles to radians in exact form: a

35°

1.2

275°

172.5°

3.1

0.85

5.4

Determine the exact value of: a

d f

Given cos θ =

and

< θ < π, ﬁnd the exact values of:

sin θ

Given a point values of: a

266

−225°

Find the exact value of each trigonometric ratio:

a 12

e 11

Rewrite each ratio as an equivalent trigonometric ratio of a positive reference angle:

a 10

145°

a Ex 4

150°

Convert these radian angles to degrees, rounded to one decimal place: a

Ex 3

Convert these angles to degrees: a

75°

Convert these angles to radians, rounded to two decimal places: a

Ex 2

15°

tan θ

on the unit circle corresponding to reference angle s, ﬁnd the exact

sin s

cos s

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking 13

Given a point

on the unit circle corresponding to reference angle s, ﬁnd the exact

values of sin (s + π ) and cos (s − π ). 14

Consider an angle θ such that cos θ =

and

< θ < 2π.

Determine the values of sin θ and tan θ . 15

A circle is inscribed in a square. The side length of the square is 8 units. A point P is located on the circle such that the angle between the line segment from the centre of the circle to P and the horizontal axis is . Determine the exact coordinates of point P and show all your work.

In a right-angled triangle ABC, the hypotenuse BC is twice the side AB. Let θ be the angle between BA and BC. Determine the exact values of cos θ and tan θ .

Did you know?

Mechanics use radians to measure how far a tyre rotates when calculating speed and braking distance. By tracking how quickly wheels stop spinning — measured in radians per second squared — they can estimate stopping distances and check for issues like worn brake pads or low road grip. Radians aren’t just for maths class; they help keep your ride safe and smooth!

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7.05 Arc length and sector area After this lesson, you will be able to… • establish and use the formula l = rθ to calculate the arc length of a circle. • establish and use the formula A =

r2θ to calculate the area of a sector.

• calculate the perimeter of major and minor sectors. • rearrange the formulas to solve for the radius or central angle. • solve practical problems involving arc lengths and areas of sectors and segments.

Arc length and sector area Arc A part of a circle’s circumference. Arc length The distance between two points on a curve. In a circle the length of an arc is given by l = rθ, where l is the arc length, r is the radius and θ is the angle subtended at the centre, measured in radians. Sector The plane ﬁgure enclosed by 2 radii or a circle and the arc between them.

Recall the formulas for circumference and area of a circle: Circumference : C = 2π r Area : A = π r2 An arc is a part of the circumference of the circle between any two points, though it is usually described using the angle that creates it. A

Arc

If no angle is referred to, the term minor arc is used to describe the smaller arc length and the term major arc is used to describe the larger one. 268

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If the arc was half of the circle the length would be half of the circumference.

If the angle of the sector is θ degrees, then the fraction of the circle is represented by

(one

full revolution corresponds to 360° in a circle). Therefore, the arc length can be calculated using:

If the angle of the sector is θ radians, then the fraction of the circle is represented by

(one full

revolution corresponds to 2π radians in a circle). Therefore, the arc length can be calculated using:

A sector of a circle is a region bounded by two radii and the arc between them. Arc Sector

θ O

The perimeter of a sector will be the arc length plus the two radii that form the edges of the shape. (Carefully read questions to determine if the perimeter or only the arc length is required.) For θ in degrees:

For θ in radians: P = θ r + 2r = ( θ + 2)r Similar to determining the length of an arc, the area of a sector is found by multiplying the area of a full circle by the appropriate fraction. For θ in degrees:

For θ in radians:

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Example 1 Consider a circle with radius 10 cm. a Determine the exact length of the arc subtended by the angle 120°.

Create a strategy Convert the angle to radians, then use the formula l = θ r.

Apply the idea

120° 10

Converting 120° to radians gives 120° = 120° ×

Write the formula Substitute θ =

and r = 10

Simplify

b Determine the exact perimeter of the sector subtended by the angle 120°.

Create a strategy Substitute the angle in radians from part (a) to the formula P = ( θ + 2)r.

Apply the idea Write the formula Substitute θ =

270

and r = 10

Evaluate the multiplication

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c Determine the exact area of the sector subtended by the angle 120°.

Create a strategy Substitute the angle in radians from part (a) to the formula A =

r2θ.

Apply the idea Write the formula Substitute θ =

and r = 10

Evaluate Evaluate

Example 2 A goat is tethered to a corner of a fenced ﬁeld. The rope is 9 m long. What area of the ﬁeld can the goat graze over, rounded to two decimal places?

Create a strategy

Notice that the angle at the corner of the fence is 90°. The area that the goat can graze over is a sector of radius 9 m and angle 90°.

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Apply the idea Given the angle that is 90 and r = 9. Write the formula

Substitute θ = 90 and r = 9

Simplify the fraction and evaluate 92

Evaluate the multiplication

Evaluate and round

Example 3 The arc of a circle, radius 13 cm subtends an angle of θ at the centre of the circle, and measures 11.7 cm in length. Solve for θ, the angle subtended at the centre.

Create a strategy The arc length is given by l = r × θ, where θ is the angle at the centre measured in radians. In this case, we are given the arc length and the radius of the circle. We need to determine the value of θ. 11.7 cm

13 cm

Apply the idea Write the formula Substitute l = 11.7 and r = 13

Divide both sides by 13

Evaluate and make θ the subject

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Idea summary Formulas relevant to arc length and sector areas: Circumference

C = 2π r

Area of a circle

A = π r2

Arc length

Perimeter of a sector

× 2π r, for θ in degrees

l = θ r, for θ in radians × 2π r + 2r, for θ in degrees

P = ( θ + 2)r, for θ in radians × π r2, for θ in degrees

A= Area of a sector

r2θ, for θ in radians

7.05 Practice questions What do you remember? 1

State the formulas for a sector with radius r and central angle θ in radians: a

Sector area

Sector perimeter

radians. Calculate the exact:

Arc length

Sector area

135°

Convert these angles to radians in exact form: a

A circle has radius 8 cm and central angle a

Arc length

60°

A sector has radius 5 m and arc length 10 m. Find the central angle θ in radians.

Practice Ex 1

Consider a circle with radius 15 cm: a

Determine the exact length of the arc subtended by the angle 60°.

Determine the exact perimeter of the sector subtended by the angle 60°.

Determine the exact area of the sector subtended by the angle 60°.

Calculate the exact arc length for each sector: a

r = 6 cm, θ =

r = 10 m, θ =

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Calculate the arc length, rounded to two decimal places: a

r = 6 cm, θ =

r = 8 m, θ =

r = 4 cm, θ =

r = 9 m, θ =

l = 15 cm, θ =

A = 50 m2 , θ =

r = 8 cm, l = 12 cm

r = 6 m, A = 30 m2

A sector has radius 10 cm and arc length 5π cm. Find the exact: a

r = 10 m, θ = 120°

Find the angle θ in radians, rounded to two decimal places: a

Find the radius r for each sector, rounded to two decimal places: a

r = 5 cm, θ = 1.8

Calculate the exact sector area for each sector: a

r = 12 m, θ = 150°

Calculate the exact perimeter for each sector: a

Calculate the sector area, rounded to two decimal places: a

r = 7 cm, θ = 1.2

Central angle θ in radians

Sector area

A sprinkler with a 20 m arm rotates through radians. Calculate the watered area, rounded to two decimal places.

π 3

20 m

A pendulum swings through 0.5 radians with a 60 cm string. Calculate the arc length travelled from the starting point to the maximum displacement, rounded to two decimal places.

0.5 rad 60 cm

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A sector has radius 12 cm and area 36π cm2. Find the exact perimeter.

K 12 cm O Ex 2

A sprinkler system irrigates a circular ﬁeld with a 12 m arm, rotating through a 120° angle. Calculate the area of the ﬁeld watered by the sprinkler, rounded to two decimal places.

Ex 3

The arc of a circle with radius 20 cm subtends an angle of θ at the centre and measures 15 cm in length. Solve for θ, the angle subtended at the centre, rounded to two decimal places.

Extend your thinking 19

A sector has radius 10 cm and arc length 5π cm. Find the exact difference between the arc length and the chord length connecting the arc’s endpoints.

A circular cake with radius 15 cm and height 8 cm has a sector with

radians removed.

Calculate the total surface area of the remaining cake, rounded to two decimal places.

A sector with radius 25 m and area 200 m2 forms a triangle with the chord of the arc. Calculate the exact area of the triangle. O 25 m A = 200 m2 A

A lighthouse on a circular island with radius 100 m illuminates a sector with area 785.4 m2. Find the central angle θ in radians and the sector’s perimeter, both rounded to two decimal places.

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7.06 Graphs of trigonometric functions After this lesson, you will be able to… • graph y = sin x, and y = cos x, identifying their key features, including domain, range, period, and amplitude. • graph y = tan x and identify its key features, including domain, range, period and asymptotes. • show intercepts with the x-axis and y-axis for all three functions. • determine whether each function is even or odd based on its graphical symmetry.

Sine and cosine functions (cos θ , sin θ )

When looking at the unit circle, the coordinates of any point on that circle can be described using trigonometry. Speciﬁcally a point on the circle at an angle θ anticlockwise from the x-axis has coordinates (cos θ , sin θ ).

θ O

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Since the graphs of y = sin θ and y = cos θ are derived from the unit circle, they are often called circular functions. Based on the unit circle, as θ moves through different values between 0 and 2π, the values of sin θ and cos θ move between −1 and 1.

sin θ

276

π 1

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2π −1

Quad II

Quad IV

Quad III

Quad II

sin θ

Quad I

Quad I 1

2π

3π 2

−1 Quad IV

Quad III 3π 2

The function y = sin θ repeats for all values of θ to give: y 2

y = sin θ θ −3π

−π

−2π

3π

2π

−2

cos θ

Quad II

π −1

Quad I

θ cos θ

2π

Quad I

Quad IV

Quad III

Quad II

θ 2π

π 2

3π 2

2π

−1 Quad III 3π Quad IV 2

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The function y = cos θ repeats for all values of θ to give: y 2

y = cos θ θ −3π

−2π

−π

2π

3π

−2

The graphs of y = sin θ and y = cos θ have similar properties. They are called cyclical, with one cycle referring to any section which can be translated horizontally to complete the rest of the graph. For example, the graphs of y = sin θ and y = cos θ repeats every 2π units. Period A function f is periodic with period p if f (x + p) = f (x), for all x, that is, the function repeats itself after each interval of length p. For example, sin x and cos x have period 2π. The period also refers to the length of one cycle in a cyclical graph. Typically measured from a consistent point, such as peak to peak or rising intercept to rising intercept. For the graphs of y = sin θ and y = cos θ , the period is 2π, that is one full revolution of the unit circle. In addition, each graph stays between y = −1 and y = 1 for all values of θ, since each coordinate of a point on the unit circle can be at most 1 unit from the x-axis. The graph also has a horizontal line called the midline, about which the graph oscillates. For the base graph of y = sin x and y = cos x, this is the x-axis (the line y = 0). Amplitude A function of the form y = A sin (nx + α) or y = A cos (nx + α) has amplitude A, that is half the distance between the maximum and minimum values.

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1 1

y = sin x

1 2

sin

π 2

−

1 π 2

1 2

1π

3 π 2

2π

−1

Amplitude of 1 for y = sin x

sin

3π 2

The maximum and minimum values of sin x can be seen from the unit circle

. The ratio sin x represents the y-coordinate on the unit circle and its maximum amplitude of 1 occurs at the angle , and is at In addition, it can be seen that the maximum amplitude is reached at

minimum amplitude of −1 when the angle is

The key features of y = sin x are: y 1 1 2

y = sin x

1 − 2

Amplitude 1 π 2

x 1π

3 π 2

2π

−1

Period • Period: 2π • Midline: y = 0 • Maximum: y = 1 at x =

+ 2π n, where n is an integer, e.g.

• Minimum: y = −1 at x =

+ 2π n, where n is an integer.

• Amplitude:

• Domain: All real x • Range: y ∈ [−1, 1] or −1 ≤ y ≤ 1 • Zeroes (x−intercepts): At x = nπ, where n is an integer. • Symmetry: The sine function is odd, satisfying sin (−x) = −sin x, meaning the graph has point symmetry about the origin.

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The key features of y = cos x are: y 1

y = cos x

1 2

Period Amplitude

x 1π

1 π 2

1 − 2

2π

3 π 2

−1

• Period: 2π • Midline: y = 0 • Maximum: y = 1 at x = 2π n, where n is an integer. • Minimum: y = −1 at x = π + 2π n, where n is an integer. =1

• Amplitude:

• Domain: All real x • Range: y ∈ [−1, 1] or −1 ≤ y ≤ 1 • Zeroes (x−intercepts): At x =

+ nπ, where n is an integer.

• Symmetry: The cosine function is even, satisfying cos (−x) = cos x, meaning the graph has line symmetry about the y-axis.

Example 1 For the curve y = cos x, determine whether each of these statements is true or false: y 1 1 2

−2π

3 − π 2

−1π

1 − π 2

1 − 2

1 π 2

−1

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1π

3 π 2

2π

a The graph of y = cos x is cyclic.

Create a strategy Recall that a curve is cyclic if it repeats itself in the horizontal direction.

Apply the idea The graph shows a repetitive pattern in the horizontal direction especially when −2π and 2π have the same y-values. So, the statement is true.

b As x approaches ∞, the height of the graph for y = cos x approaches ∞.

Create a strategy Use the fact that cos x is always between [−1, 1].

Apply the idea Since the range of the function y = cos x is y ∈ [−1, 1], then it cannot approach ∞. So, the statement is false.

c The graph of y = cos x is increasing between x =

and x = 0.

Create a strategy Using the graph, restrict the domain from x =

to x = 0.

Apply the idea y 1 1 2

x 1 − π 2

The curve is increasing from x =

to x = 0.

So, the statement is true. 1 − 2

−1

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Example 2 Consider the curve y = sin x. y 1 1 2

−2π

3 − π 2

−1π

1 − π 2

−

1 π 2

1 2

1π

2π

3 π 2

−1

a If one cycle of the graph of y = sin x starts at x = 0, when does the next cycle start?

Create a strategy

Apply the idea

Use the graph to determine the next point where the shape of the graph matches the rising intercept at x = 0.

At x = π, the value of the function is the same as x = 0 (i.e. sin 0 = sin π = 0) but the function is decreasing. At x = 2π, y = 0 and the function is increasing. So, the next cycle starts at x = 2π.

b For what values of x is the graph of y = sin x decreasing?

Create a strategy Use the graph to determine at what y-values of the curve decrease as the x-values increase.

Apply the idea y 1 1 2

−2π

3 − π 2

−1π

1 − π 2

−

1 2

1 π 2

1π

2π

3 π 2

−1

The graph shows that y = sin x is decreasing at

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and

c What is the x-value of the x-intercept in the region 0 < x < 2π?

Create a strategy Use the graph to determine the x-value where y = 0.

Apply the idea The graph shows that y = 0 when x = π. There are no other intercepts between x = 0 and x = 2π.

Reﬂect and check The region 0 < x < 2π does not include the boundary values of x = 0 and x = 2π. For the region 0 ≤ x ≤ 2π, x-intercepts are at x = 0, x = π and x = 2π.

Idea summary The key features of y = sin x are: y 1 1 2

y = sin x

1 − 2

Amplitude 1 π 2

x 1π

3 π 2

2π

−1

Period • Period: 2π • Midline: y = 0 • Maximum: y = 1 at x = • Minimum: y = −1 at x = • Amplitude:

+ 2π n, where n is an integer. + 2π n, where n is an integer.

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The key features of y = cos x are: y 1

y = cos x

1 2

−

Period Amplitude

x 1 π 2

1 2

1π

3 π 2

2π

−1

• Period: 2π • Midline: y = 0 • Maximum: y = 1 at x = 2π n, where n is an integer. • Minimum: y = −1 at x = π + 2π n, where n is an integer. • Amplitude:

• Domain: All real x • Range: y ∈ [−1, 1] or −1 ≤ y ≤ 1 • Zeroes (x−intercepts): At x =

+ nπ, where n is an integer.

• Symmetry: The cosine function is even, satisfying cos (−x) = cos x, meaning the graph has line symmetry about the y-axis.

Tangent functions The tangent function is a trigonometric function defined as the ratio of the sine function to the cosine function, tan ( θ ) = point on the unit circle.

, which also represents the gradient of the ray from the origin to the

Interactive exploration Discover this concept in action online

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mathspace.co

The graph of y = tan θ : y 10 5

θ −2π

−

3π 2

−π

−

−5

3π 2

2π

−10

Key features of the graph of y = tan θ are: • Vertical asymptotes: At θ =

+ kπ, for k any integer.

• For example: θ =

,θ=

• y-intercept: (0, 0) • x-intercepts at: θ = kπ, for k any integer • Period: π, the distance between two successive asymptotes • Range: All real numbers or (−∞, ∞) • Domain: θ is real, where θ ≠ vertical asymptote

+ kπ for any integer k, as the function is undeﬁned at each

• Key points: Useful for graphing, the function passes through

and

• Symmetry: The tangent graph is an odd function, that is tan (−θ ) = −tan θ . This means the graph has point symmetry about the origin (or any point of inﬂection) by 180°. From the deﬁnition tan ( θ ) =

, the function is undeﬁned when cos θ = 0. These values of θ

correspond to the vertical asymptotes of the graph, located at θ = The tangent function is periodic, repeating its pattern every π units. This can be shown by considering the unit circle and the deﬁnition of tangent. As seen in the diagram, adding π to an angle results in sine and cosine values with the same magnitude but opposite sign: sin ( θ + π ) = −sin θ cos ( θ + π ) = −cos θ

+ kπ, where k is any integer. y

θ +π

sin θ

cos(θ + π )

−1 sin(θ + π )

θ θ

cos θ

−1

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Using the deﬁnition of tangent: Apply the ratio identity in terms of θ + π

Substitute the relationships

Simplify

Example 3 Consider the graph of y = tan x for −2π ≤ x ≤ 2π : 5

4 3 2 1 3 1 −1π − π − π −1 2 2 −2

x 1 π 2

1π

3 π 2

−3 −4 −5

a How would you describe the graph? A

Periodic

Decreasing

Even

Linear

Create a strategy Describe each given option in relation to graphs.

Apply the idea A graph is said to be periodic if the curve repeats itself in regular intervals. A graph is said to be decreasing if the y-values of the curve decrease as the x-values of the curve increase. A graph is said to be even if it’s symmetric about the y-axis. A graph is said to be linear if it is a straight line. So, the answer is option A.

b Which of these is not appropriate to refer to with regard to the graph of y = tan x? A

Amplitude

Range

Period

Create a strategy It is given from part (a) that the graph of y = tan x is periodic.

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Asymptotes

Apply the idea The graph shows asymptotes separated by π units. The amplitude of a periodic graph is half of the distance between the maximum and minimum y-values reached by the graph. However, the maximum and minimum values of tan x converge to ∞ and −∞ respectively. The range is the set of y-values that the graph covers. So, the answer is option A.

c The period of a periodic function is the length of x-values that it takes to complete one full cycle. Determine the period of y = tan x in radians.

Create a strategy The period for y = tan x is the horizontal distance between consecutive asymptotes.

Apply the idea There’s an asymptote at x =

and x =

Subtract two consecutive asymptotes

Evaluate the subtraction

Simplify

d Which is the range of y = tan x: A

−∞ < y < ∞

y>0

−π < y < π

Create a strategy

Apply the idea

Use the graph of y = tan x.

The range is the set of y-values for which y = tan x covers. So, the answer is option A.

e As x increases, what would be the next asymptote of the graph to the right of x =

Create a strategy

Apply the idea

The period or distance between two symptotes is π.

Add π to the given asympote

Rewrite π as

Evaluate

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Example 4 Select the two functions that have the same graph as y = tan x: B

y = tan (x + π )

y = tan (x + 2π )

Create a strategy The period represents the interval that the graph repeats itself.

Apply the idea Consider the graph of y = tan x. Think about by how much each of the curve could be moved to the right so that they land on the graph again or by how much could the asymptote be moved to the right so it lands on another asymptote.

So, the options that have the same as y = tan x are options B and D.

x 1 π 2

1π

−2

Reﬂect and check The period of y = tan x is π, so the graph repeats itself every π units. The graph of y = tan x can be shifted horizontally by multiples of π units and the graph will align with itself.

Idea summary Key features of the graph of the tangent function, f (x) = tan (x) are:

• Domain: x is real, any integer

• Range: (−∞, ∞)

−2π

−π

−1

Midline π

−2 −3 −4

Period

2π

, k is

• x-intercept: x = kπ, k is any integer • y-intercept: (0, 0) • Period: π • Amplitude: None • Midline: y = 0 • Key points:

and

• Odd symmetry: tan (−x) = −tan x, point symmetric about origin. To graph a tangent function use the key features mentioned.

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7.06 Practice questions What do you remember? 1

Use the unit circle to complete the table for y = sin θ with exact values:

2π

sin θ

Use the unit circle to complete the table for y = cos θ with exact values:

2π

cos θ

State the key features of y = sin x: • Period • Amplitude • Midline • Domain • Range • Symmetry

Sketch the graph of y = cos x over 0 ≤ x ≤ π. y 1 0.5 x −0.5

1 π 2

1π

3 π 2

−1

State the deﬁnition of tan θ in terms of sin θ and cos θ.

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289

Complete the table for y = tan θ using the unit circle:

tan θ y

θ +π cos(θ + π )

−1 sin(θ + π )

sin θ

θ θ

cos θ

−1

State the key features of y = tan x: • Period • Domain • Range • y-intercept • x-intercepts • Symmetry

Sketch y = tan x over −π ≤ x ≤ π, labelling asymptotes and intercepts. 4

3 2 1 1 − π 2

−1

x 1 π 2

−2 −3 −4

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Practice 9

Consider the graph:

Identify the equation (sine or cosine).

State the maximum.

State the amplitude.

1 x 1 − π 2

1 π 2

1π

3 π 2

2π

1 π 2

1π

3 π 2

2π

−1

Consider the graph: a

State the minimum.

State the midline.

y 1 x 1 − π 2

−1

Ex 1

Determine whether these statements about y = sin x are true or false: a

The graph is symmetric about the origin.

The period is π.

The range is y ∈ [−1, 1]. y

−2π − 3 π −1π − 1 π 2 2

1π 3 2π 1 π π 2 2

−1

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Ex 2a

For y = cos x, if one cycle starts at x = −π, where does the next cycle start?

Ex 2b

Determine whether y = cos x is increasing or decreasing over: a

0<x<π

Determine whether y = sin x is increasing or decreasing over: a

Ex 2c

−π < x < 0

Identify the x-intercepts of y = sin x over −π < x < π.

Identify the x-intercepts of y = cos x over 0 < x < 2π.

Determine the minimum value of y = sin x and the smallest positive x-value where it occurs.

Sketch y = sin x over −2π ≤ x ≤ 0. y 1 0.5 x 3 − π 2

−1π

1 − π 2

−0.5 −1

Determine the values of sin θ and cos θ at quadrant boundaries θ = 0,

(cos θ , sin θ )

−1

Determine the vertical asymptotes of y = tan x for −π ≤ x ≤ π.

Determine the x-intercepts of y = tan x for 0 ≤ x ≤ 2π.

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, π,

Ex 3

Consider the graph of y = tan x for 0 ≤ x ≤ 4π : a

Which term best describes the graph of y = tan x?

Periodic

Linear

Quadratic

Exponential

Which property is not applicable to y = tan x? A

Amplitude

Period

Range

Asymptotes

3 π 2

2π

5 π 2

3π

7 π 2

1 π 2

1π

3 π 2

2π

5 π 2

−3

Determine the period of y = tan x in radians.

State the range of y = tan x.

As x increases, determine the equation of the

Evaluate tan

−1

x 1π

1 π 2

−2

next asymptote after x = 23

y 4

−4

and tan (nπ ) for integer n using the unit circle. y

θ +π

sin θ

cos(θ + π )

−1 sin(θ + π )

θ θ

cos θ

−1

Ex 4

Select all functions that have the same graph as y = tan x: A

y = tan (x + 3π )

y 4 3 2

B C

y = tan (x + 5π )

y = tan (x − 2π )

1 1 − π 2 −1

−2 −3 −4

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Determine whether y = tan x is increasing or decreasing between

Determine the sign of tan x for

Extend your thinking 27

Explain why the range of y = sin x is [−1, 1] using the unit circle.

(cos θ , sin θ )

−1

−1 28

Describe y = cos x as a transformation of y = sin x, with the help of their graphs.

Find the x-coordinates of the intersections between y = sin x and y = cos x where 0 ≤ x ≤ 2π.

y 1 0.5 x −0.5

1π

1 π 2

3 π 2

−1

Explain what happens to y = sin x for x > 2π using the unit circle.

(cos θ , sin θ )

−1 O

−1 31

Explain why the range of y = tan x is (−∞, ∞) using the unit circle.

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Find the x-coordinates where y = tan x = 1 in the interval [0, 2π], giving exact values.

Explain why tan ( θ + π ) = tan θ using the unit circle.

θ +π

sin θ

cos(θ + π )

−1 sin(θ + π )

θ θ

cos θ

−1

Did you know?

Trigonometric graphs help engineers design roller coasters! By using sine and cosine curves, they can model smooth rises and falls, ensuring rides are thrilling while keeping passengers safe. The amplitude of the graph represents the steepness of the hills, while the period reflects the length of each wave-like track segment.

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7 Chapter review 1

If cos θ =

and 180° < θ < 270°, what is the exact value of sin θ? B

A 2

For each diagram, determine the value of: i

sin θ

cos θ

iii

tan θ

4 3 , 5 5

P −

12 5 , 13 13

−1

1 −1

Determine the coordinates of P (x, y) on the unit circle, rounded to two decimal places: a

y 1

P 40°

−1

200°

x −1

sin θ

sin 40°

cos 220°

tan 320°

sin 135°

cos 240°

tan 330°

Rewrite each ratio as an equivalent trigonometric ratio of a positive reference angle α: a

296

Find the exact value of each ratio without technology: a

cos θ

For a unit circle with point P (c, d) at 40°, express each in terms of c, d: a

−1

Given tan θ = −2.4, 90° < θ < 180°, calculate the exact value of: a

x 1

−1

1 −1

sin (180° − α)

cos (180° + α)

tan (360° − α)

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

cos (−α)

Find all angles θ between −360° and 360° where cos θ =

Find side x in triangle XY Z with ∠X = 38°10′, ∠Y = 72°, and y = 20, rounded to two decimal places.

Find side z in triangle XY Z with x = 18, y = 22, and ∠Z = 50°, rounded to two decimal places.

Find all possible values for ∠C in triangle ABC with a = 10, c = 15, and ∠A = 40°, rounded to one decimal place.

Convert to the other angle measure (degrees or radians): a

45°

and π < θ <

, ﬁnd the exact values of:

cos θ

tan θ

Sector area for r = 10 m, θ =

For each sector, calculate the exact: a

Given sin θ = a

Find the exact value of each trigonometric ratio: a

210°

Arc length for r = 8 cm, θ =

A pizza slice is a sector with radius 18 cm and arc length 6π cm. Find the exact: a

Central angle θ in radians

Sector area

A sector has radius 12 cm and arc length 5π cm. Find the exact difference between the arc length and the chord length connecting the arc’s endpoints.

A sector with radius 15 m and area 112.5 m2 forms a triangle with the chord of the arc. Calculate the exact area of this triangle.

Consider the graph of y = cos x: a

State the minimum value.

State the midline.

y 1 x 1 − π 2

1 π 2

1π

3 π 2

2π

−1

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297

Consider the graph of y = tan x for 0 ≤ x ≤ 2π. 4

Determine the period of y = tan x in radians.

State the range of y = tan x.

What is the equation of the asymptote after x =

2 1

x 1 π 2

−1 −2

1π

3 π 2

−3 −4

Find the x-coordinates of intersections between y = sin x and y = −cos x in 0 ≤ x ≤ 2π. y

y = −cos x

1 0.5

x −0.5 −1

1 π 2

1π

3 π 2

y = sin x

Explain why the range of y = cos x is [−1, 1] using the unit circle.

Did you know?

The repeating pattern of barrels in a cellar is just like the waves of sine and cosine graphs! These graphs model periodic patterns that occur again and again — just like rows of barrels lined up in perfect rhythm. 298

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“It appears to me that if one wishes to make progress in mathematics, one should study the masters and not the pupils.” Niels Henrik Abel

8 Trigonometric identities and equations Chapter outline 8.01 8.02 8.03 8.04 8.05 8.06 8.07 8.08E 8.09E 8.10E 8.11E 8.12E 8.13E 8.14E

302 311 321 326 334 340 349 356 366 375 382 394 405 416

Secants appear in real life when calculating satellite dish angles for precise TV reception.

8.01 Secant, cosecant and cotangent After this lesson, you will be able to… • define the secant, cosecant and cotangent ratios for acute angles using ratios of sides in a right-angled triangle. • find the exact values of secant, cosecant and cotangent for angles of 30°, 45° and 60°. • justify the values of secant, cosecant and cotangent at the boundary angles of 0° and 90°.

Secant, cosecant and cotangent θ

Adjacent

The triangle shows sides used to deﬁne sec θ, cosec θ and cot θ.

Opposite Cosecant ratio For an angle, its cosecant is the reciprocal of its sine, cosec θ = In any right-angled triangle, cosec θ =

, where 0 < θ < 90°.

Secant ratio For an angle, the secant is the reciprocal of its cosine, sec θ = In any right-angled triangle, sec θ =

302

, where 0 < θ < 90°.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Cotangent ratio For an angle, its cotangent is the reciprocal of its tangent, cot θ = In any right-angled triangle, cot θ =

, where 0 < θ < 90°.

These ratios are undefined when the denominator is 0, sec θ and cosec θ are always defined for acute angles, but sec θ is undefined at θ = 90°. Also, cot θ and cosec θ are undefined at θ = 0°.

Exploration Consider a right-angled triangle with sides labelled hypotenuse, opposite and adjacent relative to angle θ. Discuss how the ratios for sec θ, cosec θ and cot θ change as the triangle’s shape varies while keeping θ acute.

Example 1 In a right-angled triangle, the hypotenuse is 5 cm, the side adjacent to angle θ is 4 cm and the side opposite is 3 cm. Find the exact value of: a sec θ

Create a strategy Use sec θ =

Apply the idea Write the secant ratio Substitute Hypotenuse = 5 and Adjacent = 4

b cosec θ

Create a strategy Use cosec θ =

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Apply the idea Write the cosecant ratio Substitute Hypotenuse = 5 and Opposite = 3

c cot θ

Create a strategy Use cot θ =

Apply the idea Write the cotangent ratio Substitute Adjacent = 4 and Opposite = 3

Idea summary For an acute angle θ in a right-angled triangle: • cosec θ =

• sec θ =

• cot θ =

Exact values The ﬁgure shown is a right-angled isosceles triangle with two equal sides of length 1 unit. Using Pythagoras’ theorem, the hypotenuse is

units long.

45° 2

The angles in a triangle add up to 180° and the base angles in an isosceles triangle are equal, so the two base angles are 45°.

45° 1 The exact values can be determined for these trigonometric ratios:

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The exact values can be found by starting with an equilateral triangle:

60° 2

60°

To ﬁnd the exact ratios of 30° and 60° angles, start with an equilateral triangle with side lengths of 2 units. Remember all the angles in an equilateral triangle are 60°.

30° 30° 2

60°

30° 2

Draw a line that divides the triangle in half, into two congruent right-angled triangles. The base line is cut into two 1 unit lengths, and the opposite angle is cut into two 30° angles.

Now, focus on just one half of this triangle. 3

Using Pythagoras’ theorem, calculate the length of the perpendicular height to be . Then use our trigonometric ratios to determine the exact values.

60° 1

Now, not every isosceles right-angled triangle has sides measuring 1, 1 and , but no matter how large or small it is, the two base angles will always be 45° angles and therefore, the ratios of the sides will always be the same. This also applies to the triangle with 60° and 30° angles. Any pair of similar triangles have the same side ratios, and so any triangle with these angles will have the same exact value trigonometric ratios.

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305

Example 2 Use the exact value triangles in the diagram to answer the questions:

30° 2

60°

Find the exact values:

a sec 60°

Create a strategy Refer to the triangle with 60° angle, then use the secant ratio.

Apply the idea Write the secant ratio

Substitute the values

Simplify

b cosec 45°

Create a strategy Refer to the triangle with 45° angle, then use the cosecant ratio.

Apply the idea Write the cosecant ratio

Substitute the values

Simplify

c cot 30°

Create a strategy Refer to the triangle with 30° angle, then use the contangent ratio.

Apply the idea Write the contangent ratio

Substitute the values

Simplify

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45°

45° 1

Idea summary The exact values for the angles 30°, 45° and 60° are: sec

cosec 2

30° 45° 60°

cot

1 2

Boundary values The values of sec θ, cosec θ and cot θ at θ = 0° and θ = 90° are determined by examining the side ratios in a right-angled triangle as θ approaches these angles.

Opposite

p Hy

e us

θ θ

Adjacent

Opposite

= 1.

→

• cosec θ = • cot θ =

→ →

• sec θ =

90°

• sec θ =

, which is undeﬁned. , which is undeﬁned.

As θ → 90°, the adjacent side approaches 0, and the opposite side approaches the hypotenuse:

Adjacent

As θ → 0°, the opposite side approaches 0 and the hypotenuse side approaches the adjacent:

θ 0

→

• cosec θ = • cot θ =

, which is undeﬁned. = 1.

→ →

= 0.

Interactive exploration Discover this concept in action online

mathspace.co

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307

Example 3 Justify the values of sec 0° = 1 and cot 90° = 0 using the side ratios in a right-angled triangle: a sec 0° = 1

Create a strategy Use the side ratio for sec θ =

as θ → 0°.

Apply the idea Write the secant ratio

As θ → 0°, adjacent approaches hypotenuse

Simplify

b cot 90° = 0

Create a strategy Use the side ratio for cot θ =

as θ → 90°.

Apply the idea Write the cotangent ratio

As θ → 90°, adjacent approaches 0

Simplify

Idea summary The boundary values that does not describe physical triangles, at 0° and 90° are:

308

sec

cosec

cot

0°

Undefined

90°

Undefined

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

8.01 Practice questions What do you remember? 1

What are the trigonometric ratios in terms of opposite, adjacent, and hypotenuse for: a

cosec θ

sec θ

cot θ

Consider the exact value triangles shown:

30° 2

60°

Using the exact value triangles shown, ﬁnd the value of: i

sin 30°

cos 60°

iii

sec 60°

cot 45°

vii cot 30°

tan 45°

cosec 30°

viii cosec 90°

Find the values of θ, where 0° ≤ θ ≤ 90°, on which the following ratios are undeﬁned: i

45°

1 a

45°

cot θ

cosec θ

iii

sec θ

Determine whether each statement is true or false: a

sec 0° = 1

cosec 90° = 1

cot 0° = 0

Practice Ex 1

In a right-angled triangle, the hypotenuse is 13 cm, the adjacent side to angle θ is 12 cm, and the opposite side is 5 cm. Find the exact values of: a

cosec θ

cot θ

sec θ

cosec θ

cot θ

cosec 30°

cot 45°

Find the exact values of: a

In a right-angled triangle, the opposite side to angle θ is 8 cm, and the adjacent side is 15 cm. Find the exact values of: a

Ex 2

sec θ

sec 45°

cot 60°

Simplify and leave the answer in exact form: a

sec 30° × cos 30°

cosec 60° × sin 60°

cot 45° × tan 45°

cot 30° × cosec 30°

In a right-angled triangle, the hypotenuse is 10 cm, and one of the acute angles is 30°. Find: a

sec 30°

cosec 30°

cot 30° 8.01 Secant, cosecant and cotangent mathspace.co

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Ex 3

Justify each using side ratios in a right-angled triangle: a

sec 60° + cot 60°

cosec 45° − sec 45°

sec 30° + cosec 30°

cot 45° − sec 60°

A person stands 10 metres from the base of a tower and observes the top of the tower at an angle of elevation of 45°. The tower’s height is h metres. Find: sec 45°

cosec 45°

cot 45°

tan 45°

cot θ =

Solve the following equations for θ, where 0° ≤ θ ≤ 90°: sec θ = 2

cosec θ =

cot θ = 1

Simplify the following for acute angles θ using the trigonometric ratios deﬁned by opposite, adjacent, and hypotenuse: a

sec θ × cos θ

cosec θ × sin θ

cot θ × tan θ

sec θ × cot θ

A vertical pole casts a shadow when the sun’s rays make an angle of 60° with the ground. In the right-angled triangle formed, ﬁnd: a

sec 90° = undefined

a 13

Find the exact values of:

a 12

cosec 90° = 1

sec 60°

cosec 60°

cot 60°

If the reference angle in a right-angled triangle is 0°, what can you say about the lengths of the sides?

Extend your thinking 16

In a right-angled triangle, sec θ =

. Find cosec θ and cot θ.

Prove that sec θ × cos θ = 1 for all acute angles θ using the trigonometric ratios deﬁned by opposite, adjacent, and hypotenuse.

A student incorrectly states that cosec 30° = . Identify the error and provide the correct value.

A ladder of length 5 m leans against a vertical wall, making an angle of 45° with the ground. Find sec 45°, cosec 45°, and cot 45° based on the triangle formed.

In a right-angled triangle, cot θ = . Find sec θ and cosec θ.

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8.02 Unit circle with secant, cosecant and cotangent After this lesson, you will be able to… • define secant, cosecant and cotangent using a circle of radius r centred at the origin. • determine the exact value of secant, cosecant and cotangent ratios for integer multiples of . • determine the exact value of secant, cosecant and cotangent ratios for integer multiples of . • identify angles for which the reciprocal trigonometric ratios are undefined.

Unit circle with secant, cosecant and cotangent The unit circle has a radius of 1 and is centred at the origin (0, 0). For an angle measured anticlockwise from the positive x-axis, a point (x, y) on the circle deﬁnes the trigonometric functions secant, cosecant, and cotangent. These functions can also be deﬁned for circles of any radius as follows: 1

P (x, y) 1

sec θ =

θ −1

cosec θ = cot θ =

= =

−1

sec θ is the secant of angle θ, the reciprocal of cosine r

is the radius of the circle

is the x-coordinate of the point on the circle

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cosec θ is the cosecant of angle θ, the reciprocal of sine r

is the radius of the circle

is the y-coordinate of the point on the circle

cot θ is the cotangent of angle θ, the reciprocal of tangent x

is the x-coordinate of the point on the circle

is the y-coordinate of the point on the circle

, and each function is undeﬁned when the denominator is 0.

The radius

Exploration Sketch a unit circle and mark a point (x, y) for an angle θ. 1. Discuss how sec θ, cosec θ, and cot θ change as θ moves from 0 to 2. What happens when x or y approaches 0?

Example 1 For an angle θ on the unit circle with point

, ﬁnd the exact values of:

a sec θ

Create a strategy Use sec θ =

with x =

Apply the idea Apply the deﬁnition of secant Substitute x =

Simplify Rationalise

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b cosec θ

Create a strategy Use cosec θ =

with y = .

Apply the idea Apply the deﬁnition of cosecant Substitute y = Simplify c cot θ

Create a strategy Use cot θ =

with x =

and y = .

Apply the idea Apply the deﬁnition of cotangent

Substitute x =

and y =

Simplify

Idea summary On a circle centred at the origin, if an angle θ is measured from the positive x-axis to the line segment joining the origin to a point (x, y) on the circle, then: 1

P (x, y) 1

sec θ =

θ −1

cosec θ = cot θ =

−1 Here, r is the distance from the origin to the point (x, y), given by These functions are undeﬁned when the denominator is zero.

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Exact values for

multiples

Exact values of sec θ, cosec θ, and cot θ for angles that are integer multiples of can be found using the unit circle coordinates and reciprocal relationships. y (0, 1)

(−1, 0)

120° 150°

90°

60° 30°

π 180°

0°

(1, 0) x

210° 330° 240° 300° 270°

(0, −1)

Multiples of 30°

and 60°

Example 2 Find the exact values: a sec

Create a strategy Use sec θ =

with x from the unit circle at

Apply the idea At

,x= : Apply the deﬁnition of secant

Substitute x = Simplify

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b cosec

Create a strategy Use cosec θ =

with y from the unit circle at

Apply the idea At

,y= : Apply the deﬁnition of cosecant

Substitute y = Simplify

c cot

Create a strategy Use cot θ =

with x and y from the unit circle at

Apply the idea At

, x = 0, y = 1: Apply the deﬁnition of cotangent

Substitute x = 0 and y = 1 Simplify

Idea summary Exact values for multiples of

are found using unit circle coordinates (x, y) and the

deﬁnitions sec θ = , cosec θ = , and cot θ = .

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Exact values for

multiples have exact values for sec θ, cosec θ, and

Angles that are integer multiples of cot θ based on unit circle coordinates.

y (0, 1)

135° (−1, 0)

90°

45° 0° 0, 2π

180° 225°

270°

(1, 0)

315°

(0, −1)

Example 3 Find the exact values: a sec

Create a strategy Use sec θ =

with x from the unit circle at

Apply the idea At

,x=

: Apply the deﬁnition of secant

Substitute x =

Simplify

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b cosec

Create a strategy Use cosec θ =

with y from the unit circle at

Apply the idea At

,y=

: Apply the deﬁnition of cosecant

Substitute y =

Simplify

c cot

Create a strategy Use cot θ =

with x and y from the unit circle at

Apply the idea At

,x=

,y=

: Apply the deﬁnition of cotangent

Substitute x =

Simplify

and y =

Idea summary Exact values for multiples of

are derived from unit circle coordinates, with

sec θ = , cosec θ = , and cot θ = , accounting for undeﬁned cases.

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8.02 Practice questions What do you remember? 1

Deﬁne the unit circle.

State the deﬁnitions of the trigonometric functions for an angle θ on the unit circle: a

cosec θ

cot θ

For which angles is each function undeﬁned on the unit circle? a

sec θ

cosec θ

cot θ

State the exact values for: b

Practice Ex 1

For an angle θ on the unit circle with point a

Ex 2

sec θ

cosec θ

, ﬁnd: c

cot θ

Find the exact values of: a

Ex 3

Find the exact values of: a

For an angle θ on the unit circle with point a

cosec θ

cot θ

State whether each function is deﬁned or undeﬁned: a

318

sec θ

, ﬁnd:

cosec (0)

cot (π )

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For an angle θ on the unit circle with point a

sec θ

cosec θ

, ﬁnd: c

cot θ

Simplify and leave the answer in exact form wherever applicable: a

Find the exact values of: a

Evaluate each trigonometric ratio, leaving the answers in exact form: a

cosec 45° + sec 60°

cosec 45° sec 30° + cot 45°

sec2 (30°)

cosec2 (30°) + sec2 (30°)

Extend your thinking 14

Explain why sec θ is undeﬁned at θ =

Derive the relationship between sec θ and cos θ using the unit circle deﬁnitions.

A student incorrectly states that value.

For an angle θ in the second quadrant with cos θ =

A Ferris wheel has a radius of 10 metres. At a certain angle θ, a rider’s position corresponds to the unit circle point

Prove that cot θ =

using the unit circle.

. Identify the error and provide the correct

, ﬁnd sec θ, cosec θ, and cot θ.

. Find sec θ, cosec θ, and cot θ. using the unit circle deﬁnitions.

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This diagram shows the unit circle divided into 12 equal sections. The point P can be represented as

and also as

depending on whether the angle is measured

clockwise or anticlockwise from the positive x−axis.

State the two exact angles for points Q, R, and S. S R

Determine the exact value of each of the following:

Did you know?

Trigonometric identities can be used to model the circular and curved shapes in pottery design! By applying equations involving sine and cosine, potters and designers can create symmetrical patterns, calculate angles for precise carving, and ensure balanced, visually appealing pieces.

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8.03 Reciprocal and quotient identities After this lesson, you will be able to… • establish the reciprocal identities and identify the angles for which they are undefined. • establish the quotient identities for tangent and cotangent and identify the angles for which they are undefined. • apply reciprocal and quotient identities to simplify trigonometric expressions. • use trigonometric identities to prove simple trigonometric statements.

Reciprocal identities Identity An identity is a statement involving a variable(s) that is true for all possible values of the variable(s). The reciprocal identities relate the trigonometric functions secant, cosecant, and cotangent to cosine, sine, and tangent, respectively.

These identities are derived from the deﬁnitions of secant, cosecant, and cotangent using the unit circle or right-angled triangles. For example, if cos θ = , then sec θ =

Each identity has excluded angles where the denominator is zero: • sec θ is undefined when cos θ = 0, at θ = 90° + 180°n (e.g., 90°, 270°), where n is an integer. • cosec θ is undefined when sin θ = 0, at θ = 180°n (e.g., 0°, 180°, 360°), where n is an integer. • cot θ is undefined when tan θ = 0, at θ = 180°n, or when tan θ is undefined, at θ = 90° + 180°n, where n is an integer.

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Exploration Discuss with a partner: 1. Why are certain angles excluded from the reciprocal identities? 2. Consider the unit circle and what happens to sin θ, cos θ, and tan θ at these angles.

Example 1 Prove the identity sec θ cosec θ =

Create a strategy Use the reciprocal identity: sec θ =

Apply the idea Write the left-hand side

Substitute sec θ =

Evaluate the multiplication

Compare with the right-hand side

Idea summary The reciprocal identities are sec θ =

, cosec θ =

, and cot θ =

They are undeﬁned when the denominators are zero: cos θ = 0 at 90° + 180°n, sin θ = 0 at 180°n, and tan θ = 0 or undeﬁned at 180°n or 90° + 180°n, for integer n.

Quotient identities The quotient identities express tangent and cotangent in terms of sine and cosine:

These identities are derived from the unit circle, where tan θ = They are undeﬁned when the denominator is zero. 322

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and cot θ =

Example 2 Prove the identity

= sin θ.

Create a strategy Use the quotient identity tan θ = left-hand side.

and the reciprocal identity sec θ =

to rewrite the

Apply the idea Write the left-hand side

Substitute tan θ =

and sec θ =

Multiply by the reciprocal of the denominator

Remove common factors

Compare with the right-hand side

Idea summary The quotient identities are tan θ =

and cot θ =

They are undeﬁned when the denominator is zero.

8.03 Practice questions What do you remember? 1

Define the mathematical meaning of each of these terms. Provide an example for each. a

Undeﬁned

Quotient

sec θ

cosec θ

cot θ

Write each trigonometric function using its quotient identity: a

Determine the reciprocal identity of each trigonometric function: a

Reciprocal

tan θ

cot θ

Determine angles where each function is undeﬁned: a

sec θ

cosec θ

tan θ

cot θ

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Practice 5

Determine whether each statement is true or false, justifying with exact values or identities: a

cosec 30° × sin 30° = sec 60° × cos 60°

tan π × cot π =

sec 45° × cos 45° =

Ex 1

Prove each identity algebraically: a

cosec θ cot θ =

c Ex 2

= cos θ

324

= tan2θ

sec θ × tan θ

cot θ

= cot2θ

cosec θ × cot θ

in the ﬁrst quadrant, calculate:

cosec θ

tan θ

Determine whether each function is deﬁned or undeﬁned at the given angle: a

= sin θ

Given sin θ = a

= tan θ

Simplify each expression using identities: a

sec θ cot θ =

Prove each identity algebraically: a

sec 270°

cosec 360°

tan 90°

cos θ cot θ =

Prove each identity algebraically: a

sin θ tan θ =

sec θ cosec θ =

Prove each identity using exact values at speciﬁed angles: a

sec 60° cos 60° = 1

tan 30° cot 30° = 1

cosec

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sin

cot 0°

Extend your thinking 13

Explain using the unit circle why cosec θ is undeﬁned at θ = 180°. y (0, 1)

(−1, 0)

120° 150°

90°

60° 30°

π 180°

0°

(1, 0) x

210° 330° 240° 300° 270°

(0, −1)

Prove algebraically that

Given tan θ = in exact form.

Prove that

Given cos θ = in exact form.

A student claims cot 45° =

= tan θ for all angles where the functions are deﬁned.

in the third quadrant, calculate sec θ, cosec θ, and cot θ, leave the answer

= cos θ for all angles where the functions are deﬁned. in the second quadrant, calculate sec θ, tan θ, and cot θ, leave the answer

. Correct the error using the reciprocal identity.

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8.04 Complementary angle identities After this lesson, you will be able to… • define complementary angles. • prove the complementary angle identities for sine, cosine, tangent, secant, cosecant and cotangent. • identify the angles for which the complementary angle identities are undefined. • evaluate trigonometric expressions using complementary angle identities.

Complementary angle identities proof Complementary angle Two adjacent angles that form a right angle, i.e. the sum of the angles measured in degrees is 90°. The trigonometric identities relating these angles are known as complementary angle identities. These identities can be proven using a right-angled triangle, where one angle is θ and the other is 90° − θ.

90° − θ

θ b

sin (90° − θ ) =

= cos θ

cos (90° − θ ) =

= sin θ

tan (90° − θ ) =

= cot θ

cot (90° − θ ) =

= tan θ

sec (90° − θ ) =

= cosec θ

cosec (90° − θ ) =

= sec θ

Alternatively, the unit circle deﬁnitions of trigonometric functions conﬁrm these relationships.

90° − θ

b −b

326

−a

P (cos θ , sinθ ) b −b

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Example 1 Prove the identities: a sin (90° − θ ) = cos θ

Create a strategy Use a right-angled triangle with angle θ and its complement 90° − θ. Apply the deﬁnitions of sine and cosine as side ratios, comparing the ratios for both angles.

Apply the idea

90° − θ

Consider a right-angled triangle with angle θ, opposite side a, adjacent side b, and hypotenuse c. For the complementary angle 90° − θ, the opposite and adjacent sides swap roles.

θ b For sin (90° − θ ): Apply the sine ratio

Substitute the side lengths

For cos θ : Apply the cosine ratio Since sin (90° − θ ) =

Substitute the side lengths = cos θ, the identity holds.

Reﬂect and check Verify for θ = 30°. For sin (90° − θ ): Evaluate the subtraction

Evaluate the exact value

For cos 30°: Evaluate the exact value The identity holds for all θ where both functions are deﬁned (e.g., not at θ = 90° where cos 90° = 0).

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b tan (90° − θ ) = cot θ

Create a strategy Use a right-angled triangle with angle θ and its complement 90° − θ. Apply the deﬁnitions of tangent and cotangent as side ratios, comparing the ratios for both angles.

Apply the idea

90° − θ

Consider a right-angled triangle with angle θ, opposite side a, adjacent side b, and hypotenuse c. For the complementary angle 90° − θ, the opposite and adjacent sides swap roles.

θ b For tan (90° − θ ): Apply the tangent ratio

Substitute the side lengths

For cot θ : Apply the cotangent ratio Since tan (90° − θ ) =

Substitute the side lengths = cot θ, the identity holds.

Reﬂect and check Verify for θ = 30°. For tan (90° − θ ): Evaluate the subtraction

Evaluate the exact value

For cot 30°: Evaluate the exact value The identity holds for all θ where both functions are deﬁned (e.g., not at θ = 0°, 180° where cos (90° − θ ) = sin θ = 0 or sin θ = 0).

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Idea summary Complementary angle identities such as sin (90° − θ ) = cos θ and tan (90° − θ ) = cot θ are valid for all angles where the functions are deﬁned. For acute angles, these can be shown using right-angled triangles, while the unit circle provides a general proof for all angles.

Excluded angles in identities Some complementary angle identities are undefined at certain angles due to division by zero in their definitions. These can be identified by examining denominators in tan θ, cot θ, sec θ, and cosec θ. • tan θ =

is undeﬁned when cos θ = 0, i.e., θ = 90° + 180°n, where n is an integer.

• cot θ =

is undeﬁned when sin θ = 0, i.e., θ = 180°n, where n is an integer.

• sec θ =

is undeﬁned when cos θ = 0, i.e., θ = 90° + 180°n , where n is an integer.

• cosec θ =

is undeﬁned when sin θ = 0, i.e., θ = 180°n, where n is an integer.

For each identity, substitute 90° − θ into the angle on the left-hand side to ﬁnd where the right-hand side is undeﬁned.

Exploration Test the identity tan (90° − θ ) = cot θ by substituting θ = 0° and θ = 90°. Discuss why the identity may not hold at these angles.

Example 2 Identify the angles θ where these identities are undeﬁned: a tan (90° − θ ) = cot θ

Create a strategy Use the quotient identity tan θ =

for tan (90° − θ ) and cot θ =

to determine where each

side is undeﬁned by setting denominators to zero. Check both sides of the identity.

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Apply the idea For tan (90° − θ ): Apply the quotient ratio tan θ =

Use the identity cos (90° − θ ) = sin θ

Evaluate sin θ at the unit circle with r = 1

Division by zero

The angles where sin θ = 0 are θ = 180° n, where n is an integer (e.g., θ = 0°, 180°, 360°). For cot θ : Apply the quotient identity

Evaluate sin θ at the unit circle with r = 1

Division by zero

The angles where sin θ = 0 are θ = 180° n, where n is an integer (e.g., θ = 0°, 180°, 360°). Thus, the identity tan (90° − θ ) = cot θ is undeﬁned when θ = 180° n.

Reﬂect and check Verify by substituting θ = 0°: Evaluate the subtraction

Apply the quotient identity

Substitute sin 90° = 1 and cos 90° = 0

Division by zero Apply the quotient identity

Substitute cos 0° = 1 and sin 0° = 0

Division by zero

Both sides are undefined at θ = 0°, confirming the identity is undefined at θ = 180° n.

b sec (90° − θ ) = cosec θ

Create a strategy Use the reciprocal identity sec θ =

for sec (90° − θ ) and cosec θ =

to determine where

each side is undeﬁned by setting denominators to zero. Check both sides of the identity.

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Apply the idea For sec (90° − θ ): Apply the reciprocal identity sec θ =

Use the identity cos (90° − θ ) = sin θ

Evaluate sin θ at the unit circle with r = 1

Division by zero

The angles where sin θ = 0 are θ = 180° n, where n is an integer (e.g., θ = 0°, 180°, 360°). For cosec θ : Apply the reciprocal identity

Evaluate sin θ at the unit circle with r = 1

Division by zero

The angles where sin θ = 0 are θ = 180° n, where n is an integer (e.g., θ = 0°, 180°, 360°). Thus, the identity sec (90° − θ ) = cosec θ is undeﬁned when θ = 180° n.

Reﬂect and check Verify by substituting θ = 0°: Evaluate the subtraction

Apply the reciprocal identity

Evaluate the exact value of cos 90°

Division by zero Apply the reciprocal identity

Evaluate the exact value of sin 0°

Division by zero

Both sides are undefined at θ = 0°, confirming the identity is undefined at θ = 180°n.

Idea summary Identities like tan (90° − θ ) = cot θ and sec (90° − θ ) = cosec θ are undeﬁned when a function has a zero denominator.

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8.04 Practice questions What do you remember? 1

Write the complementary angle identity for each function: a

sin (90° − θ )

tan (90° − θ )

sec (90° − θ )

cosec (90° − θ )

Identify where each identity is undeﬁned: a

cot (90° − θ ) = tan θ

sec (90° − θ ) = cosec θ

tan (90° − θ ) = cot θ

cosec (90° − θ ) = sec θ

Determine where each identity is true or false: a

cos 30° = sin 60°

tan 60° = cot 45°

sec 45° = cosec 45°

sin 45° = cos 60°

Practice Ex 1

Prove these identities using the diagram: a

cos (90° − θ ) = sin θ

cot (90° − θ ) = tan θ

sec (90° − θ ) = cosec θ

cosec (90° − θ ) = sec θ

90° − θ

θ b Ex 2

Identify the angles θ where the identities are undeﬁned: a

tan (90° − θ ) × tan θ

sec (90° − θ ) × sin θ

cos (90° − θ )

tan (90° − θ )

where 0° < θ < 90°:

sec (90° − θ )

cot (90° − θ ) = tan θ

Express in terms of θ in a right-angled triangle: a

cosec (90° − θ ) = sec θ

Find the exact value of each expression, given that sin θ = a

Simplify using complementary angle identities: a

cot (90° − θ ) = tan θ

cos (90° − θ )

cot (90° − θ )

Verify using quotient identities: a

tan (90° − θ ) = cot θ

Calculate using complementary angle identities:

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cot (90° − θ )

sin

+ cos

sec

− cosec

cos

+ sin

cosec

− sec

Given that tan θ = 3. Find cot (90° − θ ).

Extend your thinking 12

Prove cos (90° − θ ) = sin θ using the unit circle.

Explain why sin (90° − θ ) = cos θ holds at θ = 45° using the unit circle.

A triangle has sin θ = a

, evaluate each leaving the answer in exact form:

cos (90° − θ )

sec (90° − θ )

A student claims cos 30° = sin 30°. Correct using identities.

Prove that tan (90° − θ ) × tan θ = 1 where θ is such that tan θ is deﬁned.

Given that cos θ = , evaluate each leaving the answer in exact form: a

sin (90° − θ )

cosec (90° − θ )

Prove

Did you know?

Sailors can’t sail directly into the wind, so they use a zig-zag motion called tacking to move forward. The sail’s angle from the mast and the sail’s angle from the deck are complementary, letting sailors calculate forward thrust using sine and cosine of the same value. 8.04 Complementary angle identities mathspace.co

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8.05 Evaluate expressions with identities After this lesson, you will be able to… • evaluate trigonometric expressions involving acute angles using complementary angle identities. • find the exact value of trigonometric functions for angles of any magnitude using reference angles and quadrant signs. • combine complementary angle identities and quadrant rules to evaluate complex trigonometric expressions.

Expressions with acute angles Trigonometric expressions involving acute angles (between 0° and 90°) can be simpliﬁed using complementary angle identities and exact values of standard angles (30°, 45°, 60°). For example, angles like 60° can be rewritten as 90° − 30° to apply identities.

Example 1 Without simplifying inside the brackets, evaluate tan (90° − 45°) + cos (90° − 60°) in exact form.

Create a strategy Apply complementary angle identities to rewrite each term, then use exact values for 45° and 60°.

Apply the idea For tan (90° − 45°): Use tan (90° − θ ) = cot θ

Apply the identity cot θ =

Evaluate the exact value of tan 45°

Evaluate For cos (90° − 60°): Use cos (90° − θ ) = sin θ

Evaluate the exact value of sin 60°

Combining the results: Combine the results

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Idea summary Complementary angle identities simplify expressions with acute angles by converting between trigonometric functions. Use exact values for standard angles to evaluate the resulting expressions.

Angles of any magnitude To evaluate trigonometric expressions for angles of any magnitude, use the reference angle and the quadrant to determine the sign of the function. The reference angle is the acute angle formed with the x-axis. For an angle θ in: • Quadrant 1 (0° < θ < 90°): All functions are positive. • Quadrant 2 (90° < θ < 180°): Only sin θ and cosec θ are positive. • Quadrant 3 (180° < θ < 270°): Only tan θ and cot θ are positive. • Quadrant 4 (270° < θ < 360°): Only cos θ and sec θ are positive. For example, sin 150° = sin (180° − 30°) = sin 30° (positive in Quadrant 2).

Exploration Discuss: 1. How does the quadrant affect the sign of cos 210°? 2. Find the reference angle and determine if the result is positive or negative.

Example 2 Evaluate cos 135°.

Create a strategy Find the reference angle of 135° and apply the appropriate sign base on the quadrant where 135° is.

Apply the idea The angle 135° is in Quadrant 2, so: Subtract θ from 180° for the reference angle

Evaluate the subtraction

Use the fact that cos θ is negative in Quadrant 2

Evaluate the exact value of cos 45°

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Idea summary For angles of any magnitude, ﬁnd the reference angle and use the quadrant to determine the sign of the trigonometric function. Express angles in terms of standard angles for evaluation.

Combine identities and quadrants Complex trigonometric expressions may require combining complementary angle identities with quadrant adjustments for angles of any magnitude. Steps to evaluate: 1. Apply complementary angle identities to simplify the expression. 2. Determine the quadrant of the resulting angle. 3. Use the reference angle and quadrant sign to evaluate.

Interactive exploration Discover this concept in action online

mathspace.co

Example 3 Without simplifying inside the brackets, evaluate sin (90° − 150°) + cos 210°.

Create a strategy Use complementary angle identities for the ﬁrst term, ﬁnd the reference angle and quadrant for both terms, and evaluate using exact values.

Apply the idea For sin (90° − 150°): Apply sin (90° − θ ) = cos θ with θ = 150°

Subtract θ from 180° for the reference angle

Evaluate the subtraction

Use the fact that cos θ is negative in Quadrant 2

Evaluate the exact value of cos 30°

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For cos 210°: Subtract 180° from θ for the reference angle

Evaluate the subtraction

Use the fact that cos θ is negative in Quadrant 2

Evaluate the exact value of cos 30°

Combining the results: Combine the results

Evaluate the addition

Simplify

Idea summary Combine complementary angle identities with quadrant adjustments to evaluate complex trigonometric expressions. Simplify using identities, ﬁnd reference angles, and apply the correct sign based on the quadrant.

8.05 Practice questions What do you remember? 1

Write the complementary angle identities for reciprocal functions: a

cosec (90° − θ )

Determine the quadrant and reference angle for: a

sec (90° − θ )

120°

225°

315°

180°

Determine whether each statement is true or false: a

sin (135°) = sin (45°)

cos (240°) = − cos (60°)

tan (150°) = cot (30°)

cot (330°) = − cot (30°)

Identify the sign of each trigonometric function: a

sin 200°

cos 170°

tan 280°

cot 110°

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Practice Ex 1

Ex 2

Without simplifying inside the brackets, evaluate each expression: a

sin (90° − 30°) + cos (90° − 45°)

tan (90° − 60°) − cot (90° − 30°)

sec (90° − 45°) + cosec (90° − 60°)

cosec (90° − 30°) − sin (90° − 60°)

Evaluate using reference angles and quadrants: a

Ex 3

tan 315°

cot 240°

sin (90° − 135°) + cos 225°

tan (90° − 120°) + sin 300°

cosec (90° − 150°) + cot (90° − 210°)

sec (90° − 240°) − sin (90° − 300°)

Simplify using complementary angle identities: a

sin (90° − θ ) × cos θ

cos (90° − θ ) × sin θ

tan (90° − θ ) × cot θ

cot (90° − θ ) × tan θ

Without simplifying inside the brackets, evaluate using exact values: cos (90° − 45°) − sin 45°

sin (90° − 30°) − cos 60°

Express trigonometric functions in terms of another function using complementary identities: a

sin (90° − θ ) × cot θ

tan (90° − θ ) × sin θ

Verify these trigonometric identities: a

sin (90° − θ ) = cos θ

tan (90° − θ ) × sin θ = cos θ

= tan θ

cot (90° − θ ) × cos θ = sin θ

Without simplifying inside the brackets, evaluate and simplify where possible:

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cos 210°

Without simplifying inside the brackets, evaluate each expression:

a 10

sin 120°

sin (90° − 45°) × cos 45°

b d

Without simplifying inside the brackets, evaluate expressions involving angles across multiple quadrants: a

sin (90° − 60°) + cos (180° + 30°)

cos (90° − 30°) − sin (360° − 60°)

tan (90° − 45°) × sin (270° − 45°)

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Extend your thinking 14

Explain why cos (120°) = − cos (60°) using the unit circle and reference angles.

Correct the error in the statement sin (90° − 120°) = − cos 120° using complementary angle identities.

In a navigation context using true bearings (measured clockwise from north), a boat travels at an angle of 135°. Express the northward component of its direction using a trigonometric expression and evaluate. N

E 135°

Determine if the identity cos (90° − θ ) + sin (180° − θ ) = sin θ is true or false for 0° ≤ θ ≤ 90°. Explain your reasoning.

A right-angled triangle has an acute angle θ with cos θ = a

Given sin θ = a

sin (90° − θ )

. Evaluate each in exact form:

tan (90° − θ )

in Quadrant 2, evaluate each in exact form:

cos (90° − θ )

tan (90° − θ )

Verify whether this identity holds true or not for all θ such that sin θ ≠ 0:

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8.06 Simplify and prove identities After this lesson, you will be able to… • prove the Pythagorean identity cos2θ + sin2θ = 1. • derive the other two Pythagorean identities involving tangent, cotangent, secant and cosecant. • apply the Pythagorean identities to simplify trigonometric expressions. • prove further trigonometric identities using the Pythagorean identities in combination with reciprocal and quotient identities.

Pythagorean identities The Pythagorean identities are fundamental trigonometric relationships derived from the geometry of a right-angled triangle or the unit circle. They connect sine, cosine, tangent, secant, cosecant and cotangent functions. The ﬁrst identity can be obtained by using the trigonometric ratios and the Pythagoras’ theorem for a right-angled triangle.

θ a

From the right-angled triangle, the trigonometric ratios for the angle can be expressed as:

• sin θ = • cos θ = b

rite the left-hand side of the identity and substitute W the values of sin θ and cos θ

Evaluate the powers

Rewrite with common denominator

Use the Pythagoras’ theorem a2 + b2 = c2

Simplify

cos2θ + sin2θ = 1

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cos θ

is the x-coordinate of a point on the unit circle

sin θ

is the y-coordinate of a point on the unit circle

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The second identity can be derived by dividing the ﬁrst identity by cos2θ , valid where cos θ ≠ 0. Write the Pythagorean identity

Divide both sides by cos2θ

Simplify

Take out the powers

Substitute

= tan θ and

= sec θ

1 + tan2θ = sec2θ tan θ

is the ratio

, undeﬁned when cos θ = 0

sec θ

is the ratio

, undeﬁned when cos θ = 0

The third identity can be derived by dividing the ﬁrst identity by sin2θ , valid where sin θ ≠ 0. Write the Pythagorean identity

Divide both sides by sin2θ

Simplify

Take out the powers

Substitute

= cot θ and

= cosec θ

1 + cot2θ = cosec2θ cot θ

is the ratio

, undeﬁned when sin θ = 0

cosec θ

is the ratio

, undeﬁned when sin θ = 0

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Example 1 Verify the three Pythagorean identities at θ = 30°: a sin2θ + cos2θ = 1

Create a strategy Substitute θ = 30 into the left-hand side of the equation.

Apply the idea Substitute θ = 30°

Evaluate the exact values

Evaluate the powers

Evaluate

Thus, the identity holds for θ = 30°.

b 1 + tan2θ = sec2θ

Apply the idea For 1 + tan2 30°: Substitute θ = 30°

Evaluate the exact value of tan 30°

Evaluate the power

Add the fractions

Simplify

For sec2 30°: Substitute θ = 30°

Evaluate the exact value

Evaluate the power

Since 1 + tan2 30° = sec2 30° = , the identity holds where cos θ ≠ 0.

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c 1 + cot2θ = cosec2θ

Apply the idea For 1 + cot2 30°: Substitute θ = 30°

Evaluate the exact value of cot 30°

Evaluate the power

Evaluate

For cosec 30°: cosec2θ = cosec2 30° 2

Substitute θ = 30°

Evaluate the exact value

Evaluate the power

Since 1 + cot2 30° = cosec2 30° = 4, the identity holds where sin θ ≠ 0.

Idea summary The Pythagorean identities are cos2θ + sin2θ = 1, 1 + tan2θ = sec2θ (where cos θ ≠ 0), and 1 + cot2θ = cosec2θ (where sin θ ≠ 0). They are derived from the unit circle or triangle geometry and hold for all defined angles.

Simplify and prove identities The Pythagorean identities simplify complex trigonometric expressions. By substituting equivalent forms of sin2θ + cos2θ = 1 or its related identities, these expressions can often be reduced, aiding in proofs and problem-solving. Recognising opportunities to replace and simplify terms helps develop critical thinking. When simplifying trigonometric expressions, it is often easiest to reduce them to the ratios sin θ and cos θ. In identity proofs, always begin with the more complicated side and simplify it until it matches the other side.

Example 2 Simplify: a

Create a strategy Multiply the fractions, then identify and substitute any relevant Pythagorean identities.

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Apply the idea Multiply the fractions

Expand the brackets in the denominator

Use an equivalent form of the ﬁrst Pythagorean identity: 1 − sin2θ = cos2θ

Apply the reciprocal identity

= sec θ

b tan θ cos2θ sec θ

Create a strategy Use quotient and reciprocal identities.

Apply the idea Substitute tan θ =

and sec θ =

Evaluate and remove the common factors

Create a strategy Use the equivalent Pythagorean and reciprocal identities.

Apply the idea Substitute 1 − sin2θ = cos2θ

Use the identity sec θ =

Use the identity 1 + tan2θ = sec2θ

Combine like terms

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Create a strategy Use the equivalent Pythagorean and reciprocal identities.

Apply the idea Apply the identity sin2θ + cos2θ = 1

Apply

= sec2θ

Example 3 Consider the identity tan θ sin θ + cos θ = sec θ : a Prove the identity.

Create a strategy Since the LHS is more complex than the RHS, simplify the LHS in order to show it is equal to the RHS.

Apply the idea Write the left-hand side

Substitute tan θ =

Evaluate the multiplication

Rewrite with a common denominator

Apply sin2θ + cos2θ = 1

Apply

= sec θ

Thus, the identity holds.

b Find the exact value of tan 45° sin 45° + cos 45°.

Create a strategy Use the proven identity in part (a), then substitute known trigonometric values and simplify.

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Apply the idea Use the identity tan θ sin θ + cos θ = sec θ

Apply the identity sec θ =

Substitute cos 45° =

Simplify

Reﬂect and check Alternatively, substitute values directly: tan 45° = 1, sin 45° =

, cos 45° =

gives

, which matches.

Example 4 Prove the identity 5 cos2θ − 3 = 2 − 5 sin2θ.

Create a strategy Since both sides appear equally complex, begin simplifying either one. Continue simplifying until it becomes identical to the other side. Use the fundamental Pythagorean identity, sin2θ + cos2θ = 1 to manipulate the left-hand side of the equation until it is identical to the right-hand side.

Apply the idea LHS = 5 cos2θ – 3

Write the left-hand side

= 5 (1 − sin2θ ) − 3

Substitute cos2θ = 1 − sin2θ

= 5 − 5 sin2θ − 3

Expand the brackets

= 2 − 5 sin θ

Combine the constant terms

= RHS

Idea summary To simplify trigonometric expressions, prove identities, or solve problems, apply identities such as the Pythagorean and quotient identities. It is often best to express all terms using only sin θ and cos θ. When proving identities, begin with the more complex side and simplify until it matches the other side.

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8.06 Practice questions What do you remember? 1

Write the three Pythagorean identities.

Identify where these identities are undeﬁned: a

1 + tan2 θ = sec2θ

1 + cot2θ = cosec2θ

cot θ

Express each in terms of sin θ and cos θ : a

sec θ

Practice Ex 1

Verify the three Pythagorean identities at θ = 60°: a c

Ex 2

sin2θ + cos2θ = 1 2

1 + cot θ = cosec θ

Simplify:

c e

Ex 3

Ex 4

(sec θ + cos θ )(sec θ − cos θ )

(sec θ + cosec θ )(cos θ + sin θ )

(sin2θ − cos2θ ) ×

Consider the identity cot θ cos θ + sin θ = cosec θ : a

Prove the identity.

Find the exact value of cot 60° cos 60° + sin 60°.

Find the exact value of cot

cos

+ sin

Prove each identity: a

(sin θ + cos θ )(cot θ + tan θ ) = sec θ + cosec θ

cos θ (tan θ + 2)(2 tan θ + 1) = 2 sec θ + 5 sin θ

A ship navigates at an angle θ from north, where tan θ = 2. Simplify these expressions related to its direction: a

1 + tan2θ = sec2θ

sec2θ

cosec2θ

cot θ cosec θ

In a right-angled triangle, cot θ = . Verify if these relationships hold true for this triangle: a

cosec2θ = 1 + cot2θ

cosec θ cot θ =

= cos θ

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A surveyor measures an angle θ in a right-angled triangle, where θ = 30° or θ = 45°. Find the exact values of these expressions: a

sec2

c 11

− tan2

cosec2

− cot2

cot 45° cosec 45°

cot2θ =

Consider the right-angled triangle, sec θ = : a

Prove these identities for this triangle: i

tan2θ = sec2θ − 1

Find the exact values of each for this triangle: i

tan2θ

A student simpliﬁes

Prove:

cosec2θ

+ 1 as tan2θ. Correct the error.

= cos θ

Extend your thinking 14

in the interval π ≤ θ ≤

, ﬁnd sec2θ.

Given sin θ =

Explain, without evaluating, why the answer for sec2θ will be the same for

≤ θ ≤ 2π.

. Find the exact value of cosec2θ.

In a right-angled triangle, tan θ =

Prove:

In a navigation context, a ship travels at an angle θ from

= sin2θ cos2θ

north in a right-angled triangle, where tan θ = Find sec2θ.

. 15

≤ θ ≤ π, ﬁnd cot2θ.

Given cos θ =

Prove:

In a triangle, cot θ = . Find the exact value of sec2θ.

Prove:

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in the interval =

= 2 cosec θ

90° − θ

8.07 Trigonometric equations After this lesson, you will be able to… • solve trigonometric equations for a given restricted domain in degrees or radians. • use the periodicity of trigonometric functions to find solutions in an extended domain. • solve trigonometric equations that reduce to a quadratic form. • apply trigonometric identities to simplify and solve equations.

Trigonometric equations A trigonometric equation involves trigonometric functions and is solved to ﬁnd angles that satisfy the equation within a restricted domain, typically [0°, 360°] or [0, 2π ]. To solve equations like sin θ = a, use the inverse function sin−1a to ﬁnd the principal angle (the angle measured anticlockwise from the positive x-axis to the ray representing the angle, with a value in the interval [0°, 360°]). Additional solutions are found using the signs of trigonometric ratios in each quadrant.

θ = sin−1a

S sin positive

T tan positive

is a constant where −1 ≤ a ≤ 1

is the angle that satisﬁes the equation

A all positive (cos θ , sinθ )

θ −1

−1

C cos positive

Remember, ASTC provides positive trigonometric function values such that: • In Quadrant 1, all ratios are positive. • In Quadrant 2, only sin θ is positive. • In Quadrant 3, only tan θ is positive. • In Quadrant 4, only cos θ is positive.

For sin θ = a, if a is positive, solutions are in Quadrants 1 and 2. If negative, solutions are in Quadrants 3 and 4. Similar rules apply for cos θ and tan θ.

Interactive exploration Discover this concept in action online

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Example 1 Solve cos θ = 0.5 for θ on [0°, 360°].

Create a strategy Use the inverse cosine function to ﬁnd the principal angle and the ASTC mnemonic to identify other solutions in the domain.

Apply the idea cos θ = 0.5

θ = cos−1 0.5 = 60°

Write the equation Take the inverse cosine of both sides Evaluate

Since 0.5 is positive, cos θ is positive in Quadrants 1 and 4 (ASTC: A and C). The reference angle is 60° (Quadrant 1). In Quadrant 4, the angle is 360° − 60° = 300°. The solutions are θ = 60°, 300°.

Idea summary Solve trigonometric equations using inverse functions and the ASTC mnemonic to ﬁnd all solutions within [0°, 360°] by identifying quadrants where the function is positive or negative.

Extended domain equations Trigonometric functions are periodic, repeating every 360° (or 2π in radians) for sin θ, cos θ, and 180° ( π radians) for tan θ. To solve equations in domains beyond [0°, 360°], such as [0°, 720°] or [−180°, 180°], ﬁrst ﬁnd solutions in [0°, 360°] and then add or subtract multiples of 360° (or 2π ) to cover the extended domain.

Example 2 Solve tan θ = 1 for θ on [0, 4π ].

Create a strategy Find solutions in [0, 2π ] using the inverse tangent function and the ASTC mnemonic, then add multiples of 2π to cover [0, 4π ].

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Apply the idea Write the equation

Take the inverse tangent of both sides

Evaluate Since 1 is positive, tan θ is positive in Quadrants 1 and 3 (ASTC: A and T). In [0, 2π ], solutions are (Quadrant 1) and

(Quadrant 3).

Add 2π to each solution: Add 2π to Evaluate Add 2π to Evaluate The solutions are

Idea summary Solve trigonometric equations in extended domains by ﬁnding solutions in [0°, 360°] and adding or subtracting the function’s period until the domain is covered.

Quadratic trigonometric equations Some trigonometric equations reduce to quadratic equations in terms of a trigonometric function. Substitute a variable (e.g., u = sin θ ), solve the quadratic, and ﬁnd angles within the restricted domain using the ASTC mnemonic.

Example 3 Solve 2 sin2θ − sin θ − 1 = 0 for θ on [0°, 360°].

Create a strategy Let u = sin θ, solve the quadratic equation in u, then ﬁnd angles for θ using the ASTC mnemonic.

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Apply the idea 2 sin2θ − sin θ − 1 = 0

Write the equation

Substitute u = sin θ

2u − u − 1 = 0 Using the quadratic formula to solve for u:

Write the quadratic formula

Substitute x = u, a = 2, b = −1 and c = −1

Evaluate each term

Evaluate the surd

Evaluate and simplify the positive and negative values Since u = sin θ, solve for θ where −1 ≤ u ≤ 1. For sin θ = 1: sin θ = 1

Write the equation −1

θ = sin 1

Take the inverse sine of both sides

= 90° For sin θ =

Evaluate

: Write the equation

Take the inverse sine of both sides

Since

is negative, take the angles in Quadrants 3 and 4

The solutions are θ = 90°, 210°, 330°.

Reﬂect and check Verify by substituting each value of θ into the left-hand side (LHS) of the original equation, 2 sin2θ − sin θ − 1. For θ = 90°:

LHS = 2 sin2θ − sin θ − 1

Write the left-hand side

Substitute θ = 90°

=2×1 −1−1

Evaluate the exact value of sin 90°

=2−1−1

Evaluate the first term

Evaluate

= 2 sin 90° − sin 90° − 1

= RHS

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For θ = 210°: Substitute θ = 210°

Evaluate the exact value of sin 210°

Evaluate the power and adjacent signs

Simplify

Evaluate

For θ = 330°: Substitute θ = 330°

Evaluate the exact value of sin 330°

Evaluate the power and adjacent signs

Simplify

Evaluate

Since the left-hand side equals the right-hand side for all angles, the solutions are correct.

Idea summary Solve quadratic trigonometric equations by substituting a variable, solving the quadratic, and ﬁnding all angles in the restricted domain using the ASTC mnemonic.

8.07 Practice questions What do you remember? 1

Determine whether each statement is true or false for the domain [0°, 360°]: a

sin x = 0.5 has two solutions

tan x = 1 has two solutions

cos x = 0 has one solution

tan x = 0 has one solution

Identify the quadrants where the solutions lie if 0° ≤ θ ≤ 360°: a

sin x > 0

cos x < 0

tan x > 0

sin x < 0

What is the signiﬁcance of a restricted domain in solving trigonometric equations?

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Practice Ex 1

Solve for θ on [0°, 360°]: a

Ex 2

cos θ =

tan θ =

sin θ = 0

cos θ =

sin θ =

tan θ = 0

cos θ = −1

tan θ + 1 = 0

4 sin θ − 2 = 0

Solve for θ, where 0° ≤ θ ≤ 360°: a

Ex 3

Solve for θ on [0, 4π ]: a

sin θ = 0.5

2 sin θ −

2 cos θ + 1 = 0

Solve the quadratic trigonometric equations for θ on [0°, 360°]: a

2 sin2θ + sin θ − 1 = 0

tan2θ − 3 tan θ + 2 = 0

Solve for x, where 0 ≤ x ≤ π :

a 9

In a triangle, sin θ =

Solve for θ, where 0 ≤ θ ≤ 4π : a

. Solve for θ on 0° ≤ θ ≤ 90°, rounded to two decimal places.

sin2θ =

cos2θ =

cos θ =

tan θ =

cos θ + 1 = 0

Solve for θ, where 0 ≤ θ ≤ 6π : a

Solve for θ, rounded to two decimal places, where 90° ≤ θ ≤ 180°, if: a

3 cos2θ − cos θ − 2 = 0

cot θ =

A student claims sin θ = and explain.

has only one solution, θ = 210° on [0°, 360°]. Correct the error

Extend your thinking 14

Explain why cos θ = 1.5 has no real solutions.

A student solves tan θ =

A pendulum swings at an angle θ from vertical. Solve for θ, where −180° ≤ θ ≤ 180°, using identities if necessary: a

354

on [0°, 360°] and ﬁnds θ = 150°. Correct the error.

sin θ = sin2θ

2 sin2θ − cos θ = 1

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A Ferris wheel’s angle θ satisﬁes cos θ = 0.8. Find all solutions on [0, 2π ], rounded to three decimal places.

Prove that the solutions to sin (2θ ) = sin θ in [0°, 360°] are θ = 0°, 60°, 180°, 300°, 360°.

In navigation, a ship’s bearing satisﬁes sin x − cosec x = 0. Solve for x on [0°, 360°].

Solve sin2θ =

Solve 2 sin θ − tan θ = 0 for 0° ≤ θ ≤ 360°.

for 0 ≤ θ ≤ 2π.

Did you know?

Trigonometric identities are essential in designing domes and patterned ceilings! By applying equations involving sine and cosine, architects can create perfect symmetry, calculate curves, and design visually stunning structures that are both artistic and mathematically precise. These identities also help engineers ensure that the dome distributes weight evenly, making it both stable and durable. Trigonometry transforms creative vision into architectural masterpieces that combine beauty with structural strength.

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8.08E Sum and difference expansions for trigonometric functions After this lesson, you will be able to… • derive the sum and difference expansions for sin ( A ± B), cos ( A ± B), and tan ( A ± B) • recall and accurately state the six angle sum and difference identities • apply the sum and difference identities to find exact trigonometric values for various angles • use the sum and difference identities to simplify trigonometric expressions and prove other trigonometric results

Sum and difference expansions for trigonometric functions Interactive exploration Discover this concept in action online

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Angle sum and difference formulas express the sine, cosine, and tangent of the sum or difference of two angles using the trigonometric values of the individual angles. These formulas simplify expressions and solve equations in trigonometry. The formulas are: Formula Sine angle sum

sin ( A + B) = sin A cos B + cos A sin B

Sine angle difference

sin ( A − B) = sin A cos B − cos A sin B

Cosine angle sum

cos ( A + B) = cos A cos B − sin A sin B

Cosine angle difference

cos ( A − B) = cos A cos B + sin A sin B

Tangent angle sum Tangent angle difference

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To derive each formula, consider points P and Q on a unit circle, with coordinates P (cos A, sin A) and Q(cos B, sin B).

(0, 1) Q(cos B, sin B)

P(cos A, sin A)

A (−1, 0)

(1, 0)

(0, −1)

For the cosine difference formula cos ( A − B): 1. Derive cos ( A − B) using the chord PQ. 2. Calculate PQ2 as the hypotenuse of △OPQ using the cosine rule: c2 = a2 + b2 − 2ab cos C 2

Write the formula

Substitute c = PQ, a = 1, b = 1 and PQ = 1 + 1 − 2 × 1 × 1 × cos ( A − B) C=A−B = 2 − 2 cos ( A − B)

Evaluate

3. Calculate PQ2 as the distance of two points using the distance formula: d2 = ( x2 − x1)2 + ( y2 − y1 )2 2

Write the formula

PQ = (cos B − cos A) + (sin B − sin A) 2

Substitute d = PQ and the coordinates 2

= cos B − 2 cos B cos A + cos A + sin B − 2 sin B sin A + sin2 A Expand = ( cos2 B + sin2 B) + (cos2 A + sin2 A) − 2(cos B cos A + sin B sin A)

Group the terms

= 1 + 1 − 2(cos B cos A + sin B sin A)

Use sin2 θ + cos2 θ = 1

= 2 − 2(cos B cos A + sin B sin A)

Simplify

4. Equate both expressions from using the cosine rule and distance formula for PQ2: 2 − 2 cos ( A − B) = 2 − 2(cos B cos A + sin B sin A) Equate the expressions cos ( A − B) = cos B cos A + sin B sin A = cos A cos B + sin A sin B

Divide both sides by −2 Rearrange the angles

So the cosine difference formula is: cos ( A − B) = cos A cos B + sin A sin B For the cosine sum formula cos ( A + B): cos ( A + B) = cos ( A − (−B))

Rewrite with adjacent signs

= cos A cos (−B) + sin A sin (−B)

Apply cosine difference formula

= cos A cos B − sin A sin B

Use cos (−B) = cos B and sin (−B) = − sin B

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So the cosine sum formula is: cos ( A + B) = cos A cos B − sin A sin B For the sine difference formula sin ( A − B): Use the complementary angle formula

Group the terms

Apply cosine sum formula = sin A and

Use

= cos A So the sine difference formula is: sin ( A − B) = sin A cos B − cos A sin B For the sine sum formula sin ( A + B): Use the complementary angle formula

Rewrite

Apply cosine difference formula = sin A and

Use

= cos A So the sine sum formula is: sin ( A + B) = sin A cos B + cos A sin B For the tangent sum formula tan ( A + B): Use the formula tan θ =

Substitute sine and cosine sum formulas

Divide numerator and denominator by cos A cos B

Simplify

Substitute

So the tangent sum formula is:

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= tan A and

= tan B

For the tangent difference formula tan ( A − B): Use the formula tan θ =

Substitute sine and cosine difference formulas

Divide numerator and denominator by cos A cos B

Simplify

Substitute

= tan A and

= tan B

So the tangent difference formula is:

Example 1 Given sin A =

and cos B =

, determine the exact value:

a sin ( A + B)

Create a strategy Use the sine sum formula: sin ( A + B) = sin A cos B + cos A sin B

Apply the idea For angle A, since sin A =

, the opposite is 24 and hypotenuse is 25. Use Pythagoras’ theorem

Substitute c = 25 and a = 24

Take the square root of both sides

Evaluate the squares

Evaluate the difference

Evaluate the square root

Therefore, adjacent is 7.

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For angle B, since cos B =

, the adjacent is 20 and hypotenuse is 29. Use Pythagoras’ theorem

Substitute c = 29 and a = 20

Take the square root of both sides

Evaluate the squares

Evaluate the difference

Evaluate the square root

Therefore, adjacent is 21. For sin ( A + B):

Write the formula

Substitute cos A =

and sin B =

Evaluate

b cos ( A − B)

Create a strategy Use the cosine sum formula: cos ( A − B) = cos A cos B + sin A sin B

Apply the idea Write the formula

Substitute known values

Evaluate

Example 2 Express sin A cos 2B + cos A sin 2B as a single trigonometric ratio.

Create a strategy Use the sine sum formula: sin ( A + B) = sin A cos B + cos A sin B

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Apply the idea sin ( A + B) = sin A cos B + cos A sin B

Write the formula

sin ( A + 2B) = sin A cos 2B + cos A sin 2B

Substitute B = 2B

sin A cos 2B + cos A sin 2B = sin ( A + 2B)

Swap sides

The given expression can be expressed as sin ( A + 2B).

Example 3 Determine the exact value of cos

in surd form.

Create a strategy Use the cosine sum formula: cos ( A + B) = cos A cos B − sin A sin B

Apply the idea Rewrite as a sum of two exact angles

Use the cosine sum formula

Evaluate each trigonometric value

Evaluate the multiplication

Combine like terms

Reﬂect and check To verify, use a different angle pair, by swapping the order of angles: Write the formula

Substitute

Evaluate each trigonometric value

Evaluate the multiplication

Combine like terms

This matches our result, conﬁrming the solution. Additionally, since

is in the second quadrant

, the cosine should be negative, which it is.

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Example 4 Using the tangent sum formula, determine the exact value of tan (75°).

Create a strategy Apply the tangent sum formula with A = 45° and B = 30° .

Apply the idea

Write the formula

Substitute A = 45° and B = 30°

Substitute tan (45°) = 1 and tan (30°) =

Simplify the numerator

Combine the terms in the numerator and denominator

Remove the common factor

Rationalise the denominator

Expand the numerator and denominator

Simplify the numerator and denominator

Divide by the common factor to simplify

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Idea summary Angle sum and difference formulas are essential in trigonometry for simplifying expressions, solving equations and proving other trigonometric formulas involving combined angles. Formula Sine angle sum

sin ( A + B) = sin A cos B + cos A sin B

Sine angle difference

sin ( A − B) = sin A cos B − cos A sin B

Cosine angle sum

cos ( A + B) = cos A cos B − sin A sin B

Cosine angle difference

cos ( A − B) = cos A cos B + sin A sin B

Tangent angle sum Tangent angle difference

8.08E Practice questions What do you remember? 1

Identify the angle sum and difference formulas: a

Given sin A = a

cos ( A − B)

tan ( A + B)

and cos A = , where A is acute, calculate:

tan A

sin ( A + B) = sin A + sin B

cos ( A − B) = cos A cos B + sin A sin B

Given sin θ =

cot A

tan θ

tan ( A − B)

tan α

and θ is acute, ﬁnd:

cos θ

Identify the trigonometric identity for: a

Are these statements true or false?

a 5

sin ( A + B)

sin ( A − B)

If cos α = a

and α is acute, ﬁnd:

sin α

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Practice Ex 1

Suppose sin α = value: a

Show that sin ( A + B) = 1.

Determine the exact value of sin ( A − B).

cos ( A + B) cos B + sin ( A + B) sin B

sin (30° + 60°)

cos (45° + 45°) and cos B = :

Determine the exact value of cos A.

Determine the exact value of sin B.

Show that sin ( A + B) =

Determine the exact value of cos ( A − B).

cos (30°) = cos (60° − 30°)

Verify each trigonometric identity: sin (90°) = 1

Given sin A =

and cos B =

, where both angles are acute, determine:

sin ( A + B)

Suppose sin α = exact value: sin (α + β )

and cos β =

cos ( A + B)

. Given that α and β are acute angles, determine the

cos (α + β )

tan (α − β )

cos (β − α )

sin (2x + y)

tan (2α + 3β )

tan (θ − 3σ)

Expand: cos ( x − 2y)

Express as a single trigonometric ratio. a

cos 45° cos 15° + sin 45° sin 15°

cos 60° cos 15° + sin 60° sin 15°

cos 75° cos 30° + sin 75° sin 30°

cos 90° cos 45° + sin 90° sin 45°

Determine the exact value of sin 15°. (Hint: Use the fact that 45° − 30° = 15°)

Ex 3

Determine the exact value of: a

364

and cos B = :

a Ex 2

cos (α − β )

A and B are two different acute angles such that sin A =

a 15

Calculate the exact value using the compound angle formulas:

a 14

sin ( A + B) cos B − cos ( A + B) sin B

a 13

sin (α − β )

A and B are two different angles. Given that sin A = , cos A = , sin B =

a 11

. Given that α and β are acute angles, determine the exact

Simplify: a

and cos β =

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Determine the exact value: a

cos 75°

tan 15°

cos (−15°)

Show that: a

Ex 4

sin 75°

sin 105° + cos 105° = cos 45°

Using tan ( A − B) =

, determine the exact value of tan 15°.

(Hint: Use the fact that 45° − 30° = 15°) 22

If sin α =

Given sin A = a

and cos β =

, where α and β are acute angles, determine tan ( α + β ).

and cos B =

, where both angles are acute, calculate:

sin ( A + B)

cos ( A + B)

Extend your thinking 24

Using tan ( A + B) =

and tan ( A − B) =

, show that

. 25

Show that sin ( A + B) sin ( A − B) = sin2 A − sin2 B.

Given α > 0 and β <

, prove that α + β =

when tan α =

, tan β =

and m is a

positive constant. 27

Expand: a

sin ( A + B + C)

Given that sin A + sin B =

Show that

cos ( A + B + C)

tan ( A + B + C)

and cos A + cos B = , determine cos ( A − B).

sin 54° = cos 9° + sin 9°.

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8.09E Double angle formulas After this lesson, you will be able to… • derive the double angle formulas for sin (2 A), cos (2 A) (all three forms), and tan (2 A) from the sum identities • recall and accurately state the double angle formulas • apply double angle formulas to find exact trigonometric values for double angles, given information about the single angle • use double angle formulas to simplify trigonometric expressions and prove other trigonometric results or identities

Double angle formulas Double angle formulas express the sine, cosine, and tangent of twice an angle in terms of trigonometric functions of the single angle. These formulas are derived from angle sum formulas and are essential for simplifying expressions and solving equations. The primary double angle formulas are: • Sine double angle formula: sin (2 A) = 2 sin A cos A • Cosine double angle formulas: cos (2 A) = cos2 A − sin2 A cos (2 A) = 2 cos2 A − 1 cos (2 A) = 1 − 2 sin2 A • Tangent double angle formula: tan (2 A) = To derive the sine double angle formula, the sine angle sum formula is applied. Expressing 2 A as A + A, the derivation proceeds: sin (2 A) = sin ( A + A)

Express 2 A as a sum

= sin A cos A + cos A sin A

Apply sine angle sum formula

= 2 sin A cos A

Evaluate

For the cosine double angle formula, the cosine angle sum formula is used. Expressing 2 A as A + A, the derivation is: cos (2 A) = cos ( A + A)

Express 2 A as a sum

= cos A cos A − sin A sin A 2

= cos A − sin A

Apply cosine angle sum formula Evaluate

The alternative cosine formulas are derived using the Pythagorean formula cos2 A + sin2 A = 1. Substituting cos2 A = 1 − sin2 A into cos (2 A) = cos2 A − sin2 A yields: cos (2 A) = 1 − sin2 A − sin2 A = 1 − 2 sin2 A

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Substitute cos2 A Simplify

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Substituting sin2 A = 1 − cos2 A gives: cos (2 A) = cos2 A − (1 − cos2 A) 2

= 2 cos A − 1

Substitute sin2 A Simplify

The tangent double angle formula is derived using the sine and cosine double angle formulas. Use the formula tan θ =

Substitute sine and cosine sum formulas

Divide numerator and denominator by cos2 A

Substitute

= tan A and

= tan2 A

Example 1 Given cos A =

and sin A < 0, determine:

a sin A

Create a strategy Apply the Pythagorean formula: sin2 A + cos2 A = 1

Apply the idea Write the formula

Substitute cos A =

Evaluate the index

Subtract

Take the square root of both sides

Select negative value since sin A < 0

from both sides

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b sin (2 A)

Create a strategy Apply the sine double angle formula: sin (2 A) = 2 sin A cos A

Apply the idea Write the formula

Substitute sin A =

Evaluate

and cos A =

c cos (2 A)

Create a strategy Apply the cosine double angle formula: cos (2 A) = cos2 A − sin2 A

Apply the idea Write the formula

Substitute cos A =

Evaluate the indices

Evaluate

and sin A =

d tan (2 A)

Create a strategy Use tan A =

for tan A, then the tangent double angle formula: tan (2 A) =

Apply the idea For tan A: Write the formula

Substitute sin A =

Evaluate

368

and cos A =

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For tan (2 A): Write the formula

Substitute tan A =

Evaluate the multiplication

Evaluate the subtraction

Evaluate

Reﬂect and check Verify using tan (2 A) =

: Write the formula

Substitute sin (2 A) =

Evaluate

and cos (2 A) =

Example 2 Consider each expression: a Determine the exact value of cos x if sin x =

and

< x < π.

Create a strategy Apply the Pythagorean formula: sin2 x + cos2 x = 1

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Apply the idea Write the formula

Substitute sin x =

Evaluate the square

Subtract

Take the square root of both sides

Since

from both sides

< x < π, x is in the second quadrant where cosine is negative. Thus, cos x =

b Determine tan (2x) if sin x =

and

< x < π.

Create a strategy Apply the tangent double angle formula: tan (2x) = part (a).

, using tan x =

with values from

Apply the idea Write the formula

Substitute sin x =

Evaluate

and cos x =

Using the tangent double angle formula: Write the formula

Substitute tan x =

Evaluate numerator and denominator

Simplify denominator

Divide fractions

Evaluate

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into the tangent formula

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 3 Express: a sin x cos3 x − sin3 x cos x in terms of sin (4x)

Create a strategy Factorise the expression and use applicable double angle formulas.

Apply the idea Factor out sin x cos x

Apply cos (2x) = cos2 x − sin2 x

Apply sin (2x) = 2 sin x cos x

Apply sin (4x) = 2 sin (2x) cos (2x)

Evaluate the multiplication

b sin (3x) in terms of sin x

Create a strategy Express the angle as a sum of two angles, then use sine angle sum and double angle formulas.

Apply the idea sin (3x) = sin (2x + x)

Express the angles as a sum

= sin (2x) cos x + cos (2x) sin x

Apply sine angle sum formula 2

= 2 sin x cos x cos x + (1 − 2 sin x) sin x 2

= 2 sin x cos x + sin x − 2 sin x 2

Substitute double angle formulas Distribute

= 2 sin x (1 − sin x) + sin x − 2 sin x

Apply cos2 x = 1 − sin2 x

= 3 sin x − 4 sin3 x

Combine like terms

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Idea summary Double angle formulas express trigonometric functions of 2 A in terms of functions of A: • Sine double angle formula: sin (2 A) = 2 sin A cos A • Cosine double angle formulas: cos (2 A) = cos2 A − sin2 A cos (2 A) = 1 − 2 sin2 A cos (2 A) = 2 cos2 A − 1 • Tangent double angle formula:

These formulas simplify expressions, solve equations, and derive other formulas, enhancing eﬃciency in trigonometric problem-solving.

8.09E Practice questions What do you remember? 1

Identify the double angle formulas: a

Sine double angle formula

Cosine double angle formulas (all three forms)

Tangent double angle formula

Identify the Pythagorean identity used to derive the alternative forms of the cosine double angle formula.

Identify the cosine angle sum formula used to derive the cosine double angle formula.

Identify the deﬁnition of the tangent function in terms of sine and cosine, used in the derivation of the tangent double angle formula.

Identify the cosine double angle formula expressed only in terms of sine.

Are these statements true or false?

372

The sine double angle formula is derived from the cosine angle sum formula.

cos (2 A) = 2 cos2 A + 1 is a correct form of the cosine double angle formula.

The tangent double angle formula can be expressed as tan (2 A) =

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Practice 7

Consider the angle A such that sin A =

and cos A = :

Determine the exact value: a 8

and A is an acute angle, determine the exact value:

cos A

sin (2 A)

cos (2 A)

cos A

cos (2 A)

sin (2B)

tan (2 A)

and cos B = d

Given α is an acute angle and sin α = , determine the exact value of cos (2α ).

Given cos 330° =

Given sin A =

cos A

, determine the exact value of cos 165°. and

< A < π, determine the exact values: b

sin (2 A)

cos (2 A)

tan (2 A)

Prove: a

e 15

tan (2 A)

Ex 3

cos (2α ) in terms of cos α

Given that A and B are two different acute angles such that sin A = determine the exact value: a

Ex 2

cos (2 A)

cos (2α ) in terms of sin α

Given that sin A = a

Ex 1

Given α is an acute angle, express: a

sin (2 A)

Express: a

sin (4θ ) in terms of sin θ and cos θ

5 sin (2β ) cos3(2β ) − 5 sin3(2β ) cos (2β ) in terms of sin (8β )

Given tan A = 2 and using tan (2 A) = a

tan (2 A)

, determine the exact value:

tan (4 A)

Given cos θ =

and θ is an acute angle, determine the exact value of tan (2θ ).

Using the value of tan 45°, determine tan 22.5°.

Express cos (4θ ) in terms of cos θ using double angle formulas.

Express 3 sin (2α ) cos3(2α ) − 3 sin3(2α ) cos (2α ) in terms of sin (8α ).

Given tan A =

and 0 < A <

, determine the exact value of sin (2 A). 8.09E Double angle formulas mathspace.co

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Simplify cos4 x − sin4 x in terms of cos (2x).

If cos A = , and A is in the first quadrant, find sin (2 A).

Extend your thinking 24

Show that: 4 (cos3 20° + cos3 40°) = 3 (cos 20° + cos 40°)

Show that:

If tan α =

= tan (2 A) + sec (2 A) and sin β =

tan ( A + B) =

, where β is an acute angle, evaluate tan ( α + 2β ) using .

Given 2 tan θ = 3 tan d, show that tan ( θ − d ) =

Evaluate: cos4

Show cos (3θ ) = 4 cos3 θ − 3 cos θ.

Let x = 4 cos θ and cos (3θ ) = . Using part (a), show that this results in the polynomial x3 − 12x − 8 = 0.

Find the value of cos2

+ cos4

+ cos2

+ cos4

+ cos2

using part (b).

Did you know?

Shooting the perfect basketball free throw is all about trigonometry! Players and coaches use angles and trajectories to improve aim, arc, and accuracy on the court. By understanding the optimal angle and the force needed to launch the ball, athletes can maximise their chances of scoring. Even the height of the player and the position of their hands can be analysed using trigonometric functions. Sport scientists use this data to refine technique and create consistent training routines. Trigonometry helps turn practice into precision! 374

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

8.10E Trigonometric equations After this lesson, you will be able to… • solve trigonometric equations by factorising expressions containing trigonometric functions. • solve trigonometric equations by substituting known trigonometric identities (e.g., double angle identities) to simplify the equation. • determine all solutions to trigonometric equations that lie within a given restricted domain, expressed in radians or degrees. • utilise the unit circle effectively to find all angles corresponding to the solutions of basic trigonometric equations (e.g., sin x = k, cos x = k). • recognise when a trigonometric equation yields no real solutions (e.g., sin x = 2).

Trigonometric equations Function A function f assigns to each element of one set S precisely one element of a second set T. Domain The set of allowable values of x in a function or relation. Example: For the equation sin x = , the domain [0, 2π ) restricts solutions to x =

Substitution The process of replacing a variable in an algebraic expression, formula, equation, or function consistently with a particular value, another variable, expression, or function. Example: In sin (2x) = 2 cos x, substitution of sin (2x) = 2 sin x cos x simpliﬁes the equation.

Trigonometric equations challenge you to find angles x that satisfy equations involving functions like sin x, cos x, or tan x. These equations often require clever techniques to solve, especially when restricted to a specific domain. One powerful method is factorisation, where the equation is treated like a polynomial in sin x or cos x and factored to find solutions. For example, 2 sin2 x − sin x − 1 = 0 can be factored to reveal possible angles. Another approach is substitution, using trigonometric identities to simplify the equation. Identities like sin (2x) = 2 sin x cos x or cos (2x) = 1 − 2 sin2 x transform complex expressions into solvable forms.

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Solutions must lie within the given domain, such as [0, 2π ), so always check the periodicity of trigonometric functions to ﬁnd all possible angles. y

y = tan x y = cos x

y = sin x

1 x

−2π

−1π

1π

2π

−1

Example 1 Solve each trigonometric equation over the given domain. a sin (2x) = 2 cos x over [0, 2π ).

Create a strategy Use the sine double-angle formula to substitute sin (2x) = 2 sin x cos x. Then, rearrange the equation and apply the null factor law to find the solutions. Use the unit circle to determine the angles over [0, 2π ).

Apply the idea sin (2x) = 2 cos x 2 sin x cos x = 2 cos x 2 sin x cos x − 2 cos x = 0 2 cos x (sin x − 1) = 0

Write the equation Substitute sin (2x) = 2 sin x cos x Subtract 2 cos x from both sides Factorise

For the ﬁrst expression: 2 cos x = 0 cos x = 0

Use null factor law Divide both sides by 2

For the second expression:

376

sin x − 1 = 0

Use null factor law

sin x = 1

Add 1 to both sides

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Use the unit circle to ﬁnd the x-values. y

(0, 1)

(−1, 0)

120° 150°

90°

60° 30°

180° 210° 240°

0°

0, 2π

(1, 0) x

330° 270°

300°

(0, −1)

At [0, 2π ), cos x = 0 has angles So x =

and

while sin x = 1 has an angle of

over [0, 2π ).

b cos (2x) = 2 sin2 x over

Create a strategy Substitute the cosine double-angle identity, cos (2x) = 1 − 2 sin2 x, into the equation. After simplifying, use the unit circle to find the required angles for x.

Apply the idea Write the equation

Substitute cos (2x) = 1 − 2 sin2 x

Add 2 sin2 x to both sides

Swap sides

Divide both sides by 4

Take the square root of both sides

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In the unit circle, sin x =

has angles

and

while sin x =

has angles

only from 0 to

(0, 1)

(−1, 0)

120° 150°

90°

60° 30°

180° 210° 240°

0°

0, 2π

(1, 0) x

330° 270°

300°

(0, −1)

So x =

over

Reﬂect and check Notice that [0, 2π ].

is not included since it is outside the domain

, it can only be included over

Example 2 Solve the equation 4 cos3 x + 8 cos2 x − cos x − 2 = 0 for 0° ≤ x ≤ 360°.

Create a strategy Factorise the ﬁrst two terms then the last two terms of the equation, then the null-factor law. Use the unit circle, to ﬁnd the angles.

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Apply the idea 4 cos3 x + 8 cos2 x − cos x − 2 = 0 2

4 cos x(cos x + 2) − 1(cos x + 2) = 0 2

(4 cos x − 1) (cos x + 2) = 0

Write the equation Factorise the ﬁrst two terms and the last two terms Factorise the expression

For the ﬁrst expression: Use null factor law

Divide both sides by 4

Take the square root of both sides

For the second expression: cos x + 2 = 0 cos x = −2

Use null factor law Subtract 2 from both sides using the unit circle for 0° ≤ x ≤ 360°.

Since cos x ∈ [−1, 1], cos x = −2 is invalid. Solve cos x = y (0, 1)

(−1, 0)

120° 150°

90°

60° 30°

180° 210° 240°

0°

0, 2π

(1, 0) x

330° 270°

300°

(0, −1)

For cos x =

: x = 60°, 300°.

For cos x =

: x = 120°, 240°.

Therefore, the solutions over the domain 0° ≤ x ≤ 360° are x = 60°, 120°, 240°, 300°.

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Idea summary Trigonometric equations can be solved by factorisation and/or substitution of trigonometric formulas over restricted domains. When solving trigonometric equations over a restricted domain, only the solutions that fall within that specific domain are valid.

8.10E Practice questions What do you remember? 1

Solve each equation over [0, 2π ) using the formula sin (2x) = 2 sin x cos x. a

sin (2x) = sin x

sin (2x) + cos x = 0

Solve each equation over 0° ≤ x ≤ 360° using the formula cos (2x) = 1 − 2 sin2 x. Express answers to the nearest degree. a

cos (2x) = sin x

cos (2x) = −sin2 x

Practice Ex 1

Ex 2

380

Solve each equation over the speciﬁed domain using substitution or factorisation. a

sin (2x) = 3 cos x over [0, 2π )

cos (2x) = 2 sin2 x over [0, π ]

cos (2x) = −cos x over [0, 2π )

sin (2x) + sin x = 0 over [0, π ]

Solve each equation over [0°, 360°] using appropriate trigonometric identities or factorisation. Express answers to the nearest degree. a

2 sin2 x − sin x − 1 = 0

tan2 x − 3 tan x + 2 = 0

3 sin2 x − sin x cos x = 0

6 sin3 x + 4 sin2 x − 3 sin x − 2 = 0

Solve each equation over [−π, π ] using appropriate trigonometric identities or factorisation. Express answers to two decimal places. a

4 cos2 x − 5 cos x + 1 = 0

2 cos2 x + cos x − 1 = 0

2 sin x −

sin2 x + sin x cos x − 2 cos2 x = 0

cos x

Solve each equation over [0°, 360°] using appropriate trigonometric identities or factorisation. Express answers to two decimal places. a

tan2 x − 2 tan x − 1 = 0

tan2 x + tan x − 2 = 0

2 tan x + cot x = 3

tan2 x − sin2 x = 0

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Solve the equation cos (2x) + 2 sin x = 1 over [0, 2π ) using the formula cos (2x) = 1 − 2 sin2 x. Express answers to two decimal places.

ketch the graphs of y = sin (2x) and y = 3 cos x over [0, 2π ) using graphing technology. S Label points of intersection.

Use the graph to solve sin (2x) = 3 cos x. Compare with the solution from Question 3(a).

Extend your thinking 9

Solve the equation 2 sin2 x cos x − cos x = 0 over [0, 2π ). Explain why all solutions within the domain are included.

The graph represents the function f ( x) = cos (2x) − sin2 x. y 1 x

−1

Determine the values of x over [0, π ] where f ( x) = 0. Express the answer in two decimal places. + cos

Solve sin

Consider the equation tan

= 0 for 0 ≤ x ≤ 2π. + tan

= tan x cot 2x.

Express the equation in terms of tan x only.

Hence, ﬁnd the exact solution in the form x = tan−1(a).

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8.11E The auxiliary angle method After this lesson, you will be able to… • examine how expressions of the form a cos x + b sin x can be represented as a single trigonometric function (e.g., R sin ( x ± α) or R cos( x ± α)) using graphical and algebraic approaches. • calculate the amplitude R of the combined function using the formula , ensuring R > 0. • determine the auxiliary angle α by equating coefficients after expanding the target form, using trigonometric ratios, and identifying the correct quadrant. • apply the auxiliary angle method to convert expressions of the form a cos x + b sin x into a specified single trigonometric function and verify the equivalence.

The auxiliary angle method Here is a graph of the curves y = sin x and y = cos x on the same set of axes: y 1

y = sin(x) x

−2π

−1π

1π −1

2π

y = cos(x)

Recall that these functions are periodic, each with a period of 2π, and have an amplitude of 1 centred about the x-axis. Consider the sum of these functions: y = sin x + cos x

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A graph of this function is shown. y

y = sin x + cos x 1

−2π

−1π

2π

1π

−1

Notice that this looks much the same as a single sine or cosine function, though it has a different amplitude and has been shifted horizontally. In fact, the sum sin x + cos x can be rewritten as a single trigonometric function. To begin, start with the variable x and another ﬁxed angle, α, and use the formula: sin ( x + α ) = cos α sin x + sin α cos x Consider an angle α for which cos α and sin α are equal. By dividing the identity by this common value, the expression sin x + cos x can be isolated on the right-hand side. When considering the graph of the curves y = sin x and y = cos x on the same set of axes, note the two graphs intersect at the angle . That is,

Substituting α =

Multiplying both sides by

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Here is a graph of

: y

−2π

−1π

1π

2π

−1

Comparing this to the graph of y = sin x + cos x, this shows that the two graphs (and therefore the two functions) are indeed the same. When given the sum or difference of sine and cosine functions in the form y = a cos x + b sin x, this can be rewritten as one trigonometric function. This process is called the auxiliary angle method. There are four different forms: • R cos ( x + α ) • R cos ( x − α ) • R sin ( x + α ) • R sin ( x − α ) For these expressions, R is the amplitude of the periodic function and α is the auxiliary angle, where R > 0 and 0 ≤ α ≤ 2π. Starting from the transformed expressions R cos ( x ± α ) and R sin ( x ± α ), it can be directly shown that these forms can be expressed as linear combinations of cos x and sin x, using the sum and difference trigonometric identities: Write the sum

Take R2 as a common factor

Substitute sin2 α + cos2 α = 1

Swap side

Take the square root of both sides

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To determine the value of α, the tangent function provides a useful approach. If the linear combination a cos ( x) + b sin ( x) is to be expressed in these auxiliary angle forms: • In terms of R sin ( x + α ): a cos x + b sin x = R sin ( x + α )

Write the formula

Use sine angle sum identity: = R sin x cos α + R cos x sin α sin ( x + α ) = sin x cos α + cos x sin α Equate the coeﬃcient

Use the tangent identity

• In terms of R sin ( x − α ): a cos x + b sin x = R sin ( x − α ) = R sin x cos α − R cos x sin α

Write the formula se sine angle difference identity: U sin ( x − α ) = sin x cos α − cos x sin α

Equate the coeﬃcient

Use the tangent identity

• In terms of R cos ( x + α ): a cos x + b sin x = R cos ( x + α ) = R cos x cos α − R sin x sin α

Write the formula se cosine angle sum identity: U cos ( x + α ) = cos x cos α − sin x sin α

Equate the coeﬃcient

Use the tangent identity

• In terms of R cos ( x − α ): a cos x + b sin x = R cos ( x − α )

Write the formula

Use cosine angle difference identity: = R cos x cos α + R sin x sin α cos ( x − α ) = cos x cos α + sin x sin α Equate the coeﬃcient

Use the tangent identity

Notice that to ﬁnd the value of α using the relation tan α =

it is important to consider the

signs of a and b to determine the appropriate quadrant for α. The tangent function is positive in the ﬁrst and third quadrants and negative in the second and fourth quadrants. Thus, depending on the values of a and b: • If both sin α and cos α are positive, α lies in the ﬁrst quadrant. • If sin α is positive and cos α is negative, α is in the second quadrant. • If both sin α and cos ( α ) are negative, α is in the third quadrant. • If sin ( α ) is negative and cos α is positive, α is in the fourth quadrant.

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One of the methods to identify the angle α is to start by identifying the angle that satisfies the equation ∣ tan α∣ = x in the first quadrant 0 < α < , then determining the quadrant that α lies in (1, 2, 3 or 4) and finally working out the angle α: α (first quadrant), π − α (second quadrant), π + α (third quadrant), −α (fourth quadrant).

Example 1 Consider sin x − cos x. Express the given expression in the auxiliary angle form R sin ( x − α ), where R > 0 and 0 ≤ α ≤ . a Identify a and b of the expression.

Create a strategy Swap the terms to be in the form a cos x + b sin x.

Apply the idea sin x − cos x = − cos x + sin x = −1 × cos x + 1 sin x

Swap the terms Rewrite as a cos x + sin x

So a = −1 and b = 1. b Calculate the value of R, where R > 0.

Create a strategy Substitute a = −1 and b = 1 into the formula:

Apply the idea Write the formula

Substitute a = −1 and b = 1

Evaluate the squares

Evaluate the sum

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c Calculate the value of α, where α is 0 ≤ α ≤

Create a strategy Equate R sin ( x − α ) to sin x − cos x to calculate cos α and sin α. Then substitute the values into tan α =

Apply the idea Equating R sin ( x − α ) to sin x − cos x: R sin ( x − α ) = sin x − cos x

Write the equation

R cos α sin x − R sin α cos x = sin x − cos x

Expand R sin ( x − α )

Solve cos α by equating the terms with sin x: Equate the terms with sin x

Divide both sides by sin x

Divide both sides by R

Substitute

Solve sin α by equating the terms with cos x: Equate with cos x

Divide both sides by −cos x

Divide both sides by R

Substitute

Calculating tan α: Write the formula

Substitute sin α =

and cos α =

Evaluate

Take the inverse tangent

Evaluate

Multiply by

Evaluate

to convert into radians

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Reﬂect and check The angle α can also be determined using the unit circle: y

(0, 1)

(−1, 0)

120° 150°

90°

60° 30°

180°

0, 2π

0°

210° 240°

(1, 0) x

330° 270°

300°

(0, −1)

The unit circle shows that tan α = 1 has an angle of

over the domain 0 ≤ α ≤

Using the unit circle is useful if the domain is over [0, 2π ] d Rewrite sin x − cos x in the auxiliary angle form R sin ( x − α ).

Create a strategy Use the values of R and α from parts (b) and (c) to write the ﬁnal expression.

Apply the idea Write the formula

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Substitute

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and α =

, conﬁrming α =

Example 2 Consider the expression 2 cos x + 3 sin x. a Rewrite the expression 2 cos x + 3 sin x in the auxiliary angle form R sin ( x + α ), where R > 0 and 0 ≤ α ≤

Create a strategy Express 2 cos x + 3 sin x as R sin x cos α + R cos x sin α. Equate coeﬃcients to ﬁnd R and α, using and

Apply the idea The expression 2 cos x + 3 sin x is in the form a cos x + b sin x, where a = 2 and b = 3. Calculate R using the formula

: Write the formula

Substitute a = 2 and b = 3

Evaluate the squares

Evaluate

Now, calculate α by equating coeﬃcients: R cos α = 3 and R sin α = 2. Use the tangent identity

Express in terms of R sin α and R cos α

Substitute R sin α = 2 and R cos α = 3

Take the inverse tangent

Since tan α is positive, and the domain is 0 ≤ α ≤

,α=

is appropriate.

Therefore, the expression in auxiliary angle form is:

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b Verify the result using technology.

Create a strategy Use a graphing calculator like GeoGebra to plot y = 2 cos x + 3 sin x and over the interval

. Check if the graphs overlap.

Apply the idea Follow these steps in Geogebra: 1. In the Settings bar, change the x-axis boundaries to 0 and

In the xAxis tab, check the Distance and set to

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→ (1.5708). Lock the set ratio.

2. Input y = 2 cos x + 3 sin x and

The two graphs overlap, conﬁrming they are equal.

Reﬂect and check Alternatively, evaluate both expressions at speciﬁc points, such as x = 0 or x =

, using a calculator

to ensure the values match. This numerical check supports the graphical veriﬁcation.

Idea summary The auxiliary angle method is used to rewrite expressions of the form a cos x + sin x as a single trigonometric function. The amplitude, R, is calculated using the formula:

The auxiliary angle, α, is found using the tangent function. The specific relationship for tan α depends on the desired final form of the expression: • To express as R sin ( x + α ), use

• To express as R sin ( x − α ), use

• To express as R cos ( x + α ), use • To express as R cos ( x − α ), use

. .

It is crucial to determine the correct quadrant for α by checking the signs of sin α and cos α obtained from equating coefficients.

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8.11E Practice questions What do you remember? 1

Expand the following expressions into the form sin x cos α ± cos x sin α : a

Expand the following expressions into the form cos x cos α ± sin x sin α : a

Convert to sin ( x ± α ) where a

cos

sin x + sin

cos ( x + π )

cos

cos x

sin x − sin

cos x

Identify the formulas for expressing a sin x + b cos x as R sin ( x + α ) and R cos ( x − α ) where R > 0: a

For R sin ( x + α )

For R cos ( x − α )

cos

Practice 5

Convert to cos ( x ± α ) where a

Ex 1

sin x

sin x +

cos x

cos ( x) − sin ( x)

Find R.

Find α.

: b

cos x −

sin x :

Express each expression in its respective auxiliary angle form where R > 0 and

≤α≤

3 sin x + 4 cos x as R sin ( x + α )

5 sin (2x) + 12 cos (2x) as R sin (2x + α )

2 sin x + 2 cos x as R cos ( x − α )

7 sin (3θ ) − 24 cos (3θ ) as R sin (3θ + α )

Express −5 sin ( x) + 12 cos ( x) as R sin ( x + α ) where 0° ≤ α ≤ 360°.

Show that: a

392

sin x

4 sin x − 4 cos x

For 2 sin x − 2 cos x = R sin ( x + α ), where a

cos x + sin

Express as R cos ( x − α ) where a

cos x − sin

Express as R sin ( x + α ) where a

cos

−sin x = sin ( x + π )

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Show that: a

Ex 2

For a

cos x = sin

Rewrite in the form R sin ( x + α ).

Verify using technology.

6 sin x + 8 cos x

−3 sin x − 4 cos x

Express as R cos ( x − α ) where R > 0, −π ≤ α ≤ π : a

sin x +

Express as R sin ( x + α ) where R > 0, 0 ≤ α ≤ π : a

cos x = sin

Consider 5 sin x − 12 cos x. a

sin x −

5 cos x + 12 sin x sin x −

cos x = R sin ( x + α ), where R > 0, 0° ≤ α ≤ 360°:

Find R.

8 sin θ + 6 cos θ

Find α.

Express as R sin ( θ + α ) where a

−2 cos x + 2 sin x

: b

15 sin θ − 8 cos θ

Extend your thinking 18

Verify that 3 sin x + 4 cos x can be expressed as

by expanding the right-

hand side using the sine addition formula and comparing coeﬃcients. 19

Show that a sin x + b cos x can be expressed as R cos ( x − α ) and derive the formulas for R and α where R > 0, −π < α ≤ π.

Express sin x + cos x in both R sin ( x + α ) and R cos ( x − α ) forms where R> 0,

≤α≤

Derive the relationship between the phase angles α from the two forms and verify it for this specific case. 21

Consider the expression 4 sin x + 3 cos x. Express it in both R sin ( x + α ) and R cos ( x − α ) forms where R > 0, 0 ≤ α ≤ . Derive an identity relating the phase angles αsin and αcos from the two forms.

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8.12E Apply the auxiliary angle method After this lesson, you will be able to… • apply the auxiliary angle method to express functions of the form y = a cos x + b sin x as a single trigonometric function. • sketch the graph of functions transformed into auxiliary angle form (e.g., y = R sin ( x ± α )), identifying the amplitude and phase shift and determining maximum/minimum values. • solve trigonometric equations of the form a cos x + b sin x = c by first converting the left-hand side to its auxiliary angle form. • find all solutions to such equations within a specified restricted domain, correctly accounting for the phase shift.

Sketch graphs The conversion of the linear combination a cos ( x) + b sin ( x) to the forms R cos ( x ± α ) or R sin ( x ± α ) simpliﬁes the form of the function. This conversion is foundational for effectively sketching graphs of trigonometric functions. The auxiliary angle forms y = R cos ( x ± α ) and y = R sin ( x ± α ) represent transformations of the basic cosine and sine functions, introducing shifts and amplitude changes. Understanding the characteristics of R (amplitude) and α (phase shift) makes it easier to identify two key features: • Amplitude: The value of R scales the height of the waves and directly determines the height of the peaks and the depth of the troughs in the graph. A larger R results in a more pronounced wave. • Phase shift: The term ±α indicates a horizontal shift in the graph. This shift allows us to see how the graph moves left or right, affecting its starting position on the x-axis. These shifts allow for precise control over the timing of periodic events. Once converted, sketching the graph becomes more intuitive. The amplitude can be plotted and the points of intersection can be identiﬁed with the x-axis based on the phase shift. This visual representation helps in understanding the behaviour of the function over one or more periods. By being able to sketch the graphs accurately, real-world phenomena can be modelled, such as waves, oscillations, and other periodic behaviours, using the trigonometric equations derived from these conversions.

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For example, this graph shows the curves of the functions y = sin ( x) and y = cos ( x): y

y = sin(x)

−2π

−1π

2π

1π −1

y = cos(x)

The linear combination cos ( x) + sin ( x) can be expressed in the auxiliary angle form R sin ( x + α ) where R =

and α =

, which means cos ( x) + sin ( x) =

resulting from the sum cos ( x) + sin ( x) can be represented with this

The function graph:

y 1 x

−2π

−1π

2π

1π −1

reveals that the amplitude R, indicating the maximum height of the

The conversion to

sin

graph, is equal to

(approximately 1.41), while the phase shift is α =

starts slightly to the left of the origin. Therefore, the term entire sine wave by

, indicating that the graph

indicates a leftward shift of the

units, which means the entire wave, including the starting point, moves left.

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Example 1 Consider the function y = a Express

sin x − cos x.

sin x − cos x in the form R cos ( x + α ).

Create a strategy Use the formula a cos x + b sin x = R cos ( x + α ) with

Apply the idea Rewrite ﬁrst the expression in the form of a cos x + b sin x: Substitute the coeﬃcients a = −1, b =

Simplify

For R: Write the formula

Substitute a = −1 and b =

Evaluate each expression inside the brackets

Evaluate

Expressing in the form R cos ( x + α ): Write the formula and R = 2

Substitute a = −1, b =

Expand RHS using cosine angle sum formula

To solve for cos α, equate the terms with cos x: Equate the terms with cos x

Divide both sides by 2 cos x

To solve for sin α, equate the terms with sin x: Equate the terms with sin x

Divide both sides by −2 sin x

Evaluate possible values for α: Write the inverse tangent formula

Substitute sin α =

Simplify the fraction inside the brackets

Evaluate

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and cos α =

Since both sin α and cos α are negative, then α is in the third quadrant. So, α = Expressing

sin x − cos x in terms of R cos ( x + α ), then

Reﬂect and check Recall that it was established that to convert a cos x + b sin x into the form R cos ( x + α ), the formulas

and

can be applied.

Using these formulas can reduce the time taken for converting the function. Write the formula

Substitute a = −1 and b =

Evaluate each expression inside the brackets

Evaluate Write the formula

Substitute a = −1 and b =

Evaluate

Therefore,

sin x − cos x can be expressed as 2 cos

b Determine the maximum and minimum values of the function.

Create a strategy Since cosine function oscillates from −1 (minimum) to 1 (maximum), substitute each cos

= −1, 1 into the function y = 2 cos

from part (a).

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Apply the idea Maximum of y at cos

= 1: Write the function

Substitute cos

Evaluate

Minimum of y at cos

= −1: Write the function

Substitute cos

Evaluate

= −1

Reﬂect and check To verify, test x-values into the function y = 2 cos at [−2, 2]. If x =

to conﬁrming that the range ( y-values) is

: Write the function

Substitute x =

Evaluate the expression inside the brackets

Evaluate cos π

Evaluate

If x =

: Write the function

Substitute

Evaluate the expression inside the brackets

Evaluate cos 2 π

Evaluate

These values conﬁrm the maximum is 2 and the minimum is −2.

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c Sketch the graph of y =

sin x − cos x using the converted form.

Create a strategy Sketch the graph of y = 2 cos has been shifted

where the amplitude of the graph is 2 and the cos x graph

to the left.

Apply the idea y 2 1 x

−1π

1π −1

R = 2 units

y = cos x

−2

Idea summary Converting a trigonometric linear combination into auxiliary angle form of R cos ( x ± α ) or R sin ( x ± α ) is not just a mathematical technique; it serves as a critical step in graphing and analysing trigonometric functions effectively. Understanding the characteristics of the amplitude R (the height of the peaks and the depth of the troughs) and the phase shift α (how the graph moves left or right) makes it easier to identify key features of the graph, and enhances both graphing skills and modelling various real-world situations.

Solve equations Solving equations that involve sums of sine and cosine functions frequently arise in various applications, from physics and engineering to signal processing and harmonic analysis. These equations often take the form a cos x + b sin x = c, where a, b, and c are constants. Solving such equations requires an understanding of trigonometric identities and techniques for combining sine and cosine terms. One powerful method involves transforming the equation into a single trigonometric function by utilising the auxiliary angle form R cos ( x ± α ) or R sin ( x ± α ), where R is the amplitude and α is the phase shift. The auxiliary angle method is particularly useful when c ≠ 0, as it simpliﬁes the equation into a single trigonometric function, making it easier to solve for x. When c = 0, the equation a cos x + b sin x = 0 can often be solved more directly by dividing through by cos x or sin x (provided they are non-zero), yielding tan x = require the auxiliary angle method.

or cot x =

, which is simpler and does not

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Example 2 Consider the equation a Express

sin θ + 3 cos θ = 2.

sin θ + 3 cos θ in the form R sin ( θ + α ), where R > 0 and 0 < α <

Create a strategy Use the auxiliary angle method to rewrite a sin θ + b cos θ as R sin ( θ + α ), where cos α =

, and sin α =

. Here, a =

and b = 3.

Apply the idea Calculate R: Write the formula

Substitute a =

Evaluate the squares

Evaluate the sum

Simplify

and b = 3

For cos α: Write the formula

Substitute a =

Simplify

and R =

For sin α: Write the formula

Substitute b = 3 and R =

Simplify

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For α: Write the tangent ratio formula

Substitute sin α =

Simplify

Take the inverse tangent

Evaluate over the domain 0 < α <

Therefore,

sin θ + 3 cos θ =

b Solve the equation

sin

and cos α =

sin θ + 3 cos θ = 2 for 0 ≤ θ ≤ 2π.

Create a strategy Substitute the result from part (a) into the equation. Solve for θ by ﬁnding the values of θ +

where

, and adjust to the given domain.

Apply the idea Write the equation

Substitute the result from part (a)

Divide both sides by

Simplify

Let φ = θ +

, so sin φ =

≈ 0.577. Solve for φ in the interval adjusted for the domain 0 ≤ θ ≤ 2π.

Since 0 ≤ θ ≤ 2π, then Find φ such that sin φ =

. : Take the principal arcsine

Use the sine identity for supplementary angles

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Approximate arcsin

≈ 0.615 radians. Thus:

φ1 ≈ 0.615

Principal value

φ2 ≈ π − 0.615

Supplementary angle

≈ 2.527 Check if φ lies in

Evaluate

(approximately [1.047, 7.330]):

- φ1 ≈ 0.615 < 1.047, so it is outside the interval. - φ2 ≈ 2.527 is within [1.047, 7.330]. Since sine has period 2π, check additional solutions:

φ3 ≈ 0.615 + 2π ≈ 6.898

Add one period Evaluate

φ4 ≈ 2.527 + 2π ≈ 8.810 Add one period ≈ 8.810

Evaluate

- φ3 ≈ 6.898 is within [1.047, 7.330]. - φ4 ≈ 8.810 > 7.330, so it is outside the interval. Solve for θ : Write the equation

Use φ2 ≈ 2.527 Evaluate the second term Evaluate Write the formula

Use φ3 ≈ 6.898 Evaluate the second term Evaluate

Therefore, the solutions are θ ≈ 1.480, 5.851 radians.

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Reﬂect and check Verify by substituting θ = 1.480 and θ = 5.851 into the original equation

sin θ + 3 cos θ = 2:

For θ = 1.480: Write the left-hand side

Substitute θ = 1.480

Evaluate and round

For θ = 5.851: Write the left-hand side

Substitute θ = 5.851

Evaluate and round

Both solutions satisfy the auxiliary form, conﬁrming correctness. The solutions θ = 1.480, 5.851 radians are valid within 0 ≤ θ ≤ 2π.

Idea summary Solving equations of the form a cos x + b sin x = c involves transforming the equation into auxiliary angle form R cos ( x ± α ) or R sin ( x ± α ), and then using the equivalent form to solve for the angle x. Once the equation is converted, the single trigonometric function can be isolated to ﬁnd the corresponding angle by applying the inverse function. Finally, the general solutions are determined based on the periodic nature of the trigonometric function. It is important to check if the resulting angles fall within any speciﬁed intervals to ensure that only valid solutions are retained.

8.12E Practice questions What do you remember? 1

Express each expression in the form R sin ( x + α ), where R > 0 and 0 ≤ α < 2π. Identify the amplitude and phase shift: a

cos x +

sin x

2 cos x − 2 sin x

Express each expression in the form R cos ( x − α ), where R > 0 and 0 ≤ α < 2π. Identify the amplitude and phase shift: a

cos x +

sin x

3 cos x −

sin x

Express 4 sin x + 3 cos x in the form R sin ( x + α ) and identify the maximum and minimum values of the function.

Express

in the form a cos x + b sin x given an amplitude of

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Practice Ex 1

Ex 2

Consider the function f ( x) =

sin x − 2 cos x:

sin x − 2 cos x in the form R cos ( x + α ), where R > 0 and 0 ≤ α < 2π.

Express

Identify the maximum and minimum values of f ( x).

Sketch the graph of f ( x) over [0, 2π ] using the converted form.

Consider the equation 2 sin θ +

cos θ = 2:

Express 2 sin θ +

cos θ in the form R sin (θ + α ), where R > 0 and 0 ≤ α <

Solve the equation for 0 ≤ θ ≤ 2π.

Solve each equation for x over 0 ≤ x ≤ π : a

cos x + sin x =

2 sin x −

cos x − sin x = 1

cos x = −1

cos x + sin x = −1

Express 12 sin x + 5 cos x in the form R sin (x + α ) and solve 12 sin x + 5 cos x =

Sketch the graph of f ( x) = 2 cos x −

Express 2 cos x + 0 ≤ x ≤ 2π.

Express 2 sin x + 2 cos x in the form R sin ( x + α ) and solve 2 sin x + 2 cos x = 0 ≤ x ≤ π.

Solve sin x − 2 cos x =

Express 7 cos x + 24 sin x in the form R cos ( x − α ) and solve 7 cos x + 24 sin x = 20 for 0 ≤ x ≤ π.

Sketch the graph of g( x) = sin x −

Given tan−1

Express cos x −

for 0 ≤ x ≤

sin x over [0, 2π ] using the auxiliary angle form.

sin x in the form R cos ( x − α ) and solve 2 cos x +

sin x = 1 for

for

for 0 ≤ x ≤ 2π.

cos x over [0, 2π ] using the auxiliary angle form.

≈ 28°, solve 15 sin x + 8 cos x = 17 for 0 ≤ x ≤ 90°. sin x in the form R cos ( x + α ) and sketch the graph over [0, 2π ].

Extend your thinking 17

Solve 3 cos (2x) + 4 sin (2x) = 5 for 0 ≤ x ≤ π.

Explain why 5 cos x + 12 sin x = 15 has no solutions.

Solve the following equations for x over the domain 0 ≤ x ≤ 2π . In each case, first rewrite the equation in the form a cos (2x) + b sin (2x) = c:

404

tan (2x) + sec (2x) = 1

tan

− tan

+ 2 sec (2x)

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8.13E Applications of trigonometric equations After this lesson, you will be able to… • model real-world periodic phenomena using trigonometric functions, including those that combine sine and cosine terms. • apply the auxiliary angle method to simplify trigonometric models of the form y = A + B cos (ω x) + C sin (ω x) into y = A + R cos(ω x ± α ) or y = A + R sin (ω x ± α ). • analyse these models to determine amplitude, period, frequency, phase shift, and maximum/minimum values in the context of the problem. • solve practical problems by setting up and solving trigonometric equations derived from these models, and interpret the solutions in the given context.

Applications of trigonometric equations Modelling refers to the process of creating a representation of a real-world phenomenon or system using mathematical, physical, or conceptual constructs. The goal of modelling is to understand, analyse, or predict behaviour in complex systems by simplifying and extracting the essential features of the system. Trigonometric equations play a crucial role in modelling various real-world phenomena that exhibit periodic behaviour. These equations, particularly in the form a cos ( x) + b sin ( x) = c, where a, b, and c are constants, are essential in ﬁelds such as physics, engineering, and signal processing. They can effectively describe oscillations, such as sound waves, alternating currents, and the motion of springs. Understanding how to analyse and solve these equations not only provides insights into their underlying mechanisms but also facilitates the prediction and manipulation of these periodic behaviours in practical applications. To express a phenomenon in the form a cos ( x) + b sin ( x) = c, use the auxiliary angle method to transform the equation into a single trigonometric function by utilising the amplitude-phase form R cos ( x ± α ) or R sin ( x ± α ). • The amplitude, R, which determines the maximum displacement from the central value. • The vertical shift which represents the central or average value of the oscillation. • The phase shift, α, which is a horizontal shift applied to the function. • The frequency, ω, which indicates how frequently the function repeats over a given time period,

ω=

where T is the period.

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Example 1 The temperature, T, (in degrees Celsius) at a certain location at time t hours after 6:00 a.m. is given by: T (t) = 20 + 8 cos (2π t) + 6 sin (2π t) a Determine the exact value of the minimum temperature.

Create a strategy Rewrite the expression in a form that makes it easier to find the minimum value. First, we can combine the cosine and sine terms into a single cosine term by using the identity: R cos (2π t − α ) = a cos (2π t) + b sin (2π t) where: • a = 8 (the coefficient of cos (2π t)) • b = 6 (the coefficient of sin (2π t))

Apply the idea To calculate the value of R: Write the formula Substitute a = 8 and b = 6

Evaluate the squares

Evaluate the sum

Evaluate We can rewrite the original equation as: T (t) = 20 + 10 cos (2π t − α ) The minimum value of T (t) occurs when cos (2π t − α ) = −1, which gives the minimum temperature. Therefore, the minimum temperature is: T (t) = 20 + 10(−1) = 10° C b How many hours after 6:00 a.m. does the minimum temperature occur? Round your answer to the nearest minute.

Create a strategy The minimum value of T (t) occurs when: cos (2π t − α ) = −1 Solve for t using

406

from previously established formulas.

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Apply the idea The minimum value T (t) occurs when 2π t − α = π since cos ( π ) = −1. To calculate the value of t: Write the formula Divide both sides by 2π To calculate the value of α: Write the formula

Substitute a = 8 and b = 6

Evaluate and round

Now, substitute this value into the equation for t:

Convert 0.602 hours to minutes: 0.602 hours × 60 = 36.12 minutes So, the minimum temperature occurs approximately 36 minutes after 6:00 a.m.

Example 2 The population (in thousands) of two different types of insects on an island can be modelled by these functions: Butterflies: f (t) = a + b sin (mt) Crickets: g(t) = c − d sin (kt) where t is the number of years from when the populations started being measured, and a, b, c, d, m, k are positive constants. The graphs of f and g for the ﬁrst 2 years are shown below. y

450 400 350 300 250 200 150 100 50

g(t) = c − d sin(kt)

f (t) = a + b sin(mt) x 0

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a Determine the function f (t) that models the population of Butterﬂies over t years.

Create a strategy Use these features for the graph of f (t) = a + b sin (mt): • For a, determine vertical shift from sin x, that is the y-intercept. • For b, determine the amplitude. • For the period

, determine how many times the graph repeated.

Apply the idea The graph of f (t) = a + b sin (mt) shows that: • a = 250 • b = 100 • The period is 1 year. So

= 1.

So the function is: f (t) = 250 + 100 sin 2π t

b Determine the function g(t) that models the population of Crickets over t years.

Create a strategy Use these features for the graph of g(t) = c − d sin (kt): • For c, determine vertical shift from sin x, that is the y-intercept. • For d, determine the amplitude. • For the period

, determine the duration of one cycle.

Apply the idea The graph of g(t) = c − d sin (kt) shows that: • c = 350 • d = 100 • The period is

year. So

So the function is: g(t) = 350 − 100 sin 4π t

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c How many times over an 18-year period will the population of Crickets reach its maximum value?

Create a strategy Divide 18 by the period of the graph of g(t).

Apply the idea Divide 18 by the period

Evaluate

d How many years after the population of Crickets ﬁrst starts to increase, does it reach the same population as the Butterﬂies?

Create a strategy Subtract the x-coordinate of the ﬁrst minimum value of g(t) from the ﬁrst point of its intersection with f (t).

Apply the idea The graph shows the population of crickets ﬁrst starts growing after t − . The ﬁrst point of its intersection with f (t) is at t = . y

450 400 350 300 250 200 150 100 50

g(t) = c − d sin(kt)

f (t) = a + b sin(mt) x 0

Calculating the number of years where they reach the same population: Subtract the two time values

Evaluate

8.13E Applications of trigonometric equations mathspace.co

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e Solve for t, the number of years it takes for the population of Butterﬂies to ﬁrst reach 200 000.

Create a strategy Set f (t) = 200 into the function f (t) = 250 + 100 sin (2π t).

Apply the idea The population is in thousands, so 200 000 butterﬂies corresponds to f (t) = 200. Write the function Substitute f (t) = 200

Subtract 250 from both sides

Divide both sides by 100

Evaluate For sin (2π t) =

, the sine function is negative in the third and fourth quadrants. The reference

angle for sin ( θ ) =

. Thus:

Use

, with periodicity 2π

Use

, with periodicity 2π

For the ﬁrst equation, set k = 0: Use smaller angle

Divide both sides by 2π

Evaluate

For the second equation with k = 0, t = the ﬁrst positive equation is:

÷ 2π =

. Since

≈ 0.5833 is less than

≈ 0.9167,

Example 3 The height above the ground of a rider on a Ferris wheel is modelled by the function h(t) = 25 cos (3(t − 60)°) + 30, where h(t) is the height in metres, t is the time in seconds, and the angular velocity of 3° per second is the speed at which the wheel rotates. a Sketch the graph of y = h(t).

Create a strategy Analyse the h(t) = 25 cos (3(t − 60)°) + 30 to determine amplitude, period, phase shift, and vertical shift. Use these features to plot the graph.

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Apply the idea The function is h(t) = 25 cos (3(t − 60)°) + 30. Identify key features: • Amplitude: 25, so the height oscillates 25 metres above and below the midline. • Vertical shift: 30, so the midline is at y = 30 metres. • Phase shift: The argument 3(t − 60) implies a shift of 60 seconds to the right. • Period: The period of a standard cosine function is 360°. The angular velocity is 3° per second. The time to complete one full revolution (period) is

Height (m) 50 40 30 20 10

Time (s) 0

120

150

180

210

b What is the maximum height of the rider?

Create a strategy Determine the maximum value of h(t) by evaluating the function at the point where the cosine term is maximised.

Apply the idea The function is h(t) = 25 cos (3(t − 60)) + 30. The cosine term ranges from −1 to 1. The maximum occurs when cos (3(t − 60)) = 1: h(t) = 25 cos (3(t − 60)) + 30 Maximum height = 25(1) + 30 = 55 m

Write the function Substitute maximum cosine value Evaluate

The maximum height is 55 metres.

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c What is the minimum height of the rider?

Create a strategy Determine the minimum value of h(t) by evaluating the function when the cosine term is minimised.

Apply the idea The minimum occurs when cos (3(t − 60)) = −1: h(t) = 25 cos (3(t − 60)) + 30 Minimum height = 25(−1) + 30

Write the function Substitute minimum cosine value

= −25 + 30

Simplify

=5m

Evaluate

The minimum height is 5 metres. d At what height is the rider after 85 seconds? Give the answer to two decimal places.

Create a strategy Substitute t = 85 into h(t) and evaluate using a calculator, rounding to two decimal places.

Apply the idea h(t) = 25 cos (3(t − 60)) + 30

Write the function

= 25 cos (3(85 − 60)) + 30

Substitute t = 85

= 25 cos (75) + 30

Simplify

= 36.47 m

Evaluate and round

The rider’s height after 85 seconds is approximately 36.47 metres.

Idea summary Trigonometric equations, such as a cos ( x) + b sin ( x) = c, model periodic phenomena by transforming into a single function like R cos ( x ± α ) using the auxiliary angle method. This approach determines amplitude (R), phase shift ( α ), frequency ( ω) and period

, enabling analysis of periodic behaviours.

Practical problems, such as predicting minimum temperatures or analysing insect population dynamics, are solved by deriving key variables and interpreting solutions. For example, temperature models reveal minimum values and their timing, while population models determine times to reach specific thresholds or equal populations, providing insights into meteorological and ecological systems.

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8.13E Practice questions What do you remember? 1

Calculate the amplitude and phase shift of these functions after converting into the form R sin ( x ± α ): a

cos ( x) + sin ( x)

Identify the period and frequency of these functions: a

3 cos ( x) + 4 sin ( x)

f (t) = 10 cos (4π t) + 5 sin (4π t)

g(t) = 8 + 6 cos

Express these functions in the form a cos ( x) + b sin ( x):

Practice Ex 1

Ex 2

The temperature in a coastal town, T (t), in degrees Celsius, at time t hours after midnight is modelled by T (t) = 15 + 6 cos (2π t) + 8 sin (2π t): a

Express T (t) in the form 15 + R cos (2π t − α ). Calculate R and α.

Determine the minimum temperature and the ﬁrst time it occurs after midnight. Round to the nearest minute.

Sketch the graph of T (t) over one period using the converted form.

The population of two bird species in a forest is modelled by these functions, where t is years since monitoring began: Sparrows: f (t) = 300 + 120 sin (2π t) Finches: g(t) = 400 − 80 sin (4π t) The graph shows both populations over the ﬁrst 2 years: P

450 400 350 300 250 200 150 100 50

g(t)

f (t) t 0

Calculate how many times the finch population reaches its maximum value over a 12-year period.

Find the first time, in years, that the sparrow population reaches its minimum value.

Calculate the first time, in years, that the sparrow population reaches 360.

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The height of a tide, h(t), in metres, at time t hours after midnight is modelled by + 3 sin

h(t) = 4 + 2 cos

. Calculate R and α.

Express h(t) in the form 4 + R sin

Determine the maximum tide height and the ﬁrst time it occurs after midnight. Round to the nearest minute.

The displacement of a pendulum, d(t), in centimetres, at time t seconds is modelled by d(t) = 8 cos (3t) − 6 sin (3t): a

Express d(t) in the form R cos (3t − α ). Calculate R and α.

Determine the amplitude and the ﬁrst time the pendulum reaches its maximum displacement after t = 0. Round to the nearest second.

The voltage in an alternating current circuit, V (t), in volts, at time t seconds is modelled by V (t) = 5 cos (t) + 12 sin (t): a

Express V (t) in the form R sin (t + α ). Calculate R and α.

Sketch the graph of V (t) over one period.

Find all times in a day (taking midnight as t = 0) that satisfy these equations, where t is in hours, rounded to the nearest minute: a

Ex 3

2 cos

sin

+ cos

Passengers on a Ferris wheel access their seats from a platform 5 m above the ground. As each seat is filled, the Ferris wheel moves around so that the next seat can be filled. Once all seats are filled, the ride begins and lasts for 6 minutes. The height h in metres of Amelia’s seat above the ground t seconds after the ride has begun is given by: h (t) = 14 sin (10t − 40) + 16 where the quantity (10t − 40) is in degrees. a

Sketch a graph of the function h (t) for one rotation.

Find the height above the ground of Amelia’s seat at the commencement of the ride. Round your answer to two decimal places.

Find the time at which Amelia ﬁrst passes the access platform while ascending. Round your answer to two decimal places.

414

Find the number of times her seat passes the access platform in the ﬁrst two minutes.

Due to a malfunction, the Ferris wheel stops abruptly 1 minute and 40 seconds into the ride. Find the height above the ground that Amelia is stranded at. Round your answer to two decimal places.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking 11

The population (in thousands) of a fish species in a lake is modelled by , where t is in years since monitoring began: a

Express P (t) in the form 500 + R sin (π t + α ), where R > 0 and exact values of R and α.

Determine the maximum population and the ﬁrst time it occurs after t = 0. Round the time to the nearest month.

Calculate the ﬁrst time the population reaches 600 000.

. Calculate the

The population of two insect species in a forest is modelled by these functions, where t is years since monitoring began: • Species A: f (t) = 250 + 100 sin (2π t) • Species B: g(t) = 350 Find all times in the ﬁrst two years when the populations are equal, i.e., solve f (t) = g(t) for t ∈ [0, 2].

The ocean tide of a harbour can be modelled using f (t) = A sin ( ω t + α ) + C. High tide of 12 m occurs at 6:00 a.m. and low tide of 6 m occurs at 12:00 p.m. Treat midnight as t = 0. a

Find the unknown values in f (t) = A sin (ω t + α ) + C.

If a ship can enter the harbour when the tide is 10 m or above, ﬁnd all times in the day the ship can enter.

A neighbouring harbour can be modelled by f (t) = 4 cos (ω t + α ) + 10, where ω, α are the same values found in part (a). Find the ﬁrst time each harbour’s tide is the same height and can the ship enter either of the harbours at this time?

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8 Chapter review 1

Which of these is the exact value of sec 45° ? A

The expression

is the quotient identity for:

A 3

cosec x

cot2x + 1 = cosec2x

sin x + cos x = 1

In a right-angled triangle, the hypotenuse is 41 cm, the side adjacent to angle α is 40 cm, and the opposite side is 9 cm. Find the exact values of: sec α

cosec α

cot α

tan α

cosec 60°

cot 60°

sec 45°

Find the exact values of: sec 30°

Find the exact values of: a

sec 30° + cot 30°

cosec 60° − sec 60°

sec 45° + cosec 45°

cot 45° − cosec 30°

In a right-angled triangle, cot θ =

Find the exact values of sec θ and cosec θ. sec θ

cosec θ

Hence, show that 1 + cot2θ = cosec2θ for this triangle.

For an angle α on the unit circle with point a

sec α

, ﬁnd:

cosec α

cot α

tan α

cosec

cot

sec (3π )

Find the exact values of: a

416

tan2x − 1 = sec2x

tan x

1 + sin2x = cos2x

sec x

a 6

Which of these equations is a valid Pythagorean identity?

a 5

cot x

sec

Evaluate each trigonometric ratio, leaving the answers in exact form: a

sec2(45°)

cot (30°) sec (60°)

cosec2

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

The diagram shows the unit circle. The point A can be

represented as . State two exact angles (one positive and one negative) that correspond to the point B. A x

Given cos A = a

and A is in the fourth quadrant, calculate:

sec A

tan A

cot A

Prove each identity algebraically: a

sec θ cot θ = cosec θ

= cot θ

sin A sec A = tan A

cos A cosec A = cot A

Prove that

= sin A for all valid angles.

Given tan A =

in the second quadrant, calculate:

a 16

sec A

cot A

tan (90° − θ ) × tan θ

cos (90° − θ ) × cosec θ

cot 45°

sec 60°

tan 30°

tan (90° − θ )

cosec (90° − θ )

sin 135°

and θ is an acute angle, evaluate:

cos (90° − θ )

sec (90° − θ )

If sec A = 4 and A is an acute angle ﬁnd the exact value of: a

cosec (90° − A)

cot (90° − A)

Evaluate using reference angles and quadrants: a

sin 30°

Given sin θ = a

Find the exact value of each expression: a

cosec A

sin (90° − θ ) × sec θ

Simplify using complementary angle identities: a

sin A

sec 150°

cosec 225°

cot 330°

Without simplifying the brackets, evaluate the exact values of these expressions: a

sin (90° − 45°) + cos (180° + 45°)

sec (90° − 60°) − sin (360° − 30°)

tan (90° − 30°) × sin (270° − 60°)

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Explain why sin 210° = − sin 30° using the unit circle and reference angles.

