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ADVANCED

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 1: 2026 ISBN: 978-1-963022-91-9 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

Algebraic techniques 0.01

Index laws

0.02

Negative and fractional indices

0.03

Expand binomial products

0.04

Factorise algebraic expressions

0.05

Factorise quadratics

0.06

Complete the square

0.07

Algebraic fractions

0.08

Further algebraic fractions

0.09

Simplify surds

0.10

Operations with surds

0.11

Further operations with surds Chapter 0 review

Solve equations 1.01

Systems of linear equations

1.02

Solve quadratic equations

Investigation: Derive the quadratic formula

The discriminant

Chapter 1 review

1.03

Introduction to functions and relations

2.01

Functions and relations

2.02

Variables and substitution

2.03

Characteristics of functions

Chapter 2 review

Contents mathspace.co

iii

Linear functions

3.01

Characteristics of linear graphs

3.02

Equations of lines

3.03

Linear inequalities

3.04

Linear models

102

3.05

Simultaneous equations

120

Chapter 3 review

139

Quadratic and cubic functions

144

4.01

Characteristics of quadratics

146

4.02

Completed square form

159

4.03

Graph parabolas

167

4.04

Equations of parabolas

175

4.05

Solve quadratic systems

182

4.06

Quadratic inequalities

188

4.07

Quadratic models

194

4.08

Cubic functions

201

Chapter 4 review

209

Function properties

214

5.01

Further domain and range

216

5.02

Even and odd functions

224

5.03

Composite functions

232

5.04

Piecewise functions

241

Chapter 5 review

253

Functions and relations

256

6.01

Graphs of reciprocal functions

258

6.02

Direct variation models

265

6.03

Inverse variation models

272

6.04

Introduction to absolute value functions

278

6.05

Absolute value functions

283

6.06

Circles and semicircles

291

Chapter 6 review

299

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Trigonometry 304 7.01

Exact trigonometric values

306

7.02

Angles of elevation, depression and bearings

313

7.03

Unit circle

330

7.04

Related angles and identities

343

7.05

Sine and cosine rules

350

7.06

Radians

361

7.07

Arc length and sector area

370

7.08

Graphs of trigonometric functions

378

Chapter 7 review

398

Trigonometric identities and equations

402

8.01

Secant, cosecant and cotangent

404

8.02

Unit circle with secant, cosecant and cotangent

413

8.03

Reciprocal and quotient identities

423

8.04

Complementary angle identities

428

8.05

Evaluate expressions with identities

436

8.06

Simplify and prove identities

442

8.07

Trigonometric equations

451

Chapter 8 review

458

Probability 462 9.01

Sets and notation

464

9.02

Set operations and complements

470

9.03

Venn diagrams

476

9.04

Probability and events

484

9.05

Mutually exclusive events

491

9.06

Multistage events and conditional probability

499

9.07

Conditional probability formulas

508

9.08

Independent events

515

Chapter 9 review

522

Contents mathspace.co

Data 526 10.01

Random variables

528

10.02

Organise and graph datasets

532

10.03

Analyse data

541

Chapter 10 review

546

Introduction to rates of change

550

11.01

Average rate of change

552

11.02

Speed as a rate of change

557

11.03

Instantaneous vs. average speed

563

11.04

Instantaneous speed and tangents

568

11.05

Linear and quadratic rates of change

575

Chapter 11 review

584

The derivative

588

12.01

Gradient of a curve

590

12.02

Derivatives of basic functions

602

12.03

First principles for derivatives

606

12.04

Derivative notation and basic rules

613

12.05

Tangents and normals

623

12.06

Chain rule

636

Extension: Proof of chain rule 12.07

Product rule

643

Extension: Proof of product rule 12.08

Quotient rule

648

12.09

Apply differentiation rules

655

12.10

Graphical behaviour of functions

662

12.11

Derivatives as rates of change

675

Chapter 12 review

684

Extension: Proof of quotient rule

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Exponential functions

688

13.01

Exponential graphs

690

13.02

Tangent gradient at y-intercept

702

13.03

Euler’s number and derivatives

706

Investigation: Differentiation of exponential functions

712

Applications

714

Chapter 13 review

720

Logarithmic functions

724

13.04

14.01

Logarithms

726

14.02

Exponential-logarithmic equivalence

735

14.03

Logarithm laws and properties

742

14.04

Logarithm expressions and equations

748

14.05

Logarithmic graphs

755

Investigation: Logarithms

763

Chapter 14 review

767

Transformations 770 15.01

Reflections in axes

772

15.02

Horizontal and vertical translations

778

15.03

Dilations

784

Chapter 15 review

793

Advanced applications of transformations

796

16.01

Linear, quadratic and cubic functions

798

16.02

Exponential and logarithmic functions

811

16.03

Reciprocal and absolute value functions

824

16.04

Circles and translations

832

16.05

Order of transformations

839

16.06

Multiple transformations

850

Chapter 16 review

858

Answers

862

Contents mathspace.co

vii

Big ideas • Linear equations in one variable form the basis for systems of two linear equations, which represent two lines on a Cartesian plane; their solution is the intersection point, found using methods such as substitution and elimination. • The discriminant reveals the number and nature of a quadratic equation’s real roots, which correspond graphically to the x-intercepts of its parabola, while methods such as factorisation and the quadratic formula are used to find their exact values.

1 Solve equations Chapter outline 1.01 1.02 1.03

Systems of linear equations Solve quadratic equations Investigation: Derive the quadratic formula The discriminant Chapter 1 review

4 15 26 30 36

Fast food combos hide equations – one burger + one drink = $12, two drinks + one burger = $15. Solve for lunch!

1.01 Systems of linear equations After this lesson, you will be able to… • solve linear equations in one variable. • solve systems of linear equations using the substitution method or elimination method.

Linear equations Linear equation An equation involving linear expressions. The general form of a linear equation in one variable is ax + b = c, where a, b, and c are constants. A linear equation is an equation of the first degree, meaning the highest power of any variable is 1. To solve linear equations in one variable, apply inverse operations systematically to find the value of the variable. For equations involving fractions, it is often easiest to first multiply every term by the lowest common multiple of the denominators to eliminate them.

Example 1 Solve the linear equations: a 3(x − 2) + 5 = 2x + 7

Create a strategy Expand the brackets, collect like terms, and use inverse operations to isolate x.

Apply the idea 3(x − 2) + 5 = 2x + 7

Write the equation

3x − 6 + 5 = 2x + 7

Expand the brackets

3x − 1 = 2x + 7

Simplify the left-hand side

x−1=7

Subtract 2x from both sides

x=8 The solution is x = 8.

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Add 1 to both sides

Create a strategy Eliminate fractions by multiplying all terms by the lowest common multiple of the denominators, then use inverse operations to isolate x.

Apply the idea Write the equation Multiply both sides by the lowest common denominator 4

Expand the brackets

Collect like terms

Subtract 5 from both sides

Multiply both sides by −1

The solution is x = −5.

Reﬂect and check To verify, substitute x = −5 into the left-hand side to show that it is equal to the right-hand side: Write the left-hand side

Substitute x = −5

Simplify the numerators

Write the fractions with common denominator

Combine the fractions

Simplify the numerator

Simplify

Compare with the right-hand side

Since LHS = RHS, the solution is veriﬁed.

Idea summary Linear equations in one variable are solved using inverse operations to isolate the variable. Solving results in a unique solution.

1.01 Systems of linear equations mathspace.co

Solve simultaneous equations by substitution 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. Simultaneous equations can be solved using substitution by isolating one variable in one equation and using its value to replace that variable in another equation. A system of equations involves two or more equations containing two or more variables. These equations can be solved simultaneously, meaning the solution must satisfy all equations in the system at the same time. When at least one equation in a system has a variable that is already isolated (or can be easily isolated), the system can be solved eﬃciently by substituting the isolated variable into another equation. This method is called “substitution”. The substitution method involves making one variable the subject of one equation, then substituting its equivalent expression into the other equation. This produces an equation in a single variable, which can then be solved.

Example 2 Solve this system of equations using the substitution method: x−y=5 2x + 3y = 6

Create a strategy First, number the equations to make them easier to work with: 1

x−y=5

2x + 3y = 6

Isolate x in equation 1 , then substitute the x-value to equation 2 to solve for y. Finally, substitute the y-value into equation 1 to solve for x.

Apply the idea First, isolate x in equation 1 : x−y=5 x=5+y

Write equation 1 Add y to both sides

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Then, substitute x = 5 + y in equation 2 to solve for y: Write equation 2 Substitute x = 5 + y

Expand the brackets

Collect like terms

Subtract 10 from both sides

Divide both sides by 5

Finally, substitute y =

to equation 1 to solve for x: Write equation 1

Substitute y =

Evaluate the adjacent signs

Subtract The solution is

from both sides

Reﬂect and check Substitute the solutions in both equations to verify: Verifying for 1 x − y = 5: Write the left-hand side of 1 and y =

Substitute x =

Evaluate the adjacent signs

Add

Simplify

Compare with the right-hand side

Verifying for 2 2x + 3y = 6: Write the left-hand side of 2

Substitute x =

Evaluate the multiplication

Subtract

Simplify

Compare with the right-hand side

and y =

Since LHS = RHS, the solution is verified.

1.01 Systems of linear equations mathspace.co

Idea summary The substitution method solves a system of equations by expressing one variable in terms of the other, substituting this expression into the second equation, solving for the remaining variable, and finally substituting back to find the first.

Solve by elimination The elimination method is another technique for solving a system of equations. It can often be quicker, depending on how the equations are presented. First, align like terms in columns: For example, in this system, the terms are aligned: 3x + 4y = 12 x + y = 11 In this system, they are not aligned: 3x = 12 − 4y x + y = 11 Next, adjust the equations so that the coeﬃcients of one variable are equal in magnitude; that is, their absolute values are the same. Then, if the coeﬃcients have the same sign, subtract the equations; if they have opposite signs, add the equations.

Exploration Consider the system of equations: x + y = 10 x−y=4 1. Add the two equations together. (Add the left-hand sides, and add the right-hand sides). What is the resulting equation? 2. Can you solve this new equation for x? 3. Now, subtract the second original equation from the ﬁrst. What is the resulting equation this time? 4. Can you solve this new equation for y? 5. Discuss why adding or subtracting the full equations helps to ﬁnd the solution.

The goal is to eliminate one of the variables, leaving an equation with only one variable.

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Example 3 Solve each system of equations using the elimination method. a 9x + y = 62 5x + y = 38

Create a strategy Number the equations to help describe the process: 1

9x + y = 62

5x + y = 38

The goal is to combine the two equations to eliminate one variable. In this problem, the y-coeﬃcients are the same. Subtracting equation 2 from equation 1 will eliminate the y-term. Note that subtracting equation 1 from equation 2 would also eliminate the y-term. However, this would produce a negative constant and a negative x-term, making the resulting equation less convenient to work with. Subtracting equation 2 from equation 1 avoids this, resulting in a simpler equation.

Apply the idea Subtract equation 2 from equation 1 :

4x = 24

Eliminate the y-term

x=6

Divide both sides by 4

Now that x is known, substitute it back into either of the original equations to ﬁnd y: 5x + y = 38

Write equation 2

5(6) + y = 38

Substitute x = 6

30 + y = 38

Evaluate the multiplication

y=8

Subtract 30 from both sides

The solution to this system is the ordered pair (6, 8).

1.01 Systems of linear equations mathspace.co

b 2x + 3y = 19 4x − y = 10

Create a strategy Number the equations to help describe the process: 1

2x + 3y = 19

4x − y = 10

There are two ways to eliminate a variable: 1. Multiply equation 1 by 2, so that the x-coeﬃcients of both equations will be 4x. Then, subtract equation 2 from equation 1 to eliminate x. Multiply equation 1 by 2: 2x + 3y = 19

Write equation 1

4x + 6y = 38

Multiply both sides by 2

The modiﬁed system is: 1

4x + 6y = 38

4x − y = 10

2. Multiply 2 by 3, so that the y-coeﬃcients will be 3y and − 3y. Then, add the modiﬁed equations to eliminate y. Multiply equation 2 by 3: 4x − y = 10

Write equation 2

12x − 3y = 30

Multiply both sides by 3

The modiﬁed system is: 1

2x + 3y = 19

12x − 3y = 30

Since the ﬁrst method is much simpler, use this method to solve the system of the equations.

Apply the idea 1

4x + 6y = 38

4x − y = 10

Subtract equation 2 from equation 1 :

7y = 28

Eliminate the x-term

y=4

Divide both sides by 7

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Now substitute y = 4 back into one of the original equations: Write equation 2 Substitute y = 4

Add 4 to both sides

Divide both sides by 4

Simplify The solution to this system is the ordered pair

Reﬂect and check Substitute the solutions in both equations to verify: Verifying for 1 2x + 3y = 19: Write the left-hand side of 1 Substitute x =

and y = 4

Evaluate the multiplication

Evaluate the addition

Compare with the right-hand side

Since LHS = RHS, the solution is verified. Verifying for 2 4x − y = 10: Write the left-hand side of 2 Substitute x =

and y = 4

Evaluate the multiplication

Evaluate the subtraction

Compare with the right-hand side

Since LHS = RHS, the solution is verified.

Idea summary The elimination method aims to remove one variable by adding or subtracting the equations in a system. To do this, multiply one or both equations by a constant so that the coeﬃcients of one variable are equal in magnitude.

1.01 Systems of linear equations mathspace.co

1.01 Practice questions What do you remember? 1

State the inverse operation for each of the following: a

Addition

Multiplication

Subtraction

Division

Identify the operation applied to these equations: a

3x + 5 = 11 to 3x = 6

= 2 to x = 8

Verify that x = 3 is a solution to the equation 2x − 1 = 5.

Brieﬂy deﬁne the substitution method for solving simultaneous equations.

Brieﬂy explain the elimination method for solving simultaneous equations.

Practice 6

Solve the following one-step linear equations: a

x − 4 = −1

3x + 2 = 11

2x − 5 = 3

+1=3

4x = 2x + 7

3(x + 2) = 12

2(2x − 3) = 6

4(2x + 1) = 3x − 1

5(x − 1) = 3x + 1

Solve the following linear equations: a

2(x − 3) + 4 = x + 5

3(x + 1) − 2x = x + 5

Solve the following linear equations involving fractions: a

x+5=8

Solve the following equations with brackets: a

Ex 1

Solve the following two-step linear equations: a

4x = 16

Consider the system of equations: 2x + 5y = 13 x−y=4

Which variable is easier to isolate in the second equation to solve by substitution?

How would you eliminate y using the elimination method?

What does the solution (x, y) represent graphically?

Mathspace New South Wales – Year 11 Advanced mathspace.co

Consider the system of equations: 3x + 7y = −24 4x − 5y = 11

Ex 2

To eliminate the variable y, what constants should the ﬁrst and second equations be multiplied by, respectively?

After multiplying, what is the new equivalent system of equations?

By adding the two new equations, solve for x and then ﬁnd y.

Solve the following systems of equations using the substitution method: a

y = 6x + 14

a=b+5

y = 2x + 2 a + 2b = 23 c

m = 7 − 2n

z = −5y − 22

3m + 4n = 11 z = 7y + 38 e

y = 5x + 37

q = −3p + 22

5y = 35x + 265 p + q = 10 g

y = − 4x – 27

y = 3x + 14

2y = 10x + 54 x + y = 2 14

Solve the system y = x + 11 y = 3x + 19 using substitution, and verify the solution in both equations.

Ex 3

Solve the following systems of equations using the elimination method: a

7x + y = 49

3a + 5b = 3

3x + y = 25 6a − 5b = −84 c

j + 4k = 34

7d + 9e = 92

−j + 3k = 22 −7d + 5e = −36 e g

5p + q = 5

−3s − 7t = −69

5s − 7t = 3

45p + 2q = 10

0.5x + 3y = 14

p + 4q = 2.8

6x − 4y = 32 0.1p + q = 6.4 i

−6w − 5x = −12

4x − 9y = 7

−30w − 10x = 30

−16x + 3y = 71

Stella bought a novel and a magazine for $59. The novel costs $10 less than twice the magazine price. The system is represented by: m + n = 59 n = 2m − 10 Solve the system to ﬁnd the cost of the magazine (m) and the novel (n).

1.01 Systems of linear equations mathspace.co

Extend your thinking 17

Two cyclists start at the same point and travel in opposite directions. One travels at 15 km/h and the other at 20 km/h. How long will it take for them to be 70 km apart?

Solve

Two investments yield a total return of $1200. One investment earns $x, and the other earns $200 less than three times the ﬁrst.

= 1 for x, expressing the solution in terms of p, q, r, and s.

Form an equation and solve for x.

Calculate the amount earned by the second investment.

Write simultaneous equations with solution (1, 4) and solve using substitution.

A store sells apples at $2 each and oranges at $1.50 each. A customer buys 10 fruits for $17.50.

Write the system of equations.

Solve to ﬁnd the number of apples and oranges.

A business has ﬁxed costs of $500 and variable costs of $10 per unit. Revenue is $20 per unit. a

Write the cost and revenue equations.

Find the break-even point where total revenue equals total costs, resulting in no proﬁt or loss and interpret.

Solve the following systems of equations using any algebraic method. State if there is one solution, no solution, or inﬁnitely many solutions. a

4x + 3y = 10

y = 3x − 1

2x − y = 8 6x − 2y = 4 c

5x − 2y = 7

x = 2y + 1

10x − 4y = 14 3x − 6y = 3 24

Consider the system: 2x − 1.8y = 5.6 0.65x + 0.7y = −0.75

Rewrite with integer coeﬃcients.

Solve the system.

Mathspace New South Wales – Year 11 Advanced mathspace.co

1.02 Solve quadratic equations After this lesson, you will be able to… • solve quadratic equations by factorisation using the null factor law. • solve quadratic equations by completing the square and using the quadratic formula. • apply various methods to solve practical problems involving quadratic equations.

Solve by factorisation Quadratic equation An equation of the form ax2 + bx + c = 0, where a ≠ 0, b and c are constants.

Exploration What values of the variables make each of these equations true? • z−7=0 • 13a = 0 • 3(d + 4) = 0 • x×y=0 • x2 = 4

The null factor law states that if the product of two or more factors is equal to 0, then at least one of the factors must be equal to 0. That is, if ab = 0, then either a = 0 or b = 0 (or both). This property can be used to solve quadratic equations by first writing the equation in the general factored form: a(x − α )(x − β ) = 0. Once a quadratic equation is in factored form, the null factor law implies that either x − α = 0 or x − β = 0. This means the solutions (or roots) of the quadratic equation are x = α and x = β. Not all quadratic equations can be efficiently solved by factorisation, but it is usually best to attempt this method first, as it is often the fastest. Quadratic language: • Monic: A quadratic where the coefficient of x2 equals 1, that is, when a = 1 in ax2 + bx + c = 0. • Non-monic: A quadratic where the coefficient of x2 does not equal 1(or 0). • Trinomial: Exactly three non-zero terms. • Constant: A fixed number with no variable factor, that is, in ax2 + bx + c = 0, c is the constant term.

1.02 Solve quadratic equations mathspace.co

Example 1 Solve these equations by factorising: a x2 + 6x − 55 = 0

Create a strategy The equation is a monic quadratic trinomial. To factorise it, ﬁnd two integers that multiply to give the constant term c = −55 and add to give the coeﬃcient of x, which is b = 6. Then, use these integers to write the equation in factored form and apply the null factor law.

Apply the idea The two integers that multiply to − 55 and add to 6 are 11 and − 5. Rewrite the equation in factored form: x2 + 6x − 55 = 0

Write the equation

(x + 11)(x − 5) = 0

Factorise the quadratic

x + 11

0, x − 5 = 0

Use null factor law

x + 11 = 0

Write the equation

For x + 11 = 0: x = −11

Subtract 11 from both sides

For x − 5 = 0: x−5=0

Write the equation

x=5

Add 5 to both sides

b 2x2 + 7x − 4 = 0

Create a strategy This is a non-monic quadratic equation. To factorise, ﬁnd two numbers that multiply to a × c = 2 × (− 4) = −8, and add to the coeﬃcient of x, which is b = 7. Use these numbers to split the middle term, factorise by grouping, and apply the null factor law to solve for x.

Apply the idea Find two numbers that multiply to − 8 and add to 7. The numbers are 8 and − 1. 2x2 + 7x − 4 = 0 2

2x + 8x − x − 4 = 0

Split the middle term using 8x − x

(2x + 8x) − (x + 4) = 0

Group the terms

2x(x + 4) − (x + 4) = 0

Factorise each group

(2x − 1)(x + 4) = 0 2x − 1

Write the equation

0, x + 4 = 0

Mathspace New South Wales – Year 11 Advanced mathspace.co

Factorise the common binomial (x + 4) Use null factor law

For 2x − 1 = 0: Write the equation

Add 1 to both sides

Divide both sides by 2

For x + 4 = 0: x+4=0 x = −4 The solutions are x =

Write the equation Subtract 4 from both sides

and x = − 4.

Reﬂect and check To verify, substitute the solutions on the left-hand side of the equation to show that it is equal to the right-hand side of the equation. For x = : Write the left-hand side

Substitute x =

Evaluate the power

Multiply

Simplify the ﬁrst fraction

Add the fractions

Simplify the fraction

Evaluate

Compare with the right-hand side

Since LHS = RHS, the solution is verified. For x = − 4: LHS = 2x2 + 7x – 4

Write the left-hand side

= 2 × (− 4) + 7 × (− 4) – 4

Substitute x = − 4

= 2 × 16 + 7 × (− 4) – 4

Evaluate the power

= 32 − 28 – 4

Multiply

Evaluate

= RHS

Compare with the right-hand side

Since LHS = RHS, the solution is verified.

1.02 Solve quadratic equations mathspace.co

Idea summary To solve a quadratic equation by factorisation, set the equation equal to 0. Then, write the quadratic expression in factored form a(x − α)(x − β ) = 0 and use the null factor law to ﬁnd the solutions x = α and x = β.

Solve by completing the square Completing the square is a method used to solve any quadratic equation by converting it into a perfect square trinomial. A perfect square trinomial is one that can be factorised as (x + k)2 or (x − k)2. The process involves rearranging the equation ax2 + bx + c = 0 into the form (x + k)2 = d and then solving by taking the square root of both sides. The key steps are: 1. Move the constant term c to the right-hand side of the equation. 2. If the equation is non-monic (i.e., a ≠ 1), divide all terms by a. 3. Take the coeﬃcient of the x term, divide it by 2, and square the result. Add this value to both sides of the equation. 4. Factorise the left-hand side as a perfect square and simplify the right-hand side. 5. Solve for x by taking the square root of both sides (remembering the ± symbol) and isolating x.

Example 2 Solve these quadratic equations by completing the square: a x2 + 6x − 7 = 0

Create a strategy The equation is a monic quadratic. Move the constant term to the right-hand side, complete the square on the left-hand side by adding

to both sides, and then solve by taking the square root.

Apply the idea Write the equation

Add 7 to both sides

Add

Factorise the left-hand side and simplify the right-hand side

Take the square root of both sides

Subtract 3 from both sides and evaluate the square root

= 9 to both sides to complete the square

The two solutions are x = −3 + 4 = 1 and x = −3 − 4 = −7.

Mathspace New South Wales – Year 11 Advanced mathspace.co

b 2x2 − 8x + 6 = 0

Create a strategy This is a non-monic quadratic. First, divide the entire equation by the leading coeﬃcient 2 to make it monic. Then, proceed with the standard steps for completing the square.

Apply the idea Write the equation

Divide all terms by 2

Subtract 3 from both sides

Add

Factorise the left-hand side and simplify the right-hand side

Take the square root of both sides

Add 2 to both sides

= 4 to both sides

The two solutions are x = 2 + 1 = 3 and x = 2 − 1 = 1.

Idea summary Completing the square solves any quadratic equation ax2 + bx + c = 0 by transforming it into the form (x + k)2 = d. This is achieved by making the quadratic monic (dividing by a if necessary), and then adding the square of half the new x-coeﬃcient to both sides.

Solve by the quadratic formula Quadratic formula The roots of a quadratic equation ax2 + bx + c = 0 where a ≠ 0 are given by the quadratic formula:

The quadratic formula provides a direct method for ﬁnding the solutions to any quadratic equation in the standard form ax2 + bx + c = 0. It is derived by applying the method of completing the square to this general form:

are the solutions (roots) of the equation

a, b, c are the coeﬃcients of the quadratic equation ax2 + bx + c = 0, where a ≠ 0 1.02 Solve quadratic equations mathspace.co

Example 3 Use the quadratic formula to solve these equations: a −x2 + 6x − 8 = 0

Create a strategy The equation is in standard form ax2 + bx + c = 0. Identify the coeﬃcients a, b, and c, and substitute them into the quadratic formula to solve for x.

Apply the idea For − x2 + 6x − 8 = 0, the coeﬃcients are a = −1, b = 6, and c = −8. Write the formula

Substitute a = −1, b = 6 and c = −8

Evaluate the terms inside the square root

Simplify the expression inside the square root

Evaluate the square root

For x with the positive value: Write the equation with the positive value

Simplify the numerator

Evaluate For x with the negative value: Write the equation with the negative value

Simplify the numerator

Evaluate The solutions are x = 2 and x = 4.

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b 5x2 = 8x + 1

Create a strategy First, rewrite the equation in the standard form ax2 + bx + c = 0. Then, identify the values of a, b, and c and substitute them into the quadratic formula. Simplify the resulting expression.

Apply the idea 5x2 − 8x − 1 = 0

Rearrange into standard form

The coeﬃcients are a = 5, b = −8, and c = −1. Write the formula

Substitute a = 5, b = −8 and c = −1

Evaluate the terms inside the square root

Simplify the expression inside the square root

Factorise 84

Simplify the surd

Factorise the numerator

Simplify

The solutions are x =

and x =

Idea summary Any quadratic equation in the form ax2 + bx + c = 0 can be solved by substituting the coeﬃcients a, b, and c into the quadratic formula:

1.02 Solve quadratic equations mathspace.co

1.02 Practice questions What do you remember? 1

Solve these equations using the null factor law: a

x (x − 7) = 0

(k + 4) (k − 2) = 0

(2y − 1) (3y + 5) = 0

5 (m + 6) (m − 6) = 0

The standard form of a quadratic equation is ax2 + bx + c = 0. Find the values of a, b and c for these equations: a

2x2 − 5x + 3 = 0

x2 = 7x − 4

−3x2 + 9 = 0

State the quadratic formula used to solve equations of the form ax2 + bx + c = 0.

What value must be added to each expression to create a perfect square trinomial?

x2 + 8x

x2 − 12x

x2 + 5x

x = −6, x = −2

x = −6, x = 2

x = 6, x = −2

x = 6, x = 2

Any quadratic equation that can be solved by completing the square can also be solved by the quadratic formula.

Using completing the square to find the solution of a non-monic quadratic, the first step should be dividing the quadratic by the coefficient of x2.

The equation x2 = −9 has no real solutions.

Completing the square can only be used on monic quadratic equations.

Solve these monic quadratic equations by factorising: a

x2 − 5x − 14 = 0

f 2 + 6f − 55 = 0

h2 + 19h + 88 = 0

x2 − 20x + 100 = 0

a2 + 15a + 56 = 0

x2 − x − 72 = 0

Solve these non-monic quadratic equations by factorising: a

2y2 − 9y − 5 = 0

3x2 − 14x + 8 = 0

5x2 + 19x − 4 = 0

6k2 − 5k − 6 = 0

9p2 + 12p + 4 = 0

4m − 25 = 0

Solve these equations by ﬁrst taking out a common factor: a

e 9

x2 − x

Is each statement true or false?

Practice 7

Consider the quadratic equation x2 − 4x − 12 = 0. Which of these are the correct solutions? A

Ex 1b

5x2 + x = 0

Ex 1a

6x2 + 54x = 0 2

4y − 8y2 = 0

3x − 27x = 0

5y2 + 5y − 30 = 0

−2a2 + 16a − 30 = 0

x3 − 7x2 + 10x = 0

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Ex 2

Solve these equations by ﬁrst rearranging into standard form and then factorising: a

m2 = 3m + 10

−6y = y2 + 8

x (x + 1) = 42

(x + 1) (x + 2) = 20

Solve by completing the square. Leave answers in exact form: a

x2 − 8x + 13 = 0

x2 + 10x − 3 = 0

x2 − 6x − 11 = 0

x2 + 3x − 1 = 0

3x2 − 12x + 5 = 0

e Ex 3

x2 − 12x = −20

2x + 12x − 10 = 0

Solve these equations using the quadratic formula. Provide exact answers unless otherwise speciﬁed: a

−x2 + 10x − 24 = 0

x2 − 7x + 9 = 0

3x2 + 9x − 4 = 0

4x2 + 20x + 25 = 0

−2x2 + 8x + 5 = 0

1.8x2 + 5.2x − 2.3 = 0 (Round to two decimal places)

Solve these equations involving fractions by ﬁrst multiplying by the lowest common denominator: a

Solve these equations by factorising: a

x2 − 81 = 0

9x2 − 1 = 0

(x − 1)2 − 16 = 0

49 − (x + 2)2 = 0

2x2 − 18 = 0

x − 5 = 0

A rectangle has a perimeter of 34 cm and an area of 60 cm2. Find the dimensions of the rectangle.

The area of a rectangle is 60 cm2. If the length is x + 7 cm and the width is x cm, ﬁnd the dimensions.