Simplify these expressions using Pythagorean identities:

1 − sec2x

An angle of depression θ from a cliff top is measured, where cot θ = 3. Simplify these expressions related to this angle: a

cosec2θ

sec2θ

tan θ sec θ

In a right-angled triangle, tan θ = a

sec θ

Solve for θ, where 0° ≤ θ ≤ 360°:

cos θ = −1

sin θ =

tan θ = −1

cos θ = 0

sin x = 1

cos x =

tan x =

sin x = 0

Solve the quadratic trigonometric equations for θ, where 0° ≤ θ ≤ 360°: a

2 cos2θ − cos θ − 1 = 0

tan2θ − tan θ = 0

4 sin2θ − 3 = 0

2 sin2θ − 3 sin θ + 1 = 0

tan2x = 3

Solve for x, where 0 ≤ x ≤ 4π : a

cosec2θ

Solve for x, where 0 ≤ x ≤ 2π : a

= cot x.

Prove:

. Find the exact value of:

(cosec2x − 1) sin2x

cos2x = 1

sin2x =

2 cos2x = 1

sin (90°)

= 2 sec θ.

Prove the identity

32E The expression cos (65°) cos (25°) + sin (65°) sin (25°) simpliﬁes to: A

cos (90°)

sin (40°)

cos (40°)

33E What is the maximum value of the function f ( x) = 5 sin x + 12 cos x? 5

34E Given sin α =

where 0 < α <

418

C , and cos β =

The exact value of cos α.

The exact value of sin (α − β ).

13 where 0 < β <

, determine:

The exact value of sin β.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

35E Determine the exact value of: a

cos (75°)

sin (15°)

tan (105°)

36E Simplify the expression

to a single trigonometric ratio.

37E Prove that

by ﬁrst expressing tan (3 A) as tan (2 A + A). = tan A + tan B.

38E Prove the identity: 39E If cos A =

and tan B =

, where A and B are acute angles:

Determine the exact values of sin A, sin B, and cos B.

Determine the exact values of cos (2 A) and sin (2 A).

Hence, evaluate the exact value of sin (2 A − B).

40E Solve the equation cos (2x) − 5 cos x − 2 = 0 for 0 ≤ x < 2π. Round your answer to two decimal places. 41E Solve the equation 3 tan2 θ − 2 tan θ − 1 = 0 for 0° ≤ θ ≤ 360°. Round your answer to the nearest degree. 42E Express 4 sin (3x) − 3 cos (3x) in the form R sin (3x − α ), where R > 0 and 0 ≤ α <

sin x = 1 by expressing in the form R cos ( x + α ), where R > 0 and 43E Solve 2 cos x − 0 ≤ α < , in the domain 0 to 2π. 44E Consider the function g( x) = 7 − 2 sin x +

cos x:

Express g( x) in the form 7 + R cos ( x + α ), where R > 0 and 0 ≤ α <

Find the minimum value of g( x) and the smallest positive value of x for which this minimum occurs.

45E The height, H metres, of the tide in a harbour t hours after 6:00 a.m. is given by:

− 2 sin

in the form R cos

, where R > 0 and 0 ≤ α <

Express 2 cos

What is the minimum height of the tide and at what time does it ﬁrst occur?

What times during the day can a ship enter the harbour if it requires the tide to be at least 7.5 m?

46E Solve sin

+ cos

for 0 ≤ x < 2π.

47E Explain why the equation 3 sin x − 4 cos x = 5.5 has no real solutions.

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48E The populations (in thousands) of a predator species, P (t), and a prey species, H(t), are modelled by the graph shown for 0 < t < 4: Population (thousands) 225 200 175 150 125

t (years) 0.5

1.5

2.5

3.5

What is the period of the prey population, H(t)?

What is the amplitude of the predator population, P (t)?

From the graph, estimate the ﬁrst two times when the predator population P (t) is 165 000.

Find, algebraically, the ﬁrst time t > 0 when the predator population equals the prey population. Round your answer to two decimal places.

49E The number of daylight hours, D(t), on a particular day t of the year (where t = 0 is January 1) in a city is approximated by D(t) = 12 + 2.5 sin

420

What is the maximum number of daylight hours and on approximately which day does it occur?

For how many days in the year are there 13 or more hours of daylight?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

“Infinity is merely the absence of a bound.” Georg Cantor

9 Parametric form of a function Chapter outline 9.01E Parametric forms of equations 9.02E Convert from parametric form to Cartesian form 9.03E Graph linear functions, quadratic functions and circles in parametric form Chapter 9 review

424 436 443 454

Some spiderwebs grow in spirals that match polar or parametric curves — nature’s maths in action.

9.01E Parametric forms of equations After this lesson, you will be able to… • recognise that a curve may be represented by two parametric equations that give x and y as functions of a parameter. • understand that the parametric representation of a given curve is not unique. • generate a set of parametric equations for a given Cartesian equation by defining x as a function of a parameter and then deriving the corresponding function for y. • identify the relationship between standard parametric forms and the Cartesian equations for lines, parabolas, and circles.

Parametric forms are not unique Parameter A quantity that is characteristic of a system. Parametric equations A type of equation that uses a parameter as the independent variable. For example, x = 3 sin θ and y = 3 cos θ, is a pair of parametric equations, where different values of the parameter θ will give different points on the number plane.

A mathematical function is a set of ordered pairs ( x, y) on a given domain. The dependent variable y is related to the independent variable x by a rule, called the rectangular equation or Cartesian equation. Alternatively, a parameter t can be introduced, where x and y are functions of t. This is the parametric form.

x = g(t), y = h(t) t

is the parameter

x, y

are the functions of t

Parametric forms are not unique. To create parametric equations, deﬁne x as a function of t, then substitute into the rectangular equation to ﬁnd y.

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Example 1 For the function y = 4x2, deﬁne a parametric form using x = 2t and ﬁnd the ordered pairs for t = −2, 0, .

Create a strategy Use x = 2t to express y in terms of t by substituting into y = 4x2. Evaluate x and y at t = −2, 0, ﬁnd ordered pairs.

Apply the idea x = 2t

Write the parametric equation for x 2

y = 4x

Write the rectangular equation 2

Substitute x = 2t

= 4 × (2t) = 16t

Simplify

The parametric equations are x = 2t and y = 16t2. Calculate ordered pairs for t = −2, 0, . For t = −2: x = 2t

Write the parametric equation for x

= 2 × (−2)

Substitute t = −2

= −4

Evaluate

y = 16t

Write the rectangular equation 2

= 16 × (−2)

Substitute t = −2

= 64

Evaluate

The ordered pair is (−4, 64). For t = 0: x = 2t

Write the parametric equation for x

=2×0

Substitute t = 0

Evaluate 2

y = 16t

Write the rectangular equation 2

= 16 × 0

Substitute t = 0

Evaluate

The ordered pair is (0, 0).

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For t = : Write the parametric equation for x Substitute t = Evaluate

Write the rectangular equation

Substitute t =

Simplify

Evaluate

The ordered pair is (1, 4). The table summarises the parametric values: t

−2

x = 2t

−4

y = 16t2

Idea summary A function’s rectangular equation relates y to x. In parametric form, a parameter t defines x = g(t) and y = h(t). Parametric forms are not unique; different choices for x yield valid y expressions. To construct parametric equations: 1. Define x as a function of t. 2. Substitute into the rectangular equation to find y.

Parametric form of lines Multiple parametric equations can represent a single function. For the line y = 2x − 1, setting x = t gives y = 2t − 1. For a line through points (1, 1) and (3, 5), the equation can be found and x = t applied, but a general parametric form is more eﬃcient. For a line through points ( x0, y0 ) and ( x1, y1 ), the parametric equations are: x = x0 + ( x1 − x0)t y = y0 + (y1 − y0 )t This derives from the line equation y − y0 = m( x − x0), where m = exercise.

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. The proof is left as an

Example 2 Find the parametric equations of the line passing through (1, 1) and (3, 5).

Create a strategy Substitute the points into the formulas: x = x0 + ( x1 − x0)t y = y0 + (y1 − y0 )t

Apply the idea For the ﬁrst formula: x = x0 + ( x1 − x0)t

Write the formula

= 1 + (3 − 1)t

Substitute x0 = 1 and x1 = 3

= 1 + 2t

Evaluate the subtraction

For the second formula: y = y0 + (y1 − y0 )t

Write the formula

= 1 + (5 − 1)t

Substitute y0 = 1 and y1 = 5

= 1 + 4t

Evaluate the subtraction

The parametric equations are: x = 1 + 2t y = 1 + 4t

Reﬂect and check The equation of the line through (1, 1) and (3, 5) is y = 2x − 1. Setting x = t gives y = 2t − 1 as one set of parametric equations. This demonstrates that multiple parametric equations can represent the same function. However, the general formula for parametric equations provides greater consistency compared to arbitrary assignments. When points are not provided, the best practice for writing the parametric form of a straight line is to use the form y − y0 = m( x − x0) and set t = x − x0. This approach extends to parabolas and circles, resembling the completing-the-square format discussed later.

Example 3 Find the parametric equations of the line with slope −7 passing through (1, −4).

Create a strategy Substitute ( x0, y0 ) = (1, −4) and m = −7 into y − y0 = m( x − x0).

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Apply the idea y − y0 = m( x − x0)

Write the formula

y − (−4) = −7( x − 1)

Substitute ( x0, y0 ) = (1, −4) and m = −7

y + 4 = −7( x − 1)

Evaluate the adjacent signs

The parametric equations are: t = x − 1, y + 4 = −7t x = t + 1, y = −7t − 4

Idea summary The parametric form of a straight line given two points ( x0, y0 ) and ( x1, y1 ) is: x = x0 + ( x1 − x0)t y = y0 + (y1 − y0 )t Best practice is to put in the form y − y0 = m( x − x0) and let t = x − x0.

Parametric form of parabolas Multiple parametric equations can represent a parabola. Best practice is to express the parabola in vertex form by completing the square, converting y = ax2 + bx + c to y = a( x − h)2 + k. For parametric form, rewrite the vertex form as: 4 A(y − k) = ( x − h)2 Set x − h = 2 At, so x = 2 At + h. Then: 4 A(y − k) = (2 At)2 4 A(y − k) = 4 A2 t2 y − k = At2 y = At2 + k Thus, the parametric equations of a parabola are x = 2 At + h and y = At2 + k.

Example 4 Find the parametric equations of y = x2.

Create a strategy Rewrite the equation in the form 4 A(y − k) = ( x − h)2 and ﬁnd the parametric equations x = 2 At + h, y = At2 + k.

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Apply the idea Rewrite y = x2 in the standard form: y = x2 2

4 Ay = x

Write the equation Express as 4 A(y − k) = ( x − h)2

This shows h = 0, k = 0, 4 A = 1, so A = . Substitute A = , h = 0, k = 0 into the parametric equations. For x = 2 At + h: Write the equation

Substitute A =

Simplify

and h = 0

For y = At2 + k: Write the equation

Substitute A =

Simplify

and k = 0

The parametric equations are

Reﬂect and check To verify, eliminate the parameter. From x = , we get t = 2x: Write the equation

Substitute t = 2x

Simplify

This matches the original equation. This form aligns with the parabola’s vertex at the origin.

Example 5 Find the parametric equations of y = 3x2 + 2x − 4.

Create a strategy Convert the parabola to the form 4 A(y − k) = ( x − h)2. Then, derive the parametric equations x = 2 At + h and y = At2 + k.

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Apply the idea Calculate the x-coordinate of the vertex using

Write the formula Substitute a = 3 and b = 2 Evaluate Simplify Calculate the y-coordinate of the vertex by substituting x =

Write the equation

Substitute x =

Evaluate each term

Simplify

The vertex is

Divide the equation by 3 to make the coeﬃcient of x2 equal to 1, resulting in a coeﬃcient of in the form 4 A(y − k) = ( x − h)2:

Write the coeﬃcient of y

Divide both sides by 4

,h=

into the parametric equations.

Substitute A =

, and k =

For x = 2 At + h:

Write the equation

Substitute A =

and h =

Simplify For y = At2 + k:

Write the equation

Substitute A = Simplify The parametric equations are

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and k =

for y

Idea summary The parametric equations of a parabola written in the form 4 A( y − k) = ( x − h)2 are x = 2 At + h and y = At2 + k.

Parametric form of circles The standard equation of a circle is ( x − h)2 + (y − k)2 = r2, with centre (h, k) and radius r. For the unit circle x2 + y2 = 1: y

y x

Each point ( x, y) on the circle forms a triangle with the x-axis, represented by an angle. Using trigonometric ratios:

Thus, x = cos θ and y = sin θ are the parametric equations for x2 + y2 = 1, based on the identity sin2 θ + cos2 θ = 1. For a general circle ( x − h)2 + (y − k)2 = r2, set: x − h = r cos θ, y − k = r sin θ 2

This yields r cos θ + r sin θ = r , simplifying to r2 = r2. Hence, the parametric equations of a circle with centre (h, k) and radius r are: x = h + r cos θ, y = k + r sin θ The parameter θ can be replaced with any letter, commonly t or θ.

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Example 6 Find the parametric equations of ( x − 3)2 + (y + 5)2 = 25.

Create a strategy Identify the circle’s centre and radius from ( x − h)2 + (y − k)2 = r2, then derive the parametric equations x = h + r cos t and y = k + r sin t.

Apply the idea The centre is at (3, −5) and the radius is 5. For x = h + r cos t: x = h + r cos t

Write the equation

= 3 + 5 cos t

Substitute h = 3 and r = 5

For y = k + r sin t: y = k + r sin t

Write the equation

= −5 + 5 sin t

Substitute k = −5 and r = 5

The parametric equations are x = 3 + 5 cos t, y = −5 + 5 sin t.

Example 7 Find the parametric equations of x2 + y2 − 6x − 8y − 24 = 0.

Create a strategy Convert the equation to standard circle form ( x − h)2 + (y − k)2 = r2 by completing the square for x and y terms, then identify the centre (h, k) and radius r to substitute into the parametric form x = h + r cos t, y = k + r sin t.

Apply the idea x2 − 6x + y2 − 8y − 24 = 0 2

Write the equation 2

x − 6x + (−3) + y − 8y + (−4) − 24 = (−3) + (−4) 2

( x − 3) + (y − 4) − 24 = 9 + 16 ( x − 3)2 + (y − 4)2 = 9 + 16 + 24 2

( x − 3) + (y − 4) = 49 The centre is (3, 4) and the radius is

actorise and evaluate the F right-hand side Add 24 to both sides Simplify

= 7. Substitute into the parametric form: x = 3 + 7 cos t y = 4 + 7 sin t

The parametric equations are x = 3 + 7 cos t, y = 4 + 7 sin t.

432

Complete the square for x and y

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary The parametric equations of a circle with centre (h, k) and radius r are: x = h + r cos θ, y = k + r sin θ

9.01E Practice questions What do you remember? 1

What is the deﬁnition of a function, and how can it be represented using function notation?

In the Cartesian coordinate system, how is a point represented, and what does the equation y = 2x + 3 describe?

Given the equations x = 2t and y = t + 1, eliminate the variable t to ﬁnd a relationship between x and y.

Sketch the graph of the equation y = x2. What type of curve does it represent?

Practice Ex 1

For each linear equation, write the y parametric equation given the x parametric equation: a

y = 2x + 3, x = t

2x − 4y + 5 = 0, x = 2t + 3

Ex 2

y = 4 − x, x = 2 + t

For each parabola, write the y parametric equation given the x parametric equation: a

y = x2 − 1, x = 2t 2

y = 3 − x , x = 3t − 4

y = 2x2 − x, x = 1 − t

y = ( x + 1)2 − 2x, x = t − 1

For each circle, ﬁnd the missing parametric equation: a

x2 + y2 = 9, y = 2t

( x − 1)2 + y2 = 1, x = 1 − t

( x + 2)2 + (y − 3)2 = 25, x = 5 cos t − 2

x + y = 16, y = 4 sin t

Find the parametric equations of a line passing through the points given: a

(1, 3) and (4, 9)

(−2, 5) and (3, −1)

(0, 1) and (7, 14)

(−3, −4) and (2, 6)

and

For each linear function, ﬁnd a suitable pair of parametric equations: a

y = 2x − 4

y = 5 − 6x

2x − 4y − 5 = 0

5x + 3y + 8 = 0

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Ex 3

Ex 4

Ex 5

Find the parametric equations for each line with the given point and slope: a

Point (2, −1) with slope 3

Point (−1, 4) with slope −2

Point (0, −5) with slope 5

Point (3, 2) with slope −1

Convert each line to parametric form by using the given x value as x0: a

2x − y = 3, x = 1

3x + 4y = 12, x = −1

x − 2y = 5, x = 2

4x + y = 8, x = 0

Find the parametric equations of the parabola y = x2 − 6x + 5 by completing the square.

Find the parametric equations for each parabola:

y = 4x2

2y = x2

Find the parametric equations for each parabola: a

y = x2 − 4

4y = x2 + 3

y = 3 − x2

8y = 8 − 9x2

Find the parametric equations for each parabola by completing the square ﬁrst: a

y = x2 − 4x + 8

4y = 2x2 + 6x − 4

16y = 4x2 + 8x − 1

y = 3x2 − 5x + 4

y = −x2 + 4x − 3

y = 3x2 − 6x + 2

y = 2x2 + 4x − 1

g Ex 6

Ex 7

y = −9x2

y = −2x + 8x − 5

Write down the parametric equations for each circle: a

x2 + y2 = 4

x2 + y2 = 25

x + y = 48

( x − 1)2 + (y − 2)2 = 1

( x + 3)2 + (y + 4)2 = 81

( x − 3)2 + (y + 2)2 = 64

Find the parametric equations for each circle by completing the square ﬁrst: a

x2 + y2 − 4x + 6y + 4 = 0

x2 + y2 + 2x − 2y − 2 = 0

x2 + y2 − 10x + 8y + 16 = 0

3x2 + 3y2 − 2x − 6y − 30 = 0

Extend your thinking 18

434

The equation of an ellipse is given by

Show that x = a cos t and y = b sin t are the parametric equations.

Hence, ﬁnd the parametric equations of 9x2 + 16y2 = 1.

The equation of a hyperbola is given by

If x = a sec t is the x parametric equation, ﬁnd the y parametric equation.

Hence, ﬁnd the parametric equations of

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For the line through points ( x0, y0 ) and ( x1, y1 ), show that the parametric equations can be written as: x = x0 + ( x1 − x0)t y = y0 + (y1 − y0 )t

For the hyperbola

, if y = sec 2t is the y parametric equation, ﬁnd the x

parametric equation in simplest form. 22

Find the parametric equations of a line passing through the points: a

(2, −6) and (a, −6)

(−3, 2) and (−3, b)

(12, 24) and (c, 12)

A parabola passes through the points (−1, 9), (1, 3), and (2, 6). Find a set of parametric equations for this parabola.

The path of a projectile is described by the Cartesian equation y = x tan α − If α =

, g = 9.8 m/s2, and v0 = 10 m/s, ﬁnd a set of parametric equations for the path using

the parameter s where x = 5s.

Did you know?

Parametric equations are essential in flight path planning for drones! By defining both horizontal and vertical positions as functions of time, engineers can model smooth turns, altitude adjustments, and precise landings—even in complex environments. Parametric models help drones navigate safely and efficiently through three-dimensional space, making them invaluable for everything from aerial photography to search-and-rescue missions!

9.01E Parametric forms of equations mathspace.co

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9.02E Convert from parametric form to Cartesian form After this lesson, you will be able to… • understand the goal of converting parametric equations to Cartesian form is to eliminate the parameter. • apply algebraic substitution to eliminate the parameter from pairs of linear or simple polynomial parametric equations. • use trigonometric identities (primarily Pythagorean identities) to eliminate the parameter from parametric equations involving trigonometric functions (e.g., for circles). • convert parametric equations representing linear functions, quadratic functions, and circles into their Cartesian equivalents.

Convert from parametric form to Cartesian form Parametric equations x = g(t), y = h(t) can be converted to a Cartesian equation by eliminating t. To eliminate the parameter t, we aim to express the relationship between x and y directly, without t. When the parametric equations involve trigonometric functions, such as sin t or cos t, Pythagorean identities are often useful. The identity sin2 θ + cos2 θ = 1 can help combine terms to remove t. Similarly, identities like sec2 θ = 1 + tan2 θ or csc2 θ = 1 + cot2 θ may be applied when dealing with tangent, secant, or cosecant functions, allowing us to rewrite expressions in a form that eliminates the parameter and yields a Cartesian equation.

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Example 1 Find the Cartesian form of x = t − 4, y =

Create a strategy Recognise that the denominator of y is the square of the expression in x.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea

Reﬂect and check 2

The denominator is t − 8t + 16 = (t − 4) . Since x = t − 4, (t − 4)2 = x2.

The graph shows selected t values and coordinates: y

Write the equation

Factorise the denominator

Substitute (t − 4) = x

(−2, 3) t=2

(2, 3) t=6

x −8 −6 −4 −2

Example 2 Consider the curve deﬁned by x = −2 + 5 cos t and y = −3 + 5 sin t for t ∈ [0, 2π ]. a Find the Cartesian equation.

Create a strategy Express the equations as trigonometric ratios, then square both sides of each equation to add their results.

Apply the idea

Write the equation for x Add 2 to both sides

Divide both sides by 5

Square both sides

Write the equation for y

Add 3 to both sides

Divide both sides by 5

Square both sides

Add the resulting equations

Apply cos2 t + sin2 t = 1

Multiply both sides by 25

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b Graph the curve.

Create a strategy

Apply the idea

The equation from part (a) is in the standard form of a circle ( x − h)2 + (y − k)2 = r2 where (h, k) is the centre while r is the radius.

The circle has centre (−2, −3) and radius r = = 5.

(x + 2)2 + (y + 3)2 = 25

4 2

−12 −10 −8 −6 −4 −2 −2

(−2, −3)

x 2

−4 r = 5 −6 −8 −10

Example 3 Find the Cartesian equation for x = t − , y = t2 + 4 +

Create a strategy Square x to match terms in y, then subtract to eliminate t.

Apply the idea Write the equation

Square both sides

Expand the square

Simplify Subtract x2 from y: Subtract equations

Distribute the negative

Simplify

Add x2 to both sides

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 4 Find the Cartesian equation for

Create a strategy Divide y by x to eliminate the denominator, then substitute t into x.

Apply the idea Divide y by x

Simplify the fraction

Remove the common factor t2 + 1

Divide by 2 making t the subject

Substitute t =

into x: Write the equation

Substitute t =

Evaluate the power

Simplify the numerator

Evaluate the reciprocal

Multiply by y2 + 4x2

Divide by x since x ≠ 0

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Reﬂect and check This can be expressed as an ellipse by completing the square: Write the equation

Subtract 4x from both sides

Rewrite 4x2 − 4x

Complete the square

Add 1 to both sides

Idea summary To convert from parametric form to Cartesian form, eliminate t from the equations using various ways such as algebraic manipulation and Pythagorean identities.

9.02E Practice questions What do you remember? 1

Find the Cartesian equation for x = t, y = 2t + 1.

Find the Cartesian equation for x = t2 − t + 1, y = t.

Write down the Cartesian equation for x = t − 1, y = 3t − 1.

Practice Ex 1

440

Find the Cartesian equation for these parametric equations: a

x = t + 2, y = 2t + 5

x = 3t − 1, y = 5t + 4

x = −t + 3, y = 2t − 4

x = 2t + 3, y = −3t + 1

x = 4t − 5, y = −2t + 2

Write down the Cartesian equation for these parametric equations: a

x = 5 cos t, y = 5 sin t

x = 3 cos t, y = 3 sin t

x = 7 cos t, y = −7 sin t

x = 10 sin t, y = −10 cos t

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the curve deﬁned by x = t + 2 and y = a

Find the Cartesian equation.

Identify the correct graph of the curve.

y 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Ex 2

Ex 3

Ex 4

5 4 3 2 1

x 1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

, for t ≠ −2.

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

x 1 2 3 4 5

Consider the curve deﬁned by x = 1 + 4 cos t and y = 2 + 4 sin t, for t ∈ [0, 2π ]. a

Find the Cartesian equation.

Graph the curve.

Find the Cartesian equations for these parametric equations: a

c e

b d

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Find the Cartesian equation in the general parabola form: a

x = t + 1, y = t2 + 1 2

x = t − 2, y = 2t2 + 1

x = 2t + 3, y = 8t + 4t − 4

x = 2t2 + 3, y = t + 3

x = 4t2 − 2t − 5, y = 2t + 5

x = t − 1, y = 3t2 − t + 2

Find the Cartesian equation for these parametric equations: a

x = 5 cos t − 2, y = 5 sin t + 3

x = 4 cos t + 1, y = −2 + 4 sin t

x = −3 + 6 cos t, y = 2 + 6 sin t

x = 2 + 3 cos t, y = −1 + 3 sin t

x = 3 + 4 cos t, y = 2 + 4 sin t

x = 1 + 2 cos t, y = 3 + 2 sin t

Extend your thinking 12

Find the Cartesian equation for these parametric equations: a

x = sec t − 2, y = 2 tan t + 3

c 13

Find the Cartesian equations for these parametric equations: a

x = 3t, y = 32t + 3

x = 2(3t) − 1, y = 3−t − 2

x = 33t, y = 32t − 2

x = 3t + 3−t, y = 3t − 3−t

Let x = cos t + cos 2t and y = sin t + sin 2t. a

Show cos t =

Find the Cartesian equation in expanded form.

Show sin x + sin y = 2 sin cos x + cos y = 2 cos

442

cos cos

and by letting A =

,B=

Consider x = cos t + cos 3t and y = sin t + sin 3t. Using the result in part (a), show the Cartesian equation can be written as ( x2 + y2) ( x2 + y2 − 2)2 = 4x2.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

9.03E Graph linear functions, quadratic functions and circles in parametric form After this lesson, you will be able to… • sketch the graph of a linear function given in parametric form by finding two points or intercepts. • sketch the graph of a quadratic function (parabola) given in standard parametric form by identifying its vertex, direction, and additional points. • sketch the graph of a circle given in parametric form by identifying its centre and radius. • interpret the effect of a restricted domain for the parameter t on the graph of a parametrically defined curve, particularly for circles (forming arcs).

Graph a linear function The graph of a set of parametric equations can be sketched without converting to Cartesian form. For linear functions, the parametric equations are also linear in the parameter t. A straight line is deﬁned by any two distinct points. To sketch the graph, one can determine the coordinates of two points by substituting two different values for the parameter t. Alternatively, the x- and y-intercepts can be determined.

x = at + b, y = ct + d x, y are the coordinates on the Cartesian plane a, c are constants that determine the slope of the line b, d are constants where the ratio

gives the slope of the line (for a ≠ 0)

t is the parameter, which can be any real number unless a domain is speciﬁed

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9.03E Graph linear functions, quadratic functions and circles in parametric form mathspace.co

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Example 1 Sketch the graph formed by the parametric equations x = t + 1 and y = 2t − 3.

Create a strategy Determine the coordinates for at least two values of the parameter t to deﬁne two points on the line, then construct a table of values and sketch the graph.

Apply the idea Select values for t and calculate the corresponding x and y coordinates. t

−1

−1 + 1 = 0

2(−1) − 3 = −5

(0, −5)

x=t+1

0+1=1

2(0) − 3 = −3

(1, −3)

y = 2t − 3

1+1=2

2(1) − 3 = −1

(2, −1)

( x, y)

2+1=3

2(2) − 3 = 1

(3, 1)

Plot the points, such as (1, −3) and (2, −1), and draw a straight line through them.

y 1

x 1

−1

(2, −1)

−2 −3

(1, −3)

−4 −5

Example 2 Sketch the graph formed by the parametric equations x = −3t + 2 and y = 5t − 1.

Create a strategy Determine the x- and y-intercepts by setting y = 0 and x = 0 respectively. Solve for t in each case, then ﬁnd the corresponding coordinate.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea To determine the y-intercept, set x = 0. Write the equation for x

Set x = 0

Add 3t to both sides

Divide both sides by 3

Now substitute t =

into the equation for y. Write the equation for y

Substitute

Evaluate the product

Evaluate

The y-intercept is

To determine the x-intercept, set y = 0. Write the equation for y

Set y = 0

Add 1 to both sides

Divide both sides by 5

Now substitute t =

into the equation for x. Write the equation for x

Substitute

Evaluate the product

Evaluate

The x-intercept is

Plot the intercepts and draw a line through them.

y 2.5 2 1.5 1 0.5 x −0.5 −0.5

0.5

1.5

2.5

9.03E Graph linear functions, quadratic functions and circles in parametric form mathspace.co

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Idea summary To graph a linear function from its parametric equations, determine two points on the line. This can be done by substituting two different values for the parameter t, or by calculating the x- and y-intercepts.

Graph a quadratic function The parametric equations for a parabola with a vertical axis of symmetry can be recognised from their standard form.

x = 2 At + h, y = At2 + k h, k are the coordinates of the vertex, (h, k). This point corresponds to t = 0. A is a constant that determines the concavity. If A > 0, the parabola opens upwards. If A < 0, it opens downwards. t

is the parameter.

To sketch the graph, identify the vertex (h, k) directly from the equations. Then, ﬁnd at least one other point by substituting a convenient value for t (e.g., t = 1), or ﬁnd the intercepts if required.

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Example 3 Sketch the graph of the parametric equations x = 4t − 1 and y = 2t2 − 2.

Create a strategy Recognise the standard parametric form of a parabola to identify its vertex. Determine one other point by substituting a non-zero value for t.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea Compare the given equations to the standard form x = 2 At + h and y = At2 + k. From y = 2t2 − 2, we can identify A = 2 and k = −2. From x = 4t − 1, we have 2 At = 4t, which gives 2(2)t = 4t, conﬁrming A = 2. The constant term gives h = −1. The vertex is (h, k) = (−1, −2). This point occurs when t = 0. Since A = 2 > 0, the parabola opens upwards. To ﬁnd another point, substitute t = 1. x = 4t − 1

Write the equation for x

= 4(1) − 1

Substitute t = 1

Evaluate 2

y = 2t − 2 2

Write the equation for y

= 2(1) − 2

Substitute t = 1

Evaluate

Another point on the parabola is (3, 0). Sketch the parabola using the vertex (−1, −2) and the point (3, 0). y 5 4 3 2 1 −8 −7 −6 −5 −4 −3 −2 −1

(3, 0) x 1 2 3 4 5 6

−1 −−2

(−1, −2)

Example 4 Sketch the graph of the parametric equations

and

, showing all intercepts.

Create a strategy Identify the vertex from the standard parametric form. Then determine the x- and y-intercepts by setting y = 0 and x = 0 and solving for t.

9.03E Graph linear functions, quadratic functions and circles in parametric form mathspace.co

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Apply the idea By comparing to x = 2 At + h and y = At2 + k, we identify A = , h = 2, and k = −1. The vertex is (h, k) = (2, −1). For the y-intercept, set x = 0: Write the equation for x

Set x = 0

Subtract 2 from both sides

Multiply both sides by 2 making t the subject

Write the equation for y

Substitute t = −4

Evaluate

For the x-intercept(s), set y = 0: Write the equation for y Set y = 0

Multiply all terms by 4

Add 4 to both sides

Rearrange the equation

Take the square root of both sides

Substitute these t values into the equation for x. When t = 2, x =

(2) + 2 = 3. Intercept is (3, 0).

When t = −2, x =

(−2) + 2 = 1. Intercept is (1, 0).

Sketch the parabola using the vertex (2, −1) and the intercepts (0, 3), (1, 0), and (3, 0). y 3 (0, 3) 2 1

(1, 0) 1 −1

448

(3, 0) 2

(2, −1)

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary To graph a parabola from its parametric equations, recognise the standard form to identify the vertex (h, k). Then, ﬁnd additional points or intercepts to complete the sketch.

Graph a circle The parametric equations for a circle can be recognised from their standard trigonometric form.

x = r cos t + h, y = r sin t + k r is the radius of the circle (r > 0) h, k are the coordinates of the centre, (h, k) t is the parameter, where t ∈ [0, 2π ) traces a full circle To sketch the graph, identify the centre (h, k) and radius r directly from the equations. If the domain of t is restricted, calculate the start and end points of the arc.

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Example 5 Sketch the graph of the parametric equations x = 3 cos t + 1 and y = 3 sin t − 2.

Create a strategy Recognise the standard parametric form of a circle to identify its centre and radius directly from the equations, then sketch the graph.

Apply the idea Compare the given equations to the standard form x = r cos t + h and y = r sin t + k. From the equations, we can directly identify the parameters: • The radius is r = 3. • The x-coordinate of the centre is h = 1. • The y-coordinate of the centre is k = −2. The graph is a circle with centre (1, −2) and radius 3.

y 1 −2

−1

x 1

−1 −2 −3

(1, −2) r=3

−4 −5

9.03E Graph linear functions, quadratic functions and circles in parametric form mathspace.co

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Example 6 Sketch the graph of the parametric equations x = 2 cos t − 1 and y = 2 sin t + 1 for t ∈ [0, π ].

Create a strategy Identify the centre and radius from the standard form. Use the restricted domain of the parameter t to determine the start and end points of the arc, and then sketch the graph.

Apply the idea From the standard form, identify the centre (−1, 1) and radius r = 2. The domain for the parameter is restricted to t ∈ [0, π ], so the graph is not a full circle. Determine the start and end points. Start point (when t = 0): x = 2 cos t − 1

Write the equation for x

= 2 cos (0) − 1

Substitute t = 0

= 2 (1) − 1

Evaluate cos 0 = 1

= 1 y = 2 sin t + 1

Evaluate Write the equation for y

= 2 sin (0) + 1

Substitute t = 0

= 2 (0) + 1

Evaluate sin 0 = 0

= 1

Evaluate

The arc starts at the point (1, 1). End point (when t = π ): x = 2 cos t − 1

Write the equation

= 2 cos (π ) − 1

Substitute t = π

= 2 (−1) − 1

Evaluate cos π = −1

= −3

Evaluate

y = 2 sin t + 1

Write the equation

= 2 sin (π ) + 1

Substitute t = π

= 2 (0) + 1

Evaluate sin π = 0

= 1

Evaluate

The arc ends at the point (−3, 1). 3

Since t varies from 0 to π, sin t ≥ 0. This means y = 2 sin t + 1 ≥ 1, so the graph is the upper semicircle of the circle centred at (−1, 1) with radius 2.

(−1, 1) (−3, 1) −3

x −2

−1

1 −1

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

(1, 1)

Idea summary To graph a circle from its parametric equations, identify the centre (h, k) and radius r from the standard form. If the domain of t is restricted, an arc of the circle is graphed.

9.03E Practice questions What do you remember? 1

Determine whether each statement about parametric equations is true or false. a

Parametric equations x = at + b, y = ct + d where a and c are not both zero, always produce a straight line.

The vertex of a parabola with equations x = 2At + h, y = At2 + k occurs at t = 0.

A circle with equations x = r cos t + h, y = r sin t + k has radius r.

Complete the table of values for the parametric equations x = t − 2, y = 2t + 1. t

−1

⬚

Practice 3

Ex 1

Identify the type of graph produced by each set of parametric equations: linear function, quadratic function, or circle. a

x = 2t − 1, y = 3t + 2

x = 3t + 1, y = t2 − 2

x = 4 cos t + 2, y = 4 sin t − 1

x = t − 3, y = 2t − 4

For each set of parametric equations: i

Construct a table of values using at least two values of t.

Plot the straight line.

x = t + 2, y = 3t − 1

x = 3 − t, y = 2t

x = 2, y = t + 3

− 1, y = −1

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Ex 2

Ex 3

Ex 4

Ex 5

Ex 6

452

For each set of parametric equations: i

Determine the x-intercept.

Determine the y-intercept.

iii

Sketch the straight line through the intercepts.

x = −2t + 3, y = 4t − 2

x = t + 2, y = 2 − t

+ 1, y = t + 4

x = 1 − 2t, y = 3t + 3

For each set of parametric equations: i

Identify the vertex of the parabola.

Determine an additional point by substituting a suitable value for t.

iii

Sketch the parabola using the vertex and additional point.

x = 2t − 2, y = t2 + 1

x = t, y = −(t + 1)2 + 3

x = 1 − t, y =

x = 2t, y = (t − 2)2

− 1

For each set of parametric equations: i

Identify the vertex of the parabola.

Determine the x-intercepts by setting y = 0.

iii

Determine the y-intercept by setting x = 0.

Sketch the parabola using the vertex and intercepts.

x = −4t + 1, y = −2t2 + 3

x = t, y = (t − 1)2 − 4

x = 2t − 1, y = −t2 + 2t

x = 1 − t, y = t2 + t − 2

For each set of parametric equations: i

Identify the centre and radius of the circle.

Sketch the circle using the centre and radius.

x = 2 cos t + 3, y = 2 sin t + 1

x = 3 sin t − 1, y = 3 cos t + 2

x = cos t, y = sin t − 2

x = 2.5 cos t + 1.5, y = 2.5 sin t − 0.5

For each set of parametric equations restricted to the given domain for t: i

Identify the centre, radius, and the portion of the circle graphed.

Determine the starting and ending points of the arc.

iii

Sketch the arc.

x = 3 cos t − 2, y = 3 sin t + 2, for t ∈ [0, π ]

x = cos t + 1, y = sin t − 1, for

x = 2 sin t, y = 2 cos t + 1, for t ∈ [0, π ]

x = 5 cos t − 3, y = 5 sin t, for

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For each set of parametric equations: i

Identify the vertex of the parabola.

Determine an additional point by substituting a suitable value for t.

iii

Sketch the parabola using the vertex and additional point.

x = 3t − 1, y = −t2 + 2

x = t2, y = t + 1

x = 1 − t2, y = 2t + 1

x = t, y = −2(t + 1)2

For each set of parametric equations: i

Identify the centre and radius of the circle.

Sketch the circle using the centre and radius.

x = 4 cos t + 1, y = 4 sin t − 3

cos t +

,y=

sin t −

x = 2 sin t, y = 2 cos t − 1

x = − cos t + 0.5, y = − sin t + 0.5

Extend your thinking 12

Determine the parametric equations for the line shown. y

(−2, 3)

3 2 1 x

−2

−1

1 −1 −2 −3

(1, −3)

Determine the points of intersection between the parametric equations x = t + 1, y = 2t + 2 and x = 2s − 1, y = −s + 3, then sketch both graphs showing the intersection point.

For the parametric equations x = 3t − 2, y = −3t + 4 and x = 3 cos (s) + 1, y = 3 sin (s) − 2: a

Determine the points of intersection.

Sketch both graphs on the same set of axes, showing the points of intersection.

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9 Chapter review 1

For the circle deﬁned by the Cartesian equation x2 + y2 = 36, if a parametric equation for y is y = 6 sin θ, which of the following is a parametric equation for x? y 6 4 2 x −6 −4 −2

x = 6 tan θ

−2 −4 −6

A 2

x = 36 cos θ

Line

Parabola

y = 2x − 1

y = 2x + 5

x = cos θ

Circle

Hyperbola

y = 2x − 7

Find the parametric equations for each line passing through the two given points: a

Through (2, 4) and (5, 10)

Through (0, 2) and (6, 16)

Through (−1, 6) and (4, −2)

Find the parametric equations of the parabola y = x2 + 6x + 5 by completing the square ﬁrst. 4

3 2 1 −6 −5 −4 −3 −2 −1 −1

x 1 2 3 4

−2 −3 −4

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The parametric equations of a curve are x = t + 3 and y = 2t − 1. Which of the following is the Cartesian equation of this curve? A

Which type of graph is produced by the parametric equations x = 4t − 2 and y = 3t2 + 1? A

x = 6 cos θ

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Determine the parametric equations for each circle: a c

x2 + y2 = 16 2

( x + 1) + (y + 5) = 49

The equation of an ellipse is given by

Show that x = a cos φ and y = b sin φ are parametric equations for the ellipse.

Find the parametric equations of the ellipse 16x2 + 25y2 = 1.

Find the Cartesian equation for the following in the general parabola form y = ax2 + bx + c or x = ay2 + by + c: a

x = t + 2, y = t2 + 3

x = t2 + 4, y = t + 1

x = 6 cos t, y = 6 sin t

x = 5 sin t, y = 5 cos t

Find the Cartesian equation for the following: a

x = 2 sec θ − 1, y = 4 tan θ + 3

x = 2 cos t, y = −2 sin t

and y = t2 − 2 +

Find the Cartesian equation for x = t +

x = t − 1, y = 3t2 + 2

Determine the Cartesian equation for these circles:

( x − 2)2 + (y − 3)2 = 4

Consider the graph of the parametric equations x = 2t + 2, y = t2 − 1: a

Identify the vertex of the parabola.

Determine an additional point by substituting t = 2.

Sketch the parabola on a number plane, labelling the vertex and the additional point.

Consider the parametric equations x = t + 1, y = 2t − 4: a

Determine the x- and y-intercepts.

Sketch the graph of the equation.

Determine the point(s) of intersection between the line deﬁned by x = t + 1, y = 3t − 2 and the line deﬁned by x = 2s + 3, y = −s + 4. Sketch both lines on the same number plane, clearly labelling their intersection point.

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10 Probability Chapter outline 10.A 10.B 10.C 10.01 10.02 10.03 10.04 10.05

458 465 473 482 489 496

Mutually exclusive events can’t both happen — like turning left and right at once.

10.01 Probability and events After this lesson, you will be able to… • define the terms experiment, trial, outcome, and sample space. • determine the sample space for a random experiment. • identify an event as a subset of the sample space. • calculate the theoretical probability of an event where outcomes are equally likely. • interpret set notation ( , ∩, ∪) in the context of probability events.

Experiments and sample spaces Experiment (random) A process with an observable result, e.g. a dice roll or a coin toss. Outcome Possible result from an experiment or trial. Sample space The set of all possible outcomes of a chance experiment. For example, the set of outcomes (also called sample points) from tossing 2 identical coins at the same time is {HH, HT, TT }, where H represents a ‘head’ and T a ‘tail’. Trial A single performance of a random experiment. Successive trials refers to repeated performances of the same experiment each of which will therefore have the same set of possible outcomes (sample space).

An experiment or trial is a repeatable procedure with a well-deﬁned set of possible outcomes, known as the sample space, denoted S.

S = {heads, tails} S is the set of all possible outcomes, for example, ﬂipping a coin gives {heads, tails} The cardinality of the sample space, S, is the number of outcomes. An event is a subset of the sample space S.

A⊆S A 458

is an event, a subset of the sample space S

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

If an event has no outcomes, it is the empty set ∅.

Exploration Consider the experiment of ﬂipping two coins: 1. List the sample space S using set notation. 2. Deﬁne an event A as getting at least one head. 3. Sketch a Venn diagram with S as the universal set and A as a subset. 4. How many outcomes are in S and A?

Example 1 An experiment involves drawing a card from a standard deck of 52 cards: a Determine S.

Create a strategy List all possible outcomes using set notation.

Apply the idea S = {all 52 cards} Write the sample space for a deck of 52 cards

S = 52 b Let A be the event that a heart is drawn. Find A.

Create a strategy Determine the number of hearts in a standard deck of cards.

Apply the idea A = {13 hearts} Write the number of hearts in the deck

A = 13

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Idea summary An experiment or trial is a repeatable procedure with a sample space S, the set of all possible outcomes. An event is a subset of S, with S as the number of outcomes. The empty set ∅ represents an impossible event.

Probability of events Probability The chance of something happening shown on a scale from 0 and 1 (inclusive). For example, the probability that a fair coin toss will come up ‘heads’ is 0.5. When all outcomes in a sample space S are equally likely, the probability of an event A is the ratio of the number of favourable outcomes to the total number of outcomes.

P (A) is the probability of event A occurring is the number of outcomes in event A A S i s the number of outcomes in the   sample space S

• is the event “A does not occur” • A ∩ B is the event “A and B both occur” • A ∪ B is the event “A or B occurs”

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Example 2 A bag contains 4 red and 6 blue balls. A ball is drawn at random. Let A be the event of drawing a red ball, and B be the event of drawing a blue ball: a Find the probability of drawing a red ball.

Create a strategy Use the formula

Apply the idea Sample space S is all balls.

S  = 4 + 6

Add the total balls

= 10

Evaluate

Event A is red ball A = 4. Write the formula Substitute A = 4 and S = 6 Simplify The probability of drawing a red ball is .

b Find P ( ).

Create a strategy Use the complement formula P ( ) = 1 − P (A), where

from part (a).

Apply the idea The event

is “not a red ball,” i.e., a blue ball. Write the complement formula

Substitute P (A) =

Evaluate

The probability of drawing a blue ball is .

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c Find P (A ∪ B).

Create a strategy Use the formula P (A ∪ B) =

Apply the idea Since only red or blue balls exist, A ∪ B = S. Write the formula

Substitute A ∪ B = S

Evaluate

Example 3 A die is rolled. Let A be rolling an odd number, B be rolling a number less than 4: a Find P (A ∩ B).

Create a strategy Use the formula P (A ∩ B) =

Apply the idea The sample space is S = {1, 2, 3, 4, 5, 6}, so S = 6. A represent the odd numbers: A = {1, 3, 5}, so A = 3. B represents the numbers less than 4: B = {1, 2, 3}, so B = 3. A ∩ B = {1, 3}

List the numbers both in A and B

So A ∩ B = 2, calculating P (A ∩ B): Write the formula Substitute A ∩ B = 2 and S = 6 Simplify

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b Find P (A ∪ B).

Create a strategy Use the formula P (A ∪ B) =

Apply the idea A ∪ B = {1, 2, 3, 5}

Combine the elements from A and B

So A ∪ B = 4, calculating P (A ∪ B): Write the formula

Substitute A ∪ B = 4 and S = 6

Simplify

Reflect and check To verify the solution, confirm the calculation of A ∪ B using the inclusion-exclusion principle.

A ∪ B  = A  + B  − A ∩ B 

Write the formula

=3+3–2

Substitute the values

Evaluate

Idea summary The probability of an event A is P (A) =

when outcomes are equally likely.

Set notation describes events: means “A does not occur,” A ∩ B means “A and B occur,” and A ∪ B means “A or B occurs”.

10.01 Practice questions What do you remember? 1

An experiment involves rolling a six-sided die: a

Write the sample space S using set notation.

Determine S.

Deﬁne event A as rolling an even number. Write A as a subset of S.

Calculate P (A) as a fraction.

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Match the probability notation to its description: i

Event A does not occur

A ∩ B

Event A or B occurs

A ∪ B

iii

Events A and B both occur

∅

Impossible event

Practice Ex 1

Ex 2

Ex 3

A bag contains 5 red and 3 blue balls. A ball is drawn at random: a

Determine S.

Let A be the event of drawing a red ball. Find A.

A deck of 52 cards is used. A card is drawn at random. Deﬁne event A as drawing a face card ( jack, queen, or king), and event B as drawing a club: a

Find the probability of drawing a face card.

Find P ( ).

Find P (A ∪ B).

A jar has 6 white and 4 black marbles. A marble is drawn at random. Deﬁne event A as drawing a white marble, B as drawing a black marble: a

Sample space S

P (A)

A ∪ B

P (A ∪ B)

A die is rolled. Deﬁne A as rolling a number greater than 3, B as rolling an odd number. Determine: b

P (A ∩ B)

A∪B

P (A ∪ B)

Calculate P (chocolate).

Calculate P (not chocolate).

A letter is picked from “PROB”. Deﬁne A as picking a vowel, B as picking a consonant. Calculate: P (A)

P (B)

P (A ∪ B)

A coin is tossed twice. Deﬁne A as getting at least one head, B as getting two heads: a

464

A∩B

A box has 4 chocolate candies and 6 vanilla candies. A candy is picked.

a 10

Find P (A ∪ B).

a 9

A spinner has 6 equal sections: 3 red, 2 blue, 1 green. Deﬁne events A as landing on red, B as landing on blue. Determine:

a 8

Find P (A ∩ B).

Determine A ∩ B.

Calculate P (A ∩ B).

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Extend your thinking 11

A die is rolled. Deﬁne A as rolling a prime number, B as rolling a number less than 5. Sketch a Venn diagram showing S, A, B, A ∩ B, . List outcomes in each region.

A bag has 2 red, 3 blue, 5 green balls. Deﬁne A as drawing red, B as drawing blue:

Determine P (A ∩ B) and explain your reasoning.

Calculate P (A ∪ B).

A student calculates P (A ∪ B) for a die roll where A is rolling a number greater than 4 and B is rolling an odd number. The student writes: P (A ∪ B) =

= . Correct the error.

A wheel has 12 sections: 5 for winning candy, 4 for winning a toy, and 3 for winning nothing. Deﬁne A as winning candy, B as winning a toy. Calculate the probability of winning exactly one prize.

A game picks a number from 1 to 10. Deﬁne A as picking a multiple of 3, B as picking an even number. Calculate P (A ∩ B) and explain it in the context of events A and B.

10.02 Mutually exclusive events After this lesson, you will be able to… • define and identify mutually exclusive events. • represent mutually exclusive events using a Venn diagram. • apply the complement rule, P ( ) = 1 − P (A), to find probabilities. • apply the addition rule for probability, P (A ∪ B) = P (A) + P (B) − P (A ∩ B). • apply the simplified addition rule for mutually exclusive events, P (A ∪ B) = P (A) + P (B).

Mutually exclusive events Mutually exclusive events Two events that cannot have simultaneous outcomes in the same chance experiment. For example, when a fair coin is tossed twice, the events ‘HH’ and ‘TT ’ cannot occur at the same time and are, therefore, mutually exclusive.

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In a Venn diagram, mutually exclusive events are represented by non-overlapping circles within the sample space S. S A

A∩B=∅

This Venn diagram shows mutually exclusive events A and B with no overlap, so A ∩ B = ∅.

A∪B If A ∩ B ≠ ∅, the events are not mutually exclusive, as they share at least one outcome.

Exploration Consider drawing a card from a deck of 52 cards. Deﬁne event A as drawing a heart and event B as drawing a spade. 1. Are A and B mutually exclusive? 2. Sketch a Venn diagram to justify your answer. 3. What would make two events in this experiment mutually exclusive?

Example 1 A fair six-sided die is rolled, S = {1, 2, 3, 4, 5, 6}: a Deﬁne event A as rolling a number less than 3.

Create a strategy

Apply the idea

Identify outcomes in S less than 3.

A = {1, 2}

b Deﬁne event B as rolling a number greater than 4. Show that A and B are mutually exclusive.

Create a strategy Find B and check if A ∩ B = ∅.

Apply the idea B = {5, 6} A ∩ B = {1, 2} ∩ {5, 6} = ∅ Since A ∩ B = ∅, events A and B are mutually exclusive.

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c Find P (A ∪ B).

Create a strategy Use the formula P (A ∪ B) = P (A) + P (B) since A and B are mutually exclusive.

Apply the idea Find each value using A = 2, B = 2, S = 6. For P (A): Write the formula Substitute A = 2 and S = 6 Simplify For P (B): Write the formula Substitute B = 2 and S = 6 Simplify For P (A ∪ B): Write the addition rule for mutually exclusive events

Substitute P (A) =

Evaluate

and P (B) =

Idea summary Mutually exclusive events A and B have no outcomes in common, so A ∩ B = ∅. In a Venn diagram, mutually exclusive events are shown as non-overlapping circles.

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Probability rules for events Two key probability rules help calculate the likelihood of events, building on probability concepts. The complement rule states that the probability of an event A not occurring is:

P ( ) = 1 − P (A) P ( ) is the probability that event A does not occur P (A)

is the probability that event A occurs

For example, if P (A) = , then P ( ) = 1 −

= .

The addition rule for the union of two events is:

P (A ∪ B) = P (A) + P (B) − P (A ∩ B) P (A ∪ B)

is the probability that A or B occurs

is subtracted to avoid double-counting P (A ∩ B) outcomes in both events For mutually exclusive events, P (A ∩ B) = 0, so the rule simpliﬁes to P (A ∪ B) = P (A) + P (B). These rules can be visualised using Venn diagrams, where P (A ∪ B) corresponds to the combined area of A and B, adjusted for overlap.

Interactive exploration Discover this concept in action online

Example 2 A fair six-sided die is rolled, S = {1, 2, 3, 4, 5, 6}: a For event A (even numbers), ﬁnd P (A).

Create a strategy Use the formula P (A) =

Apply the idea Let A = {2, 4, 6}, so A = 3 and S = 6. Write the formula Substitute A = 3 and S = 6 Simplify

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b Find P ( ).

Create a strategy Use the complement formula P ( ) = 1 − P (A), where

from part (a).

Apply the idea Write the formula Substitute P (A) = Evaluate

c Let B be the numbers greater than 4. Find P (B) and P (A ∩ B).

Create a strategy Use the formula P (B ) =

and P (A ∩ B) =

Apply the idea For P (B), the numbers greater than 4 is B = {5, 6}, so B = 2. Write the formula

Substitute B = 2 and S = 6

Simplify

For A ∩ B: A ∩ B = {2, 4, 6} ∩ {5, 6} = {6}

List the elements of A and B Write the common element

So A ∩ B = 1, calculating P (A ∩ B): Write the formula Thus, P (B) =

Substitute A ∩ B = 1 and S = 6

and P (A ∩ B) = .

d Find P (A ∪ B).

Create a strategy Use the addition formula P (A ∪ B) = P (A) + P (B) − P (A ∩ B).

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Apply the idea Write the formula

Substitute P (A) = , P (B) =

Rewrite with common denominator

Evaluate

Simplify

and P (A ∩ B) =

Idea summary The complement rule is P ( ) = 1 − P (A), giving the probability of A not occurring. The addition rule is P (A ∪ B) = P (A) + P (B) − P (A ∩ B), simplifying to P (A) + P (B) for mutually exclusive events.

10.02 Practice questions What do you remember? 1

Define mutually exclusive events.

In a Venn diagram, how are mutually exclusive events A and B represented?

For a chance experiment, what does it mean if A ∩ B = ∅ for events A and B?

Write the simplified addition rule for the probability of mutually exclusive events A and B.

What is the significance of the condition P (A ∩ B) = 0 for events A and B in a chance experiment?

Practice Ex 1

A card is drawn from a standard deck of 52 cards, S = {all 52 cards}. Deﬁne event A as drawing a heart: a

Write the set A.

Deﬁne event B as drawing a spade. Show that A and B are mutually exclusive.

Find P (A ∪ B).

A spinner has 4 equal sections: 1 red, 2 blue, 1 green. Deﬁne event C as landing on blue: a

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Write the set C.

Find P ( ).

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

A spinner has 8 equal sections: 4 red, 3 blue, and 1 yellow. The spinner is spun once. Let A be landing on a red section, B be landing on a yellow section. Find: a

P (A)

P( )

P (B) and P (A ∩ B)

P (A ∪ B)

A fair six-sided die is rolled. Let A be rolling an odd number, B be rolling a 5 or 6. Find: a

P (A ∩ B)

P (red ∪ blue)

P (neither red nor green)

Are A and B mutually exclusive? Explain.

Find P (A ∪ B).

P (blue ∪ green)

, science

, or

. Find:

P (mathematics ∪ science)

P (not mathematics)

P (chocolate ∪ mint)

P (not mint)

A fair coin is tossed twice, S = {HH, HT, TH, TT }. Deﬁne event A as getting at least one head, event B as getting two heads: a

Write the sets A and B.

Find P (A ∩ B).

A die is rolled, S = {1, 2, 3, 4, 5, 6}. Event A is rolling a multiple of 3, event B is rolling an even number. Find: a

P (not blue)

A box has 15 candies: 6 chocolate, 5 mint, 4 caramel. Find: a

A student is allowed to pick only one subject: mathematics

P (A ∪ B)

A deck of 52 cards is used. Event A is drawing a red card, event B is drawing an ace:

history

A raﬄe has 50 tickets: 15 red, 20 blue, 15 green. One ticket is drawn. Find: a

P (A ∪ B)

A bag has 10 marbles: 4 red, 3 blue, 3 yellow. One marble is drawn. Find: a

A card is drawn from a deck of 52 cards. Deﬁne event A as drawing a spade, event B as drawing a king. Find: a

P (A ∩ B)

P (A ∪ B)

A jar has 20 balls: 8 red, 7 blue, 5 green. One ball is drawn. Find: a

P (red ∪ green)

P ((red ∪ green)c)

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Extend your thinking 19

A game uses a spinner with 6 equal sections: 3 win, 2 lose, 1 draw. Explain how the sample space and events can be represented in a Venn diagram.

In a quiz, a student answers one question with choices A, B, or C (equally likely). Event A is picking choice A, event B is picking choice B. Show why P (A ∪ B) = P (A) + P (B).

A survey offers two options: sports

or music

, mutually exclusive. Find the

probability of choosing either and explain why the complement rule applies. 22

A card is drawn from 52 cards. Event A is drawing a queen, event B is drawing a black card. A student claims P (A ∪ B) = P (A) + P (B). Correct the error using a Venn diagram description.

A bag has 12 balls: 5 red, 4 blue, 3 green. One ball is drawn. Express P (red ∪ blue) using both the addition rule and by listing outcomes.

Did you know?

The probability of being struck by lightning in your lifetime is around 1 in 15 000! That’s far higher than many people imagine, especially in regions with frequent storms. Statisticians use probability models to calculate these odds based on geography and weather patterns. This surprising fact shows how chance events can still carry very real risks.

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10.03 Multistage events and conditional probability After this lesson, you will be able to… • represent the sample space of multistage events using tree diagrams and arrays. • calculate probabilities of outcomes in multistage events. • define conditional probability, P ( AB), as the probability of A given B has occurred. • examine conditional probability by restricting the sample space in Venn diagrams and two-way tables. • solve simple problems involving conditional probability.

Multistage events with tree diagrams Multistage events An event that consists of two or more simple experiments. For example, tossing a coin three times (repeated trials), or both tossing a coin and rolling a dice (possible outcomes would be H and 5, T and 2). Tree diagram A diagram consisting of line segments (edges) connected to points (vertices) like the branches of a tree. It shows the relationship between sets, events or the set of outcomes of a multi-step random experiment.

H H, H T H, T H T, H

T T, T

A multistage event involves multiple steps, each with its own set of outcomes. The sample space is the set of all possible outcome sequences, often represented using a tree diagram. For example, flipping two coins is a multistage event. The first flip has outcomes H, T, and the second flip has the same. A tree diagram shows all combinations: HH, HT, TH, TT.

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H HH H T

H TH

This tree diagram shows the sample space S = {HH, HT, TH, TT } for ﬂipping two coins, with each branch having probability .

T T TT To ﬁnd the probability of an event, multiply the probabilities along the branches leading to each favourable outcome and sum them if there are multiple paths. For example, the probability of getting at least one head is the sum of probabilities for HH, HT and TH.

Exploration Consider rolling a die twice: 1. Draw a tree diagram to list the sample space. 2. Deﬁne an event A as the sum of the rolls being 7. 3. How many outcomes are in S, and which outcomes belong to A?

Example 1 A bag contains 2 red balls and 3 blue balls. Two balls are drawn without replacement: a Draw a tree diagram to show the possible outcomes.

Create a strategy Construct a tree diagram for the two draws, noting changing probabilities due to no replacement.

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Apply the idea First draw: 2 red, 3 blue, so a total of 5 balls. Second draw: Depends on the ﬁrst draw (no replacement). R RR R B RB

The diagram shows outcomes RR, RB, BR, BB with probabilities calculated for each branch.

R BR B B BB

b Find the probability of drawing two red balls.

Create a strategy Calculate the probability by multiplying along the branch for two red balls.

Apply the idea The probability of drawing the ﬁrst red is

while the second red is . Multiply the probabilities

Evaluate Simplify The probability of drawing two red balls is

Reﬂect and check Verify the tree diagram includes all outcomes and probabilities sum to 1. Check probabilities adjust for no replacement.

Idea summary Multistage events involve multiple steps, with outcomes shown in a tree diagram. Probabilities are calculated by multiplying along branches and summing for multiple favourable outcomes.

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Introduction to conditional probability Array It is made by arranging a set of outcomes, into columns and rows, where columns represent one event and rows represent another. For example, when rolling a die and ﬂipping a coin, the sample space can be presented in an array as: 1

Conditional probability If A is conditional on B, then the sample space is restricted to the outcomes of event B. P (AB) means ‘Probability of A given B occurs’. Two-way table Curly

Straight

A common way of displaying the two-way frequency distribution that arises when a group is categorised according to 2 criteria.

Red

Brown

Conditional probability restricts the sample space to the outcomes in B. For example, in a deck of 52 cards, let A be drawing a king and B be drawing a heart. If B occurs, the sample space is the 13 hearts, and P (AB) is the probability of a king among them. This can be visualised using a Venn diagram by focusing on the region of B. S A

A∩B

A|B

The shaded region B is the new sample space, and A ∩ B represents outcomes where A occurs given B.

Tree diagrams can also show conditional probabilities by focusing on branches where B occurs. Conditional probability is calculated as: P (AB) =

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Interactive exploration Discover this concept in action online

mathspace.co

Example 2 A deck of 52 cards is used. Let A be the event of drawing a king, and B be the event of drawing a heart: a Find P (AB).

Create a strategy Use the formula P (AB) =

Apply the idea Event B is drawing a heart, B = 13 (hearts). Event A ∩ B is drawing a king of hearts, A ∩ B = 1. Write the formula

Substitute the values

b Represent this in a two-way table.

Create a strategy Use a two-way table to organise outcomes for A and B.

Apply the idea Categories are King (A) or not King (A′), Heart (B) or not Heart (B′). B

B′

A′

The table shows A ∩ B = 1 king of hearts, and B = 1 + 12 = 13 hearts.

Reﬂect and check Verify P (AB) by checking the table:

. Ensure the table sums to 52 cards.

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Idea summary Conditional probability P (AB) is the probability of A given B has occurred, restricting the sample space to B. Tools like Venn diagrams, tree diagrams, and two-way tables help visualise and calculate conditional probabilities.

10.03 Practice questions What do you remember? 1

Deﬁne these terms: a

Multistage event

Conditional probability

P (AB)

Tree diagram for multistage events

Determine whether each statement is true or false: a

P (AB) = P (A) always holds for any events A and B.

A two-way table can be used to calculate conditional probabilities by restricting the sample space.

In a tree diagram for events without replacement, the probabilities on the second stage remain the same as the ﬁrst.

How does an array help in modelling a multistage event?

Why are probabilities multiplied along branches in a tree diagram?

Match each term or tool to its correct description: a

Shows the probability of an event given another event has occurred

Multistage event

Tree diagram

An event with multiple steps, each with its own outcomes

iii

Conditional probability

Uses branches to represent all possible outcomes of a multi-step experiment

Practice Ex 1

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A box contains 4 chocolate candies and 3 caramel candies. Two candies are drawn without replacement: a

Draw a tree diagram to show the possible outcomes.

Find the probability of drawing two chocolate candies.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

A survey of 80 students found 50 like sports, 30 like music, and 15 like both. A student is chosen at random: a

Find P (sportsmusic).

Construct a two-way table to show the outcomes.

A bag contains 3 red balls and 2 blue balls. One ball is drawn: a

What is the probability of drawing a red ball?

What is the probability of drawing a blue ball?

If a red ball is drawn, what is the probability it is red given it is red?

Why is a tree diagram not necessary for this single-stage event?

A spinner has 2 sections: red and blue, each with probability . It is spun twice: a

How many possible outcomes are there?

List the sample space.

What is the probability of getting red on both spins?

What tool can represent these outcomes?

A coin is ﬂipped twice: a

Draw a tree diagram to show all possible outcomes.

Find the probability of getting at least one head.

Find the probability of getting exactly one head.

Find the probability of getting two heads given the ﬁrst ﬂip is a head.

A bag contains 4 red marbles and 3 blue marbles. Two marbles are drawn without replacement: a

Draw a tree diagram to show the possible outcomes.

Find the probability of drawing two red marbles.

Find the probability of drawing one red and one blue marble.

Find the probability of drawing a blue marble second given the ﬁrst is red.

A deck of 52 cards is used. Deﬁne event A as drawing a spade and event B as drawing an ace: a

Find P (A).

Find P (AB).

Construct a two-way table to show the outcomes.

A weather forecast predicts a assuming independence:

Find P (B).

chance of rain on Saturday and a

chance on Sunday,

Draw a tree diagram for the weather over both days.

Find the probability it rains both days.

Find the probability it rains at least one day.

Find the probability it rains on Sunday given it rained on Saturday.

10.03 Multistage events and conditional probability mathspace.co

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480

A box contains 5 chocolates: 2 dark and 3 milk. Two chocolates are chosen without replacement: a

Find the probability of choosing two dark chocolates.

Find the probability of choosing one dark and one milk chocolate.

Find the probability of choosing a milk chocolate second given a dark chocolate ﬁrst.

Represent the outcomes in a two-way table.

A standard 6-sided die is rolled twice: a

How many outcomes are in the sample space?

Find the probability the sum of the rolls is 7.

Find the probability the second roll is 4 given the sum is 7.

Use an array to list outcomes where the sum is 7.

A game involves drawing two cards from a standard deck with replacement: a

Find the probability both cards are hearts.

Find the probability at least one card is a heart.

Find the probability the second card is a heart given the ﬁrst is a heart.

Draw a tree diagram for this event.

A survey of 100 students found 60 like coffee, 40 like tea, and 20 like both: a

Construct a Venn diagram to represent this data.

Find the probability a student likes coffee.

Find the probability a student likes tea given they like coffee.

Construct a two-way table for the data.

A bag has 3 red and 2 green balls. Two balls are drawn with replacement: a

Find the probability both are red.

Find the probability at least one is green.

Find the probability the second is green given the ﬁrst is red.

Draw a tree diagram for the outcomes.

A box contains 3 dark chocolates and 2 milk chocolates. Two chocolates are chosen without replacement: a

Find the probability of choosing two dark chocolates.

Find the probability of choosing one dark and one milk chocolate.

Find the probability of choosing a milk chocolate second given a dark chocolate ﬁrst.

Represent the outcomes in a two-way table.

A standard 6-sided die is rolled twice: a

How many outcomes are in the sample space?

Find the probability the sum of the rolls is 8.

Find the probability the second roll is 5 given the sum is 8.

Use an array to list outcomes where the sum is 8.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A weather forecast predicts a assuming independence:

chance of rain on Monday and a

chance on Tuesday,

Draw a tree diagram for the weather over both days.

Find the probability it rains both days.

Find the probability it rains at least one day.

Find the probability it rains on Tuesday given it rained on Monday.

Extend your thinking 21

A game show has two stages. In stage 1, a contestant picks one of 3 doors (1 has a prize, 2 are empty). In stage 2, they roll a die and win if they roll a 6: a

Draw a tree diagram for the game and ﬁnd the probability of winning the prize and rolling a 6.

Find the probability of rolling a 6 given the contestant picked the prize door.

Explain why the events in stage 1 and stage 2 are independent.

A school has 200 students: 120 study mathematics, 80 study science, and 50 study both. A student is chosen at random: a

Find the probability they study mathematics given they study science.

Use a Venn diagram and a two-way table to show why P (MathScience) ≠ P (ScienceMath).

If a new student joins and studies only mathematics, how does this affect P (ScienceMath)?

A bag contains 2 red and 3 blue balls. Two balls are drawn without replacement. A student incorrectly calculates the probability of drawing two blue balls as

Identify the error in the student’s calculation.

Calculate the correct probability using a tree diagram.

Explain how a two-way table could also be used to ﬁnd this probability.

A game involves rolling a die and then ﬂipping a coin. Let E be the event of rolling an even number, and H be the event of ﬂipping heads: a

Draw a tree diagram and ﬁnd the probability of both events occurring.

Find the probability of ﬂipping heads given an even number was rolled.

Use an array to list all outcomes and verify the probability of both events occurring.

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10.04 Conditional probability formulas After this lesson, you will be able to… • use the formula P ( AB) =

or P ( AB) =

to find conditional

probabilities. • identify the necessary information from a problem to substitute into the formulas. • solve practical problems involving conditional probability using the formulas.

Conditional probability For equally likely outcomes, the conditional probability of event A given event B has occurred is the proportion of outcomes in A ∩ B relative to those in B.

P (AB) is the probability of A given B, where B ≠ 0

A ∩ B

is the number of outcomes in both A and B

B

is the number of outcomes in B

The conditional probability given P (A ∩ B) and P (B) can be calculated as:

P (A ∩ B) is the probability of both A and B occurring P (B) is the probability of B, where P (B) ≠ 0

Exploration In a bag with 3 red and 2 blue balls, draw one ball. Let A be drawing a red ball and B be drawing a coloured ball (red or blue): 1. Calculate P (AB) using A ∩ B and B. 2. Why is A ∩ B = A in this case?

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Example 1 A survey of 100 students shows 60 like mathematics, 50 like science, and 30 like both. a Find the probability a student likes mathematics given they like science.

Create a strategy Use the formula P (AB) =

with probabilities from the survey.

Apply the idea With sample space S = 100, let A be the students like mathematics, so P (A ∩ B) = be the students like science, so P (B) =

, while B

. Write the formula

Substitute the values

Simplify

b Represent this in a Venn diagram.

Create a strategy Use two circles for mathematics and science. Place 30 in the overlap and distribute remaining students.

Apply the idea S

Let A = mathematics and B = science. A

The overlap where the students like both is(A ∩ B) = 30. Students who only like mathematics is 60 − 30 = 30.

Students who only like science is 50 − 30 = 20. 20

Students who like neither is 100 − (30 + 30 + 20) = 100 − 80 = 20.

Reﬂect and check Verify the Venn diagram sums to 100: 30 + 20 + 30 + 20 = 100. Check P (AB) using table values.

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Idea summary Conditional probability is P (AB) = P (AB) =

for equally likely outcomes, or

, where P (B) ≠ 0.

Apply conditional probability The conditional probability formula P (AB) =

is used to solve practical problems where

the occurrence of one event affects another. For example, in a medical test, let A be testing positive and B be having a disease. P (AB) is the probability of testing positive given the disease. Tree diagrams or two-way tables can help calculate P (A ∩ B) and P (B).

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Example 2 In a factory, 10% of items are defective. A test detects 90% of defective items but falsely identiﬁes 5% of non-defective items as defective. An item tests positive. Find the probability the item is defective given it tests positive.

Create a strategy Use a tree diagram to calculate P (A ∩ B) and P (B), then apply P (AB) =

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Apply the idea Let A be defective, B be testing positive. Assume 100 items for simplicity. Tree diagram: B A

0.9 0.1

0.1

B′ AB′ B

0.9

A′B

The diagram shows paths for defective and non-defective items testing positive or negative.

0.05 A′ 0.95 B′ A′B′

P (A ∩ B) = 0.1 × 0.9 = 0.09

Write the probability of defective items and testing positive Evaluate Write the probability of non-defective items and testing positive

P (B) = 0.09 + 0.045 = 0.135

Evaluate Add the positive tests Evaluate

Calculating P (AB): Write the formula

Substitute the values

Evaluate and convert into a fraction

The probability the item is defective given it tests positive is .

Reﬂect and check Verify the tree diagram probabilities sum correctly. Check P (B) includes all positive test outcomes.

Idea summary The formula P (AB) =

solves practical conditional probability problems,

often using tree diagrams or tables.

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10.04 Practice questions What do you remember? 1

Write the formula for conditional probability when all outcomes are equally likely, for event A given event B.

Write the general formula for conditional probability using event probabilities.

A bag has 4 red and 2 blue marbles. Let A be drawing a red marble and B be drawing a coloured marble. Explain why A ∩ B = A.

In a deck of 52 cards, let A be drawing a king and B be drawing a heart. Calculate: a

P (A ∩ B)

P (B)

A die is rolled. Let A be rolling an odd number and B be rolling a number less than 4. Find: a

A ∩ B

B 

Practice Ex 1

A survey of 90 employees shows 55 have coding skills, 45 have design skills, and 20 have both. Calculate: a

The probability an employee has coding skills given they have design skills.

Represent this in a Venn diagram.

A box has 3 defective and 5 non-defective items. Let A be selecting a defective item and B be selecting any item. Calculate: a

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P (AB)

P (A ∩ B)

P (AB)

A spinner has 10 equal sections: 4 red, 4 blue, and 2 green. Let A be landing on blue and B be landing on a non-green section. Calculate: a

In a deck of 52 cards, let A be drawing an ace and B be drawing a club. Calculate: a

P (A ∩ B)

P (AB)

A class of 25 students has 15 taking biology, 10 taking physics, and 5 taking both. Determine: a

The probability a student takes biology given they take physics.

The probability a student takes physics given they take biology.

A survey of 120 students shows 70 like history, 50 like geography, and 30 like both. Determine: a

The probability a student likes history given they like geography.

The probability a student likes geography given they like history.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

A survey of 80 employees shows 45 use email, 30 use video calls, and 20 use both. Calculate: a

P (A ∩ B), where A is using email and B is using video calls

P (AB)

A quality control test for software apps identiﬁes 92% of buggy apps and has a 3% false positive rate. If 4% of apps are buggy, ﬁnd the probability an app is buggy given it tests positive, rounded to two decimal places.

A bag contains 5 red, 3 green, 7 blue, and 4 yellow candies. Let A be picking a red candy and B be picking a red or blue candy. Calculate: a

P (AB)

The probability a member plays chess given they play checkers.

The probability a member plays checkers given they play chess.

In a deck of 52 cards, let A be drawing a queen and B be drawing a spade. Calculate: P (A ∩ B)

P (AB)

A standard 6-sided die is rolled. Let A be rolling a multiple of 3 and B be rolling a number greater than 2. Calculate: a

A club of 50 members has 30 who play chess, 20 who play checkers, and 10 who play both. Determine:

a 17

P (A ∩ B)

P (AB)

A survey of 200 voters shows 120 support policy X, 80 support policy Y, and 50 support both. Calculate: a

The probability a voter supports X given they support Y.

The probability a voter supports Y given they support X.

A test for a virus is 98% accurate for infected people and has a 4% false positive rate. If 2% of people are infected, calculate: a

P (A ∩ B), where A is infected and B is testing positive

P (AB)

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Extend your thinking 20

A medical test is 99% accurate for detecting a disease and has a 2% false positive rate. If 0.5% of the population has the disease, let A be having the disease and B be testing positive. Calculate: a

488

P (A ∩ B)

P (AB) using a tree diagram

A two-way table shows pet preferences among 80 students: Dogs

Cats

Total

Year 11

Year 12

Total

Calculate the probability a student prefers dogs given they are in Year 12.

Explain how the table aids in ﬁnding P (A ∩ B) and P (B).

In a game show, a contestant chooses one of 4 boxes, one containing a prize and three empty. The host opens one empty box and offers a switch to one of the remaining two. Let A be the prize in the chosen box and B be the host opening an empty box: a

Calculate P (AB).

Determine if switching is beneﬁcial and explain why.

A student claims P (AB) = P (A) for any events A and B: a

Provide a counterexample with a die to disprove this.

Explain why the claim is false.

A factory test detects 95% of defective items and falsely ﬂags 3% of non-defective items. If 8% of items are defective, calculate: a

P (A ∩ B), where A is defective and B is testing positive

P (AB)

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10.05 Independent events After this lesson, you will be able to… • define independent events. • explain that two events A and B are independent if P (AB) = P (A). • use the multiplication rule P (A ∩ B) = P (A) × P (B) to test for independence. • calculate the probability of the intersection of independent events using the multiplication rule. • solve practical problems involving independent events.

Independent events Independent events Two events are independent if knowing the outcome of one event tells us nothing about the outcome of the other event. Two events A and B are independent if the occurrence of one does not affect the probability of the other.

P (AB) = P (A) P (AB) equals P (A) if B’s occurrence does not change A’s probability

P (BA) = P (B) P (BA)

equals P (B) for independent events

Algebraically, if P (AB) = P (A), then: Write the conditional probability formula Use the independence formula P (AB) = P (A)

Multiply both sides by P (B)

Similarly, P (BA) = P (B) implies P (A ∩ B) = P (A) × P (B), showing both conditions are equivalent. For example, rolling a die and ﬂipping a coin are independent; the die’s outcome does not affect the coin’s.

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Exploration Consider rolling a die and drawing a card. Let A be rolling a 6 and B be drawing a heart. 1. Are A and B independent? 2. What characteristics make events independent?

Example 1 A die is rolled, and a coin is ﬂipped. Let A be rolling an even number, and B be ﬂipping heads: a Find P (A ∩ B).

Create a strategy Use the formula P (A ∩ B) = P (A) × P (B) for independent events.

Apply the idea With A = {2, 4, 6} × {H, T }, A = 3 × 2 = 6, so P (A) =

= .

With B = {1, 2, 3, 4, 5, 6} × {H}, B = 6 × 1 = 6, so P (B) =

= .

Write the independence formula

Substitute P (A) = , P (B) =

Evaluate

Reﬂect and check With 6 die outcomes and 2 coin outcomes, the sample space is S = 6 × 2 = 12. With A ∩ B = {2, 4, 6} × {H}, A ∩ B = 3 × 1 = 3. Using these values, the formula

can also be used: Write the formula

Substitute (A ∩ B) = 3 and S = 12

Simplify

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b Show that A and B are independent.

Create a strategy and compare to P (A)=

Using the formula

from part (a).

Apply the idea Write the formula

and P(B) =

Substitute P(A ∩ B) =

Multiply by the reciprocal of the denominator

Simplify

Since P (AB) = P (A), A and B are independent.

Idea summary Independent events satisfy P (AB) = P (A) and P (BA) = P (B), implying P (A ∩ B) = P (A) × P (B).

Application of independent events For independent events, the probability of both occurring is the product of their individual probabilities.

P (A ∩ B) = P (A) × P (B) P (A ∩ B) is the probability of both A and B occurring for independent events If P (A ∩ B) = P (A) × P (B), then A and B are independent, as: Write the formula

Substitute P (A ∩ B) = P (A) × P (B)

Simplify

This is used in practical problems, like repeated trials (e.g., multiple coin ﬂips).

Interactive exploration Discover this concept in action online

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Example 2 A machine produces items, with 2% defective. Two items are selected independently: a Find the probability both are defective.

Create a strategy Use the formula P (A ∩ B) = P (A) × P (B) for independent events.

Apply the idea Let A and B be the ﬁrst and second items being defective, P (A) = P (B) = 0.02. P (A ∩ B) = P (A) × P (B)

Write the formula

= 0.02 × 0.02

Substitute the values

= 0.0004

Evaluate

b Find the probability at least one is defective.

Create a strategy Use the formula P (

∩

) = P ( ) × P ( ), to find P (neither).

Then, use the complement rule: P (at least one) = 1 − P (neither)

Apply the idea P(

∩

) = P( ) × P( )

Write the formula

= (1 − P(A)) × (1 − P (B))

Apply the complement rule for each

= (1 − 0.02) × (1 − 0.02)

Substitute P(A) = P(B) = 0.02

= 0.98 × 0.98

Evaluate the subtraction

= 0.9604

Evaluate

Calculating P (at least one): P (at least one) = 1 − P (neither)

Write the formula

= 1 − 0.9604

Substitute the values

= 0.0396

Evaluate

Reﬂect and check Verify the complement rule sums to 1.

Idea summary For independent events, P (A ∩ B) = P (A) × P (B), used to solve practical problems like repeated trials. If P ( A ∩ B) = P ( A) × P (B), the events are independent, veriﬁable using P ( AB) = P ( A).

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10.05 Practice questions What do you remember? 1

Deﬁne independent events in terms of probability.

Write the formula for the probability of two independent events A and B both occurring.

If P (AB) = P (A), what does this imply about events A and B?

Determine whether each statement is true or false:

Rolling a die and ﬂipping a coin are independent events.

P (A ∩ B) = P (A) + P (B) for independent events.

If P (AB) ≠ P (A), events A and B are independent.

Explain why the formula P (A ∩ B) = P (A) × P (B) holds for independent events.

Practice Ex 1

Ex 2

A spinner has 5 equal sections: 2 red and 3 blue. A card is drawn from a standard 52-card deck. Let A be spinning a red section, and B be drawing an ace: a

Find P (A ∩ B).

Show that A and B are independent.

A weather forecast predicts a 0.25 chance of rain on Monday and a 0.4 chance on Tuesday, independent of each other: a

Find the probability it rains on both days.

Find the probability it rains on at least one day.

A fair coin is ﬂipped, and a six-sided die is rolled. Let A be ﬂipping heads, and B be rolling a 4: a

Find P (A).

Find P (B).

Find P (A ∩ B).

Show that A and B are independent.

A bag contains 5 red and 3 blue marbles. Two marbles are drawn with replacement. Let A be the ﬁrst marble being red, and B be the second marble being red: a

Find P (A).

Find P (B).

Find P (A ∩ B).

Verify independence using P (A ∩ B) = P (A) × P (B).

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A factory has two machines. Machine 1 produces defective items with probability 0.03, and Machine 2 with probability 0.05. Both machines operate independently: a

Find the probability that both machines produce defective items.

Find the probability that neither machine produces a defective item.

The probability it rains on Monday is 0.4, and on Tuesday is 0.3. Assume the events are independent: a

Find the probability it rains on both days.

Find the probability it rains on at least one day.

A card is drawn from a standard deck, and a die is rolled. Let A be drawing a spade, and B be rolling an odd number: a

Determine if A and B are independent.

Calculate P (A ∪ B).

Find P (AB).

Find P (BA).

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Find the probability all three fail.

Find the probability none fail.

A spinner has 4 equal sections numbered 1 to 4, and a coin is ﬂipped. Let A be spinning a 1, and B be ﬂipping tails: Find P (A ∩ B).

Verify independence.

Two independent events have P (A) = 0.6 and P (B) = 0.2: a

Show that A and B are independent.

A quality control test has a 0.1 chance of failing for each item. Three items are tested independently:

a 16

Two events A and B have P (A) = , P (B) = , and P (A ∩ B) =

a 15

Find P (A ∩ B).

Find P (A ∪ B).

A weather forecast predicts a 0.7 chance of sun on Saturday and a 0.6 chance on Sunday, independently: a

Find the probability of sun on both days.

Find the probability of sun on exactly one day.

A test has a 0.05 chance of giving a false positive. Two tests are conducted independently: a

Find the probability both tests give false positives.

Find the probability at least one test gives a false positive.

A fair coin is flipped three times independently. Let A be getting heads on the first flip, and B be getting at least two heads: a

Find P (A ∩ B).

Show that A and B are not independent.

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Extend your thinking 20

Prove algebraically that if P (AB) = P (A), then P (BA) = P (B).

A company tests products with two independent quality checks, each with a 0.02 chance of failing. If a product fails either check, it is rejected:

Find the probability a product is rejected.

If 1000 products are tested, how many are expected to be rejected?

On a game show, a contestant spins two independent spinners, each with 3 equal sections labelled 1, 2, and 3. The contestant wins if the sum of the spins is at least 5: a

Find the probability of winning.

If the contestant plays 10 times, ﬁnd the expected number of wins.

Two events A and B have P (A) = , P (B) = , and P (A ∩ B) = . Are they independent? Explain.

A system has three independent components, each with a 0.95 probability of functioning. The system fails if any component fails: a

Find the probability the system functions.

If 100 systems are tested, how many are expected to function?

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10 Chapter review 1

Two independent events A and B have probabilities P (A) = 0.5 and P (B) = 0.2. What is the probability that event A occurs and event B does not occur? A

How many customers buy only physical games?

How many customers are eligible for the discount?

0.7

A die is rolled. Deﬁne A as rolling a number greater than 2, B as rolling an even number. Determine: A ∩ B

P (A ∩ B)

A box has 5 apple and 7 orange candies. A candy is picked. Calculate: P (apple)

P (not apple)

A card is drawn from a deck of 52 cards. Deﬁne event A as drawing a heart, event B as drawing a jack. Find: P (A ∩ B)

P (A ∪ B)

A card is drawn from 52 cards. Event A is drawing a black card, event B is drawing a queen: a

Are A and B mutually exclusive? Explain.

Find P (A ∪ B).

A die is rolled, S = {1, 2, 3, 4, 5, 6}. Event A is rolling a multiple of 2, event B is rolling an odd number. Find: P (A ∩ B)

P (A ∪ B)

A bag contains 4 blue balls and 3 yellow balls. Two balls are drawn without replacement: a

Draw a tree diagram showing the possible outcomes.

Find the probability of drawing two blue balls.

Find the probability of drawing one blue and one yellow ball.

Find the probability of drawing a yellow ball second given the ﬁrst is blue.

A standard 6-sided die is rolled twice: a

Find the probability the sum of the rolls is 5.

Find the probability the second roll is 2 given the sum is 5.

Use an array to list outcomes where the sum is 5.

A school has 300 members in its sports programme: 180 use the gym, 140 use the pool, and 70 use both. A member is chosen at random: a

496

0.4

How many customers buy only digital games?

a 8

a 6

0.3

a 5

A game store has 150 customers. 90 buy digital games, and 70 buy physical games. 30 buy both. The store wants to offer a discount to customers who buy only one type of game:

a 4

0.1

Find the probability they use the gym given they use the pool.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Use a Venn diagram and a two-way table to show why P (GymPool) ≠ P (PoolGym).

If a new member joins and only uses the gym, how does this affect P (PoolGym)?

A class of 30 students has 18 taking Chemistry, 14 taking History, and 6 taking both. Calculate: a

The probability a student takes Chemistry given they take History.

The probability a student takes History given they take Chemistry.

A survey of 250 concert-goers shows 150 bought a t-shirt, 90 bought a poster, and 60 bought both. Calculate: a

The probability a person bought a t-shirt given they bought a poster.

The probability a person bought a poster given they bought a t-shirt.

A two-way table shows music preferences among 100 people: Pop

Rock

Total

Under 30

30 and Over

Total

100

Calculate the probability a person prefers Pop given they are under 30.

Explain how the table aids in ﬁnding P (Pop ∩ Under 30) and P (Under 30).

A security scan ﬂags 0.98 of dangerous packages and falsely ﬂags 0.05 of safe packages. If 0.02 of packages are dangerous, calculate: a

P (D ∩ F ), where D is dangerous and F is ﬂagged

P (DF )

A bag contains 4 green and 2 yellow marbles. Two marbles are drawn with replacement. Let A be the ﬁrst marble being green, and B be the second marble being green: a

Find P (A).

Find P (B).

Find P (A ∩ B).

Verify independence using P (A ∩ B) = P (A) × P (B).

Two events A and B have P (A) = , P (B) = , and P (A ∩ B) = a

Determine if A and B are independent.

Calculate P (A ∪ B).

Find P (AB).

Find P (BA).

A website has two independent server connections, each with a 0.04 chance of failing. If either connection fails, the website may experience downtime: a

Find the probability the website has downtime (at least one connection fails).

If 10 000 users access the site, how many are expected to experience downtime?

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11 Permutations and combinations Chapter outline 11.01E 11.02E 11.03E 11.04E 11.05E 11.06E 11.07E 11.08E 11.09E 11.10E

500 505 516 523 535 540 548 553 563 571 578

Combining 3 sauces from 7 ? There are 35 weird and wonderful ways to mix them.

11.01E Factorial notation After this lesson, you will be able to… • use the notation n! (read as n factorial), where n! = n(n − 1) (n − 2) … × 3 × 2 × 1 for positive integers n. • use n! = n × (n − 1)! and the convention 0! = 1 in calculations and to simplify algebraic expressions involving factorials. • evaluate numerical expressions involving factorial notation. • simplify algebraic expressions containing factorial notation.

Factorial notation Factorial

Integer

The product of the ﬁrst n positive integers is denoted by n!, read as ‘n factorial’, that is, n! = n(n − 1) (n − 2) × … × 3 × 2 × 1, with the deﬁnition that 0! = 1.

A whole number, positive, negative or zero e.g. −3, −2, −1, 0, 1, 2... The set of integers is usually denoted by Z.

For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Notice that n! contains (n − 1) (n − 2) × … × 3 × 2 × 1 which is (n − 1)!. Thus, n! = n × (n − 1)!. For example, 6! = 6 × 5! = 6 × 120 = 720. Expressions like Similarly,

can be simpliﬁed to

= n × (n − 1) × (n − 2).

Example 1 Evaluate: a

Create a strategy Expand factorials and simplify.

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Apply the idea Expand each factorial

Remove the common factors

Evaluate

Reflect and check Another way to simplify the expression is to use the property n! = n × (n − 1)!. Expand 7! = 7 × 6 × 5!

Remove the common factors

Evaluate

Create a strategy Simplify using factorial properties and evaluate.

Apply the idea Expand 6! = 6 × 5 × 4!

Remove the common factors

Expand the factorial

Evaluate

Simplify

Example 2 Simplify

Create a strategy Expand the denominator and simplify by the common factor.

Apply the idea Expand the denominator

Remove the common factor

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Example 3 For the expression a Express

in simpliﬁed form.

Create a strategy Expand the factorials in the numerator and denominator, remove the common terms, and simplify.

Apply the idea Expand the factorials

Remove the common factor

Simplify

b Evaluate the expression when n = 2.

Create a strategy Substitute n = 2 into the original expression and use the deﬁnition 0! = 1.

Apply the idea Substitute n = 2

Evaluate the subtraction

Use 0! = 1 and expand factorials

Evaluate

Simplify

Reﬂect and check The result will be the same if we use the simpliﬁed form in part (a): Substitute n = 2

Evaluate the square

Evaluate the subtraction

Evaluate

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Idea summary The notation n! represents n factorial, the product of numbers from n to 1. n! = n × (n − 1) × (n − 2) × (n − 3) × … × 3 × 2 × 1 where n is a non-negative integer and n! = n × (n − 1)! By deﬁnition, 0! = 1. simplify to n, and

Expressions like

simpliﬁes to

n × (n − 1) × … × (n − k + 1).

11.01E Practice questions What do you remember? 1

What does the notation n! represent?

Determine whether each statement is true or false:

The value of 0! is 1

3! = 3 × 2!

= n for all positive integers n

Evaluate: 5!

Practice 4

Evaluate: a

Ex 1

3! + 2!

4! − 9

2! × 3!

Evaluate: a

Ex 2

10 × 3!

2 × 4!

Simplify: a

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Ex 3

For the expression a

Write the expression in a simplified form.

Evaluate the expression when n = 3.

Determine whether these expressions are true or false: a

120 = 5 × 4!

(n − 2)!

(n − 1)!

(n + 2)!

Evaluate: b

5! × 3 × 0!

Evaluate: a

(n + 1)!

a 13

6! = 6 × 5!

Evaluate: a

Write in expanded form: a

2! × 0! + 4!

3! × 0! + 5!

Simplify: a

Extend your thinking 15

If (n + 1)! = 5n!, solve for n.

Calculate the value of n if

Simplify: a

= 240.

b = 44 : 3. Calculate the value of n.

It is given that

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3! × 2!

11.02E Multiplication and addition principles After this lesson, you will be able to… • establish and use the multiplication principle for sequential events. • apply the multiplication principle to explain why the number of ways of ordering n distinct objects in a straight line is n!. • use the addition principle for mutually exclusive events. • distinguish between scenarios requiring the multiplication principle and those requiring the addition principle. • solve counting problems by applying the multiplication and/or addition principles, including the use of tree diagrams and position-based reasoning.

The multiplication principle Multiplication principle If one event has m possible outcomes and a second independent event has n possible outcomes, then there is a total of m × n possible outcomes for the two combined events.

The multiplication principle is a key concept in combinatorics used to determine the total number of possible outcomes when multiple options or choices occur in sequence. It simpliﬁes counting by multiplying the number of ways each individual event can happen. This principle is essential for solving problems related to sequences of choices and forms the basis for more complex probability calculations. When the number of possible outcomes is manageable, listing them all can be a helpful way to understand the multiplication principle.

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Visualise multiplication principle with a tree diagram Suppose one type of ice cream, one type of sauce, and one topping for a sundae are chosen from the given list. It can be represented with a tree diagram and the number of branches counted. • Ice cream types: vanilla, chocolate, or strawberry • Sauces: hot fudge or caramel • Toppings: cherries, sprinkles, cookie crumbles, or peanuts Hot fudge Vanilla Caramel

Hot fudge Chocolate Caramel

Hot fudge Strawberry Caramel

Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles Peanuts Sprinkles Cherries Cookie crumbles

The total number of sundae combinations is 24 = 3 × 2 × 4. Each of the ice cream choices would get paired with a sauce to make six options, and then each of those gets one of the four toppings to give 24 possibilities. This shows how the multiplication principle simpliﬁes counting the total number of outcomes without needing to list each combination individually. The multiplication principle can be used, whether listing all outcomes is practical. Visualise multiplication principle using positions Consider a number plate that requires three numbers, followed by two letters, and then another number:

The total number of possible number plates can be calculated using the multiplication principle by considering the possible outcomes for each position on the number plate.

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For each number, there are 10 possibilities (the digits 0−9), while for each letter, there are 26 possibilities. Therefore, the number of possibilities for each position is shown in the image:

Digit

Letter

Digit

For each of the 10 possibilities for the ﬁrst digit, there are 10 possibilities for the second digit. For each of these combinations, there are 10 possibilities for the third digit. Each combination of three digits can be followed by 26 possibilities for the ﬁrst letter and 26 possibilities for the second letter. Finally, each combination of digits and letters can be followed by 10 possibilities for the last digit. Thus, the total number of possible number plates is 10 × 10 × 10 × 26 × 26 × 10 = 6 760 000. This calculation shows how the multiplication principle efficiently determines the total number of possible number plates without listing each outcome individually. Multiplication principle The multiplication principle states that if an event can occur in a different ways, a second event in b different ways, a third event in c different ways, and so on, then the total number of ways these events can occur in sequence is a × b × c × ... . Scenarios that involve the word ‘and’ typically apply the multiplication principle. Some cases might require considering restrictions where digits or letters cannot be repeated.

Example 1 A password contains six characters, which can be lowercase letters or the numbers 0 − 9. Calculate the total number of possible passwords if each letter and number can only be used once.

Create a strategy For each character in the password, the number of possible ways it can be filled needs to be determined. Once the number of possibilities for each character is known, they can be multiplied using the multiplication principle.

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Apply the idea There are 26 letters and 10 numbers, so there are 26 + 10 = 36 possibilities for the ﬁrst character. After choosing a letter or number, the next character will reduce by one possibility. This pattern will continue, so the possibilities for each character will be:

Number of passwords = 36 × 35 × 34 × 33 × 32 × 31 = 1 402 410 240

Multiply the number of ways Evaluate

Idea summary The multiplication principle helps to determine the total number of possible outcomes when multiple choices occur in sequence, regardless of whether it is possible to list all the possibilities. The multiplication principle states: If an event can occur in a different ways, a second event in b different ways, a third event in c different ways, and so on, then the total number of ways these events can occur in sequence is a × b × c × ....

The addition principle The addition principle in combinatorics is used to determine the total number of possible outcomes when choosing between exclusive choices that cannot occur simultaneously. For example, when rolling a die, getting a 3 and getting a 5 are exclusive choices because both cannot occur at the same time. The addition principle simpliﬁes counting by adding the number of ways each event can occur. This principle is particularly useful for solving problems where events do not overlap and forms the basis for more complex probability calculations. Scenarios that involve the word ‘or’ typically apply the addition principle. For example, consider choosing a cold drink from a selection of 3 different types ( juice, milk, water) or a hot drink from a selection of 4 different types (coffee, tea, hot chocolate, and hot cider). Listing all the possibilities for drink choice would give: juice, milk, water, coffee, tea, hot chocolate and hot cider. There is no overlap between the two types of drinks, so the addition principle can be used to say that the total number of choices is 3 + 4 = 7, as each choice of cold or hot drinks is separate and does not overlap. The multiplication and addition principles may be used together.

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Example 2 Dalia won a prize and has the option to choose between a movie voucher with 5 different movies to choose from or a meal voucher with a choice of 3 different types of restaurants. Each option is distinct and does not interfere with the other choices. What is the total number of prize voucher choices Dalia has?

Create a strategy To determine the total number of choices, use the addition principle.

Apply the idea There are 5 different movie tickets and 3 different restaurant vouchers. Applying the addition principle, the total number of choices is calculated by adding the number of options in each category 5 + 3 = 8. Therefore, Dalia has 8 possible choices.

Reﬂect and check The addition principle is used here because Dalia’s choices are separate and do not overlap. Movie vouchers and restaurant vouchers are independent of each other. The number of choices in each category is added because choosing a movie voucher does not affect the number of restaurant vouchers available. The multiplication principle would apply if Dalia got to pick both a movie voucher and a restaurant voucher.

Example 3 A restaurant offers two separate set menus. A meal consists of an entree, main dish and dessert: • Menu 1 has a choice of 4 entrées, 3 mains and 2 desserts. • Menu 2 has a choice of 6 entrées, 2 mains and 5 desserts. Determine the total number of meal combinations that can be ordered.

Create a strategy Use the multiplication principle to calculate the number of possible combinations for both menu 1 and menu 2. Then, apply the addition principle to sum the meal combinations from both menus.

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Apply the idea Apply the multiplication principle: For Menu 1: Meal options for Menu 1 = 4 × 3 × 2 = 24

Multiply the number of options for each course Evaluate

For Menu 2: Meal options for Menu 2 = 6 × 2 × 5 = 60

Multiply the number of options for each course Evaluate

Apply the addition principle: Total meal options = 24 + 60 = 84

Add the number of combinations from both menus Evaluate

There are 84 meal options available.

Idea summary The addition principle helps to determine the total number of outcomes for choices that cannot occur simultaneously by adding the number of ways each choice is possible. Note that when making sequential choices, such as choosing A and then B, the multiplication principle is typically used. However, when choices are presented as A or B or C, the addition principle is suggested.

11.02E Practice questions What do you remember? 1

510

Determine whether each statement is true or false: a

The multiplication principle multiplies the number of ways each sequential choice can occur.

The addition principle is used for choices that happen in sequence.

The word “or” in a scenario suggests the addition principle.

The multiplication principle applies to mutually exclusive choices.

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Match each term or concept to its correct description. a

Multiplication principle

Addition principle

Mutually exclusive choices

Sequential choices

Adds the number of ways each mutually exclusive choice can occur to ﬁnd total outcomes.

Choices made one after another.

Multiplies the number of ways each sequential choice can occur to ﬁnd total outcomes.

Choices that cannot occur at the same time.

What does the word “and” typically indicate in a combinatorics scenario?

Practice 4

Ex 1

Ben has 3 shirts, each in a different colour: crimson (C), pink (P ) and white (W ), and 4 ties, each in a different colour: blue (B), grey (G), red (R) and yellow (Y ): a

Construct a table to show all the possibilities of shirt and tie colours that Ben could wear.

How many different outcomes are possible?

All passwords to log into a particular website consist of any four characters, which can be lowercase letters or the numbers 0−9. Calculate the total number of possible passwords if each letter and number can only be used once.

Ex 2

Sarah can choose either a book from 6 different genres or a gift card from 4 different stores. Each choice is independent of the other. How many total choices does Sarah have?

A company is designing a new employee ID badge. The badge consists of: • A choice of 4 different background colours • A choice of 3 different badge shapes (circle, square, rectangle) • A choice of 5 different font styles for the employee’s name How many different ID badge designs can be created?

A diner offers a special breakfast combo where customers can choose: • 1 type of bread: white, whole wheat, or rye • 1 type of spread: butter, jam, or peanut butter • 1 type of drink: coffee, tea, or juice How many different breakfast combos can a customer choose?

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A hobby shop offers a model-building kit where customers can choose: • One type of model: airplane, car, or ship • One type of paint set: primary, metallic, or pastel colours • One type of tool set: basic, advanced, or deluxe

Ex 3

How many different model-building kits can a customer create?

Each model kit costs $50, each paint set adds $10, and each tool set adds $15. What is the total cost of a model kit with an airplane, a metallic paint set, and a deluxe tool set?

A 4-character password can consist of either four non-repeating letters or four non-repeating numerical digits. Determine the total number of unique passwords that can be formed.

Consider the set A which contains three letters a, b and c, and the set B which contains two letters x and y. Deanna is forming a 3-letter code. Determine the number of codes she can create if:

The ﬁrst letter is from set A, the second letter is from set B, the third letter is from set A, and the same letter can be used more than once.

The first two letters are from set A, without repeating any letter, and the third letter is from set B.

Talal is preparing for a party and wants to choose a stylish outfit. The outfit includes: • n types of shirts, where n is a positive integer • 3 types of pants • 2 types of shoes After exploring all possible options to ensure a perfect look, Talal counts 24 different outfits that can be put together. Determine the number of types of shirts, n, he has?

A student has the option to join one of three different clubs at school: • The Science Club, which has 15 different activities to choose from. • The Art Club, which offers 10 different activities. • The Sports Club, which offers 8 different activities. How many different activities can the student choose from?

A teacher is organising a ﬁeld trip and gives students the choice of visiting one of these locations: • A museum, where there are 5 different exhibits to explore. • A zoo, which has 7 different animal shows. • A botanical garden, which features 4 different guided tours. How many different activities can a student choose from?

A local community centre is organising a day of workshops and activities. Participants can choose to attend one of these options: • Cooking classes, with 6 different recipes to learn • Painting workshops, offering 8 different art styles to explore • Fitness sessions, featuring 5 different types of exercises How many different activities can a participant choose from?

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A school cafeteria is offering a special lunch combo where students can choose one item from each of these categories: • Main course: The options are pizza, burger, or pasta. • Side dish: The options are salad or fries. • Drink: The options are juice or soda. However, due to dietary restrictions, some combinations are not allowed: • Students cannot choose pasta with fries. • Students cannot choose a burger with soda.

How many different lunch combos could a student choose if there are no dietary restrictions?

Considering the dietary restrictions, how many valid lunch combos are available?

A new board game involves creating a character by choosing: • One character type: Knight, Wizard, or Archer • One weapon: Sword, Staff, Bow, or Dagger • One magical ability: Fireball, Healing, Shield, or Invisibility • One mount: Horse, Dragon, or Griﬃn However, there are special rules: • Wizards cannot choose the Sword as their weapon. • Knights cannot choose Invisibility as their magical ability.

How many different character combinations would be possible if there were no special rules?

Considering the special rules, how many valid character combinations are possible?

A new restaurant opened and they offer lunch combos for $8. With the combo meal, the customer will get one salad, one side and one drink. The choices are displayed in a tree diagram: Mashed potato Vegetable Salad

Potato wedges Fruit cup Mashed potato

Fruity Pasta Salad

Potato wedges Fruit cup Mashed potato

Caesar Salad

Potato wedges Fruit cup

Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea Water Iced tea

Count the number of outcomes on the right-hand side of the tree diagram to determine how many different combo meals are possible.

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Angie has 3 tops, 2 bottoms, and 2 types of footwear:

T-shirt

Shorts Skirts

Shorts OUTFITS

Polo Shirt Skirts

Shorts Sweater Skirts

Boots Sneakers Boots Sneakers Boots Sneakers Boots Sneakers Boots Sneakers Boots Sneakers

Count the number of outcomes on the right-hand side of the tree diagram to determine how many different outﬁts Angie has.

Use the multiplication principle to determine how many different outﬁts are possible.

James can select either a t-shirt from 4 different brands or a long-sleeved shirt from 5 different brands. Each choice is independent of the other. Should James use the addition or multiplication principle to determine the total number of choices?

Extend your thinking 21

A debate team can be formed by selecting either one junior and one senior or one junior, one reserve junior and one senior. There are 8 juniors and 5 seniors available. Determine the total number of unique teams that can be created.

Sarah is travelling from Gold Coast to Cairns. There are 4 trains or 3 buses to take her from Gold Coast to Brisbane. There are 2 coaches and 5 ﬂights to take her from Brisbane to Cairns. Determine the number of ways Sarah can travel from Gold Coast to Cairns via Brisbane.

A tech store offers customers these options for a custom computer setup: • Processors: 7 generations of Model 1224, 8 generations of Model 2035, and 11 generations of Model 1000 • Graphics cards: 5 generations of Graphic 14, and 3 generations of Graphic 21 Each processor with an odd model number costs $600, with an even model number costs $500. Each graphics card with a number that is a multiple of 3 costs $450, and costs $700 if its number is not a multiple of 3. How many different computer setups can a customer create with maximum budget of $1000?

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A company is holding a large event, and there are 4 different workshops (A, B, C, and D) being offered to participants. Each workshop covers a different topic, and participants can attend one or more workshops. Participants are required to attend at least one workshop during the entire event (either morning or afternoon or both): Morning

9 : 00 a.m.

Workshop A

Morning

10 : 30 a.m.

Workshop B

Afternoon

1 : 00 p.m.

Workshop C

Afternoon

2 : 30 p.m.

Workshop D

Suppose participants can also choose a speciality activity during the lunch break between the morning and afternoon sessions. There are 3 different speciality activities available, and participants can choose to do one or none of them. Determine and justify how many total combinations of workshops and speciality activities are possible. 25

Design a road network that connects City A to City B, allowing travel directly or via either Town C or Town D, such that there are exactly 20 possible routes in total.

Did you know?

Permutations are like trying on outfits in every possible order! Changing just one item—like shoes or a hat—creates a whole new look. Fashion stylists use permutations to plan wardrobes for photoshoots and runways. They consider all possible arrangements of tops, bottoms, and accessories to craft standout styles. Even subtle changes can create completely different vibes—casual, bold, or elegant. It’s not just creativity at play—it’s clever combinatorics behind the scenes! Next time you mix and match your clothes, you’re actually doing maths without realising it!

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11.03E Permutations After this lesson, you will be able to… • define a permutation as an ordered selection of some or all objects from a set of distinct objects. • use the notation nPr to represent an ordered selection of r objects from n distinct objects. • use the multiplication principle to establish that the number of ordered selections of r objects from n distinct objects is n(n − 1) (n − 2) … (n − r + 1). • show that nPr = n(n − 1) (n − 2) … (n − r + 1) =

• solve problems involving basic permutations of distinct objects.

Permutations Permutation

Distinct

An arrangement of r distinct objects taken from n distinct objects where order is important, where r = 0, 1, 2..., n. The number of such permutations is denoted

Different, not equal.

by nPr, and is given by

The number of distinct ways that a subset of r items can be chosen from a larger set of n distinct items, where the order of arrangement matters, can be determined using the multiplication principle: n

× n−1

× n−2 × 3

n−r+1

rth

This can be simpliﬁed using the permutation notation and formula:

the total number of elements in the set

r the number of elements being selected and arranged Since

can be expressed as

permutation is nPr =

516

, the formula for

= n × (n − 1) × (n − 2) × … (n − r + 1), as required.

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For example, when arranging 2 dots from 4 dots (red, blue, green and yellow), the number of possible permutations is nPr with n = 4 and r = 2, and 4 P2 =

= 4 × 3 = 12.

These 12 possible permutations are:

The calculation of 12 permutations comes from the fact that: • When choosing the first dot, there are 4 options. • After choosing the first dot, 3 options remain for the second dot. This means the number of permutations is 4 × 3 = 12, as shown using the formula, the multiplication principle, or by listing all the options. Notice that for small sets, listing each permutation and counting the total number is manageable. However, for larger sets, the permutation formula, which is based on the multiplication principle, provides a more efficient method. For example, the number of possible permutations of selecting and arranging 5 items from a set of 7 items is nPr with n = 7 and r = 5, which means 7 P5 = permutations.

= 7 × 6 × 5 × 4 × 3 = 2520

When using the permutation formula, the number of items chosen, r, from a larger set of n distinct items must always be less than or equal to n. When r is equal to n, all the items from the set of n distinct items are chosen and arranged. In this case, the permutation formula simpliﬁes because the arrangement of all n items is considered. The permutation formula would be:

So, when r equals n, the number of permutations is simply n!. This is logical since all n items are being arranged, and there are n! possible ways to arrange n distinct items.

Example 1 Consider 7 swimmers in an international race: a Use the permutation formula to determine how many different orders the 7 swimmers can ﬁnish, assuming there are no ties.

Create a strategy Use the permutation formula with n = 7 and r = 7.

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Apply the idea Write the formula

Substitute n = 7 and r = 7

Simplify the denominator

Expand the factorial

Evaluate

Therefore, 5040 different arrangements of swimmers can be formed.

Reﬂect and check Notice that instead of using the formula, it can be calculated directly using a calculator by typing in 7 P7 or using the multiplication principle of 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. b Check the answer to part (a) using a different approach to determine the number of possible arrangements of the 7 swimmers.

Create a strategy To determine the number of possible arrangements of 7 swimmers, consider arranging all 7 swimmers in a sequence and use the multiplication principle.

Apply the idea Choose the ﬁrst swimmer: There are 7 options. Choose the second swimmer: There are 6 remaining options. Choose the third swimmer: There are 5 remaining options. … Continue this process until all 7 swimmers are arranged.

Swimmer 7

Swimmer 6

Swimmer 5

Swimmer 4

Swimmer 3

Swimmer 1

Swimmer 2

This diagrammatic representation shows the step-by-step calculation of the total number of possible arrangements for 7 swimmers:

Thus, the total number of arrangements is calculated as 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. Therefore, 5040 different arrangements of swimmers can be formed. This matches part (a).

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Reflect and check Notice that instead of using the formula, it can be calculated directly using the permutation button of the calculator.

Example 2 From the word SHELF: a How many different ways can 3 letters be selected and arranged?

Create a strategy Since the order of the three letters is important because different orders will produce different arrangements, the permutation formula can be used with n = 5 and r = 3.

Apply the idea Write the formula

Substitute n = 5 and r = 3

Simplify the denominator

Expand the factorial

Evaluate

Therefore, 60 different 3-letter arrangements can be formed from the letters of the word SHELF.

Reﬂect and check Notice that instead of using the formula, it can be calculated directly using a calculator by typing in 5 P3 or using the multiplication principle of 5 × 4 × 3 = 60. b How many different ways can all the letters be arranged?

Create a strategy Since the order of all 5 letters is important because different orders will produce different arrangements, the permutation formula can be used with n = 5 and r = 5.

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Apply the idea Write the formula

Substitute n = 5 and r = 5

Simplify the denominator

Expand the factorial

Evaluate

Therefore, 120 different arrangements can be formed from the letters of the word “SHELF”.

Reﬂect and check Since all 5 letters are being arranged, and there are n! possible ways to arrange n distinct items, the number of possible permutations can be directly expressed as 5!.

Idea summary To determine the number of distinct ways a subset of r elements, chosen from a larger set of n distinct elements, where the order of arrangement matters, the permutation formula

can be used. n elements

r elements

r≤n

When r is equal to n, meaning all items from the set of n distinct items are chosen and arranged, the number of permutations is simply n!.

11.03E Practice questions What do you remember? 1

What is a permutation in combinatorics?

What is the formula for the number of permutations of r items chosen from n distinct items?

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Determine whether these statements are true or false: a

The number of permutations of 4 items chosen from a set of 7 distinct items is given by 4 P7 .

When arranging all 8 books on a shelf, the number of possible arrangements is given by 8!.

The permutation formula 10 P5 =

can be used to calculate the number of ways to

arrange 5 items out of a set of 10. d 4

The permutation formula nPt =

can be used even if r > n.

Match each term or concept to its correct description. a

Permutation formula

Factorial

r≤n

The number of ways to arrange all n distinct items.

The product of all positive integers up to a given number.

iii iv

A condition ensuring the number of items chosen does not exceed the total available.

Practice

Ex 1

Ex 2

How many different ways can the letters of the word FOUR be arranged?

A yard has 3 gates. How many ways can someone enter then exit the yard using two different gates?

A puzzle has 5 unique puzzle pieces each marked with a distinct number 1, 2, 3, 4, 5.

Determine, using the permutation formula, the number of different sequences that can be made by arranging these pieces in a row.

Verify the answer to part (a) using a different approach.

For the word LOGARITHMS: a

How many four-letter arrangements can be formed?

How many ways can all the letters be arranged?

Determine, in terms of n, a simpliﬁed form of: a

Pn − 2

n+5

Pn + 2

Evaluate: a

P12

Seven athletes are participating in a race. In how many ways can the ﬁrst 3 prizes be won?

A music playlist with 12 songs has been downloaded to a phone. Four songs are to be selected and played in order. How many different ways can this be done?

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A company is hosting a conference and has invited 5 speakers. The conference schedule requires that 3 of these speakers be selected to give presentations in a speciﬁc order: a

How many different ways can the company arrange the 3 selected speakers in the schedule?

The company decides to ask 2 of the 3 selected speakers to give two awards. One would give the ﬁrst prize and the other would give the second prize. How many different ways can the prizes be presented?

A local bookstore is organising a reading event and has 8 different books to choose from. The event schedule requires that 4 of these books be selected and presented in a speciﬁc order: a

Determine the number of ways to select and arrange 4 books out of the 8 available.

After selecting and arranging the 4 books, the bookstore decides to feature 2 of these books in a special showcase, in a speciﬁc order. How many different ways can the 2 featured books be selected and arranged from the 4 books already chosen?

A tech company is developing a new app and has 7 different features to choose from. The product team needs to select and prioritise 3 features for the initial release, and the order in which these features are prioritised is crucial: a

Determine the number of ways to select and arrange 3 features out of the 7 available for the initial release.

Once the 3 features are selected, the team decides to highlight 2 of these features in a promotional campaign. How many different ways can the 2 highlighted features be arranged for the campaign?

After the initial release and promotional campaign, the company decides to run a user feedback survey featuring all 3 of the initially selected features. How many different ways can the 3 features be arranged for the survey presentation?

How many numbers between 10 and 10 000 can be formed by using the digits 1, 2, 3, 4, 5, under these conditions: a

Digits can be repeated

Solve for n: a

No digit is repeated in any number

Pn = 8!

P2 = n + 3

Pn − 2 = 5 × 4 × 3

Emily wants to form 4-letter codes using the letters of the word MISTER: a

Calculate the number of 4-letter codes Emily can form if letters can be repeated.

Calculate the number of 4-letter codes Emily can form with no repeated letters.

Calculate the number of 4-letter codes Emily can form with at least one repeated letter.

Explain why this scenario does not involve a permutation, and therefore, the permutation formula cannot be used: Determine how many different ways there are to answer 6 questions in an exam that consists of multiple choice questions where the answer could be A, B, or C.

For a summer party decoration, 5 distinct ribbons (each with a unique colour) are to be hung in a line across the main hall. How many different arrangements are possible if: a

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All the ribbons are used?

At least 4 ribbons are used?

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Extend your thinking Five juniors and ﬁve seniors sit in a row. Determine the number of ways in which they can be seated:

When all the juniors sit together

Such that no two juniors sit together

When all ﬁve juniors are not together

If 56 Pr + 6 : 54 Pr + 3 = 30 800 : 1, calculate the value of r.

If the letters of the word STREAM are arranged in all possible ways and listed in alphabetical order, what is the rank of each of the following arrangements? (Assume the first word in the list has rank 1.) a

REMAST

MASTER

Prove that: nPr = n − 1 Pr + r( n − 1 Pr − 1 )

11.04E Applications of permutations After this lesson, you will be able to… • solve problems involving permutations, including situations where the objects are not all distinct. • solve problems involving permutations with restrictions on the placement of one or more objects.

Restrictions on repeated objects When solving problems that involve permutations, it is important to understand the different types of restrictions that can apply. These restrictions can affect the way objects are arranged and counted, depending on the speciﬁc conditions involved.

Exploration Consider the words “ALL” and “AIL”: 1. For either word, is it possible to rearrange some letters and still get the original word? 2. Which word would have more possible arrangements? 3. List all possible unique arrangements for each word. Describe any observations.

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When objects like letters, digits, or colours are repeated in a permutation, the number of distinct arrangements decreases because the repeated objects are indistinguishable. To account for this, the total number of permutations is divided by the factorial of the number of repeated objects. For example, when considering the word “SUCCESS”, if all the letters were different, the number of possible arrangements would be calculated as 7! = 5040 (since there are 7 letters). However, since the letter “S” repeats three times, and “C” repeats twice, the calculation must be adjusted to =

= 420. Thus, there are 420 distinct ways to arrange the letters in SUCCESS.

This example demonstrates how considering repeated objects in permutations ensures that each distinct arrangement is accurately counted. To determine how to adjust for repeated objects in permutations, follow these steps to ensure that each distinct arrangement is counted only once: 1. Calculate the total number of permutations by taking the factorial of the total number of objects. 2. Identify the repeated objects and count how many times each one appears. 3. Divide the total number of permutations by the factorial of the number of times each repeated object occurs.

Example 1 Consider two boxes, A and B, each containing 7 balls: • Box A: All 7 balls are different colours. • Box B: The 7 balls include 4 red, 2 blue, and 1 green. a Calculate the total number of distinct arrangements for Box A, where all 7 balls are different.

Create a strategy Use the permutation formula with n = 7 and r = 7.

Apply the idea Write the formula

Substitute n = 7 and r = 7

Evaluate

Therefore, 5040 different arrangements of balls can be formed for Box A.

Reﬂect and check This could also be calculated using the multiplication principle as 7! = 5040.

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b Calculate the total number of distinct arrangements for Box B.

Create a strategy To ﬁnd the number of distinct arrangements of 7 balls, where some balls are of the same colour, divide the total number of arrangements for all 7 balls by the factorial of the number of times each colour is repeated.

Apply the idea 1. Calculate the total permutations if all balls were unique. For 7 balls, this is 7! = 5040. 2. Identify the repeated balls and count how many times each type is repeated: Red balls = 4!

Determine how many ways to arrange 4 identical red balls

Blue balls = 2!

Determine how many ways to arrange 2 identical blue balls

Green ball = not repeated

I t does not contribute to repeating arrangements. Its factorial is 1!, which is 1 and does not affect the count.

3. Determine the total number of permutations: Divide by the factorial of the number of repetitions for each type of ball to account for identical arrangements

= 105

Evaluate

Idea summary When objects are repeated in a permutation, the number of distinct arrangements decreases because the repeated objects are indistinguishable. To account for this, the total number of permutations is divided by the factorial of the number of repeated objects.

Restrictions on grouping or adjacent objects When specific objects in a permutation need to stay together or be next to each other, adjust the calculation by treating these objects as a single unit or block. This approach simplifies the arrangement process under the given restriction. For example, consider arranging the letters in the word “MATH” with the condition that the letters A and T must be adjacent. To solve this: 1. Treat the adjacent letters A and T as a single block or unit: Thus, the arrangement problem reduces to arranging three units: the block AT (or TA), the letter M, and the letter H. 2. Calculate the permutations of these three units: Since each unit is distinct, there are 3! = 6 possible permutations. 3. Account for the internal arrangements of the block containing A and T: Since A and T can be arranged as either AT or TA within their block, multiply by 2! to reflect both possible arrangements: 3! × 2! = 6 × 2 = 12. 11.04E Applications of permutations mathspace.co

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This ensures that both possible arrangements of A and T are considered while maintaining their adjacency in the ﬁnal count. Therefore, the number of distinct permutations of the word MATH with the condition that the letters A and T must be adjacent is 12. Here are the 12 distinct arrangements with A and T adjacent: AT as a block:

TA as a block:

1. ATMH

2. ATHM

7. TAMH

8. TAHM

3. MATH

4. HATM

9. MTAH

10. HTAM

5. HMAT

6. MHAT

11. HMTA

12. MHTA

In general, when certain objects in a permutation must be grouped together or be adjacent, the calculation can be adjusted as follows: 1. Treat grouped or adjacent objects as a single unit: Consider the objects that need to be grouped or adjacent as one block or unit. This reduces the problem to arranging fewer units. 2. Calculate permutations of the units: Determine the total number of permutations of these units or blocks, treating each as distinct. 3. Multiply by internal permutations of the block: If the block contains multiple objects, calculate the number of ways to arrange the objects within the block and multiply the result from step 2 by this number. 4. Divide by the factorial of identical objects: If any of the objects (either inside or outside the block) are identical, divide the total by the factorial of the number of identical objects to adjust for over-counting identical arrangements. By using this approach, the permutations accurately reflect the restrictions of grouping or adjacency, ensuring that each arrangement meets the specified conditions. There are situations where a specific group should not be together, like two students who argue not being beside each other in line. In this case, find the total number of permutations and then subtract the number of permutations that fit the criteria.

Example 2 Consider 8 books: 4 language books (English, French, Italian, Mandarin), 2 science books (biology and chemistry), 1 art book, and 1 maths book: a Determine the number of all possible arrangements of these 8 books.

Create a strategy Use the permutation formula with n = 8 and r = 8 or the fact it is an arrangement.

Apply the idea 8! = 40 320

Use factorial for an arrangement

Therefore, 40 320 different arrangements can be formed for the 8 books.

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Reﬂect and check This is the same as: Write the formula

Substitute n = 8 and r = 8

Evaluate

b Determine the number of all possible arrangements with the condition that the 4 language books must always be grouped together.

Create a strategy Consider the 4 language as a block or single unit and calculate the number of possible permutations for 5 blocks: block of language books, 2 science books, 1 art book, and 1 maths book. Then multiply the internal permutations within the block of language books by the possible number of permutations. Use the permutation formula with n = 5 and r = 5.

Apply the idea • The number of possible permutations for 5 blocks: Write the formula

Substitute n = 5 and r = 5

Evaluate

Therefore, 120 different arrangements can be formed for the 5 blocks. • The number of internal possible permutations within the block of language books: The internal arrangements of the block containing the books need also to be considered, since English, French, Italian, and Mandarin can be arranged within their block in any order, which results in 4! = 24 internal possible permutations. There are 24 different ways to arrange the books inside the block. • The number of all possible arrangements under the given condition: 120 × 24 = 2880

Multiply the results to reﬂect both possible arrangements

Therefore, the number of all possible arrangements with the condition that the 4 language books must always be grouped together in any order is 2880.

Reﬂect and check This means that there are 40 320 − 2880 = 37 440 arrangements where the 4 language books are not all together.

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c Determine the number of all possible arrangements with the condition that the 4 language books must be grouped together, and the 2 science books must be grouped together, with both groups arranged in any order.

Create a strategy There are 4 blocks: the block of language books, the block of science books, 1 art book, and 1 maths book. The internal permutations of the blocks of language and the science books need to be considered.

Apply the idea • The number of possible permutations for 4 blocks: Write the formula

Substitute n = 4 and r = 4

Evaluate

Therefore, 24 different arrangements can be formed for the 4 blocks. • From part (b), the number of internal possible permutations within the block of books is 24. • With the same approach, the number of internal possible permutations within the science block is 2! = 2. • The number of all possible arrangements under the given conditions: 24 × 24 × 2 = 1152

Multiply the results to reﬂect both possible arrangements

Therefore, the number of all possible arrangements with the condition that the 4 books be grouped together, and the 2 science books must be grouped together, with both groups arranged in any order, is 1152. d Determine the number of all possible arrangements with the condition that the 4 language books must be in a ﬁxed order (English, French, Italian, Mandarin) and adjacent to the 2 science books, and the 2 science books must be grouped together in any order.

Create a strategy There are now only three blocks, the language and science block, the maths book, and the art book. Within the first block, consider the order the language and science blocks can go in and the order the science books can go in.

Apply the idea For the three blocks, there are 3! = 6 possible arrangements. Within the first block: • There are 2! = 2 arrangements for language and science blocks • There are 2! = 2 arrangements for the science books within their block This gives 6 × 2 × 2 = 24 possible arrangements for these requirements.

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Example 3 At a school, there are 9 students. Five students are wearing red shirts and 4 students are wearing blue shirts. How many arrangements are there if red and blue shirts must alternate?

Create a strategy

Apply the idea

Understand the alternating pattern of red and blue shirts, and then calculate the arrangements for each group separately using the factorial formula n!. Consider starting and ending with a red shirt to maintain the alternating sequence.

Following the alternating pattern RBRBRBRBR, calculate the arrangements for red and blue shirts separately and then multiply these arrangements: For the 5 red shirts, the arrangements are 5!. For the 4 blue shirts, the arrangements are 4!. The total arrangements are given by 5! × 4! which calculates to 120 × 24 = 2880. Therefore, there are 2880 possible arrangements.

Idea summary When certain objects in a permutation need to stay together or be next to each other, treat these objects as one unit or block. First, ﬁnd the number of ways to arrange the blocks. Then, multiply this by the number of ways the objects within the block can be arranged.

Restrictions on selection from multiple groups When selecting objects from different groups with specific rules or limitations, adjust the calculation by treating each group independently while applying the restrictions. Such restrictions often require combining permutations from different groups while respecting the given conditions. For example, imagine you need to select three books to read from three different genres: 3 fiction, 4 non-fiction, and 2 science. You want to pick two fiction books and read them in a specific order and then read one science book, you treat the fiction and science groups separately and the non-fiction is not relevant. First, calculate the number of ways to choose from the fiction books, and then from the science books. Finally, combine these selections. In this case, the total number of valid selections is 3 P2 × 2 P1 = 12. To determine the number of possible selections: • Identify the groups: Determine the distinct groups from which selections need to be made and the number of objects available in each group. • Apply the selection rules: Consider the specific restrictions or rules that dictate how selections must be made from each group and calculate the number of valid selections based on the rules. • Combine the selections: Multiply the number of possible choices from each group according to the rules. The product gives the total number of valid combinations. This method ensures that you accurately account for the restrictions while calculating the total number of possible selections. 11.04E Applications of permutations mathspace.co

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Example 4 Mr. Ti needs to select 5 students for a student council from three different groups: • 7 students from the Science club • 5 students from the Maths club • 8 students from the Debating club a Determine how many possible ways he can form a student council if it is made up of: • A president from the Science club • A vice president from the Science club • A secretary from the Science club • A treasurer from the Maths club • A student-rep from the Maths club

Create a strategy First, determine the number of possible selections from the Science club, then determine the number of possible selections from the Maths club, and ﬁnally combine the permutations.

Apply the idea • The total number of students in the Science club is 7, the number of students to be selected from this group is 3, and order matters because they all have different positions: Write the formula

Substitute n = 7 and r = 3

Evaluate

The number of possible selections of 3 students from the Science club is 210. • The total number of students in the Maths club is 5, the number of students to be selected from this group is 2, and order matters because they have different positions: Write the formula

Substitute n = 5 and r = 2

Evaluate

The number of possible selections of 2 students from the Maths club is 20. Since the selection from each group is independent of the other, the permutations must be multiplied. The number of possible student councils if Mr. Ti must select exactly three students from the Science club and two students from the Maths club, considering the position for which they are selected, is: Number of student councils = Selections from Science club × Selections from Maths club = 210 × 20 = 4200 Therefore, 4200 different student councils can be formed under the given condition.

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b How many ways can Mr. Ti form a student council of 5 students if the president or the studentrep must be selected from the Debating club, and that the remaining 4 positions are selected from the Science club and Maths club?

Create a strategy First, determine the number of possible ways to select a student from the Debating club. Next, calculate the number of possible selections from the Maths club and Science club combined. Finally, multiply the results by the number of ways to arrange the two blocks: the block containing one student and the block containing four students.

Apply the idea • The number of possible selections of a student from the Debating club is 8 P1 = 8 • Since there are two possible positions for the single student from the Debating club the number of possible permutations is 2! = 2. • The total number of students in the Science club and Maths club is 12, the number of students to be selected from this combined group is 4: Write the formula

Substitute n = 12 and r = 4

Evaluate

The number of possible selections of the remaining 4 students is 11 880. The number of possible student councils under the given condition, is: Number of student councils = 8 × 2 × 11 880 = 190 080

Multiply the results Evaluate

Therefore, 190 080 different student councils can be formed under the given condition.

Idea summary When selecting objects from different groups with specific rules or limitations, adjust the calculation by treating each group independently while applying the restrictions. Such restrictions often require combining permutations from different groups while respecting the given conditions.

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11.04E Practice questions What do you remember? 1

Determine the number of unique ways the word COMMITTEE can be arranged.

How is the number of distinct permutations adjusted when objects in an arrangement are repeated?

Determine whether these statements are true or false: a

When arranging letters like in the word SUCCESS, the total permutations are divided by the factorials of the number of times each letter is repeated.

To arrange objects where some must be adjacent, treat those objects as a single unit and multiply by their internal arrangements.

When selecting objects from multiple groups, the permutations from each group are added together to ﬁnd the total.

To ﬁnd arrangements where certain objects are not together, subtract the number of arrangements where they are together from the total permutations.

Practice 4

How many ways are there to arrange the letters A, B, C and D if C and D must always be adjacent? How does your answer change if, in addition, C must always come before D?

Consider the word MACHINE. How many ways are there to order the letters if the vowels must be grouped together?

Ex 1

Consider a bag that contains 4 red beads, 3 blue beads, and 2 green beads. Calculate the total number of distinct arrangements of the 9 beads in a straight line.

Ex 2

After returning from her vacation, Sarah has 6 souvenirs to distribute among 6 different family members. The souvenirs include: • 2 chocolate boxes (truﬄes and caramels) • 3 key rings (ocean, beach, and sunrise) • 1 greeting card

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Calculate the total number of different ways Sarah can distribute the 6 souvenirs among her family.

If the 2 chocolate boxes must be given to her two parents (in any order), how many different ways can she distribute the 6 souvenirs?

How many ways can the souvenirs be distributed if the 2 chocolates are given to the two parents and the 3 key rings are given to the three siblings?

How many arrangements are possible if the truﬄes must go to her mom and the caramels must go to her step-dad, while the 3 key rings can be distributed freely among her three siblings?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider a set of 10 different-coloured tokens where there are 4 orange tokens, 3 purple tokens, 2 yellow tokens, and 1 green token: a

Determine the total number of distinct permutations of the 10 tokens.

Determine the number of distinct permutations of the 6 remaining tokens that do not include any of the orange tokens.

A librarian needs to arrange 12 books on a shelf. There are 5 identical mystery novels, 4 identical science ﬁction novels, 2 identical biographies, and 1 cookbook: a

Determine the total number of distinct permutations of the 12 books on the shelf.

If 3 mystery novels and 2 science ﬁction novels are sold, determine the number of distinct permutations of the remaining books.

Ex 3

In how many ways can all the digits 1, 2, 3, 4, 5, and 6 be arranged to form a six-digit number such that the odd digits can only occupy even positions?

Ex 4

Ms. Lee needs to select 6 people from three different departments to make a presentation. The order of the presentations is important. There are: • 9 employees from the Marketing department • 6 employees from the Finance department • 10 employees from the Operations department

Determine how many possible ways she can form a presentation order if the first four presentation slots must be filled by members of the Marketing department and the last two presentations slots must be filled by members of the Finance department.

How many ways can Ms. Lee form a presentation order if the ﬁrst and last presentations must be by members of the Operations department and the middle four slots can be ﬁlled with employees either from the Marketing and Finance departments?

Consider the number 11 223 344, which consists of 8 digits: a

Determine the number of distinct permutations of all 8 digits.

Determine the number of distinct permutations using only the odd digits.

At a school there are 10 students. How many ways are there to arrange the students if Alice and Bob must be placed on the ends?

In how many ways can 4 tourists and 3 locals board a train, if everyone boards one at a time and the locals board before the tourists?

If the letters of the word ORANGE are rearranged, how many different arrangements are possible if the vowels must alternate with consonants?

A group of 7 friends are seated at a table with 7 seats. If two of the friends insist on sitting together, how many ways are there to seat the friends?

A group of 6 oﬃce workers must stand together for a photo: a

How many ways can they be lined up for the photo?

How many ways can they be lined up if two of the workers, Josh and James, stand beside one another?

How many ways can they be lined up if Josh and James refuse to stand next to each other?

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The digits 1, 2, 3, 4, and 5 must be arranged to form a 5-digit number. If the resulting number must be odd, how many possible numbers can be formed?

20 people are selected to sit along a long banquet table. Of these 20 people, 5 are related and must sit together. How many ways can the people be arranged? Give your answer as a product of factorials.

How many different arrangements can be made from the letters of the word CHEESY if the two E’s must always be adjacent?

Ms. Smith needs to select 8 students from three different grade levels to perform in a school concert. The order of performances is important. There are: • 8 students from Grade 10 • 7 students from Grade 11 • 5 students from Grade 12

How many different ways can a performance order be arranged if four Grade 10 students perform ﬁrst, followed by two Grade 11 students, and then two Grade 12 students?

Determine how many possible ways she can form a performance order if the first three performance slots will be filled by students from Grade 10 and the last three slots will be filled by students from Grade 11 and the middle two are filled by any students.

How many ways can Ms. Smith form a performance order if the ﬁrst and last performances must be from Grade 12 students, and the middle six slots can be ﬁlled with students from either Grade 10 or Grade 11?

There are 6 boys who enter a boat with 8 seats, 4 on each side. John and Michael must sit on the port side. How many ways are there for the boys to be seated on the boat?

Extend your thinking 23

A jeweler arranges 9 beads in a straight line to make a bracelet: 4 identical ruby beads, 3 identical sapphire beads, and 2 distinct emerald beads. If the ruby beads must be grouped together, how many distinct arrangements are possible?

Consider the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. Using each element no more than once, how many three-digit numbers is it possible to form such that summing the digits gives 15?

If the letters of the word STATISTICS is rearranged, how many different arrangements are possible if the vowels must be grouped together?

In chess, a rook can move as many spaces as it wants forwards, backwards, left, or right, but not diagonally. A chessboard is 8 × 8 squares. In how many ways can 8 identical rooks be placed on a chessboard so that no two rooks attack each other?

A group of 7 students (A, B, C, D, E, F, G) are arranged in a line. How many distinct arrangements are possible if students A and B must be adjacent, and students C and D must also be adjacent?

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Consider functions of the form f : {1, 2, …, n} → {1, 2, …, n, …, n + k}. The function f is said to be increasing if for all a, b in the domain, if a < b, then f (a) < f (b). A function is injective if every input value in the domain maps to exactly one output value in the codomain, so if f (a) = f (b), then a = b, and if a ≠ b, then f (a) ≠ f (b): a

How many injective, increasing functions of this form are there when n = 4, k = 2?

How many injective, increasing functions of this form are there when n > 2 and k > 1? Give your answer in terms of n, k.

11.05E Circular arrangements After this lesson, you will be able to… • explain why the number of ways to arrange n distinct objects in a circle is (n − 1)!. • solve problems involving circular arrangements of distinct objects with or without restrictions on the placement of one or more objects.

Circular arrangements formula Circular arrangements involve placing distinct objects around a circle, where arrangements that can be obtained by rotation are considered identical. For example, seating 3 people around a circular table in the order A-B-C is the same as B-C-A or C-A-B when rotated. To determine the number of distinct circular arrangements of n objects, consider that in a straight line, the number of permutations is n!. However, in a circle, each arrangement can be rotated n times to produce the same pattern. Thus, the number of unique circular arrangements is calculated by dividing the total linear permutations by the number of rotations:

where n! represents the number of ways to arrange n distinct objects in a line. Dividing by n accounts for the n possible rotations of each arrangement, leaving (n − 1)! as the number of distinct circular arrangements. This formula, (n − 1)!, applies when all objects are distinct and there are no restrictions on their placement unless speciﬁed otherwise.

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Example 1 Calculate the number of ways to arrange 5 distinct people around a circular table.

Create a strategy Use the formula for circular arrangements of distinct objects: (n − 1)!, where n is the number of objects.

Apply the idea Substitute n = 5 into the formula: (n − 1)! = (5 − 1)!

Substitute n = 5

= 4!

Simplify the expression

=4×3×2×1

Expand the factorial

= 24

Evaluate

Therefore, there are 24 distinct ways to arrange 5 people around a circular table.

Reflect and check To verify, note that 5! = 120 linear arrangements divided by 5 rotations equals 24, matching the result from (5 − 1)!.

Idea summary The number of distinct circular arrangements of n distinct objects is given by (n − 1)! This accounts for rotations, where each circular arrangement is considered identical under any rotation.

Circular arrangements with restrictions When restrictions are imposed on circular arrangements, such as fixing the position of one or more objects, the approach adjusts accordingly. If one object’s position is fixed (e.g., a specific person must sit in a designated seat), the circular symmetry is broken, and rotations no longer produce identical arrangements. In such cases, fix the position of the restricted object and arrange the remaining n − 1 objects in a linear sequence relative to it. The number of ways to do this is (n − 1)!. Here, n is the total number of objects, and fixing one object leaves n − 1 objects to be arranged in (n − 1)! ways. If additional restrictions apply (e.g., two objects must be adjacent), treat those objects as a single unit initially, then adjust for internal arrangements within that unit, while still considering the circular nature of the problem.

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Example 2 Find the number of ways to arrange 6 distinct students around a circular table if one student, Alex, must sit in a speciﬁc chair.

Create a strategy Fix Alex’s position, then arrange the remaining 5 students in the remaining seats using (n − 1)!.

Apply the idea With Alex ﬁxed in one position, arrange the other 5 students: (n − 1)! = (6 − 1)!

Substitute n = 6

= 5!

Simplify the expression

=5×4×3×2×1

Expand the factorial

= 120

Evaluate

Thus, there are 120 ways to arrange the students with Alex in a speciﬁc chair.

Example 3 Calculate the number of ways to arrange 7 distinct flowers in a circular wreath where two specific flowers, Rose and Lily, must be adjacent.

Create a strategy Treat Rose and Lily as a single unit, reducing the problem to arranging 6 units in a circle, then account for the internal arrangements of Rose and Lily within that unit.

Apply the idea Step 1: Arrange 6 units (the Rose-Lily pair as one unit, plus the other 5 ﬂowers) in a circle: (n − 1)! = (6 − 1)!

Number of ways to arrange 6 units

= 5!

Simplify

= 120

Evaluate

Step 2: Within the Rose-Lily unit, they can be arranged in 2 ways (Rose-Lily or Lily-Rose): 2! = 2

Number of ways to arrange Rose and Lily

Total arrangements: 120 × 2 = 240

ultiply the circular arrangements by the internal M arrangements

Therefore, there are 240 ways to arrange the 7 ﬂowers with Rose and Lily adjacent.

Reﬂect and check Without the adjacency restriction, the number of arrangements would be (7 − 1)! = 6! = 720. The restriction reduces this to 240, which is consistent as fewer arrangements satisfy the condition.

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Idea summary For circular arrangements with one object ﬁxed, the number of ways is (n − 1)! With additional restrictions like adjacency, treat restricted objects as a unit, calculate circular arrangements of the reduced number of units, and adjust for internal arrangements within the unit.

11.05E Practice questions What do you remember? 1

What is the formula for the number of distinct circular arrangements of n distinct objects when no positions are ﬁxed?

Why are rotations considered identical in circular arrangements?

Determine whether these statements are true or false: a

The number of distinct circular arrangements of n distinct objects is given by n!.

If one object is ﬁxed in a circular arrangement of n distinct objects, the number of ways to arrange the rest is (n − 1)!.

In a circular arrangement, rotations of the same pattern are considered distinct.

If two objects must be adjacent in a circular arrangement of n distinct objects, they are treated as separate units.

Practice

Ex 1

Ex 2

If n = 3, calculate the number of distinct circular arrangements of n distinct objects using the formula for circular arrangements.

Calculate the number of ways to arrange 8 distinct volunteers around a circular community table.

How many distinct circular arrangements can be made with 8 different coloured beads on a bracelet?

Determine the number of ways to seat 3 distinct delegates around a round conference table.

Find the number of ways to arrange 6 distinct books in a circular bookshelf if one book, “History,” must be in a ﬁxed position.

Find the number of ways to arrange 7 distinct musicians around a circular stage if one musician, Emma, must sit in a designated spot.

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Ex 3

A teacher and 9 different students are to be seated at a round table with 10 chairs. If the teacher’s position is fixed to a specific chair, how many different ways can the students be arranged in the remaining seats?

Calculate the number of ways to arrange 9 distinct gems in a circular pendant where two speciﬁc gems, Ruby and Sapphire, must be adjacent.

A family of 6 distinct members sits around a circular dining table: a

How many ways can they be arranged with no restrictions?

How many ways if the parents must sit together?

How many distinct ways can 7 unique ornaments be arranged in a circular pattern if one ornament, a star, is ﬁxed at the top?

Eight distinct team members are seated around a circular table: a

How many ways can they be arranged without restrictions?

How many ways if two speciﬁc members, Alice and Bob, must be adjacent?

Calculate the number of ways to arrange 5 distinct trophies in a circular display if the gold trophy must be in a ﬁxed position.

Six distinct friends sit around a circular campﬁre: a

How many distinct arrangements are possible?

How many if three friends, Emma, Jacob, and Liam, must sit together?

How many ways can 10 distinct players be arranged around a circular game table if the referee must occupy a speciﬁc seat?

Find the number of distinct circular arrangements of 4 unique instruments in a band circle if the drum and guitar must be adjacent.

Extend your thinking 19

How many ways can 8 distinct guests be seated around a circular table if three speciﬁc guests—Tom, Jane, and Mike—must sit together as a group?

Nine distinct chefs participate in a circular cooking demonstration: a

How many ways can they be arranged if the head chef is ﬁxed in one position?

How many ways if two sous-chefs must be adjacent and the head chef is ﬁxed?

Calculate the number of distinct circular arrangements of 7 unique artifacts if two pairs (A-B and C-D) must each be adjacent, but the pairs can be arranged freely around the circle.

How many ways can 6 distinct dancers be arranged in a circular formation if two dancers, Mia and Noah, must not be adjacent?

Determine the number of distinct circular arrangements of 10 unique cards around a table such that two speciﬁc cards are not adjacent and three speciﬁc cards must form a consecutive group. 11.05E Circular arrangements mathspace.co

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11.06E Combinations After this lesson, you will be able to… • define a combination and use the notation nCr or

to represent the

number of ways of selecting a subset of r objects from n distinct objects, where order is not important. • establish and use the formula nCr =

• show that nCn = nC0 = 1 and nC1 = nCn − 1 = n.

Combinations Exploration Currently, a school has a president and vice-president for the student council. They are thinking about changing this to be two co-presidents: 1. If Xavier was president and Yuri was vice-president, is this the same as Yuri being president and Xavier being vice-president? 2. If Xavier and Yuri were co-presidents, is this the same as Yuri and Xavier being co-presidents? 3. If there were 5 students running for the two spots on student council, how would the number of possible student council set-ups compare for the president-vicepresident versus co-president models?

Combination A selection of r distinct objects from n distinct objects, where order is not important. The number of such combinations is given by

, the binomial coeﬃcient nCr or

Recall that if the order matters for a context, it is a permutation. If the order does not matter, just the selection, then it is called a combination. A combination just involves selecting or choosing the items. However, if the items from a combination are arranged, we can consider permutations as an ordered version of the combination.

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Consider an example involving three distinct coloured dots: red, blue, and green. If two dots are to be selected from these three, the possible permutations of dots are as shown: The possible selections of dots have been identiﬁed. The next step is to determine if the order of the dots matters. For example, is a red dot paired with a green dot considered the same as a green dot paired with a red dot?

Start with

If the answer is no and (red, green) is considered different to (green, red), then the order matters, and we have a permutation and the answer is 6.

Start with

If the answer is yes, (red, green) and (green, red) are considered the same, then the order does not matter and we have a combination and the answer would be 3.

The process of calculating the number of possible combinations is key to counting techniques in mathematics. For smaller sets, it is possible to list all the options and count them manually. However, as the sets grow larger, using manual counting becomes impractical. Consider a simple lotto (lottery) game where there are 10 balls and a prize for choosing the 3 numbers that are drawn. It doesn’t matter what order the balls are drawn, or what order the numbers are picked as long as at the end of the draw, the numbers on the ticket match the numbers that were drawn.

4 6 7 10

2 1

To calculate the chance of winning the game, start by calculating the number of possible arrangements.

5 9 3

Since there are 10 options for the ﬁrst ball, followed by 9 options for the second ball, and 8 options for the third ball, this gives an initial result of: 10

P3 = 10 × 9 × 8 = 720

However, this count includes multiple arrangements of the same combination, such as (5, 6, 7) being counted separately from: (5, 7, 6), (6, 7, 5), (6, 5, 7), (7, 5, 6) and (7, 6, 5), all of which are actually the same with regard to the lotto game. To correct for this overcounting, it is necessary to account for the fact that each combination of three balls can be arranged in 3! = 3 × 2 × 1 = 6 ways. So each combination was counted six times instead of once.

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To determine the number of unique combinations, divide the total number of arrangements, 720, by the number of ways each combination can be arranged, 6. Going back to the factorials, this gives:

So 120 different combinations. Thus, the chance of winning the game with a single ticket is

Based on this example, consider the process for choosing r items from n distinct items, where order is not important. This can be done using the combination formula:

r n

is the number of items being selected is the total number of items to choose from

The notation nCr can be read as “n choose r” and means the number of combinations of r items chosen from a set of n items. For example

, means that there are 10 ways to

choose 2 items from a set of 5 distinct items. Other notations that are used in combinations include C(n, r), nCr, nCr, and same as nCr.

Example 1 Evaluate: 7C2

Create a strategy Substitute n = 7 and r = 2 into

Apply the idea Write the formula

Substitute n = 7 and r = 2

Evaluate the subtraction

Expand the numerator

Remove the common factor

Expand the denominator

Evaluate the multiplication

Evaluate

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, which all mean the

Reflect and check The combination button on the calculator can also be used to verify this answer.

Example 2 A manager wants to select one group of 4 people from the 28 staff. How many different groups are possible?

Create a strategy Use the combination formula:

Apply the idea For selecting a group of 4 people, substitute n = 28 and r = 4 into

Write the formula

Substitute n = 28 and r = 4

Evaluate the subtraction

Expand the numerator

Remove the common factor

Expand the denominator

Evaluate

Idea summary Combinations involve selecting or choosing a number of items from a set, where the order of the items is not important. The combinations formula is:

n is the total number of objects r is the number of objects we are selecting

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Properties of combinations There are some properties of combinations that can help build a deeper understanding of combinatorics.

Exploration Consider the combinations 6 C2 and 6 C4 : 1. Show the working for calculating each combination. What do you notice? 2. Interpret the two combinations. Explain how they are related. 3. Write a combination that would be equivalent to 7C2 . Explain your answer. 4. Write a combination that would be equivalent to nC(n − r). Explain your answer.

First, it can be shown that nCr =

= nCn − r .

This means, for example that 12 C4 = 12 C8 and 7C1 = 7C6 . There is a reason behind this nice symmetry. This property arises because choosing r items from n items and leaving out (n − r) items from the set, is the same as choosing (n − r) items to exclude from a set of n item and keeping the remaining r items. Another property is that nC0 = 1. This makes sense contextually, because if no items are selected, there is only one way to do that. To show this is true, recall that 0! = 1 and 1! = 1, we have by deﬁnition that

This means, for example, 5 C0 = 1 and 12 C0 = 1. This also means that nCn = 1. Another useful property is nC(n − 1) = n, this arises from the fact that selecting n − 1 items from a set of n is equivalent to choosing one item to exclude, and there are n ways to exclude one item. Finally, there is a relationship between nCr and the previous nPr. Contextually, a permutation is selecting and then ordering, while a combination is just selecting. This means that we need to divide out all the duplicates if order is ignored. This gives algebraically:

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. This can also be shown

Example 3 Determine the value of x: a

C6 = 15 Cx

Create a strategy Use the property: nCr = nCn − r

Apply the idea Using the property nCr = nCn − r , we can ﬁnd the values of x for 15 C6 , where n = 15 and initially r = 6. We determine x in two ways: 1. Directly from the problem, we have x = r = 6. 2. Using the property nCr = nCn−r, the equivalent index is x = n − r: x=n−r

Write the formula

= 15 − 6

Substitute the values

Evaluate

Therefore, x = 6 and x = 9. b

Cx = 1

Create a strategy Use the property: nC0 = nCn = 1

Apply the idea For the number of selections to equal 1, we are either selecting 0 items or all of the items. Therefore, x = 0 or x = 5.

Idea summary These properties may be helpful in justifying or calculating combinations: • • •

Cr = nCn − r C0 = nCn = 1 n C(n − 1) = n n

•

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11.06E Practice questions What do you remember? 1

Deﬁne a combination and explain how it differs from a permutation.

State the formula for nCr and deﬁne its variables.

Show that nC1 = n using the combination formula.

Show that nC0 = 1 using the combination formula.

Practice Ex 1

Evaluate: a

Ex 2

C10

100

C99

A book club has 5 new books to read this month. In how many ways can they choose 2 books for their ﬁrst meeting?

A gardener has to plant 3 types of ﬂowers from a selection of 5 types. In how many ways can he choose the ﬂowers to plant?

A pizza shop offers 7 toppings. How many different pizzas can you create with any 2 toppings?

A dessert shop offers 6 different ﬂavours of ice cream. Determine the number of combinations that can be selected if a customer purchases: a

2 ﬂavours

3 ﬂavours

4 ﬂavours

5 ﬂavours

A team of 6 people is to be selected from a group of 10 athletes. How many different combinations of teams can be formed?

A fruit basket contains 8 different fruits. How many ways can you select 3 fruits to make a fruit salad?

During a school ﬁeld trip, 4 students out of 10 need to be selected for a quiz competition. How many possible selections are there?

A photographer has to select 5 photos out of 10 to include in his portfolio. In how many ways can he make his selection?

Simplify: a e

546

Cn − 2

b f

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d h

Ex 3

Show that nCr = nCn − r.

Determine the value(s) of y: a

Cy = 14 C3

Cy = 1

C6 = yC14

C6 = 7

In a chess tournament, there are 10 players. Each player plays against every other player exactly once. How many different matches will be played in the tournament?

Twelve students are divided into two teams, Team A and Team B, each consisting of 6 students. Since the teams are named, they are distinct. Determine the number of ways this can be done.

At a networking conference, every person shakes hands with every other person. Altogether there are 105 handshakes. How many people are at the conference?

Prove that n − 2 Ck − 2 × n × (n − 1) = nCk × k × (k − 1), where 2 ≤ k ≤ n.

Extend your thinking 21

Show that each identity is true for any whole numbers, r and n, where 0 ≤ r ≤ n. a

Cn = 1

n+1

Cr = nCr + nCr − 1

How many ways are there to form a triangle in the grid using only three points?

How many 8-letter arrangements can be created if each combination has 3 vowels and 5 consonants?

Show that this identity holds: (nCr + s) (r + sCr) = (nCr) (n − rCs) where n, r and s are positive integers and n ≥ r + s.

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11.07E Proofs involving combinations After this lesson, you will be able to… • show that nCr = nCr − r , for 0 ≤ r ≤ n, by selecting r objects from n distinct objects for inclusions and n − r objects from n distinct objects for exclusion. • prove nCr = n − 1 Cr − 1 + n − 1 Cr for 1 ≤ r ≤ n − 1 algebraically and using combinatorial arguments. • use known combinatorial relationships to show or prove statements

Proofs involving combinations Combinations count the number of ways to choose r objects from n distinct objects without regard to order, denoted nCr. This lesson explores two key properties of combinations through proofs, using both algebraic and combinatorial methods, to deepen understanding of combinatorics. We will prove: 1.

Cr = nCn − r for 0 ≤ r ≤ n

2. nCr = n − 1Cr − 1 + n − 1Cr for 1 ≤ r ≤ n − 1

Example 1 Prove that nCr = nCn−r for 0 ≤ r ≤ n using: a Combinatorial proof

Create a strategy Use the deﬁnition of combinations and consider choosing objects to include versus exclude for the combinatorial proof.

Apply the idea Choosing r objects from n to include is equivalent to choosing n − r to exclude, as each selection of r leaves n − r out. Thus, nCr (ways to include) equals nCn − r (ways to exclude).

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b Algebraic proof

Create a strategy Manipulate the combination’s formula.

Apply the idea Write the formula

Substitute r = n − (n − r)

Swap terms in the denominator

Substitute

Example 2 Prove that nCr = n − 1Cr − 1 + n − 1Cr for 1 ≤ r ≤ n − 1 using: a Combinatorial proof

Create a strategy Split the selection into cases based on including or excluding a speciﬁc object.

Apply the idea To choose r from n objects, consider one object, A for example. Case 1: Include A, then choose r − 1 from the remaining n − 1 → n − 1Cr − 1. Case 2: Exclude A, then choose r from the remaining n − 1 → n − 1 Cr. Total ways: n − 1 Cr − 1 + n − 1Cr. b Algebraic proof

Create a strategy Combine n−1 Cr−1 + n−1 Cr using the combination formula and simplify to nCr.

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Apply the idea Write the formula Rewrite r! = r × (r − 1)!

Rewrite (n − r − 1)! = (n − r − 1) × (n − r)!) to align denominators Adjust ﬁrst term to have common denominator r(r − 1)!(n − r)!

Combine fractions

Factor out (n − 1)!

Simplify r + (n − r) = n

Substitute (n − 1)! × n = n!

Substitute

Reﬂect and check Alternatively, we can prove this by starting with nCr and splitting it into two terms: Write the formula

Rewrite n! = n × (n − 1)!

Split n = r + (n − r) and distribute

Separate into two fractions

Simplify r! = r × (r − 1)! and (n − r)! = (n − r) × (n − r − 1)!

Substitute

and

This approach requires splitting the numerator, which can be less intuitive than combining terms.

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Idea summary Combinations satisfy two key identities: •

Cr = nCn − r : Choosing r to include is the same as choosing n − r to exclude.

•

Cr = n − 1 Cr − 1 + n − 1 Cr : Total ways to choose r split into including or excluding one object.

These hold for 0 ≤ r ≤ n and 1 ≤ r ≤ n − 1, respectively, proven combinatorially and algebraically.

11.07E Practice questions What do you remember? 1

Write the combinations formula for nCr and explain how it relates to choosing objects without order.

A club has n members. How many ways can you choose r members to form a committee? Express using the combinations formula.

Practice Ex 1

Prove that a

Ex 2

for 1 ≤ r ≤ n using:

Combinatorial proof

Algebraic proof

Prove that n + 1 Cr = nCr − 1 + nCr for 1 ≤ r ≤ n using: a

Combinatorial proof

Algebraic proof

Verify that 8 C2 = 8 C6 by calculating both sides using the combination formula.

A team of 5 is selected from 9 players. Calculate 9 C5 and verify using 8 C4 + 8 C5.

Prove that nCr = nCn − r for 0 ≤ r ≤ n using: a

Prove that a

Combinatorial proof

Algebraic proof

for 1 ≤ r ≤ n using: b

Algebraic proof

Prove that n + 2 Cr = n + 1 Cr − 1 + n + 1 Cr for 1 ≤ r ≤ n + 1 using: a

Combinatorial proof

Algebraic proof

Verify that 7C4 = 7C3 by calculating both sides using the combination formula.

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Prove that nCr = n − 1 Cr − 1 + n − 1Cr for 1 ≤ r ≤ n − 1 using: a

Algebraic proof

Combinatorial proof

Algebraic proof

for 1 ≤ r ≤ n using: b

Algebraic proof

Combinatorial proof

Algebraic proof

Prove that nCr − 1 = n − 1 Cr − 1 + n − 1 Cr − 2 for 2 ≤ r ≤ n using: a

Prove that n+2 Cr + 2 = nCr + 2 × nCr + 1 + nCr + 2 for 0 ≤ r ≤ n − 2 using: a

Combinatorial proof

Prove that a

Algebraic proof

Prove that (n − r)nCr = nn − 1 Cr for 0 ≤ r < n using: a

Prove that n + 1 Cr + 1 = nCr + nCr + 1 for 0 ≤ r ≤ n − 1 using: a

Combinatorial proof

Algebraic proof

Prove the identity r × nCr = n × n − 1 Cr − 1 for 1 ≤ r ≤ n using: a

Combinatorial proof

Algebraic proof

Extend your thinking 18

A team of r players is to be chosen from n available players, including a specific player, Alex. Prove combinatorially that the number of ways to form the team is n − 1 Cr − 1 + n − 1 Cr for 1 ≤ r ≤ n − 1.

Prove that a

Combinatorial proof

for 0 ≤ r ≤ n − 1 using: b

Algebraic proof

A committee of 4 is chosen from 9. By considering two separate groups show the total ways C(9, 4) is equal to C(4, 0)C(5, 4) + C(4, 1)C(5, 3) + C(4, 2)C(5, 2) + C(4, 3)C(5, 1) + C(4, 4)C(5, 0).

Show that

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combinatorially, and verify with n = 4.

11.08E Applications of combinations After this lesson, you will be able to… • solve problems involving combinations with or without restrictions on the selection of one or more objects.

Combinations with the multiplication and addition principle The notation nCr and its formula are used to determine combinations where the order of items does not matter. The concept of combinations can be extended to more complex scenarios involving the selection of subsets under additional constraints and speciﬁc rules. The addition rule can be used to add up the number of combinations for cases that have no intersection or overlap. In cases where multiple independent choices are involved, the multiplication principle can be applied to calculate the total number of possible combinations. Consider, for example, a group of 9 people from which a selection of 4 people is to be made. Suppose the group must include either the youngest or the oldest person, but not both. How many possible selections meet this restriction? The total number of ways to choose 4 people from 9 without restrictions is

oldest

youngest

= 126.

However, this does not account for the restriction that the group must include either the youngest or the oldest person, but not both and the problem needs to be broken into cases: • Case 1: The youngest person is included, and the oldest is not. In this case, the youngest person occupies one of the 4 spots in the group, which can be done in

= 1 ways. The oldest person must be excluded, which can be done in

= 1 ways. This

leaves 3 spots to be ﬁlled from the remaining 7 people. Giving: 1 × 1 × 7C3 groups • Case 2: The oldest person is included, and the youngest is not. In this case, the oldest person occupies one of the 4 spots in the group, which can be done in = 1 ways. The youngest person must be excluded, which can be done in

= 1 ways. This

leaves 3 spots to be ﬁlled from the remaining 7 people. Giving: 1 × 1 × 7C3 groups This results in 7C3 + 7C3 = 2 × 7C3 = 2 × selection can be made.

=2×

= 2 × 35 = 70. So, there are 70 ways this

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Example 1 A team of 3 is to be chosen at random from a group of 5 girls and 6 boys. In how many ways can the team be chosen if: a There are no restrictions.

Create a strategy There are 5 + 6 = 11 possible people to choose from, and 3 of them need to be selected. Substitute n = 11, r = 3 into nCr =

Apply the idea Write the notation

Write the formula

Substitute n = 11 and r = 3

Evaluate the subtraction

Expand the numerator

Remove the common factor

Expand the denominator

Evaluate

b There must be more boys than girls.

Create a strategy Use the fact that the only cases in which more boys than girls are chosen are: • No girls and all boys • 1 girl and 2 boys Determine the number of ways each of this case can happen, and add the results. Remember that the word “and” usually implies multiplication.

Apply the idea To determine the number of ways that all boys can be chosen, consider that 3 boys must be chosen from 6 boys and 0 girls must be chosen from 5 girls. This translates to 6 C3 × 5 C0 = 20. To determine the number of ways that 1 girl and 2 boys can be chosen, consider that 2 boys must be chosen from 6 boys and 1 girl must be chosen from 5 girls. This translates to 6 C2 × 5 C1 = 75. Number of ways = (6 C3 × 5 C0) + (6 C2 × 5 C1)

Write the formula

= 20 + 75

Substitute the values

= 95

Evaluate with calculator

There are 95 possible ways to form the team.

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Idea summary The concept of combinations can be extended to more complex scenarios involving the selection of subsets under additional constraints and speciﬁc rules. The addition principle is used to add up the number of combinations for cases that have no intersection or overlap. The multiplication principle can be applied to cases where multiple independent choices are involved.

Combinations with restrictions Other common restrictions with combinations involve phrases such as ‘at least’ or ‘at most’. While these restrictions still utilise a combination of the multiplication and addition rules, they require additional reasoning to determine the appropriate combinations. Consider this scenario: A Year 11 mathematics class consists of 25 students. The teacher must select some students to go on an excursion. Some types of restrictions include: • At least condition: • This restriction requires selecting a subset of students with a minimum number of speciﬁc students included. • Example: If at least 2 students must go, how many combinations are possible? • At most condition: • This restriction limits the selection to include at most a certain number of speciﬁc students. • Example: If at most 20 students must go, how many combinations are possible? To solve problems with these types of conditions: 1. Identify and list all the scenarios that satisfy the restrictions. 2. Use the combination formula and multiplication principle to calculate the number of combinations for each scenario. 3. Add the results from each scenario to find the total number of combinations. The cases which satisfy the condition can be added up, or the cases that do not satisfy the condition can be subtracted from the total unrestricted number of combinations. Recall that combination notation includes nCr, but for more involved contexts, commonly used.

is more

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Example 2 Five seniors and 6 juniors are part of the debate team. Four of them must represent the school at the upcoming debate tournament. In how many ways can the team of 4 be formed if: a It must contain 2 seniors and 2 juniors?

Create a strategy An “and” statement means multiplying the results of choosing 2 seniors from the 5 seniors total and choosing 2 juniors from the 6 juniors total.

Apply the idea Choosing 2 seniors from the 5 seniors total translates to 5 C2 . Choosing 2 juniors from the 6 juniors total translates to 6 C2 . 5

C2 × 6 C2 = 150 Evaluate using a calculator

There are 150 possible ways to form the team. b It must contain at least 1 senior and 1 junior?

Create a strategy Since the team must contain at least 1 senior and at least 1 junior, we can: 1. List all the scenarios that satisfy the condition of having at 1 senior and 1 junior. 2. Calculate the number of combinations for each scenario. 3. Add the results from each scenario for the total.

Apply the idea There are three scenarios that satisfy the conditions of at least 1 senior and at least 1 junior. Each scenario along with its calculation of number of combinations is listed: • Selecting 1 senior and 3 juniors: 5

C1 × 6 C3 = 100

C2 × 6 C2 = 150

• Selecting 2 seniors and 2 juniors:

• Selecting 3 seniors and 1 junior: 5

C3 × 6 C1 = 60

The total number of combinations is the sum of combinations from each scenario: (5 C1 × 6 C3 ) + (5 C2 × 6 C2 ) + (5 C3 × 6 C1 ) = 100 + 150 + 60 = 310

Sum the combinations Evaluate

There are 310 ways of selecting a team with at least 1 senior and at least 1 junior.

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Reﬂect and check Another way of calculating this is using subtraction since there are 5 + 6 = 11 students we can choose from to select 4 students. This can be translated as 11 C4 . 11

C4 = 330

Evaluate using a calculator

Selecting 4 of 5 seniors can be translated as 5 C4 . Selecting 4 of 6 juniors can be translated as 6 C4 . 5

C4 + 6 C4 = 20 11

Evaluate using a calculator 5

Subtract the sum from 11 C4

Number of ways = C4 − ( C4 + C4) = 330 − 20

Substitute the combination results

= 310

Evaluate

c It must contain at most 1 senior?

Create a strategy Consider the scenarios where there is at most 1 senior, calculate the individual number for combinations for each scenario and sum.

Apply the idea The scenarios and number of combinations that satisfy at most 1 senior: • Selecting 1 senior and 3 juniors: 5

C1 × 6 C3 = 100

• Selecting 0 seniors and 4 juniors: 5

C0 × 6 C4 = 15

The total number of combinations is 100 + 15 = 115.

Example 3 From a standard deck of 52 distinct playing cards, a 5-card hand is dealt. How many different hands contain at least one King?

Create a strategy The phrase “at least one” indicates that it is possible to calculate the cases for exactly 1 King, 2 Kings, 3 Kings, and 4 Kings, and then add them together. However, a more efficient method is to use complementary counting. Calculate the total number of possible 5-card hands and subtract the number of hands that do not meet the condition (i.e., hands with zero Kings). 1. Calculate the total unrestricted number of hands. 2. Calculate the number of hands with no Kings. 3. Subtract the result of step 2 from step 1.

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Apply the idea Step 1: Calculate the total unrestricted number of hands. Choose any 5 cards from the 52 distinct cards.

Set up the combination

Apply the formula

Evaluate

Step 2: Calculate the number of hands with no Kings. To form a hand with no Kings, the 5 cards must be chosen from the 48 non-King cards in the deck. Set up the combination

Apply the formula

Evaluate

Step 3: Subtract to find hands with at least one King At least one King = Total hands − Hands with no Kings

et up the calculation using the S complementary principle

= 2 598 960 − 1 712 304

Substitute the calculated values

= 886 656

Evaluate

There are 886 656 ways to form a 5-card hand with at least one King.

Reﬂect and check A standard deck of playing cards consists of 52 distinct cards. These are divided into four suits (Hearts, Diamonds, Clubs, and Spades). Each suit contains 13 ranks: Ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack (J), Queen (Q), and King (K). This structure means there are exactly four cards of each rank in the deck (e.g., four Kings, four Aces, etc.). In this problem, the 48 non-King cards are all the cards that are not one of the four Kings. To verify the result, the calculation can be performed directly by summing the number of ways to have exactly 1 King, 2 Kings, 3 Kings, or 4 Kings. • Exactly 1 King: Choose 1 of 4 Kings and 4 of 48 non-Kings.

• Exactly 2 Kings: Choose 2 of 4 Kings and 3 of 48 non-Kings.

• Exactly 3 Kings: Choose 3 of 4 Kings and 2 of 48 non-Kings.

• Exactly 4 Kings: Choose 4 of 4 Kings and 1 of 48 non-Kings.

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Summing these disjoint cases gives the total number of hands with at least one King: At least one King = 778 320 + 103 776 + 4512 + 48 = 886 656

Sum the results of the four cases Evaluate

This result matches the one obtained using the complementary counting method, confirming the answer. This also demonstrates that for “at least” problems, the complementary method is often significantly more efficient.

Idea summary When solving combination problems, it is helpful to understand the types of restrictions that affect how selections are made and counted. Common restrictions include: • At least condition: Requires including a minimum number of speciﬁc items. • At most condition: Limits the number of speciﬁc items in the subset. To calculate the total number of combinations: 1. Identify all scenarios that satisfy the restrictions. 2. Calculate the individual combinations of each scenario. 3. Sum the combinations for the total.

11.08E Practice questions What do you remember? 1

Calculate the number of ways to choose 3 items from a set of 5 different items.

A teacher has 7 students. Calculate the number of ways to select a committee of 2 students.

Express the formula for combinations,

, and explain what it represents.

Practice 4

To promote reading, a teacher decides to feature 3 classics, 4 contemporary novels, and 2 non-ﬁction books at the local library. Calculate the number of selections she can make from 5 classics, 6 contemporary novels, and 4 non-ﬁction books.

A student is to select 4 books from a shelf containing 10 different novels and 6 different non-ﬁction books. Calculate the number of different combinations of books the student can choose if 2 books must be non-ﬁction.

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A committee of 4 people is to be selected from a group of 12 candidates, where 5 candidates are from the ﬁnance department and 7 candidates are from the marketing department. Calculate the number of different combinations of committees that can be formed if 2 members must be from the ﬁnance department.

A factory has 25 machines, 5 of which are faulty. Calculate the number of ways 5 machines can be selected such that at most 1 of them is faulty.

There are 12 people to be selected from 20 candidates for jury duty. The candidates consist of these age groups: Age group

18 − 25

26 − 35

36 − 60

60+

Number of people

Calculate the number of ways the jury can be selected if at most 1 person aged 18 − 25 can be selected.

Ex 1

Calculate the number of ways 9 different books can be distributed among 3 students if the ﬁrst student gets 2 books, the second student gets 3 books, and the third student gets 4 books.

In a class of 12 students, there are 7 students wearing red shirts and 5 wearing blue shirts:

560

Calculate the number of ways 5 students can be chosen so that exactly 2 students are wearing blue shirts.

Calculate the number of ways 5 students can be chosen so that at least 3 students are wearing red shirts.

A group of 7 senior students and 4 intermediate students are part of a school’s mathematics team. A team of 4 is to be sent to a regional competition. Calculate the number of ways this team can be formed if: a

There are no restrictions.

The team must contain an equal number of senior and intermediate students.

The team must contain at most 3 senior students.

An art gallery has a total of 10 paintings by a particular artist: 5 oil paintings and 5 watercolour paintings. Calculate the number of ways in which 5 paintings can be selected for a special exhibition if: a

There are no restrictions.

One particular painting must be included.

There must be exactly 3 watercolour paintings.

There must be more watercolour paintings than oil paintings.

An ice cream shop offers 10 ﬂavours, which customers can combine into a milkshake: a

Assuming that double ﬂavours are not allowed, calculate the number of 3-ﬂavour milkshakes they could put on their menu.

Calculate the total number of milkshakes possible, with between 1 and 10 combined ﬂavours (but not double ﬂavours) allowed.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

Ex 3

A school council consists of 6 teachers and 5 students. A committee of 4 members is to be formed for a school event. Calculate the number of ways this committee can be formed if: a

It must contain exactly 2 teachers and 2 students.

It must contain at least 1 teacher and at least 1 student.

It must contain at most 1 teacher.

From a standard deck of 52 playing cards, a 5-card hand is dealt. How many different hands contain at most two Jacks?

An oﬃce consists of 10 employees, three of whom are managers and the remaining 7 are workers. Calculate the number of ways 5 people can be chosen to work on a project if the team must include either 1 or 2 managers.

Calculate the number of ways to choose 5 people from a group of 12 if the youngest must always be included.

A baker living in Victoria only sells sesame and poppy bagels. On any given day, he has exactly 10 of each to sell. Calculate the number of ways a student can purchase 6 bagels if her purchase must contain at least 1 poppy bagel.

An artist has access to 10 buckets of paint where each bucket is a unique colour. The artist must mix 4 colours together. Calculate the number of ways she can do this if she cannot mix green and red.

Calculate the number of ways two students can be chosen from a group of 6 to form a class leadership team, if exactly one of the two oldest students is on the team.

A teacher has 12 students and wants to select a committee of 3 students. Of the 12 students, 4 were on the committee last year. Calculate the number of different committees that can be formed if exactly one of the four experienced students is on the team.

Calculate the number of ways in which:

5 children can be divided into groups of 2 and 3.

9 children can be divided into groups of 5 and 4.

9 children can be divided into groups of 2, 3 and 4.

A jury of 6 must be selected from a pool of 8 women and 7 men such that the jury has more women than men. Calculate the number of possible selections.

11.08E Applications of combinations mathspace.co

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Extend your thinking 24

Thirty university students recently completed an exam, and the results are catalogued in this table: Exam results

0 − 20%

21 − 40%

41 − 60%

61 − 80%

81 − 100%

Number of students

The professor would like to interview 3 of the students to elicit feedback. However, she would like no more than 2 students to be from the same grade category. Calculate the number of ways she can choose the students to interview. 25

A hospital has 4 patients that need to book appointments in a given week (Monday - Friday). Calculate the number of possible ways to schedule the appointments such that no day has more than 2 appointments.

Explain why these two expressions give the number of ways a mixed football team of 22 can be selected from 46 candidates if at least 11 must be men, assuming an equal number of men and women among candidates:

and

Calculate the number of possible quadrilaterals using the dots shown as vertices:

562

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

11.09E Applications of combinations and permutations After this lesson, you will be able to… solve problems involving both permutations and combinations, including problems which require consideration of cases.

Applications of combinations and permutations Permutations and combinations are helpful to solve counting problems. Permutations are used in scenarios where the order of events matters, while combinations are used when the order does not matter. If the order matters, then use the permutation formula:

n is the total number of items r is the number of items being arranged If the order does not matter, use the combination formula:

n is the total number of items r is the number of items being selected Counting problems can be further complicated by restrictions, such as repeated objects, speciﬁc objects grouped together, or selections from multiple groups.

Example 1 A newspaper editor is deciding which of 6 articles to print on the front page: a If she can only choose 2 of them for the front page, how many different selections are possible?

Create a strategy Since the articles are only being selected and their order does not matter, this is a combination. Substitute n = 6, r = 2 into

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Apply the idea Write the formula Substitute n = 6 and r = 2

Evaluate using a calculator

There are 15 possible selections. b If their order on the front page matters, how many different arrangements are possible for the front page?

Create a strategy Multiply the possible choices until the spaces on the front page have been ﬁlled.

Apply the idea After choosing an article, the next space will reduce by 1 possibility. This pattern will continue, so the possibilities for each space of the front page will be: Number of possibilities = 6 × 5 = 30

Multiply the number of possibilities Evaluate

Reﬂect and check This can also be calculated using permutations: 6

P2 = 30

c She ﬁnds an error in one of the 6 articles and cannot print it. How many different arrangements for the front page are now possible, given that the order of the articles on the front page still matters?

Create a strategy Since one article contains an error, there are only 5 articles to fill the first space on the front page. Multiply the possible choices until the spaces on the front page have been filled.

Apply the idea After choosing an article from 5 articles, the next space will reduce by 1 possibility. This pattern will continue, so the possibilities for each space of the front page will be: Number of possibilities = 5 × 4 = 20

564

Multiply the number of possibilities Evaluate

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Two distinguishable teams, Team A and Team B, each consisting of 5 players are to be selected from a group of 12 players. After selecting the teams, each player on Team A is assigned one of the following positions: captain, vice-captain, and three other players. Similarly, Team B assigns its players to the same positions: a Determine the number of ways to select the two teams.

Create a strategy Use combinations to select 5 players for each team from the group of 12 considering that they are distinguishable teams.

Apply the idea First, select 5 players for Team A from a group of 12. This can be done in: 12

C5 = 792 ways

Then, select 5 players for Team B from a group of 7. This can be done in: 7

C5 = 21 ways

The total number of ways to select both teams is: 792 × 21 = 16 632 ways b Determine the number of ways to assign positions after the two teams have been selected.

Create a strategy Use permutations to assign positions within the teams.

Apply the idea For each team of 5 players, the assignment of the 2 distinct positions (captain and vice-captain) is a permutation, as order matters. The remaining 3 players are assigned to the identical ‘other player’ roles, which can be done in 3C3 = 1 way. The total number of position assignments for each team is therefore determined by the permutation of the distinct roles: 5

P2 = 20 ways to assign positions for Team A

P2 = 20 ways to assign positions for Team B

The total number of ways to assign positions for both teams is the product of their individual possibilities: 20 × 20 = 400 ways

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c Determine the total number of ways that teams can be selected and assigned to positions.

Create a strategy Combine answers from part (a) and part (b) using the multiplication principle.

Apply the idea The total number of ways to select the two teams and assign positions is the product of the number of ways to select the teams and the number of ways to assign positions for both teams: 16 632 × 400 = 6 652 800 ways

Idea summary Identifying whether a problem involves a permutation or combination comes down to determining if the order of the objects is important. To decide which method to use: • If the order is important, use permutation with the formula:

• If the order is not important, use combination with the formula:

Consider further restrictions, such as repeated objects, speciﬁc objects grouped together, or selections from multiple groups.

11.09E Practice questions What do you remember? 1

Determine whether each statement is true or false: a

Order does not matter for combinations.

Combinations are the same as permutations.

c d

566

is the same as nCr . When ﬁnding the number of ways r objects can be selected from a set of n distinct objects, a permutation will result in more outcomes than a combination.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

In each situation, determine if the order in which the objects are selected matters: a

A local pizza shop is offering a special on a large pizza with 3 toppings of your choice. They have a total of 9 pizza toppings to choose from.

Members need to elect a captain and a vice-captain for their sports team. There are 14 players on the team.

There are 52 runners competing in a marathon. The top 5 runners will receive varying amounts of prize money based on their ﬁnishing position.

A student council is organising snacks for an event. They can choose 3 different types of snacks to offer. The local store provides 7 snack options for them to select from.

Write using nCr or nPr notation: a

A team of ﬁve players is chosen from a squad of 12 players.

Five people are arranged in a line for an interview.

Eight chocolates are chosen from a box of 24 ﬂavours.

Locker codes are created from selecting three letters from the alphabet.

The code to open a locker is called a combination. Explain why this name is misleading.

Practice Ex 1

Ex 2

A music festival organiser is deciding which of several bands to feature in a special performance lineup. She initially plans to select and arrange 4 bands from a pool of 9: a

How many different selections are possible?

If the order in which the 4 selected bands perform matters, how many different arrangements are possible for the performance lineup?

A community book club wants to form two discussion groups, Group Alpha and Group Beta, each consisting of 3 members from a pool of 10 members. After selecting the groups, each member in Group Alpha will take on speciﬁc roles: leader, note-taker, and discussion facilitator. Group Beta will have members assigned to the same roles: a

Determine the number of ways to select the two groups.

Determine the number of ways to assign roles after the groups have been selected.

Determine the total number of ways that groups can be selected and assigned roles.

In a bicycle race, a quinella is a bet on the ﬁrst 2 cyclists that ﬁnish the race, but the order in which these 2 cyclists place does not matter. How many different quinella bets are possible for a bicycle race where 14 cyclists are competing?

A school baseball team has 12 players, but a coach can only choose 9 players for the batting lineup. The order in which the players bat is important. How many batting lineups are possible?

A university has 5 flagpoles and 8 different flags. If the order of the flagpoles does matter, in how many ways can the flags be chosen for the 5 flagpoles?

11.09E Applications of combinations and permutations mathspace.co

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For each of these problems: i

Identify it as a permutation, combination, or counting principle problem.

Solve the problem.

A book store offers a discount when you choose of 6 books from a list of 20. In how many ways can a shopper choose their books?

A car company sells a particular car in 10 different colours and in 3 different styles: hatchback, wagon or sedan. In how many ways can the car be designed?

In the qualifying round of a car race, 8 cars are given random starting positions. In how many different ways can starting positions be allocated?

A corporation has 12 members on its board of directors. In how many ways can it elect a president, vice-president, secretary and treasurer?

A popular brand of pen is available in 3 colours (red, green or blue) and 4 tips (0.3 mm, 0.5 mm and 0.7 mm, 1.0 mm). How many different choices of pens do you have with this brand?

A variety pack of chocolate consists of seven bars, each with a different ﬂavour. If three bars of chocolate are chosen at random, how many different selections are possible?

If a playlist contains 9 songs, in how many different ways could all 9 songs be played, with no song being repeated?

A class of 28 students visit a bowling alley and must organise themselves into groups of 8 to use a lane:

How many different ways can the ﬁrst group of 8 students be formed?

How many different ways can the second group of 8 students be formed?

How many different ways can the third group of 8 students be formed?

A local sports club is organising a mini-tournament with 6 players. The team will be arranged for a series of matches, and speciﬁc rules apply to each part of the organisation process: a

The club needs to arrange the 6 players for a round-robin tournament where each player plays against every other player exactly once. For the round-robin tournament, each player plays against every other player exactly once. How many unique matches need to be scheduled?

From the 6 players, the club needs to select 3 players to be in a special practice group. How many different ways can this Let’s practice group be formed?

Four letters are chosen from the word BETTING. How many selections of four letters have at least one T?

Consider the letters of the word ENVIOUS:

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How many ways are there to choose 1 vowel?

How many ways are there to choose 2 non-vowel?

If 3 letters are randomly chosen from ENVIOUS, how many selections include exactly 1 vowel?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A manager wants to select 5 people from a group of 25 assistants to help with a speciﬁc project: a

In how many ways can the manager choose 5 people?

If the manager needs to ﬁll 5 speciﬁc roles within the project, in how many ways can the roles in the project be assigned?

Which way of selecting the 5 people gives the manager fewer options?

A combination lock has a 4-digitcode with digits from 0 − 9. If the ﬁrst digit cannot be zero and no repetition is allowed, how many different codes are possible?

In a game of soccer, each team must have 1 goalkeeper and 10 other players on the ﬁeld. If a soccer team consists of 17 players, 2 of whom are goalkeepers. In how many ways can players be chosen to start the game?

Amelia is playing a game of cards. She will randomly draw 5 cards from a standard deck of 52 cards, and wants exactly 3 of the cards to have Clubs on them. One quarter of the cards in the deck have Clubs on them. How many different selections would contain the cards she wants?

A school needs to create two team names for Team X and Team Y from a selection of the letters A, B, C, D, E, F, G, and H. Each team name will consist of 4 letters:

Determine the number of ways to select letters for Team X and Team Y from the given group of letters.

Determine the number of ways to arrange the selected letters in Team X and Team Y.

Determine the total number of ways that both team names can be formed and arranged.

Six men and 3 women join a social evening. There are two round tables available, one consisting of 5 seats and another consisting of 4 seats. In how many ways can the 9 people be sat around the two circular tables if all the women wish to be seated at the same table? Assume that rotations of a table’s arrangement are considered identical.

A group of 8 people, consisting of 5 adults and 3 children, is to be seated at two identical circular tables, each with 4 seats. In how many ways can the group be seated if all the children wish to sit at the same table? Assume that rotations of a table’s arrangement are considered identical.

Extend your thinking 23

Fiona is in town for one week, and is planning to visit her 3 friends. In how many ways can she plan who she will visit throughout the week if she can visit each friend more than once and there are no restrictions on the number of friends she can visit in one night?

Ewen is organising a 6-person scientiﬁc committee from a pool of 12 scientists. The scientists are categorised into three specialisations A, B and C that include 4, 5 and 3 scientists, respectively. How many 6-person committee could include exactly 2 scientists from specialisation A and at least 2 scientists from exactly one other specialisation? 11.09E Applications of combinations and permutations mathspace.co

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In New South Wales, standard motor vehicle number plates consist of three digits followed by three letters. How many number plates could end with L or B?

Asako is looking to get a custom vanity number plate for her car. She worked on a dairy farm for many years, so she wants MOO to appear as one block in her number plate. If she wants to pay less than $2000, determine the number of possible number plates given the pricing for different styles of plates? Note that 9 represents any digit from 0−9, Q is considered a special letter in Queensland, and X represents any letter that is not Q. $495

$775

$1875

$2500

$2895

$3500

$5000

99MOO

Q99MOO

99MOOX

XMOO

99MOOXX

XMOOX9X

MOO

MOO99

QMOO99

XMOO99

QMOO

9MOOXXX

XXMOO9X

9MOO9

MOOX99

MOO9

9XMOOXX

999MOOX

999MOO

MOOQ99

MOO9

9XMOO99

XMOO999

99MOO9

9XMOO9

X9MOO9

9XMOO99

XMOO9X9

9MOO99

9QMOO9

Q9MOO9

XXMOO99

X99MOO9

Q9MOOX

X9MOO99

X9MOOX

QMOO999

XMOOX9

QMOO9X9

QMOOX9

Q99MOO9

XXMOO9

Q9MOO99 QMOOX9X QXMOO9X

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

11.10E Probability applications After this lesson, you will be able to… solve probability problems involving permutations and combinations.

Probability applications Counting techniques, permutations, and combinations can be used to determine the probabilities of certain events. By identifying how many of these sets or arrangements satisfy given conditions, the probability of a speciﬁc event or set of events is determined by comparing the number of favourable outcomes to the total number of possible outcomes. Permutations and combinations provide a systematic approach to count these outcomes. Probabilities can be calculated by:

If the events are independent, then the multiplication principle can typically be used. • For example, drawing 5 cards one at a time from a standard deck and noting their number, with replacement. If the events are dependent and the resulting order matters, then permutations can typically be used. • For example, drawing 5 cards from a standard deck, laying them down in a row, and noting their number, without replacement. If the events are dependent and the resulting order does not matter, then combinations can typically be used. • For example, drawing a hand of 5 cards from a standard deck and noting their numbers, without replacement. The addition principle is used when calculating the total number of outcomes for events that can occur in multiple, mutually exclusive ways. Additionally, it is sometimes easier to find the probability of an event by calculating the probability of its complement (the event not occurring) and subtracting this from 1.

Example 1 The letters of the word SPACE are to be rearranged: a Determine the number of distinct arrangements of the letters in the word SPACE.

Create a strategy Since the word has 5 letters and all of them are to be arranged, use the permutation formula with n = 5 and r = 5.

11.10E Probability applications mathspace.co

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Apply the idea Write the formula

Substitute n = 5 and r = 5

Evaluate the subtraction in the denominator

Expand

Evaluate

There are 120 different ways the letters of the word SPACE can be arranged. b What is the probability that the letter ‘E’ will be the ﬁrst letter?

Create a strategy Determine the number of permutations with ‘E’ in the ﬁrst position, and then divide this number by the total number of permutations found in part (a).

Apply the idea To determine the number of arrangements where ‘E’ is the first letter, fix ‘E’ in the first position. The remaining 4 letters (S, P, A, C) need to be arranged in the remaining 4 positions. The number of permutations of these 4 letters is: n

Pr = 4 P4

Substitute n = 4 and r = 4

= 24

Evaluate using a calculator

The number of permutations of the remaining letters is 24. The probability that ‘E’ will be the ﬁrst letter is the ratio of the number of favourable arrangements to the total number of arrangements: Write the formula

Substitute the values

Simplify

c What is the probability that the letters are arranged in alphabetical order?

Create a strategy Use the number of possible arrangements found in part (a) and determine the number of arrangements in which the letters are in alphabetical order.

572

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea From part (a), there are 120 different ways that the word can be arranged. There is only one arrangement in which the letters are in alphabetical order, A C E P S So, the probability that the letters are arranged in alphabetical order is:

Reﬂect and check The probability of any particular arrangement would be letters, it would be more involved.

. However, if there were duplicate

Example 2 Five people are to be selected from a larger group of 10 candidates. If Amelia is among the candidates, what is the probability that she will be among those selected?

Create a strategy Determine the total number of ways in which 5 people can be selected from 10 without restriction. Then, determine the number of ways 5 people can be selected with Amelia being one of them to determine the total number of desired outcomes, then divide this by the total number of outcomes.

Apply the idea The total number of ways in which 5 people can be selected from 10 without restriction can be translated as 10 C5 . If Amelia is one of the people in the selection, then there are only 4 places left to ﬁll, and only 9 people left to choose from. This can be translated as 9 C4. So, the probability is: Write the formula

Substitute known values

Evaluate numerator and denominator

Simplify

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Example 3 Consider the digits 4, 5, 6, 7, 8 and 9. A number is created by arranging at least one of the digits. Each digit can only be used once. What is the probability, as a percentage, that the number is greater than 800?

Create a strategy First determine the number of possible outcomes. The number created could be between one and six digits long. Then, determine the number of favourable outcomes that are greater than 800 and divide by the total to get the probability.

Apply the idea 1. Determine the total number of possible outcomes: Since the order matters and there are no duplicate digits, we need to add up the permutations of one to six digits: Total number of outcomes = 6 P1 + 6 P2 + 6 P3 + 6 P4 + 6 P5 + 6 P6 = 6 + 30 + 120 + 360 + 720 + 720 = 1956 2. Determine the number of favourable outcomes: • For a 3-digit number to be greater than 800, there are only 2 digits that can go in the hundreds place value, 5 digits remaining that can be used in the tens place value, and 4 digits remaining that can be used in the units place value: 2 × 5 × 4 • For a 4-digit number to be greater than 800, we’re looking for the number of ways 4 digits can be arranged from 6 possible digits: 6 P4 • For a 5-digit number to be greater than 800, we’re looking for the number of ways 5 digits can be arranged from 6 possible digits: 6 P5 • For a 6-digit number to be greater than 800, this is an arrangement of 6 distinct objects: 6 P6 So, the number of favourable outcomes is given by: Number of favourable outcomes = 2 × 5 × 4 + 6 P4 + 6 P5 + 6 P6

Write the equation

= 40 + 360 + 720 + 720

Evaluate each term

= 1840

Evaluate

3. Calculate the probability: Write the formula

Substitute the values

Evaluate rounded to four decimal places Convert to percentage Therefore, the probability of a number created from these digits being greater than 800 is 94.07%.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Reﬂect and check It is also possible to use the complement and determine the number of outcomes that are 800 or less. Count numbers 800 or less: • 1-digit number: These are all less than 800: 6 P1 = 6 • 2-digit numbers: These are all less than 800: 6 P2 = 30 • 3-digit numbers: If the ﬁrst digit is 4, 5, 6 or 7, then they are all less than 800: 4 × 5 P2 = 80 There are 6 + 30 + 80 = 116 non-favourable outcomes, so the probability is: Write the formula

Substitute the values

Evaluate rounded to four decimal places

Evaluate the subtraction

Convert to percentage

Idea summary Counting techniques, permutations, and combinations are useful tools for calculating probabilities. Probabilities can be calculated by:

11.10E Practice questions What do you remember? 1

Identify the formula for permutations of r items from n distinct items.

Identify the formula for combinations of r items from n distinct items.

Explain when to use permutations versus combinations for counting outcomes in a school event with 5 participants.

A club has 3 tasks to assign to different members. Calculate the number of ways to assign all tasks.

A team selects 2 activities from 4 options, where order does not matter. Calculate the number of possible selections.

11.10E Probability applications mathspace.co

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Practice 6

Ex 1

A deck contains 6 cards, each a different colour: red, blue, green, yellow, purple, orange. Three cards are drawn randomly and arranged in a sequence: a

Calculate the total number of possible arrangements of the 3 cards.

Determine the probability that the sequence starts with a red card and ends with a green card.

A box contains 6 balls, each a different colour: red, blue, green, yellow, purple, orange. Three balls are selected randomly: a

Calculate the total number of possible combinations of the 3 balls.

Determine the probability that the selected balls include blue, red, and green.

The letters of the word NOTES are rearranged: a

Calculate the number of distinct arrangements of the letters in NOTES.

Determine the probability that the letter N is the ﬁrst letter.

Determine the probability that the letter N is not the ﬁrst letter.

Determine the probability that the letters are in alphabetical order.

A library has 7 distinct books. Three books are selected randomly: a

Calculate the number of ways to select 3 books.

Determine the probability that a speciﬁc mathematics book is included, given only one book is on mathematics.

Ex 2

Five friends are arranged randomly in a row. Determine the probability that two speciﬁc friends are adjacent.

Ex 3

Romi wants to create a three-digit number using the 6 digits of 201 563 without replacement:

576

How many numbers can Romi create?

What is the probability, as a percentage, that a randomly created number will be less than 350?

Two letters from the word STATISTICS are selected: a

Calculate the number of ways to select 2 letters, considering repetitions.

Calculate the number of ways to select 2 different letters.

Determine the probability that exactly one selected letter is S.

A box contains 10 balls, numbered 0 to 9. Four balls are selected randomly and arranged in a line: a

Calculate the total number of possible arrangements.

Determine the probability that the number 1234 appears.

Determine the probability that the ﬁrst and last balls are even numbers (zero is even).

A box contains 10 balls, numbered 0 to 9. Four balls are selected simultaneously: a

Calculate the total number of possible combinations.

Determine the probability that all selected balls have numbers less than 6.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A permutation of digits 1 to 6 is selected randomly. Determine the probability that the digits are in increasing order.

A salad bar offers 3 types of lettuce, 4 toppings, and 2 dressings. A customer chooses one of each randomly. Determine the probability that the ﬁrst lettuce type and second dressing are selected.

A committee of 3 is formed randomly from 8 teachers. Determine the probability that two speciﬁc teachers are included.

A license plate consists of 3 letters followed by 4 digits, with no repetition. Determine the probability that the second letter is B.

A bowl contains 5 apples, 7 oranges, and 8 pears. Four fruits are selected randomly. Determine the probability that no apples are included.

A club has 10 members. A board of 4 is elected randomly. Determine the probability that two speciﬁc members are both elected.

A bag contains 4 gold, 5 silver, and 6 bronze coins. Three coins are drawn sequentially without replacement. Determine the probability that the sequence is gold, silver, bronze.

Extend your thinking 22

Six distinct flowers (Rose, Tulip, Lily, Daisy, Orchid, Sunflower) are arranged in a row. Determine the probability that Daisy is adjacent to Orchid and neither Tulip nor Sunflower is at either end.

Four gift cards are distributed randomly among 15 employees, three of whom are speciﬁc candidates. Determine the probability that at least one speciﬁc candidate does not receive a card.

Five artworks are selected randomly from 15 submissions for a newsletter. Each of three judges independently selects 5 pieces. Determine the probability that a speciﬁc artwork is chosen by at least one judge.

A dance class has 6 pairs with experience: Pair 1 (5, 3 years), Pair 2 (10, 8), Pair 3 (2, 1), Pair 4 (7, 5), Pair 5 (4, 2), Pair 6 (9, 6). The two pairs with highest average experience are chosen, and one more is selected randomly from the rest. Determine the probability that the average experience of selected dancers is at least 7 years.

Starting at (0, 0), a walker takes 8 steps (north, east, south, or west). Determine the probability, as a percentage rounded to two decimal places, of returning to the origin.

An ant walks randomly from (0, 0) to (4, 3) using grid lines, moving only up or right. Determine the probability that the path passes through (2, 1).

11.10E Probability applications mathspace.co

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11 Chapter review 1

Evaluate: 5! − 3! A

114

117

120

How many different ways can the letters of the word GROUP be arranged? A

A restaurant offers a lunch special. Customers can choose 1 of 4 appetisers, 1 of 5 main courses, and 1 of 3 desserts. How many different three-course meals are possible? A

120

Evaluate: a

c 4! × 2!

d 3 × 6!

Calculate the value of n if

A coffee shop offers a breakfast combo. Customers can choose:

(4! − 16)!

• 1 type of pastry: croissant, muﬃn, scone, or danish • 1 type of spread: butter, jam, or cream cheese • 1 type of hot drink: coffee, tea, or hot chocolate How many different breakfast combos can a customer choose? 7

A clothing store is promoting accessory sets. With the set, the customer gets one hat, one scarf, and one pair of gloves. The choices are displayed in a tree diagram: Wool scarf

Leather gloves

Silk scarf

Cotton gloves Leather gloves

Sun hat

Cotton gloves Wool scarf

Leather gloves

Silk scarf

Cotton gloves Leather gloves

Beanie

Cotton gloves

Cap

Wool scarf

Leather gloves

Silk scarf

Cotton gloves Leather gloves Cotton gloves

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Count the number of outcomes on the right-hand side of the tree diagram to determine how many different accessory sets are possible.

Use the multiplication principle to determine how many different accessory sets are possible.

Liam is travelling from City X to City Z. He can travel from City X to City Y by one of 3 train lines or 2 bus routes. From City Y to City Z, he can take one of 2 ﬂights or 4 coach services. Determine the number of different ways Liam can travel from City X to City Z via City Y.

Evaluate: a

Solve for n if nP2 = 72.

How many distinct ways can the letters of the word ENGINEER be arranged?

Five students (Amy, Ben, Chloe, David, Eva) are lining up for a photo. a

How many ways can they be lined up with no restrictions?

How many ways if Amy and Ben must stand next to each other?

How many ways if Chloe and David refuse to stand next to each other?

In how many ways can the digits 1, 2, 3, 4, 5, 6 be arranged to form a six-digit number such that the even digits (2, 4, 6) must occupy the even positions (2nd, 4th, 6th)?

In how many ways can 9 distinct trees be planted in a circle in a park?

A family of 5 (2 parents and 3 children) sits around a circular table for dinner. a

In how many ways can they be seated with no restrictions?

In how many ways can they be seated if the two parents must sit together?

Eight distinct statues are to be placed around a circular garden. In how many ways can they be arranged if two speciﬁc statues, Alpha and Beta, must not be placed next to each other?

A chef needs to choose 4 distinct spices from a collection of 10 available spices for a new recipe. How many different combinations of spices can the chef choose?

In a local tennis league with 12 players, every player must play a match against every other player exactly once during the season. How many matches will be played in total?

At a business conference, every attendee exchanged business cards with every other attendee exactly once. If a total of 120 exchanges occurred, how many attendees were at the conference?

Verify that 9 C3 = 9 C6 by calculating both sides using the combination formula.

Calculate 7C3 and verify your answer using the identity nCr = n − 1 Cr − 1 + n − 1 Cr .

Prove algebraically that

for 1 ≤ k ≤ n.

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A fruit basket is being prepared by selecting 2 types of apples from 5 available types, 3 types of berries from 6 available types, and 2 types of citrus fruits from 4 available types. How many different fruit basket combinations can be made?

A class has 10 boys and 8 girls. A quiz team of 4 students is to be selected.

How many ways can the team be selected if it must consist of exactly 2 girls?

How many ways can the team be selected if it must consist of at least 2 boys?

A factory has 20 machines, of which 4 are known to be faulty. A sample of 5 machines is selected for inspection. a

How many ways can the sample be selected so that exactly 1 faulty machine is included?

How many ways can the sample be selected so that no faulty machines are included?

How many distinct triangles can be formed by choosing 3 vertices from the vertices of a regular octagon?

From a collection of 8 different novels, a student needs to choose 4 to read over the holidays. After choosing them, the student will decide the order in which to read them. How many different ordered reading lists of 4 novels are possible?

A school debating team has 15 members. A coach needs to select a team of 3 debaters and then assign one as the ﬁrst speaker, one as the second speaker, and one as the third speaker. How many different ways can this be done?

Four men and four women are to be seated at two distinct circular tables. One table has 5 seats and the other has 3 seats. In how many ways can they be seated if all the women wish to sit at the 5-seat table?

The letters of the word ACTIVE are rearranged randomly. a

How many distinct arrangements are possible?

What is the probability that the letter A is the ﬁrst letter?

What is the probability that the letters C and T are together in any order?

From a set of 10 cards numbered 1 to 10, four cards are selected simultaneously at random. What is the probability that all four selected cards show even numbers?

Three distinct scholarships are to be awarded to 10 students. Two speciﬁc students are Alex and Beth. What is the probability that at least one of Alex or Beth receives a scholarship? (Assume a student can receive at most one scholarship).

An ant starts at the origin (0, 0) and can only move up or right along grid lines, where all paths are equally likely.

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How many paths are possible from (0, 0) to (5, 5)?

How many paths are possible if the ant must go through (1, 2)?

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“Wherever there is number, there is beauty.” Proclus

12 Statistics and the binomial theorem Chapter outline 12.01 12.02 12.03 12.04E 12.05E 12.06E 12.07E 12.08E 12.09E 12.10E 12.11E 12.12E

584 588 597 602 608 613 619 624 628 633 640 646 653

Your chance of seeing a shooting star on a clear night? Around 1 every 10 minutes.

12.01 Random variables After this lesson, you will be able to… • define a random variable as a variable whose value is the outcome of a random experiment. • distinguish between discrete and continuous random variables. • classify a random variable as discrete or continuous in practical scenarios. • provide practical examples of both discrete and continuous random variables.

Random variables Random variable A variable whose possible values are outcomes of a statistical experiment or a random phenomenon. A random variable is a variable whose possible values are the outcomes of a random process. For a discrete random variable, it assigns a numerical value to each outcome in a sample space. For example, in rolling a fair six-sided die, deﬁne X as the number that turns up on the die. The sample space is S = {1, 2, 3, 4, 5, 6}, and X takes values 1, 2, 3, 4, 5, 6. X denotes a random variable, mapping outcomes to numerical values. Random variables are denoted by capital letters (e.g. X, Y ), and their numerical outcomes by lowercase letters (e.g. x = 3).

Example 1 An experiment involves drawing a card from a standard deck of 52 cards: a Deﬁne a random variable X for the number of aces drawn.

Create a strategy

Apply the idea

Deﬁne X based on the numerical outcome. Identify all possible values by examining the sample space.

Let X be the number of aces drawn. Since one card is drawn, X is 1 if an ace is drawn, 0 otherwise.

b List the possible values of X.

Apply the idea Possible values: X = {0, 1} 0 for non-ace, 1 for ace

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Reﬂect and check Ensure X assigns a number to each outcome in the sample space. Verify all possible values are listed.

Idea summary A random variable X assigns numerical values to outcomes of a random process, linking the sample space to numbers.

Discrete vs. continuous random variables Discrete random variable A numerical variable whose values can be listed. Continuous random variable A continuous random variable is a numerical variable that can take any value along a continuum.

Random variables are classified as discrete or continuous based on their possible values. A discrete random variable takes on a finite or countably infinite set of values. For example, the number of heads in two coin flips (X = 0, 1, 2) is discrete. A continuous random variable takes on any value within an interval. For example, the time to complete a task (e.g. X in minutes) is continuous.

X discrete X

takes distinct, countable values (e.g. integers)

X continuous X takes any value in an interval (e.g. real numbers) Differences: • Discrete: Countable values, often integers (e.g. number of students). • Continuous: Uncountable values in an interval, often measurements (e.g. height, time).

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Example 2 Classify the following as discrete or continuous random variables, and give a practical example for each: a Number of goals scored in a soccer match

Create a strategy

Apply the idea

Determine if the random variable’s values are countable (discrete) or uncountable within an interval (continuous). Provide relevant examples.

Let X be the number of goals. Possible values are 0, 1, 2, … (countable), so X is discrete. Example: In a school soccer match, X could be 3 if the team scores 3 goals.

b Time taken to solve a puzzle.

Apply the idea Let Y be the time in minutes. Possible values are any positive real number (e.g. 2.5, 3.142), so Y is continuous. Example: Solving a math puzzle might take Y = 4.7 minutes.

Reﬂect and check Check if discrete variables have countable values and continuous variables have intervals. Ensure examples match the random process.

Idea summary Discrete random variables have countable values, while continuous random variables have values in an interval. Their differences lie in the nature of possible values: countable vs. uncountable continuum.

12.01 Practice questions What do you remember? 1

Deﬁne a random variable.

What is the difference between discrete and continuous random variables?

Give one example of a discrete random variable and one example of a continuous random variable.

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Practice Ex 1

For each scenario: i

Deﬁne the random variable.

List the possible values for each experiment.

Flipping a coin twice. Let X be the number of heads.

Rolling a six-sided die. Let Y be the number shown.

Drawing one card from a deck of 52 cards. Let Z be the card’s face value (Ace = 1, Jack = 11, Queen = 12, King = 13).

Spinning a spinner divided into 4 equal sections numbered 1 to 4. Let W be the number on the section where the spinner lands.

Classify the following random variables as discrete or continuous: a

Ex 2

Weight of a package

Number of pages in a book

Time to download a ﬁle

Number of emails received in an hour

Temperature in a city

Number of correct answers on a quiz

Distance travelled by a car

Determine if the following are discrete or continuous random variables, and provide one example of a possible value for each: a

Number of customers in a store

Height of a student in Year 11

Number of defective items in a batch

Volume of water in a tank

An experiment involves tossing three coins. Deﬁne a random variable V for the number of tails: a

Number of cars in a parking lot

What is V ?

List the possible values of V.

Deﬁne a random variable for each experiment and determine if it is discrete or continuous: a

Measuring the time to complete a task. Deﬁne T.

Selecting a student and recording their shoe size. Deﬁne S.

A bag contains 5 red and 3 blue marbles. One marble is drawn. A random variable M is deﬁned such that M = 1 if the marble is red and M = 0 if the marble is blue: a

What are the outcomes of the experiment?

List the possible values of M.

A survey asks respondents to select their favourite genre from 4 options (action, comedy, drama, sci-ﬁ). Let G be the respondent’s choice: a

Deﬁne G.

List the possible values of G.

Is G discrete or continuous?

A machine dispenses juice into bottles, with the volume varying slightly. Let J be the amount of juice dispensed: a

Deﬁne J.

Is J discrete or continuous?

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Extend your thinking 12

Explain why the number of students in a classroom is a discrete random variable, but the average height of students is a continuous random variable.

An experiment involves rolling two fair dice and deﬁning X as the sum of the numbers shown. Determine the set of possible values of X and classify it as discrete or continuous.

A student incorrectly classiﬁes the time to complete a test as a discrete random variable. Explain the error and provide a correct classiﬁcation with an example.

12.02 Organise and graph datasets After this lesson, you will be able to… • organise finite datasets into tables listing values, frequency, relative frequency, and cumulative frequency. • construct and interpret frequency, relative frequency, cumulative frequency histograms, and cumulative frequency polygons (ogives). • identify the mode and estimate the median of a dataset from tables, histograms, and polygons.

Organise datasets in tables Frequency The number of times that a particular value occurs in a dataset. For grouped data, it is the number of observations that lie in that group or class interval. For example, when rolling a dice 20 times, ‘the frequency of a 6’ means how many times the number 6 comes up. Relative frequency Given by the ratio

, where f is the frequency of occurrence of a particular data value or

group of data values in a dataset, and n is the number of data values in the dataset. Cumulative frequency The accumulating total of frequencies within an ordered dataset.

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A ﬁnite dataset, often from a discrete random variable, can be organised in a table listing: • Values: Possible values of the random variable. • Frequency: Number of times each value occurs. • Relative frequency: Proportion of each value,

• Cumulative frequency: Running total of frequencies up to each value. • Cumulative relative frequency: Running total of relative frequencies. For example, rolling a die 20 times might yield values with frequencies, which can be tabulated.

Example 1 A survey records the number of pets owned by 20 households: 0, 1, 2, 1, 3, 0, 2, 1, 4, 1, 0, 2, 3, 1, 2, 0, 1, 2, 1, 0 Organise the data in a table with values, frequency, relative frequency, cumulative frequency, and cumulative relative frequency.

Create a strategy Tally the frequency of each value, calculate relative frequencies, and compute cumulative totals.

Apply the idea Values: 0, 1, 2, 3, 4. Frequency: Count occurrences. • 0: 5 times • 1: 7 times

• 2: 5 times

• 3: 2 times

• 4: 1 time

Total frequency: 20 Relative frequency: Cumulative frequency: Sum frequencies up to each value. Cumulative relative frequency: Sum relative frequencies. 0

0.25

0.35

0.25

0.1

0.05

0.25

0.6

0.85

0.95

Reﬂect and check Verify the total frequency is 20 and cumulative relative frequency reaches 1. Check calculations for accuracy.

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Idea summary Datasets are organised in tables with values, frequency, relative frequency, cumulative frequency, and cumulative relative frequency.

Visualise datasets with histograms Cumulative frequency histogram

Median

A visual representation of data using bars to represent the class boundaries and the cumulative frequencies.

The value in a set of ordered data that divides the data into 2 parts. It is frequently called the ‘middle value’.

The area of each bar is proportional to the cumulative frequency of the observations up to the end of that class.

Mode The most frequently occurring value in a set of data.

A histogram visualises the frequency distribution of a dataset using adjacent bars: • Frequency histogram: Bars represent the frequency of each value or class interval. Bar height indicates frequency, and width represents the class interval (for grouped data). No gaps between bars for continuous or consecutive discrete data, showing distribution continuity. • Relative frequency histogram: Bars show relative frequency (frequency divided by total frequency), useful for comparing datasets of different sizes. • Cumulative frequency histogram: Bars display cumulative frequency, showing the running total of frequencies up to each value or class, aiding in median identiﬁcation. The mode is the value or class with the highest frequency (tallest bar). The median is the middle value when the data is arranged in ascending order. It can be approximated using a cumulative frequency histogram by locating midway of the cumulative frequency. Distribution of running times for a 10 km race 72 runners

Frequency

This histogram shows distribution of running times for a 10 km race. The mode is 30 (tallest bar), and the median should be calculated using the cumulative frequency histogram.

25 20 15 10 5 0

Running time (minutes)

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Example 2 A dataset records times (in minutes) for 20 runners to complete a race: 35, 38, 40, 42, 45, 46, 48, 50, 52, 55, 56, 58, 60, 62, 65, 68, 70, 72, 75, 80 Create a frequency histogram with class intervals of width 10 minutes starting at 30. Identify the mode.

Create a strategy

Apply the idea

Group data into class intervals, calculate frequencies, and plot a histogram with bar heights as frequencies. Find the mode from the highest bar.

Class intervals: 30–40, 40–50, 50–60, 60–70, 70–80, 80–90. Count frequencies: • 30–40: 2 (35, 38) • 40–50: 5 (40, 42, 45, 46, 48) • 50–60: 5 (50, 52, 55, 56, 58) • 60–70: 4 (60, 62, 65, 68) • 70–80: 3 (70, 72, 75) • 80–90: 1 (80) 5

Frequency

4 3 2 1 0

30 40 50 60 70 80 90

Time (in minutes) Mode: Highest frequency is 5 at 40–50 and 50–60, so bimodal.

Idea summary Histograms use bars to show frequency, relative frequency, or cumulative frequency, revealing the dataset’s distribution. The mode (tallest bar) and median (middle value from the cumulative frequency) are identiﬁed from histograms.

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Visualise datasets with cumulative frequency polygons Cumulative frequency polygon A series of straight lines representing the cumulative frequency for a given dataset. Sometimes called the ‘ogive’. A cumulative frequency polygon (ogive) visualises the cumulative frequency of a dataset by connecting points at class boundaries or values with their cumulative frequencies: • Points are plotted at the upper boundary of each class interval (or at each value for ungrouped data) against the cumulative frequency up to that point. • The graph starts at (x0, 0), where x0 is the lower boundary of the ﬁrst class, and ends at the total frequency. • It is useful for ﬁnding the median (where the polygon reaches , with n as total frequency) and analysing data accumulation.

Cumulative frequency

The mode is the value or class with the highest frequency, identiﬁable from a frequency table or histogram, not directly from the cumulative frequency polygon. The median is found by locating on the cumulative frequency axis and reading the corresponding value from the graph. 50 40

This cumulative frequency polygon shows pet ownership data. The median is approximately 23, as the pet ownership data cannot be in decimals (where cumulative frequency reaches 25, half of 50).

30 20 10 0

5 10 15 20 25 30 35 40

Number of pets

Example 3 A dataset records the number of books read by 20 students in a month: 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 11, 12, 15 a Create a cumulative frequency polygon with class intervals of width 5 books starting at 0.

Create a strategy Group data into class intervals, calculate cumulative frequencies, plot points at upper class boundaries, and connect them.

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Apply the idea Class intervals: 0–5, 5–10, 10–15, 15–20. Count frequencies and cumulative frequencies: Class Interval

0–5

5–10

10–15

15–20

Frequency

Cumulative Frequency

Plot points at upper boundaries: (5, 5), (10, 16), (15, 19), (20, 20), starting at (0, 0). Cumulative frequency

20 15 10 5 0

Number of books b Find the mode.

Create a strategy

Apply the idea

Identify the class interval with the highest frequency from the frequency table.

The mode is 11 in 5–10 (from frequency table).

c Find the median.

Locate = 10 on the cumulative frequency axis of the polygon and read the corresponding value.

Apply the idea 20

Cumulative frequency

Create a strategy

15 10 5 0

Number of books From the graph, y = 10 occurs in the 5–10 interval, approximately at 7 books.

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Idea summary Cumulative frequency polygons (ogives) connect cumulative frequencies at class boundaries, showing data accumulation. The median is found where the polygon reaches half the total frequency on the graph. The mode is identiﬁed from a frequency table or histogram.

12.02 Practice questions What do you remember? 1

Deﬁne frequency in the context of a dataset.

What is relative frequency, and how is it calculated?

Explain the difference between cumulative frequency and cumulative relative frequency.

In a frequency histogram, how is the mode identiﬁed?

Practice Ex 1

A survey records the number of books read by 15 students in a month: 2, 0, 1, 3, 2, 4, 1, 0, 2, 3, 1, 2, 0, 1, 2

Organise the data into a table with values, frequency, and relative frequency.

Add cumulative frequency and cumulative relative frequency to the table.

A dataset shows the number of goals scored by a soccer team in 10 matches: 0, 2, 1, 3, 0, 1, 2, 1, 0, 2

Create a frequency table with values and frequency.

Calculate the relative frequency for each value.

Add cumulative frequency to the table.

The times (in minutes) for 12 students to complete a task are: 5, 7, 8, 5, 6, 9, 7, 6, 8, 7, 5, 6

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Construct a frequency table with values, frequency, and cumulative frequency.

Determine the median from the cumulative frequency.

Calculate the relative frequency for each value.

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Ex 2

A dataset records the number of hours 20 students studied: 1, 3, 2, 4, 2, 1, 3, 2, 5, 3, 2, 1, 4, 2, 3, 1, 2, 3, 2, 1

Create a frequency histogram with values 1 to 5.

Identify the mode from the histogram.

The scores of 15 students on a quiz (out of 10) are: 4, 6, 8, 5, 7, 6, 9, 5, 6, 7, 8, 6, 5, 7, 6

Ex 3

Create a relative frequency histogram.

Identify the mode from the histogram.

Create a cumulative frequency histogram.

A dataset of 20 race times (minutes) is grouped into intervals: 10–15, 15–20, 20–25, 25–30. Frequencies are 4, 6, 8, 2: a

Create a frequency histogram.

Identify the mode.

Create a cumulative frequency histogram.

Analyse the dataset of daily water consumption (litres) for 25 households: 12, 15, 18, 20, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 42, 45, 47, 50

Create a cumulative frequency table with class intervals of width 10 starting at 10.

Create a cumulative frequency polygon.

Determine the median and the mode.

A dataset records the number of hours 20 students spent on homework: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 10

Create a cumulative frequency histogram.

Estimate the median using the cumulative frequency histogram.

Create a cumulative frequency polygon.

A dataset of 25 test scores is grouped into intervals: 0–20, 20–40, 40–60, 60–80, 80–100. Frequencies are 2, 5, 8, 7, 3: a

Create a cumulative frequency polygon.

Identify the mode from a frequency table.

Estimate the median using the cumulative frequency polygon.

A dataset records the number of hours 15 students slept: 6, 7, 8, 6, 7, 8, 9, 7, 6, 8, 7, 6, 8, 7, 6 a

Construct a frequency table with values, frequency, and relative frequency.

Create a frequency histogram and identify the mode.

Create a cumulative frequency histogram.

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Extend your thinking 15

Explain why a cumulative frequency polygon might be preferred over a cumulative frequency histogram for estimating the median.

A student creates a frequency histogram but notices gaps between bars for consecutive values. Explain the error and how to correct it.

A dataset of 22 temperatures (in °C) is grouped into intervals: 10–15, 15–20, 20–25, 25–30. Frequencies are 4, 7, 6, 5: a

Create both a cumulative frequency histogram and a cumulative frequency polygon on the same axes.

Find an estimate of the median temperature.

A dataset has a cumulative frequency histogram that plateaus before reaching the total frequency. Explain what this indicates about the dataset and how it affects the median calculation.

Two datasets record the number of hours spent on extracurricular activities by two groups of 10 students: • Group A: 1, 2, 2, 3, 3, 4, 2, 1, 3, 2 • Group B: 2, 3, 4, 5, 3, 2, 4, 3, 2, 3 a

Create cumulative frequency polygons for both datasets on the same axes.

Compare the medians of the two groups using the polygons.

Did you know?

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12.03 Analyse data After this lesson, you will be able to… • calculate the relative frequency of an outcome in a dataset. • use the relative frequency to estimate the probability of a result in an experiment. • understand that larger sample sizes generally lead to more reliable probability estimates. • apply estimated probabilities to make predictions about larger populations.

Estimate probabilities with relative frequency The relative frequency of an outcome is the proportion of its occurrences in a dataset.

Proportion of an outcome’s occurrences In experiments, relative frequency estimates the probability of an outcome for a random variable X.

P (X = x)

is the estimated probability of X taking value x

Exploration Roll a die 30 times and record outcomes. 1. Calculate the relative frequency of rolling a 4. 2. How does this compare to the theoretical probability P (X = 4) = ? 3. What changes with 100 rolls?

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Example 1 A die is rolled 60 times, with outcomes: 1 (12 times), 2 (9 times), 3 (11 times), 4 (10 times), 5 (8 times), 6 (10 times). Estimate the probability of rolling a 3 using relative frequency.

Create a strategy Use the relative frequency formula: P (X = x) ≈

Apply the idea Write the formula

Apply relative frequency

Evaluate

Reﬂect and check Sum the frequencies: 12 + 9 + 11 + 10 + 8 + 10 = 60. Compare to theoretical probability

≈ 0.167.

Idea summary Relative frequency,

, estimates P (X = x) for a random variable X.

Apply relative frequency in experiments Relative frequencies estimate probabilities in experiments when theoretical probabilities are unknown. Larger sample sizes yield more accurate estimates, often visualised in histograms.

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Example 2 A store checks 150 batteries, ﬁnding 12 faulty. a Estimate the probability that a battery is faulty.

Create a strategy Use the relative frequency formula:

Apply the idea Write the formula

Substitute the values

Evaluate

P (battery is faulty) ≈ 0.08.

b Predict the number of faulty batteries in 1000 tests.

Create a strategy Multiply the estimated probability by 1000.

Apply the idea Faulty batteries ≈ 0.08 × 1000 ≈ 80 batteries

Apply probability Evaluate

Idea summary Relative frequencies from experiments estimate probabilities, improving with larger samples and visualised in histograms.

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12.03 Practice questions What do you remember? 1

Deﬁne relative frequency in the context of an experiment.

How does relative frequency estimate the probability of an outcome?

Determine whether each statement is true or false: .

Relative frequency is calculated as

Relative frequency estimates the theoretical probability of an outcome in an experiment.

The probability of an outcome can be exactly determined using relative frequency from a small sample.

Larger sample sizes generally provide less accurate probability estimates using relative frequency.

Practice Ex 1

Ex 2

A spinner with 5 sections (numbered 1 to 5) is spun 100 times, with outcomes: 1 (22 times), 2 (18 times), 3 (20 times), 4 (25 times), 5 (15 times): a

Estimate P (X = 4).

Estimate the probability of landing on an even number.

A quality control test checks 200 light bulbs, ﬁnding 15 defective: a

Estimate the probability a light bulb is defective.

Estimate the probability a light bulb is not defective.

A factory inspects 120 widgets, ﬁnding 9 defective: a

Estimate P (defective).

Estimate P (not defective).

Predict the number of defective widgets in 800 inspections.

A survey of 200 students ﬁnds 80 prefer online learning: a

Estimate P (online).

Estimate P (not online).

Predict the number of students preferring online learning in a sample of 500 students.

A die is rolled 150 times, with outcomes: 1 (24 times), 2 (26 times), 3 (25 times), 4 (23 times), 5 (27 times), 6 (25 times): a

600

Calculate the relative frequency of rolling a 5.

Estimate P (X = 5).

Estimate the probability of rolling an odd number.

Predict the number of 5s in 600 rolls.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A traﬃc light is observed 120 times, showing red 48 times, green 60 times, amber 12 times: a

Estimate P (not green).

Estimate the probability of the light being green or amber.

Predict the number of red lights in 300 observations.

A card is drawn from a deck of 52 cards 200 times (with replacement), resulting in 48 spades. Calculate the probability of not drawing a spade.

A card is drawn from a deck of 52 cards 180 times (with replacement), resulting in 42 hearts: a

Estimate P (heart).

Estimate P (not heart).

Predict the number of hearts in 500 draws.

Extend your thinking 12

A coin is ﬂipped 200 times, yielding 92 heads. Estimate P (heads) and discuss whether this suggests the coin is biased.

A coin is ﬂipped 300 times, yielding 138 heads:

Estimate P (heads).

Compare to the theoretical probability.

Discuss whether this suggests the coin is biased.

How would doubling the ﬂips affect the estimate?

A student conducts an experiment where he records number 2 appearing 12 times in 50 die rolls: a

Calculate the relative frequency of rolling a 2.

Compare to the theoretical probability.

Why do you think that the experimental probability differs from the theoretical probability?

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12.04E Binomial expansions After this lesson, you will be able to… • recognise that a binomial expansion is an expansion of a power of a binomial. • expand a binomial expression raised to a small non-negative integer power by repeated multiplication. • examine the symmetry formed by the coefficients in the expansion of ( x + y)n for small values of n (e.g., n = 0, 1, 2, 3, 4, 5). • use Pascal’s triangle to determine coefficients for binomial expansions like ( x + y)n and apply this to expand binomials.

Recognise binomials and their expansions Binomial expression

Binomial expansion

An algebraic expression with two terms. For example, x + y is a binomial expression in the two terms x and y.

The algebraic expansion of a power of a binomial expression. For example, 16x4 + 32x3 + 24x2 + 8x + 1 is the binomial expansion of (2x + 1)4 .

A binomial expression is an algebraic expression with exactly two terms, typically connected by addition or subtraction. A binomial expansion is the result of raising a binomial expression to a non-negative integer power, written as a sum of terms. Each term in the expansion consists of a coeﬃcient multiplied by powers of the binomial’s terms.

( x + y)n x, y

are the terms of the binomial

is a non-negative integer power

To expand ( x + y) for small n, we can multiply the binomial repeatedly. For example, for n = 2: ( x + y)2 = ( x + y) × ( x + y) 2

Write the expression as a product 2

= x + xy + yx + y 2

= x + 2xy + y

Expand using the distributive property Combine like terms

The coeﬃcients in this expansion are 1, 2, 1. These patterns become more evident as we expand for higher powers.

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Example 1 Expand ( x + y)3.

Create a strategy Write ( x + y)3 as ( x + y)2 × ( x + y), expand ( x + y)2 ﬁrst, then multiply by ( x + y).

Apply the idea ( x + y)3 = ( x + y) ( x + y) ( x + y) 2

Express as linear factors

= ( x + 2xy + y ) × ( x + y) 3

Perform the ﬁrst multiplication

= x + x y + 2x y + 2xy + y x + y 3

= x + 3x y + 3xy + y

Distribute each term Combine like terms

The expansion is x + 3x y + 3xy + y .

Reﬂect and check The coeﬃcients are 1, 3, 3, 1. We can check by substituting x = 1, y = 1 into the left-hand side and right-hand side of the equation: ( x + y)3 3

(1 + 1)3 = 8

Left-hand side

Right-hand side

1 +3×1 ×1+3×1×1 +1 =8 The values match, conﬁrming the expansion.

Idea summary A binomial expression has two terms, and its expansion for a power n is a sum of terms with coeﬃcients and powers of the variables.

Explore symmetry in binomial coefficients Pascal’s triangle A triangular ﬁgure with rows of numbers starting and ending with 1, where each interior number is equal to the sum of the two numbers immediately above it.

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

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The coefficients in a binomial expansion of ( x + y)n form a symmetric pattern. For example, in ( x + y)3 = x3 + 3x2 y + 3xy2 + y3 , the coefficients are 1, 3, 3, 1, which are symmetric about the middle. The rows of Pascal’s triangle are numbered starting from row 0 at the top. Therefore, row n contains the coefficients for the expansion of ( x + y)n. For example, the coefficients for ( x + y)2 are 1, 2, 1, which are found in row 2. These coefficients can be arranged in Pascal’s triangle, where each row corresponds to the coefficients of ( x + y)n for n = 0, 1, 2, …. Each row of Pascal’s triangle gives the coefficients for ( x + y)n. For example, the third row (1, 3, 3, 1) matches ( x + y)3. 0th row

1 1 1 1 1 1 1 1

5 6

3rd row

3 6

2nd row

2 3

1st row

4th row

15 20 15 6

7 21 35 35 21

5th row

10 10 5

6th row

1 1

7th row

Observe, for example, how row 4 of Pascal’s triangle is formed by adding adjacent pairs of numbers from row 3, such as 1 + 3 = 4 and 3 + 3 = 6. Similarly, to form row 5, start with row 4: 1, 4, 6, 4, 1. Begin and end with 1. The middle numbers are calculated as follows: 1 + 4 = 5, 4 + 6 = 10, 6 + 4 = 10 and 4 + 1 = 5. Thus, row 5 is 1, 5, 10, 10, 5, 1.

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Example 2 List the coeﬃcients of ( x + y)4 and identify the corresponding row in Pascal’s triangle.

Create a strategy Expand ( x + y)4 by repeated multiplication or recall the coeﬃcients from Pascal’s triangle row 4.

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Apply the idea ( x + y)4 = ( x + y)2 × ( x + y)2 2

Express as a product of squares

= ( x + 2xy + y ) × ( x + 2xy + y )

Use the expansion of ( x + y)2

= x4 + 2x3 y + x2 y2 + 2x3 y + 4x2 y2 + 2xy3 + x2 y2 + 2xy3 + y4

Distribute terms

= x + 4x y + 6x y + 4xy + y

Combine like terms

The coeﬃcients are 1, 4, 6, 4, 1, which correspond to row 4 of Pascal’s triangle.

Reﬂect and check Notice the symmetry: 1, 4, 6, 4, 1 reads the same forwards and backwards. This matches the pattern in Pascal’s triangle.

Example 3 Expand (3x − 1)4 using Pascal’s triangle.

Create a strategy Use Pascal’s triangle to figure out the coefficients for each term. Sub in the first and second term from the binomial expression, noting the power of the first term (3x) starts at 4 and decreases to 0, while the power of the second term (−1) starts at 0 and increases to 4.

Apply the idea For power of 4, the coeﬃcients are 1, 4, 6, 4, 1 from Pascal’s triangle, hence: (3x − 1)4 = (3x)4(−1)0 + 4(3x)3(−1)1 + 6(3x)2(−1)2 + 4(3x)1(−1)3 + (3x)0(−1)4 Expand = 81x4 − 108x3 + 54x2 − 12x + 1

Evaluate

Reflect and check The sum of the powers in each term equals n = 4, as seen in the exponents: 4 + 0, 3 + 1, 2 + 2, 1 + 3, 0 + 4 For clarity, terms with a power of 0 or a coefficient of 1 are often simplified, so the expansion can be written as: (3x)4 + 4(3x)3(−1) + 6(3x)2(−1)2 + 4(3x) (−1)3 + (−1)4

Idea summary The coeﬃcients of ( x + y)n are symmetric and form the nth row of Pascal’s triangle.

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12.04E Practice questions What do you remember? 1

What is a binomial expression?

Expand ( x + y)1 and identify the coeﬃcients.

Write the ﬁrst four rows of Pascal’s triangle (for n = 0, 1, 2, 3).

True or false? a

The expansion of ( x + y)2 is x2 + xy + y2.

The coeﬃcients of ( x + y)3 are 1, 2, 2, 1.

Practice Ex 1

Expand ( x + y)5.

Ex 2

List the coeﬃcients of these expansions using Pascal’s triangle. a

Ex 3

( x + y)6

( x + y)7

( x + y)8

(3x + 2)3

( x + 2)4

(2x − 3)6

Expand using Pascal’s triangle: a

( x + 2)3

(2x − 1)

( x + 3)4

( x − 2)

(2x + 1)

Identify the speciﬁed term in the expansion of ( x + y)4. a c

( x + y)5

The term containing x2 y2 3

The term containing xy

The term containing x3 y

The term containing y4

Identify the speciﬁed term in the expansion of ( x + y)5 a

The term containing x4 y

The term containing x3 y2

2 3

The term containing xy4

The term containing x y

Expand (2x + 3y)3 and simplify. Then identify the coeﬃcient of x2 y.

Expand (2x − 1)4 and simplify. Then identify the coeﬃcient of x3.

Extend your thinking 12

Explain why the coeﬃcients in the expansion of ( x + y)n are symmetric, using ( x + y)3 as an example.

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13 A company models its proﬁt as (100 + x)2, where x is the increase in sales (in thousands of dollars). a

Expand the expression to determine the proﬁt components.

Interpret the terms in the expanded expression and explain why a company might choose to model proﬁt in this way.

Show that the sum of the coeﬃcients in the expansion of ( x + y)n is 2n by substituting x = 1, y = 1. Use n = 4 as an example.

The expansion of ( x − y)5 is written incorrectly as x5 − 5x4 y + 10x3 y2 − 15x2 y3 + 5xy4 − y5. Identify and correct the error.

Expand and simplify: a

Find the term independent of x in (3x − 2)3

b .

Did you know?

The binomial theorem helps us expand expressions like (a+b)n without having to multiply them out one step at a time! This powerful shortcut is used in everything from computer algorithms to probability modelling. Graphic designers even use it to calculate gradients and smooth curves in digital art. It’s all about patterns, symmetry, and coefficients—just like in nature! Next time you admire a perfectly curved shape or a digital design, there might be a little binomial maths behind the magic!

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12.05E Coefficients in a binomial expansion After this lesson, you will be able to… (or nCr) as the coefficient of xn − r yr in the

• define a binomial coefficient expansion of ( x + y)n. =

• recall and use the formula

to calculate binomial coefficients.

• understand and use factorial notation n! in calculations, including 0! = 1. • recognise the equivalence between the coefficient of xn − r yr in the expansion of ( x + y)n and

• calculate the coefficient of a specific term in a binomial expansion. • interpret

as the number of ways to choose r items from n distinct items.

Coefficients in a binomial expansion Binomial coeﬃcient The coeﬃcient of the term xn − r yr in the expansion of ( x + y)n. It is written as nCr or where r = 0, 1, 2, … , n and is given by

In Pascal’s triangle, the uppermost row is row 0. The entries in row n are the binomial coeﬃcients

where r = 0, 1, 2, … , n of the expansion of ( x + y)n.

In the expansion of ( x + y)n , each term has the form coeﬃcient.

, where

is the coeﬃcient of the term n r

608

is the power of the binomial is the power of y in the term

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

is the binomial

The binomial coeﬃcient

is calculated using the formula:

is the factorial of n, i.e., n × (n − 1) × … × 1 is the factorial of r is the factorial of n − r

n! r! (n − r)! Combinatorially,

represents the number of ways to choose r items from n items, which

corresponds to selecting r factors of y from n factors in the expansion.

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Example 1 Find the coeﬃcient of x4 y5 in the expansion of ( x + y)9.

Create a strategy Identify the power n from the binomial ( x + y)9 and the value r from the power of y in the term x4 y5. Then, calculate the coefficient

Apply the idea From the expression ( x + y)9, the power is n = 9. For the term x4 y5, the power of y provides the value r = 5. The coeﬃcient is

. Write the formula

Substitute n = 9 and r = 5

Evaluate the subtraction

Expand the factorials

Remove the common factor

Evaluate the multiplication

Evaluate the division

The coeﬃcient of x4 y5 is 126.

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Reﬂect and check Verify using

, since

: Write the formula

Substitute n = 9 and r = 4

Evaluate the subtraction

Expand the factorials

Remove the common factor

Evaluate the multiplication

Evaluate the division

Same as

, conﬁrming symmetry.

Example 2 Calculate the binomial coeﬃcient of x2 y3 in the expansion of ( x + y)5.

Create a strategy

Apply the idea

Identify the power n from the binomial ( x + y)5 and the value r from the power of y in the term

From the expression (x + y)5, the power is n = 5.

x2 y3. Then, calculate the coefficient

For the term x2 y3, the power of y provides the value r = 3. The coeﬃcient is

. Write the formula

Substitute n = 5 and r = 3

Evaluate the subtraction

Expand the factorials

Remove the common factor

Evaluate the multiplication

Evaluate the division

The coeﬃcient of x2 y3 is 10.

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Reﬂect and check Verify using

, since

: Write the formula

Substitute n = 5 and r = 2

Evaluate the subtraction

Expand the factorials

Remove the common factor

Evaluate the multiplication

Evaluate the division

Same as

, conﬁrming symmetry.

Idea summary The coeﬃcient of xn − r yr in the expansion of ( x + y)n is

, which also

represents the number of ways to choose r items from n.

12.05E Practice questions What do you remember? 1

What is a binomial coeﬃcient?

Calculate

In the expansion of ( x + y)n, write the general form of any term.

True or false? a

The coeﬃcient of x3 y3 in ( x + y)6 is

The statement

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Practice Ex 1

Ex 2

Find the coeﬃcient of the speciﬁed term in the expansion of (2x + y)6. a

The term containing x4 y2

The term containing x3 y3

The term containing x2 y4

The term containing xy5

Calculate the coeﬃcient for the speciﬁed term in the expansion of ( x + y)6. a

The term containing x4 y2

The term containing x3 y3

The term containing x2 y4

The term containing y6

Calculate the coeﬃcient for the speciﬁed term in the expansion of ( x + y)7. a

The term containing x5 y2

The term containing x4 y3

The term containing x3 y4

The term containing x2 y5

Find the coeﬃcient of the speciﬁed term in the expansion of ( x − 3y)5. a

The term containing x3 y2

The term containing x2 y3

The term containing xy4

The term containing y5

Calculate the coeﬃcient of x6 y4 in the expansion of ( x + y)10.

Calculate the coeﬃcient of x3 y3 in the expansion of (2x + 3y)6.

Calculate the coeﬃcient of x4 in the expansion of ( x − 2)6.

If the coeﬃcient of x3 y5 in ( x + y)n is 56, determine n.

Extend your thinking 13

A student incorrectly calculates the coeﬃcient of x5 y3 in ( x + y)8 as correct the error.

Calculate the coeﬃcient of the x4 term in the expansion of (1 + x + x2)3.

Show that

= 56. Identify and

equals the coeﬃcient of xn − r yr in ( x + y)n by calculating both for n = 5, r = 2

and comparing. 16

In the expansion of ( x + y)n, determine the value of n such that the coefficient of x6 y4 equals the coefficient of x4 y6, and calculate this common coefficient.

Express the sum of the coeﬃcients of all terms in the expansion of ( x + y)8 where the exponent of x is even, and calculate its value.

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12.06E Properties of Pascal’s triangle After this lesson, you will be able to… • recall and understand the recursive identity of Pascal’s triangle: . • recall and understand the symmetry identity of Pascal’s triangle:

• verify the recursive identity for specific values of n and r using calculations or by referencing Pascal’s triangle. • verify the symmetry identity for specific values of n and r using calculations or by referencing Pascal’s triangle. • use patterns and symmetry in Pascal’s triangle to confirm these identities.

Recursive property of Pascal’s triangle In Pascal’s triangle, each entry is the sum of the two entries directly above it. This corresponds to the recursive identity for binomial coeﬃcients:

is the binomial coefficient for row n, position r is the coefficient from the previous row, position r − 1 is the coefficient from the previous row, position r This identity holds for 1 ≤ r ≤ n − 1 and reflects how each coefficient is formed by combining coefficients from the row above.

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Example 1 Verify the identity

using Pascal’s triangle.

Create a strategy Calculate

, and

using the binomial coeﬃcient formula and conﬁrm their sum.

Apply the idea For

: Write the formula

Substitute n = 5 and r = 2

Evaluate the subtraction

Expand the factorials

Remove the common factors

Evaluate the multiplication

Evaluate

For

: Write the formula

Substitute n = 4 and r = 1

Evaluate the subtraction

Expand the factorials

Remove the common factors

Evaluate

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For

: Write the formula

Substitute n = 4 and r = 2

Evaluate the subtraction

Expand the factorials

Remove the common factors

Evaluate the multiplication

Evaluate

Substituting the obtained values in the identity: Write the identity

Substitute

Evaluate

The identity

= 10,

= 4,

holds.

Reﬂect and check In Pascal’s triangle, the entry in row 5, position

is the sum of row 4, positions 1 and

Idea summary The recursive identity

shows that each binomial coeﬃcient in

Pascal’s triangle is the sum of the two coeﬃcients above it.

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Symmetry in Pascal’s triangle Pascal’s triangle exhibits symmetry, where coeﬃcients at opposite ends of a row are equal. This is expressed by the identity:

is the binomial coefficient for row n, position r is the coefficient for row n, position n − r This identity holds for 0 ≤ r ≤ n and reflects the mirror-like structure of each row in Pascal’s triangle.

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Example 2 Verify the identity

algebraically.

Create a strategy Calculate

and

using the binomial coeﬃcient formula and compare their values.

Apply the idea For

: Write the formula

Substitute n = 6 and r = 2

Evaluate the subtraction

Expand the factorials

Remove the common factors

Evaluate the multiplication

Evaluate

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For

: Write the formula

Substitute n = 6 and r = 4

Expand the factorials

Remove the common factors

Evaluate the multiplication

Evaluate

holds, as 15 = 15.

The identity

Reflect and check In Pascal’s triangle, row 6 has coefficients 1, 6, 15, 20, 15, 6, 1. Positions 2 and 4 both have 15, confirming symmetry.

Idea summary The symmetry identity

shows that coeﬃcients at opposite ends of a

row in Pascal’s triangle are equal.

12.06E Practice questions What do you remember? 1

How are the numbers in each row of Pascal’s triangle related?

State the recursive identity for binomial coeﬃcients in Pascal’s triangle.

State the symmetry identity for binomial coeﬃcients in Pascal’s triangle.

Determine whether these statements are true or false: a

In Pascal’s triangle,

The coeﬃcient

equals equals

. due to symmetry.

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Practice Ex 1

Verify the recursive identity a

Ex 2

n = 5, r = 1

n = 4, r = 2

n = 5, r = 4

n = 6, r = 2

n = 6, r = 3

n = 7, r = 2

n = 4, r = 2

n = 5, r = 2

n = 6, r = 3

n = 6, r = 4

Conﬁrm the symmetry identity triangle row.

Calculate

Verify algebraically that

n = 8, r = 4

n = 7, r = 3

, then conﬁrm using the binomial coeﬃcient c

n = 7, r = 5

n = 8, r = 6

by listing the relevant Pascal’s triangle

n = 5, r = 3

n = 7, r = 4

, then conﬁrm using the binomial coeﬃcient

Conﬁrm the recursive identity rows. a

for these values algebraically.

Use the symmetry identity to ﬁnd the value of formula. a

n = 5, r = 3

Use the recursive identity to ﬁnd the value of formula. a

Verify the symmetry identity a

n = 4, r = 2

for these values using Pascal’s triangle.

n = 6, r = 4

n = 7, r = 2

for n = 6, r = 2 by listing the relevant Pascal’s

using the recursive identity twice, starting from known values in row 4. using the symmetry identity.

Extend your thinking 13

A student claims that

but calculates

= 10,

= 10, and concludes

10 + 10 = 20. Identify and correct the error. 14

Explain why the symmetry identity structure of Pascal’s triangle.

Prove algebraically that formula.

Show

holds for all 0 ≤ r ≤ n by considering the

for 1 ≤ r ≤ n − 1 using the binomial coeﬃcient

= 32 by manually calculating each component of

the LHS.

618

Hence, using the symmetry identity, show

What does the result in part (b) mean for the binomial expansion (1 + x)5?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

12.07E The binomial theorem After this lesson, you will be able to… • state the binomial theorem for ( x + y)n using summation notation and in expanded form. • identify the components of the binomial theorem formula, including binomial coefficients and powers of terms. • understand that the binomial theorem generalises the process of expanding binomials for positive integer powers n. • derive the binomial theorem for small specific values of n (e.g., n = 3) by generalising from manual expansion. • verify the binomial theorem for specific cases by substitution.

The binomial theorem Binomial theorem The formula for the expansion of a power of a binomial expression: ( x + y)n = nC0 xn + nC1 xn − 1 y + nC2 xn − 2 y2 + … + nCn − 1 xyn − 1 + nCnyn when n is a positive integer and nC0 , nC1 , ..., nCn are binomial coeﬃcients. The binomial theorem provides a general formula for expanding ( x + y)n for a positive integer n. It states:

∑

is the sum over a range of values (from r = 0 to r = n) is the binomial coeﬃcient,

xn − r yr is the term with x to the power n − r and y to the power r r

is the index ranging from 0 to n

This can be written explicitly as:

are the binomial coeﬃcients n

is a positive integer power

x, y

are the terms of the binomial 12.07E The binomial theorem mathspace.co

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The theorem generalises the pattern observed in expansions. For example, for n = 2: Expand ( x + y)2 manually

Apply the binomial theorem to ( x + y)2

Equate coeﬃcients: come from Pascal’s triangle, and the powers of x and y sum to n in each term.

The coeﬃcients

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In a binomial expansion (a + b)n, setting a = 1 and b = 1 gives the sum of all coeﬃcients: (1 + 1)n = 2n

Example 1 Derive the binomial theorem for n = 3 by expanding ( x + y)3 and generalising the pattern.

Create a strategy Expand ( x + y)3 manually, identify the coeﬃcients and terms, and express the result using

Apply the idea ( x + y)3 = ( x + y) ( x + y) ( x + y) 2

Express as linear factors

= ( x + 2xy + y ) × ( x + y) 3

Perform the ﬁrst multiplication 2

= x + x y + 2x y + 2xy + y x + y 3

= x + 3x y + 3xy + y

Distribute each term Combine like terms

The coeﬃcients are 1, 3, 3, 1. Substitute n = 3 and r = 0, 1, 2, 3 in the binomial coeﬃcient formula to prove.

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For

Write the formula

Substitute n = 3 and r = 0

Substitute n = 3 and r = 1

Evaluate the subtraction

Expand the factorial 0!

Expand the factorials

Remove the common factors

Evaluate

The coeﬃcient of x2 y is 3.

The coeﬃcient of x3 is 1. For

For

: Write the formula

Substitute n = 3 and r = 2

Substitute n = 3 and r = 3

Evaluate the subtraction

Expand the factorials

Expand the factorial 0!

Remove the common factors

Evaluate

The coeﬃcient of y3 is 1.

The coeﬃcient of xy is 3. Thus, Generalising,

. .

Reﬂect and check The coeﬃcients match row 3 of Pascal’s triangle: 1, 3, 3, 1. The pattern of the general form.

and powers suggests

Example 2 Verify the binomial theorem for n = 4 by substituting x = 1, y = 1 in ( x + y)4.

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Create a strategy Apply the binomial theorem to expand ( x + y)4, substitute x = 1, y = 1, and compare with (1 + 1)4.

Apply the idea Substituting n = 4 into the right-hand side of the formula: Substitute n = 4

Expand

Evaluate the coeﬃcients

Adding the coeﬃcients obtained: Add the coeﬃcients Evaluate Substituting x = 1, y = 1 into the left-hand side of the formula: ( x + y)4 = (1 + 1)4 4

Substitute x = 1, y = 1

Evaluate the addition

= 16

Evaluate

The values match, verifying the theorem for n = 4.

Reﬂect and check The sum of coeﬃcients 1 + 4 + 6 + 4 + 1 = 16 equals 24, consistent with the binomial theorem’s structure.

Idea summary The binomial theorem states that

, providing a general

formula for expanding binomials using binomial coeﬃcients.

12.07E Practice questions What do you remember? 1

State the binomial theorem for expanding ( x + y)n.

In the expansion of ( x + y)n, what is the role of the binomial coeﬃcient

Write the ﬁrst three terms of the expansion of ( x + y)4 using the binomial theorem.

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Determine whether these statements are true or false: a

The binomial theorem applies only to positive integer powers of n.

The sum of the exponents in each term of ( x + y)n is n − 1.

Practice Ex 1

In the expansion of ( x + y)5, write the binomial coeﬃcient for each term: a

Ex 2

x4 y

x 3 y2

x 2 y3

Complete the expansion of (a + b)4 by ﬁlling in the missing coeﬃcients and terms. a4 + ⬚ + 6a2 b2 + ⬚ + b4

For the binomials ( x + y)3 and( x − y)3: a

Expand and determine the coeﬃcients of these binomials using the binomial theorem.

Identify the pattern in their coeﬃcients.

Verify the pattern for the coefficient of the term where r = 2.

Use the binomial theorem to ﬁnd the sum of the coeﬃcients in each expansion by substituting x = 1 and y = 1. a

( x + y)9

( x + y)6

( x − y)4

( x + 2y)3

Use the binomial theorem to expand (1 + x)n for these values of n and simplify. a

n=3

n=4

n=5

n=6

Expand ( x + 2)5.

Use the binomial theorem to ﬁnd the value of (1.1)4 by expanding (1 + 0.1)4 and simplifying.

Use the binomial theorem to expand (2x + y)6 and identify the coeﬃcient of x4 y2.

Extend your thinking 13

A student claims the expansion of ( x + y)3 is x3 + 2x2 y + 2xy2 + y3. Use the binomial theorem to correct the error and explain the mistake.

Use the binomial theorem to ﬁnd the values of: a

0.984

Find the term independent of x in each of the following by using the binomial theorem: a

973

Simple algebra exponential rules show that (1 + x)8 = (1 + x)3(1 + x)5. By expanding both sides using the binomial theorem, show that:

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12.08E Apply the binomial theorem After this lesson, you will be able to… • apply the binomial theorem to expand expressions of the form ( x + a)n, where a is a constant. • apply the binomial theorem to expand expressions of the form (ax + b)n, where a and b are constants. • apply the binomial theorem to expand expressions of the form (ax + by)n. • simplify terms in a binomial expansion by evaluating powers of constants and combining numerical coefficients. • correctly handle negative signs within binomial terms during expansion and simplification.

Apply the binomial theorem The binomial theorem applies to various forms, such as ( x + y)n, ( x + a)n (with a constant) or (ax + by)n (with coefficients). After expansion, terms are simplified by combining coefficients and variables.

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Example 1 Use the binomial theorem to expand and simplify ( x + 2)4.

Create a strategy Apply the binomial theorem to express ( x + 2)4 as a sum of terms. Substitute y = 2 and n = 4 into the formula, compute the binomial coeﬃcients polynomial.

624

, evaluate each term and simplify the resulting

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea Write the formula

Substitute y = 2 and n = 4

Expand the sum

Evaluate the coeﬃcients and powers

Simplify

The simpliﬁed expansion is x4 + 8x3 + 24x2 + 32x + 16.

Reflect and check The coefficients 1, 4, 6, 4, 1 correspond to row 4 of Pascal’s triangle. To verify, substitute x = 1: (1 + 2)4 = 34 = 81 and 1 + 8 + 24 + 32 + 16 = 81, confirming the expansion is correct.

Example 2 Use the binomial theorem to expand and simplify ( x + y)5.

Create a strategy Apply the binomial theorem to write the expansion as a sum, compute each terms.

and simplify the

Apply the idea Write the formula

Substitute n = 5

Expand the sum

Evaluate the coeﬃcients

The simpliﬁed expansion is x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5.

Reflect and check The coefficients 1, 5, 10, 10, 5, 1 match row 5 of Pascal’s triangle, and the exponents sum to 5 in each term, confirming the expansion.

12.08E Apply the binomial theorem mathspace.co

625

Example 3 Use the binomial theorem to expand and simplify (2x + 3)4.

Create a strategy Apply the binomial theorem with x = 2x, y = 3, compute each term and simplify by combining coeﬃcients and variables.

Apply the idea Write the formula:

Substitute y = 3 and n = 4:

Expand the sum:

Evaluate the coeﬃcients and powers: (2x + 3)4 = 16x4 + 96x3 + 216x2 + 216x + 81 The simpliﬁed expansion is 16x4 + 96x3 + 216x2 + 216x + 81.

Reﬂect and check Substituting x = 1 gives (2 × 1 + 3)4 = 54 = 625, and 16 + 96 + 216 + 216 + 81 = 625, conﬁrming the expansion.

Idea summary The binomial theorem expands and simpliﬁes expressions like ( x + y)n or (ax + by)n by computing

and combining terms into a simpliﬁed polynomial.

12.08E Practice questions What do you remember? 1

626

What is the purpose of the binomial theorem in expanding expressions? Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Write the binomial theorem formula for expanding ( x + y)n.

Expand ( x + y)3 using the binomial theorem and simplify.

Determine if each statement is true or false. If a statement is false, explain why. a

The binomial theorem can be used to expand (2x + 3y)4.

The expansion of ( x + y)n always has n terms.

Practice Ex 1

Expand and simplify using binomial theorem: a

Ex 2

(z + 4)3

( x + y)4

(a + b)5

( p + q)3

(u + v)6

( x − y)4

( x − 2)5

( y − 1)3

(z − 3)4

(2x + y)3

(3x + 2y)4

(2x + 3y)3

( x + 2y)5

( x + y)4

( x + y)5

( x − y)3

( x + 3y)4

(3x + 2y)9

( x − 2y)7

Expand and simplify using the binomial theorem: a

( y + 3)5

Verify the expansion of each binomial by substituting x = 1, y = 1 and comparing both sides of the equation. a

Use the binomial theorem to expand and simplify each binomial: a

( x + 2)3

Expand and simplify using binomial theorem: a

Ex 3

Expand and simplify using binomial theorem: a

( x + 1)4

( x + 2)6

(2x − y)8

Expand ( x + 1)4 and ( x − 1)4 using the binomial theorem, then ﬁnd their sum and simplify.

Extend your thinking 12

Show that the sum of the coeﬃcients in the expansion of ( x + y)5 is 25 by substituting x = 1, y = 1.

A student expands ( x + y)4 as x4 + 3x3 y + 6x2 y2 + 3xy3 + y4. Identify and correct the error using the binomial theorem.

Determine the value of n such that the coeﬃcient of the speciﬁed term in the expansion of ( x + 1)n matches the given value: a

x4, coeﬃcient 126

x3, coeﬃcient 35

The coeﬃcient of a3 b3 in the expansion of (a + 2b)n is 160. Find n and conﬁrm the term.

The coeﬃcients of the 6th and 8th term of (2x + 3y)n have a ratio 14 : 9. Find n.

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12.09E Specific terms in a binomial expression After this lesson, you will be able to… • recall and use the general term formula Tr + 1 = x n − r y r for the binomial n expansion of ( x + y) . • determine the value of r corresponding to a specific term in a binomial expansion (e.g., the term containing xk). • calculate the coefficient of a specific term in a binomial expansion, including all numerical factors. • find the constant term (term independent of x) in a binomial expansion. • apply these techniques to binomials where terms may include coefficients or variables in the denominator.

Specific terms in a binomial expression The binomial theorem provides a method to expand expressions of the form( x + y)n, where n is a positive integer. It allows identification of specific terms or coefficients in the expansion without computing the entire expression. The binomial theorem states that:

Here,

is the binomial coeﬃcient, calculated as

, representing the number of ways to

choose r items from n. The general term in the expansion is:

To find the coefficient of a specific term, identify the value of r that gives the desired powers of x and y, then compute

and any numerical coeﬃcients.

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mathspace.co

Example 1 Consider the binomial expansion of

a Given that the powers of x in the expansion are in decreasing order, write an expression for the (r + 1)th term. Fully simplify the answer.

Create a strategy Apply the binomial theorem to ﬁnd the general term, where a = x and b = . Simplify the expression by combining the exponents of x.

Apply the idea The general term in the expansion of

is:

Simplify the expression by combining the exponents of x: Rewrite

as x

Combine the exponents

Simplify the exponent

The (r + 1)th term is

b What must be the value of n so that the 8th term of the expansion is a constant?

Create a strategy Use the general term from part (a). The 8th term corresponds to r = 7 (since Tr + 1 = T8 when r = 7). Set the exponent of x to 0 (since the term is constant) and solve for n.

Apply the idea From part (a), the general term is

The 8th term corresponds to r = 7:

For the term to be constant, the exponent of x must be 0: n − 14 = 0 n = 14

Set the exponent to 0 Solve for n

The value of n must be 14.

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Reﬂect and check Substitute n = 14 into the general term to conﬁrm: constant.

, which is indeed a

Example 2 Find the coeﬃcient of x3 in the expansion of

Create a strategy Use the general term

to ﬁnd the term where the power of x is 3.

Compute the binomial coeﬃcient and any numerical factors.

Apply the idea The general term in the expansion of

is:

Simplify the term: Expand the powers

Combine exponents of x

Simplify the exponent

Set the exponent of x to 3: 7 − 2r = 3 Equate the exponent to 3 −2r = −4 Subtract 7 from both sides r = 2 Divide both sides by −2 Since r = 2, the term containing x3 is the third term, T3. Substitute r = 2 into the simplified general term to find its value: Write the simplified term

Substitute r = 2

Evaluate the binomial coefficient and powers

Evaluate the final product

The term is 20 412x3, so the coeﬃcient of x3 is 20 412.

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Idea summary The binomial theorem expands ( x + y)n to:

To ﬁnd a speciﬁc term: .

• Use the general term

• Set the exponent of the variable to match the desired term and solve for r • Compute

and numerical coeﬃcients.

12.09E Practice questions What do you remember? 1

What is the general term Tr + 1 in the binomial expansion of ( x + y)n?

In the expansion of (a + b)5, what are the powers of a and b in the term corresponding to r = 3?

Determine whether these statements are true or false:

The general term of ( x + y)n is

The binomial coeﬃcient

In the expansion of ( x + y)4, the sum of the exponents in each term equals 4.

The constant term in

is calculated as

occurs when r = 3.

Write the binomial coeﬃcients for the expansion of ( x + y)4 using Pascal’s triangle.

Practice Ex 1

Ex 2

Consider the binomial expansion of

Write an expression for the (r + 1)th term, fully simpliﬁed.

Find the value of n so that the 6th term is a constant.

Find the coeﬃcient of x2 in the expansion of

Find the coeﬃcient of x2 in the expansion of ( x + 3)4.

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Find the constant term in the expansion of

Determine:

The coeﬃcient of x3 in (2x + 1)5

The coeﬃcient of x2 in (3x − 2)4

The constant term in

The term independent of x in

Find the coeﬃcient of x5 in the expansion of

Determine the coeﬃcients or constant terms:

In ( x + 2)6, the coeﬃcient of x4

, the constant term

In (2x + 3)5, the coeﬃcient of x3

, the coeﬃcient of x

For what value of r does the term x4 y3 appear in the expansion of ( x + y)7?

In a binomial experiment with 6 trials and probability of success p, the probability of exactly r successes is given by

. What is the probability of exactly 4 successes?

Find the coeﬃcient of x9 in the expansion of

Extend your thinking 15

Determine the constant term in the expansion of

Consider the expansion of

The coeﬃcient of x6 is claimed to be 11 520. Verify this by calculating the coeﬃcient.

Explain why the coeﬃcient must be positive.

If the expression were

Generalise how changing the sign of the second term affects coeﬃcients in binomial expansions.

, how would the coeﬃcient of x6 change?

is 160. If a and b are positive integers, ﬁnd

The constant term in the expansion of their possible values.

A student incorrectly assumes the coeﬃcient of x3 in compute the correct coeﬃcient.

Determine the values of n and r such that the term x3 in ( x + 2)n has a coeﬃcient of 40.

Find the term independent of x in the following expansion

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is 1008. Identify the error and

12.10E Simplify expressions with binomial coefficients After this lesson, you will be able to… • recall the identities

= 1 and

= 1.

• recall and apply the symmetry identity

to simplify expressions.

= + to simplify sums of • recall and apply Pascal’s identity binomial coefficients. • simplify expressions involving sums, differences or combinations of binomial coefficients using these identities. • verify simplifications by substituting numerical values where appropriate.

Simplify expressions with binomial coefficients Identity An identity is a statement involving a variable(s) that is true for all possible values of the variable(s).

Binomial coeﬃcients, denoted

, arise in the binomial theorem and represent the number of

ways to choose r items from n items. Simplifying expressions involving binomial coefficients often requires the use of specific identities. The following identities are useful for simplification: • • •

= 1 and

= 1 for all positive integers n for 0 ≤ r ≤ n (symmetry identity) for 1 ≤ r ≤ n − 1 (Pascal’s identity)

These identities allow expressions involving binomial coeﬃcients to be rewritten in simpler forms or evaluated directly.

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Example 1 Consider the expression

a Simplify the expression.

Create a strategy Apply Pascal’s identity, which is given in the theory as expression

. Match the given

to the right-hand side of this identity form.

Apply the idea Pascal’s identity is stated in the theory as:

Our expression is

. , identify the corresponding parts:

To match the right-hand side of Pascal’s identity,

• Let n − 1 (from the identity) correspond to x − 1 in our expression. This implies that the n in the identity corresponds to x in our problem. • Let r − 1 (from the identity) correspond to 3. This implies that the r in the identity corresponds to 4. . Substituting our correspondences, this becomes

• The second term in the identity is

, which matches the second term in our expression. So, with the n of the identity being our x, and the r of the identity being 4, the expression is the right-hand side of Pascal’s identity. Therefore, it simpliﬁes to the left-hand side of the identity, correspondences.

, which becomes

using our

Apply Pascal’s identity The simpliﬁed expression is

b Verify the result from part (a) by making the substitution x = 9.

Create a strategy Substitute x = 9 into the original expression x = 9 into the simpliﬁed expression are equal.

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and calculate its value. Then, substitute

from part (a) and calculate its value. Conﬁrm both values

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea Substitute x = 9 into the original expression: Substitute x = 9 Calculate

Simplify superscripts

: Write the formula

Substitute n = 8 and r = 3

Evaluate the subtraction

Expand

Remove the common factors

Evaluate Calculate

: Write the formula

Substitute n = 8 and r = 4

Evaluate the subtraction

Expand

Remove the common factors

Evaluate So,

= 56 + 70 = 126.

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Now, substitute x = 9 into the simpliﬁed expression

from part (a): Write the formula

Substitute x = 9

Evaluate the subtraction

Expand

Remove the common factors

Evaluate Both calculations yield 126.

Example 2 , assuming x ≥ 2.

Consider the expression a Simplify the terms

and

separately.

Create a strategy = 1,

Apply the identities

= 1 and

to simplify the pairs of terms. Here, our general

power is x instead of n.

Apply the idea First, simplify

: Apply the identity

= 1 (with n = x)

Apply the identity

= 1 (with n = x)

Substitute the values

Evaluate

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Next, simplify

(with n = x):

using

Apply the identity

Simplify the subscript

Further simplify the subscript

Substitute

Evaluate

b Use the results from part (a) to simplify the entire expression.

Create a strategy Combine the simpliﬁed results from part (a) to evaluate the expression.

Apply the idea From part (a): •

•

Substitute these into the original expression: Group the terms

Substitute the simpliﬁed results

Evaluate

The simplified expression is 2.

Idea summary Binomial coefficients can be simplified using the following identities: • • •

= 1 and

=1 for 0 ≤ r ≤n for 1 ≤ r ≤ n − 1

Apply these identities to rewrite or combine terms in an expression, then evaluate the resulting coefficients.

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12.10E Practice questions What do you remember? 1

Identify whether these statements are true or false: a

= 1 for all positive integers n.

for all r.

Pascal’s identity is

is always zero.

d 2

What is the symmetry identity for binomial coeﬃcients?

Write Pascal’s identity for binomial coeﬃcients.

Write an equivalent expression using the symmetry identity: a

Practice Ex 1

Simplify the expression

Ex 2

Consider the expression

. , where n = 12:

Simplify the terms

Use the result in part (a) to simplify the entire expression.

and

separately.

Write an equivalent expression for

Simplify each of the following sums or differences to a single binomial coeﬃcient. Do not evaluate the expressions. a

using the symmetry identity (do not evaluate).

Verify Pascal’s identity for n = 5 and r = 3 by computing both sides.

Calculate the value of n such that

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= 84.

In a lottery, the number of ways to choose r winning numbers from 6 is

. Calculate the

total number of ways to choose either 2 or 3 winning numbers. 12

Simplify: a

Use Pascal’s identity to express

as a sum of two binomial coeﬃcients.

Extend your thinking 14

for n = 4.

Show that

Explain why this identity holds for any n using the binomial theorem.

Compute

How could this identity be used in probability?

A student simpliﬁes

Prove that

How big must a squad be so that the total number of 3 and 4-player teams is equal to the number of 3-player teams from 7 players?

Verify that

Prove

For the result in part (a), prove it using combinatorial methods.

Deduce

to 28. Identify the error and provide the correct simpliﬁcation. using factorials.

for n = 5. .

for all non-negative n.

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12.11E Proofs with binomial expansions After this lesson, you will be able to… • understand the concept of a mathematical identity involving binomial coefficients. • prove identities involving binomial. • verify given identities for specific values of n.

Proofs with binomial expansions Binomial expansions, derived from the binomial theorem, allow us to express ( x + y)n as a sum of terms involving binomial coeﬃcients

. The theorem states:

represents the number of ways to choose r items from n. These coeﬃcients

Here,

are central to proving identities, which are equations that hold true for all valid inputs. Prove binomial identities by: • Substituting values: Assign specific values to x and y to simplify the expansion and verify the identity. • Comparing coefficients: Equate coefficients of corresponding powers of x in expansions on both sides of the identity. • Combinatorial arguments: Interpret the coefficients as counting the same quantity in different ways. These methods leverage the structure of binomial expansions to establish relationships like or

Interactive exploration Discover this concept in action online

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mathspace.co

Example 1 Prove that

by substituting values into the binomial theorem.

Create a strategy Substitute x = 1 and y = 1 into the binomial expansion of ( x + y)n. Evaluate both sides to verify the identity.

Apply the idea Write the formula

Substitute x = 1 and y = 1

Simplify exponents

Simplify the terms

The identity is proven.

Reﬂect and check This result makes sense combinatorially:

counts subsets of size r from n items, and summing

over all r gives the total number of subsets, which is 2n.

Example 2 Prove that

using a combinatorial argument.

Create a strategy Interpret the left-hand side as counting the number of subsets of a set with n elements, and compare it to the right-hand side.

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Apply the idea Consider a set with n elements. The number of subsets of this set is 2n, since each element can either be included or excluded (2 choices per element). Now, count the subsets by their size: • Subsets with 0 elements: • Subsets with 1 element: • Subsets with n elements: The total number of subsets is the sum of subsets of all possible sizes:

Since this counts all subsets, it equals the total number of subsets:

Thus, the identity holds.

Example 3 Prove that

by comparing coeﬃcients in the binomial expansion.

Create a strategy Consider the binomial expansion of ( x + y)n. The coefficient of xn − r yr is the expansion to find the coefficient of xr yn − r and compare.

. Use the symmetry of

Apply the idea The binomial expansion is:

The general term is:

So, the coeﬃcient of xn − r yr is

Now, consider the term where the power of x is r and the power of y is n − r. Let the index be k:

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We need n − k = r and k = n − r: n−k=r

Set the exponent of x

k=n−r

Solve for k

Substitute k = n − r:

The coeﬃcient of xr yn − r is

Since ( x + y)n = ( y + x)n, the coeﬃcient of xr yn − r in ( y + x)n is

(from the term

Thus:

The identity is proven.

Reflect and check This symmetry reflects Pascal’s triangle, where the rth and (n − r)th entries in row n are equal, confirming our result.

Idea summary Binomial expansions can be used to prove identities involving binomial coeﬃcients. By substituting values, comparing coeﬃcients or using combinatorial arguments, relationships like

and

can be verified.

12.11E Practice questions What do you remember? 1

Identify the binomial theorem for ( x + y)n.

Deﬁne

List three methods to prove binomial coeﬃcient identities.

Write the general term in the expansion of ( x + y)n.

and its combinatorial meaning.

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Practice Ex 1

Verify the identity a

Ex 2

Ex 3

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for the following values of n:

n = 2

n=3

for n ≥ 1:

Consider the identity a

Prove the identity using substitution.

Verify for n = 4.

Verify for n = 2.

for 0 ≤ r < n:

Consider the identity a

Prove the identity using algebraic manipulation.

Verify for n = 5, r = 1.

Verify for n = 6, r = 2. for n ≥ 1:

Consider the identity a

Prove the identity combinatorially.

Verify for n = 4.

Consider the identity

Verify for n = 3.

Verify for n = 1.

Prove the identity using substitution.

Verify for n = 2.

Consider the identity

Prove the identity using a symmetry argument.

Verify for n = 4, r = 1.

Verify for n = 5, r = 3.

Consider the identity

Prove the identity using substitution.

Verify for n = 2.

Consider the identity

Verify for n = 1.

for k ≥ 1:

Prove the identity using comparing coeﬃcients.

Verify for n = 4, k = 2.

Verify for n = 5, k = 1.

Consider the identity

Prove the identity using substitution.

Verify for n = 2.

Verify for n = 1.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the identity a

Prove the identity combinatorially.

Verify for n = 3.

Consider the identity

Verify for n = 2.

Verify for n = 1.

Prove the identity using substitution.

Verify for n = 2.

Extend your thinking 16

Consider the identity

. Provide a combinatorial interpretation.

Consider the identity

Prove using comparing coeﬃcients.

Verify for n = 2.

Verify for n = 3.

Provide a combinatorial interpretation.

Consider the identity a

Prove combinatorially.

Verify for n = 1, m = 3.

Consider the identity a

Prove using comparing coeﬃcients. Verify for n = 3.

Prove using comparing coeﬃcients.

Verify for n = 4.

Prove combinatorially.

Verify for n = 4, k = 2.

Consider the identity

Verify for n = 2.

for even n:

Consider the identity

Verify for n = 2, m = 2.

Consider the identity

Verify for n = 2.

for 0 ≤ k ≤ n: b

Verify for n = 3, k = 1.

Verify for n = 2.

Prove the identity using differentiation.

Verify for n = 3.

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12.12E Prove further identities with binomial coefficients After this lesson, you will be able to… • understand and apply the convention that

= 0 if r < 0 or r > n.

• use the expansion of (1 + x)n to prove identities by substituting values for x. • prove identities by comparing coefficients in products of binomial expansions (e.g., Vandermonde’s identity). • apply previously proven identities to derive or prove further identities. • construct proofs for identities involving sums or combinations of binomial coefficients without using calculus.

Prove further identities with binomial coefficients Binomial coeﬃcients, denoted

, arise in the binomial theorem and combinatorial problems. They

are deﬁned for non-negative integers n and integers r as follows:

The condition

= 0 for cases where r < 0 or r > n is a standard and important convention.

Combinatorially, this means: • If r > n, it signifies that you cannot choose more items than are available (so there are 0 ways). • If r < 0, it signifies that you cannot choose a negative number of items (so there are 0 ways). This convention is crucial as it simplifies the handling of sums involving binomial coefficients, allowing terms outside the primary range 0 ≤ r ≤ n to naturally become zero. It also aligns with the visual representation of Pascal’s triangle, which can be thought of as being surrounded by zeros. These coefficients satisfy various identities that can be used to prove further relationships. Some key identities include: •

(symmetry)

•

(Pascal’s identity)

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•

(sum of binomial coefficients)

Another powerful technique for proving identities involving binomial coeﬃcients is to use the binomial expansion of (1 + x)n:

By substituting speciﬁc values for x (for example, x = 1, x = −1 or x = 2) into this expansion, or by comparing coeﬃcients in products of such expansions, various identities can be derived. For instance: • Substituting x = 1 gives

, which simpliﬁes to

This proves the identity for the sum of all binomial coeﬃcients. • Substituting x = −1 gives

, which simpliﬁes to

This proves the identity for the alternating sum of binomial coeﬃcients. These identities can be combined or manipulated to prove more complex relationships without calculus, relying on algebraic manipulation or combinatorial reasoning. Let’s explore how binomial coeﬃcients relate to each other through Pascal’s identity. Consider constructing Pascal’s triangle and observing how each entry is the sum of the two entries above it.

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 Prove the identity

for non-negative integers n and m.

Create a strategy Recognise the left-hand side as the sum of products of binomial coefficients. Use the binomial theorem by considering the product of two binomial expansions, (1 + x)n(1 + x)m, and find the coefficient of xn in the resulting expansion.

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Apply the idea Consider the product (1 + x)n(1 + x)m = (1 + x)n + m. The binomial expansion of (1 + x)n is:

Similarly, for (1 + x)m:

Their product is: Write the product of the expansions To find the coefficient of xn in this product, consider pairs (r, s) such that r + s = n. Rewrite the second sum with s = n − r: Sum the coefficients where the exponents add to n Now, the right-hand side is (1 + x)n + m, which expands as:

The coeﬃcient of xn is

Equating the coeﬃcients of xn from both sides: Match coeﬃcients Thus, the identity holds.

Reﬂect and check This identity is known as Vandermonde’s identity. It has a combinatorial interpretation: the left-hand side counts ways to choose n items from two groups of sizes n and m, which equals choosing n items from a total of n + m items.

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Example 2 Consider the sum

for positive integers n.

a Prove that

Create a strategy Use the deﬁnition of binomial coeﬃcients to rewrite

and simplify to match

Apply the idea Start with the left-hand side: Expand

Rewrite the factorial

Simplify

Recognise the binomial coeﬃcient

Thus,

b Hence, prove that

Create a strategy Use the property

to transform the sum, then apply the known identity for the sum of

binomial coeﬃcients.

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Apply the idea Start with the left-hand side. Note that when r = 0, the term is

, so we can start the sum from r = 1: Exclude the zero term

Use the identity

: Apply the identity

Factor out n

Reindex the sum by letting k = r − 1. When r = 1, k = 0; when r = n, k = n − 1: Change the index and divide by n noting r = n This is the sum of binomial coeﬃcients for n − 1: Apply Thus: Combine results

Reflect and check Verify for n = 2: Left-hand side is

. Right-hand side is

2 × (22 − 1) = 2 × 2 = 4, confirming the result. This suggests a need to recheck, but our derivation uses a standard identity, confirming correctness for general n.

Idea summary Binomial coefficients satisfy identities like symmetry, Pascal’s identity, and the sum to 2n. These can be used to prove further identities, such as Vandermonde’s identity or weighted sums, by algebraic manipulation or binomial theorem applications, without calculus. The binomial expansion of (1 + x)n is often helpful in proofs involving binomial coefficients.

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12.12E Practice questions What do you remember? 1

State the symmetry identity for binomial coeﬃcients and explain what it means combinatorially.

Write down Pascal’s identity for binomial coeﬃcients and give its bounds for r.

State the identity for the sum of binomial coeﬃcients in a row of Pascal’s triangle.

Explain how the binomial theorem can be used to prove identities involving binomial coeﬃcients.

Practice 5

Verify the following identities using known binomial identities:

a 6

Use the symmetry identity to ﬁnd the value of the following pairs: a

Ex 1

given

= 28

Prove the identity

Use the binomial theorem to verify the identity

Prove the identity

= 21

by calculating each term. for n = 3.

for n = 2, m = 3 by direct calculation.

Compute the left-hand side.

Use the identity

given

Ex 2

Compute the right-hand side.

to evaluate:

Prove the identity

Use the binomial theorem to prove ( x + 1)n.

by direct calculation. by setting an appropriate value for x in

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Use the identity

to compute:

Prove the identity

Extend your thinking 17

Prove the identity

Prove the identity signiﬁcance.

Using the identity

and explain why this identity holds combinatorially.

and explain its combinatorial

Prove that

Verify the result from part (a) for the case where n = 3.

for even n by simplifying the sum, and verify for n = 4.

Prove that

By considering the expansions of (1 + 1)n and (1 − 1)n, prove that the sum of the binomial coefficients with even indices is 2n − 1. That is, for n ≥ 1, prove:

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12 Chapter review 1

Which of the following describes a continuous random variable? A

The number of students present in a classroom.

The number of goals scored in a soccer match.

The exact time it takes for a student to run 100 metres.

The result of rolling a standard six-sided die.

An experiment involves rolling two fair six-sided dice and deﬁning X as the sum of the numbers shown. Which set represents all possible values of X ? A

{1, 2, 3, 4, 5, 6}

{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

{2, 12}

An experiment involves drawing one card from a standard deck of 52 cards. The random variable Z is deﬁned as the card’s face value, with Ace = 1, Jack = 11, Queen = 12, and King = 13. What are the possible values for Z? A

{1, 2, …, 10}

{1, 2, …, 13}

{Hearts, Diamonds, Clubs, Spades}

{1, 11, 12, 13}

An experiment involves tossing four coins. Deﬁne a random variable H for the number of heads: a

What is H ?

List the possible values of H.

A bag contains 6 green and 4 yellow marbles. One marble is drawn. A random variable C is deﬁned as 1 if the marble is green and 0 if it is yellow: a

What does the random variable C represent?

List the possible values of C.

Explain why the number of books on a shelf is a discrete random variable, but the total weight of the books is a continuous random variable.

A survey records the number of beyblades owned by 20 students: 1, 0, 2, 3, 1, 0, 0, 2, 4, 1, 2, 1, 0, 1, 2, 3, 1, 0, 2, 1

Organise the data into a table with values, frequency, and relative frequency.

Add cumulative frequency and cumulative relative frequency to the table.

The number of smoothies sold by a cafe each hour for 12 hours are: 8, 10, 11, 8, 9, 12, 10, 9, 11, 10, 8, 9 a

Construct a frequency table with values, frequency, and cumulative frequency.

Determine the median from the cumulative frequency.

Calculate the relative frequency for each value.

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A dataset records the number of hours 20 students spent gaming in a week: 5, 8, 6, 9, 6, 5, 8, 7, 10, 8, 7, 5, 9, 7, 8, 5, 7, 8, 7, 5

Create a frequency histogram.

Identify the mode(s) from the histogram.

Create a cumulative frequency histogram.

Estimate the median using cumulative frequency histogram.

A dataset of 30 commute times (minutes) is grouped into intervals: 10–20, 20–30, 30–40, 40–50, 50–60 Frequencies are 5, 9, 8, 3, 5:

Create a frequency histogram.

Identify the modal class.

Create a cumulative frequency histogram.

Estimate the median using cumulative frequency histogram.

A dataset of 50 plant heights (in cm) is grouped into intervals: 0–10, 10–20, 20–30, 30–40, 40–50 Frequencies are 4, 11, 18, 12, 5:

Create a cumulative frequency polygon.

Identify the modal class.

Determine the median using the cumulative frequency polygon.

A dataset records the number of hours 15 students slept: 7, 8, 9, 7, 8, 9, 10, 8, 7, 9, 8, 7, 9, 8, 7

Construct a frequency table with values, frequency, and relative frequency.

Create a frequency histogram and identify the mode.

Create a cumulative frequency polygon.

Estimate the median using the cumulative frequency polygon.

Two datasets record the quiz scores (out of 10) for two classes of 15 students: • Class A: 5, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 10, 10 • Class B: 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10

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Create cumulative frequency polygons for both datasets on the same axes.

Compare the medians of the two classes using the polygons.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A spinner with 8 sections (numbered 1 to 8) is spun 200 times. The table shows the outcomes: Outcome

Frequency

Estimate P ( X = 4).

Estimate the probability of landing on a number greater than 5.

Predict the number of times a 4 appears in 500 spins.

A quality control test checks 300 smartphones, ﬁnding 12 defective: a

Estimate the probability a smartphone is defective.

Estimate the probability a smartphone is not defective.

A factory tests 400 laptops, ﬁnding 16 faulty: a

Estimate P (faulty).

Estimate P (not faulty).

Predict the number of faulty laptops in a new batch of 1500, keeping manufacturing conditions consistent.

A card is drawn from a deck of 52 cards 250 times (with replacement), resulting in 55 clubs: a

Estimate P (club).

Estimate P (not a club).

Predict the number of clubs in 800 draws.

A coin is ﬂipped 300 times, yielding 165 tails. Estimate P (tails) and discuss whether this suggests the coin is biased.

From 60 die rolls, it was found that a number greater than 4 (a 5 or a 6) appeared 24 times: a

Based on this experiment, what is the relative frequency of rolling a number greater than 4?

Compare this to the theoretical probability of the same event.

Why might the experimental probability differ from the theoretical probability in this case?

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20E Which of the following lists the coeﬃcients for the expansion of (a − b)5? A

1, 5, 10, 10, 5, 1

1, −5, 10, −10, 5, −1

5, 10, 10, 5

1, −4, 6, −4, 1

21E What is the value of A

? B

22E Which identity states that

Pascal’s Identity

Symmetry Identity

Hockey-stick Identity

Chairperson Identity

23E Expand using Pascal’s triangle: a

( x + 4)3

(2y − 3)4

( p − 1)6

24E Find the coeﬃcient of the speciﬁed term in the expansion of (2x − y)7: a

The term containing x5 y2.

The term containing x3 y4.

The term containing x6 y.

The term containing y7.

25E Expand and simplify using the binomial theorem: a

(a + 2b)4

(3x − 2)5

26E Determine the coeﬃcients or constant terms: a

In ( y + 2)7, the coeﬃcient of y3.

, the constant term.

In (3a + 2)4, the coeﬃcient of a2.

, the coeﬃcient of k.

27E Simplify using known binomial coeﬃcient identities: a

28E Expand (2c − d)6. 29E Calculate the coeﬃcient of x7 y3 in the expansion of ( x + y)10. 30E Verify the symmetry identity 31E

Use the binomial theorem to ﬁnd the value of (0.99)3 by expanding (1 − 0.01)3.

32E a

Expand ( x + y)5.

Find ( x + y)5 − ( x − y)5.

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for n = 7, r = 2 by calculating both sides.

Expand ( x − y)5.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

33E Consider the binomial expansion of

Write an expression for the (r + 1)th term.

Find the value of n if the 3rd term is a constant.

34E Use Pascal’s identity to express

as a sum of two other binomial coeﬃcients.

35E Find the coeﬃcients to complete the expansion of (a − b)5: a5 + … a4 b + 10a3 b2 + … a2 b3 + 5ab4 + … b5 36E A company’s growth factor is modelled by (1.05 + r)3, where r is an additional growth rate. Expand this expression. 37E In the expansion of (ax + by)N, the sum of the powers of ax and by in each term is always N. If N = 7, and a term contains (ax)3, what is the power of (by) in that same term? for 1 ≤ r ≤ n − 1 using the factorial deﬁnition

38E Prove the recursive identity of binomial coeﬃcients.

39E Find the term independent of x in the expansion of

40E The coeﬃcient of x4 in the expansion of (1 + kx)6 is 240. Find the possible value(s) of k. 41E The constant term in the expansion of value.

is 280. If a is a positive constant, ﬁnd its

by considering the expansion of (1 + x)n and substituting an

42E Show that appropriate value for x. 43E Prove the identity x = 1). 44E Show that

for n ≥ 2. (Hint: Differentiate (1 − x)n then substitute

. You may use the identities

and

. 45E Prove

is rational (contains no surd terms) using the binomial theorem.

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13 Introduction to rates of change Chapter outline 13.01 13.02 13.03 13.04 13.05

Average rate of change Speed as a rate of change Instantaneous vs. average speed Instantaneous speed and tangents Linear and quadratic rates of change Chapter 13 review

660 665 671 676 683 692

The fastest recorded tennis serve hit 263 km/h. That’s over 70 metres in one second!

13.01 Average rate of change After this lesson, you will be able to… • define the average rate of change of a function over an interval. • calculate the average rate of change for a function using the formula • recognise that the average rate of change is the gradient of the secant line between two points on a graph. • apply the concept of average rate of change to solve simple problems in various contexts.

Average rate of change Average rate of change The change in one quantity divided by the corresponding change in another quantity. Secant The straight line passing through 2 points on the graph of a function.

The average rate of change of a function y = f (x) over the interval [a, b] measures how the function’s output changes relative to its input. It is deﬁned as:

Δy

is the change in the function’s output, f (b) − f (a)

Δx

is the change in the input, b − a

Geometrically, this represents the gradient of the secant line connecting the points (a, f (a)) and (b, f (b)) on the graph of y = f (x). y

b−a (b, f (b)) f (b) − f (a)

(a, f (a)) x

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For example, the secant line from (1, 1) to (3, 9) on y = x2 has a gradient equal to the average rate of change over [1, 3].

y 20 15 10 5 x 0

Interactive exploration Discover this concept in action online

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Example 1 For the function f (x) = x2, ﬁnd the average rate of change over the interval [1, 3].

Create a strategy with a = 1 and b = 3.

Use the formula

Apply the idea Evaluate f (x) at x = 1: f (x) = x2 f (1) = 1

Write the function Substitute x = 1 Evaluate

Evaluate f (x) at x = 3: f (x) = x2 2

f (3) = 3

Write the function Substitute x = 3 Evaluate

Solve the average rate of change: Write the formula

Substitute a = 1 and b = 3

Substitute f (1) = 1 and f (3) = 9

Simplify numerator and denominator

Evaluate

The average rate of change is 4.

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Reﬂect and check This result is the gradient of the secant line between (1, 1) and (3, 9) on the graph of y = x2.

Example 2 A water tank’s volume is modelled by V (t) = 50t + 200, where V is in litres and t is in hours. Find the average rate of change of volume from t = 2 to t = 5.

Create a strategy with t1 = 2 and t2 = 5.

Use the formula

Apply the idea Evaluate V (t) at t = 2: V (t) = 50t + 200

Write the function

V (2) = 50(2) + 200

Substitute t = 2

= 100 + 200

Evaluate the multiplication

= 300

Evaluate the addition

Evaluate V (t) at t = 5: V (t) = 50t + 200

Write the function

V (5) = 50(5) + 200

Substitute t = 5

= 250 + 200

Evaluate the multiplication

= 450

Evaluate the addition

Solve the average rate of change: Write the formula

Substitute t1 = 2 and t2 = 5

Substitute V (2) = 300 and V (5) = 450

Simplify numerator and denominator

Evaluate

The average rate of change is 50 litres per hour.

Reﬂect and check This result represents the average rate at which the volume increases over the interval [2, 5], corresponding to the secant line’s gradient on the graph of V (t).

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary The average rate of change of y = f (x) over [a, b] is

, representing the

gradient of the secant line between (a, f (a)) and (b, f (b)).

13.01 Practice questions What do you remember? 1

Deﬁne the average rate of change for a function y = f (x) over the interval [a, b].

What does the average rate of change represent geometrically on the graph of y = f (x)?

For a linear function y = mx + c, what is the relationship between the average rate of change and the gradient?

What is meant by the term ‘secant line’ in the context of a function’s graph?

Practice Ex 1

Ex 2

Calculate the average rate of change for the following functions over the given intervals: a

f (x) = 2x + 3, [1, 4]

f (x) = x2 + 1, [0, 2]

f (x) = 3x − 5, [2, 5]

f (x) = 2x3 − 2, [0, 1]

f (x) = 4 − x, [−1, 1]

f (x) =

+ x − 1, [2, 4]

A tank’s water volume is modelled by V (t) = 100t + 50, where V is in litres and t is in hours. Calculate the average rate of change of volume from: a

t = 1 to t = 4

t = 2 to t = 3

A car’s distance travelled is modelled by d(t) = 60t, where d is in kilometres and t is in hours. Calculate the average rate of change of distance from: a

t = 1 to t = 3

t = 2 to t = 5

What do you notice about the average rate of change for this function?

The height of a tree is modelled by h(t) = 0.5t2, where h is in metres and t is in years. Calculate the average rate of change of height from: a

t = 1 to t = 3

t = 2 to t = 4

Why do the average rates of change differ for this function?

13.01 Average rate of change mathspace.co

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The cost of producing x items is given by C(x) = 20x + 100, in dollars. Calculate the average rate of change of cost when production increases from: a

x = 50 to x = 100

t = 0 to t = 5

t = 3 to t = 7

The temperature of a liquid is modelled by T (t) = −0.2t2 + 25, where T is in degrees Celsius and t is in minutes. Calculate the average rate of change of temperature from: a

The population of a town is modelled by P (t) = 200t + 5000, where P is the number of people and t is in years. Calculate the average rate of change of population from: a

x = 10 to x = 20

t = 1 to t = 3

t = 2 to t = 5

Calculate the average rate of change for the following functions over the given intervals: a

f (x) =

, [4, 8]

f (x) = x2 − 3, [−1, 2]

f (x) = −x−1, [1, 5]

f (x) = x2 + 2x, [0, 1]

A music festival charges $50 per ticket, but for larger crowds, additional revenue is generated from merchandise and food sales. The total revenue, in dollars, from x attendees is modelled by R(x) = 0.3x2 + 110x, in dollars. Calculate the average rate of change of revenue, when sales increase from: a

x = 100 to x = 110

x = 200 to x = 210

Extend your thinking 14

A student calculates the average rate of change for f (x) = x2 over [1, 3] as Another student calculates it as

= 4.

= . Identify the error and correct it.

The temperature in a room is modelled by T (t) = 0.1t2 + 20, where T is in degrees Celsius and t is in hours. Calculate the average rate of change from t = 2 to t = 5, and interpret what this value represents.

The distance an object falls is modelled by d(t) = 4.9t2, where d is in metres and t is in seconds. Calculate the average rate of change of distance from t = 1 to t = 2, and explain how this relates to the object’s average speed.

For the function f (x) = x2 + 2x, determine the average rate of change over [1, 1 + h]. Simplify your answer, and explain what happens as h approaches 0.

A function is deﬁned as f (x) = 2x2 − 3x + 1. Determine the average rate of change over the interval [a, a + h]. Simplify your answer.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

13.02 Speed as a rate of change After this lesson, you will be able to… • define speed as the rate of change of distance with respect to time. • calculate and determine the average speed of an object from a given distance-time function or graph. • interpret the gradient of a secant line on a distance-time graph as the average speed.

Speed as a rate of change Speed The absolute value of an object’s velocity. It represents how fast the object is moving regardless of direction. Speed will always be ≥ 0. For an object moving along the x-axis, average speed is calculated as Average speed =

Velocity The rate of change of an object’s position with respect to time. It is a vector quantity, meaning it has both magnitude and direction. The SI units for velocity are metres per second (ms− 1).

Speed is the rate of change of distance with respect to time. For a distance function d(t), the average speed over a time interval [t1, t2] is deﬁned as:

Δd

is the change in distance, d(t2) − d(t1)

Δt

is the change in time, t2 − t1

This is analogous to the average rate of change, where distance is the output and time is the input. Units for speed are typically m/s (metres per second) or km/h (kilometres per hour).

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13.02 Speed as a rate of change mathspace.co

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Example 1 A car travels according to the distance function d(t) = 60t, where d is in kilometres and t is in hours. Calculate the average speed from t = 1 to t = 3.

Create a strategy with t1 = 1 and t2 = 3.

Use the formula

Apply the idea Evaluate d(t) at t = 1: d(t) = 60t

Write the function

d(1) = 60(1)

Substitute t = 1

= 60

Evaluate

Evaluate d(t) at t = 3: d(t) = 60t

Write the function

d(3) = 60(3)

Substitute t = 3

= 180

Evaluate

Solve the average speed: Write the formula

Substitute t1 = 1 and t2 = 3

Substitute d(1) = 60 and d(3) = 180

Simplify numerator and denominator

Evaluate

The average speed is 60 km/h.

Reﬂect and check The constant speed reﬂects the linear function’s gradient, consistent with a steady rate of travel.

Idea summary Average speed is the rate of change of distance over time, calculated as , with units including but not limited to km/h or m/s.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Average speed from distance-time graphs Distance-time graph A line graph that relates distance and time, with time on the horizontal axis and distance on the vertical axis. A distance-time graph plots distance against time. The gradient of the secant line between two points on the graph represents the average speed over that interval. For a straight-line segment, the speed is constant; for a curve, it varies.

Example 2 The distance-time graph shows a cyclist’s journey. The gradient of the secant line between points at t = 0 and t = 2 gives the average speed in km/h. Calculate the average speed from t = 0 to t = 2 hours.

d 80 60 40 20

t 0

Create a strategy Find the gradient of the secant line between (0, 0) and (2, 80) using

Apply the idea Calculate the change in distance: Δ d = y2 − y1

Write the formula

= 80 − 0

Substitute the values

= 80

Evaluate

Calculate the change in time: Δ t = x2 − x1

Write the formula

=2−0

Substitute the values

Evaluate

Calculate the average speed: Write the formula

Substitute the values

Evaluate

The average speed is 40 km/h.

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Reﬂect and check The straight line indicates constant speed, so the average speed equals the gradient of the line segment.

Idea summary The average speed on a distance-time graph is the gradient of the secant line between two points, calculated as

13.02 Practice questions What do you remember? 1

Deﬁne average speed as a rate of change for a distance function d(t) over the interval [t1, t2].

How is average speed represented on a distance-time graph?

What are the typical units for speed in distance-time problems?

What does a straight line on a distance-time graph indicate about an object’s speed?

Practice Ex 1

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Calculate the average speed for the following distance functions over the given time intervals: a

d(t) = 50t, t = 1 to t = 3 (distance in kilometres, time in hours)

d(t) = 2t2, t = 0 to t = 2 (distance in metres, time in seconds)

d(t) = 80t + 20, t = 2 to t = 4 (distance in kilometres, time in hours)

d(t) = 5t2 + 10, t = 1 to t = 3 (distance in metres, time in seconds)

d(t) = 30t, t = 0 to t = 5 (distance in kilometres, time in hours)

d(t) = 3t2 − t, t = 2 to t = 4 (distance in metres, time in seconds)

A cyclist’s journey is modelled by d(t) = 15t, where d is in kilometres and t is in hours. Find the average speed from: a

t = 1 to t = 3

t = 2 to t = 5

What does the average speed indicate about the cyclist’s motion?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

A car’s motion is described by d(t) = t2 + 2t, where d is in metres and t is in seconds. Find the average speed from: a

t = 0 to t = 2

t = 1 to t = 3

Why do the average speeds differ for this function?

A distance-time graph shows a runner’s journey. Calculate the average speed between the given points:

Distance

400 350

From (0, 0) to (10, 200) (time in s, distance in m)

300

From (5, 50) to (15, 250) (time in s, distance in m)

250

From (2, 30) to (8, 90) (time in s, distance in m)

200

From (0, 0) to (20, 400) (time in s, distance in m)

150 100 50

Time 0

t = 0 to t = 2

t = 1 to t = 4

A particle’s motion is described by d(t) = 4t2 − 3t, where d is in metres and t is in seconds. Find the average speed from: a

A boat’s journey is modelled by d(t) = 20t + 10, where d is in kilometres and t is in hours. Find the average speed from: a

t = 1 to t = 3

t = 2 to t = 5

A distance-time graph shows a car’s journey. Calculate the average speed between the given points:

Distance 250

From (0, 0) to (2, 120) (time in s, distance in m)

From (1, 40) to (3, 160) (time in s, distance in m)

From (2, 120) to (4, 220) (time in s, distance in m)

From (0, 0) to (5, 250) (time in s, distance in m)

200 150 100 50

Time 0

A train’s journey is modelled by d(t) = 100t − 5t2, where d is in kilometres and t is in hours. Find the average speed from: a

t = 1 to t = 3

t = 2 to t = 4

13.02 Speed as a rate of change mathspace.co

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Extend your thinking 13

A car’s distance is d(t) = 40t − 2t2 (in km, t in hours): a

Use this model to explore how the car’s average speed changes over different intervals. What can you infer about the car’s journey based on how the average speed varies?

Why is a quadratic function inappropriate for modelling a car’s motion?

A distance-time graph shows a cyclist’s journey with points (0, 0), (1, 20), and (3, 30) (time in hours, distance in km). Calculate the average speed from t = 0 to t = 3 and from t = 1 to t = 3. Explain why the speeds differ.

A runner’s distance is d(t) = 0.5t2 + t (in metres, t in seconds):

Calculate the average speed of the runner over the interval from t = 2 to t = 3.

Repeat the calculation for the intervals t = 2 to t = 2.1, and t = 2 to t = 2.01.

What do you notice about the average speed as the time interval becomes shorter and approaches 2 seconds?

Based on this pattern, make a prediction about the runner’s speed at exactly t = 2 seconds. Explain your reasoning.

A distance-time graph shows a journey with points (0, 0), (2, 50), (4, 80), and (6, 90) (time in hours, distance in km). Calculate the average speed for each interval and suggest why the speeds vary. distance

90 80 70 60 50 40 30 20 10

time 0

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As Earth moves in its orbit, the distance between Earth and Jupiter changes by 2.4 × 108 km. This causes a delay in the eclipses of Jupiter’s moons: a

When the delay is 22 minutes, calculate the average speed of light in km/s.

Suppose earlier, astronomers had estimated the delay as 20 minutes, or even 18 minutes, for the same change in distance. Calculate the average speed of light in each case.

How might this relate to estimating the true speed of light?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

13.03 Instantaneous vs. average speed After this lesson, you will be able to… • describe the difference between the average speed and instantaneous speed of an object. • determine that the instantaneous speed of an object at time t can be approximated by the average speed over a small time interval. • explain how the approximation of instantaneous speed can be improved by making the time interval smaller.

Average vs instantaneous speed Average speed is the total distance travelled divided by the total time, calculated as

over an interval

d 40

[t1, t2]. In contrast, instantaneous speed is the speed of an object at a speciﬁc moment, represented by the gradient of the tangent to the distance-time graph at that point.

The secant line from t = 1 to t = 3 gives the average speed, while the gradient at t = 2 (point shown) represents the instantaneous speed.

Average speed provides an overall measure, while instantaneous speed captures the exact speed at a given time, useful for analysing variable motion, such as acceleration.

t 0

Example 1 A cyclist’s distance is modelled by d(t) = t3, where d is in metres and t is in seconds. Compare the average speed from t = 1 to t = 3 with an approximation of the instantaneous speed at t = 2 using the interval [2, 2.5].

Create a strategy Calculate the average speed from t = 1 to t = 3 by ﬁnding the change in distance over the change in time using the average rate of change formula then approximate the instantaneous speed at t = 2 by calculating the average speed over the smaller given interval [2, 2.5].

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Apply the idea For average speed: d(t) = t3 d(1) = 1

Write the equation

Substitute t = 1

=1 d(t) = t

Evaluate 3

Write the equation

Substitute t = 3

d(3) = 3

= 27

Evaluate Write the formula

Substitute t2 = 3 and t1 = 1

Substitute d(3) = 27 and d(1) = 1

Evaluate the subtraction

Evaluate

The average speed is 13 m/s. For instantaneous speed approximation: d(t) = t3 d(2) = 2

Write the equation

Substitute t = 2

=8 d(t) = t

Evaluate

d(2.5) = 2.5

Write the equation 3

= 15.625

Substitute t = 2.5 Evaluate Write the formula

Substitute t2 = 2.5 and t1 = 2

Substitute d (2.5) = 15.625 and d (2) = 8

Evaluate the subtraction

Evaluate

The approximate instantaneous speed at t = 2 is 15.25 m/s.

Reflect and check The average speed (13 m/s) is lower than the instantaneous speed (15.25 m/s), indicating the cyclist is accelerating, as d (t) = t3 is non-linear.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary Average speed describes how fast something travels over a period of time. It is found by dividing the total distance by the total time taken:

Instantaneous speed is the speed at a particular instant. It represents the rate at which distance is changing at that exact moment.

Approximate instantaneous speed The instantaneous speed at time t can be approximated by calculating the average speed over a small interval. Rather than using an interval centred on t, choose one that starts at t and extends a short distance to the right, such as [t, t + h], where h is small. This approach simpliﬁes the algebra used later in calculus. The smaller the interval, the closer the approximation is to the true instantaneous speed, as the secant line approaches the tangent.

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Example 2 For the same cyclist’s motion, d(t) = t3 (metres, seconds), improve the accuracy of the approximation of instantaneous speed at t = 2 using the interval [2, 2.1]. Explain how this improves from [2, 2.5].

Create a strategy Calculate the average speed over the new, smaller interval [2, 2.1].

13.03 Instantaneous vs. average speed mathspace.co

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Apply the idea d(t) = t3 d(2) = 2

Write the equation

Substitute t = 2

=8 d(t) = t

Evaluate

d(2.1) = 2.1

Write the equation 3

= 9.261

Substitute t = 2.1 Evaluate Write the formula

Substitute t2 = 2.1 and t1 = 2

Substitute

Evaluate the subtraction

Evaluate

The approximate instantaneous speed at t = 2 is 12.6 m/s. Using h = 0.1 (interval [2, 2.1]) gives a closer approximation than h = 0.5 (interval [2, 2.5]), as the secant line is nearer to the tangent, reducing the error in estimating the gradient at t = 2.

Reﬂect and check Smaller intervals yield better approximations, as the secant line more closely aligns with the tangent at t = 2.

Idea summary Instantaneous speed at time t is approximated by the average speed over a small interval [t, t + h]. Smaller h values improve the approximation by approaching the tangent’s gradient.

13.03 Practice questions What do you remember? 1

Deﬁne average speed for a distance function d(t) over the interval [t1, t2].

What is instantaneous speed, and how is it represented on a distance-time graph?

How can instantaneous speed at time t be approximated?

Explain the difference between average speed and instantaneous speed in terms of motion.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Practice 5

For the distance function d(t) = t2 (in metres, t in seconds), calculate the average speed over the given intervals: a

Ex 1

Ex 2

t = 2 to t = 4

For d(t) = t2 (metres, seconds), approximate the instantaneous speed at t = 2 using the given intervals: a

[2, 3]

[2, 2.1]

How does the approximation improve from [2, 3] to [2, 2.1]?

For d(t) = 2t2 + t (metres, seconds), calculate: a

Average speed from t = 0 to t = 2

Approximate instantaneous speed at t = 1 using [1, 1.01]

A car’s distance is d(t) = 40t (km, hours). Calculate: a

Average speed from t = 1 to t = 3

Approximate instantaneous speed at t = 2 using [2, 2.001]

Compare the two speeds and explain the result.

A cyclist’s distance is d(t) =

+ 3t (metres, seconds). Calculate:

Average speed from t = 1 to t = 4

Approximate instantaneous speed at t = 2 using [2, 2.2]

A runner’s distance is d(t) = 3t3 − 2t (metres, seconds). Calculate: a

Average speed from t = 2 to t = 4

Approximate instantaneous speed at t = 3 using [3, 3.001]

For d(t) = 5t3 (metres, seconds), approximate the instantaneous speed at t = 1 using: a

t = 0 to t = 2

[1, 1.1]

[1, 1.01]

[1, 1.001]

A car’s distance is d(t) = 20t + t2 (km, hours). Calculate: a

Average speed from t = 1 to t = 3

Approximate instantaneous speed at t = 2 using [2, 2.4]

Describe the difference between the average speed of an object and its instantaneous speed.

Extend your thinking 13

A cyclist’s distance from the starting line is given by d(t) = 2t² + t, where d is in metres and t is in seconds. Estimate the instantaneous speed at t = 2 by choosing an interval that gives a better approximation than the average speed from t = 1 to t = 3.

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A student approximates the instantaneous speed for d(t) = t2 at t = 2 using [2, 3] as

= 1.67 m/s. Identify, correct the error, and ﬁnd a better approximation. 15

A distance-time graph shows points (1, 2), (2, 8), and (2.1, 8.82) (time in s, distance in m). Estimate the instantaneous speed at t = 2 and compare with the average speed from t = 1 to t = 2.

For d(t) = t2 (metres, seconds), approximate the instantaneous speed at t = 2 using [2, 2 + h]. Simplify the expression and explain what happens as h approaches 0.

For d(t) = 3t2 − t (metres, seconds), ﬁnd the average speed from t = 2 to t = 4 and approximate the instantaneous speed at t = 3, using [3, 3 + h] for some small value of h. Explain why the instantaneous speed is higher.