A ball is launched from a height of 20 m with an initial velocity of 25 m/s. Its height, h metres, after t seconds is given by h = − 4.9t2 + 25t + 20. Determine the number of seconds it will take the ball to reach the ground. Round your answer to two decimal places.

1.02 Solve quadratic equations mathspace.co

Extend your thinking 18

Solve these equations by making a suitable substitution: a

x4 − 13x2 + 36 = 0

x6 + 7x3 − 8 = 0

(x + 1)2 − 5 (x + 1) + 4 = 0

2 (x2 + 1)2 + 9 (x2 + 1) − 5 = 0

A school football field is a rectangle measuring 100 metres by 50 metres. It is surrounded by stadium seating of a uniform width, x. If the total area of the field including the stadium seating is 8400 m2, find the width of the stadium seating. x

PANTHERS

50 m

100 m

A = 8400 m2

The sum of the ﬁrst n positive integers 1 + 2 + 3 + … + n is given by the formula Sn = How many consecutive integers, starting from 1, must be added to get a sum of 66?

A quadratic equation is x2 − 8x + k = 0. If one of the roots is x = 4 − other root and the constant k.

A right-angled triangle has side lengths x, x + 7, and x + 8. Find the value of x and the lengths of the three sides.

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, ﬁnd the value of the

Eddie was asked to solve the quadratic equation below. Identify their mistake and provide the correct solution.

Solutions: 24

and

. His working is shown

A rectangular garden has an area of 24 m2. The length and width satisfy the equation , where L and W are the length and width in metres. Determine the dimensions of the garden.

Did you know?

Equations help engineers design water slides! By solving equations involving angles, water flow, and speed, they ensure the ride is thrilling while keeping everyone safe. Every twist and splash is carefully planned with maths.

1.02 Solve quadratic equations mathspace.co

INVESTIGATION

Derive the quadratic formula Investigate a method for deriving the quadratic formula.

Objectives • Derive the quadratic formula. • Find solutions to any quadratic equation of the form ax2 + bx + c = 0.

A new way to solve quadratic equations We will derive a method for solving quadratic equations. First, we will use a numerical example, then generalise to form the quadratic formula. x2 + 2x − 8 = 0 Start by using completing the square to solve x2 + 2x − 8 = 0. This could be solved by factorising, but we will use completing the square, with colour-coded terms, to clarify the process. Print the downloadable worksheet to complete the steps in this investigation. Starting with x2 + 2x − 8 = 0, move the constant term to the right-hand side:

x2 + 2x = 8 Represent this visually: the area of a square with side length x units plus a rectangle with side lengths 2 units and x units equals a square with area 8 square units. 1. Fill in the missing side lengths:

2x 8

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2. Split the green rectangle in half and place it on either side of the blue square. Fill in the missing side lengths and areas: x

x 8

3. Complete the square on the left by adding the missing corner, balancing both sides. Fill in the missing areas:

x 8

4. Combine the right-hand side terms to form a new square. Fill in the missing area for the new square: x

Thus:

(x + 1)2 = 9

The side length x + 1 units equals the side length of the grey square, 3 units, so a solution is x = 2. For generality, consider x + 1 = −3, giving a second solution x = − 4.

Investigation: Derive the quadratic formula mathspace.co

Repeat the process to derive the quadratic formula. Start with ax2 + bx + c = 0, divide by a, and move the constant term to the right:

Represent this visually: a square with side length x units plus a rectangle with side lengths units and x units equals a square with area

square units.

1. Fill in the missing side lengths and area:

2. Split the green rectangle in half and place it on either side of the blue square. Fill in the missing side lengths and areas: x

3. Complete the square on the left by adding the missing corner, balancing both sides. Fill in the missing areas: x x x2

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4. Combine the right-hand side terms to form a new square. Simplify the area of the new grey square: Combine fractions using a common denominator.

x x

Thus:

Rearrange algebraically: 1. Take the square root of both sides, indicating two solutions. 2. Simplify the fraction using

3. Isolate x on the left. 4. Write the right-hand side as a single fraction. For a quadratic equation ax2 + bx + c = 0, the solutions are:

Investigate 1.

Solve x2 − 5x + 6 = 0. Verify the solution(s) using the quadratic formula.

Solve x2 − x − 1 = 0. Why is the quadratic formula more eﬃcient than factorising or completing the square for this equation?

Solve x2 + 4x + 4 = 0. How does using the quadratic formula differ from the previous two problems?

Discussion 1.

How does the quadratic formula relate to the axis of symmetry in the standard form of a quadratic equation?

When is the quadratic formula the best approach? When might other methods be preferable?

Investigation: Derive the quadratic formula mathspace.co

1.03 The discriminant After this lesson, you will be able to… • calculate the discriminant of a quadratic equation. • use the discriminant to determine the number and nature of the roots. • relate the value of the discriminant to the graphical features of a parabola. • solve for unknown variables in quadratic equations by applying conditions for equal, distinct, or real roots.

The discriminant Discriminant In the quadratic expression ax2 + bx + c = 0, the discriminant is Δ = b2 − 4ac. The discriminant Δ = b2 − 4ac determines root properties: • Δ > 0 for two distinct real roots which are rational if Δ is a perfect square and irrational otherwise.

y 3 2 1 x 1

−1

• Δ = 0 for one real root, called a repeated root, double root, or equal roots, because it has multiplicity 2.

1 x

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• Δ < 0 for no real roots

1 x 1

Example 1 For these quadratic equations, compute the discriminant to determine the nature of the roots, then solve using an eﬃcient method. Justify your choice of method. a x2 − 6x + 9 = 0

Create a strategy Compute the discriminant to assess the roots. Check if the equation is easily factorisable or suits another method.

Apply the idea For x2 − 6x + 9 = 0, the values are a = 1, b = −6, c = 9. Using discriminant formula: Δ = b2 − 4ac

Write the formula

= (−6) − 4 × 1 × 9

Substitute a = 1, b = −6 and c = 9

= 36 − 36

Evaluate each term

Evaluate

This indicates one real, repeated root. Since Δ = 0, factorising or square roots is eﬃcient. The equation resembles a perfect square trinomial. x2 − 6x + 9 = 0 2

(x − 3) = 0 x−3=0 x=3

Write the equation Factorise as a perfect square Take the square root of both sides Add 3 to both sides

The solution is x = 3 (repeated root).

Reflect and check The discriminant indicated one root, and the perfect square form made factorising efficient. Alternatively, the quadratic formula would confirm

1.03 The discriminant mathspace.co

b x2 − 4x − 5 = 0

Create a strategy Compute the discriminant to determine the nature of the roots. Check if factorising is straightforward due to integer coeﬃcients.

Apply the idea For x2 − 4x − 5 = 0, the values are a = 1, b = − 4, c = −5. Using discriminant formula: Δ = b2 − 4ac

Write the formula

= (− 4) − 4 × 1 × (−5)

Substitute a = 1, b = − 4 and c = −5

= 16 + 20

Evaluate each term

= 36

Evaluate

Since Δ = 36 is a perfect square, there are two real, rational roots, and factorising is eﬃcient. x2 − 4x − 5 = 0

Write the equation

(x − 5)(x + 1) = 0

Factorise by ﬁnding factors of −5 that sum to − 4

x−5

0, x + 1 = 0

Use null factor law

x−5=0

Write the equation

x=5

Add 5 to both sides

x+1=0

Write the equation

For x − 5 = 0:

For x + 1 = 0: x = −1

Subtract 1 from both sides

The solutions are x = 5, x = −1.

Reflect and check The discriminant Δ = 36 confirmed two rational roots, making factorising efficient. Verify x = 5: LHS = x2 − 4x – 5 2

= (5) − 4 × 5 – 5

Write the left-hand side Substitute x = 5

= 25 − 20 – 5

Evaluate each term

Evaluate

= RHS

Compare with the right-hand side

Since LHS = RHS, the solution is verified. Verify x = −1: LHS = x2 − 4x – 5 2

Write the left-hand side

= (−1) − 4 × (−1) – 5

Substitute x = −1 into the left-hand side

=1+4–5

Evaluate each term

Evaluate

= RHS

Compare with the right-hand side

Since LHS = RHS, the solution is verified.

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Example 2 Find k if (2k + 3) x2 − 4kx + 4 = 0 has equal roots.

Create a strategy First, identify a, b, and c to ﬁnd the discriminant Δ. Then, set the discriminant equal to 0 to ﬁnd the values of k.

Apply the idea In the equation, a = 2k + 3, b = − 4k and c = 4. Δ = b2 − 4ac

Write the formula

= (− 4k) − 4 × (2k + 3) × 4

Substitute the values

Evaluate each term

Expand the brackets

= 16k − 16(2k + 3) = 16k − 32k − 48 Equating Δ to 0: 16k2 − 32k − 48 = 0 2

Equate Δ to 0

k − 2k − 3 = 0

Divide both sides by 16

(k − 3)(k + 1) = 0

Factorise

k−3

0, k + 1 = 0

Use null factor law

k − 3 = 0

Write the equation

k = 3

Add 3 to both sides

k + 1 = 0

Write the equation

For k − 3 = 0:

For k + 1 = 0: k = −1

Subtract 1 from both sides

The solutions are k = 3 or k = −1.

Idea summary The discriminant Δ = b2 − 4ac determines root nature: • Δ > 0 for two distinct real roots. The roots are rational if Δ is a perfect square and irrational otherwise. • Δ = 0 for one real root called a double root, repeated root, or equal roots • Δ < 0 for no real roots

1.03 The discriminant mathspace.co

1.03 Practice questions What do you remember? 1

State the discriminant formula for a quadratic equation ax2 + bx + c = 0 and describe how it determines the nature of the roots.

For the quadratic equation 3x2 − 6x + 2 = 0, identify the values of a, b, and c, and compute the discriminant.

True or false: a

If the discriminant is a perfect square and the coeﬃcients a, b, and c are rational, the roots of the quadratic equation are rational.

Any quadratic equation that can be solved by factorising can also be solved by the quadratic formula.

All quadratic equations can be solved by factorising over the integers.

The discriminant of a quadratic equation determines the number of real solutions.

Why is the discriminant useful before solving a quadratic equation?

Can the quadratic formula solve any quadratic equation? Do all quadratic equations have real solutions?

Practice Ex 1

Ex 2

Compute the discriminant and determine the number and nature of the roots for each quadratic equation: a

x2 + 6x + 9 = 0

2x2 − 4x − 6 = 0

x2 + 2x + 3 = 0

4x2 + 4x + 1 = 0

x − 8x + 16 = 0

x2 + 2x − 3 = 0

2x2 + 3x + 2 = 0

For the quadratic equation (2k + 1) x2 − 4x + 3 = 0, use the discriminant to ﬁnd: a

The values of k for which the equation has equal roots.

The values of k for which the equation has no real solutions.

Find the values of p for which (3p − 2)x2 − 5x + 4 = 0 has one real root.

A projectile’s height is modelled by h = − 4.9t2 + 25t + 2 (in metres). Determine if the projectile reaches a height of 30 m using the discriminant and at what time. Round to two decimal places.

The stopping distance of a car is modelled by s = 0.06u2 + 0.7u (in metres, u in km/h). Determine if a stopping distance of 40 m is possible, and if so, ﬁnd the speed to the nearest km/h.

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Determine the nature of the roots for each quadratic equation using the discriminant: a

x2 − 3x + 1 = 0

3x2 − 6x + 3 = 0

Solve x2 − 10x + 25 = 0 using an eﬃcient method after computing the discriminant. Justify your choice of method.

Solve 2x2 − 4x − 3 = 0 using an eﬃcient method after computing the discriminant. Justify your choice of method.

Extend your thinking 14

Find the values of m for which x2 + mx + 16 = 0 has real and rational roots.

A quadratic equation has no real roots. If a = 2 and c = 5, ﬁnd the possible values of b.

A quadratic equation x2 + kx + 4 = 0 has one root 2 + of k.

Find the values of m for which mx2 + 3x + 2 = 0 has two distinct real roots.

A right-angled triangle has sides x, x + p, and x + q (shortest to longest, no two equal): a

Write a quadratic equation in standard form for the sides.

Given q = 2p, ﬁnd x in terms of p.

. Find the other root and the value

x+q

x+p

x 19

A quadratic equation x2 + bx + c = 0 has roots r and s. Express b and c in terms of r and s.

For which values of k is the function y = kx2 − 4x + k + 3 negative for all x?

1.03 The discriminant mathspace.co

1 Chapter review 1

What is the solution to the system of equations 3x + y = 10 and x − y = 2? A

(4, −2)

(1, 7)

x = −5, x = − 4

x = 5, x = − 4

x = −5, x = 4

x = 5, x = 4

The equation 2x2 − 3x + 4 = 0 has: A

one real root

two distinct real roots

no real roots

two rational roots

Solve the following systems of equations using the substitution method: a

y = 4x + 11

m + 3n = 15

p = 8 − 3q 2p + 5q = 19

m=n+3

y = x + 2

(3, 1)

The solutions to the quadratic equation x2 + x − 20 = 0 are: A

(2, 4)

y = 2x + 9

x+y=3

Solve the following systems of equations using the elimination method: a

3x + 2y = 16

5a + b = 22

5x − 2y = 16 2a + b = 10 c

2p − 3q = 1

3m + 4n = 25

4p + q = 23 4m − 5n = − 4 6

Solve the following linear equations involving fractions: a

A laptop and a mouse cost $1250. The laptop costs $50 more than 3 times the mouse price. The system is represented by: L + M = 1250 L = 3M + 50 Solve the system to ﬁnd the cost of the laptop (L) and the mouse (M ).

A café sells coffees at $4.50 each and teas at $3.50 each. A customer buys 12 drinks for a total of $49: a

Write a system of equations to model this situation.

Solve the system to ﬁnd the number of coffees and teas purchased.

A company has ﬁxed costs of $800 and variable costs of $15 per item. The revenue from selling the items is $35 per item: a

Write the cost (C) and revenue (R) equations in terms of the number of items, x.

Find the break-even point, where total revenue equals total costs, and interpret the result.

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Solve these monic quadratic equations by factorising: a c

p + 14p + 48 = 0

a2 + 4a − 21 = 0

m2 − 16m + 64 = 0

Solve these non-monic quadratic equations by factorising: a

x2 − 3x − 28 = 0

2x2 + 9x − 5 = 0 2

4p + 11p − 3 = 0

3a2 − 10a + 8 = 0

9k2 − 49 = 0

Solve these equations by ﬁrst rearranging into standard form and then factorising: a

x(x − 2) = 35

a2 = 2a + 24

( p − 1)( p + 3) = 12

− 2x = 6

Solve by completing the square. Leave your answers in exact form: a

x2 − 8x + 5 = 0

a2 + 6a − 2 = 0

p2 − 5p + 1 = 0

2m2 − 8m − 6 = 0

Solve these equations using the quadratic formula. Leave your answers in exact form: a

3x2 − 5x − 1 = 0

x2 − 9x + 10 = 0

−x2 + 6x + 2 = 0

9x2 − 6x + 1 = 0

A right-angled triangle has side lengths x, x + 3, and x + 6. Find the value of x and the lengths of the three sides.

A rectangular sports field measures 80 metres by 40 metres. It is surrounded by a path of uniform width, x. If the total area of the field and the path is 5200 m2, find the width of the path. x

PANTHERS

40 m

80 m

A = 5200 m2

Chapter 1 review mathspace.co

Solve these equations by making a suitable substitution: a

x4 − 10x2 + 9 = 0

c 18

(x − 2)2 − 2(x − 2) − 15 = 0

Compute the discriminant and determine the number and nature of the roots for each quadratic equation: a

x2 − 12x + 36 = 0

2x2 − 7x − 4 = 0

3x2 + 5x − 1 = 0

x + 3x + 5 = 0

Find the value(s) of k for which the quadratic equation x2 − 5x + (k + 1) = 0 has equal roots.

Solve x2 − 10x + 25 = 0 using an eﬃcient method after computing the discriminant. Justify your choice of method.

Solve 3x2 + 2x − 4 = 0 using an eﬃcient method after computing the discriminant. Justify your choice of method.

Find the values of m for which mx2 − 4x + 1 = 0 has two distinct real roots.

A quadratic equation has no real roots. If the coefficient a = 3 and the constant term c = 5, find the possible range of values for the coefficient b.

A right-angled triangle has sides of length x, x + a, and x + 2a: a

By using Pythagoras’ theorem, show that the side lengths are related by the equation x2 − 2ax − 3a2 = 0.

Solve this equation to ﬁnd x in terms of a, noting that x must be a positive length.

x + 2a

x+a

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“The science of operations, as derived from mathematics more especially, is a science of itself, and has its own abstract truth and value.” Ada Lovelace

Big ideas The concept of a function formalises the relationship between variables, defining a rule where each input corresponds to a unique output; this relationship is communicated using function notation and analysed by its key characteristics, such as domain, range, and intercepts.

2 Introduction to functions and relations Chapter outline 2.01 2.02 2.03

Functions and relations Variables and substitution Characteristics of functions Chapter 2 review

42 53 60 66

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

2.01 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

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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.

2.01 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.

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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

2.01 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 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.

2.01 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 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.

2.01 Functions and relations mathspace.co

2.01 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 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

⬚ ⬚ ⬚

2.01 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 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.

2.02 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

2.02 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 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.

2.02 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 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.

2.02 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’.

2.02 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 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).

2.02 Variables and substitution mathspace.co

2.03 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 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.

2.03 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 mathspace.co

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

2.03 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

2.03 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 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

2.03 Characteristics of functions mathspace.co

2 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)

Mathspace New South Wales – Year 11 Advanced mathspace.co

h(0.2)

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 2 review mathspace.co

Big ideas • The algebraic form of a linear function directly describes its geometric properties, such as gradient and intercepts. Key formulas allow for the determination of a line’s equation from given information, including its relationship (parallel or perpendicular) to other lines. • The solution to a linear inequality is a set of values, not a single point, found by using inverse operations with the critical rule of reversing the inequality symbol when multiplying or dividing by a negative number. • Linear functions and systems of equations are powerful tools for modelling real-world scenarios involving constant rates of change. The solution to such a system represents a point where multiple conditions are met simultaneously, such as a break-even point in business analysis.

3 Linear functions Chapter outline 3.01 3.02 3.03 3.04 3.05

Characteristics of linear graphs Equations of lines Linear inequalities Linear models Simultaneous equations Chapter 3 review

70 81 95 102 120 139

Ever noticed escalators follow perfect straight lines? They’re built on a constant slope —that’s a linear function!

3.01 Characteristics of linear graphs After this lesson, you will be able to… • identify the gradient and y-intercept from an equation in gradient-intercept form. • determine the x- and y-intercepts of a line from its equation. • graph a straight line using its gradient and y-intercept. • graph a straight line using its x- and y-intercepts. • convert a linear equation between gradient-intercept form and general form.

Equation in the gradient-intercept form Gradient The slope of a line. It is calculated as the gradient of a line segment. 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 .

The gradient represents the rate of change in the vertical direction relative to the horizontal direction. The equation of a straight line can be expressed in a form that highlights both its steepness and its point of intersection with the y-axis. This is known as the gradient-intercept form and is useful for describing and graphing linear relationships. Gradient-intercept form y = mx + c is the gradient-intercept form of a straight line, where m is the gradient of the line and (0, c) is the point at which the line intersects the y-axis. c is called the y-intercept.

The gradient-intercept form of a straight-line equation:

y = mx + c

m c

represents the gradient of the line is the y-intercept, which is the value of y when x = 0

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The graph of a linear function is always a straight line. 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. To determine the equation of a line in gradient-intercept form, both the gradient and the y-intercept must be identified. These values may be provided directly, extracted from a graph, or calculated from given coordinates. If two points on the line are known, the gradient can be found using the formula of m as given earlier, and then one point can be substituted into the equation to determine the value of c. Note that while the gradient-intercept form y = mx + c makes it easy to identify the gradient and y-intercept, straight-line equations can also be expressed in other forms, such as the general form ax + by + c = 0.

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Example 1 A straight line has gradient 3 and y-intercept 4. Determine its equation in gradient-intercept form.

Create a strategy Use the gradient-intercept form y = mx + c, then identify the gradient m and the y-intercept, c, and finally substitute the given values for m and c.

Apply the idea y = mx + c

Write the formula

y = 3x + 4

Substitute m = 3 and c = 4

Example 2 A straight line passes through the point (2, 3) with a gradient of gradient-intercept form.

. Determine its equation in

Create a strategy Substitute the coordinates of the point (2, 3), and the value of the gradient, m, into the gradient-intercept form y = mx + c to determine the y-intercept, c.

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Apply the idea y = mx + c

Write the formula Substitute y = 3,

,x=2

3 = –1 + c

Evaluate the multiplication

c=4

Add 1 to both sides making c the subject

Substituting the values of m and c, the equation is: y = mx + c

Write the formula Substitute

and c = 4

Reflect and check Verify that the equation y = −

x + 4 is for a straight line that passes through the point (2, 3) by

substituting and x = 2 and y = 3 into

: Write the formula Substitute x = 2, y = 3

3 = –1 + 4

Evaluate the multiplication

3=3

Evaluate the right-hand side

The left-hand side equals the right-hand side, so the point (2, 3) lies on the line. This confirms that the equation is correct.

Example 3 Determine the equation of the straight line that passes through the points (2, 3) and (−1, 6), in the gradient-intercept form y = mx + c.

Create a strategy Substitute the points into the formula

, then use a point to determine the y-intercept, c.

Apply the idea Calculating the gradient: Write the formula Substitute (x1, y1 ) = (2, 3) and (x2, y2 ) = (−1, 6) Evaluate the numerator and denominator = −1

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Evaluate

Using the point (2, 3) to determine the y-intercept, c: y = mx + c

Write the gradient-intercept formula

3 = (−1) × 2 + c

Substitute x = 2, y = 3 and m = −1

3 = −2 + c

Evaluate the multiplication

c=5

Add 2 to both sides making c the subject

Substituting back m = −1 and c = 5: y = mx + c

Write the gradient-intercept formula

y = (−1) × x + 5

Substitute m = −1 and c = 5

y = −x + 5

Simplify

The equation is y = −x + 5.

Idea summary The gradient-intercept form of a linear equation is written as y = mx + c, where m is the gradient and c is the y-intercept. A function is described as linear function if it can be expressed in this form. The gradient represents the rate of change in y with respect to x, and the y-intercept indicates the value of y when x = 0. This form, y = mx + c, is useful for both interpreting and graphing straight-line equations. To determine the equation of a line in this form, identify the gradient and y-intercept from given information, or use two known points to calculate the gradient m and substitute one point to find the y-intercept c.

Graph with gradient-intercept form The gradient-intercept form of a straight-line equation, y = mx + c, is particularly useful for graphing because both key features of the line, the gradient m and the y-intercept c, are immediately identifiable. To graph a straight line using this form, begin by identifying the y-intercept c. The y-intercept corresponds to the point (0, c), where the line crosses the y-axis. This point should be plotted first. Next, identify the gradient m, which represents the rate of change in y with respect to x, and which can be used to locate a second point on the line. The gradient can be interpreted as a ratio of rise to run. From the y-intercept, move horizontally and vertically according to the gradient to plot a second point. For example, a gradient of 2 units up.

indicates a movement of 3 units to the right and

After plotting two points, draw a straight line that extends through them in both directions. Alternatively, if the gradient is not easily interpreted as a ratio, or if additional precision is needed, a second point can be found by substituting any convenient value of x into the equation to calculate the corresponding y-value.

3.01 Characteristics of linear graphs mathspace.co

Note that if the gradient is negative, the direction of movement changes. For example, a gradient can be interpreted as moving 3 units to the right and 2 units down, or 3 units to the

left and 2 units up. Either interpretation will correctly position a second point, as long as the ratio between rise and run is maintained.

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Example 4 Graph the straight line represented by the equation y = 2x − 3.

Create a strategy Identify and plot the y-intercept, then use the gradient m to determine a second point. Draw a straight line through the two points to complete the graph.

Apply the idea y 3

y = 2x–3

The y-intercept is c = −3, so the point (0, −3) should be plotted first.

x –3

–2

–1

1 –1 –2

The gradient is

move 1 unit right and 2 units up to locate a second point at (1, −1). Draw a line through the points (0, −3) and (1, −1).

(0, –3)

–3

. From (0, −3),

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Idea summary To graph a straight line given in gradient-intercept form, y = mx + c, begin by identifying and plotting the y-intercept at (0, c). The gradient m, interpreted as a rise over run, is then used to determine a second point on the line. A straight line is drawn through both points to complete the graph.

Equation in the general form and graph General form (of straight line) A straight line in general form is given as ax + by + c = 0, where a, b and c are constants. The general form does not immediately display the gradient or the y-intercept, but it remains suitable for graphing, particularly when the coefficients are integers and the intercepts are easy to identify. To graph a straight line from general form, begin by finding the intercepts. The x-intercept is determined by setting y = 0 and solving the resulting equation for x. This gives the point where the line crosses the x-axis. Similarly, the y-intercept is found by setting x = 0 and solving for y, identifying where the line crosses the y-axis. Once both intercepts have been determined, plot the corresponding points on the Cartesian plane. A straight line is then drawn through them and extended in both directions. If either intercept is fractional or inconvenient to work with, the equation can be rearranged into gradient-intercept form y = mx + c, or a second point may be obtained by substituting a convenient value for x and solving for the corresponding y-value. This method provides a practical approach for graphing straight lines when equations are not initially presented in gradient-intercept form. Note that it is often helpful to make the coefficient a positive in the general form. If necessary, use algebraic techniques (such as multiplying both sides of the equation by −1) to achieve this, as it can make the intercepts easier to identify and the equation simpler to interpret when graphing.

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Example 5 Express the straight line equation y = 2x − 5 in general form.

Create a strategy Rearrange all terms to one side, ensuring the coefficient of x is positive.

Apply the idea y = 2x − 5

Write the gradient-intercept form

−2x + 5 + y = 0

Subtract 2x − 5 from both sides

2x − 5 − y = 0

Multiply both sides by −1

2x − y − 5 = 0

Rewrite in ax + by + c = 0

Example 6 Consider the straight line equation 3x + 4y − 12 = 0: a Determine the x-intercept and y-intercept.

Create a strategy Set y = 0 to determine the x-intercept, and x = 0 to determine the y-intercept.

Apply the idea To determine the x-intercept, substitute y = 0: 3x + 4y − 12 = 0

Write the equation

3x + 4 × 0 − 12 = 0

Substitute y = 0

3x − 12 = 0

Evaluate the multiplication

3x = 12

Add 12 to both sides

x=4

Divide both sides by 3

Therefore, the x-intercept is at (4, 0). To determine the y-intercept, substitute x = 0: 3x + 4y − 12 = 0

Write the equation

3 × 0 + 4y − 12 = 0

Substitute x = 0

4y − 12 = 0 4y = 12

Add 12 to both sides

y=3

Divide both sides by 4

Therefore, the y-intercept is at (0, 3).

Evaluate the multiplication

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Reflect and check Verify by converting to gradient-intercept form: 3x + 4y − 12 = 0

Write the equation

4y = −3x + 12

Subtract 3x − 12 from both sides

y=− x+3

Divide both sides by 4

The y-intercept is (0, 3) and setting y = 0 gives x = 4. This confirms that intercepts are correct.

b Graph the straight line represented by the equation.

Create a strategy Plot the x-intercept and y-intercept from part (a), then draw a straight line through the two points.

Apply the idea Plot the points (4, 0) and (0, 3), and then draw a straight line through them. y 4 3

(0, 3)

2 1

(4, 0) –2

–1

–1 –2 –3 –4

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Reflect and check Alternatively, turn the equation into the gradient-intercept form to plot the y-intercept and use the gradient to find and plot another point: 3x + 4y − 12 = 0

Write the equation

4y = − 3x + 12

Add −3x + 12 to both sides

y=− x+3 Divide both sides by 4 So, the gradient is m = −

and the y-intercept is c = 3.

y 4 3

(0, 3)

2 1 –2

–1

Plot the y-intercept at (0, 3), then move 4 units right and 3 units down to locate and plot another point.

(4, 0) 6

–1 –2 –3 –4

Idea summary o graph a straight line given in general form ax + by + c = 0, begin by determining T the intercepts. The x-intercept is found by setting y = 0, and the y-intercept is found by setting x = 0. Plotting these two points and drawing a straight line through them provides an efficient method for graphing when the equation is not written in gradient-intercept form.

3.01 Practice questions What do you remember? 1

Determine whether each statement is true or false: a

The equation y = mx + c is in gradient-intercept form.

The value of x when y = 0 is called the y-intercept.

A linear function can be represented by any line, curved or straight.

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d 2

Identify whether each equation is in gradient-intercept form y = mx + c or general form ax + by + c = 0: a

y = −4x + 2

3x + 5y − 15 = 0

Its general form

x−3

For the equation y = 3x − 6, determine: a

Gradient, slope, and rate of change are all related to the same concept.

The gradient

y-intercept

Group these descriptions under the correct heading: Gradient or y-intercept: a

Rate of change in y with respect to x

Where the line crosses the y-axis

Represented by the constant m in y = mx + c

Appears as the point (0, c)

Affects the steepness of the line

Found by setting x = 0

Gradient

y-intercept

A student says: “In the equation y = x + 2, 2 is the y-intercept and there is no gradient.” What is incorrect about this statement?