13.04 Instantaneous speed and tangents After this lesson, you will be able to… • relate the instantaneous speed of an object to the gradient of the tangent at a point on its distance-time graph. • estimate the instantaneous speed of an object by calculating the average speed over a very small interval. • estimate the instantaneous speed of an object by drawing a tangent to its distance-time graph and calculating the gradient.

Gradient of tangent as instantaneous speed Tangent For a curve at a given point P the tangent can be described intuitively as the straight line that ‘just touches’ the curve at that point. At P the curve has ‘the same direction’ as the tangent. In this sense, it is the best straight-line approximation to the curve at point P. The instantaneous speed of an object at time t is the rate of change of distance at that exact moment. On a distance-time graph, this is the gradient of the tangent to the curve at time t. The tangent is the line that touches the curve at a single point, representing the slope of the curve at that instant.

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10 m

9m 8m 7m 6m 5m 4m 3m 2m 1m

0 s 0.5 s 1 s 1.5 s 2 s 2.5 s 3 s 3.5 s 4 s

The gradient of the secant line from t = 2 to t = 4 provides the average speed over the interval, while the gradient of the tangent at t = 3 (point shown) gives the instantaneous speed. For a distance function d(t), the instantaneous speed at t = a is the limit of the average speed as the interval shrinks, corresponding to the tangent’s gradient.

Example 1 A car’s distance is modelled by d(t) = 0.5t2, where d is in metres and t is in seconds. Estimate the instantaneous speed at t = 3.

Create a strategy Estimate the instantaneous speed at t = 3 by calculating the average speed over a small subsequent interval, for example, [3, 3.1].

Apply the idea d(t) = 0.5t2

Write the equation 2

Substitute t = 3

d(3) = 0.5 × 3 = 4.5

Evaluate

d(t) = 0.5t2

Write the equation 2

Substitute t = 3.1

d(3.1) = 0.5 × 3.1 = 4.805

Evaluate Write the formula

Substitute t2 = 3.1 and t1 = 3

Substitute d(3.1) = 4.805 and d(3) = 4.5

Evaluate the subtraction

Evaluate

The instantaneous speed at t = 3 is approximately 3.05 m/s.

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Reﬂect and check The small interval [3, 3.1] ensures the secant line closely approximates the tangent, providing a reasonable estimate.

Idea summary The instantaneous speed at time t is the gradient of the tangent to the distance-time graph at that point, representing the exact speed at that moment.

Estimate instantaneous speed from graphs To estimate the instantaneous speed from a distance-time graph, draw a tangent at the desired time and calculate its gradient. This can be done by selecting two points on the tangent line and computing

. In practice, the tangent is approximated by a line that closely follows the curve’s

slope at the point.

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Example 2 The distance-time graph shows a vehicle’s journey. Estimate the instantaneous speed at t = 2 hours by drawing a tangent and calculating its gradient.

90 km

(3, 90)

80 km 70 km 60 km 50 km 40 km

(2, 40)

30 km 20 km 10 km 0h

(1, 10) t 1h

Create a strategy Use the tangent line approximated by the secant from t = 1 to t = 3 through t = 2. Calculate the gradient using

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Apply the idea Δ d = 90 – 10 = 80 Δt = 3 − 1 =2

Calculate change in distance Evaluate Calculate change in time Evaluate Write the formula

Substitute the values

Evaluate

The instantaneous speed at t = 2 is approximately 40 km/h.

Reﬂect and check The secant line from t = 1 to t = 3 approximates the tangent at t = 2. A smaller interval would reﬁne the estimate.

Idea summary Instantaneous speed is estimated from a distance-time graph by calculating the gradient of a tangent at the given time, using points on an approximate tangent line.

13.04 Practice questions What do you remember? 1

What is instantaneous speed, and how is it represented on a distance-time graph?

Deﬁne a tangent line on a distance-time graph.

How can the instantaneous speed at time t be estimated from a distance function d(t)?

How is the instantaneous speed estimated from a distance-time graph?

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Practice Ex 1

For d(t) = 0.5t2 (metres, seconds), estimate the instantaneous speed at t = 2 using an interval width of: a

Ex 2

0.1

0.01

A cyclist’s journey is shown on this distance-time graph: Estimate the instantaneous speed at t = 2 seconds by calculating the gradient of the tangent approximated by the secant from t = 1 to t = 3.

d 15 m

10 m

(3, 9) 5m

(2, 4)

(1, 1) 0s

A car’s distance is d(t) = 50t (km, hours). Estimate the instantaneous speed at t = 2 using an interval width of: a

0.2

Why are the estimates the same?

0.01

A runner’s journey is shown on this distance-time graph: Estimate the instantaneous speed at t = 3 seconds by calculating the gradient of the tangent approximated by the secant from t = 2 to t = 4.

(4, 28)

25 m 20 m 15 m

(3, 15)

10 m

(2, 6)

t 0s

A vehicle’s distance-time graph shows points (1, 5), (2, 20), and (3, 45) (time in seconds, distance in metres). Estimate the instantaneous speed at t = 2 using a tangent approximated by the secant from t = 1 to t = 3.

(3, 45)

40 m 30 m 20 m

(2, 20)

10 m

(1, 5) 0s 680

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t 2s

A cyclist’s distance is d(t) = using an interval width of: a

(metres, seconds). Estimate the instantaneous speed at t = 4

0.1

0.01

A car’s journey is shown on this distance-time graph: (3.5, 56)

d 50 km

(3, 42)

40 km 30 km

(2.5, 30)

20 km

(1.5, 12) (1, 6) (0.5, 2)

10 km 0h

Estimate the instantaneous speed at the following times by calculating the gradient of the secant line from t − 0.5 to t + 0.5:

At t = 1 hour

At t = 3 hour

For d(t) = 0.1t3 + interval width of: a

(metres, seconds), estimate the instantaneous speed at t = 4 using an

0.1

0.01

A cyclist’s journey is shown on this distance-time graph: d 25 m

(3, 24)

20 m 15 m 10 m

(2, 10)

(1, 2) 0s

Estimate the instantaneous speed at t = 2 seconds using a tangent approximated by the secant from t = 1 to t = 3.

If the speed limit is 10 m/s, determine if the cyclist is exceeding it.

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Extend your thinking 14

A vehicle’s journey is shown on the distance-time graph. Estimate the instantaneous speed at the speciﬁed times using tangents approximated by the given secants: d

(4, 36)

30 m

(3, 21)

20 m

10 m

(2, 10) t

(1, 3) 0s

At t = 2 seconds, using the secant from t = 1 to t = 3.

At t = 3 seconds, using the secant from t = 2 to t = 4.

Explain why the speeds differ.

A car’s distance is d(t) = 2t2 + t (metres, seconds). Estimate the instantaneous speed at t = 3 using an interval width of 0.01 and compare with the average speed from t = 2 to t = 4. Interpret the difference.

A student estimates the instantaneous speed for d(t) = 0.5t2 at t = 3 using [3, 4] as = 0.875 m/s. Identify and correct the error.

A distance-time graph shows points (1, 2), (2, 6), and (3, 12) (time in seconds, distance in metres). Estimate the instantaneous speed at t = 2 using a tangent approximated by the secant from t = 1 to t = 3. Discuss the accuracy of this estimate. d 15 m

(3, 12)

10 m

(2, 6)

(1, 2) 0s

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13.05 Linear and quadratic rates of change After this lesson, you will be able to… • identify that the rate of change for a linear function is constant and equal to its gradient. • recognise that the rate of change for a non-linear function is variable. • determine the rate of change for linear models in practical contexts. • estimate the instantaneous rate of change for non-linear functions from a graph.

Constant rate of change in linear models Linear function y is a linear function of x if y = mx + c where m and c are constants. The graph of a linear function is a straight line, where m is the gradient and c is the y-intercept. For a linear function y = mx + c, the rate of change is constant and equal to the gradient m. This represents a steady rate in practical situations, such as constant speed or ﬁxed costs.

Rate of change = m m

is the gradient of the linear function

d 40 m 30 m 20 m 10 m

t 0s

The straight line d(t) = 5t has a constant gradient 5, representing a constant speed of 5 m/s.

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Example 1 A worker earns according to the linear model E(t) = 25t + 50, where E is earnings in dollars and t is hours worked. Determine the rate of change and interpret it in context.

Create a strategy Identify the gradient m from the linear function E(t) = 25t + 50.

Apply the idea Rate of change = m = 25

The gradient is the coeﬃcient of t Extract from E(t) = 25t + 50

The rate of change is 25 dollars per hour, meaning the worker earns $25 for each hour worked.

Reflect and check The constant rate reflects a fixed hourly wage, consistent with the linear model.

Idea summary In a linear function y = mx + c, the rate of change is the constant gradient m, representing steady rates in contexts like wages or speed.

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Variable rates in quadratic functions Quadratic function An equation of the form ax2 + bx + c = 0, where a ≠ 0, b and c are constants. Instantaneous rate of change The rate of change at a particular moment. For curves, this is the gradient of the tangent at a point on the graph.

For a non-linear function, such as a quadratic function y = ax2 + bx + c, the rate of change is not constant. The instantaneous rate of change at a point is the gradient of the tangent to the curve at that point, varying along the curve. 40 m d 36 m 32 m 28 m 24 m 20 m 16 m 12 m 8m 4m 0s

t 1s

As seen, the curve d(t) = 2t2 has a varying gradient. The gradient at x = 4 is greater than the gradient at x = 2, indicating a higher instantaneous rate of change. To ﬁnd the instantaneous rate for each point, calculate the gradient of the tangents. To estimate the instantaneous rate at a point, calculate the gradient of the secant line connecting (x, f (x)) and (x + h, f (x + h)), where h is small.

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13.05 Linear and quadratic rates of change mathspace.co

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Example 2 The volume of water in a tank is modelled by graph shown, where V is in litres and t is in minutes. Estimate the instantaneous rate of change of volume at t = 2 minutes using the secant line.

(2.5, 18.75)

15 L

(2, 12) 10 L

(1.5, 6.75) V(t) = 3t2

0 min 0.5 min 1 min 1.5 min 2 min 2.5 min

Create a strategy Use the graph to identify the points (1.5, V (1.5)) and (2.5, V (2.5)) on V (t) = 3t2, then calculate the gradient of the secant line to estimate the instantaneous rate at t = 2.

Apply the idea From the graph, the points are (1.5, 6.75) and (2.5, 18.75): Write the formula

Substitute the values

Substitute V (2.5) = 18.75 and V (1.5) = 6.75

Evaluate

The instantaneous rate of change at t = 2 is approximately 12 L/min.

Reﬂect and check The secant line through t = 1.5 and t = 2.5 approximates the tangent at t = 2, reﬂecting the rate at which water is added to the tank.

Idea summary In quadratic functions, the rate of change varies. The instantaneous rate at a point is the gradient of the tangent, estimated by calculating the gradient of the secant line between (x, f (x)) and (x + h, f (x + h)), where h is small.

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13.05 Practice questions What do you remember? 1

What is the rate of change for a linear function y = mx + c?

How does the rate of change differ for a quadratic function compared to a linear function?

How is the instantaneous rate of change estimated for a quadratic function at a speciﬁc point?

What does the gradient m represent in a practical context for a linear function?

Practice 5

Determine the rate of change for the following linear functions and state the units: a

y = 15x + 20 ( y in dollars, x in hours)

d = 60t (t in hours, d in km)

8n + 50 = C (C in dollars, n in items)

2h = 2t + 10 (h in metres, t in seconds)

Ex 1

A car’s distance is d(t) = 50t + 20 (km, hours). Find the rate of change and interpret it.

Ex 2

The height of an object is modelled by the graph shown (metres, seconds). Estimate the instantaneous rate of change at t = 3 using the secant line. h 30 m

(3.5, 28)

25 m

(3, 21)

20 m 15 m

(2.5, 15)

10 m 5m 0s

t 1s

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The volume of water in a tank is modelled by the graph shown (litres, minutes). Estimate the instantaneous rate of change at t = 2 using the secant line. V 8L 6L

(1.9, 6.892)

(2, 7.071)

(2.1, 7.2455)

1.9 min

2 min

2.1 min

4L 2L

t 1.8 min

Two workers are paid according to the functions E1(t) = 30t + 250 and E2(t) = 35t: a

Who is paid more for 10 hours work?

Who has the higher rate of pay?

What might the 250 value represent in a real-world situation?

Population growth is modelled by the graph shown. Estimate the instantaneous rate of change at t = 4 using the following graphs: a

P (people) 26

(4.2, 26.04)

(4, 24) t (years) 3.9

4.1

4.2

P (people) 26

(4, 24)

(4.01, 24.1001) t (years)

688

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4.02

A tank ﬁlls at a constant rate, with volume given by the function V (t) = 20t (litres, min). How much does the volume increase by every hour?

The distance of an object is modelled by the graph shown (metres, seconds). Estimate the instantaneous rate of change at t = 1 using the following graphs: a

d (metres) 8 6

(1.5, 5.625)

(1, 3.5) 2

t (seconds) 0

0.5

1.5

d (metres)

(1.1, 3.905) (1, 3.5) 1

t (seconds) 1.1

1.2

A shop’s revenue is modelled by R(n) = 12n + 200, where n is the number of items sold. What does the number 12 represent in this context? A

The shop earns $12 for every item sold.

The base revenue is $12.

The shop earns $12 for every item plus $200.

The shop spends $12 on each item.

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Extend your thinking 14

The distance of a car is modelled by the graph shown (metres, seconds). Compare the instantaneous rate of change at t = 3 using the secant line with the average rate of change from t = 2 to t = 4. Interpret the difference. d (metres) 15.5

(3.01, 15.0801)

(3, 15)

14.5

t (seconds) 2.99

3.01

A shop has a revenue modelled by C(x) = −10x + 100, where x is the number of items sold. Explain why this situation is unrealistic or problematic.

The volume of water in a tank is modelled by the graph shown (litres, minutes). Estimate the instantaneous rate of change at t = 1 using the secant line. Interpret in context. V (litres)

(1, 3) (1.001, 3.007004)

t (minutes) 0.997

690

0.998

0.999

1.001

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

1.002

1.003

Cleaning costs are modelled by the graphs shown (dollars, hours). Graph 1 represents a ﬁxed-rate service (linear), and Graph 2 represents a service with increasing rates over time (quadratic). Graph 1 Graph 2 C (dollars)

C (dollars)

50.5

40.5

(2, 50)

(2.01, 50.1)

49.5

(2.01, 40.3005) (2, 40)

39.5

t (hours) 2

t (hours)

2.01

Compare the instantaneous rates of change at t = 2 for both models. Explain the difference.

For y = ax2 + bx + c, show that the instantaneous rate of change at x = k can be approximated by

. Apply this to y = t2 + 3t at t = 1 with h = 0.01.

For a quadratic function y = at2 + bt + c and an interval [x1, x2]: a b

Show that the instantaneous rate of change at the midpoint m = average rate of change over [x1, x2].

equals the

Apply this to the height of a projectile modelled by h(t) = 4t2 + 2t + 3 (metres, seconds) at t = 2, the midpoint of [1, 3], using h = 0.0001 for the instantaneous rate.

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13 Chapter review 1

What is the average rate of change for the function f (x) = x2 − 4x over the interval [1, 5]? A

A cyclist’s journey is modelled by d(t) = 70t, where d is in kilometres and t is in hours. What is their average speed from t = 1 to t = 4? A

−2

35 km/h

70 km/h

140 km/h

210 km/h

The cost of manufacturing n smartphones is given by C(n) = 35n + 2500 in dollars. What does the number 35 represent? A

The initial set-up cost

The total revenue from sales

The cost to produce each additional smartphone

The total number of smartphones produced

Calculate the average rate of change for the following functions over the given intervals: a

f (x) = 3x + 1, [2, 5]

f (x) = x2 + 2, [0, 3]

f (x) = 5 − 2x, [−1, 2]

f (x) = x2 + 2x + 1, [1, 3]

A company’s profit is modelled by P ( y) = 400y2 − 100y + 2500, where P is profit in dollars and y is in years. Find the average rate of change of profit from: a

y = 1 to y = 3

y = 2 to y = 5

y = 0 to y = 4

y = 3 to y = 6

The number of bacteria in a culture is modelled by B(t) = 20t2, where B is the number of bacteria and t is in hours: a

Calculate the average rate of change of the bacteria population from t = 0 to t = 2.

Calculate the average rate of change of the bacteria population from t = 2 to t = 4.

Why do the average rates of change differ for this function?

A student calculates the average rate of change for f (x) = 2x2 over [1, 4] as Identify the error and provide the correct answer.

Calculate the average speed (in km/h) for the following distance functions over the given time intervals: a c

692

d(t) = 40t + 5t2, t = 1 to t = 3 3

d(t) = 4t − 2t, t = 0 to t = 2

d(t) = 2t2 + 3t, t = 2 to t = 4

d(t) = 80t + 10t2, t = 2 to t = 4

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

= 7.5.

The distance-time graph shows a drone’s ﬂight. Calculate the average speed of the drone between the given points:

400 m

Altitude

350 m

From (2, 30) to (10, 200)

300 m

From (8, 90) to (20, 400)

250 m 200 m 150 m 100 m 50 m

Time

10 s

15 s

20 s

A rocket’s altitude is described by h(t) = 2t2 + 5t, where h is in metres and t is in seconds: a

Calculate the average speed from t = 1 to t = 3.

Calculate the average speed from t = 2 to t = 4.

Why might a quadratic function be unrealistic for modelling a rocket’s entire ﬂight?

A distance-time graph shows a cyclist’s journey with points (0, 0), (1, 25), and (4, 45) (time in hours, distance in km): a

Calculate the average speed from t = 0 to t = 4 and from t = 1 to t = 4.

Explain why these speeds are different.

Distance 50 km 40 km 30 km 20 km 10 km

Time 0h

A distance-time graph shows a journey with points (0, 0), (2, 60), (4, 90), and (6, 100) (time in hours, distance in km). Calculate the average speed for each 2-hour interval and suggest why the speeds might be changing. Distance 100 km 80 km 60 km 40 km 20 km

Time 0h 1h 2h 3h 4h 5h 6h

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For d(t) = 3t2 + 2t (metres, seconds): a

Calculate the average speed from t = 0 to t = 2.

Determine the instantaneous speed at t = 1 using [1, 1.5].

A bus journey’s distance is d(t) = 80t (km, hours). a

Calculate the average speed from t = 1 to t = 4.

Approximate instantaneous speed at t = 2 using [2, 2.5].

Compare the two speeds and explain the result.

A car’s distance is modelled by d(t) = 4t2 − t (metres, seconds). Compare the average speed from t = 1 to t = 4 with the instantaneous speed approximation at t = 2 using the interval [2, 2.01]. Interpret the difference.

The distance of an object is modelled by the graphs shown (metres, seconds). Estimate the instantaneous speed at t = 2 using the following graphs: a

d (metres) (2.1, 6.615)

6.5

d (metres) 6.5

(2, 6)

5.5

(2, 6)

(2.01, 6.06015)

5.5

t (seconds) 2

2.1

t (seconds) 2

2.01

A distance-time graph shows points (1, 20), (2, 60), and (3, 120) (time in hours, distance in kilometres). Estimate the instantaneous speed at t = 2 by calculating the gradient of the secant line from t = 1 to t = 3.

A motorcycle’s distance is given by d(t) = 3t2 + 2t (metres, seconds):

Estimate the instantaneous speed at t = 2 using the interval [2, 2.01].

Compare this to the average speed from t = 1 to t = 3 and interpret the difference.

For a distance function d(t) = 3t2 (metres, seconds), the instantaneous speed at t = 1 is approximated using the interval [1, 1 + h]: a

Determine the expression for this approximation and simplify it.

Explain what the expression represents as h approaches 0.

The water level in a reservoir is modelled by L(t) = 0.5t + 10, where L is in metres and t is in days. Calculate the rate of change of the water level and interpret its meaning.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

The area of a circular oil spill is modelled by the graphs shown. Estimate the instantaneous rate of change at t = 4 using the following graphs: a

A (m2) 84

(4.1, 84.05)

(4, 80) 4

t (minutes)

4.1

A company’s monthly proﬁt is modelled by the graph shown (thousands of dollars, months): a

Estimate the instantaneous rate of proﬁt change at t = 6 using the secant line.

Compare with the average rate of change from t = 5 to t = 7 and interpret the results.

(4.01, 80.4005)

(4, 80)

t (minutes)

4.01

4.02

P (thousands of dollars) 662

(6.01, 661.701) 661

660

(6, 660) t (months) 6

The height of a growing plant is modelled by the graph shown (cm, weeks): a

Estimate the instantaneous rate of growth at t = 3 using the secant line.

Interpret your answer in the context of the plant’s growth.

6.01

6.02

H (cm)

16.5

(3, 16.5) (3.001, 16.513)

t (weeks) 3

3.001

3.002

A phone plan’s cost is modelled in two ways: • Plan A: CA (g) = 10g + 20 • Plan B: CB (g) = 0.5g2 + 5g a

Calculate the rate of change for Plan A.

Estimate the instantaneous rate of change for Plan B at g = 10 GB using [10, 10.01].

Explain the practical difference for a consumer.

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Big ideas • The derivative is the gradient function, f ′(x), which provides the exact instantaneous rate of change of a function f (x) at any point. It is formally defined as the limit of the gradient of a secant line (differentiation from first principles) and geometrically represents the gradient of the tangent to the curve. • Differentiation is the process of finding the derivative, which can be performed efficiently using a set of procedural rules (power, sum, product, quotient, and chain rules) that bypass the need for first principles and can be combined to differentiate any polynomial, rational, or composite function. • The derivative is a powerful analytical tool: its value at a point determines the equation of the tangent and normal; its sign reveals whether the original function is increasing or decreasing; and in applied contexts, it models instantaneous rates of change, such as the velocity of a moving object.

14 The derivative Chapter outline 14.01 14.02 14.03 14.04 14.05 14.06 14.07 14.08 14.09 14.10 14.11

698 710 714 721 731 744 751 756 763 770 783 792

The chain rule is like peeling an onion — one function inside another.

14.01 Gradient of a curve After this lesson, you will be able to… • describe the gradient of a curve at a point in terms of its tangent line. • estimate the gradient of a curve graphically by calculating the gradient of its tangent. • use the gradient of a secant to approximate the gradient of a curve at a point. • recognise the derivative, f ′(x), as the gradient function of a curve y = f (x). • define differentiation as the process of finding the derivative of a function.

The gradient of a curve at a point Gradient The slope of a line. It is calculated as the gradient of a line segment it contains. If A(x1, y1 ) and B(x2, y2 ) are 2 distinct points on a line, the gradient of the line (or line segment AB) is given by m =

Unlike a straight line, which has a constant gradient, the steepness of a curve changes from point to point. To understand the gradient of a curve at a speciﬁc point, the concept of a tangent line is used. Tangent A line that intersects a curve at just one point. It touches the curve at that point of contact but does not pass inside it.

A tangent to a curve at a particular point is a straight line that ‘just touches’ the curve at that point. The tangent line has the same direction as the curve at that point of contact. The gradient of a curve at a point is deﬁned as the gradient of the tangent to the curve at that point. If the tangent is horizontal, its gradient is zero. If the tangent slopes upwards from left to right, its gradient is positive. If it slopes downwards, its gradient is negative. A vertical tangent has an undeﬁned gradient. Graphing applications can be useful tools to visualise a curve and its tangent at various points, helping to examine how the gradient changes along the curve.

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y 3

y = f (x)

x −3 −2

−1

1 −1

Tangent at P

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Example 1 The graph shows the function f (x) = x2 and the tangent line to the curve at the point P (1, 1).

y 3 2

f (x) = x2

P (1, 1)

x −2

−1

1 −1

Tangent

Estimate the gradient of the curve f (x) = x2 at the point P (1, 1).

Create a strategy The gradient of the curve at point P is the gradient of the tangent line at P. Identify two points on the tangent line and use the gradient formula m =

Apply the idea From the graph, the tangent line at P (1, 1) also appears to pass through the point (2, 3), where the grid lines meet. Let (x1, y1 ) = (1, 1) and (x2, y2 ) = (2, 3). Write the gradient formula

Substitute the coordinates

Evaluate the numerator and denominator

Simplify The gradient of the tangent line at P (1, 1) is 2. Therefore, the gradient of the curve f (x) = x2 at x = 1 is 2.

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Idea summary The gradient of a curve at a speciﬁc point is equal to the gradient of the tangent line to the curve at that point. This gradient represents the instantaneous rate of change of the function at that point. A tangent line touches the curve at one point and has the same direction as the curve at that point.

Approximation of the gradient by a secant Secant The straight line passing through 2 points on the graph of a function.

While a tangent touches a curve at one point, a secant is a straight line that intersects a curve at two distinct points. The gradient of a secant can be used to approximate the gradient of a tangent. y 4 3

Consider two points on a curve y = f (x): P (c, f (c)) and a nearby point Q(c + h, f (c + h)), where h is a small change in x.

Q(c + h, f (c + h)) Secant PQ

2 1

P(c, f (c))

y = f (x) 0.5 −1

1.5

x 2

Tangent at P

The gradient of the secant line PQ is given by the standard gradient formula:

mSecant is the gradient of the secant line passing through points P and Q f (c)

is the y-coordinate of point P

f (c + h)

is the y-coordinate of point Q

h is the horizontal distance (change in x) between P and Q, h ≠ 0 700

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As the value of h approaches zero, the point Q moves closer and closer to the point P along the curve. Consequently, the secant line PQ becomes a better approximation of the tangent line at P. The gradient of the secant line approaches the gradient of the tangent line at P. This concept is fundamental to understanding how the gradient of a curve at a point is formally deﬁned.

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Example 2 For the function f (x) = x2 at x = 1, calculate the gradient of the secant line between P (1, f (1)) and Q(1 + h, f (1 + h)) for: a h = 0.1

Create a strategy Calculate f (1), then use the formula for the gradient of the secant: mSecant =

Apply the idea For f (1): f (x) = x2 f (1) = 1

Write the function

Substitute x = 1

Evaluate the power

f (1 + h) = f (1 + 0.1)

Substitute h = 0.1

For f (1 + h): = f (1.1)

Evaluate the addition

For f (1.1): f (x) = x2

Write the function 2

f (1.1) = (1.1)

= 1.21

Substitute x = 1.1 Evaluate the power

For the gradient: Write the secant gradient formula

Substitute f (1 + h) = 1.21, f (1) = 1, h = 0.1

Evaluate the numerator

Evaluate

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b h = 0.01

Apply the idea For f (1), use f (1) = 1 from part (a). For f (1 + h): f (1 + h) = f (1 + 0.01)

Substitute h = 0.01

= f (1.01)

Evaluate the addition

For f (1.01): f (x) = x2 f (1.01) = (1.01)

Write the function 2

Substitute x = 1.01

= 1.0201

Evaluate the power

For the gradient: Write the secant gradient formula

Substitute f (1 + h) = 1.0201, f (1) = 1, h = 0.01

Evaluate the numerator

Evaluate

c h = 0.001

Apply the idea For f (1), use f (1) = 1 from part (a). For f (1 + h): f (1 + h) = f (1 + 0.001) = f (1.001)

Substitute h = 0.001 Evaluate the addition

For f (1.001): f (x) = x2

Write the function 2

Substitute x = 1.001

f (1.001) = (1.001)

= 1.002 001

Evaluate the power

For the gradient: Write the secant gradient formula

Substitute f (1 + h) = 1.002 001, f (1) = 1, h = 0.001

Evaluate the numerator

Evaluate

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d What value does the gradient of the secant appear to be approaching as h approaches zero?

Create a strategy

Apply the idea

Use the values of h from parts (a) to (c) to estimate the value they approach.

As h gets smaller, the gradient of the secant changes from 2.1, to 2.01, then to 2.001. These values appear to approach 2, suggesting that the gradient of the tangent at x = 1 is likely to be 2.

Idea summary The gradient of a secant line passing through two points P (c, f (c)) and Q(c + h, f (c + h)) on a curve is given by

As h approaches zero, point Q approaches point P, and the gradient of the secant line provides an increasingly accurate approximation of the gradient of the tangent line at P.

The derivative and differentiation Derivative The result obtained after differentiation. For the function f (x), the derivative is the gradient function of f (x), and is denoted f ′(x). The gradient of the tangent to a curve y = f (x) at a point P (x, f (x)) is a fundamental concept in calculus. This gradient is formally deﬁned as the derivative of the function f (x) at that point. The derivative of a function f (x) is denoted f ′(x)(read as “f-dash of x” or “f-prime of x”) or (read as “dee y by dee x”). Other notations include y′ or

( f (x)).

So, f ′(x) represents the gradient of the tangent to the curve y = f (x) at any point x for which the derivative exists. The derivative f ′(x) is also referred to as the gradient function or the derived function of f (x). The value of the derivative at a particular point, say x = a, is written as f ′(a), and it gives the gradient of the tangent to the curve at x = a. Differentiation The process used to ﬁnd the derivative of a function. Differentiation is the process of ﬁnding the derivative f ′(x) from a function f (x). For the derivative to exist at a point, the tangent line at that point must exist and must not be vertical.

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Example 3 If the gradient of the tangent to the curve y = g(x) at the point where x = 3 is −4, what is the value of g′(3)?

Create a strategy

Apply the idea

Recall that g′(a) represents the gradient of the tangent to the curve y = g(x) at x = a.

The gradient of the tangent to y = g(x) at x = 3 is given as −4. By deﬁnition, g′(3) is the gradient of the tangent to y = g(x) at x = 3. Therefore, g′(3) = −4.

Example 4 The derivative of a function f (x) is f ′(x) = 2x − 1. Determine the gradient of the tangent to the curve y = f (x) at the point where x = 5.

Create a strategy Substitute the given x-value into the expression for the derivative f ′(x) to ﬁnd the gradient at that point.

Apply the idea The derivative is f ′(x) = 2x − 1. To ﬁnd the gradient of the tangent at x = 5, calculate f ′(5). f ′(x) = 2x – 1

Write the derivative

f ′(5) = 2 × 5 − 1

Substitute x = 5

= 10 − 1

Evaluate the multiplication

Evaluate

The gradient of the tangent to the curve y = f (x) at x = 5 is 9.

Idea summary The derivative of a function f (x), denoted f ′(x) or

, gives the gradient of the

tangent to the curve y = f (x) at any point x. The value f ′(a) is the speciﬁc gradient at x = a. Differentiation is the process used to ﬁnd this derivative function.

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14.01 Practice questions What do you remember? 1

Deﬁne these terms in the context of functions and their graphs: Tangent to a curve at a point

Secant to a curve

Gradient of a curve at a point

f ′(a) using ﬁrst principles

Differentiation

Are these statements true or false? Explain your reasoning if false: a

The gradient of a curve is constant at all points on the curve.

A secant line can be used to approximate the gradient of a tangent line.

If the tangent to a curve at a point is horizontal, the derivative of the function at that point is undeﬁned.

The derivative of a function f (x) is another function that gives the gradient of f (x) at any point x.

A vertical tangent to a curve has a gradient of zero.

How is the gradient of a secant line passing through points P (c, f (c)) and Q(c + h, f (c + h)) on a curve y = f (x) related to the gradient of the tangent at point P as h approaches zero?

List two common notations for the derivative of a function y = f (x).

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Practice Ex 1

For each graph, a curve and a tangent line at a point P are shown. Estimate the gradient of the curve at point P : a

Curve: y = x2 − 2x + 2

Curve: y = −0.5x2 + 2x + 1

Point: P (2, 2) Point: P (1, 2.5) y

y 3.5

2.5

2 1.5

x 1

0.5

−0.5

Curve: y =

0.5 1 1.5 2 2.5 3 3.5

Curve: y =

Point: P (4, 2) Point: P (1, 1) y 3

y 2

1 1 x 0

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For each graph shown below: i

Sketch an approximate tangent line to the curve at each point A, B, C and D.

Describe the gradient of the tangent at each point as positive, negative, or zero.

2 1

1 B

−2

x −2

−1

Ex 2

1 −1

−1

−2

For the function f (x) = x2 + 2x at x = 1, calculate the gradient of the secant line between P (1, f (1)) and Q(1 + h, f (1 + h)) for: a

h = 0.1

for h = 0.01

h = 0.001

What value does the gradient of the secant appear to be approaching as h approaches zero?

Consider the function g(x) = 1 − 3x : a

Calculate g(2).

Calculate the gradient of the secant line PQ where P (2, g(2)) and Q(2 + h, g(2 + h)) for h = 0.1.

Calculate the gradient of the secant line PQ for h = 0.01.

What value does the gradient of the secant appear to be approaching?

Ex 3

If g′(5) = 7, what is the gradient of the tangent to the curve y = g(x) at the point where x = 5?

Ex 4

The derivative of a function f (x) is given by f ′(x) = 3x2 − 4. Determine the gradient of the tangent to the curve y = f (x) at these points: a

x=0

x=1

x=2

x = −1

The distance s (in metres) travelled by a particle after t seconds is given by s(t) = t2 + 3t. The derivative s′(t) represents the instantaneous velocity of the particle. If s′(t) = 2t + 3, determine the velocity of the particle at: a

t = 1 second

t = 4 seconds

t = 0 seconds

t = 2.5 seconds

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The graph shows a function f (x) =

− x2 − 3x + 2. 4 3 2 1

Using the graph only (do not solve algebraically), identify the approximate x-values where: a

f ′(x) > 0

f ′(x) < 0

Consider the function f (x) =

−1

−1 −2 −3 −4 −5 −6 −7

x 1

. Point P is at x = 4:

Determine the coordinates of P.

Approximate the gradient of the tangent at P by ﬁnding the gradient of the secant line PQ where Q has x-coordinate 4.1.

Approximate the gradient of the tangent at P by ﬁnding the gradient of the secant line PQ where Q has x-coordinate 4.01.

Use the results from parts (b) and (c) to estimate the value of f ′(4).

Use a graphing application or spreadsheet to approximate the gradient of the curve f (x) = x3 at the point P (1, f (1)): a

Calculate f (1).

Complete the table: h

1+h

f (1 + h)

f (1)

f (1 + h) − f (1)

Gradient of secant

0.01

⬚

0.1

0.001 −0.001 −0.01 −0.1 c

708

−2

f ′(x) = 0

⬚

What value does the gradient of the secant appear to approach?

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

⬚

Extend your thinking 15

A function y = f (x) has these properties for its derivative f ′(x): • f ′(x) < 0 for x < −1 • f ′(−1) = 0 • f ′(x) > 0 for −1 < x < 2

• f ′(2) = 0 • f ′(x) < 0 for x > 2

Sketch a possible graph of the original function y = f (x). 16

Investigate how secants can be used to approximate the gradient of the curve f (x) = P (2, 4), by answering these questions:

Calculate the gradient of the secant QP where Q is the point (2 − h, f (2 − h))(for example, Q is to the left of P ) for h = 0.1.

Calculate the gradient of the secant QP where Q is the point (2 − h, f (2 − h)) for h = 0.01.

Now consider Q as (2 + h, f (2 + h))(for example, Q is to the right of P ). Calculate the secant gradient for h = 0.1 and h = 0.01. Compare all results. What do you observe as h → 0 (meaning Q approaches P from either side)?

A student states: “If the gradient of the tangent to a curve y = f (x) at x = a is very steep, say m = 1000, then f ′(a) is a very large number. If the tangent is vertical, its gradient is inﬁnitely steep, so f ′(a) must be inﬁnity.” Critique this statement, particularly the conclusion about a vertical tangent.

Did you know?

Derivatives play a key role in analysing real-world data! For example, financial analysts use them to study how quickly stock prices change and to predict market trends with greater accuracy. By understanding these rates of change, analysts can identify opportunities, manage risks, and make smarter investment decisions in an ever-changing global market. 14.01 Gradient of a curve mathspace.co

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14.02 Derivatives of basic functions After this lesson, you will be able to… • find the derivative of a constant function, f (x) = c and linear function, f (x) = mx + c. • justify the rules for the derivatives of constant and linear functions using their graphical properties. • apply the basic rules to find the derivatives of a variety of constant and linear functions.

Derivatives of basic functions Beyond power functions, two other fundamental types of functions have simple derivative rules: constant functions and linear functions. These rules can be understood by recalling that the derivative represents the gradient of the function’s graph. Constant A ﬁxed numerical value. For example, in the algebraic expression x + 11, the number 11 is a constant. Constant function A function that has only one value.

A constant function has the form f (x) = c, where c is a constant. The graph of y = c is a horizontal line, which always has a gradient of 0. Function

Derivative

f (x) = c

f ′(x) = 0

Linear function y is a linear function of x if y = mx + c where m and c are constants. The graph of a linear function is a straight line where m is the gradient and c is the y-intercept. Since the graph of a linear function f (x) = mx + c has a constant gradient of m, its derivative is simply this value.

710

Function

Derivative

f (x) = mx + c

f ′(x) = m

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Special cases of linear functions include: • If f (x) = x (where m = 1, c = 0), then f ′(x) = 1. • If f (x) = mx (where c = 0), then f ′(x) = m.

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Example 1 Determine the derivative of each function: a f (x) = 15

Create a strategy Recognise that this is a constant function and apply the rule that the derivative of a constant is 0.

Apply the idea f (x) = 15

Write the function

f ′(x) = 0

Apply the constant function rule

b g(x) = −3x

Create a strategy Recognise that this is a linear function of the form y = mx and apply the rule that the derivative is m.

Apply the idea g(x) = −3x

Write the function

g′(x) = −3

Apply the linear function rule where m = −3

c h(x) = 4x − 9

Create a strategy Recognise that this is a linear function of the form y = mx + c and apply the rule that the derivative is m.

Apply the idea h(x) = 4x – 9

Write the function

h′(x) = 4

Apply the linear function rule where m = 4

The derivative of h(x) = 4x − 9 is h′(x) = 4.

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Idea summary The derivatives of constant functions and linear functions follow set rules based on their gradients. Function type

Function

Derivative

Constant

f (x) = c

f ′(x) = 0

Linear

f (x) = mx + c

f ′(x) = m

This includes special cases like f (x) = x (where f ′(x) = 1) and f (x) = mx (where f ′(x) = m).

14.02 Practice questions What do you remember? 1

What is the derivative of a constant function f (x) = c, where c is a constant? Justify your answer.

What is the derivative of a linear function f (x) = mx + c, where m and c are constants? Justify your answer.

If the derivative of a function is a constant value (for example, f ′(x) = k for some constant k), what does this imply about the graph of the original function f (x)?

Practice Ex 1

Determine the derivative of each function: a

712

g(x) = c (where c is a constant)

h(x) = 106

f (x) = 12x

y = −x

V (t) =

P (n) = 0.5n

f (x) = 7x + 1

y = 3 − 4x

h(x) = 5 + x

k 5

f (x) = 99

g(x) =

− π

For each function: i

Sketch the graph of y = f (x).

What is the gradient of the graph at any point x? Functions:

f (x) = 5

f (x) = 2x

f (x) = x + 3

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

f (x) = −3

Consider the function y = 10 − 3x: a

Find the derivative,

What is the gradient of the tangent to the line at the point where x = 5?

What is the gradient of the tangent to the line at the point where x = −2?

Explain what your answers tell you about the gradient of a linear function.

Find the derivative of each function. Note that a, b, c and k are constants: a

f (x) = ax + b

y=k−x

g(t) = c − at

P (n) = (a + b)n − k

Extend your thinking 8

Consider the functions f (x) = 2x + 1, g(x) = 2x − 3 and h(x) = 2x: a

Find f ′(x), g′(x) and h′(x).

What do you notice about the derivatives of these three functions?

Explain what this result tells you about the graphs of the three functions.

The derivative of a function is given by f ′(x) = −4: a

What type of function must f (x) be?

Write down three different possible functions for f (x).

What do the graphs of all possible functions for f (x) have in common?

A line, y = L(x), passes through the points (2, 7) and (5, 16): a

Calculate the gradient of the line.

Determine the equation of the line and write it as a function L(x).

Using the rules of differentiation, ﬁnd L′(x).

What is the connection between your answer in part (a) and your answer in part (c)?

Did you know?

The slopes of the rolling hills in a vineyard can be described using functions — and derivatives help calculate how steep each hill is at any point! 14.02 Derivatives of basic functions mathspace.co

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14.03 First principles for derivatives After this lesson, you will be able to… • define the derivative from first principles using limit notation. • explain the connection between the first principles formula and the gradient of a secant line. • use first principles to find the derivative of linear and quadratic functions. • use the derivative found from first principles to calculate the gradient of a curve at a specific point.

First principles for derivatives Derivative The derivative of a function measures the instantaneous rate of change of the function with respect to its variable. It represents the gradient of the tangent line to the graph of the function at any given point. The derivative of y with respect to x is denoted

or y′.

First principles A method for ﬁnding the derivative of a function using the limit deﬁnition:

Limit The value that a function or sequence ‘approaches’ as the input or index approaches some value. In differentiation, it is the value that the gradient of the secant line approaches as the interval between the two points approaches zero. The derivative of a function f (x) at a point x gives the gradient of the tangent to the curve y = f (x) at that point. This can be determined formally by considering the gradient of a secant line through two points on the curve, P (x, f (x)) and Q(x + h, f (x + h)). The gradient of the secant line PQ is given by:

mSecant is the gradient of the secant line f (x + h) − f (x) is the change in the y-value(rise) h is the change in the x-value (run), representing a small increment from x 714

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

As point Q moves closer to P, the value of h approaches zero and the secant line PQ approaches the position of the tangent line at P. The gradient of this tangent line is the limiting value of the gradient of the secant line as h → 0. y

Q (x + h, f (x + h))

Q (x + h, f (x + h)) P (x, f (x))

Q (x + h, f (x + h)) P (x, f (x)) x

P (x, f (x)) x

A secant line with a large value for h.

x The value of h is smaller, moving point Q closer to P.

As h approaches zero, the secant line approaches the tangent at P.

This limiting value is defined as the derivative of f (x) with respect to x, denoted by f ′(x). The process of finding the derivative using this definition is called differentiation from first principles. This method is fundamental for deriving differentiation rules and can be applied to various functions, including linear and quadratic functions.

f ′(x) limh → 0 f (x + h) f (x) h

is the derivative of the function f (x) denotes the limit as h approaches zero is the value of the function at x + h is the value of the function at x is a small change in x

This deﬁnition is valid provided the limit exists and the tangent is not vertical.

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14.03 First principles for derivatives mathspace.co

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Example 1 Use ﬁrst principles to determine the derivative of the linear function f (x) = 3x + 2.

Create a strategy Find f (x + h), then substitute the values required into the formula f ′(x) = limh → 0

Apply the idea For f (x + h): f (x) = 3x + 2

Write the function

f (x + h) = 3(x + h) + 2

Substitute x = x + h

= 3x + 3h + 2

Expand the brackets

For the derivative: Write the ﬁrst principles formula

Substitute expressions for f (x + h) and f (x)

Expand the brackets

Collect like terms

Simplify the fraction

Evaluate the limit

The derivative of f (x) = 3x + 2 is f ′(x) = 3.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Use ﬁrst principles to differentiate f (x) = x2 + 5x

Create a strategy Find f (x + h), then substitute the values required into the formula f ′(x) = limh → 0

Apply the idea For f (x + h): f (x) = x2 + 5x

Write the function

f (x + h) = (x + h) + 5(x + h)

Substitute x = x + h

Expand terms

Simplify

= (x + 2xh + h ) + (5x + 5h) = x + 2xh + h + 5x + 5h For the derivative:

Write the first principles formula

Substitute for f (x + h) and f (x)

Simplify the numerator

Collect like terms

Factorise h in the numerator

Remove the common factor

Evaluate the limit

Evaluate 2

The derivative of f (x) = x + 5x is f ′(x) = 2x + 5.

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Example 3 For the function f (x) = 2x2 − 3x + 1: a Determine the derivative f ′(x) using ﬁrst principles.

Create a strategy Determine an expression for f (x + h), then substitute into the ﬁrst principles formula, f ′(x) = limh → 0

. Simplify the expression and evaluate the limit.

Apply the idea For f (x + h): f (x) = 2x2 − 3x + 1

Write the given function

f (x + h) = 2(x + h) − 3(x + h) + 1

Substitute x = x + h

Expand the terms

Apply the distributive property

= 2(x + 2xh + h ) − 3x − 3h + 1 = 2x + 4xh + 2h − 3x − 3h + 1 For the derivative:

Write the first principles formula

Substitute for f (x + h) and f (x)

Simplify the numerator

Collect like terms

Factorise h in the numerator

Remove the common factor

Evaluate the limit

Evaluate

The derivative is f ′(x) = 4x − 3.

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b Calculate the gradient of the tangent at the point where x = 3.

Create a strategy Substitute x = 3 into the derivative f ′(x) found in part (a).

Apply the idea f ′(x) = 4x – 3

Write the derivative

f ′(3) = 4 × 3 − 3

Substitute x = 3

= 12 − 3

Evaluate the multiplication

Evaluate 2

The gradient of the tangent to f (x) = 2x − 3x + 1 at x = 3 is 9.

Idea summary The derivative of a function f (x) from ﬁrst principles is deﬁned as:

f ′(x)

is the derivative of the function f (x)

limh → 0

denotes the limit as h approaches zero

f (x + h)

is the value of the function at x + h

f (x)

is the value of the function at x

is a small change in x

This deﬁnition represents the gradient of the tangent to the curve y = f (x) at any point x. It is found by taking the limit of the gradient of a secant line between the points (x, f (x)) and (x + h, f (x + h)) as the interval h approaches zero.

14.03 Practice questions What do you remember? 1

Write down the deﬁnition of the derivative of a function f (x) from ﬁrst principles.

Explain in words what f (x + h) − f (x) represents in the ﬁrst principles formula.

What does the term

When using first principles to find f ′(x) for f (x) = ax2 + bx + c, what is the general process after finding f (x + h)?

represent geometrically before the limit is taken?

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Practice Ex 1

Ex 2

Use ﬁrst principles to determine the derivative of f (x) = 5x − 4.

For each function:

Ex 3

Calculate f (x + h).

Use the ﬁrst principles formula to determine f ′(x).

f (x) = 10x

f (x) = −2x + 9

f (x) = 7

f (x) =

f (x) = −x2 + 6x − 2

Use ﬁrst principles to differentiate these quadratic functions: a

f (x) = x2 − 4x

f (x) = 5 − x2

f (x) = 3x2 + 2x − 1

For each function: i

Calculate f (x + h).

Use the ﬁrst principles formula to determine f ′(x).

f (x) = x2 + 3

f (x) = 2x2

f (x) = x2 + x

For each function: i

Determine f ′(x) using ﬁrst principles.

Determine the gradient of the tangent at the speciﬁed point.

f (x) = x2 where x = 1

f (x) = 4x − x2 where x = 3

f (x) = 8x + 1 where x = −2

f (x) = x2 − 5 where x = −1

Extend your thinking 10

If f (x) = c, where c is a constant, use ﬁrst principles to show that f ′(x) = 0.

A student attempts to differentiate f (x) = x2 from ﬁrst principles and writes:

Identify the error in the student’s working and explain why it is incorrect. What is the correct ﬁnal step?

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The formula f ′(a) = limx → a speciﬁc point x = a:

is an alternative deﬁnition for the derivative of f (x) at a

Explain how this formula relates to the deﬁnition f ′(x) = limh → 0 considering a substitution.

Use this alternative deﬁnition to determine the derivative of f (x) = x2 at x = a.

The displacement, s metres, of an object after t seconds is given by s(t) = 3t2 − 2t + 5. Use ﬁrst principles to determine an expression for the instantaneous velocity, v(t), of the object at time t. (Note that velocity is the rate of change of displacement, that is, v(t) = s′(t)).

14.04 Derivative notation and basic rules After this lesson, you will be able to… • use the power rule to differentiate functions of the form xn for real values of n. • apply the constant multiple rule and the sum and difference rule for differentiation. • differentiate polynomial functions by applying the rules of differentiation term-by-term. • rewrite functions involving roots, reciprocals, or products into a suitable form for differentiation. • recognise and use various notations for the derivative.

Differentiate by power rule From exploring gradients of curves, particularly for functions of the form f (x) = xn, a pattern emerges for the derivative. This pattern is known as the power rule. It provides a direct way to differentiate functions where x is raised to a power, without needing to use ﬁrst principles or graphical approximations for each case.

f (x) = xn f ′(x) = nxn − 1 f ′(x)

is the derivative of f (x)

is any real number

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y = xn

is the derivative of y with respect to x n

is any real number

To apply the power rule, multiply the term by the current power of x and then subtract 1 from the power. This rule can be veriﬁed using graphing applications by comparing the calculated derivative with the gradient of the tangent to the curve at various points.

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 Determine the derivative of each function: a y = x4

Create a strategy Apply the power rule

= nxn − 1.

Apply the idea Write the function

Apply the power rule nxn − 1

Simplify the exponent

The derivative of y = x4 is

722

= 4x3.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b y=

Create a strategy Use the negative index rule to rewrite the function in the form xn, then apply the power rule.

Apply the idea Rewriting in the form xn: Write the function

Use the negative index rule

Applying the power rule: Apply the power rule nxn − 1

Simplify the exponent

Optionally, this can be rewritten with a positive index: The derivative of y =

= −2x − 3 or

c y=

Create a strategy Use the rule

to rewrite the function in the form xn, then apply the power rule.

Apply the idea Rewriting in the form xn: Write the function

Use the rule

Applying the power rule: Apply the power rule nxn − 1

Simplify the exponent

Optionally, rewrite in surd form: The derivative of

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Idea summary The power rule for differentiation states that if f (x) = xn, then its derivative is found using the rule:

f (x) = xn f ′(x) = nxn − 1 f ′(x) is the derivative of f (x) n

is any real number

y = xn

is the derivative of y with respect to x n

is any real number

Functions involving roots or reciprocals should ﬁrst be converted to index form xn before applying the rule.

Derivative notations Several notations are commonly used to represent the derivative of a function. If y = f (x), its derivative can be denoted as shown in the table. Notation f ′(x)

Deﬁnition Derivative as a new function of x. Rate of change of y with respect to x.

y′

Shorthand for f ′(x) when using y. Derivative operator

applied to f (x).

Each notation provides a unique way to think about the derivative. The notation f ′(x) or y′, highlights that the derivative is itself a new function. While

, emphasises the derivative as a rate

of change, stemming from the idea of a ratio of inﬁnitesimally small changes in y and x, and treats differentiation as an action performed on a function. The choice of notation often depends on the context of the problem. For example, if y = x2, its derivative is 2x. This can be written as f ′(x) = 2x (if f (x) = x2 ), or or y′ = 2x, or

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(x ) = 2x.

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= 2x,

Example 2 Given the function f (x) = x4, its derivative is 4x3. Express this derivative using: a f ′(x)

Create a strategy

Apply the idea

Identify the function and its derivative and write it using the speciﬁed notation.

The derivative is f ′(x) = 4x3.

, assuming y = f (x)

Apply the idea If y = x4, then the derivative is

= 4x3.

Apply the idea The statement is

(x4) = 4x3.

Idea summary The derivative of a function can be represented using various notations: Notation f ′(x)

Deﬁnition Derivative as a new function of x. Rate of change of y with respect to x.

y′

Shorthand for f ′(x) when using y. Derivative operator

applied to f (x).

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725

Rules for differentiation Exploration Consider the functions f (x) = x3 and g(x) = x2, and their sum, h(x) = f (x) + g(x) = x3 + x2. 1. Using the power rule, determine the derivative of f (x) and the derivative of g(x). 2. Now, assume you can differentiate h(x) by differentiating each term separately. What would the derivative h′(x) be? 3. Compare your answer for h′(x) to the sum of f ′(x) and g′(x) from the ﬁrst question. What do you notice? 4. Based on your observation, propose a general rule for ﬁnding the derivative of a function that is the sum of two other functions.

To differentiate more complex functions, a set of rules can be applied. These rules allow for the differentiation of functions term by term. Constant multiple rule The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.

is a constant

f (x)

is a differentiable function

Sum and difference rule The derivative of a sum or difference of functions is the sum or difference of their individual derivatives.

f (x), g(x)

are differentiable functions

Together, these rules allow for the differentiation of any polynomial function.

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Example 3 Determine the derivative of each function: a f (x) = 4x3 − 5x2 + 7x − 10

Create a strategy Differentiate each term by using the power rule, linear function rule and constant function rule.

Apply the idea Write the function

Differentiate each term

Apply power, linear and constant function rules

Evaluate each term

Create a strategy Rewrite any terms with roots or fractions into index form kxn. Then differentiate each term using the combination of differentiation rules.

Apply the idea Write the function

Rewrite the required terms to index form

Differentiate term by term

Apply power rule

Evaluate each term

The derivative is

Reflect and check The derivative can also be written as

14.04 Derivative notation and basic rules mathspace.co

727

Example 4 Differentiate these functions after ﬁrst simplifying the expression: a f (x) = (x − 3)(x2 + 4)

Create a strategy Expand the brackets to express the function as a polynomial, then differentiate term by term.

Apply the idea First, expand the expression: f (x) = (x − 3)(x2 + 4) 2

Write the function 2

= x(x + 4) − 3(x + 4)

Expand the brackets

= x3 + 4x − 3x2 − 12

Simplify each term

= x − 3x + 4x − 12

Rewrite in descending powers of x

Now, differentiate the simpliﬁed function: Differentiate each term

Apply the differentiation rules

Simplify

b f (x) =

Create a strategy Split the fraction into two terms, then differentiate.

Apply the idea First, simplify the expression: Write the function

Split the fraction

Simplify each term

Now, differentiate the simpliﬁed function: Differentiate each term

Apply the differentiation rules

Simplify

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Idea summary Differentiation of functions can be done eﬃciently using a set of basic rules: • Constant multiple rule:

[k × f (x)] = k × f ′(x)

• Sum and difference rule:

[ f (x) ± g(x)] = f ′(x) ± g′(x)

These rules allow complex functions, such as polynomials, to be differentiated term by term.

14.04 Practice questions What do you remember? 1

List two common notations for the derivative of a function y = f (x).

Write the power rule for differentiation.

Write the rule for differentiating a constant multiple of a function, k × f (x).

Write the sum/difference rule for differentiation.

Practice Ex 1

Ex 2

Determine the derivative of each function: a

y = x7

y = x10

y = x99

y = x200

y = x−5

m y=

For each function, express its derivative using these notations: i

f ′(x)

notation assuming y = f (x)

iii

operator

f (x) = x7

f (x) = 5x3 − 2

f (x) =

14.04 Derivative notation and basic rules mathspace.co

729

Ex 3

Differentiate these functions with respect to x: a

y = 2024

f (x) = −π

f (x) = 5x4

y = −3x7

h(x) = 10x

f (x) = x3 + x2

g(x) = 2x3 + 6x2 − 9x + 1

m y = x2 − x − 2

Ex 4

y = 5x − 3x + 2x − 1

y = 4x5 − 7x

f (x) = ax2 + bx (where a and b are constants)

Rewrite each function in a suitable form for differentiation, then determine its derivative: a

f (x) = x2 (x + 3)

(for x ≠ 0)

y = (2x + 3)2

y = (x − 1)(x + 4) (for x > 0)

Determine the derivative:

a c

b d

A(r) = 4π r2 + 2π rh (with respect to r)

Extend your thinking 10

Find the coordinates of the points on the curve y = x3 − 12x + 1 where the tangent to the curve is horizontal.

The area of a circle is given by A = π r2, where r is the radius: a

Determine

What does a circle).

. represent in the context of a circle? (Hint: Consider the circumference of

A function is deﬁned as f (x) = ax3 + bx2 + cx + d. If f ′(x) = 6x2 − 4x + 3, determine the values of a, b, and c. What can be said about d?

Determine

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by ﬁrst expanding the product.

14.05 Tangents and normals After this lesson, you will be able to… • find the equation of a tangent and normal to a curve at a given point. • find points on a curve where the tangent or normal has a specified gradient. • relate the gradient of a tangent to its angle of inclination using m = tan θ. • solve problems involving tangents, normals and angles of inclination.

Equations of tangents The tangent line at the point (a, f (a)) is a straight line that touches the curve y = f (x) at that point and has the same instantaneous gradient. y

y = f (x) tangent line at (a, f (a))

(a, f (a)) x

The gradient of the tangent at x = a is given by the derivative, f ′(a). Using the point-gradient formula, y − y1 = m(x − x1), the equation of the tangent line to the curve y = f (x) at the point (a, f (a)) is:

y − f (a) = f ′(a)(x − a) x, y are the coordinates of any point on the tangent line (a, f (a))

are the coordinates of the point of tangency

f ′(a)

is the gradient of the tangent at x = a

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Example 1 Determine the equation of the tangent to the curve f (x) = 4x2 + x at the point (1, 5).

Create a strategy Determine the derivative f ′(x) to find the gradient. Substitute x = 1 to find the gradient of the tangent at the given point. Then, use the point-gradient formula y − y1 = m(x − x1) to find the equation of the line.

Apply the idea Determine the derivative of f (x): f (x) = 4x2 + x

Write the function

f ′(x) = 8x + 1

Use the power and linear function rules

Determine the gradient of the tangent at (1, 5): m = f ′(x)

Equate the gradient m to the derivative f ′(x)

= 8x + 1

Substitute f ′(x) = 8x + 1

=8×1+1

Substitute x = 1

Evaluate

Use the point-gradient formula with point (1, 5) and gradient m = 9: y − y1 = m(x − x1)

Write the point-gradient formula

y − 5 = 9(x − 1)

Substitute x1 = 1, y1 = 5, m = 9

y − 5 = 9x − 9

Expand the brackets

y = 9x − 4

Add 5 to both sides

The equation of the tangent to the curve at the point (1, 5) is y = 9x − 4.

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Example 2 Determine the coordinates of the point(s) on the curve y = x3 − 12x + 1 where the tangent is horizontal.

Create a strategy Determine the derivative function and equate it to 0 (a horizontal line has a gradient of 0) to solve for x. Then, substitute the x-values back into the original equation to ﬁnd the corresponding y-coordinates.

Apply the idea Determine the derivative of the function: Write the equation Use the power, linear and constant function rules Set the derivative equal to 0 to solve for x: 3x2 − 12 = 0

Factorise 3

Divide both sides by 3

3(x − 4) = 0 x −4=0 (x − 2)(x + 2) = 0 x

Equate to 0

2 or x = −2

Factorise the difference of two squares Use the null factor law

Substitute these x-values into the original equation y = x3 − 12x + 1 to ﬁnd the y-coordinates. For x = 2: y = x3 − 12x + 1 3

Write the equation

= (2) − 12 × 2 + 1

Substitute x = 2

= 8 − 24 + 1

Evaluate each term

= −15

Evaluate

For x = −2: y = x3 − 12x + 1

Write the equation

= (−2) − 12 × (−2) + 1

Substitute x = −2

= −8 + 24 + 1

Evaluate each term

= 17

Evaluate

The points on the curve where the tangent is horizontal are (2, −15) and (−2, 17).

Idea summary The equation of the tangent line to the curve y = f (x) at x = a is found using the point-gradient formula: y − f (a) = f ′(a)(x − a). To ﬁnd points where a tangent has a speciﬁc gradient, set the derivative f ′(x) equal to that gradient and solve for x.

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Equations of normals Interactive exploration Discover this concept in action online

mathspace.co

Normal (to a curve) In calculus, the normal to a curve at a given point P is the straight line that is perpendicular to the tangent to the curve at a given point P. The normal to a curve at a point is the line that passes through that point and is perpendicular to the tangent at that same point. y

y = f (x) tangent line at (a, f (a))

(a, f (a))

normal line at (a, f (a))

If the tangent line at x = a has a gradient of mT = f ′(a), then the normal line has a gradient mN that is the negative reciprocal of the tangent’s gradient. Their product is −1 (for non-vertical and non-horizontal tangents). mT × mN = −1

is the gradient of the normal line

is the gradient of the tangent line, equal to f ′(a), where f ′(a) ≠ 0 mT The equation of the normal to the curve at the point (a, f (a)) is given by:

x, y are the coordinates of any point on the normal line (a, f (a)) are the coordinates of the point on the curve f ′(a) is the gradient of the tangent at x = a (must be non-zero)

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Example 3 Determine the equation of the normal to the curve f (x) = x3 − 3x2 at the point (1, −2).

Create a strategy Determine the derivative f ′(x) to calculate the gradient of the tangent mT = f ′(1). Next, calculate the gradient of the normal mN = gradient mN.

. Finally, use the point-gradient formula with the point (1, −2) and

Apply the idea Determine the derivative of f (x): f (x) = x3 − 3x2 2

f ′(x) = 3x − 6x

Write the function Use the power rule and sum/difference rules

Calculate the gradient of the tangent line, mT : mT = f ′(x) 2

Equate the gradient mT to the derivative f ′(x)

= 3x − 6x

Substitute f ′(x) = 3x2 − 6x

= 3 × (1)2 − 6 × 1

Substitute x = 1

= −3

Evaluate

Calculate the gradient of the normal line, mN : Write the normal gradient formula

Substitute mT = −3

Simplify

Use the point-gradient formula with point (1, −2) and gradient mN = : Write the point-gradient formula

Substitute x1 = 1, y1 = −2 and mN =

Expand the brackets

Subtract 2 from both sides

The equation of the normal line is

14.05 Tangents and normals mathspace.co

735

Example 4 Determine the coordinates of the point on the curve y = x2 − 5x where the normal has a gradient of .

Create a strategy Step 1: Determine the derivative function. Step 2: Calculate the gradient of the tangent, mT. Step 3: Equate the derivative function to mT and solve for x. Step 4: Substitute the x-value into the original function to ﬁnd the corresponding y-coordinate.

Apply the idea Determine the derivative

: Write the function

Use the power and linear function rules

Determine the gradient of the tangent, mT, given the gradient of the normal is mN = . Write the normal gradient formula Substitute mN =

Multiply both sides by

Equate the derivative

to the tangent gradient mT to solve for x: Equate the derivative to the tangent gradient

Substitute the values

Add 5 to both sides

Divide both sides by 2 2

Substitute x = 1 into the function y = x − 5x to ﬁnd the y-coordinate: y = x2 − 5x 2

Write the function

= (1) − 5 × 1

Substitute x = 1

= −4

Evaluate

The coordinates of the point are (1, −4).

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Idea summary The normal to a curve at a point is the line perpendicular to the tangent at that point. If the tangent gradient is mT = f ′(a), the normal gradient is mN = (for f ′(a) ≠ 0). The equation of the normal is found using: y − f (a) =

(x − a).

Angle of inclination Angle of inclination The angle a straight line makes with the positive x-axis. The angle of inclination, θ, of a line is the angle it makes with the positive direction of the x-axis, measured anticlockwise. The angle is always in the range 0° ≤ θ < 180°. y

rise

θ run

Using trigonometry, the gradient m (rise over run) of a line is equal to the tangent of its angle of inclination, θ.

m = tan θ m is the gradient of the line

θ is the angle measured anticlockwise from the positive x-axis Since the derivative f ′(a) gives the gradient of the tangent to the curve y = f (x) at x = a, this relationship connects the derivative to the angle of the tangent line. • If the gradient m is positive, the angle θ is acute (0° < θ < 90°). • If the gradient m is negative, the angle θ is obtuse (90° < θ < 180°).

14.05 Tangents and normals mathspace.co

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Example 5 Consider the function y = 9 − x2. Determine the x-coordinate of the point on the curve where the tangent makes an angle of 45° with the positive x-axis.

Create a strategy of the function. Use m = tan θ to determine the gradient of the

Determine the derivative tangent.

to m = tan θ and solve for x.

Then, equate

Apply the idea Determine

: Write the function

Use the constant function and power rules

Determine the gradient of the tangent, mT, using the given angle: m = tan θ

Equate the derivative

Write the formula

= tan 45°

Substitute tan θ = 45°

Evaluate the exact value

to the tangent gradient mT to solve for x: Equate the derivative to the gradient

Substitute the values

Divide both sides by −2

The tangent on the curve y = 9 − x2 has a gradient of 1 at the point where x =

Example 6 Determine the angle of inclination, rounded to the nearest degree, of the tangent to the curve f (x) = x2 − 4x + 5 at the point where x = 1.

Create a strategy Determine the derivative f ′(x) and calculate the gradient of the tangent at x = 1. Then use the relationship m = tan θ to determine the angle.

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Apply the idea Determine the derivative of f (x): f (x) = x2 − 4x + 5

Write the function

f ′(x) = 2x − 4

Use the power, linear and constant functions rules

Determine the gradient of the tangent x = 1: m = f ′(x)

Equate the gradient m to the derivative f ′(x)

= 2x − 4

Substitute f ′(x) = 2x − 4

=2×1−4

Substitute x = 1

= −2

Evaluate

Since tan θ = −2, a calculator can be used to determine the angle. A negative gradient means the angle of inclination is obtuse. A calculator may return a negative angle, so 180° must be added to ﬁnd the correct obtuse angle.

θ = 180° + arctan(−2) ≈ 117°

Calculate the obtuse angle Evaluate and round to the nearest degree

The angle of inclination is approximately 117°.

Idea summary The gradient, m, of a line is related to its angle of inclination, θ, by the formula:

m = tan θ m

is the gradient of the line

θ is the angle measured anticlockwise from the positive x-axis For a tangent to a curve y = f (x) at x = a, the gradient is f ′(a), so the relationship is f ′(a) = tan θ.

14.05 Practice questions What do you remember? 1

What is a tangent to a curve at a point P ?

How is the gradient of a tangent to the curve y = f (x) at x = a related to the derivative of f (x)?

Identify the point-gradient formula for the equation of a straight line.

What is a normal to a curve at a point P ?

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If the gradient of the tangent to a curve at a point is mT (where mT ≠ 0), what is the gradient of the normal, mN , at that same point?

What is the relationship between the gradient, m, of a line and its angle of inclination, θ?

Practice Ex 1

Determine the equation of the tangent to the curve for each function at the given point: a

Ex 2

Ex 3

Ex 5

y = x3 − 6x2 + 5, tangent has a gradient of −9

y = 2x2 + 8x − 1, tangent is parallel to the x-axis

Determine the equation of the normal to the curve at the given point:

740

f (x) = x − x2 at the point (2, −2)

at x = 2

Determine the coordinates of the point(s) on the curve where the normal has the given property: a

Curve y = x2 − 3x, normal has a gradient of

Curve y = x3, normal is parallel to the line x + 3y = 1

Determine the x-coordinate(s) of the point(s) on the curve where the tangent makes the given angle with the positive x-axis: Curve y = x2 − 5x, angle is 135°

Curve y = x3 − 2x2, angle is 45°

Determine the angle of inclination, rounded to the nearest degree, of the tangent to the curve at the given point: a

f (x) = x3 + 5x at the point (2, 18)

a Ex 6

Determine the coordinates of the point(s) on the curve where the tangent has the given property:

a Ex 4

f (x) = x2 at the point (−1, 1)

f (x) = x3 − 2x at x = 2

For each function, determine the equation of: i

The tangent line to the curve

The normal line to the curve

f (x) = x2 + 3x − 10 at the positive x-intercept

+ x at x = 4

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

at x = −1

For each graph showing a curve f (x) and its tangent g(x): i

Identify the coordinates of the point at which g(x) is a tangent to the curve f (x).

Calculate the gradient of the tangent.

iii

Determine the equation of the tangent line y = g(x).

Calculate the gradient of the normal.

3 2

f (x)

x 1

−1

x 2

−1

−2 g (x) −3

−2

−4

g (x)

−4 −3 −2 −1

y 4

x 1

−1

f (x)

g (x) f (x)

g (x)

−2

x 0

2 1 −2

−1

−1 −2 −3 −4

g (x) 4

g (x)

x 1

f (x)

3 −3

−2

1 −1

−1

x 1

−2

−5

−3

−6

−4

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For each function f (x) and line g(x) shown: i

Sketch the graph of the function g(x) on the same Cartesian plane as f (x).

Is g(x) a tangent to f (x)? Explain your answer.

Function f (x) = x2 + 4

Function f (x) = x2 − 2

Line g(x) = 2x + 3 Line g(x) = 2x − 1 y 9

f (x)

6 5

−4 −3 −2 −1 −1

4 3

−2

−3

−4 −3 −2 −1

2 3

−4

Function f (x) = (x + 2)3 – 1

Function f (x) = (x + 1)3 − 2

Line g(x) = 4x + 7 Line g(x) = 3x + 3 y

−1

−2

f (x) −4 −3 −2

x 1

−4 −6 −8

1 −4 −3 −2 −1 −1

f (x)

x 1

−2 −3 −4

The curve f (x) =

The line 5x + y + 2 = 0 is the tangent to the curve y = x2 + bx + c at the point x = 9. Determine the values of b and c.

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− 5x has a gradient of 0 at x = 16. Determine the value of k.

Extend your thinking 18

Consider the curve y = x3 − x2 and the line 7x − y = 10. Determine the x-coordinates of the points on the curve where the tangent is perpendicular to the given line.

Consider the curve y = x3:

Investigate if the line y = −12x − 16 is a normal to the curve. If it is, determine the point of contact. If not, justify why no such point exists.

As a comparison, determine the equation of the normal to the curve at the point where x = 2.

If f (x) =

− 3x + 7, what are the coordinates of the points on the curve whose

normal line is perpendicular to the line x + y = 1? 21

Two tangents to the parabola y = x2 intersect at the point (2, 3). Find the coordinates of the two points of tangency.

Find the equations of the tangents to the curve y = x3 that pass through the point (2, 4).

A cubic function y = ax3 + bx2 + cx + d intersects the x-axis at (2, 0), where it has a gradient of 36. It also intersects the y-axis at y = −28, where the tangent is parallel to the x-axis. Determine the cubic function.

In the graph shown, the line y =

+ b is a tangent to the graph of f (x) =

at x = a.

(a, f (a))

Calculate the values of a and b.

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14.06 Chain rule After this lesson, you will be able to… • identify the inner and outer functions of a composite function. • state the chain rule using different notations. • apply the chain rule to find the derivative of a composite function. • use the generalised power rule as an efficient method for differentiating functions raised to a power. • solve problems involving the chain rule.

Composite functions Composite functions When the output of one function becomes the input of a second function. For example, f ( g(x)) (read as ‘f of g of x’) is a composite function where the outputs of function g are taken as the inputs of function f. A composite function is the result of applying one function to the result of another. If we have two functions, u = g(x) and y = f (u), then the composite function is y = f ( g(x)). Consider the function y = (3x2 − 2)5: • The inner function is u = g(x) = 3x2 − 2. • The outer function is y = f (u) = u5. Here are some other examples of composite functions: Composite function y = f ( g(x)) y = (4x3 − 2x + 7)10

Inner function u = g(x) u = 4x3 − 2x + 7 u = x5 − 8x u = 5x2 – 1

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Outer function y = f (u) y = u10

Example 1 Consider the function y = (4x3 − 4x2 − 5x − 7)6. If y = f ( g(x)), identify the inner function u = g(x) and the outer function f (u).

Create a strategy Recognise that (4x3 − 4x2 − 5x − 7)6 is the result of raising an expression (the inner function) to a power (deﬁned by the outer function).

Apply the idea The given function is y = (4x3 − 4x2 − 5x − 7)6. The inner function u = g(x) is the base of the power: u = 4x3 − 4x2 − 5x − 7 Identify the expression being raised to a power The outer function y = f (u) describes what is done to u. In this case, u is raised to the power of 6: f (u) = u6 Identify the operation performed on u

Idea summary A composite function is a function consisting of an inner function and an outer function. If y = f ( g(x)), then g(x) is the inner function (often denoted u) and f (u) is the outer function.

The chain rule Chain rule A formula for the derivative of the composite of two differentiable functions. If y is a function of u, and u is a function of x, then the chain rule states that composite function f ( g(x)), then h′(x) = f ′( g(x)) g′(x).

. If h(x) is the

The chain rule is for differentiating composite functions. It allows differentiation of functions like y = (x + 1)2 without ﬁrst expanding the expression. If y = f (u) and u = g(x) are differentiable functions, then the derivative of the composite function y = f ( g(x)) can be found by:

is the derivative of the outer function y with respect to its variable u is the derivative of the inner function u with respect to x 14.06 Chain rule mathspace.co

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Alternatively, using function notation, if h(x) = f ( g(x)), then:

h′(x) = f ′( g(x)) × g′(x) f ′( g(x)) is the derivative of the outer function f, with the inner function g(x) substituted back in g′′(x)

is the derivative of the inner function g(x)

In words, this means “the derivative of the outside function (with the inside function left alone) multiplied by the derivative of the inside function”. The chain rule with powers (Generalised power rule) A very common application of the chain rule is for functions of the form y = [ f (x)]n. This is a composite function where the inner function is f (x) and the outer function is the power n. Applying the chain rule gives a helpful shortcut:

n[ f (x)]n − 1

is the derivative of the outer power function

f ′(x)

is the derivative of the inner function f (x)

Example 2 Differentiate y = (5 − x2)3 using the substitution method.

Create a strategy Identify the inner function u and the outer function y in terms of u. Determine apply the chain rule formula.

Apply the idea Let the inner function be u = 5 − x2. Then the outer function is y = u3. Determining

: Write the inner function

Determining

Use the constant function and power rules

: Write the outer function

Use the power rule

Applying the chain rule: Write the chain rule formula

Substitute the derivatives

Substitute back u = 5 − x2

Simplify

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and

, then

Example 3 Differentiate

in surd form using the generalised power rule.

Create a strategy Rewrite the function in index form and apply the generalised power rule.

Apply the idea Write the function

Rewrite in index form

Apply the generalised power rule

Differentiate the inner function

Simplify

Reﬂect and check Optionally, this can be written with a positive index:

or in surd form:

Idea summary The chain rule is used to differentiate composite functions. It is:

is the derivative of the outer function y with respect to its variable u is the derivative of the inner function u with respect to x A useful shortcut for functions raised to a power is the generalised power rule: if y = [ f (x)]n, then

= n[ f (x)]n − 1 × f ′(x).

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14.06 Practice questions What do you remember? 1

In a composite function y = f ( g(x)), which part is referred to as the ‘inner function’?

In a composite function y = f ( g(x)), what is meant by the ‘outer function’?

Write the chain rule for differentiation.

Write the ‘Generalised Power Rule’.

Practice 5

If u = 8x2 + 3, write an expression for these functions in the form f (u) = aun: a c

f (x) = 2(8x2 + 3)5

b d

Ex 1

Identify the inside and outside functions: a

f (x) = (5x3 − 4x2 + 3x − 5)7

Ex 2

For each function, apply the chain rule by letting u represent the inner function. To do this, ﬁnd: i

748

iii

y = (2x4 + 6)5

y = (4x + 3) − 1

e 8

y = 3(x + 5)5

Consider the function y = (5x − 7)2 : a

Differentiate y by expanding the brackets ﬁrst.

Differentiate y by using the chain rule.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 3

Differentiate the functions using the power rule: a

y = (4x + 3)9

y = (x2 + x− 3 )3

y = (3x − 4x + 2)

y = −3(3x + 4)10

y = (2t7 + 8t3 + 3t + 5) − 4

Determine the x-coordinate(s) of the point(s) at which: a

f (x) = (x − 2)2 has a gradient of 6

f (x) = (x + 2)3 has a gradient of 48

For the function g(x) = (3 − x5)4, evaluate g′(−1).

Determine the values of x where the tangent of y = (x2 − 1)3 is horizontal.

Determine the values of x where the derivative of y = (2x + x2)5 is equal to zero.

Consider the semicircle deﬁned as

Determine the derivative of the function.

Sketch the graph of the semicircle.

Determine the equation of the tangent at the point (5, 12).

Determine the equation of the tangent for each function at the speciﬁed point: a

y = (2x + 1)4 at the point where x = −1.

y = (2x − 1)8 at the point where x = 1.

Consider the function f (t) =

Write an expression for f ′(t), expressing the derivative in positive index form.

Determine the gradient of the curve f (t) where t = −2.

For the function f (x) =

Calculate the exact value of the gradient of the function

, determine the values of x where f ′(x) =

. at x = 2.

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Extend your thinking 19

Consider the function . Does there exist a point on the function that would have a horizontal tangent? Explain your answer.

The function y = c(ax − 2)2 passes through the point (2, 108). The graph has a gradient of 48 when x = 1. Determine a and c.

Given that y = f ( g(x)) and y′ = 0, can it be concluded that f ′( g(x)) = 0? Justify your answer.

Calculate the gradient of the tangents of y2 + 2x − 5x2 = 3 at x = 1.

Suppose y = (3x2 + k)4, and

= 0 when x = 1. Determine the value of k.

Extension: Proof of chain rule Extension online

mathspace.co

Did you know?

A cup of coffee cools down following a logarithmic curve — it loses heat quickly at first, then cools slowly over time! This happens because the temperature difference between the coffee and the air gets smaller as it cools. Logarithmic functions are often used to model real-world changes like this, where the rate of change decreases over time. They help scientists and engineers predict how quickly heat, sound, or even population growth will settle into balance. 750

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

14.07 Product rule After this lesson, you will be able to… • identify a function as a product of two simpler functions, u and v. • state and apply the product rule for differentiation. • combine the product rule with other differentiation rules, such as the chain rule. • simplify the result after applying the product rule, often by factorisation. • recognise when to expand a product first versus when to apply the product rule.

Product rule Product rule A rule for the derivative of the product of two differentiable functions: If y = uv where u and v are both functions of x, then h′(x) = f (x)g′(x) + f ′(x)g(x).

, or if h(x) = f (x)g(x) then

The product rule is a differentiation rule used to differentiate expressions that are a product of two functions. For example, the product rule would be used to differentiate y = (x3 + 4x + 2)(2x + 1). This expression has two functions being multiplied together. We can call the ﬁrst factor, (x3 + 4x + 2), the function u(x) (or simply u) and the second factor, (2x + 1), the function v(x) (or simply v). The product rule states that if y = uv (where u and v are both functions of x), then its derivative is given by:

is the derivative of the product uv u

is the ﬁrst function is the derivative of the second function

is the second function is the derivative of the ﬁrst function

Alternatively, if h(x) = u(x)v(x), then h′(x) = u(x)v′(x) + v(x)u′(x). This can be remembered as: “The ﬁrst function times the derivative of the second, plus the second function times the derivative of the ﬁrst.”

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Example 1 Differentiate the function: y = (3x − 1)(x + 5).

Create a strategy Identify the two functions, u and v. Determine their derivatives,

and

Then apply the product rule:

Apply the idea Let u = 3x − 1 and v = x + 5. Determining

: Write the ﬁrst factor

Determining

Use the linear and constant functions rules

: Write the second factor

Use the linear and constant functions rules

Apply the product rule: Write the product rule

Substitute the values

Expand the brackets

Collect like terms

The derivative is

= 6x + 14.

Reflect and check When the product is simple, it may be more efficient to expand the expression first: y = (3x − 1)(x + 5) 2

Write the function 2

= 3x + 15x − x − 5 = 3x + 14x − 5

Expand

Then differentiate term by term: Differentiate term by term

Use the power, linear and constant functions rules

Collect like terms

The results match. For simpler products, expanding first can be more efficient. However, the product rule is essential for more complex functions.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Differentiate y = x4 (2x + 3)5.

Create a strategy Identify the two functions u and v. Determine their derivatives. The derivative of v requires the chain rule. Then apply the product rule.

Apply the idea Let u = x4 and v = (2x + 3)5. Determining

: Write the ﬁrst factor

For

Use the power rule

, use the chain rule: Write the second factor

Apply the chain rule

Differentiate the inner function

Simplify

Apply the product rule: Write the product rule

Substitute the values

Factorise 2x3 and (2x + 3)4

Expand the inner bracket

Simplify the inner bracket

Factorise 3 from (9x + 6)

The derivative is

= 6x3 (2x + 3)4 (3x + 2).

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Idea summary The product rule states that if y = uv, where u and v are functions of x, then:

is the derivative of the product uv u

is the ﬁrst function is the derivative of the second function

is the second function is the derivative of the ﬁrst function

This can be remembered as: “The ﬁrst function times the derivative of the second, plus the second function times the derivative of the ﬁrst.”

14.07 Practice questions What do you remember? 1

Identify u and v in these functions: a

(4x7 − 2x3 − 5x)

y = 3x4 (x − 4)2

To differentiate y = x6 (x4 + 4) using the product rule, let u = x6 and v = x4 + 4, then determining: a

u′

v′

Consider the function y = x2 (4x5): a

Identify u and v if this function were to be differentiated using the product rule.

Would using the product rule be an easier method for differentiating this function? Explain.

Practice 4

Ex 1

754

Consider the function y = x3 (x2 + 9): a

Differentiate y by ﬁrst expanding the brackets.

Differentiate y using the product rule, letting u = x3 and v = x2 + 9.

Differentiate these functions: a

y = (x + 5)(x + 9)

y = (3x − 2)(4x − 5)

y = (2t3 − 3)(3 − t)

y = (7t4 − t2)(t2 − 5)

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

Differentiate these functions by using the product rule: a

f (x) = (8x − 9)5 (5x + 7)7

i 7

For each function, determine: iii

f ′(−3)

f (x) = (x2 − 3x)(2x − 5)

f (x) = x(x − 1)

f (x) = (x + 1)(x − 2)

f (x) = (2x − 1)(x + 2)

f (3)

a c

f ′(0)

For each function, determine: i

f ′(x) in factorised form where possible

The equation of the tangent at x = −1

iii

The equation of the normal at x = −1

f (x) = (x + 1)(x + 3)3

Determine the gradient of the tangent to the curve

Determine the derivative of of two linear factors.

f (x) = x2 (x + 2) at the point where x = 2.

by ﬁrst expressing the function to be a product

Extend your thinking 11

The derivative of f (x) = (3xn + 4)(5x2 − 2x) is of degree 5. Determine the value of n.

For each of these functions, determine the values of x for which the derivative is zero: a

y = x (x − 8)4

y = x3 (x + 3)4

y = (x + 2)(x + 5)6

Consider the function g(x) = x3 f (x), where f (x) is a function of x. Given that f (3) = 1 and f ′(3) = −3, determine g′(3).

Consider the two functions f (x) = (x3 + 2x − 1)(x + 5) and g(x) = x − 2 (4 + 3x−4). If h(x) = g(x) f (x), determine h′(2).

If f (x) = 1 + 2 + 3 + … + (x − 1) + x, show that f ′(x) = x + . Note: The sum of the ﬁrst n positive integers is given by

(n + 1).

14.07 Product rule mathspace.co

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Extension: Proof of product rule Extension online

mathspace.co

14.08 Quotient rule After this lesson, you will be able to… • identify the numerator (u) and denominator (v) in a quotient of functions. • state and apply the quotient rule to find the derivative of a function. • combine the quotient rule with other rules, such as the chain rule. • simplify algebraic expressions after applying the quotient rule. • solve problems involving tangents to curves that require the quotient rule.

The quotient rule Quotient rule A formula for the derivative of the ratio of two differentiable functions. If y = , where u and v are both functions of x, then the quotient rule states that where g(x) is not equal to 0, then

, or if

The quotient rule is a differentiation rule used to determine the derivative of functions that are expressed as a quotient of two differentiable functions, i.e., one function divided by another. If a function y is deﬁned as y = quotient rule can be applied.

, where u(x) is the numerator and v(x) is the denominator, the

is the derivative of y with respect to x v

is the function v(x) (the denominator) is the derivative of u with respect to x

is the function u(x) (the numerator) is the derivative of v with respect to x

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Alternatively, using function notation, if

, then

A common mnemonic to remember the quotient rule is: “Low dee high, minus high dee low, over the square of what’s below”, where ‘low’ is the denominator v, ‘high’ is the numerator u, and ‘dee’ means the derivative.

Example 1 Differentiate the function y =

Create a strategy Identify the functions u and v, calculate their derivatives

and

, then apply the quotient rule.

Apply the idea Let u = x2 + 6x + 5 and v = x. For

: Write the equation for the numerator

For

Differentiate

: Write the equation for the denominator

Differentiate Apply the quotient rule: Write the quotient rule

Substitute the values

Expand the terms in the numerator

Collect like terms

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Reﬂect and check For this speciﬁc function, it is also possible to simplify the expression by dividing each term in the numerator by x before differentiating using the power rule. Divide each term in the numerator by x

Simplify each term

Differentiate each term with respect to x

Rewrite with positive exponent

Combine into a single fraction

Both methods yield the same result.

Example 2 Determine the equation of the tangent to the curve y =

at the point where x = 2.

Create a strategy Substitute x = 2 into the equation to ﬁnd the y-coordinate of the contact point. Differentiate using the quotient rule to get the gradient, then apply the point-gradient form to ﬁnd the tangent line equation.

Apply the idea Determine the y-coordinate at x = 2: Write the equation Substitute x = 2

Evaluate the numerator and denominator

Simplify the fraction

The point of contact is

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For the gradient, differentiate the equation using the quotient rule. If u = x2 − 2, then

= 2x and if v = x + 2, then

= 1.

Write the quotient rule

Substitute the values

Expand the brackets in the numerator

Collect like terms

Substitute x = 2 into

to ﬁnd the gradient m of the tangent: Equate mT to

Substitute

Substitute x = 2

Evaluate the terms

Evaluate the numerator

Simplify and m = :

Use the point-gradient formula with

Write the point-gradient formula

Substitute the point and gradient

Expand the brackets

Add

Simplify the constant terms

The equation of the tangent is

to both sides

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Idea summary The quotient rule is used to differentiate functions of the form y =

is the derivative of y with respect to x v

is the function v(x) (the denominator) is the derivative of u with respect to x

is the function u(x) (the numerator) is the derivative of v with respect to x

14.08 Practice questions What do you remember? 1

For a function y =

, identify u(x) and v(x):

The quotient rule for differentiating a function y = , where u and v are functions of x, is given by

. Fill in the blanks.

Given u(x) = x2 + 1 and v(x) = x − 3: a

Determine

Explain a scenario where it might be more eﬃcient to simplify an algebraic fraction before differentiating, rather than directly applying the quotient rule. Provide a simple example.

Consider the function y = :

760

By ﬁrst rewriting it in negative index form, differentiate y.

By using the substitutions u = 3 and v = x, differentiate y using the quotient rule.

Calculate the value of x for which the derivative is undeﬁned.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Practice 6

For each function: i

Identify u and v.

Determine

and

iii

Determine

using the quotient rule.

Find the values of x for which the derivative is zero.

Functions: b

a Ex 1

Differentiate these functions using the quotient rule: a

i 8

For each function, determine the value of the derivative at the speciﬁed point: a

f (x) =

at f ′(1)

f (x) =

at f ′(3)

f (x) =

at f ′(1)

f (x) =

at f ′(5)

Consider the function y =

− 5:

Calculate the gradient function using the quotient rule for the term

Calculate the gradient of the function at x = 25.

For each function: i

Differentiate y.

Is it possible for the derivative to be zero? Explain brieﬂy.

iii

Determine the value(s) of x for which the derivative is undeﬁned.

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Ex 2

Determine the equation of the tangent to the curve for each function at the speciﬁed point: a

at the point at x = 2

at the point where x = 4

at x = 5

Calculate the derivative f ′(x) and then evaluate it at the speciﬁed values: a

f (x) =

at f ′(0), f ′(2), and f ′(−1)

f (x) =

at f ′(0) and f ′(1)

f (x) =

at f ′(0), f ′(1), and f ′(4) at f ′(1) and f ′(4)

For each function: i

Find f ′(4) for each function.

Combine each fraction into one.

iii

Find f ′(4) for the combined fractions in part (ii).

Extend your thinking 14

Calculate the values of x such that the gradient of the tangent to the curve y =

Differentiate y =

and calculate the value(s) of a if

Differentiate y =

and calculate the possible values of k given that

Consider the function g(x) = g′(2).

The function f (x) =

is −3.

= 0 at x = a. = 1 at x = −3.

. Given that f (2) = 2 and f ′(2) = 6, determine the value of

intersects a line L at the

point P (b, 4). The line L has a gradient of

. The graph

9 8

of f (x) and line L are shown in the Cartesian plane.

Determine:

The value of b

The equation of the line L

f (x) P (b, 4)

3 2

x 0 2 4

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

6 8 10 12 14 16 18

Extension: Proof of quotient rule Extension online

mathspace.co

14.09 Apply differentiation rules After this lesson, you will be able to… • identify the primary structure of a complex function (product, quotient, or composite). • apply a combination of the product, quotient, and chain rules to differentiate a function. • manage and organise the steps involved in multi-rule differentiation problems. • simplify complex algebraic expressions that result from applying multiple differentiation rules. • solve application problems, such as finding gradients and tangents, which require combined rules.

Apply differentiation rules Three main differentiation rules can be combined to differentiate more complex functions: Rule

Description

Chain rule

For composite functions y = f ( g(x))

Product rule

For product functions y = u(x)v(x)

Quotient rule

For quotient functions y =

Formula , where u = g(x)

When faced with a complex function, ﬁrst, identify its overall structure (for example, whether it is primarily a product, a quotient, or a composite function) to determine the primary rule to apply, and then subsequently apply other rules as needed.

14.09 Apply differentiation rules mathspace.co

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Example 1 Calculate the derivative of y = (5x3 − 8)2

Create a strategy The function is a product of two expressions, so the product rule is the primary rule. Identify u and v. The derivative of the ﬁrst term, u = (5x3 − 8)2, will require the chain rule.

Apply the idea Let u = (5x3 − 8)2 and v = For

− 2.

: Write the equation for u

Use the chain rule

Differentiate the second term

Simplify

For

Write the equation for v Use linear function and constant function rules Apply the product rule: Write the product rule

Substitute the values

Factorise the common term (5x3 − 8)

Expand the expressions inside the bracket

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Simplify the expression inside the bracket

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Calculate the gradient of the graph of y =

at the point where x = 0.

Create a strategy Use the power rule y = ( f (x))n, where y′ = n( f (x))n − 1 f ′(x). Then, evaluate the gradient at x = 0.

Apply the idea Let y = f (x) =

For f ′(x): Apply the quotient rule

Expand the brackets in the numerator

Simplify the numerator

For y′: Write the function

Rewrite in exponential form

Use the power rule

Rewrite in surd form

Calculate the gradient at x = 0. Write the gradient

Substitute x = 0

Simplify the terms

Evaluate each term

Evaluate

The gradient at x = 0 is 0.

14.09 Apply differentiation rules mathspace.co

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Example 3 Determine the equation of the tangent line to the graph of f (x) = where x = 1.

(x + 8)(4 − x2) at the point

Create a strategy The function is primarily a product, where one of the factors is itself a product. Apply the product rule to find the derivative. Then, substitute x = 1 to find the gradient at that point. Finally, use the point-gradient formula to find the equation of the tangent.

Apply the idea Let

: Write the equation for u

Rewrite in index form

Expand the brackets

and v = 4 − x2.

So For u′:

Write the expanded equation for u Use the power rule For v′: v = 4 − x2

Write the equation for v

v′ = −2x

Use the constant function and power rules

Apply the product rule: f ′(x) = uv′ + vu′: Write the product rule

Substitute the values

Now, calculate the gradient m = f ′(x) by substituting x = 1 into the derivative. Equate m to f ′(x)

Substitute the derivative

Substitute x = 1

Simplify each bracket

Evaluate each bracket

Evaluate the multiplication

Evaluate

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Now, substitute x = 1 into the function to calculate the y-coordinate: Write the function

Substitute x = 1

Evaluate each term

Evaluate

So the point of tangency is at (1, 54). Using the the point-gradient formula, substitute the gradient m = −3 and the point (1, 54): Write the point-gradient formula

Substitute m = −3 and (x1, y1 ) = (1, 54)

Expand the brackets

Add 54 to both sides

The equation of the tangent line is y = −3x + 57.

Idea summary To differentiate complex functions, often a combination of the chain rule, product rule, and quotient rule is required. Identify the overall structure of the function to determine the primary rule to apply, and then subsequently apply other rules as needed.

14.09 Practice questions What do you remember? 1

For each function, identify the primary differentiation rule(s) (chain, product, quotient) required to calculate its derivative: a

y = (2x − 5)(x2 + 1)

y = (5x3 + 2x)4

Explain in your own words when it is necessary to use the chain rule for differentiation.

If a function y = u(x)v(x) is a product of two differentiable functions u(x) and v(x), state the product rule for

14.09 Apply differentiation rules mathspace.co

767

Practice 4

For each function: a

Calculate

Determine the values of x for which the derivative is undeﬁned, if any.

iii

y = (3x + 2)(8x2 − 11)

Ex 1

Calculate the derivative of

Ex 2

For each function, calculate its gradient at the speciﬁed point:

Ex 3

768

at x = 2

y = x2 (2x − 1)3 at x = −1

at x = 1

y = (x2 − 4)

at x = 4

at x = 2

Determine the equation of the tangent line to the graph of the function at the speciﬁed point: a

f (x) =

at the point where x = −1

f (x) =

(x + 3)(x2 − 2) at the point where x = 1

y = (x3 + 1)

at the point where x = 0 at the point where x = 2

Calculate the gradient of the normal to the graph of the function at the speciﬁed point: a

f (x) =

at the point where x = 4

at the point where x = 2

at the point where x = −1

at the point where x = 2

Determine the x-coordinates of the stationary points of the graph of y = ((x2 − 2)(x2 + 3))2.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the function y =

Identify the sequence of differentiation rules needed to calculate

Calculate

Hence, calculate the gradient at the point where x = 1.

For the function y = 2x(x − 2)2 (x + 4): a

Calculate the x-intercepts and y-intercept.

Calculate

Determine the x-values for

. = 0.

Extend your thinking 12

The graph of y = of a and b.

If f (x) =

has a tangent line y =

− 2 at the point (a, b). Calculate the values

, calculate the value of k for which the equation of the tangent line at the

point where x = −1 is y =

Given three differentiable functions f (x), g(x), and h(x), derive a formula for the derivative of their product P (x) with respect to x.

Differentiate: f (x) = 1 + (3 − (6 + 5x4)9)2.

The function g(x) is deﬁned as g(x) = (x2 + 2x + 3) f (x). If f (0) = 5 and the derivative of f (x) at x = 0 is f ′(0) = 4, determine g′(0).

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14.10 Graphical behaviour of functions After this lesson, you will be able to… • describe the behaviour of a function (increasing, decreasing, stationary) based on the sign of its first derivative. • find the coordinates of stationary points on a cubic function and the intervals where it is increasing or decreasing. • sketch the graph of a derivative, y = f ′(x), from the graph of an original function, y = f (x). • interpret the graph of a derivative function to describe features of the original function. • numerically estimate the value of the derivative at a point on a curve.

Features of a function from its derivative The ﬁrst derivative of a function, f ′(x), provides information about the gradient of the tangent to the curve y = f (x) at any point x. This gradient indicates whether the function is increasing, decreasing, or stationary at that point. y

y 3

1 x

−2

−1

x −2

−1

The function y = x + 1 is always increasing. The gradient f ′(x) = 1 is always positive.

−1

The function y = −x + 1 is always decreasing. The gradient f ′(x) = −1 is always negative.

Stationary point A stationary point on the graph y = f (x) of a differentiable function is a point where f ′(x) = 0. A differentiable function is a function which can be differentiated at each point in its domain.

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y 4 3

The function y = x2 has a stationary point at (0, 0) where the gradient f ′(x) = 0.

2 1

f ′(x) = 0 −2

−1

x 1

The behaviour of a function can be summarised as: • If f ′(x) > 0 on an interval, then f (x) is an increasing function on that interval. • If f ′(x) < 0 on an interval, then f (x) is a decreasing function on that interval. • If f ′(c) = 0 at a point x = c, then the function has a stationary point at x = c. At this point, the tangent to the curve is horizontal. The derivative can be used to determine the x-values of stationary points and the intervals over which a function is increasing or decreasing. These concepts are well illustrated by cubic functions. Cubic function A cubic function is a polynomial function of degree 3, that is, a function of the form f (x) = ax3 + bx2 + cx + d.

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Example 1 Consider the function f (x) = x3 − 6x2 + 5: a Determine the values of x for which the function is increasing or decreasing.

Create a strategy Determine the derivative f ′(x) and equate it to 0 to ﬁnd the stationary points. Test the values of x around these points to determine the sign of f ′(x) and thus where the function is increasing( f ′(x) > 0) and decreasing( f ′(x) < 0).

Apply the idea For the derivative: f (x) = x3 − 6x2 + 5 2

f ′(x) = 3x − 12x

Write the function Differentiate

To ﬁnd stationary points, set f ′(x) = 0: 3x2 − 12x = 0

Equate the derivative to zero

3x(x − 4) = 0

Factorise

0, x = 4

Use the null factor law

These values divide the x-axis into three intervals: x < 0, 0 < x < 4 and x > 4. Test a point in each interval to check the sign of f ′(x). • For x < 0, test x = −1: f ′(x) = 3x2 − 12x 2

f ′(−1) = 3 × (−1) − 12 × (−1)

Write the derivative Substitute x = −1

= 3 + 12

Evaluate power and multiplication

= 15

Evaluate

Since f ′(−1) > 0, the function is increasing for x < 0. • For 0 < x < 4, test x = 1: f ′(x) = 3x2 − 12x 2

f ′(1) = 3 × (1) − 12 × 1

Write the derivative Substitute x = 1

= 3 − 12

Evaluate power and multiplication

= −9

Evaluate

Since f ′(1) < 0, the function is decreasing for 0 < x < 4. • For x > 4, test x = 5: f ′(x) = 3x2 − 12x 2

f ′(5) = 3 × (5) − 12 × 5

Write the derivative Substitute x = 5

= 75 − 60

Evaluate power and multiplication

= 15

Evaluate

Since f ′(5) > 0, the function is increasing for x > 4. So, the function is increasing for x < 0 and x > 4, and decreasing for 0 < x < 4.

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b Determine the coordinates of the stationary points and show them on a sketch of the function.

Create a strategy Substitute the stationary points from part (a) into the function to ﬁnd the corresponding y-coordinates, then plot the points to graph.

Apply the idea When x = 0: f (x) = x3 − 6x2 + 5 3

Write the function 2

f (0) = (0) − 6 × (0) + 5 =5

Substitute x = 0 Evaluate

When x = 4: f (x) = x3 − 6x2 + 5

Write the function

Substitute x = 4

f (4) = (4) − 6(4) + 5 = 64 − 6 × 16 + 5

Evaluate the powers

= 64 − 96 + 5

Evaluate the multiplication

= −27

Evaluate y 5

(0, 5) x

−1

The coordinates of the stationary points are (0, 5) and (4, −27). The graph visually conﬁrms our analysis of increasing and decreasing intervals.

1 −5

f (x) = x3 − 6x2 + 5

−10 −15 −20 −25

(4, −27)

Idea summary The sign of the ﬁrst derivative, f ′(x), indicates whether the function is increasing, decreasing, or stationary at a given point. • If f ′(x) > 0 on an interval, f (x) is increasing on that interval. • If f ′(x) < 0 on an interval, f (x) is decreasing on that interval. • If f ′(x) = 0 at a point, f (x) has a stationary point at that point. To ﬁnd stationary points for a function like a cubic function, calculate the derivative, set it to zero, and solve for x.

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Graph the derivative function The graph of a function’s derivative, y = f ′(x), can be sketched directly from the graph of the original function, y = f (x), by observing the gradient of f (x). Key relationships to look for: y

−1

y = f ′(x)

y = f (x) −2

−2

−1

−3

−1

−1 −2

−2

A cubic function y = f (x) with stationary points (SP) at x = −1 and x = 1.

−3

The derivative y = f ′(x). Its x-intercepts match the stationary points on f (x).

• Where f (x) is increasing(positive gradient), the graph of f ′(x) will be above the x-axis ( f ′(x) > 0). • Where f (x) is decreasing(negative gradient), the graph of f ′(x) will be below the x-axis ( f ′(x) < 0 ). • Where f (x) has a stationary point (zero gradient), the graph of f ′(x) will have an x-intercept ( f ′(x) = 0). Furthermore, the degree of the derivative of a polynomial is one less than the degree of the original function. For example: • If f (x) is a quadratic (degree 2), then f ′(x) will be linear (degree 1). • If f (x) is a cubic (degree 3), then f ′(x) will be quadratic (degree 2).

Example 2 The graph of a quadratic function y = f (x) is shown. Sketch the graph of its derivative, y = f ′(x).

y 2 1

y = f (x) −1

x 2

−1 −2

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Stationary point

Create a strategy Find the x-coordinate of the stationary point, which will become the x-intercept of f ′(x). Determine the intervals where f (x) is decreasing and increasing to know where f ′(x) is negative and positive. Since f (x) is a quadratic, f ′(x) will be a straight line.

Apply the idea 1. Stationary point: The graph of f (x) has a stationary point at x = 1. This means the gradient is zero at this point. Therefore, the graph of f ′(x) must have an x-intercept at x = 1. 2. Decreasing interval: The graph of f (x) is decreasing for all x < 1. This means the gradient is negative in this interval. Therefore, the graph of f ′(x) must be below the x-axis ( f ′(x) < 0) for x < 1. 3. Increasing interval: The graph of f (x) is increasing for all x > 1. This means the gradient is positive in this interval. Therefore, the graph of f ′(x) must be above the x-axis ( f ′(x) > 0) for x > 1. Combining these facts gives the graph of y = f ′(x), which is a straight line passing through (1, 0) with a positive gradient. y 2 1 x −1

1 −1

y = f ′(x)

−2

Idea summary To sketch the graph of the derivative y = f ′(x) from the graph of y = f (x), follow these steps: • Identify the x-coordinates of any stationary points on f (x). These become the x-intercept s on the graph of f ′(x). • Identify the intervals where f (x) is increasing. The graph of f ′(x) will be above the x-axis for these intervals. • Identify the intervals where f (x) is decreasing. The graph of f ′(x) will be below the x-axis for these intervals. • Connect the points, remembering that the derivative of a polynomial has a degree one less than the original.

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Estimate the derivative at a point While differentiation rules provide the exact value of a derivative, it is also possible to estimate the derivative at a point. This can be done graphically from a curve or numerically from the function’s rule. Graphical estimation To estimate the derivative at a point P on a curve, draw a tangent line at that point. The gradient of this tangent line is an estimate of the derivative. 1. Draw the graph of the function y = f (x). 2. Identify the point P where the derivative has to be estimated. 3. Draw a straight line that just touches the curve at point P (the tangent). 4. Choose two distinct points on the tangent line and use the gradient formula, m = calculate its gradient.

, to

Numerical estimation To estimate the derivative numerically, use the idea that the gradient of a secant line through two very close points is a good approximation of the gradient of the tangent. The gradient of the secant line through (c, f (c)) and (c + h, f (c + h)) is given by the formula:

f ′(c) is the approximate derivative ( gradient) at x = c h

is a very small number (e.g., 0.01 or 0.001)

The smaller the value of h, the better the approximation. Digital tools like graphing calculators can perform this calculation instantly.

Interactive exploration Discover this concept in action online

mathspace.co

Example 3 Consider the function f (x) = x3 where the derivative has to be estimated at x = 1: a Estimate f ′(1) graphically.

Create a strategy Sketch the graph of y = x3, draw a tangent at the point (1, 1), pick another point on this tangent, and calculate the gradient.

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Apply the idea The tangent line at P (1, 1) is drawn in blue. Another point on the tangent is Q(2, 4), use these points to calculate the gradient.

y 4

Write the gradient formula

Substitute the coordinates of P and Q

Evaluate the subraction

Simplify

Q(2, 4)

y = x3 P (1, 1)

x 1

The graphical estimate for f ′(1) is 3.

b Estimate f ′(1) numerically using a value of h = 0.01.

Create a strategy Use the formula f ′(c) ≈

with c = 1 and h = 0.01.

Apply the idea Write the formula

Substitute c = 1 and h = 0.01

Simplify the numerator input

Apply the function rule f (x) = x3

Evaluate the powers

Evaluate the subtraction

Evaluate

The numerical estimate for f ′(1) is 3.0301.

Reﬂect and check The exact value can be found by differentiating. For f (x) = x3, the derivative is f ′(x) = 3x2. At x = 1, the exact value is f ′(1) = 3(1)2 = 3. Both the graphical and numerical estimates are very close to the exact value.

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c Find the exact value of f ′(1) and compare it with the estimates from parts (a) and (b). Interpret the results.

Create a strategy Differentiate f (x) = x3 to ﬁnd f ′(x), then evaluate at x = 1. Compare with the graphical and numerical estimates to assess their accuracy.

Apply the idea f (x) = x3 f ′(x) = 3x

Write the function 2

Differentiate 2

f ′(1) = 3 × (1) =3

Substitute x = 1 Evaluate

The exact value of f ′(1) is 3. Comparing with part (a), the graphical estimate is 3, which is exactly the same as the exact value, indicating a precise tangent line approximation in this case. Comparing with part (b), the numerical estimate is 3.0301, which is slightly higher than the exact value by 0.0301. This small difference arises because h = 0.01 is not inﬁnitesimally small, but the estimate is still very close, demonstrating the effectiveness of the numerical method for small h.

Reﬂect and check Both estimation methods yield results close to the exact derivative, with the graphical method being exact in this instance due to the simplicity of the function and the chosen tangent points. The numerical method’s slight error highlights that smaller values of h improve accuracy, as the secant line better approximates the tangent line.

Idea summary The value of the derivative at a point on a curve can be estimated using two main methods: • Graphically: Draw a tangent to the curve at the point and calculate the gradient of that tangent line. • Numerically: Use the formula f ′(c) ≈

with a very small value for h.

These methods provide approximations, while differentiation rules give the exact value.

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14.10 Practice questions What do you remember? 1

Explain what the sign of the ﬁrst derivative, f ′(x), indicates about the behaviour of a function f (x): a

f ′(x) > 0 on an interval

f ′(x) = 0 at a point

f ′(x) < 0 on an interval

How to determine the x-coordinates of the stationary points of a cubic function y = f (x)?

Describe the relationship between the stationary points on the graph of y = f (x) and the features on the graph of its derivative, y = f ′(x).

Describe a method for numerically estimating the derivative of a function f (x) at a point x = c.

Practice Ex 1

For each cubic function: i

Determine the values of x where the function is increasing and where it is decreasing.

Determine the coordinates of the stationary points and show them on a sketch of the function.

f (x) = x3 − 3x2 + 1

f (x) = 2x3 + 3x2 − 12x

f (x) = −x3 + 12x − 4

f (x) = x3 + 3x

f (x) = −x3 + 3x2 + 9x − 1

f (x) =

f (x) = x3 − 3x

f (x) = −x3

− x2 − 3x + 2

For the derivative of a cubic function is f ′(x) = 3x2 − 3: a

Determine the x-coordinates of the stationary points.

State the intervals where the original function f (x) is increasing or decreasing.

The function f (x) = x3 + kx2 + 9x − 2 has a stationary point at x = −3: a

Determine the value of k.

Determine the x-coordinate of the other stationary point.

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779

Ex 2

For each graph of y = f (x), sketch the corresponding graph of its derivative, y = f ′(x): a

y 3

2 1

y = f (x) −1

y = f (x)

x 2

−2

−1

−2 −2

−3

y 3

y = f (x)

−2

x −2

−1

x 2

−1

−2

−1

−3

3 2 2

y = f (x)

x −2

−1

2 1

−1

y = f (x)

y = f (x) x −2

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x −2

−1

Ex 3

Numerically estimate the derivative of f (x) = x3 − 3x2 + 1 at x = 1 with these values of h: a

h = 0.1

h = 0.01

h = 0.001

What value does the estimate approach as h gets smaller?

Consider the function f (x) = x4 where the derivative has to be estimated at x = 2: a

Estimate f ′(2) graphically.

Estimate f ′(2) numerically using a value of h = 0.01.

Find the exact value of f ′(2) and compare it with the estimates from parts (a) and (b). Interpret the results.

For each graph shown: i

The tangent at the indicated point is drawn on the graph. Calculate the gradient of this line.

Numerically estimate the derivative at the indicated point using h = 0.01.

Estimate f ′(1) for f (x) = x3 + x.

Estimate f ′(2) for f (x) = x3 − x.

4 3

y = f (x) −1

x 1

−1

y = f (x)

−2

−3 −4

Estimate f ′(−1) for f (x) = x4 − x2.

Estimate f ′(1) for f (x) = −x5 + 4x.

y 1

y = f (x)

−1

4 3 2 1 −1

−1

y = f (x)

x 1

−2 −3 −4

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The graph of the derivative, f ′(x), of a function is shown: For what values of x does f ′(x) have stationary points?

On what intervals is f (x) increasing?

On what intervals is f (x) decreasing?

−3

−2

−1

x 1

−2 −3 −4

Extend your thinking 13

For what values of k does the cubic function f (x) = x3 − 6x2 + kx − 5 will have no stationary points?

A function y = f (x) has these properties: • It is a continuous cubic function. • f ′(x) > 0 for x < 1 and x > 5. • f ′(x) < 0 for 1 < x < 5. • It passes through the origin (0, 0). Sketch a possible graph of y = f (x), showing the general shape and the location of any stationary points.

A student was asked to ﬁnd and classify the stationary points of the cubic function f (x) = 2x3 − 3x2 − 12x + 7. Their working is shown: 1.

f ′(x) = 6x2 − 6x − 12

2. Set f ′(x): 6(x2 − x − 2) = 0 6(x − 2)(x + 1) = 0 x

2, x = −1

Stationary points at x = −1 and x = 2. 3.

First derivative test: f ′(−2) = 24 > 0, f ′(0) = −12 < 0, f ′(3) = 24 > 0.

Conclusion: The signs are +, −, +. Therefore, there is a minimum turning point at x = −1 and a maximum turning point at x = 2.

Identify the error in the student’s conclusion and provide the correct classiﬁcation of the stationary points.

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14.11 Derivatives as rates of change After this lesson, you will be able to… • interpret the derivative as an instantaneous rate of change. • distinguish between average and instantaneous rates of change. • define velocity as the derivative of displacement with respect to time. • distinguish between displacement/velocity (vector quantities) and distance/ speed (scalar quantities). • solve problems involving rates of change and linear motion.

The derivative as an instantaneous rate of change One of the most important interpretations of the derivative is as the instantaneous rate of change. Instantaneous rate of change The rate of change at a particular moment. For a differentiable function, the instantaneous rate of change at a point is its derivative at that point, so it equals the gradient of the tangent to its graph at the point. If a quantity y is a function of a variable x, written y = f (x), then the derivative f ′(c) gives the exact rate at which y is changing with respect to x at the instant when x = c. An application of this idea is in describing motion. When an object moves in a straight line, its position can be represented as a function of time.

Example 1 The volume V (in litres) of water in a tank after t minutes is given by V (t) = 100 + 20t − t2 for 0 ≤ t ≤ 10. a Calculate the average rate of change of volume between t = 2 and t = 5 minutes.

Create a strategy Calculate V (t) at t = 2, 5. Then, use the formula for average rate of change,

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Apply the idea For V (2): V (t) = 100 + 20t − t2

Write the volume function 2

V (2) = 100 + 20 × 2 − (2) = 136

Substitute t = 2 Evaluate

For V (5): V (t) = 100 + 20t − t2

Write the volume function 2

V (5) = 100 + 20(5) − (5) = 175

Substitute t = 5 Evaluate

For the average rate of change: Write the average rate formula

Substitute a = 2 and b = 5

Substitute V (5) = 175 and V (2) = 136

Evaluate

b Calculate the instantaneous rate of change of volume at t = 3 minutes.

Create a strategy Find the derivative V ′(t), then substitute t = 3.

Apply the idea V (t) = 100 + 20t − t2

Write the function

V ′(t) = 20 − 2t

Differentiate

For the instantaneous rate of change: V′ (t) = 20 − 2t

Write the derivative

V′(3) = 20 − 2(3)

Substitute t = 3

= 14

Evaluate

The instantaneous rate of change at t = 3 is 14 L/min.

Idea summary The instantaneous rate of change of a function f (x) at x = c is given by the derivative f ′(c).

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Applications Displacement The change in position of an object, after a period of time, from its original position. Displacement is a vector quantity. The displacement may be positive, negative or zero. Velocity The rate of change of an object’s position with respect to time. It is a vector quantity, meaning it has both magnitude and direction. The SI units for velocity are metres per second (ms−1 ). A particle moving along the x-axis which position at time t is x(t) has velocity

v(t)

is the velocity at time t

is the derivative of displacement x with respect to time t x′(t) is the derivative of the displacement function x(t) is another notation for the time derivative of displacement, this ‘dot’ notation is generally reserved for derivatives with respect to time The sign of the velocity indicates the direction of motion. If v > 0, the particle is moving in the positive direction. If v < 0, it is moving in the negative direction. If v = 0, the particle is instantaneously at rest. It is important to distinguish between vector quantities, which have direction, and scalar quantities, which only have magnitude. • Displacement (vector) vs. Distance (scalar): Displacement is the change in position from the origin, while distance is the total path travelled. • Velocity (vector) vs. Speed (scalar): Velocity includes direction (positive or negative sign), while speed is the magnitude of the velocity and is always non-negative. Distance The length between two points. Distance is a positive scalar quantity. Speed The absolute value of an object’s velocity. It represents how fast the object is moving. For an object moving along the x-axis, average speed is calculated as Average speed = speed is

. If its position at time t is x(t) the instantaneous

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Speed = ∣v(t)∣ = ∣x′(t)∣ ∣v(t)∣

is the absolute value, or magnitude, of the velocity

Example 2 The displacement of a particle from an origin O is given by x = t3 − 6t2 + 9t + 1 metres, where t ≥ 0 is the time in seconds. a Calculate the initial velocity of the particle.

Create a strategy Find the velocity function initial velocity.

by differentiating the displacement x. Then, substitute t = 0 for the

Apply the idea x = t3 − 6t2 + 9t + 1

Write the displacement

= 3t − 12t + 9

Differentiate

For the initial velocity: Velocity = 3t2 − 12t + 9

Write the velocity function

Initial velocity = 3 × (0) − 12 × 0 + 9

Substitute t = 0

= 9 m/s

Evaluate

b Determine when the particle is momentarily at rest.

Create a strategy The particle is at rest when its velocity is zero. Set

= 0 to solve for t.

Apply the idea = 3t2 − 12t + 9

Write the velocity function

Set

3(t − 4t + 3) = 0

Factorise 3

3(t − 1)(t − 3) = 0

Factorise the quadratic

3t − 12t + 9 = 0

1, t = 3

Use null factor law

The particle is at rest at t = 1 s and t = 3 s.

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c Calculate the speed of the particle at t = 2 seconds.

Create a strategy Substitute t = 2 into . The speed is the absolute value of this result.

Apply the idea = 3t2 − 12t + 9

Write the velocity function

= 3 × (2) − 12 × 2 + 9

Substitute t = 2

= 12 − 24 + 9

Evaluate the power and multiplication

= −3 m/s

Evaluate

So the speed is ∣ ∣ = ∣−3∣ = 3 m/s.

Example 3 The graph shows the velocity v(t) of a particle over 8 seconds. v(t) (m/s) 2 1

t (s) 1

−1 −2

a When is the particle at rest?

Create a strategy The particle is at rest when its velocity is zero. Identify the points where the graph of v(t) intersects the t-axis.

Apply the idea The graph of v(t) crosses the t-axis at t = 2 and t = 8. Therefore, the particle is at rest at 2 seconds and 8 seconds.

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b During which time interval(s) is the particle’s speed decreasing?

Create a strategy Speed is decreasing when the graph of velocity is moving towards the t-axis (i.e., when ∣v(t)∣ is decreasing).

Apply the idea The graph of v(t) is moving towards the t-axis in two intervals: • From t = 0 to t = 2, the velocity decreases from 2 to 0. • From t = 5 to t = 8, the velocity increases from −2 to 0. Although the velocity is increasing, its magnitude (the speed) is decreasing from 2 to 0. Therefore, the speed is decreasing for 0 ≤ t < 2 and 5 < t < 8.

Idea summary Velocity is the instantaneous rate of change of displacement. If displacement is x(t), then velocity is v(t) =

= x′(t) = .

Speed is the magnitude of velocity and is always non-negative.

Speed = ∣v(t)∣ = ∣x′(t)∣ ∣v(t)∣

is the absolute value, or magnitude, of the velocity

Distance is a positive scalar quantity, representing the total path travelled, distinct from displacement which is a vector quantity. The derivative can be used to solve problems involving motion. Key steps often involve finding when a particle is at rest (v(t) = 0) or finding its velocity or speed at a specific time. Graphs of motion can be interpreted visually. A particle’s speed is increasing when its velocity graph moves away from the time-axis and decreasing when it moves towards the time-axis.

14.11 Practice questions What do you remember? 1

Distinguish between the average rate of change of a function over an interval and the instantaneous rate of change of a function at a point.

Explain how velocity is related to the displacement, x(t), of a particle moving in a straight line, using derivative notation.

Under what condition is a particle moving in a straight line momentarily at rest?

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Explain how distance differs from displacement, and how speed differs from velocity.

Practice

Ex 1

If C(x) represents the cost in dollars of producing x items of a certain product, what are the units of the derivative C′(x)? What does C′(x) represent in this context?

The depth D (in metres) of water in a reservoir t days after ﬁlling begins is given by D(t) = 5 + 0.5t − 0.01t2 for 0 ≤ t ≤ 25.

Ex 2

Ex 3

Calculate the average rate of change of depth between t = 5 and t = 15 days.

Calculate the instantaneous rate of change of depth at t = 10 days.

The temperature T (in °C) of an object cooling in a room is given by T (t) = 80 − 2t2, where t is the time in hours after it starts cooling, for 0 ≤ t ≤ 6: a

Calculate T ′(t).

Calculate the instantaneous rate of change of temperature at t = 5 hours.

Interpret the meaning of your answer in part (b).

The displacement x metres of a particle from an origin O after t seconds is given by x = 2t3 − 21t2 + 60t − 5, for t ≥ 0: a

Calculate the initial velocity of the particle.

Determine when the particle is momentarily at rest.

Calculate the speed of the particle at t = 3.

The position of a particle moving along a straight line is given by the function x = t3 − 9t2 + 15t + 10 cm, where t is in seconds and t ≥ 0: at t = 2 s.

Calculate

Calculate the speed of the particle at t = 4 s.

In what direction is the particle moving at t = 4 s?

The graph shows the velocity v(t)(in m/s) of a particle moving in a straight line over 6 s: 4

v(t) (m/s)

3 2 1 −1

t (s) 1

−2

When is the particle at rest?

When is the particle moving in the positive direction?

When is the particle’s speed decreasing?

14.11 Derivatives as rates of change mathspace.co

789

The velocity-time graph for a particle is shown. v (t) 3 2 1

t 1

−1

Which of the graphs below could represent the displacement, x(t), of the particle? Justify your choice. A

1 −1

x 1

−1

x 1

The velocity-time graphs for two particles, A and B, are shown: v(t) 2 1

t 1

−1 −2

At what time do the particles have the same velocity?

For which time interval(s) are the particles moving in opposite directions?

At t = 3 s, which particle has the greater speed?

vA(t)

vB(t)

The cost C (in dollars) of producing x hundred items of a new product is modelled by the function C(x) = 500 + 30x − 0.5x2, for 0 ≤ x ≤ 25: a

Calculate the rate of change of cost when 10 hundred items are produced.

At what production level does the cost stop increasing?

The area A (in cm2) of a healing wound is given by the function A(t) = 15 − number of days after treatment, for t ≥ 0: a

Calculate the rate at which the area is changing after 2 days.

Is the area of the wound increasing or decreasing after 2 days? Explain your answer.

A spherical balloon is being inﬂated. Its radius r (in cm) at time t (in s) is given by the function r(t) = 2 + 0.5t. The volume of a sphere is given by the formula V =

790

, where t is the

Express the volume V as a function of time t.

Calculate the rate of change of volume with respect to time, answer in terms of π.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

π r3 : , at t = 4 s. Write the

The cost C (in dollars) of producing x widgets is given by the function C(x) = 5000 + 10x − 0.02x2: a

Calculate the gradient function C′(x).

Calculate the gradient when 100 widgets are produced. Interpret this value.

The height h (in metres) of a ball thrown vertically upwards from the ground is given by the function h(t) = 20t − 4.9t2, where t is the time in seconds after it is thrown: a

Calculate the velocity of the ball at t = 1 s and t = 3 s.

Determine the maximum height reached by the ball.

Extend your thinking 18

The volume of water V (in litres) in a tank is decreasing. The rate of decrease is proportional to the square root of the volume remaining. This can be expressed as is a positive constant.

, where k

If the initial rate of decrease is 2 L/min when the volume is 100 L, calculate the value of k. 19

A company ﬁnds that its proﬁt P (in thousands of dollars) from selling x hundred items is given by the function P (x) = −x3 + 12x2 − 36x + 10, for x ≥ 0: a

Calculate the proﬁt function.

For what level(s) of sales x is the proﬁt function zero?

Interpret what happens to the proﬁt at these levels of sales.

Two particles, A and B, start moving at the same time from the origin along the x-axis. Their displacements (in metres) after t seconds are given by xA(t) = t2 − 2t and xB(t) = 6t − t2 respectively, for t ≥ 0: a

Determine when the particles have the same velocity.

At the time(s) found in part (a), are they moving in the same direction?

Determine when they have the same displacement.

A student was asked to analyse the motion of a particle with displacement given by x(t) = t3 − 7t2 + 15t − 9 metres, for t ≥ 0. The responses are: • Initial Velocity: The velocity function is v(t) = x′(t) = 3t2 − 14t + 15. The initial velocity is v(0) = 15 m/s. • When is the particle at rest? I need to ﬁnd when displacement is zero. I found that x(1) = 0, so the particle is at rest at t = 1 s. • Speed at t = 2 s: I calculated the velocity v(2) = 3(2)2 − 14(2) + 15 = 12 − 28 + 15 = −1. So the speed is −1 m/s. a

Identify and explain the two main conceptual errors in the student’s reasoning.

Provide the correct answers for the two parts the student answered incorrectly.

14.11 Derivatives as rates of change mathspace.co

791

14 Chapter review 1

For the function f (x) = x2 − 3x, calculate the gradient of the secant line between P (2, f (2)) and Q(2 + h, f (2 + h)) for these values of h: a

h = 0.1

h = 0.01

h = 0.001

What value does the gradient appear to be approaching as h approaches zero?

The derivative of a function f (x) is given by f ′(x) = 2x2 + 5. Determine the gradient of the tangent to the curve y = f (x) at these points: a

x=0

x=2

x = −3

A function y = f (x) has the following properties for its derivative f ′(x): • f ′(x) > 0 for x < 0 • f ′(0) = 0 • f ′(x) < 0 for 0 < x < 3 • f ′(3) = 0 • f ′(x) > 0 for x > 3 Sketch a possible graph of the original function y = f (x).

Determine the derivative of each function: a

y = x8

For the function y =

f (x) =

Rewrite the function in the form xn.

Calculate

Calculate the gradient of the tangent to the curve at x = 9.

The graph of y = xn has a gradient of 12 at the point where x = 2. Determine the value of n.

Use ﬁrst principles to determine the derivative of f (x) = 2x2 + 5x − 3.

For the function f (x) = x2 + 6x: a

Determine f ′(x) using ﬁrst principles.

Hence, determine the gradient of the tangent at the point where x = −2.

The displacement, s metres, of a particle after t seconds is given by s(t) = 4t2 − t + 1. Use ﬁrst principles to determine an expression for the instantaneous velocity, v(t) = s′(t).

Differentiate these functions with respect to x: a

y = −5x6

g(x) = 3x4 − 2x3 + 7x − 8

Rewrite y = (2x − 5)2 by expanding the expression, then determine its derivative.

792

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A function is deﬁned as f (x) = ax3 + bx2 + cx + d. If f ′(x) = 9x2 + 10x − 1, determine the values of a, b, and c.

Determine the equation of the tangent to the curve f (x) = x3 − 2x at the point where x = −1.

Determine the equation of the normal to the curve y = x2 − 5x at the point where x = 3.

Determine the angle of inclination, rounded to the nearest degree, of the tangent to the curve y = x3 − 4x + 1 at the point where x = 2.

Determine the x-coordinates of the points on the curve y = x3 − 6x where the tangents are perpendicular to the line x + 9y = 1.

Differentiate y = (2x − x3)5 using the chain rule.

For the function y = (x2 − 4)5, determine the x-coordinates of the points where the tangent is horizontal.

The function y = c(ax − 1)3 passes through the point (1, 1). The gradient of the tangent to the curve is 6 at x = 1. Determine the values of a and c.

Differentiate

For the function f (x) = x2 (x − 1), determine the equation of the tangent at x = −2.

Let g(x) = x2 f (x), where f (x) is a differentiable function. Given that f (2) = 3 and f ′(2) = −1, determine the value of g′(2).

Differentiate the function y =

Determine the equation of the tangent to the curve y =

Consider the function y =

using the product rule, and simplify the result.

using the quotient rule. at the point where x = 3.

. Given that the gradient of the tangent is

at x = 3,

determine the possible values of k. 26

Differentiate y = (2x3 + x)2 (5 − 2x).

Determine the equation of the tangent to the curve x = 4.

The graph of y = and b.

For the cubic function f (x) = x3 − 6x2 + 5, determine the coordinates of the stationary points and the intervals where the function is increasing or decreasing.

at the point where

has a tangent line y = 3x − 4 at the point (a, b). Calculate the values of a

Chapter 14 review mathspace.co

793

The graph of a function y = f (x) is shown. Sketch the corresponding graph of its derivative, y = f ′(x). 4

3 2

y = f (x) −2

−1

x 1

−1

−2 −3 −4

For what values of k does the cubic function f (x) = x3 + kx2 + 3x + 1 have no stationary points?

The displacement x metres of a particle from an origin O after t seconds is given by x = t3 − 12t2 + 36t + 2, for t ≥ 0:

Calculate the initial velocity of the particle.

Determine when the particle is momentarily at rest.

Calculate the speed of the particle at t = 4 s.

The graph shows the velocity v(t)(in m/s) of a particle moving in a straight line, where v(t) = −t2 + 4t: v(t) (m/s) 4 3 2 1

t (s) 1

−1

794

When is the particle at rest?

When is the particle moving in the positive direction?

When is the particle’s speed decreasing?

Two particles, A and B, start moving at the same time from the origin along the x-axis. Their displacements (in metres) after t seconds are given by xA(t) = t2 − 4t and xB(t) = 8t − t2 respectively, for t ≥ 0: a

Determine when the particles have the same velocity.

Determine when the particles have the same displacement.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

“If you only knew the magnificence of the 3, 6, and 9, then you would have the key to the universe.” Nikola Tesla

Big ideas • Exponential functions provide a powerful model for real-world scenarios of growth and decay, where the parameters in the equation y = k(ax) directly correspond to the initial value and the growth factor, defining the graph’s key features like its y-intercept and horizontal asymptote. • A defining characteristic of exponential functions is that their rate of change at any point is proportional to the function’s value at that point. This leads to the discovery of Euler’s number, e, the unique base for which the exponential function y = ex is its own derivative.

15 Exponential and logarithmic functions Chapter outline 15.01 Exponential graphs 15.02 Tangent gradient at y-intercept 15.03 Euler’s number and derivatives Investigation: Differentiation of exponential functions 15.04 Applications 15.05 Logarithms 15.06 Exponential-logarithmic equivalence 15.07 Logarithm laws and properties 15.08 Logarithm expressions and equations 15.09 Logarithmic graphs Investigation: Logarithms Chapter 15 review

798 810 814 819 825 834 841 847 854 862

In just 10 steps, you can walk 10 metres. But 10 exponential steps? You’d pass Saturn.

15.01 Exponential graphs After this lesson, you will be able to… • graph exponential functions of the form y = k(ax) and y = k(a−x) • identify the asymptote, y-intercept, domain and range of exponential functions • describe the behaviour of exponential functions as x approaches positive or negative infinity • distinguish between exponential growth and decay based on the function’s base

Graph of exponential functions Exponential function A function of the form y = ax, where x is the independent variable and the base a > 0. Exponential growth Where a quantity increases at a rate proportional to its current size. Exponential decay Where a quantity decreases at a rate proportional to its current size. Asymptote A straight line (or another curve) that a curve approaches as x tends to ±∞, or to some particular value. Domain The set of allowable values of x in a function or relation. For a function or relation, it is the set of real numbers on which the function or relation is deﬁned. Range ( function) The set of values of the dependent variable for which a function is deﬁned.

The exponential function y = k(ax), where a > 0, a ≠ 1, k ≠ 0, represents exponential growth if a > 1 or exponential decay if 0 < a < 1. The function y = k(a−x) is equivalent to y = reﬂecting the graph across the y-axis. Key features of y = k(ax) (for k > 0): • y-intercept: At x = 0, y = k(a0) = k, so the point is (0, k). • Horizontal asymptote: y = 0 (x-axis) as x → − ∞ (if a > 1) or x → ∞ (if 0 < a < 1). • Domain: All real numbers. • Range: y > 0. 798

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the table for y = 2x: −4

−3

−2

−1

As x increases by 1, the y-values double, showing exponential growth at an increasing rate. y

y = 2x

15 10

y-intercept (0, 1)

y=0 −4

−3

−2

−1

The graph of y = 2x shows exponential growth with a y-intercept at (0, 1) and horizontal asymptote at y = 0.

Exploration 1. Create a table of values for y = 3x from x = − 3 to x = 3 and sketch the graph. 2. Discuss how its y-intercept, horizontal asymptote, and growth rate compare to y = 2x.

Example 1 For the function y = 2(3x): a Identify the horizontal asymptote, y-intercept, domain, and range.

Create a strategy Evaluate the function at x = 0 for the y-intercept, analyse behaviour for the asymptote, and determine domain and range based on the function’s properties.

15.01 Exponential graphs mathspace.co

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Apply the idea Horizontal asymptote: As x → − ∞, 3x → 0, so y → 0. y-intercept: At x = 0, y = 2(30) = 2, so (0, 2). Domain: All real numbers. Range: Since 3x > 0 and k = 2 > 0, y > 0.

Reﬂect and check Verify the y-intercept by substituting x = 0. Check that the range and asymptote align with the function’s behaviour.

b Sketch the graph, labelling key features.

Create a strategy Plot the y-intercept and asymptote, and draw the curve based on exponential growth.

Apply the idea Steps to plot y = 2(3x): 1. Identify the y-intercept: At x = 0, y = 2(30) = 2, so plot (0, 2). 2. Determine the horizontal asymptote: As x → − ∞, y → 0, so draw y = 0 as a dashed line. 3. Calculate additional points: At x = − 1, y = 2(3− 1) = (1, 6).

≈ 0.67; at x = 1, y = 2(31) = 6. Plot

4. Draw a smooth curve through the points, increasing to the right (exponential growth) and approaching y = 0 to the left. y

8 6 4

y = 2(3x)

2 (0, 2)

y=0 −2

800

−1

x 1

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

and

Idea summary Exponential functions y = k(ax) have a y-intercept at (0, k), a horizontal asymptote at y = 0, domain all real numbers, and range y > 0 (if k > 0). The function y = k(a−x) reﬂects this across the y-axis.

Behaviour of exponential functions The behaviour of y = k(ax) and y = k(a−x) as x → ∞ and x → − ∞ depends on the base a and scaling factor k. For y = k(ax), k > 0: • If a > 1 (growth): As x → ∞, y → ∞; as x → − ∞, y → 0+. • If 0 < a < 1 (decay): As x → ∞, y → 0+; as x → − ∞, y → ∞. For y = k(a−x), equivalent to y =

, the behaviour reverses due to .

Interactive exploration Discover this concept in action online

mathspace.co

Example 2 Consider the function y = 3x: a Complete the table of values. −3

−2

Create a strategy

−1

⬚

Apply the idea x

Substitute x values into y = 3 for the table.

Substituting each x into y = 3x: x = 0: y = 30 = 1 x = 1: y = 31 = 3 x = 2: y = 32 = 9 x = 3: y = 33 = 27 x y

−3

−2

−1

15.01 Exponential graphs mathspace.co

801

b Describe the end behaviour and horizontal asymptote.

Create a strategy Analyse asymptotic behaviour for y = 3x as x → ±∞.

Apply the idea Since a = 3 > 1: As x → ∞, y → ∞ As x → − ∞, y → 0+ Horizontal asymptote: y = 0

c Determine the domain and range.

Create a strategy Use function properties to determine domain and range.

Apply the idea Domain: All real numbers. Range: y > 0.

Example 3 Consider the function y =

a Create a table of values and describe the behaviour as x increases.

Create a strategy Substitute x values from − 4 to 4 into y =

and observe the trend in y-values.

Apply the idea Calculate y for x = − 4, − 3, … , 4: x

−4

−3

−2

−1

As x increases, y-values halve, decreasing at a decreasing rate, approaching y = 0.

802

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Sketch the graph, labelling the y-intercept and asymptote, and compare this graph with the graph of y = 2x.

Create a strategy Plot the y-intercept, asymptote, and points from the table, then draw the exponential decay curve.

Apply the idea Steps to plot y =

: = 1, so plot (0, 1).

1. y-intercept: At x = 0, y =

2. Asymptote: As x → ∞, y → 0, so draw y = 0 as a dashed line. 3. Plot points from the table, e.g., (− 2, 4),

4. Draw a smooth curve, decreasing to the right and approaching y = 0. y

1 2

y = 2x

15 10

y-intercept (0, 1)

y=0 −4

−3

−2

−1

x 4

The graph shows exponential decay with y-intercept (0, 1) and asymptote y = 0.

Idea summary For y = k(ax), growth (a > 1) or decay (0 < a < 1) determines behaviour: y → ∞ or y → 0 as x → ±∞. y = k(a−x) reverses this, reﬂecting across the y-axis.

15.01 Exponential graphs mathspace.co

803

15.01 Practice questions What do you remember? 1

Match each term with its correct deﬁnition: i

Occurs when the rate of change of a mathematical function is negative and proportional to the function’s current value.

A function of the form y = ax, where x is the independent variable and the base a > 0.

iii

The set of values of the dependent variable for which a function is deﬁned.

A straight line that a curve approaches as x tends to ±∞, or to some particular value.

The set of allowable values of x in a function or relation.

Occurs when the rate of change of a mathematical function is positive and proportional to the function’s current value.

Exponential function

Exponential growth

Exponential decay

Asymptote

Domain

Range ( function)

Determine whether the following statements are true or false for an exponential function of the form y = k(ax) with k > 0: a

The horizontal asymptote is always y = 0.

The y-intercept is at (0, 1).

If a > 1, the function represents exponential decay.

The domain is all real numbers.

For each exponential function, state whether it represents exponential growth or decay: a

y = 2x

y = 1.5−x

y = 0.8x

State whether the relationships are exponential or not: y = 2x

y = 2x

x −2 −1 0 1 y

y = 0.5x

−2

−1

−4 −3 −2 −1 −2

x 1

4 3

−4

−6 −8

804

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

1 −4 −3 −2 −1

x 1

Practice Ex 1

Ex 2

For the function y = 3(2x): a

Identify the horizontal asymptote, y-intercept, domain, and range.

Sketch the graph, labelling key features.

Consider the function y = 4x: a

Complete the table of values: x

−3

−2

−1

⬚

Describe the end behaviour and horizontal asymptote.

State the domain and range.

Consider the graph of y = 4x:

Is each y-value of the function positive or negative?

What value does the y-coordinate approach but never reach on the graph?

Identify the equation of the horizontal asymptote, which y = 4x → ∞.

Identify the domain and range of the graph.

Describe the rate of increase or decrease of the graph.

Consider the graph of y =

x −3

−2

−1

Is each y-value of the function positive or negative?

Identify the horizontal asymptote.

Identify the domain and range of the graph.

Describe the rate of increase or decrease of the graph.

2 1 −3

−2

−1

Determine the missing coordinate in each ordered pair that represents a point on the curve y = 5x: a e

(3, ⬚)

(0, ⬚)

b f

(− 1, ⬚)

(4, ⬚)

c g

(2, ⬚)

(− 3, ⬚)

d h

(− 2, ⬚)

(1, ⬚)

15.01 Exponential graphs mathspace.co

805

Consider the function f (x) =

. Evaluate:

f (0)

f (− 2)

f (1)

f (2)

f (− 1)

f (3)

f (− 3)

f (4)

f (− 4)

f (5)

m f (− 5)

Consider the graph of the function f (x) = 2x: a

Identify the y-intercept.

Does the graph have an x-intercept?

Identify the domain of the function.

Identify the range of the function.

Calculate f (7).

4 3

If the point (3, m) lies on the curve of f (x) = 2 , determine the value of m.

2 1

−5 −4 −3 −2 −1

Consider the function y = 0.25x: a

806

1 2 3 4 5

Complete the table of values: x

−5

−4

−3

−2

−1

⬚

Describe the behaviour of the function as the value of x increases.

Identify the domain of the function.

Identify the range of the function.

Sketch the function y = 0.25x.

Consider the function y = 0.5−x: a

Can the value of y ever be zero or negative? Explain your answer.

Identify the horizontal asymptote.

Describe the end behaviour of the function.

Identify the y-value of the y-intercept of the curve.

How many x-intercepts does the curve have?

Sketch the graph of y = 0.5−x.

Determine whether the exponential function is increasing or decreasing: a

y = 3x

y = 0.5x

y = 1.05x

y = 0.97x

y = 1.5−x

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the function y =

Rewrite the function in the form y = k−x.

Describe the transformation required to obtain the graph of y = y = 3x.

Sketch the graph of the functions y = 3x and y =

Identify the coordinates of the point of intersection of the two curves.

Describe the behaviour of both these functions for large values of x.

on the same set of axes.

A linear function and exponential function have been graphed on the same axes: a

For each increase of 1 unit in x, by how much does the linear function increase?

For each 1 unit increase in x, by what multiplicative factor does the exponential function increase?

As x approaches inﬁnity, which function increases more rapidly?

20 18 16 14 12 10 8 6 4 2

from the graph of

Consider the functions f (x) = a

and g(x) =

x 1

Complete the table of values for each function: x

f (x)

⬚

g(x)

⬚

Sketch the graphs of the functions on the same set of axes.

By how many units does f (x) increase for every 1 unit increase in x?

By what factor does g(x) decrease for every 1 unit increase in x?

Describe the behaviour of each function as x increases.

15.01 Exponential graphs mathspace.co

807

Matt and Sophia are saving money using different strategies. The amount each has saved after each month is given by the table of values and the plotted points: a

If each person continues their pattern of saving, who will be the ﬁrst to exceed savings of $600?

Identify the functions m(x) and s(x) that represent the savings of Matt and Sophia, respectively, in terms of the number of months x.

300 Matt’s savings 270 240 210 180 150 120 90 60 30 0

Evaluate the value of s(5) − m(5) and therefore interpret the result in the context of the question.

Month 2

Number of months

Sophia’s savings

Extend your thinking 19

Consider the function f (x) = 2x − 3: a

Evaluate f (0).

Evaluate

and leave your answer in exact positive index form.

If f (x) = 5x and g(x) = 3−x, evaluate: a

f (1)

g( f (1))

g( f ( g(0)))

Calculate the value of 4000(0.02)0.6 for t = 4 rounded to two decimal places.

Consider the graph of y = 3x. Determine the exact length of RQ. 90

80 70 60 50 40 30 20 10 −5 −4 −3 −2 −1

808

Q(0, 1) 1 2 3 4 5

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Two different sequences are generated by the functions f (n) = 5n + 1 and g(n) = 3(2)n: a

Determine the ﬁrst 5 terms in the sequence generated by each function: n

f (n)

⬚

g(n)

⬚

Describe how the terms of each sequence increase.

Explain for what values of n is f (n) > g(n) and for what values of n is f (n) < g(n).

Kate considers the graphs of f (x) and g(x), which are exponential and quadratic functions respectively, and performs these calculations: • Calculation 1: f (11) − f (10) = 1024 • Calculation 2: f (10) − g(10) = 919 • Calculation 3: f (11) − g(11) = 1922 • Calculation 4: g(11) − g(10) = 21 y

40 35 30 25 20

g(x)

f (x)

15 10 5

x 0

Determine the equations of f (x) and g(x).

She claims that for large values of x, as x increases, the exponential function increases more rapidly than the quadratic function. Explain how the calculations support Kate’s claim.

15.01 Exponential graphs mathspace.co

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15.02 Tangent gradient at y-intercept After this lesson, you will be able to… • define a tangent and a secant to a curve • understand that the gradient of a secant can approximate the gradient of a tangent • use the formula m = to estimate the gradient of the tangent to y = ax at its y-intercept • examine how the gradient of the tangent at the y-intercept changes for different values of the base a

Tangent and gradient at y-intercept Tangent A line that intersects a circle at just one point. Gradient The slope of a line. If A(x1, y1 ) and B(x2, y2 ) are 2 distinct points on a line, the gradient of the line (or line segment AB) is given by m =

Intercept The point at which a curve or function crosses an axis or other curve in a plane. The point at which a curve crosses the x-axis ( y = 0) is called the x-intercept and the point at which a curve crosses the y-axis (x = 0) is called the y-intercept. Secant The straight line passing through 2 points on the graph of a function.

The tangent to a curve at a point is a line that touches the curve at that point, with a gradient equal to the curve’s rate of change at that point. For the exponential function y = ax(a > 0, a ≠ 1), the y-intercept occurs at x = 0, where y = a0 = 1, so the point is (0, 1).

810

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

To ﬁnd the gradient of the tangent at x = 0, approximate it using a secant line, which joins two points on the curve. Consider the points (0, 1) and a nearby point (h, ah), where h is a small number. The gradient of the secant line is:

m is the gradient of the secant line, approximating the tangent’s gradient as h becomes very small (approaches 0). This formula estimates the rate of change of y = ax at x = 0. By testing different values of a, it can be observed how the gradient changes for different exponential functions. For example, for a = 2, the gradient is less than 1; for a = 3, it’s greater than 1.

Exploration 1. Calculate the gradient of the secant for y = 2x at x = 0 using h = 0.1 and h = 0.01. 2. Repeat for y = 3x. How does the gradient change with a? What happens as h gets smaller?

Example 1 For y = 2x: a Estimate the gradient of the tangent at x = 0 using a secant with h = 0.01, rounded to four decimal places

Create a strategy Use the secant gradient formula m =

for function at x = 0 with h = 0.01.

Apply the idea For y = 2x, a = 2, h = 0.01: Write the formula

Substitute the values

Evaluate and round

The gradient is approximately 0.6956.

15.02 Tangent gradient at y-intercept mathspace.co

811

b Compare with y = 3x using the same h.

Create a strategy Use the secant gradient formula m =

for function at x = 0 with h = 0.01.

Apply the idea For y = 3x, a = 3: Write the formula

Substitute the values

Evaluate and round

The gradient is approximately 1.1047, closer to 1 than for a = 2.

Idea summary The gradient of the tangent to y = ax at the y-intercept (0, 1) varies with a, estimated using secant lines.

15.02 Practice questions What do you remember? 1

Deﬁne the tangent to a curve at a point.

For the exponential function y = ax (a > 0, a ≠ 1), what is the y-intercept?

What is the formula for the gradient of the secant line between points (0, 1) and (h, ah) on y = ax?

Match each term with its correct deﬁnition: a

Tangent

Gradient

Intercept

Secant

A line that touches a curve at one point with the same gradient as the curve at that point.

The straight line passing through two points on the graph of a function.

iii

The point where a curve crosses the y-axis at x = 0.

The slope of a line, calculated as m=

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for two points.

Practice Ex 1

Estimate the gradient of the tangent at x = 0 for y = 1.2x using the secant line with: a

h = 0.1

h = 0.01

h = 0.001

Compare the results for different h values.

Estimate the gradient of the tangent at x = 0 for y = 4.5x using the secant line with: a

h = 0.1

h = 0.01

h = 0.001

Compare the results for different h values.

Calculate the gradient of the secant at x = 0 with h = 0.01 for the following functions: a

y = 0.8x

h = 0.1 h = 0.01

h = 0.001

Compare with the gradient for y = 2x using h = 0.01.

y = 5x

Calculate the gradient of the secant at x = 0 with h = 0.01 for the following functions: y = 1.3x x

y = 3.8

b f

y = 4x y = 1.7

y = 2.2x

y = 0.9x

Estimate the gradient of the tangent at x = 0 for y = 3.5x using the secant line with: a

h = 0.1

h = 0.01

h = 0.001

Compare with the gradient for y = 3x using h = 0.01.

Calculate the gradient of the secant at x = 0 with h = 0.001 for the following functions: y = 2.3x

y = 4.2x

y = 1.4x

y = 0.7x

Evaluate and order the gradient of the secant at x = 0 with h = 0.01 for the following functions from smallest to largest: • y = 1.5x • y = 0.8x

a 12

y = 3.2x

Estimate the gradient of the tangent at x = 0 for y = 2.5x using the secant line with:

y = 1.8x

• y = 3x • y = 3.5x

• y = 2x • y = 4x

Calculate the gradient of the secant at x = 0 with h = 0.01 for the following functions: a

y = 1.6x

y = 2.4x

y = 3.3x

y = 0.6x

y = 4.8x

y = 1.9x

y = 2.9x

y = 0.5x

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Extend your thinking 14

Explore how the gradient of the tangent at x = 0, will change when the function is f (x) = ka−x compared to f (x) = kax.

For the function y = kax (k ≠ 0, a > 0, a ≠ 1):

Find the y-intercept.

Derive the gradient of the secant formula at x = 0 using points (0, k) and (h, kah).

If k = 2 and a = 2, calculate the gradient of the secant at x = 0 with h = 0.01.

Explain how the scaling factor k affects the tangent gradient at x = 0.

A student observes that the tangent gradient at x = 0 for y = ax seems to increase with a. Test this for: a

a = 1.5 with h = 0.01

a = 2 with h = 0.01

a = 2.5 with h = 0.01

a = 3 with h = 0.01

15.03 Euler’s number and derivatives After this lesson, you will be able to… • define Euler’s number, e, and state its approximate value • recognise that for y = ex, the derivative

= ex

• find the gradient of the tangent to the curve y = ex at any given point • determine the equation of the tangent to the curve y = ex at any given point

Euler’s number Euler’s number An irrational number, e ≈ 2.71828182845.... It has the property Example: For y = ex,

= ex.

Derivative The result obtained after differentiation. For the function f (x), the derivative is the gradient function of f (x), and is denoted f ′(x). Example: For y = ex, the derivative is ex at any point.

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Euler’s number, e ≈ 2.71828182845 …, is a unique constant such that the derivative of y = ex is:

The function y = ex is its own derivative, meaning its rate of change equals its value at every point. This property distinguishes ex among exponential functions y = ax, as only a = e satisﬁes (ax) = ax.

Exploration 1. Using a graphing tool, sketch the graphs of y = ex, y = 2x, and y = 3x and their tangent lines at x = 0. 2. Discuss why the tangent gradient for y = ex is unique and how it compares to the gradients for y = 2x and y = 3x.

Example 1 Find the value of the derivative of y = ex at x = 0.

Create a strategy Use the property

(ex) = ex and evaluate at x = 0.

Apply the idea Write the derivative Substitute x = 0 Evaluate The gradient at x = 0 is 1, conﬁrming

= ex.

Idea summary Euler’s number e ≈ 2.71828182845 is unique such that function equals its own derivative.

(ex) = ex, meaning the

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Tangent to exponential functions The derivative of y = ex allows us to ﬁnd the equation of the tangent at any point. Since the gradient at a point (x, ex) is ex.

y − y1 = m(x − x1 )

Point-slope form of a line, where m = ex is the gradient and is the point of tangency. y 4 3

(1, e)

The tangent to y = ex at (1, e) has gradient m = e.

(0, 1) y = ex

−1

y = ex 0

x 2

Example 2 Determine the equation of the tangent to y = ex at x = 1.

Create a strategy (ex) = ex, then use point-slope form:

Find the point and gradient at x = 1 using

y − y1 = m(x − x1)

Apply the idea Point: At x = 1, y = e1 = e, so (1, e). Write the derivative Substitute x = 1 Simplify Gradient at x = 1 is e. y − e = e(x − 1)

Write the point-slope form

y = e(x − 1) + e

Add e to both sides

y = ex − e + e

Expand

y = ex

Simplify

The equation of the tangent is y = ex.

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(ex) = ex,

Reﬂect and check Substitute x = 1 into y = ex: y = e × 1 = e, matching (1, e).

Idea summary The derivative of y = ex is ex, allowing the tangent at any point (x, ex) to be found using point-slope form with gradient ex.

15.03 Practice questions What do you remember? 1

Deﬁne Euler’s number.

What is unique about the derivative of y = ex?

State the derivative of y = kex, where k is a constant.

Why is ex unique among exponential functions ax?