Practice Ex 1

Ex 2

Ex 3

Determine the equation of each line in gradient-intercept form y = mx + c given the gradient and y-intercept: a

Gradient =

c e

and y-intercept = −2

Gradient = −2 and y-intercept = 5

Gradient = 1 and y-intercept =

Gradient =

Gradient = −1 and y-intercept =

and y-intercept = −4

and y-intercept = 3

Determine the equation of the straight line, in the gradient-intercept form y = mx + c, that passes through the given point with the specified gradient: a

Gradient = 2 and point (0, −7)

Gradient = −1 and point (0, 4)

Gradient =

and point (0, 5)

Gradient = 4 and point

Gradient =

and point (0, −1)

and point (0, −2)

Determine the equation of the line passing through each pair of points, in gradient-intercept form y = mx + c: a

(−3, 4) and (1, −2)

(0, 5) and (4, −1)

(−4, 5) and (2, 4)

(1, −4) and (−5, 2)

(−4, −1) and (2, 3)

and (5, 2)

3.01 Characteristics of linear graphs mathspace.co

Ex 4

Graph the straight line represented by each gradient-intercept equation: a

y = −3x + 4

c e 10

Ex 5

x −

Rewrite each equation in gradient-intercept form y = mx + c: a

x + 2y = 6

3x − 2y − 4 = 0

2x + 5y − 10 = 0

4x − 3y = 12

5x + 2y = 8

x = 4y − 16

Convert these equations from gradient-intercept form to general form: a

Ex 6

y = 3x − 7

y = −2x + 4

y = −x + 1

−4

For each straight-line equation: i

Determine the x-intercept and the y-intercept.

Graph the straight line.

5x + 2y − 10 = 0

2x + 3y − 6 = 0

4x − 3y − 12 = 0

x + 4y − 8 = 0

2x + 7y − 14 = 0

5x − y − 5 = 0

Extend your thinking 13

A parking fee in Sydney charges a flat fee of $5, plus an additional $3 for every hour parked: a

Form the equation for the total cost y, in terms of the number of hours parked x.

Identify the y-intercept and explain what it represents in this context.

A phone plan charges a fixed fee plus a per-minute rate. The total cost is given by the equation y = 2x + 5, where x is the number of minutes used, and y is the cost in dollars: a

Determine the y-intercept and interpret its meaning in the context of the phone plan.

Determine the x-intercept and explain why it may not be meaningful in this context.

Graph the straight line for 0 ≤ x ≤ 3, using the y-intercept and the gradient.

A bakery makes batches of muffins and cookies. Each muffin weighs 5 kg and each cookie weighs 3 kg. A batch has a total weight of 15 kg: a

Write an equation in the form ax + by + c = 0 to represent the total weight, where x is the number of muffins and y is the number of cookies.

Determine the x-intercept and y-intercept. Interpret what each intercept means in the context of the bakery.

Sketch the straight line for 0 ≤ x ≤ 3, labelling the intercepts.

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A straight line passes through the points

Determine the equation of the straight line in gradient-intercept form y = mx + c.

Find the x-intercept and y-intercept of this line.

Find the point that lies exactly halfway between the given points and show that this point lies on the line by substituting it into the equation from part (a).

3.02

Equations of lines

After this lesson, you will be able to… • use the point-gradient formula to find the equation of a line. • find the equation of a line passing through two given points. • find the equation of a line that is parallel or perpendicular to a given line. • express linear equations in gradient-intercept or general form as required.

Point-gradient form Point-gradient form The equation of a line given its gradient and a point, written as y − y1 = m(x − x1). The point-gradient formula is used to find the equation of a line when given its gradient m and a point (x1, y1 ).

y − y1 = m(x − x1 ) m is the gradient of the line (x1, y1) is a point on the line To use the formula, substitute the given gradient and coordinates, then rearrange into gradientintercept form y = mx + c if required.

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3.02 Equations of lines mathspace.co

Example 1 Find the equation of a line with gradient 3 passing through (1, 4) in gradient-intercept form.

Create a strategy Substitute m = 3, x1 = 1 and y1 = 4 into the point-gradient formula y − y1 = m(x − x1).

Apply the idea y − y1 = m(x − x1)

Write the point-gradient formula

y − 4 = 3(x − 1)

Substitute m = 3, x1 = 1 and y1 = 4

y − 4 = 3x − 3

Expand the brackets

y = 3x + 1

Add 4 to both sides

Idea summary The point-gradient formula y − y1 = m(x − x1) determines a line’s equation using its gradient m and a point (x1, y1 ). Rearrange to y = mx + c for gradient-intercept form.

Line through two points To find the equation of a line passing through two points (x1, y1 ) and (x2, y2 ), calculate the gradient using:

m is the gradient of the line (x1, y1) are the coordinates of the first point (x2, y2) are the coordinates of the second point Then, use the point-gradient formula y − y1 = m(x − x1) with one point and the calculated gradient, rearranging to y = mx + c.

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(x2 , y2)

4 3 2

(x1 , y1) 1 –4

–3

–2

–1

For points (−2, 1) and (3, 4), the gradient is

–1 –2 –3 –4

Example 2 Find the equation of the line through (1, 2) and (3, 6) in gradient-intercept form.

Create a strategy Calculate the gradient using rearrange to y = mx + c.

, then use the point-gradient formula with one point and

Apply the idea Calculate the gradient: Write the formula Substitute (x1, y1) = (1, 2) and (x2, y2) = (3,6)

Evaluate the subtraction = 2

Simplify

Use the point-gradient formula with m = 2 and (1, 2): y − y1 = m(x − x1)

Write the formula

y − 2 = 2(x − 1)

Substitute into m = 2, x1 = 1 and y1 = 2

y − 2 = 2x − 2

Expand the brackets

y = 2x

Add 2 to both sides

3.02 Equations of lines mathspace.co

Idea summary For two points (x1 − y1) and calculate the gradient

, then use the

point-gradient formula with one point to find the equation, rearranging to y = mx + c.

Parallel lines Parallel Two distinct lines, rays or line segments in the same plane that have no points of intersection and so necessarily have the same gradient (slope). Two lines with equations y = m1 x + c1 and y = m2 x + c2 are parallel if their gradients are equal:

m1 = m2 If c1 = c2 as well, the two lines are identical rather than distinct. To determine the equation of a straight line parallel to a given line and passing through a specific point (x1, y1 ): 1. Identify the gradient m from the given line. 2. Use the point-gradient form y − y1 = m(x − x1) with the given point (x1, y1 ) and the same gradient m. 3. Rearrange into gradient-intercept form y = mx + c or general form ax + by + c = 0 as required. y 4

y = 2x + 3

y = 2x – 1

2 1

–4

–3

–2

–1

x 1

–1

For example, both lines have gradient m = 2 but different y-intercepts (c = −1, c = 3), so they are parallel.

–2 –3 –4

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Example 3 Determine the equation of the line parallel to y = 3x − 2 and passing through (1, 4) in gradientintercept form.

Create a strategy Use the gradient of the given line in the point-gradient form with the given point, then rearrange to gradient-intercept form.

Apply the idea The gradient of y = 3x − 2 is m = 3. For a parallel line, the gradient remains m = 3. Using the point-gradient form with (x1, y1 ) = (1, 4): y − y1 = m(x − x1)

Write the point-gradient formula

y − 4 = 3(x − 1)

Substitute m = 3, x1 = 1 and y1 = 4

y − 4 = 3x − 3

Expand the brackets

y = 3x + 1

Add 4 to both sides

The equation of the straight line parallel to y = 3x − 2 and passing through (1, 4) is y = 3x + 1.

Reflect and check The gradient of both lines is m = 3, and their y-intercepts differ (c = −2 and c = 1), which means they are distinct parallel lines. A sketch or graph can also be used to confirm this visually: y 6 5 4 3

y = 3x – 2

2 1 –4

–3

–2

y = 3x + 1

–1

x 1

–2 –3 –4

3.02 Equations of lines mathspace.co

Idea summary Two straight lines y = m1 x+c1 and y = m2x + c2 are parallel if they have equal gradients (m1 = m2) but different y-intercepts. To determine the equation of a line that is parallel to a given line, use the gradient of the given line in the point-gradient form y − y1 = m(x − x1) with the given point, then rearrange into the required form.

Perpendicular lines Perpendicular Two straight lines that intersect at a 90° angle (a right angle). If the gradients of the two lines are m1 and m2, they are perpendicular if: m1 × m2 = −1 This means the gradient of one line is the negative reciprocal of the other: and Note that a vertical line (undefined gradient) is always perpendicular to a horizontal line (gradient 0). To determine the equation of a line perpendicular to a given line through a specific point: 1.

Determine the gradient m of the given line. .

2. Calculate the gradient of a line perpendicular to it using

3. Use the point-gradient form with the given point (x1, y1 ): y − y1 = m2(x − x1) 4. Rearrange into gradient-intercept or general form as required. y 4 3

y=–

1 x+2 2

For example, on this graph, the gradients of the

–4 –3 –2 –1

1 –1

y = 2x – 1

x 2

two straight lines are m1 = 2 and product is 2 × are perpendicular.

–2 –3 –4

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. Their

= −1, confirming the two lines

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Example 4 Determine the equation of the line perpendicular to in gradient-intercept form.

and passing through (3, 1),

Create a strategy Calculate the negative reciprocal of the given gradient, use the point-gradient form with the given point, then rearrange to gradient-intercept form.

Apply the idea The gradient of

, so the gradient of the perpendicular line is: Write the formula

= − 3

Substitute Evaluate

Use the point-gradient form with m2 = −3, (x1, y1 ) = (3, 1): y − y1 = m2(x − x1) Write the formula y − 1 = −3(x − 3)

Substitute m2 = −3 and (x1, y1 ) = (3, 1)

y − 1 = −3x + 9

Expand the brackets

y = −3x + 10

Add 1 to both sides is y = −3x + 10.

The equation of the line perpendicular to

Reflect and check Verify: Multiply the gradients = −1

Evaluate

3.02 Equations of lines mathspace.co

Gradients product is −1, confirming the lines are perpendicular. This can also be verified by graphing the two lines: y 10

y = –3x + 10

8 6 4

1 y= x+2 3 –4

2 x

–2

–2 –4

The lines intersect at a 90° angle, confirming the lines are perpendicular. Point (3, 1) lies on y = −3x + 10, satisfying the equation.

Idea summary wo lines with gradients m1 and m2 are perpendicular if the product of their T gradients is m1 × m2 = −1. To determine the equation of a line perpendicular to a given line, and passing through a point (x1, y1 ), take the negative reciprocal of the given gradient, substitute it into the point-gradient form with the given point, and rearrange into the required form.

Applications of parallel and perpendicular lines Parallel and perpendicular lines are used in real-world contexts, such as engineering, architecture, and physics, to model relationships like parallel roads, perpendicular walls, or reflected paths in optics. These applications often involve constraints, such as fixed points or specific angles. To solve problems involving parallel or perpendicular lines: 1. Identify the given line’s equation and determine its gradient m. 2. For parallel lines, use the same gradient; for perpendicular lines, use the negative reciprocal . 3. Apply the point-gradient form with given points or constraints, and consider the context (e.g., horizontal/vertical lines). 4. Verify the solution geometrically (e.g., angles, intercepts) or algebraically.

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Example 5 A laser beam follows the path y = −2x + 6 and reflects off a mirror along the x-axis. The reflection passes through (4, 2). a Find the point of reflection on the mirror.

Create a strategy The mirror is horizontal, so it’s represented by the x-axis (y = 0). Find the point where the laser beam’s path intersects y = 0.

Apply the idea Set y = 0 in y = −2x + 6: y = −2x + 6

Write the equation

0 = −2x + 6

Substitute y = 0

−6 = −2x

Subtract 6 from both sides

x =3

Divide both sides by −2 making x the subject

The point of reflection is (3, 0).

b Determine the equation of the reflected path in gradient-intercept form.

Create a strategy Use the two points (3, 0) (point of reflection) and (4, 2) to determine the gradient, then apply the point-gradient form and rearrange to y = mx + c.

Apply the idea Calculate the gradient of the reflected path using (3, 0) and (4, 2): Write the formula

Substitute (x1, y1 ) = (3, 0) and (x2, y2 ) = (4, 2) Evaluate the subtraction

= 2

Simplify

Use point-gradient form with m = 2 and (3, 0): y − y1 = m(x − x1)

Write the formula

y − 0 = 2(x − 3)

Substitute m = 2, x1 = 3 and y1 = 0

y = 2x − 6

Expand the brackets and simplify

The reflected path is y = 2x − 6.

3.02 Equations of lines mathspace.co

Reflect and check This can also be confirmed by graphing the two equations in gradient-intercept form. y 6 4 2

y = –2x + 6 –1

y = 2x – 6 3

–2 –4 –6

The laser reflects at (3,0), passing through (4, 2).

Idea summary arallel and perpendicular lines model real-world scenarios like reflections or P structural alignments. Parallel lines have the same gradient, perpendicular lines have gradients that are negative reciprocals (m1 × m2 = −1), and equations are found using point-gradient form, applying any given constraints such as fixed points or specified angles.

3.02 Practice questions What do you remember? 1

In the equation expressed in gradient-intercept form y = mx + c: a

What does m describe on the graph?

What does c describe on the graph?

In the equation expressed in general form ax + by + c = 0: a

How is the x-intercept found?

How is the y-intercept found?

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Using the graph, determine if the lines are parallel or perpendicular: a

a and b

b and c

a and d

b and e

3 2 1

c –4

x –3

–2

–1

a –2 –3 –4

Consider two straight lines with equation y = m1 x + c1 and y = m2 x + c2. a

What must be true about the gradients m1 and m2: i

If the two lines are distinct parallel

If the two lines are perpendicular

What must be true about the gradients m1 and m2, and the y-intercepts c1 and c2 if the two lines are identical? y Given the lines:

• y=x+5 • y=x+2 • y=x−3 Determine whether each statement is true or false: a

They all have the same gradient.

They are all parallel to y = x.

They all have positive y-intercepts.

y=x+5 x –5

y=x+2 –5

y=x–3

Give one real-world example where parallel or perpendicular lines are used.

3.02 Equations of lines mathspace.co

Practice Ex 1

Ex 2

Ex 3

Ex 4

Write the equation of each line in gradient-intercept form: a

Gradient = 2 and point (1, 3)

Gradient = −1 and point (2, 5)

Gradient =

and point (0, 4)

Gradient =

and point

Gradient =

and point

Gradient =

and point (−1, −2)

Write the equation of each line in general form: a

Gradient =

and point

Gradient = −2 and point (−3, 4)

Gradient =

and point

Gradient =

and point

Gradient = 1 and point

Gradient =

and point

Graph the straight line that: a

passes through the points (1, 2) and (2, 3)

has a gradient of 3 and passes through the point (1, 1)

has a gradient of −2 and a y-intercept of 2

is parallel to the x-axis and passes through the point (3, −1)

Determine, in gradient-intercept form, the equation of the line that is: a

Parallel to y = 2x − 5 and passes through (1, 4)

Parallel to

and passes through (−2, 1)

Parallel to

and passes through (5, 0)

Parallel to

and passes through

Parallel to

and passes through (−1, 2)

Parallel to y = −x + 3 and passes through

Determine, in gradient-intercept form, the equation of the line that is: a

Perpendicular to

Perpendicular to y = −2x + 5 and passes through (0, −4)

Perpendicular to

and passes through (1, 2)

Perpendicular to

and passes through

Perpendicular to y = −3x + 2 and passes through (4, −1)

Perpendicular to

and passes through (0, 6)

and passes through

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Determine the equation, in general form, of the line parallel to the given line in the diagram and passing through point P : a

7 6 5 4 3 2 1

7 6 5 4 l 3 2 1

–7–6–5–4–3–2 –1 1 2 3 4 5 6 7 –1 –2 –3 –4 P –5 –6 –7

P x 1 2 3 4 5 6 7

Determine the equation, in general form, of the line perpendicular to the given line through point P: a

Ex 5

–7–6–5–4–3–2 –1 –1 –2 l –3 –4 –5 –6 –7

1 2 3 4 5 6 7

–2 –3 –4 –5 –6 –7

7 6 5 4 3 2 1

–7–6–5–4–3–2 –1–1

7 6 5 4 3 2 1

–7–6–5–4–3–2 –1–1 P

x 1 2 3 4 5 6 7

–2 –3 –4 –5 –6 –7

A sonar signal off NSW’s coast follows y = −x + 2, reflecting off a barrier at y = 0: a

Find the point of reflection.

Determine the equation of the reflected path in gradient-intercept form.

Consider the three equations: • First equation: • Second equation: 4x + y − 12 = 0 • Third equation: y − 4 = 3(x + 2) a

For each, state the fastest graphing method (intercepts, rise/run, or convert forms).

Graph the three straight lines on the same Cartesian plane.

3.02 Equations of lines mathspace.co

Extend your thinking 16

A straight line L has equation

. A second line Lp is parallel to the line L and must

pass through a point in the first quadrant with integer coordinates and y-intercept ≥ −1. Determine an example equation for Lp. 17

A surveyor maps a straight path between two markers at coordinates (in kilometres): a

Determine the equation of the path in gradient-intercept form.

Convert the equation to general form.

Verify if the point

and

lies on the path.

A fence follows the line

. A perpendicular fence passes through the intersection of

this line and 4x − 5y + 10 = 0:

Find the point of intersection of the two lines.

Determine the equation of the perpendicular fence in general form.

Verify that the two fences are perpendicular.

A farmer measures crop height y (in centimetres) at two points in time. On day 4, plants are cm tall; on day 10, they are

cm tall:

Write the equation in gradient-intercept form.

Convert to general form.

Interpret the gradient in the context of the crop height.

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3.03

Linear inequalities

After this lesson, you will be able to… • solve linear inequalities using algebraic operations. • recognise the condition under which the inequality symbol must be reversed. • represent the solution of a linear inequality on a number line. • distinguish between inclusive (≤, ≥) and strict (<, >) inequalities in graphical representations.

Solve linear 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

Inequalities are important because in real life, many conditions are not exact—they set limits or thresholds. For example, “You must be at least 16 to drive” is age ≥ 16, and “Speed must be under 60 km/h” is speed < 60. Linear inequalities are solved in a similar way to equations, by isolating the variable and applying these properties to keep the inequality true: • Addition or subtraction: Adding or subtracting the same number on both sides keeps the inequality direction. If a < b, then a + c < b + c and a − c < b − c. • Multiplication or division by a positive number: Multiplying or dividing both sides by a positive number keeps the inequality direction the same. If a < b and c > 0, then a × c < b × c and

• Multiplication or division by a negative number: Multiplying or dividing both sides by a negative number reverses the inequality direction. If a < b and c < 0, then a × c > b × c and (inequality reverses). To solve a linear inequality: 1. Simplify both sides (e.g. expand brackets, collect like terms). 2. Isolate the variable using addition, subtraction, multiplication, or division. 3. Reverse the inequality symbol if multiplying or dividing by a negative number. 4. Write the solution set as an inequality (e.g. x > 4). 5. Check your solution by testing a number inside the solution set and one outside.

3.03 Linear inequalities mathspace.co

Interactive exploration Discover this concept in action online

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Example 1 Solve the inequality 3x − 7 > 5.

Create a strategy Use inverse operations to isolate the variable in the inequality.

Apply the idea 3x − 7 > 5

Write the inequality

3x > 12

Add 7 to both sides

x >4

Divide both sides by 3

The solution set is x > 4.

Reflect and check Check two values: one inside the solution set (x = 5) and one outside (x = 4) Test with x = 5: 3x − 7 > 5 3×5−7 >5

Write the inequality Substitute x = 5

15 − 7 > 5

Evaluate

8 >5

Simplify

Since this is a true statement, this confirms the number 5 is in the solution set. Test with x = 4: 3x − 7 > 5 3×4−7 >5

Write the inequality Substitute x = 4

12 − 7 > 5

Evaluate

5 >5

Simplify

Since this is a false statement, this confirms x = 4 is not in the solution set.

Example 2 Solve the inequality −2(x + 3) ≤ 4.

Create a strategy Expand the bracket, then apply the inverse operations to isolate the variable in the inequality.

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Apply the idea −2(x + 3) ≤ 4

Write the inequality

−2x − 6 ≤ 4

Expand the brackets

−2x ≤ ≤ 10

Add 6 to both sides

x ≥ −5

Divide both sides by −2 reversing the inequality

The solution set is x ≥ −5.

Reflect and check Check two values: one inside the solution set (x = −5) and one outside x = −6) Test with x = −5: −2(x + 3) ≤ 4

Write the inequality

−2((−5) ≤ 4 + 3)

Substitute x = −5

−2 × (−2) ≤ 4 4 ≤4

Evaluate the expression inside the brackets Evaluate

Since this is a true statement, this confirms the number −5 is in the solution set. Test with x = −6: −2(x + 3) ≤ 4

Write the inequality

−2((−6) ≤ 4 + 3)

Substitute x = −6

−2 × (−3) ≤ 4 6 ≤4

Evaluate the expression inside the brackets Evaluate

Since this is a false statement, this confirms x = −6 is not in the solution set.

Idea summary An inequality is a mathematical statement that compares two quantities, showing that one is less than, greater than, or equal to the other. Linear inequalities are solved in a similar way to equations by isolating the variable and applying the properties of inequalities. The key difference is that when multiplying or dividing both sides by a negative number, the inequality sign must be reversed. The solution set contains all values that make the inequality true and can be written in inequality form (e.g. x ≥ −5).

3.03 Linear inequalities mathspace.co

Graph linear inequalities A number line is a straight, horizontal line where each point represents a real number. It is often used to show the solution set of an inequality visually. Graphing an inequality makes it easy to see which numbers satisfy the condition and whether the boundary value is included. To graph the solution set of a linear inequality on a number line: 1. Identify the boundary point from the solution (e.g. x = 4 in x ≥ 4). 2. Mark the boundary point with: • Open circle for strict inequalities < or > (boundary not included). • Closed circle for inclusive inequalities ≤ or ≥ (boundary included). 3. Draw an arrow to indicate the direction of all values that satisfy the inequality: • Left for < or ≤ • Right for > or ≥ 4. Label the number line with key values, especially the boundary point, so it is clear where the solution starts or ends.

–5 –4 –3 –2 –1 0

For x < 2, the open circle at 2 indicates 2 is not included in the solution set, and the arrow to the left shows the values that satisfy the inequality. For x ≥ −3, the closed circle at −3 indicates −3 is included in the solution set, and the arrow to the right shows all numbers greater than or equal to −3.

Interactive exploration Discover this concept in action online

Example 3 Consider the inequality

a Solve the inequality.

Create a strategy Apply the inverse operations to isolate x.

Apply the idea Write the inequality Subtract 1 from both sides x ≥ 4 Multiply both sides by 2 The solution set is x ≥ 4.

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b Graph the solution on a number line.

Create a strategy Mark the boundary point at x = 4 with a closed circle (filled) to show it is included in the solution set, then draw an arrow to the right for all values greater than or equal to 4.

Apply the idea

–5 –4 –3 –2 –1 0

Idea summary number line is used to clearly represent the solution set of a linear inequality. A The boundary point is shown with a closed circle (≤, ≥) if it is included in the solution set, or with an open circle (<, >) if it is not. An arrow is drawn to show all values that satisfy the inequality: left for < or ≤, and right for > or ≥.

3.03 Practice questions What do you remember? 1

Explain the differences between the inequalities: 0 ≤ x < 4 and 3 ≤ x ≤ 7.

Which of these number lines represent the solution set for x < 3? A

B –5 –4 –3 –2 –1 0

2 3 4 5

–5 –4 –3 –2 –1 0

2 3 4 5

C 3

–5 –4 –3 –2 –1 0

2 3 4 5

Which of these inequalities represents the solution set for r in “5 more than 2r is less than 39”? A

–5 –4 –3 –2 –1 0 1 2 3 4 5

r > 17

r < 17

r > 22

r < 22

Write each statement as an inequality using mathematical symbols: a

The sum of 3 groups of p and 9 is less than 24.

The sum of 5 times x and 3 is at least 23.

Six more than the value of x is at least seven.

Half of x is no more than five.

The product of negative four and x is at most three.

3.03 Linear inequalities mathspace.co

Practice Ex 1

Ex 2

Solve: a

3x − 7 < 8

2(x − 3) < −16

4x + 1 > 9

−5 + 2x < 7

−6x − 7 ≤ 5

Solve: a

5x − 40 ≥ 50

−8(x + 8) ≥ −40

e 7

x > −9

x ≤ 15

44 46 48 50 52 54 56 58 60 62 64 66

B 44 46 48 50 52 54 56 58 60 62 64 66

D 44 46 48 50 52 54 56 58 60 62 64 66

44 46 48 50 52 54 56 58 60 62 64 66

For each inequality: i

Solve for x.

Represent the solution set on a number line.

4 < 6x − 2

c 10

To ride the Ferris wheel in Fantasy World, a child must be at least 47 inches tall and less than 64 inches tall. Which of these number lines represent these conditions? A

Ex 3

2 − 3.6x ≤ 20.9

Represent the solution set of each inequality on a number line: a

−8 − m > 3

1.5x + 8 > 12.5

Consider the situation “3 less than 3 groups of p is no more than 24”: a

Write the situation as an inequality.

Solve the inequality.

Find the largest value p can take.

Skye has a budget for school stationery of $43, but has already spent $21.14 on books and folders. Let p represent the amount that Skye can spend on other stationery. Write the budget constraint as an inequality and solve for p.

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James is saving up to buy a laptop that costs $550. He has $410 in his bank account and expects to receive some money for his birthday next month: a

Let x represent the amount he is to receive for his birthday. Write an inequality to model the situation.

Represent the solution set of the inequality on a number line.

Extend your thinking 13

Ryan wants to save enough money to buy a new sports equipment set, which costs $40.00. He has already saved $22.10 from his birthday. To make more money, he plans to wash neighbours’ windows for $2 per window: a

Let w be the number of windows that Ryan washes. Write an inequality to represent the situation.

Solve for w, rounded to two decimal places.

State whether each statement is correct. Explain your thinking. i

Ryan must wash more than 9 windows to be able to afford the equipment.

Ryan must wash at least 8 windows to be able to afford the equipment.

iii

If Ryan washes 8 windows and 95% of another window, he can afford the equipment.

The number of windows Ryan must wash to be able to afford the equipment must be greater than or equal to 9.

Rochelle tried to solve the following inequality but made a mistake in her work: Step 0: − 4 − 2x > 10 Step 1:

−2x > 14

Step 2:

x>−7

Determine which step is incorrect and explain the error. 15

Percy tried to represent the solution set of the inequality 4x + 28 ≥ −8 on a number line. However, his answer is incorrect. –10 –9 –8 –7 –6 –5 –4 –3 –2

–1

Identify the error(s) in Percy’s work and explain how to correct them.

3.03 Linear inequalities mathspace.co

101

3.04 Linear models After this lesson, you will be able to… • identify the independent and dependent variables in a real-world context. • construct linear functions to model practical situations. • interpret the gradient as a rate of change and the vertical intercept as an initial value. • use linear inequalities to model constraints in real-world problems. • solve problems using linear models and justify conclusions within the context of the problem.

Model with linear functions Many real-world situations involve relationships that can be described using a straight-line equation or inequality. These situations can be modelled mathematically using linear functions, solved, and interpreted in the context of the problem. Model A mathematical, conceptual or physical representation that describes, simplifies, clarifies or provides an explanation of the structure, workings or relationships within an object, system or idea. Models can provide a means of testing and predicting behaviour within limited conditions. Models may be physical or exist in digital form. Independent variable A variable used to represent values in the domain (input values) of a function. Generally represented on the horizontal axis of a graph. Dependent variable The variable used to represent the output values of a function. A dependent variable is generally represented on the vertical axis of a graph.

The independent and dependent variables in a straight-line equation often represent physical quantities such as time, distance, cost, mass, or temperature.

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Fuel consumption of a mid-sized SUV Volume of fuel in tank V (L) 60 50

For example, if modelling the rate of fuel being used in a car: • The independent variable may represent the distance travelled, in kilometres • The dependent variable may represent the volume of fuel, in litres, in the car’s fuel tank

40 30

Instead of x and y, the variables d and V can be used to better represent the context. Each variable has appropriate units of measurement.

20 10 Distance travelled d (km) 0

200 400 600 800 1000

Recall, the two key features of a linear function, or straight-line graph, are the gradient and the y-intercept. In linear modelling situations, the y-intercept is often called the vertical intercept, because the vertical axis may be labelled with a variable other than y. Fuel consumption of a mid-sized SUV Volume of fuel in tank V (L) lnitial value

60 50

rate of change

40 30

The vertical intercept represents an initial value. In the example shown, the vertical intercept is 63 litres. It represents the volume of fuel in a full tank, before the car began its journey. The gradient represents a rate of change. Using the example shown, the gradient is a measure of the car’s fuel consumption in L/km or L/ 100 km.

20 10 Distance travelled d (km) 0

200 400 600 800 1000

Rate of change For a function, the relative change in a function’s value f (x) to a change in x. Linear graphs are often extended indefinitely in both directions. However, most physical quantities such as distance, volume, or time, cannot be negative, so many linear models exist primarily in the first quadrant of the Cartesian plane. This is not always the case; for example, temperature can have negative values. When analysing a linear model, consider which values are appropriate in the given context.

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Example 1 This graph shows the amount of water remaining in a bucket that was initially full before a hole was made in its side: Quantity (L)

40 36 32 28 24 20 16 12 8 4

Time (mins) 0

a Determine the gradient of the function.