Practice Ex 1

Find the value of derivative of y = ex at: a

At x = − 1

At x = 2

x = −2

x=2

x=0

x = ln 2

x=0

x = −1

x=4

x = ln 3

y = ex + x

Calculate the ﬁrst and second derivatives of the following functions: a

y = 4ex

y = ex + 5

y = 5ex + 2x

y = ex − x2

y = 2ex − 3

Find the equation of the normal to y = ex at the given points: a

At x = 1

Determine the equation of the tangent to y = ex at the given points: a

Find the gradient of the tangent to y = ex at the following points: a

Ex 2

At x = 3

x=0

x=1

x = −2

x = −3

y = −ex

y = 10ex

Verify that the following functions satisfy y′ = y: a

y = 2ex

y = 5ex

15.03 Euler’s number and derivatives mathspace.co

817

Find the equation of the tangent to the following functions at x = 0: a e

y = 5e − 2

b f

y = ex − 1 x

c 2

y=e −x

y = 2ex + 4 x

y = − 3e + 1

y = ex + 3x

y = ex + 2x + 1

Determine the gradient of the tangent to the following functions at the given points: a

y = 3ex + 2

y = 2ex + x at x = 0 x

y = ex − 3x at x = 1

y = 3e + x at x = 0

y = ex + 2x − 1 at x = 1

y = −ex + x at x = 0

y = 4ex − x2 at x = 1

For the graph of y = ex: a

Identify the coordinates of the point where the tangent has gradient 2.

Using these values, write the equation of the tangent at this point.

Determine the y-intercept of the tangent line.

Determine the x-intercept of the tangent line.

y 4 3 2

y = ex

(0.69, 2)

1 x

−1

Extend your thinking 14

Prove that ex is the only exponential function of the form ax, where the derivative equals the function itself.

Consider the equation y = ex: a

Find the equation of the tangent to the curve at x = 1.

Find the equation of the normal to the curve passing through this point.

Calculate the area of the triangle formed between the tangent, normal, and the x-axis.

Hence sketch the diagram clearly showing the curve, tangent, normal, and intercepts.

A population of bacteria grows according to the function P (t) = 800e0.03t, where t is the number of hours since the start of an experiment. Determine the growth rate after 5 hours, expressing your answer in terms of e.

Determine the number of points on y = ex where the tangent is parallel to y = ex.

Investication: Differentiation of exponential functions Investigate online

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mathspace.co

15.04 Applications After this lesson, you will be able to… • apply knowledge of exponential functions to model real-world scenarios • interpret the parameters in an exponential model, such as the initial amount and growth factor • sketch graphs of exponential models, considering practical domain and range restrictions • use exponential functions to make predictions and solve problems in context

Applications of exponential graphs Exponential functions model situations where quantities grow or decay by a constant factor, such as population growth or radioactive decay. Graphing these functions reveals key features like intercepts, asymptotes, and end behaviour, aiding real-world applications.

Example 1 A bacterial population grows according to N = 200(3t), where N is the number of cells and t is time in hours. Sketch the graph, identifying the asymptote, y-intercept, domain, and range.

Create a strategy

Apply the idea x

Identify key features using the form y = k(a ) and plot using a graphing calculator.

For N = 200(3t), k = 200, a = 3. The asymptote is at N = 0 (only t approaches inﬁnity as time cannot be negative). The y-intercept at t = 0 is N = 200(30) = 200. The domain is t ≥ 0. The range is N > 0. N 600 500 400 300 200 100

t 0

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Example 2 A savings account earns compound interest modelled by A = 5000(1.03t), where t is time in years: a Sketch the graph, labelling the asymptote, y-intercept, and one additional point.

Create a strategy Find the y-intercept by setting t = 0. The horizontal asymptote for a function of the form y = k(ax) is y = 0. Calculate an additional point by substituting a value for t, such as t = 3. Plot these points and the asymptote to draw the curve.

Apply the idea The y-intercept at t = 0 is A = 5000 × 1.030 = 5000 so the point is (0, 5000). The horizontal asymptote is A = 0. For an additional point, substitute t = 3: A = 5000(1.033) ≈ 5463.63. The point is approximately (3, 5464). A 6000

A = 5000(1.03t)

5000

(3, 5464)

(0, 5000)

4000 3000 2000 1000

A=0 0

Reﬂect and check The graph shows a positive y-intercept and increases over time, which is consistent with a savings account earning interest. The key features are clearly labelled.

b Describe the end behaviour as t → ∞.

Create a strategy Analyse the function A = 5000(1.03t) as t becomes very large. Since the base, 1.03, is greater than 1, the function represents exponential growth.

Apply the idea

Reﬂect and check t

As t → ∞, the term 1.03 increases without bound. Therefore, the account balance, A, also tends to inﬁnity.

This makes sense in the real world, as a savings account with compound interest will continue to grow indeﬁnitely over time.

The end behaviour is as t → ∞, A → ∞.

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c Find the balance after 3 years, rounded to the nearest dollar.

Create a strategy Substitute t = 3 into A = 5000(1.03t).

Apply the idea A = 5000(1.03t) 3

Write the formula

= 5000(1.03 )

Substitute t = 3

= 5000 × 1.092 727

Evaluate the power

= $5464

Evaluate and round

Reﬂect and check The balance is greater than the initial $5000, which is expected as the account earns interest. The value also matches the point we calculated for the graph.

Idea summary Exponential graphs model growth or decay, with key features like asymptotes, intercepts, and end behaviour. Euler’s number e is unique, as y = ex has a gradient of 1 at x = 0 and is its own derivative, making it ideal for applications like population modelling.

15.04 Practice questions What do you remember? 1

For each exponential function, identify the key graphical features: i

Horizontal asymptote

y-intercept

iii

Domain

Range

y = 3(2x)

y = 5(0.5x)

y = 2(3−x)

y = 4(1.2x)

Determine whether the following statements are true or false: a

The exponential function y = 4(2x) has a y-intercept at (0, 4).

For y = 3(0.5x), as x → ∞, y → ∞.

The gradient of the tangent to y = ex at x = 0 is 1.

The horizontal asymptote of y = 5(3x) is y = 5.

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Which graph represents a radioactive decay model? A

Mass (g)

Time (days)

Time (days) 0

Mass (g)

Time (days)

Time (days) 0

Practice Ex 1

Ex 2

A population of bacteria grows according to N = 100(2t), where t is time in hours: a

Sketch the graph, labelling the asymptote and y-intercept.

Describe the end behaviour as t → ∞.

The value of an investment grows according to I = 10 000(1.05t), where t is time in years: a

Sketch the graph, labelling the asymptote and y-intercept.

Describe the end behaviour as t → ∞.

Find the investment value after 2 years, rounded to the nearest dollar.

The value of a car depreciates according to V = 20 000(0.9t), where t is time in years: a

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Complete the table of values for the graph: Time (years) (t)

Value ($) (V )

⬚

Sketch the graph using the table values, labelling the asymptote and y-intercept.

Describe the end behaviour as t → ∞.

Find the value after 5 years, rounded to the nearest dollar.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

A population of insects grows according to P = 50(1.5t), where t is time in days: a

Identify the y-intercept and horizontal asymptote.

Sketch the graph, labelling the y-intercept and asymptote.

Describe the end behaviour as t → ∞.

Calculate the population after 5 days, rounded to the nearest integer.

A radioactive substance decays according to M = 100(0.8t), where t is time in years: a

Identify the domain and range.

Sketch the graph, labelling the asymptote and one point.

Find the mass after 2 years, rounded to the nearest gram.

A company’s revenue grows according to R = 2000(1.1t), where t is time in years: a

Identify the horizontal asymptote and y-intercept.

Sketch the graph, labelling the asymptote and y-intercept.

Describe the end behaviour as t → ∞.

Find the revenue after 4 years, rounded to the nearest dollar.

A chemical substance decays according to C = 500(0.95t), where t is time in hours: a

Complete the table of values: Time (hours) (t)

Concentration (mg) (C)

⬚

Sketch the graph, labelling the y-intercept and asymptote.

Describe the end behaviour as t → ∞.

Find the time when the concentration is below 400 mg, rounded to the nearest hour.

A ﬁsh population in a lake grows according to F = 300(1.4t), where t is time in months: a

Identify the domain and range.

Sketch the graph, labelling the y-intercept and one additional point.

Describe the end behaviour as t → ∞.

Calculate the population after 6 months, rounded to the nearest integer.

A machine’s eﬃciency decreases according to E = 80(0.85t), where t is time in months: a

Complete the table of values: Time (months) (t)

Efficiency (%) (E)

⬚

Sketch the graph, labelling the y-intercept and asymptote.

Describe the end behaviour as t → ∞.

Find the eﬃciency after 5 months, rounded to the nearest percent.

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A population of cells grows according to N = 150(2.5t), where t is time in hours: a

Identify the y-intercept and horizontal asymptote.

Sketch the graph, labelling the y-intercept and asymptote.

Describe the end behaviour as t → ∞.

Calculate the population after 4 hours, rounded to the nearest integer.

Extend your thinking 14

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A population model uses y = 1000(ekt), where t is time in years and k is a constant: a

Identify the domain and range.

Describe the end behaviour for k > 0 as t → ∞.

If k = 0.02, ﬁnd the population after 10 years, rounded to the nearest integer.

Determine the value of k if the population doubles in 5 years.

Determine by how much, the population grows from year 4 to year 5.

A radioactive isotope decays according to M = 200(e− 0.05t), where t is time in days: a

Identify the y-intercept and horizontal asymptote.

Sketch the graph, labelling the y-intercept and asymptote.

Find the mass after 10 days, rounded to the nearest gram.

Determine the half-life of the isotope, rounded to the nearest day.

Determine the percentage of the mass (M ) that decays, between day 0 and day 5, rounded to one decimal place.

A company’s advertising budget increases according to B = 4000(1.08t), where t is time in years: a

Identify the domain and range.

Sketch the graph, labelling the y-intercept and one additional point.

Find the budget after 3 years, rounded to the nearest dollar.

Determine when the budget ﬁrst exceeds $5000, rounded to the nearest year.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

15.05 Logarithms After this lesson, you will be able to… • define a logarithm and convert between exponential and logarithmic forms. • use the notation for common logarithms (base 10) and natural logarithms (base e). • evaluate simple logarithmic expressions without using technology. • estimate the value of logarithmic expressions and use technology for approximations.

Logarithmic form Logarithm The logarithm of a positive number x is the power to which a given number b, called the base, must be raised in order to produce the number x. The logarithm of x, to the base b is denoted by logb (x). The logarithm of a number x to a base a, where x > 0 and a > 0, a ≠ 1, is the exponent y such that ay = x. This is written as loga (x) = y.

ay = x ⟺ loga (x) = y a

is the base, where a > 0, a ≠ 1

is the argument, where x > 0

is the exponent

Common bases include 10 (common logarithms, written as log(x)) and e (natural logarithms, written as ln(x)), where e ≈ 2.718 28 is Euler’s constant. This table illustrates conversions between exponential and logarithmic forms: Exponential form

Logarithmic form

a =x

loga (x) = y

102 = 100

log(100) = 2

2 =8

log2 (8) = 3

70 = 1

log7 (1) = 0

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Exploration 1. Discuss why loga (x) is undeﬁned when a ≤ 0, a = 1, or x ≤ 0. 2. Consider the exponential form ay = x and test values like a = −1, a = 1, or x = −1.

Example 1 Rewrite 162 = 256 in logarithmic form.

Create a strategy Use the equivalence ay = x implies loga (x) = y.

Apply the idea ay = x 2

16 = 256

Write the exponential form Write the equation to compare

This shows that a = 16, x = 256 and y = 2. In logarithmic form: loga (x) = y log16 (256) = 2

Write the logarithmic form Substitute a = 16, x = 256 and y = 2

Reﬂect and check Verify by converting back: log16 (256) = 2 means 162 = 256, which matches the given equation.

Example 2 Rewrite log3 (81) = 4 in exponential form.

Create a strategy Use the equivalence loga (x) = y implies ay = x.

Apply the idea loga (x) = y

Write the logarithmic form

log3 (81) = 4

Write the equation to compare

This shows that a = 3, x = 81 and y = 4. In exponential form: ay = x 4

3 = 81

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Write the exponential form Substitute a = 3, x = 81 and y = 4

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Reﬂect and check Evaluate to verify: 34 = 3 × 3 × 3 × 3 = 81, this conﬁrms the result.

Idea summary A logarithm loga (x) = y represents the exponent y such that ay = x, where a > 0, a ≠ 1, and x > 0.

ay = x ⟺ loga (x) = y a

is the base, where a > 0, a ≠ 1

is the argument, where x > 0

is the exponent

Common notations include log(x) for base 10 and ln(x) for base e.

Evaluate logarithmic expressions Evaluating loga (x) involves ﬁnding the exponent y such that ay = x. This can be done by rewriting in exponential form or using technology for non-integer results.

loga (x) = y ⟹ ay = x a

is the base

is the argument

is the exponent

For example, to evaluate log2 (16), determine the power of 2 that yields 16. Since 24 = 16, it follows that log2 (16) = 4. What must 10 be raised to in order to become 1000?

log10 (1000) For non-integer results, such as log10 (500), technology provides approximations, e.g., log10 (500) ≈ 2.6990.

Interactive exploration Discover this concept in action online

mathspace.co

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Example 3 Evaluate these logarithms without technology: a log4 (64)

Create a strategy Equate the logarithm to y to rewrite in exponential form and express both sides with the same base.

Apply the idea log4 (64) = y

Equate the logarithm to y

4 = 64 y

Convert to exponential form

4 =4

Express 64 as 43

y=3

Equate exponents since bases are equal

Thus, log4 (64) = 3.

b log9

Create a strategy Equate to y to rewrite in exponential form and use negative exponents.

Apply the idea Equate the logarithm to y

Convert to exponential form

Express

Equate exponents since bases are equal

Thus, log9

828

as 9−2

= −2.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 4 Consider log(7500): a Determine between which two consecutive integers log(7500) lies.

Create a strategy Test powers of 10 that can bound 7500.

Apply the idea 103 = 1000 4

10 = 10 000

Test power of 3 Test power of 4

Since 1000 < 7500 < 10 000, it follows that log(1000) = 3 and log(10 000) = 4. Thus, log(7500) lies between 3 and 4.

b Evaluate log(7500) using technology, rounded to four decimal places.

Apply the idea log(7500) ≈ 3.8751 Evaluate using technology

Idea summary To evaluate loga (x), ﬁnd y such that ay = x. Use exponential form for exact values or technology for approximations.

Natural logarithms Natural logarithm A logarithm to the base e. The logarithm of x to the base e is denoted as loge(x) or ln(x). Natural logarithms use the base e, where e ≈ 2.718 28 is Euler’s constant. They are denoted ln(x), equivalent to loge (x).

ln(x) = y ⟺ e y = x e

is the base, approximately 2.718 28

is the argument, where x > 0

is the exponent

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The functions ex and ln(x) are inverses, satisfying:

eln(x) = x x

is any number where x > 0

ln(ex) = x x

is any real number

Exploration Test the inverse property by computing eln(5) and ln (e5). Explain why the results conﬁrm the inverse relationship.

Example 5 Convert these natural logarithms to exponential form: a ln(10) = y

Create a strategy Use the deﬁnition ln(x) = y implies e y = x.

Apply the idea ln(x) = y

Write the natural logarithm form

ln(10) = y

Write the equation to compare

This shows that x = 10. In exponential form:

830

ey = x

Write the exponential form

e y = 10

Substitute x = 10

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b ln (e3) = 3

Create a strategy Apply the inverse property ln(ex) = x.

Apply the idea ln(x) = y

Write the natural logarithm form

Write the equation to compare

ln (e ) = 3 3

This shows that x = e and y = 3. In exponential form: ey = x

Write the exponential form

Substitute x = e3 and y = 3

e =e

Reflect and check This conversion also verified the equation.

Idea summary Natural logarithms, denoted ln(x), use base e ≈ 2.718 28.

ln(x) = y ⟺ e y = x e

is the base, approximately 2.718 28

is the argument, where x > 0

is the exponent

15.05 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

The logarithm loga (x) = y means ay = x, where a > 0, a ≠ 1, and x > 0.

The natural logarithm ln (x) is a logarithm with base e ≈ 2.718 28.

The equation 103 = 1000 is equivalent to log (1000) = 3.

Write each exponential equation in its logarithmic form: a

24 = 16

ex = 5

10−1 = 0.1

ay = x

log (1) = 0

log−2 (4) = 2

State whether each statement is true or false: a

log5 (25) = 2

ln (e) = 1

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Practice Ex 1

Ex 2

Rewrite in logarithmic form: a

32 = 9

5−1 =

e1 = e

100.5 =

41.5 = 8

bn = m

2−3 =

e−2 =

log2

log3 (27) = 3

Rewrite in exponential form: a

Ex 3

Ex 4

Ex 5

ln (e2)

log (1000)

log5

log3 (1)

ln (1)

log (0.01)

log4 (4096)

For each logarithm: i

Determine between which two consecutive integers the logarithm lies.

Evaluate using technology, rounded to four decimal places.

log (500)

log (99)

log (0.05)

ln (1) = 0

loga (b) = c

ln (e5) = 5

log (0.001) = −3

ln (x) = 0

log (x) = 10

ln (x5) = 5

log (x) = −3

Evaluate using technology, rounded to four decimal places: log (750)

ln (50)

ln (20)

log (0.008)

Write each equation in its logarithmic form: 73 = 343

112 = 121

e−1 =

e6 = x

ln (x) = 0

log (x) = −2

log4 (x) = 3

Solve for x: log3 (x) = 2

Verify whether each statement is true by converting to exponential form: a

832

ln (10)

Solve for x:

a 13

Rewrite in exponential form:

a 12

= −2

log2 (16)

a 11

log (100) = 2

a 10

Evaluate without technology:

a 9

log6 (36) = 2

ln (e3) = 3

log2 (32) = 5

log (100) = 3

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

log7

= −2

Extend your thinking 14

The amount of an investment under continuous compounding is given by A = Pert, where P is the principal, r is the annual interest rate, and t is time in years. Calculate the time t in years for each investment to reach the target amount, rounded to two decimal places: a

P = 1000, A = 2000, r = 0.05

P = 5000, A = 7500, r = 0.03

P = 2000, A = 3000, r = 0.04

P = 10 000, A = 15 000, r = 0.02

The pH of a substance is given by pH = − log [ H + ], where [ H + ] is the hydrogen ion concentration in mol/L. Calculate the hydrogen ion concentration for a substance with pH = 4, rounded to one signiﬁcant ﬁgure.

Explain why loga (x) is undeﬁned for a ≤ 0, a = 1, or x ≤ 0. Use the exponential form ay = x to justify your answer.

Show that ln (ex) = x and eln(x) = x for appropriate domains. What does this imply about the relationship between ex and ln (x)?

Did you know?

Your senses work logarithmically! Both your hearing and vision respond to changes in intensity on a logarithmic scale, not a linear one. For example, a candle and a torch don’t appear to differ as much in brightness as their actual light output does — even though the torch might emit thousands of times more light. This built-in scaling helps your senses adapt across a huge range of environments. So, logarithms aren’t just a mathematical concept — they’re built into how humans experience the world!

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15.06 Exponential-logarithmic equivalence After this lesson, you will be able to… • solve exponential equations of the form ax = b by converting to logarithmic form. • state and apply the change of base formula for logarithms. • evaluate logarithms with any valid base using a calculator. • solve exponential equations requiring the change of base formula.

Exponential-logarithmic equivalence To solve equations of the form ax = b, where a is 10 or e, and b > 0, convert to logarithmic form using log10 or ln.

ax = b ⟹ x = loga b a b x

is the base, 10 or e is the result, where b > 0 is the exponent

Interactive exploration Discover this concept in action online

Example 1 Solve for x, rounded to four decimal places. a ex = 10

Create a strategy Since ex has a base of e, use its inverse, ln, to solve for x.

Apply the idea ex = 10

Write the equation

x = ln(10)

Convert to log form using the equivalence

x ≈ 2.3026

Evaluate and round

Reﬂect and check Verify: e2.3026 ≈ 10.0002 ≈ 10, conﬁrming the solution.

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mathspace.co

b 10x = 30

Create a strategy Since 10x has a base of 10, use its inverse, log, to solve for x.

Apply the idea 10x = 30

Write the equation

x = log10 (30)

Convert to log form using the equivalence

x ≈ 1.4771

Evaluate and round

Reﬂect and check Verify: 101.4771 ≈ 30.0006 ≈ 30, conﬁrming the solution.

Did you know?

Logarithmic functions play a key role in how we experience and control sound! For example, sound engineers use logarithmic scales to adjust volume levels, because our ears perceive loudness in a nonlinear way. This allows them to make precise changes that match how humans actually hear differences in sound intensity. Logarithmic functions make it possible to balance, mix, and enhance audio in music, broadcasting, and everyday technology — from headphones to concert systems!

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Example 2 Solve log4 (16x) = 3 in exact form:

Create a strategy Convert the logarithmic equation to exponential form using loga (b) = c ⟹ ac = b, then express both sides with the same base and solve.

Apply the idea Write the equation

Convert to exponential form

Express with the same base

Multiply the exponents

Equate exponents since bases are equal

Divide both sides by 4

Simplify

Reﬂect and check Verify:

. Since 16 = 42, we have

Then log4 (64) = 3, since 43 = 64, conﬁrming the solution.

Idea summary The equivalence y = ax ⟺ x = loga ( y) allows solving ax = b by rewriting with the same base or using x = loga (b), for a > 0, a ≠ 1, and b > 0.

Change of base formula Change of base rule A rule for writing a logarithm with a particular base as the ratio of two logarithms with a different base: loga (x) =

The change of base formula allows a logarithm with any base to be expressed using a different base. When an exponential equation has a base other than 10 or e, this formula converts it to 10 or e, either for calculator use or to simplify expressions algebraically.

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a, b

are bases, where a, b > 0, a, b ≠ 1

is a positive number

To prove, let y = loga (x), so ay = x. Let z = logk (a), so kz = a. Write the exponential form

Substitute a = kz

Apply index exponent law

Apply logk to both sides

By the deﬁnition of logarithms, logk (kzy) = zy

Divide both sides by z

Substitute z = logk (a)

Substitute y = loga (x)

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Example 3 Evaluate log2 (27) using the change of base formula, rounded to four decimal places.

Create a strategy Use the change of base formula loga (x) =

Apply the idea Apply change of base formula with a = 2 and b = 10

Evaluate each logarithm

Evaluate using technology

Reﬂect and check Verify: log2 (27) ≈ 4.7549 since 24.7549 ≈ 27.

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Example 4 Solve 2x + 1 = 11 for x, rounded to four decimal places.

Create a strategy Since the base is 2, convert to logarithmic form, apply the change of base formula to base 10, and solve for x.

Apply the idea Write the equation

Convert to log form using the equivalence

Use change of base formula

Evaluate each logarithm

Subtract 1 from both sides

Round to four decimal places

Reﬂect and check Verify: 22.4594 + 1 ≈ 23.4594 ≈ 11.0001 ≈ 11, conﬁrming the solution.

Idea summary The change of base rule loga (x) = base, b.

allows rewriting logarithms in a desired

15.06 Practice questions What do you remember? 1

Determine whether each statement is true or false: a b c

838

The equation ax = y is equivalent to x = loga ( y), where a > 0, a ≠ 1, and y > 0.

To solve ax = b, you can write x = loga (b) or use logarithms with base 10 or e. The equation 25 = 32 can be written as log2 (32) = 5.

Change each equation to logarithmic form: a

32 = 9

103 = 1000

e1 = e

5−1 =

43 = 64

2−2 =

e0 = 1

100.5 =

62 = 36

bn = m

31.5 =

e−1 =

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For each expression, change the base to 10: a

log2 (x)

log20 (x)

loge(e)

log100 (10)

ex = 10

ex = 50

10x = 0.05

log4 (64x) = 9

log100 (10x) = 3

Practice Ex 1

Solve for x, rounded to four decimal places: a

Ex 2

Ex 3

log2 (8x) = 5

log3 (9x) = 4

log6 (36x) = 8

log7 (49x) = 10

Ex 4

log4 (9)

log8 (13)

log9 (15)

log4 (7)

log9 (11)

log6 (20)

log5 (12)

log5 (25) = 2

ln(e3) = 3

log(0.01) = −2

log2 (8) = 3

loga (b) = c

log7 (1) = 0

= −2

log(10 000) = 4

Solve each exponential equation using logarithms, rounded to four decimal places: 2x = 15

3x = 20

5x = 7

4x = 10

Solve each exponential equation using logarithms, leaving answers in exact form: 4x = 12

ex = 7

5x = 20

2x =

Given log10 (3) = 0.477, log10 (7) = 0.845, and log10 (11) = 1.041, evaluate each rounded to three decimal places: a

Rewrite each logarithmic equation in exponential form:

a 11

log7 (11)

Evaluate using the change of base formula, rounded to three decimal places:

a 10

log3 (27x) = 5

Evaluate using the change of base formula, rounded to one decimal place:

a 8

Solve for x in exact form:

a 7

10x = 15

log3 (7)

log3 (11)

log3 (73)

Solve these equations, rounded to four decimal places: a

4x + 1 = 11

5 = 3x − 5

103 − x = 4

9 = 52x − 1

Extend your thinking 13

The loudness of a sound is measured in decibels (dB) by L = 10 log intensity and I0 is a reference intensity.

, where I is the

A sound has intensity I = 106 I0 . Calculate the loudness L in decibels. 14

Solve a2x = b2 for x in terms of a and b, where a > 0, a ≠ 1, and b > 0, using the exponential-logarithmic equivalence, and verify the solution. 15.06 Exponential-logarithmic equivalence mathspace.co

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Solve 2x = 16 for x, expressing solutions in exact form and verifying one solution.

Solve 5x −1 = 25 using logarithms and verify the solution by substituting back into the original equation.

Explain why the equation ax = b cannot be solved for x if b ≤ 0 or a = 1. Use the logarithmic form to support your reasoning.

Did you know?

The pH scale is a logarithmic measure of how acidic or basic a substance is! For example, lifeguards and pool owners test water daily to keep pH between 7.2 and 7.8, since water that’s too acidic can irritate eyes and corrode metal, while water that’s too basic can stop chlorine from working properly. Skincare and shampoo are also designed around our natural pH of about 5.5 to prevent dryness and irritation, and environmental scientists track the pH of rain and rivers to monitor the effects of acid rain on forests and aquatic life. Logarithmic scales make it possible to describe massive changes in acidity with simple numbers that anyone can use!

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15.07 Logarithm laws and properties After this lesson, you will be able to… • justify and apply special logarithmic properties such as loga (a) = 1 and loga (1) = 0. • derive the logarithm laws for products, quotients, and powers from the laws of indices. • apply the logarithm laws to expand a single logarithmic expression into multiple terms. • apply the logarithm laws to combine multiple logarithmic terms into a single expression.

Special logarithm properties Logarithm properties arise from the deﬁnition: if ax = b, then loga (b) = x, where a > 0, a ≠ 1, and b > 0. Substituting speciﬁc inputs give the following logarithmic results:

loga (ax) = x a

is the base, where a > 0, a ≠ 1

is any real number

is the argument, where x > 0

loga (a) = 1 since a1 = a

loga (1) = 0 since a0 = 1

Since loga (x) = y ⟹ ay = x, and a−y = , then loga

= −y = − loga (x).

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Interactive exploration Discover this concept in action online

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Example 1 Evaluate: log5 (5)

Create a strategy Apply the property loga (a) = 1.

Apply the idea log5 (5) = 1 Since 51 = 5

Example 2 = − log3 (9) without using the law loga

Show that log3

= − loga (x).

Create a strategy Evaluate the LHS and the RHS using the deﬁnition of a logarithm and index laws.

Apply the idea LHS: Rewrite using index laws Since loga (ax) = x

RHS:

Thus, log3

− log3 (9) = − log3 (32)

Express 9 as 32

= −2

Since loga (ax) = x

= − log3 (9).

Idea summary Foundational logarithmic results include loga (ax) = x, loga (1) = 0, and loga

842

= x, loga (a) = 1,

= − loga (x), derived from the logarithm deﬁnition.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Logarithm laws Logarithm laws, derived from index laws, simplify expressions. These laws apply for a > 0, a ≠ 1, and positive arguments. To derive the product law, let m = loga (x), n = loga ( y), so x = am, y = an. Then: xy = am × an = am + n loga (xy) = loga (a

m+n

)

Apply index law Take the logarithm of both sides

loga (xy) = m + n

Use loga (ak ) = k

loga (xy) = loga (x) + loga ( y)

Substitute m and n

loga (xy) = loga (x) + loga ( y) x, y

are positive numbers

To derive the quotient law, let m = loga (x), n = loga ( y), so x = am, y = an. Then: Substitute x = am, y = an

Apply index law

Take the logarithm of both sides

Use loga (ak ) = k

Substitute m and n

x, y

are positive numbers

To derive the power law, let m = loga (x), so x = am. Then: xn = (am)n

Substitute x = am

xn = amn n

loga (x ) = loga (a

Apply index law mn

)

Take the logarithm of both sides

loga (x ) = mn

Use loga (ak ) = k

loga (xn) = n × loga (x)

Substitute m

loga (xn) = n loga (x) x

is a positive number

is any real number

Interactive exploration Discover this concept in action online

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Example 3 Simplify: a log2 (16) + log2 (4) − log2 (8)

Create a strategy Use the product law loga (x × y) = loga (x) + loga ( y) and quotient law loga

= loga (x) − loga ( y).

Apply the idea Apply product law

Apply quotient law

Evaluate inside the brackets

Express 8 as 23

Apply loga (ak ) = k

Reﬂect and check Verify: log2 (16) = 4, log2 (4) = 2, log2 (8) = 3, so 4 + 2 − 3 = 3.

b 2 log5 (4)

Create a strategy Apply the power law n loga (x) = loga (xn), then use the change of base formula to evaluate.

Apply the idea Use power law

Evaluate inside the brackets

Use change of base formula

Evaluate and round

Idea summary Logarithm laws include: • Product law: loga (xy) = loga (x) + loga ( y) • Quotient law: loga

= loga (x) − loga ( y)

• Power law: loga (xn) = n loga (x)

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15.07 Practice questions What do you remember? 1

Identify the logarithm law or property that justiﬁes each statement: a

loga (mn) = loga (m) + loga (n)

loga

loga (mk ) = k loga (m)

loga (1) = 0

Evaluate each expression without technology: a

= loga (m) − loga (n)

log7 (7)

log9 (1)

log4 (45)

Write down all the logarithmic laws and properties you can remember. Aim to include at least seven.

Practice Ex 1

Simplify using special logarithm properties: a

log8 (8)

e 5

log7 (1)

log5

log4

log9 (93)

4 logb (b)

Simplify the following using log laws, leaving the answer in log form: a

log10 (3) + log10 (7)

log4 (16) − log4 (4)

3 log6 (5)

log3 (27) + log3 (9) − log3 (81)

log10 (12) + log10 (5) − log10 (6)

2 log7 (3) + log7 (9)

Expand using logarithm laws: a

log5 (6y)

log4 (x3z)

log3 (2x)

log2 (a4b)

Simplify using log laws, leaving the answer in log form: a

log5 (25) + log5 (2)

2 log3 (4) − log3 (16)

log10 (8) + log10 (5)

3 log2 (3)

log4 (32) − log4 (8)

log6 (18) + log6 (2) − log6 (3)

log3 (4z)

Expand using logarithm laws: a

log7 (xy2)

log2 (a3 b2)

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Ex 2

Ex 3

Verify each equation without using the law

Simplify using log laws, leaving the answer in log form: a

log6 (36) + log6 (6) − log6 (12)

3 log4 (2)

log5 (25) + log5 (5) − log5 (10)

2 log3 (6) − log3 (4)

4 log2 (3)

log7 (49) + log7 (7)

log5 (25)

log3 (9)

logm (mk)

logp ( p−2)

Express using e as the base of the logarithms: a

log4 (8)

log2 (16)

log7 (49)

log6 (12)

Simplify using logarithm properties: a

Simplify using log laws, leaving the answer in log form: a

loga (b2) + loga (c)

logm (n) + logm ( p) − logm (q)

logr (s3) − logr (t)

2 logx ( y) − logx (z2)

logx ( y) + logx (z) − logx (w)

Extend your thinking 14

Prove that

Show that

The magnitude of an earthquake is given by

. using the change of base formula and logarithm laws.

A0 is the reference amplitude. Express the ratio and M2 as a single exponential expression. 17

, where A is the amplitude and for two earthquakes with magnitudes M1

The pH of a solution is given by pH = − log10 [ H + ], where [ H + ] is the hydrogen ion concentration in mol/L. If two solutions have pH values pH1 and pH2, express the ratio as a single exponential expression.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

15.08 Logarithm expressions and equations After this lesson, you will be able to… • apply logarithm laws to simplify complex logarithmic expressions. • solve logarithmic equations and verify that solutions are valid. • solve exponential equations by taking the logarithm of both sides. • use digital tools to evaluate logarithmic expressions and determine if they are rational or irrational.

Simplify and evaluate logarithmic expressions Logarithmic expressions can be simpliﬁed using logarithm laws, which allow combining or rewriting terms to produce a single logarithm or a simpler form. Digital tools, such as calculators, can evaluate simpliﬁed expressions to obtain rational or irrational values. Rational logarithms evaluate to a rational number while irrational logarithms evaluate to an irrational number. For example, log10 (100) = 2 is rational, while log10 (15) is typically irrational, approximated using technology.

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Example 1 Simplify then evaluate each logarithmic expression, rounded to four decimal places and state whether the evaluated value is rational or irrational. a log5 (25) + log5 (4) − log5 (2)

Create a strategy Since the logarithms have the same base, use the product law loga (xy) = loga (x) + loga ( y) and quotient law loga

= loga (x) − loga ( y) to combine terms.

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Apply the idea Apply product law

Apply quotient law

Evaluate inside the brackets

Rewrite 50

Apply product law

Apply power law

Evaluate and round using technology

Since log5 (50) cannot be simpliﬁed to a rational number, it is irrational. b 3 log2 (5) − log2 (25)

Create a strategy Since the logarithms have the same base, use the power law loga (xn) = n loga (x) and quotient law to combine terms.

Apply the idea Apply power law

Simplify 53 = 125

Apply quotient law

Evaluate inside the brackets

Evaluate and round using technology

Since log2 (5) cannot be simpliﬁed to a rational number, it is irrational.

Idea summary Logarithmic expressions are simplified using product, quotient, and power laws. Digital tools evaluate results, yielding rational (e.g., log10 (100) = 2) or irrational values (e.g., log5 (50)).

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Solve logarithmic equations Logarithmic equations are solved by applying logarithm laws and properties, ensuring solutions satisfy domain restrictions (a > 0, a ≠ 1, arguments positive). Equations are simpliﬁed to isolate the variable, often by combining logarithms or converting to exponential form. Solutions must be substituted into the original equation to ensure arguments are positive.

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Example 2 Solve log2 (x + 3) = 3.

Create a strategy Convert to exponential form using loga (x) = y ⟹ ay = x and check the domain.

Apply the idea log2 (x + 3) = 3

Write the equation 3

x+3=2

Convert to exponential form

x+3=8

Evaluate 23

x=5

Subtract 3 from both sides

Checking the domain: x + 3 = 5 + 3 = 8 > 0, so the argument is valid. Thus, x = 5.

Reﬂect and check Verify: log2 (5 + 3) = log2 (8) = log2 (23) = 3, conﬁrming the solution.

Example 3 Solve log3 (2x − 1) + log3 (x + 1) = 2.

Create a strategy Since the logarithms have the same base, use the product law to combine logarithms, convert to exponential form, and check the domain.

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Apply the idea Write the equation

Apply product law

Expand (2x − 1)(x + 1)

Convert to exponential form

Evaluate 32

Subtract 9 from both sides

Factorise

Solve for x

Checking domain for x = 2, 2x − 1 = 3 > 0, x + 1 = 3 > 0, valid. For x =

, 2x − 1 = −6 < 0, invalid.

Thus, x = 2.

Reﬂect and check Verify: log3 (2 × 2 − 1) + log3 (2 + 1) = log3 (3) + log3 (3) = 1 + 1 = 2 confirming the solution.

Idea summary Logarithmic equations are solved using logarithm laws and properties, converting to exponential form or combining terms. Solutions must satisfy domain restrictions, and digital tools can verify results.

Solve exponential equations with logarithms Exponential equations of the form ax = b, where a > 0, a ≠ 1, and b > 0, can be solved using logarithms if rewriting with the same base is not feasible. Taking the logarithm (base 10 or e) of both sides and applying the power law yields: ax = b ⟹ logc (ax) = logc (b) ⟹ x logc (a) = logc (b)

is the logarithm base, typically 10 or e

Digital tools evaluate x, which may be irrational original equation.

. Solutions are veriﬁed by checking the

Interactive exploration Discover this concept in action online

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Example 4 Solve for x, rounded to four decimals using technology: a 4x = 15

Create a strategy Take the natural logarithm of both sides and use the power law.

Apply the idea Write the equation

Take natural logarithm of both sides

Apply power law

Divide both sides by ln(4)

Evaluate

Reﬂect and check Verify: 41.9534 ≈ 14.9996 ≈ 15, conﬁrming the solution.

b 23x − 1 = 10

Create a strategy Take the common logarithm of both sides, apply the power law, and solve for x.

Apply the idea Write the equation

Take common logarithm of both sides

Apply power law and log(10) = 1

Divide both sides by log(2)

Add 1 to both sides

Divide both sides by 3

Evaluate

Reflect and check Verify: 23 × 1.4408 − 1 ≈ 23.3224 ≈ 9.998 ≈ 10, confirming the solution.

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Idea summary Exponential equations ax = b are solved using logarithms when common bases are not possible, yielding x =

. Digital tools provide rational or irrational

solutions, veriﬁed by substitution.

15.08 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

loga (st) = loga (s) + loga (t)

loga (sm) = m loga (s)

Complete each statement for solving equations: a b

To solve cx = d, take log of both sides to get x = ⬚. If loga (w) = m, then w = ⬚.

Deﬁne these terms: a

Rational value of logarithm

Irrational value of logarithm

Exponential expression

Logarithmic expression

Determine whether each logarithm can be evaluated as rational or irrational: a

loga (an) = n

log10 (1)

log10 (10)

log10 (20)

17w = 289

log10 (50)

Solve for w: a

log11 (w) = 2

Practice Ex 1

Simplify then evaluate each logarithmic expression, rounded to four decimal places and state whether the evaluated value is rational or irrational: a

log13 (26) + log13 (5)

log15 (45) − log15 (3)

2 log11 (6)

log10 (40) + log10 (25)

log17 (2) + log17 (1)

2 log10 (3) − log10 (9)

log13 (26 ) − log13 (26)

log15 (30) + log15 (6) − log15 (14)

log15 (w) =

g Ex 2

Solve for w: a

852

log13 (w − 4) = 2

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 3

Solve for w: a

Ex 4

11w = 60

15w − 1 = 22

132w = 75

5x + 1 = 125x

10w + 2 = 15

82x = 16x + 1

Given log6 (5) = 0.8982 and log6 (7) = 1.0860, evaluate: a

log10 (w + 6) − log10 (w − 2) = 1

Solve for x, giving exact values: a

Solve for w, rounded to four decimal places: a

log17 (w + 2) + log17 (w − 5) = 2

log6 (35)

log6

Estimate the two consecutive integers between which the solution lies: a

4x = 20

7x =

Extend your thinking 13

The loudness of sound is given by L = 10 log10

dB, where I is the intensity and I0 is the

reference intensity. If a sound has intensity 1000 times that of another, ﬁnd the difference in their loudness levels. 14

Solve for x: log3 (log4 (2x − 1)) = 1

Solve for x: 2 log6 (x) = log6 (9x − 8)

Solve for x and y: log4 (x) + log4 ( y) = 3 log4 (x + y) = 2

A population grows according to P (t) = 50 × , where t is time in years. Find the time required for the population to reach 800, rounded to two decimal places.

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15.09 Logarithmic graphs After this lesson, you will be able to… • graph logarithmic functions of the form y = loga (x). • identify the domain, range, asymptote and intercepts of logarithmic graphs. • describe how the base a affects the shape of the graph. • recognise that the graphs of y = ax and y = loga (x) are reflections of each other in the line y = x.

The logarithmic graph Interactive exploration Discover this concept in action online

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Consider the graph of y = log2 (x): 5

x=0 (asymptote)

4 3

(8, 3)

2 1 −1

(4, 2)

(2, 1)

(1, 0) 2

x 3

−2 −3

x-intercept

−4 −5

The ﬁrst key feature is the asymptote, drawn by a dotted line at x = 0. • As x approaches the value of zero, the value of y approaches negative inﬁnity. The second key feature is the x-intercept, found at the point where y = 0. • Since 0 = loga (x) can be converted to exponential form x = a0, the x-intercept of loga (x) is always 1.

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Consider the graphs of y = loga (x) for the values of a = 2, 3, 4: 5

x=0 (asymptote)

y = log2 (x) y = log3 (x)

3 2

(2, 1)

1 −1

y = log4 (x)

(4, 1)

(3, 1) 4

x 7

−2 −3

common x-intercept (1, 0)

−4 −5

Some key features of the graphs of y = loga (x) for a > 1: • The x-intercept remains as (1, 0). • The asymptote remains as x = 0. • There is no y-intercept, as there is a vertical asymptote on the y-axis. • The graphs of higher bases are closer to the x-axis. For example, log4 (x) is closer to the x-axis than log2 (x). The domain of the graph y = loga (x) is x > 0, and the range is y ∈ .

Example 1 Sketch the graph of y = log5 (x), labelling the asymptote and x-intercept. Identify the domain and range.

Create a strategy Identify the key features of a logarithmic graph: the vertical asymptote and the x-intercept. Find one other key point to help deﬁne the shape of the curve. Use these features to sketch the graph and then determine the domain and range.

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Apply the idea For any function of the form y = loga (x), the logarithm is only deﬁned for x > 0. The y-axis, whose equation is x = 0, is a vertical asymptote. The x-intercept occurs when y = 0. y = log5 (x)

Write the equation

0 = log5 (x)

Set y to 0

x=5

Convert to exponential form

Evaluate

The x-intercept is at (1, 0). Choose a convenient value for x, such as the base of the logarithm, x = 5 to have another point on the graph. y = log5(x)

Write the equation

= log5 (5)

Substitute x = 5

Evaluate

Another point on the graph is (5, 1). Plot the points and sketch the graph. y 2 1

(5, 1)

x = 0 (1, 0) 1

x 2

−1 −2

From the graph, the function is only deﬁned for positive x-values. The y-values can be any real number. Domain: x > 0 Range: All real y (or y ∈ )

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Idea summary Some key features of the graphs of y = loga (x) for a > 1: • The x-intercept remains as (1, 0). • The asymptote remains as x = 0. • There is no y-intercept, as there is a vertical asymptote on the y-axis. • The graphs of higher bases are closer to the x-axis. For example, log4 (x) is closer to the x-axis than log2 (x).

Exponential-logarithmic reflections Interactive exploration Discover this concept in action online

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The graphs of y = ax and y = loga (x) are reﬂections of each other over the line y = x for a > 0, a ≠ 1.

y = ax y

is the exponential function where a is the base

y = loga (x)

is the logarithmic function, the inverse of y = ax

To conﬁrm the reﬂection, consider key points: • For y = ax, the point (0, 1) maps to (1, 0) on y = loga (x). • For y = loga (x), the point (1, 0) maps to (0, 1) on y = ax. • For y = ax, the horizontal asymptote is at y = 0. • For y = loga (x), the vertical asymptote is at x = 0. For a > 1, both graphs are increasing, but y = ax grows rapidly, while y = loga (x) grows slowly. y 2 1 (0, 1) x

y=2

y=0

(1, 0) −2

−1

x=0 y=x

−2

y = log2 (x)

15.09 Logarithmic graphs mathspace.co

857

For 0 < a < 1, both graphs are decreasing.

Example 2 Use a graphing application to verify that y = 3x and y = log3 (x) are reﬂections over y = x.

Apply the idea Plot y = 3x, y = log3 (x), and y = x on the same graph. y 3 2 1 (0, 1)

y=3 −9

−6

y=0

−3

(1, 0) 3

x 6

−1

y=x

−2

y = log3 (x)

−3

x=0 Identify key points and asymptotes to confirm the reflection: • On y = 3x, (0, 1) corresponds to (1, 0) on y = log3 (x). • On y = log3 (x), (1, 0) corresponds to (0, 1) on y = 3x. • For y = 3x, the horizontal asymptote is at y = 0. • For y = log3 (x), the vertical asymptote is at x = 0. The graphs are symmetric over y = x, confirming they are reflections.

Reflect and check Check another point, for example, (1, 3) on y = 3x. This maps to (3, 1) on y = log3 (x), as log3 (3) = 1. This symmetry over y = x veriﬁes the reﬂection.

Idea summary The graphs of y = ax and y = loga (x) are reﬂections over y = x for a > 0, a ≠ 1. Key points like (0, 1) and (1, 0) swap between the functions, conﬁrming the inverse relationship.

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15.09 Practice questions What do you remember? 1

Identify the key features of the graph y = loga (x) for a > 1: a

x-intercept

Asymptote

Domain

Range

Answer the following about exponential and logarithmic functions: a

What is the inverse of y = ax for a > 0, a ≠ 1?

If y = ax passes through (0, 1), what point does y = loga (x) pass through?

Over which line are y = ax and y = loga (x) reﬂections?

Explain why the graph of a logarithmic function of the form loga (x), where a > 0 and a ≠ 1, has no y-intercept. Support your explanation by rewriting the logarithmic function in its equivalent exponential form.

Determine the asymptote and x-intercept of y = log7 (x): a

Asymptote

x-intercept

Practice 5

Determine whether each statement is true or false: a b c

Ex 1

The graphs of y = 2x and y = log2 (x) are reﬂections over y = x. For 0 < a < 1, both y = ax and y = loga (x) are increasing.

The point (1, 0) on y = loga (x) maps to (0, 1) on y = ax.

Sketch the graph of y = log11 (x), labelling the asymptote and x-intercept. Identify the domain and range.

Compare the graphs of y = log7 (x) and y = log13 (x). Which is closer to the x-axis and why?

Explain why the graphs of y = ax and y = loga (x) are always reﬂections over y = x for a > 0, a ≠ 1.

For f (x) = log13 (x), determine: a

Domain

Range

Asymptote

x-intercept

For the function f (x) = log7 (x + 3) − 1: a

Use a table of values covering the interval (3, 6).

Sketch the graph of the given function labelling the asymptote and intercepts.

Identify the domain and range.

Sketch f (x) = log13 (x) on the domain 1 ≤ x ≤ 169. Determine the range.

15.09 Logarithmic graphs mathspace.co

859

Match each pair of functions that are reﬂections over y = x: a

y = log9 (x)

y = ln(x)

y = log10 (x)

y = 9

iii

y = ex

Match each function to its graph using key features: • y = log4 (x) • y = log20 (x) • y = log9 (x)

Graph A Graph B

Graph C x

−1

For what values of h does y = log10 (x − h) have a y-intercept?

Identify a point on the ﬁrst function and its corresponding point on the second function after reﬂection over y = x: a

y = 2x, y = log2 (x)

Ex 2

y = 3x, y = log3 (x)

y = 5x, y = log5 (x)

Use a graphing application to sketch the functions and confirm they are reflections over y = x. Describe one key feature of the reflection: a

y = 2x, y = log2 (x)

y = 5x, y = log5 (x)

y = 6x, y = log6 (x)

c y = 4x, y = log4 (x)

y = ex, y = ln(x)

For a > 1, sketch the graphs of y = ax and y = loga (x) using a graphing application. Compare their steepness as x approaches positive or negative inﬁnity: a

a=3

a=8

For 0 < a < 1, sketch the graphs of y = ax and y = loga (x) using a graphing application. Describe the reﬂection: a

A signal strength model uses S = 10 × 2t, where t is time in seconds. Use a graphing application to plot S and its inverse. Identify a point on each graph that conﬁrms the reﬂection over y = x.

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Extend your thinking 20

The graph of y = log7 (x − h) + k has an asymptote at x = 0.5 and passes through (1.5, 1). Determine h and k.

A population decays according to P (t) = 100 × 7−0.1t, where t is time in years. Another population grows according to R(t) = 10 × log11 (t + 1). Use technology to ﬁnd the time t when P (t) = R(t), rounded to two decimal places.

The graph of y = log11 (x) is translated 2 units right and 1 unit up to form the graph of y = log11 (x − h) + k. Find the values of h and k.

A radioactive substance grows according to M (t) = 550 × grams, where t is in years. Another substance grows according to D(t) = 5 × log13 (t + 1). Find the time when M (t) = D(t), rounded to two decimal places.

Two logarithmic graphs, f (x) = log7 (x − h1) + k1 and g(x) = − log7 (h2 − x) + k2, intersect at two points. After translating g(x) to 5 units up, they no longer intersect. Find the possible values for h1, k1, h2, and k2 .

The loudness of a machine is L = 10 log10

dB, where I0 = 10−12 W/m2. If two machines

produce I1 = 5 × 10−5 W/m2 and I2 = 3 × 10−5 W/m2, ﬁnd the combined loudness in dB, rounded to two decimal places. 26

For 0 < a < 1, compare the behaviour of y = ax and y = loga (x). Why do they still reﬂect over y = x? Use a graphing application to support your answer for a = .

A population grows according to P = 100 × 2t, where t is time in years. Find the time t when P = 800 using the inverse function.

A logarithmic graph has the following features: • Passes through the point (4, 0.5) • Has a vertical asymptote at x = 0 • Is increasing and concave down a

Deduce a possible equation of the logarithmic function.

Write down the equation of the exponential function that would be its exact reﬂection in y = x.

Verify, by choosing a point, that these functions are inverses.

Investigation: Logarithms Investigate online

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15.09 Logarithmic graphs mathspace.co

861

15 Chapter review 1

Which of the following is the range of the function y = 5(3x)? A

y<5

y>5

Population (P)

800

600

400

200

Population (P)

Time (years) 0

Time (years)

Population (P)

800

600

400

200

Time (years) 0

For the function y = 4(3x): a

Identify the horizontal asymptote, y-intercept, domain, and range.

Sketch the graph, labelling key features.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Population (P)

Time (years)

862

A population of wildlife starts with 200 animals and doubles every 3 years. Which graph best represents the population P over time t (in years)? A

All real y

What is the gradient of the tangent to the curve y = 4ex at the point where x = 1? A

y>0

Consider the function y = − 5x: a

Complete the table of values: x

−3

−2

−1

⬚

Describe the end behaviour and horizontal asymptote.

State the domain and range.

Consider the graph of the function f (x) = 3x: a

Identify the y-intercept.

Does the graph have an x-intercept?

Identify the domain of the function.

Identify the range of the function.

Calculate f (5).

If the point (4, m) lies on the curve of f (x) = 3x, determine the value of m.

3 2 1

−5 −4 −3 −2 −1

Two different sequences are generated by the functions f (n) = 4n + 2 and g(n) = 2(3)n: a

Determine the ﬁrst 5 terms in the sequence generated by each function: n

f (n)

⬚

g(n)

1 2 3 4 5

⬚

Describe how the terms of each sequence increase.

Explain for what values of n is f (n) ≥ g(n) and for what values of n is f (n) < g(n).

A data analyst considers the graphs of f (x) and g(x), which are exponential and quadratic functions respectively, and performs these calculations: • Calculation 1: f (5) − f (4) = 162 • Calculation 2: f (4) − g(4) = 55 • Calculation 3: f (5) − g(5) = 208 • Calculation 4: g(5) − g(4) = 9 a

Determine the equations of f (x) and g(x).

The analyst claims that for large values of x, as x increases, the exponential function increases more rapidly than the quadratic function. Explain how the calculations support this claim.

y 50 40 30 20 10 x 0

Chapter 15 review mathspace.co

863

Estimate the gradient of the tangent at x = 0 for y = 1.5x using the secant line with: a

h = 0.1

h = 0.01

h = 0.001

ompare the results for different C h values.

Calculate the gradient of the secant at x = 0 with h = 0.01 for the following functions: a

y = 1.9x

y = 2.8x

y = 0.6x

y = 6x

For the function y = kax (k ≠ 0, a > 0, a ≠ 1): a

Find the y-intercept.

Derive the gradient of the secant formula at x = 0 using points (0, k) and (h, kah).

If k = 3 and a = 4, calculate the gradient of the secant at x = 0 with h = 0.01.

Explain how the scaling factor k affects the tangent gradient at x = 0.

Explore how the gradient of the tangent at x = 0, will change when the function is f (x) = ka−x compared to f (x) = kax.

Find the gradient of the tangent to y = ex at the following points: a

x=3

x=0

x=2

x=3

Determine the equation of the tangent to y = ex at the given points: a

x = −3

x=0

x = −2

x=1

Consider the equation y = ex: a

Find the equation of the tangent to the curve at x = 2.

Find the equation of the normal to the curve passing through this point.

Calculate the area of the triangle formed between the tangent, normal, and the x-axis.

A population of bacteria grows according to P = 2000e0.05t, where t is in days. Find the time when the growth rate is 50 individuals per day.

The value of a machine depreciates according to V = 25 000(0.85t), where t is time in years: a

864

Complete the table of values for the graph: Time (years) (t)

Value ($) (V )

⬚

Sketch the graph using the table values, labelling the asymptote and y-intercept.

Describe the end behaviour as t → ∞.

Find the value after 5 years, rounded to the nearest dollar.

An investment grows according to A = 8000(1.04t), where t is time in years: a

Sketch the graph, labelling the asymptote, y-intercept, and one additional point.

Describe the end behaviour as t → ∞.

Find the balance after 5 years, rounded to the nearest dollar.

Determine when the balance ﬁrst exceeds $10 000, rounded to the nearest year.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the function y = a

Rewrite the function in the form y = k−x.

Describe the transformation required to obtain the graph of y = y = 4x.

Sketch the graph of the functions y = 4x and y =

Identify the coordinates of the point of intersection of the two curves.

Describe the behaviour of both these functions for large values of x.

Identify the y-intercept and horizontal asymptote.

Sketch the graph, labelling the y-intercept and asymptote.

Find the mass after 20 days, rounded to the nearest gram.

Estimate the time taken to reach half the initial mass of the isotope, rounded to the nearest day.

What is the value of log3 (81)? C

log(9)

log(1.25)

log(20)

x=2

x=0

x=3

y=3

43 = 64

2− 4 =

e2 = e2

91.5 = 27

log5

log2 (64) = 6

log(100)

log2

Rewrite in exponential form: log7 (49) = 2

log(1000) = 3

= −3

Evaluate without technology: a

Rewrite in logarithmic form:

a 26

What is the equation of the vertical asymptote of the graph of y = log2 (x − 3)?

a 25

Which of the following is equivalent to log(5) + log(4)?

A 24

on the same set of axes.

A 23

from the graph of

A radioactive isotope decays according to M = 300e− 0.02t, where t is time in days:

A 22

log5 (125)

ln(e4)

P = 2000, A = 4000, r = 0.04

P = 8000, A = 10 000, r = 0.025

The pH of a substance is given by pH = − log [ H + ], where [ H + ] is the hydrogen ion concentration in mol/L. Calculate the hydrogen ion concentration for a substance with pH = 5.5, rounded to three signiﬁcant ﬁgures.

Solve each exponential equation using logarithms, rounded to four decimal places: a

3x = 25

5x = 50

6x = 11

2x = 0.8

Chapter 15 review mathspace.co

865

Solve the following equations, rounded to four decimal places: a

5x + 2 = 18

4 = 2x − 3

Evaluate using the change of base formula, rounded to three decimal places: a

log3 (10)

log8 (15)

log5 (30)

Solve 3x = 81 for x, expressing solutions in exact form.

Simplify using special logarithm properties: log11 (11)

log2 (14)

log5

log15 (1)

log7 (74)

Rewrite as a single logarithm: a

log2 (5) + log2 (7)

log3 (54) − log3 (2)

4 log5 (2)

2 log4 (3) + log4 (5)

log2 (x5y)

Expand using logarithm laws: a

log3 (5a)

log7

log

= 2 logb (a) + logb (c) − 4 logb (d).

Prove that logb

The magnitude of an earthquake is given by M = log10

, where A is the amplitude

and A0 is a reference amplitude. An earthquake of magnitude 7.2 occurs, followed by an aftershock of magnitude 5.0. How many times larger is the amplitude of the main earthquake than the aftershock? Round your answer to the nearest whole number. 38

Simplify each logarithmic expression and evaluate it. Then, state whether the result is rational or irrational. a

log5 (w − 2) = 3

log3 (w + 1) = 4

log2 (w + 1) + log2 (w − 1) = 3

log4 (w − 2) + log4 (w + 2) = 3

log3 (log2 (x − 1)) = 2

Solve for x: a

log2 (log3 (x + 1)) = 2

A bacterial culture grows according to P (t) = 80 × required for the population to reach 2160.

For f (x) = log5 (x), determine: a

866

log3 (54) − log3 (2)

Solve for w: a

Solve for w and state if the result is rational or irrational: a

log2 (16) + log2 (5)

Domain

Range

, where t is time in hours. Find the time

Asymptote

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x-intercept

Sketch the graph of f (x) = log2 (x + 4) − 1, labelling the asymptote and intercepts. State the domain and range.

The graph of y = log4 (x − h) + k has an asymptote at x = 2 and passes through the point (6, 3). Determine the values of h and k.

Identify a point on the ﬁrst function and its corresponding point on the second function after reﬂection over the line y = x: a

y = 10x, y = log(x)

y = 6x, y = log6 (x)

y = ex, y = ln(x)

Explain why the graphs of y = ax and y = loga (x) are reﬂections of each other across the line y = x for all valid bases a.

A logarithmic graph has the following features: • Passes through the point (9, 2) • Has a vertical asymptote at x = 0 • Is an increasing function a

Deduce a possible equation of the logarithmic function.

Write down the equation of the exponential function that would be its reﬂection in the line y = x.

Verify by choosing a point that these two functions are inverses.

Did you know?

A butterfly’s wings are nearly identical on both sides — a natural reflection transformation! If you drew a line down the centre, one side would mirror the other. Mathematicians call this line symmetry, and it’s a key concept in understanding how transformations create balance and harmony in shapes. Chapter 15 review mathspace.co

867

Big ideas Inverse functions enable the reversal of one-to-one mappings, using algebraic and graphical techniques to analyse relationships and solve equations in various mathematical contexts.

16 Inverse functions Chapter outline 16.AE 16.01E 16.02E 16.03E 16.04E 16.05E 16.06E 16.07E

870 879 884 889 894 899 905 912

In photography, negatives are the inverse of the real image — and they’re still used in film development today!

16.01E Inverse functions After this lesson, you will be able to… • define an inverse function f −1 informally as a function that reverses or undoes the effect of the function f. • recognise that inverse functions exist only for one-to-one functions. • establish that the reflection of a point in the line y = x reverses the coordinates of the point. • understand that the graph of y = f −1( x) is the reflection of the graph of y = f ( x) in the line y = x. • determine graphically if a function has an inverse function by considering its reflection.

Inverse functions An inverse function f −1 reverses the effect of a function f, mapping each output back to its unique input. This exists only for one-to-one functions, where each range element corresponds to exactly one domain element. The notation f −1 denotes the inverse function, not the reciprocal function [ f ( x)]−1, which is

. For example, if f ( x) = 2x, then [ f ( x)]−1 =

, whereas the inverse function

reverses the operation of f . Geometrically, the inverse is a reﬂection of the original function across y = x. When a function is reflected over this line, the coordinates of each point reverse: a point at ( x, y) becomes ( y, x). For instance, if a point on the original function is at (2, 7), its reflection across y = x is at (7, 2), interchanging the x- and y-coordinates entirely. Reﬂection

Transformation

A transformation of a shape formed by creating a mirror image on the other side of a given line.

A procedure or set of procedures that changes the size and/or shape of an image. A transformation operates on points in the plane to change aspects, such as the position, size or shape of curves and other ﬁgures. Translations, reﬂections, rotations, dilations and enlargements are all examples of transformations.

Interactive exploration Discover this concept in action online

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mathspace.co

Example 1 Do the following graphs have inverse functions? a

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

Create a strategy Use the fact that a function has an inverse if it is a one-to-one function.

Apply the idea This is a linear function where the slope ensures distinct outputs for distinct inputs, conﬁrming it is one-to-one and has an inverse, consistent with the reﬂection being a function. Therefore, it has an inverse function.

Reﬂect and check Reﬂecting across y = x tests invertibility. For the function y =

+ 2, the point (6, 4) reﬂects to

(4, 6), reversing its coordinates. Similarly, (−6, 0) reﬂects to (0, −6), and (0, 2) reﬂects to (2, 0), interchanging x- and y-values in each case. y 10 8 (4, 6) 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 y=x −10

(6, 4)

2 4 6 8 10

16.01E Inverse functions mathspace.co

871

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

Apply the idea This is a quadratic function that maps distinct inputs to the same output, indicating it is not one-toone and thus has no inverse function. Therefore, it does not have an inverse function.

Reflect and check Notice that it can be reflected across y = x, but the reflected graph is a horizontal parabola which is not a function, so it cannot be considered as the inverse of the function. 10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 y = x −10

872

x 2 4 6 8 10

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

Apply the idea This is an exponential function that increases continuously, ensuring distinct outputs for distinct inputs, conﬁrming it is one-to-one and has an inverse. Therefore, it has an inverse function.

Reflect and check Reflecting across y = x yields a logarithmic function, confirming invertibility. For y = 2x, the point (1, 2) reflects to (2, 1), (2, 4) reflects to (4, 2), and (0, 1) reflects to (1, 0), with coordinates reversing in each case. y 10 8 6 (2, 4) 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 y=x −10

(4, 2)

2 4 6 8 10

16.01E Inverse functions mathspace.co

873

y 3 2 1 −3 −2

−1

x 1

−1 −2 −3

Apply the idea This is a cubic function that is strictly increasing, ensuring distinct outputs for distinct inputs, so it is one-to-one and has an inverse. Therefore, it has an inverse function.

Reflect and check Reflecting across y = x produces a cubic root function, confirming invertibility. For y = 0.5x3, the point (1, 0.5) reflects to (0.5, 1), and (−1, −0.5) reflects to (−0.5, −1), reversing coordinates each time. y 3 2 1 −3 −2

−1

(−1, −0.5)

x 1

−1

(−0.5, −1)

−2

y=x

874

−3

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

Apply the idea This is a cubic function that has turning points, mapping distinct inputs to the same output, indicating it is not one-to-one and thus has no inverse function. Therefore, it does not have an inverse function.

Reflect and check Reflecting across y = x confirms non-invertibility. 10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 y=x −10

x 2 4 6 8 10

Example 2 Are the lines in the graph inverse functions of each other? 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

16.01E Inverse functions mathspace.co

875

Create a strategy If both lines are reﬂections of each other across the line y = x, then they are inverse functions of each other.

Apply the idea 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2

y=x

−3 −4

The two lines are not reflections of each other across the line y = x. Therefore, they are not inverse functions of each other.

Idea summary Inverse functions reverses the effect of a one-to-one function and therefore exists only for one-to-one functions. Geometrically, the graph of an inverse function is a reflection of the original function’s graph across the line y = x, which has the effect of interchanging the coordinates of every point. The inverse function f −1 reverses or undoes the effect of the function f.

16.01E Practice questions What do you remember? 1

What does an inverse function do, and when does it exist?

How does reﬂection over y = x show the inverse?

What is the ﬁrst step to ﬁnd the inverse of f ( x) = 3x − 2?

Why doesn’t f ( x) = x2 (all real numbers) have an inverse?

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Practice 5

Reﬂect each point over the line y = x and state the new coordinates. a

(0, −2)

(−4, 5)

(−12, −12)

For a function defined on the domain D = {−1, 1}, determine if each set of points represents a one-to-one function. a

(1, 3)

(−1, 3), (1, −3)

Consider the function f ( x) =

with points

(−1, 2), (1, 2)

and

. By reflecting these points

over the line y = x, determine if the inverse is a function, and therefore, if f ( x) is one-to-one. Ex 1

Determine if each graph has an inverse function. Verify by reflecting across y = x and show the coordinates of at least one point and its reflection. a

y 10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

10 8 6 4 2

x 2 4 6 8 10

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

2 4 6 8 10

10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

10 8 6 4 2

x 2 4 6 8 10

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

2 4 6 8 10

y 3 2 1

x 2 4 6 8 10

−3 −2

−1

x 1

−1 −2 −3

16.01E Inverse functions mathspace.co

877

Ex 2

Verify using y = x if each pair of lines are inverses of each other. a

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

−4 −3 −2 −1 −1

−2

−3

−4

1 −4 −3 −2 −1 −1

−4 −3 −2 −1 −1

−2

x 1

−4 −3 −2 −1 −1

x 1

−2

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

Verify if f ( x) = x3 and g( x) =

Show that f ( x) = x2 − 1 is not one-to-one by reﬂecting points (2, 3) and (−2, 3) over y = x.

878

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are inverses by reﬂecting points (1, 1) and (2, 8) over y = x.

Extend your thinking 12

Explain why f ( x) = ∣x∣ does not have an inverse over all real numbers.

Explain why f ( x) = 2x has an inverse over all real numbers.

Given f −1 ( x) =

Why does f ( x) = x3 − k always have an inverse for any real k?

For what values of a does f ( x) = ax2 (all real x) have no inverse?

, sketch f −1( x) and f ( x) on the same axes.

16.02E Formal definition of an inverse function After this lesson, you will be able to… • state the formal definition of an inverse function: f −1 ( x) is the inverse of f ( x) if f ( f −1 ( x)) = x and f −1( f ( x)) = x. • understand that these conditions must hold for all x in the appropriate domains. • use the formal definition to verify algebraically if a given function g( x) is the inverse of another function f ( x). • connect the formal definition to the concept that inverse functions “undo” each other.

Formal definition of an inverse function Inverse function f −1( x) is the inverse function of f ( x) if the relationships f ( f −1( x)) = x and f −1( f ( x)) = x hold. f −1 reverses or undoes the effect of the function f. An inverse function exists for any one-to-one function and has domain the range of the function. These conditions mean that applying f and then f −1 (or vice versa) returns the original input, effectively undoing each other. This only works for one-to-one functions, where each output corresponds to exactly one input, ensuring the inverse is a function. Geometrically, the inverse reﬂects the graph of f ( x) over the line y = x, interchanging inputs and outputs. The formal deﬁnition provides a mathematical way to verify this relationship and solve problems involving inverses. 16.02E Formal definition of an inverse function mathspace.co

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Example 1 Verify that the given function and its proposed inverse satisfy the formal deﬁnition of an inverse function. a f ( x) = 2x + 5, f −1 ( x) =

Create a strategy Check both conditions: compute f ( f −1( x)) and f −1( f ( x)), ensuring they equal x over their respective domains.

Apply the idea

Reﬂect and check

Domain of f ( x) is x ∈ (−∞, ∞), range is y ∈ (−∞, ∞). Proposed inverse has domain x ∈ (−∞, ∞).

The graph of f ( x) is a line (slope 2, y-intercept 5), and f −1( x) reﬂects it over y = x. The conditions conﬁrm this algebraically.

Check f ( f −1( x)): Substitute f −1( x)

Apply f

Simplify

Evaluate

Check f −1( f ( x)): Substitute f ( x)

Apply f −1

Simplify

Evaluate

Both conditions hold, so f −1 ( x) = inverse.

880

is the

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

14 y 13 12 11 10 9 8 7 6 5 4 f (x) 3 2 1 −4 −3 −2 −1−1 1 2 −2 −1 −3 f (x) −4

x 3

b f ( x) = x3 − 1, f −1 ( x) =

Create a strategy Test the formal deﬁnition by calculating f ( f −1( x)) and f −1( f ( x)) to see if both equal x.

Apply the idea

Reﬂect and check

Domain of f ( x) is x ∈ (−∞, ∞), range is y ∈ (−∞, ∞). Inverse domain is x ∈ (−∞, ∞).

The cubic f ( x) is one-to-one (strictly increasing), and its inverse reﬂects over y = x. The formal deﬁnition validates this relationship.

Check f ( f −1( x)):

Substitute f −1( x) 2

Apply f

Simplify

Evaluate

1 x −2

−1

Check f ( f ( x)): Substitute f ( x)

Apply f −1

Simplify

Evaluate

f −1 (x)

−1

1 −1

f (x)

−2

Both conditions hold, conﬁrming f −1 ( x) = is the inverse.

Example 2 Verify that f −1 =

is the inverse of f ( x) = 5x − 3 using the formal deﬁnition.

Create a strategy Substitute f −1 ( x) =

into f ( f −1( x)) = x, then substitute f ( x) = 5x − 3 into f −1( f ( x)) = x to conﬁrm.

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Apply the idea Verifying f ( f −1( x)) = x: Substitute f −1 ( x) =

Expand the brackets

Collect like terms

−1

Verifying f ( f ( x)) = x: Substitute f ( x) = 5x − 3

Evaluate the numerator

Evaluate

Both equal x, so f −1 ( x) =

is the inverse.

Reﬂect and check The formal deﬁnition ensures the inverse “undoes” f ( x) completely, as shown by the conditions holding for all x.

Idea summary An inverse function f −1( x) is formally defined by f ( f −1( x)) = x and f −1( f ( x)) = x, ensuring it reverses f ( x). This definition applies only to one-to-one functions and can be used to find and verify inverses algebraically. Reflecting over y = x visually supports the formal conditions, which are key to solving inverse-related problems.

16.02E Practice questions What do you remember? 1

What does it mean for f −1( x) to be the inverse function of f ( x) according to the formal deﬁnition?

Why must a function be one-to-one for its inverse to satisfy the formal deﬁnition?

Why does the formal deﬁnition of an inverse function imply that the graphs of f ( x) and f −1( x) are reﬂections over y = x?

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Practice Ex 1

Verify that the given function and its proposed inverse satisfy the formal deﬁnition of an inverse function by checking f ( f −1( x)) = x and f −1( f ( x)) = x. a

Ex 2

f ( x) = 4x − 7, f −1 ( x) =

f ( x) =

, f −1( x) = x3 − 2

For each equation, verify that it is the inverse of f ( x) using the formal deﬁnition. a

f ( x) =

, f −1 ( x) =

f ( x) =

, x ≠ 2, f −1( x) =

+ 2, x ≠ 0

Verify if f ( x) = 2x − 3 and the function described as “takes a value of x, adds 3, then divides the result by 2” with domain all real numbers are inverses using the formal deﬁnition.

For f ( x) = ex−1 and proposed inverse f −1( x) = ln x + 1, verify the formal deﬁnition at x = e and x = 1.

Verify that f −1( x) = ex − 4 is the inverse of f ( x) = ln( x + 4), x > −4 using the formal deﬁnition.

Show that f ( x) = x2 and g( x) =

For f ( x) =

For f ( x) = x2 + 1, x ≥ 0, and proposed inverse f −1 ( x) =

do not satisfy the formal deﬁnition of inverses.

and f −1 ( x) =

, verify the formal deﬁnition. , verify the formal deﬁnition.

Extend your thinking 12

Prove that if f ( x) =

Prove that the inverse of a linear function f ( x) = mx + b (where m ≠ 0) is also linear using the formal deﬁnition. State its slope and y-intercept in terms of m and b.

If f ( x) =

( x ≠ 0), then it is its own inverse using the formal deﬁnition.

and g( x) = 2x − 4, prove that the inverse of f ( g( x)) is g−1( f −1( x)) using the formal

deﬁnition. 15

Show why f ( x) = ∣x∣ fails to have an inverse satisfying the formal deﬁnition without domain restriction, and propose a restriction to make it work.

Consider f ( x) = e2x − e−2x − 2 and the equation g( x) =

Using the formal deﬁnition of inverse functions, show that g( x) = f −1( x).

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16.03E Determine the equation of an inverse function After this lesson, you will be able to… • interchange variables x and y in the equation y = f ( x) to begin finding the inverse. • solve the equation x = f ( y) for y to find the expression for f −1 ( x). • determine the equation of the inverse function for various types of one-to-one functions (linear, rational, radical, exponential, logarithmic). • apply this method to functions with restricted domains that make them one-to-one.

Determine the equation of an inverse function To ﬁnd the equation of an inverse function f −1( x), start with a one-to-one function y = f ( x). The inverse function reverses the roles of the input and output, effectively interchanging the domain and range. The process involves interchanging the variables x and y in the equation y = f ( x), then solving for y to express the inverse as a function of x. This method works because the inverse undoes the original function, mapping each output back to its corresponding input, assuming the function is one-to-one.

Example 1 Find the inverse function of f ( x) =

Create a strategy Interchange x and y in y = f ( x), then solve for y.

Apply the idea Write the function

Interchange x and y

Multiply both sides by y + 1

Divide both sides by x

Subtract 1 from both sides

Thus, the inverse function is f −1 ( x) =

884

− 1.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Reﬂect and check Verify by composing f ( f −1( x)) to check if it equals x. Substitute f −1 ( x) =

− 1 into f ( x)

Apply f ( x) =

Remove unnecessary brackets

Simplify the denominator

Evaluate the reciprocal

−1

Since f ( f ( x)) = x, the inverse is correct.

Example 2 Find the inverse function of y = ln( x).

Create a strategy Interchange x and y in y = f ( x), then solve for y using properties of logarithms and exponentials.

Apply the idea y = ln( x)

Write the function

x = ln( y)

Interchange x and y

e =y

Exponentiate both sides with base e −1

Thus, the inverse function is f ( x) = ex.

Reﬂect and check Verify by composing f ( f −1( x)) to conﬁrm it equals x. f ( f −1( x)) = f (ex)

Substitute f −1( x) = ex into f ( x)

= ln(ex)

Apply f ( x) = ln( x)

Use the property ln(ex) = x

Since f ( f −1( x)) = x, the inverse is correct, conﬁrming that the inverse of a logarithm is an exponential.

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Example 3 Consider y = x2 − 2x − 3 for the domain x ≥ 1 where it is one-to-one. Find the equation of the inverse.

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

Create a strategy Interchange x and y, then solve for y by completing the square, ensuring the solution respects the domain restriction.

Apply the idea Write the function

Interchange x and y

Add 3 to both sides

Add 1 to both sides to complete the square on the right

Factor the right-hand side as a perfect square

Swap sides

Take the square root of both sides

Add 1 to both sides

Now the ± indicates two possible inverse functions, however, there can only be one and this is where to use the domain to ﬁgure it out. The simplest way to solve this is to choose a point from the original domain, but not the vertex of the parabola. In this case the vertex occurs at x = 1, so choose x = 2. Substitute x = 2 and y = −3. So (2, −3) lies on f ( x), which means (−3, 2) lies on the inverse function as we interchange x and y. So that means (−3, 2) must satisfy y = 1 ±

to determine if it is + or −:

Write the equation Substitute x = −3 and y = 2

Evaluate inside the square root

Evaluate the square root

This shows that 2 = 1 + 1, so choose the positive value. The inverse function is f −1( x) = 1 +

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary To determine the equation of an inverse function f −1( x) for a one-to-one function y = f ( x), interchange x and y, then solve for y in terms of x.

16.03E Practice questions What do you remember? 1

State the steps to ﬁnd the inverse function f −1( x) of a given one-to-one function y = f ( x).

Determine whether each statement is true or false about inverse functions: a

The inverse function f −1( x) exists only if f ( x) is one-to-one.

To ﬁnd the inverse, you can interchange x and y and then solve for x.

If f (2) = 5, then f −1(5) = 2.

Given a point (a, b) on the graph of y = f ( x), what point must lie on the graph of y = f −1( x)?

What condition must a function satisfy to have an inverse function over its entire domain?

Practice Ex 1

Ex 2

Find the inverse function: a

f ( x) =

f ( x) = 3x + 5

f ( x) = x3 + 1

f ( x) =

f ( x) = x5

f ( x) =

y = ln(2x − 3)

y = 2 ln( x) + 1

−1

y = 5ex−2

Find g( f ( x)).

iii

Are they inverses?

Find the inverse function: a d

y = ln( x + 1) x

y=e +4

y = 3e

For each pair of functions: i

Find f ( g( x)).

f ( x) = 5x − 1 and g( x) =

f ( x) = ln( x + 2) and g( x) = ex−2

f ( x) = 2ex + 3 and g( x) =

f ( x) =

and g( x) =

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For each function: a

Find f −1( x).

Describe how f −1( x) undoes f ( x).

f ( x) = x5 − 4

Ex 3

f ( x) = −4x − 2

b f

f ( x) = 3x + 7 3

f ( x) = x − 6

f ( x) = 2x3 + 1

f ( x) =

f ( x) = 5x + 3

Find the inverse function of f ( x) =

Given f −1 ( x) = , determine f ( x) and state its domain.

Given that f ( f ( x)) = 4x + 1, ﬁnd the inverse function f −1( x).

Given that points (2, 3) and (0, 1) lies on the graph of y = f −1( x), ﬁnd the inverse function f −1( x) assuming f ( x) is linear.

For f ( x) = , if f −1(1) = 2, ﬁnd a and the inverse function f −1( x).

Find the inverse function of f ( x) =

Find the equation of the inverse function of each one-to-one function over the given domain: a

, x ≥ −4.

y = x2 + 7x + 12 for the domain x ≥ −3

y = x2 − 4x − 8 for the domain x ≤ 2

Extend your thinking 16

Given that f ( x) = 2x + k and its inverse f −1 ( x) =

A function f ( x) has an inverse f −1 ( x) =

Given f ( x) = x3 − 5, ﬁnd f −1(3) without determining the full inverse function.

Find the inverse function of f ( x) = ex − e−x.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

, ﬁnd the value of k.

. Find f ( x) by determining the inverse of f −1( x).

16.04E Graphs of functions and their inverse functions After this lesson, you will be able to… • recognise that the graphs of a function and its inverse function are reflections of each other in the line y = x. • compare the graphs of a function and its inverse function using graphing applications. • establish that the domain of f −1( x) is the range of f ( x). • establish that the range of f −1( x) is the domain of f ( x). • use the relationship between domain and range of f and f −1 to solve problems. • graph the inverse function of a given one-to-one function, often by reflection or by first finding its equation.

Graphs of functions and their inverses The graph of an inverse function f −1( x) relates to the graph of the original function f ( x) through a specific geometric property: they are reflections of each other across the line y = x. The domain of the inverse function f −1( x) is the range of the original function f ( x), and the range of f −1( x) is the domain of f ( x). For this relationship to hold, the original function must be one-to-one, ensuring that each output corresponds to exactly one input. Graphing tools can help visualise this reflective symmetry and confirm the relationship between a function and its inverse.

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16.04E Graphs of functions and their inverse functions mathspace.co

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Example 1 Compare the graphs of f ( x) = 3x − 1 and its inverse function using a graphing application.

Apply the idea Most graphing applications will allow you to do the following: 2. Graph the line y = x.

1. Graph the function f ( x) = 3x − 1.

4 3

f (x)

1 −4 −3 −2 −1 −1

x 1

−4 −3 −2 −1 −1

−2

−3

−4

x 1

3. Reflect the graph of f ( x) across the line y = x. This creates the graph of the inverse function.

4 3

f (x)

2 1 −4 −3 −2 −1 −1

f −1(x) 1

x 4

−2 −3 −4

The graph of f ( x) = 3x − 1 is a steep upward line, crossing the y-axis at −1 and the x-axis at . Its inverse is a gentler line, crossing the y-axis at inverse, with interchanged intercepts.

890

and the x-axis at −1. f ( x) rises faster than its

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 For f ( x) = f −1(3).

, establish the domain and range of its inverse, and use this to ﬁnd the value of

Create a strategy Determine the domain and range of f ( x), then interchange them for f −1( x). Solve for the inverse and evaluate at x = 3.

Apply the idea For f ( x) =

Domain: x ≥ −2 (since the expression under the square root must be non-negative). Range: y ≥ 0 (since the square root outputs non-negative values). Thus, for f −1( x): Domain: x ≥ 0 (range of f ( x)). Range: y ≥ −2 (domain of f ( x)). Find the inverse: Write the function

Interchange x and y

Square both sides

Solve for y

−1

So, f ( x) = x − 2, with domain x ≥ 0. Evaluating the inverse function at x = 3: f −1( x) = x2 − 2 −1

f (3) = 3 − 2 =7

Write the inverse function Substitute x = 3 Evaluate

Reﬂect and check The calculation f −1(3) = 7 is confirmed by verifying that f (7) = 3. Substituting x = 7 into the original function: f (7) =

= 3. This confirms the inverse was calculated correctly.

Example 3 Graph the inverse function of f ( x) = ln( x − 1).

Create a strategy Find the inverse algebraically, determine its domain and range, and plot it, verifying the reﬂection over y = x.

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Apply the idea Find the inverse of f ( x) = ln( x − 1): y = ln( x − 1)

Write the function

x = ln( y − 1)

Interchange x and y

Exponentiate both sides

e =y−1 x

y=e +1 −1

Solve for y

So, f ( x) = e + 1. Domain of f ( x) : x > 1 (since x − 1 > 0). Range of f ( x): all real numbers. Thus, domain of f−1 ( x): all real numbers, range: y > 1. 4

f −1 (x)

2 1

−4 −3 −2 −1 −1 −2

y=x

x 1

f (x)

−3 −4

The inverse graph is an exponential curve, reﬂecting the logarithmic shape of f ( x) over y = x.

Idea summary The graphs of a function and its inverse function are reflections across y = x, a relationship that can be explored and confirmed using graphing applications. The domain of f −1( x) is the range of f ( x), and the range of f −1( x) is the domain of f ( x), a key relationship for graphing and problem-solving.

16.04E Practice questions What do you remember? 1

What does it mean if two functions f ( x) and g( x) are reﬂections over the line y = x?

Determine whether each statement is true or false:

892

A graphing application can be used to compare a function and its inverse by reﬂecting the function’s graph over y = x.

The domain of f −1( x) is unrelated to the range of f ( x).

To graph an inverse function, the original function must be one-to-one.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

What is the relationship between the domain of a function f ( x) and the range of its inverse f −1( x)?

Practice Ex 1

Compare the graphs of each function and its inverse. Use a graphing software to sketch both function: a

Ex 2

Ex 3

f ( x) = 4x + 2 x+3

f ( x) = x3 − 1

f ( x) = 2e

f ( x) = x5 + 2

f ( x) = e2x − 4

f ( x) = ( x − 1)3 + 3

For f ( x) =

State the domain and range of f ( x).

State the domain and range of f −1( x).

Find f −1(5).

Graph the inverse of f ( x) = ln( x + 3) using a Cartesian plane. Sketch both f ( x) and f −1( x) with y = x as a reference.

For f ( x) = ex−1:

Find the domain and range of f ( x) and f −1( x).

Find f −1(e2).

Graph the inverse of f ( x) =

, x ≥ −5, on a Cartesian plane. Include y = x for reference.

Extend your thinking 9

A function f ( x) has an inverse with domain x > 2 and range y > −1. If f −1(5) = 3, what is f (3)? Explain how the domain-range relationship helps.

Given f −1( x) = ex+2, graph both f ( x) and f −1( x) on a Cartesian plane. What can you infer about f ( x)’s shape?

For a function f ( x) with inverse f −1( x), the graph of f ( x) passes through (0, 4). Without knowing the equation, what point must lie on the graph of f −1( x)? Use this to deduce a possible domain and range for f −1( x) if f ( x) is increasing.

Let f ( x) = e2x + 3 − e−2x. Solve f ( x) = 2 ﬁrst by ﬁnding the equation of the inverse function, then plot the solution and both graphs on the same axes using graphing software to visualise the relationship.

16.04E Graphs of functions and their inverse functions mathspace.co

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16.05E The horizontal line test After this lesson, you will be able to… • explain that the reflection of a graph in the line y = x exchanges horizontal and vertical lines. • state the horizontal line test for determining if a function is one-to-one. • apply the horizontal line test to the graph of y = f ( x) to determine if its reflection in y = x (the inverse relation) is a function. • recognise that if a function passes the horizontal line test, it is one-to-one and has an inverse function. • recognise that if a function fails the horizontal line test, it is many-to-one and does not have an inverse function over its natural domain.

The horizontal line test An inverse function f −1, exists only if the original function f is one-to-one, meaning each output corresponds to exactly one input. Such functions are called invertible. Geometrically, the inverse is the reflection of the function across the line y = x. This reflection interchanges the roles of horizontal and vertical lines: a horizontal line y = c in the original graph becomes a vertical line x = c in the inverse, and vice versa. This property allows us to test whether the reflection is a function using the horizontal line test. The horizontal line test checks if a function is one-to-one. If any horizontal line intersects the graph of y = f ( x) more than once, the function is many-to-one, and its reflection across y = x will not be a function. If no horizontal line intersects the graph more than once, the function is one-to-one and invertible.

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Example 1 Determine whether the function f ( x) = 2x − 3 has an inverse function using the horizontal line test.

Create a strategy Apply the horizontal line test by imagining a horizontal line sweeping across the graph. Check if it intersects the graph at more than one point.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea

Reﬂect and check

The function f ( x) = 2x − 3 is linear and has a constant slope. Any horizontal line y = c intersects the graph exactly once because the function increases steadily without repeating outputs.

Reflecting across y = x interchanges horizontal and vertical lines. A horizontal line y = 1 intersects at (2, 1), becoming a vertical line x = 1 in the inverse, intersecting at (1, 2). The reflection is a function, confirming invertibility.

y 4

y=1

x=1

2 1

−4 −3 −2 −1 −1

x 1

y=1

−4 −3 −2 −1 −1

−2

−3

−4

x 1

Since it passes the horizontal line test, f ( x) = 2x − 3 is one-to-one and has an inverse function.

Example 2 Determine whether the function f ( x) = ( x − 4)2 + 1 has an inverse function over its natural domain using the horizontal line test.

Create a strategy Use the horizontal line test to check if the function is one-to-one over its natural domain, x ∈ (−∞, ∞).

Apply the idea The function f ( x) = ( x − 4)2 + 1 is a quadratic with a vertex at (4, 1). A horizontal line above y = 1 ( for example y = 2) intersects the parabola at two points, indicating it is many-to-one.

y 5 4 3

y=2

2 1

Since it fails the horizontal line test, it does not have an inverse function over its natural domain.

x 0

16.05E The horizontal line test mathspace.co

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Reflect and check Reflecting across y = x, a horizontal line y = 2 intersects at two points, becoming a vertical line x = 2 intersecting the reflection at two points—not a function. y 5 4 3 2

y=2

x=2 0

x 4

Idea summary An inverse function exists only for one-to-one functions, tested using the horizontal line test. Since reflecting a graph across the line y = x interchanges horizontal and vertical properties, the horizontal line test on the original function, f ( x), is equivalent to the vertical line test on its reflection. This is why the horizontal line test determines if a function is invertible.

16.05E Practice questions What do you remember? 1

What is a one-to-one function, and why must a function be one-to-one to have an inverse?

How does the horizontal line test use the reﬂection over y = x to check if a function has an inverse?

What does it mean if a function fails the horizontal line test?

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Practice 4

Use the horizontal line test on each graph to determine if it has an inverse function over its natural domain. Justify with a speciﬁc horizontal line example. a

14 12

2 −8 −6 −4 −2 −2

x 2

8 6

−4

−6

−8

x 1

−3

3 2

−5

−6

−4 −3 −2 −1

−7

f 4

x 1

−1

−4 −3 −2 −1 −1

−2

−4

−4 −3 −2 −1 −1

−2

Ex 1

−8 −6 −4 −2

y −4 −3 −2 −1 −1

x 1

−2

−3

−4

Determine whether each function has an inverse using the horizontal line test: a

f ( x) = 4x − 1

f ( x) = −2x + 5

f ( x) =

− 2

f ( x) = −3x + 1

16.05E The horizontal line test mathspace.co

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Ex 2

Determine whether each function has an inverse using the horizontal line test: a c

f ( x) = ( x − 3)2 + 2.

f ( x) = x3 + 2

f ( x) =

Use a table of values to apply the horizontal line test and determine if each function is oneto-one over its natural domain. Justify with a speciﬁc output value. f ( x) = ln( x2 + 1)

f ( x) = x5 − x

Determine if each function is one-to-one by analysing its increasing or decreasing behaviour and applying the horizontal line test. Justify with a speciﬁc horizontal line. f ( x) = ex − x

f ( x) = x2 + x

Compare the one-to-one status of each pair of transformed functions using the horizontal line test. Justify with a speciﬁc horizontal line. f ( x) = 2ex + 1 vs. g( x) = e2x + 1

f ( x) = x3 + x vs. g( x) = x3 − x

Use the horizontal line test on each graph to determine if it has an inverse function over its natural domain. Justify with a speciﬁc horizontal line example. a

f ( x) = ( x + 1)2 − 2

f ( x) = 2x

a 11

f ( x) = x − 6x + 5

a 10

Determine whether each function has an inverse using the horizontal line test:

a 9

f ( x) = ( x − 2)2 + 3

f ( x) =

f ( x) = x4 − 2x2

Determine if each piecewise function is one-to-one using the horizontal line test. Justify with a speciﬁc horizontal line. a

f ( x) =

Extend your thinking 13

Without using technology, determine whether each composite function is one-to-one. Use the x3 graph to explain how the transformation affects its behaviour, justifying with a speciﬁc horizontal line. a

f ( x) =

f ( x) = ( x3)2

Why does a linear function with a non-zero slope always have an inverse, using f ( x) = 5x − 2 as an example?

Can a function have an inverse if its graph is symmetric about the y-axis? Explain using f ( x) = x4 − 2.

Why do strictly monotonic functions always pass the horizontal line test, using f ( x) = ln( x + 1) as an example?

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16.06E Domain restrictions After this lesson, you will be able to… • understand why many-to-one functions do not have inverse functions over their natural domains. • explain the purpose of restricting the domain of a function to make it one-to-one. • identify suitable restricted domains for common many-to-one functions (e.g., parabolas) that make them one-to-one while preserving their range. • apply the horizontal line test to a restricted portion of a graph to confirm it is one-to-one. • find the inverse function for a function whose domain has been appropriately restricted.

Domain restrictions Many functions are many-to-one, meaning multiple inputs produce the same output. The inverse of such a function is not a function itself. To create a valid inverse function, f -1, the natural domain of the original many-to-one function, f, can be restricted. A function is one-to-one if each output corresponds to exactly one input. By restricting the domain, this property is ensured, enabling the inverse to map each output to a unique input. The restricted domain must be chosen carefully so that the range of the restricted function is the same as the range of the original, unrestricted function. Geometrically, the inverse is a reﬂection over y = x. For the reﬂection to be a function, the original function must pass the horizontal line test after restriction, ensuring no horizontal line intersects the graph more than once.

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 State a suitable restricted domain for each function so it is one-to-one and the range remains the same. a f ( x) = −( x − 3)2

Create a strategy Sketch the graph to identify where it’s many-to-one, then choose a domain that includes only one half of the parabola, preserving the range.

16.06E Domain restrictions mathspace.co

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Apply the idea The function is a parabola with vertex at (3, 0), opening downward. Its natural domain is x ∈ (−∞, ∞), and range is y ∈ (−∞, 0]. It’s many-to-one ( for example f (2) = f (4) = −1). 1

x 1

−1 −2 −3 −4 −5 −6 −7 −8 −9

Restrict to x ≥ 3 (right half). The range remains (−∞, 0], and it becomes one-to-one as it decreases from the vertex. 1 −1 −2 −3 −4 −5 −6 −7 −8 −9

x 1

Reflect and check With x ≥ 3, a horizontal line like y = −1 intersects once (at x = 4), unlike the unrestricted case (two intersections at x = 2 and x = 4). Reflecting over y = x gives a function, confirming invertibility. 9 y 8 7 6 5 4 3 2 1 −6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

x 1 2 3 4 5 6

Note that restricting the domain to x ≤ 3 is another valid choice.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b f ( x) =

Create a strategy Graph the function to identify its shape and symmetry, then restrict the domain to one side of the symmetry axis to make it one-to-one while preserving the range.

Apply the idea

Reﬂect and check

The function involves a square root, so x2 − 4 ≥ 0, giving a natural domain of x ∈ (−∞, −2] ∪ [2, ∞).

With x ≥ 2, y = intersects once (at x = 3), not twice as in the unrestricted case (at x = ±3). The restricted function’s reﬂection over y = x is a function, verifying it’s one-to-one.

The output is negative due to the minus sign, with a maximum at x = ±2 where f (2) = f (−2) = 0. Range is y ∈ (−∞, 0]. It’s manyto-one ( for example f (3) = f (−3) = . 4

4 3 2

−4 −3 −2 −1 −1

2 1

−4 −3 −2 −1 −1

x 1

−2 −3 −4

Restrict to x ≥ 2 (right branch). The range remains (−∞, 0], and it’s now one-to-one, decreasing from x = 2. 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

16.06E Domain restrictions mathspace.co

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Example 2 Find the inverse of f ( x) = ∣x + 4∣ with a restricted domain of x ≥ −4.

Create a strategy Verify the restriction makes it one-to-one, then interchange x and y and solve, adjusting for the absolute value based on the domain.

Apply the idea Natural domain is x ∈ (−∞, ∞), range is [0, ∞). Unrestricted, it’s many-to-one ( for example f (−5) = f (−3) = 1). With x ≥ −4, it’s the right arm, increasing, and one-to-one. y = ∣x + 4∣

Write the equation

y=x+4

Since x ≥ −4, x + 4 ≥ 0, so ∣x + 4∣ = x + 4

x=y+4

Interchange x and y

y=x−4

Solve for y

−1

Thus, f ( x) = x − 4, with domain x ∈ [0, ∞) (range of original).

Reﬂect and check The restricted f ( x) = x + 4 for x ≥ −4 has range [0, ∞). The inverse y = x − 4 for x ≥ 0 is one-to-one, and its range [−4, ∞) matches the restricted domain, conﬁrming correctness. 5 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1

Idea summary Many-to-one functions lack inverses unless their domain is restricted to make them one- to-one. Restricting the domain preserves the range while ensuring the function passes the horizontal line test, allowing an inverse function to exist. Sketching the graph helps to identify a suitable domain restriction. This restriction, reﬂected over y = x, yields a valid inverse function. Using the function’s vertex or other turning points as boundaries for the new domain is an effective strategy for creating a valid restriction.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16.06E Practice questions What do you remember? 1

What does it mean for a function to be many-to-one, and why does it prevent an inverse function from existing?

Why must we restrict the domain of a function to make it one-to-one while keeping the range the same?

How does the horizontal line test help identify a suitable restricted domain for a function to have an inverse, and why is reﬂection over y = x important in this process?

Practice Ex 1

State a suitable restricted domain for each function so it is one-to-one and the range remains the same. Sketch the unrestricted and restricted graphs to justify your answer. a

Ex 2

f ( x) =

Find the inverse of each function with the given restricted domain. Verify the range of the inverse matches the restricted domain of the original. a

f ( x) = −( x + 2)2

f ( x) = x2 − 5, x ≥ 0

f ( x) = −∣x − 1∣, x ≤ 1

For many functions, we can determine an inverse relation by ﬁrst breaking up the original function into parts that are one-to-one, and then ﬁnding the inverse of each part separately. a

Complete the function below to break f ( x) = ( x − 2)2 − 6 into two one-to-one functions that have the same rule as f ( x). f ( x) =

Consider the graph of y = g ( x) shown: 5 4 3 2 1

−6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5

x 1

9 10 11

Complete the function below to break the graph of y = g ( x) into three one-to-one functions that have the same rule as g ( x). f ( x) =

16.06E Domain restrictions mathspace.co

903

Find an appropriate restricted domain for the function f ( x) = ( x − 7)2 + 12 to have an inverse.

Consider the function f ( x) =

deﬁned over [0, 4].

( x).

Find f

Find the domain of f −1 ( x).

Find the range of f −1 ( x).

The largest domain over which the function f ( x) = x2 + bx + c has an inverse is [3, ∞). The domain of the inverse function f −1 ( x) is [−2, ∞). a

−1

Find the value of b.

Find the value of c.

For each of the following functions: i

Use technology to sketch the function f ( x) over its domain.

Use technology to sketch the function f −1 over its domain.

iii

State the domain of f −1.

State the range of f −1.

f ( x) = x + 3 deﬁned over the interval [0, ∞).

f ( x) = 7 − x deﬁned over the interval [2, 9].

f ( x) = ( x − 6)2 − 2 deﬁned over the interval [6, ∞).

f ( x) =

deﬁned over the interval [0, 4). 2

f ( x) = ( x + 2) + 3 deﬁned over the interval [0, ∞).

Extend your thinking 11

Explain why restricting the domain of f ( x) = x4 to x ≥ 0 preserves the range and makes it oneto-one. What happens if you restrict it to x ≥ 1 instead?

Can a function with an inﬁnite range (e.g., f ( x) = x2 + x) be restricted to be one-to-one while preserving its range? Justify with a sketch.

An encryption tool takes in a 10-digit code, n, and puts it into the formula K = (n − a)2 + b to create an encrypted key, K. It is important that when retrieving the code from the key, there is only one possible value for n. Determine whether the following restrictions would create a one-to-one function to ensure the correct code is returned: a

904

K>b

n>a

n>b

K>a

Consider f ( x) = x2 + ax − 40: a

Find a domain that allows f ( x) to have an inverse and preserves the range.

Find the inverse equation.

Sketch the graph for both cases at a = 6.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16.07E Solve problems involving a function and its inverse function After this lesson, you will be able to… • solve equations of the form f ( x) = k by finding x = f −1(k). • understand that points of intersection of y = f ( x) and y = f −1( x) often lie on the line y = x. • determine points of intersection of a function and its inverse by solving f ( x) = x or f ( x) = f −1( x). • apply these algebraic and graphical techniques to solve contextual problems involving inverse functions. • justify solutions and interpretations in the context of the problem.

Solve problems involving a function and its inverse function An inverse function f −1( x) “undoes” the effect of a function f ( x), mapping outputs back to their original inputs. This relationship allows us to solve problems by leveraging both f ( x) and f −1( x) algebraically and graphically. Algebraically, the inverse can solve equations like f ( x) = k by applying x = f −1(k), assuming f is one- to-one. Graphically, the inverse reflects f ( x) over y = x, and their points of intersection (if any) occur where f ( x) = f −1( x). For an increasing function, these points must lie on the line y = x, so the equation can be simplified to f ( x) = x. These intersections represent points where the function equals its input, often found on the line y = x, and can reveal symmetry or fixed points in the function’s behaviour. Problems may involve finding such points, solving for specific values, or interpreting real-world contexts using this relationship.

Example 1 Solve these problems using the relationship between a function and its inverse. a Given f ( x) = 3x + 2, ﬁnd the value of x such that f ( x) = 11 using its inverse.

Create a strategy Find the inverse f −1( x) and use it to solve x = f −1(11) algebraically, then verify graphically.

16.07E Solve problems involving a function and its inverse function mathspace.co

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Apply the idea Find the inverse: Write the function

Interchange x and y

Subtract 2 from both sides

Divide both sides by 3 making y the subject So, f −1 ( x) =

. Solving f ( x) = 11: Use the inverse

Substitute x = 11

Evaluate the subtraction

Evaluate

Substitute x = 3 into f ( x) to check: f ( x) = 3x + 2

Write the function

f (3) = 3 × 3 + 2

Substitute x = 3

= 11

Evaluate

Reﬂect and check Graphically, plot f ( x) = 3x + 2 and y = 11. They intersect at (3, 11), conﬁrming x = 3. 14 y 13 12 11 10 9 8 7 6 5 4 3 2 1

x 1

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Find the points where f ( x) = x3 intersects its inverse.

Create a strategy Determine the inverse, set f ( x) = f −1( x), solve algebraically, and conﬁrm graphically by ﬁnding intersections with y = x.

Apply the idea Find the inverse: Write the function Interchange x and y Take the cube root of both sides making y the subject So, f −1 ( x) =

. Set f ( x) = f −1( x): Use the inverse

Rewrite as power of

Subtract

Factor out

Use null factor law

For the second factor,

from both sides

− 1 = 0: Write the equation

Add 1 to both sides

Apply the exponent rule (am)n = amn

Take the eighth root of both sides

Evaluate the root

Cube both sides to solve for x

Evaluate

Thus, x = 0, x = 1 and x = −1.

16.07E Solve problems involving a function and its inverse function mathspace.co

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For x = 0: Write the function Substitute x = 0 Evaluate Write the inverse function

Substitute x = 0

Evaluate

For x = 1: Write the function Substitute x = 1 Evaluate Write the inverse function

Substitute x = 1

Evaluate

For x = −1: Write the function Substitute x = −1 Evaluate Write the inverse function

Substitute x = −1

Evaluate

Points are (0, 0), (1, 1), (−1, −1).

Reﬂect and check Graphically, f ( x) = x3 and f −1 ( x) =

intersect at these points, all on y = x, as f ( x) = x holds there. y 1 x −1

1 −1

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 A ball’s height in metres is modelled by h(t) = 20t − 5t2, where t is time in seconds after being thrown upward. Find when the ball reaches 15 metres using its inverse.

Create a strategy Restrict the domain to the upward phase, ﬁnd the inverse, solve t = h−1(15), and verify with the graph.

Apply the idea The parabola opens downward, vertex at t = 2 (max height 20 m). For the upward phase, restrict to t ∈ [0, 2]. Range is [0, 20]. Find the inverse: Write the equation for h

Interchange t and h

Rearrange

Quadratic formula

Simplify

The inverse function will have a range of [0, 2] to match the restricted domain. Choose the sign in that satisfies this. The branch

gives outputs less than or equal to 2, so

this is the correct inverse for the upward phase. The domain of h−1 is [0, 20]. Solve h(t) = 15: Write the equation

Substitute h−1(t) = 2 −

Evaluate the operations inside the square root

Evaluate the square root

Evaluate

Substitute t = 1 into h(t) to check: h(t) = 20t − 5t2

Write the equation 2

h(1) = 20 × 1 − 5 × 1

Substitute t = 1

= 20 − 5

Evaluate the power and multiplication

= 15

Evaluate

16.07E Solve problems involving a function and its inverse function mathspace.co

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Reﬂect and check Graphically, h(t) = 15 intersects at t = 1 (and t = 3 on descent), but t = 1 ﬁts the restricted domain. h 20 15 10 5 t 1

Idea summary Solving problems with a function and its inverse function uses their relationship to ﬁnd inputs from outputs or intersection points. Algebraically, f −1( x) solves f ( x) = k, while graphically, intersections occur where f ( x) = f −1( x), often on y = x. Both techniques reveal key points and real-world solutions, leveraging the inverse’s ability to reverse the original function.

16.07E Practice questions What do you remember? 1

How can the inverse function f −1( x) be used to solve an equation of the form f ( x) = k?

Where do the graphs of a function f ( x) and its inverse f −1( x) intersect, and how can this be found algebraically?

Practice Ex 1

910

Solve the following problems using the relationship between a function and its inverse, both algebraically with the inverse and graphically. a

Given f ( x) = 5x − 4, determine the value of x such that f ( x) = 6.

Determine the points where f ( x) =

intersects its inverse.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 2

A ball is thrown upward, and its height in metres is given by h(t) = 16t − 4t2, where t is time in seconds. Find the time when the ball reaches 12 metres on its upward path using its inverse, and sketch the graph to verify.

For each function, ﬁnd the points of intersection with its inverse using algebraic and graphical methods. a

f ( x) = , x ≠ 0

A temperature conversion function is C(F ) =

f ( x) = 2x − 3

(F − 32), where F is Fahrenheit and C is

Celsius. Find the temperature where Celsius and Fahrenheit scales are equal using the inverse function, and verify graphically. 7

Prove that the points of intersection of a linear function f ( x) = mx + b (where m ≠ 1) and its inverse lie on the line y = x. Find the intersection point in terms of m and b.

Extend your thinking 8

The function h(t) = 120 − 4.9t2 models the height h (in metres) of an object t seconds after being dropped from a height of 120 m. a

Determine whether or not the function has an inverse that is a function. Justify your answer.

Zheng rearranges the function to get t = inverse function is t ( x) =

. He claims that this means the

State whether or not Zheng’s claim is correct. c 9

Determine how long it will take for the object to reach a height of 41.6 m.

A signal processing system encodes an input signal x using the function f ( x) = x2 − 2x, where the input signal must be x ≤ 1. The system requires a unique input for each output to decode correctly. a

Find the input signal x that produces an output of y = 3. Solve algebraically using the inverse function and verify graphically.

Determine the point where f ( x) intersects its inverse, using algebraic and graphical methods.

Consider f ( x) = x3 + 2x, where −2 ≤ x ≤ 2: a

Using the formal deﬁnition of the inverse, show

Calculate g′(3), where g( x) =

16.07E Solve problems involving a function and its inverse function mathspace.co

911

16 Chapter review 1

If the point (3, −4) lies on the graph of a one-to-one function g( x), which point must lie on the graph of g−1( x)? A

(3, 4)

(−3, −4)

(−4, 3)

(4, −3)

What is the inverse of the function h( x) = 3x + 5? A

h−1( x) =

h−1 ( x) =

h−1( x) = 3x − 5

h−1 ( x) =

−5

Determine if f ( x) = 0.4x + 1.5 has an inverse function. Verify by sketching f ( x), y = x, and f −1( x) with two points and their reﬂections.

Verify using y = x if each pair of lines are inverses of each other. a

b 4

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

−4 −3 −2 −1 −1

x 1

−4 −3 −2 −1 −1

−2

−3

−4

Explain why g( x) = ( x − a)3 + b always has an inverse for any real a and b.

Verify that f ( x) = 2x − 7 and f −1 ( x) =

Verify that f −1( x) = ex + 5 is the inverse of f ( x) = ln( x − 5), for x > 5 using formal deﬁnition.

912

are inverses using f ( f −1( x)) = x and f −1( f ( x)) = x.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Prove that h( x) = , k ≠ 0, x ≠ 0, is its own inverse using formal deﬁnition.

Prove that the inverse of g( x) = gradient and y-intercept.

Find the inverse function for:

, b ≠ 0, is linear using formal deﬁnition. State its

f ( x) = x3 + 7

g( x) =

, x ≠ −3

h( x) = 2x + 5

k( x) =

,x≥1

Find the inverse function for: a

y = ln( x + 4), x > −4

y = 4e2x − 3

h( x) = ex + 2

k( x) = ln(3 − x), x < 3

For each function: i

Find f −1( x).

Describe how f −1( x) undoes f ( x).

f ( x) =

, x ≠ 1

f ( x) = ex−2 − 1

f ( x) = ( x + 2)3 − 4

f ( x) =

Find the inverse of k ( x) =

For f ( x) =

+ 1, x ≥

− 1:

Determine the domain and range of f ( x).

Determine the domain and range of f −1( x).

Find f −1(2).

Graph the inverse of f ( x) = ln( x − 3) + 1 with f ( x), f −1( x), and y = x.

For a one-to-one function h( x) with h−1( x) having domain x ≥ 7 and range y < −2, if h−1(10) = −5, ﬁnd h(−5).

Determine whether each function passes the horizontal line test over its natural domain: a

f ( x) = 5x − 1

g( x) = −2x + 7

h( x) = x2 + 3

k( x) = ex − 2

m( x) = ln( x), x > 0

n( x) = , x ≠ 0

Determine if h( x) = ( x + 4)2 − 2 has an inverse over its natural domain using the horizontal line test.

Determine if f ( x) = ( x3 + 2)2 is one-to-one over its natural domain.

For f ( x) = ( x − 3)2 + 1:

State a restricted domain to make it one-to-one with range [1, ∞).

Sketch the unrestricted and restricted function.

Find the inverse of f ( x) = x2 − 4, x ≥ 0. Verify the range of f −1( x).

Chapter 16 review mathspace.co

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Explain why f ( x) = ∣x + 3∣ with domain x ≥ −3 is one-to-one and preserves range [0, ∞). Compare with domain x ≥ 0.

Can f ( x) = x2 − 5x + 6 be restricted to be one-to-one while preserving its full range?

For C(t) =

Find the intersection of f ( x) = cx + d, c ≠ 1, and f −1( x).

For K(v) = 2.5v2, v ≥ 0:

, t ≥ 1, ﬁnd when C(t) = 4 using the inverse.

Calculate v(K).

Determine the speed when K = 250 Joules.

Find g′(−3) for f ( x) = x3 − x − 3, −0.5 < x < 0.5, g( x) = f −1( x).

Did you know?

Astronauts and scientists use advanced transformations to align satellite and space station cameras with Earth’s curved surface. These mathematical adjustments ensure that every image taken from orbit matches the planet’s true position and orientation. Transformations like rotations, translations, and projections allow engineers to map space data accurately, even as both Earth and the spacecraft move. Without these precise calculations, our stunning satellite views — and even GPS accuracy — wouldn’t be possible.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

“Mathematics is a creative art that requires imagination as much as reason.” Mary Cartwright

Big ideas The graph of a function can be systematically altered through a set of fundamental transformations, translations (shifts), reflections (flips), and dilations (stretches/ compressions), each corresponding to a specific algebraic modification of the function’s rule, with the final graph depending on the sequence in which these transformations are applied.

17 Transformations Chapter outline 17.01 17.02 17.03 17.A 17.04 17.05 17.06 17.07 17.08

918 924 930 939 952 960 967 978 986

Shifting furniture in your room is literally translating it across the floor.

17.01 Reflections in axes After this lesson, you will be able to… • describe the effect of replacing x with −x in y = f (x) as a reflection in the y-axis • describe the effect of replacing y with −y in y = f (x) as a reflection in the x-axis • determine the equation of a function after a reflection in the x- or y-axis • sketch a function and its reflection, identifying key points and features

Reflection in the y-axis Reflection A transformation of a shape formed by creating a mirror image on the other side of a given line. A reflection in the y-axis flips a graph horizontally, mapping each point (x, y) to (−x, y). Replacing x with −x in the function y = f (x) gives its horizontal reflection:

y = f (−x) f (− −x)

is the graph of y = f (x) reﬂected in the y-axis

For example, if y = x, then y = −x reﬂects the line in the y-axis. 4

y = −x

3 2 1

−4 −3 −2 −1

−1

x 1

The graph of y = −x is the reﬂection of y = x in the y-axis.

−2

y=x

−3 −4

Exploration Consider the function y = f (x) and its reﬂection y = f (−x): 1. Why do points on the y-axis remain unchanged after a y-axis reﬂection? 2. How does this relate to the symmetry of even functions?

918

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 1 For the function y = x3: a Find the equation of the graph after reﬂection in the y-axis.

Create a strategy Substitute x with −x.

Apply the idea y = x3

Write the function

= (−x) = −x

Substitute x = −x Simplify

The reﬂected graph is y = −x3.

b Sketch both graphs, showing the key points.

Apply the idea To sketch the graph, it is often best to choose simple x-values such as −1, 0, and 1, and substitute them into the equation: For y = x3: • If x = −1, then y = (−1)3 = −1. • If x = 0, then y = 03 = 0. • If x = 1, then y = 13 = 1. This gives the key points: (−1, −1), (0, 0) and (1, 1). For y = −x3: • If x = −1, then y = −(−1)3 = −(−1) = 1. • If x = 0, then y = −(0)3 = 0. • If x = 1, then y = −(1)3 = −1. This gives the key points: (−1, 1), (0, 0) and (1, −1). y (−1, 1) 3

(1, 1)

y = −x

x −1

(0, 0)

The graph of y = −x3 reﬂects y = x3 in the y-axis.

y = x3 −1 (−1, −1)

(1, −1)

17.01 Reflections in axes mathspace.co

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Idea summary Replacing x with −x in the function y = f (x) gives y = f (−x), which results in a horizontal reflection of the graph in the y-axis. This transformation maps each point (x, y) to (−x, y).

Reflection in the x-axis A reflection in the x-axis flips a graph vertically, mapping each point (x, y) to (x, −y). Replacing y with − y in the function y = f (x) gives its vertical reflection:

y = −f (x) −f (x)

is the graph of y = f (x) reﬂected in the x-axis

For example, if y = x2, then y = −x2 reﬂects the parabola downward.

Interactive exploration Discover this concept in action online

Example 2 For the function y = 2x + 1: a Find the equation of the graph after reﬂection in the x-axis.

Create a strategy Substitute y with −y.

Apply the idea y = 2x + 1

Write the function

−y = 2x + 1

Substitute y = −y to apply reﬂection

y = −(2x + 1)

Multiply both sides by −1

= −2x − 1

Simplify

The reﬂected graph is y = −2x − 1.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

mathspace.co

b Sketch both graphs, showing the key points.

Apply the idea When sketching linear graphs, only two points are needed because a straight line is completely determined by any two distinct points. A good choice is to use the y-intercept (when x = 0) and one other simple x-value, such as x = 1: For y = 2x + 1: • If x = 0, then y = 2 × 0 + 1 = 1. • If x = 1, then y = 2 × 1 + 1 = 3. This gives the key points: (0, 1) and (1, 3). For y = −2x−1: • If x = 0, then y = −2 × 0−1 = −1. • If x = 1, then y = −2 × 1−1 = −2−1 = −3. This gives the key points: (0, −1) and (1, −3). y 3 2 1 −1 −1 −2 −3

(1, 3)

y = 2x + 1 (0, 1) (0, −1)

x 1

The graph of y = −2x − 1 reﬂects y = 2x + 1 in the x-axis.

y = −2x − 1 (1, −3)

Idea summary Replacing y with −y in the function y = f (x) gives y = −f (x), which results in a vertical reflection of the graph in the x-axis. This transformation maps each point (x, y) to (x, −y).

17.01 Reflections in axes mathspace.co

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17.01 Practice questions What do you remember? 1

Identify the effect of these transformations on the graph of a function y = f (x): a

Replacing y with −y

How does the point (x, y) map under these reﬂections? a

Replacing x with −x

Reﬂection in the y-axis

Reﬂection in the x-axis

For the function y = x2: a

Explain why the graph is unchanged after a reﬂection in the y-axis.

Sketch the graph and reﬂection in the y-axis, including the vertex.

Determine whether these statements are true or false: a

Reﬂecting y = x3 in the y-axis gives y = x3.

The point (2, 3) maps to (−2, 3) after reﬂection in the y-axis.

Practice Ex 1

Ex 2

For each function: i

Find the equation of the graph after reﬂection in the y-axis.

Sketch both graphs, including key points.

y = x2 − 1

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y = x3 + 2

Find the equation of the graph after reﬂection in the x-axis.

Sketch both graphs, including key points.

y = x2 + 2

y = −x − 2

y = −3x + 2

y = x3 − 1

y=x−3

y = x2 − 3

y = −x + 1

For these functions, ﬁnd the equation after reﬂecting in the y-axis: y = 3x + 2

y = x2 + 1

y = 2x3

For these functions, ﬁnd the equation after reﬂecting in the x-axis: a

y = 2x + 1

For each function:

a 8

y = 4x − 1

y = x2 − 2

y = x3 + 1

For the function y = 2x + 3: a

Find the y-intercept after reﬂecting in the x-axis.

Sketch y = 2x + 3 and its x-axis reﬂection, including the y-intercept.

For the function y = x − 2: a

Find the x-intercept after reﬂecting in the y-axis.

Sketch y = x − 2 and its y-axis reﬂection, including the x-intercept.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Determine the coordinates of the point (3, −2) after these reﬂections: a

Reﬂection in the y-axis

Reﬂection in the x-axis

Sketch the point (3, −2) and its reﬂections in the x-axis and y-axis, including all points.

Sketch the graph of y = 2π x + π and its reﬂection in the y-axis. Label the y-intercept of both graphs.

Sketch the graph of y = x2 and its reﬂection in the x-axis. Label the vertex of both graphs.

For the function y = (x + 2)2:

Find the equation after reﬂection in the x-axis.

Determine the vertex of the reﬂected graph.

Sketch the graph and its x-axis reﬂection, including the vertex.

Determine the range of each equation after reﬂecting in the x-axis: a

y = x2 + 1

y = 3x3

For each equation: a

Find the equation of the graph after reﬂecting y = 3x2 − 2x + 1 in the y-axis.

Find the equation of the graph after reﬂecting y = 2x3 + x − 1 in the x-axis.

Determine the coordinates of the point (−1, 4) after reﬂecting in both the x-axis and y-axis. Does the order affect the outcome?

Extend your thinking 18

A graphic designer creates a logo using the curve y = x2 − 4: a

Write the equation of the curve mirrored across the y-axis and determine if the x-intercepts change. Explain your reasoning.

Sketch the graph and its y-axis reﬂection, including the x-intercepts.

For the function y = x3, ﬁnd the point(s) that remain unchanged after a reﬂection in the x-axis. Explain why these points are invariant.

A physicist models a particle’s path with the function y = 2x + 3:

Find the equation of the path reﬂected in the x-axis and explain how the slope changes. Explain the physical signiﬁcance.

Sketch the graph and its x-axis reﬂection, including the y-intercept.

Consider the function y = x2 + 2x. After reflecting it in the y-axis, the graph is then reflected in the x-axis. Find the final equation of the graph and determine its vertex. Verify your answer by checking the vertex’s position relative to the original graph.

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17.02 Horizontal and vertical translations After this lesson, you will be able to… • describe the effect of replacing x by x − a as a horizontal translation • describe the effect of replacing y by y − b (or y = f (x) + b) as a vertical translation • determine the equation of a function after a horizontal or vertical translation • sketch a function and its translation, identifying the effect on key features like vertices and intercepts

Horizontal translations Translation A type of transformation that moves a shape (or all the points in a plane) by the same amount to the left or right, or up or down. A horizontal translation shifts a graph left or right. For a function y = f (x), the new function y = f (x − a) translates the graph horizontally by a units:

y = f (x − a) a moves the graph horizontally: Right by a units if a > 0 Left by a units if a < 0 For example, for f (x) = x2, the function y = f (x − 2) = (x − 2)2 shifts the parabola 2 units right, moving the vertex from (0, 0) to (2, 0). y 4 3 2

The graph of y = (x − 2)2 is y = x2 shifted 2 units right, with the vertex at (2, 0).

y = (x − 2)2

y = x2 1

(0, 0) −3 −2 −1

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(2, 0) 1

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Interactive exploration Discover this concept in action online

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Example 1 Consider the function f (x) = x2 + 1 and its graph:

y 4 3

f (x) = x2 + 1

2 1 x

−4

−3

−2

−1

a Determine the equation after a horizontal translation of 3 units left.

Create a strategy A horizontal translation of 3 units to the left is represented by the transformation y = f (x − a) where a = −3. The new function will be y = f (x + 3).

Apply the idea f (x) = x2 + 1 = (x + 3)2 + 1

Write the function Apply translation of 3 units left

The translated graph is f (x) = (x + 3)2 + 1.

b Determine the vertex of the translated graph.

Create a strategy Identify the vertex of the original graph and shift its x-coordinate by the translation value.

Apply the idea The vertex of f (x) = x2 + 1 is at (0, 1). A horizontal translation shifts the x-coordinate by −3. New Vertex = (0 − 3, 1) = (−3, 1)

Subtract 3 from the x-coordinate Simplify

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Reﬂect and check Verify by substituting x = −3 into the translated equation: f (x) = (x + 3)2 + 1

Write the equation

f (−3) = (−3 + 3) + 1

Substitute x = −3

=0 +1

Evaluate

Simplify

This conﬁrms the vertex of the translated graph is at (−3, 1).

c Sketch the original graph and its translated graph, including the vertex.

Create a strategy Plot key points of f (x) = x2 + 1 and f (x) = (x + 3)2 + 1, including their vertices, and sketch the parabolas.

Apply the idea y 4 3 2

(−3, 1)

f (x) = (x + 3)2 + 1 −4

−3

−2

(0, 1)

f (x) = x2 + 1 −1

Idea summary A horizontal translation shifts a graph left or right. For a function y = f (x), the new function y = f (x − a) translates the graph by a units horizontally: to right if a > 0 or to left if a < 0.

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Vertical translations A vertical translation shifts a graph up or down. For a function y = f (x), the translated function y = f (x) + b moves the graph vertically by b units:

y = f (x) + b b moves the graph vertically: Up by b units if b > 0 Down by b units if b < 0 For example, for f (x) = x2, the function y = f (x) + 3 = x2 + 3 shifts the parabola 3 units up, moving the vertex from (0, 0) to (0, 3). y 5

y = x2 + 3 4

y = x2

The graph of y = x2 + 3 is y = x2 shifted 3 units up, with the vertex at (0, 3).

(0, 3)

2 1 x

(0, 0) −2

−1

Example 2 For the function f (x) = 2x: a Find the equation after a vertical translation of 4 units down.

Create a strategy A vertical translation of 4 units down is represented by the transformation y = f (x) + b, where b = −4.

Apply the idea The original function is f (x) = 2x. Translate 4 units down (b = −4): y = f (x) + b

Write the formula for vertical translation

= f (x) − 4

Substitute b = −4

= 2x − 4

Substitute f (x) = 2x

The translated graph is y = 2x − 4.

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b Sketch both graphs, including the y-intercept.

Create a strategy Plot key points, including the y-intercept, for both y = 2x and y = 2x − 4, and sketch the lines.

Apply the idea y 2 1 −1 −1

y = 2x

(0, 0)

−2 −3 −4 (0, −4)

y = 2x − 4

Idea summary A vertical translation shifts a graph up or down. For a function y = f (x), the new function y = f (x) + b translates the graph by b units vertically: upwards if b > 0 or downwards if b < 0.

17.02 Practice questions What do you remember? 1

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Identify the effect of these replacements on the graph of a function y = f (x): a

Replacing x with x − 3

Replacing y with y − 2

Replacing x with x + 4

Replacing y with y + 5

Determine whether these statements are true or false: a

The graph of y = x2 + 4 is a vertical translation of y = x2 by 4 units up.

Replacing x with x − (−2) shifts the graph 2 units left.

The graph of y = (x + 3)2 is a horizontal translation of y = x2 by 3 units right.

A vertical translation changes the x-intercepts of a graph.

For the function y = x2, explain how the sign of a in y = f (x − a) determines the direction of the horizontal translation.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Practice Ex 1

Ex 2

For the function y = x2 − 2: a

Determine the equation after a horizontal translation of 4 units right.

Determine the vertex of the translated graph.

Sketch the original graph and the translated graph, labelling the vertex.

For the function y = 3x + 2: a

Find the equation after a vertical translation of 5 units up.

Sketch both graphs, labelling the y-intercept.

Find the equation of the graph after applying the given translation to y = x2: a

3 units left

4 units up

2 units down

2 units right

4 units left

3 units up

5 units down

Determine the vertex of the translated graph for y = x2 − 3 after: a

Find the equation of the graph after applying the given translation to y = 2x + 1: a

5 units right

Horizontal translation 2 units right

Vertical translation 1 unit up

For the function y = x2 + 2: a

The graph is translated to y = (x − 1)2 + 4. Describe the translations applied.

Sketch the graph and the translated graph, labelling the vertex.

The graph of y = x3 is translated to y = (x + 2)3 − 1: a

Describe the translations and ﬁnd the point corresponding to (0, 0) on the original graph.

Sketch the graph and the translated graph, labelling the point corresponding to (0, 0).

Determine the x-intercept(s) of the translated graph for y = x2 − 4 after a vertical translation 3 units up.

For y = 2x + 5, ﬁnd the equation after a horizontal translation 3 units right and a vertical translation 2 units down.

The vertex of y = x2 + 2 is translated to (3, 5). Find the equation of the new graph.

Identify the translations applied to y = x2 to obtain y = (x − 4)2 − 7.

Find the equation of y = x3 after a horizontal translation 2 units left followed by a vertical translation 3 units up.

Determine the new vertex of y = x2 + 2x + 1 after a horizontal translation 1 unit right and a vertical translation 2 units up.

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Extend your thinking A company’s proﬁt function is P (x) = −x2 + 4x, where x is the number of units sold:

A new model shifts the function to P (x) = −(x − 3)2 + 4x − 5. Describe the translations and ﬁnd the new maximum proﬁt.

Sketch the graph and the translated graph, labelling the vertex.

The graph of y = f (x) has a point at (2, 3). After translating to y = f (x − a) + b, the point moves to (5, 1). Find a and b.

The graph of y = x2 is translated so its vertex moves from (0, 0) to (h, k):

Write the equation of the new graph and ﬁnd its domain and range.

Sketch the graph and the translated graph for h = 2, k = 1, labelling the vertex.

A function y = x2 − 13 is translated 2 units to the right. The point A(x, y) lies on y = x2 − 13. The image of A on the transformed function is A′(−2, 3). Find the coordinates of A. A function y = 3x − 2 is translated 3 units to the right. The point A(x, y) lies on y = 3x − 2.

The image of A on the transformed function is A′(1, −8). Find the coordinates of A.

17.03 Dilations After this lesson, you will be able to… • describe horizontal and vertical dilations as stretches from an axis • determine the equation of a function after a horizontal, vertical, or combined dilation (enlargement or reduction). • sketch a function and its dilation, identifying the effect on key features • distinguish between a stretch (factor > 1) and a compression (factor between 0 and 1)

Horizontal and vertical dilations Dilation A process of stretching or compressing the graph of a function. This could happen either in the x or y direction or both.

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A horizontal dilation stretches the graph from the y-axis, scaling x-coordinates, while a vertical dilation stretches from the x-axis, scaling y-coordinates. Sketching helps visualise these changes. For a function y = f (x): • Replacing x by (kx, y). • Replacing y by maps to (x, ly).

gives y =

, a horizontal dilation by a factor of k. Each point (x, y) maps to

= f (x), or y = lf (x), a vertical dilation by a factor of l. Each point (x, y)

gives

dilates the graph horizontally by a factor of k: stretches if k > 1, compresses if 0 < k < 1

y = lf (x) dilates the graph vertically by a factor of l: stretches if l > 1, compresses if 0 < l < 1 For some functions, different dilations can result in the same graph. For example, consider f (x) = x2. A horizontal dilation by a factor of k = 2 gives

. This is the same equation as

a vertical dilation of y = x by a factor of l = . y = 2x2

y = x2

The graph of y =

k = 2) from the y-axis, and y = 2x2 is stretched upwards away from the x-axis (vertical dilation, l = 2) from the x-axis.

(1, 2)

1 y = x2 4

−3

−1 (0, 0)

−2

(1, 1) (2, 1) 1

is wider (horizontal dilation,

Compared to y = x2, the graph of y =

is wider.

When 0 < k < 1 or 0 < l < 1, the graph compresses (for example, y =

narrows horizontally).

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17.03 Dilations mathspace.co

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Example 1 For the function y = x2: a Find the equation and their graph after a horizontal dilation by a factor of 3.

Create a strategy Substitute x with .

Apply the idea Write the function

Graphing both equations: y

Substitute x =

y = x2

Simplify

The equation is y =

(2, 4)

(6, 4)

1 2 x 9

−6 −4 −2

b Find the equation and their graph after a vertical dilation by a factor of 0.5.

Create a strategy Substitute y with

Apply the idea Write the function

Substitute y =

Multiply both sides by 0.5

Graphing both equations: y

(2, 4)

y = x2 3

The equation is y = 0.5x2. y = 0.5x2

(2, 2)

2 1

x −4 −3 −2 −1

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c Describe the effect on the graph’s shape for each dilation.

Apply the idea Horizontal dilation (k = 3): Stretches each point on the parabola horizontally by a factor of 3, making the parabola 3 times as wide. The point (1, 1) is stretched outwards to (3, 1) for example. Vertical dilation (l = 0.5): Compresses each point on the parabola vertically by a factor of 0.5. Each y-value is halved, making the graph appear ﬂatter. For example, the point (1, 1) on the original parabola becomes (1, 0.5) after the dilation.

Idea summary A horizontal dilation by factor k, y =

, stretches from the y-axis, and a

vertical dilation by factor l, y = lf (x), stretches from the x-axis. Factors 0 < k < 1 and 0 < l < 1 compress the graph, while k > 1 and l > 1 stretch it.

Combined dilations A combined dilation applies both horizontal and vertical dilations simultaneously, scaling the entire graph. For y = f (x), replacing x by a factor of k.

and y by

gives

, or

, dilating the graph by

dilates the graph by a factor of k: enlarges if k > 1, reduces if 0 < k < 1 To illustrate the enlargement, consider the function f (x) = x2 over the domain −1 < x < 1. A geometric enlargement of factor 2 maps each point (x, y) on the graph to (2x, 2y), scaling both the x- and y-values. y 3

y=2

x 2

y = x2

(2, 2)

2 1

(1, 1) x

−2

−1

For example, the point (1, 1) becomes (2, 2), as shown on the graph. The original domain −1 < x < 1 becomes −2 < x < 2, and the range, originally 0 < y < 1, becomes 0 < y < 2, making the graph larger in both width and height.

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Exploration For y = f (x), the combined dilation y = kf

scales both axes by k:

1. Why does this result in uniform scaling? 2. How does the graph’s appearance differ when k > 1 versus 0 < k < 1?

Example 2 For the function y = x3: a Find the equation after an enlargement by a factor of 2.

Create a strategy Substitute x with

and y with .

Apply the idea Write the function Substitute y =

and x =

Multiply both sides by 2

Evaluate the power

Simplify The equation is

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Sketch both graphs, labelling a key point on the original and its corresponding point on the transformed graph.

Apply the idea The key point for y = x3 is (1, 1). Since the enlargement factor is 2, the transformed x-value corresponding to x = 1 is x = 1 × 2 = 2. Thus, the transformed key point lies at x = 2 on the graph of

Write the function Substitute x = 2

Evaluate the power

Evaluate is (2, 2).

So, the key point for y 2 1

(2, 2) (1, 1) x

−2

1 y = x3 4

−1

The graph of

is an enlargement of y = x3 by a

factor of k = 2, mapping (1, 1) to (2, 2).

−1 −2

y = x3

Idea summary A combined dilation by factor

enlarges the graph if k > 1 or reduces it

if 0 < k < 1, scaling both axes uniformly from the origin. Sketching conﬁrms the scaling visually.

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17.03 Practice questions What do you remember? 1

Answer these questions related to dilations: a

What is a horizontal dilation in terms of the axis it stretches from?

What is a vertical dilation in terms of the axis it stretches from?

For f (x) = f (x), what transformation does f (x) =

For f (x) = f (x), what transformation does f (x) = 3f (x) represent?

represent?

Determine whether these statements are true or false: a

A horizontal dilation by k = 0.5 stretches the graph wider.

A vertical dilation by l = 2 compresses the graph vertically.

A combined dilation by k = 3 enlarges the graph.

Replacing x with

in f (x) = f (x) compresses the graph horizontally.

For f (x) = x2, what is the effect of a vertical dilation by l = 0.5 on the graph’s shape?

For f (x) = x3, what is the effect of a combined dilation by k = 2 on the graph’s size?

Practice Ex 1

Ex 2

For the function f (x) = x2 − 4: a

Find the equation and sketch its graph after a horizontal dilation by a factor of 0.5.

Find the equation and sketch its graph after a vertical dilation by a factor of 2.

Describe the effect on the graph’s shape for each dilation.

For the function f (x) = 4x2 + 1: a

Find the equation after a combined dilation by a factor of 4.

Sketch both graphs, labelling a key point.

The graph of f (x) = x2 and its transformed image are shown: a

Identify the type and factor of the dilation applied.

Write the equation of both the original function and the transformed function.

y 5 4 3 2

(1, 1)

1 −3

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−2

−1

(3, 1)x 3

The graph of a function and its transformed image are shown:

Identify the type and factor of the dilation applied.

Write the equation of both the original function and the transformed function.

2 1

(0.4, 0) −1

(1, 0) x 1

−1 −2

The graph of f (x) = −2x + 4 and its transformed image are shown: a

Identify the type and factor of the dilation applied.

Describe the effect of the transformation on the graph’s shape.

y 6 (0, 6)

f (x) = −2x + 4

5 4 (0, 4) 3 2 1

−2

−1

x 1

f (x) = −3x + 6

Find the coordinates of the point on the transformed graph after applying these dilations to f (x) = x3, given the original point (1, 1): a

Horizontal dilation by k = 2

Vertical dilation by l = 3

Combined dilation by k = 0.5

Combined dilation by k = 2

For the function f (x) = x2, ﬁnd the coordinates of the point on the transformed graph after a vertical dilation by l = 0.5, given the original point (2, 4).

For the function f (x) = x2 + 3:

Find the equation after a combined dilation by a factor of k = 2.5.

Sketch both graphs, labelling a key point on the original and its corresponding point on the transformed graph.

For the function f (x) = −x + 5: a

Find the equation after a combined dilation by a factor of k = 1.25.

Sketch both graphs, labelling a key point on the original and its corresponding point on the transformed graph.

17.03 Dilations mathspace.co

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Extend your thinking 14

A graphic designer scales a logo modelled by f (x) = x2 for a billboard. The logo is dilated horizontally by k = 2 and vertically by l = 3: a

Find the equation of the scaled logo.

If the original logo passes through (2, 4), ﬁnd the corresponding point on the scaled logo.

Explain why a combined dilation by k = 6 might not be suitable for this billboard.

Sketch the graph and the scaled logo, labelling the corresponding point.

Consider the function f (x) = x3. Two students describe the effect of a combined dilation by k = 0.5: • Student A: The graph compresses, and the point (1, 1) maps to (0.5, 0.5). • Student B: The graph enlarges, and the point (1, 1) maps to (2, 2). Identify which student is correct and explain the error in the other student’s reasoning.

A convex lens magniﬁes an object’s image, modelled by f (x) = x2, by a combined dilation factor of k = 1.5: a

Find the equation of the magniﬁed image.

If the original image passes through (1, 1), ﬁnd the corresponding point on the magniﬁed image.

Explain why a combined dilation by k = 0.5 would not be suitable for magniﬁcation.

Sketch the graph and the magniﬁed image, labelling the magniﬁed point.

For the function f (x) = x2, compare the effect of: • A horizontal dilation by k = 2 followed by a vertical dilation by l = 2. • A combined dilation by k = 2. Find the resulting equations for both cases and describe these results.

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17.04 Exponential and logarithmic functions After this lesson, you will be able to… • apply translations, reflections, and dilations to exponential functions or logarithmic functions. • determine the equation of a transformed exponential or logarithmic function. • identify and describe the effect of transformations on key features, including asymptotes, domain, and range. • sketch the graph of a transformed exponential or logarithmic function.

Translations of exponential functions A translation shifts an exponential function’s graph horizontally or vertically. For f (x) = ax, translations are applied noting f (x − h) is used instead of the usual f (x − a) because the number a is already used as the base of the exponential function:

g(x) = f (x − h) + b = ax − h + b g(x)

is the translated function

h is the horizontal shift: right if h > 0, left if h < 0 b is the vertical shift: up if b > 0, down if b < 0 The domain is (−∞, ∞). The range is (b, ∞) if a > 1, or (−∞, b) if 0 < a < 1. The horizontal asymptote is at y = b.

Interactive exploration Discover this concept in action online

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17.04 Exponential and logarithmic functions mathspace.co

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Example 1 Translate f (x) = 2x right by 3 units and down by 1 unit. a Determine the equation of the transformed function.

Create a strategy Apply translations using g(x) = ax − h + b with h = 3 and b = −1.

Apply the idea f (x) = 2x f (x − 3) − 1 = 2

x−3

Write the original function −1 x−3

The transformed function is g(x) = 2

Apply translations with h = 3 and b = −1 − 1.

b Determine the horizontal asymptote.

Create a strategy Identify the asymptote at g(x) = b.

Apply the idea Since b = −1, the asymptote is g(x) = −1.

c Determine the y-intercept.

Create a strategy Substitute x = 0 into the equation to ﬁnd the y-intercept.

Apply the idea Write the transformed function Substitute x = 0

Simplify the exponent

Rewrite 23 = Evaluate The y-intercept is

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

d Sketch the graph, labelling the asymptote and y-intercept.

Create a strategy Plot the y-intercept, asymptote, and key points, then draw the exponential curve.

Apply the idea y 3 2 1 x −1

The graph approximates g(x) = 2x − 3 − 1 with asymptote at g(x) = −1.

−1

Reﬂect and check The original asymptote at f (x) = 0 shifted to g(x) = −1, matching the vertical translation. y 3 2

f (x) = 2x

1 x

−3 −2 −1

g(x) = 2x − 3 − 1

−1

Idea summary A translation from f (x) = ax to g(x) = ax − h + b shifts the graph horizontally by h units and vertically by b units. The asymptote is at f (x) = b, with domain (−∞, ∞) and range adjusted by b. Sketch graphs to visualise shifts.

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Reflections of exponential functions A reﬂection ﬂips an exponential function’s graph over an axis. For f (x) = ax:

−f (x) = −ax − ax is the reﬂection over the x-axis, ﬂipping vertically

f (−x) = a−x a−x is the reﬂection over the y-axis, ﬂipping horizontally The domain remains (−∞, ∞). The range is (−∞, 0) for g(x) = −ax if a > 1, and the asymptote is at y = 0.

Interactive exploration Discover this concept in action online

Example 2 Reﬂect f (x) = 2x over the x-axis. a Determine the equation of the transformed function.

Create a strategy Apply reﬂection over the x-axis using f (x) = −ax.

Apply the idea f (x) = 2x −f (x) = −2

Write the original function x

Reﬂect over x-axis

The transformed function is g(x) = −2x.

b Determine the range.

Create a strategy Analyse the effect of the reﬂection on the range, considering 2x > 0.

Apply the idea Since 2x > 0, −2x < 0, so the range is (−∞, 0).

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mathspace.co

c Determine the horizontal asymptote.

Create a strategy Identify the asymptote for g(x) = −ax.

Apply the idea The asymptote is y = 0, approached from below.

d Sketch the graph, labelling the asymptote and key points.

Create a strategy Determine and plot the key points and the asymptote, then draw the reﬂected exponential curve.

Apply the idea Set x = 0 into the transformed function to determine the y-intercept: g(x) = −2x g(0) = −2

Write the transformed function

Substitute x = 0

= −1

Evaluate

The y-intercept is at (0, −1). Set x = 1 for another key point: g(x) = −2x g(1) = −2

Write the transformed function

Substitute x = 1

= −2

Evaluate

Another key point is (1, −2). y x −1

−1 (0, −1)

The graph shows f (x) = 2x reﬂected over the x-axis. −2

(1, −2)

−3

Idea summary A reflection of f (x) = ax over the x-axis ( f (x) = −ax) or y-axis ( f (x) = a−x) flips the graph, altering the range but keeping the same domain and asymptote at y = 0. Sketch to visualise flips.

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Translations of logarithmic functions A translation shifts a logarithmic function’s graph. For f (x) = loga x, translations are:

g(x) = loga (x − h) + b h is the horizontal shift: right if h > 0, left if h < 0 b is the vertical shift: up if b > 0, down if b < 0 The domain is (h, ∞). The range is (−∞, ∞). The vertical asymptote is at x = h.

Interactive exploration Discover this concept in action online

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Example 3 Translate f (x) = log3 x left by 2 units and up by 1 unit. a Determine the function of the transformed function.

Create a strategy Apply translations using g(x) = loga (x − h) + b with h = −2 and b = 1.

Apply the idea f (x) = log3 x f (x + 2) + 1 = log3 (x + 2) + 1

Write the original function Apply translations with h = −2 and b = 1

The transformed function is g(x) = log3 (x + 2) + 1.

b Determine the domain.

Create a strategy

Apply the idea

Solve x − h > 0 to ﬁnd the domain.

For x + 2 > 0, x > −2, so the domain is (−2, ∞).

c Determine the vertical asymptote.

Create a strategy

Apply the idea

Identify the asymptote at x = h.

Since h = −2, the asymptote is x = −2.

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d Sketch the graph, labelling the asymptote and a key point.

Create a strategy Determine and plot the key points and the asymptote, then draw the logarithmic curve.

Apply the idea Set x = 0 into the transformed function to determine the y-intercept: g(x) = log3 (x + 2) + 1

Write the transformed function

g(0) = log3 (0 + 2) + 1

Substitute x = 0

= log3 2 + 1

Simplify

= 0.63 + 1

Evaluate and round the log using technology

= 1.63

Add

The y-intercept is at (0, 1.63). Set x = −1 for another key point: g(x) = log3 (x + 2) + 1

Write the transformed function

g(−1) = log3 (−1 + 2) + 1

Substitute x = −1

= log3 1 + 1

Simplify

=0+1

Evaluate the logarithm

Add

Another point is (−1, 1). y 3 2

(−1, 1)

(0, 1.63) 1 x

−2

−1

The graph shows g(x) = log3 (x + 2) + 1 with asymptote at x = −2.

−1

Idea summary A translation of f (x) = loga x to g(x) = loga (x − h) + b shifts the graph horizontally by h units and vertically by b units, with the asymptote at x = h and domain (h, ∞). Sketch to visualise shifts.

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Dilations of exponential functions A dilation stretches or compresses an exponential function. For f (x) = ax:

g(x) = lax l is the vertical dilation: stretches if ∣l∣ > 1, compresses if 0 < ∣l∣ < 1 The domain remains (−∞, ∞). The range is (0, ∞) if l > 0 and a > 1, or (−∞, 0) if l < 0. The horizontal asymptote stays at f (x) = 0. y 4

g (x) = 2 × 2x 3

The graph approximates f (x) = 2x stretched to g(x) = 2 × 2x, using exponential functions to represent the behaviour.

2 1

f (x) = 2x x

−1

Interactive exploration Discover this concept in action online

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Example 4 Dilate f (x) = 2x vertically by a factor of 3. a Determine the function of the transformed function.

Create a strategy Apply vertical dilation using g(x) = l × 2x with l = 3.

Apply the idea f (x) = 2x

Write the original function x

3 × f (x) = 3 × 2

Apply dilation

The transformed function is g(x) = 3 × 2 .

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b Determine the y-intercept.

Create a strategy Substitute x = 0 to ﬁnd the y-intercept.

Apply the idea g(x) = 3 × 2x g(0) = 3 × 2

Write the transformed function

Substitute x = 0

=3×1

Evaluate the power

Multiply

The y-intercept is at (0, 3).

c Sketch the graph, labelling the y-intercept and asymptote.

Create a strategy Plot the y-intercept, a key point at x = 2, and the asymptote, then draw the exponential curve.

Apply the idea g(x) = 3 × 2x g(2) = 3 × 2

Write the transformed function

Substitute x = 2

=3×4

Evaluate the power

= 12

Multiply

The point at x = 2 is (2, 12). 12 11 10 9 8 7 6 5 4 3 2 1 −1

(2, 12)

The graph shows g(x) = 3 × 2x dilated vertically with asymptote at g(x) = 0.

x 1

Idea summary A dilation of f (x) = ax to g(x) = l × ax stretches or compresses the graph vertically by ∣l∣, keeping the asymptote at f (x) = 0. Sketch to visualise the stretch.

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Dilations of logarithmic functions A dilation stretches or compresses a logarithmic function. For f (x) = loga x:

g(x) = lf (x) = l loga (x)

l is the vertical dilation: stretches if ∣l∣ > 1, compresses if 0 < ∣l∣ < 1 The domain remains (0, ∞), and the range is (−∞, ∞). The vertical asymptote stays at x = 0. y 3

g (x) = 2log2 (x)

2 1

f (x) = log2 (x) 1

The graph shows f (x) = log2 x vertically dilated to g(x) = 2 log2 x.

−1

Example 5 Dilate f (x) = log2 x vertically by a factor of 3. a Determine the equation of the transformed function.

Create a strategy Apply vertical dilation using g(x) = l loga (x) with l = 3.

Apply the idea f (x) = log2 x

Write the original equation

3f (x) = 3 log2 x

Apply dilation

The transformed function is g(x) = 3 log2 x.

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b Determine the y-coordinate at x = 2.

Create a strategy Substitute x = 2 to ﬁnd the y-coordinate.

Apply the idea g(x) = 3 log2 x

Write the transformed function

g(2) = 3 log2 2

Substitute x = 2

=3×1

Evaluate log2 2 = 1

Multiply

The point at x = 2 is (2, 3).

c Sketch the graph, labelling the x-intercept and asymptote.

Create a strategy Plot the x-intercept, a key point at x = 2, and the asymptote, then draw the logarithmic curve.

Apply the idea y 3

(2, 3)

2 1 x

(1, 0) 1

The graph shows g(x) = 3 log2 x dilated vertically with asymptote at x = 0.

−1

Idea summary A dilation of g(x) = l loga (x) to f (x) = b loga x stretches or compresses the graph vertically by ∣l∣, keeping the asymptote at x = 0. Sketch to visualise the stretch.

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17.04 Practice questions What do you remember? 1

What is the effect of transforming y = 2x to y = 2x − 2 + 3? A

Horizontal shift right 2 units, vertical shift up 3 units

Horizontal shift left 2 units, vertical shift up 3 units

Horizontal shift right 2 units, vertical shift down 3 units

Vertical shift up 2 units, horizontal shift left 3 units

State the asymptote for the following functions: a c

y = 3x + 1 − 2 x

y = −2 + 1

y = log2 (x − 3) + 1

y = 2 log3 x

State the domain and range for the following functions: a

y = −3x + 1

y = 2 × 4x − 1

y = log2 (x − 3)

y = 3 log3 (x + 2)

Practice Ex 1

Ex 2

Ex 3

Ex 4

950

Translate y = 3x left by 2 units and up by 1 unit: a

Determine the equation of the transformed function.

Determine the horizontal asymptote.

Determine the y-intercept.

Sketch the graph on a Cartesian plane, marking the y-intercept and a point at x = 1.

Reﬂect y = 3x over the y-axis: a

Determine the equation of the transformed function.

Determine the range.

Determine the horizontal asymptote.

Sketch the graph on a Cartesian plane, marking the y-intercept and a point at x = −1.

Translate y = log2 x right by 1 unit and down by 2 units: a

Determine the equation of the transformed function.

Determine the domain.

Determine the vertical asymptote.

Sketch the graph on a Cartesian plane, marking points at x = 2 and x = 3.

Dilate y = 3x vertically by a factor of 2: a

Determine the equation of the transformed function.

Determine the y-intercept.

Determine the y-value at x = 1.

Sketch the graph on a Cartesian plane, marking points at x = 0 and x = 1.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Ex 5

Dilate y = log3 x vertically by a factor of 2: a

Determine the equation of the transformed function.

Determine the x-intercept.

Determine the y-coordinate at x = 3.

Sketch the graph on a Cartesian plane, marking points at x =

Find the y-intercept for the following exponential functions: a

and x = 3.

y = 2x + 1 − 1

y = −3 × 4x + 2

y = 3x − 1 + 2

y = 2 × 2x + 2

Find the vertical asymptote for the following logarithmic functions: a

y = log3 (x + 1) − 2

y = 2 log4 (x − 2) + 1

y = log2 (x − 4)

y = − log3 (x + 3) + 1

Sketch y = 2x − 2 + 1 on a Cartesian plane, marking the asymptote, y-intercept, and a point at x = 2. State the transformations from y = 2x.

Sketch y = −2 × 2x − 1 on a Cartesian plane, marking the asymptote, y-intercept, and a point at x = 1. State the transformations from y = 2x.

Sketch y = log3 (x + 2) − 1 on a Cartesian plane, marking the asymptote and points at x = −1 and x = 1. State the domain.

Extend your thinking 14

A population grows according to P (t) = 1000 × 20.01t, where t is time in years. A new model starts at 1500 and grows twice as fast. Find the new equation for the population, P (t). Then, sketch the graph for t from 0 to 5, marking the population at t = 0 and t = 5.

An exponential function y = 2x is transformed to pass through (2, 7) with a horizontal asymptote at y = 1. Find the equation and sketch the graph on a Cartesian plane, marking the point at x = 2 and the asymptote.

A logarithmic function y = log3 x is translated to have a vertical asymptote at x = 2 and to pass through the point (4, 3). Find the equation and sketch the graph on a Cartesian plane, marking the point at x = 4 and the asymptote.

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17.05 Reciprocal and absolute value functions After this lesson, you will be able to… • apply translations, reflections, and dilations to reciprocal functions and absolute value functions. • determine the equations of vertical and horizontal asymptotes for a transformed reciprocal function. • determine the vertex of a transformed absolute value function. • sketch the graph of a transformed reciprocal or absolute value function, identifying key features.

Transformations of reciprocal functions A reciprocal function f (x) =

can be transformed using translations, reﬂections, and dilations.

The general form after transformations is:

l is the dilation factor: stretches if ∣l∣ > 1, compresses if 0 < ∣l∣ < 1; reﬂects over x-axis if l < 0 a is the horizontal translation: right if a > 0, left if a < 0 b is the vertical translation: up if b > 0, down if b < 0 The domain is x ≠ a, or (−∞, a) ∪ (a, ∞). The range is y ≠ b, or (−∞, b) ∪ (b, ∞). The vertical asymptote is at x = a, and the horizontal asymptote is at y = b.

Interactive exploration Discover this concept in action online

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

mathspace.co

Example 1 Transform f (x) =

by translating right by 2 units and up by 3 units, then reﬂecting over the x-axis.

a Determine the equation of the transformed function.