Create a strategy Use the formula:

Apply the idea Quantity (L)

40 36 32 28 24 20 16 12 8 4

Time (mins) 0

Using the intercepts (0, 30) and (60, 0), the rise is 30 units down and the run is 60 units across. Write the formula for the gradient Substitute the values Simplify

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b Determine the vertical intercept.

Create a strategy Determine the value at which the line intersects the y-axis.

Apply the idea The line intersects the y-axis at y = 30.

c Explain what the gradient represents in this context.

Create a strategy The gradient represents a rate of change between the given physical quantities.

Apply the idea The gradient represents the rate of change of the amount of water in the bucket for a change in time. The gradient of

means the amount of water in the bucket decreases by 1 litre every 2 minutes.

Reflect and check This also means that 1 litre of water is lost every 2 minutes.

d Explain what the vertical intercept represents in this context.

Create a strategy The y-intercept or vertical intercept represents the initial value.

Apply the idea The vertical intercept represents the initial amount of water in the bucket, before water began flowing out through the hole. The bucket initially contained 30 litres.

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e Determine an equation using appropriate variables to represent the situation.

Create a strategy Use the variable V to represent volume (dependent variable) and t to represent time (independent variable), then construct an equation in a similar format to y = mx + c.

Apply the idea y = mx + c

Write the equation in gradient-intercept form

V = mt + c

Substitute the variables y = V and x = t and the vertical intercept c = 30

Substitute the gradient

where V is the amount of water in the bucket in litres, and

The equation of the line is t is the time in minutes.

Reflect and check The letters used for variables can be uppercase or lowercase. What matters is that each variable is clearly defined, along with its units, when modelling straight-line graphs in context.

Determine the amount of water remaining in the bucket after 20 minutes.

Create a strategy Substitute t = 20 into the equation from part (e) and solve for V.

Apply the idea Write the equation Substitute t = 20

= −10 + 30

Evaluate the multiplication

= 20

Evaluate

The amount of water remaining in the bucket after 20 minutes is 20 L.

Example 2 Let the height of a candle be h cm. When the candle is lit, the height decreases according to the equation h = −2t + 8, where t is the elapsed time in minutes: a Complete this table of values: Time (t min)

Height of candle (h cm)

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Create a strategy Substitute each value of t into the equation h = −2t + 8.

Apply the idea Substitute t = 0, which gives h = −2 × (0) + 8 = 8. Substitute t = 1, which gives h = −2 × (1) + 8 = 6. Substituting t = 2 and t = 3, the complete table of values is: Time (t min)

Height of candle (h cm)

b Sketch the graph of h = −2t + 8.

Create a strategy Plot the four points from the table of values completed in part (a).

Apply the idea Plot the points (0, 8) , (1, 6) , (2, 4), and (3, 2). Then draw a straight line that passes through them. h 12 10 8 6 4 2 –8 –6 –4 –2 –2

t 2

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c The height of the candle and the time elapsed cannot be negative. Using part (b), for which values of t does the equation h = −2t + 8 accurately model the height of the candle?

Create a strategy Given that h ≥ 0 and t ≥ 0, the equation only models the height of the candle in the first quadrant.

Apply the idea h 12 10 8 6 4 2 –8 –6 –4 –2 –2

t 2

By observing the graph in the first quadrant, the graph is only valid for 0 ≤ t ≤ 4. At t = 4, h = −2(4) + 8 = 0, indicating the candle has completely burned.

Example 3 A carpenter charges a call-out fee of $150 plus $45 per hour: a Determine an equation to represent the total amount charged, C, by the carpenter as a function of the number of hours worked, h.

Create a strategy Express the relationship Total amount charged = $45 × Number of hours + $150 in mathematical form using the assigned variables.

Apply the idea

108

y = mx + c

Write the equation in gradient-intercept form

C = mh + c

Substitute the variables y = C and x = h

C = 45h + 150

Substitute the gradient m = 45 and the vertical intercept c = 150

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b Determine the total amount charged by the carpenter for 6 hours of work.

Create a strategy Substitute h = 6 into the equation from part (a) and solve for C.

Apply the idea C = 45h + 150

Write the equation

= 45 × 6 + 150 Substitute h = 6 = 270 + 150 Evaluate the multiplication = $420 Evaluate The total amount charged is $420.

c Determine the number of hours worked if the total amount charged is $735.

Create a strategy Substitute C = 735 into the equation from part (a) and solve for h.

Apply the idea C = 45h + 150

Write the equation

735 = 45h + 150

Substitute C = 735

585 = 45h

Subtract 150 from both sides

h = 13

Divide both sides by 45

The number of hours worked is 13 hours.

Idea summary A model is a representation that helps describe or explain a real situation. In a linear model, the independent variable (input) is usually on the horizontal axis, and the dependent variable (output) is usually on the vertical axis. In linear modelling situations: • The y-intercept or vertical intercept represents an initial value. • The gradient represents a rate of change. When analysing a linear model, consider carefully what values are appropriate in the given context.

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Model with linear inequalities A linear inequality shows all values that satisfy certain constraints. Many real-world situations can be modelled using linear inequalities, but the solutions must be interpreted in context. A viable solution satisfies the inequality and is practical for the situation, while a non-viable solution is mathematically correct but not realistic (e.g., negative or fractional items). To model real-world situations using linear inequalities and solve and interpret them in context: 1. Define a variable (e.g., x for number of items) and write an inequality based on the problem’s constraints. 2. Solve the inequality algebraically to find the solution set. 3. Interpret the solution in context, identifying viable solutions (e.g., whole numbers for items) and non-viable solutions (e.g., negative quantities). 4. Graph the solution on a number line, highlighting viable solutions. When analysing a linear inequality, consider both the mathematical result and what is meaningful in the given situation.

–1 0

2 3 4 5 6 7 8 9 10

For a budget inequality 5x ≤ 40 (where x is items costing $5 each), the solution x ≤ 8 is graphed. Viable solutions are whole numbers 0 ≤ x ≤ 8, as items cannot be negative or fractional.

Exploration Complete the chart by performing the indicated operations: Consider the inequality

Perform the operation on the inequality

1<4

Add 2 to both sides

6 > −2

Subtract 2 from both sides

3 < 10

Multiply by 2 on both sides

1 > −7

Multiply by − 2 on both sides

4>2

Divide by 2 on both sides

−8 < 12

Divide by − 2 on both sides

Write the new inequality

True or false?

1. Did any operations cause the given inequality to become a false inequality? 2. Can you think of something to change about a false inequality without changing the operation performed? 3. What general rule must be applied when multiplying or dividing both sides of an inequality by a negative number to maintain its truth?

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Example 4 Calandra charges $54.00 to style hair, as well as an additional $8.50 per foil. Pauline would like the total cost for her styling to be no more than $137.00: a Write an inequality that represents the number of foils Pauline could get.

Create a strategy Pauline has no more than $137.00 to spend. “No more than” means “less than or equal to.”

Apply the idea Write an inequality in words that represents the cost to style Pauline’s hair: Cost of styling + Cost per foil × Number of foils ≤ Total Pauline can spend Translating that into an algebraic expression: 54.00 + 8.50N ≤ 137.00 where N represents the number of foils.

b How many foils could Pauline get and still afford the styling?

Create a strategy Solve the inequality and then write the solution set.

Apply the idea 54.00 + 8.50N ≤ 137.00 8.50N ≤ 83.00 N ≤ 9.76

Write the inequality Subtract 54.00 from both sides Divide both sides by 8.50

According to the solution, Pauline could get 9.76 foils or fewer. However, since she can’t get a partial foil, a more realistic solution is that she can get 9 foils or fewer.

c Determine whether N = −2 is a viable solution to the inequality in the context of the question.

Create a strategy Keep in mind, it is not realistic to get part of a foil or a negative number of foils.

Apply the idea Pauline can get a maximum of 9 foils and a minimum of 0 foils, so while −2 is mathematically part of the solution set for the inequality N ≤ 9.76, it is not a viable solution in this context.

Reflect and check This is an example of a non-viable solution: it satisfies the inequality mathematically but is not possible in the real-world context.

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Idea summary Many real-world situations can be modelled using linear inequalities. A viable solution satisfies the inequality and makes sense in context, while a non-viable solution may be mathematically correct but is not practical. When solving, interpret the results carefully and keep only the values that are meaningful in the situation. To model and interpret real-world situations using linear inequalities: 1. Define a variable and write an inequality based on the problem’s constraints. 2. Solve the inequality algebraically to find the solution set. 3. Interpret the solution in context, identifying viable and non-viable solutions. 4. Graph the solution on a number line, highlighting viable solutions.

3.04 Practice questions What do you remember? 1

For the given graph, determine the: a

y-intercept

x-intercept

gradient

6 5 4 3 2 1 –1

–1

x 1

–2 2

Consider the pattern of blue boxes:

Complete the table of values: Number of columns (c)

Number of blue boxes (b) b

112

Determine an equation that describes the relationship between the number of blue boxes (b) and the number of columns (c).

Mathspace New South Wales – Year 11 Advanced mathspace.co

The table shows Mae’s heart rate at different times after she starts running. Her heart rate increases at a constant rate: Number of minutes passed, t

Heart rate, H

Determine Mae’s heart rate after 12 minutes.

Calculate the constant rate of change in heart rate, in beats per minute.

Determine an equation that describes the relationship between the number of minutes passed, t, and Mae’s heart rate, H.

Practice 4

Ex 1

The amount of medication M (in milligrams) in a patient’s body gradually decreases over time t (in hours) according to the equation M = 1050 − 15t. a

After 61 hours, how many milligrams of medication are left in the body?

Calculate the number of hours it will take for the medication to be completely removed from the body.

The graph shows the amount of water remaining in a bucket that was initially full before a hole was made in its side: a

Determine the gradient of the line.

Determine the y-intercept.

Explain the meaning of the gradient in this context.

Explain the meaning of the y-intercept in this context.

Determine an equation to represent the amount of water remaining in the bucket, W, as a function of time, t.

Determine the amount of water remaining in the bucket after 20 minutes.

32 Water (L) 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

Time (min) 48 56

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The graph shows the temperature of a room after the heater has been turned on: a

Determine the gradient of the line.

Determine the y-intercept.

Express the equation in gradient-intercept form.

Ex 2

Explain the meaning of the gradient in this context.

Explain the meaning of the y-intercept in this context.

Determine the temperature of the room after the heater has been turned on for 30 minutes.

Temp (0C)

4 2 1

Time (min) 0

1 2 3 Discuss the usefulness and limitations of this model (e.g., how long it is valid, what assumptions it makes).

5 6

8 9

The cost of a taxi ride C is given by C = 2.50t + 3 where t is the duration of the trip in minutes: a

Complete the table of values: Time in minutes (t)

Cost in dollars (C)

Sketch the graph of C = 2.50t + 3.

For which values of t does the graph accurately model the cost of the taxi ride?

A racing car starts the race with 250 litres of fuel. From there, it uses fuel at a rate of 5 litres per minute: a

Complete the table of values: Number of minutes passed, x

Amount of fuel left in tank, y

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Determine an equation relating the number of minutes passed, x, and the amount of fuel left in the tank, y.

Describe how the amount of fuel in the car is changing over time.

For which values of x does the graph accurately model the amount of fuel left in the tank?

A diver starts at the surface of the water and begins to descend below the surface at a constant rate. The table shows the depth of the diver over 4 minutes: Number of minutes passed, x

Depth of diver in metres, y

1.4

2.8

4.2

5.6

Calculate the increase in depth each minute.

Determine a linear equation for the relationship between the number of minutes passed, x, and the depth, y, of the diver.

Calculate the depth of the diver after 6 minutes.

Calculate the time it takes for the diver to reach a depth of 12.6 metres.

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Ex 3

Ex 4

The number of fish in a river is approximated over a five-year period. The results are shown in the table: Time in years, t

Number of fish, F

4800

4600

4400

4200

4000

3800

Sketch a graph that corresponds to this information.

Calculate the gradient of the line.

Explain the meaning of the gradient in this context.

Determine the value of F when the line crosses the vertical axis.

Determine an equation for the line, using the given values.

Determine the number of fish remaining in the river after 13 years.

Determine the number of years, t, when 2000 fish remain in the river.

Beth’s income is based solely on the number of hours she works, and she is paid a fixed hourly wage. She earns $25 per hour. Let I represent Beth’s income after working h hours: a

Determine an equation relating h and I.

Calculate Beth’s income when she works 25 hours.

Calculate the number of hours that Beth must work to earn $125.

A plumber charges a $75 call-out fee plus $60 per hour for repairs. Let C represent the total cost and h the number of hours worked: a

Determine an equation for C in terms of h.

Calculate the total cost for 5 hours of work.

Calculate the number of hours worked when the total cost is $315.

A landscaper charges $85.00 for a consultation plus $45.50 per hour for labour. A client’s budget is no more than $220.00. Let h represent the number of labour hours: a

Write an inequality for the number of hours the client can afford.

Solve the inequality to find the maximum number of hours the client can afford.

Determine whether h = −1 is a viable solution in the context of the question.

The graph shows the conversion between temperatures in Celsius and Fahrenheit: a

Use the graph to convert 10°C into Fahrenheit.

0°C is 32°F. For every 1°C increase, by how much does the Fahrenheit temperature increase?

Would 80°F be above or below normal body temperature (approximately 37°C)?

Determine the equation for conversion between Celsius (°C) and Fahrenheit (°F).

Convert 65°C into Fahrenheit.

100 90 80 70 60 50 40 30 20 10

C 0

5 10 15 20 25 30 35 40

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The graph shows the amount of Euros that can be bought with Australian dollars on a particular date:

Euros

12 10

How many Euros can 20 AUD buy?

How much Australian currency is required to buy 6 Euros?

Calculate the number of Euros that 1 AUD buys.

Determine the equation for conversion between AUD (A) and Euros (E).

How many Euros can 465 AUD buy?

4 2 AUD 0

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2 4 6 8 10 12 14 16 18 20

The cost, C, for a business to operate can be expressed in terms of h, the total number of hours it has operated for. The cost is $120 an hour: a

Sketch a graph that displays the cost against time.

Determine the gradient of the line.

Determine an equation relating h and C.

Calculate the total cost for the business to operate for 28 hours.

Determine the number of hours that the business operates for to incur a total cost of $3840.

A mobile phone salesperson earned $600 in a particular week during which he sold 26 phones, and $540 in another week during which he sold 20 phones. a

Determine a linear equation to represent the weekly earnings of the salesperson, E, as a function of the number of phones sold, n.

Determine how much the salesperson will earn in a week during which he sells 36 phones.

Gardening Company A charges a fixed call-out fee plus a certain amount for every hour of work performed on any particular garden. One customer pays $380 for 4 hours of work. Another customer pays $490 for 6 hours of work: a

Using a linear equation as a model, determine an expression for the amount of money, P, the company charges in terms of t, the number of hours worked.

How much would the company charge for a job involving 7 hours of work?

Gardening Company B charges a fixed fee of $130 for every job, plus $65 for every hour of labour. David usually takes 5.5 hours to complete his garden work but injured his hand and is unable to work. Which company should he call to work on his garden?

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A café notices that for every 5°C increase in temperature outside, they sell 20 fewer cups of coffee a day. On a certain day when the temperature was 15°C, they sold 250 cups of coffee: a

If the temperature outside is 25°C, predict how many cups of coffee the café will sell.

Estimate the temperature at which the café would sell no coffee.

The café decides to implement a marketing strategy that increases their cups of coffee sales by two cups per degree Celsius increase in temperature. On a 15°C day, they sell 275 cups of coffee. Determine a new equation for the model.

Consider the inequality: 5 (x + 3) ≤ 35. a

Solve for x.

State whether each solution is viable or non-viable: i

x = −4

x=8

iii

x=4

x=0

x=2

Consider the inequality: 4 − 2x < 3x − 2. a

Solve for x.

State whether each solution is viable or non-viable: ii

i 22

x=1

iii

When breeding certain types of fish it is recommended that the number of female fish is more than double the number of male fish: a

Write the inequality for the recommended relationship. Let f be the number of female fish and m be the number of male fish.

State whether each combination is viable or non-viable: i

f = 10, m = 8

f = 23, m = 9

iii

f = 13, m = 10

f = 7, m = 4

To get a grade of C or better, Uther must obtain an average score of at least 75 over his four exams. So far he has taken the first three exams and achieved scores of 68, 60, and 86: a

Write the inequality that models the situation for the score, x, Uther must obtain on his last exam to get a C or better.

Solve for x.

Interpret the solution in the context of Uther’s score.

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Extend your thinking 24

The table shows the maximum amount of a harmful gas to which the body can be exposed without adverse effects, based on the total duration of exposure: Use a linear model to express the amount of gas exposed, G, in terms of the duration of exposure, t.

Duration (seconds)

Amount (grams)

What is the maximum safest duration that 30 g of the gas can be exposed to the body?

Peter found out that he has been exposed to 40 g of the gas over 17 seconds. Is he safe or at risk from the effects of the gas?

Anyone who has been exposed to the gas for longer than the safest duration can be treated using a particular medicine A. For every gram of gas above the recommended amount exposed to the body, 4 grams of medicine A must be administered. John was exposed to 28 g of the gas over 26 seconds. How many grams of medicine A should he be administered?

Mario wants to determine which of two slow-release pain medications is more rapidly absorbed by the body. Consider the given graph and table, which show the amount of medication in the bloodstream for the liquid and capsule form of the medication: Liquid form A

Capsule form Time (min), t

Amount in blood (mg), A

24.6

42.3

77.7

35 30 25 20 15 10 5

t 0

5 6

8 9

In which form is the medication absorbed more rapidly?

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Luke purchases 30 L of mixed leaded and unleaded petrol solution, which costs him a total of $103.20. The price of leaded petrol is $3.54/L, while the price of unleaded petrol is $2.94/L. Let x and y be the number of litres of leaded and unleaded petrol that make up the solution respectively:

Determine the amount of leaded petrol that was used in the solution.

Determine the amount of unleaded petrol that was used in the solution.

The population of City A can be modelled by the equation P = 80 000 + 1500t, where t is the number of years since 2020. The population of City B is modelled by P = 60 000 + 2000t: a

In how many years will the populations of the two cities be equal?

If the population growth of City A continues at this rate, in what year will its population reach 150 000?

The graph shows the number of customers inside a department store t hours after it opens at 8:00 a.m.: a b c

How many customers are inside the store initially?

occur? 29

Number of customers

100 90 How many customers left the store from 80 10:00 a.m. to 2:00 p.m.? 70 The department store decides to close 60 for the day when there are fewer than 50 35 customers. When did they close? At one point in the day, customers bought 40 30 a total of $2475 worth of clothing. If, on 20 average, each customer bought $45 worth 10 of clothes, when during the day did this

Time (hours) 0

2 4 6 8 10 12 14 16 18 20 22 24

An investment Account A has a starting balance of $5000 and earns $200 per month. Another investment Account B starts with $7000 and earns $150 per month: a

Calculate the number of months required for both accounts to have the same balance.

After 15 months, $1500 is withdrawn from Account A, and $1750 is withdrawn from Account B. If the investments continue to grow as they did before, after how long from the start will the two accounts contain the same amount of money again?

James, who owns Account A, begins making monthly withdrawals of $300 after 12 months. If no further interest is earned after the withdrawals begin, how long after this will he use up all the money in the account?

A bakery sells cakes for $30 each and cupcakes for $5 each. The bakery has a budget of at most $300 for ingredients to make cakes and cupcakes. Let c represent the number of cakes and k the number of cupcakes: a

Write an inequality to model the bakery’s budget constraint.

Solve the inequality for k when c = 6.

Determine whether the combination c = 8, k = 15 is a viable solution in the context of the problem. 3.04 Linear models mathspace.co

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3.05

Simultaneous equations

After this lesson, you will be able to… • solve simultaneous linear equations by finding the point of intersection on a graph. • construct a pair of simultaneous linear equations from a word problem. • interpret the point of intersection in the context of a real-world problem. • model cost and revenue scenarios to find the break-even point. • analyse the viability of solutions in practical contexts.

Solve simultaneous equations graphically Simultaneous equation Also known as system of linear equations, is a set of 2 or more equations considered at the same time. The aim is to find values for the variables that make all the equations true together. In most cases at this level, there are two equations with two variables, usually written as x and y. Each equation can be represented by a straight line on a coordinate plane. The solution is the point of intersection, where the lines meet, and its coordinates satisfy both equations at the same time. A system of linear equations can have no solutions, infinitely many solutions, or one unique solution. y

120

When two lines are parallel and distinct, they have no points of intersection. The corresponding system of equations has no solutions.

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When two lines are identical, they overlap at each other. The corresponding system of equations has infinitely many solutions.

When two lines are not parallel, they have exactly one point of intersection. The corresponding system of equations has one solution.

A solution to a system of equations in a given context is said to be viable if the solution makes sense in the context and non-viable if it does not make sense within the context, even if it would otherwise be algebraically valid.

Example 1 Solve each system of equations graphically: a

1 2

4y = 3x − 24

Create a strategy Convert 2 in the system to gradient-intercept form and graph both equations using the y-intercept and gradient.

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Apply the idea 4y = 3x − 24

2 Divide both sides by 4

Since 2 is equivalent to 1 in the system, the lines will be the same, and therefore, the graph will show that the solution to the system of equations is infinite solutions. y

x –5

–5

6x + 3y = 18

12x + 6y = 24

Create a strategy Calculate the intercepts of each equation in the system and graph them.

Apply the idea Find the x-intercept of 1 : 6x + 3y = 18 6x + 3 × 0 = 18

122

1 Substitute y = 0

6x = 18

Evaluate the multiplication

x=3

Divide both sides by 6

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Find the y-intercept of 1 : 6x + 3y = 18

6 × 0 + 3y = 18

Substitute x = 0

3y = 18

Evaluate the multiplication

y=6

Divide both sides by 3

The intercepts of 1 are (3, 0) and (0, 6). y

x –5

–5

Find the x-intercept of 2 : 12x + 6y = 24 12x + 6 × 0 = 24

2 Substitute y = 0

12x = 24

Evaluate the multiplication

x=2

Divide both sides by 12

Find the y-intercept of 2 : 12x + 6y = 24 12 × 0 + 6y = 24

2 Substitute x = 0

6y = 24

Evaluate the multiplication

y=4

Divide both sides by 6

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The intercepts of 2 are (2, 0) and (0, 4). y

x –5

–5

The graphs of the equations never intersect, so there is no solution to the system of equations.

Reflect and check Converting the system of equations to gradient-intercept form before graphing shows that the lines have the same gradient but different y-intercepts, indicating that the system has no solution even before drawing the graphs.

Example 2 Gordiano made two trips to a flower shop to purchase roses and sunflowers. On his first trip, he purchased 4 roses and 4 sunflowers and paid $12. The following day, Gordiano went back to the flower shop and purchased 12 roses and 8 sunflowers for $16. The situation can be represented by this system of equations:

4x + 4y = 12

12x + 8y = 16

a Graph the system of equations.

Create a strategy First, define the variables. Based on the problem, each trip to the flower shop is represented in its own equation. Gordiano buys 4 roses, then 12, so the expressions 4x and 12x indicate the cost for the roses. Gordiano buys 4 sunflowers, then 8 sunflowers, so the terms 4y and 8y indicate the cost of the sunflowers. Let x = the cost per rose, and let y = the cost per sunflower. Now that the variables have been defined, the equations can be converted to gradient-intercept form and graphed on the same coordinate plane.

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Apply the idea Convert 1 to gradient-intercept form: 4x + 4y = 12

4x + 4y − 4x = 12 − 4x 4y = 12 − 4x y = −x + 3

Subtract 4x from both sides Combine like terms Divide both sides by 4

Convert 2 to gradient-intercept form: 12x + 8y = 16

12x + 8y − 12x = 16 − 12x 8y = 16 − 12x

Subtract 12x from both sides Combine like terms Divide both sides by 8

Flower shop purchases Cost per sunflower (dollars) 8 7 6 5 4 3 2 1 –5 –4 –3 –2 –1

Cost per rose (dollars) 1 2 3 4 5 6 7

b Interpret the solution.

Apply the idea Each equation models the relationship between the total cost and the quantities of each type of flower purchased on a given day. The solution to the system is (−3, 6), which suggests a cost of −$3 per rose and $6 per sunflower. A negative cost is not realistic in this context, so the solution is non-viable. This indicates that at least one of the per-item costs must have been different on the two days.

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Idea summary The solution to simultaneous equations (a system of linear equations) is the ordered pair that satisfies all equations in the system at the same time. On a coordinate plane, this solution is the point where the graphs of the equations intersect: the point of intersection. A system of two linear equations can have: • One solution: the lines intersect at exactly one point. • No solution: the lines are parallel and do not meet. • Infinitely many solutions: the lines are identical and overlap completely. Graphing both equations on the same set of axes makes it possible to determine how many solutions the system has and to locate the solution when it exists. In real-world problems, a solution is viable if it makes sense in the context and non-viable if it does not.

Applications of simultaneous equations Simultaneous linear equations model practical situations with two unknowns, such as quantities, prices, or rates. The equations are based on the relationships described in the problem. Solving the system gives values that satisfy all conditions and often represent quantities such as total sales, production costs, or distances travelled. To solve problems using simultaneous equations: 1.

Define variables for the unknowns (e.g. x for students, y for adults).

2. Form two equations from the information in the context (e.g. total attendees, total revenue). 3. Solve the system using a graphical method, substitution, elimination, or technology. 4. Interpret the solution in context and determine whether it is viable (for example, quantities must be non-negative integers). Solution methods: • Graphical: Model the relationships as straight-line graphs and identify their point of intersection. This shows the solution visually, though manual plotting may be less precise. • Substitution: Rewrite one equation to express one variable in terms of the other, then substitute into the second equation. This is efficient when a variable already has a coefficient of 1 or –1. • Elimination: Add or subtract the equations to remove one variable, then solve for the other. This works well when coefficients match or can be made to match. • Technology: Use graphing calculators or computer software to model and solve the system quickly and accurately. While efficient, algebraic methods develop stronger understanding.

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Example 3 Bixia is saving up her 20 cent and 5 cent pieces in a jar. She has a total of $10.50 in 90 coins: a Write a system of equations that models this situation.

Create a strategy A system of equations can be written using the information provided in the problem, which contains only two numerical totals. One equation is based on the dollar amount, while the other is based on the total number of coins. The unknown variables are: • x = the number of 20 cent pieces • y = the number of 5 cent pieces

Apply the idea x + y = 90

1 2

0.20x + 0.05y = 10.50

Reflect and check The simultaneous equations can be confirmed by examining the units in each equation. Since the total of 1 is the total number of coins, it should make sense that the number of 20 cent pieces added to the number of 5 cent pieces is equal to the total number of coins, which is what is being represented in this equation. The total of 2 represents the fact that Bixia has $10.50. The expression 0.20x represents the value of one 20 cent piece multiplied by how many 20 cent pieces are in the jar. This term is the dollar amount of all the 20 cent pieces combined. The expression 0.05y represents the value of one 5 cent piece multiplied by how many 5 cent pieces are in the jar. This term is the dollar amount of all the 5 cent pieces combined. When these two expressions are added together, the result is the total value of all the coins, or $10.50.

b Graph the system of equations. Use appropriate axes, labels, and scales.

Create a strategy

Apply the idea

In the equations, x represents the number of 20 cent pieces, since a 20 cent piece is worth $0.20, and y represents the number of 5 cent pieces, since a 5 cent piece is worth $0.05. The intercepts of both equations can be calculated and used to determine suitable maximum values for the x-axis and y-axis.

First, calculate the x-intercept of 1 : x + y = 90

x + 0 = 90

Substitute y = 0

x = 90

Simplify

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Now, calculate the y-intercept of 1 : x + y = 90

0 + y = 90

Substitute x = 0

y = 90

Simplify

The x-intercept of 1 is (90, 0) and the y-intercept is (0, 90). Next, calculate the x-intercept of 2 : 0.20x + 0.05y = 10.50 0.20x + 0.05 × 0 = 10.50

2 Substitute y = 0

0.20x = 10.50

Evaluate the multiplication

x = 52.5

Divide both sides by 0.20

Finally, calculate the y-intercept of 2 : 0.20x + 0.05y = 10.50 0.20 × 0 + 0.05y = 10.50

2 Substitute x = 0

0.05y = 10.50

Evaluate the multiplication

y = 210

Divide both sides by 0.05

The x-intercept of 2 is (52.5, 0) and the y-intercept is (0, 210). Since the x-intercepts have a maximum value of 90, the x-axis can be extended to 100 and marked in intervals of 10. Since the y-intercepts have a maximum value of 210, the y-axis can be extended to 220 and marked in intervals of 50. Bixia’s coin jar Number of 5 cent pieces

150

100

Number of 20 cent pieces 0

128

10 20 30 40 50 60 70 80 90

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c Interpret the solution to the system of equations.

Create a strategy

Apply the idea

The graph can be used to identify the solution to the system, and the axes labels to interpret the meaning of the point of intersection. If the solution is not clear from the graph, technology can be used to produce a more precise graph of the system.

The solution to the system of equations is (40, 50). Since x represents the number of 20 cent pieces, the number of 20 cent pieces she has is 40. Since y represents the number of 5 cent pieces, the number of 5 cent pieces she has is 50 in the jar.

Example 4 The length of a rectangle is 3 metres less than twice its width. If the perimeter of the rectangle is 48 metres, determine the length.

Create a strategy Start by defining the variables, then build a system of equations to model the situation. The length of the rectangle in metres is unknown, so it can be assigned the variable l. Since the length and width make up the perimeter of the rectangle and the width in metres is also unknown, the variable w can be assigned to the width.

Apply the idea Based on the first sentence in the problem, the length is equal to 3 metres less than twice its width. The equation that models this relationship is l = 2w − 3. As the perimeter of a rectangle is equal to the sum of twice the length and twice the width, so the equation that models this relationship is 2l + 2w = 48. The system of equations can be written as: 1

l = 2w − 3 2l + 2w = 48

Solve for w: 2l + 2w = 48 2(2w – 3) + 2w = 48 4w – 6 + 2w = 48

2 Substitute l = 2w − 3 Apply distributive property

6w – 6 = 48

Combine like terms

6w = 54

Add 6 to both sides

w=9

Divide both sides by 6

Solve for l: l = 2w − 3

= 2 × 9 − 3 Substitute w = 9 = 15 Simplify The length of the rectangle is 15 metres.

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Example 5 When comparing test results, Verna noticed that the sum of her Chemistry and English test scores was 128 and that their difference was 16. She scored higher on her Chemistry test: a Write a system of equations for this scenario, where x represents Verna’s Chemistry test score, and y represents her English test score.

Create a strategy Since Verna’s Chemistry test score is higher than her English test score, then the difference of the two scores can be expressed as x − y to get a positive result.

Apply the idea

Reflect and check

1 x + y = 128

2 x − y = 16

There are two other constraints based on typical test scores that must be applied to this problem: 0 ≤ x ≤ 100 and 0 ≤ y ≤ 100.

b Solve the system of equations to determine her test scores.

Create a strategy Since the equations have the same coefficient of y with opposite sign, the two equations can be added to eliminate y and solve for x first.

Apply the idea x + y = 128 +

x−y =

2x − 0y = 144 2x = 144

Eliminate the y term

x = 72

Divide both sides by 2

To solve for y, substitute x = 72 into 1 : x + y = 128

72 + y = 128

Substitute x = 72

y = 56

Subtract 72 from both sides

So Verna scored 72 on her Chemistry test and 56 on her English test.

Reflect and check In this case, the two equations also had the same coefficient of x with the same sign. This means the system could also be solved by subtracting one equation from the other to find y first.

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c Does the solution make sense in terms of the context? Explain your answer.

Apply the idea Yes. Assuming that the tests were out of 100, then 72 and 56 are both valid test scores to obtain.

Idea summary Simultaneous linear equations model real-world situations involving two unknowns, such as quantities, prices, or costs. The process begins by defining variables and forming two equations from the information in the context. These equations describe the relationships between the unknowns and represent them mathematically. The system can be solved using graphical, substitution, elimination methods, or using technology. Once a solution is found, it is interpreted in the context of the problem, and its viability is checked to ensure it makes sense in the situation. For example, a solution may be non-viable if it gives negative values for quantities that must be positive.

Break-even analysis Another common application of simultaneous equations in business is break-even analysis, used to identify the sales volume at which total revenue equals total cost. Break-even point The point at which income from production and cost of production are equal. To find the break-even point: 1.

Write the cost and the revenue equations (e.g., C = fixed + variable × x, R = price × x).

2. Equate cost and revenue (C = R) and solve for x (units sold). 3. Find the corresponding cost/revenue by substituting x. 4. Interpret the break-even point, considering constraints (e.g., integer units). Break-even analysis also compares pricing plans (e.g., subscription models) by finding where their cost equations intersect.

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Example 6 A bakery produces cakes at a cost of $25 each with a daily fixed cost of $200. Each cake sells for $30: a Write expressions for the daily cost and revenue for x cakes.

Create a strategy Define x as the number of cakes. Write cost as fixed plus variable cost per cake, and revenue as price per cake times quantity.

Apply the idea For the cost, the fixed or constant value is $200 plus $25 per cake. For the revenue, the value is $30 per cake. C = 200 + 25x

Write the cost equation

R = 30x

Write the revenue equation

Reflect and check The constraint is x ≥ 0, which are integers, as cakes are sold in whole units.

b Determine the break-even point algebraically.

Create a strategy Equate cost and revenue (C = R), solve for x, and determine the corresponding cost and revenue, adjusting the value for whole cakes.

Apply the idea C=R 200 + 25x = 30x 200 = 5x x = 40

Equate C to R Substitute the values Subtract 25x from both sides Divide both sides by 5

Substitute x = 40, into R for the revenue: R = 30x

Write the revenue equation

= 30 × 40

Substitute x = 40

= $1200

Evaluate

The break-even point is (40, 1200), meaning selling 40 cakes results in cost and revenue both equal to $1200.

Reflect and check Since x = 40 is an integer and satisfies x ≥ 0, it is viable and confirms the break-even point.

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c Calculate the profit if 50 cakes are sold.

Create a strategy Compute revenue and cost for x = 50, then subtract the cost from the revenue.

Apply the idea For the revenue: R = 30x

Write the revenue equation

= 30 × 50 Substitute x = 50 = $1500 Evaluate For the cost: C = 200 + 25x

Write the cost equation

= 200 + 25 × 50

Substitute x = 50

= 200 + 1250

Evaluate the multiplication

= $1450

Evaluate

For the profit: Profit = Revenue − Cost

Subtract the cost from the revenue

= 1500 − 1450 Substitute the revenue and cost values = $50 Evaluate

Idea summary Break-even analysis identifies the break-even point, which is the point where cost equals revenue (C = R), solved either algebraically or graphically, representing no profit or loss (e.g., x units sold). It is used in business applications such as sales and pricing plan comparisons, with solutions interpreted to ensure they are viable in context.

3.05 Practice questions What do you remember? 1

Define simultaneous linear equations and their use in real-world applications.

Explain how to set up simultaneous equations for a sales scenario.

What is a break-even point, and how is it found?

List three methods to solve simultaneous equations and their advantages in applications.

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Select the solution to the system of equations on the given graph: A

(−1, −7)

(−6, −7)

(−7, −6)

(1, −6)

No solution

4 2 x –8

–6

–4

–2

–2 –4 –6 –8

Select the solution to the system of equations on the given graph:

(−3, 0)

(−3, 5)

(3, 5)

Infinitely many solutions

x –8 –6 –4 –2 –2

No solution

–4 –6 –8

Solve each system of equations graphed below: a

–6

8 7 6 5 4 3 2 1

–8

–5 –4 –3 –2 –1–1

4 2 –4 –2 –2

x 2

–4

–10 –12 –14 –16 –18

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–2 –3 –4 –5 –6 –7 –8

x 1

9 8 7 6 5 4 3 2 1 –3 –2 –1–1 –2 –3 –4 –5 –6 –7 –8 –9

9 8 7 6 5 4 3 2 1

–9 –8 –7 –6 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 –6 –7 –8 –9

1 2 3 4 5 6 7 8 9

x 1 2 3 4

Practice Ex 1

For each system of equations: i

Sketch the two lines representing the equations on the coordinate plane.

Solve the system of equations using the graph.

y = −x − 2 y = 2x − 5 y=x−2 y = 2x − 1

Ex 2

b d

A baker makes two purchases of flour and sugar. In the first, 3 kg of flour and 2 kg of sugar cost $11. In the second, 5 kg of flour and 6 kg of sugar cost $23. Let x be the cost per kg of flour and y the cost per kg of sugar. Use appropriate labels and scales for the axes:

1 3x + 2y = 11

2 5x + 6y = 23

y = −4x + 5 y = −4x − 4 y = −x y = −2x + 1

Graph the system of equations.

Interpret the solution.

The sum of two mystery numbers is 4. The difference of the two numbers is −2: a

Write a system of equations to represent the situation.

Graph the model to represent the equations.

Find the solution to the system of equations.

Rochelle and Mohamad are sister and brother. Rochelle’s age is 11 more than 4 times the age of Mohamad. The sum of their ages is 21: a

Write a system of equations to represent the situation.

Graph the model to represent the equations.

Determine how old Rochelle and Mohamad are.

3.05 Simultaneous equations mathspace.co

135

Ex 3

Ex 4

Write a system of equations to represent the situation.

Graph the model to represent the equations.

Solve the system of equations and interpret the result in terms of the context.

A shop sells apples and oranges. In one purchase, 5 apples and 3 oranges cost $14. In another, 2 apples and 6 oranges cost $13. Let x be the cost per apple and y the cost per orange: a

Write a system of equations that models this situation.

Graph the system of equations.

Interpret the solution.

Miji saves $200 and adds $75 each month to buy a laptop costing $650, to be released in 6 months. Let x represent months and y represent savings in dollars: a

Write a system of equations to represent the situation.

Sketch the two lines representing these equations on the coordinate plane.

Determine if the student can afford the laptop by the release date.

The total cost of 5 rulers and 3 books is $15.30. If the cost of a ruler is a and a book is $2.70 more expensive, find a.

Valentina has $2000 to invest, and wants to split it up between two accounts: Account A earns 8% annual interest, while Account B earns 9% annual interest. Her target is to earn $177 total interest from the two accounts in one year:

Ex 5

Kang and Elvia are working on an assignment together. They have broken down the work into 9 parts of the same size. Kang works at 2 times the speed of Elvia but has 9 pieces of work to do for another subject:

Write a system of equations to represent the context.

Solve the system of equations.

State if you would make the same investment as Valentina. Explain your answer.

The function f (x) = 0.47x + 8.9 represents the US annual bottled water consumption (in billions of gallons) and the function g (x) = −0.17x + 14.2 represents the US annual soda consumption (in billions of gallons). For both functions, x is the number of years since 2009: a

Determine the year in which the bottled water and soda consumption in the US is the same.

Determine whether the solution is viable in terms of the context. Explain your answer.

A school concert sells student tickets for $2.50 and adult tickets for $5.00. A total of 150 people attended, raising $550. Determine the number of students and adults who attended.

When comparing her test, results Judy noticed that the sum of her Geography test score and Maths test score was 137, and that their difference was 29. Judy scored higher on her Geography test than her maths test:

136

Write a system of equations where x represents Judy’s Geography score and y represents her Maths score.

Solve for Judy’s Geography score using the elimination method.

Now, solve for Judy’s Maths score.

Mathspace New South Wales – Year 11 Advanced mathspace.co

Ex 6

12 pens and 5 rulers cost $70 while 3 pens and 25 rulers cost $65: a

Write a system of equations where x represents the price of a pen and y represents the cost of a ruler.

Solve for the price of a pen using the elimination method.

Now, solve for the price of a ruler.

A small coffee cart has a fixed daily cost of $150 and a variable cost of $2.50 per cup of coffee. Each cup sells for $5: a

Write equations for the total daily cost C and total daily revenue R when x cups are sold.

Find the break-even point, giving the number of cups and the corresponding cost/ revenue.

Interpret the result in context.

Fred spent $55.25 to purchase 9 flowers. He bought rhododendrons which cost $6.45 each and chrysanthemums which cost $5.75 each: a

If R is the number of rhododendrons and C is the number of chrysanthemums that Fred bought, construct two equations describing the total number of flowers bought, and the total amount spent in dollars.

Solve for the number of rhododendrons and chrysanthemums that Fred purchased.

Extend your thinking 23

Two equations, y1 and y2 represent the growth of two different house plants over time. Use the graph of y1 and y2 to support the claim that the two plants will never reach the same height on the same day: y 25 20 15 10 5 –40 –30 –20 –10 y1

–5

x 10

–10 –15 –20 –25

Describe a situation where it would be unfeasible to solve a system of equations by graphing.

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137

Write a scenario to represent the system of equations and its solution. Explain what the solution to the system means in terms of the scenario:

18 16 14 12 10 8 6 4 2 0

x 2 4 6 8 10 12 14 16 18 20 22 24

At Soul Food Express you can buy two orders of oxtail stew and a slice of sweet potato pie for $38.49, or you can get five orders of oxtail stew and three slices of sweet potato pie for $99.22. Construct a system of equations to model the situation and use it to determine the cost of 4 oxtail stews and 2 slices of sweet potato pie.

Kwabena bought some fresh produce. He picked up 2 oranges and 3 bananas. The cost of the Kwabena’s shopping was $18.30. Amra also went to the same shop and bought 5 oranges and 7 bananas. The cost of the Amra’s shopping was $44.03:

138

Construct a system of equations to model the scenario.

Explain two different ways to approach finding the cost per orange and the cost per banana.

Omeida’s piggy bank contains 80 pieces of 10 cent and 5 cent pieces with a total value of $5.50: a

Construct a system of equations to model the scenario and use it to determine how many of each coin Omeida has.

Suppose the coins in the piggy bank were only 5 cent pieces and 20 cent pieces. Revise the model in part (a) for this change. Explain how this changes the solution.

Mathspace New South Wales – Year 11 Advanced mathspace.co

3 Chapter review 1

Which of these inequalities represents the solution for n in “ 7 more than 3 times n is less than 40”? A

n > 11

n < 11

To enter a theme park ride, a person must be at least 120 cm tall and less than 190 cm tall. Which of these number lines represents these conditions? A 110

120

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140

150

160

170

180

190

200

110

120

130

140

150

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200

110

120

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110

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Which of these equations is the gradient-intercept form of the equation 3x + 4y = 12?

A C 4

y = 3x − 12

Rewrite each equation in gradient-intercept form y = mx + c: a

y = −3x + 12

4x + 2y = 8

5x − 2y = 6

3x + 6y = −12

−2x + 3y = 9

6x − 4y + 8 = 0

9x + 3y = −15

2x − 5y = 10

7x + 4y = −20

Sketch the graph of the following equations: a

3x − 2y = 6

4x + 2y = 8

5x − y = 10

2x + 3y = −6

6x − 4y + 8 = 0

9x + 3y = −18

A gym membership includes a fixed charge of $10 plus $5 per class: a

Form the equation for the total fee y in terms of the number of classes x.

Find the y-intercept and explain its meaning in the context of the problem.

A streaming service plan charges a fixed fee plus a per-movie rate. The total cost is given by y = 3x + 8, where x is the number of movies rented and y is the cost in dollars:

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Find the y-intercept and interpret its meaning in the context of the plan.

Find the x-intercept and explain why it may not be meaningful in this context.

Graph the line for 0 ≤ x ≤ 4 by plotting the y-intercept and using the gradient.

A farmer packs apples and oranges into a crate. The total weight of a full crate is 20 kg. Each apple weighs 0.2 kg and each orange weighs 0.25 kg: a

Write an equation in general form ax + by + c = 0 to represent the total weight, where x is the number of apples and y is the number of oranges.

Find the x-intercept and y-intercept. Interpret what each intercept means in the context of the problem.

Sketch the line for x ≥ 0, y ≥ 0, labelling the intercepts.

Convert each equation from gradient-intercept form to general form: a

y = 4x − 5

y = −2x + 3

y = x + 4

y = −3x − 1

Determine the equation of the line in gradient-intercept form given the gradient and a point: a

Gradient −3, point (2, 4)

Gradient 2, point (1, 5)

Gradient −1, point (3, −2)

Gradient

point (4, 3)

Gradient

, point (2, −1)

Gradient

point

Determine the equation in gradient-intercept form of the line that is parallel to y = 3x + 1 and passes through the point (2, 5).

Determine the equation in gradient-intercept form of the line that is perpendicular to y=

and passes through the point (1, 1).

A pipeline follows the line

. A perpendicular pipeline must be built that passes

through the intersection of the first pipeline and the line 5x − 2y + 1 = 0:

140

Find the point of intersection of the two existing lines.

Determine the equation of the new perpendicular pipeline in general form.

Show that the new pipeline is perpendicular to the first pipeline.

Solve: a

−4(x + 2) ≥ −20

3x − 5 < 7

2(x − 3) ≤ 8

−x + 4 > 1

e 15

For each inequality: i

Solve for x.

Plot the solution on a number line.

Mathspace New South Wales – Year 11 Advanced mathspace.co

3 < 4x − 5

2x + 3 ≤ 7

−3x + 4 > 10

5 − 2x ≥ 1

Sarah has a gift card with a balance of $50. She has already spent $18.50. Let p represent the amount that Sarah can still spend. Write the budget constraint as an inequality and solve for p.

Ben wants to save up enough money to buy a new video game, which costs $50.00. Ben has $15.00 that he saved from his allowance. To make more money, he plans to wash cars for $5 per car:

Let w be the number of cars that Ben washes. Write an inequality to represent the situation.

Solve for w.

Explain what the solution means in the context of the problem.

A student tried to solve the following inequality but made a mistake in their work: Problem: − 3x −6 > 12 Step 1:

−3x > 18

Step 2:

x > −6

Determine which step is incorrect and explain the error. 19

The graph shows the amount of water remaining in a tank that was initially full before a tap was opened at its base:

Water (L) 30 (0, 30)

Determine the gradient of the line.

Determine the y-intercept.

Explain the meaning of the gradient in this context.

Explain the meaning of the y-intercept in this context.

Determine an equation to represent the amount of water remaining in the tank, W, as a function of time, t.

A cyclist starts at the top of a hill and 0 10 begins to descend at a constant rate. The table shows the altitude of the cyclist over 4 minutes:

Time (min) (60, 0) 20

Number of minutes passed, x

Altitude of cyclist in metres, y

500

480

460

440

420

Calculate the decrease in altitude each minute.

Determine a linear equation for the relationship between the number of minutes passed, x, and the altitude, y, of the cyclist.

Calculate the altitude of the cyclist after 7 minutes.

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An electrician charges a $90 call-out fee plus $75 per hour for repairs. Let C represent the total cost and h the number of hours worked: a

Determine an equation for C in terms of h.

Calculate the total cost for 3.5 hours of work.

Find the number of hours worked if the total cost is $465.

A catering company charges $150.00 for setup plus $35.50 per person. A client’s budget is no more than $1200.00. Let p represent the number of people attending: a

Write an inequality for the number of people the client can afford.

Solve the inequality to find the maximum number of people the client can afford.

A salesperson earned $700 in a week when they sold 15 computers, and $900 in a week when they sold 25 computers. Their earnings consist of a base salary and a commission per computer sold: a

Determine a linear equation to represent the weekly earnings of the salesperson, E, as a function of the number of computers sold, n.

Determine how much the salesperson will earn in a week during which they sell 32 computers.

The table shows the maximum amount of a chemical that can be present in a laboratory environment without adverse effects, based on the total duration of exposure: a

Amount (mg)

Use a linear model to express the amount of chemical, C, in mg, in terms of the duration of exposure, t, in hours.

What is the maximum safest duration that 50 mg of the chemical can be present?

A scientist finds they have been exposed to 40 mg of the chemical over 18 hours. Are they safe or at risk?

For each simultaneous equation: i

Sketch the two lines representing the equations on the coordinate plane.

Solve the simultaneous equations using the graph.

y = x − 3

y = 3x − 2

y = −2x + 6

y = −x + 6

y = −2x + 5

y = x − 1

142

Duration (hours)

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y = −x + 4

A cinema sells child tickets for $12.00 and adult tickets for $18.00. On a particular evening, a total of 200 people attended, and the total revenue was $2760.00. Determine the number of children and adults who attended.

The sum of a student’s scores on a physics test and a chemistry test was 150. The difference between the scores was 18, with the Physics score being higher: a

Write a system of equations where p represents the physics score and c represents the chemistry score.

Solve for the student’s physics and chemistry scores.

At a café, you can buy two sandwiches and one coffee for $25.00. An order of five sandwiches and three coffees costs $65.00. Construct a system of equations to model the situation and use it to determine the cost of 4 sandwiches and 2 coffees.

A piggy bank contains 50 coins, which are a mix of 10 cent and 20 cent pieces. The total value of the coins is $7.50: a

Construct a system of equations to model the scenario and use it to determine how many of each coin there are.

Suppose the total value was $12.00 with the same number of coins. Revise the model and explain why this situation is impossible.

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Big ideas • The different algebraic forms of a quadratic function, general, factored, and vertex form, are strategic tools that each reveal specific geometric features of its parabolic graph, such as intercepts, vertex, and concavity, enabling both systematic sketching and the determination of its equation from given information. • The analysis of quadratic equations extends to solving problems of interaction and optimisation; the discriminant determines the number of intersection points between curves, inequalities define regions on a plane, and the vertex provides solutions to real-world modelling problems involving maximum or minimum values. • Cubic functions introduce a higher-order polynomial graph whose key features, such as intercepts and end behaviour, are determined using principles analogous to those for quadratics, primarily through an analysis of the factored form and the leading coefficient.

4 Quadratic and cubic functions Chapter outline 4.01 4.02 4.03 4.04 4.05 4.06 4.07 4.08

Characteristics of quadratics Completed square form Graph parabolas Equations of parabolas Solve quadratic systems Quadratic inequalities Quadratic models Cubic functions Chapter 4 review

146 159 167 175 182 188 194 201 209

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

4.01 Characteristics of quadratics After this lesson, you will be able to… • identify key features of a parabola, including the vertex, axis of symmetry, intercepts and concavity. • identify the x-intercepts of a parabola from its factored form. • use the discriminant to determine the number and nature of roots of a quadratic equation. • determine the domain and range of a quadratic function. • sketch the graph of a parabola showing its key features.

Key features of quadratic functions Quadratic function A function of the form f (x) = ax2 + bx + c, where a ≠ 0, b and c are constants. Parabola The graph of y = x2. The point (0, 0) is called the vertex of the parabola and the y-axis is the axis of symmetry of the parabola. Some other parabolas are the graphs of y = ax2 + bx + c where a ≠ 0.

9 8 7 6 5 4 3 2 1 −4 −3 −2 −1

The simplest quadratic is y = x2. Consider its values:

146

−4

−3

−2

−1

Mathspace New South Wales – Year 11 Advanced mathspace.co

x 1 2 3 4

Plotting these points yields: 16

14 12 10 8 6 4 2 −7

−6

−5

−4

−3

−2

−1

Turning point (0, 0)4

Vertex The turning point of a parabola. Concave up A function for which all points on a graph between any two given points on the graph lie on or below the chord joining the given points. Concave down A function in which all points on a graph between any two given points on the graph lie on or above the chord joining the given points.

The turning point or vertex at (0, 0) is the minimum for y = x2, where the parabola is concave up (positive a). It also serves as the x-intercept and y-intercept. 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. Axis of symmetry A straight line that divides a shape or curve into two so that one side is a reﬂection of the other in the given line.

The axis of symmetry is a vertical line through the vertex, here x = 0. For y = ax2 + bx + c, it is given by:

is the coeﬃcient of x2

is the coeﬃcient of x 4.01 Characteristics of quadratics mathspace.co

147

Parabolas may have zero, one, or two x-intercepts (where y = 0) and exactly one y-intercept (where x = 0). y 15

x-intercepts

10 5

−15 −10 −5 −5 −10

x 5

y-intercept

−15

Interactive exploration Discover this concept in action online

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Example 1 Identify the key features of the parabola y = x2 − 4x + 3: a Find the y-intercept.

Create a strategy Substitute x = 0 into the equation.

Apply the idea y = x2 − 4x + 3 2

=0 −4×0+3

Substitute x = 0

Evaluate

The y-intercept is (0, 3).

b Find the axis of symmetry.

Create a strategy Use

148

Write the equation

with a = 1, b = −4.

Mathspace New South Wales – Year 11 Advanced mathspace.co

Apply the idea Write the formula Substitute a = 1 and b = −4 Evaluate The axis of symmetry is x = 2.

Idea summary A parabola’s key features include its vertex, x- and y- intercepts, and axis of . The sign of a determines concavity.

symmetry at

Factored form and x-intercepts A quadratic function in factored form is y = a(x − r)(x − s), where r and s are the x-intercepts (roots, zeroes, solutions), and a determines the parabola’s stretch and orientation.

y = a(x − r)(x − s) r, s

are the x-intercepts

is the stretch factor and orientation

To ﬁnd the x-intercepts, set y = 0: y = a(x − r)(x − s)

Write the formula

0 = a(x − r)(x − s)

Set y = 0

0 = (x − r)(x − s)

Divide both sides by a

Set each factor to 0: x−r=0 x=r x−s=0 x=s

Write the equation of the ﬁrst factor Add r to both sides Write the equation for the second factor Add s to both sides

Thus, the x-intercepts are at x = r and x = s.

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Example 2 For the quadratic y = 2(x − 3)(x + 1), ﬁnd: a The x-intercepts

Create a strategy Set y = 0 and solve for x.

Apply the idea y = 2(x − 3)(x + 1)

Write the formula

0 = 2(x − 3)(x + 1)

Set y = 0

0 = (x − 3)(x + 1)

Divide both sides by 2

Set each factor to 0: x−3=0 x=3 x+1=0 x = −1

Write the equation of the ﬁrst factor Add 3 to both sides Write the equation for the second factor Subtract 1 from both sides

The x-intercepts are x = 3 and x = −1.

b The y-intercept

Create a strategy Substitute x = 0.

Apply the idea y = 2(x − 3)(x + 1)

Write the formula

= 2(0 − 3)(0 + 1)

Substitute x = 0

= 2 × (−3) × 1

Evaluate each expression inside the brackets

= −6

Evaluate

The y-intercept is (0, −6).

Idea summary In factored form, y = a(x − r)(x − s), the x-intercepts are x = r and x = s. Set x = 0 for the y-intercept.

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The discriminant and number of roots Discriminant In the quadratic expression ax2 + bx + c, the discriminant is Δ = b2 − 4ac. For a quadratic function f (x) = ax2 + bx + c, the discriminant determines the number and nature of x-intercepts (roots) where f (x) = 0.

Δ = b2 — 4ac Δ a b c

is the discriminant is the coeﬃcient of x2 is the coefficient of x is the constant term

Quadratic formula The roots of a quadratic equation ax2 + bx + c = 0 where a ≠ 0 are given by the quadratic formula:

The discriminant appears in the quadratic formula:

x a b c

are the roots of the quadratic is the coeﬃcient of x2 is the coefficient of x is the constant term

The discriminant’s value indicates the number and nature of real roots: • Δ > 0: Two distinct real roots (two x-intercepts). If Δ is a perfect square, roots are rational; otherwise, irrational. • Δ = 0: One real root (repeated, rational), and the parabola touches the x-axis at the vertex. • Δ < 0: No real roots, and the parabola does not intersect the x-axis. y 4 2 −2

−1

∆<0 ∆=0 1

−2

x 2

∆>0

−4

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Three parabolas show Δ > 0 (two intercepts), Δ = 0 (one intercept), and Δ < 0 (no intercepts) for a > 0.

Interactive exploration Discover this concept in action online

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Example 3 For f (x) = x2 − 6x + 8: a Calculate the discriminant.

Create a strategy Use Δ = b2 − 4ac with coeﬃcients from f (x).

Apply the idea Δ = b2 − 4ac 2

Write the discriminant formula

= (−6) − 4 × 1 × 8

Substitute a = 1, b = −6 and c = 8

= 36 − 32

Evaluate the power and multiplication

Evaluate

b Determine the number and nature of x-intercepts.

Create a strategy Analyse Δ’s value and whether it is a perfect square.

Apply the idea Since Δ = 4 > 0, there are two distinct real roots. As 4 is a perfect square rational.

c Justify the parabola’s position relative to the x-axis.

Create a strategy Find the vertex’s coordinate and concavity to determine position.

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, the roots are

Apply the idea Find the x-coordinate of the vertex: Write the formula

Substitute a = 1 and b = −6

Evaluate

Find the y-coordinate of the vertex: f (x) = x2 − 6x+8

Write the function

Substitute x = 3

f (3) = 3 − 6 × 3 + 8 = 9 − 18 + 8

Evaluate the power and multiplication

= −1

Evaluate

The vertex is (3, −1). Since a = 1 > 0, the parabola is concave up, with its minimum below the x-axis, but crossing it at two points due to Δ > 0.

Idea summary The discriminant in the quadratic expression ax2 + bx + c is given by:

Δ = b2 − 4ac Δ

is the discriminant

is the coeﬃcient of x2

is the coefficient of x

is the constant term

The discriminant determines the number and nature of x-intercepts: • two distinct (rational if Δ is a perfect square, irrational otherwise) if Δ > 0 • one rational if Δ = 0 • none if Δ < 0 The vertex and concavity clarify the parabola’s position.

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Graph and nature of roots Root The solution to an equation. Example: (x − 1)(x + 3) = 0 has roots x = 1 and x = −3 The discriminant aids graphing by indicating x-intercepts. To graph f (x) = ax2 + bx + c: • Calculate Δ to determine x-intercepts (if Δ ≥ 0), or by factorising. • Find the y-intercept: f (0) = c • Find the vertex at

, then compute

• Determine concavity: a > 0 for concave up, a < 0 for concave down. The domain is . The 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

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Example 4 For f (x) = 2x2 + 3x + 4: a Calculate the discriminant.

Create a strategy Use Δ = b2 − 4ac.

Apply the idea Δ = b2 − 4ac

Write the discriminant formula

=3 −4×2×4

Substitute a = 2, b = 3 and c = 4

= 9 − 32

Evaluate the power and multiplication

= −23

Evaluate

b Determine the number and nature of x-intercepts.

Create a strategy

Apply the idea

Analyse Δ’s value from part (a).

Since Δ = −23 < 0, there are no real roots, so no x-intercepts.

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c Determine the domain and range.

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

Apply the idea The domain is . For the vertex: Write the formula Substitute a = 2 and b = 3 Evaluate Substitute

into the function: Write the function

Substitute x =

Evaluate the power and multiplication

Rewrite with common denominator

Evaluate

Simplify

The vertex is

. Since a = 2 > 0, the range is

d Sketch the graph, showing key features.