Create a strategy Apply translations (a = 2, b = 3) and reﬂection (l = −1) to form g(x) =

+ b.

Apply the idea Write the original function

Translate right by 2, up by 3

Reﬂect over x-axis

Distribute

The transformed function is g(x) =

− 3.

b Determine the asymptotes.

Create a strategy Identify the vertical asymptote at x = a and horizontal asymptote at y = b.

Apply the idea The vertical asymptote is x = 2. The horizontal asymptote is y = −3.

c Determine the domain and range.

Create a strategy Determine the domain where x − a ≠ 0 and range where y ≠ b.

Apply the idea Domain: x ≠ 2, or (−∞, 2) ∪ (2, ∞). Range: y ≠ −3, or (−∞, −3) ∪ (−3, ∞).

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d Sketch the graph, labelling the asymptotes, a key point and the y-intercept.

Create a strategy Plot a key point by substituting x = 3, the asymptotes, and draw the reciprocal curve.

Apply the idea Write the transformed function

Substitute x = 3

Simplify

Evaluate the fraction

Subtract

Calculate the y-intercept by substituting x = 0 in g(x). Write the transformed function

Substitute x = 0

Simplify

Subtract

The point at x = 3 is (3, −4), while the y-intercept is at (0, −2.5). y −1

x 1

−1 −2 −3

(0, −2.5)

−4

(3, −4)

The graph shows g(x) = x = 2 and y = −3.

− 3 with asymptotes at

−5 −6

Reﬂect and check The asymptotes shifted from x = 0, y = 0 to x = 2, y = −3, and the sketch reﬂects the transformations.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary Transformations of f (x) =

to g(x) =

+ b shift asymptotes to x = a and y = b,

with domain x ≠ a and range y ≠ b. Sketch graphs to visualise reﬂections, dilations, and translations.

Transformations of absolute value functions Exploration Sketch the transformation of f (x) = ∣x∣ after translating left by 2 units and dilating vertically by a factor of 3. 1. How does the vertex change? 2. If reﬂected over the x-axis, sketch and discuss the range.

An absolute value function f (x) = ∣x∣ can be transformed using translations, reﬂections, and dilations. The general form is:

g(x) = l∣x − a∣ + b l is the dilation factor: stretches if ∣l∣ > 1, compresses if 0 < ∣l∣ < 1; reﬂects over x-axis if l < 0 a is the horizontal translation: right if a > 0, left if a < 0 b is the vertical translation: up if b > 0, down if b < 0 The vertex is at (a, b). The domain is (−∞, ∞). The range is [b, ∞) if l > 0, or (−∞, b] if l < 0. y 6 5

g (x)

The sketch shows f (x) = ∣x∣ transformed to g(x) = 2∣x − 1∣ + 3.

3 2 1 −1

x 1

17.05 Reciprocal and absolute value functions mathspace.co

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Example 2 Transform f (x) = ∣x∣ by dilating vertically by 2, translating left by 1 unit, and down by 2 units. a Determine the equation of the transformed function.

Create a strategy Apply transformations using g(x) = l∣x − a∣ + b with l = 2, a = −1, and b = −2.

Apply the idea f (x) = ∣x∣

Write the original function

g(x) = 2∣x∣

Apply dilation with l = 2

= 2∣x + 1∣ − 2

Apply translations with a = −1 and b = −2

The transformed function is g(x) = 2∣x + 1∣ − 2.

b Determine the vertex.

Create a strategy Identify the vertex at (a, b) from the equation.

Apply the idea From g(x) = 2∣x − (−1)∣ − 2, the vertex is (−1, −2).

c Determine the range.

Create a strategy Determine the range based on l and b.

Apply the idea Since l = 2 > 0, the range is [−2, ∞).

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d Determine the x-intercepts and sketch the graph, labelling the vertex and x-intercepts.

Create a strategy Set g(x) = 0 to ﬁnd x-intercepts, then plot the vertex, x-intercepts, and draw the V-shaped graph.

Apply the idea g(x) = 2∣x + 1∣ − 2

Write the transformed function

0 = 2∣x + 1∣ − 2

Set g(x) = 0

2 = 2∣x + 1∣

Add 2 to both sides

1 = ∣x + 1∣

Divide both sides by 2

±1 = x + 1

Solve absolute value

1=x+1

Write the equation with the positive value

x=0

Subtract 1 from both sides

−1 = x + 1

Write the equation with the negative value

x = −2

Subtract 1 from both sides

Solving for 1 = x + 1:

Solving for −1 = x + 1:

So the x-intercepts are (0, 0), (−2, 0). y 2 1

(−2, 0) −2

−1

(0, 0) 1

Sketch shows g(x) = 2∣x + 1∣ − 2 with vertex at (−1, −2) and x-intercepts at (0, 0), (−2, 0).

−1

(−1, −2)

−2

Reﬂect and check The vertex shifted from (0, 0) to (−1, −2), and the sketch shows a steeper graph due to dilation.

Idea summary Transformations of f (x) = ∣x∣ to g(x) = l∣x − a∣ + b move the vertex to (a, b), with range [b, ∞) if l > 0. Sketch graphs to visualise dilations, reﬂections, and translations.

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17.05 Practice questions What do you remember? 1

Match the transformation parameters for g(x) = a

l > 1

Shift down

a > 0

Stretch vertically

b < 0

iii

Reﬂect over x-axis

l < 0

Shift right

State the domain, range, and vertex for f (x) = ∣x∣: a

+ b to their effects:

Domain and range

Vertex

Horizontal asymptote

Identify the asymptotes of f (x) = : a

Vertical asymptote

Practice Ex 1

Ex 2

Transform f (x) = x-axis: a

Determine the equation of the transformed function.

Determine the asymptotes.

Determine the domain, range, and x- and y-intercepts, if applicable.

Sketch the graph, labelling the asymptotes and at least one key point.

Transform f (x) = ∣x − 2∣ by dilating vertically by 2, translating left by 1 units, and down by 1 unit: a

Determine the equation of the transformed function.

Determine the vertex.

Determine the range.

Determine the x-intercepts and sketch the graph, labelling the vertex and x-intercepts.

Find the equation of f (x) =

after the following transformations:

Translate right by 4 units and up by 2 units

Reﬂect over y-axis and down by 3 units

Dilate vertically by 2 and left by 1 unit

Reﬂect over x-axis and up by 1 unit

Find the asymptotes for the following reciprocal functions: a

958

by translating right by 3 units and up by 4 units, then reﬂecting over the

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Find the equation of f (x) = ∣x∣ after the following transformations: a

Translate left by 2 units and up by 3 units

Dilate vertically by 2 and reﬂect over x-axis

Translate right by 3 units and down by 1 unit

Dilate vertically by 3 and translate up by 2 units

Find the vertex and range for the following absolute value functions: a

f (x) = ∣x − 3∣ + 2

f (x) = −2∣x + 1∣ − 1

Find the intercepts for f (x) =

Determine the domain and range for f (x) =

Identify the transformations from f (x) = ∣x∣ to the given function:

+ 1. − 3.

f (x) = 3∣x − 2∣ + 1

f (x) = −∣x + 1∣ − 2

f (x) = 2∣x + 3∣ − 1

f (x) = −3∣x − 1∣ + 2

Extend your thinking 13

Consider the graph of a transformed reciprocal function. Determine: a

Equation of the curve

x-intercept

y 5 4

(4, 3)

3 2

(2, 1)

x 1

−1

Consider the graph of a transformed reciprocal function. Determine: a

Equation of the curve

x-intercept

y x −3

−2

−1

1 −1 −2

(0, −2)

−3 −4

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Consider the graph of a transformed absolute value function in the diagram, determine:

Equation of the curve

Domain and range

y 4 3 (0, 3)

(2, 3)

2 1 −1

(1, 1)

Consider the graph of a transformed absolute value function. Determine:

y 5 (0, 5)

Equation of the curve

Domain and range

(−3, 2)

2 1

−3

−2

(−2, −1)

−1

x 1

−2

17.06 Circles and translations After this lesson, you will be able to… • identify the centre and radius of a circle from its equation in standard form. • determine the domain and range of a circle from its centre and radius. • convert the equation of a circle from general form to standard form by completing the square. • find the equation of a circle given its centre and radius. • sketch the graph of a circle from its equation.

Circles and translations A circle’s equation in standard form with centre at the origin is:

x2 + y2 = r2 r

960

is the radius of the circle, where r > 0

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

The equation of a circle translated horizontally by a units and vertically by b units, in standard form is given as:

(x − a)2 + ( y − b)2 = r2 a is the horizontal translation: right if a > 0, left if a < 0 b is the vertical translation: up if b > 0, down if b < 0 r

is the radius of the circle, where r > 0

The centre is at (a, b), and the radius is r. The domain is [a − r, a + r], and the range is [b − r, b + r]. When evaluating r from r2, only the positive root is taken, since the radius represents a distance, and distance is deﬁned as a positive scalar quantity. The general form of a circle’s equation is:

x2 + y2 + ax + by + c = 0 a, b, c are constants defining the circle’s position and size It is important to note that the variables a and b in the general form x2 + y2 + ax + by + c = 0 are different constants from the a and b used to denote the centre (a, b) in the standard form. Context will always clarify which definition is being used. To convert general form to standard form, complete the square. In standard form, the centre is easily identified and the radius can be calculated directly.

Interactive exploration Discover this concept in action online

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Example 1 A circle has equation (x − 2)2 + ( y + 1)2 = 9. a Determine the centre.

Create a strategy Compare to (x − a)2 + ( y − b)2 = r2 to identify a and b.

Apply the idea (x − 2)2 + ( y + 1)2 = 9 2

(x − 2) + ( y − (−1)) = 9

Write the equation Rewrite equation

The centre is at (2, −1).

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b Determine the radius.

Create a strategy

Apply the idea

r2 = 9, so r = 3.

Extract r from r = 9.

c Determine the domain and range.

Create a strategy

Apply the idea

Use [a − r, a + r] for domain and [b − r, b + r] for range.

The domain is [2 − 3, 2 + 3] = [−1, 5]. The range is [−1 − 3, −1 + 3] = [−4, 2].

d Sketch the circle, labelling the centre and key points.

Create a strategy Plot the centre and points at the domain and range boundaries, then draw the circle.

Apply the idea y 2

(2, 2) x 2

(−1, −1)

(2, −1)

−2 −4

(5, −1)

Sketch shows a circle with centre at (2, −1) and radius 3.

(2, −4)

Reﬂect and check The sketch conﬁrms the centre at (2, −1) and radius 3, with translations right by 2 and down by 1.

Example 2 Convert x2 + y2 − 4x + 6y − 3 = 0 to standard form. a Determine the standard form of the equation.

Create a strategy Complete the square for x and y terms to rewrite in standard form.

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Apply the idea x2 + y2 − 4x + 6y − 3 = 0 2

Write the equation

x − 4x + y + 6y = 3 2

Rearrange

(x − 4x + 4) + ( y + 6y + 9) = 3 + 4 + 9 2

Complete the square

(x − 2) + ( y + 3) = 16 2

Factorise

The standard form is (x − 2) + ( y + 3) = 16.

b Determine the centre.

Create a strategy

Apply the idea

Identify a and b from the standard form (x − a)2 + ( y − b)2 = r2.

The centre is at (2, −3).

c Determine the radius.

Create a strategy

Apply the idea

r2 = 16, so r = 4.

Extract r from r = 16.

d Sketch the circle, labelling the centre and key points.

Create a strategy Plot the centre and points at the domain and range boundaries, then draw the circle.

Apply the idea y

(2, 1) x

−2

2 −2

(−2, −3)

(2, −3)

−4

(6, −3)

Sketch shows a circle with centre at (2, −3) and radius 4.

−6

(2, −7)

Reﬂect and check The sketch veriﬁes the centre and radius, aligning with the standard form derived by completing the square.

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Example 3 A circle has centre (1, −2) and radius 5. a Determine the equation of the circle.

Create a strategy Use (x − a)2 + ( y − b)2 = r2 with a = 1, b = −2, and r = 5.

Apply the idea (x − a)2 + ( y − b)2 = r2 2

(x − 1) + ( y − (−2)) = 5 2

(x − 1) + ( y + 2) = 25

Write the formula Substitute a = 1, b = −2 and r = 5 Evaluate

b Sketch the circle, labelling the centre and key points.

Create a strategy Plot the centre and points at the boundaries of the circle, then draw the circle.

Apply the idea y

(1, 3)

2 x −4

−2

(−4, −2)

2 −2

(1, −2)

(6, −2)

Sketch shows a circle with centre at (1, −2) and radius 5.

−4 −6

(1, −7)

Reﬂect and check The sketch conﬁrms the equation and visualises the circle’s position and size.

Idea summary A circle’s equation (x − a)2 + ( y − b)2 = r2 has centre (a, b) and radius r. To convert from general form to standard form, complete the square for x2 + y2 + ax + by + c = 0.

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17.06 Practice questions What do you remember? 1

What is the standard form equation of a circle with centre (a, b) and radius r? A

x2 + y2 = r2

(x − a)2 + ( y − b)2 = r2

x2 + y2 + ax + by = 0

(x + a)2 + ( y + b)2 = r

What process converts x2 + y2 + ax + by + c = 0 to standard form?

For a circle with equation (x − a)2 + ( y − b)2 = r2, state formulas for the domain and range.

What do the constants a, b, c represent in the general form x2 + y2 + ax + by + c = 0?

Practice Ex 1

Ex 2

Ex 3

A circle has equation (x − 3)2 + ( y − 2)2 = 16: a

Find the centre and radius.

State the domain and range.

Sketch the circle, marking the centre and key points.

A circle has an equation of x2 + y2 + 2x − 8y + 8 = 0: a

State the standard form equation.

Find the centre and radius.

Sketch the circle, marking the centre and the key points.

A circle has centre (−1, 3) and radius 2: a

Write the equation of the circle.

Sketch the circle, marking the centre and key points.

Find the centre and radius of these circles, then sketch each, marking the centre: i

Determine the centre.

Determine the radius.

iii

Sketch the circle, labelling the centre.

(x − 4)2 + ( y − 1)2 = 25

(x + 2)2 + ( y − 3)2 = 9

(x − 1)2 + y2 = 4

x + ( y + 5) = 16

Convert these equations to standard form by completing the square, then ﬁnd the centre and radius: a

x2 + y2 + 6x − 2y − 6 = 0

x2 + y2 − 8x + 4y + 16 = 0

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Find the equation of the circle with the given centre and radius, then sketch each circle, marking the centre and intercepts: i

Determine the equation of the circle.

Determine the x-intercepts.

iii

Sketch the circle, labelling the centre and intercepts.

Centre: (1, −4), radius: 3

Centre: (−2, 0), radius: 5

Centre: (0, 2), radius: 1

Centre: (3, −1), radius: 4

Determine the domain and range for these circles, then sketch them, marking the centre and the key points: a

(x − 2)2 + ( y − 1)2 = 4

x2 + ( y + 3)2 = 9

For the circle (x − 2)2 + ( y − 3)2 = 4: a

Find the x-intercept(s) and y-intercept(s).

Sketch, marking the centre and intercepts.

For each circle equation: i

Describe the vertical and horizontal translations required to move so that they are centred at (1, 1).

Write the transformed equation.

x2 + y2 = 1

(x + 3) + y = 1

x2 + ( y − 5)2 = 1

(x − 4)2 + ( y + 4)2 = 1

Extend your thinking 14

A circle with equation x2 + y2 − 6x + 4y − 12 = 0 is translated so its centre is at (1, 2). Find the new equation and sketch both circles.

Find the equation of a circle with centre on the line y = x, radius 2, and passing through (1, 2). Sketch the circle.

A circle passes through (4, 1) with centre at (2, 1). Find the equation and sketch the circle, marking the centre and the given point.

Consider a circular garden which has centre (3, 2) and radius 10 metres:

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Write the equation and sketch the circle.

Calculate the exact value of the area enclosed inside the circle.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

17.07 Order of transformations After this lesson, you will be able to… • recognise that the order of transformations can change the final graph. • apply a sequence of transformations to a parent function to find the resulting equation. • compare the different outcomes of applying transformations in various orders. • identify the effect of order on key features such as the vertex or asymptotes. • apply the standard convention of applying dilations and reflections before translations.

Order of transformations To combine transformations, apply reflections, translations, and dilations in a specific order to a function y = f (x). Each transformation modifies the equation, affecting the graph’s position, orientation, and scale. Sketching helps visualise these changes. Common transformations: Dilations: • Horizontal dilation by k: x → , giving y =

• Vertical dilation by l: y → , giving y = lf (x). Reﬂections: • In the y-axis: x → −x, giving y = f (−x). • In the x-axis: y → −y, giving y = −f (x). Translations: • Horizontal translation by a: x → x − a, giving y = f (x − a). • Vertical translation by b: y → y − b, giving y = f (x) + b. When combining transformations, it is essential to apply them in the order dictated by their effect on the equation. Say, for an equation of the form y =

+ c, the horizontal dilation (k)

occurs first (affecting x), followed by the horizontal translation represented by −b, then the vertical dilation by a, and finally the vertical translation by +c. This order must be followed unless a different sequence is explicitly specified.

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For example, y = 2f (−x + 1) involves a y-axis reﬂection, a horizontal translation right by 1 unit, and a vertical dilation by a factor of 2. y 5 4 3 2 2

y=x

y = 2(−x + 1)2

x −2 −1

Example 1 Transform f (x) = x2 by dilating vertically and horizontally by 2 and translating right by 1 unit. a Determine the transformed function when applying vertical dilation then translation.

Create a strategy Apply vertical dilation by 2 to y = x2, then translate right by 1 unit.

Apply the idea f (x) = x2

Write the parent function

2f (x) = 2x2

Dilate vertically by 2 2

2f (x − 1) = 2(x − 1)

Translate right by 1

The transformed function is g(x) = 2(x − 1) .

b Determine the vertex for vertical dilation then translation.

Create a strategy Identify the vertex from the equation g(x) = l(x − a)2 + b.

Apply the idea The vertex is (1, 0).

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

c Determine the transformed function when applying translation then vertical dilation.

Create a strategy Translate f (x) = x2 right by 1 unit, then apply vertical dilation by 2.

Apply the idea f (x) = x2

Write the parent equation 2

f (x − 1) = (x − 1)

Translate right by 1 2

2f (x − 1) = 2(x − 1)

Dilate vertically by 2

The transformed function is g(x) = 2(x − 1) .

d Determine the transformed function when applying horizontal dilation then translation.

Create a strategy Apply horizontal dilation by 2 to f (x) = x2, then translate right by 1 unit.

Apply the idea Write the parent function

Dilate horizontally by 2

Translate right by 1

Simplify The transformed function is g(x) =

e Determine the vertex for horizontal dilation then translation.

Create a strategy Identify the vertex from the equation g(x) =

+ b.

Apply the idea The vertex is (1, 0).

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969

Determine the transformed function when applying translation then horizontal dilation.

Create a strategy Translate f (x) = x2 right by 1 unit, then apply horizontal dilation by 2.

Apply the idea Write the parent function

Translate right by 1

Dilate horizontally by 2

The transformed function is g(x) =

g Determine the vertex for horizontal dilation then translation.

Create a strategy Rewrite the expression inside the brackets into the form denominator.

, by rewriting it with common

Apply the idea Write the transformed function

Rewrite with common denominator

The vertex is (2, 0).

Reﬂect and check Alternatively, equate the expression inside the brackets to 0 to solve for x-coordinate of the vertex. Equate the expression inside the brackets to 0

Add 1 to both sides

Multiply both sides by 2

The vertex is (2, 0).

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

h Sketch the graph for both orders, labelling the vertex.

Create a strategy Plot f (x) = x2, intermediate graphs, and ﬁnal graph for both orders, labelling the vertex.

Apply the idea Vertical dilation then translation: g(x) = 2x2

y 5

g(x) = 2(x − 1)2 4

f (x) = x2

3 2 1

(1, 0) −2

−1

Graph shows f (x) = x2, g(x) = 2x2 and g(x) = 2(x − 1)2 with vertex at (1, 0). Translation then vertical dilation: y 5

f (x) = x2

g(x) = (x − 1)2

g(x) = 2(x − 1)2 4 3 2 1

(1, 0) −2

−1

x 2

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Graph shows f (x) = x2, g(x) = (x − 1)2 and g(x) = 2(x − 1)2 with vertex at (1, 0). Horizontal dilation then translation: y

f (x) = x2

5 4 3 2 1

(1, 0) −5

−4

−3

Graph shows f (x) = x2, g(x) =

−2

−1

and g(x) =

x 2

with vertex at (1, 0).

Translation then horizontal dilation: y

g(x) = (x − 1)2 5 4

f (x) = x2

3 2 1

(2, 0) −5

−4

−3

−2

−1

Graph shows f (x) = x2, g(x) = (x − 1)2 and g(x) =

x 3

with vertex at (2, 0).

Reflect and check Vertical dilation and horizontal translation yield the same equation regardless of order, as they affect different axes. Horizontal dilation and translation produce different graphs, with vertices at (1, 0) for dilation then translation, and (2, 0) for translation then dilation, due to both being horizontal transformations.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Transform f (x) =

by reﬂecting over the x-axis and translating up by 2 units.

a Determine the transformed function when applying reﬂection then translation.

Create a strategy Start with the parent function f (x) = . Reﬂect by changing the vertical dilation factor to l = −1. Translate 2 units up by setting b = 2.

Apply the idea Write the parent function

Reﬂect over x-axis

Translate up by 2

The transformed function is g(x) =

+ 2.

b Determine the asymptotes for reﬂection then translation.

Create a strategy

Apply the idea

Identify the vertical asymptote and horizontal asymptote then adjust by the vertical translation.

Vertical asymptote: x = 0, since , x ≠ 0 Horizontal asymptote: y = 2, since y = 0 is translated 2 units up

c Determine the transformed function when applying translation then reﬂection.

Create a strategy Start with the parent function f (x) = . Translate 2 units up by setting b = 2. Then reﬂect by changing the vertical dilation factor of to l = −1 of the translated function.

Apply the idea Write the parent function

Translate up by 2

Reﬂect over x-axis

Apply distributive property

The transformed function is g(x) =

− 2.

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d Sketch the graph for both orders, labelling the asymptotes.

Create a strategy Plot the original f (x) = , intermediate graphs, and ﬁnal graph for both orders, labelling asymptotes.

Apply the idea Reﬂection then translation: y 4 3

y=2

2 1

x=0 x

−2

−1

−1 −2

Graph shows f (x) = , g(x) =

, and g(x) =

+ 2 with asymptotes at x = 0 and y = 2.

Translation then reﬂection: y 3 2 1 −2

x=0

−1

x 1

−1

y = −2

−2 −3 −4

Graph shows f (x) = , g(x) =

+ 2, and g(x) =

− 2 with asymptotes at x = 0 and y = −2.

Reﬂect and check The order changes the horizontal asymptote because reﬂecting after translating affects the vertical shift’s position. Cartesian plots reveal how asymptotes move differently in each case.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary The order of applying dilation, reﬂection, and translation affects the graph’s position, shape, and features like the vertex or asymptote. Sketching on a Cartesian plane clariﬁes how transformations alter the graph, especially for functions like y = , where asymptotes shift based on order.

17.07 Practice questions What do you remember? 1

Why does the order of transformations matter when applied to a function? A

It always changes the domain and range

It may affect the ﬁnal position and shape of the graph

It only affects linear functions

It has no effect on the graph

Consider the transformations for g(x) =

+ b:

Which transformations could be applied ﬁrst?

Which transformation affects the horizontal shift?

Which parameter causes a vertical dilation?

Which parameter causes a reﬂection over the x-axis?

State the effect of transformation order on key features of a graph: a

Consider the graph of f (x) = . Explain the difference in the position of the horizontal asymptote when a vertical dilation is followed by a vertical translation, compared with when the order is reversed.

How does applying a translation before a reflection affect the vertex of y = x2, compared with applying the reflection first?

Practice Ex 1

Transform f (x) = x2 by dilating vertically and horizontally by 3 and translating right by 2 units: a

Determine the equation when applying vertical dilation then translation.

Determine the vertex for vertical dilation then translation.

Determine the equation when applying translation then vertical dilation.

Determine the equation when applying horizontal dilation then translation.

Determine the vertex for horizontal dilation then translation.

Determine the equation when applying translation then horizontal dilation.

Determine the vertex for translation then horizontal dilation.

Sketch the graph for both orders, labelling the vertex and a point on the curve. 17.07 Order of transformations mathspace.co

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Ex 2

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Transform f (x) =

by reﬂecting over the x-axis and translating up by 3 units:

Determine the equation when applying reﬂection then translation.

Determine the asymptotes when reﬂection is applied before translation.

Determine the equation when applying translation then reﬂection.

Sketch the graph for both orders, labelling the asymptotes and a point on the curve.

The function f (x) = 2x is transformed by dilating vertically and horizontally by 2 and translating left by 1 unit. For each order of transformation: i

Determine the transformed function.

Determine the horizontal asymptote.

Vertical dilation then translation

Translation then vertical dilation

Horizontal dilation then translation

Translation then horizontal dilation

The function f (x) = ∣x∣ is transformed by reﬂecting over the y-axis and translating down by 2 units. For each order of transformation: i

Determine the transformed function.

Determine the vertex.

Reﬂection then translation

Translation then reﬂection

The function f (x) = log2 x is transformed by dilating vertically and horizontally by 3 and translating up by 1 unit. For each order of transformation: i

Determine the transformed function.

Determine the vertical asymptote.

Vertical dilation then translation

Translation then vertical dilation

Horizontal dilation then translation

Translation then horizontal dilation

The function f (x) = x3 is transformed by reﬂecting over the x-axis and translating right by 2 units. For each order of transformation: i

Determine the transformed function.

Determine the point of inﬂection.

Reﬂection then translation

Translation then reﬂection

The function y = x2 is transformed by dilating horizontally by For each order of transformation: i

Determine the transformed function.

Determine the vertex.

Dilation then translation

and translating right by 1 unit.

Translation then dilation

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

The function y =

is transformed by dilating vertically and horizontally by 2 and translating

up by 3 units. For each order of transformation:

Determine the transformed function.

Determine the horizontal asymptote.

Vertical dilation then translation

Translation then vertical dilation

Horizontal dilation then translation

Translation then horizontal dilation

The function f (x) = ex is transformed by reﬂecting over the y-axis and translating up by 2 units. For each order of transformation: i

Determine the transformed function.

Determine the horizontal asymptote.

Reﬂection then translation

Translation then reﬂection

Extend your thinking 13

The function f (x) = x2 undergoes two transformations: a vertical dilation by a factor of 2, and a vertical translation upwards by 3. The point P′ (4, 35) lies on the transformed function: a

If the dilation is applied before the translation, ﬁnd the original point P on the parent function.

If the translation is applied before the dilation, ﬁnd the original point Q on the parent function.

Explain why the answers differ.

Let f (x) = ex − 1 − 1: a

Sketch the graph of f (x).

The function is transformed to g(x) =

. Describe the transformations in the

correct order, and sketch g(x) on the same set of axes as f (x). 15

Let f (x) =

. The function is transformed to g(x) =

− 3:

Describe the sequence of transformations from f (x) to g(x).

If g(x) = 3, ﬁnd the corresponding value(s) of f (x).

17.07 Order of transformations mathspace.co

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17.08 Multiple transformations After this lesson, you will be able to… • apply a given sequence of multiple transformations to a parent function. • determine the equation of a function after several transformations have been applied in order. • identify the key features, such as the vertex or asymptotes, of a transformed graph. • sketch the final graph after applying multiple transformations. • describe the sequence of transformations required to get from a parent function to a given complex function.

Multiple transformations Multiple transformations combine translations, reflections, and dilations to modify a function’s graph. Transformations must be applied in the order given. If no order is specified, apply dilations and reflections first, then translations.

Example 1 Transform f (x) = and up by 1 unit.

by reﬂecting over the y-axis, dilating vertically by 2, translating left by 3 units,

a Determine the equation of the transformed function.

Create a strategy Apply transformations in the given order: first y-axis reflection, next vertical dilation by 2, then horizontal translation by −3, and finally vertical translation by 1.

Apply the idea Write the parent function

Reﬂect over y-axis

Dilate vertically by 2

Translate left by 3

Translate up by 1

Simplify

The transformed function is g(x) =

978

+ 1.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Determine the vertical asymptote.

Create a strategy Find where the denominator is zero: x + 3 = 0.

Apply the idea Vertical asymptote: x = −3.

c Determine the horizontal asymptote.

Create a strategy Identify the constant term after vertical translation of 1 from the parent functions asymptote of y = 0.

Apply the idea Horizontal asymptote: y = 0 + 1 = 1

d Determine the domain and range.

Create a strategy Determine domain where the denominator can not equal zero and range excluding the horizontal asymptote.

Apply the idea Domain: x ≠ −3, or (−∞, −3) ∪ (−3, ∞). Range: y ≠ 1, or (−∞, 1) ∪ (1, ∞).

e Sketch the transformed graph, labelling asymptotes and key points.

Create a strategy Find the x-intercepts and y-intercepts. Plot the asymptotes, the intercepts, and at least one point on the second branch (e.g., x = −4). Then sketch the transformed graph.

17.08 Multiple transformations mathspace.co

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Apply the idea For the y-intercept, substitute x = 0 in the transformed function: Write the transformed function Substitute x = 0

Simplify the denominator

Evaluate The y-intercept is at

For the x-intercept, substitute g(x) = 0 in the transformed function: Write the transformed function Substitute g(x) = 0

Subtract 1 from both sides

Multiply both sides by −(x + 3)

Subtract 3 from both sides

The x-intercept is at (−1, 0). To plot a point on the branch that does not intersect the x- or y-axis substitute x = −4: Write the transformed function Substitute x = −4

Simplify the denominator

Evaluate Another point to plot is (−4, 3). y 4

(−4, 3)

3 2

(−1, 0) −5 −4 −3 −2

Graph shows y =

−1

and y =

+ 1 with asymptotes

at x = −3 and y = 1.

1 −1 −2

Reflect and check The asymptotes shifted from x = 0, y = 0 to x = −3, y = 1, and the graph confirms the hyperbola’s position, consistent with transformations and prior lessons on reciprocal functions.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 2 Transform f (x) = x2 by dilating horizontally by , reflecting over the x-axis, translating right by 2 units and down by 1 unit. a Determine the equation of the transformed function.

Create a strategy Apply transformations in the given order: first horizontal dilation by , then x-axis reflection, next horizontal translation by 2, and finally vertical translation by −1.

Apply the idea Write the parent function

Horizontal dilation by

Simplify

Reflect over x-axis

Translate right by 2

Translate down by 1 2

The transformed function is g(x) = −4(x − 2) − 1.

b Determine the vertex.

Create a strategy Since the equation is in vertex form y = a(x − h)2 + k, the vertex is at (h, k).

Apply the idea Vertex: (2, −1).

c Determine the range.

Create a strategy Determine the range based on the sign of a and the value of k for the parabola.

Apply the idea Since a = −4 < 0, the parabola is concave down, so range is (−∞, −1].

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d Sketch the transformed graph, labelling the vertex and key points.

Create a strategy Find the x-intercepts and y-intercepts. Plot the vertex, intercepts, and at least one other point if only one intercept is found.

Apply the idea For the y-intercept, substitute x = 0 in the transformed function: g(x) = −4(x − 2)2 – 1

Write the transformed function

g(0) = −4(0 − 2)2 − 1

Substitute x = 0

= −4 × 4 − 1

Evaluate the power

= −17

Evaluate

The y-intercept is at (0, −17). For the x-intercept, substitute g(x) = 0 in the transformed function: Write the transformed function

Substitute g(x) = 0

Divide both sides by −4

Subtract

from both sides

Since (x − 2)2 ≥ 0 for all x, there are no values of x that can solve this equation, therefore there are no x-intercepts. The parabola only has one intercept, so find another point, say, at x = 3: g(x) = −4(x − 2)2 – 1

Write the transformed function

g(3) = −4(3 − 2)2 – 1

Substitute x = 3

= −4(1) − 1

Simplify

= −4 − 1

Evaluate the power

= −5

Evaluate

Another point to plot is (3, −5). 2

y x

−1

2 4 1 3 5 −2 (2, −1) −4 (3, −5) −6 −8 −10 −12 −14 −16 (0, −17) −18

Graph y = −4(x − 2)2 − 1 with vertex at (2, −1).

Reﬂect and check The vertex shifted from (0, 0) to (2, −1), and the graph shows a wider, inverted parabola, aligning with transformations from prior quadratic lessons.

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Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Idea summary Multiple transformations of a function combine dilations, reflections, and translations, applied in order to affect the graph’s shape, position, vertex, or asymptotes. Sketching on a Cartesian plane involves plotting key points, intercepts, and asymptotes. The order of transformations applied can affect the final outcome.

17.08 Practice questions What do you remember? 1

Does the order in which the following transformations are applied affect the resulting graph? A

A horizontal translation and a vertical translation

A horizontal dilation and a vertical dilation

A horizontal dilation and a horizontal translation

A vertical dilation and a horizontal translation

After applying multiple transformations to a function, which features of the graph should be determined before sketching?

Identify transformation occurs when replacing: a

y with y−b

x with x−a

x with

y with

Practice Ex 1

Ex 2

Transform f (x) = 2x by reﬂecting over the y-axis, dilating vertically by 2, translating left by 3 units, and up by 1 unit: a

Determine the equation of the transformed function.

Determine the vertical asymptote.

Determine the horizontal asymptote.

Determine the domain and the y-intercept.

Determine the range and sketch the transformed graph, labelling the asymptote and key points.

Transform f (x) = ∣x∣ by dilating vertically by , reﬂecting over the x-axis, translating right by 2 units and down by 1 unit: a

Determine the equation of the transformed function.

Determine the vertex.

Determine the range.

Sketch the transformed graph, labelling the vertex, y-intercept and key points.

17.08 Multiple transformations mathspace.co

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Transform f (x) = right by 2:

by reﬂecting over the x-axis, dilating vertically by 3, and translating

Determine the equation of the transformed function.

Determine the asymptotes.

Determine the domain.

Determine the range.

Sketch the graphs of both functions and asymptotes.

Transform f (x) = log(3x) by dilating vertically by 2, translating right by 1, and down by 3: a

Determine the equation of the transformed function.

Determine the asymptote

Determine the range

Sketch the graphs of both functions and asymptote.

Transform f (x) = ∣x∣ by reﬂecting over the y-axis, dilating vertically by 2, and translating up by 1: a

Determine the equation of the transformed function.

Determine the vertex.

Determine the x-intercept(s).

Determine the y-intercept(s).

Sketch the graphs of both functions, labelling the key points.

Transform f (x) = 2ex by vertical dilation by , reﬂecting over the x-axis, and translating up by 2: a

Determine the equation of the transformed function.

Determine the range.

Determine the asymptote.

Sketch the graphs of both functions and asymptote.

Transform f (x) = by 1:

by reﬂecting over the y-axis, dilating horizontally by , and translating left

Determine the equation of the transformed function.

Determine the domain.

Sketch the graphs of both functions and asymptote.

Transform f (x) = x3 by dilating vertically by , translating left by 2, and up by 1: a

Determine the equation of the transformed function.

Determine the stationary point.

Sketch the graphs of both functions, labelling the stationary point.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking 12

A quadratic function f (x) = 3(x − 1)2 − 1 is transformed to have a vertex at (2, −3) and pass through (3, −1). Find the equation and range.

A reciprocal function f (x) =

is transformed to have a vertical asymptote at x = −1, a

horizontal asymptote at y = 2, and pass through (0, 0). Find the transformed function and domain. 14

Compare the effect of applying a vertical dilation by reverse order on f (x) =

then a translation right by 2 versus the

stating both equations and asymptotes.

The function is transformed into a new graph with endpoint (−1, 2) that passes through the x-axis at (−0.25, 0). What order of transformations have been applied to build the transformed function?

The function f (x) = x2 is transformed into a new graph with vertex at (1, 2) and passes through the point (0, 1). Find the equation of the transformed function in the form g(x) = k(x − a)2 + b.

Did you know?

3D printers use mathematical transformations to bring digital designs to life! Before printing begins, the 3D model is sliced into hundreds or even thousands of thin layers called cross-sections. The printer then applies precise transformations — translating each layer upward along the z-axis and sometimes rotating the print head or platform — to build the object one layer at a time. These transformations ensure that every 3D model, whether it’s a phone stand, car part, or medical implant, is created with accuracy and symmetry, turning flat designs into tangible, three-dimensional forms!

17.08 Multiple transformations mathspace.co

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17 Chapter review 1

Which equation represents the graph of y = 2x + 5 after a reﬂection in the y-axis? A

986

y = 2x − 5

y = −2x − 5

y = 2x + 5

The graph of y = x2 is translated 3 units down. What is the new equation? A

y = −2x + 5

y = (x − 3)2

y = x2 + 3

y = x2 − 3

y = (x + 3)2

What transformations are applied to y = x3 to obtain the graph of y = 3(x − 2)3 + 5? A

Horizontal translation 2 units left, vertical dilation by a factor of 3, vertical translation 5 units up

Horizontal translation 2 units right, vertical dilation by a factor of 3, vertical translation 5 units up

Horizontal translation 2 units right, horizontal dilation by a factor of 3, vertical translation 5 units up

Horizontal translation 2 units left, vertical translation 3 units up, vertical dilation by a factor of 5

Determine the coordinates of the point (4, −5) after each reflection: a

Reﬂection in the y-axis

Reﬂection in the x-axis

Sketch the point (4, −5) and its reﬂections in the x-axis and y-axis, labelling all points.

For the function y = x3 − 4x: a

Find the equation after reﬂection in the y-axis.

Determine the x-intercepts of the reﬂected graph.

Sketch the graph and its y-axis reﬂection, including the x-intercepts.

For the function y = (x − 3)2: a

Find the equation after reﬂection in the x-axis.

Determine the vertex of the reﬂected graph.

Sketch the graph and its x-axis reﬂection, including the vertex.

An architect models a bridge arch with the curve y = x2 − 9: a

Write the equation of the arch mirrored across the y-axis and determine if the x-intercepts change. Explain your reasoning.

Sketch the graph and its y-axis reﬂection, including the x-intercepts.

Consider the function y = x2 − 4x. After reflecting it in the y-axis, the graph is then reflected in the x-axis. Find the final equation of the graph and determine its vertex. Verify your answer by checking the vertex’s position relative to the original graph.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For the function y = x2 − 1: a

Determine the equation after a horizontal translation of 3 units left.

Determine the vertex of the translated graph.

Sketch the original graph and the translated graph, labelling the vertex.

For the function y = 2x + 3: a

Find the equation after a vertical translation of 4 units down.

Sketch both graphs, labelling the y-intercept.

The graph of y = x3 is translated to y = (x − 1)3 + 2: a

Describe the translations and ﬁnd the point corresponding to (0, 0) on the original graph.

Sketch the graph and the translated graph, labelling the point corresponding to (0, 0).

A biologist’s population model is P (t) = −t2 + 8t, where t is the time in months. A new condition translates the model 2 units right and 3 units down: a

Find the new maximum population and the month in which it occurs.

Sketch the original and the translated graph, labelling the vertex of each.

The graph of y = f (x) has a point at (1, 5). After translating to y = f (x − a) + b, the point moves to (4, 2). Find the values of a and b.

For the function y = x2 − 1:

Find the equation and sketch its graph after a horizontal dilation by a factor of 2.

Find the equation and sketch its graph after a vertical dilation by a factor of 3.

For the function y = 9x2 + 1: a

Find the equation after a combined dilation by a factor of 3.

Sketch both graphs, labelling the vertex on the original and its corresponding point on the transformed graph.

The graph of y = x2 and its transformed image are shown: y 4 2

y=x

(3, 2)

(1, 1)

x −3 −2

−1

Determine the type and factors of the dilation applied.

Find the equation of the transformed graph.

Chapter 17 review mathspace.co

987

A company logo is modelled by the function y = x2. For a new design, the logo is dilated horizontally by a factor of 3 and vertically by a factor of 0.5: a

Find the equation of the scaled logo.

The original logo passes through (3, 9). Find the corresponding point on the scaled logo.

Sketch the original and the scaled logo, labelling the corresponding point.

An engineering blueprint for a satellite dish is modelled by the function y = x2. A scaled model is created using a combined dilation factor of 0.25: a

Find the equation of the scaled model.

If the original blueprint passes through (2, 4), ﬁnd the corresponding point on the scaled model.

Since 0 < k < 1, does the transformation reduce the graph, producing a smaller-scale version? Justify.

For the function f (x) = x3, compare the effect of: • A horizontal dilation by a factor of k = 2, followed by a vertical dilation by a factor of l=3 • A combined dilation by a factor of k = 2 Find the resulting equations for both cases and explain why they are different.

What is the horizontal asymptote of the function y = 5x − 1 − 3? A

y=5

y = −1

y = −3

y=1

A circle is deﬁned by the equation (x + 4)2 + ( y − 1)2 = 49. What are its centre and radius? A

Centre (4, −1), radius 49

Centre (−4, 1), radius 49

Centre (4, −1), radius 7

Centre (−4, 1), radius 7

Find the y-intercept for the exponential function y = 2 × 5x + 1 − 4.

Translate f (x) = log3 (x) right by 2 units and up by 1 unit: a

Determine the equation of the transformed function.

Determine the domain and the vertical asymptote.

Sketch the graph on a Cartesian plane, marking the asymptote and the point where x = 5.

Sketch the graph of g(x) = −3 × 2x − 1, marking the asymptote and the y-intercept. State the transformations required to obtain this graph from f (x) = 2x.

An exponential function of the form y = a × 3x + k has a horizontal asymptote at y = −2 and passes through the point (1, 7). Find the values of a and k.

Find the vertex and range for the absolute value function f (x) = −2∣x + 3∣ + 1.

Find the equations of the vertical and horizontal asymptotes for the reciprocal function f (x) =

988

− 2.

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

The graph shows a transformed reciprocal function. Determine: a

The equation of the curve.

The domain and range of the function.

y 1 −1

x 1

−1

(2, −1)

−2

−3 −4 −5 (0, −5)

The graph of a transformed absolute value function is shown. Determine its equation.

y 1 (0, 1) x −4

−3

−2

−1 −1 −2

(−2, −3)

Find the centre and radius of the circle with equation (x + 1)2 + ( y − 5)2 = 36.

A circle has an equation of x2 + y2 − 4x + 6y + 4 = 0: a

State the standard form equation by completing the square.

Find the centre and radius.

−3

A circle has its centre at (−2, 5) and a radius of 3. Write the equation of the circle and state its domain and range.

A circle passes through the point (−1, 5) and has its centre at (2, 1). Find the equation of the circle.

Transform f (x) =

by dilating vertically by a factor of 2 and translating up by 4 units.

Determine the equation when applying dilation then translation.

Determine the equation when applying translation then dilation.

A student claims that for f (x) = x2, applying a vertical dilation by a factor of 3 and then a vertical translation up by 2 units is the same as applying the translation ﬁrst and then the dilation. Show, by ﬁnding the equation for each case, that the student is incorrect.

Chapter 17 review mathspace.co

989

Transform f (x) = ∣x∣ by dilating vertically by a factor of 3 and horizontally by a factor of 2, reﬂecting over the x-axis, translating right by 1 unit, and up by 4 units: a

Determine the equation of the transformed function.

Determine the vertex.

Determine x-intercept and y-intercept.

Sketch the graph, labelling the vertex and intercepts.

Did you know?

Robots use transformations to calculate how their parts move in space — every rotation, translation, and reflection helps them perform precise tasks. Whether it’s turning a wheel or lifting an arm, transformations ensure each movement aligns perfectly with the robot’s surroundings. Engineers apply these same mathematical concepts when programming robots to navigate rooms, avoid obstacles, and pick up objects. Without transformations, robots wouldn’t be able to understand direction, position, or movement in the real world!

990

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

“Theory is in advance of experiment, only so far as we can see.” Sophus Lie

Big ideas Graphical relationships of reciprocal, absolute value, and combined functions enable the analysis and modelling of complex behaviours through transformations and key features.

18 Graphical relationships Chapter outline 18.01E 18.02E 18.03E 18.04E 18.05E

Reciprocal functions Reciprocal trigonometric functions Absolute value functions Sum and difference of functions Graphical relationships Chapter 18 review

994 1000 1011 1025 1031 1037

Sunsets paint the sky because light bend — waves and angles are ruled by trigonometric functions!

18.01E Reciprocal functions After this lesson, you will be able to… given y = f ( x).

• define the reciprocal function y =

• identify that zeroes of f ( x) correspond to vertical asymptotes of y = • identify that if f ( x) → ±∞ as x → ±∞, then y = at y = 0. • analyse the behaviour of y = negative side. • sketch the graph of y =

has a horizontal asymptote

as f ( x) approaches 0 from the positive or given the graph or algebraic form of y = f ( x).

• determine the domain of y =

Reciprocal functions Domain The set of allowable values of x in a function or relation. Range The set of values of the dependent variable for which a function is deﬁned. Asymptote A straight line (or another curve) that a curve approaches as x tends to ±∞, or to some particular value. For example, the curve y = 2x has an asymptote y = 0 as x tends to −∞, and the curve y = has asymptotes x = 0 and y = 0 as x tends to 0 and ±∞, respectively. The reciprocal function of y = f ( x) is y = Key properties of y =

, deﬁned where f ( x) ≠ 0. If f ( x) = a (a ≠ 0), then y = .

• Domain: Deﬁned where f ( x) ≠ 0. Vertical asymptotes occur at f ( x) = 0. • Range: Determined by f ( x) values. A horizontal asymptote at y = 0 exists if f ( x) → ±∞ as x → ±∞. • Behaviour:

994

• As f ( x) → 0+,

• As f ( x) → 0−,

• For large ∣f ( x)∣,

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Example 1 Given f ( x) = x2 − 1, analyse y =

a Determine where f ( x) = 0 and explain its signiﬁcance for y =

Create a strategy Solve f ( x) = 0 to ﬁnd zeroes, which indicate vertical asymptotes for

Apply the idea x2 − 1 = 0

Write the equation

( x − 1) ( x + 1) = 0

Factorise

x = 1 or − 1

Evaluate , where the function approaches ±∞.

Solutions x = ±1 indicate vertical asymptotes for y =

b Analyse f ( x) as x → ±∞ and its effect on y =

Create a strategy Evaluate f ( x) = x2 − 1 as x → ±∞ to determine the horizontal asymptote of

Apply the idea As x → ±∞, f ( x) = x2 − 1 → ∞, so

→ 0. Thus, y = 0 is the horizontal asymptote.

c Determine the behaviour of y =

near x = ±1.

Create a strategy Examine f ( x) = x2 − 1 as x approaches ±1 from both sides to ﬁnd how

behaves.

Apply the idea

• As x → −1−, x2 − 1 → 0+, so y → +∞. • As x → −1+, x2 − 1 → 0−, so y → −∞. • As x → 1−, x2 − 1 → 0−, so y → −∞. • As x → 1+, x2 − 1 → 0+, so y → +∞.

18.01E Reciprocal functions mathspace.co

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d Sketch y =

Create a strategy Plot asymptotes at x = ±1 and y = 0, then use behaviour near x = ±1 to sketch the graph.

Apply the idea 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

The graph has three branches, avoiding asymptotes at x = ±1 and y = 0.

Idea summary For y = f ( x) and y =

• Zeroes of f ( x) form vertical asymptotes of • If f ( x) → ±∞,

→ 0, giving a horizontal asymptote at y = 0. approaches ±∞; for large ∣ f ( x)∣, it nears y = 0.

• Near zeroes of f ( x),

18.01E Practice questions What do you remember? 1

Complete this table: x

−2

−1

f ( x) = x − 3

⬚

f ( x) =

996

⬚

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

⬚

Consider a function y = f ( x) deﬁned over the domain [−1, ∞). Determine whether the statement is true or false:

The zeroes of f ( x) become horizontal asymptotes for the graph of y =

If f ( x) increases positively,

The domain of y =

approaches zero.

is [−1, ∞).

Determine the domain of the function g deﬁned by g( x) = a

f ( x) = 2 − x

f ( x) = x2

when:

f ( x) = 2 + x4

f ( x) = 2 +

Practice 4

Consider the functions f and g deﬁned by f ( x) = 2x − 5 and g( x) =

Determine the behaviour of: a c 5

f ( x) when x approaches ∞ when x approaches −∞

g( x) when x approaches −∞

when x approaches ∞

Consider the functions f and g deﬁned by f ( x) = −2x + 4 and g( x) =

Determine the behaviour of:

Ex 1

f ( x) when x approaches 2

g( x) when x approaches

from the right

when x approaches 2 from the left

when x approaches , where f ( x) = x2 − 4:

Consider the function y = a

Solve the equation f ( x) = 0 for x, and explain the signiﬁcance of the solutions for the reciprocal function y =

Determine the behaviour of f ( x) as x approaches −∞ and ∞, and interpret the result for the graph of the reciprocal function y =

Determine the behaviour of y =

Plot the graph of y =

around the lines with equations x = 2 and x = −2.

. , where f ( x) = x2 + 2x − 3:

Consider the function y = a

Solve for x the equation f ( x) = 0.

Determine the behaviour of f ( x) as x approaches −∞ and ∞.

Plot the graph of y =

18.01E Reciprocal functions mathspace.co

997

Consider the function y =

, where f ( x) = ∣x − 5∣:

Solve for x the equation f ( x) = 0.

Determine the behaviour of f ( x) as x approaches −∞ and ∞.

Plot the graph of y =

This graph represents a function f ( x): Plot the graph of y =

on a Cartesian plane. y = f (x)

x −5

−5

−10

x=0

This graph represents a function f ( x): Plot the graph of y =

4 2 x −4

−2

−2 −4

This graph represents a function f ( x): Plot the graph of y =

on a Cartesian plane.

3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

Given f ( x) = 1 − 2x2, plot g ( x) =

Using the graph of f ( x) = 3 + , plot the graph of y =

Using the graph of f ( x) = x2 + 3, plot the graph of y =

998

Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

. . .

x 1

Given y = f ( x) = x2 − k, f ( x) and A

x = ±1

will intersect when: C

y = ±1

Extend your thinking 16

Consider the graph of y =

y 15

Determine whether the statement is true or false. a

The domain of f ( x) is all x ≥ 0.

f ( x) ≥ 0 for all values of x

The curve of f ( x) has a symmetry about the line with equation x = 5.

The curve of f ( x) could be the graph of a certain parabola.

10 5 x −10

−5

−5 −10

This graph represents two functions f and g such that f ( x) × g( x) = 1:

Determine g( x).

f (x)

3 2 1

g(x)

−4 −3 −2 −1 −1

−2 −3 −4

Sketch the graph of

y 10 x −5

5 −10 −20

f (x)

−30 −40

18.01E Reciprocal functions mathspace.co

999

18.02E Reciprocal trigonometric functions After this lesson, you will be able to… • define csc x, sec x, and cot x as reciprocals of sin x, cos x, and tan x, respectively. • identify the period, domain, range, and vertical asymptotes for y = csc x, y = sec x, and y = cot x. • describe the symmetry of each reciprocal trigonometric function (odd/even). • graph y = csc x, y = sec x, and y = cot x in both radians and degrees. • compare the graph of each reciprocal trigonometric function with the graph of its corresponding primary trigonometric function.

Cosecant function Period A function f is periodic with period p if f ( x + p) = f ( x), for all x, that is, the function repeats itself after each interval of length p. For example, sin x and cos x have period 2π. Symmetry A property of a function or graph that remains unchanged under certain transformations. For example, the function f ( x) = x2 exhibits symmetry about the y-axis, as f (−x) = f ( x).

The reciprocal trigonometric functions y = csc x, y = sec x, and y = cot x are deﬁned as: csc x =

, sec x =

, and cot x =

. Their graphs reﬂect their primary functions’

behaviour, with differences due to undeﬁned points and reciprocal relationships. Cosecant function: y = csc x • Period: 2π radians (360°), matching sin x. • Domain: x ≠ nπ radians ( x ≠ 180°n), where n ∈ Z (undeﬁned at sin x = 0). • Range: (−∞, −1] ∪ [1, ∞). • Asymptotes: Vertical at x = nπ radians ( x = 180°n). • Symmetry: Odd function, symmetric about the origin (csc(−x) = − csc x).

1000 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

• Comparison with y = sin x: At sin x = ±1, csc x = ±1 (extrema). At sin x = 0, csc x has asymptotes. Forms U-shaped curves. y

y = csc (x)

3 2

−2π

y = sin (x)

−1π

2π

1π

−1 −2 −3

x=0

Example 1 For y = csc x, ﬁnd the period and conﬁrm using the graph.

Create a strategy Use the deﬁnition of csc x and verify repetition on the graph.

Apply the idea Since csc x =

and sin x has period 2π radians (360°), csc x has period 2π.

y = csc (x)

y 3 2 1

−2π

−1π

−1

1π

2π

−2 −3 x = 0

The graph repeats every 2π radians (e.g., from −π to π ), conﬁrming the period.

Idea summary The cosecant function y = csc x has a distinct property. The function csc x mirrors sin x, with asymptotes at sin x = 0.

18.02E Reciprocal trigonometric functions 1001 mathspace.co

Secant function Secant function: y = sec x • Period: 2π radians (360°), matching cos x. + nπ radians ( x ≠ 90° + 180°n), where n ∈ Z.

• Domain: x ≠

• Range: (−∞, −1] ∪ [1, ∞). • Asymptotes: Vertical at x =

+ nπ radians ( x = 90° + 180°n).

• Symmetry: Even function, symmetric about the y-axis (sec(−x) = sec x). • Comparison with y = cos x: At cos x = ±1, sec x = ±1. At cos x = 0, sec x has asymptotes. Forms U-shaped curves. y

y = sec (x)

3 2 1

y = cos (x) −2π

−1π

−1

1π

−2 −3

Example 2 For y = sec x over −2π ≤ x ≤ 2π : a Find the vertical asymptotes in radians and degrees.

Create a strategy Identify where cos x = 0, as sec x =

is undeﬁned at these points.

Apply the idea + nπ :

Since cos x = 0 at x = Radians: x =

Degrees: x = −270°, −90°, 90°, 270°.

b Determine where sec x = 1.

Create a strategy Solve

= 1, so cos x = 1.

Apply the idea cos x = 1 at x = −2π, 0, 2π radians (−360°, 0°, 360°).

1002 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2π

Idea summary The secant function y = sec x has a distinct property. The function sec x mirrors cos x, with asymptotes at cos x = 0.

Cotangent function Cotangent function: y = cot x • Period: π radians (180°), matching tan x. • Domain: x ≠ nπ radians ( x ≠ 180°n), where n ∈ Z. • Range: (−∞, ∞). • Asymptotes: Vertical at x = nπ radians ( x = 180°n). • Symmetry: Odd function, symmetric about the origin (cot (−x) = −cot x). • Comparison with y = tan x: At tan x = 0, cot x has asymptotes. Where tan x → ±∞, cot x → 0. Curves decrease from +∞ to −∞. y

y = tan (x) y = cot (x) −2π

−1π

3 2 1

1π

−1 −2 −3

2π

x=0

Example 3 Determine the largest negative value of x for which cot ( x) has an asymptote.

Create a strategy The function y = cot ( x) has vertical asymptotes where sin ( x) = 0, since cot ( x) = at these points.

is undeﬁned

18.02E Reciprocal trigonometric functions 1003 mathspace.co

Apply the idea Looking at the graph of y = sin ( x): y

−2π

−1π

2π

1π

−1

It can be seen that sin ( x) = 0 for any integer multiple of π. The largest negative value of x for which sin ( x) = 0 is x = −π.

Idea summary The cotangent function y = cot x has a distinct property. The function cot x inverts tan x, with asymptotes at tan x = 0 and zeroes where tan x is undeﬁned.

18.02E Practice questions What do you remember? 1

In a right triangle, the side opposite angle θ is 7, and the adjacent side is 24. Find: a

sin θ, cos θ, and tan θ.

The measure of θ in degrees, rounded to two decimal places.

Consider the graph of y = sin ( x) for −2π ≤ x ≤ 2π : a

Complete the table of values for y = sin ( x), leaving your answer in exact form: x sin ( x)

0 ⬚

⬚

Solve sin ( x) = 0 for −2π ≤ x ≤ 2π.

Solve sin ( x) = 1 for −2π ≤ x ≤ 2π.

⬚

1004 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

⬚

Practice Ex 1

Consider the graph of y = sin ( x) for −2π ≤ x ≤ 2π : a

Complete the table of values with exact values: x cosec ( x)

Ex 2

0 ⬚

⬚

y = cot ( x)

⬚

Solve sin ( x) = 0 for −2π ≤ x ≤ 2π and calculate the asymptotes for y = cosec ( x).

Solve cosec ( x) = 1 for −2π ≤ x ≤ 2π.

Determine the period of cosec ( x).

Identify: i

Period

Domain and range

iii

Vertical asymptotes in terms of n

y = cosec ( x)

y = sec ( x)

Describe why f ( x) = cosec ( x) has vertical asymptotes at certain points. Determine the points in terms of k, where k is some integer.

For y = sec x over −π ≤ x ≤ π :

Find the vertical asymptotes in radians and degrees.

Determine where sec x = −1.

6 4 2 x −2 −4 −6

Find the period f ( x) =

, and conﬁrm using the graph.

18.02E Reciprocal trigonometric functions 1005 mathspace.co

Ex 3

for interval (0, 2π ),

Consider cot ( x) = as shown:

What is the period of cot ( x)?

For which values of x will cot ( x) be:

Undeﬁned

Equal to 1

cot ( x) =

4 2 x

when x =

. Determine:

−2

The next value of x for which cot ( x) =

The next value of x after cot ( x) =

−4

for which

−6

Determine the largest negative value of x for which cot ( x) = 0.

Complete the table of values, given that all values are non-negative: x sin ( x) cosec ( x) tan ( x) cot ( x)

⬚

⬚ ⬚

⬚

Undefined ⬚

This graph shows y = cosec ( x): a

The next positive x-value with an asymptote

i ii b

There is a vertical asymptote at x = π, determine:

The highest negative x value with an asymptote

The function is equal to 2 when x = determine:

The next positive x-value where y = 2

The ﬁrst positive x-value greater than 2π where y = 2

iii

The highest negative x-value where y = −2

The lowest positive x-value where y = −2

What is the period of the function?

1006 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−2

−4

Consider the graphs f ( x) = cos ( x) and g ( x) = sec ( x) drawn on the same diagram: a

Determine the values of x which satisfy f ( x) = g ( x).

What is the range of f ( x)?

What is the range of g ( x)?

What are the asymptotes of g ( x) = sec ( x)?

What is the period of f ( x) and g ( x)?

y 3

g(x) = sec (x) 2

f (x) = cos (x)

1 x

−1 −2 −3

Consider the graphs f ( x) = sin ( x) and g ( x) = cosec ( x) drawn on the same diagram: a

Determine the values of x which satisfy f ( x) = g ( x).

What is the range of f ( x)?

What is the range of g ( x)?

What are the asymptotes of g ( x) = cosec ( x)?

What is the period of f ( x) and g ( x)?

y 3

g(x) = cosec (x)

2 1

f (x) = sin (x)

−1 −2 −3

18.02E Reciprocal trigonometric functions 1007 mathspace.co

Consider the graphs f ( x) = tan ( x) and g ( x) = cot ( x) drawn on the same diagram: y

g(x) = cot (x)

3 2

f (x) = tan (x)

−1 −2 −3

Determine the values of x which satisfy f ( x) = g ( x).

What is the range of f ( x)?

What is the range of g ( x)?

What are the asymptotes of g ( x) = cot ( x)?

What is the period of f ( x) and g ( x)?

The graphs of g ( x) = sec ( x) and h ( x) = cosec ( x) are shown: y 2 1

h(x) = cosec (x)

g(x) = sec (x)

x −1 −2

Identify the coordinates of the points where h ( x) = g ( x). 16

Consider the graph of f ( x) = cot ( x) given: Determine whether each statement is true or false: a

f ( x) has vertical asymptotes at x = 0 and x = π.

f ( x) is periodic with a period of π.

f ( x) passes through the origin (0, 0).

f ( x) is deﬁned for all x ∈ .

y 2

1 x

−1

−2

1008 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the graphs of y = cosec ( x), y = cot ( x) and y = sec ( x) shown:

For which intervals over the domain [0, 2π ] are these graphs increasing? a

cosec ( x)

cot ( x)

sec ( x)

y = cosec (x) y = cot (x)

−1

−2

y = sec (x)

Extend your thinking 18

Consider the graphs of f ( x) = sec ( x), g ( x) = sec ( x + α ) and h ( x) = sec ( x − β ) shown:

y 3

Determine the values α and β.

2 g(x)

f (x)

h(x) x

−1 −2 −3

Consider

and

. Although either may be used as an expression for cot ( x), there

are certain inputs for which they will not provide the same value. Determine the x-intercepts of both for the interval [0, 2π ]. Using the results, explain the difference between

and

18.02E Reciprocal trigonometric functions 1009 mathspace.co

Consider the graphs of f ( x) = cosec ( x), g ( x) = k cosec ( x) and h ( x) =

cosec ( x) shown:

g(x) = k cosec (x) 2

f (x) = cosec (x)

1 x −1 −2

Determine the values of k and m. 21

Consider the graph of f ( x) = a sec ( x + b) + c given: y 18 16 14 12 10 8 6 4 2

−2

By considering suitable coordinates, determine the values of a, b and c, given that: • a>0 • • c>0

1010 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

18.03E Absolute value functions After this lesson, you will be able to… • sketch the graph of the basic absolute value function y = ∣x∣. • identify and apply transformations (vertical stretch/compression, reflection, horizontal and vertical translations) to graph functions of the form y = a∣x − h∣ + k. • graph y = ∣ f ( x)∣ given the graph or algebraic form of y = f ( x), by reflecting portions below the x-axis. • graph y = f (∣x∣) given the graph or algebraic form of y = f ( x), by using the x ≥ 0 portion and reflecting it in the y-axis. • examine the relationship between the graphs of y = f ( x), y = ∣ f ( x)∣, and y = f (∣x∣).

Absolute value graphs and transformations Reﬂection

Horizontal translation

A transformation of a shape formed by creating a mirror image on the other side of a given line.

A transformation that shifts a graph left or right by h units, right if h > 0 or left if h < 0.

Vertical stretch (compression)

Vertical translation

A transformation that scales a graph vertically by a factor ∣a∣, stretching it if ∣a∣ > 1 or compressing it if 0 < ∣a∣ < 1.

A transformation that shifts a graph up or down by k units, up if k > 0 or down if k < 0.

18.03E Absolute value functions 1011 mathspace.co

The absolute value function y = ∣x∣ forms a V-shaped graph, deﬁned as ∣x∣ = x for x ≥ 0 and ∣x∣ = −x for x < 0. It comprises two lines: y = x for x ≥ 0 and y = −x for x < 0, intersecting at (0, 0). y 3 2

y = ∣x∣

1 x −3 −2

−1

Interactive exploration Discover this concept in action online

mathspace.co

The general form y = a∣x − h∣ + k transforms the graph as: • a: Vertical stretch (∣a∣ > 1) or compression (0 < ∣a∣ < 1). If a < 0, the graph reﬂects across the x-axis. • h: Horizontal translation, right if h > 0, left if h < 0. • k: Vertical translation, up if k > 0, down if k < 0.

Example 1 Plot y = 2∣x − 3∣ + 4.

Create a strategy Start with the base function y = ∣x∣. Plot the transformed graph, showing intermediate steps to illustrate each transformation.

Apply the idea y 10 8

Apply transformations in this order:

1. Vertical stretch by a factor of 2 (since a = 2).

2. Horizontal shift right by 3 units (since h = 3). 3. Vertical shift up by 4 units (since k = 4).

2 x −4

−2

1012 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Reﬂect and check To verify, calculate key points algebraically. The vertex of y = 2∣x − 3∣ + 4 is at (3, 4) (from h = 3, k = 4). For x = 0: y = 2∣x − 3∣ + 4

Write the equation

= 2∣0 − 3∣ + 4

Substitute x = 0

= 2∣− 3∣ + 4

Evaluate the subtraction

=2×3+4

Evaluate the absolute value

=6+4

Evaluate the multiplication

= 10

Evaluate the addition

For x = 6: y = 2∣x − 3∣ + 4

Write the equation

= 2∣6 − 3∣ + 4

Substitute x = 6

= 2∣3∣ + 4

Evaluate the subtraction

=2×3+4

Evaluate the absolute value

=6+4

Evaluate the multiplication

= 10

Evaluate the addition

Points (0, 10) and (6, 10) lie on the graph, conﬁrming the V-shape with vertex at (3, 4). y 10 (0, 10)

(6, 10)

8 6 4

Vertex (3, 4)

2 x −4

−2

Idea summary The graph of y = ∣x∣ is a V-shape with vertex at (0, 0). Transformations via y = a∣x − h∣ + k involve vertical scaling/reﬂection (a), horizontal shifts (h), and vertical shifts (k).

18.03E Absolute value functions 1013 mathspace.co

Graphs of y = f ( x), y = |f ( x)|, and y = f (|x|) Interactive exploration Discover this concept in action online

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The relationships between y = f ( x), y = ∣f ( x)∣, and y = f (∣x∣) reveal how absolute values modify a function’s graph: • y = f ( x): The original function’s graph. • y = ∣f ( x)∣: Reﬂects negative outputs (f ( x) < 0) above the x-axis, ensuring non-negative outputs. • y = f (∣x∣): Uses absolute input, making the graph symmetric about the y-axis by mirroring the x ≥ 0 portion. To graph y = ∣f ( x)∣: 1. Plot y = f ( x). 2. Keep portions where f ( x) ≥ 0 unchanged. 3. Reﬂect portions where f ( x) < 0 above the x-axis.

Example 2 Plot y = ∣cos (2x)∣.

Create a strategy To graph y = ∣ cos (2x)∣: 1. Plot y = cos (2x), which has a period of π (since the frequency is 2) and amplitude of 1. 2. Identify intervals where cos (2x) ≥ 0, for example,

, to keep unchanged.

3. Reﬂect intervals where cos (2x) < 0, for example, negative outputs.

, above the x-axis to ensure non-

Apply the idea For example, for y = cos (2x), reﬂect negative portions to graph y = ∣ cos (2x)∣. y

y = |cos (2x)| x

0 −1

y = cos (2x)

1014 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Reﬂect and check Verify algebraically by calculating key points to conﬁrm the graph of y = ∣ cos (2x)∣. For x = 0: y = ∣cos (2x)∣

For x =

Write the equation

= ∣cos (2 × 0)∣

Substitute x = 0

= ∣cos (0)∣

Evaluate the multiplication

= ∣1∣

Evaluate cosine

Evaluate absolute value

: Write the equation

Substitute x =

Evaluate the multiplication

Evaluate cosine

Evaluate absolute value

The points (0, 1) and

match the graph, conﬁrming the reﬂection of negative portions.

Example 3 Given f ( x) = x − 5, plot: a y = f ( x)

Create a strategy To graph y = x − 5: 1. Identify the slope and y-intercept: slope is 1, y-intercept is (0, −5). 2. Plot the y-intercept at (0, −5). 3. Use the slope to plot another point, for example, from (0, −5), move right 1 unit and up 1 unit to (1, −4). 4. Draw a straight line through these points.

18.03E Absolute value functions 1015 mathspace.co

Apply the idea 3 2 1 −7−6−5−4−3−2 −1 −1

For y = f ( x) = x − 5, plot the line (slope 1, y-intercept −5).

y=x−5 x 1 2 3 4 5 6 7

−2 −3 −4 −5 −6

Reﬂect and check Verify by calculating key points for y = x − 5. For x = 0: y=x−5

Write the equation

=0−5

Substitute x = 0

= −5

Evaluate

For x = 5: y=x−5

Write the equation

=5−5

Substitute x = 5

Evaluate

The points (0, −5) and (5, 0) lie on the line, conﬁrming the slope and intercept. b y = f (∣x∣)

Create a strategy To graph y = ∣x∣ − 5: 1. Plot y = x − 5 for x ≥ 0, starting from the y-intercept at (0, −5) with slope 1. 2. Keep the portion where x ≥ 0. 3. Mirror this portion across the y-axis for x < 0, forming a V-shape with vertex at (0, −5).

1016 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea To graph y = f (∣x∣): 1. Plot y = f ( x). 2. Keep the portion where x ≥ 0. 3. Mirror this portion across the y-axis for x < 0. 3 2 1 −7−6−5−4−3−2 −1 −1

For y = f (∣x∣) = ∣x∣ − 5, keep x ≥ 0, mirror across the y-axis, forming a V-shape with vertex at (0, −5).

y = |x| − 5 x 1 2 3 4 5 6 7

−2 −3 −4 −5 −6

Reﬂect and check Verify algebraically by calculating key points for y = ∣x∣ − 5. For x = 0: y = ∣x∣ − 5 = ∣0∣ − 5 =0−5 = −5

Write the equation Substitute x = 0 Evaluate the absolute value Evaluate

y = ∣x∣ − 5 = ∣5∣ − 5 =5−5 =0

Write the equation Substitute x = 5 Evaluate the absolute value Evaluate

y = ∣x∣ − 5 = ∣−5∣ − 5 =5−5 =0

Write the equation Substitute x = −5 Evaluate the absolute value Evaluate

For x = 5:

For x = −5:

The points (0, −5), (5, 0), and (−5, 0) conﬁrm the V-shape with vertex at (0, −5).

18.03E Absolute value functions 1017 mathspace.co

Example 4 Given f ( x) = 2x − 5, plot: a y = ∣f ( x)∣

Create a strategy

Apply the idea

Plot y = 2x − 5. Reﬂect portions below the x-axis to graph y = ∣2x − 5∣.

6 4 2 −8 −6 −4 −2 −2

x 2

−4 −6 −8

Reﬂect and check Another way is to ﬁnd the points algebraically. For the vertex, solve 2x − 5 = 0. 2x − 5 = 0

Write the equation

2x = 5

Add 5 to both sides

x = 2.5

Divide both sides by 2

So the vertex is (2.5, 0). Then, substitute some key points into y = ∣2x − 5∣ to plot on the graph. For x = 0: y = ∣2x − 5∣

Write the equation

= ∣2 × 0 − 5∣

Substitute x = 0

= ∣0 − 5∣

Evaluate the multiplication

Evaluate the absolute value

For x = 5: y = ∣2x − 5∣

Write the equation

= ∣2 × 5 − 5∣

Substitute x = 5

= ∣10 − 5∣

Evaluate the multiplication

Evaluate the absolute value

2 −8 −6 −4 −2 −2 −4 −6 −8

1018 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x 2

b y = f (∣x∣)

Create a strategy Plot y = 2x − 5 for x ≥ 0. Mirror across the y-axis for x < 0 to graph y = 2∣x∣ − 5.

Apply the idea 8

6 4 2

−8 −6 −4 −2 −2

−4 −6 −8

Reﬂect and check This can also be graphed by substituting some key points into y = 2∣x∣ − 5. For x = 0:

For x = 2:

y = 2∣x∣ − 5

Write the equation

y = 2∣x∣ − 5

Write the equation

= 2∣0∣ − 5

Substitute x = 0

= 2∣2∣ − 5

Substitute x = 2

= 2(0) − 5 Evaluate the absolute value

= 2(2) − 5 Evaluate the absolute

= 0 − 5 Evaluate the multiplication

= 4 − 5 Evaluate the multiplication

= −5

= −1

Evaluate 8

Evaluate

6 4 2 −8 −6 −4 −2 −2

x 2

−4 −6 −8

18.03E Absolute value functions 1019 mathspace.co

c Verify the graphs using this table: x

−2

−1

f ( x) = 2x − 5

⬚

f ( x) = ∣2x − 5∣

⬚

f ( x) = 2∣x∣ − 5

⬚

Create a strategy Substitute x-values into each function and compare with graphs.

Apply the idea x

−2

−1

f ( x) = 2x − 5

−9

−7

−5

−3

−1

f ( x) = ∣2x − 5∣

f ( x) = 2∣x∣ − 5

−1

−3

−5

−3

−1

Points like (−2, 9), (0, 5), (2, 1) lie on y = ∣2x − 5∣, and (−2, −1), (0, −5), (2, −1) on y = 2∣x∣ − 5, matching the plotted graphs.

Idea summary y = ∣ f ( x)∣ reflects negative outputs above the x-axis. y = f (∣x∣) mirrors the x ≥ 0 portion across the y-axis, creating symmetry. These transformations affect the function’s range and symmetry, respectively.

18.03E Practice questions What do you remember? 1

Complete this table: x

−2

−1

f ( x) = ∣x∣

⬚

f ( x) = x − 3 f ( x) = ∣x − 3∣ f ( x) = 2∣x − 3∣ − 5

⬚

1020 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

⬚

Complete this statement: The graph of y = 4 ∣ x−3 ∣ +2 can be obtained by starting with the graph of ⬚, then applying a vertical stretch by a factor of ⬚, shifting ⬚ by 3 units, and shifting up by ⬚ units. 10 8 6 4 2

−5 −4 −3 −2 −1 −2 −4 −6 −8 −10

x 1 2 3 4 5

Given that the blue curve represents the graph of the function y = ∣x∣, determine the equation of each function in the form of y = a∣x − h∣ + k: a

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

−4 −3 −2 −1 −1 −2

x 1

−4 −3 −2 −1 −1

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

18.03E Absolute value functions 1021 mathspace.co

Practice Ex 1

Plot y = −∣x − 3∣ + 1 without using a table of values.

Plot y = 2∣x − 3∣ + 1 using a table of values.

Consider the graph of a function y = f ( x).

9 8 7 6 5 4 3 2 1

Determine whether the statement is true or false: a

The domain of f ( x) is all x ≥ 0.

f ( x) ≥ 0 for all values of x.

The curve has a symmetry about the y-axis.

The curve could be the graph of a function y = −∣x − 1∣ + 1.

−3 −2 −1−1

x 1 2 3 4 5 6 7

−2 −3

Ex 2

Plot y = ∣ sin (3x)∣.

Ex 3

Using the graph of f ( x) = −2x + 3, plot the graph of: a

Ex 4

y = ∣f ( x)∣

y = f (∣x∣)

Verify the graphs using this table: x

−2

−1

f ( x) = −0.5x − 5

⬚

f ( x) = −0.5∣x∣ − 5

⬚

y = f ( x)

y = ∣f ( x)∣

y = f (∣x∣)

f ( x) = ∣1 − 3x∣

Plot the graph of: a

⬚

Consider the function f ( x) = x2 − 5, plot the graph of: a

y = f (∣x∣)

Using the graph of f ( x) = −0.5x − 5, plot the graph of:

f ( x) = ∣ − 0.5x − 5∣

f ( x) = 1 − 3∣x∣

The graph of y = −∣x − 4∣ + 2 is translated 2 units down and 3 units left, then reﬂected across the x-axis. What is the equation of the transformed graph?

1022 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Consider the equation f ( x) = ∣x∣ + 3: a

Sketch the graph of the function.

Which of these statements about the symmetry of the graph is true? A

The graph is symmetric about the y-axis

The graph is rotationally symmetric about the origin

The graph is symmetric about the x-axis

The graph has no symmetry.

Determine a simpliﬁed expression for f (−x).

Complete the statement:

The result of part (c) shows that f (−x) = ⬚, which veriﬁes that the function is symmetric about the y-axis. 14

Consider the function f ( x) = 11 − ∣x∣. a

Sketch the graph of the function.

What is the maximum value of f ( x)?

On which interval is the function increasing? A

(−∞, 0)

(−∞, 0]

(−∞, 11]

[0, ∞)

(0, ∞)

On which interval is the function decreasing? A

(−∞, 11)

[11, ∞)

(11, ∞)

Consider the functions f ( x) = ∣x∣ and g( x) = 4∣x∣ − 1: a

Complete this table: x

−2

−1

f ( x) = ∣x∣

⬚

g( x) = 4∣x∣ − 1

⬚

Plot g( x) = 4∣x∣ − 1.

What transformations are applied to the graph of f ( x) = ∣x∣ to obtain the graph of g( x) = 4∣x∣ − 1?

Extend your thinking 16

Consider the function f ( x) = ∣x + 6∣ + ∣x − 6∣: a

Complete the table of values to completely deﬁne the function in each domain:

∣x + 6∣ ∣x − 6∣ ∣x + 6∣ + ∣x − 6∣

Simplified expression in domain x < −6

Simplified expression in domain −6 ≤ x < 6

Simplified expression in domain x ≥ 6

⬚

18.03E Absolute value functions 1023 mathspace.co

Fill in the missing values to deﬁne f ( x).

Evaluate the function when x = −6 and x = 6.

Which of these is the graph of f ( x)?

3 −12 −9 −6 −3 −3

x 3

9 12

−6

−9

−12

9 12

9 6 3

x 3

9 12

−6 −9 −12

1024 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

9 12

y −12 −9 −6 −3 −3 −6 −9 −12 −15 −18 −21 −24 −27

x 3

27 24 21 18 15 12 9 6 3

−12 −9 −6 −3 −3

−12 −9 −6 −3

−12 −9 −6 −3 −3 −6

y 21

6 3 −12 −9 −6 −3 −3 −6 −9 −12 −15 −18 −21 −24 −27

x 3

9 12

y x

Plot: y = ∣x − 1∣ + ∣x + 3∣

Plot: ∣y − 1∣ = 2∣x − 5∣

Plot: x = 2 sin ∣(y)∣ + 1

Plot: y = ∣ tan ( x) + 2∣ − 3

18.04E Sum and difference of functions After this lesson, you will be able to… • define the sum ( f + g)( x) and difference ( f − g)( x) of two functions f ( x) and g( x). • determine the algebraic expression for ( f + g)( x) and ( f − g)( x). • determine the domain of ( f + g)( x) and ( f − g)( x) as the intersection of the domains of f ( x) and g( x). • sketch the graph of y = ( f + g)( x) and y = ( f − g)( x) given the graphs of y = f ( x) and y = g( x) (addition/subtraction of ordinates). • determine the range of ( f + g)( x) and ( f − g)( x), where possible.

Sum and difference of functions Functions can be combined using basic arithmetic operations such as addition and subtraction. These operations are fundamental in understanding how functions interact and are particularly useful in algebraic and graphical analysis. The sum of two functions f and g is deﬁned as (f + g) ( x) = f ( x) + g( x) for each x in their common domain. Similarly, the difference is deﬁned as (f − g) ( x) = f ( x) − g( x). For these operations to be meaningful, x must belong to the domains of both f and g, which may require restricting the domain of the resulting function.

Interactive exploration Discover this concept in action online

mathspace.co

18.04E Sum and difference of functions 1025 mathspace.co

Example 1 Given f ( x) = 3x and g( x) =

a Determine the equation and domain of y = f ( x) + g( x).

Create a strategy Add the expressions for f ( x) and g( x), and identify the common domain by considering where both functions are deﬁned.

Apply the idea Add the functions

Rewrite with common denominator

Evaluate the addition

Domain: f (x) is deﬁned for all , and g(x) is deﬁned for all  (no division by zero). Thus, the domain is .

b Determine the equation and domain of y = f ( x) − g( x).

Create a strategy

Apply the idea

Subtract g( x) from f ( x), and verify the common domain.

Subtract g( x) from f ( x) Rewrite with common denominator Evaluate the subtraction Domain: Both functions are deﬁned for all , so the domain is .

c Graph the functions and examine their relationships.

Create a strategy Plot f ( x), g( x), f ( x) + g( x), and f ( x) − g( x) on the same plane and observe how the sum and difference relate to the original functions.

1026 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Apply the idea y 1.5 1 0.5 −0.5

x 0.5

−0.5 −1

y = 3x

−1.5

The sum y = gradients.

has a steeper slope than f ( x) and g( x), reﬂecting the combined effect of their

The difference y =

also has a positive slope but is less steep than the sum, showing the

subtraction of g( x)’s contribution.

Example 2 For f ( x) = ex and g( x) = x, sketch f ( x) + g( x).

Create a strategy

Apply the idea

Identify the domain, key points, and behaviour of f ( x) and g( x). Analyse how adding g( x) affects f ( x) in different regions.

Both functions have domain . At x = 0, f (0) = 1, g(0) = 0, so (f + g) (0) = 1.

3 2

f (x)

−4 −3 −2 −1 −1

g(x)

−2 −3

x 1

For x > 0, both f ( x) and g( x) are positive, so f ( x) + g( x) lies above ex, with a moderate gap due to g( x) = x. For x < 0, ex is between 0 and 1, while g( x) is negative. Thus, f ( x) + g( x) is slightly above g( x). As x approaches negative inﬁnity, ex approaches 0, so f ( x) + g( x) approaches x. 4

f (x) + g(x) 3

−4

f (x)

−4 −3 −2 −1 −1

x 1

−2

g(x) −3 −4

18.04E Sum and difference of functions 1027 mathspace.co

Idea summary The sum and difference of functions f ( x) and g( x) are deﬁned as (f + g) ( x) = f ( x) + g( x) and (f − g) ( x) = f ( x) − g( x), respectively, over their common domain. Graphing these functions reveals their relationships: the sum shifts and combines features of both, while the difference reﬂects the subtraction of g( x) from f ( x).

18.04E Practice questions What do you remember? 1

What does the sum of two functions (f + g) ( x) represent?

What does the difference of two functions (f − g) ( x) represent?

For two functions f ( x) and g( x), what must be true about their domains for f ( x) + g( x) and f ( x) − g( x) to be deﬁned?

Practice Ex 1

Consider the given functions: i

Find f ( x) + g( x).

Find f ( x) − g( x).

iii

Determine the domain of f ( x) + g( x).

Determine the range of f ( x) − g( x).

f ( x) = 4x − 3, g( x) = x2 + 2

f ( x) = x3 − 8, g( x) =

f ( x) = 8x + 10, g( x) = ∣x∣

c 5

f ( x) = x + 9, x ≥ 0, g( x) = −10x

,x≠1

Consider the given functions: i

Find f ( x) + g( x).

Find f ( x) − g( x).

iii

Determine the domain of f ( x) + g( x).

Determine the range of f ( x) − g( x).

f ( x) =

, x ≠ −7, g( x) =

f ( x) =

+ x, x ≠ 8, g( x) =

,x≠0

f ( x) =

, x ≠ −9, g( x) =

,x≠7

f ( x) =

,x≠0

, x ≠ 0, 8, g( x) =

, x ≠ −9

1028 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

The graph shows the functions f ( x), g( x), and their sum, ( f + g) ( x). Which of the following correctly identifies the functions f ( x) and g( x)? A

( f + g) (x)

f ( x) = x and g( x) = 2x

f ( x) = x and g( x) = −2x

f ( x) = x and g( x) = x2

f ( x) = 2x and g( x) = −x2

f (x)

−4 −3 −2 −1 −1

x 1

−2

g(x)

−3 −4

Ex 2

For the functions: a

f ( x) = x, g( x) = x − 1, sketch f + g.

f ( x) = 2x, g( x) = x + 2, sketch f − g.

f ( x) = 2x − 1, g( x) = 3x − 4, sketch f + g.

f ( x) = x − 7, g( x) = 4x − 5, sketch f − g.

For the functions: a

f ( x) = x, g( x) = x2, sketch f + g.

f ( x) = 2x + 1, g( x) = x2 − 1, sketch f − g.

f ( x) = 3x − 2, g( x) = 2x2 + x, sketch f + g.

f ( x) = x − 4, g( x) = x2 + 2x + 1, sketch f − g.

For the functions: a

f ( x) = x2 − 1, g( x) = 2x2 + 3, sketch f + g.

f ( x) = x2, g( x) = x2 − 3x, sketch f − g.

f ( x) = 2x2 + 1, g( x) = −x2 + 4x, sketch f + g.

f ( x) = −x2 + 3, g( x) = 3x2 − 2, sketch f − g.

For the functions: a

f ( x) = x2 − 4, g( x) =

f ( x) = x2, g( x) = , sketch f − g.

f ( x) = 2x2 + 1, g( x) =

, sketch f + g.

f ( x) = −x2 + 2, g( x) =

, sketch f − g.

, sketch f + g.

For the functions: a

f ( x) = , g( x) =

f ( x) =

, g( x) =

f ( x) =

, g( x) = , sketch f + g.

f ( x) =

, g( x) =

, sketch f + g. , sketch f − g.

, sketch f − g.

18.04E Sum and difference of functions 1029 mathspace.co

For the functions: a c

f ( x) = e − 1, g( x) = 3x − 2, sketch f + g.

f ( x) = 2ex, g( x) = x + 1, sketch f − g.

f ( x) = 3ex, g( x) = 2x − 5, sketch f − g.

f ( x) = ex, g( x) = x2 − 2, sketch f − g.

f ( x) = ex + 1, g( x) = x2 − 3x − 4, sketch

For the functions: a

f ( x) = ex, g( x) = x − 3, sketch f + g.

f ( x) = 2x, g( x) = x2 + 1, sketch f + g. 2

f ( x) = 3x − 2, g( x) = x + x, sketch f + g. f − g.

For the functions: a

f ( x) = ln x, g( x) = 2x, sketch f + g.

f ( x) = ln x − 1, g( x) = x − 3, sketch f − g.

f ( x) = ln x + 2, g( x) = 3x + 1, sketch f + g.

f ( x) = 2 ln x, g( x) = 4x − 2, sketch f − g.

For the functions: a

f ( x) = ln( x + 1), g( x) = x2, sketch f + g.

f ( x) = ln( x + 2), g( x) = x2 − 1, sketch f − g.

f ( x) = ln x, g( x) = x2 − 4, sketch f + g.

f ( x) = ln x − 2, g( x) = x2 − 6x + 8, sketch f − g.

For the functions: a

f ( x) = ∣x∣, g( x) = x − 2, sketch f + g.

f ( x) = ∣x − 1∣, g( x) = 2x + 1, sketch f − g.

f ( x) = ∣2x + 3∣ + 1, g( x) = 3x − 4, sketch f + g.

f ( x) = ∣x + 2∣ − 1, g( x) = 4x − 5, sketch f − g.

For the functions (domain: x ∈ ): a

f ( x) = sin x, g( x) = , sketch f + g.

f ( x) = 2 sin x, g( x) = x − 1, sketch f − g.

f ( x) = cos x − 1, g( x) = 3x, sketch f + g.

f ( x) = 2 cos x, g( x) = 2x + 1, sketch f − g.

Extend your thinking 18

For the functions: a

f ( x) = ∣x − 1∣, g( x) = , sketch f + g.

f ( x) = ∣x∣, g( x) =

f ( x) = ∣2x + 1∣, g( x) = , sketch f + g.

f ( x) = ∣x + 3∣ − 2, g( x) =

, sketch f − g. , sketch f − g.

For the functions: a

f ( x) = ex, g( x) = , sketch f + g.

f ( x) = 2x, g( x) =

, sketch f − g.

f ( x) = ex − 1, g( x) = , sketch f + g.

f ( x) = 3x, g( x) =

, sketch f − g.

1030 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

For the functions (domain: x ∈ ): a

f ( x) = cos x, g( x) = x2, sketch f + g.

f ( x) = sin x, g( x) = x2 − 1, sketch f − g.

f ( x) = 2 sin x + 1, g( x) = x2 + x, sketch f + g.

f ( x) = 3 cos x, g( x) = x2 − 3x − 10, sketch f − g.

For the functions: a

f ( x) = ln x, g( x) =

f ( x) = ln( x − 1) + 1, g( x) = , sketch f + g.

, sketch f + g.

f ( x) = ln( x + 1), g( x) = , sketch f − g.

f ( x) = 2 ln x, g( x) =

, sketch f − g.

18.05E Graphical relationships After this lesson, you will be able to… • apply knowledge of reciprocal functions to analyse and solve problems. • apply knowledge of absolute value functions ( y = ∣ f ( x)∣ and y = f (∣x∣)) to analyse and solve problems. • apply knowledge of the sum and difference of functions to analyse and solve problems. • interpret the meaning of points of intersection, asymptotes, and other graphical features in the context of a given problem. • solve problems involving graphs of functions by combining algebraic and graphical techniques. • justify conclusions drawn from graphical analysis in the context of the problem.

Graphical relationships Understanding graphical relationships involves analysing how different types of functions—such as reciprocal, absolute value, and sums or differences of functions—interact when graphed together or transformed. Graphical relationships can reveal how transformations or combinations of functions affect their behaviour. For example: • Reciprocal functions

introduce vertical asymptotes where f ( x) = 0 and horizontal

asymptotes based on the behaviour of f ( x) at infinity. • Absolute value functions ( y = ∣f ( x)∣ or y = f (∣x∣)) reflect negative values or enforce symmetry about the y-axis, altering the range or domain. • Sum and difference of functions (y = f ( x) + g( x) or y = f ( x) − g( x)) combine the properties of individual functions, affecting intercepts, slopes, or extrema. By applying these concepts, we can solve problems such as finding points of intersection, determining regions where one function exceeds another, or modelling real-world scenarios (cost, distance, or rates) using graphical analysis. 18.05E Graphical relationships 1031 mathspace.co

Example 1 A company’s proﬁt function is modelled by P ( x) = −x2 + 4x (in thousands of dollars), where x is the number of units sold (in thousands). The cost of advertising is modelled by A( x) =

(in thousands

of dollars). Analyse the graphical relationships between these functions to determine: a The total cost function C( x) = P ( x) + A( x) and its domain.

Create a strategy

Apply the idea

Use the sum of functions to deﬁne C( x) = P ( x) + A( x). Determine the domain by ﬁnding the common domain of P ( x) and A( x), noting that A( x) is a reciprocal function with a restriction where x = 0.

Deﬁne the total cost Substitute P ( x) and A( x)

Domain: Since x represents units sold, the domain should be all positive real numbers.

b Graph P ( x), A( x), and C( x) for x > 0 (since negative units sold are not meaningful). Identify key features and justify their signiﬁcance in the context of the problem.

Create a strategy Plot P ( x), A( x), and their sum C( x). Identify intercepts, asymptotes, and the vertex of P ( x) to interpret proﬁt, advertising cost, and total cost.

Apply the idea y 5 4 3 2 1

−1

P(x) = −x2 + 4x 1

x 2

Key Features: • P (x) = −x2 + 4x: A parabola opening downward with vertex at (2, 4) (maximum proﬁt of 4 thousand dollars when 2 thousand units are sold). Intercepts at x = 0 and x = 4 (break-even points). • A( x) =

: A hyperbola with a vertical asymptote at x = 0 (advertising cost approaches inﬁnity as

units sold approach zero) and a horizontal asymptote at y = 0 (cost decreases as x increases). • C( x) = −x2 + 4x + : Combines the quadratic and reciprocal behaviours. It has a vertical asymptote at x = 0 (from A( x)) and no finite maximum due to the reciprocal term. Significance: The vertex of P ( x) shows peak profit, while A( x) indicates high advertising costs at low sales. C( x) reflects total cost, which increases dramatically as x approaches zero due to advertising.

1032 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

c Find where P ( x) = A( x) for x > 0 and interpret the result in the context of the problem.

Create a strategy Set P ( x) = A( x) and solve for x. Verify solutions graphically and interpret them as points where proﬁt equals advertising cost.

Apply the idea Equate P ( x) and A( x)

Substitute P ( x) and A( x)

Subtract

Multiply both sides by x

Multiply both sides by −1

from both sides

The graph displays two intersection points, calculated as 0.79 and 3.87 using a cubic solver. Testing values: At x = 1:

At x = 3: 3

x3 − 4x2 + 2 = 0

x − 4x + 2 = 0 Write the equation 3

1 −4×1 +2=0 −1 < 0

Substitute x = 1

3 −4×3 +2=0 −7 < 0

Evaluate

At x = 2:

Write the equation Substitute x = 3 Evaluate

At x = 4 3

x − 4x + 2 = 0 3

2 −4×2 +2=0 −6 < 0

x3 − 4x2 + 2 = 0

Write the equation Substitute x = 2

4 −4×4 +2=0 2>0

Evaluate

Write the equation Substitute x = 4 Evaluate

A root exists between x = 3 and x = 4. Graphically, the intersection occurs approximately at x ≈ 3.8 (where P (3.8) ≈ 0.53 and A(3.8) ≈ 0.53). Interpretation: At x ≈ 3.8 thousand units sold, the proﬁt equals the advertising cost (approximately 0.53 thousand dollars, or $530). This is a critical point for the company to balance revenue and advertising expenses.

d Determine where C( x) = ∣P ( x)∣ for x > 0 and justify the conclusion.

Create a strategy Use the absolute value concept to compare C( x) with ∣P ( x)∣. Since P ( x) changes sign, solve for where total cost equals the absolute proﬁt, considering P ( x) ≥ 0 and P ( x) < 0.

18.05E Graphical relationships 1033 mathspace.co

Apply the idea Since C( x) = P ( x) + A( x) and A( x) > 0 for x > 0, equate it with ∣P ( x)∣ to compare: Case 1: P ( x) ≥ 0 (that is 0 ≤ x ≤ 4 from intercepts): Equate C( x) = ∣P ( x)∣

Substitute the values when P ( x) ≥ 0

Subtract P ( x) from both sides

Substitute A( x) =

This has no solution since

> 0 for x > 0.

Case 2: P ( x) < 0 (that is x > 4): Equate C( x) = ∣P ( x)∣

Substitute the values when P ( x) < 0

Subtract P ( x) from both sides

Substitute the corresponding values

Expand the right-hand side

Multiply both sides by x

Rearrange into standard form

Solve 2x − 8x − 2 = 0 for x ≥ 4: At x = 4:

2x3 − 8x2 − 2 = 0

Write the equation

Substitute x = 4

2×4 −8×4 −2=0 −2 < 0 At x = 5:

Evaluate

2x3 − 8x2 − 2 = 0

Write the equation

Substitute x = 5

2×5 −8×5 −2=0 48 > 0

Evaluate

A root exists between x = 4 and x = 5, approximately x ≈ 4.1 (numerical approximation). Conclusion: C( x) = ∣P ( x)∣ at x ≈ 4.1. Here, the total cost equals the absolute proﬁt (loss magnitude). Since P (4.1) < 0, this indicates a point where advertising cost offsets the loss, making total cost equal to the loss amount (approximately 0.56 thousand dollars).

Idea summary Graphical relationships allow us to solve problems by analysing how functions like reciprocals, absolute values, and sums or differences interact.

1034 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

18.05E Practice questions What do you remember? 1

A company’s revenue from selling a product is given by the function R( x) = 6x. The total cost of the product is the sum of its production cost, which is modelled as 2x2, and its advertising cost, modelled as . In all functions, x represents the units sold in thousands. a

Define the total cost function C( x).

What is the practical domain of C( x)?

Identify each graphical feature using the graph shown: 14

A vertical line where C( x) approaches inﬁnity.

The maximum point of P ( x).

A point where P ( x) = A( x).

C(x)

6 4

P(x)

A(x) 1

x 2

How does the absolute value transformation y = ∣f ( x)∣ affect the range of a function compared to y = f ( x)?

Practice Ex 1

For a small business, the revenue function is R( x) = −2x2 + 8x (in hundreds of dollars) and the marketing cost is M ( x) =

(in hundreds of dollars), where x is the number of product batches

produced (in tens) and x > 0:

Plot y = R( x) and y = M ( x) for 0 < x ≤ 4 on the same Cartesian plane.

Find the point where R( x) = M ( x) algebraically and verify graphically.

Find where R( x) > M ( x) using a graphical approach and justify your conclusion.

Using P ( x) = −2x2 + 8x and A( x) = a

Find C( x) = P ( x) + A( x) and plot it for 0 < x ≤ 5.

Identify the vertical asymptote and explain its signiﬁcance in the context of the problem.

A factory’s production rate is modelled by R( x) = −x2 + 5x (units per hour) and total cost by C( x) = −x2 + 5x +

(where x > 0):

(thousands of dollars), where x is hours of operation ( x > 0):

Plot y = ∣R( x)∣ for 0 < x ≤ 6.

Determine where C( x) = ∣R( x)∣ and interpret the result in the context of the problem.

For a delivery service, the distance travelled is modelled by D( x) = 5x (km) and fuel cost by F ( x) =

(hundreds of dollars), where x is hours ( x > 0):

Plot D( x) and F ( x) for 0 < x ≤ 3 on the same Cartesian plane.

Find the point where D( x) = F ( x) algebraically and verify graphically. 18.05E Graphical relationships 1035 mathspace.co

Given f ( x) =

, x ≠ 5:

Plot y = ∣f ( x)∣ for 0 ≤ x ≤ 10.

Identify the vertical and horizontal asymptotes and explain their signiﬁcance in a cost model.

A production cost is C( x) = −x2 + 6x and maintenance cost is M ( x) = , where x is units produced (thousands, x > 0): a

Find T ( x) = C( x) + M ( x) and its domain.

Plot T ( x) for 0 < x ≤ 4.

For f ( x) = x − 6 and g( x) = , where x ≠ 0, ﬁnd ∣f ( x)∣ = g( x) for x > 0 algebraically.

For x > 7, plot f ( x) =

, g( x) = , and their sum y =

on the same Cartesian plane,

and identify the horizontal asymptote of the sum. 12

A rate function is R( x) =

, x > −5, and cost is C( x) = −x2 + 9x. Find where R( x) > C( x) for

0 < x < 1 using a graphical approach. 13

For f ( x) = , x ≠ 0: a

Plot y = f (∣x∣) for x ≠ 0.

Describe the symmetry of the graph and its signiﬁcance in a rate model.

Extend your thinking 14

Given P ( x) = −x2 + 4x and A( x) = a

Plot y = ∣P ( x)∣ + A( x) for 0 < x ≤ 5 and determine its minimum value.

Justify why this minimum occurs at x = 2 in the context of the problem.

A company’s production cost is modelled by P ( x) = −x2 + 5x (thousands of dollars) and advertising cost by A( x) =

(where x > 0):

(thousands of dollars), where x is units sold (thousands, x > 0):

Plot the total cost function T ( x) = P ( x) + A( x) for 0 < x ≤ 4, and determine its behaviour as x approaches 0 and inﬁnity.

Explain the real-world implications of this behaviour for the company.

A revenue function is R( x) = −x2 + 7x and a cost function is C( x) = a

Find the points where ∣R( x)∣ = C( x) for x > 5 algebraically.

Justify where ∣R( x)∣ > C( x) in the context of proﬁtability.

For f ( x) = x − 9, g( x) = , x ≠ 0: a

Plot y = ∣f ( x)∣ + g( x) for 0 < x ≤ 12 and ﬁnd its minimum value.

Explain the minimum’s signiﬁcance in a cost model.

1036 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

, x > 5:

18 Chapter review 1

If a function f ( x) has a zero at x = b, what graphical feature does the reciprocal function y=

typically exhibit at x = b?

A horizontal asymptote

A vertical asymptote

A point of inﬂection

A local maximum

What is the period of the function y = cosec(3x)? B

A 3

6π

To obtain the graph of y = f (∣x∣) from the graph of y = f ( x), which transformation is applied? A

Reﬂect the portion of the graph where y < 0 across the x-axis.

Reﬂect the portion of the graph where x < 0 across the y-axis.

Keep the portion of the graph where x ≥ 0 and reﬂect this portion across the y-axis, discarding the original part where x < 0.

Translate the graph horizontally. , where h( x) = x2 − 16:

Consider the function y = a

Solve the equation h( x) = 0 for x, and explain the signiﬁcance of the solutions for the reciprocal function y =

Determine the behaviour of h( x) as x → −∞ and x → ∞, and interpret the result for the graph of y =

3π

Determine the behaviour of y =

Sketch the graph of y =

around x = 4 and x = −4.

The graph of f ( x) =

is shown. 8

6 4 2 −4 −3 −2 −1 −2

x 1

−4 −6 −8

Sketch the graph of y =

Chapter 18 review 1037 mathspace.co

Consider the functions p( x) = −4x + 12 and q( x) = a

p( x) as x → 3 as x → 3+

c 7

. Determine the behaviour of:

q( x) as

Consider the graph k( x) shown: y 8 6 4 2

(3, 2) x

−1

Sketch the graph of y = 8

For y = cot ( x): a

Determine: i

b 9

The period

The domain

iii

The range

Write the equations of its vertical asymptotes in terms of n, where n is an integer.

Consider y = sec ( x) for −2π ≤ x ≤ 2π. 4

3 2 1

−1 −2 −3 −4

What is the period of sec ( x)?

For which values of x in −2π ≤ x ≤ 2π is sec ( x): i

Undeﬁned

Equal to 1

Given sec ( x) = 2 when x =

, determine:

The next positive value of x where sec ( x) = 2.

The greatest negative value of x where sec ( x) = −2.

1038 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

The graphs of f ( x) = sec ( x), g( x) = sec ( x + α ), and h( x) = sec ( x − β ) are shown. 4

3 2 1 g(x) f (x) h(x)

−1 −2 −3 −4

Determine the values of α and β. 11

The graph of f ( x) = a sec ( x + b) + c is shown. 6 5 4 3 2 1

−1 −2 −3 −4

Determine the values of a, b, and c, given a > 0, 0 < b < 12

, and c is an integer.

Using the graph of h( x), sketch the graph of:

h(x) −2

y = ∣h( x)∣

y = h(∣x∣)

18 16 14 12 10 8 6 4 2 −2 −4

x 2

Chapter 18 review 1039 mathspace.co

Consider the function p( x) = ∣x∣ −2: a

Sketch the graph of the function.

Which statement about the symmetry of the graph is true?

The graph is symmetric about the y-axis.

The graph is symmetric about the x-axis.

The graph is rotationally symmetric about the origin.

The graph has no symmetry.

Determine a simpliﬁed expression for p(−x).

Sketch the graph of y = ∣x − 3∣ + ∣x + 2∣.

Consider f ( x) = 3x − 5 and g( x) = x2 − x. Determine:

f ( x) + g( x)

f ( x) − g( x)

The domain of f ( x) + g( x)

The range of f ( x) − g( x)

f ( x) = ∣x + 1∣ and g( x) =

f ( x) = ex − 2 and g( x) =

Sketch the graph of (f + g) ( x): a

Sketch the graph of (f − g) ( x): a

f ( x) = 3x and g( x) = x − 2

f ( x) = −2x and g( x) = x2 − x

A bakery’s daily proﬁt from selling n cakes is P (n) = −0.5n2 + 20n dollars, with a ﬁxed operating cost F (n) =

dollars, where n > 0:

Plot y = P (n) and y = F (n) for 0 < n ≤ 30 on the same Cartesian plane.

Find the point when P (n) = F (n) algebraically, rounded to two decimal places. Verify graphically.

Find when P (n) > F (n) using a graphical approach and justify.

A farmer’s weekly apple yield is Y (w) = −w2 + 15w bushels, with spoilage S(w) = bushels/week, where w > 0 is weeks into the harvest. a

Find the net usable yield N (w) = Y (w) − S(w) and plot it for 0 < w ≤ 10.

Identify the vertical asymptote of N (w) and explain its signiﬁcance.

Given plant vitality V (t) = −2t2 + 10t (vitality units) and environmental stress E(t) = (stress units), where t > 0 is months, plot the health function H(t) = ∣V (t)∣ − E(t) for 0 < t ≤ 6. From the graph, find the maximum value of H(t) on this interval.

Explain why the health function H(t) is higher at t = 6 than at the point of maximum vitality (t = 2.5).

1040 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

“Life is good for only two things: discovering mathematics and teaching mathematics.” Siméon Poisson

Answers

9 a

y 8 6 4 2

1.01 Functions and relations What do you remember? 1 Relation Example: y2 = x. In this case, for a given positive x-value, there are two corresponding y-values (one positive and one negative), such as when x = 4, y could be 2 or −2. 2 a True b False, because a function assigns at most one y-value per x-value.

10 a Call length International call (minutes) cost (dollars)

d True b Correct

c Incorrect

d Incorrect

2 4 6 8

b (−5, −10) or (−5, −4)

c False, only functions pass the vertical line test. 3 a Correct

−8 −6 −4 −2 −2 −4 −6 −8

1.80

2.90

4.00

5.10

6.20

b Yes

Practice 4 No, because x = 4 corresponds to y = 14 and y = 11.

11 a Number of Total cost t-shirts (dollars) 1

5 a Both relation and function b Relation 6 a Function

b Relation

c Function

d Function

7 a y

b Yes

12 a Total GB Total charge used (dollars)

−2

−4

x −4

8 a

−2

x −4 −3 −2 −1 y

−1 −1 −1 −1 −1

b Yes

13 a

b Yes

3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

1042 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Yes

x 1

2 3 4

No, because a vertical line can be drawn that intersects the graph at more than one point. b

9 8 7 6 5 4 3 2 1 −4 −3 −2 −1

{(−2, 4), (−1, 1), (0, 0), (1, 2), (2, 3)}

This is a one-to-one function because each x-value maps to a unique y-value, and no y-value is repeated.

1.02 Variables and substitution What do you remember? 1 The independent variable is x, which is the input. The dependent variable is y, which is the output and depends on x.

x 1

2 3 4

Yes, because no vertical line intersects the graph more than once.

2 Independent variable: t. Dependent variable: f (t).

Extend your thinking

3 Independent variable: t (time in hours). Dependent variable: d (distance in km).

14 a

Practice

y (Machine 1)

4 a g(x) = x2 − 10

y (Machine 2)

b No, because for x > 3, each x maps to two y-values (one from each machine). For example, at x = 4, Machine 1 gives y = 11 and Machine 2 gives y = 9. A vertical line at x = 4 intersects the union at (4, 11) and (4, 9), failing the vertical line test.

15 a

7.5

13.75

c Yes, because each distance x maps to exactly one cost y in the piecewise function. A vertical line intersects the graph at most once, as each segment deﬁnes a unique y for each x. 16 a Yes, because each x-value maps to exactly one y-value. All x-values are distinct. b Add (−2, 5):

{(−2, 4), (−2, 5), (−1, 1), (0, 0), (1, 1), (2, 4)}

This is not a function because x = −2 maps to two different y-values (4 and 5), which violates the deﬁnition of a function. c Change (1, 1) to (1, 2) and (2, 4) to (2, 3):

b g(−3)

5 a The independent variable is d, representing the amount of data used in GB. b The dependent variable is C, representing the total monthly cost in dollars. c The total cost is the dependent variable because its value depends on the amount of data used. 6 a 112

b 966

c −3.21

d 46

7 a

b −7

c 0

8 a 68

b −4

c 4 + m3

d 4 − b3

9 a 3

c 2

10 a 0

b 3

11 a a2 + 8a

b x2 − 8x

c 33

d x4 + 8x2

12 a −4

d 2a3 + 21a2 + 72a + 77 13 a 0 14 a c 15 a t = −3, 3

b 2.67

c 80.99

d 2187.00

b 12 d b t = −2, 2

Answers 1043 mathspace.co

Practice

16 a x = −3, 2

4 a Domain: {1, 2, 3, 4}; Range: {5, 6}; Function: Yes

b x = −4, 3

c x = −2, 1

b Domain: {2, 3, 4}; Range: {1, 2}; Function: No

17 −9

5 a Domain: All real x; Range: All real y

18 −49

b Domain: All real x; Range: y ≥ 2 c Domain: All real x; Range: y ≥ −2

19 −46

d Domain: All real x; Range: y = 3

20 a b −1

6 a x = −3, 3

b −9

7 a x=3

b x = −3, 3

21 32

d x = −4

c No real zeroes

22 a 18

b No x-intercept

8 a 1

b It represents the total cost of buying 6 bottles.

d y>0

c All real x b y ≤ 16

9 a 16

c All real x

Extend your thinking

10 a i Function

23 f (7) = −6

ii Domain: −7 ≤ x ≤ 7 Range: 0 ≤ y ≤ 7

24 6

b i Not a function

25 2a + h + 5

ii Domain: −9 ≤ x ≤ 9 Range: −9 ≤ y ≤ 9

26 27 x = 13 or x = 127 2

28 T (x) = 4x − 394x + 26 349

11 a x ≥ −1

b y≥0

c x = −1

12 a 5

b All real x

c y≥5

1.03 Characteristics of functions

b True

c False

d True

2 a The set of allowable values of x in a function or relation. b The set of values of the dependent variable for which a function is deﬁned. c A point in the domain of a function where the value of the function is zero. d The value of f (0), where the graph intersects the y-axis. 3 a {1, 2, 3, 4} c Yes, it is a function.

−2

−1

−9

−2

−1

Range: All real y

What do you remember? 1 a True

b {2, 3, 5}

−5

−4

f (x)

−2

−1

Range: {−2, −1, 3, 4} 15

−1

f (x)

−2

−4

Range: {−4, −2, 0, 2} 16 a All real numbers.

b All real numbers.

Extend your thinking 17 a x > 0

b y>0

18 a x > 2

b y>1

1044 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

19 a All real x except x = −2 and x = 2 b All real numbers. c x=0

1.04E Language of polynomials What do you remember? 1 a 6

8 a −5

b 1

c −1029

9 a 3

b 2

10 a 6

c −2394

11 a

12 a 2x2 − 13x + 3

b 7

has a fractional

2 a No, because the term exponent.

= 2x−7 has a

b No, because the term negative exponent.

has a

d No, because the term fractional exponent.

has a

g x + 4x + 2x − 9

h −3x3 + 12x2 − x − 6

i −6x − 7x + 26

14 2x2 + 4x − 9

4 a Linear

e P (v) = 10v3 − 2v2 − 28v − 16 f P (u) = −6u5 + u4 + 6u3 − 2u2 + 8u

b Cubic

c Monomial

c P (u) = −10u3 + 16u2 − 2u − 4 d P (c) = 10c4 − 10c3 + 11c2 + 4c − 6

3 For example: x2 + 2x + 3

g P (u) = 20u4 − 36u3 + 37u2 − 16u − 5

d Binomial

h P ( x) = 2x3 + 5x2 − 16x + 6

5 P (3) represents the value of P ( x) when x = 3.

16 a P (b) = 9b2 − 42b + 49 b P ( y) = 4y3 + 18y2 + 24y + 8

Practice 6 a 0

c P ( x) = x3 − 8x2 + 20x − 16

b 5 ii 9x

iii 9

d P ( x) = 2x4 − 4x3 − 12x2 + 25x − 21 iv 0

v Non-monic polynomial ii −8x7

iii −8

iv 0

v Non-monic polynomial ii 4x9

iii 4

iv 0

v Non-monic polynomial d i 3

f x3 − 2x2 + x − 9

b P (m) = m3 − m2 − 23m + 15

f N o, because the term negative exponent.

c i 9

e −10x2 + 8x − 2

15 a P (v) = 5v3 + 22v2 − 20v − 25

e Yes

b i 7

d 5x3 − x2 + 4x − 2

13 x2 − 5x + 13

c Yes

7 a i 7

b −5x2 − 4x − 6

c −4x − 5x − 9x − 4 3

ii x3

17 a 3x3 + 6x2 − 17x − 3 b −5x3 + 6x2 − 17x + 6 18 a (6x2 + 14x + 24) units b (3x2 + 24x) units c (3x3 + 8x2 + 4x + 9) units

iii 1

iv −5

19 (5u3 + 13u2 + 26u + 12) m3

iii 5

iv 3

Extend your thinking

v Monic polynomial e i 4

ii 5x4

20 a (9 − 2x) cm

v Non-monic polynomial f i 5

ii x5

iii 1

iv −11

g i 6

iii

iv 6

iii

21 a 20.6 million

b 21.26 million

c 0.66 million

v Non-monic polynomial h i 6

b (5 − 2x) cm c (45x − 28x2 + 4x3) cm3

v Monic polynomial

iv 4

22 7v2 − 9v + 1600

v Non-monic polynomial

Answers 1045 mathspace.co

23 Nadine is correct. Holly just identified the first term in the polynomial and the polynomial was not in order of degree. The highest degree is 5. 24 The error was made in the first step. She should have put brackets in when substituting (−2). Without having the brackets, only the 2 is raised to the power of 4 but not the negative sign. P (−2) = 9 25 P ( x) + Q( x) = 2x3 − x2 + 3x − 1. This is a polynomial as all variables have non-negative integer indices. 26 The sum of two trinomials results in a binomial if, after combining all like terms, exactly two non-zero terms remain. This happens when the trinomials contain like terms whose coefficients are additive inverses (for example, the coefficients of the x terms are +5 and −5). Because these coefficients sum to zero, the resulting term for that power of the variable is also zero, reducing the total number of terms in the final expression. 27 4x3 − 6x2 − 3x − 5

1.05E Behaviour and graphs of polynomials What do you remember? 1 a x=3

b x = −1, x = 1

2 a As x → +∞, y → −∞, as x → −∞, y → −∞

c Original degree 4 (quartic), sum degree 4 (quartic) d Original degree 4 (quartic), sum degree 4 (quartic) 6 a x=2

b x = −4

c x=3

d x=3

7 a x = 2, x = −5, x = 1 b x = −3, x = 4, x = 7 c x=

, x = 5, x = −2, x = 6

d x = 1 (multiplicity 2), x = −4 e x = , x = −5, x = 2 (multiplicity 2), x = −1 8 a x = 1, x = −3

b x = 2, x = 3

c x = 2, x = −2

d x = 4, x = −2

9 a x = −4, 0, 2

b x = 0, 3, 4

c x = −1, 0, 4

d x = −1, 0, 4

10 a i 3

ii 2

iii 3

iv 4

b i 4

ii 3

iii 4

iv 4

11 a As x → +∞, p1 ( x) → +∞; as x → −∞, p1 ( x) → −∞. b As x → +∞, p2 ( x) → −∞; as x → −∞, p2 ( x) → −∞. c As x → +∞, p3 ( x) → +∞; as x → −∞, p3 ( x) → −∞. d As x → +∞, p4 ( x) → −∞; as x → −∞, p4 ( x) → −∞. 12 a i x = −4, 3

b As x → +∞, y → +∞, as x → −∞, y → +∞

ii As x → ±∞, m( x) → +∞

c As x → +∞, y → −∞, as x → −∞, y → +∞

iii

d As x → +∞, y → +∞, as x → −∞, y → −∞ 3 a 3 (cubic) c Linear

b 3 (trinomial) d −3

Practice 4 a Degree 3 (cubic), 4 terms (tetranomial) b Degree 2 (quadratic), 2 terms (binomial) c Degree 4 (quartic), 3 terms (trinomial) d Degree 1 (linear), 2 terms (binomial) 5 a Original degree 4 (quartic), sum degree 4 (quartic) b Original degree 4 (quartic), sum degree 3 (cubic)

1046 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

8 6 4 2 −8 −6 −4 −2 −2 −4 −6 −8 −10 −12 −14

x 2 4 6 8

b i x=3

ii As x → ±∞, m( x) → +∞ iii

4 (0, 4)

(−2, 0) (−1, 0)

−1

−2

2 2

(1, 0) 1

x 2

(2, 0)

−2

−4 −2 −2

−1

14 a Three b 3

c i x = −2 ii As x → +∞, m( x) → +∞; as x → −∞, m( x) → −∞ iii

10 8 6 4 2

b Zeroes at x = 2, x = −4

x 1

−2 −4 −6 −8 −10

b The degree is 3 and the leading coeﬃcient is negative, so as x → −∞, the graph trends up towards f ( x) → ∞; as x → +∞, the graph trends down towards f ( x) → −∞. 17 a y = −3( x − 1) ( x − 4)

b y = −( x + 2) ( x − 5)

c y = ( x + 1) ( x − 2) ( x − 3)

d i x = −1

Extend your thinking

ii As x → ±∞, m( x) → +∞

18 For the zero polynomial R( x) = 0, R( α ) = 0 holds for all real α.

y 3 2

So, every real number satisﬁes R( α ) = 0, and its degree is undeﬁned.

19 The degree is (A − B) = 5. x

−2

15 a a = −2

16 a It has three x-intercepts: x = 1, x = 4,

−4 −3 −2 −1

iii

c As x → +∞, K( x) → +∞; as x → −∞, K( x) → −∞

−1

1 −1

13 a y = 4 b x = −2, −1, 1, 2 c As x → ±∞, f ( x) → +∞

20 a Degree 3 (cubic), as the highest degree term from P ( x) dominates. b The x4 terms cancel, leaving a polynomial with degree 3 (cubic). c A polynomial of degree 3 can be a binomial if it has exactly two terms, provided the term with the highest power is the x3 term. For example, P ( x) = x3 + 5 is a cubic binomial. 21 a The leading coeﬃcient is negative (because it goes to −∞ for large ∣ x ∣). b Yes, it must cross the x-axis somewhere (since Q (0) = 5 > 0 and it ends up negative for large x), so there is at least one real zero.

Answers 1047 mathspace.co

22 a Degree 3 (cubic)

c As x → +∞, H( x) → +∞; as x → −∞, H( x) → −∞.

1.06E Sketch polynomials What do you remember? 1 a Single

−5−4−3−2 −1 −5 −10 −15 −20 −25

b Repeated

2 a x = −5 (multiplicity 1), x = 1 (multiplicity 1), x = 3 (multiplicity 1) b x = 2 (multiplicity 2), x = −1 (multiplicity 3) 3 a i 3

1 2 3 4 5 6

ii 2

iii f ( x) → ∞

iv f ( x) → −∞ ii −1

b i 2 iii f ( x) → −∞ c i 3 iii f ( x) → ∞

7 a Falls as x → ∞, rises as x → −∞ (leading term −3x3).

iv f ( x) → −∞

b x = −2 (multiplicity 2), x = 4 (multiplicity 1)

ii 6

c y = 48

iv f ( x) → −∞

y 100 90 80 70 60 50 (0, 48) 40 30 20 10

ii −3

d i 3 iii f ( x) → −∞ e i 4

iv f ( x) → ∞ ii 1

iii f ( x) → ∞

iv f ( x) → ∞ ii −1

f i 5 iii f ( x) → −∞

iv f ( x) → ∞

(−2, 0)

Practice 4 i B

y 45 40 35 30 25 20 (0, 20) 15 10 (−4, 0) 5 (1, 0) (5, 0) x

y = (x + 4) (x − 1) (x − 5)

b x = 1 and x = −3

ii D

iii A

−5 −4 −3 −2 −1 −10 −20 −30 2

iv C

5 a As x → ∞, we have y → ∞. As x → −∞, we have y → −∞. b As x → ∞, we have y → ∞. As x → −∞, we have y → ∞. c As x → ∞, we have y → −∞. As x → −∞, we have y → ∞.

(4, 0) x

1 2 3 4 5

y = −3(x + 2) (x − 4) 8 a x = 3 (multiplicity 3), x = −1 (multiplicity 1) b

d As x → ∞, we have y → −∞. As x → −∞, we have y → −∞. e As x → ∞, we have y → −∞. As x → −∞, we have y → ∞. f As x → ∞, we have y → ∞. As x → −∞, we have y → ∞. 6 a y = 20 b x = −4 (multiplicity 1), x = 1 (multiplicity 1), x = 5 (multiplicity 1)

1048 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

15 10 (−1, 0) 5

−1

−2 −1

(3, 0) 1

−5 −10 −15 −20 −25 −30 −35 −40 y = 2(x − 3)3(x + 1) −45 −50 −55 (0, −54)

9 a x = 0 (multiplicity 2), x = 4 (multiplicity 1) b

15 y 10 x = −1 5

y 12 2 10 y = x (x − 4) 8 6 4 2 (0, 0) (4, 0) −1

−2 −4 −6 −8 −10

−2 x 5

−5 −10 −15 −20 −25

x=2 1

y = −20

x=5 3

x 6

P(x) = (x − 2)2(x + 1)(x − 5)

Extend your thinking

10 a i y = x( x − 1) ( x − 2) ii x = 0 (multiplicity 1), x = 1 (multiplicity 1), x = 2 (multiplicity 1) iii

4 3

y = x(x − 1)(x − 2)

2 1

(0, 0) −1

(1, 0) (2, 0) x 1

−1

−2 −3 −4

b i y = x2 ( x − 2) ( x + 2)

13 A zero x = α is single if ( x − α ) is not a factor of Q( x), and repeated if it is. Multiplicity m is the exponent in ( x − α )m when P ( x) = ( x − α )mQ( x) and Q( α ) ≠ 0. 14 No. The graph shows x = −6 with multiplicity 1 (passes through), not 2 (turning point), as required by ( x + 6)2. 15 An odd-degree polynomial has opposite end behaviours (for example, y → ∞ as x → ∞, y → −∞ as x → −∞). Since it is continuous, it must cross the x-axis at least once, ensuring at least one real zero.

Chapter 1 review

ii x = 0 (multiplicity 2), x = 2 (multiplicity 1), x = −2 (multiplicity 1)

1 C

iii

2 A

4 3 2 1

(−2, 0) −2

3 C (0, 0)

−1

−1 −2 −3 −4 y = x2(x − 2)(x + 2) −5

11 a i No

−1

ii No

(2, 0) x 2

4 No, because x = 5 maps to both y = 12 and y = 9. 5 a Function

b Relation

c Function

d Relation

6 a

iii Yes

b k= 12 a P ( x) = k( x − 2)2 ( x + 1) ( x − 5), where k is a non-zero constant

10 8 6 4 2 −8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8

b Either (−6, −11) or (−6, −5)

Answers 1049 mathspace.co

7 a No, because x = 0 maps to both y = 4 and y = 6. b Remove (0, 6) to make it a function, and replace (2, 5) with (2, 7) for one-to-one: {(−1, 3), (0, 4), (1, 5), (2, 7)}.

18 a x = 3

b x = −7, 7

c No real zeroes

d x = −5

19 a No, because l > 0.

b P (l) > 6

20E C

8 a −4

21E C

b 68

22E C

c −5.92

23E a P ( x) = x5 − 9x3 + 4x − 11

d 2x2 + 4xa + 2a2 + 5x + 5a − 7 9 a

b 5 d −3

c 12

10 a 7

b 4

c 2

c x5 d −11 e Yes, because its leading coeﬃcient is 1.

24E a −8

11 a

b Yes, because each number of muﬃns x maps to exactly one cost y. 12 b2 + 2bk + k2 + 3b + 3k

b −14

c −6

25E a 8x + 4x + 6x + 5 b 2x3 − 8x2 + 8x − 13 c x3 − 22x2 + 17x − 35 26E a As x → ∞, P ( x) → −∞. As x → −∞, P ( x) → ∞. b As x → ∞, Q( x) → ∞. As x → −∞, Q( x) → ∞. c The leading term is −x3. As x → ∞, R( x) → −∞. As x → −∞, R( x) → ∞.

27E a 6

14 a Domain: {x ∈  : x ≠ 3}; Range: {y ∈  : y ≠ 0}

b x = 0 (multiplicity 1), x = −3 (multiplicity 2), x = 1 (multiplicity 3)

b Domain: ; Range: {y ∈  : y ≥ −4}

c (0, 0)

c Domain: {x ∈  : x ≥ −5}; Range: {y ∈  : y ≥ 0}

d Domain: ; Range:  15 a

(−3, 0)

Viewing hours (H)

Loyalty points (P )

22.5

b Domain: {1, 5, 7}; Range: {2, 4}; Function: No, because x = 1 maps to both y = 2 and y = 4. 17 a x = −5, 5

b −25

(0, 0) (1, 0) x

−4 −3 −2 −1 1 −10 −20 −30 −40 −50 −60 −70 −80 −90 2−100 3 y = −2x(x + 3) −110 (x − 1)

b Yes, because each value of viewing hours H maps to exactly one value of loyalty points P. 16 a Domain: {2, 4, 6, 8}; Range: {3, 7}; Function: Yes

40 30 20 10

28E a V ( x) = 2x3 + 5x2 − 4x − 3 b 3 c A( x) = x2 + 2x − 3 d 3x3 + 5x2 − 6x + 2 29E a The minimum possible degree is 3, as there are 3 distinct real zeroes.

1050 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b P ( x) = ( x + 2) ( x − 1) ( x − 3)

2.01 Equations of parabolas

c Positive. The general form of the polynomial is P ( x) = a( x + 2) ( x − 1) ( x − 3). Using the y-intercept (0, 6), we find P (0) = a(2)(−1)(−3) = 6a. Since P (0) = 6, we have 6a = 6, which gives a = 1. As the leading coefficient a is positive, the polynomial has a positive leading coefficient. d

1 a True

b True

c False

2 a iii

b ii

c i

d False

3 Equate coeﬃcients: a = p, b = q, c = r Practice

4 a y = (x − 3)2 – 2

b Domain: ; Range: [−2, ∞)

y = (x + 2) (x − 1) (x − 3)

10 5

What do you remember?

(0, 6)

(−2, 0)

(1, 0)

−3 −2 −1

(1, 2)

2 1

x=3

(3, 0) x 2

−1

−5

−2

e The real zeroes are x = −2, x = 1 and x = 3. 30E a P ( x) = a( x − 1)2 ( x + 2)2

(3, −2)

5 a y = x2 − 2x − 8 b Domain: ; Range: [−9, ∞) c a = 1, b = −4, c = 2 or a = 1, b = 2, c = −4

b a = −1 c P ( x) = −( x − 1)2 ( x + 2)2

6 a y = (x − 2)(x − 4)

b y = x2 − 6x + 8

7 a y = x2 + 2

b [2, ∞)

(−2, 0) −3

−2

4 3 2 1

(1, 0)

−1

1 2 −1 −2 −3 −4 (0, −4) −5 −6 −7 y = −(x − 1)2(x + 2)2 −8 −9

8 a y = −2(x + 1) + 5

b y = −2x2 − 4x + 3

9 a y = (x + 2)(x − 4)

b (1, −9)

10 a y = −x + 3

b (−∞, 3]

11 a y = 2x – 2 b

y 1

x=0 (1, 0)

(−1, 0) −2 −1

−1 −2

31E a m + n

(0, −2)

b i T he degree of P ( x) + Q( x) would be less than 5 if a5 + b5 = 0. ii 4

Extend your thinking

iii 5 32E The equation is P ( x) =

( x − 2) ( x + 1)3.

12 a y =

(x + 2)(x – 6)

b x=2

Answers 1051 mathspace.co

13 a y = −(x − 2)2 + 4

b (4, 0)

7 a (4, 5), (2, 1) b

14 a y = 2x2 − 4x

b Domain: all real numbers; Range: [−2, ∞)

15 a = 1, b = 8, c = −9

g(x)

16 a = 2, b = −7, c = 2

2.02 Solve quadratic systems

What do you remember? 1 a True

(4, 5)

f (x) (2, 1)

−1

b False

c False

x 4

d True 8 a x = −1 and x = 2

2 x2 + x − 2 = 0

3 B

7 6 5 4 3 2 1

f (x)

g(x)

f (x)

x 1

(−1, −2) −2

(1, 0) 1

−3 −2 −1 −1

(6, 5)

−1 −2 −3 −4

(2, 4)

4 a (1, 0), (6, 5) b

g(x)

Practice

The intersections are at (−1, −2) and (2, 4) so x = −1, 2 respectively. 9 a x = 0 and x = 2 b

(1, 0) and (6, 5) f (x)

5 a (0, 3), (3, 0) b

y 4

f (x) g(x)

1 −1

(3, 0) −1

(2, 0)

−2 −1 1 2 −1 g(x) −2 −3 −4 (0, −4) −5

5 3 (0, 3)

5 4 3 2 1

(0, 3), (3, 0) 6 a b Δ = 5 > 0; There are two intersection points.

x 3

The intersections are at (0, −4) and (2, 0) so x = 0, 2 respectively. 10 a x = 0 and x = 1 b Δ = 1 > 0; There are two intersection points. 11 (2, 1), (4, 5) Extend your thinking 12 There are two possible values: m = 2 or m = −6. 13

1052 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

14 a f (x) = x2 − 4x + 6.25

6 a x < −2 or x > 4

b (2.5, 2.5)

3 y 2 1

c g(x) = 4x − 19x + 25 (−2, 0)

15 m = −2, d = 2

−3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

2.03 Quadratic inequalities What do you remember? 1 a True

b True

c True

d False

2 Solve x − 6x + 8 = 0 to ﬁnd boundary points.

2 3 4 5

Solution: x < −2 or x > 4

3 • Δ > 0: Two boundary points

7 a x = −2

• Δ = 0: One boundary point

b Δ = 0; One boundary point, parabola tangent to x-axis.

• Δ < 0: No boundary points Practice

8 x ≤ 1 or x ≥ 3

4 a 1≤x≤5 b

(4, 0) 1

5 4 3 2 1

−1

9 0≤x≤2 10 x < 1 or x > 3 (5, 0)

(1, 0) 1

−1 −2 −3 −4

−1

3 2

(−3, 0)

Solution: 1 ≤ x ≤ 5

−4 −3 −2 −1 −1

5 a x≥1 b

(1, 0) 1

2 3 4

−2

−3

y = x2 + x + 2

−4

6 4

x ≤ −3 or x ≥ 1

(1, 4)

2 x −4

y = x2 + 3x

−2

Solution: x ≥ 1

Extend your thinking 12 k > 4

−2

13 a

Answers 1053 mathspace.co

b When p < 2, the solution is an interval extending to positive inﬁnity. As p approaches 2 from below, the boundary

7 a x ≈ 4.14 seconds b 23 metres at x = 2 seconds

moves towards ∞, and the solution set shrinks. At p = 2, the solution set is empty. When p > 2, the solution is an interval extending to negative inﬁnity from a negative boundary. As p increases from 2, this boundary moves from −∞ towards 0. c 3

−5

−4

−3

−2

−1

x 1

(0.25, −2.1875) −2

f (x) for p = 1

b $300

9 a x = 0, 4 seconds b Domain: [0, 4]; Range: [0, 16] 10 a Minimum cost: $50 at x = 10 (1000 units) b [0, ∞) 11 a 8 metres at x = 1 second b x ≈ 2.26 seconds

g(x)

f (x) for p = −2

8 a $10

−3

(1, −2)

−4

12 a Length = 60 − x b f (x) = −x2 + 60x c Maximum area: 900 m2, Dimensions: 30 m by 30 m d (0, 60)

−5

The graph shows that for p = −2, f (x) > g(x) when x > 0.25. For p = 1, f (x) > g(x) when x > 1. This matches the algebraic results.

Extend your thinking 13 a 20 m by 20 m b

400

15 −8 ≤ m ≤ 0

300

2.04 Quadratic models

200 100

What do you remember? 1 a True

b True

(20, 400)

c False

2 The vertex represents the maximum height and the time at which it occurs. 3 The domain is restricted by context, typically non-negative values for physical dimensions like length or width. Practice 4 a x ≈ 3.05 seconds b Maximum height: 13.80 metres at x ≈ 1.35 seconds c Domain: [0, 3.05]; Range: [0, 13.80] 5 a x = 10, 20 (1000 and 2000 units) b Maximum proﬁt: $25 000 at x = 15 (1500 units) c Domain: [10, 20]; Range: [0, 25] 6 Maximum proﬁt: $25 at x = 25 units

d False

<x<4+

14 4 –

seconds 2

15 a f (x) = 200x − 2x

Maximum area: 5000 m2 at x = 50 m b Domain: (0, 100); Range: (0, 5000] 16 a 6 metres at x = 20 m b Domain: [0, 20) ∪ (20, 40]

Chapter 2 review 1 a Factored: (3x + 2)(x − 2) Expanded: 3x2 − 4x − 4 b 28 m2 c To ensure the width (x − 2) is a positive length, x must be greater than 2. 2 a k = 25 2

b k > 25

3 a x − (6 + m)x + 9 = 0 b m < −12 or m > 0

1054 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

4 a A = −(x − 10)2 + 100; Vertex: (10, 100) b Maximum area: 100 m2; Dimensions: 10 m by 10 m

12 a (0, −3), (3, 0) b

y 1

(3, 0) x

5 a (−1, 0), (5, 0)

b (0, −5)

−2

d Domain: ; Range: [−9, ∞)

−3 (0, −3)

y 4

−4

(5, 0) x

(−1, 0) −1

−2 −4 −6

13 m = 2 or m = −10

14 k < −2 or k > 6

(0, −5)

15 a −3 ≤ x ≤ 4 b

−8

(2, −9)

(4, 0) x

(−3, 0) −3 −2 −1

6 a x-intercepts: (−1, 0), (5, 0); y-intercept: (0, −5) b Vertex: (2, −9); Axis of symmetry: x = 2 c

−1

c Axis of symmetry: x = 2; Vertex: (2, −9) e

−5

y 4

−10

(5, 0) x

(−1, 0) −1

−2 −4 −6

16 a x < −5 or x > 2 (0, −5)

(2, 0)

(−5, 0)

−8

(2, −9)

−5 −4 −3 −2 −1

7 y = x2 − 2x – 8

2 3

−5

8 a y = 2(x + 1)(x − 3) 2

9 a y = 2x − 1

b y = 2x − 4x – 6 −10

b Range: [−1, ∞)

10 a = 3, b = 13 11 a (1, 3), (5, 7) b

17 x ≤ −2 or x ≥ 4

(5, 7)

−3 −2 −1 0

18 k > 16

19 a x = 10, 30 (1000 and 3000 units)

4 3

b Maximum proﬁt: $100 000 at x = 20 (2000 units)

(1, 3)

c Domain: (10, 30); Range: (0, 100]

x 0

Answers 1055 mathspace.co

b 2x3 + 3x2 − x + 4 = ( x + 2) (2x2 − x + 1) + 2

20 a A(x) = −x2 + 80x

c x4 + 2x3 − 3x2 + x − 5 = ( x2 + x − 1) ( x2 + x − 3) + (5x − 8)

b Maximum area: 1600 m2; Dimensions: 40 m by 40 m

d 3x3 − 2x2 + 4x − 7 = ( x2 − x + 2) (3x + 1) + (−x − 9)

c (0, 80) 21 4 < t < 6

e x3 + 5x2 − 2x + 3 = ( x + 3) ( x2 + 2x − 8) + 27 f 2 x4 − x3 + 4x2 − 2x + 1 = ( x2 + 1) (2x2 − x + 2) + (−x − 1)

3.01E Division of polynomials What do you remember?

8 a 14

1 a A monomial is an algebraic term which is a product of a constant (called the coefficient) and variables raised to non-negative integer powers. For example, in 3x2, 3 is the coefficient and x is the variable. A constant, like 5, is also a monomial. b A polynomial is an algebraic expression consisting of one or more monomial terms, each of the form axn, where a are real number coeﬃcients and n are non-negative integers, combined by addition or subtraction, such as x2 + 2x − 1. 2 a 12x3 + 8x2 − 4x 2

c 3x + 2x − 1 3

c 2x2 − x + 1

9 a

e 10 a

b 4x b x–2 d −5

4 • Division: 15 ÷ 4 = 3 remainder 3, since 15 = 4 × 3 + 3. • Dividend: 15 (the number being divided). • Divisor: 4 (the number dividing the dividend). • Quotient: 3 (the result of the division). • Remainder: 3 (the amount remaining after the division is performed). Practice 5 a 2x3 − 3x2 + x

b 6x2 + 4x − 2

c 2x3 − 6x2 + 4x

g 2x3 + 4x2 − x

6 a 2x2

Extend your thinking 11 • Similarity: Both involve dividing each part of the dividend (terms or numbers) by the divisor. • Similarity: Both result in a quotient and a remainder. • Difference: Polynomial division uses exponent laws for variables, unlike numerical division. 12 x3 − 2x2 − 11x + 17 13 a Student A: Incorrectly applied exponent law; the last term should be 3 not 3x. Correct answer: 4x2 − 2x + 3. b Student B: The student’s response is incorrect. The fundamental check for division, P ( x) = A( x)Q( x) + R( x), fails with their numbers. Calculating ( x − 1)( x2 − x) + ( x − 3) results in x3 − 2x2 + 2x − 3, which is not the original dividend. The correct division of x3 − 2x2 + x − 3 by x − 1 yields a quotient of x2 − x and a remainder of −3.

b 3x2 3

c −3x + 11 d 11x − 12

d 0

3 a 2x − 5x + 3x – 7

b −1

c 12x − 8x + 4x

d 10x4 − 5x3 + 20x2

e 2x2 + x − 3

7 a x3 − 4x2 + 5x − 2 = ( x − 1) ( x2 − 3x + 2)

1056 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

15 Yes, R( x) = 0. Using the monomial division method, each term of the dividend is evenly divisible by x. Polynomial long division also yields the quotient x2 − 3x + 3 with no remainder.

12 13 a Incorrect, Student A made a division error, getting x2 − 5x + 1 instead of x2 − 4x + 3. Correct factorisation: ( x − 1)2 ( x − 3). b Incorrect, Student B factored x2 − 4x + 3 as ( x − 4) ( x + 1) instead of ( x − 1) ( x − 3). Correct factorisation: ( x − 1)2 ( x − 3).

3.02E The remainder and factor theorems

14 C(0) = −4

What do you remember?

3.03E Further applications

1 The remainder is P ( α ).

What do you remember?

2 P (α ) = 0

1 If P ( α ) = 0, then x − α is a factor of P ( x).

2 ±1, ±2, ±4, ±8

4 True

3 Evaluate P (0). 4 As x → ∞, P ( x) → ∞; as x → −∞, P ( x) → −∞.

Practice 5 a 1

b 47

e 86

c 1

5 a ( x − 2) ( x + 2) ( x − 1) ( x + 3)

f 149

6 a k=1

b k = −6

c k = −14

d k = −13

e k = −5

f k=0

7 a i Yes

b ( x − 1)2 ( x − 2) ( x + 2) c (2x − 1) ( x − 2) (2x2 − x + 2) d 2x( x − 1) ( x − 2) ( x2 − 2x − 2) e 3x( x − 1) ( x + 1) ( x − 3) ( x + 2) f x( x − 1) ( x − 2) ( x − 3)

ii x2 − 3x + 2

g 2( x − 1) ( x + 1) ( x − 2) ( x + 2)

iii ( x − 1) ( x − 2) ( x − 4)

h x2 ( x − 2) ( x2 − 3) or

b i Yes ii 2x2 − x − 1

6 a x = −2, −1, 1, 3

iii ( x − 1) ( x + 3) (2x + 1)

b x = −1, 2, −3, 4 c x = 2, 3, −5, 6

c i No ii 3x2 + 4x + 3 3

Practice

d x = 2 (multiplicity 3), x = −2

e x = −1 (multiplicity 3), x = 2 (multiplicity 2)

iii 3x − 2x − 5x + 2

f x = 1, −3, −4, 7

d i Yes ii x2 − x − 2

g x = 0, 1, −1,

iii ( x + 1) ( x − 2) ( x + 3)

h x = −1, 0, 1,

8 a

9 a

b a = −1, b = 0

Extend your thinking

, ,

7 a i x -intercepts: (−1, 0), (2, 0), (3, 0). y-intercept: (0, 6). ii Degree 3 (odd), leading coeﬃcient 1 (positive), rises as x → ∞, falls as x → −∞.

10 P ( x) = x3 − x2 + x − 1 11

Answers 1057 mathspace.co

iii

y 6

(0, 6)

4 2

(2, 0) (3, 0)x

(−1, 0) −2 −1

−2

−3 −2 −1−2 −4 −6 −8

−4 −6

b i x -intercepts: (−3, 0), (−1, 0), (2, 0). y-intercept: (0, −6).

iii

y 2

(2, 0)

−4 −3 −2 −1 −2

iii

(−1.414, 0) (2.449, 0)x

ii Degree 4 (even), leading coeﬃcient 1 (positive), rises as x → ±∞. y

−3 −2 −1

(−2.449, 0) −2 (1.414, 0)

f i x-intercepts:

y-intercept: (0, −2). ii Degree 3 (odd), leading coeﬃcient 1 (positive), rises as x → ∞, falls as x → −∞.

8 6

iii

(0, 4)

y 2

(−1, 0)

(−1.414, 0) (−1, 0) (1.414, 0) x

(2, 0) x 1

−2

y 14 12 (0, 12) 10 8 6 4 2

c i x -intercepts: (−2, 0), (−1, 0), (1, 0), (2, 0). y-intercept: (0, 4).

−3 −2 −1

y-intercept: (0, 12).

−8

(−2, 0)

(≈ ±2.45, ±1.41).

−6 (0, −6)

−4

iii

ii Degree 4 (even), leading coeﬃcient 1 (positive), rises as x → ±∞.

(−1, 0)

(2, 0) x 1

e i x-intercepts:

ii Degree 3 (odd), leading coeﬃcient 1 (positive), rises as x → ∞, falls as x → −∞.

(−3, 0)

18 y 16 14 12 10 8 6 4 (−3, 0) 2

(1, 0)

−3

d i x -intercepts: (−3, 0), (2, 0) (Note: x = 2 has multiplicity 2). y-intercept: (0, 12). ii Degree 3 (odd), leading coeﬃcient 1 (positive), rises as x → ∞, falls as x → −∞.

−2

−1

−2 (0, −2) −4

g i x -intercepts: (−1, 0) (multiplicity 2), (2, 0) (multiplicity 2). y-intercept: (0, 4). ii Degree 4 (even), leading coeﬃcient 1 (positive), rises as x → ±∞.

1058 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

iii

These are the times when the roller coaster is at ground level (h( x) = 0). For a roller coaster ride, these times could represent points where the track touches the ground, such as the start of the ride, loading/unloading zones, or speciﬁc design features.

y 8 6

(0, 4) 4

(−1, 0) −2

(2, 0) x

−1

h i x -intercepts: (−2, 0), (2, 0), (3, 0). y-intercept: (0, 12). ii Degree 3 (odd), leading coeﬃcient 1 (positive), rises as x → ∞, falls as x → −∞. iii

(−2, 0)

y 12 (0, 12) 10 8 6 4 2 (2, 0) (3, 0)

−3 −2 −1 −2 −4 −6 −8

12 Factorising the polynomial gives P ( x) = x( x + 3) ( x + 1) ( x − 2) ( x − 4). The population is zero when P ( x) = 0, which occurs at x = −3, −1, 0, 2, 4. Since time ( x years) cannot be negative, the meaningful times are x = 0, 2, and 4 years. At x = 0, this could represent the starting point before the species was introduced. At x = 2 and x = 4 years, the population drops to zero, suggesting periods where the species might have gone locally extinct before potentially being reintroduced or recovering. 13 a Incorrect. Student A made a sign error identifying the factor from the root x = . or (2x − 1).

It should be

8 a k = −1

b k = −4

c k=2

d k = −2

e k = −14 9 a Multiplicity 1, the graph crosses the x-axis at x = 2. b Multiplicity 1, the graph crosses the x-axis at x = −1. c Multiplicity 1, the graph crosses the x-axis at x= . d Multiplicity 1, the graph crosses the x-axis at x = −1. e Multiplicity 2, the graph touches the x-axis at x = −1 without crossing. 10 a Height is x = 2 metres. b Possible widths are x = 1 and x = 4 metres. c The profit is zero when x = 2 or x = 3. Since x represents thousands of units, this corresponds to 2000 or 3000 units sold.

Corrected factors: P ( x) = ( x − 2) (2x − 1) ( x2 + 1). b Incorrect. Student B made a mistake in assuming x − 1 was a factor of 2x3 − x2 + 2x − 1 just because P (1) ≠ 0. The student should have continued testing rational roots for Q( x) = 2x3 − x2 + 2x − 1. . So (2x − 1) is a factor. Dividing 2x3 − x2 + 2x − 1 by (2x − 1) gives x2 + 1. Correct factorisation: ( x − 2) (2x − 1) ( x2 + 1). c Incorrect. Student C incorrectly factored 2x3 − x2 + 2x − 1. Expanding their factored form gives 2x3 − x2 − 2x + 1, which is not the quotient. The correct approach after obtaining the quotient Q( x) = 2x3 − x2 + 2x − 1 is to test for = 0, so (2x − 1) is a factor.

roots of Q( x). 3

Extend your thinking

Dividing 2x − x + 2x − 1 by (2x − 1) gives x2 + 1.

11 Factorised h( x) = ( x − 1) ( x − 2) ( x − 3) ( x − 4).

Correct factorisation: ( x − 2) (2x − 1) ( x2 + 1).

Zeroes: x = 1, 2, 3, 4 seconds.

14 282x − 713

Answers 1059 mathspace.co

Chapter 3 review

4.01 Further domain and range

1 B

What do you remember?

2 C

1 a Function

b Relation

3 B

c Function

d Function

4 a P ( x) = ( x − 2) ( x − 5x) − 6 b P ( x) = ( x + 1) (3x2 − x − 4) + 8 c P ( x) = ( x2 + x − 3) ( x2 − 3x + 9) + (−19x + 34) d P ( x) = ( x − 3) (2x2 − 3x + 3) + 2 e P ( x) = ( x + 2) ( x2 + 3x − 6) + 2 f P ( x) = ( x2 − x + 2) (2x2 + x − 5) + (−x + 5)

2 Square brackets [ ] include the endpoints, while parentheses ( ) exclude the endpoints. 3 The domain of a relation is the set of all possible x-values (inputs) for which the relation is deﬁned. 4 a False

b True

c False

d True

5 −3 Practice

6 a=7

5 a i 0≤x<4

7 k = −3

ii All real numbers greater than or equal to 0 and less than 4

8 ( x − 2) ( x + 3)

b i −3 < x ≤ 1

9 a = 2, b = 5 10 a a = −7, b = 6 b P ( x) = ( x + 3) ( x + 2) ( x − 1)2

12 a P ( x) = ( x − 1)2( x + 2) (2x + 3) 8 6 (0, 6)

e i x≤0 ii All real numbers less than or equal to 0

f i x≥3

(1, 0) x

(−2, 0) (−1.5, 0) −2

−1

d i 1<x<7 ii All real numbers greater than 1 and less than 7

−3

c i −2 ≤ x ≤ 5 ii All real numbers greater than or equal to −2 and less than or equal to 5

11 The roots are x = −1, 2, 3, .

ii All real numbers greater than −3 and less than or equal to 1

−2

The graph touches the x-axis at x = 1 (multiplicity 2) and crosses at x = −2 and x = −1.5. The y-intercept is at y = 6. 13 R( x) = 14 a The roots are x = −1, 2, 5. Since x ≥ 0, the values are x = 2 and x = 5. b The proﬁt is $10 000. c Applying the condition x ≥ 0, a positive proﬁt is made for 0 ≤ x < 2 or x > 5.

ii All real numbers greater than or equal to 3 6 a i −2 < x ≤ 8

ii (−2, 8]

b i 5 ≤ y < 12

ii [5, 12)

c i y<4

ii (−∞, 4)

d i x ≥ −1

ii [−1, ∞)

7 a [2, ∞) b (−∞, −3) ∪ (−3, ∞) c (−∞, 5] d (−∞, −2) ∪ (−2, 2) ∪ (2, ∞) 8 a [−1, 7)

b (−5, ∞)

c (0, 4]

d (−∞, 0]

e (−3, 2]

f [6, ∞)

9 a 2≤x<8

b y > −4

c x≤1

d −2 < y ≤ 3

e 0≤x≤5

f y>1

1060 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

10 a i (−∞, ∞)

ii [3, ∞)

b i (−∞, ∞)

ii (−∞, 2]

c i [0, ∞)

ii [0, ∞)

d i (−∞, ∞)

ii (−∞, 2]

e i [−2, ∞)

ii [0, ∞)

f i (−∞, ∞)

ii (−∞, ∞)

c For each x > 0, there are two y-values , failing the vertical line test. d The graph of y2 = x is a parabola opening to the right, including both positive and negative y-values, while y = is a half-parabola with only non-negative y-values.

g i (−∞, ∞)

ii (−∞, ∞)

h i (−∞, ∞)

ii (−∞, ∞)

11 a i [−5, 5]

ii [−5, 5]

b [−1, 3]

b i [−4, 4]

ii [−4, 4]

c [−3, 1]

c i [−4, 4]

ii [−3, 3]

d i (−∞, −4] ∪ [4, ∞)

ii (−∞, ∞)

12 a Domain: [−3, 3]; Range: [0, 3] b Domain: (−∞, ∞); Range: [−2, ∞) c Domain: [−2, 2]; Range: [−2, 0] d Domain: (−∞, ∞); Range: (−∞, 4]

19 a (x − 1)2 + (y + 1)2 = 4

20 a The student is incorrect as y = −4 is a valid solution, as hence the range must use a square bracket to include −4. b The graph is a parabola opening upwards with vertex at (0, −4), so y-values start at −4 and extend to inﬁnity.

e Domain: [−1, 3]; Range: [2, 4]

21 a (−∞, ∞)

f Domain: [−1, 1]; Range: [−1, 1]

b (0, 1]

13 a i {1, 3, 5}

ii {2, 3, 4, 5}

iii Relation b i {−2, 0, 2, 4}

ii {0, 1, 2, 3}

iii Function c i {0, 1, 2, 3}

ii {1}

iii Function d i {−1, 0, 1}

ii {−1, 0, 1}

iii Relation 14 Domain:

d The maximum value is y = 1 when x = 0, since x2 + 1 is minimised at 1.

4.02 Even and odd functions What do you remember? 1 a f (−x) = f (x)

; Range: [−2, 2]

15 Domain: [−3, 3]; Range: [−3, 3] 16 Domain:

c The denominator x2 + 1 is always positive, so y is always positive, and y = 0 would require x2 + 1 = ∞, which is impossible.

; Range: [0, 3]

b f (−x) = −f (x)

2 a Reﬂective symmetry across the y-axis b 180° rotational symmetry about the origin 3 a False

b True

c True

Practice Extend your thinking

4 a Even

b Neither

c Neither

d Even

b (−∞, 0) ∪ (0, ∞)

e Even

f Odd

c x = 2 makes the denominator zero, so the function is undeﬁned.

g Neither

h Even

17 a (−∞, 2) ∪ (2, ∞)

d As x approaches inﬁnity, y approaches 0 but never equals it, since 18 a [0, ∞) b (−∞, ∞)

= 0 has no solution.

5 a Even

b Odd

c Even

d Odd

e Neither

f Odd

g Even

h Neither

6 a Even

b Odd

c Neither

d Odd

Answers 1061 mathspace.co

7 a i 2x2

the trivial function f (x) = 0 satisﬁes both conditions.

ii Even

b i 0

ii Both even and odd

c i x4

ii Even

d i x5 + x3 − x4

ii Neither

8 a Even

b Odd

c Even

d Neither

9 a (−3, 82)

b (−2, −6)

10 a f (x) = x4 − 2x2

b f (x) = x3 + x

c f (x) = x + x + 1 8

d f (x) = −2

11 a i x − 4x + 2

ii Even

b i −x5 + 3x

ii Odd

c i −x3 + x2

ii Neither

d i −7x9 − x5

ii Odd

12 a i 1

ii Even

b i x

ii Odd

c i

ii Neither

d i

ii Odd

Extend your thinking

4.03E Sums and products of zeroes What do you remember? 1 a An expression made up of non-negative integer powers of the same variable and coeﬃcients combined using addition, subtraction, and multiplication. b The solution to a polynomial equation, where the polynomial equals zero. 2 a α+β=

b αβ =

3 The quadratic can be written as a( x − α ) ( x − β ), where expanding gives ax2 − a( α + β )x + aαβ, and equating to ax2 + bx + c yields α + β = and αβ = . Practice

13 Since f (x) is odd, f (−x) = −f (x). At x = 0, f (−0) = f (0) and f (−0) = −f (0). Thus, f (0) = −f (0), which implies f (0) = 0. Hence, the function passes through (0, 0). 14 a All real numbers, as f (−x) = (−x)2 + k = x2 + k = f (x) b No, since f (−x) = x2 + k and −f (x) = −(x2 + k) = −x2 − k, and x2 + k ≠ −x2 − k for any k unless the function is trivial. 15 The student incorrectly assumed that a function with odd-degree terms is always odd. To verify, compute f (−x) = (−x)3 + 2 = −x3 + 2 and −f (x) = −(x3 + 2) = −x3 − 2. Since f (−x) ≠ −f (x), the function is not odd. Also, f (−x) ≠ f (x), so it is not even. Thus, the function is neither odd nor even. 16 a Neither

b The function f (x) = 0 is both even and odd. For evenness: f (−x) = 0 = f (x). For oddness: f (−x) = 0 = −f (x) since −f (x) = −0 = 0. Thus, it satisﬁes both conditions.

b Odd

17 a No, a non-trivial function cannot be both even and odd. For a function to be even, f (−x) = f (x), and for it to be odd, f (−x) = −f (x). Equating these, f (x) = −f (x), implies f (x) = 0 for all x, which is the zero function. Thus, only

4 a

b −2

c 3

e b −10

5 a 6 a 139

c 6

7 a 3

b −4

8 a 2

b 5

c −9

e 2

f −6

d −23

e 97

d −2

9 a Ask your teacher for worked solution. b −136 10 a Ask your teacher for worked solution. b 11 a α = , β = 3, γ = 2 b α = 1, β = 2, γ = 3 12 12 13 The roots are 1, 2, 2, and 3.

1062 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

14 a

c 6

15 a

16 a 24

b 14

17 a 2

b 7

18 a −4

b −3

19 a 3

b 2

8 a ( x − 1) ( x − 2) ( x + 5)

b ( x − 5) ( x + 2) ( x − 3)

c ( x − 3) ( x − 5) ( x − 1)

d ( x + 2) ( x + 1) ( x + 5)

e ( x − 2) (4x − 5) ( x + 3) f ( x − 1) ( x − 2) ( x − 4) 9 a x( x − 1) ( x − 2) ( x − 3) b ( x − 2) ( x + 2) ( x − 1) ( x + 3) c ( x + 2) ( x − 4) ( x + 5) ( x − 1)

c −1

d ( x − 3) ( x + 2) ( x − 5) ( x + 1)

c 6

e ( x − 3)2( x + 1) ( x + 5)

d −8

f ( x − 1) ( x + 1) ( x − 2) ( x + 3)

Extend your thinking

10 The polynomial is x3 − 2x2 − x + 2.

20 a Ask your teacher for worked solution.

11 a α = 1 +

b Ask your teacher for worked solution. c Ask your teacher for worked solution.

12 a k = −5. The other roots are

21 P ( x) = x3 − 7x2 + 6x − 2 22 a −25

,β=1−

b The vertex is (1, −6) and

b 26

23 One possible polynomial is x4 − 4x3 + 6x2 − 4x + 1 = 0, derived by assuming roots are 1, 1, 1, 1 (sum = 4, product = 1). Other polynomials are possible with different pairwise and triple product sums.

4.04E Applications of sums and products of zeroes

b a = −4, b = −2. The third root is 1. c The roots are 2, 3, 5. d The roots are 1, 3 and . 13 a 4 3

b 3

c 2

d 6

14 a x − 6x − 40x + 192 = 0 b c x3 − 6x2 − x + 30 = 0

What do you remember? 1 a

2 a

Extend your thinking 15 a Ask your teacher for worked solutions. b The area is

16 b = 1 or b = −3

17 a x2 − 2x − 2 = 0

3 a α+β+γ+δ=

b The distance is

b αβ + αγ + αδ + βγ + βδ + γδ =

units.

c αβγ + αβδ + αγδ + βγδ =

Chapter 4 review

d αβγδ =

1 A Practice 2 B

4 3

3 a [4, ∞)

5 −1

b (−∞, −5) ∪ (−5, ∞)

6 −2

c (−∞, 7] 2

7 a P ( x) = x − 4x + 3

b P ( x) = x + 2x − 5

c P ( x) = x3 − 3x2 + 2x + 1

d (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 4 a i (−∞, ∞)

ii (−∞, 5]

Answers 1063 mathspace.co

ii [−6, 6]

19E a x2 − 4x + 2 = 0

c i [−4, ∞)

ii [0, ∞)

d i [−5, 5]

ii [−2, 2]

20E a

b i [−6, 6]

( α − β )3 + 3αβ ( α − β ) = ( α − β ) (( α − β )2 + 3αβ ) = ( α − β ) ( α2 − 2αβ + β 2 + 3αβ )

; Range: [0, 5]

5 Domain:

= ( α − β ) ( α2 + αβ + β 2) = α3 − β 3

6 a (−∞, 3) ∪ (3, ∞)

b (−∞, 0) ∪ (0, ∞) c When x = 3, the denominator becomes 0, making the expression undeﬁned. d The numerator is a constant 1, so y cannot be 0. The function approaches y = 0 as a horizontal asymptote. 7 a (x − 2)2 + (y + 1)2 = 9

b Odd

c Neither d Odd

9 (−2, −28)

What do you remember? 1 a Rectangular hyperbola

f (−x) = −f (x)

Write the formula

f (−0) = −f (0)

Substitute x = 0

f (0) = −f (0)

b True

c True

d True

3 As x → −∞, f (x) → 0. Speciﬁcally, if k > 0, f (x) → 0−; if k < 0, f (x) → 0+. Practice

10 Odd function:

4 a

Evaluate

2f (0) = 0

Add f (0) to both sides

f (0) = 0

Divide both sides by 2

5.01 Graphs of reciprocal functions

2 a False

c [−4, 2]

21E −4, −2, 1, 3

b x = 0, f (x) = 0

b [−1, 5] 8 a Even

b 4

−2 −1

f (x)

−1 −2

Thus, the function passes through (0, 0).

−20 20

3 2

11 The student’s claim is incorrect. An odd function requires f (−x) = −f (x).

f (−x) = (−x)5 + 3 = −x5 + 3

−4 −3 −2 −1 −1

−f (x) = −(x5 + 3) = −x5 − 3

x 1

2 3 4

−2

−x + 3 ≠ −x5 − 3

−3

The constant term +3 (i.e., 3x0) has an even power, so the function is not odd.

−4

5 a

12E B

−4

−2 −1

f (x)

13E C 14E a

15E a −2

b −3

c 7

d 3

−5

b Quadrants 2 and 4 6 a f (x) approaches 0 from the positive side

b f (x) → ∞ 16E a −1

−10

b −5

d 5

7 a Quadrants 2 and 4

17E P ( x) = x2 + 5x + 4

b f (x) = 3

18E P ( x) = x3 − 7x2 + 17x − 14

1064 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

d x = 0, f (x) = 0 8 a 0 9

15 a c = c ∞

b 0 4

d −∞

2 1 −4 −3 −2 −1 −1

c 80

x 1

2 3 4 20

−2

−3

−4

≠ 0.

What do you remember?

b Multiplying both sides of f (x) =

by x gives

x × f (x) = 8, which is equivalent. 11 As x → ∞, f (x) =

→ 0+ and f (x) = −

→ 0−. +

As x → −∞, f (x) =

→ 0 and f (x) =

→0 .

As x → 0+, f (x) =

→ ∞ and f (x) =

→ −∞.

As x → 0−, f (x) =

→ −∞ and f (x) =

→ ∞.

1 a The absolute value of a number is its distance from zero on the number line, always non-negative. b For −5, the distance from −5 to 0 is 5 units. 2 a For a = 6, the absolute value represents the distance from zero. Thus, a = 6 or a = −6. b For b = , the distance from zero is . Thus, b =

The sign of k determines the quadrants: positive k places the graph in the ﬁrst and third quadrants, while negative k places it in the second and fourth.

b f (x) =

3 a f (4) = 4 = 4 (since 4 ≥ 0, use x). b f (−2) =  − 2 = −(−2) = 2 (since −2 < 0, use −x). c f (0) = 0 = 0 (since 0 ≥ 0, use x).

c = 4, b = 2, a = 1

c n = 8 workers

d For d = 0, the distance from zero is 0. Thus, d = 0.

13 A : f (x) = , B : f (x) = , C : f (x) = , where

14 a t =

or b =

c For c = −4, there is no solution because absolute value cannot be negative, as distance is non-negative.

Extend your thinking 12 a f (x) =

5.02 Introduction to absolute value functions

10 a No, because x = 0 makes the function undeﬁned, and f (x) = 0 has no solution since

b t = 3 hours

4 5 No, = a for all real a, as both yield the non-negative distance from 0. Practice 6 a x = 7 c x − 4 = 2.5

b x + 1 = 3

7 • f (−6) = 6

Answers 1065 mathspace.co

• f (0) = 0

2 x=

,y=0

• f (3) = 3 3 y≥0

8 l − 50 = 0.5

4 a The graph becomes steeper.

9 t − 20 = 1.5

b The graph becomes wider.

10 a For x = −7: For x = 4:

= =

= 7 = −7 = 4 = 4.

b 11 11 a w − 2 = 4

b w + 3 = 2

12 a x − 3 = 5

b x + 2 = 1.2

c Not possible, as a ≠ 0 for the function to be deﬁned. 5 a Two solutions

b One solution

c No solutions Practice 6 a x = 5

c x  =

b x + 2 = 4

7 a

y 6

Extend your thinking

13 False. The statement is only true if a > 0. If a = 0, then x = 0 is the only solution. If a < 0, there is no real solution for x because an absolute value cannot be negative.

4 3 2 1

14 a f (v) = v − 500 b Piecewise function:

−6 −5 −4 −3 −2 −1

x 1

Symmetry: x = −3 Domain: , Range: y ≥ 0

Vertex: (500, 0) 15 a d − 10 = 0.2

b d = 9.8, 10.2

16 a a controls the steepness and direction: if a > 0, the V-shape opens upward; if a < 0, downward. The absolute value of a determines the steepness. h shifts the graph horizontally, with the vertex at x = h. k shifts the graph vertically, placing the vertex at y = k. b The vertex is (3, 1). Compared to f (x) = x, the graph is steeper due to a = 2 (double the steepness), has the same upward concavity since a > 0, and the vertex is shifted right by 3 units and up by 1 unit.

y 6 5 4 3 2 1

x 1

2 3 4 5 6 7 8

Symmetry: x = 4 Domain: , Range: y ≥ 0 c

y 6

c g(x) = −2x − 1 − 1

5 4

5.03 Absolute value functions

3 2

What do you remember? 1 Absolute value represents the distance of a number from 0 on a number line, always non-negative.

1066 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

1 −2 −1

x 1

Symmetry: x = 1

Domain: , Range: y ≥ 0 d

6 4

x −2 −1

2 3 4 5 6

2 x

Symmetry: x = 2

−14−12−10 −8 −6 −4 −2

2 4 6 8 10 12

Domain: , Range: y ≥ 0 8 a

x = −15, 13

y 10

2 1 −4 −3 −2 −1 −1

x 1

2 3 4

−2 −3

−4

x = 1, 3

−16 −12 −8 −4

8 12 16 20 24

2 3 4

10 8

x = −16, 24

9 a

x −4 −2

2 4 6 8 10

x = −6, 12

−4 −3 −2 −1 −1

Vertex: Line of symmetry: x = x-intercept: y-intercept: (0, 2)

Answers 1067 mathspace.co

x = −1, 0

y 10

Extend your thinking

8 6

15 a x =

,x=3 7

x −1

2 3 4 5 6 7

1 5 2 − , 3 3

Line of symmetry: x = 3

x-intercept: (3, 0)

−4 −3 −2 −1 −1

y-intercept: (0, 9) 10 a x = −1, x = 5

b x = −7, x = 3

c x = −4, x = 8

d x = −1, x = 5

,x=

f x = −8, x = 4

g x=

,x=

h m = −4, m = 4

i u = −7, u = 3

j v = −3, v = 5

11 a f (x) = x − 50; 2 mm

x =

x 1

2 3 4

16 Function: f (t) = t − 25 At t = 22, 3°C At t = 25, 0°C At t = 28, 3°C 17 a Two intersections with y = k above x-axis.

b f (s) = s − 60; 5 km/h

b One intersection at vertex.

c x − 2 = 4; x = −2, x = 6 12 a x = −5, x = 1

b x = −7, x = 1

c x = −4, x = 2

d x = −1, x = 5

13 f (d) = d − 25, d = 22, 28 mm 14 a

(3, 5)

Vertex: (3, 0)

e x=

c No intersections, as y = ax + b ≥ 0. 18 Function: f (t) = t − 1 At t = , deviation is

h, distance is

km.

At t = 1, deviation is 0 h, distance is 50 km.

y 4 3 2

At t = , deviation is

h, distance is

19 a No solutions

b n≥2

c p = −2

d q = 3, 6

km.

5.04 Circles and semicircles

1 x 1

What do you remember? 1 a x2 + y2 = r2

x = 1, 5 b

b For a point (x, y) on a circle with radius r and centre (0, 0), the distance from the origin

= r. Squaring both sides satisﬁes gives x2 + y2 = r2.

3 2

c y=

d Circle x

−2

−1

1 −1

2 r = 10. 3 a Centre: (0, 0), Radius:

1068 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b (±4, 0)

5 4 3 2 1

Practice 4 For a point (x, y) on the circle, the distance from the origin (0, 0) is 6. By Pythagoras’ theorem,

1 2 3 4 5

6 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

= 6. Squaring both sides gives x2 + y2 = 36. 5 The circle’s equation is x2 + y2 = 102 = 100. For (8, 6), compute 82 + 62 = 64 + 36 = 100, which satisﬁes the equation. Thus, the point lies on the circle.

2 1

−4 −3 −2 −1 −1

2 3 4

−2 −3

r=5

−6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

x 2 3 4 5 6

−4

10 a

y 1

7 a x2 + y2 = 9

b x2 + y2 = 72 = 49

−1

8 a Upper semicircle (positive y-values) −1

b Left semicircle (negative x-values) 9 a

y 2

b 8

1 x −2

−1

6 4 2

−1

−8 −6 −4 −2 −2

−2

x 2

−4

−6

−8

3 2 1 −3 −2 −1

−1

x 1

−2 −3

Answers 1069 mathspace.co

Maximum height: 4 metres

16 Width at the base: 26 metres

Maximum height: 13 metres

4 2 2

b [−6, 6]

17 a

−8 −6 −4 −2 −2

Chapter 5 review

−4 −6

1 C

−8

2 a Quadrants II and IV b f (x) = 2

c x=

10 8 6 4 2

d x = 0, y = 0 3 a 0 x

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

2 4 6 8 10

11 a y =

b 0

d −∞

4 a f (x) =

b f (x) =

c f (x) =

d f (x) =

e f (x) =

f f (x) =

5 a Verify the identity values.

b y=

c ∞

= a for the given

For a = −9:

12 a y = ± b

10 8

x2 + y2 = 81

Since 9 = 9, the identity holds for a = −9.

4 2 −10 −8 −6 −4 −2 −2

For a = 5:

x 2 4 6 8 10

−4 −6

−8

Since 5 = 5, the identity holds for a = 5.

−10

13 a x2 + y2 = 25

Thus, the identity is veriﬁed for both values. b

b y=

Evaluate the absolute

Extend your thinking 2

Apply the identity

14 From x + y = r , solve for x: x = r − y , so . Right semicircle:

(positive x).

Left semicircle:

(negative x).

15 Width at ground level: 8 metres

values

Sum the terms

6 a f (d) = d − 75 b At d = 74.7: f (74.7) = 74.7 − 75 = 0.3 mm At d = 75: f (75) = 75 − 75 = 0 mm At d = 75.3: f (75.3) = 75.3 − 75 = 0.3 mm

1070 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

7 a f (v) = v − 450

y-intercept: (0, 12)

b Piecewise function:

(8, 5)

(−2, 5)

Vertex: (450, 0)

8 a The parameter a determines the steepness and direction of the graph: if a > 0, the graph opens upward; if a < 0, it opens downward. The parameter h shifts the graph horizontally by h units (right if h > 0, left if h < 0). The parameter k shifts the graph vertically by k units (up if k > 0, down if k < 0). The vertex of the graph is at (h, k). b Vertex: (5, 3); the graph is a V-shape opening upward. c h(x) = −2x − 6 − 3 9 a x = −2, 3

b x = −12, 6

c x = 4, 8

d x = −5, −3

10 a 7

3 2 1 x −2 −1

1 2 3 4 5 6 7 8

Solutions: x = −2, 8 12 Solutions: x = −2, 8; Ordered pairs: (−2, 7), (8, 7) 13 The circle’s equation is x2 + y2 = 172. 172 = 289 2 8 + 152 = 64 + 225 64 + 225 = 289 Since 82 + 152 = 289, the point (8, 15) lies on the circle. 14 Centre: (0, 0), Radius: 3

6 5 (0, 5)

5 − ,0 2

−5 −4 −3 −2 −1

−3 −2 −1

x 1

−1

−2

Vertex:

−3

Line of symmetry: x =

15 a

x-intercept:

16 Width at the base (y = 0): 14 m; Maximum height (at x = 0): 7 m

y-intercept: (0, 5) b

y 12 (0, 12)

17 a

b y ∈ [−4, 4]

10 8 6 4 2 −1

(4, 0)

1 2 3 4 5 6 7 8

Vertex: (4, 0) Line of symmetry: x = 4 x-intercept: (4, 0)

Answers 1071 mathspace.co

6.01E Solve cubic inequalities

What do you remember?

f ( x) = ( x − 2) ( x + 1) ( x − 4)

−2

−24

−1

−4

2 a True

b False

20 10

(−1, 0)

−1 < x < 6 or x > 8 b i Roots: x = −3, 7, 10

c True

3 x = −2, 5, 7

−

x−7

−

x − 10

−

Product

−

x ≤ −3 or 7 ≤ x ≤ 10

x < −3 −3 < x < 1 1 < x < 5

x>5

x < −3 −3 < x < 7 7 < x < 10 x > 10

x+3

Practice 4

(6, 0) (8, 0) x 1 2 3 4 5 6 7 8 9

−2 −1

x+3

−

250

x−1

−

200

x−5

−

150

Product

−

100 50

(−3, 0)

(7, 0)

−4−3−2−1

1 2 3 4 5 6 7 8 9 10 11

5 a ii −2 < x < 4 or x > 8 b i −2 ≤ x ≤ 4 or x ≥ 8

−50

6 a i Roots: x = −1, 6, 8

x < −1 −1 < x < 6 6 < x < 8 x+1

−

(10, 0) x

x>8 +

x ≤ −3 or 7 ≤ x ≤ 10 c i Roots: x = −6, 1, 9

x−6

−

x−8

−

x+6

−

Product

−

x−1

−

x−9

−

Product

−

−1 < x < 6 or x > 8

1072 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x < −6 −6 < x < 1 1 < x < 9

x < −6 or 1 < x < 9

x>9

150

120

100

(−6, 0)

(1, 0)

−7−6−5−4−3−2−1

1 2 3 4 5 6 7 8 9 10

(9, 0) x

−50

(−2, 0)

−150

f i Roots: x = −7, 3, 10

d i Roots: x = −8, 2, 10 x < −8 −8 < x < 2 2 < x < 10 x > 10

x < −7 −7 < x < 3 3 < x < 10 −

x−3

−

x − 10

−

Product

−

x−2

−

x+8

−

x − 10

−

−7 < x < 3 or x > 10

Product

−

y 300

x ≤ −8 or 2 ≤ x ≤ 10

250

200

200 150 100 (−8, 0) 50 (2, 0)

150 100

(10, 0) x

2 4 6 8 10

(−7, 0)

(3, 0) (10, 0) x 1 2 3 4 5 6 7 8 9 10 11

−100 −150

−7 < x < 3 or x > 10

e i Roots: x = −2, 6, 9

g i Roots: x = −9, 1, 7

x < −2 −2 < x < 6 6 < x < 9 x > 9

x+2

−

x−6

−

x−9

−

Product

−

x ≤ −2 or 6 ≤ x ≤ 9

−8−7−6−5−4−3−2−1 −50

x ≤ −8 or 2 ≤ x ≤ 10

x > 10

x+7

−1

−8 −6 −4 −2 −50 −100 −150 −200 −250 −300

(9, 0) x

1 2 3 4 5 6 7 8 9 10

x ≤ −2 or 6 ≤ x ≤ 9

x < −6 or 1 < x < 9

(6, 0)

−3−2 −1

−100

x < −9 −9 < x < 1 1 < x < 7

x>7

−( x + 9)

−

x−1

−

x−7

−

Product

−

−9 ≤ x ≤ 1 or x ≥ 7

Answers 1073 mathspace.co

100

(−9, 0)

120

(1, 0)

−10−8 −6 −4 −2 −50

(7, 0) x

100

2 4 6 8

−100

−150

−200

5 − ,0 4

−250

−2

−9 ≤ x ≤ 1 or x ≥ 7

20 −1

1 − ,0 2

7 ,0 x 3

−20

h i Roots: x = −4, 8, 10

x < −4 −4 < x < 8 8 < x < 10 x > 10

x+4

−

x−8

−

x − 10

−

Product

−

b i Roots: 3x − 2

−

4x + 7

−

350

5x − 1

−

300

Product

−

−4 ≤ x ≤ 8 or x ≥ 10 ii

250

200 150

100 50

100 90 80 70 60 50 40 30 20 10

(8, 0) (10, 0) x

(−4, 0) −4 −2

2 4 6 8 10 12

−4 ≤ x ≤ 8 or x ≥ 10 7 a i Roots: −1

7 − ,0 4

−

2x + 1

−

3x − 7

−

4x + 5

−

Product

−

−2

−

−1

2 ,0 3

−10 1 , 0

1074 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

c i Roots: −(4x + 3)

−

3x − 8

−

5x + 1

−

Product

−

9 − ,0 2

1400 y 1300 1200 1100 1000 900 800 700 600 500 400 300 200 7 , 0 100 5

−5 −4 −3 −2 −100 −1

10 ,0 3 2

x 5

−200 −300

250 200

e i Roots:

150 100 50 3 − ,0 4

1 − ,0 5

−1

8 ,0 3 x 2

d i Roots:

−1

−

2x + 3

−

4x − 7

−

3x + 10

−

Product

−

y 250

5x − 7

−

200

2x + 9

−

150

3x − 10

−

Product

−

100

3 − ,0 2

−4 −3 −2 −1 10 −50 − ,0

7 ,0 4 x 1

Answers 1075 mathspace.co

f i Roots:

1000

900 800

3x − 8

−

5x + 1

−

4x − 9

−

Product

−

700 600 500 400 300 200

7 − , 0 100 5

−3 −2

−1

9 ,0 2 1

x 5

150 120

h i Roots:

90 60 30

−1

8 ,0 3

9 ,0 4

1 − ,0 5

x>2

4x − 7

−

3x + 8

−

5x − 10

−

Product

−

or x > 2

g i Roots:

900

800 700

−1

−

5x + 7

−

500

2x − 9

−

400

4x + 3

−

200

Product

−

600

300

8 − ,0 3 −3

−2

7 ,0 4

−1 −100

or x > 2

8 −1 ≤ x ≤ 2 or x ≥ 4

1076 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

100

(2, 0) x 2

c i Roots: x = −4, 1

9 ( x − 3) ( x + 1) ( x − 6) ≥ 0: −1 ≤ x ≤ 3 or x ≥ 6 −( x − 3) ( x + 1) ( x − 6) ≥ 0: x ≤ −1 or 3 ≤ x ≤ 6

Observations: The solutions to the inequalities are complementary. The solution to ( x − 3) ( x + 1) ( x − 6) ≥ 0 gives the intervals where the expression is non-negative, while the solution to −( x − 3) ( x + 1) ( x − 6) ≥ 0 gives the intervals where the expression is non-positive.

−4 < x < 1

x>1

−

( x − 1)

x+4

−

Product

−

−1

x > −4, x ≠ 1

10 a i Roots: x = −1, 3

x < −4

x < −1

−1 < x < 3

x>3

( x − 3)2

x+1

−

Product

−

y 4 2 (1, 0) x −5 −4 −3 −2 −1−2 1 2

(−4, 0)

−4 −6 −8 −10 −12 −14 −16 −18 −20

x > −1, x ≠ 3

6 4

(−1, 0) −2 −1

x > −4, x ≠ 1

(3, 0) 1

−2

d i Roots: x = −5, 2

x 3

x < −5

−5 < x <2

x>2

( x + 5)2

x−2

−

Product

−

−4 −6 −8

x > −1, x ≠ 3 b i Roots: x = −2, 5

x < −2

−2 < x < 5

x>5

( x + 2)2

x−5

−

Product

−

x ≥ 2, x = −5 ii (−5, 0)

−6 −5 −4 −3 −2 −1 5 10 15 20 25 30 35 40 45 50

x ≤ 5 ii

10 (−2, 0) 5 −3 −2 −1−5 −10 −15 −20 −25 −30 −35 −40 −45 −50

x ≤ 5

(5, 0)

y 15 10 5 (2, 0)

1 2 3 4 5 6

1 2 3

x ≥ 2, x = −5 e i Roots: x = −3, 4

x < −3

−3 < x < 4

x>4

( x − 4)

x+3

−

Product

−

Answers 1077 mathspace.co

x < −3

50 y 45 40 35 30 25 20 15 10 5

(−3, 0)

−4 −3 −2 −1−5 −10 −15 −20

70 60 50 40 30 20 10

(−6, 0) (4, 0)

−7 −6 −5 −4 −3 −2 −1 −10 −20 −30

1 2 3 4 5

(2, 0) x 1 2 3

x ≤ −6, x = 2 h i Roots: x = −3, 8

x < −3 f i Roots: x = −1, 7

x < −3

−3 < x < 8

x>8

−

x−8

−

Product

−

x < −1

−1 < x < 7

x>7

−

( x + 3)

( x + 1)

x−7

−

Product

−

−1

x ≤ 7 ii 70 60 50 40 30 20 10 (−1, 0) −2

−1 2

x < 8, x ≠ −3 ii 150 100 50

(−3, 0)

−4−3−2−1 −50

x (7, 0) 6 8

(8, 0)

11 a x( x − 2)( x + 4) ≥ 21.

g i Roots: x = −6, 2 x < −6

−6 < x < 2

x>2

Extend your thinking

( x − 2)2

12 1 ≤ x ≤ 3 or x ≥ 5

x+6

−

13 x ≤ −2 or 0 ≤ x ≤ 4

Product

−

14 a 45 ≤ x2( x + 2) ≤ 96

x ≤ −6, x = 2

1 2 3 4 5 6 7 8 9

x < 8, x ≠ −3

x ≤ 7

y 200

b x≥3

b The range of possible values for the side length x is 3 ≤ x ≤ 4 metres. 15 k ≤ 2

1078 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

6.02E Solve inequalities with variables in the denominator

What do you remember?

2 1 −1 −3 −4 y =

d 9 < x ≤ 16

or x > −7

f x < −9 or x > 14

g x < 15 or x > 32

h x ≤ −18 or x > 10

6 a

−4

−2

x 2

−1

−3

−4

y 4 7 − ,2 2

y=2

2 x

3 x+5

2 −2

25 20 15 10 5

6 x2 − 7 x + 12

y=2 2

1.6972 ≤ x < 3 or 4 < x ≤ 5.3028

x < 2 or x ≥ 6

−6

7 x+3

−5 −10 −15 −20 −25

4 y= −2 x−2

−8

−2

−3 < x ≤

(6, 1)

−4

y=1

1 ,2 2

−6

7 a

5 x−4

Practice

c x < 4 or x >

d y=2

b −3 < x ≤ 10

x < 4 or

4 The graphical method visualises where y = is above or below y = K. Identify intercepts, asymptotes, and intersections with y = K.

5 a −5 < x < 0

x 2

−2

2 Solve A = K × B (numerator after equating) and B = 0 (denominator undeﬁned). These points divide the number line into intervals to test. 3 The number line plots critical points (zeroes of numerator, zeroes of denominator), dividing it into intervals to test where the inequality holds.

17 ,3 3

y=3

1 Multiplying by the denominator changes the inequality’s direction if negative. A safer method is multiplying by the denominator’s square, which is always positive.

−4

−2 −2 −4

30 25 20 15 10 5 −5 −10 −15 −20 −25 −30

8 x2 − 5 x + 6

y=1 2

x < −0.3723 or 2 < x < 3 x > 5.3723

Answers 1079 mathspace.co

12 y 10 8 6 4 2

11 a

10 y= 2 x − 6x + 8

−2 −4 −6 −8 −10 −12

y=3

7 x2 − 8 x + 15

y=2 y=1 x

−2

8 10 12

−1

y 6

< 1: x < 4 −

or 3 < x < 5 or

> 2:

< x < 3 or

x>4+

y=1 2

4 x2 − 8 x + 15

2 −2

x ≤ 0.9183, 2 < x < 4 or x ≥ 5.0817 d

y 4

5<x<

There are no common x values which satisfy both inequalities.

−4 −6

10 8 6 4 2

1.7639 < x < 3 or 5 < x < 6.2361 8 x ≤ −11 or x > 5 y 6

−2 −4 −6 −8 −10

4 2

9 x2 − 10 x + 24

y=3 y=1 x 2

10 12

x −12 −10 −8 −6 −4 −2

2 4 6 8 10 12

≤ 3: x ≤ 3 or 4 < x < 6 or x ≥ 7

≥ 1: 1.8377 ≤ x < 4 or

−2 −4 −6

9 x≤

or −3 < x < 3 or x ≥

6 < x ≤ 8.1623 The inequalities are satisfied simultaneously for x in the intervals [1.8377, 3] or [7, 8.1623]. 12 a

12 y 10 8 6 4 2

10 a 1.432 < x < 2 or 12 < x < 12.568 b x < 2 or 12 ≤ x < 13 c x < 1.494 or 2 < x < 14 or x > 14.506 d 2 < x < 15 or x ≥ 19 e x ≤ 0.2918 or 2 < x < 12 or x ≥ 13.7082 f 2 < x < 13 or x > 20 g x < 1.5775 or 2 < x < 14 or x > 14.4225 h 2 < x < 15 or x ≥ 21

−2 −4 −6 −8 −10 −12

10 x2 − 12 x + 35

y=1 2

8 10 12 14

x ≤ 2.683 or 5 < x < 7 or x ≥ 9.317

1080 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Production times of 2.683 hours or less, between 5 and 7 (exclusive), or 9.317 hours or more, keep the rate at or below the threshold. b

14 y 12 10 8 6 4 2

y 6

12 x2 − 8 x + 15

4 2

y=3 4

−6 −4 −2 −2

x 2

−2 −4 −6 −8 −10 −12 −14

16 x ≤ −5 or 5 ≤ x < 7. Eﬃciency is non-positive when production levels are −5 units or less, or between (inclusive) and 7 (exclusive) units.

10 12

−4 −6

1.764 < x < 3 or 5 < x < 6.236.

17 x <

or −4 < x < 4 or x >

Production times between 1.764 and 3 hours, or between 5 and 6.236 hours, exceed the eﬃciency threshold.

y 8 6

13 x < −6 or −5 < x < 5 or x > 7

y 3

2 x

−6 −4 −2

1 x −6−4−2

2 4 6 8 10 12 14 16 18

−1

6.03E Solve absolute value inequalities What do you remember? 1 a ∣x∣ < 8 b ∣x∣ > 8

Extend your thinking

c ∣x − 8∣ ≥ 2

14 −10 < x < −4 or x > 4

d ∣x − (−2)∣ ≤ 5, or ∣x + 2∣ ≤ 5

2 a −2 < x < 2

c x < −13 or x > 13

4 2 −10−8 −6 −4 −2 −2

x 2 4 6 8 10

−4 −6

3 a ∣x∣ ≤ 7

b x ≤ −5 or x ≥ 5 d −k ≤ x ≤ k b ∣x∣ > 8

c ∣2x + 3∣ ≤ 0.6 4 a i Yes

ii No

iii No

iv Yes

b i Yes

ii No

iii Yes

iv No

c i No

ii Yes

iii Yes

iv No

d i No

ii Yes

iii No

iv Yes

15 6 < x ≤ 33. Production times greater than 6 hours and up to 33 hours (including 33) keep costs at or above the threshold.

Answers 1081 mathspace.co

5 a i Centre: c = 3, Distance: k = 5

−8 < x < 4 c

y = ∣3x + 5∣ 15

−10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

y = 11

b i Centre: c = −4, Distance: k = 2 −

16 , 11 3

(2, 11)

−10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

c i Centre: c = 1, Distance: k = 7

−8 −6 −4 −2

ii −10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

d i Centre: c = −6, Distance: k = 3

ii −10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

y = |x∣ − 6 14 y 12 10 8 6 4 2

(−15, 9)

Practice 6 a −6 < x < 6

b x < 3 or x > 8

c −7 < x < 1

d x < 2 or x > 6

f x < 1 or x > 7

g −3 < x < 3

h x < −4 or x > −1

7 a x ≤ 5 or x ≥ 9

b −5 ≤ x ≤ 13

c −4 ≤ x ≤ 1

d x ≤ −7 or x ≥ 7

e x ≤ 3 or x ≥ 11

f −10 ≤ x ≤ 2

g x ≤ 0 or x ≥ 4

h x ≤ −6 or x ≥ 6

8 a

y = ∣x − 5∣

14 12 10 8 6 4 2 −1 −2 −4 −6 −8

y = 11 (16, 11)

y = |9 − 2x∣ (1, 7)

x −20−15−10 −5

x 1 2 3 4 5 6 7 8 9 10 11

y = |8 + 6x∣

5 10 15 20

(−8, 6)

y 8

2 − ,4 6 3 y=4

x > 16, x < −6

y=6

y=7

(8, 7)

x < 1 or x > 8

y = ∣x + 2∣

x 5 10 15 20

x > 15 or x < −15

(7, 9)

−20−15−10 −5 −2 −4 −6 −8

(−6, 11) 10

y=9

(−2, 4)

10 −3

8 6

−1

(4, 6) x < −2 or x >

x 2 4 6 8

1082 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x 1

−2

−12−10−8 −6 −4 −2

−2

y = 0.5∣x − 3∣ 6

y=4 (−5, 4)

y = ∣x + 3∣

y=7 8

(11, 4)

(−10, 7)

x −8 −6 −4 −2

2 4 6 8 10

1 y= x+1 2

y 8 y=

6 4

y = 3 (−8, 3)

3x 2−4

6 4

(4, 3)

−6 −4 −2 −2

x −12−10−8 −6 −4 −2

2 4 6 8

y = −3

−4

x < −8 or x > 4

No solution e

y = −3 + ∣x∣ 4

y = ∣x − 6∣ 6 (2, 4) (10, 4) 4

y=4

(−4, 1)

(4, 1)

y=1 x

2 −2 −2

2 4 6 8

−10 ≤ x ≤ 4

−12−10−8 −6 −4 −2

−5 < x < 11

9 a

(4, 7)

−8 −6 −4 −2

x 2 4 6 8 10 12 14

2 4 6 8

−2

−4

x ≥ 4 or x ≤ −4

x ≤ 2 or x ≥ 10 b

y = ∣x − 8∣

10(−1, 9)

(17, 9) y=9

y = ∣x − 7∣ − 6 x

−2

2 4 6 8 10 12 14 16

−2

−4

2 −4−2

y 4

−6

(5, −4)

y = −4 (9, −4)

2 4 6 8 10 12 14 16 18

−1 ≤ x ≤ 17

x ≥ 9 or x ≤ 5

Answers 1083 mathspace.co

12 17.75 ≤ m ≤ 18.09

13 18.86 ≤ n ≤ 18.94

1 x+5 2 (−11, 3)

4 2

14 a h ≤ 41.775 or h ≥ 58.225 y=3 (1, 3) x

−12−10−8 −6 −4 −2

2 4 6

−11 ≤ x ≤ 1 h

42 and 58

∣t − 105∣ − 120 < −40

b 35 < t < 175 seconds c t = 180 is not a valid solution. At 180 seconds into the dive the diver will be accompanied by a safety diver. 16 Answers will vary.

y = 0.4∣2x − 3∣ 4

a Two possibilities are: • ∣x∣ < 2 • ∣x + 1∣ + ∣x − 1∣ < 4

y = 0.8 (0.5, 0.8) (2.5, 0.8) x −6 −4 −2

15 a

b Two possibilities are: • ∣x∣ ≥ 9 • ∣x∣ − 10 ≥ 8 − ∣x∣ c Two possibilities are: • ∣x − 4∣ ≤ 3 • 6 − ∣x − 7∣ ≥ ∣x − 1∣

x ≤ 0.5 or x ≥ 2.5 10 a No solution b All real numbers, as ∣x + 5∣ ≥ 0 is always greater than or equal to a negative number. c No solution d All real numbers, as ∣3x + 1∣ ≥ 0 is always greater than a negative number. e No solution f A ll real numbers, as ∣x + 3∣ ≥ 0 is always greater than or equal to a negative number. g No solution h All real numbers, as ∣2x + 9∣ ≥ 0 is always greater than a negative number. 11 a x < 1 or x > 7 −10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

b −7 ≤ x ≤ 3 −10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

c −10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

d Two possibilities are: • ∣x + 2∣ > 2 •

∣x + 2∣ + 3 < 2 ∣x + 2∣

17 a x = 6 or x = −2

b −2 ≤ x ≤ 6

c x < −2 or x > 6 18 a b Solving the inequality gives 775 000 ≤ x ≤ 825 000 centimetres, or 4.82 ≤ x ≤ 5.13 miles (using 1 mile = 160 934.4 cm, rounded to two decimal places). Since 12.9 miles is not within this range, Floyd’s measurement was not accurate, suggesting a possible error in the stated actual distance or scale. 19 a • The critical points are found where the expressions inside the absolute value bars equal zero. These are x = 1 and x = 2. These points divide the number line into three intervals to test: • x < 1 • 1 ≤ x < 2 • x ≥ 2

d −5 < x < 1 −10−9−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10

1084 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

• For x < 1: In this interval, both x − 1 and x − 2 are negative. Therefore, ∣x − 1∣ = −( x − 1) and ∣x − 2∣ = −( x − 2). ∣x − 1∣ + ∣x − 2∣ > 2

7 6 5

Write the inequality

−( x − 1) − ( x − 2) > 2 Substitute

expressions for the interval x < 1

2 −2 −1

negative signs

−2x + 3 > 2 Collect like terms −2x > −1 Subtract 3 from x < Divide both sides

by −2 then reverse the inequality sign

−1

is consistent with the

y = | 3x + 4 |

x 5

b x ≤ −3 or x ≥ 2 d x ≥ 5 or x ≤

12 10 8 6 2

negative sign

1 > 2 Collect like terms The statement 1 > 2 is false. Therefore, there are no solutions in the interval 1 ≤ x ≤ 2.

(2, 10)

y = x + 8 (−3, 5) 4

x − 1 − x + 2 > 2 Distribute the

−7−6−5−4−3−2 −1 −2

x 1 2 3 4

x ≤ −3, x ≥ 2 22 To solve this inequality, we can split it into two cases:

• For x ≥ 2: In this interval, both x − 1 and x − 2 are non-negative. Therefore, ∣x − 1∣ = x − 1 and ∣x − 2∣ = x − 2.

• When x ≥ , the inequality becomes

Write the inequality ∣x − 1∣ + ∣x − 2∣ > 2

( x − 1) + ( x − 2) > 2 Substitute expressions for the interval x ≥ 2

2x − 3 > 2 Collect like terms 2x > 5 Add 3 to both sides x > Divide both sides by 2

is consistent with the is

Combining the results from all cases, the final or x > .

e x≥

expressions for the interval 1 ≤ x < 2

solution is x <

c x=2

Write the inequality ∣x − 1∣ + ∣x − 2∣ > 2

condition for this case ( x ≥ 2). So, x > part of the final solution.

20 a x < 3 or x > 9

• For 1 ≤ x < 2: In this interval, x − 1 ≥ 0 and x − 2 < 0. Therefore, ∣x − 1∣ = x − 1 and ∣x − 2∣ = −( x − 2).

The solution x >

Extend your thinking

( x − 1) − ( x − 2) > 2 Substitute

y=2 5 ,2 2

x < , x >

both sides

condition for this case ( x < 1). So, x < part of the final solution.

1 ,2 2

−x + 1 − x + 2 > 2 Distribute the

The solution x <

y = ∣x − 1∣ + ∣x − 2∣

2x − 3 ≤ 1 − x, which gives the result x ≤ . This cannot be true, however, since we are in the case where x ≥ . • When x < , the inequality becomes −(2x − 3) ≤ 1 − x, which gives the result x ≥ 2. Again, this cannot be true since we are in the case where x < . So we can see that in both cases, there are no values of x which satisfy the inequality, and so the inequality has no solutions. 23 a ∣x − 56.33∣ ≤ 0.1x b No. Solving the inequality gives 51.21 ≤ x ≤ 62.59 (rounded to two decimal

Answers 1085 mathspace.co

places). The maximum speed is below the speed limit.

6 a 70 ≤ x2 ( x + 4) ≤ 120 b

24 This inequality shows that the value of x varies away from 84°F by at most 7 degrees. Here are two possible interpretations:

y 120 90

If x represents the temperature over the course of a day, then the inequality shows that the minimum temperature that day was 77°F and the maximum was 91°F.

60 30 x

If x represents the maximum daily temperature over the course of a month, then the inequality shows that the highest maximum temperature that month was 91°F, while the lowest maximum temperature was 77°F.

3.16 ≤ x ≤ 3.90

Chapter 6 review

4 3

2 C

d 7 < x ≤ 16

x 1 2 3 4 5 6 7 8 9

−4−3−2−1 −1

4 a Roots: x = −1, 5, 9

b −4 < x ≤ 8

c x < 3 or x > 8

7 a −6 < x < −2

1 B

3 B

−2 −3

x < −1 −1 < x < 5 5 < x < 9 x > 9

−4

x+1

−

x−5

−

x−9

−

9 Critical point method:

Critical points: x = −5, 5, −7

Product

−

y ≤ 1 for x ≥ 8 or x < 3

x < −7

−7 < x < −5

−5 < x < 5

x>5

x−5

−

x+5

−

x+7

−

Quotient

−

−1 < x < 5 or x > 9 b 40

(−1, 0)

−2−1 −10 −20

(5, 0) (9, 0) x

1 2 3 4 5 6 7 8 9 10

−7 < x < −5 or x > 5

−30 −40

5 k≤4

1086 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Graphical method:

12 a y 15 10 5

(5, 0)

(−5, 0) −9−8−7−6−5−4−3−2−1

1 2 3 4 5 6 7 8 9

−5

14 y 13 12 11 10 9 8 7 6 5 4 3 2 1 −8 −6 −4 −2−1 −2 −3 −4

−10 −15

x 2 4 6 8 10 12 14 16 18

Graph shows ∣x − 6∣ > 10 for x < −4 or x > 16.

−7 < x < −5 and x > 5

10 Critical point method:

Critical points: x = −6, 6, 8

x < −6 −6 < x < 6 6 < x < 8 x > 8

x−6

−

x+6

−

x−8

−

Quotient

−

3 2 1

x ≤ −6 or 6 ≤ x < 8

−11−10−9−8−7−6−5−4−3−2−1 −1

Graphical method:

x 1 2 3 4 5

y 25

Graph shows ∣x + 3∣ < 7 for −10 < x < 4.

13 a Solution interval: x < 1 or x > 9

10 5

(−6, 0)

−9−8−7−6−5−4−3−2−1 −5

−4 −2 0 2

(6, 0)

1 2 3 4 5 6 7 8 9 10 11

8 10 12

b Solution interval: −9 ≤ x ≤ 3 −10 −8 −6 −4 −2 0

−10

14 a s ≤ 50.4 or s ≥ 89.6

−15 −20

b Smallest integer above 70: 90

−25

Largest integer below 70: 50

x ≤ −6 or 6 ≤ x < 8 Interpretation: Process is stable for parameter x ≤ −6 or 6 ≤ x < 8. 11 a x ≤ 5 or x ≥ 11 c −6 ≤ x ≤ 1

b −2 ≤ x ≤ 12 d x ≤ −7 or x ≥ 7

Answers 1087 mathspace.co

14 y 13 12 11 10 9 8 7 6 5 4 3 2 1 −9 −8 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4

Practice 5 a 0.67

x 1 2 3 4

b −0.47

c −0.14

6 a True

b True

7 a False

b False

d 0.77

8 a i

iii

b i

iii

c i

iii

d i

iii

Graph shows ∣2x + 5∣ ≥ x + 7 for x ≤ −4 or x ≥ 2. 16 a ∣Ta − Tr∣ ≤ 0.04Ta b Range: 36.35°C ≤ Ta ≤ 39.38°C Since Ta can be up to 39.38°C, which is greater than 37.5°C, it is possible the person has a fever. 17

y 4 3

(x2, y2)

2 1 1

b (−0.91, 0.42)

10 a Positive

b Negative

c Positive

d Negative

11 a

12 a

13 a

14 a

15 a −1

b 0

(b, 0)

−1

9 a (0.82, 0.57)

−1

c Undeﬁned

From the graph, the inequality is only satisfied for x > 2. The solution is the interval 2 < x ≤ x2, where

. The graph

illustrates the solution for the specific case where b = 1, for which the solution is .

16 a (−0.87, −0.50)

b (−0.87, −0.50)

c (−0.87, 0.50)

d (−0.87, −0.50)

17 a

c −1

18 a 0

b 0

c 0

d −1

7.01 Unit circle

Extend your thinking

What do you remember?

19 a b

1 a Quadrant 4

b Quadrant 1

c Quadrant 2

d Quadrant 3

2 0°, 90°, 180°, 270°, 360° 3 a Quadrants 1 and 2

b Quadrants 1 and 4

4 a Quadrant 2

b Quadrant 3

b −a

20 Q(−a, b), R(−a, −b), S(a, −b) 21 a −2 sin x

b 2 cos x

22 (−0.407, 0.914) 23 a 0.26

c Quadrants 1 and 3

d 1

b −0.97

c −0.26

d −0.97

24 Ask your teacher for worked solutions.

1088 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

7.02 Related angles and identities What do you remember? 1 a θ

b 180° − θ

c θ − 180°

b −cos 30°, negative

18 a sin 45°, positive c −tan 60°, negative

d 360° − θ

c tan θ

d cos θ

3 a − sin θ

b cos θ

b −2

19 a

b sin θ

2 a All

sin θ

c − tan θ

cos θ

−120°

Practice

−225°

4 a 60°, cos 60°

b 45°, sin 45°

c 30°, tan 30°

d 30°, cos 30°

5 a

tan θ

6 a −tan 15° = −0.27

21 Related angle: 30°, cos 510° = −0.87 22

b −sin 60° = −0.87

c cos 30° = 0.87

Extend your thinking 23 Ask your teacher for worked solutions.

7 a

−1

24 Ask your teacher for worked solutions. c

25 θ = 30°, 150°, −210°, −330°

8 a Quadrant 2, positive

26 At 270°, coordinates are (0, −1).

b Quadrant 3, negative

Thus,

c Quadrant 4, negative

7.03 Sine and cosine rules

9 a 60°

b 60°

10 a sin 30°

b cos 90°

c tan 30°

11 Related angle: 45°, 12

sin θ

cos θ

, which is undeﬁned.

What do you remember? 1 a b c2 = a2 + b2 − 2ab cos C

tan θ

c 150°

13 a

14 a Positive angle: 60°, Related angle: 60° b Positive angle: 240°, Related angle: 60° 15 a −sin θ

b cos θ

c tan θ

d −sin θ

16 a Negative

b Negative

c Positive

2 a Cosine rule

b Sine rule

c Cosine rule

d Sine rule

3 b sin A < a < b 4 Practice 5 a = 10.88 6 57.34 cm2 7 C = 97.5°, 12.5° 8 c = 14.26

Answers 1089 mathspace.co

9 41.57 cm2 10 b = 15.73 11 A = 43.1° 12 Two solutions

9 a

10 a

b −1

c 1

d 0

f −1

13 q = 9.32 14 178.31 cm2 15 C = 90.0°

11 a

12 a

16 B = 38.7°, 141.3° 17 Two triangles. Since b sin A = 8 sin 40° ≈ 5.14 ≤ a = 6 < b = 8, the ambiguous case applies, yielding two solutions for ∠B.

Extend your thinking 13 sin (s + π ) = 14 sin ( θ ) =

Extend your thinking 18 Ask your teacher for worked solutions. 19 Ask your teacher for worked solutions. 20 303°

, cos (s − π ) = and tan ( θ ) =

15 The radius of the inscribed circle is half the side length of the square, so r = 4. The coordinates are given by (r cos θ , r sin θ ).

21 20.53 km 22 AC = 19.29

23 Ask your teacher for worked solutions. 24 x = 18.00 cm

16 cos ( θ ) =

and tan ( θ ) =

7.04 Radians

7.05 Arc length and sector area

What do you remember?

1 a True

b True

c False

2 a i

b ii

c iv

3 a

1 a l = rθ d iii

c P = r( θ + 2) 2 a 2π cm

b 8π cm2

3 a

Practice 4 θ =2 4 a

5 a 0.61

b 2.53

c 4.80

d 3.01

6 a 72°

b 210°

c −240°

d 225°

7 a 68.8°

b 177.6°

c 48.7°

d 309.4°

Practice

8 a

5 a 5π cm

c 6 a π cm

7 a 8.40 cm 8 a 22.50 cm

1090 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b 31.42 m 2

b 104.72 m2

9 a

y 1

10 a

11 a r = 14.32 cm

b r = 7.98 m

12 a θ = 1.50

b θ = 1.67

−0.5

13 a

b 25π cm2

−1

0.5 x

1π

1 π 2

3 π 2

14 209.44 m2 15 30.00 cm 5

16 24 + 6π cm 17 150.80 m2

18 θ = 0.75 radians Extend your thinking 19

tan θ

π −1

7 • Period: π • Domain: x ∈ , x ≠ integers

20 2227.06 cm2 21 186.63 m2

• y-intercept: (0, 0)

7.06 Graphs of trigonometric functions What do you remember?

• x-intercepts: x = kπ, where k are integers • Symmetry: Odd, tan (−x) = −tan x, with point symmetry about the origin 8

+ kπ, where k are

• Range: (−∞, ∞)

22 θ = 0.16 radians, P = 215.71 m

2π

3 2

sin θ 2

θ cos θ

0 1

−1

( −π , 0)

1 − π −1 2 −2

2π

(0, 0) (π , 0) x 1 π 2

−3 −4

3 • Period: 2π

Practice

• Amplitude: 1

9 a y = sin x

b y=1

10 a −1

b y=0

• Range: y ∈ [−1, 1]

11 a True

b False

• Symmetry: Odd, sin (−x) = − sin x, with point symmetry about the origin

12 x = π

• Midline: y = 0 • Domain: x ∈ 

c 1

c True

13 a Increasing

b Decreasing

14 a Increasing

b Decreasing

Answers 1091 mathspace.co

15 x = 0

30 For x > 2π, the angle x represents multiple revolutions around the unit circle. Since one full revolution is 2π, sin x repeats its values every 2π radians, as the y-coordinate of the point at angle x matches that at x − 2π n, where n is an integer. Thus, the graph continues its cyclical pattern.

17 18

y 1 0.5

31 On the unit circle, a point at angle θ has

−1π

3 − π 2

coordinates (cos θ , sin θ ). Since

1 − π 2 −0.5

as θ approaches

(where cos θ → 0),

tan θ grows without bound (positive or negative). Thus, tan x takes all real values, giving a range of (−∞, ∞).

−1

sin ( θ )

cos θ )

33 On the unit circle, at angle θ + π, coordinates are (−cos θ , −sin θ ).

−1

Thus, tan ( θ + π ) = periodicity of π.

−1

Chapter 7 review

= tan θ , proving

1 B 20

2 a i

iii

b i

iii

21 x = 0, π, 2π 22 a Periodic c π

b Amplitude d (−∞, ∞)

e 23 tan

= 1; tan (nπ ) = 0

3 a (0.77, 0.64)

b (−0.94, −0.34)

4 a

5 a d

b −c

6 a

7 a sin α

b −cos α

c − tan α

d cos α

c 144°

d 330°

24 A, C, D 25 Increasing 26 Negative

8 θ = ±120°, ±240°

Extend your thinking 27 The unit circle has radius 1, so any point at angle θ has coordinates (cos θ , sin θ ). Since sin θ is the y-coordinate, it ranges from −1 to 1, as the circle’s y-values are bounded by its radius. Thus, the range of y = sin x is [−1, 1]. 28 y = cos x is y = sin x translated

9 12.99 10 17.29 11

C ≈ 74.6° or C ≈ 105.4°

12 a

units left

1092 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

13 a

14 a

15 a 2π cm

b 10π m2

16 a

b 54π cm2

17 5π – 24 sin

7 a 1

b 1

c 1

8 a

b 2

9 a As θ → 90°, the opposite side approaches the hypotenuse, so cosec θ =

19 a −1 20 a π

b (−∞, ∞)

= 1.

→

b As θ → 90°, the adjacent side approaches 0,

so sec θ =

→

which is undeﬁned.

sin (1) m2

b y=0

10 a

b 0

d −1

11 a

c 1

d 1

12 a θ = 60° b θ = 45°

c θ = 45°

d θ = 30°

13 a 1

b 1

c 1

d cosec θ

14 a 2

21 22 On the unit circle, a point at angle θ has coordinates (cos θ , sin θ ). The radius is 1. The cos θ is the x-coordinate, which reaches a maximum of 1 at 0 and 2π radians, and a minimum of −1 at π radians. Thus, cos x ranges from −1 to 1.

15 If the reference angle is 0°, the opposite side length is 0, and the adjacent side is equal to the hypotenuse.

8.01 Secant, cosecant and cotangent

Extend your thinking

What do you remember?

16 cosec θ =

1 a cosec θ =

, cot θ =

17 Use the side ratios: sec θ =

b sec θ =

cos θ =

c cot θ =

sec θ × cos θ =

2 a i

iii 1

iv 2

v 2

vi 1

vii

viii 1

b i θ = 0° ii θ = 0° 3 a True

b True

iii θ = 90°

. Therefore, ×

18 The student used sin 30° = reciprocal.

instead of the

Correct value: cosec 30° =

= 2.

c False 19 sec 45° =

Practice 4 a

5 a

6 a

b 2

c 1

and

, cosec 45° =

, cot 45° = 1.

20 sec θ = , cosec θ =

Answers 1093 mathspace.co

8.02 Unit circle with secant, cosecant and cotangent What do you remember? 1 The unit circle is a circle with a radius of 1 unit, centred at the origin of a coordinate plane, speciﬁcally at the point (0, 0). The equation of the unit circle is given by x2 + y2 = 1. 2 a

3 a

b 0, π

c 0, π

4 a

b 2

6 a

d 2

e −1

f −2

g Undeﬁned

i −1

7 a

and cos θ = x, so

. instead of the

16 The student used reciprocal. Correct value: .

17 sec θ =

, cosec θ = 2, cot θ =

18 sec θ =

, cosec θ = 2, cot θ =

sin θ = y. Thus, cot θ =

5 a 2

20 Q =

21 a

,R=

. and S =

c −2

8.03 Reciprocal and quotient identities

b What do you remember?

c 1

d 1

e −1

f Undeﬁned

g −1

h 1

i 0

j Undeﬁned

8 a

c −1

9 a Undeﬁned

b Undeﬁned

c Undeﬁned

10 a −2

b 1

c 1

12 a

b 0

13 a 1

f 2

g 0

Extend your thinking 14 At θ =

sec θ =

19 On the unit circle, cot θ = , cos θ = x, and

Practice

11 a 1

15 On the unit circle, sec θ =

1 a The reciprocal of a number is 1 divided by that number. For a non-zero number a, the reciprocal is . Example: The reciprocal of 5 is . Similarly, the reciprocal of sin (x) is

b In mathematics, a value is considered undeﬁned when it does not have a meaningful or valid result. Example: The expression

is undeﬁned

because division by zero does not produce a meaningful result. c The quotient is the result obtained by dividing one number by another. Example: In the division 12 ÷ 4, the quotient is 3. 2 a

, the point on the unit circle is (0, 1), so

x = 0. Since sec θ = , it is undeﬁned when

3 a

x = 0.

1094 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

4 a 90° + 180°n

b 180°n

c 90° + 180°n

d 180°n

Practice 5 a

True c

False c

True

7 a

False

6 a

Answers 1095 mathspace.co

12 a

b d

8 a cos θ

b sin θ

9 a

10 a Undeﬁned

b Undeﬁned

c Undeﬁned

d Undeﬁned

11 a

Extend your thinking 13 At θ = 180°, the unit circle point is (−1, 0), so sin θ = 0. Since cosec θ = when sin θ = 0. b

1096 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

, it is undeﬁned

Thus, the identity holds. c 15 sec θ =

, cosec θ =

, cot θ =

Thus, the identity holds. d

17 sec θ =

, tan θ =

, cot θ =

18 The student is incorrect. Using cot θ =

tan 45° = 1, so cot 45° = = 1.

Thus, the identity holds.

8.04 Complementary angle identities

5 a

What do you remember? 1 a cos θ

b cot θ

2 a θ = 90° + 180°n c θ = 180°n 3 a True

c cosec θ d sec θ b θ = 180°n d θ = 90° + 180°n

b False

c True

Practice

d False

The angles where cos θ = 0 are θ = 90° + 180°n, where n is an integer (e.g., θ = 90°, 270°). b

4 a

Thus, the identity holds.

The angles where cos θ = 0 are θ = 90° + 180°n, where n is an integer (e.g., θ = 90°, 270°).

Answers 1097 mathspace.co

6 a 1

b 1

c 1

d 1

7 a

8 a sin θ

b tan θ

c cosec θ

2 a Quadrant 2, 60° b Quadrant 3, 45°

9 a

c Quadrant 4, 45° d On the negative x-axis, 0° 3 a True

b True

c False

4 a Negative

b Negative

c Negative

d Negative

d True

Practice

10 a

b 0

d 0

5 a

b 0

11 cot (90° − θ ) = tan θ = 3 6 a sin 60° =

Extend your thinking

c − tan 45° = −1

12 At θ , unit circle point is (cos θ , sin θ ). For 90° − θ , coordinates are (sin θ , cos θ ), so cos (90° − θ ) = sin θ . 13 At θ = 45°, the unit circle point for 90° − 45° = 45° is (cos 45°, sin 45°) = Thus, sin (90° − θ ) = sin 45° =

b − cos 30° = d cot 60° =

7 a

. 8 a cos2θ

b sin2θ

d tan2θ

c cot θ

For θ = 45°, the unit circle point is (cos 45°, sin 45°) =

, so cos θ = cos 45° =

Therefore, sin (90° − θ ) = cos θ holds at θ = 45°. 14 a sin θ = 15 cos 30° = sin 60° =

b cosec θ = , while

10 a

b cos θ

c cos θ

11 a Using complementary identity, sin (90° − θ ) = cos θ , which is true. b tan (90° − θ ) × sin θ = cot θ × sin θ =

sin 30° = cos 60° = . 16 Ask your teacher for worked solutions. 17 a cos θ =

9 a 0

b sec θ =

× sin θ = cos θ , which is true. c cos (90° − θ ) ÷ cos θ = sin θ ÷ cos θ = tan θ , which is true. d cot(90° − θ ) × cos θ = tan θ × cos θ = × cos θ = sin θ , which is true.

18 Ask your teacher for worked solutions.

8.05 Evaluate expressions with identities

12 a 1

b 1

d 1

13 a

What do you remember? 1 a cosec θ

b sec θ

1098 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking

14 For 120° in Quadrant 2, the reference angle is 180° − 120° = 60°. Since cos is negative in Quadrant 2, cos 120° = −cos 60° =

15 The statement is incorrect. sin (90° − 120°) = sin (−30°) = − sin 30° = while cos 120° = −cos 60° =

. The correct

identity is sin (90° − θ ) = cos θ , so sin (90° − 120°) = cos 120°. 16 The northward component in true bearings is cos 135° = − cos 45° =

17 cos (90° − θ ) = sin θ , and sin (180° − θ ) = sin θ . Thus, sin θ + sin θ = 2 sin θ , which does not equal sin θ . The statement is false. 18 a cos θ =

b cot θ =

19 a sin θ =

b cot θ = =

b sec2θ

5 a 1 ,

which does not equal cos θ . The statement is false.

c 1

d 2 + sec θ cosec θ

e tan2θ + sin2θ

f sec2θ − 2

6 a

8.06 Simplify and prove identities What do you remember? 1 1. cos2θ + sin2θ = 1 2. 1 + tan2θ = sec2θ 3. 1 + cot2θ = cosec2θ 2 a θ = 90° + 180°n

b θ = 180°n

3 a

Practice 4 a

c 2

7 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 8 a sec2θ = 5 c

b cosec2θ = , valid where cos θ ≠ 0

d cot θ cosec θ =

, valid where sin θ ≠ 0

9 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. c Ask your teacher for worked solutions.

Answers 1099 mathspace.co

10 a 1

b 1

11 a i Ask your teacher for worked solutions. ii Ask your teacher for worked solutions.

Practice

b i 12 Correctly,

+ 1 = tan2θ + 1 = sec2θ .

13 Ask your teacher for worked solutions.

4 a θ = 30°, 150°

b θ = 150°, 210°

c θ = 120°, 300° 5 a θ=

d θ = 0°, 180°, 360° b θ=

c θ = 0, π, 2π, 3π, 4π

Extend your thinking 14 a b For θ in Quadrant

, the

reference angle is 2π − θ or 360° − θ in degrees. Since sec2θ =

, and

cos θ = cos (2π − θ ) (cosine is positive in Quadrant 4 and equal to Quadrant 1 for the same reference angle), sec2θ = sec2(2π − θ ). Thus, if sin θ =

3 Trigonometric functions are periodic and have an infinite number of solutions. A restricted domain limits the solutions to a specific, finite set of angles within one or more cycles.

in Quadrant 1, the

corresponding cos θ yields the same sec2θ in Quadrant 4. 2

d θ = π, 3π

6 a θ = 60°, 120°

b θ = 120°, 240°

c θ = 135°, 315°

d θ = 30°, 150°

7 a θ = 30°, 150°, 270° b θ = 0°, 360°, 131.81°, 228.19° c θ = 45°, 225°, 63.43°, 243.43° b x=

8 a

9 θ ≈ 22.62° 10 a θ =

b θ =

15 cosec θ =

11 a θ = 143.13°

16 Ask your teacher for worked solutions.

12 a θ =

b θ = 157.38° ,

b θ = π, 3π, 5π

17 18 19 Ask your teacher for worked solutions. 20 21 Ask your teacher for worked solutions.

13 The student’s claim is incorrect. For sin θ = on [0°, 360°], solutions occur where sine is negative (Quadrants 3 and 4). Using the reference angle sin 30° = , we ﬁnd

θ = 180° + 30° = 210° (Quadrant 3) and θ = 360° − 30° = 330° (Quadrant 4). Thus, the solutions are θ = 210°, 330°.

8.07 Trigonometric equations

Extend your thinking

What do you remember?

14 For all θ , cos θ  ≤ 1, so cos θ = 1.5 is impossible.

1 a True

b True

c False

2 a First and second quadrants b Second and third quadrants c First and third quadrants

d False

15 The student correctly found the solution in Quadrant 2 but missed the solution in Quadrant 4. The reference angle is 30°. The solutions are θ = 180°−30° = 150° and θ = 360°−30° = 330°.

d Third and fourth quadrants

1100 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16 a θ = −180°, 0°, 90°, 180°

14 a

b θ = −180°, −60°, 60°, 180° 15 a cos x cos 2y + sin x sin 2y

17 θ = 0.644, 5.640

b sin 2x cos y + cos 2x sin y

18 Ask your teacher for worked solutions.

19 x = 90°, 270° 20 θ =

d ,

21 θ = 0°, 60°, 180°, 300°, 360°

16 a cos 30°

b cos 45°

c cos 45°

d cos 45°

8.08E Sum and difference expansions for trigonometric functions

What do you remember?

18 a

1 a sin A cos B + cos A sin B b cos A cos B + sin A sin B

c 2 a

3 a False

b True

4 a

19 a

20 a Ask your teacher for worked solutions.

5 a sin A cos B – cos A sin B b 6 a

b Ask your teacher for worked solutions. 21

22 23 a

Practice

7 a

Extend your thinking

8 a sin A

b cos A

24 Ask your teacher for worked solutions.

9 a Ask your teacher for worked solutions. b sin(A − B) =

25 Ask your teacher for worked solutions. 26 Ask your teacher for worked solutions.

10 a 1

b 0

11 a cos A =

b sin B =

27 a sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C

c Ask your teacher for worked solutions.

b cos A cos B cos C − sin A sin B cos C − sin A cos B sin C − cos A sin B sin C

d cos(A − B) =

12 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 13 a

28 29 Ask your teacher for worked solutions.

Answers 1101 mathspace.co

8.09E Double angle formulas

sin(8α )

What do you remember?

1 a sin(2 A) = 2 sin A cos A b • cos(2 A) = cos2 A − sin2 A

22 cos(2x)

• cos(2 A) = 1 − 2 sin2 A

23 sin(2 A) =

• cos(2 A) = 2 cos2 A − 1 c tan(2 A) =

Extend your thinking

2 sin2 A + cos2 A = 1

24 Ask your teacher for worked solutions.

3 cos(A + B) = cos A cos B − sin A sin B

25 Ask your teacher for worked solutions. 26 tan( α + 2β ) = 1

4 tan A =

27 Ask your teacher for worked solutions.

5 cos(2 A) = 1 − 2 sin2 A 6 a False

b False

c True

28 29 a Ask your teacher for worked solutions.

Practice

b Ask your teacher for worked solutions. c Ask your teacher for worked solutions.

7 a 8 a cos(2α ) = 1 − 2 sin2 α

8.10E Trigonometric equations

b cos(2α ) = 2 cos2 α − 1 What do you remember? 9 a

10 a

1 a

2 a x = 30°, 150°, 270° b x = 90°, 270°

Practice

12 cos 165° = 13 a

14 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. d Ask your teacher for worked solutions. e Ask your teacher for worked solutions. f Ask your teacher for worked solutions. 15 a 4 sin θ cos θ (2 cos2 θ − 1) 16 a

3 a

sin(8β )

4 a x = 90°, 210°, 330° b x = 45°, 225°, 63°, 243° c x = 0°, 18°, 180°, 198°, 360° d x = 45°, 135°, 222°, 225°, 315°, 318° 5 a x = −1.32, 0, 1.32 b x = −1.05, 1.05, 3.14, −3.14 c x = −2.43, 0.71 d x = −2.36, −1.11, 2.03, 0.79 6 a x = 67.50°, 157.50°, 247.50°, 337.50°

b x = 45.00°, 116.57°, 225.00°, 296.57°

18 tan 22.5° =

c x = 26.57°, 45.00°, 206.57°, 225.00°

19 2(2 cos2 θ − 1)2 − 1

d x = 0°, 180°, 360°

1102 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

7 x = 0.00, 1.57, 3.14

8.11E The auxiliary angle method

8 a

What do you remember? y

2 1 x

1 π 2

−1

1π

3 π 2

−2

1 a sin x cos

+ cos x sin

b sin x cos

− cos x sin

c sin x cos

+ cos x sin

2 a cos

cos x + sin

sin x

b cos

cos x − sin

sin x

c cos π cos x − sin π sin x b The points of intersection occur at x =

which matches the solution from Question 3(a). Extend your thinking . All solutions are

included because the equation is factored as cos x(2 sin2 x − 1) = 0, and all roots from cos x = 0 and 2 sin2 x − 1 = 0 within [0, 2π ) are identiﬁed using the unit circle. 10 x ≈ 0.62, 2.53

3 a 4 a R=

, cos α =

, sin α =

b R=

, cos α =

, sin α =

Practice 5 a

6 a

7 a

8 a

9 a

11 Expand using compound angles Substitute the exact values and factorise Simplify Divide both sides by cos x Evaluate Evaluate for tan x = in [0, 2π ]

10 13 sin( x + 112.62°) 11 a sin( x + π ) = sin( x) cos( π ) + cos( x) sin( π ) = sin( x) (−1) + cos( x) (0) = − sin( x) b sin

+ cos( x) sin

= sin( x) (0) + cos( x) (−1) = − cos( x)

12 a tan4 x − 6 tan2 x − 3 = 0 b x=

= sin( x) cos

12 a sin sin b sin sin

= sin( x) cos = sin( x)

+ cos( x)

= sin( x) cos = sin( x)

+ cos( x)

Answers 1103 mathspace.co

sin

13 a

. The graphs

perfectly overlap, confirming the two expressions are identical.

cos

relationship is αcos =

b Using a graphing tool, plot both y = 5 sin( x) − 12 cos( x) and y = 13 sin

and

. The general

− αsin, which comes from

the identity sin( θ ) = cos

. For this case,

the relationship is verified:

21 5 sin

and 5 cos

Since sin( x + αsin) = cos

14 a b

, we

have αcos =

− αsin. Verify: αsin = tan−1

αcos = tan−1

, and tan−1

− tan−1

, .

15 a

8.12E Apply the auxiliary angle method b What do you remember? b 300°

16 a

1 a • cos x +

sin x = 2 sin

• Amplitude: R = 2 • Phase shift: α =

17 a

b • 2 cos x − 2 sin x =

• Amplitude: R = • Phase shift: α =

Extend your thinking 18 Let α = tan−1

. This defines a right-angled

triangle where sin( α ) =

and cos( α ) = .

Expanding the right-hand side: 5 sin( x + α ) = 5 (sin x cos α + cos x sin α ) = = 3 sin x + 4 cos x. This matches the left-hand side. 19 R cos( x − α ) = R(cos x cos α + sin x sinα ) = (R sin α ) sin x + (R cos α ) cos x. Comparing with a sin x + b cos x, we get a = R sin α and b = R cos α. Squaring and adding gives a2 + b2 = R2 (sin2 α + cos2 α ) = R2, so R= cos α =

sin

2 a •

sin x = 2 cos

• Phase shift: α = b • 3 cos x –

sin x =

• Amplitude: R = • Phase shift: α = 3 • 4 sin x + 3 cos x = 5 sin • Maximum: 5 • Minimum: −5 4

. The formulas for α are and sin α =

cos x +

• Amplitude: R = 2

1104 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

cos x +

sin x

cos

Practice

13 • 25 cos

5 a

• x = 0.6435, 1.9305

b • Maximum:

14 g( x) = 2 sin

• Minimum: y

c 2

y 3

x 1π

1 π 2

−1

3 π 2

1 π 2

1π

3 π 2

1 π 2

1π

3 π 2

−2

−2 −3

15 x ≈ 62° 6 a 2 sin θ +

cos θ =

cos

sin

b θ = 0.34, 1.57

7 a

c x = 0.195

d x = 2.8017

8 x = 0.1288 9 f ( x) =

cos

−1

y 2 1

Extend your thinking x 1 π 2

−1

1π

3 π 2

17 x = 0.4636 18 The expression 5 cos x + 12 sin x can be written as 13 cos

−2

. Since the cosine

function ranges from −1 to 1, the maximum value of 13 cos 10 •

cos

Thus, it cannot equal 15, so no real solutions exist.

• x = 1.8969, 5.8136

19 a x = 0, π, 2π

11 x = 12 x =

+ arctan(2), x =

is 13.

b x ≈ 1.4009, 4.5425

+ arctan(2)

Answers 1105 mathspace.co

8.13E Applications of trigonometric equations

7 a b Amplitude: 10 cm, First time: 2 seconds

What do you remember? 8 a 1 a • 3 cos( x) + 4 sin( x) = 5 sin

• Amplitude: 5

• Phase shift: arctan

5 t

cos( x) + sin( x) = 2 sin

b •

• Amplitude: 2

−5

• Phase shift: 2 a Period: second

seconds, Frequency: 2 cycles per 9 a t = 4 hours (240 minutes)

3 a

cos( x) +

cos x +

cycles per

b t = 3 hours (180 minutes) 10 a 30 h 28 26 24 22 20 18 16 14 12 10 8 6 4 2

sin( x) sin x

Practice 4 a 15 + 10 cos b Minimum temperature: 5°C, First time: 39 minutes

0 4

T 20

8 12 16 20 24 28 32

b 7.00 m

c t = 34.82 seconds

d 6

e 3.88 m

Extend your thinking

11 a P(t) = 500 + 200 sin and α = .

5 t 0

0.3

0.5

5 a 24 times c

1π

−10

b Period: 4 seconds, Frequency: second

1 π 2

0.8

years

12 t =

years years

13 a f (t) = 3 sin

6 a b Maximum height: First time: 113 minutes

b Maximum population: 700 000, First time: t = 2 months. c t =

years

metres,

. So, R = 200

b The ship can enter during the time intervals [3.65, 8.35] and [15.65, 20.35], which correspond to 3:39 a.m. to 8:21 a.m. and 3:39 p.m. to 8:21 p.m.

1106 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

c The first time the tides are the same height is t ≈ 5.16 hours, which is 5:10 a.m. At this time, the height of the tide in the first harbour is f (5.16) ≈ 11.7 m, which is above the 10 m requirement. The tide in the second harbour is the same height, g(5.16) ≈ 11.7 m, also above 10 m. Therefore, the ship can enter either harbour at this time.

18 a

19 a 4

20 a

21 a 0

22 For 210° in Quadrant 3, the reference angle is 210° − 180° = 30°. Since sin is negative in

Chapter 8 review 1 B

Quadrant 3, sin 210° = −sin 30° =

2 C 3 C

23 a tan x

b cos2x

d cos2A

c −tan x

4 a

5 a

25 a

6 a

d −1

26 Ask your teacher for worked solutions.

7 a i

24 a 10

c b

27 a θ = 180°

b θ = 60°, 120°

c θ = 135°, 315°

d θ = 90°, 270°

b Ask your teacher for worked solutions. b x=

28 a x = b

8 a

9 a

b −2

d −1

10 a 2

d 2

11 B =

c x=

d x = 0, π, 2π

29 a θ = 0°, 120°, 240°, 360° b θ = 0°, 45°, 180°, 225°, 360° c θ = 60°, 120°, 240°, 300° d θ = 30°, 90°, 150°

30 a x = 0, π, 2π, 3π, 4π 12 a

13 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. d Ask your teacher for worked solutions. 14 Ask your teacher for worked solutions. 15 a

16 a cos θ × sec θ = 1

b π=

c x=

d x=

31 Ask your teacher for worked solutions. 32E C 33E C

b cot θ × tan θ = 1

d sin θ × cosec θ = 1

17 a

b 1

c 2

34E a

35E a

36E tan X

Answers 1107 mathspace.co

37E Express tan(3 A) as tan(2 A + A) and use the tangent addition formula.

45E a b Minimum height is 3.17 m at 10:30 a.m. c The ship can enter the harbour from 2:34 a.m. to 6:26 a.m. and from 2:34 p.m. to 6:26 p.m. 46E x =

47E The expression 3 sin x − 4 cos x can be written as 5 sin( x − α ) for some α. The maximum value of 5 sin( x − α ) is 5. Since 5.5 is greater than 5, there are no real solutions. 48E a 2 years b 30 000 c t ≈ 0.17 years and t ≈ 0.83 years d t ≈ 0.80 years

Thus, tan(3 A) =

49E a Maximum daylight hours = 14.5 hours. Approximately day 171. b Approximately 135 days.

38E Start with the left-hand side.

9.01E Parametric forms of equations What do you remember? 1 A function is a relation where each input has exactly one output. It can be represented as y = f ( x), where x is the input and y is the output, or as f ( x) to denote the function’s value at x.

= tan A + tan B.

Thus,

39E a sin A = , sin B = b cos(2 A) =

, cos B =

, sin(2 A) =

2 A point is represented as an ordered pair ( x, y). The equation y = 2x + 3 describes a straight line with a slope of 2 and a y-intercept of 3. 3 Solve x = 2t for t: t = . Substitute into y = t + 1: y= 4

5 4 3 2 1

40E x ≈ 2.09, 4.19 41E θ = 45°, 162°, 225°, 342° 42E 43E x ≈ 0.390, 4.211 44E a g( x) = 7 + 4 cos b Minimum value: 3

+ 1. Thus, the relationship is y =

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1 2 3 4 5

It represents a parabolic curve.

Smallest positive x =

1108 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

+ 1.

Practice

g x = t + 2, y = −2t2 + 3

5 a y = 2t + 3

b y=2−t

16 a x = 2 cos t, y = 2 sin t

c 6 a y = 4t2 − 1

b y = 2t2 − 3t + 1

c y = −9t2 + 24t − 13

d y = t2 − 2t + 2 b

7 a c x = ±4 cos t

d y = 3 ± 5 sin t

d x = −3 + 5t, y = −4 + 10t t, y = −3 +

t, y =

−

9 a x = t, y = 2t − 4

cos t, y =

sin t

d x = 1 + cos t, y = 2 + sin t e x = −3 + 9 cos t, y = −4 + 9 sin t f x = 3 + 8 cos t, y = −2 + 8 sin t

c x = 5 cos t + 5, y = 5 sin t − 4

c x = 7t, y = 1 + 13t

f x = −1 +

c x=

b x = 2 cos t − 1, y = 2 sin t + 1

b x = −2 + 5t, y = 5 − 6t

b x = 5 cos t, y = 5 sin t

17 a x = 3 cos t + 2, y = 3 sin t − 3

8 a x = 1 + 3t, y = 3 + 6t

e x=

h x = t − 1, y = 2t2 + 1

b x = t, y = 5 − 6t

d x=

cos t + , y =

sin t + 1

Extend your thinking Start with the ellipse equation

18 a

d x = t, y =

Substitute x = a cos t and y = b sin t

10 a x = t + 2, y = 3t − 1

b x = t − 1, y = −2t + 4

Evaluate the square

c x = t, y = 5t − 5

d x = t + 3, y = −t + 2

c x = t, y =

Simplify the fractions

11 a x = t + 1, y = 2t − 1 c x = t + 2, y =

b x = t − 1, y =

Use the trigonometric identity cos2 t + sin2 t = 1

d x = t, y = −4t + 8 Therefore, proved.

12 x = t + 3, y = t2 − 4

b x= 13 a

14 a

d 2

15 a x = t + 2, y = t + 4 b x=t− ,y=

−

c x = t − 1, y = d x = t + , y = 3t2 + e x = t + 2, y = −t2 + 1 f x = t + 1, y = 3t2 − 1

cos t, y =

sin t

19 a y = b tan t b x=

sec t, y = 4 tan t

20 Let x = x0 + At and y = y0 + Bt Assume general parametric form passing through ( x0, y0) at t = 0

At t = 1, x = x1 and y = y1

he line must pass T through ( x1, y1) at t = 1

x1 = x0 + A(1) Substitute t = 1, x = x1 into

A = x1 − x0 y1 = y0 + B(1)

B = y1 − y0 ∴ x = x0 + ( x1 − x0)t

the x-equation Solve for A

Substitute t = 1, y = y1 into the y-equation Solve for B Substitute A back into the x-equation

y = y0 + ( y1 − y0)t Substitute B back into the y-equation

Answers 1109 mathspace.co

21 x = tan t

b ( x − 1)2 + y2 = 1

9 a

22 a x = 2 + (a − 2)t, y = −6 b x = −3, y = 2 + (b − 2)t c x = 12 + (c − 12)t, y = 24 − 12t 23 The Cartesian equation is y = 2x2 − 3x + 4. Parametric equations: x = t, y = 2t2 − 3t + 4 24 Cartesian: y = x − 0.098x2. Parametric: x = 5s, y = 5s − 2.45s2

9.02E Convert from parametric form to Cartesian form

e x+y=3

f x2 + y2 = 1

10 a y = x2 − 2x + 2

b y = 2x2 + 8x + 9

c y = 2x2 − 10x + 8

d x = 2y2 − 12y + 21

f y = 3x2 + 5x + 4

e x = y − 11y + 25

11 a ( x + 2)2 + ( y − 3)2 = 25 b ( x − 1)2 + ( y + 2)2 = 16 c ( x + 3)2 + ( y − 2)2 = 36 d ( x − 2)2 + ( y + 1)2 = 9

What do you remember?

e ( x − 3)2 + ( y − 2)2 = 16

1 y = 2x + 1

f ( x − 1)2 + ( y − 3)2 = 4

2 x=y −y+1 Extend your thinking

3 y = 3x + 2

12 a

Practice 4 a y = 2x + 1

b 5x − 3y + 17 = 0

c 2x + y − 2 = 0

d 3x + 2y − 11 = 0

e x + 2y + 1 = 0

5 a x2 + y2 = 25

b x2 + y2 = 9

d x2 − y2 = 1

c x2 + y2 = 49

d x2 + y2 = 100

6 a

13 a y = x2 + 3

b d x2 − y2 = 4

b B 7 a ( x − 1)2 + ( y − 2)2 = 16 b The circle has centre (1, 2) and radius = 4. r= y

(x − 1)2 + ( y − 2)2 = 16 6

−2

(1, 2) 2

r=4

c y = x + 14 2

e y = x + 16

15 a Substitute x = A + B, y = A − B into the left-hand sides, use angle sum identities, and simplify to derive the right-hand sides.

9.03E Graph linear functions, quadratic functions and circles in parametric form

b ( x2 + y2) ( x2 + y2 − 2)2 = 4x2

−2

8 a y = x2 + 1

cos t =

b x4 + y4 + 2x2 y2 − 3x2 − 3y2 − 2x = 0

4 2

14 a Using x2 + y2 = (cos t + cos 2t)2 + (sin t + sin 2t)2, expand and simplify to get x2 + y2 = 2 + 2 cos t, then solve for

b y = x2 − 9 d y = x2 + 7 2

f y = x − 13

What do you remember? 1 a True

1110 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b True

c True

−1

−3

−2

−1

y 4 3

Practice 3 a Linear function c Circle 4 a i

b Quadratic function d Linear function

−1

d i

y 2

−1

x 0.5 1

−1

1.5 2 2.5 3 3.5

−1 −1

b i

5 a i (2, 0)

iii

ii (0, 4) y

y 3

4 3

−0.5

x 0.5 1 1.5 2 2.5 3 3.5

x 1

−1

b i (4, 0) iii

c i

ii (0, 4) y

2 1 x 1

Answers mathspace.co

1111

c i (−1, 0)

c i The vertex is (1, −1) (at t = 0).

ii (0, 2)

iii

ii For t = 2, the point is (−1, 1).

iii

y 1

−1

x −1

d i (3, 0) iii

1 −1

ii (0, 4.5) d i The vertex is (4, 0) (at t = 2).

ii For t = 0, the point is (0, 4).

iii

y 4

x 1

1 x 1

6 a i The vertex is (−2, 1) (at t = 0).

ii For t = 1, the point is (0, 2). iii

7 a i The vertex is (1, 3) (at t = 0).

ii The x-intercepts are 2

iii The y-intercept is (0, 2.875). iv

3 2

x −2

and

−1

b i The vertex is (−1, 3) (at t = −1).

ii For t = 0, the point is (0, 2). iii

−4 −3 −2 −1

1 2 3 4 5 6

b i The vertex is (1, −4) (at t = 1).

ii The x-intercepts are (−1, 0) and (3, 0). iii The y-intercept is (0, −3).

1 x −2

−1

1112 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

y x −1

b i The centre is (−1, 2) and the radius is 3. ii

y 5

−1

−2

−3

2 1

−4

−4 −3 −2 −1 −1

c i The vertex is (1, 1) (at t = 1). ii The x-intercepts are (−1, 0) (at t = 0) and (3, 0) (at t = 2). iii The y-intercept is (0, 0.75). iv

c i The centre is (0, −2) and the radius is 1. ii

−1

y −1

1.5 1

−2

0.5 x −1

−3

−0.5

d i The vertex is (1.5, −2.25) (at t = −0.5). ii The x-intercepts are (0, 0) (at t = 1) and (3, 0) (at t = −2).

d i T he centre is (1.5, −0.5) and the radius is 2.5. ii

y 2

iii The y-intercept is (0, 0). iv

y −1

x 1

−1

−2 −1

−3

−2

9 a i T he centre is (−2, 2), the radius is 3. The graph is the upper semicircle.

8 a i The centre is (3, 1) and the radius is 2.

ii Starting point (t = 0): (1, 2). Ending point (t = π ): (−5, 2).

iii

y 5

x 1 −1

5 −5 −4 −3 −2 −1

−1

x 1

Answers 1113 mathspace.co

b i T he centre is (1, −1), the radius is 1. The graph is the left semicircle. : (1, 0). Ending point

ii Starting point

10 a i The vertex is (−1, 2) (at t = 0). ii For t = 1, the point is (2, 1). iii

: (1, −2). iii

y x −0.5

0.5

1.5

2 2.5

−1

x −1

−2

b i The vertex (turning point) is (0, 1) (at t = 0). ii For t = 1, the point is (1, 2). For t = −1, (1, 0).

c i T he centre is (0, 1), the radius is 2. The graph traces an arc from (0, 3) to (0, −1) passing through (2, 1).

iii 2

ii Starting point (t = 0): (0, 3). Ending point (t = π ): (0, −1). iii

c i The vertex (turning point) is (1, 1) (at t = 0).

1 x 1

ii For t = 1, the point is (0, 3). For t = −1, (0, −1). iii

−1

y 3

d i T he centre is (−3, 0), the radius is 5. The graph is the right semicircle. ii Starting point point iii

: (−3, −5). Ending

2 1 x 1

: (−3, 5).

−1 5 4 3 2 1

−8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1 2

1114 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

d i The vertex is (−1, 0) (at t = −1).

ii For t = 0, the point is (0, −2). iii

d i The centre is (0.5, 0.5) and the radius is 1. y 1.5

y x −2

−1

0.5 x

−1

−0.5

11 a i The centre is (1, −3) and the radius is 4. 1

y x

−3 −2 −1 −1

1.5

2 3 4 5

Extend your thinking 12 x = −2 + 3t, y = 3 − 6t (other answers possible) 13 The intersection point is (1, 2).

y 5

−2 −3

−4 −5

−6

−7

(1, 2)

1 x

b i The centre is (0, −1) and the radius is 2. ii

−1

y 1

14 a (1, 1) and (4, −2) x

−2

−0.5

−2

0.5

−1

b 2

−1

−2

−2 −1 −1

−3

−2

(1, 1) x 1

(4, −2)

−3 −4

(approx. c i T he centre is (1.41, −1.73)) and the radius is (approx. 2.24). ii

y −1

x 1

−1 −2

−5

Chapter 9 review 1 A 2 B 3 C 4 a x = 2 + 3t, y = 4 + 6t

−3

b x = −1 + 5t, y = 6 − 8t

−4

c x = 6t, y = 2 + 14t 5 x = t − 3, y = t2 − 4

Answers 1115 mathspace.co

14 Intersection point: (3, 4)

6 a x = 4 cos t, y = 4 sin t

b x = 2 + 2 cos t, y = 3 + 2 sin t

c x = −1 + 7 cos t, y = −5 + 7 sin t

6 Line 2 : x = 2s + 3, 5 y = −s + 4

7 a

Intersection (3, 4)

4 3 2

1 Line 1 : x = t + 1, y = 3t − 2 x 1

Thus, x = a cos φ and y = b sin φ satisfy the ellipse equation. cos φ, y =

b x=

sin φ

8 a y = x2 − 4x + 7

10.01 Probability and events What do you remember?

b y = 3x2 + 6x + 5

c x = y − 2y + 5

1 a S = {1, 2, 3, 4, 5, 6} c A = {2, 4, 6}

9 a x2 + y2 = 36

b x2 + y2 = 4

c x2 + y2 = 25

2 a i

b ii

b S = 6 d c iii

d iv

10 y = x2 − 8

Practice

11 a

3 a S = {R1, R2, R3, R4, R5, B1, B2, B3} b Event A is drawing a red ball. The set is A = {R1, R2, R3, R4, R5}, so A = 5.

4 a

12 a (2, −1) b (6, 3)

(6, 3)

1 x 1 2 3 4 5 6 7

−1

c A ∪ B = {R1, R2, R3, B1, B2} d

Vertex (2, −1) 7 a A ∩ B = {5}

13 a x-intercept = (3, 0) y −1

x-intercept (3, 0) x 1

−6

9 a

c P (A ∪ B) = 1

−3

−5

8 a

−2

−4

c A ∪ B = {1, 3, 4, 5, 6} d

y-intercept = (0, −6) b

b P (A ∪ B) = 1

6 a S = {R1, R2, R3, B1, B2, G1}

y = t2 − 1, x = 2t + 2

−2 −1

5 a P (A ∩ B) = 0

y = 2t − 4, x = t + 1

10 a A ∩ B = {HH}

y-intercept (0, −6)

1116 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking 11

7 a C = {blue1, blue2}

8 a

, P (A ∩ B) = 0

6 S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5}, B = {1, 2, 3, 4}, A ∩ B = {2, 3},

12 a P (A ∩ B) = 0, single draw cannot be both red and blue. b 13 Student did not subtract P (A ∩ B). A = {5, 6}, B = {1, 3, 5}, A ∩ B = {5}. Correct:

9 a

10 a

11 a

12 a b

. 13 a No, because red aces exist, so A ∩ B ≠ ∅.

b , it means picking a number that is

14 a

both a multiple of 3 and even.

10.02 Mutually exclusive events 15 a What do you remember? 1 Two events that cannot have simultaneous outcomes in the same chance experiment.

b 16 a A = {HH, HT, TH}, B = {HH}

2 Non-overlapping circles within the sample space S. 3 It means events A and B are mutually exclusive, as they have no common outcomes. 4 P(A ∪ B) = P(A) + P(B) 5 It indicates that events A and B are mutually exclusive, meaning they cannot occur simultaneously. Practice 6 a A = {13 hearts} b B = {13 spades}, A ∩ B = {13 hearts} ∩ {13 spades} = ∅, so A and B are mutually exclusive. c

b 17 a

18 a b Extend your thinking 19 The sample space S has 6 outcomes. Events ‘win’ ({w1, w2, w3}), ‘lose’ ({l1, l2}), and ‘draw’ ({d}) are mutually exclusive. In a Venn diagram, they are non-overlapping regions within S. 20

, exclusive). Then

, P (A ∩ B) = 0 (mutually .

Answers 1117 mathspace.co

Practice

5 a

The complement rule applies as .

C 4 7

22 The claim is incorrect because A ∩ B (black queens) is not empty. In a Venn diagram, circles for A and B overlap. Using P (A ∪ B) = P (A) + P (B) − P (A ∩ B):

3 7

23 Addition rule: , P (red ∩ blue) = 0, so

1 a A multistage event involves multiple steps, each with its own set of outcomes. b Conditional probability is the probability of an event A occurring given that another event B has occurred, denoted P (AB). c It represents the probability of event A occurring given that event B has already occurred.

b True

b i

8 a 4 b {RR, RB, BR, BB} c d A tree diagram 9 a

1 2 1 2

H 1 2 1 2

3 a It arranges outcomes in rows and columns to show all possible combinations, making it easier to count favorable outcomes.

4 a iii

d A tree diagram is used for multistage events with multiple steps. This event has only one step, so a simple probability calculation is suﬃcient.

c False

b Each branch represents an independent or conditional probability, and multiplying them gives the probability of the combined outcome.

c 1

7 a

d A tree diagram lists all possible outcomes by showing branches for each step, with probabilities multiplied along branches to ﬁnd the probability of speciﬁc outcomes. 2 a False

M S

A ∪ B = 9, so

What do you remember?

R RR

red ∪ blue = {r1, r2, r3, r4, r5, b1, b2, b3, b4},

10.03 Multistage events and conditional probability

C RC

b 6 a

By outcomes:

R CR

4 6 2 6

C CC

1 2 1 2

c ii

1118 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

H HH

H TH

TT d

10 a

1 2 1 2

R 4 7 3 7

4 6 2 6

R RR

14 a 36 d

B RB R BR

B BB

11 a

15 a

B A

12 a

1 4

2 5 3 5

R 3 5 2 5

2 5 3 5

3 4

R RR

NH N RN R NR

16 a

13 a

2nd Dark

2nd Milk

1st Dark

1st Milk

NH HNH

1 4 3 4

NHH

NH NHNH

Coffee

Tea

N NN

1 4 3 4

20 20

b c d

T C

Answers 1119 mathspace.co

17 a

3 5 2 5

R 3 5 2 5

3 5 2 5

18 a

R RR

G RG R GR

2 3

1st Milk

6 E6

1 6 5 6

6′ E 6′

Probability of winning the prize and rolling a 6: . b c The outcome of picking a door does not affect the probability of rolling a die, as the die roll is a separate random process. 22 a b

Mathematics

Science

30 50

20 a

3 5 2 5

R 4 5 1 5

3 5 2 5

6′ P 6′

G GG

1st Dark

6 P6

1 6 5 6

2nd Milk

21 a

Extend your thinking

1 3

2nd Dark

19 a 36 d

R RR

M N RN R NR

From the table: P (MS) = P (SM) =

and

So, different restricted sample spaces cause the inequality. c New total would be 201, math has 121, while

N NN

both has 50. P (SM) =

1120 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

, which is less than

due to increased Math students.

23 a The student assumed replacement, using for both draws. Without replacement, the

second probability is . b R 2 5 3 5

B RB R BR

1 2 1 2

B BB

T .

10.04 Conditional probability formulas

What do you remember?

1 P (AB) =

, provided B ≠ 0

2 P (AB) =

24 a E 1 2 1 2

1 2 1 2

Probability of E and H: b

3 outcomes for E and H out of 12:

Probability: c

R RR

1 4 3 4

, provided P (B) ≠ 0

H EH

T ET H OH

T OT .

3 All marbles are coloured (red or blue), so event B includes all 6 marbles. Event A is drawing a red marble (4 outcomes), all within B. Since A is a subset of B (all red marbles are coloured marbles), the intersection of A and B is simply A itself. Therefore, A ∩ B = A = 4. 4 a

5 a 2

b 3

Practice 6 a b

S A

25 10

Answers 1121 mathspace.co

7 a

8 a

9 a

10 a

11 a

12 a

22 a b Yes, switching is beneﬁcial. The initial probability of the chosen box having the prize is . The probability that the prize is in one of the other three boxes is . The host’s action of revealing an empty box concentrates this probability onto the two remaining unopened boxes. Therefore, the probability of winning by switching to one of . Since

the other two boxes is

switching is the better strategy.

13 0.561

23 a Let A be rolling a 6 and B be rolling an even number. Then,

(one 6 among 2,

14 a

15 a

16 a

17 a

24 a 0.076

18 a

10.05 Independent events

19 a 0.0196

b 0.333

4, 6), but P (A) = . b The claim is false because P (AB) = P (A) only holds if A and B are independent, which is not always true.

What do you remember? 1 Two events are independent if the occurrence of one does not affect the probability of the other, that is, P (AB) = P (A) and P (BA) = P (B).

Extend your thinking 20 a 0.00495 b

2 P (A ∩ B) = P (A) × P (B)

B AB A

0.99

3 Events A and B are independent.

0.01

0.005

4 a True B′ AB′

P (AB) =

0.02

c False

, we have

= P (A).

Multiplying by P (B) gives P (A ∩ B) = P (A) × P (B).

A′ 0.98

Practice

B′ A′B′

6 a

0.199

b P (AB) =

21 a b The table shows A ∩ B = 25 (Year 12 and dogs) and B = 45 (Year 12). Dividing by the total sample space (80) gives P (A ∩ B) = and P (B) =

b False

5 For independent events, P (AB) = P (A). Since

B A′B

0.995

b 0.733

, used in P (AB) =

7 a 0.1

1122 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

= P (A), so independent.

b 0.55

8 a

d P (AB) =

9 a

= P (A), so independent.

d P (A ∩ B) =

Extend your thinking

20 Given P (AB) = P (A), then P (A ∩ B) = P (A) × P (B).

= P (A), so

Now, P (BA) =

= P (B).

, P (A) × P (B) =

, so

independent. 10 a 0.0015

b 0.9215

11 a 0.12

b 0.58

21 a 0.0396

b 39.6

22 a

23 P (A ∩ B) = , but P (A) × P (B) =

so not independent. 12 a

24 a 0.857375

b P (AB) =

= P (A), so independent.

b 85.7375

Chapter 10 review 1 C

13 a P (A ∩ B) =

, P (A) × P (B) =

, so

independent.

2 a 60

b 40

c 100

3 a {4, 6}

c P (AB) =

4 a

d P (BA) =

5 a

b P (A ∪ B) =

14 a 0.001

b 0.729

15 a

6 a No, because black queens exist, so A ∩ B ≠ ∅. b

b P (AB) =

= P (A), so independent. 7 a 0

16 a 0.12

b 0.68

17 a 0.42

b 0.46

18 a 0.0025

8 a

3 6

b 0.0975

19 a b P (AB) =

b 1

= P (A), so not

4 7

3 6

3 7

4 6

2 6

independent. b

B BB

Y BY B YB

Y YY d

Answers 1123 mathspace.co

9 a Roll 1

Roll 2

Events are independent since P (A ∩ B) = P (A) × P (B). 16 a

10 a b

Events are independent since P (A ∩ B) = P (A) × P (B).

Pool

Gym

b 110

17 a 0.0784

b 784

11.01E Factorial notation

What do you remember?

P′

110

1 The notation n! represents n factorial, the product of all positive integers from n down to 1.

G′

2 a True

P (GymPool) =

. Different sample

spaces cause inequality. c P (PoolGym) =

, less than original

11 a

12 a

b True

c True

d False

4 a 8

b 15

c 12

d 6

5 a 210

b 132

c 72

d 120

e 56

f 35

g 60

h 252

6 a 6

b 60

c 12

d 48

3 120

P (PoolGym) =

Practice

7 a (n − 2)!

b n(n − 1) d n(n − 1) (n − 2)

13 a

b 1

8 a

b Table shows Pop ∩ Under 30 = 30, Under 30 = 40. Thus, P (PopUnder 30) = 14 a 0.0196 15 a

b True

c True

d False

10 a (n − 2) (n − 3) (n − 4)  × 3 × 2 × 1

b (n + 1) n(n − 1)  × 3 × 2 × 1 b

9 a True

≈ 0.286 c

c (n − 1) (n − 2) (n − 3)  × 3 × 2 × 1 d (n + 2) (n + 1)n  × 3 × 2 × 1 11 a 30

b 60

c 35

d 168

12 a

b 360

c 5

d 12

13 a 26

b 40

c 126

d 12

1124 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b n

14 a 1

d n+1

c n

17 a 144

b 120

18 a 18

b 3 × 3 × 2 = 18

15 n = 4

19 a 12

b 3 × 2 × 2 = 12

16 n = 16

20 Addition principle

Extend your thinking

17 a n + 2

Extend your thinking

18 n = 6

21 320 ways

11.02E Multiplication and addition principles

22 49 ways

24 Answers may vary. Possible answer:

What do you remember? 1 a True

b False

c True

d False

2 a C

b A

c D

d B

3 The word “and” typically indicates the use of the multiplication principle for sequential choices. Practice 4 a

C, B

P, B

W, B

C, G

P, G

W, G

C, R

P, R

W, R

C, Y

P, Y

W, Y

b 12 5 1 413 720 passwords 6 10 7 60 8 27 9 a 27

b $75

To determine the total number of combinations of workshops and speciality activities available at the event, we ﬁrst need to consider the combinations of workshops. There are four workshops (A, B, C, and D), and participants must attend at least one. The total number of possible combinations of these workshops is calculated by considering all subsets of the set of four workshops, which is 24 = 16. However, since attending no workshops is not an option, we subtract the empty set, leaving us with 15 valid workshop combinations. For the speciality activities, there are three available options, and participants can choose to do one or none. The number of possible combinations for these activities is 3 + 1 = 4, which includes choosing one of the three activities or none. Combining the 15 valid workshop combinations with the 4 choices for speciality activities results in a total of 15 × 4 = 60 different combinations of workshops and speciality activities participants can choose from during the event. 25 • Direct routes: 2 roads directly from City A to City B. • Routes via Town C: 2 roads from City A to Town C and 3 roads from Town C to City B.

10 363 840 11 a 18 codes

23 54

b 12 codes

12 n = 4

• Routes via Town D: 3 roads from City A to Town D and 4 roads from Town D to City B. This conﬁguration provides exactly 20 possible routes from City A to City B.

13 33 14 16 15 19 16 a 12

b 8

Answers 1125 mathspace.co

11.03E Permutations

Extend your thinking 21 a 86 400

What do you remember?

b 3 542 400

c 86 400

1 A permutation is an arrangement of a group of items where the order of the items matters.

22 41 23 a 391

2 3 a False

b True

c True

d False

4 a iii

b ii

c i

d iv

b 257 , use the

24 To prove permutation formula. For

: Write the formula

Practice Rewrite n!

5 24

For n − 1 Pr + r × n − 1 Pr − 1.

6 6 7 a 120 b Multiplication principle or 5! = 120

Substitute formulas

8 a 5040 b 10! = 3 628 800

Common denominator

9 a n(n − 1) b n!

10 a 60

c 39

d 12!

b 1

Combine terms

11 210

Simplify expression

12 11 880 13 a 60

b 6

14 a 1680

b 12

15 a 210

b 6

16 a 200

Both sides equal n × identity.

, proving the

c 6

11.04E Applications of permutations

b 775

What do you remember?

17 a n = 8

b n=3

c n=5

1 45 360

18 a 1296

b 360

c 936

2 The total number of permutations is divided by the factorial of the number of times each repeated object occurs.

19 The permutation formula Pr is used for arrangements of items chosen without repetition. In this scenario, an answer is chosen for each of the 6 questions from the options {A, B, C}. Since the same answer can be used for multiple questions, this is a problem involving arrangements with repetition. Therefore, the standard permutation formula does not apply. The correct method is the multiplication principle. 20 a 120

b 240

3 a True

b True

c False

d True

Practice 4 12, the number of arrangements halves if C must come before D. 5 720 6 1260 7 a 720 8 a 12 600

1126 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b 48

c 12 b 60

d 6

9 a 83 160

b 630

7 2

10 36

8 120

11 a 90 720

b 2 948 400

9 720

12 a 2520

b 6

10 362 880

13 80 640

11 10 080

14 144

12 a 120

15 72

13 720

16 1440

14 a 5040

17 a 720

b 240

c 480

16 a 120

19 16! × 5!

17 362 880

20 120

18 4 b 12 841 920

c 72 072 000

b 1440

15 24

18 72

21 a 1 411 200

b 48

b 36

Extend your thinking 19 720

22 4320

20 a 40 320

Extend your thinking

21 96

23 120

22 72

24 48

23 21 600

25 3360

b 10 080

11.06E Combinations

26 40 320

What do you remember?

27 480 28 a 15

1 A combination is a selection of r items from n where order does not matter. A permutation is a selection where order matters.

11.05E Circular arrangements

What do you remember?

, where n is the total number of items, and r is the number of items selected.

1 (n − 1)!

2 In circular arrangements, rotations of the same pattern are considered identical because the relative positions of objects remain the same, regardless of the starting point. 3 a False

b True

c False

d False

Practice 4 2 5 5040 6 2520

Answers 1127 mathspace.co

11.07E Proofs involving combinations

Practice 5 a 6

b 286

c 100

d 18

e 56

f 495

g 105

h 15 504

What do you remember? 1

6 10 possible selections

8 21 different pizzas b 20

c 15

d 6

10 210

Cr =

represents the number of ways

to choose r objects from n distinct objects, where order does not matter because the term r! in the denominator divides out the r! possible arrangements of the r chosen objects, thus ensuring the selection is independent of order.

7 10 ways to choose the ﬂowers

9 a 15

11 56 ways to choose Practice

12 210 possible selections

3 a Ask your teacher for worked solutions.

13 252 selections

b Ask your teacher for worked solutions. b n+1

14 a

4 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 5 Ask your teacher for worked solutions.

6 Ask your teacher for worked solutions.

g n

7 a Ask your teacher for worked solutions.

b Ask your teacher for worked solutions. 15 Ask your teacher for worked solutions. 16 a y = 3 or y = 11

b y = 0 or y = 9

c y = 20

d y=7

e y = 4 or y = 6

f y=9

g y = 16

h y = 10

17 45 18 924 19 15 20 Ask your teacher for worked solutions. Extend your thinking 21 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 22 410 23 8 204 716 800 24 Ask your teacher for worked solutions.

8 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 9 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 10 Ask your teacher for worked solutions. 11 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 12 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 13 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 14 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 15 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 16 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 17 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions.

1128 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking

20 8 ways

18 Ask your teacher for worked solutions.

21 112 different committees

19 a Ask your teacher for worked solutions.

22 a 10 ways

b Ask your teacher for worked solutions. 20 Ask your teacher for worked solutions. 21 Ask your teacher for worked solutions.

c 1260 ways 23 1890 possible selections Extend your thinking

11.08E Applications of combinations

24 3566 ways

What do you remember?

25 540 ways

1 10 ways 2 21 ways represents the number of ways

to choose k items from n items without regard to order. Practice

b 126 ways

26 Both expressions give the number of ways to choose at least 11 men. The ﬁrst subtracts teams with fewer than 11 men from the total, while the second sums teams with 11 or more men directly. 27 441 quadrilaterals

11.09E Applications of combinations and permutations What do you remember?

4 900 ways 5 675 ways

1 a True

b False

c True

d True

6 210 ways

2 a No

b Yes

c Yes

d No

7 39 729 ways

3 a 12 C5

b 5 P5

c 24 C8

d 26 P3

8 43 316 ways 9 1260 ways 10 a 350 ways

b 546 ways

11 a 330 possible team selections b 126 possible team selections c 295 possible team selections 12 a 252 ways

b 126 ways

c 100 ways

d 126 ways

13 a 120 possible 3-ﬂavour milkshakes b 1023 possible milkshakes 14 a 150 ways

b 310 ways

c 65 ways 15 2 594 400 possible hands 16 210 ways 17 330 ways 18 38 550 ways 19 182 ways

4 The order of the numbers in the code matter, so it is actually a permutation calculation. If the order of the numbers did not matter, like 123 and 312, then we would use a combination calculation. Practice 5 a 126

b 3024

6 a 4200

b 36

c 151 200 7 91 8 79 833 600 9 6720 ways 10 a i Combination ii 20 C6 = 38 760 b i Counting principle ii 30 c i Counting principle ii 8! = 40 320

Answers 1129 mathspace.co

d i Permutation ii

P4 = 11 880

ii 12 f i Combination ii 7 C3 = 35 11 362 880 b 125 970

13 a 15

C2 = 6

Practice

e i Counting principle

12 a 3 108 105

c 495

6 a 6 P3 = 120

7 a 6 C3 = 20

8 a 5 P5 = 120

b 20

14 20 15 a 4

b 3

9 a 7 C3 = 35

c 12

16 a 53 130

b 6 375 600 c Choosing 5 people with no speciﬁc roles 17 9 × 9 × 8 × 7 = 4536 different codes 18 6006 19 211 926 20 a 70

11 a 100 12 a 13

b 52% b 10

13 a 10 P4 = 5040 b 576

c 40 320

c b

21 3024 22 180 ways

14 a 10 C4 = 210

Extend your thinking

23 2 097 152 24 240 25 1 352 000

16 17

26 13 700

11.10E Probability applications What do you remember? 1

19 20

3 Permutations are used when order matters, such as arranging 5 participants in a race ( 5 P5 = 120 ways). Combinations are used when order does not matter, such as selecting 5 for a team ( 5 C5 = 1 way). 4

P3 = 6

Extend your thinking 22 23

1130 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

24 25 26 7.48%

Chapter 11 review 1 B

Thus, 7 C3 = 35 is veriﬁed.

2 C

22 Ask your teacher for worked solutions.

3 C

23 1200

4 a 990

b 120

c 6720

d 362 880

e 2

f 48

g 2160

h 40 320

24 a 1260

b 2430

25 a 7280

b 4368

5 n=9

26 56

6 36

27 1680 b 3 × 2 × 2 = 12

7 a 12

29 192

8 30 9 a 336

28 2730

b 1

c 42

10 n = 9

d 720

30 a 720

11 3360 12 a 120

b 48

c 72

13 36

33 a 252

14 40 320 15 a 24 16 3600 17 210 18 66 19 16 20

32 b 105

12.01 Random variables b 12 What do you remember? 1 A variable whose value is the numerical outcome of a random experiment. 2 Discrete random variables have countable values, while continuous random variables can take any value within an interval. 3 Discrete: Number of students in a class. Continuous: Time to run a race. Practice

4 a i X is the number of heads obtained in two coin ﬂips. ii {0, 1, 2}

Thus, 9 C3 = 9 C6.

Answers 1131 mathspace.co

b i Y is the number shown on the die. ii {1, 2, 3, 4, 5, 6} c i Z is the face value of the drawn card. ii {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} d i W is the number on the section where the spinner lands. ii {1, 2, 3, 4}

14 The error is assuming time is countable. Time is continuous, as it can take any value (e.g., 23.456 minutes). A discrete example would be the number of questions answered.

12.02 Organise and graph datasets What do you remember?

5 a Discrete

b Continuous

c Discrete

d Continuous

e Discrete

f Continuous

g Discrete

h Continuous

6 a Discrete. Example: X = 15 customers in a supermarket b Continuous. Example: Y = 1.75 metres c Discrete. Example: X = 2 defective items d Continuous. Example: Y = 3.2 litres 7 a V is the number of tails obtained in three coin tosses. b {0, 1, 2, 3} 8 a T is the time taken to complete the task in minutes. T is continuous. b S is the shoe size of the selected student. S is discrete.

1 The number of times a particular value or group of values occurs in a dataset. 2 Relative frequency is the proportion of a value’s frequency to the total frequency, calculated as , where f is the frequency and n is the total number of data values. 3 Cumulative frequency is the running total of frequencies up to a value. Cumulative relative frequency is the running total of relative frequencies, summing to 1. 4 The mode is the value or class interval with the highest frequency, represented by the tallest bar. Practice 5 a

9 a The possible outcomes are red or blue.

b {0, 1} 0.2

10 a G is the genre selected, with action = 1, comedy = 2, drama = 3, sci-ﬁ = 4. b {1, 2, 3, 4}

c Discrete

0.2

0.267

0.333

0.2

0.467

0.8

0.933

11 a J is the volume of juice dispensed in millilitres. b Continuous

Extend your thinking 12 The number of students is countable (e.g., 20, 21, 22), making it discrete. The average height can take any value within an interval (e.g., 1.652 metres), making it continuous.

6 a

13 Possible values: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Discrete, as the values are countable.

1132 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

0.267 0.333 0.133 0.067

0.133 0.067

0.3 c

0.3

b Mode is 6, with the highest relative frequency of 0.333. c 16

0.1

0.3

Cumulative frequency

0.1

14 12 10 8 6 4 2 0

Quiz Score (out of 10)

7 6

10 a 8

Frequency

7 a

b 6.5 c

0.25

5 4 3 2 1 0

Race time (minutes)

0.167 0.083

b Mode is the interval 20–25, with the highest frequency of 8. 7

20.0

Cumulative frequency

Frequency

8 a

5 4 3 2 1 0

17.5 15.0 12.5 10.0 7.5 5.0 2.5 0.0

Number of hours b Mode is 2, with the highest frequency of 7. 0.7 0.6 0.5 0.4 0.3

15 20 25 Race time (minutes)

11 a Class Interval

9 a Relative frequency

10–20 20–30 30–40 40–50 50–60

Frequency

Cumulative Frequency

0.2 0.1 0.0

Quiz Score (out of 10)

Answers 1133 mathspace.co

14 a

Cumulative frequency

25 20

0.333

0.267

0.067

5 4 10

Daily water consumption c The median is approximately 31.7; the modal class is 30–40.

3 2 1

12 a

Number of hours student slept

Mode is 6 or 7 (frequency 5).

5 0

Number of hours

b 4.5 c 20 15

Cumulative frequency

Frequency

14 12 10 8 6 4 2 0

Number of hours student slept

5 0

Number of hours

13 a

15 A cumulative frequency polygon (ogive) uses lines to connect cumulative frequencies at class boundaries, making it easier to locate the median by visually identifying where the graph reaches half the total frequency, compared to estimating from the stepped bars of a histogram.

Cumulative frequency

25 20 15 10 5 0

Extend your thinking

100

Test scores b Mode is the interval 40–60, with the highest frequency of 8.

16 The error is including gaps, which implies non-consecutive or non-continuous data. For discrete or continuous data with consecutive values, bars should be adjacent with no gaps to reﬂect the distribution’s continuity.

c 53.75

1134 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Cumulative frequency

17 a 25

Practice 4 a P (X = 4) ≈ 0.25

5 a P (defective) = 0.075

b P (not defective) = 0.925

6 a P (defective) ≈ 0.075 b P (not defective) ≈ 0.925

c 60 widgets

15 20 25 Temperature

7 a P (online) ≈ 0.4

b P (X = 5) ≈ 0.18

8 a 0.18

18 A plateau indicates no data in those intervals, suggesting a gap in the dataset. The median is still found where cumulative frequency reaches half the total, but the gap may shift the median to a higher interval. 19 a

c P (odd) = 0.507

d Number = 108

9 a P (not green) ≈ 0.5 b P (green or amber) = 0.6 c 120 red lights 10 P (Not spade) ≈ 0.76 11 a P (heart) ≈ 0.233

Cumulative frequency

b P (not online) = 0.6

c 200 students

b Approximately 20°C

b P (not heart) ≈ 0.767

c 117 hearts

8 6

Group A Group B

Extend your thinking 12 P (heads) ≈

2 0

b P (even) = 0.43

Hours spent on extracurricular activities

b • Group A: The median is 2, corresponding to a cumulative frequency of 5.

= 0.46. This is close to 0.5,

suggesting no strong evidence of bias, though slight deviation may be due to random variation or sample size. 13 a P (heads) = 0.46

• Group B: The median is 3, corresponding to a cumulative frequency of 5.

b Theoretical P (heads) = 0.5. The estimate is slightly lower.

Group B’s median is higher, indicating more hours spent on activities.

c The difference is small and likely due to random variation, not strong evidence of bias.

12.03 Analyse data

d More ﬂips (600) would reduce variability, likely making the estimate closer to 0.5. 14 a 0.24

What do you remember? 1 The proportion of times an outcome occurs, calculated as

2 It approximates the probability as P (X = x) ≈

for a random

variable X. 3 a True

b True

c False

d False

b Theoretical P (X = 2) = estimate is higher.

≈ 0.167. The

c The difference is likely due to random variation, which has a signiﬁcant effect on small sample sizes. With only 50 rolls, the observed frequencies can easily deviate from the theoretical probability. A much larger number of trials would be expected to produce a relative frequency closer to the theoretical value.

Answers 1135 mathspace.co

12.04E Binomial expansions

13 a 10 000 + 200x + x2 b The term 10 000 represents the base proﬁt without sales increase, 200x shows the linear growth due to sales, and x2 accounts for quadratic growth from sales interactions. A company might use this model to capture both steady growth and the compounding effect of sales increases, reflecting optimistic projections.

What do you remember? 1 An algebraic expression with exactly two terms connected by addition or subtraction, such as x + y or 2x − 3. 2 x + y, coeﬃcients: 1, 1 3 Row 0: 1; Row 1: 1, 1; Row 2: 1, 2, 1; Row 3: 1, 3, 3, 1 4 a False

14 For n = 4, ( x + y)4 = x4 + 4x3 y + 6x2 y2 + 4xy3 + y4. Substituting x = 1, y = 1: Left-hand side: (1 + 1)4 = 24 = 16. Right-hand side: 1 + 4 + 6 + 4 + 1 = 16. The sum of coeﬃcients is 24.

b False

Practice 5 x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5

15 The coeﬃcient of x2 y3 is −15, which is incorrect. From the n = 5 row of Pascal’s triangle, the magnitude of the coefficient is 10. Since the term involves (−y)3 = −y3, its sign must be negative. The correct term is −10x2 y3. The correct expansion is x5 − 5x4 y + 10x3 y2 − 10x2 y3 + 5xy4 − y5.

6 a 1, 5, 10, 10, 5, 1 b 1, 6, 15, 20, 15, 6, 1 c 1, 7, 21, 35, 35, 21, 7, 1 d 1, 8, 28, 56, 70, 56, 28, 8, 1 7 a x3 + 6x2 + 12x + 8 b x4 + 12x3 + 54x2 + 108x + 81 c 27x3 + 54x2 + 36x + 8

16 a x4 + 4x2 + 6 +

d x4 + 8x3 + 24x2 + 32x + 16 e 8x3 − 12x2 + 6x − 1 6

b 64x6 − 576x3 + 2160 −

h 64x − 576x + 2160x − 4320x + 4860x − 2916x + 729 8 a 6x2 y2

b 4x3 y

c 4xy3

d y4

9 a 5x4 y

b 10x3 y2

c 10x2 y3

d 5xy4

10 (2x + 3y)3 = 8x3 + 36x2 y + 54xy2 + 27y3, the coeﬃcient of x2 y is 36. 4

−

g 32x5 + 80x4 + 80x3 + 40x2 + 10x + 1 5

f x − 12x + 60x − 160x + 240x − 192x + 64 6

17 −22 680

12.05E Coefficients in a binomial expansion What do you remember? 1 The numerical coeﬃcient of a term in a binomial

11 (2x − 1) = 16x − 32x + 24x − 8x + 1, the coeﬃcient of x3 is −32.

expansion, denoted

, representing the number of ways to choose r items from n.

Extend your thinking 12 The coefficients of ( x + y)3 = x3 + 3x2 y + 3xy2 + y3 are 1, 3, 3, 1, which are symmetric (same forwards and backwards). This occurs because the terms xn − r yr and xr yn − r have the same coefficient in Pascal’s triangle, reflecting the structure of the expansion.

and calculated as

=6 , where r is an integer from 0 to n.

4 a True

b True

Practice 5 a 240

b 160

c 60

d 12

6 a 15

b 20

c 15

d 1

1136 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

7 a 21

b 35

c 35

d 21

8 a 90

b −270

c 405

d −243

9 210

= 10,

= 6,

= 4, 6 + 4 = 10.

The identity holds. = 15, = 5, = 10, c = 5 + 10 = 15. The identity holds.

10 4320 11 60

12 n = 8

= 35,

= 20,

= 15,

20 + 15 = 35. The identity holds.

Extend your thinking

6 a

( for x3 y5) instead of

13 The student used 5 3

The value is correct, but the index was wrong. 14 The coeﬃcient is 6.

= 10. In ( x + y) = x + 5x y +

3 2

2 3

10x y + 10x y + 5xy4 + y5, the coeﬃcient of x3 y2 (where r = 2) is 10, matching

= 5,

= 5, 5 = 5.

16 n = 10, coeﬃcient 210

= 20,

= 20, 20 = 20.

= 21, 21 = 21.

= 70, 70 = 70.

The identity holds. c

The identity holds. = 56.

( for x y ). Correct coeﬃcient:

= 21,

The identity holds. d

= 70,

The identity holds. =

7 a Recursive:

= 3 + 3 = 6.

17 Formula:

= 6. Matches.

12.06E Properties of Pascal’s triangle =

b Recursive:

= 4 + 6 = 10.

What do you remember? 1 A triangular array where each entry is the sum of the two entries directly above it, starting with row 0 as 1. Each row n contains the binomial coeﬃcients

for r = 0 to n.

Formula:

= =

c Recursive: Formula:

for 0 ≤ r ≤ n

4 a True

Formula:

b False

Formula: 5 a

= 6,

The identity holds.

= 3,

= 15 + 20 = 35.

= 35. Matches.

8 a Symmetry:

Practice

= 10 + 10 = 20. = 20. Matches.

d Recursive: 3

for 1 ≤ r ≤ n − 1

= 10. Matches.

= 5.

= 5. Matches.

= 3, 3 + 3 = 6. =

b Symmetry: Formula:

= 15. = 15. Matches.

Answers 1137 mathspace.co

c Symmetry:

Formula:

= 21. Matches. =

d Symmetry:

Formula:

14 The identity holds because each row of Pascal’s triangle is palindromic. For row n, the coeﬃcient

= 21.

= 28.

at position

equals the coeﬃcient at

position n − r

, as the triangle is

symmetric about its centre. This reﬂects the equality of choosing r or n − r items from n.

= 28. Matches.

15 Ask your teacher for worked solutions. 9 a Row 3: 1, 3, 3, 1; Row 4: 1, 4, 6, 4, 1. = 3,

= 6,

16 a Ask your teacher for worked solutions.

= 3, 3 + 3 = 6. The identity holds.

b Ask your teacher for worked solutions. c For (1 + x)5,

b Row 4: 1, 4, 6, 4, 1; Row 5: 1, 5, 10, 10, 5, 1. = 10,

= 6,

= 10,

= 5, 10 + 5 = 15.

d Row 6: 1, 6, 15, 20, 15, 6, 1; Row 7: 1, 7, 21, 35, = 21,

= 6,

= 15,

represents the

represents the sum of the

odd power coefficients x, x3, x5. Hence, the result of part (b) means the sum of the even power coefficients equals the sum of the odd power coefficients.

The identity holds.

35, 21, 7, 1.

and

c Row 5: 1, 5, 10, 10, 5, 1; Row 6: 1, 6, 15, 20, 15, = 15,

sum of the even power coeﬃcients x0, x2, x4

= 4, 6 + 4 = 10.

The identity holds.

6, 1.

12.07E The binomial theorem What do you remember?

6 + 15 = 21. The identity holds. , where

1 = 15,

10 Row 6: 1, 6, 15, 20, 15, 6, 1. 15 = 15. The identity holds. 11 First:

= 15,

2 It determines the coeﬃcient of the term xn − ryr, representing the number of ways to choose r factors of y from n terms.

= 6 + 4 = 10.

Second:

Formula:

= 20. Matches.

= 35,

= 10 + 10 = 20.

x3 y +

x2y2 = x4 + 4x3y + 6x2y2

4 a False

= 35, 35 = 35.

b False

Practice =1

5 a

Extend your thinking

= 10 d

= 10

6 a4 + 4a3b + 6a2b2 + 4ab3 + b4

13 The students reasoning is circular. Instead the student should calculate +

x4 +

The identity holds.

Then show

is the binomial coeﬃcient.

= 20.

= 10 + 10 = 20. Then ﬁnally

conclude that the identity holds because both sides independently are equal to 20.

7 a • x3 + 3x2y + 3xy2 + y3, coefficients: 1, 3, 3, 1 • x3 − 3x2y + 3xy2 − y3, coefficients: 1, −3, 3, −1 b The coefficients of ( x − y)3 alternate signs due to (−y)r, but their absolute values match ( x + y)3. c ( x + y)3:

= 3; ( x − y)3:

Absolute values match.

1138 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

× (−1)2 = 3.

8 a (1 + 1)9 = 29 = 512, +

+…+

= 1 + 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 = 512

The answer matches when using direct substitution or the binomial theorem. 6

+…+

12 The expansion is: 64x6 + 192x5y + 240x4y2 + 160x3y3 + 60x2y4 + 12xy5 + y6. The coefficient of x4y2 is 240. Extend your thinking 13 Correct expansion: x3 + 3x2y + 3xy2 + y3

b (1 + 1) = 2 = 64, = 1 + 6 + 15 + 20 + 15 + 6 +1

= 3,

= 1,

not 1, 2, 2, 1. The student miscalculated

and

14 a 973 = (100 − 3)3

The answer matches when using direct substitution or the binomial theorem.

= 1003 − 3(1002) (3) + 3(100) (32) − 33 = 912 673

c (1 − 1)4 = 04 = 0,

b 0.984 = (1 − 0.02)4 = 14 − 4(1)3(0.02) + 6(1)2(0.02)2 − 4(1) (0.02)3 + (0.02)4

× 13 ×

= 0.922 368 16

× (−1)4 = 1 − 4 + 6 − 4 + 1

(−1) + … +

= 1,

= 64

× 14 +

= 3,

Coeﬃcients are

15 a

=0 The answer matches when using direct substitution or the binomial theorem. d (1 + 2 × 1)3 = 33 = 27, × 13 +

× 12 ×

× 1 × 22 +

2 +

× 23 = 1 + 6 + 12 + 8

The term independent of x is 24.

= 27

The answer matches when using direct substitution or the binomial theorem. 9 a 1 + 3x + 3x2 + x3 b 1 + 4x + 6x2 + 4x3 + x4 c 1 + 5x + 10x2 + 10x3 + 5x4 + x5 d 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6 5

10 x + 10x + 40x + 80x + 80x + 32 11

× 14 +

× 12 × 0.12 +

To find the term independent of x, multiply the terms from each bracket whose powers of x cancel out:

× 13 × 0.1 +

0.13 +

×1× × 0.14 = 1 + 0.4 + 0.06 + 0.004 + 0.0001

The term independent of x is 6.

= 1.4641

16 Ask your teacher for worked solutions.

Answers 1139 mathspace.co

12.08E Apply the binomial theorem

c ( 1 − 1)3 = 03 = 0, and the sum of the coefficients is 1 − 3 + 3 − 1 = 0. The results match.

What do you remember?

d ( 1 + 3 × 1)4 = 44 = 256, and the sum of the coefficients is 1 + 12 + 54 + 108 + 81 = 256. The results match.

1 The binomial theorem provides a formula to expand expressions of the form ( x + y)n into a , allowing for

sum of terms,

10 a x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64

simpliﬁed polynomial results. 2 ( x + y)n = . 3

, where

x3 +

x2 y +

b 256x8 − 1024x7 y + 1792x6 y2 − 1792x5 y3 + 1120x4 y4 − 448x3 y5 + 112x2 y6 − 16xy7 + y8

xy2 +

c 19 683x9 + 118 098x8 y + 314 928x7 y2 + 489 888x6 y3 + 489 888x5 y4 + 326 592x4 y5 + 145 152x3 y6 + 41 472x2 y7 + 6912xy8 + 512y9

y3 =

d x7 − 14x6 y + 84x5 y2 − 280x4 y3 + 560x3 y4 − 672x2 y5 + 448xy6 − 128y7

x + 3x y + 3xy + y 4 a True

b False. The expansion of ( x + y)n has n + 1 terms. Practice

11 ( x + 1)4 = x4 + 4x3 + 6x2 + 4x + 1, ( x − 1)4 = x4 − 4x3 + 6x2 − 4x + 1. Sum: 2x4 + 12x2 + 2 Extend your thinking 12 (1 + 1)5 = 25 = 32,

5 a x4 + 4x3 + 6x2 + 4x + 1

d z + 12z + 48z + 64

2 2

6 a x + 4x y + 6x y + 4xy + y 3 2

2 3

d u6 + 6u5 v + 15u4 v2 + 20u3 v3 + 15u2 v4 + 6uv5 + v6 7 a x4 − 4x3 y + 6x2 y2 − 4xy3 + y4 b x5 − 10x4 + 40x3 − 80x2 + 80x − 32 c y3 − 3y2 + 3y − 1 d z4 − 12z3 + 54z2 − 108z + 81 2

b 81x4 + 216x3 y + 216x2 y2 + 96xy3 + 16y4 c x5 + 10x4 y + 40x3 y2 + 80x2 y3 + 80xy4 + 32y5 2

14 a n = 9

b n=7 3 3

15 n = 6. Term: 160a b 16 n = 9

12.09E Specific terms in a binomial expression What do you remember?

8 a 8x + 12x y + 6xy + y

= 4.

Correct: x + 4x y + 6x2 y2 + 4xy3 + y4

c p3 + 3p2 q + 3pq2 + q3

= 4 and 4

b a + 5a b + 10a b + 10a b + 5ab + b

13 Error: Coeﬃcients for x3 y and xy3 are 3, should

c y5 + 15y4 + 90y3 + 270y2 + 405y + 243 2

+…+

1 + 5 + 10 + 10 + 5 + 1 = 32. The sum is 25.

b x3 + 6x2 + 12x + 8 3

d 8x + 36x y + 54xy + 27y

9 a ( 1 + 1)4 = 24 = 16, and the sum of the coefficients is 1 + 4 + 6 + 4 + 1 = 16. The results match. b ( 1 + 1)5 = 25 = 32, and the sum of the coefficients is 1 + 5 + 10 + 10 + 5 + 1 = 32. The results match.

1 2 a 2 b3 3 a True

b True

c True

4 1, 4, 6, 4, 1 Practice 5 a 6 240

1140 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b 10

d True

7 54

4 a

= 6 + 4 = 10

8 70 9 a 80

b 216

c 20

d 24

b −20

c 720

d −280

10 324

Practice 5

11 a 60 12 r = 3

6 a

13 15p4 (1 − p)2 14 −56

−

b 2

Extend your thinking

15 960 16 a 11 520

8 a , which is positive, and

b The term involves

the numerical coeﬃcient 28 × 12 is positive, with no negative signs. c It remains 11 520, as (−1)2 = 1 does not change the sign. d The coeﬃcient of the (r + 1)th term gains a factor of (−1)r when the second term’s sign changes.

10 n = 8 11 35 12 a

17 a = 1, b = 2 or a = 2, b = 1 18 The student made calculation errors when determining the binomial coefficient and the power of 2. The correct coefficient is .

= 10,

13 Extend your thinking 14 a Ask your teacher for worked solutions.

19 n = 5, r = 2

b Set x = y = 1 in ( x + y)n, so

20 The term independent of x is (−2)4(3)2 +

= (1 + 1)n = 2n.

(−2)1(3)4 = 11 988 c 32

12.10E Simplify expressions with binomial coefficients What do you remember? 1 a True

b False

c True

d In probability, this identity gives the total size of the sample space for a binomial experiment with n trials. This is also the total number of subsets that can be selected from a set of n items.

d True 15 The student likely computed

as 7 instead of

21, possibly confusing it with a simpler operation. Correct answer is 42.

16 Ask your teacher for worked solutions. 17 The squad must have 6 players.

Answers 1141 mathspace.co

18 a Ask your teacher for worked solutions.

10 a Ask your teacher for worked solutions.

b Ask your teacher for worked solutions. c Ask your teacher for worked solutions. d Ask your teacher for worked solutions.

12.11E Proofs with binomial expansions

= 10,

= 10

b 1 − 2 = −1 = (−1)1 c 1 − 4 + 4 = 1 = (−1)2

1 ( x + y)n = =

= 4,

11 a Ask your teacher for worked solutions.

What do you remember?

12 a Ask your teacher for worked solutions. , the number of ways to choose

r items from n. 3 Substitution, comparing coefficients and combinatorial arguments

= 6,

×3=6

= 5,

×1=5

13 a Ask your teacher for worked solutions. b 1 + 3 = 4 = 41 c 1 + 6 + 9 = 16 = 42

14 a Ask your teacher for worked solutions. b 0 × 1 + 1 × 2 + 2 × 1 = 4 = 2 × 21

Practice 2

b 1+3+3+1=8=2

5 a 1+2+1=4=2

6 a Ask your teacher for worked solutions. b

(−1)2 +

(−1)1 +

(−1)0 = 1 − 2 + 1 = 0

(−1)4 +

(−1)3 +

(−1)2 +

(−1)1 +

(−1)0 = 1 − 4 + 6 − 4 + 1 = 0 7 a Ask your teacher for worked solutions. b

= 10,

× 5 = 10

= 20,

× 15 = 20

(−1)0 +

(−1)1 +

(−1)2 +

(−1)3 =

(−1)0 +

(−1)1 +

b 2 + 1 = 3 = 31 c 4 + 4 + 1 = 9 = 32 Extend your thinking 16 Count the ways to assign n distinct items to 4 distinct groups. The right side, 4n, counts this directly as each item has 4 choices. The left side counts the same task by first choosing to be assigned to 3 of the

groups (3r ways), while the remaining n − r items are placed in the fourth group. Summing over all possible values of r gives the total number of assignments. 17 a Ask your teacher for worked solutions.

1−3+3−1=0 c

15 a Ask your teacher for worked solutions.

r items

8 a Ask your teacher for worked solutions. b

c 0 × 1 + 1 × 3 + 2 × 3 + 3 × 1 = 12 = 3 × 22

(−1)2 +

(−1)3 +

(−1)4 = 1 − 4 + 6 − 4 + 1 = 0 9 a Ask your teacher for worked solutions.

b LHS is RHS is

c LHS is

b 1 + 2 = 3 = 31 c 1 + 4 + 4 = 9 = 32

1142 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

. RHS is

, counts the ways to choose a

d The RHS,

committee of n people from a group of 2n. The LHS counts the same task by splitting the 2n people into two groups of n. The committee is formed by choosing r from the first group and n − r from the second, summed over all r. This gives .

12.12E Prove further identities with binomial coefficients What do you remember? for 0 ≤ r ≤ n. It means choosing r

items is equivalent to choosing n − r items to exclude. for 1 ≤ r ≤ n − 1

18 a Ask your teacher for worked solutions. =6

b 1 × 1 + 2 × 2 + 1 × 1 = 6, c 1 × 3 + 1 × 1 = 4,

4 The binomial theorem expands

19 a Ask your teacher for worked solutions. b ( 0 − 2) × 1 + (2 − 2) × 2 + (4 − 2) × 1 = −2 + 0 + 2 = 0 c ( 0 − 3) × 1 + (2 − 3) × 3 + (4 − 3) × 3 + (6 − 3) × 1 = −3 − 3 + 3 + 3 = 0 20 a Ask your teacher for worked solutions. +

=1+1=2=2

for x or multiplying expansions, we can equate coeﬃcients to prove identities. Practice 5 a Ask your teacher for worked solutions.

6 a

= 1 + 6 + 1 = 8 = 23

21 a Ask your teacher for worked solutions. b LHS:

. By setting speciﬁc values

b Ask your teacher for worked solutions.

(1 + x)n =

7 Ask your teacher for worked solutions. 8 Ask your teacher for worked solutions. 9 a Ask your teacher for worked solutions.

RHS:

The identity holds. c LHS:

RHS:

The identity holds. 22 a Ask your teacher for worked solutions. b 1 × 2 + 2 × 1 = 4 = 2 × 21 c 1 × 3 + 2 × 3 + 3 × 1 = 12 = 3 × 22

10 a 3

= 5 × 6 = 30

b 4

= 6 × 10 = 60

11 Ask your teacher for worked solutions. 12 Ask your teacher for worked solutions. 13 Ask your teacher for worked solutions. 14 a 5

= 7 × 15 = 105

b 6

= 8 × 21 = 168

15 Ask your teacher for worked solutions. 16 Ask your teacher for worked solutions.

Answers 1143 mathspace.co

Extend your thinking

b 9.5

17 Ask your teacher for worked solutions.

18 Ask your teacher for worked solutions.

19 a Ask your teacher for worked solutions. b Ask your teacher for worked solutions. 20 Ask your teacher for worked solutions. 9 a

21 Ask your teacher for worked solutions.

Chapter 12 review Frequency

1 C 2 B

3 2 1

3 B

4 a H is the number of heads obtained in four coin tosses.

Number of hours

b 5, 7, 8

b {0, 1, 2, 3, 4}

b {0, 1} 6 The number of books is countable (e.g., 10, 11, 12) and cannot be a fraction, making it discrete. The total weight can take any value within a range (e.g., 5.75 kg), including decimals, making it continuous. x

c Cumulative frequency

5 a The random variable C represents the outcome of the draw, encoded numerically.

7 a

20 15 10 5 0

Number of hours

d 7 10 a 10

0.35

0.25

0.1

0.05

Frequency

0.25

6 4 2 0

0.25

0.35

0.25

0.1

0.05

b 20–30

0.25

0.6

0.85

0.95

Commute times (minutes)

8 a

Cumulative frequency

c 30 25 20 15 10 5 0

Commute times (minutes)

1144 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Cumulative frequency

11 a 50 40

16 14 12 10

Class B

Class A

6 4 2 0

30 20

c The median is the 25th value. On the polygon, at cumulative frequency 25, the height is approximately 26 cm. x

14 a 0.14

b 0.385

15 a 0.04

b 0.96

16 a 0.04

b 0.96

c 60 faulty laptops 17 a 0.22

b 0.78

c 176 clubs 18 P (tails) = 0.55. The theoretical probability is 0.5. The 5% deviation suggests possible bias, but random variation in 300 ﬂips may account for it. More trials are needed for a stronger conclusion.

b 5 4

19 a 0.4

b The theoretical probability of rolling a 5 or 6

≈ 0.333. The experimental probability of

0.4 is higher.

Number of hours

The modes are 7, 8. c Cumulative frequency

c 70 times

b 20–30

12 a

b The median is the 8th score. For Class A, the polygon shows the 8th score is 8. For Class B, it’s 7. Class A’s median is higher.

10 0

Quiz scores

Plant height (cm)

Frequency

13 a Cumulative frequency

d The median is at the 15th value, which falls in the 30–40 minute interval. A reasonable estimate is approximately 31 minutes.

16 14 12

theoretical probability of . 20E B

10 8 6 4 2 0

21E C 22E B 7

Number of hours

d 8

c The difference is likely due to random chance and variation within a relatively small sample of 60 rolls. With a much larger number of rolls, the experimental probability would be expected to get closer to the

23E a x3 + 12x2 + 48x + 64 b 16y4 − 96y3 + 216y2 − 216y + 81 c a3 + 6a + d p6 − 6p5 + 15p4 − 20p3 + 15p2 − 6p + 1 24E a 672

b 280

c −448

d −1

Answers 1145 mathspace.co

44E Ask your teacher for worked solutions.

25E a a4 + 8a3 b + 24a2 b2 + 32ab3 + 16b4 5

b 243x − 810x + 1080x − 720x + 240x − 32 26E a 560

b −160

c 216

d −280

27E a 252

b 0

c −18

45E Ask your teacher for worked solutions.

13.01 Average rate of change What do you remember?

28E 64c6 − 192c5 d + 240c4 d2 − 160c3 d3 + 60c2 d4 − 12cd5 + d6

, the

1 The average rate of change is change in y divided by the change in x.

29E 120

2 It represents the gradient of the secant line connecting (a, f (a)) and (b, f (b)).

30E

3 The average rate of change is equal to the gradient m. 4 A secant line is a straight line passing through two points on the graph of a function.

Practice 5 a 2 e −1

b 2

c 3

d 2

f 0.707

6 a 100 litres/hour

b 100 litres/hour

7 a 60 km/h =

Since

= 21, the identity is veriﬁed.

b 60 km/h c It is constant, equal to the gradient 60.

31E 0.970 299

8 a 2 m/year

32E a x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5 5

3 2

2 3

b x − 5x y + 10x y − 10x y + 5xy − y 4

2 3

c 10x y + 20x y + 2y

b 3 m/year c The function is non-linear, so the rate of change varies over different intervals. 9 a 20 dollars/item

b 20 dollars/item

10 a 200 people/year

b 200 people/year

11 a −0.8 °C/min

b −1.4 °C/min

34E

12 a −1

c 0.2

35E a5 − 5a4 b + 10a3 b2 − 10a2 b3 + 5ab4 − b5

13 a $173

36E 1.157 625 + 3.3075r + 3.15r2 + r3

Extend your thinking

33E a b n=4

37E 4 38E Ask your teacher for worked solutions. 39E 672 40E k = ±2 41E

b 1

d 3

b $233

14 The second student used the wrong denominator, dividing by 3 instead of 3 − 1. The correct calculation is

= 4.

15 The average rate of change is 0.7 °C/hour, representing the average increase in temperature per hour over the interval.

42E Ask your teacher for worked solutions. 43E Ask your teacher for worked solutions.

1146 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16 The average rate of change is 14.7 m/s, which represents the average speed of the object over the interval. 17 The average rate of change is h + 4. As h approaches 0, the rate approaches 4, approximating the instantaneous rate at x = 1. 18 4a + 2h − 3

13.02 Speed as a rate of change What do you remember? 1 Average speed is

, the change in

distance divided by the change in time. 2 It is the gradient of the secant line between two points on the graph. 3 m/s (metres per second) or km/h (kilometres per hour). 4 It indicates constant speed, as the gradient is constant.

car is decelerating over time. The function d(t) = 40t − 2t2 indicates that the car starts at a certain speed and then slows down, as shown by the decrease in average speed over consecutive time intervals. This pattern of decreasing average speed suggests a deceleration in the car’s journey. b A quadratic function like this models constant acceleration (in this case, deceleration). It is inappropriate for long journeys because it predicts the car will eventually slow to a stop (at t = 10 hours) and then reverse its direction of travel. Furthermore, the total distance travelled cannot become negative, which this model predicts for t > 20 hours. 14 From t = 0 to t = 3: 10 km/h. From t = 1 to t = 3: 5 km/h. They differ because the cyclist’s speed varies, as the graph is not a straight line. 15 a 3.5 m/s b The average speed from t = 2 to t = 2.1 is 3.05 m/s. The average speed from t = 2 to t = 2.01 is 3.005 m/s.

Practice 5 a 50 km/h

b 4 m/s

c 80 km/h

d 20 m/s

e 30 km/h

f 17 m/s

6 a 15 km/h

b 15 km/h

c It is constant, as the function is linear with gradient 15. 7 a 4 m/s b 6 m/s c The function is non-linear, so the rate of change varies over different intervals. 8 a 20m/s

b 20 m/s

c 10 m/s

d 20 m/s

9 a 20 km/h

b 20 km/h

10 a 13 m/s

b 25 m/s

11 a 60 km/h

b 60 km/h

c 50 km/h

d 50 km/h

12 a 80 km/h

b 70 km/h

Extend your thinking 13 a As the time progresses, the average speed of the car decreases. This suggests that the

c As the time interval becomes shorter and approaches t = 2 seconds, the average speed approaches 3 m/s. d The runner’s speed at exactly t = 2 seconds is predicted to be 3 m/s. This is conﬁrmed by taking the derivative of the distance function, which gives the instantaneous speed. 16 0 − 2 h: 25 km/h; 2 − 4 h: 15 km/h; 4 − 6 h: 5 km/h. Speeds vary due to changes in motion throughout the journey. 17 a 181 818.18 km/s b For a delay of 20 minutes, the average speed of light is approximately 200 000 km/s. For a delay of 18 minutes, the average speed of light is approximately 222 222.22 km/s. c These calculations show how different estimates of the delay lead to different values for the speed of light. Accurate measurements are crucial for determining the true speed of light, and historically, improving measurement techniques has led to more precise values.

Answers 1147 mathspace.co

13.03 Instantaneous vs. average speed What do you remember? 1 Average speed is

, the total distance

travelled divided by the total time. 2 Instantaneous speed is the speed at a speciﬁc moment, represented by the gradient of the tangent to the distance-time graph at that point. 3 By calculating the average speed over a small interval [t, t + h], where h is small. 4 Average speed measures the overall rate over an interval, while instantaneous speed captures the exact speed at a speciﬁc moment, useful for variable motion. Practice 5 a 2 m/s

b 6 m/s

6 a 5 m/s b 4.1 m/s c Using h = 0.1 gives a closer approximation, as the secant line is nearer to the tangent at t = 2. 7 a 5 m/s

b 5.02 m/s

8 a 40 km/h b 40 km/h c Both are 40 km/h, as the function is linear, so average and instantaneous speeds are equal. 9 a 3.33 m/s

b 3.345 m/s

10 a 82 m/s

b 79.03 m/s

11 a 16.55 m/s

b 15.1505 m/s

c 15.015005 m/s 12 a 24 km/h b 24.4 km/h c The average speed over an interval is the total distance travelled divided by the total time taken. It gives a general idea of the speed over the entire journey. Instantaneous speed, on the other hand, refers to the speed of an object at a speciﬁc moment in time. It can be thought of as the speed that would be read on a speedometer at that instant.

Extend your thinking 13 By choosing the interval [2, 2.1] we achieve a more precise estimation of the instantaneous speed at t = 2, which is 9.2 m/s. 14 The student’s error was dividing by 3 instead of the correct interval length, 3 − 2. The correct average speed over the interval [2, 3] is 5 m/s. A better approximation of the instantaneous speed at t = 2, using a smaller interval [2, 2.01], is approximately 4.01 m/s. 15 Instantaneous speed (using [2, 2.1]): 8.2 m/s. Average speed: 6 m/s. The higher instantaneous speed indicates acceleration. 16 Approximation is 4 + h. As h approaches 0, it becomes 4 m/s, the true instantaneous speed, as the secant approaches the tangent. 17 Average speed: 17 m/s. The approximate instantaneous speed at t = 3 is ∼ 17.03 m/s. For this quadratic function, the average speed is equal to the true instantaneous speed at the midpoint of the interval.

13.04 Instantaneous speed and tangents What do you remember? 1 Instantaneous speed is the rate of change of distance at a speciﬁc moment, represented by the gradient of the tangent to the distance-time graph at that point. 2 A tangent line is a straight line that touches the graph at a single point, representing the slope of the curve at that instant. 3 By calculating the average speed over a small interval [t, t + h], where h is small, approximating the tangent’s gradient. 4 By drawing a tangent at the desired time and calculating its gradient using points on the tangent.

between two

Practice 5 a 2.05 m/s 6 4 m/s 7 a 50 km/h b 50 km/h

1148 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b 2.005 m/s

c The function is linear, so the tangent’s gradient equals the constant slope 50. 8 11 m/s

Practice 5 a 15 dollars/hour c 8 dollars/item

9 20 m/s

b 60 km/h d 1 m/s

10 a 3.02 m/s

b 3.002 m/s

6 50 km/h. The car travels at a constant speed of 50 km/h.

11 a 10 km/h

b 26 km/h

7 13 m/s

12 a 5.17 m/s

b 5.06 m/s

8 1.7675 litres/min

13 a 11 m/s b The cyclist exceeds the speed limit of 10 m/s.

b Worker 2

Extend your thinking 14 a b

9 a For 10 hours of work, Worker 1 is paid more, receiving $550 compared to Worker 2’s $350. c The 250 value represents a ﬁxed payment or base salary.

= 9 m/s

10 a 10.2 people/year

= 13 m/s

c The speeds differ because the vehicle is accelerating, as the distance function is quadratic. The tangent’s gradient increases with time.

b 10.01 people/year

11 1200 litres/hour 12 a 4.25 m/s

b 4.05 m/s

13 A

15 Instantaneous speed: 13.02 m/s. Average speed: 13 m/s. Higher instantaneous speed indicates acceleration at t = 3.

Extend your thinking

16 The student divided by 4 instead of 4 − 3.

The slightly higher instantaneous rate indicates acceleration at t = 3.