Create a strategy Plot the vertex, y-intercept, without x-intercepts. Draw the parabola concave up.

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Apply the idea For the y-intercept, substitute x = 0 into f (x): f (x) = 2x2 + 3x + 4 2

f (0) = 2(0) + 3 × 0 + 4 =4

Write the function Substitute x = 0 Evaluate

So the y-intercept is at (0, 4). y 5 4 (0, 4) 3 2

Plot the vertex

and y-intercept (0, 4). Draw the

axis of symmetry at x =

. Sketch a concave up parabola

above the x-axis.

1 x −2

−1

Idea summary The discriminant informs the graphing of quadratics by indicating if there are two, one or no x-intercepts. Domain is , the range depends on the vertex and concavity. Rational roots occur if Δ is a perfect square, irrational otherwise.

4.01 Practice questions What do you remember? 1

156

Select the correct terms to complete the statements about quadratic functions: a

The graph of y = x2 is concave (up/down) with a (maximum/minimum) turning point.

The graph of y = −2x2 is concave (up/down) with a (maximum/minimum) turning point.

Match each feature of a quadratic graph with its description: a

Axis of symmetry

Vertex

x-intercept

y-intercept

Mathspace New South Wales – Year 11 Advanced mathspace.co

Point(s) where the graph crosses the x-axis

Point representing the maximum or minimum value

iii

Vertical line dividing the graph symmetrically

Point where the graph crosses the y-axis

Deﬁne these terms: a

Quadratic formula

Identify the coeﬃcients a, b, and c in the quadratic equation ax2 + bx + c = 0 for: a

Discriminant

3x2 − 5x + 2 = 0

x2 + 4 = 2x

Match each discriminant condition to the number of real roots: a

No real roots

Δ>0

One real root

Δ=0

Two distinct real roots

iii

Δ<0

What conditions must be satisﬁed for a quadratic equation to have rational solutions?

Practice Ex 1

For the quadratic function y = x2 − 6x + 8, ﬁnd: a

Ex 2

Ex 3

Ex 4

The axis of symmetry

For the quadratic function y = 2(x − 1)(x − 5), ﬁnd: a

The y-intercept

x-intercepts

y-intercept

For the quadratic function y = −(x + 2)(x − 3): a

Find the x-intercepts.

Find the y-intercept.

Find the axis of symmetry.

For the quadratic function y = x2 − 4x + 3: a

Calculate the discriminant.

Determine the number and nature of x-intercepts.

Justify the parabola’s position relative to the x-axis.

For the quadratic function y = 3x2 − 6x + 5: a

Calculate the discriminant.

Determine the number and nature of x-intercepts.

Determine the domain and range.

Sketch the graph, showing the vertex, y-intercept, and axis of symmetry.

For each quadratic equation: i

Calculate the discriminant.

Determine the number and nature of roots:

x2 − 4x + 4 = 0

2x2 − x = 0

−3x2 + 2x − 5 = 0

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For the quadratic functions y = x2 − 2x − 3 and y = x2 + 2x + 2: a

Calculate the discriminant for each and identify the number of x-intercepts.

Sketch both graphs on the same axes, including key features.

For the quadratic equation kx2 − 4x + 1 = 0, determine the values of k for which there are no real roots.

For each quadratic, determine the number of x-intercepts by calculating the discriminant: a

y = 4x2 − 12x + 9

For the equation x2 − 8x + k = 0, ﬁnd the values of k for which it has: a

y = −x2 + 3x − 4

One real solution

Two real solutions

A ball is thrown upward with an initial velocity of 15 m/s from a height of 1 metre. Its height is given by h = −4.9t2 + 15t + 1: a

At what time does the ball hit the ground?

What is the maximum height achieved by the ball?

Sketch the graph, including key features and the axis of symmetry.

Extend your thinking 18

A parabola is concave up with its vertex in Quadrant 1. How many x-intercepts does it have? Explain.

Two quadratic functions, A and B, share the same x-intercepts at x = 1 and x = 5. Function A is concave down with a y-intercept at (0, −5). Function B is concave up with a y-intercept at (0, 10). Find the vertical distance between their vertices.

A rectangular garden has a length of (2x + 3) m and a width of (x − 1) m: a

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

Find the area when x = 2 m.

Explain why x must be greater than 1.

2x + 3 x−1

A company’s profit from selling x units is P = −x2 + 100x − 1500 dollars. Find the number of units to maximise profit and the maximum profit.

A parabola y = x2 − 4x + 5 intersects a line y = mx + 2: a

Form the quadratic equation for the points of intersection.

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

For m = −8, sketch the parabola and line, including intersection points.

The quadratic equation x2 + (k + 3)x − k = 0 has two distinct real roots. Find the possible values of k.

For what values of k does the quadratic Explain why no value of k gives two rational roots.

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have two real solutions?

4.02 Completed square form After this lesson, you will be able to… • convert a quadratic function from general form to completed square (vertex) form. • identify the vertex and axis of symmetry of a parabola from its equation in completed square form. • show that the axis of symmetry of y = ax2 + bx + c is by completing the square. • determine the domain and range of a quadratic function using its vertex. • sketch a parabola using the key features derived from its completed square form.

Completed square form and key features A quadratic function in completed square form (or vertex form) is f (x) = a(x − h)2 + k, where (h, k) is the vertex, and a ≠ 0 determines concavity and stretch.

f (x) = a(x − h)2 + k h k a

is the x-coordinate of the vertex is the y-coordinate of the vertex determines concavity (positive for concave up, negative for concave down) and stretch

The axis of symmetry is the vertical line through the vertex:

x=h h

is the x-coordinate of the vertex

For a quadratic in general form f (x) = ax2 + bx + c, the axis of symmetry is derived by completing the square: Write the function

Factorise a from the ﬁrst two terms

Add and subtract

Factorise the perfect square trinomial and distribute a

Simplify the constant terms

Thus, the vertex is

, and the axis of symmetry is

inside the brackets

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Example 1 For the quadratic function f (x) = x2 + 6x + 5, use the completed square form to identify key features and graph it: a Express in completed square form.

Create a strategy Complete the square for x2 + 6x + 5 by adding and subtracting

Apply the idea f (x) = x2 + 6x + 5

Write the function

= x2 + 6x + 9 − 9 + 5

Add and subtract

= (x + 3)2 − 4

Factorise and simplify 2

The completed square form is f (x) = (x + 3) − 4.

b Identify the vertex and axis of symmetry.

Create a strategy

Apply the idea 2

Use the form f (x) = a(x − h) + k to read the vertex (h, k) and axis of symmetry x = h.

From f (x) = (x + 3)2 − 4, rewrite as f (x) = (x − (−3))2 − 4. Thus, h = −3, k = −4. The vertex is (−3, −4), and the axis of symmetry is x = −3.

c Find the y-intercept.

Create a strategy Substitute x = 0 into f (x).

Apply the idea f (x) = (x + 3)2 – 4 2

f (0) = (0 + 3) − 4

Substitute x = 0

=9−4

Evaluate the power

Evaluate

The y-intercept is at (0, 5).

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d Find the x-intercepts.

Create a strategy Set f (x) = 0 and solve for x using the completed square form.

Apply the idea Set the function equal to zero

Add 4 to both sides

Take the square root of both sides

Evaluate the square root

Solve for the two possible values of x: x = −3 + 2

First solution

= −1

Evaluate

x = −3 − 2

Second solution

= −5

Evaluate

The x-intercepts are at (−5, 0) and (−1, 0).

e Sketch the graph, showing key features.

Create a strategy Plot the vertex, y-intercept, and axis of symmetry. Draw a smooth curve, noting concavity from a.

Apply the idea Since a = 1 > 0, the parabola is concave up. y 4

x = −3

(0, 5)

2 x −6 −5 −4 −3 −2 −1

−2

(−3, −4)

Plot the vertex (−3, −4), the x-intercepts (−5, 0) and (−1, 0), and y-intercept at (0, 5). Draw the axis of symmetry at x = −3. Sketch a concave up parabola through these points.

−4

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Idea summary The completed square form f (x) = a(x − h)2 + k directly gives the vertex (h, k) and axis of symmetry x = h. Completed square form is also known as vertex form.

f (x) = a(x − h)2 + k h

is the x-coordinate of the vertex

is the y-coordinate of the vertex

a determines concavity (positive for concave up, negative for concave down) and stretch For f (x) = ax2 + bx + c, completing the square yields symmetry.

for the axis of

Domain, range, and non-monic quadratics The domain of a quadratic function f (x) = a(x − h)2 + k is , unless restricted by context. The range depends on concavity: for a > 0, range is [k, ∞); for a < 0, range is (−∞, k], where k is the vertex’s y-coordinate. For non-monic quadratics (f (x) = ax2 + bx + c, a ≠ 1), factorise a before completing the square: Write the function

Factorise a

Add and subtract

Factorise the perfect square trinomial and distribute a

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Example 2 For f (x) = 2x2 − 8x + 3: a Express in vertex form.

Create a strategy Factorise 2 and complete the square for the resulting monic quadratic.

Apply the idea f (x) = 2x2 − 8x + 3

Write the function

= 2 (x − 4x) + 3

Factorise 2

= 2 (x2 − 4x + 4 − 4) + 3

Add and subtract

= 2 ((x − 2)2 − 4) + 3

Factorise the perfect square trinomial

= 2(x − 2) − 8 + 3 2

= 2(x − 2) − 5

Distribute and simplify Collect like terms

The vertex form is f (x) = 2(x − 2) − 5.

b Find the vertex and axis of symmetry.

Create a strategy Use f (x) = 2(x − 2)2 − 5 to identify (h, k) and x = h.

Apply the idea From f (x) = 2(x − 2)2 − 5, h = 2, k = −5. The vertex is (2, −5), and the axis of symmetry is x = 2.

c Determine the domain and range.

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

Apply the idea The domain is . Since a = 2 > 0, the parabola is concave up, with a minimum at y = −5. Thus, the range is [−5, ∞).

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d Find the y-intercept.

Create a strategy Substitute x = 0 into f (x).

Apply the idea f (x) = 2(x − 2)2 – 5

Write the function

Substitute x = 0

= 2 × (−2) − 5

Evaluate the expression inside the brackets

=2×4−5

Evaluate the power

Evaluate

f (0) = 2(0 − 2) − 5

The y-intercept is at (0, 3).

e Sketch the graph, showing key features.

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

Apply the idea y 4

(0, 3) 2 −1

x=2 1

x 4

−2

Plot the vertex (2, −5) and y-intercept (0, 3). Draw the axis of symmetry at x = 2. Sketch a concave up parabola through these points.

−4

(2, −5)

Idea summary Non-monic quadratics require factorising a before completing the square. The domain is , and the range is determined by the vertex and concavity.

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4.02 Practice questions What do you remember? 1

State whether each statement about the quadratic function y = a(x − h)2 + k is true or false: a

The vertex is at (h, k).

The axis of symmetry is x = k.

If a > 0, the parabola is concave down.

The range is [k, ∞) if a > 0.

For the quadratic y = 2(x − 3)2 + 5, identify: a

The vertex

The axis of symmetry

The concavity

The range

What term must be added to x2 + 8x to form a perfect square trinomial?

Practice Ex 1

Ex 2

For the quadratic function y = x2 + 4x + 3: a

Express in vertex form.

Identify the vertex and axis of symmetry.

Find the point of the y-intercept.

Sketch the graph, labelling key features.

For the quadratic function y = 3x2 − 12x + 5: a

Express in vertex form.

Identify the vertex and axis of symmetry.

Determine the domain and range.

Sketch the graph, labelling key features.

Express each quadratic in vertex form and identify the vertex: a

y = x2 − 4x + 7

y = 3x2 + 12x − 5

A quadratic function is deﬁned as y = ax2 − 2x + 3. The vertex of the parabola is at (1, 2): a

Determine the value of a.

Express the quadratic in vertex form.

For the quadratic y = −x2 − 6x + 2: a

Express in vertex form.

Determine the vertex, axis of symmetry, and range.

Find the x-intercepts in exact form.

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For the quadratic y = 2x2 + 10x − 3: a

Express in vertex form.

Determine the vertex and concavity.

Sketch the graph, labelling the vertex, y-intercept, and axis of symmetry.

Complete the square to ﬁnd the vertex and y-intercept of y = 3x2 − 6x + 4.

For the quadratic y = −x2 − 4x + 1:

Express in vertex form.

Sketch the parabola, labelling the vertex, axis of symmetry, and y-intercept.

A parabola has a vertex at (−1, 4) and a y-intercept at (0, 7): a

Find the equation of the parabola in vertex form, assuming a is an integer.

State the axis of symmetry and range.

Express the quadratic y = x2 + 10x + 13 in vertex form.

Extend your thinking 14

A rectangular garden has a perimeter of 20 metres and area A = x(10 − x) m2, where x is the width in metres: a

Express the area in vertex form and ﬁnd the vertex.

Determine the maximum area and corresponding dimensions.

For the quadratic y = kx2 + 4x + 3, ﬁnd the values of k such that the vertex lies on the x-axis.

A projectile’s height is given by h = −5t2 + 20t + 1 metres after t seconds: a

Express in vertex form and identify the vertex.

Determine the maximum height and when it occurs.

A quadratic function has vertex (2, −3) and passes through (4, 1). Find its equation in vertex form.

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4.03 Graph parabolas After this lesson, you will be able to… • identify all key features of a parabola from its equation in standard form. • use the quadratic formula to find the x-intercepts of a parabola, if they exist. • use the formula

to determine the axis of symmetry and the

coordinates of the vertex. • determine the domain and range of a quadratic function. • sketch an accurate graph of a parabola, labelling the vertex, axis of symmetry, and all intercepts.

Graph quadratics in standard form A quadratic function in standard form, where a ≠ 0.

f (x) = ax2 + bx + c a determines concavity (positive for concave up, negative for concave down) b

affects the vertex position

is the y-coordinate of the y-intercept (0, c)

To graph a quadratic: • Find the y-intercept by setting x = 0: f (0) = c. • Find x-intercepts (if they exist) by solving ax2 + bx + c = 0 using the quadratic formula:

While factorisation can be quicker for simple quadratics, testing for factors also takes time, so applying the quadratic formula directly in all cases is often the most eﬃcient approach. • Find the axis of symmetry: • Find the vertex by substituting

. into f (x).

• The domain is . The range is [k, ∞) for a > 0 or (−∞, k] for a < 0, where k is the vertex’s y-coordinate.

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Example 1 For f (x) = x2 − 4x + 3: a Find the x-intercepts.

Create a strategy Set f (x) = 0 and use the discriminant to determine if factorisation is possible or if the quadratic formula is required.

Apply the idea f (x) = x2 − 4x + 3

Write the function

0 = x − 4x + 3

Set f (x) = 0

Determine if the quadratic expression can be factorised by calculating the discriminant. Δ = b2 − 4ac

Write the discriminant formula

= (−4) − 4(1)(3)

Substitute a = 1, b = −4 and c = 3

= 16 − 12

Evaluate each term

Evaluate the subtraction

Since the discriminant is 4, which is a perfect square, the quadratic equation x2 − 4x + 3 = 0 can be factorised. x2 − 4x + 3 = 0

Write the equation

(x − 1)(x − 3) = 0

Factorise the quadratic expression

1, x = 3

Set each factor equal to zero and solve for x

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

b Find the y-intercept.

Create a strategy Substitute x = 0 into f (x).

Apply the idea f (x) = x2 − 4x + 3

Write the function

Substitute x = 0

f (0) = 0 − 4 × 0 + 3 =3 The y-intercept is at (0, 3).

c Find the axis of symmetry and vertex.

Create a strategy Use

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for the axis of symmetry, then substitute into f (x) for the vertex.

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Evaluate

Apply the idea For the axis of symmetry: Write the formula

Substitute a = 1 and b = −4

Evaluate

The axis of symmetry is x = 2. For the vertex, substitute x = 2: f (x) = x2 − 4x + 3

Write the function

Substitute x = 2

f (2) = 2 − 4 × 2 + 3 =4−8+3

Evaluate each term

= −1

Evaluate

The vertex is (2, −1).

d Determine the domain and range.

Create a strategy

Apply the idea

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

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

e Sketch the graph, showing key features.

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

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

y 3 (0, 3)

x=2

2 1

(1, 0) 1 −1

(3, 0) 2

(2, −1)

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Idea summary To graph f (x) = ax2 + bx + c, identify the y-intercept (0, c), x-intercepts (if any) via the quadratic formula, axis of symmetry

, and vertex

The domain is , and the range depends on concavity and vertex.

Graph non-monic quadratics For non-monic quadratics (a ≠ 1), the process is the same: ﬁnd x-intercepts, y-intercept, axis of symmetry, vertex, domain, and range. The coeﬃcient a affects the parabola’s steepness.

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Example 2 For f (x) = 2x2 + 4x − 3: a Find the x-intercepts.

Create a strategy Set f (x) = 0 and use the discriminant to determine if factorisation is possible or if the quadratic formula is required.

Apply the idea f (x) = 2x2 + 4x – 3 2

0 = 2x + 4x − 3

Write the function Set f (x) = 0

Determine if the quadratic expression can be factorised by calculating the discriminant. Δ = b2 − 4ac 2

Write the discriminant formula

= (4) − 4(2)(−3)

Substitute a = 2, b = 4 and c = −3

= 16 − (−24)

Evaluate terms

= 16 + 24

Simplify

= 40

Evaluate the sum

Since Δ = 40, which is not a perfect square, the quadratic formula is necessary.

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Use the quadratic formula to ﬁnd the x-intercepts. Write the formula

Substitute a = 2, b = 4 and c = −3

Evaluate each term

Evaluate the expression inside the square root

Simplify square root

Simplify the fraction

The x-intercepts are

and

b Find the y-intercept.

Create a strategy

Apply the idea f (x) = 2x2 + 4x – 3 Write the function

Substitute x = 0 into f (x).

f (0) = 2 × (0)2 + 4 × 0 − 3 = −3

Substitute x = 0 Evaluate

The y-intercept is at (0, −3). c Find the axis of symmetry and vertex.

Create a strategy Use

for the axis of symmetry, then substitute into f (x) for the vertex.

Apply the idea Write the formula Substitute a = 2, b = 4 Evaluate The axis of symmetry is x = −1. For the vertex, substitute x = −1: f (x) = 2x2 + 4x – 3 2

f (−1) = 2 × (−1) + 4 × (−1) − 3

Write the function Substitute x = −1

=2−4−3

Evaluate each term

= −5

Evaluate

The vertex is (−1, −5).

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d Determine the domain and range.

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

Apply the idea The domain is . Since a = 2 > 0, the parabola is concave up, with a minimum at y = −5. Thus, the range is [−5, ∞).

e Sketch the graph, showing key features.

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

Apply the idea Plot x-intercepts

, the y-intercept (0, −3) and vertex (−1, −5).

Draw the axis of symmetry at x = −1. Sketch a concave up parabola. y 4 2 x −6

−4

−2

2 −2

(0, −3)

−4

(−1, −5)

Idea summary Non-monic quadratics follow the same graphing process as monic quadratics. The coeﬃcient a affects steepness, but x-intercepts, y-intercept, axis of symmetry, vertex, domain, and range are found using the same techniques.

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4.03 Practice questions What do you remember? 1

State whether each statement about the quadratic function y = ax2 + bx + c is true or false: a

The y-intercept is (0, c).

The axis of symmetry is

If a < 0, the parabola is concave up.

The domain is all real numbers.

For y = ax2 + bx + c, identify: a

The vertex

The y-intercept

The concavity if a < 0

The range if a < 0

What is the formula for the axis of symmetry of a quadratic y = ax2 + bx + c?

Practice Ex 1

Ex 2

For the quadratic function y = x2 − 6x + 5: a

Find the points of x-intercept(s).

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 function y = 3x2 − 6x + 2: a

Find the points of x-intercept(s) in exact form.

Find the point of y-intercept.

Find the axis of symmetry and vertex.

State the domain and range.

Sketch the graph, labelling key features.

For each quadratic function, ﬁnd the y-intercept and axis of symmetry: a

y = x2 + 6x − 2

y = −3x2 + 12x + 5

For the quadratic y = −2x2 + 8x − 1: a

Find the x-intercepts in exact form.

Find the vertex and range.

Sketch the graph, labelling key features.

Find the vertex and x-intercepts of y = 4x2 − 8x + 5.

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11 12

For the quadratic y = 3x2 − 9x + 4: a

Find the axis of symmetry and vertex.

Determine the concavity and range.

Sketch the graph, labelling the vertex, y-intercept, and axis of symmetry.

For the quadratic y = −x2 + 4x − 2: a

Find the x-intercepts in exact form, if they exist.

Find the y-intercept and axis of symmetry.

Sketch the graph, labelling the vertex, axis of symmetry, and intercepts.

Sketch the parabola y = −x2 + 2x + 3, labelling the vertex, axis of symmetry, and intercepts. For the quadratic y = (x + 2)(x − 4): a

Find the x-intercepts and y-intercept.

Find the vertex and axis of symmetry.

Sketch the graph, labelling key features.

Extend your thinking 13

A rectangular garden has a perimeter of 30 metres. Let x be the width: a

Find the formula for the area, A, in ax2 + bx + c form.

Find the vertex and the maximum possible area.

Determine the dimensions for the maximum area.

A projectile’s height is given by h = −5t2 + 20t + 1 metres after t seconds: a

Find the vertex and hence the maximum height.

Determine when the projectile hits the ground.

For the quadratic y = ax2 + 6x + 2, ﬁnd the values of a such that the parabola has one x-intercept.

A quadratic function has vertex (2, −4) and passes through (3, −3). Find its equation in standard form.

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4.04 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.

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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.

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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).

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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.

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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.

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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.

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4.04 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

180

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.

Mathspace New South Wales – Year 11 Advanced 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.

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4.05 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

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.

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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

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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.

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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

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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.

4.05 Practice questions What do you remember? 1

2 3

186

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).

Mathspace New South Wales – Year 11 Advanced 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:

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.

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4.06 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

mathspace.co

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.

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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)

≤ 0?

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

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

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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.

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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)

Yes No

= −2 (2 + 2)(2 − 1) = 4 × 1

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.

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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.

4.06 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.

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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.

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4.07 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.

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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

Write the function Substitute x = 1.02 Evaluate and round

At x = 1.02 seconds, the maximum height is 7.10 metres.

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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.

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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).

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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

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

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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.

4.07 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: a

Find the time when the projectile reaches the ground.

Find the maximum height.

4.07 Quadratic models mathspace.co

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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:

200

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.

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4.08 Cubic functions After this lesson, you will be able to… • recognise and graph cubic functions of the form f (x) = kx3. • identify the intercepts of a cubic function from its factored form. • determine the end behaviour of a cubic graph from its leading coefficient. • sketch the graph of a cubic function of the form f (x) = k(x − a)(x − b)(x − c). • interpret the practical domain for a cubic function used in a real-world model.

Graph cubic functions of the form f (x) = kx3 Cubic function A polynomial function of degree 3, that is, a function of the form f (x) = ax3 + bx2 + cx + d. A cubic function f (x) = kx3, where k ≠ 0, passes through the origin at (0, 0). The coeﬃcient k affects steepness and direction. • If k > 0, the graph increases from bottom left to top right. • If k < 0, the graph decreases from top left to bottom right. • The domain is , and the range is also .

Interactive exploration Discover this concept in action online

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Example 1 Graph f (x) = 2x3 identifying key points, direction, domain, and range.

Create a strategy Substitute x = −1, 0 and 1 into f (x) and plot the corresponding points. Since k > 0, sketch the cubic so that it extends from the bottom left to the top right passing through these points.

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Apply the idea For key points, substitute x = −1, 0, 1 into the function. If x = −1: f (x) = 2x3 f (−1) = 2(−1)

Write the function 3

Substitute x = −1

= −2

Evaluate

If x = 0: f (0) = 2(0)3

Substitute x = 0

Evaluate

If x = 1: f (1) = 2(1)3

Substitute x = 1

Evaluate

Since k = 2 > 0, the graph increases from bottom left to top right. The domain is , and the range is , as the function is continuous and unbounded. y 6 4

f (x) = 2x3 (1, 2)

(−1, −2)

−1

(0, 0)

−2

The graph increases through key points (−1, −2), (0, 0), (1, 2).

−4 −6

Idea summary Cubic functions f (x) = kx3 pass through (0, 0). The sign of k determines direction; ∣k∣ affects steepness. Domain and range are .

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Graph factored cubic functions A cubic function f (x) = k(x − a)(x − b)(x − c), where k ≠ 0, has x-intercepts at (a, 0), (b, 0), (c, 0). • If k > 0, the graph increases; if k < 0, it decreases. • The domain is , and the range is , unless restricted by context.

Interactive exploration Discover this concept in action online

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Example 2 Consider the function f (x) = (x + 2)(x − 1)(x − 3): a Find the x- and y-intercepts.

Create a strategy For the x-intercepts, set f (x) = 0 and evaluate f (0) for the y-intercept.

Apply the idea For the x-intercepts: f (x) = (x + 2)(x − 1)(x − 3) (x + 2)(x − 1)(x − 3) = 0 x

−2, x

1, x = 3

Write the function Set f (x) = 0 Use null factor law

So the x-intercepts are at (−2, 0), (1, 0), and (3, 0). For the y-intercept: f (x) = (x + 2)(x − 1)(x − 3)

Write the function

f (0) = (0 + 2)(0 − 1)(0 − 3)

Substitute x = 0 into the function

Evaluate

So the y-intercept is at (0, 6).

b Determine the domain and range.

Apply the idea Since the function is not restricted, the domain is , and the range is also .

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c Graph the function.

Create a strategy Plot the intercepts from part (a), considering the domain and range from part (b).

Apply the idea 8

6 (0, 6) 4 2

−2

−1

−2

(3, 0) x

(1, 0)

(−2, 0)

The graph has x-intercepts at (−2, 0), (1, 0), (3, 0) and y-intercept at (0, 6).

−4

f (x) = (x + 2)(x − 1)(x − 3) −6

−8

Idea summary Factored cubics f (x) = k(x − a)(x − b)(x − c) have x-intercepts at a, b, c. The graph’s direction depends on k. Domain and range are typically .

Applications of cubic functions Cubic functions model real-world phenomena, including the bending of beams in engineering, speciﬁc cost and proﬁt functions in economics, smooth curves in computer graphics, population growth models, and the relationship between the side length of a cube and its volume.

Bending of beams

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Bézier curve in computer graphics

Example 3 A box is made from a 10 cm by 6 cm cardboard by cutting squares of side x cm from each corner. The volume is V (x) = 4x(3 − x)(5 − x) cm3: a Determine the practical domain for which the volume is positive.

Create a strategy Find the roots of V (x). Divide into intervals. Test the sign of V (x) on each interval between consecutive roots. Keep the intervals where V (x) > 0, then apply the context (dimensions/constraints) to obtain the practical domain.

Apply the idea Find the roots by setting V (x) = 0: V (x) = 4x(3 − x)(5 − x) 4x(3 − x)(5 − x) = 0 x

0, x

3, x = 5

Write the function Set V (x) = 0 Use null factor law

Divide into intervals: (−∞, 0), (0, 3), (3, 5), and (5, ∞). Test a point in each interval to determine where V (x) > 0. Test point (−∞, 0)

−1

(0, 3)

(3, 5)

(5, ∞)

x(3 − x)(5 − x) 4(−1)(3 − (−1))(5 − (−1)) = (−4)(4)(6) = −96 4(1)(3 − 1)(5 − 1) = (4)(2)(4) = 32 4(4)(3 − 4)(5 − 4) = (16)(−1)(1) = −16 4(6)(3 − 6)(5 − 6) = (24)(−3)(−1) = 72

V (x) > 0? No Yes No Yes

V (x) > 0 over the intervals (0, 3) and (5, ∞). However, since the shorter side of the box is 6 cm, the cut-out squares can be at most 3 cm, as 3 + 3 = 6 would leave no width remaining. Therefore, the only practical domain is (0, 3).

b Find the x- and y-intercepts.

Create a strategy For the x-intercepts, set V (x) = 0 and check against the practical domain. For the y-intercept, evaluate V (0).

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Apply the idea For the x-intercepts: V (x) = 4x(3 − x)(5 − x) 4x(3 − x)(5 − x) = 0 x

3, x = 5

0, x

Write the function Set V (x) = 0 Use null factor law

Within the domain (0, 3), no x-intercepts exist, as V (x) > 0 in this interval. The roots x = 0 and x = 3 are endpoints where V (x) = 0. For the y-intercept: V (x) = 4x(3 − x)(5 − x)

Write the function

V (0) = 4(0)(3 − 0)(5 − 0)

Substitute x = 0

Evaluate

The y-intercept is at (0, 0), which is an endpoint of the domain.

c Graph the function.

Create a strategy Calculate additional points within the practical domain (0, 3), then plot these points along with the boundary behaviour.

Apply the idea Using x = 1: V (x) = 4x(3 − x)(5 − x)

Write the function

V (1) = 4 × 1 × (3 − 1)(5 − 1)

Substitute x = 1

=4×2×4

Evaluate each term

= 32

Evaluate

Using x = 2:

206

V (x) = 4x(3 − x)(5 − x)

Write the function

V(2) = 4 × 2 × (3 − 2)(5 − 2)

Substitute x = 2

=8×1×3

Evaluate each term

= 24

Evaluate

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The additional points are (1, 32) and (2, 24). y 35

V (x) = 4x(3 − x)(5 − x)

(1, 32)

30 25

The graph shows the volume for 0 < x < 3, with points (1, 32) and (2, 24). The volume approaches zero at the boundaries x = 0 and x = 3. The range is estimated as (0, 32] based on these points.

(2, 24)

20 15 10 5

Idea summary Cubic functions can model specific real-world scenarios, but practical constraints must be taken into account when determining the domain and range of the model.

4.08 Practice questions What do you remember? 1

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

The function f (x) = kx3 has an x-intercept at (0, 0).

If k > 0 in f (x) = k(x − a)(x − b)(x − c), the graph is always increasing.

The range of f (x) = kx3 is .

The function f (x) = k(x − a)(x − b)(x − c) has exactly three x-intercepts.

For f (x) = kx3, how does the sign of k affect the end behaviour of the graph?

What are the x-intercept of f (x) = k(x − a)(x − b)(x − c)?

Practice Ex 1

Graph

Determine the domain and range.

Sketch the graph, labelling key points.

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Ex 2

Find the x-intercepts, y-intercept.

Determine the end behaviour of the function.

Determine the domain and range.

Sketch the graph, labelling key points.

Graph f (x) = −x3, labelling the key points.

For the cubic function f (x) = (x − 1)(x + 2)(x + 3):

Ex 3

Graph f (x) = (x + 1)(x − 2)(x − 4):

Determine the x-intercepts and y-intercept.

Sketch the graph, labelling intercepts.

Graph f (x) = −(x + 3)(x − 1)(x − 2): a

Identify the x-intercepts, and y-intercept.

Sketch the graph, labelling key points.

The volume of a box made from a 12 cm by 10 cm cardboard by cutting squares of side x cm from each corner is V (x) = 4x(6 − x)(5 − x) cm3: a

Determine the practical domain.

Find the x- and y-intercepts.

Graph the volume function for its practical domain.

The mass of a cube is m(x) = 0.002x3 grams, where x is the side length in cm: a

Graph the mass function for 0 < x ≤ 10.

Find the side length for a mass of 8 grams, rounded to two decimal places.

Extend your thinking 11

A cubic function has x-intercepts at (−1, 0), (2, 0), (4, 0) and y-intercept at (0, 8): a

Determine the equation in factored form.

Sketch the graph, labelling key points.

The deﬂection of a beam is modelled by D(x) = x(4 − x)(6 − x), where x is the distance (in metres) from one end of the beam. For what values of x does the deﬂection D(x) take positive values?

A company models its weekly proﬁt with the function P (x) = x(5 − x)(10 − x), where x is the number of items sold (in hundreds). Determine the values of x for which the company makes a proﬁt.

The growth rate of a population is modelled by R(t) = t(2 − t)(8 − t), where t is the number of years since the population was introduced. For what time values does the population increase?

The volume of a cylindrical tank is V (r) = π r2(10 − r), where r is the radius and 10 − r is the height in metres. Graph the volume function for the practical domain. From the graph, estimate the maximum volume of the cylinder.

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4 Chapter review 1

Which of the following is the vertex of the parabola with the equation y = 3(x + 4)2 − 1? A

(−4, −1)

(4, 1)

(−4, 1)

How many x-intercepts does the graph of the function y = 3x2 − 4x + 2 have? A

(4, −1)

Inﬁnitely many

Which of the following graphs could represent the function f (x) = −(x + 1)(x − 2)(x − 4)? A

5 x

x −3 −2 −1

−3 −2 −1

−5

−10

x −3 −2 −1

−5

−10

For each quadratic function: i

Find the x-intercepts.

Find the y-intercept.

iii

Find the axis of symmetry.

y = −(x + 1)(x − 5)

y = (x − 3)(x + 2)

y = 2(x − 1)(x − 4)

y = −(x + 3)(x − 2)

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For each quadratic function: i

Find the y-intercept.

Find the axis of symmetry.

iii

Determine the vertex.

y = 2x2 − 8x + 6

y = x2 + 4x − 5

y = −3x2 + 6x + 9

y = x2 − 2x − 3

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 quadratic function y = x2 − 8x + 12: a

Calculate the discriminant.

Determine the number and nature of x-intercepts.

Justify the parabola’s position relative to the x-axis.

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

210

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.

For the quadratic function y = x2 + 6x + 5: a

Express in completed square form.

Identify the vertex and axis of symmetry.

Find the point of the y-intercept.

Sketch the graph, labelling key features.

For the quadratic function y = 2x2 − 12x + 11: a

Express in completed square form.

Identify the vertex and axis of symmetry.

Determine the domain and range.

Sketch the graph, labelling key features.

Mathspace New South Wales – Year 11 Advanced mathspace.co

A parabola has a vertex at (2, −5) and a y-intercept at (0, 3): a

Find the equation of the parabola in completed square form.

State the axis of symmetry and range.

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):

Find the equation in factored form.

Convert to standard form.

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.

Chapter 4 review mathspace.co

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

Algebraically

Graphically

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

Algebraically

Graphically

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.

Graph f (x) = −2x3, labelling the key points.

For the equation f (x) = (x + 2)(x − 1)(x − 4):

212

Find the x-intercepts and the y-intercept.

Sketch the graph, labelling key points.

A cubic function has x-intercepts at (−2, 0), (1, 0), (3, 0) and a y-intercept at (0, 12): a

Determine the equation of the function in factored form.

Sketch the graph, labelling key points.

Mathspace New South Wales – Year 11 Advanced mathspace.co

“Mathematics is not about numbers, equations, computations, or algorithms: it is about understanding.” William Paul Thurston

Big ideas • A function is characterised by its domain and range, which define its boundaries, and by its inherent symmetry, which classifies it as even, odd, or neither through specific algebraic tests. • New functions can be constructed by combining existing functions through composition, where the output of one becomes the input for another, or by defining a function in separate pieces over different intervals, which introduces the concept of continuity at the boundaries.

5 Function properties Chapter outline 5.01 5.02 5.03 5.04

Further domain and range Even and odd functions Composite functions Piecewise functions Chapter 5 review

216 224 232 241 253

Some car engines adjust fuel delivery using piecewise functions – more power when you floor it!

5.01 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

5.01 Further domain and range mathspace.co

219

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.

5.01 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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221

Practice Ex 1

Express the given interval notation of the domain in: i

Ex 2

Ex 3

Ex 4

222

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 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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223

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.

5.02 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.

Interactive exploration Discover this concept in action online

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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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5.02 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 mathspace.co

−2

−4

−2

228

−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.

Mathspace New South Wales – Year 11 Advanced mathspace.co

x 1

−2

−4

−2

230

f (x) = x2 + x + 1

f (x) = x + x

−2

f (x) = −2

iii

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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5.03 Composite functions After this lesson, you will be able to… • use the composite function notation f ( g(x)). • determine the equations of composite functions by substitution. • evaluate composite functions for numerical and algebraic inputs. • determine the domain and range of composite functions.

g(x) = 2x + 1

(2x + 1)

f (x) = x2

(2x + 1)2

To form f ( g(x)), substitute g(x) into f (x). The order matters, as g( f (x)) and f ( g(x)) may yield different results.

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Example 1 Given f (x) = 3x − 2 and g(x) = x2 + 1, determine f ( g(x)).

Create a strategy Substitute g(x) into f (x).

Apply the idea f (x) = 3x − 2 2

f ( g(x)) = 3(x + 1) − 2 = 3x2 + 3 − 2 2

= 3x + 1

Write the function Substitute g(x) = x2 + 1 Use the distributive property Evaluate

Example 2 If f (x) = −2x − 3 and g(x) = −2x − 6, then evaluate: a f (7) and g (7)

Create a strategy Substitute x = 7 into the functions f (x) = −2x − 3 and g(x) = −2x − 6.

Apply the idea For f (7): f (x) = −2x − 3

Write the function

f (7) = −2 × 7 − 3

Substitute x = 7

= −14 − 3

Evaluate the multiplication

= −17

Evaluate

For g(7): g(x) = −2x − 6

Write the function

g(7) = −2 × 7 − 6

Substitute x = 7

= −14 − 6

Evaluate the multiplication

= −20

Evaluate

Thus, f (7) = −17 and g(7) = −20.

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b f ( g(7))

Create a strategy Use the result for g(7) from part (a) and substitute it into f (x) = −2x − 3.

Apply the idea From part (a), g(7) = −20: f (x) = −2x − 3

Write the function

f ( g(7)) = −2 × g(7) − 3

Substitute x = g(7)

= −2 × (−20) − 3

Substitute g(7) = −20

= 40 − 3

Evaluate the multiplication

= 37

Evaluate

c g( f (7))

Create a strategy Use the result for f (7) from part (a) and substitute it into g(x) = −2x − 6.

Apply the idea From part (a), f (7) = −17: g (x) = −2x − 6

Write the function

g ( f (7)) = −2 × f (7) − 6

Substitute x = f (7)

= −2 × (−17) − 6

Substitute f (7) = −17

= 34 − 6

Evaluate the multiplication

= 28

Evaluate

d Determine if g( f (x)) = f ( g(x)) for all x.

Create a strategy Compare the results of g( f (7)) and f ( g(7)) from parts (b) and (c).

Apply the idea From part (b), f ( g(7)) = 37. From part (c), g( f (7)) = 28. g( f (7)) = f ( g(7)) → 28 ≠ 37 Since g( f (7)) ≠ f ( g(7)), it is not true that g( f (x)) = f ( g(x)) for all x.

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Idea summary A composite function f ( g(x)) is formed by applying g(x) ﬁrst, then f (x) to the result. Similarly, g( f (x)) applies f (x) ﬁrst. The order of composition affects the resulting function.

Domain and range of composite functions The domain of a composite function g( f (x)) depends on the domain of f (x) and the values f (x) produces, which must be valid inputs for g(x). The range of g( f (x)) is determined by the range of g(x), restricted by the output of f (x). For example, if f (x) has no restrictions but g(x) requires non-negative inputs, the domain of g( f (x)) is restricted to values of x where f (x) ≥ 0. To ﬁnd the domain of g( f (x)), follow these steps: 1. Determine the domain of f (x). 2. Determine the range of f (x). 3. Determine the domain of g(x). 4. Restrict the range of f (x) to only those values that lie within the domain of g(x). 5. Restrict the domain of f (x) to only those inputs that produce outputs in the restricted range from Step 4. This restricted domain of f (x) is the domain of g( f (x)). To ﬁnd the range of g( f (x)), follow these steps: 1. Find the range of f (x). 2. Identify the values in the range of f (x) that are also in the domain of g(x). 3. Evaluate g( f (x)) using only the inputs x for which f (x) lies in the set from Step 2. The resulting set of g( f (x)) values is the range of g( f (x)).

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mathspace.co

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235

Example 3 Given f (x) = 3x − 1 and g(x) =

, determine:

a The domain of g( f (x))

Create a strategy Identify the values of x where f (x) is a valid input for g(x).

Apply the idea First, form the composite function g( f (x)). Write the function Substitute f (x) = 3x − 1 For the domain, g(x) =

requires x ≥ 0, so set f (x) ≥ 0. Write the inequality

Substitute f (x) = 3x − 1

Add 1 to both sides

Divide both sides by 3

The domain is

b The range of g( f (x))

Create a strategy Analyse the output of g( f (x)).

Apply the idea For the range, g( f (x)) =

produces non-negative values, as the square root function outputs

y ≥ 0. Since 3x − 1 ≥ 0 for x ≥ , the range is [0, ∞).

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Example 4 The graphs of functions f (x) and g(x) are shown: y

4 3 2

f (x)

1 −1

g (x) x

−2

x −2

−1

−3 −4

What is the range of the composite function f ( g(x))?

Create a strategy Observe the range of g(x) from its graph, intersect with the domain of f (x), and determine the output values of f (x) over this interval using its graph.

Apply the idea From the graph of g(x), the function has a minimum at (0, 0) and approaches y = 1 as x increases or decreases without bound, giving range [0, 1). The graph of f (x) shows a vertical asymptote at x = 0, so its domain is (0, ∞). Intersecting with [0, 1) gives (0, 1). On the graph of f (x), for inputs x ∈ (0, 1), the output ranges from − ∞ as x approaches 0 to 0 as x approaches 1, excluding 0. The range of f ( g(x)) is (− ∞, 0).

Reﬂect and check Ensure the range of g(x) excludes 1 due to the asymptote, and verify f (x)’s domain excludes 0. A common error is including 0 in the range, but g(x) < 1 prevents this.

Idea summary The domain of g( f (x)) requires f (x) to produce valid inputs for g(x). The range depends on the output of g(x), constrained by f (x)’s output.

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5.03 Practice questions What do you remember? 1

Deﬁne a composite function and give an example.

Given f (x) = 2x and g(x) = x − 1, what is the difference between f ( g(x)) and g( f (x))?

For a composite function g( f (x)), what determines its domain?

Identify whether each statement is true or false: a

f ( g(x)) = g( f (x)) for all functions.

The range of g( f (x)) depends on g’s output.

f (x) =

Composite functions are always linear.

has domain x ≥ 0.

Practice Ex 1

Given f (x) = 2x + 1 and g(x) = x2 − 3, determine: a

Ex 4

q( p(5))

h(k(x))

f (3)

k(h(x))

and g(x) = b

g( f (2))

f ( g(3))

p(q(−2))

q( p(−2))

h(k(0))

k(h(0))

, evaluate:

g( f (3))

Given f (x) = 4x + 5 and g(x) =

, determine the:

Domain

Range

g( f (x))

f ( g(x))

Given f (x) =

and g(x) = x − 2, determine the:

Domain

Range

f ( g(x))

g( f (x))

Domain of p(q(x))

Given p(x) = x2 − 1 and q(x) = , determine: a

238

p(q(5))

Given f (x) = 3x2 + a

Ex 3

f ( g(x))

Given h(x) = x2 and k(x) = x + 5, determine: a

Ex 2

Given p(x) = x − 4 and q(x) = 3x, evaluate: a

g( f (x))

p(q(x))

Mathspace New South Wales – Year 11 Advanced mathspace.co

The graphs of functions f (x) and g(x) are shown: y

g (x) 2

f (x) x 1

x −2

−1

Determine the range of the composite function f ( g(x)). 13

The graphs of functions f (x) and g(x) are shown: y

3 2

f (x)

−4 −3 −2 −1 −1

x −2

−1

−2 −3

−2

−4

g (x)

Determine the domain of the composite function f ( g(x)). 14

The graphs of functions f (x) and g(x) are shown: y

y 4

g (x) 2

2 1

f (x) x 1

x −2

−1

Determine the range of the composite function f ( g(x)).

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The graph shows y = f (x) and y = g(x):

y 4

Given that f (x) is a parabola, determine: a

f (1)

g( f (1))

f (x) = x2

1 −4 −3 −2 −1

x 1

−1

−2

g (x) = 2x

−3 −4

Given f (x) = x + 2 and g(x) = 3x − 1, determine x such that: a

Given h(x) = a

g( f (x)) = 8

Domain of h(k(x))

Domain of g( f (x))

and k(x) = x2, determine:

h(k(x))

Given f (x) = a

f ( g(x)) = 8

and g(x) = x − 3, determine:

g( f (x))

Given f (x) = 2x and g(x) = x2 + 1, determine the domain and range of these functions: a

f (x)

g(x)

f ( g(x))

g( f (x))

Extend your thinking 20

Given f (x) = 2x + 1 and g(x) = x2, determine all x such that f ( g(x)) = g( f (x)).

The graph shows y = f (x) and y = g(x). Sketch the graph of y = g( f (x)).

−2

y = x2 2

−2

y=x+1

A temperature conversion uses f (x) =

(Fahrenheit to Celsius) and g(x) = x + 273.15

(Celsius to Kelvin). Determine and interpret g( f (x)). 23

Given f (x) =

and g(x) = , determine and compare the domains of f ( g(x)) and g( f (x)).

Explain the reasoning for each domain based on the function constraints.

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5.04 Piecewise functions After this lesson, you will be able to… • interpret and evaluate piecewise-defined functions. • graph piecewise-defined functions involving linear and non-linear components. • identify points where a function is continuous or discontinuous. • determine if a piecewise-defined function is even or odd.

Piecewise functions Piecewise-deﬁned function A function which is deﬁned differently for different parts of the domain. For example,

When a piecewise graph has no gaps or breaks, and all the lines are connected to one another, it creates a continuous piecewise function. Continuous function The graph of a continuous function is an unbroken curve. It can be drawn without lifting the pen off the paper. If the lines are disconnected, the piecewise function has discontinuities. Discontinuity A point at which the graph of a function is broken (not continuous) and the function is undeﬁned.

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241

Consider the piecewise function:

The function uses f (x) = x2 for x ≤ 0 and f (x) = 2x + 1 for x > 0. The graph is not continuous at x = 0, as the two pieces do not meet at the same point.

y 6 5 4 3 2 1 −2

−1

x 1

To check for continuity at the boundary point x = 0, evaluate each piece of the function at this value. If the results are different, there is a discontinuity. For the piece f (x) = x2 (where x ≤ 0): f (0) = 02 =0

Substitute x = 0 Evaluate

For the piece f (x) = 2x + 1 (where x > 0): f (0) = 2(0) + 1 =1

Substitute x = 0 Evaluate

Since the two pieces result in different y-values (0 and 1) at x = 0, the function is not continuous. There is a discontinuity at x = 0. To determine if a piecewise function is even or odd, check if f (− x) = f (x) (even) or f (− x) = −f (x) (odd). In a piecewise function, f (x) and f (− x) may be deﬁned by different expressions, depending on which interval x and − x fall into. So it is a must to: 1. Identify which piece(s) deﬁne f (− x). 2. Identify which (possibly different) piece(s) define f (− x). 3. Evaluate and compare f (− x) with f (− x).

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Example 1 Consider this piecewise function:

Sketch the piecewise graph.

Create a strategy Graph each linear function over each interval.

Apply the idea For x < 0, graph the constant function y = 5. Since the function is not deﬁned for x = 0, there will be a hollow circle at (0, 5).

y 7 6 5 4 3 2 1 −2

−1

For x > 0, graph the linear function y = x + 5.

y 7 6 5 4 3 2 1 −2 −1

x 1

Reﬂect and check This piecewise function is discontinuous as the graphs are not connected at x = 0.

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Example 2 What is the function deﬁnition of the graph? y 4 3 2 1 −4 −3 −2 −1 −1

x 1

2 3 4 5

−2 −3 −4

Create a strategy Consider the three distinct parts of this graph and determine the domain for each of these parts.

Apply the idea The ﬁrst part of the graph exists for all values of x from − ∞ to 0.

y 4 3

This means this is a linear function with a gradient of

−4 −3 −2 −1 −1

1 2 3 4 5

and a y-intercept of 3.

Therefore the graph is y =

+ 3 in the domain x ≤ 0.

−2 −3 −4

The second part of the graph exists for all values of x between 0 and 3, including 3.

y 4 3

The value x = 0 is not included in this interval since x = 0 has a hollow circle for this domain.

2 1 −4 −3 −2 −1 −1 −2 −3

x 1

2 3 4 5

This means it is a linear function with a gradient of =

Therefore the graph is y = − x + 3 in the domain 0 < x ≤ 3.

−4

244

= −1 and y-intercept y = 3.

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The ﬁnal part of the graph exists for all values of x between 3 and 5.

y 4 3

The value x = 3 is not included in this interval since the graph has hollow circles at the endpoints.

2 1 −4 −3 −2 −1 −1

x 1

This means it is a linear function with a gradient of

2 3 4 5

= 2 and y-intercept − 6. Therefore the graph is

y = 2x − 6 in the domain 3 < x < 5.

−2 −3 −4

Idea summary Piecewise functions deﬁne different rules over distinct domain intervals. A function is continuous if the graph has no breaks. Discontinuities occur at points where the graph has gaps or jumps. To determine if a piecewise function is even or odd, compare f (− x) with f (x).

Non-linear piecewise functions Piecewise functions can include non-linear functions such as quadratics and cubics, deﬁned over speciﬁc domains. When graphing, identify key features like turning points, intercepts, and endpoints. Check continuity at domain boundaries to identify discontinuities. Consider the piecewise function:

To check continuity at x = 2, evaluate both pieces at the boundary point. For x ≤ 2, substitute x = 2:

For x > 2, substitute x = 2:

f (x) = x2 − 1 f (2) = 22 − 1 =3

Write the function Substitute x = 2 Evaluate

f (x) = −(x − 3)2 + 2 f (2) = −(2 − 3)2 + 2 =1

Write the function Substitute x = 2 Evaluate

Since the y-values from the two pieces are different at the boundary point x = 2 (3 ≠ 1), the two parts of the graph do not meet. Therefore, the function has a discontinuity at x = 2.

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Example 3 Sketch the graphs of: a

Create a strategy Graph each function over their corresponding intervals. Consider the key points that may be required, such as endpoints, turning points, and intercepts.

Apply the idea • First, consider y = 2 − 3x over the domain x < 0. A linear graph requires two points to sketch. To determine the coordinate of the boundary point for the domain x < 0, evaluate this piece of the function at x = 0: y = 2 − 3x

Write the equation

= 2 − 3(0)

Substitute x = 0

Evaluate

Therefore, the boundary point is at (0, 2). Use any value less than zero to identify a second point. At x = −1: y = 2 − 3x

Write the equation

y = 2 − 3(−1)

Substitute x = −1

Evaluate

So (−1, 5) is a point on the line y = 2 − 3x. • Next, consider y = (x − 2)2 − 2 over the domain x ≥ 0. Use x = 0 to determine the endpoint of this parabola: y = (x − 2)2 − 2

Write the equation

= (0 − 2) − 2

Substitute x = 0

Evaluate

This parabola has an endpoint at (0, 2). • Then, consider the turning point as this is a key feature of any parabola. The equation y = (x − 2)2 − 2 is in turning point form, so the turning point is (2, −2) which will occur over the domain x ≥ 0. Determine the x-intercepts of the parabola: Write the equation

Substitute y = 0

Add 2 to both sides

Take the square root of both sides

Add 2 to both sides making x the subject

There are two x-intercepts:

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The graph can be sketched after all key points have been determined: y 6

(−1, 5)

5 4 3 2

(0, 2)

1 −3

−2

−1

x 1

−1 −2

(2, −2)

−3

Apply the idea • First, consider y = 1 over the domain x < 2: This is a horizontal line with zero gradient. • Next, consider y = (x − 3)3 over the domain 2 ≤ x ≤ 4: Use x = 2 and x = 4 to determine the endpoints of this cubic function. At x = 2: y = (x − 3)3 Write the equation = (2 − 3)3

Substitute x = 2

= −1

Evaluate

This cubic graph has an endpoint at (2, −1). At x = 4: y = (x − 3)3

Write the equation

= (4 − 3)3

Substitute x = 4

Evaluate

This cubic graph has an endpoint at (4, 1). • Finally, consider the graph of y = 2 − (x − 5)2 over the domain x > 4. To determine the coordinate of the boundary point for the domain x > 4, evaluate this piece of the function at x = 4. At x = 4: y = 2 − (x − 5)2 Write the equation = 2 − (4 − 5)2

Substitute x = 4

Evaluate

The boundary point for this piece is at (4, 1).

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• Also consider the turning point as this is a key feature of parabolas. The parabola is in vertex form, therefore the turning point is (5, 2). Since this occurs at x = 5, this will occur over the domain x > 4. To determine the x-intercepts, let y = 0 and solve for x: Write the equation

Substitute y = 0

Multiply both sides by −1

Add 2 to both sides

Take the square root of both sides

Add 5 to both sides making x the subject

Reject x = 5 − as it is not within the interval x > 4. So, there is one x-intercept on . the right-side of the turning point: The graph can be sketched after all key points have been determined: y 4 3 2 1 −2

(5, 2) (0, 1)

−1

(2, 1) 1

−1

(4, 1)

(3, 0) 2

(2, −1)

x 6

−2 −3 −4

Idea summary Non-linear piecewise functions use functions like quadratics and cubics over speciﬁc domains. Graphing requires identifying key features and checking for discontinuities at domain boundaries.

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5.04 Practice questions What do you remember? 1

Answer these questions about piecewise functions: a

What is a piecewise function?

What does it mean for a piecewise function to be continuous?

How do you determine if a piecewise function is even?

How do you determine if a piecewise function is odd?

How to check if a piecewise function is continuous at the boundary between two intervals?

What are the key features to identify when graphing a non-linear piecewise function?

Identify whether each statement is true or false: a

A piecewise function is always continuous.

The domain of a piecewise function is always all real numbers.

A piecewise function can have non-linear components.

A piecewise function is continuous if the graph has no breaks.

Practice 5

Evaluate the piecewise function at the given points:

a Ex 1

f (0)

f (1)

f (2)

f (−1)

Sketch these piecewise functions: a

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Ex 2

Determine the function of each graph: a

4 3

1 −4 −3 −2 −1

x 1

−1

−5 −4 −3 −2 −1

−2 −3

−3

−4

−5

d (6, 3)

3 2 1 −7 −6 −5 −4 −3 −2 −1 −1

(−5, 18)

(3, 2) x 1 2 3 4 5 6 7

(−1, −2)

−2 −3

22 20 18 16 14 12 10 (0, 8) 8 6 4 2

−6 −4 −2 −2

−4

−1

−2

Ex 3

(10, 21)

(5, 6) x 4

8 10 12

Sketch these piecewise functions:

Determine the piecewise function if 2 units up.

For the piecewise function

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is shifted 3 units to the right and

, determine all values of x such that f (x) = 3.

Determine if the piecewise function is even, odd, or neither: a

A mobile phone plan charges based on data usage per month. The cost C(x) (in dollars) for x GB of data is deﬁned by:

Evaluate C(3).

Evaluate C(6).

Sketch the function for 0 ≤ x ≤ 7.

Determine the range for 0 ≤ x ≤ 7.

A taxi fare in Sydney is calculated based on the distance travelled d (in kilometres). The fare F (d) (in dollars) is deﬁned as:

Evaluate F (2).

Evaluate F (6).

Sketch the function for 0 ≤ d ≤ 7.

Determine the range for 0 ≤ d ≤ 7.

Determine the domain and range.

Sketch the graph for −3 ≤ x ≤ 3.

A piecewise function is deﬁned as:

Evaluate f (−1) and f (4).

Sketch the function for −3 ≤ x ≤ 6.

For the function f (x) = a

Identify any points of discontinuity.

Extend your thinking 16

Consider this piecewise function:

If the graph of the piecewise function is continuous, determine the value of a.

What is the y-value of the piecewise function at x = a?

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Explain why this relation is not a function:

A water tank’s drainage rate depends on the time t (in hours) since drainage began. The ﬂow rate R(t) (in litres per hour) is given by:

Sketch the function R(t) for 0 ≤ t ≤ 5 and explain why it is continuous.

With the help of the sketch of the function, explain how you would calculate the total volume of water drained between t = 0 and t = 4 hours. Hence, calculate the total volume.

Did you know?

Speed limits on roads can be modelled using piecewise functions, since the allowed speed often changes depending on the zone you’re driving in. On a drive, you might start in a school zone where the speed limit is 40 km/h, then move into a suburban street with a 60 km/h limit, and finally join a highway with a 100 km/h limit. Each stretch of road can be represented by a different “piece” of a function, with the speed staying constant within that section.

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5 Chapter review 1

What is the range of the function f (x) = x2 − 4? A

(− 4, ∞)

f (x) = 2x4 + 3

f (x) = x3 − 5x

(− ∞, ∞)

f (x) = x2 + x

f (x) = 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1 2 3 4 5

Determine the domain of each relation based on the equation: a

(− ∞, − 4]

Which piecewise function matches the graph shown? 5 4 3 2 1

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.

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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.

Given f (x) = 3x + 2 and g(x) = x2 − 1, determine: a

f ( g(x))

g( f (1))

f ( g(2))

Given f (x) = x + 5 and g(x) = 2x − 1, determine x such that: a

g( f (x))

f ( g(x)) = 10

Given f (x) =

g( f (x)) = 10

and g(x) = x − 5, determine the:

Domain

Range

f ( g(x))

g( f (x))

Given f (x) = 3x + 1 and g(x) = x2, determine all x such that f ( g(x)) = g( f (x)).

A function U (a) = 1.54a converts an amount of Australian dollars, a, to New Zealand dollars. A function E(n) = 0.56n converts an amount of New Zealand dollars, n, to Euros. Derive a composite function for converting Australian dollars directly to Euros and use it to evaluate how many Euros are equivalent to AUD $200.

Evaluate the piecewise function at the given points:

f (1)

f (2)

Sketch the piecewise function:

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f (4)

f (−2)

A streaming service charges a monthly fee based on viewing hours. The cost C(h) (in dollars) for h hours of viewing is deﬁned by:

Evaluate C(20).

Evaluate C(40).

Sketch the function for 0 ≤ h ≤ 40.

Determine the range for 0 ≤ h ≤ 40.

Consider this piecewise function:

If the graph of the piecewise function is continuous, determine the value of a.

What is the y-value of the piecewise function at x = a?

A factory’s production rate depends on the time t (in hours) into a shift. The rate P (t) (in units per hour) is given by:

Sketch the function P (t) for 0 ≤ t ≤ 6 and explain why it is continuous.

Determine the range of the function for 0 ≤ t ≤ 4.