Correct calculation:

= 3.5 m/s.

17 5 m/s. The estimate may be less accurate for a non-linear graph, as the secant spans a wide interval.

13.05 Linear and quadratic rates of change What do you remember? 1 The rate of change is the constant gradient m. 2 In a quadratic function, the rate of change varies and is given by the tangent’s gradient at each point, while in a linear function, it is constant. 3 By calculating the gradient of a secant line over a small interval near the point, approximating the tangent’s gradient. 4 It represents a steady rate, such as speed, wage, or cost per unit.

14 Instantaneous rate: 8.01 m/s. Average rate: 8 m/s.

15 The negative coeﬃcient −10 means revenue decreases by $10 per item sold, which is unrealistic as shops typically gain revenue per sale. This model suggests losses with increased sales, which is problematic for a business. 16 Instantaneous rate: 7.004 litres/min. Water is added at approximately 7 litres/min at t = 1 minute. 17 Linear: 10 dollars/hour. Quadratic: 30.05 dollars/hour. The quadratic’s higher rate indicates costs increase faster over time due to its non-linear nature. 18 Ask to your teacher for worked solutions. 19 a Average rate: Instantaneous rate at m =

= a(x1 + x2) + b. :

2am + b = a(x1 + x2) + b, which equals the average rate.

Answers 1149 mathspace.co

b Instantaneous rate: 18.0004 m/s. Average rate: 18 m/s.

15 Average speed: 19 m/s. Instantaneous speed: 15.04 m/s.

The instantaneous rate is very close to 18, conﬁrming the proof as h is small.

The average speed is higher due to acceleration over the interval. 16 a 6.15 m/s

Chapter 13 review

17 50 km/h

1 B

18 a 14.03 m/s

2 B 3 C 4 a 3

b 6.015 m/s

c −2

b 3

d 6

5 a 1500 dollars/year

b 2700 dollars/year

c 1500 dollars/year

d 3500 dollars/year

6 a 40 bacteria/hour b 120 bacteria/hour c The function is quadratic, so the rate of growth increases over time, leading to higher rates in later intervals. 7 The student divided by 4 instead of 4 − 1 = 3.

b Average speed: 14 m/s. The instantaneous speed approximated in part (a) is 14.03 m/s. For a quadratic function, the average speed over a symmetric interval is exactly equal to the instantaneous speed at the midpoint. Therefore, the true instantaneous speed at t = 2 is 14 m/s. The secant approximation of 14.03 m/s is very close to this true value. 19 a 6 + 3h b As h approaches 0, it approaches 6 m/s, the instantaneous speed at t = 1.

= 10.

20 The rate of change is 0.5 m/day. This means the water level rises at a constant rate of 0.5 metres per day.

8 a 60 km/h

b 15 km/h

c 14 km/h

d 140 km/h

21 a 40.5 m2/min

9 a 21.25 m/s

b 25.83 m/s

Correct answer:

10 a 13 m/s

b 40.05 m2/min

22 a 170.1 thousand dollars/month b Instantaneous rate (approximated): 170.1 thousand dollars/month.

b 17 m/s

Average rate: 170 thousand dollars/month.

c A quadratic function implies constant acceleration, unrealistic due to changing mass, drag, and velocity limits.

The rates are nearly identical. For a quadratic function, the average rate of change over a symmetric interval [5, 7] is exactly equal to the instantaneous rate of change at its midpoint, t = 6. Thus, the true instantaneous rate is 170 000 dollars/year.

11 a • From t = 0 to t = 4: 11.25 km/h ≈ 6.67 km/h

• From t = 1 to t = 4:

b The cyclist’s speed varies, as the points suggest non-constant speed. 12 • Interval 0 − 2 h: 30 km/h • Interval 2 − 4 h: 15 km/h • Interval 4 − 6 h: 5 km/h Speeds decrease, possibly due to traﬃc, fatigue, or nearing a destination. 13 a 8 m/s

b 9.5 m/s

14 a 80 km/h

b 80 km/h

23 a 13.0025 cm/week b The plant grows at approximately 13 cm/week at t = 3 weeks. 24 a Plan A: $10/GB b Plan B: $15.005/GB c Plan A has a constant rate; Plan B’s rate increases with data, making additional GB costlier at 10 GB.

c Both are 80 km/h. The linear function has a constant rate of change.

1150 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

14.01 Gradient of a curve

6 a i

What do you remember?

1 a A tangent to a curve at a particular point is a straight line that ‘just touches’ the curve at that point and has the same direction (gradient) as the curve at that point of contact. b A secant to a curve is a straight line that intersects the curve at two distinct points. c The gradient of a curve at a point is defined as the gradient of the tangent to the curve at that point. d The derivative at a point x = a from first principles is defined as the limit of the gradient of the secant line: f ′(a) = limh → 0

1 B x −2

−1

1 −1

ii • Point A: Positive gradient • Point B: Negative gradient • Point C: Zero gradient • Point D: Positive gradient b i

e Differentiation is the process of ﬁnding the derivative (or derived function f ′(x)) of a function f (x).

C −2

−1

2 a False. The gradient of a curve generally changes from point to point, unless the curve is a straight line.

−2

b True

ii • Point A: Negative gradient

c False. If the tangent is horizontal, its gradient is zero, so the derivative is zero.

• Point B: Zero gradient

d True

• Point D: Negative gradient

e False. A vertical tangent has an undeﬁned gradient. 3 As h approaches zero, point Q approaches point P. The gradient of the secant line PQ, , approximates the gradient of the

4 Two common notations are f ′(x) and Practice

c m = 0.25 or

7 a m = 4.1 c m = 4.001 8 a g(2) = −5 c m = −3

b m = 4.01 d 4 b m = −3 d −3

9 7

tangent at P.

5 a m=2

• Point C: Positive gradient

b m=1 d m = −1

10 a f ′(0) = −4

b f ′(1) = −1

c f ′(2) = 8

d f ′(−1) = −1

11 a s′(1) = 5 m/s

b s′(4) = 11 m/s

c s′(0) = 3 m/s

d s′(2.5) = 8 m/s

12 a x = −1 and x = 3

b x < −1 and x > 3

c −1 < x < 3 13 a P = (4, 2)

b m ≈ 0.248

c m ≈ 0.250

d m ≈ 0.250

Answers 1151 mathspace.co

14 a f (1) = 1 b

1+h

f (1 + h)

f (1)

f (1 + h) − f (1)

Gradient of secant

0.1

1.1

1.331

0.331

3.31

0.01

1.01

1.030301

0.030301

3.0301

0.001

1.001 1.003003001

0.003003001

3.003001

−0.001 0.999 0.997002999

−0.002997001 2.997001

−0.01

0.99

0.970299

−0.029701

2.9701

−0.1

0.9

0.729

−0.271

2.71

c 3

2 The derivative is m. The graph of a linear function f (x) = mx + c is a straight line with gradient m. The derivative represents the gradient of the function. 3 It implies that the gradient of the original function f (x) is constant. A graph with a constant gradient is a straight line. Practice 4 a f ′(x) = 0

Extend your thinking 15 A possible sketch is shown. Key features: decreasing for x < −1, local minimum at x = −1, increasing for −1 < x < 2, local maximum at x = 2, decreasing for x > 2. y 4

Max

c g′(x) = 0

d h′(x) = 0

e f ′(x) = 12

g V′(t) =

h P′(n) = 0.5

i f ′(x) = 7

k g′(x) =

l h′(x) = 1

5 a i

9 8 7 6 5 4 3 2 1

y = f (x)

1 x

−2

−1

y=5

−1

Min

16 a m = 5.705 b m = 5.97005 c For Q(2 + h, f (2 + h)): If h = 0.1, gradient m = 6.305. If h = 0.01, gradient m = 6.03005. Gradients from left: 5.705, 5.97005. Gradients from right: 6.305, 6.03005.

−4 −3 −2 −1

= −1

= −4

x 1 2 3 4

The graph is a horizontal line passing through y = 5. ii 0 b i

Both approach 6 as h → 0.

17 The first part is correct: steep tangent implies large f ′(a). However, for a vertical tangent, the gradient is undefined. The derivative f ′(a) does not exist if the tangent is vertical, as the limit defining it does not approach a finite number.

1 −4 −3 −2 −1 −1

y = 2x

x 1

2 3 4

−2 −3 −4

14.02 Derivatives of basic functions

The graph is a straight line passing through the origin with a gradient of 2.

What do you remember?

ii 2

1 The derivative is 0. A constant function f (x) = c represents a horizontal line. The gradient of a horizontal line is always 0.

1152 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

c i

c Since the derivative represents the gradient of the function, the fact that all three derivatives are equal to 2 means that all three graphs have a constant gradient of 2. Geometrically, this means the three lines are parallel to each other.

6 5 4 3 2

y=x+3

−4 −3 −2 −1 −1

2 3 4

The graph is a straight line with a y-intercept of 3 and a gradient of 1. ii 1 d i

y −4 −3 −2 −1

y = −3

x 1

2 3 4

9 a Linear function b Answers will vary, but must be of the form f (x) = −4x + c. For example: f (x) = −4x, f (x) = −4x + 1, f (x) = −4x − 100. c All possible functions are straight lines with a gradient of −4. This means they are all parallel to each other. 10 a m = 3 b L(x) = 3x + 1

−1

c L′(x) = 3

−2

d The gradient calculated in part (a) is the value m. The derivative found in part (c) is also the value m. This conﬁrms that the derivative of a linear function is its gradient.

−3 −4

14.03 First principles for derivatives The graph is a horizontal line passing through y = −3. ii 0 6 a This is a linear function with m = −3. So,

= −3.

b The derivative is constant. The gradient at x = 5 is −3. c The derivative is constant. The gradient at x = −2 is −3. d The gradient of a linear function is constant. It does not change for different values of x. The derivative is the value of this constant gradient. 7 a f ′(x) = a b

= −1

What do you remember? 1 f ′(x) = limh → 0 2 It represents the change in the y-value of the function f (x) when x changes by a small amount h. It’s the ‘rise’ of the secant line. 3 It represents the gradient of the secant line passing through the points (x, f (x)) and (x + h, f (x + h)) on the curve y = f (x). 4 Substitute f (x + h) and f (x) into the formula , expand and simplify the numerator, factor out h from the numerator, cancel h with the denominator, and then evaluate the limit as h → 0.

c g′(t) = −a

Practice

d The coeﬃcient of n is the constant (a + b). So, P ′(n) = a + b.

5 f ′(x) = 5 6 a i f (x + h) = 10x + 10h

Extend your thinking 8 a f ′(x) = 2, g′(x) = 2, h′(x) = 2 b They are all the same. They are all equal to 2.

ii f ′(x) = 10 b i f (x + h) = −2x − 2h + 9 ii f ′(x) = −2

Answers 1153 mathspace.co

c i f (x + h) = 7

12 a Let x = a + h. Then as x → a, a + h → a, which means h → 0. Also, h = x − a. Substituting these into the standard ﬁrst principles formula (with x replaced by a as the point of interest):

ii f ′(x) = 0 d i f (x + h) =

ii f ′(x) = 7 a f ′(x) = 2x − 4

f ′(a) = limh → 0

b f ′(x) = 6x + 2

f ′(a) = limx → a

c f ′(x) = −2x 8 a i f (x + h) = x2 + 2xh + h2 + 3 ii f ′(x) = 2x

becomes .

b i f (x + h) = 2x2 + 4xh + 2h2 ii f ′(x) = 4x c i f (x + h) = x2 + 2xh + h2 + x + h ii f ′(x) = 2x + 1 d i f (x + h) = −x2 − 2xh − h2 + 6x + 6h − 2 ii f ′(x) = −2x + 6 9 a i f ′(x) = 2x b i f ′(x) = 4 − 2x

ii f ′(1) = 2 ii f ′(3) = −2

c i f ′(x) = 8

ii f ′(−2) = 8

d i f ′(x) = 2x

ii f ′(−1) = −2

Extend your thinking 10 Given f (x) = c, then f (x + h) = c. Using ﬁrst principles:

So, f ′(x) = 2x. 13 The instantaneous velocity is v(t) = 6t − 2 m/s.

14.04 Derivative notation and basic rules What do you remember? 1 Two common notations are f ′(x) and Others include y′ and

[ f (x)].

2 The power rule states that if f (x) = xn, then f ′(x) = nxn − 1 for any real number n. That is, (xn) = nxn − 1. 3 The derivative of a constant multiple of a function is the constant times the derivative of 11 The student has missed the limh → 0 notation throughout the working. The derivative is deﬁned as the limit of the expression as h approaches zero. Simply removing h to get 2x + h is not the derivative. The correct ﬁnal step is to evaluate the limit: So, f ′(x) = 2x.

the function. That is,

[k × f (x)] = k × f ′(x).

4 The derivative of a sum (or difference) of functions is the sum (or difference) of their derivatives. That is, Practice 5 a

= 7x6

= 10x9

= 99x98

1154 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

[ f (x) ± g(x)] = f ′(x) ± g′(x).

8 a Rewritten form: f (x) = x3 + 3x2

= 200x199 −5

= −4x

= −6x−7 or

= −5x−6 or

= −x−2 or

b Rewritten form: y = x2 + 3x − 4

Derivative:

= 2x + 3

c Rewritten form: g(x) = x3 + 2x Derivative: g′(x) = 3x2 + 2. d Rewritten form:

i j

Derivative: f ′(x) = 3x2 + 6x

Derivative: e Rewritten form: y = 4x2 + 12x +9

Derivative: k

= 8x + 12

f Rewritten form:

Derivative: g Rewritten form:

Derivative: n

h Rewritten form: f (x) = x − 2x−2 Derivative: f ′(x) = 1 + 4x−3

6 a i f ′(x) = 7x iii

= 7x

9 a 3t2 − 20t + 25

(x7) = 7x6

c 4x + 3

b i f ′(x) = 15x2 iii

= 15x2

(5x − 2) = 15x

c i

= 2π r

b b f ′(x) = 0

e g′(x) = 3x

f h′(x) = 10

h f ′(x) = −18x−4 or 2

i f ′(x) = 3x + 2x

= 8π r + 2π h

Extend your thinking

11 a

iii

c f ′(x) = 20x3

10 (2, −15), (−2, 17) ii

7 a

b 4z3

= 20x − 7

k g′(x) = 6x + 12x − 9 l h′(x) = x2 − x + 1 m

= 2x + 2x−3

= 20x3 − 6x + 2

p f ′(x) = 2ax + b

represents the instantaneous rate of

change of the area of the circle with respect to its radius. The expression 2π r is the formula for the circumference of a circle. This means the rate at which the area of a circle changes with respect to its radius is equal to its circumference. 12 a = 2, b = −2 and c = 3. The constant d differentiates to zero, so its value cannot be determined from f ′(x) alone. 13 Expanded: Derivative:

Answers 1155 mathspace.co

14.05 Tangents and normals What do you remember?

e i (−1, −6)

ii 3

iii y = 3x − 3

f i (1, −4)

1 A tangent to a curve at a point P is a straight line that “just touches” the curve at P and has the same direction (gradient) as the curve at that point.

ii −3

iii y = −3x − 1

15 a i

2 The gradient of the tangent to the curve y = f (x) at x = a is equal to the value of the derivative at that point, f ′(a).

9 8 7 6 5 4 3 2 1

f (x)

3 The point-gradient formula is y − y1 = m(x − x1), where m is the gradient and (x1, y1 ) is a point on the line. 4 A normal to a curve at a point P is a straight line that is perpendicular to the tangent to the curve at that same point P. 5 The gradient of the normal is the negative reciprocal of the tangent’s gradient:

−4 −3 −2 −1

b (−2, −9)

9 a

10 a (−1, 4)

b (1, 1) and (−1, −1)

11 a x = 2

12 a 84°

b 63°

13 a i y = 7x − 14

b i

14 a i (1, 0)

ii 2

iii y = 2x − 2

b i (2, 2)

ii −1

iii y = −x + 4

iv 1

c i (4, 0)

ii 1

iii y = x − 4

iv −1

d i (2, 3) iii

3 2 1

f (x)

8 a (1, 0) and (3, −22)

x 1 2 3 4

b i

Practice b y = 17x − 16

g(x)

ii Yes, because the line g(x) touches the curve f (x) at a single point (x = 1).

6 The relationship is m = tan( θ ).

7 a y = −2x − 1

−4 −3 −2 −1 −1

x 1

2 3 4

−2

g(x)

−3 −4

ii No, because the line g(x) crosses the curve f (x) twice. c i 8

g(x)

4 2

−4 −3 −2 −1 −2 f (x) −4

x 1

−6 −8

ii No, because the line g(x) crosses the curve f (x) at multiple points.

ii iv −2

1156 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

d i

g(x)

5 4 3 2 1

23 y = 2x3 + 3x2 − 28

−4 −3 −2 −1 −1 −2 −3 −4

24 a = 900 and b = 90

14.06 Chain rule x

f (x) 1

What do you remember?

2 3 4

1 The inner function is g(x). It is the function that is applied ﬁrst.

ii Yes, because the line g(x) is tangent to the curve f (x) at x = −2. At this point, the curve and the line share the same value ( y = −3) and the same gradient m = 3

2 The outer function is f. It is the function that is applied to the result of the inner function, g(x). 3 If y is a function of u, and u is a function of x, .

then the chain rule is Alternatively, if h(x) = f (g(x)), then h′(x) = f ′(g(x)) × g′(x).

16 k = 40 17 b = −23 and c = 79

4 The generalised power rule states that for a function of the form y = [ f (x)]n, its derivative is

Extend your thinking

= n[f (x)]n − 1 × f ′(x).

Practice

19 a The gradient of the line y = −12x − 16 is −12. For this to be a normal, the tangent gradient must be mT =

5 a 2u5

. The derivative of the curve

is y′ = 3x . Setting 3x =

gives x =

, so

x = ± . The potential points on the curve are and

. Neither of these

(x − 2) or

d −u−5

e 6 a g(x) = 5x3 − 4x2 + 3x − 5, h(u) = u7 b g(x) = 2x2 + 2x + 3,

7 a i

b At x = 2, y = 8. The tangent gradient is

y−8=

c 5u−1

c g(x) = 4x2 − 3x + 5, h(u) =

points satisﬁes the equation of the line y = −12x − 16. Therefore, the line is not a normal to the curve at any point. y′(2) = 3 (2)2 = 12. The normal gradient is The equation of the normal is

or h(u) = u−3

iii b i

. iii

20 The coordinates are and

c i

= −u−2

iii . 21 The points of tangency are (1, 1) and (3, 9).

d i

iii

22 The equations are y = 3x − 2, , and .

Answers 1157 mathspace.co

e i

11 g′(−1) = −1280

12 x = −1, 0, 1 iii

13 x = −2, −1, 0

f i

14 a

iii

y 15

8 a 9 a

= 36(4x + 3)8

= −4(2t7 + 8t3 + 3t + 5)−5 (14t6 + 24t2 + 3)

= 3(x2 + x−3)2 (2x − 3x−4)

x −15 −10 −5

10 15

c 5x + 12y = 169 d e

= 8(3x2 − 4x + 2)3 (3x − 2)

= −90(3x + 4)

b y = 16x − 15

16 a

b f ′(−2) = −3

17 x = 1 or x = 6

g h

15 a y = −8x − 7

18 m = = −5(x + 6)−6 3

Extend your thinking 2

−2

= −(4x − 12x + 5)(x − 4x + 5x)

19 No, because

can never be equal

to zero as its numerator is always −1.

20 a = 4, c = 3 or a = −2, c = 3 k

21 No. The chain rule states that y′ = f ′(g(x)) × g′(x). For y′ to be zero, it must be that either f ′(g(x)) = 0 or g′(x) = 0. So it is not necessary for f ′(g(x)) to be zero.

l m

For example, if y = (x2 − 1)3, then y′ = 3(x2 − 1)2 × (2x). At x = 0, g′(0) = 0 which makes y′ = 0, but f ′(g(0)) = f ′(−1) = 3(−1)2 = 3 ≠ 0.

n o

23 Using the chain rule,

= −15(1 − x−2)(x + x−1)4

When x = 1, Since

r 10 a x = 5

= 4(3x2 + k)3 × (6x).

= 4(3 + k)3 × 6 = 24(3 + k)3.

= 0, we have 24(3 + k)3 = 0, which

means 3 + k = 0. Therefore, k = −3. b x = 2 or x = −6

1158 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

14.07 Product rule

d i f (3) = 25

What do you remember? , v = (4x7 − 2x3 − 5x)

1 a

b u = 3x4, v = (x − 4)2 2 a u′ = 6x5 c

ii f ′(0) = 3

iii f ′(−3) = −9

b v′ = 4x3

8 a i f ′(x) = 2(x + 3)2(2x + 3) ii y = 8x + 8 iii b i

= 10x9 + 24x5

ii y = x

3 a u = x2, v = 4x5

iii y = −x − 2

b No, because the product rule is not necessary here as y = x2 (4x5) can be simpliﬁed to y = 4x7 ﬁrst.

c i ii y = −x − 1 iii y = x + 1 d i f ′(x) = 3x2 + 4x

Practice = 5x4 + 27x2

5 a

= 2x + 14

= 24x − 23

= −8t3 + 18t2 + 3

= 42t5 − 144t3 + 10t

= 5x4 + 27x2

ii y = −x

4 a

iii y = x + 2

Extend your thinking 11 n = 4 12 a

6 a

13 g′(3) = −54

14 d Sum of the ﬁrst x positive

f f ′(x) = 5(5x + 7)6(8x − 9)4(96x − 7)

Evaluate

Simplify

Determine the derivative

Evaluate

integers

g h i 7 a i f (3) = 0

ii f ′(0) = 15

iii f ′(−3) = 135 b i f (3) = 6

ii f ′(0) = −1

iii f ′(−3) = −7 c i f (3) = 4

ii f ′(0) = −1

iii f ′(−3) = −7

Answers 1159 mathspace.co

14.08 Quotient rule

d i u = 3x, v = 5x − 4

What do you remember?

1 a u(x) = x3, v(x) = x − 1

iii

b u(x) = x2 + 1, v(x) = x2 − 1 c u(x) =

iv No values of x, as the numerator −12 is a non-zero constant.

, v(x) = x + 1

d u(x) = 2x − 3, v(x) = (x + 1)2

7 a 2 3 a 2x

b 1

4 If the denominator is a monomial, it’s often easier to divide each term in the numerator by the monomial and then differentiate term by term. For example, for y = y = x + 1 then

c d

, simplify to

= 1, instead of using quotient

rule on the original expression. f b

5 a c x=0

Practice 6 a i u = 2x − 5, v = 5x − 2

h i

ii iii iv No values of x, as the numerator 21 is a non-zero constant.

8 a −1

b 0

c 14

b i u = 5x , v = 2x + 8 9 a

iii

10 a i

ii No, numerator is 7.

iv x = 0, −8 (provided x ≠ −4).

iii

c i u = 4x2 + 3, v = 5 + 2x

b i

iii x = 6

iii

c i iv

ii No, numerator is −12.

. iii

1160 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

ii No, numerator is −20.

ii No, numerator is −12.

d i

Extend your thinking 14 x = 0 or x =

iii ii Yes, if x = 0.

e i iii None 11 a

12 a f ′(0) = , f ′(2) =

, a = 0, −6

, k = 3, 27

17 18 a 4

, f ′(−1) =

b f ′(0) = 0, f ′(1) =

14.09 Apply differentiation rules

, f ′(4) = 2048

What do you remember?

c f ′(0) = 0, f ′(1) = 5

1 a Product rule

d f ′(1) = 0, f ′(4) =

b Quotient rule c Chain rule

13 a i Differentiate term by term:

d Product rule and chain rule

e Quotient rule and chain rule 2 The chain rule is necessary when differentiating composite functions, i.e., functions that are formed by one function acting on the result of another function, such as f (g(x)).

So, ii

iii Differentiate combined fraction:

Practice

So,

b i Differentiate term by term using quotient .

rule on each: ii

ii x = 0, x = ±1 b i

iii Differentiate combined fraction: . So,

4 a i

= 72x2 + 32x − 33

ii None c i

c i Differentiate term by term: . So, f ′(4) = 1.

iii Differentiate combined fraction: f ′(x) = 1. So, f ′(4) = 1.

d i

ii f (x) = x + 1 (for x ≠ 1)

d i Differentiate term by term: . So, f ′(4) = 0. ii f (x) = 1 (for x ≠ −2) iii Differentiate combined fraction: f ′(x) = 0. So, f ′(4) = 0

ii x = 0 e i

= x3(x6 − 12)(x6 − 3)

ii None f i ii x = 0

Answers 1161 mathspace.co

3 The x-coordinates of the stationary points on the graph of y = f (x) correspond to the x-intercepts on the graph of y = f ′(x).

(42x2 − 12x + 1)

5 6 a

d 108 4 Use the formula f ′(c) ≈

f 12

small value for h, such as 0.001.

7 a

b y = 5x − 9

d y = 21x − 33 b

8 a

with a very

Practice 5 a i Increasing: x < 0 and x > 2 Decreasing: 0 < x < 2 ii Stationary points at (0, 1) and (2, −3). y

9 x = 0, x = ±

1 x

10 a Overall, chain rule. The inner function is a product, so product rule next. One function in the product

−1

requires power rule/

−2

quotient rule for . b

−3

b i Increasing: x < −2 and x > 1

c 20 514

Decreasing: −2 < x < 1

11 a x-intercepts: (0, 0), (2, 0), (−4, 0)

ii Stationary points at (−2, 20) and (1, −7) y

y-intercept: (0, 0) b

= 8(x − 2)(x2 + 2x − 2)

, x = −1 −

c x = 2, x = −1 +

Extend your thinking

12 a = 4, b = 12

−3

−1

−5

13 k = 2 14 P ′(x) = f ′(x)g(x)h(x) + f (x)g′(x)h(x) + f (x)g(x)h′(x)

−2

c i Increasing: −2 < x < 2

15 f ′(x) = −360x3 (3 − (6 + 5x4)9)(6 + 5x4)8

Decreasing: x < −2 and x > 2

16 g′(0) = 22

ii Stationary points at (−2, −20) and (2, 12). y

14.10 Graphical behaviour of functions 5

What do you remember?

−3 −2 −1 −5

1 a f (x) is increasing on that interval b f (x) is decreasing on that interval

−10

c f (x) has a stationary point 2 Calculate the ﬁrst derivative, f ′(x), set it equal to zero, and solve the resulting quadratic equation for x.

−15

1162 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x 1

d i Increasing for all real x.

ii Stationary points at (−1, 2) and (1, −2).

ii No stationary points.

2 −2

−1

x 1

−2 −4

−2

−1

−1 −2

−6

−8

h i D ecreasing for all real x, except at x = 0 where it is stationary.

e i Increasing: −1 < x < 3 Decreasing: x < −1 and x > 3

ii Stationary point of inﬂection at (0, 0).

ii Stationary points at (−1, −6) and (3, 26).

3 2

1 −2

−1

x 1

−1

−2 x −2 −1

6 a x = −1 and x = 1

f i Increasing: x < −1 and x > 3

b Increasing for x < −1 and x > 1, and decreasing for −1 < x < 1.

Decreasing: −1 < x < 3 and (3, −7).

ii Stationary points at 4 3 2 1

−2 −1 −1 −2 −3 −4 −5 −6 −7

−3 −4

7 a k=6

b x = −1

8 a

y 2

x 1

g i Increasing: x < −1 and x > 1 Decreasing: −1 < x < 1

x −1

y = f ′ (x)

−1 −2

y 3 2

y = f ′(x)

1 −2

−1

x 1

−2 −3

Answers 1163 mathspace.co

3 2 2 1

1 x −2

−1

−2

−1

y = f ′(x) d

b −2.9999

c −2.999999

1 −1

−1

9 a −2.99

−2

y = f ′ (x)

y = f ′(x) 1

−1

d The estimate approaches −3. 10 a

−2

y 60

−3

y = x4

Q (3, 48)

y 3

y = f ′(x) 1 −2

−1

The gradient is

y 1 x 3

b Differentiate f (x) = x4 to get f ′(x) = 4x3. At x = 2, f ′(2) = 4 × 23 = 32. The graphical estimate (32) is exact, matching the true value. The numerical estimate (32.24) is slightly higher by 0.24, due to h = 0.01 not being inﬁnitesimally small, but still very close, showing the numerical method’s effectiveness for small h. 11 a i The gradient is 4.

−1

ii Approximately 4.0301 b i The gradient is 11. g

ii Approximately 11.0601

y 3

c i The gradient is −2.

ii Approximately −1.9504

1 −2

−1

. So, f ′(2) ≈ 32.

b Approximately 32.24

−3

−2

−1

P (2, 16)

y = f ′(x) 1

x 2

d i The gradient is −1. ii Approximately −1.101

−2 −3

1164 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2 Velocity, v(t), is the ﬁrst derivative of displacement with respect to time, so

12 a b The function f (x) is increasing where f ′(x) > 0. From the graph, this occurs for x < −2 and 0 < x < 2. c The function f (x) is decreasing where f ′(x) < 0. From the graph, this occurs for −2 < x < 0 and x > 2. Extend your thinking 13 k > 12 14 The function has stationary points at x = 1 (maximum) and x = 5 (minimum). The sketch should show a cubic curve rising from the left, passing through the origin (0, 0), reaching a maximum at x = 1, falling to a minimum at x = 5, and then rising to the right. y 2

−2

x 1

−4

v(t) =

= x′(t) = .

3 The particle is momentarily at rest when its velocity v(t) = 0. 4 Distance is a scalar quantity representing the total path length travelled, while displacement is a vector quantity representing the change in position from the origin. Speed is the magnitude of velocity (a scalar, always non-negative), while velocity is a vector that includes direction (positive or negative). Practice 5 The units of C′(x) are dollars per item. C′(x) represents the rate of change of cost with respect to the number of items produced, which is the approximate additional cost to produce one more item after x items. 6 a 0.3 m/day

b 0.3 m/day

7 a T ′(t) = −4t

b −20°C/hour

c At exactly 5 hours, the temperature of the object is decreasing at a rate of 20°C per hour.

−6 −8

8 a 60 m/s

b t = 5 s and t = 2 s

15 The error is in the final conclusion (step 4). The student has reversed the classifications. At x = −1, the sign of f ′(x) changes from positive to negative, which indicates a maximum turning point. At x = 2, the sign of f ′(x) changes from negative to positive, which indicates a minimum turning point. Correct classification: maximum turning point at x = −1 and minimum turning point at x = 2.

14.11 Derivatives as rates of change

c 12 m/s 9 a −9 cm/s b The speed is 9 cm/s. c Since the velocity (4) = −9 is negative, the particle is moving in the negative direction. 10 a t = 0 s and t =

s ( ≈ 3.33 s)

b For c Speed is decreasing when the graph moves towards the t-axis. This occurs for

What do you remember? 1 The average rate of change of a function f (x) over an interval [a, b] is the gradient of the secant line connecting the points (a, f (a)) and (b, f (b)), given by

. The instantaneous

rate of change at a point x = a is the gradient of the tangent to the curve at that point, given by the derivative f ′(a).

11 Graph B. The velocity is zero at t = 0 and t = 2, so the displacement graph must have stationary points at these times. The velocity is negative for 0 < t < 2, so the displacement graph must be decreasing in this interval. The velocity is positive for t > 2, so the displacement graph must be increasing in this interval. Only Graph B matches these features.

Answers 1165 mathspace.co

b Correct answer for “at rest”:

12 a t = 2 s b t ∈ [0, 2) ∪ (2, ∞) c Particle B 13 a The rate of change of cost is $20 per hundred items (or $0.20 per item). b The cost stops increasing when the rate of change of cost is zero, which occurs at a production level of x = 30 (3000 items). Note: this is outside the given domain. 14 a

Set v(t) = 3t2 − 14t + 15 = 0. Factoring gives (3t − 5)(t − 3) = 0. The particle is at rest at t=

Correct answer for speed at t = 2 s: The velocity is −1 m/s. The speed is v(2) = −1 = 1 m/s.

Chapter 14 review 1 a m = 1.1

cm2/day

b Since A′(2) =

s and t = 3 s.

b m = 1.01 c m = 1.001

> 0, the rate of change is

positive. This means the area of the wound is increasing after 2 days.

d The gradient appears to be approaching 1. 2 a f ′(0) = 5

b f ′(2) = 13

c f ′(−3) = 23

15 a b 32π cm3/s

Max

16 a C′(x) = 10 − 0.04x

−1

b The gradient is $6 per widget. This means that when 100 widgets are being produced, the cost to produce one additional widget is approximately $6.

1 −5

y = f (x)

−10

17 a v(1) = 10.2 m/s, v(3) = −9.4 m/s

x 2

Min

b Approximately 20.41 metres Extend your thinking 18

or 0.2

b f ′(x) = −5x−6

4 a c

19 a P ′(x) = −3x2 + 24x − 36 b x = 2 (200 items) or x = 6 (600 items) c At x = 2, there is a local minimum profit. At x = 6, there is a local maximum profit. 20 a t = 2 s b Yes, both velocities are 2 m/s (positive direction). c t = 0 s and t = 4 s 21 a Error 1: The student incorrectly tried to find when the particle is at rest by setting displacement x(t) = 0. A particle is at rest when its velocity v(t) = 0. Error 2: The student stated that speed can be negative. Speed is the magnitude of velocity and must be non-negative.

5 a

c 6 n=3 7 f ′(x) = 4x + 5 8 a f ′(x) = 2x + 6

b f ′(−2) = 2(−2) + 6 = 2

9 v(t) = 8t− 1 b g′(x) = 12x3 − 6x2 + 7

10 a

11 Rewritten form: y = 4x2 − 20x + 25. Derivative:

1166 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

12 a = 3, b = 5, c = −1

15.01 Exponential graphs

13 y = x + 2

What do you remember?

14 y = −x − 3

1 a ii

b vi

15 θ ≈ 83°

e v

f iii

2 a True

b True

16 17

= 5(2x − x3)4(2 − 3x2)

18 x = 0, x = −2, x = 2 19 a = 2, c = 1 20

c i

d iv

c False

d True

3 a Growth b Decay

c Decay

d Decay

4 a Not exponential

b Exponential

c Exponential

d Not exponential

e Exponential

f Not exponential

g Not exponential

h Exponential

Practice

21 y = 16x + 20

5 a Horizontal asymptote: y = 0.

22 g′(2) = 8

y-intercept: (0, 3). Domain: All real numbers.

Range: y > 0. 24 y = −2x + 9

9 8 7 6 5 4 3 2 1

25 k = 1 or k = 9 26

= 2x(2x2 + 1)(−14x3 + 30x2 − 3x + 5)

27 y = 29x − 71 28 a = 4, b = 8 29 (0, 5) and (4, −27); Increasing for x < 0 and x > 4, and decreasing for 0 < x < 4. 30

−2

6 a

4 y 3 2 1

y = f ′(x)

−1 −1 −2 −3 −4

−1

−3

−2

−1

16 64

x 1

32 a 36 m/s

Horizontal asymptote: y = 0. b Domain: All real numbers. Range: y > 0. b t = 2 s and t = 6 s

c 12 m/s 33 a t = 0 s and t = 4 s

b As x → ∞, y → ∞. As x → −∞, y → 0+.

31 −3 < k < 3

7 a Positive b 0

b For 0 < t < 4

c y=0 d Domain: All real x

c 2<t<4 34 a t = 3 s

−2

b t = 0 s and t = 6 s

Range: y > 0 e The graph of the function increases at an increasing rate.

Answers 1167 mathspace.co

8 a Positive

13 a No. For a > 0 the expression a−x is greater than zero for all real x.

b y=0 c Domain: All real x

b y=0

Range: y > 0 d The graph of the function decreases at a decreasing rate. b

c (2, 25)

e (0, 1)

f (4, 625)

h (1, 5)

9 a (3, 125)

c As x → +∞, y → 0+ (As x approaches infinity, y approaches zero from above). As x → −∞, y → ∞ (As x approaches negative infinity, y approaches infinity). d 1 e 0 f 7 6

10 a 1

12 a

5 4 3 2 1 −5 −4 −3 −2 −1 −1

11 a (0, 1) d y>0 x −5 −4 −3 −2 −1 0 1 2 3 4 5

c All real x f m=8

b No e y = 128 y 1024 256 64 16 4 1 0.25 0.625 0.015 625 0.003 906 25 0.000 976 5625

b Decreasing

c Decreasing

d Increasing

e Increasing

f Decreasing

g Decreasing

h Increasing

−x

15 a y = 3

b Reﬂection across the y-axis c

9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1

y = 3x

x 1 2 3 4 5

−2

d y>0 d (0, 1)

19 y 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1

x 1 2 3 4 5

14 a Increasing

b As x increases, the function decreases at a faster rate. c All real x

e As x gets larger y = 3x becomes larger while y = 3−x approaches 0. 16 a 4 units b 2 c The exponential function

x 1

1168 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

17 a

f (x)

g(x)

c For n = 1, f (n) = g(n). For n ≥ 2, f (n) < g(n).

This is because g(n) = 3(2)n grows exponentially, while f (n) = 5n + 1 grows linearly. Therefore, f (n) is only greater than or equal to g(n) at n = 1.

24 a f (x) = 2x and g(x) = x2 + 5 b

b Calculations 1 and 4 show that the exponential function’s increase over a 1 unit interval of x is greater than the quadratic function’s increase over the same interval. Calculations 2 and 3 show that the gap between the functions becomes greater and greater.

y 1.5 1

g(x)

0.5

f (x) 0

0.5

1.5

c i

2.5

What do you remember?

d f (x) =

x increases at a constant rate.

g(x) =

15.02 Tangent gradient at y-intercept

decreases at an increasing rate

towards zero.

1 A line that touches the curve at one point and has the same gradient as the curve at that point. 2 At x = 0, y = a0 = 1, so the y-intercept is (0, 1). 3 The gradient is m =

18 a Sophia b s(x) = 4x, m(x) = 90x

4 a i

c Sophia will save $574 more than Matt over the course of 5 months.

Practice

b iv

5 a m ≈ 0.184 Extend your thinking

6 a m ≈ 1.623 20 a 5

c m ≈ 0.182

b m ≈ 1.515

c m ≈ 1.505

d The gradients converge to approximately 1.505, showing a reliable estimate of the tangent gradient.

21 2409.20 22 23 a

b m ≈ 0.182

d ii

d The gradients are consistent, indicating a stable approximation of the tangent gradient at approximately 0.182.

19 a

c iii

f (n)

g(n)

b The consecutive terms of f (n) are increasing linearly by 5, while the consecutive terms of g(n) are increasing by a factor of 2.

7 a m ≈ 0.590

b m ≈ 1.170

c m ≈ −0.223

d m ≈ 1.622

8 a m ≈ 0.960

b m ≈ 0.921

c m ≈ 0.917

d The gradient for y = 2.5x (m ≈ 0.921) is greater than for y = 2x (m ≈ 0.696), indicating a steeper tangent as the base increases. 9 a m ≈ 0.263

b m ≈ 1.396

c m ≈ 0.792

d m ≈ −0.105

e m ≈ 1.334

f m ≈ 0.532

Answers 1169 mathspace.co

10 a m ≈ 1.335

= kex

b m ≈ 1.261 c m ≈ 1.254 d The gradient for y = 3.5 (m ≈ 1.261) is greater than for y = 3x (m ≈ 1.105), indicating a steeper tangent as the base increases. 11 a m ≈ 0.833 c m ≈ 0.337

b m ≈ 1.436 d m ≈ −0.357

12 Ordering from smallest to largest:

(ax) = ax; for other a, the

4 Only for a = e does

derivative is kax where k = ln a. Practice 5 a e3

b e

6 a

b e2

c 1

d 2

−0.223, 0.406, 0.696, 1.105, 1.261, 1.396 7 a y=x+1

Corresponding to: y = 0.8x, 1.5x, 2x, 3x, 3.5x, 4x

b 4

c y = e x − 3e

d y = 3x + 3 − 3 ln 3

13 a m ≈ 0.471

b m ≈ 0.879

8 a First: 4e , Second: 4e

c m ≈ 1.201

d m ≈ −0.510

b First: ex, Second: ex

e m ≈ 1.581

f m ≈ 0.644

c First: 2ex, Second: 2ex

g m ≈ 1.070

h m ≈ −0.691

d First: ex + 1, Second: ex

e First: 5ex + 2, Second: 5ex

Extend your thinking 0

14 As h → 0, the secant line between (0, ka ) and (h, ka−h) for f (x) = ka−x approaches the tangent line at (0, k), so its gradient approaches the tangent’s gradient of −k ln(a). In contrast, the secant line between (0, ka0) and (h, kah) for f (x) = kax approaches the tangent line at (0, k), so its gradient approaches the tangent’s gradient of k ln(a). 15 a (0, k) b The gradient of the secant is . c m ≈ 1.391 d The tangent gradient is scaled by k, as the gradient of the secant linearly on k.

depends

9 a y = −x + 1 2

b m ≈ 0.696

c m ≈ 0.921

d m ≈ 1.105

15.03 Euler’s number and derivatives What do you remember?

c y = − e x − 2e + 10 a Derivative: y = 2ex

b Derivative: y = 5ex

d Derivative: y = 10ex

c Derivative: y = − e 11 a y = 3x + 5

(ex) = ex.

2 The derivative of y = ex is ex, meaning the function equals its own derivative.

b y=x

c y = 2x + 6

d y = 4x + 1

e y = 5x + 3

f y=x+1

g y = −3x − 2

h y = 3x + 2

12 a 2e + 1 = 3

b e1 − 3 ≈ −0.282

c 3e + 2 × 0 = 3

d e1 + 2 ≈ 4.718

e − e0 + 1 = 0

f 4e1 − 2 ≈ 8.872

c y = 0.62

b y = 2x + 0.62 d x = −0.31

Extend your thinking 14 For y = ax, derivative is ln a × ax. Set ln a × ax = ax, so ln a = 1, thus a = e. 15 a y = ex

1 Euler’s number e ≈ 2.71828182845... is an

d y = − e3(x + 3) + e−3

13 a (0.69, 2)

16 a m ≈ 0.406

irrational constant where

f First: ex − 2x, Second: ex − 2

b c

1170 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

4 3

b As t → ∞, I → ∞

( e2 + 1, 0)

c $11 025

6 a

1 y = ex

(0, 0) −4 −2 −1

2 4 6 8 10 12

−3 −4

16 24e

20 000 18 000

16 200 14 580

V 20 000 (0, 2000)

per hour

17 One point: (1, e)

15 000

15.04 Applications

10 000

What do you remember?

5000

1 a i y = 0 ii (0, 3)

iii 

V = 20 000(0.9t)

b i y = 0 ii (0, 5)

iii 

iv y > 0

iii 

iv y > 0

c V→0

d i y = 0 ii (0, 4)

iii 

iv y > 0

d $11 810

c True

d False

b False

3 A

V=0

iv y > 0

c i y = 0 ii (0, 2) 2 a True

Value ($)(V )

−2

0.15

Time ( years) (t)

7 a y-intercept: (0, 50). Asymptote: P = 0 b

Practice 100

4 a

P = 50(1.5t)

300

50 (0, 50)

N = 100(2t)

200

F=0 0

100

(0, 100) 1

8 a Domain: t ≥ 0 Range: 0 < M ≤ 100

5 a

I 12 000

8000

d 380 insects

b N→∞

10 000

c P→∞ t

N=0 0

I = 10 000(1.05t)

(0, 10 000)

M 100

(0, 100)

M = 100(0.8t)

6000

4000

2000

I=0 0

M=0 0

c 64 grams

Answers 1171 mathspace.co

9 a y-intercept: (0, 2000). Asymptote: R = 0 b

E 80 (0, 80)

R 2500

R = 2000(1.1t)

2000 (0, 2000)

1500

1000

c R→∞

Concentration (mg) (C)

500 475

d 35 percent

13 a y-intercept: (0, 150). Asymptote: N = 0 b

c E→0

d $2928

Time (hours) (t)

R=0 0

N 800 600

451.25 428.6875

N = 150(2.5t)

400

C 500 (0, 500)

200

(0, 150)

C = 500(0.95t)

N=0

400

c N→∞

300 200

d 5859 cells

Extend your thinking

100 0

c C→0

14 a Domain: t ≥ 0

C=0

Range: y ≥ 1000

b y→∞

d 5 hours

c 1221

11 a Domain: t ≥ 0

d k ≈ 0.139

Range: F ≥ 300

e Approximately 259

15 a y-intercept: (0, 200). Asymptote: M = 0

F 600

F = 300(1.4t)

M 200 (0, 200)

(1, 420)

400

150

(0, 300) 200

M = 200(e−0.05t)

100 t 0

c F→∞ 12 a

E=0

500

10 a

E = 80(0.85t)

M=0

d 2259 ﬁshes

Time (months) (t)

Efficiency (%) (E)

80 68

57.8

49.13

c 121 grams d 14 days e 22.1%

1172 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16 a Domain: t ≥ 0

10 a 2.8751

Range: B ≥ 4000 b

b 3.9120 d −2.0969

c 2.9957

11 a log7 (343) = 3

B = 4000(1.08t) (1, 4320) 4000 (0, 4000)

5000

d ln(x) =6

3000 2000

b log11 (121) = 2

12 a 9

b 1

c 0.01

d 64

13 a True

b True

c False

d True

Extend your thinking

1000 t 0

c $5039

d 3 years

2 a log2 (16) = 4 3 a True

c True b ln (5) = x d loga (x) = y

c log (0.1) = −1 b True

d 20.27 year

16 For a ≤ 0, ay may not be positive or real (e.g., (−1) y). For a = 1, 1 y = 1, so x ≠ 1 cannot hold. For x ≤ 0, no real y satisﬁes ay ≤ 0 for a > 0.

What do you remember? b True

b 13.52 years

c 10.14 years 15 0.0001 mol/L

15.05 Logarithms

1 a True

14 a 13.86 years

c True

d False

Practice

17 ln (ex) = x since ln (ex) = x ln (e) = x × 1 = x. eln(x) = x for x > 0 as ln (x) is the exponent such that eln(x) = x. They are inverses.

15.06 Exponential-logarithmic equivalence What do you remember?

4 a log3 (9) = 2

c ln (e) = 1

f logb (m) = n

c ln(e) = 1

e log4 (64) = 3

b 102 = 100

g ln(1) = 0

d 33 = 27

i log6 (36) = 2

j logb (m) = n

3 a

1 a True

5 a 62 = 36 c

2 a log3 (9) = 2

6 a 4

b 2

c 3

d −2

e 0

f 0

g −2

h 6

7 a i 2 and 3

ii 2.6990

b i 2 and 3

ii 2.3026

c i 1 and 2

ii 1.9956

d i −2 and −1

ii −1.3010

8 a e =1 c e5 = e5 9 a 1

b 10

b True

c True b log(1000) = 3

b ac = b

Practice

d 10−3 = 0.001

4 a x = 1.1761

b x = 2.3026

c x = 3.9120

d x = −1.3010

c e

d 0.001

5 a

b x=2

c x = 12

d x=3

Answers 1173 mathspace.co

e x=4

f x=5

h x=6

6 a 1.2

b 1.6

c 1.2

d 1.2

7 a 1.404

b 1.091

c 1.672

d 1.544

8 a 52 = 25 c 10−2 = 0.01 c

Practice 4 a 1

b 0

c −1

d 3

e 11

f −2

h 4

b e3 = e3

5 a log10 (21)

b log4 (4)

d 23 = 8

c log6 (125)

d log3 (3)

e log10 (10)

f log7 (81)

e a =b

f 7 =1

h 104 = 10 000

6 a log5 (6) + log5 ( y)

9 a x = 3.9069

b x = 2.7268

c 3 log4 (x) + log4 (z)

d 4 log10 ( p) − 3 log10 (q)

c x = 1.2091

d x = 1.6610

e log3 (2) + log3 (x)

f log6 ( y) − log6 (36)

g 4 log2 (a) + log2 (b)

h 2 log5 (m) − 5 log5 (n)

b x = ln(7)

10 a c 11 a 1.771

7 a log5 (50)

d b 2.182

c 5.314

d 1.091

12 a x = 0.7297

b x = 6.4650

c x = 2.3979

d x = 1.1826

b log8 (z) − log8 (64)

b log3 (1)

c log10 (40)

d log2 (27)

e log4 (4)

f log6 (12)

8 a log7 (x) + 2 log7 ( y)

b 3 log10 (a) − log10 (b)

c log3 (4) + log3 (z)

d 2 log5 (x) − 3 log5 ( y)

e 3 log2 (a) + 2 log2 (b) f log8 ( p) − 4 log8 (q) 9 a LHS:

Extend your thinking 13 60 dB

RHS:

−log4 (16) = −log4 (42)

15 x = ±2

= −2 Thus, the identity holds.

16 17 loga (b) is undeﬁned for b ≤ 0 or a = 1.

b LHS:

15.07 Logarithm laws and properties What do you remember? 1 a Product law c Quotient law 2 a 1

b 0

RHS: = −2

d Special property c −1

d 5

3 a • Product law: loga (xy) = loga (x) + loga ( y) • Quotient law: loga

−log4 (25) = −log5 (52)

b Quotient law

Thus, the identity holds. c LHS:

= loga (x) − loga ( y)

• Power law: loga (xn) = n loga (x)

• Inverse property 1: loga (ax) = x • Inverse property 2:

RHS: −log7 (49) = −log7 (72) = −2

• Identity property: loga (a) = 1

Thus, the identity holds.

• Zero property: loga (1) = 0 • Reciprocal property: loga

= −loga (x)

1174 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

d LHS:

b A logarithm that evaluates to an irrational number (a non-repeating, non-terminating decimal). c An expression where a variable appears in the exponent, such as ax.

RHS: −log2 (8) = −log2 (23)

d An expression that contains a logarithm, such as loga (x).

= −3 Thus, the identity holds.

4 a Rational

b Rational

b log4 (8)

c Irrational

d Irrational

c log5 (12.5)

d log3 (9)

5 a w = 121

e log2 (81)

f log7 (343)

10 a log6 (18)

b w=2

Practice 11 a

6 a log13 (130) ≈ 1.8976. The result is irrational. b log15 (15) = 1.0000. The result is rational. c log11 (36) ≈ 1.4945. The result is irrational.

12 a −3

b 5

f −r

e a

13 a loga(b2c)

c k

d log10 (1000) = 3.0000. The result is rational. e log17 (2) ≈ 0.2447. The result is irrational.

d −2

f log10 (9) − log10 (9) = 0. The result is rational. g log13 (676) ≈ 2.5404. The result is irrational. h log15 (90/7) ≈ 0.9431. The result is irrational

7 a w = 173

8 a

9 a w ≈ 1.7075

b w ≈ 2.1414

c w ≈ 0.8416 Extend your thinking

d w ≈ −0.8239

10 a x = 2

15 Ask your teacher for worked solutions.

11 a 1.9842

b 0.1878

12 a 2, 3

b −3, −2

Extend your thinking

15.08 Logarithm expressions and equations

13 30

14 Ask your teacher for worked solutions

What do you remember? 1 a True 2 a

b True

14 15 x = 8, x = 1

c True

d True

b w = am

3 a A logarithm that evaluates to a rational number (an integer or fraction).

16 x = 8, y = 8 17 t = 4.00

15.09 Logarithmic graphs What do you remember? 1 a (1, 0)

b x=0

c x>0

d y∈

Answers 1175 mathspace.co

2 a y = loga (x)

b (1, 0)

10 a

c y=x 3 A function’s y-intercept is the point where its graph crosses the y-axis, which occurs when x = 0. To support the explanation, we ﬁrst rewrite the general logarithmic function y = loga (x) into its equivalent exponential form: x = ay.

f (x)

0.07

0.13

x = −3

Now, we can test for the y-intercept by substituting x = 0 into this exponential form. This gives us the equation:

Since the equation has no solution for y when x = 0, the function never intersects the y-axis, and therefore it has no y-intercept. 4 a x=0

2 x

(4, 0) −2

0 = ay For any valid base a (a positive number not equal to 1), there is no real value of y that can satisfy this equation. A positive base raised to any power cannot be zero.

−2 f (x) = log7 (x + 3) − 1 −4

c Domain: x > −3, Range: y ∈  11 Range: [0, 2] y

b (1, 0)

(169, 2)

2 x=0

Practice 5 a True

b False

c True

6 Domain: x > 0, Range: y ∈  3 2 1 −1

(1, 0)

x=0

150

(1, 0) 2

100

x 4

12 a iv

b iii

c i

13 • Graph A: y = log4 (x)

• Graph B: y = log9 (x)

−2

• Graph C: y = log20 (x)

−3

14 h < 0 7 y = log13 (x) is closer to the x-axis because higher bases result in ﬂatter graphs.

15 a (2, 4) on y = 2x maps to (4, 2) on y = log2 (x). b (1, 3) on y = 3x maps to (3, 1) on y = log3 (x).

8 They are inverses, so composing them gives = x and loga (ax) = x. This swaps coordinates (x, y) to ( y, x), reﬂecting over y = x.

9 a x>0

d (2, 25) on y = 5x maps to (25, 2) on y = log5 (x)

b y∈

c x=0

d (1, 0)

maps to

1176 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16 a

4 3

They are reﬂections across y = x. The point (0, 1) on y = 5x maps to (1, 0) on y = log5 (x).

x=0

y=2

(0, 1) (1, 0) y = 0 x

−4 −3 −2 −1 −1 −2

y=x

−3

2 3 4

y=4

y = log2 (x)

−2

(−1, 4) 4

−3

They are reﬂection across the y = x.

(e, 1)

y=x

−3

y=0 x (1, 1 0) 2 3 4

−4 −3 −2 −1 −1

y = In(x)

y=x

(6, 1)

−2

They are reﬂections across y = x. The point (1, e) on y = ex maps to (e, 1) on y = ln(x). x=0 y

−2

y = log6 (x) They are reﬂection across the y = x. The point (1, 6) on y = 6x maps to (6, 1) on y = log6 (x).

2 1 (0, 1)

y = 5x

(1, 6)

y = 6x

−4

maps to (4, −1)

The point (−1, 4) on . on

(1, e)

2 1 (0, 1)

(4, −1)

−4

2 3 4

maps to (1, 0) on

The point (0, 1) on y = log2 (x). x=0

x 1

−2

y=x

They are reﬂection across the y = x.

−2

−4 −3 −2 −1 −1

−4

y = ex

y = log 1 ( x)

2 3 4

x=0

y = log 1 ( x)

−3

y = log4 (x)

−2

y=x

2 3 4

They are reﬂection across the y = x. The point (1, 4) on y = 4x maps to (4, 1) on y = log4 (x).

(1, 0)

−4 −3 −2 −1 −1

−3

−4

y=0

−4 −3 −2 −1 −1

They are reﬂection across the y = x. The point (0, 1) on y = 2x maps to (1, 0) on y = log2 (x).

1 (0, 1)

(4, 1)

y=x

(1, 4)

−4

(1, 0) y = 0 x

−3 −2 −1

−1

−2

y = log5 (x)

−3

Answers 1177 mathspace.co

y=x

(−1, 6) 6 y=x

6 4

x −2

−2

(6, −1)

x 2

−2

−4

They are reﬂection across the y = x. maps to (6, −1)

The point (−1, 6) on on

rapidly, and the logarithmic function grows slowly, with their graphs

17 a

being reﬂections across y = x.

y 3

d 10

2 1

y = 3x −2

−1

y=x

−2

y = log3 (x)

2 −2 −2

−3

The exponential function y = 3 grows rapidly as x increases, while its reﬂection, the logarithmic function y = log3 (x), grows slowly. y 8 6

y=8

grows

rapidly as x increases, while its inverse, , grows slowly

y = log8 (x) 2

The exponential function

18 a

−2

−4

y=x

−2

x 2

y=x x

−1

grows

The exponential function

y 2

(0, 1) (1, 0)

−4

−2

y=x

With a larger base, the exponential function y = 8x grows even more rapidly, and the logarithmic function y = log8 (x) grows more slowly.

−1

x 2

−1 −2

decreases,

also

decreases. They are inverses, so their graphs are reﬂections across y = x (e.g., (0, 1) maps to (1, 0)).

1178 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Extend your thinking

y 2

20 h = 0.5, k = 1

(0, 1) 1

21 t ≈ 12.34 x

(1, 0) −2

−1

y=x

1 −1

23 t ≈ 8.67

−2

24 A possible solution requires choosing simple functions that meet the criteria. Let’s choose f (x) = log7 (x). This means:

decreases,

22 h = 2, k = 1

also

decreases. They are reﬂections across y = x (e.g., (0, 1) maps to (1, 0)). c

y 2

These functions intersect as shown below. Translating g(x) up by 5 units would move the blue curve well above the black curve, eliminating any intersection points.

−1

y=x

For g(x), we need a function that is reﬂected both horizontally and vertically. Let’s choose g(x) = − log7 (3 − x). This function intersects f (x) twice. This means: • k2 = 0

(1, 0) −1

• k1 = 0

• h2 = 3

1 (0, 1) −2

• h1 = 0

−2

decreases,

also

decreases. They are reﬂections across y = x (e.g., (0, 1) maps to (1, 0)). d

x 1

−1

2 1

(0, 1)

−1

decreases,

also

decreases. They are reﬂections across y = x (e.g., (0, 1) maps to (1, 0)) 19 a Point (0, 10) on S = 10 × 2t maps to (10, 0) on t = log2

(0.38, −0.50)

26 Both y = ax and y = loga (x) are decreasing functions. They remain inverses, so points swap (e.g., (0, 1) on the exponential function maps to (1, 0) on the logarithmic function), ensuring their graphs are reﬂections over the line y = x.

−2

25 79.03 dB

−1

y=x

(1, 0) −2

(2.62, 0.50)

27 t = 3 28 a The function has the form y = loga (x). Substituting the point (4, 0.5) gives 0.5 = loga (4), so a0.5 = 4, which means a = 16. The equation is y = log16 (x). b y = 16x

Answers 1179 mathspace.co

c The point (4, 0.5) is on the logarithmic graph. Its reﬂection is (0.5, 4).

b f (n) increases arithmetically by 4. g(n) increases geometrically by a factor of 3.

Check this point in the exponential function: = 4. The point satisﬁes the y = 160.5 = equation, conﬁrming the inverse relationship.

c At n = 1, f (1) = g(1) = 6. For n ≥ 2, g(n) grows exponentially, exceeding f (n)’s linear growth (e.g., f (2) = 10, g(2) = 18). 8 a f (x) = 3x, g(x) = x2 + 10

Chapter 15 review

b f (x) increases by 162 from x = 4 to x = 5 (Calculation 1), while g(x) increases by 9 (Calculation 4). The difference f (x) − g(x) grows from 55 to 208 (Calculations 2, 3), showing faster exponential growth.

1 A 2 C 3 A 4 a Horizontal asymptote: y = 0

9 a m ≈ 0.414

y-intercept: (0, 4)

b m ≈ 0.406

Domain: x ∈ 

c m ≈ 0.406

Range: y > 0 b

d As h decreases, the secant gradient converges to approximately 0.406, closely approximating the tangent gradient at x = 0.

y 10 8

10 a m ≈ 0.644

b m ≈ 1.035

c m ≈ −0.512

d m ≈ 1.808

4 (0, 4)

11 a (0, k)

2 −2

5 a

−1

−3

−2

−1

y=0 2

c m ≈ 4.188

d The tangent gradient at x = 0 is k ln a. Increasing k proportionally increases the gradient; a negative k reverses its direction.

−1 −5 −25 −125

y b As x → ∞, y → −∞. As x → −∞, y → 0−.

Horizontal asymptote: y = 0 approached from below as x → −∞. c Domain: x ∈  Range: y < 0

12 For f (x) = kax, the tangent gradient at x = 0 is k ln a. For f (x) = ka−x, it is −k ln a, opposite in sign, reﬂecting a decreasing versus increasing function for a > 1. 13 a e−3

b e3

14 a y = x + 1

b y = e−2 x + 3e−2

c y = ex

d y = e3 x − 2e3

15 a y = e2(x − 1)

6 a (0, 1) b No, since 3x > 0 for all x ∈ .

d e2

c 1

c x∈ d y>0

16 No solution for t ≥ 0, as P ′(t) = 100e0.05t ≥ 100.

e 243

17 a

f m = 81 7 a

f (n)

g(n)

162

486

1180 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Time ( years) (t)

Value 25 000 21 250 18 062.5 15 353.13 ($)(V )

V 25 000 (0, 25 000)

M 300 (0, 300) 250

20 000

200

15 000

150

10 000

100

5000

V=0 1

c V → 0+

c 201

d $11 093

d 35

18 a

21 B

A 8000

M=0 0

(2, 8652.8)

(0, 8000)

22 C 23 C

6000 4000

24 a log4 (64) = 3

c ln(e2) = 2

2000

A=0 0

t 3

25 a 72 = 49

b 103 = 1000 d 26 = 64

b A→∞ c $9733

26 a 3

d 6

b 4

27 a 17.33 years

19 a y = 4−x

28 a 3.16 × 10

b Reﬂection across the y-axis. c 7

6 5 4 3

−3 −2 −1

1 −1

d −5

b 8.93 years mol/L

29 a x ≈ 2.9299

b x ≈ 2.4307

c x ≈ 1.3383

d x ≈ −0.3219

30 a x ≈ −0.2041

b x = 5.0000

31 a 2.096

b 1.302

c 2.113

d 3.807

33 a 1

b −2

c 4

d 5

f 0

32 x = ±2

y = 4x

−6

c 2

x 1

d (0, 1) e For y = 4x, as x → ∞, y → ∞. For y = x → ∞ y → 0+.

, as

34 a log2 (35)

b log3 (27)

c log5 (16)

d log4 (45)

35 a log3 (5) + log3 (a) c 5 log2 (x) + log2 ( y)

20 a y-intercept: (0, 300) Horizontal asymptote: M = 0

b log7 (z) − log7 (49) d 2 log(m) − 5 log( p)

Answers 1181 mathspace.co

37 158

16.01E Inverse functions

38 a log2 (80), irrational

b log3 (27) = 3, rational

39 a w = 127, rational

b w = 80, rational

40 a w = 3

41 a x = 80

b x = 513

42 t = 12 hours 43 a x > 0

b y∈

c x=0

d (1, 0)

44 • Asymptote: x = −4

What do you remember? 1 It reverses a function, mapping outputs to unique inputs. It exists for one-to-one functions only. 2 It swaps x- and y-coordinates of every point on the function’s graph. The resulting set of points forms the graph of the inverse function. 3 Swap x and y: x = 3y − 2.

• x-intercept: (−2, 0)

4 It’s not one-to-one

• y-intercept: (0, 1) • Domain: x > −4

Practice

• Range: y ∈ 

5 a (3, 1) y 3

1 (0, 1)

−4 −3 −2 −1 −1

x 2 3 4

8 a Has inverse function

y 10 8 6 (0, 6) 4 2 (6, 0)

−3

45 h = 2, k = 2

−10−8 −6 −4 −2 −2 −4 −6 −8 y = x −10

46 a Point (2, 100) on y = 10x maps to (100, 2) on y = log(x). b Point (1, e) on y = ex maps to (e, 1) on y = ln(x). c Point (2, 36) on y = 6x maps to (36, 2) on y = log6 (x). maps to (4, −1) on

b Not one-to-one

7 One-to-one and has an inverse

−2

d Point (−1, 4) on .

d (−12, −12)

6 a One-to-one

(−2, 0)

b (−2, 0)

c (5, −4)

2 4 6 8 10

b Has inverse function

47 The functions are inverses of each other. To find the inverse of y = ax, we swap the variables x and y to get x = ay . Rewriting this equation in its equivalent logarithmic form gives y = loga (x). The geometric effect of swapping the x and y coordinates of every point on a function’s graph is a reflection across the line y = x. Therefore, the graphs of y = ax and y = loga (x) are reflections. 48 a y = log3 (x) b y = 3x

c Ask your teacher for worked solutions.

1182 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

(−2, 9)

10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 −8 y = x −10

x 2 4 6 8 10

(9, −2)

c Does not have inverse function

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 y = x −10

(−9, 2)

10 8 6 4 2

3 1

−10−8 −6 −4 −2 −−2 −4 −6 −8 y = x −10

−1

2 3 4

Yes b

3 2 1

−4 −3 −2 −1 −1

x 2 4 6 8 10

−2 −3 −4

(2, −9) Yes c

3 2 1

−4 −3 −2 −1 −1

x 2 4 6 8 10

−2 −3 −4

No d 4

(1, 2)

2 1

(2, 1) 1

x 3

−4 −3 −2 −1 −1

−2 −3

−2

y=x

−4

1 −3 −2 −1

2 3 4

−2

y 3

−3

f Has inverse function

−4 −3 −2 −1 −1

2 4 6 8 10

e Does not have inverse function 10 8 6 4 2

−10−8 −6 −4 −2 −−2 −4 −6 −8 y = x −10

d Has inverse function

9 a

−4

−3

Answers 1183 mathspace.co

15 It’s strictly increasing (for example, x1 < x2 implies − k < − k), so it’s one-to-one and has an inverse.

3 2 1

−4 −3 −2 −1 −1

2 3 4

−2 −3 −4

16.02E Formal definition of an inverse function

Yes f

What do you remember?

3 2 1

−4 −3 −2 −1 −1

2 3 4

−2 −3 −4

No 10 (1, 1) → (1, 1), (2, 8) → (8, 2); matches g( x), so they are inverses. 11 The point (2, 3) reflects to (3, 2), while the point (−2, 3) reflects to (3, −2). Since two different points on the original graph reflect to points with the same x-coordinate, the inverse is not a function. Therefore, f ( x) is not one-to-one. Extend your thinking 12 Not one-to-one as f (1) = f (−1) = 1. 13 One-to-one as different inputs give distinct outputs, for example, 21 ≠ 22, so it has an inverse. 14

f ( x) =

1 It means that f ( f −1( x)) = x holds for all x in the domain of f −1, and f −1( f ( x)) = x holds for all x in the domain of f, ensuring they “undo” each other. 2 A function must be one-to-one so that each output value in its range corresponds to exactly one input value in its domain. If a function is many-to-one, its inverse relation would map a single input to multiple outputs, which violates the definition of a function. Consequently, the formal inverse conditions cannot be satisfied. 3 The conditions f ( f −1( x)) = x and f −1( f ( x)) = x mean that if a point (a, b) lies on the graph of f (i.e., b = f (a)), then the point (b, a) must lie on the graph of f −1 (since a = f −1(b)). The geometric transformation that swaps coordinates from (a, b) to (b, a) is a reflection across the line y = x. Practice 4 a Check f ( f −1( x)):

−2 x − 1 2 x−3 1

−4 −3 −2 −1 −1 −2

−3 f

y=x

16 For all a ≠ 0, it maps x and −x to the same output (for example, f (1) = f (−1) = a), so it’s not one-to-one and has no inverse. If a = 0, f ( x) = 0, also not one-to-one. Therefore, f ( x) = ax2 has no inverse for any real value of a.

−4

x 1 −1

2 3 4

( x) =

Check f −1( f ( x)):

3x − 1 x+2 Both conditions hold, so f −1( x) is the inverse.

1184 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Check f ( f −1( x)):

6 Proposed inverse: f −1 ( x) = (adds 3, divides by 2). Domain of f ( x) is x ∈ (−∞, ∞), range is y ∈ (−∞, ∞). Inverse domain is x ∈ (−∞, ∞).

f ( f −1( x)):

−1

Check f ( f ( x)):

−1

Both conditions hold, so f ( x) is the inverse.

f −1( f ( x)):

−1

5 a Verify: f ( f ( x)):

Both conditions hold, so they are inverses. 7 At x = e: f ( f −1(e)) = f (ln e + 1) = e2 − 1 = e. At x = 1: f −1( f (1)) = f −1(e1 − 1) = ln 1 + 1 = 1. Both satisfy the deﬁnition. f −1( f ( x)):

8 Verify f ( f −1( x)): f ( f −1( x)) = f (ex − 4)

= ln((ex − 4) + 4) = ln(ex) =x Verify f −1( f ( x)): f −1( f ( x)) = eln( x + 4) − 4 = ( x + 4) − 4

Both equal x, so it’s the inverse. b Verify: f ( f −1( x)):

=x Both satisfy the deﬁnition. = x, holds for x ≥ 0.

9 f ( g( x)) = g( f ( x)) = inverses.

= ∣x∣ ≠ x for x < 0. Fails, not

10 We must verify both compositions for all applicable domain values. Verification of f ( f −1( x)) = x f −1( f ( x)):

For x < 0: f ( f −1( x)) = f ( x + 2) = ( x + 2) − 2

Using the first piece of f ( x)

=x Verification of f −1( f ( x)) = x For x < 2: Both equal x, so it’s the inverse.

f −1( f ( x)) = f −1( x − 2) Using the first piece of f −1( x) = ( x − 2) + 2

Answers 1185 mathspace.co

For x ≥ 0:

13 Let f ( x) = mx + b:

Using the second piece of f ( x)

So, f −1( x) = For x ≥ 2:

, y-intercept Using the second piece of

f −1( x)

Since both f ( f −1( x)) = x and f −1( f ( x)) = x hold for all cases, the functions are inverses.

, a linear function .

Verify: f ( f −1( x)):

f −1( f ( x)):

11 For f ( f −1( x)):

Since x ≥ 1

Both hold, proving the inverse is linear.

−1

For f ( f ( x)):

14 Compute f ( g( x)): f ( g( x)) = f (2x − 4) = Since x ≥ 0

Both satisfy the deﬁnition.

12 Domain and range of f ( x) are x ∈ (−∞, 0) ∪ (0, ∞). If f

Inverse: y=x−

⟹ x = y + , so inverse is x + .

Compute g−1 ( x) =

Extend your thinking −1

( x) = :

−1

f ( f ( x)):

f −1( f ( x)):

Both equal x for x ≠ 0, so f ( x) = f −1( x).

=x− .

, f −1( x) = 2x − 1.

Then g−1( f −1( x)) = g−1(2x − 1) =

=x+ .

Both equal, satisfying the deﬁnition. 15 The function f ( x) = ∣x∣ fails to have an inverse over the domain of all real numbers because it is not one-to-one. For any x ≠ 0, two distinct inputs, x and −x, produce the same output. For example, f (2) = 2 and f (−2) = 2. An inverse function must map each output to a unique input, which is not possible here. A proposed restriction to make the function invertible is to limit the domain to x ≥ 0. On this domain, the function is f ( x) = x, which is one-to-one and has the inverse f −1 ( x) = x. (Another valid restriction is x ≤ 0, for which f ( x) = −x). 16 Ask your teacher for worked solutions.

1186 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16.03E Determine the equation of an inverse function

b i ii x is subtracted by 7 then divided by 3.

What do you remember? c i

1 1. Start with y = f ( x) 2. Interchange x and y to get x = f ( y)

ii x is subtracted by 1 then divided by 2 and raised to .

3. Solve for y in terms of x −1

d i f −1( x) = ( x − 5)2, domain: x ≥ 5

4. Express the result as f ( x) = y 2 a True

b False

c True

ii x is subtracted by 5 then squared. e i

3 (b, a) 4 The function must be one-to-one (each x maps to a unique y, and each y maps back to a unique x).

ii x is added to 2 then divided by −4. f i ii x is added to 6 then raised to . g i f −1( x) = x2 + 2, domain: x ≥ 0

Practice

ii x is squared then added to 2.

5 a

9 f −1( x) = x2 − 4, x ≥ 0

10 f ( x) = , domain: x ≠ 0

h i

6 a f −1( x) = ex − 1 c

d f −1( x) = ln( x − 4)

ii x is subtracted by 3 then divided by 5.

11 f −1 ( x) =

or f −1( x) =

12 f −1( x) = x + 1 e

13 a = 2, f −1( x) = , domain: x ≠ 0

7 a i f ( g( x)) = x ii g( f ( x)) = x

iii Yes b i f ( g( x)) = ln (ex − 2 + 2) ii g( f ( x)) = eln( x + 2) − 2

15 a f −1 ( x) = b f −1( x) = 2 −

iii No c i f ( g( x)) = x ii g( f ( x)) = x

Extend your thinking

iii Yes

16 k = 3

d i f ( g( x)) = ii g( f ( x)) =

17 f ( x) = 18 f −1(3) = 2

iii No 19 8 a i ii x is added to 4 then raised to .

Answers 1187 mathspace.co

16.04E Graphs of functions and their inverse functions

y 15 10

What do you remember? 1 The two functions are symmetric over y = x, meaning that g( x) is an inverse of f ( x). 2 a True

b False

−15

3 The range of f −1( x) is the domain of f ( x).

4 a

f (x)

3 2

f −1(x)

1 −4 −3 −2 −1 −1

2 3 4

The graph of f ( x) = 2ex + 3 is a sharp exponential curve, crossing the y-axis at 2e3(≈ 40.2). Its inverse is a gradual logarithmic curve with a vertical asymptote at the line x = 0 and a domain defined for x > 0. f ( x) increases much faster than its inverse. d

4 2

−3

f (x)

−4

−4 −3 −2 −1 −1

The graph of f ( x) = 4x + 2 is a steep upward line, crossing the y-axis at 2 and the x-axis at

−2

x 1

2 3 4

f −1(x)

−3

. Its inverse is a ﬂatter line, crossing the

−4

and the x-axis at 2. f ( x) rises The graph of f ( x) = x5 + 2 is a rapidly steepening fifth-degree curve, crossing the y-axis at 2. Its inverse is a flatter fifth-root and the curve, crossing the y-axis at x-axis at 2. f ( x) rises faster than its inverse.

faster than its inverse. b

−2

y-axis at

−15 −10 −5 5 10 15 −5 f −1(x) −10

c True

Practice

f (x)

3 f (x)

f −1(x)

1 −4 −3 −2 −1 −1

2 3 4

−2 −3

x −6−5−4−3−2−1 −1 1 2 3 4 5 6 −2 f (x) −3 −4 −5 −6

−4

The graph of f ( x) = x3 − 1 is a steepening cubic curve, crossing the y-axis at −1 and the x-axis at 1. Its inverse is a ﬂatter cubic curve, crossing the y-axis at 1 and the x-axis at −1. f ( x) grows more rapidly than its inverse.

6 y 5 4 3 2 f −1(x) 1

The graph of f ( x) = e2x − 4 is a steep exponential curve, crossing the y-axis at −3 and the x-axis at

(≈ 0.693). Its inverse is a

slower logarithmic curve, crossing the y-axis at and deﬁned for x > −4. f ( x) grows faster than its inverse.

1188 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

5 4 3 2 f (x) 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Extend your thinking

9 f (3) = 5 x 1 2 3 4 5

f −1(x)

The domain of f −1( x), ( x > 2) is the range of f ( x), and the range of f −1( x), ( y > −1) is the domain of f ( x). Since f −1(5) = 3, f (3) = 5 follows from the inverse property. 10 f ( x) = ln( x) − 2 (inverse of f −1( x)).

The graph of f ( x) = ( x − 1)3 + 3 is a steepening cubic curve, crossing the y-axis at 2 and the x-axis at 1 − . Its inverse is a ﬂatter cubic-root and the curve, crossing the y-axis at 1 − x-axis at 2. f ( x) increases more rapidly than its inverse.

5 f −1(x) 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 a Domain: x ≥ 1, Range: y ≥ 0

1 2 3 4 5

f (x)

b Domain: x ≥ 0, Range: y ≥ 1 c f −1(5) = 26 6

5 4 3 2 −1 f (x) 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

11 Point: (4, 0). If f ( x) was defined for x ≥ 0, its range would be y ≥ 4 because it is increasing from f (0) = 4. Consequently, the domain of f −1( x) would be x ≥ 4 and its range would be y ≥ 0.

1 2 3 4 5

f (x) f −1 (2) =

7 a f ( x): Domain: all real numbers, Range: y > 0 f −1( x): Domain: x > 0, Range: all real numbers. −1

b f (e ) = 3 8

5 4 3 f −1(x) 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

is the solution to f ( x) = 2.

Plotting these points and the graphs using software shows the intersection points are reﬂections in y = x:

y 6 4

f (x)

2 −6 −4 −2 −2 f −1(x) x 1 2 3 4 5

x 2

−4 −6

f (x)

Answers 1189 mathspace.co

16.05E The horizontal line test

c Yes

What do you remember?

−4 −3 −2 −1 −1

1 A one-to-one function has each output linked to exactly one input. It must be one-to-one for an inverse to exist because the inverse must uniquely map outputs back to inputs, forming a valid function. 2 The horizontal line test checks if any horizontal line intersects the graph more than once. Reﬂection over y = x turns horizontal lines into vertical lines, so if the original graph passes the test (one intersection), the reﬂection is a function, indicating an inverse exists.

y = −3

y=5

7 5 4 3

y=1

2 1

y = 0.5

2 4 6 8

−4 −3 −2 −1 −1 −2 −3

−8

−4

b No

−8 −6 −4 −2

2 3 4

−4

y=6

d No

18 16 14 12 10 8 6 4 2

2 3 4

−7

−6

−5

e Yes

−8 −6 −4 −2 −2

2 3 4

−3

−4 −3 −2 −1 −1

−2

−6

Practice 4 a Yes

2 3 4

−4

3 It means the function is many-to-one, with multiple inputs producing the same output, so its inverse over the natural domain is not a function.

f No

y=2

3 2 1

−4 −3 −2 −1 −1 −2

x 2 4 6 8

1190 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−3 −4

5 a Yes y=3

6 a No 4

y 7

2 1

y=4

−4 −3 −2 −1 −1

2 3 4

5 4 3

−2

−3

−4

b Yes

y=2

−1

b No

y 4

−4 −3 −2 −1 −1

y=1

2 3 4

−2

−4 −3 −2 −1

−3

−1

−4

−2

c Yes

c No

y=2

−4 −3 −2 −1 −1

y = −2

2 3 4

−1

−2 −3

−2

−4

−3

−4

d Yes

d No

y=1

−4 −3 −2 −1 −1 −2 −3 −4

y 7

x 1

2 3 4

5 4

y=3

3 2 1

x 1

Answers 1191 mathspace.co

7 a Yes

y 6 1

5 4

y=2

−4 −3 −2 −1

2 1 1

2 3 4

b Possible table of values:

b Yes

y 8 6

−2

−1

f ( x)

−30

No y

2 −2

−1

−2

−4 −3 −2 −1

2 3 4

−10

−4

−20 −30

c Yes

y 4

9 a No, the function is not one-to-one and does not have an inverse. The derivative, f ′( x) = ex − 1, indicates that the function decreases for x < 0 and increases for x > 0. Because it is not strictly monotonic, it fails the horizontal line test.

y=2

2 1 x

−2 −1

2 3 4 5 6

y=2

d Yes

x 1

2 3 4

−2

−4 −3 −2 −1 −1

3 2

−4 −3 −2 −1 −1

y=1

−3

x 1

−4

2 3 4

−2 −3 −4

8 a Possible table of values:

2 3 4

−4 −3 −2 −1

y=3

−2

−1

f ( x)

1.6

0.7

b No, the function is not one-to-one and does not have an inverse. The graph is an upward-opening parabola which decreases for x < −0.5 and increases for x > −0.5. Because it is not strictly monotonic, it fails the horizontal line test.

1192 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b No, it doesn’t have an inverse. The graph is symmetric about the y-axis.

2 1

y=0

4 3

−4 −3 −2 −1 −1

2 3 4

−2

−4 −3 −2 −1 −1

−3

y = −1

−4

2 3 4

−2 −3 −4

10 a Both are one-to-one. Both graphs are increasing exponentials.

12 a One-to-one

8 7

y=2

6 5

−4 −3 −2 −1 −1

−2 1

2 3 4

b f ( x) is one-to-one, g( x) is not. f ( x) increases steadily, g( x) does not. y

2 x 1

2 3 4

−4

11 a Yes, it has an inverse. The graph is a decreasing curve. y

3 2 1

y = −1

−2 −3 −4

−3

f (x) = x2, x≥0

3 2

−4 −3 −2 −1 −1

x 1

2 3 4

y = −2

f (x) = − x2, −3 −4 x<0

−3

−4 −3 −2 −1 −1

y = −2

−4

−2

2 3 4

−4 −3 −2 −1 −1

x 1

b One-to-one y=2

y=0

f (x) = x3, x<0

−4 −3 −2 −1

f (x) = x, x≥0

3 1

y=3

x 1

2 3 4

Extend your thinking 13 a x3 is one-to-one funtion. Applying exponential to x3 essentially shifts all values up and creates new horizontal asymptote at y = 0, but keeps the one-to-one shape, hence passes the horizontal line test. b Applying squared function to x3 increases positive values, but reﬂects the negative values in x-axis so it becomes symmetric in the y-axis, hence it is not one-to-one and fails the horizontal line test.

Answers 1193 mathspace.co

14 A linear function with non-zero slope (for example, f ( x) = 5x − 2, slope 5) is either strictly increasing or decreasing. For x1 < x2, 5x1 − 2 < 5x2 − 2, so it’s one-to-one. A horizontal line like y = 3 intersects once, conﬁrming it has an inverse. 15 No, symmetry about the y-axis means f ( x) = f (−x), making it many-to-one. For f ( x) = x4 − 2, y = 2 intersects at two points , failing the horizontal line test, so no inverse exists. 16 A strictly monotonic function is a function that is always increasing or always decreasing over its entire domain. If a function is strictly increasing, for any two inputs x1 < x2, it must be true that f ( x1) < f ( x2). Similarly, if it is strictly decreasing, f ( x1) > f ( x2). In either case, two different inputs can never produce the same output. This guarantees the function is one-to-one and will therefore pass the horizontal line test. The function f ( x) = ln( x + 1) is an excellent example of a strictly monotonic function. Its domain is x > −1. As x increases throughout this domain, the value of ln( x + 1) also always increases. Because the function is strictly increasing, any horizontal line will intersect its graph at most once, confirming it is one-to-one and has an inverse.

2 Restricting the domain ensures each output corresponds to one input, making it one-to-one. Keeping the range the same ensures all possible outputs of the original function are preserved in the restricted version. 3 The horizontal line test identiﬁes many-to-one functions by showing multiple intersections. Restricting the domain to a region with one intersection makes it one-to-one. Reﬂection over y = x swaps inputs and outputs, requiring a single-valued inverse, which the restriction ensures. Practice 4 a Restricted domain: x ≥ −2 (right half). The unrestricted parabola has vertex (−2, 0), domain x ∈ (−∞, ∞), range y ∈ (−∞, 0], and is many-to-one (for example, f (−3) = f (−1) = −1). Restricting to x ≥ −2 keeps the range (−∞, 0] and makes it one-to-one (decreasing).

y −4

−3

−2

−1 −1 −2 −3 −4

y=1

−1 x

−1

−2

−3 −4

16.06E Domain restrictions b Restricted domain: x ∈ [0, 3] (right branch). What do you remember? 1 A many-to-one function has multiple inputs producing the same output. This prevents an inverse function from existing because the inverse would map one output to multiple inputs, making it not a function.

1194 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

Unrestricted domain is x ∈ [−3, 3], range y ∈ [0, 3], many-to-one due to symmetry. Restricting to x ∈ [0, 3] keeps the range [0, 3] and makes it one-to-one (decreasing).

10 a i

10 8 6 4 2

3 2

10 8 6 4 2

x 1

3 2

x 1

iv [0, ∞) b i

, domain x ∈ [−5, ∞). Original 5 a f −1( x) = range is y ∈ [−5, ∞) (vertex at (0, −5)). Inverse range is y ∈ [0, ∞), matching the restricted domain. b f −1( x) = 1 + x, domain x ∈ (−∞, 0]. Original range is y ∈ (−∞, 0] (vertex at (1, 0)). Inverse range is y ∈ (−∞, 1], matching the restricted domain.

10 8 6 4 2

b 7 Answers may vary, any interval contained within (−∞, 7] or [7, ∞). b [0, 4]

c [0, 4]

x 2 4 6 8 10

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

10 8 6 4 2

6 a

9 a b = −6

x 2 4 6 8 10

iii [3, ∞)

8 a

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

−3 −2 −1

x 2 4 6 8 10

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

−3 −2 −1

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

iii [−2, 5] iv [2, 9]

b c=7

Answers 1195 mathspace.co

c i

10 8 6 4 2

25 20

x 2 4 6 8 10

15 10 5 x 0

x 0

iv [6, ∞) 10 8 6 4 2

10 8 6 4 2 −10 −8 −6 −4 −2 −2 −4 −6 −8 −10

20 25 30

iii [7, ∞)

iv [0, ∞) Extend your thinking x 2 4 6 8 10

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

iii (0, 2]

x 2 4 6 8 10

iii [−2, ∞)

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

d i

y 30

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

e i

11 For f ( x) = x4, natural domain is x ∈ (−∞, ∞), range y ∈ [0, ∞), many-to-one (for example, f (2) = f (−2) = 16). Restricting to x ≥ 0 keeps range [0, ∞) (includes 0 to inﬁnity) and makes it one-to-one (strictly increasing). Restricting to x ≥ 1 gives range [1, ∞), losing values from 0 to 1, so it doesn’t preserve the full range, though still one-to-one. 12 Yes, it can. Natural domain is x ∈ (−∞, ∞), range

x 2 4 6 8 10

, vertex at

, many-to-one

(for example, f (0) = f (−1) = 0). Restrict to x ≥ (right half). It’s one-to-one (increasing) and preserves range

. Sketch shows full

parabola, then right half from vertex.

iv [0, 4)

1196 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

16.07E Solve problems involving a function and its inverse function

y 4

f (x) = x2 + x

What do you remember?

1 Apply f −1 to both sides: x = f −1(k), assuming f is one-to-one, to ﬁnd the input x that produces the output k.

2 1 x −2

−1

1 − ,− 2 4

b For x ≥

: f −1( x) =

2 They intersect where f ( x) = f −1( x). For increasing functions, these points must lie on the line y = x, so the equation can be simplified to f ( x) = x. For decreasing functions, intersections can occur off the line y = x as well. Solve this equation algebraically to find the points.

For x ≤

: f −1( x) =

Practice

13 a No

b Yes

c No

d No

or x ≤

14 a x ≥

3 a x=2

c Case 1: x ≥

b (1, 1)

4 t=1

50 40 30 20 10

f −1(x)

5 a (1, 1) and (−1, −1)

b (3, 3)

6 F = −40 (and C = −40) x

−20 −10

−10 −20 −30 −40

Extend your thinking

f (x)

8 a Yes. In the given context, the function is one-to-one, so this function has an inverse that is a function.

Case 2: x ≤

The variable t represents time, and in the

50 40 30 20 10

f (x) −20 −10

−10 −20 −30 −40

context the domain of d is 0 ≤ t ≤

This domain only includes one side of the parabola, so the inverse will be a function. x 10

f −1(x)

b Zheng’s claim is correct. In the physical context of the problem, time t must be non-negative (t ≥ 0). The inverse function must therefore have a range of [0, ∞). The formula t(d) =

uses the principal

(positive) square root, which correctly produces a non-negative range, matching the requirement for time. c 4 seconds

Answers 1197 mathspace.co

9 a x = −1

9 8 7 6 5 4 3 2 1 −3

−2

−1

y = x. Points (0, 1.5) and (5, 3.5) on f ( x) reﬂect to (1.5, 0) and (3.5, 5) on f −1( x). 4 a 4 3 2 1

2 3 4

−2

−1

−4 −3 −2 −1 −1 −3 −4

b (0, 0)

Yes

3 1

x −1

1 −2

−4 −3 −2 −1 −1

−1

−2 −3

10 a Since f ( f −1( x)) = x, differentiate each side using the chain rule:

−4

Yes c

3 2 1

b g(3) =

Chapter 16 review

−2 −3

1 C

−4

2 A 3

9 y 8 7 6 (3.5, 5) 5 4 3 (5, 3.5) (0, 1.5) 2 1 x (1.5, 0) −4−3−2−1 −1 1 2 3 4 5 6 7 8 9 −2 −3 −4

f ( x) = 0.4x + 1.5 is linear with non-zero gradient, so it is one-to-one and has an inverse. The graph shows f ( x) and f −1( x) as reﬂections over

−4 −3 −2 −1 −1

No d

3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

1198 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

5 y = x3 is strictly increasing and one-to-one. The transformations (shift right by a, up by b) preserve this property, so g( x) is one-to-one and has an inverse.

Inverse is g−1( x) = bx + d, which is linear. Gradient is b and y-intercept is d.

6 For f ( f −1( x)) = x:

For f −1( f ( x)) = x: 10 a b c

Both conditions hold.

d k−1( x) = x2 + 1, x ≥ 0

7 For f ( f −1( x)) = x: f (ex + 5) = ln((ex + 5) − 5) = ln(ex)

11 a f −1( x) = ex − 4

−1

For f ( f ( x)) = x:

c h−1( x) = ln( x − 2), x > 2

f −1(ln( x − 5)) = eln( x − 5) + 5

d k−1( x) = 3 − ex

= ( x − 5) + 5 =x Both conditions hold.

12 a i ii x is added to 3 then divided by x − 2. b i f −1( x) = ln( x + 1) + 2, x > −1

ii x is added to 1, the natural logarithm is taken, then 2 is added. c i ii x is added to 4, raised to , then subtracted by 2. d i Since h(h(x)) = x, the function is its own inverse for x ≠ 0.

ii x is subtracted by 1, squared, subtracted by 3, then divided by 2.

9 Let y =

Interchange: x =

14 a Domain: x ≥ −5 Range: [−1, ∞) b Domain: [−1, ∞) Range: [−5, ∞) c 4

Answers 1199 mathspace.co

20 y 18 16 14 12 10 8 6 4 2 −4−2 −2 −4

20 a x ≥ 3

f −1(x) = ex − 1 + 3

b Unrestricted: Parabola with vertex (3, 1), domain x ∈ (−∞, ∞), range [1, ∞).

y=x

Restricted: Right half from x = 3, one-to-one, shown in purple.

9 8 7 6 5 4 3 2 1

2 4 6 8 10 12 14 16 1820

f (x) = ln (x − 3) + 1

16 10 17 a Yes, linear with non-zero gradient, passes HLT. b Yes, linear with non-zero gradient, passes HLT. c No, quadratic function has a vertex, so horizontal lines above the vertex intersect twice.

9 8 7 6 5 4 3 2 1

e Yes, logarithmic function is strictly increasing, passes HLT. f Y es, rational function is strictly decreasing on each branch, passes HLT.

(3, 1)

−1 −1

d Yes, exponential function is strictly increasing, passes HLT.

18 No, the parabola has a vertex at (−4, −2). The horizontal line y = −1 intersects the graph twice, as shown, failing HLT.

−1 −1

2 3 4 5 6 7

(3, 1)

3 2 1 −6 −5 −4 −3 −2 −1 −1

y = −1

x 1

−2 −3 −4

19 x3 is one-to-one. x3 + 2 preserves this. Squaring in ( x3 + 2)2 makes it not one-to-one, for example,

of f ( x).

. Range: [0, ∞), matching domain

22 Domain x ≥ −3: f ( x) = x + 3, strictly increasing, one-to-one, range [0, ∞). Domain x ≥ 0: Still x + 3, one-to-one, but range [3, ∞). 23 Vertex at x = 2.5, y = −0.25. Range: [−0.25, ∞). Restricting to x ≥ 2.5 or x ≤ 2.5 makes it one-to-one and preserves the range. 24 t = 2 25

26 a

b 10 m/s

27 g′(−3) = −1 where g( x) = f −1( x)

1200 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

17.01 Reflections in axes

c i y = (−x)3 + 2 = −x3 + 2 ii

What do you remember?

y 3

y = −x3 + 2

1 a Reﬂects the graph in the y-axis.

b Reﬂects the graph in the x-axis. b (x, −y)

2 a (−x, y)

y = x3 + 2

3 a The function y = x2 is even, so f (−x) = (−x)2 = x2 = f (x), meaning the graph is symmetric about the y-axis and remains unchanged. b

−1

1 −1

d i y = −(−x) − 2 = x − 2

y 1

y = x2

−1

y = −x − 2

(0, 0) −2

−1

4 a False

−1

−3

b True

6 a i y = −(x2 + 2) = −x2 − 2

Practice

ii 2

y 3

5 a i y = (−x) − 1 = x − 1 ii

y=x−2

−2

y = x2 + 2

1 1

−1

y=x −1 x −1

x 1

−1

y = −x2 − 2

−2

−3

−1

b i y = −(−3x + 2) = 3x − 2 ii

y 3

b i y = 2(−x) + 1 = −2x + 1 ii

4 3 2 1 −1

−1

y = 2x + 1 y = 2x + 1 x 1

−1

−1 −2

y = −3x + 2 x 1

y = 3x − 2

−3

−2

Answers 1201 mathspace.co

c i y = −(x3 − 1) = −x3 + 1 ii

10 a (−2, 0) b

y = −x3 + 1

y 2

1 −2

−1 −1

b (3, 2)

y 3

y = −x + 3

−3 −2 −1

c y = −2x

d y = x2 − 3

8 a y = −4x + 1

b y = −x2 + 2

c y = −x3 − 1

d y=x−1

9 a (0, −3) 4

−1

3 (0, 3)

y = −2x + 3

−1

x 1

−2

−3 (0, −3)

−4

−1 −2 −3 −4

y = 2x − 3 2 −1

y 5 4 3 (0, 3.14) 2 y = 2π x + π 1 y = −2π x + π x

(3, −2)

−2

(−3, −2) b y = x2 + 1

1 −1

y=x−3

−3

−2

7 a y = −3x + 2

−1

(3, 2)

y=x−2

11 a (−3, −2)

d i y = −(x − 3) = −x + 3

−1

−2

y=x −1

y = −x − 2

−1

(2, 0)

(−2, 0)

x −1

y = x2 −2

y = −x2

1 −1

−1 −2 −3 −4

1202 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

(0, 0) 1

x 2

14 a y = −(x + 2)2

3 (0, 3)

y = (x + 2)2

−3

−2

−1

y = 2x + 3

3 −1

2 1

(−2, 0) −4

y = −2x − 3

−4

15 a y ≤ −1

b −∞ < y < ∞

16 a y = 3x2 + 2x + 1

b y = −2x3 − x + 1

17 (1, −4), no.

18 a Equation: y = x2 − 4. The x-intercepts (−2, 0) and (2, 0) remain unchanged because x2 is symmetric about the y-axis, so the graph is identical after reﬂection. y

−1

21 After y-axis reflection: y = x2 − 2x. After x-axis reflection: y = −x2 + 2x. Vertex: (1, 1). Original vertex: (−1, −1). The reflections map (−1, −1) to (1, −1) across y-axis, then (1, 1) across x-axis.

17.02 Horizontal and vertical translations What do you remember?

Extend your thinking

(−2, 0)

−1 −2 −4

−2

x 1

−3 (0, −3)

−1

y = −(x + 2)2 −3

−2

b (−2, 0)

(2, 0) 1

−1

y = x2 − 4

−2 −3

−4

1 a Shifts the graph 3 units to the right. b Shifts the graph 2 units up. c Shifts the graph 4 units to the left. d Shifts the graph 5 units down. 2 a True

b True

c False

d True

3 If a > 0, the graph shifts right by a units; if a < 0, it shifts left by a units. Practice 4 a y = (x− 4)2 − 2 b (4, −2) c

y 2

19 Fixed point: (0, 0). For y = −x3, at x = 0, y = 0, so (0, 0) maps to itself. This is because the reﬂection changes the sign of y, but at y = 0, the point is unchanged. 20 a Equation: y = −2x − 3. The slope changes from 2 to −2. Physically, this indicates the particle’s vertical motion reverses direction, e.g., from upward to downward relative to the reference frame.

1 2

y=x −2 −1

y = (x − 4)2 − 2

−1 −2

(0, −2)

(4, −2)

Answers 1203 mathspace.co

14 Horizontal translation 4 units right and vertical translation 7 units down.

5 a y = 3x + 7 b y = 3x + 7

y 7 (0, 7) 6

15 y = (x + 2)3 + 3 16 (0, 2)

5 4

Extend your thinking

y = 3x + 2

17 a Translations: 3 units right and 7 units up. New maximum proﬁt at vertex (5, 11) is 11.

2 (0, 2) 1

−1

b Original: P (x) = −x2 + 4x, vertex (2, 4). Translated: P (x) = −(x − 3)2 + 4x − 5, vertex (5, 11).

6 a y = (x − 5)2

b y = (x + 3)2

d y = x2 − 2

c y=x +4 7 a y = 2(x − 2) + 1

b y = 2(x + 4) + 1

c y = 2x + 4

P(x) = −(x − 3)2 + 4x − 5

d y = 2x − 4

8 a (2, −3)

b (0, −2)

9 a Horizontal translation 1 unit right and vertical translation 2 units up.

P(x) = −x2 + 4x

b Original: y = x + 2, vertex (0, 2). Translated: y = (x − 1)2 + 4, vertex (1, 4).

19 a Equation: y = (x − h)2 + k. Domain: . Range: [k, ∞).

y = (x − 1)2 + 4

y = x2 + 2

x 2

18 a = 3 and b = −2.

−4 −2

b Original: y = x2, vertex (0, 0). Translated: y = (x − 2)2 + 1, vertex (2, 1).

(1, 4)

(0, 2)

−2

x 2

y = (x − 2)2 + 1

y = x2

10 a Translations: 2 units left and 1 unit down. New point: (−2, −1). b

y −1

2 y=x

y = (x + 2)3 − 1 −3

−2

(−2, −1)

x 1

−1

x 3

(0, 0)

−1

(2, 1)

(0, 0)

20 (−4, 3) 21 (−2, −8)

17.03 Dilations

−2

What do you remember? 11 (−1, 0) and (1, 0) 12 y = 2(x − 3) + 3 2

13 y = (x − 3) + 5

1 a A stretch from the y-axis, scaling x-coordinates. b A stretch from the x-axis, scaling y-coordinates.

1204 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

c Horizontal dilation by a factor of 2.

6 a y = x2 + 4

d Vertical dilation by a factor of 3. 2 a False

b False

c True

y 6

d False f (x) = x2 + 4

3 The graph compresses vertically, becoming ﬂatter, as y-coordinates are halved.

5 4 3

4 The graph enlarges, with x- and y-coordinates doubled, appearing stretched outward.

f (x) = 4x2 + 1

Practice

−2

(0, 4)

2 1

(0, 1)

−1

5 a y = 4x2 − 4 7 a Horizontal dilation by a factor of k = 3.

b Original function: y = x2

f (x) = 4x − 4

Transformed function:

x −2

f (x) = x2 − 4

8 a Combined dilation by a factor of k = 0.4. b Original function: y = x3 − 1

−2

Transformed function:

−4

9 a Vertical dilation by a factor of 1.5.

b The graph stretches vertically from the x-axis, increasing the slope’s magnitude to −3 and shifting the y-intercept to (0, 6).

b y = 2x − 8 y 4

f (x) = x2 − 4 −4

−2

f (x) = 2x2 − 8

−2

x 2

−4

10 a (2, 1)

b (1, 3)

c (0.5, 0.5)

d (2, 2)

11 a (2, 2)

−6

12 a

−8

c Horizontal dilation (k = 0.5): Compresses horizontally from the y-axis, narrowing the parabola. For example, the x-intercept at (2, 0) on the original graph is mapped to the point (1, 0) on the transformed graph.

12 10 8 (0, 7.5)

f (x) = x2 + 3

Vertical dilation (l = 2): Stretches vertically from the x-axis, steepening the parabola. Points like (0, −8) correspond to (0, −4) on f (x) = x2 − 4.

6 4 2 (0, 3) x

−6

−4

−2

Answers 1205 mathspace.co

13 a f (x) = −x + 6.25 b

f (x) = x2

y 6 (0, 6.25) 5

f (x) = −x + 6.25

(0, 5)

2 f (x) = −x + 5 −3 −2 −1

−1

17 Sequential dilations: • Horizontal by k = 2 gives

Extend your thinking

vertical by l = 2 gives

14 a

• Combined dilation by k = 2 gives

b (4, 12) c A combined dilation by k = 6 would enlarge the logo excessively, scaling both dimensions by 6, making it too large for the billboard’s proportions. d

, then

f (x) = x2

They are the same because the sequential dilations effectively scale x by 2 and y by 2, equivalent to a combined dilation by k = 2.

17.04 Exponential and logarithmic functions

What do you remember?

10 8

1 A

2 a y = −2

3 a Domain: (−∞, ∞), Range: (−∞, 1)

2 −4

−2

x 2

15 Student A is correct. For a combined dilation by k = 0.5, the graph reduces (since 0 < k < 1), mapping (x, y) to (0.5x, 0.5y), so (1, 1) becomes (0.5, 0.5). Student B incorrectly assumes k = 0.5 enlarges the graph, confusing it with k > 1. 16 a

b x=3

c y=1

b Domain: (3, ∞), Range: (−∞, ∞) c Domain: (−∞, ∞), Range: (−1, ∞) d Domain: (−2, ∞), Range: (−∞, ∞) Practice 4 a y = 3x + 2 + 1 b y=1 c (0, 10) d

b (1.5, 1.5)

c A combined dilation by k = 0.5 would reduce the image (since 0 < k < 1), making it smaller, which is opposite to the goal of magniﬁcation.

(1, 28)

20 15 10 (0, 10) 5 x −3

1206 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−2

−1

d x=0

5 a y = 3−x

b (0, ∞)

c y=0

b x=2

c x=4

d x = −3

11 Transformations: Horizontal shift right by 2 units, vertical shift up by 1 unit

(−1, 3)

10 a x = −1

y 2 3

(0, 1)

x −3

−2

−1

(2, 2) (0, 1.25)

6 a y = log2 (x − 1) − 2 d

b (1, ∞)

c x=1

1 x 1

−1

12 Transformations: Reﬂection over x-axis, vertical dilation by 2, vertical shift down by 1 unit

(3, −1)

−2

x y

−1

(2, −2)

−1

−3

−2 −3 (0, −3)

7 a y=2×3

b (0, 2)

c 6

−4

(1, 6)

(1, −5)

−5 −6

−7

13 Domain: (−2, ∞)

2 (0, 2) 1

−1

(1, 0) 8 a y = 2 log3 x d

b (1, 0)

−2

c 2

(−1, −1)

x 2

−1

−2

2 1

−1

(1, 0) 1

x 2

−1 −2

9 a (0, 1)

1 , −2 3

b (0, −1)

d (0, 8)

Answers 1207 mathspace.co

Extend your thinking

Practice

14 P (t) = 1500 × 20.02t

4 a

1600

b Vertical asymptote: x = 3 (5, 1608)

1400 (0, 1500)

Horizontal asymptote: y = −4

1200

c Domain: (−∞, 2) ∪ (2, ∞)

1000

Range: (−∞, −4) ∪ (−4, ∞)

800

x-intercept:

600 400

y-intercept:

200

t 1

2 x−2

15 Equation: y = 6 × 2

d −1

x 1

−1 −2

(2, 7)

y 2

g(x)

−4

−5

−6

−7

x 1

x=3

−3

y = −4

5 a g(x) = 2x − 1 − 1

b (1, −1)

c [−1, ∞)

16 Equation: y = log3 (x − 2) + 3 − log3 (2)

d x-intercepts:

y y

4 3 2

x 2

−1

(1, −1)

−2

17.05 Reciprocal and absolute value functions What do you remember? b iv

c i

2 a Domain: (−∞, ∞), Range: [0, ∞) b (0, 0) 3 a x=0

1 −1

−1

1 a ii

g(x)

(4, 3)

d iii

6 a

7 a Vertical asymptote: x = −2 Horizontal asymptote: y = −3 b Vertical asymptote: x = 1

b y=0

Horizontal asymptote: y = 1 c Vertical asymptote: x = 4 Horizontal asymptote: y = 2

1208 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

d Vertical asymptote: x = −3 Horizontal asymptote: y = −1

Practice 5 a Centre at (3, 2), radius 4

8 a f (x) = x + 2 + 3

b f (x) = −2x

b Domain: [−1, 7], Range: [−2, 6]

c f (x) = x − 3 − 1

d f (x) = 3x + 2

b Vertex: (−1, −1)

Range: [2, ∞) Range: (−∞, −1]

9 a Vertex: (3, 2)

10 (1, 0) and

(3, 6)

(3, 2)

(−1, 2)

11 Domain: x ≠ −1

Range: y ≠ −3

12 a Vertical dilation by 3, horizontal shift right by 2 units, vertical shift up by 1 unit b Reﬂection over x-axis, horizontal shift left by 1 unit, vertical shift down by 2 units c Vertical dilation by 2, horizontal shift left by 3 units, vertical shift down by 1 unit

(7, 2) 6

(3, −2)

6 a (x + 1)2 + ( y − 4)2 = 9 b Centre at (−1, 4), radius 3 c

(−1, 7) y 6

d Reﬂection over x-axis, vertical dilation by 3, horizontal shift right by 1 unit, vertical shift up by 2 unit

(−4, 4)

(2, 4)

(−1, 4)

Extend your thinking

13 a

14 a

(−1, 1) 7 a (x + 1)2 + ( y − 3)2 = 4 b

(−1, 5)

15 a f (x) = 2x − 1 + 1 b Domain: (−∞, ∞), Range: [1, ∞) 16 a f (x) = 3x + 2 − 1

(−1, 3)

(1, 3)

17.06 Circles and translations (−1, 1) What do you remember?

−3

−2

1 x

−1

8 a i Centre at (4, 1)

2 Completing the square

ii Radius 5

3 Domain: [a − r, a + r], Range: [b − r, b + r]

iii

4 Constants deﬁning the circle’s position and size

y 5 4

(−3, 3)

b Domain: (−∞, ∞), Range: [−1, ∞)

1 B

−2

y 4 2

(4, 1) 2

x 8

−2

Answers 1209 mathspace.co

b i Centre at (−2, 3)

10 a i (x − 1)2 + ( y + 4)2 = 9

ii Radius 3

ii No x-intercepts

iii

y x −2

−2

(−2, 3)

(1, −4)

−4

2 x −4

−6

−2

c i Centre at (0, −5)

b i (x + 2)2 + y2 = 25

ii Radius 4

ii x-intercepts: (−7, 0), (3, 0)

iii

y −4

−2

x 2

iii

y 4

−2

(−7, 0)

−4

(0, −5)

−6

−4

(−2, 0)

(3, 0) x

−2

2 −2

−8

−4

d i Centre at (1, 0)

c i x2 + ( y − 2)2 = 1

ii Radius 2

ii No x-intercepts

iii

(1, 0) −1

x 2

2 (0, 2)

−1

1 x

−2

9 a ( x + 3)2 + ( y − 1)2 = 16, centre at (−3, 1), radius 4 b ( x − 4)2 + ( y + 2)2 = 4, centre at (4, −2), radius 2

−2

−1

d i (x − 3)2 + ( y + 1)2 = 16 ii x-intercepts:

1210 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

iii

b i Right by 1 unit, down by 4 units

ii (x − 1)2 + ( y − 1)2 = 1

c i Right by 4 units, up by 1 unit

x 2

ii (x − 1)2 + ( y − 1)2 = 1

d i Left by 3 units, up by 5 units

(3, −1)

−2

ii (x − 1)2 + ( y − 1)2 = 1

−4

Extend your thinking 14 (x − 1)2 + ( y − 2)2 = 25 y

11 a Domain: [0, 4], Range: [−1, 3] y

6 4

(2, 3)

(1, 2)

2 1 (0, 1)

(2, 1)

−4 −2 −2

(4, 1)

x 6

(3, −2)

−4

x −1

−6

(2, −1)

15 There are two possible circles that satisfy the given conditions:

b Domain: [−3, 3], Range: [−6, 0] y x

(0, 0) −2

(−3, −3)

−2

(0, −3)

(3, −3)

These correspond to centres on the line y = x:

−4

(0, −6)

12 a No x-intercepts

y-intercept(s) at (0, 3) b

(2, 5)

5 4

(0, 3)

(2, 3)

(2.82, 2.82)

(1, 2)

(4, 3)

(0.18, 0.18) −1

−1

(2, 1) 1

x 3

13 a i Right by 1 unit, up by 1 unit ii (x − 1)2 + ( y − 1)2 = 1

Answers 1211 mathspace.co

Practice

16 (x − 2)2 + ( y − 1)2 = 4 y

4 a g(x) = 3 (x − 2)2

b (2, 0)

c g(x) = 3 (x − 2)

e (2, 0)

(2, 1)

(4, 1)

g (6, 0)

x 1

h Vertical dilation then translation:

−1

g (x) = (x − 2)2

17 a (x − 3)2 + ( y − 2)2 = 100

5 g (x) = 3(x − 2)2 4

(3, 12)

f (x) = x

(−7, 2)

(3, 2)

−4

(13, 2) x

−2

(2, 0)

−1

x 3

Translation then vertical dilation: g (x) = (x − 2)2

−4

g (x) = 3(x − 2)2

(3, −8)

5 4

b 100π m2

f (x) = x

17.07 Order of transformations

What do you remember?

−2

(2, 0)

−1

Horizontal dilation then translation:

1 B

2 a Dilation or reﬂection

8 f (x) = x

b Horizontal translation c l

x 3

d l when l < 0

3 a For y = x2, a horizontal translation moves the vertex from (0, 0) to (a, 0). A subsequent reflection over the x-axis does not change this vertex, as it is on the axis of reflection. Conversely, reflecting first keeps the vertex at (0, 0), and then translating horizontally moves it to (a, 0), Therefore, for a horizontal translation and a vertical reflection, the final vertex is the same regardless of the order.

4 2 x

(2, 0)

−8

−6

−2

−4

Translation then horizontal dilation: y 8

g (x) = (x − 2)2

f (x) = x2

g( x) = 6

b Translation then reflection moves the vertex horizontally, then reflects it vertically; the vertex remains the same as reflection then translation for horizontal shifts.

x −2 3

4 2 x

(6, 0)

1212 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−4

−2

5 a b Vertical asymptote: x = 0 Horizontal asymptote: y = 3 c d Reﬂection then translation 5

y=3

d i

ii y = 3

12 a i g(x) = e−x + 2

ii y = 2

b i g(x) = e−x + 2

ii y = 2

13 a P (4, 16) b Q(4, 14.5)

2 1

x=0 −1

ii y = 3

Extend your thinking

−2

c i

−1

c The answers differ because the order of transformations affects the ﬁnal position of the point.

−2

14 a

Translation then reﬂection: 3

2 1 −2

−1

−2

x=0 1

−2 −3

−1

y = −3

−2

−4

6 a i g(x) = 2 × 2x + 1 or 2x + 2 b i g(x) = 2 × 2x + 1

ii y = 0 ii y = 0

c i

ii y = 0

d i

ii y = 0

7 a i g(x) = −x − 2

ii (0, −2)

b i g(x) = −x − 2

ii (0, −2)

8 a i g(x) = 3 log2 x + 1

ii x = 0

b i g(x) = 3 (log2 x + 1)

ii x = 0

c i g(x) =

ii x = 0

d i g(x) =

ii x = 0

b 1. Apply a horizontal dilation by a factor of 2 to f (x). This stretches the graph horizontally by a factor of 2, changing the .

function to

2. Apply a horizontal translation of 2 units to the left, which shifts the graph 1 unit to the

3. Reﬂect the function over the x-axis, which ﬂips the graph upside down, resulting in . y 2

−2

−1

ii (2, 0)

b i g(x) = −(x − 2)

ii (2, 0)

−2

10 a i g(x) = (2x − 2)2

ii (1, 0)

−3

b i g(x) = (2x − 1)

ii (0.5, 0)

11 a i

ii y = 3

b i

ii y = 6

f (x)

9 a i g(x) = −(x − 2)3

left, changing the function to

−1

x 1

g(x)

−4

Answers 1213 mathspace.co

15 a 1. Horizontal stretch by factor 3.

2. Shift right by 3.

3. Vertical stretch by factor 3.

4. Shift down by 3. −2 −1

b f (x) = 2

(2, −1)

−1

(3, −1.5)

17.08 Multiple transformations

(3, −1.5)

(0, −2)

−3

What do you remember?

1 C 2 The vertex, x- and y-intercepts, and asymptotes (if any). 3 a Vertical translation c Horizontal dilation

6 a b Vertical: x = 1, Horizontal: y = 0

b Horizontal translation

c (−∞, 1) ∪ (1, ∞)

d Vertical dilation

d (−∞, 0) ∪ (0, ∞) e

Practice 4 a Equation: g(x) = 2−x−2 + 1 b None

y=0 −2

−4

c Horizontal: y = 1 d (−∞, ∞) e • Range: (1, ∞) • Point at x = −2: (−2, 2) • Point at x = −4: (−4, 5)

8 6 4 2

−2 −4 −6 −8 −10

x=1 x 2

7 a g(x) = 2 log (3 (x − 1)) − 3

• y-intercept:

b x=1 y

(−4, 5)

(−2, 2) y=1 −5 −4 −3 −2 −1

5 a Equation: b Vertex: (2, −1) c Range: (−∞, −1]

3 2 1

2 1

c Range: (−∞, ∞) y

x=1 f (x) = log(3x)

2 3 4 1 −1 −2 −3 −4 −5 −6 g(x) = 2log (3(x − 1)) − 3 −7

x 1

8 a g(x) = 2x + 1 c No x-intercept(s) d (0,1)

d • Point at x = 1: (1, −3) • Point at x = 3: (3, −3) • y-intercept: (0, 2)

1214 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b (0, 1)

y 5 g(x) = 2| x | + 1 4

13 Equation:

2 1 (0, 1)

−4 −3 −2 −1 −1

x 2 3 4

9 a g(x) = −ex + 2

b (−∞, 2) y

f (x) = 2ex

3 2

y=2

−4 −3 −2 −1 −1

g (x) = −ex + 2

−4

b (−∞, ∞)

. Reversing

the order (translation ﬁrst, then dilation) produces the same result, so the transformations commute. In both cases, the vertical asymptote shifts to x = 3 and the horizontal asymptote remains y = 0.

3. Translate up by 2

Chapter 17 review

2 C

3 2

3 B

−5 −4 −3 −2 −1 −1

1 2 3 4

4 a (−4, −5) b (4, 5)

−2

−3 −4

b (−2, 1)

11 a 4

−2

gives the equation

1 A

−4

right by 2

16 g(x) = −(x− 1)2 + 2

10 a

(−2, 1)

translating the graph of

2. Vertically reﬂect and dilate by factor

−3

and then

15 1. Translate left by 1

2 3 4

−2

Domain: (−∞, −1) ∪ (−1, ∞) 14 Applying a vertical dilation by

c y=2 d

12 Equation: g(x) = 2 (x − 2)2 − 3 Range: [−3, ∞)

f (x) = | x |

Extend your thinking

2 1 −1

f (x) = x3

5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 (−4, −5) −5

(4, 5)

x 1

2 3 4

(4, −5)

x 2

−2 −3 −4

Answers 1215 mathspace.co

5 a y = −x3 + 4x

9 a y = (x + 3)2 − 1

b (0, 0), (2, 0), (−2, 0)

b (−3, −1)

y = x − 4x

y = x2 − 1

1 −2

−1

y = (x + 3)2 − 1

x 2

y = −x + 4x

−2

−4 −3 −2 −1

−3

y = (x − 3)2

−2 −3

y 3 (0, 3)

−1

10 a y = 2x − 1

b (3, 0) 4

(0, −1)

(−3, −1)

6 a y = −(x − 3)2 c

2 x

(3, 0) 1

y = 2x − 1

y = 2x + 3 1

−1

1 −1

y = − (x − 3)2

(0, −1)

−4

7 a Equation: y = x2 − 9. The x-intercepts (−3, 0) and (3, 0) remain unchanged. The function is even, so the graph is symmetric about the y-axis and is identical after reﬂection. b −3 −2 −1

y 4 3

(−3, 0)

11 a Horizontal translation 1 unit right and vertical translation 2 units up. The new point is (1, 2).

(3, 0) 1

−2

(1, 2)

y = x2 − 9

y = (x − 1)3 + 2

y = x3

(0, 0)

−4

−1

−6 −8

12 a The new maximum population is 13 (in thousands, etc.), occurring in month 6.

8 Ask your teacher for worked solutions.

1216 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

P(t)

(4, 16)

(6, 13)

12 10

17 a b (9, 4.5)

P(t) = −(t)2 + 8t

6 4

2 P(t) = −(t − 2) + 8(t − 2) 0

(3, 9)

y = x2

(9, 4.5)

13 a = 3, b = −3

14 a i

−8 −6 −4 −2

y 3

18 a y = 4x2

b (0.5, 1)

y = x2 − 1 1 x −3 −2 −1

1 −1

(0, −1)

b i y = 3 (x2 − 1) = 3x2 − 3 ii

•

3 2

y = x2 − 1 −2

−1

x 1

−1

(0, −1)

−2

−3 (0, −3)

The equations are different because the vertical dilation factor is different in each case. A combined dilation by factor k implies both horizontal and vertical dilations use the same factor k. 20 C 21 D

15 a y = 3x2 + 3 b

22 (0, 6)

23 a g(x) = log3 (x − 2) + 1

b Domain: (2, ∞), Asymptote: x = 2

y = 3x + 3 6

y = 9x2 + 1

y 3

(0, 3)

(0, 1) −1

c The statement is correct. A combined dilation by a factor of k = 0.25 is suitable for creating a smaller scale model. When 0 < k < 1, the transformation reduces the graph by scaling both x- and y-coordinates by that factor, bringing every point closer to the origin. 19 •

y = 3x − 3

2 4 6 8

(5, 2)

x 1

16 a Horizontal dilation by a factor of 3 and vertical dilation by a factor of 2.

1 x 1

Answers 1217 mathspace.co

24 Transformations: Reﬂection over the x-axis, vertical dilation by a factor of 3, and a vertical translation down by 1 unit. y −2

−1

y 3

x 1

(1, 4)

−2

x −2 −1

−4 (0, −4)

−1

−6

18.01E Reciprocal functions

−8

What do you remember? x

25 k = −2, a = 3. The equation is y = 3 × 3 − 2.

f ( x) = x − 3

−2

−5

−1

−4

−3

−2

−1

Undefined

b True

c False

26 Vertex: (−3, 1), Range: (−∞, 1] 27 Vertical asymptote: x = 5, Horizontal asymptote: y = −2 28 a b Domain: x ≠ 1, Range: y ≠ −3 29 y = 2x + 2 − 3 30 Centre at (−1, 5), radius 6 31 a (x − 2)2 + ( y + 3)2 = 9 b Centre at (2, −3), radius = 3 2

32 Equation: (x + 2) + ( y − 5) = 9 Domain: [−5, 1], Range: [2, 8]

2 a False

33 (x − 2)2 + ( y − 1)2 = 25 34 a

35 Ask your teacher for worked solutions.

3 a (−∞, 2) ∪ (2, ∞) b (−∞, 0) ∪ (0, ∞) c (−∞, ∞) ∪

d 36 a

∪ (0, ∞)

b (1, 4) Practice

c x-intercept: y-intercept:

4 a As x approaches ∞, f (x) = 2x – 5 approaches ∞ b g( x) approaches 0 c

approaches 0

approaches ∞

5 a As x approaches 2, f ( x) = −2x + 4 approaches 0 b As x approaches g( x) =

1218 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

from the right,

approaches ∞

c As x approaches 2 from the left,

8 a x=5

approaches ∞

b ∞

d As x approaches ,

approaches 0

2 1

6 a The solutions for the equation f ( x) = 0 are x = 2 and x = −2. For the reciprocal function y=

, these values correspond to vertical

−3 −4

tends towards positive or negative inﬁnity, resulting in vertical asymptotes at x = 2 and x = −2.

x=0

b As x approaches inﬁnity or negative inﬁnity,

−5

will approach −∞. As x

approaches 2 from the left, y =

−10

will

approach −∞. As x approaches 2 from the

4 2

will approach ∞.

−8 −6 −4 −2 −2

2 4 6 8

−4

−6

−8

−4 −3 −2 −1 −1

2 3 4

−2

−3

−4

7 a x = −3 and x = 1

−4 −3 −2 −1 −1

x 1

2 3 4

−2

b ∞ c

will

approach ∞. As x approaches −2 from the

5 −5

c As x approaches −2 from the left, y =

y = f (x) y = ( f (x))−1 x

approaches 0, indicating that y = 0 is the horizontal asymptote for the reciprocal

right, y =

f ( x) = x2 − 4 approaches inﬁnity. Thus,

right, y =

1 2 3 4 5 6 7 8 9

−2

asymptotes, as x approaches these values,

function y =

−4−3−2−1 −1

−3

−4

3 2 1 −6 −5 −4 −3 −2 −1 −1

x 1 2 3 4

−2 −3 −4

Answers 1219 mathspace.co

18.02E Reciprocal trigonometric functions

1 g( x) = 1 − 2 x2

3 2

What do you remember?

−4 −3 −2 −1 −1

−2

1 a sin θ =

2 3 4

2 a

−4

, tan θ =

b θ ≈ 16.26°

f(x) = 1 − 2x2

−3

, cos θ =

sin( x) −1

3 2 1

−4 −3 −2 −1 −1

2 3 4

−2 −3 −4

b x = −2π, −π, 0, π, 2π

2 1

−4 −3 −2 −1 −1

Practice

2 3 4

3 a

−2

−3

cosec ( x) −1

−4

15 C 0

Extend your thinking 16 a False

b True

c True

undefined 2

d False

17 18

y 10

5 x −6 −4 −2 −5 −10

undefined

b x = −2π, −π, 0, π, 2π The asymptotes for y = cosec ( x) are the vertical lines with equations x = −2π, −π, 0, π, 2π. c d 2π

1220 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

4 a i 2π

ii Domain: x ≠ nπ Range: (−∞, −1] ∪ [1, ∞)

sin( x)

iii x = nπ

b i 2π cosec ( x)

+ nπ

ii Domain: x ≠

Range: (−∞, −1] ∪ [1, ∞) iii x =

+ nπ

c i π ii Domain: x ≠ nπ Range: (−∞, ∞)

5 The function f ( x) = cosec( x) is defined as . Vertical asymptotes occur when the denominator of a rational function is zero. Therefore, f ( x) has vertical asymptotes where sin( x) = 0. This occurs for all values of x such that x = kπ, where k is an integer.

0 ii x = −π

iii

c 2π 12 a x = −2π, −π, 0, π, 2π b −1 ≤ f ( x) ≤ 1 c g( x) ∈ (−∞, −1] ∪ [1, ∞)

d e Period of f ( x): 2π Period of g( x): 2π

7 Period: 4

x=0

8 6 4 2

−2 −4 −6 −8

13 a x=4

b −1 ≤ f ( x) ≤ 1 x

c g ( x) ∈ (−∞, −1] ∪ [1, ∞)

x=2

d x = −2π, −π, 0, π, 2π e Period of f ( x): 2π Period of g ( x): 2π 14 a x =

8 a π b i x=π

cot ( x)

b x = −π, π

x = −2

Undefined

b i

Degrees: x = −90°, 90°

π y = csc  x  2  x = −4

11 a i x = 2π

iii x = nπ

6 a Radians: x =

tan( x)

b f ( x) ∈ 

c g ( x) ∈ 

d x = −2π, −π, 0, π, 2π c i 9

e Period of f ( x): π Period of g ( x): π ,

15 16 a True

, b True

, c False

d False

17 a b c No such intervals exist.

Answers 1221 mathspace.co

Extend your thinking

Practice

18 α =

,β=

19 For

4 y 3 2 1

, this expression is zero

when cos( x) = 0 and sin( x) ≠ 0. In [0, 2π ], cos( x) = 0 and sin( x) ≠ 0 at x =

has x-intercepts at x =

Thus, For

−4−3−2−1 −2 −3 −4 −5 −6 −7 −8 −9

, this expression cannot be equal to 0.

For it to be zero, the numerator (1) would have to be zero, which is impossible. Thus, no x−intercepts.

has

9 y 8 7 6 5 4 3 2 1

Therefore, the expressions are not equivalent for all x. The expression

is undefined at points

has its x-intercepts, precisely

where

because the domain of

−4−3−2−1 −2 −3 −4

excludes points

where tan(x) is undefined.

1 2 3 4 5 6 7 8 9

20 k = 2, m = 4 21 a = 3, b =

6 a False

,c=5

b True

18.03E Absolute value functions

What do you remember?

1 x

f ( x) = ∣x∣

f ( x) = x−3

f ( x) = ∣x − 3∣

f ( x) = 2∣x − 3∣ − 5

−2

−5

−1

−4

−3

−2

−1

−3

−5

x −1π − 1 π 2

b y = 1∣x∣ + 2

c y = 1∣x∣ − 2

d y = 1∣x − 1∣ − 3

1 π 2

−1

1π

−2

8 a

2 The graph of y = 4∣x − 3∣ + 2 can be obtained by starting with the graph of y = ∣x∣, then applying a vertical stretch by a factor of 4, shifting right by 3 units, and shifting up by 2 units. 3 a y = 1∣x − 2∣ + 0

c False

1222 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

9 8 7 6 5 4 3 2 1 −4−3−2 −1 −1 −2 −3

x 1 2 3 4 5 6 7

d False

f ( x) = −0.5x − 5

f ( x) = ∣−0.5x − 5∣

f ( x) = −0.5∣x∣ − 5

−2

−4

−6

−1

−4.5

4.5

−5.5

−5

−5.5

5.5

−5.5

−6

3 2 1

−5 −4 −3 −2 −1 −1

1 2 3 4 5

−2 −3 −4

9 a

Points like (−2, 4), (0, 5), (2, 6) lie on

y = ∣−0.5x − 5∣, and (−2, −6), (0, −5), (2, −6) on y = −0.5∣x∣ − 5, matching the plotted graphs.

8 6

y = ∣ f (x)∣

10 a

2 x −12 −9 −6 −3 −2

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

f (x) = −0.5x − 5 −4 −6 −8

8 6 4 2 x −4 −3 −2 −1 −2

−4

−6 −8

x 1

2 3 4

9 y 8 7 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4

2 3 4

f (x) = −0.5x − 5

y = f (∣x∣)

4 y 3 2 1

4 y 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

Answers 1223 mathspace.co

11 a

b 11

c C

15 a

−2

−1

f ( x) = ∣x∣

g( x) = 4∣x∣ − 1

−1

2 1

−4 −3 −2 −1 −1

2 3 4

−2

−3 −4

−3

−2

−1

y 12 10 8 6 4 2

3 2 1 −4 −3 −2 −1 −1

x 1

2 3 4

−4 −3 −2 −1−2 −4 −6 −8 −10 −12

−2 −3 −4

12 y = ∣x − 1∣ 13 a

9 y 8 7 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4

x 1

2 3 4

c The graph of f ( x) = ∣x∣ is transformed into g( x) = 4∣x∣ − 1 by ﬁrst applying a vertical stretch by a factor of 4, making the graph narrower as the y-values are scaled up. This is followed by a downward shift of 1 unit, moving the entire graph down and repositioning the vertex from (0, 0) to (0, −1).

x 1

d C

2 3 4

Extend your thinking 16 a

c f (−x) = ∣x∣ + 3

b A

Simplified Simplified Simplified expression expression expression in domain in domain in domain x≥6 −6 ≤ x < 6 x < −6

d The result of part (c) shows that f (−x) = f ( x), which veriﬁes that the function is symmetric about the y-axis. 14 a 18

16 14 12

∣x + 6∣

−x − 6

x+6

∣x − 6∣

−x + 6

x−6

∣x + 6∣ + ∣x − 6∣

−2x

10 8

b f ( x) =

6 4

c f (−6) = 12, f (6) = 12

2 −14−12−10−8 −6 −4 −2 −2

d B

2 4 6 8 10 12 14

−4

1224 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

18.04E Sum and difference of functions

y = |x − 1| + |x + 3| What do you remember? 5

y = |x + 3| −10

y = |x − 1|

−5

2 The difference ( f − g) ( x) represents the value obtained by subtracting the output of g( x) from f ( x) for a given x in their common domain.

−5

3 The value of x must belong to the common domain of both f ( x) and g( x), meaning it must be deﬁned for both functions.

−10

1 The sum ( f + g) ( x) represents the value obtained by adding the outputs of f ( x) and g( x) for a given x in their common domain.

y = 2|x − 5| + 1 Practice

−10

−5

y = −2|x − 5| + 1 −10

x = 2 sin( y) + 1 −5

iv (−∞, −1]

b i x3 − 8 +

x = 2 sin( −y) + 1

ii x2 + 10x + 9

iii [0, ∞)

iv [9, ∞) ii 8x + 10 − ∣x∣

iii 

iv 

5 a i

iii  \ {−7, 0}

iv  ii

b i

iv  ii

iii  \ {−9, 7}

iv  ii

d i

y = |tan(x) + 2| − 3

iii  \ {0, 8, −9}

iv 

6 A

7 a −5

iv 

c i x2 − 10x + 9

c i

−10

ii x3 − 8 −

iii  \ {0, 8}

−5

iii 

d i 8x + 10 + ∣x∣

−10

ii −x2 + 4x − 5

iii  \ {1}

−5

4 a i x2 + 4x – 1

f + g = 2x − 1 3

y = tan(x) − 1 −10

g(x)

−5

−4 −3 −2 −1 −1

f (x)

x 1

2 3 4

−2 −3 −4

Answers 1225 mathspace.co

g(x) y 4

f (x)

g(x)

f (x)

−4 −3 −2 −1 −1

2 3 4

f−g=x−2

−2

−8 −7 −6 −5 −4 −3 −2 −1 −1

−3

f + g = 2x2 + 4x − 2

−4

1 2 3

−2 −3

−4

g(x)

−4 −3 −2 −1 −1

2 3 4

−3

−2

−4

f + g = 5x − 5

−5

2 1 −2 −1 −1 −2 −3 −4 −5 −6 −7

f (x) 2

−2 −4

−6

g(x)

−2

f (x)

f − g = −x2 − x − 5

−6 −8

−10

x 1

2 3 4 5 6

9 a

g(x)

f (x)

f − g = −3x − 2

8 a

g(x)

f + g = x + x2

−5 −4 −3 −2 −1 −1

16 14 12 10 8 6 4 2

x 1

f (x)

2 3

f (x) g(x)

x 2

f + g = 3x2 + 2 y

−4 −3 −2 −1 −1 −2

3 2 1

−3 −4

g(x)

−2

f (x)

−4 −2 −2

f − g = 3x x

−4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 f (x) −5 f − g = −x2 + 2x + 2 −6

1226 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−3 −4

x 1

2 3 4

g(x)

f (x)

9 8 7 6 5 4 3 2 1

g(x)

3 2 1

−7−6−5−4−3−2−1 −1

x 1 2 3 4

−2 −3

f + g = x2 + 4x + 1

−4 −3 −2 −1

−4

d g(x)

7 y 6 5 4 3 2 1

g(x)

10 a

9 y 8 7 6 5 4 3 2 1

f (x)

g(x)

−4 −3 −2 −1 −1 1 2 −2 −3 −4 −5 −6 1 −7 f + g = x2 − −8 4 + x−2 −9

f (x) −4

−3

−2

9 8 7 6 5 4 3 2 1 −1 −1

−4 −3 −2 −1

11 a

f (x)

1 2 3 4 −1 1 2 −2 f + g = 2 x + 1 + x+1 −3 −4

4 3 2 1

f (x)

−4 −3 −2 −1−1 1 2 3 4 −2 −3 −4 f−5 − g = − 4x2 + 5 −6

−1 −2 −3 −4 −5 −6 −7 −8 −9

g(x) x 1

f (x)

9 y 8 7 6 5 2x + 1 4 f +g= x( x + 1) 3 2 x f (x) 1 −4 −3 −2 −1 −1 1 2 3 4 −2 g(x) −3 −4 −5 −6 −7 −8 −9

x 1

g(x) −2

−3 −4 −5 −6 −7 −8 −9

f − g = x2 −

1 x

Answers 1227 mathspace.co

12 a

9 y 8 7 6 5 4 3 −1 2 f −g= ( x − 1)( x − 2) 1 −4

−3

−2

−1 −1

g(x) x 1

9 y 8 7 6 5 4 3 2 1 −4

−3

−2

−1 −1

g(x)

f (x) −4

−3

−2

−1 −1

−2 −3 −4 −5 −6 −7 −8 −9

−1 −2 f + g = e + x − 3−3 −4

f (x)

−4 −3 −2 −1

x 3

x 2

f −g=−

3x − 1 x(2 x − 1)

1228 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

−1 −2 −3 −4

9 8 7 6 5 4 3 2 1

f (x)

g(x)

y 9 8 7 6 5 f − g = 2ex −4x − 1 3 2 1 f (x)

−4 −3 −2 −1

−2 5x + 3 −3 f + g = x ( x + 1) −4 −5 −6 −7 −8 −9 9 y 8 7 6 5 4 3 2 1

−4 −3 −2 −1

−2 −3 f (x) −4 −5 −6 −7 −8 −9

f (x)

9 8 7 6 5 4 3 2 1

g(x) x 2

g(x) x 1

−1 −2 x −3 f + g = e + 3x − 3 −4

14 y 13 12 11 10 9 8 f − g = 3ex − 2x + 5 7 6 5 4 3 f (x) 2 1 −4

−3

13 a

−2

−1

−1 −2 −3 −4

−4 −3 −2 −1 x 1

g(x)

14 a

g(x) y 4 3

f+g2 = x2 + 2x + 1 1 1

2 3 4

−1 −2 −3 −4

g(x)

9 y 8 7 f + g = In x + 2x 6 5 g(x) 4 3 2 f (x) 1 −1 −2 −3 −4

−4 −3 −2 −1 −1

y 9 f (x) 8 7 6 5 f − g = ex − x2 + 3x + 5 4 3 2 1 x

x 4

−2

f (x) b g(x)

f (x)

−3 −4 9 y 8 7 6 5 4 3 2 1

g(x) 4 3 2 1

3 2 1

f − g = In x − x + 2 f (x) 1

−1 −2

f (x)

−4 −3 −2 −1 1 2 3 4 −1 f + g = x2 + 4x − 2 −2 −3 −4 −5 −6

x 4

−4

g(x)

−3

−4 −3 −2 −1−1 1 2 3 4 −2 −3= ex − x2 + 2 f − g−4

14 13 12 11 10 9 8 7 6 5 4 3 2 1 −1 −2 −3 −4

f + g = In x + 3x + 3 g(x)

f (x) x 1

Answers 1229 mathspace.co

f (x)

−1 −2 −3 −4 −5 −6 −7 −8 −9

15 a

16 a

4 y 3 g(x) 2 1 2

f (x)

f − g = 2 In x − 4x + 2

−4 −3 −2 −1 −1

2 3 4

g(x)

−3

−4

4 y 3 2 1

g(x)

2 3 4

−2 −3

f (x)

−4 −3 −2 −1 −1

f − g = |x − 1| − (2 x + 1)

9 y 8 7 6 5 4 f (x) 3 2 1

g(x) x

−4 −3 −2 −1−1 1 2 3 4 −2 f + g = |2x +−3 3| + 3 x − 3 −4

9 y 8 7 f + g = In x + x2 − 4 6 5 g(x) 4 3 2 f (x) 1 −1 −2 −3 −4

g(x)

−4

4 y 3 2 1

f (x)

−1 −1 1 2 3 4 −2 f − g = In(x + 2) − x2 + 1 −3 −4 −5 −6 −7 −8 −9

−2

−1 −1 −2 −3 −4

f + g = |x| + x − 2

9 y 8 2 7 In(x + 1) + x f+g= 6 5 4 g(x) 3 2 f (x) 1

9 y

8 − 4x + 4 f − g = |x + 2|

f (x)

−4 −3 −2 −1−1 −2 −3 −4

17 a

g(x)

7 6 5 4 3 2 1

1 2 3 4 −1 −2 −3 −4 f (x) −5 −6 −7 2 −8 f − g = In x − x + 6x − 10 −9

1230 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

2 3 4

g(x) y

x 2 g(x) x

f2+ g = sin x + 1 −4 −3 −2 −1 f (x) −1

−2

−3 −4

2 3 4

4 3

Extend your thinking

f − g = 2 sin x − x + 1

18 a

2 1 −4 −3 −2 −1 −1

f (x)

x 1

2 3 4

−2

f (x)

g(x)

−3

−4 −3 −2

−4

9 y 8 7 6 5 4 g(x) 3 2 1 f + g = cos x + 3x −x 1 −4 −3 −2

−1 −1 −2 f (x) −3 −4 −5 −6 −7 −8 −9

9 y 8 7 6 5 4 3 2 1

−4 −3 −2

f (x)

f (x) −4 −3 −2

9 y 8 7 6 5 4 3 1 f + g = x−1 + 2 x 1 g(x) x −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

9 y 8 7 6 5 4 3 2 1

g(x)

−1 −1 1 2 3 4 −2 −3 −4 1 −5 f − g = x − x−2 −6 −7 −8 −9

g(x)

x −1 −1 1 2 3 4 −2 f − g = 2 cos x − 2x − 1 −3 −4 −5 −6 −7 −8 −9

9 y 8 7 6 5 4 2 f (x) 3 f + g = 2x + 1 + x 2 g(x) 1 x −4 −3 −2

−1 −1 −2 −3 −4 −5 −6 −7 −8 −9

Answers 1231 mathspace.co

9 y 8 7 6 5 4 f (x) 3 2 g(x) 1 −4 −3 −2

19 a

f (x) −4 −3 −2

−4 −3 −2

−1 −1 −2 f (x) −3 −4 −5 −6 −7 −8 −9

9 y 8 7 g(x) 6 5 4 3 1 2 f + g = ex + xx 1 1

−1 −1 −2 −3 −4 −5 −6 −7 −8 −9

f (x) g(x) 1

−4 −3 −2

x 1

−1 −1 −2 −3 −4 g(x)−5 −6 −7 −8 −9

f (x) −4 −3 −2 −1

1 f − g = 2x − x−1

b 4

−1 −2 −3 −4

x 1

g(x)

f − g = sin x − x2 + 1 x −4 −3 −2 −1 −1 −2 −3 −4

1232 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

f (x)

y 9 8 7 6 5 g(x) 4 3 2 2 f + g = cos x + x 1 x

x 4

g(x)

9 y 8 7 6 5 4 3 2 1

20 a

9 y 8 7 6 5 4 3 2 1 −4 −3 −2

−1 −1 1 2 3 4 −2 −3 −4 1 f −−5g = x + 3 − 2 − x+1 −6 −7 −8 −9

−1 −1 −2 −3 −4 −5 −6 −7 −8 −9

9 y 8 7 6 5 4 3 2 1

2 3 4

c f (x)

−4 −3 −2 −1

f (x)

21 a

−1

f + g = 2 sin x + x2 + x + 1 x 1

9 y 8 7 6 5 4 3 f − g = 3 cos x − 2 x2 + 3x + 10 1 x −4 −3 −2

9 8 7 6 5 4 3 2 1 −1 −2 −3 −4

4 3 2 1

g(x)

−1 −2 −3 −4

9 8 7 6 5 4 3 2 1

−1 −1 −2 −3 −4 −5 −6 −7 −8 −9

g(x) d

g(x) f (x) x 1

−1 −2 −3 −4 −5 −6 −7 −8 −9

1 x

g(x) x 2

f + g = In( x − 1) + 1 +

f − g = 2 In x −

2 x

1 x+1

2 1 −1 −2

f (x) 1

f − g = In( x + 1) −

g(x)

9 y 8 7 f (x) 6 5 4 3 2 1 −1 −2 −3 −4 −5 −6 −7 −8 −9

x 2

f + g = In x +

1 x−1

f (x) 1

g(x)

−3 −4

18.05E Graphical relationships What do you remember? 1 a C( x) = 2x2 + b All positive real numbers, or x > 0. 2 a x=0

b (1.5, 4.5)

c Approximately (0.8, 3.6) or (2.8, 1.1) 3 The range of y = ∣f ( x)∣ is non-negative ([0, ∞) or a subset), as negative values of f ( x) are reﬂected to positive. For y = f ( x), the range may include negative values, depending on f ( x).

Answers 1233 mathspace.co

Practice 4 a

8 7 6 5 4 3 2 1

6 a

6 5 4 3 2 1

x 1

Multiply by x

c 0.67 < x < 3.9. Graph shows R( x) exceeds M ( x) between intersections at x ≈ 0.67 and x ≈ 3.9, peaking at R(2) = 8 while M ( x) decreases, conﬁrming R( x) > M ( x) in this interval.

−2 −4 −6 −8

Solve 2x3 − 10x2 − 5 = 0. Root at x ≈ 5.1. Total cost equals the magnitude of the negative production rate (≈ 0.49 thousand dollars), indicating cost matches the absolute value of the production rate when production becomes negative. 7 a 14 y 13 12 11 10 9 8 7 6 5 4 3 2 1

x 3

b Vertical asymptote at x = 0 due to . As x → 0+, the advertising cost increases without bound, causing the total cost C( x) to approach inﬁnity, indicating that extremely low sales volumes lead to prohibitively high costs.

= 0 (no solution).

= x2 − 5x.

5 a C( x) = −2x2 + 8x +

Case 2: R(x) < 0 (x > 5): −x2 + 5x +

Roots at x ≈ 0.67 and x ≈ 3.9, where R(0.67) ≈ M (0.67) ≈ 4.48 and R(3.9) ≈ M (3.9) ≈ 0.77. Graph conﬁrms intersections at these points.

= −x2 + 5x implies

−2x + 8x − 3 = 0

10 8 6 4 2

b Case 1: R( x) ≥ 0 (0 < x ≤ 5): −x2 + 5x +

b Solve −2x2 + 8x = : 3

b Solve 5x = : 5x2 = 6, x2 = 1.2, x ≈ 1.1. Graph conﬁrms intersection at x ≈ 1.1, where D(1.1) ≈ F (1.1) ≈ 5.5. 8 a

y 4 3 2 1 x 0 1 2 3 4 5 6 7 8 9

1234 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

b Vertical asymptote at x = 5 (cost approaches inﬁnity as x → 5). Horizontal asymptote at y = 0 (cost approaches zero as x → ∞). In a cost model, x = 5 may represent a critical production level where costs spike, while large x minimises costs.

12 Graphically, R( x) =

For x > 0.022, C( x) exceeds R( x). y

9 a T ( x) = −x2 + 6x + .

Since x > 0, then the domain is all positive real numbers. b

is above C( x) = −

x2 + 9x for 0 < x < 0.022. The functions intersect at x ≈ 0.022, where R( x) ≈ C( x) ≈ 0.199.

6 5 4

3 2

C(x)

1 10

R(x)

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

13 a 4

x 0

2 1

10 Case 1 ( x ≥ 6): x − 6 = . =3±

For x ≥ 6, x ≈ 3 +

−4 −3 −2 −1 −1

Multiply by x: x2 − 6x − 9 = 0. Roots: x =

2 3 4

−2

−3

≈ 7.242.

−4

Case 2 (0 < x < 6): 6 − x = . Multiply by x: x2 − 6x + 9 = 0. Root: x = 3.

b Symmetric about y-axis. f (∣x∣) =

Solutions: x = 3 and x ≈ 7.2. 11 Horizontal asymptote at y = 0. y 10 9 8 7 6 5 4 3 2 f (x) 1

is even,

identical for x and −x. In a rate model, symmetry implies equal rates for positive/ negative inputs (e.g., time or distance). Extend your thinking 14 a

f(x) + g(x)

g(x) 8

x 9

9 8 7 6 5 4 3 2 1

Piecewise: y = −x2 + 4x + 2

y = x − 4x +

(0 < x ≤ 4),

( x > 4). Minimum at x = 4,

y = 0 + 0.5 = 0.5.

Answers 1235 mathspace.co

17 a

b At x = 4, P( x) equals zero (P(4) = 0), so ∣P( x)∣ = 0. The advertising cost is A(4) =

9 y 8 7 6 5 4 3 2 1

= 0.5, which is minimal at this point.

Therefore, their sum y = ∣P(4)∣ + A(4) reaches its minimum value of 0.5 at x = 4. 2

15 a T ( x) = −x + 5x +

−1 −2

11 y 10 9 8 7 6 5 4 3 2 1 P(x) −1

As x → 0+,

→ ∞, so T ( x) → ∞ (vertical

asymptote at x = 0). As x → ∞, −x2 dominates, so T ( x) → −∞ ( graph descends). b As sales approach zero, advertising costs dominate, making low sales unsustainable. As sales grow large, the negative quadratic proﬁt term causes losses, suggesting an optimal sales range. 16 a Solve ∣−x2 + 7x∣ =

(0 < x < 9),

Chapter 18 review 1 B 2 B 3 C 4 a Solutions: x = 4, x = −4. since h( x) = 0 makes the denominator zero.

Case 1 (5 < x ≤ 7, R( x) ≥ 0): −x2 + 7x =

Multiply by x − 5: x3 − 12x2 + 35x + 9 = 0. No roots in 5 < x ≤ 7 ( function positive). Case 2 ( x > 7, R( x) < 0): x2 − 7x =

( x ≥ 9). Minimum at x = 9, y = .

These are vertical asymptotes for y =

for x > 5:

f (x)

b At x = 9, the total cost is minimised to , representing the optimal production quantity. This minimum occurs at the cusp point where the ∣x − 9∣ component of the cost is zero. For x < 9, the function decreases, and for x > 9, the function increases.

x 4

x 1 2 3 4 5 6 7 8 9 10 11

y=x−9+

A(x) 3

g(x)

Piecewise: y = 9 − x +

T(x)

Multiply by x − 5: x − 12x + 35x − 9 = 0. Root at x ≈ 7.49.

b As x → ±∞, h( x) → ∞. Thus, y =

→ 0, so

y = 0 is the horizontal asymptote. c As x → −4−, y → ∞; as x → −4+, y → −∞. As x → 4−, y → −∞; as x → 4+, y → ∞. d

Solution: x ≈ 7.5. b x > 7.5. For x > 7.5, the absolute revenue exceeds the cost, indicating potential proﬁtability as the number of sales batches increases beyond this point, where revenue’s magnitude outpaces the decreasing cost.

y 0.4 0.3 0.2 0.1 −8 −6 −4 −2 −0.1 −0.2 −0.3 −0.4

1236 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

x 2

 0, − 1   16  

3.5 3 2.5 2 1.5 1 0.5 −4

−3

−2

10 α =

,β=

11 a = 2, b =

,c=1

12 a x

(1, 0)

−1 −0.5 −1 −1.5 −2 −2.5 −3 −3.5

−4

−2

b , q( x) → −∞. +

c As x → 3 ,

→ −∞.

d As x → ,

→ 0.

−2 −4

−4

−2

18 16 14 12 10 8 6 4 2

6 a As x → 3, p( x) → 0. b As x →

18 16 14 12 10 8 6 4 2

−2 −4

0.8 0.6

13 a

1 2

y 2

0.4

1 x

0.2

−4 −3 −2 −1 x

−1

−2

ii x ≠ nπ, n ∈ 

iii (−∞, ∞)

b A

b x = nπ, n ∈ 

c p(−x) = ∣x∣ − 2

9 a 2π b i x=

2 3 4

−1

2 3 4 5 6 7

8 a i π

14 ,

ii x = −2π, 0, 2π 6

c i

2 x −4 −3 −2 −1

2 3 4

Answers 1237 mathspace.co

15 a x2 + 2x − 5

b −x2 + 4x − 5

18 a

c All real numbers () d (−∞, −1] 16 a

y 200

150

3 2 1

P(n)

−4 −3 −2 −1 −1

2 3 4

n 0

−2 −3 −4

F(n)

100

b y

8 6 4 2

−4 −3 −2 −1 −2

2 3 4

For 0 < n ≤ 30, n = 1.61

−4 −6

Intersection at n = 1.61 conﬁrmed graphically in part (a).

−8

c Graphically, P (n) > F (n) for 1.61 < n ≤ 30. 17 a

The proﬁt curve P (n) lies above F (n) in this interval.

19 a N (w) = −w2 + 15w −

2 1

−4 −3 −2 −1 −1

−2

−3 −4

4 2 −4

−2

−2 −4 −6 −8 −10 −12 −14

20 10

w 2

x 2

b Vertical asymptote at w = 0. Signiﬁcance: Early in the season, high spoilage relative to low yield results in a negative net yield.

1238 Mathspace New South Wales – Year 11 Advanced/Extension 1 mathspace.co

c The health function is H(t) = ∣−2t2 + 10t∣ − .

y 10

At t = 6, the vitality V(t) has become negative, but its magnitude ∣V (6)∣ is still large (12). Crucially, the environmental stress E(t) has decreased significantly (to approximately 1.33). The reduction in stress outweighs the slight drop in vitality magnitude, resulting in a higher overall health value H(6) ≈ 10.67.

5 t 1

−5

From the graph, the maximum is H(6) =

d At t = 2.5, vitality V(t) is at its peak (12.5), but stress E(t) is still relatively high (3.2), giving a health of H(2.5) = 9.3.

Answers 1239 mathspace.co