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Big ideas • Direct and inverse variation describe how quantities change in proportion to one another, a concept modelled algebraically with a constant of proportionality. The rectangular hyperbola is the graphical representation of the fundamental inverse relationship, defined by asymptotes that reflect its mathematical constraints. • The absolute value function formalises the concept of distance from zero, resulting in a characteristic ‘V’-shaped graph. Its piecewise definition is the key to both graphing and algebraically solving absolute value equations, which requires the consideration of two distinct cases. • The circle is a fundamental geometric relation whose algebraic equation, x2 + y2 = r2, is a direct application of Pythagoras’ theorem to its definition as the set of all points equidistant from a centre.

6 Functions and relations Chapter outline 6.01 6.02 6.03 6.04 6.05 6.06

Graphs of reciprocal functions Direct variation models Inverse variation models Introduction to absolute value functions Absolute value functions Circles and semicircles Chapter 6 review

258 265 272 278 283 291 299

Reciprocal graphs never touch zero — kinda like how cats always avoid water!

6.01 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).

6.01 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

262

Complete the table of values: x

−2

−1

f (x)

⬚

Sketch the graph, clearly showing the asymptotes and key points from the table.

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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

264

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.

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6.02 Direct variation models After this lesson, you will be able to… • develop models for direct variation of the form y = kxn • evaluate the constant of proportionality, k, given a pair of values • use a direct variation model to find the value of a variable • analyse and solve practical problems involving direct variation

Direct variation Direct variation A proportional relationship where one quantity directly varies with respect to a change in another quantity. This implies that if there is an increase (or decrease) in one quantity then the other quantity will experience a proportionate increase (or decrease). If a quantity y varies directly with a quantity x or a power of x, the relationship is modelled as:

y = kxn k

is the constant of proportionality, where k ≠ 0

is the power of the relationship, where n > 0

is the independent variable

is the dependent variable

Direct linear variation takes the form y = kx where n = 1. Always assume direct variation means a linear relationship unless a different power or function is speciﬁed. For example, the distance D (in metres) a car travels at a constant speed of 20 m/s varies directly with time T (in seconds). The model is D = 20T, where n = 1 and k = 20. When n = 1, the graph is a straight line through the origin, indicating that D increases proportionally with T. D (in metres)

D = 20T

100 80

The graph of D = 20T illustrates direct variation, with distance increasing linearly with time.

Direct variation relationships always pass through the origin.

40 20

T (in seconds) 0

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Idea summary Direct variation occurs when one quantity increases as another increases, modelled as y = kxn, where k ≠ 0 is the constant of proportionality and n > 0 is the power.

Constant of proportionality To apply the direct variation model y = kxn, determine the constant k using a known pair of values for x and y. Then, use the model to calculate other values. For example, if y varies directly with x (n = 1) and y = 30 when x = 5, ﬁnd k: Write the model with n = 1 Substitute y = 30 and x = 5

Divide both sides by 5

Evaluate The model is y = 6x, which can be used to ﬁnd y for other values of x.

Example 1 The cost C (in dollars) of hiring a machine varies directly with the time T (in days). If C = 150 dollars for T = 3 days: a Find the constant of linear proportionality, k.

Create a strategy Model the relationship as C = kT and substitute C = 150, T = 3 to solve for k.

Apply the idea Write the direct variation model Substitute C = 150 and T = 3

Divide both sides by 3

Evaluate The constant of proportionality is k = 50.

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b Find the cost for T = 7 days.

Create a strategy Use the model C = 50T and substitute T = 7.

Apply the idea C = 50T

Write the model with k = 50

= 50 × 7

Substitute T = 7

= $350

Evaluate

The cost is 350 dollars.

Idea summary To apply y = kxn, substitute known values to ﬁnd k, then use the model to calculate other values of y or x.

Applications of direct variation Direct variation models solve real-world problems where one quantity increases as another increases, contrasting with inverse variation, where one quantity decreases as another increases.

Example 2 The area A (in square metres) of a circle varies directly with the square of its radius R (in metres). If A = 12.56 square metres when R = 2 metres: a Find the constant of proportionality k.

Create a strategy Model the relationship as A = kR2 and substitute A = 12.56, R = 2 to solve for k.

Apply the idea Write the model with n = 2 Substitute A = 12.56 and R = 2

Evaluate 22

Divide both sides by 4

Evaluate The constant of proportionality is k = 3.14, approximately π.

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b Find the radius when A = 28.26 square metres, rounded to two decimal places.

Create a strategy Use A = 3.14R2 and substitute A = 28.26 to solve for R.

Apply the idea Write the model with k = 3.14 Substitute A = 28.26

Divide both sides by 3.14

Evaluate

Take the positive square root

Evaluate The radius is 3 metres.

Reﬂect and check Using the exact value k = π, the calculation yields R2 = result.

= 9, so R = 3 metres, confirming the

Example 3 The volume V (in cubic metres) of a sphere varies directly with the cube of its radius R (in metres). If V = 36π cubic metres when R = 3 metres: a Find the constant of proportionality k.

Create a strategy Model the relationship as V = kR3 and substitute V = 36π, R = 3 to solve for k.

Apply the idea Write the model with n = 3 Substitute V = 36π and R = 3

Evaluate 33

Divide both sides by 27

Simplify The constant of proportionality is k =

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b Find the exact volume when R = 6 metres.

Create a strategy and substitute R = 6.

Use

Apply the idea Write the model with k = Substitute R = 6 Evaluate 63

Simplify The volume is 288π cubic metres.

Reﬂect and check The volume formula for a sphere is calculations.

, which matches the model, conﬁrming the

Idea summary Direct variation models use the form y = kxn to solve problems by determining k from known values, then applying the model to ﬁnd unknown quantities.

6.02 Practice questions What do you remember? 1

State the meaning of direct variation.

If y varies directly with xn, write the general equation, using k as the constant of variation.

In the equation y = kxn, what does k represent?

Determine if these equations represent direct variation: a

y = 5x

y = 3x2

y = 2x + 1

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Practice Ex 1

Ex 2

Ex 3

270

The volume of water V (in litres) pumped by a machine varies directly with the time T (in minutes). If the volume is V = 120 litres for T = 6 minutes: a

Find the constant of proportionality k.

Find the volume for T = 9 minutes.

The distance D (in metres) travelled by a car varies directly with the time T (in seconds) at a constant speed. If D = 100 metres, for T = 5 seconds: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the distance after T = 8 seconds.

The weight W (in newtons) of an object on a planet varies directly with its mass M (in kilograms). If W = 490 newtons, for M = 50 kg: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the weight of an object with M = 75 kg.

The amount of paint P (in litres) needed to cover a wall varies directly with the area A (in square metres). If P = 4 L covers A = 20 m2: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the amount of paint needed for A = 35 square metres.

The gravitational force F (in Newtons) between two objects varies directly with their mass product M (in kg2). If F = 16 N for M = 4 kg2: a

Write the direct variation equation and ﬁnd the constant of proportionality k.

Find the force when M = 5 kg2.

The kinetic energy E (in joules) of an object varies directly with the square of its speed S (in m/s). If E = 200 joules, for S = 10 m/s): a

Write the direct variation equation and ﬁnd the constant of proportionality k.

Find the kinetic energy when S = 15 m/s.

The intensity I (in watts per square metre) of a sound wave varies directly with the square of its amplitude A (in metres). If I = 0.8 watts per square metre, for A = 0.02 metres: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the intensity when A = 0.04 metres.

The cost C (in dollars) of electricity usage varies directly with the fourth power of the appliance’s power rating P (in watts). If C = 16 dollars, for P = 2 watts: a

Find the constant of proportionality k.

Find the cost when P = 3 watts.

The volume V (in cubic metres) of a cube varies directly with the cube of its side length S (in metres). If V = 27 cubic metres, for S = 3 metres: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the volume when S = 5 metres.

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The force F (in Newtons) required to stretch a spring varies directly with the extension X (in metres). If F = 8 Newtons, for X = 2 metres: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the force when X = 3 metres.

The cost C (in dollars) of fuel for a trip varies directly with the distance D (in kilometres). If C = 45 dollars, for D = 300 kilometres: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the distance when C = 60 dollars.

The heat energy H (in joules) produced by a resistor varies directly with the square of the current I (in amperes). If H = 50 joules, for I = 5 amperes: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the current when H = 128 joules.

Extend your thinking 17

The cost C (in dollars) of running an air conditioner varies directly with the cube of its power setting P (in watts). If C = 64 dollars, for P = 4 watts: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Find the power setting when C = 125 dollars.

Explain why doubling the power setting increases the cost by a factor of 8.

The brightness B (in lumens) of a light bulb varies directly with the square of the current I (in amperes), and the power consumption P (in watts) varies directly with the cube of the current. If B = 200 lumens and P = 100 watts, for I = 2 amperes: a

Write the direct variation equations for brightness and power and ﬁnd the constant of proportionality.

Find the brightness and power when I = 3 amperes.

Sketch a graph of B against I and P against I for 0 ≤ I ≤ 1 on the same grid and describe the shape of the curves.

The distance D (in metres) a ball travels when thrown varies directly with the square of the time T (in seconds). If D = 20 metres, for T = 2 seconds: a

Write the direct variation equation and ﬁnd the constant of proportionality.

Sketch a graph of D versus T for 0 ≤ T ≤ 5, and describe its shape.

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6.03 Inverse variation models After this lesson, you will be able to… • develop models for inverse variation of the form • evaluate the constant of proportionality, k, given a pair of values • use an inverse variation model to find the value of a variable • analyse and solve practical problems involving inverse variation • distinguish between direct and inverse variation

Inverse variation Inverse variation When one variable increases as the other variable decreases. For example, if y is said to be ‘inversely proportional’ to x, the equation is of the form y = , where k is a constant of variation (or proportion). Also known as inverse proportion. The general form of inverse variation, where y decreases as x increases, is:

is the dependent variable

is the independent variable, where x ≠ 0

is the non-zero constant of proportionality

n is the positive power of the inverse relationship For example, the speed S (in km/h) varies inversely with time T (in hours) to travel a ﬁxed distance , where n = 1. Since distance equals speed times time,

of 1000 km. This is modelled as S = 1000 = T × S, so S = 500 S 450 400 350 300 250 200 150 100 50

, and the constant is k = 1000.

The graph of S =

0 10 20 30 40 50 60 70 80 90100

272

is a hyperbola, showing that as

time T increases, speed S decreases.

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Exploration Consider real-world situations where one quantity decreases as another increases, such as: • The brightness of a light decreases as the distance from the source increases. • The time to complete a task decreases as the number of workers increases. 1. What other examples of inverse variation can be identiﬁed? 2. How might these be modelled mathematically? 3. For the brightness of a light or time to complete a task, describe how doubling the distance or number of workers affects the brightness or time, respectively.

To apply the inverse variation model y = values for x and y.

, the constant k is determined using a known pair of

Example 1 The intensity of light I (in lumens per square metre) varies inversely with the square of the distance d (in metres) from the source. If I = 25 when d = 2: a Develop the inverse variation model.

Create a strategy Use the model I =

and substitute I = 25, d = 2 to ﬁnd k.

Apply the idea Write the model in terms of k Substitute I = 25, d = 2 Multiply both sides by 22 The model is I =

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b Find the distance when the intensity is I = 4 lumens per square metre.

Create a strategy Use the model I =

and substitute I = 4 to solve for d.

Apply the idea Write the model Substitute I = 4

Multiply both sides by

Take the square root of both sides

Evaluate Since distance must be positive, d = 5 metres.

Idea summary Inverse variation describes a relationship where one quantity decreases as another increases, modelled by y =

, where k ≠ 0 is the constant of

proportionality and n > 0 is the power. The constant k in y =

is found by substituting a known pair of values for x and y.

Real-world problem solving Inverse variation models solve real-world problems where one quantity decreases as another increases. The model y = questions.

is developed, k is found, and the model is applied to answer

Example 2 The time T (in days) to complete a project varies inversely with the square of the number of workers W. If T = 9 days when W = 2 workers: a Find the constant of proportionality k.

Create a strategy Use the model T =

274

and substitute T = 9, W = 2 to solve for k.

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Apply the idea Write the model in terms of k Substitute T = 9, W = 2 Multiply both sides by 22

b How many workers are needed to complete the project in T = 4 days?

Create a strategy Use the model T =

and substitute T = 4 to solve for W.

Apply the idea Write the model Substitute T = 4

Multiply both sides by

Take the square root of both sides

Evaluate Since the number of workers must be positive, W = 3 workers are needed.

Idea summary Real-world problems involving inverse variation are solved by developing the model y =

, ﬁnding k, and applying the model to ﬁnd unknown quantities.

6.03 Practice questions What do you remember? 1

State the meaning of inverse variation.

If y varies inversely with xn, write the general equation using k as the constant of variation.

Determine whether the following equations represent inverse variation: a

y = 6x + 8

xy = −7

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What does the constant k represent in inverse variation?

What is the shape of the graph of an inverse variation equation y =

Practice Ex 1

Ex 2

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The volume V (in cubic metres) of a gas varies inversely with the pressure P (in pascals). When P = 10 pascals, the volume is V = 50 cubic metres: a

Write the inverse variation model.

Find the volume when the pressure is P = 25 pascals.

The force F (in newtons) between two magnetic poles varies inversely with the square of the distance D (in metres) between them. If F = 80 newtons when D = 0.5 metres: a

Find the constant of variation k.

Find the force when D = 2 metres.

The intensity I (in watts per square metre) of sound varies inversely with the square of the distance D (in metres) from the source. If I = 16 watts per square metre at D = 1 metre: a

Find the constant of variation k.

Find the intensity at D = 4 metres.

The time T (in hours) to ﬁll a tank varies inversely with the ﬂow rate R (in litres per hour). If it takes T = 8 hours at R = 150 litres per hour: a

Write the inverse variation equation.

Find the time when the ﬂow rate is R = 200 litres per hour.

The frequency F (in hertz) of a vibrating string varies inversely with its length L (in metres). If F = 440 hertz when L = 0.5 metres: a

Find the constant of variation k.

Find the frequency when L = 0.4 metres.

The resistance R (in ohms) of a wire varies inversely with the square of its diameter D (in millimetres). If R = 10 ohms when D = 2 mm: a

Write the inverse variation equation.

Find the resistance when D = 4 mm.

The cost C (in dollars) of hiring a machine varies inversely with the number of days D it is hired for. It costs C = 150 dollars for D = 4 days: a

Write the inverse variation equation.

If the budget is $100, for how many days can the machine be hired?

The time T (in hours) to ﬁll a pool varies inversely with the pump’s ﬂow rate R (in litres per hour). If T = 6 hours when R = 200 litres per hour: a

Write the inverse variation equation.

Find the time when R = 400 litres per hour.

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Extend your thinking 14

The strength S of a radio signal varies inversely with the cube of the distance D (in kilometres) from the transmitter. At D = 2 km, the strength is S = 125 units. Find the distance where the signal strength drops to S = 8 units.

The time T (in hours) taken to paint a house varies inversely with the number of painters P. Additionally, the cost C (in dollars) of hiring P painters is proportional to P and T. Given that T = 24 hours when P = 2 painters, and the cost is C = 960 dollars. Determine the cost when P = 4 painters are hired.

Explain why the equation y =

represents inverse variation for positive values of k and n.

Use a real-world example to illustrate your explanation. 17

The gravitational force F (in newtons) between two objects varies inversely with the square of the distance D (in metres) between their centres. Given that F = 100 newtons when D = 2 metres, and the force becomes F = 25 newtons at a new distance. Find the new distance where the force is F = 25 newtons.

The time T (in minutes) to complete a task varies inversely with the number of machines M used. The cost C (in dollars) of operating the machines is directly proportional to the square of the number of machines and the time. Given that T = 60 minutes when M = 2 machines, and the cost is C = 7200 dollars: a

Find the inverse variation equation for T in terms of M.

Find the direct variation equation for C in terms of M and T.

Calculate the cost when M = 3 machines are used, and determine how many whole machines are needed to complete the task in T = 30 minutes with a cost not exceeding 18 000 dollars.

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6.04 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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Interactive exploration Discover this concept in action online

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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.

6.04 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

∣d∣ = 0

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

does not equal ∣a∣?

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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:

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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.

6.05 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

So, the y-intercept is y = 4.

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Interactive exploration Discover this concept in action online

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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.

6.05 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

4 units away from −2

f (x) = ∣2x + 6∣

f (x) = ∣x − 4∣

f (x) = ∣−2x + 2∣

∣x + 1∣ = 14

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

∣2x − 4∣ = 2

∣−x + 3∣ = 9

Use a graphing application to graph each function and identify the vertex, line of symmetry, x-intercept(s), and y-intercept: a

f (x) = ∣1.5x + 2∣

f (x) = ∣3x − 9∣

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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.

Solve each equation algebraically and identify possible solutions: a

290

hours. Then ﬁnd the

∣m + 3∣ = m + 1

∣n − 2∣ = n − 2

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∣p + 1∣ = p + 3

3∣q − 4∣ = q

6.06 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

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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

(5, 0) x 1 2 3 4 5

−2 −3

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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].

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(upper/lower) or

6.06 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

Graph the following semicircles: a

Ex 6

For the circle x2 + y2 = 64, ﬁnd the equations of the: a

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:

298

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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6 Chapter review 1

What is the radius of the circle with the equation x2 + y2 = 81? A

40.5

2.25

The cost C of painting a wall varies directly with its area A. If it costs $240 to paint a 30 m2 wall, what is the cost to paint a 45 m2 wall? A

The time taken, t, to complete a job varies inversely with the number of workers, n. If it takes 4 workers 9 hours to complete the job, what is the constant of variation, k? A

$160

For f (x) =

$285

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.

$360

$480

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?

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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

−3

−3 −4

d 4

1 1

300

−2

−3

−3 −4

−4 −3 −2 −1 −1

−4

−4 −3 −2 −1 −1

−2

−4

−4 −3 −2 −1 −1

−2

x 1

1 −4 −3 −2 −1 −1

−2

−3

−4

The time t (in hours) to complete a delivery route is inversely proportional to the number of drivers n. If 3 drivers complete the route in 8 hours: a

Write a formula relating t and n.

How long will it take with 6 drivers?

How many drivers are needed to complete the route in 2 hours?

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The distance D (in metres) travelled by a scooter varies directly with the time T (in seconds) at a constant speed. If D = 60 metres for T = 4 seconds: a

Write the direct variation equation.

Find the distance after T = 10 seconds.

The heat energy H (in joules) produced by a resistor varies directly with the square of the current I (in amperes). If H = 100 joules for I = 2 amperes: a

Find the constant of proportionality k.

Find the heat energy when I = 4 amperes.

The cost C (in dollars) of producing a decorative cube varies directly with the cube of its side length S (in cm). If C = $54 for a side length of S = 3 cm: a

Write the direct variation equation.

Find the cost for a side length of S = 5 cm.

Explain why doubling the side length increases the cost by a factor of 8.

The distance D (in metres) a ball rolls down a ramp varies directly with the square of the time T (in seconds). If D = 12 metres for T = 2 seconds: a

Write the direct variation equation.

Sketch a graph of D versus T for 0 ≤ T ≤ 4.

The force F (in newtons) between two charged particles varies inversely with the square of the distance D (in metres) between them. If F = 50 newtons when D = 0.4 metres: a

Find the constant of variation k.

Find the force when D = 2 metres.

The frequency F (in hertz) of a guitar string varies inversely with its length L (in metres). If F = 330 hertz when L = 0.6 metres: a

Find the constant of variation k.

Find the frequency when L = 0.5 metres.

The strength S of a magnetic ﬁeld varies inversely with the cube of the distance D (in metres) from the magnet. At D = 2 m, the strength is S = 160 units. Find the distance where the signal strength drops to S = 20 units.

The gravitational force F (in newtons) between two objects varies inversely with the square of the distance D (in metres) between their centres. Given that F = 240 newtons when D = 2 metres, ﬁnd the new distance where the force is F = 60 newtons.

Consider the identity

Verify 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.

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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

∣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

Lower semicircle

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:

302

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.

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“Mathematics is the art of giving the same name to different things.” Henri Poincaré

Big ideas • Trigonometry is extended beyond right-angled triangles through the unit circle, which defines the trigonometric ratios for angles of any magnitude, and the sine and cosine rules, which provide the tools to solve problems in any triangle. • Radian measure defines angles based on the intrinsic geometry of a circle, the ratio of arc length to radius, providing a natural system of measurement that simplifies formulas for arc length and sector area. • The trigonometric ratios can be viewed as functions that produce distinctive, periodic graphs, visually representing the cyclical relationship between an angle of rotation and the coordinates on a unit circle.

7 Trigonometry Chapter outline 7.01 7.02 7.03 7.04 7.05 7.06 7.07 7.08

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 Chapter 7 review

306 313 330 343 350 361 370 378 398

You can measure a tree’s height without climbing it — just use its shadow and trigonometry.

7.01 Exact trigonometric values After this lesson, you will be able to… • derive the exact trigonometric ratios for angles of 45° using an isosceles right-angled triangle. • derive the exact trigonometric ratios for angles of 30° and 60° using an equilateral triangle. • recall and use the exact values of sine, cosine and tangent for 30°, 45° and 60°. • apply exact trigonometric ratios to solve for unknown sides and angles in right-angled triangles. • recognise the relationship between trigonometric ratios of complementary angles.

Exact trigonometric values In a right-angled triangle, the ratios of the sides are given special names. For the angle θ, the main three trigonometric ratios are sine, cosine, and tangent.

sin ( θ ) = Opposite

Hypotenuse

cos ( θ ) = tan ( θ ) =

Adjacent

A common way to remember these is the mnemonic SOH CAH TOA. Pythagoras’ theorem The square of the length of the hypotenuse, c, of a right-angled triangle equals the sum of the squares of the lengths of the other 2 sides, a and b, such that c2 = a2 + b2. 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

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The exact values can be determined for these trigonometric ratios:

The exact values can be found by starting with an equilateral triangle:

60° 2

60°

To ﬁnd the exact ratios of 30 and 60 degree 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

60°

Using Pythagoras’ theorem, calculate the length of the perpendicular height to be . Then use our trigonometric ratios to determine the exact values.

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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7.01 Exact trigonometric values mathspace.co

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45° 1

45°

Notice that:

sin θ = cos (90° − θ )

45° 1

cos θ = sin (90° − θ )

45°

These angles are called complementary angles.

Example 1 Use the exact value triangles in the diagram to answer the questions: 30°

2 60° 1

Create a strategy Refer to the triangle with 45° angle, then use the cosine ratio and rationalise the value.

Apply the idea Write the cosine ratio

Substitute the values

Rationalise

Perform multiplication and simplify

Create a strategy Refer to the triangle with 60° angle, then use the cosine ratio.

Apply the idea Write the cosine ratio

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Substitute the values

45°

a What is the exact value of cos 45°?

b What is the exact value of cos 60°?

45°

Example 2 Find the angle θ in a right-angled triangle where tan θ =

Create a strategy Use the exact value triangle with a side length of

Apply the idea

To apply the tangent ratio, Opposite = 1 and Adjacent = .

30° 2

The exact value triangle shows that if that is the case, the angle is θ = 30°. So, θ = 30°

60° 1

Idea summary The exact values for the angles 30°, 45° and 60° are: sin

cos

tan

30° 45°

60° The exact values that don’t describe physical triangles, at 0° and 90° are: sin

cos

tan

0°

90°

Undefined

When answers are in surd form, always present the final result in rationalised form.

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7.01 Practice questions What do you remember? 1

Determine the exact values of the trigonometric ratios for 45° using the right-angled isosceles triangle: a

cos 45°

tan 45°

Identify the complementary angle for each angle given: a

sin 45°

36°

60°

y°

19°

Use Pythagoras’ theorem to ﬁnd the length of the adjacent side in the triangle shown. 30°

60° 1

Practice 4

Complete the table with exact trigonometric ratios for the angles given: sin

cos

⬚

30° 45° 60°

Ex 1

⬚

tan

⬚

Find the exact value of each trigonometric ratio: a

sin 60°

cos 30°

tan 45°

sin 45°

sin 30°

tan 30°

30°

60° 1

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45°

45° 1

An equilateral triangle has a side length of 4 cm: a

Find the length of exact perpendicular height.

Find the exact values of sin 60°, cos 60°, and tan 60° using a right-angled triangle.

Find the exact value of acute angle in a right-angled triangle for each equation: a

cos θ =

tan θ =

Find the exact lengths of the other two sides.

Calculate the exact area of the triangle.

sin θ =

A right-angled triangle has an angle of 45° and one of the sides of length 6 m: a

Find the exact lengths of the other side and the hypotenuse.

Calculate the exact area of the triangle.

Evaluate each expression, leaving answers in exact rationalised form: sin 30° cos 45°

tan 60° sin 45°

cos 30° + sin 60°

θ is an acute angle in a right-angled triangle. Find the value of θ : a

A right-angled triangle has one of the acute angles 30° and a hypotenuse of 8 cm:

a Ex 2

sin θ =

cos θ =

sin θ =

tan θ =

A ladder of length 5 m leans against a vertical wall, making an angle of 60° with the ground: a

Find the exact height the ladder reaches up the wall.

Find the exact distance from the base of the ladder to the wall.

Find the exact length of side h in the triangle shown. A 30°

60° B 14

An equilateral triangle has a perpendicular height of

C cm.

Find the exact side length of the triangle.

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Extend your thinking 15

Use exact values to prove that sin2 60° + cos2 60° = 1.

Use the triangle shown to demonstrate that sin 45° = cos 45°. Explain why sin θ = cos (90° − θ ) holds for any angle θ in a right-angled triangle.

45° 2

45° 1 17

A right-angled triangle has an acute angle θ where sin θ = a

Find the exact value of sin θ.

Find the exact value of tan θ.

Find the exact value of cos θ.

A ﬂagpole casts a shadow of 12 m when the sun’s rays make an angle of 30° with the ground. Find the exact height of the ﬂagpole.

312

Find the exact value of tan θ.

A right-angled triangle has an acute angle θ where tan θ = : a

A right-angled triangle has an acute angle θ where cos θ = a

Find the exact value of cos θ.

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7.02 Angles of elevation, depression and bearings After this lesson, you will be able to… • define and use angles of elevation and depression to solve for unknown lengths and angles. • interpret and represent directions using compass bearings and three-figure true bearings. • convert between compass bearings and true bearings. • solve a variety of practical problems involving bearings and right-angled trigonometry.

Angles of elevation and depression Angle of elevation The angle between horizontal and the line of sight from an observer to an object that is higher than the observer.

Obj LoS θ H

Angle of depression The angle between horizontal and the line of sight from an observer to an object that is lower than the observer.

θ LoS

Obj

Both angles are measured with respect to the horizontal. B

θ θ A

Notice that the angle of elevation between two points will always be equal to the angle of depression between those two points. This is because they are alternate angles on parallel lines (since all horizontal planes are parallel to each other).

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In practical problems with angles of elevation and depression, there are three distances between two objects: horizontal distance, vertical distance and direct distance. Horizontal

Dir

ect

dis

θ tan

Dir

ect

θ Horizontal

Angle of elevation

dis

Vertical

tan

Horizontal

Angle of depression

Example 1 Determine the value of the angle of depression from point B to point D. Use x as the angle of depression rounded to two decimal places.

9 A

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Create a strategy Identify the angle of depression and apply the appropriate trigonometric ratio.

Apply the idea Use the horizontal segment CB to identify the angle of depression, labelled ∠CBD. The adjacent side is BC = 9 and the hypotenuse BD = 19. Let x be the unknown angle. Use the cosine ratio

Substitute the values

Apply inverse cosine on both sides

Evaluate using technology and round

Example 2 From the top of a rocky ledge 188 m high, the angle of depression to a boat is 13°. If the boat is d m from the foot of the cliff, determine d rounded to two decimal places.

Create a strategy Draw a diagram to determine the trigonometric ratio that can be used.

Apply the idea Illustrating the situation gives this diagram. 13° 188 13° d

Notice how the angle opposite the 188 m side is an alternate angle to the angle of depression. That means it also measures 13°. Since the given sides are the opposite and adjacent sides, use the ratio: tan θ =

, where θ = 13°.

Write the formula

Substitute θ = 13° and

Multiply both sides by d

Divide both sides by tan 13°

Evaluate and round

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Example 3 A fighter jet flying at an altitude of 4000 m is approaching a target. At a particular time the pilot measures the angle of depression to the target to be 13°. After a minute, the pilot measures the angle of depression again and finds it to be 16°:

13° B

16°

C 4000 m

Airport a Determine the distance AC, rounded to the nearest metre.

Create a strategy With respect to the given angle 13°, the opposite side is 4000, and the adjacent side is AC, therefore, the tangent ratio can be used.

Apply the idea Use the tangent ratio

Substitute θ = 13° and

Multiply both sides by AC

Divide both sides by tan 13°

Evaluate and round

b Determine the distance BC, rounded to the nearest metre.

Create a strategy With respect to the given angle 16°, use the tangent ratio where the opposite side is 4000 and the adjacent side is BC.

Apply the idea Use the tangent ratio

Substitute θ = 16° and

Multiply both sides by BC Divide both sides by tan 16°

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Evaluate and round

c Determine the distance covered by the jet in one minute to the nearest metre.

Create a strategy The distance covered by the jet in one minute is given by the length of AB, so subtract the value of BC from the value of AC.

Apply the idea AB = AC − BC

Subtract the value of BC from the value of AC

= 17 326 − 13 950

Substitute the values

= 3376 m

Evaluate

Idea summary Trigonometric ratios can be used to determine the angles of elevation or angles of depression in practical problems. Additionally, the angles of elevation and depression with one given length can be used to determine the length of an unknown side.

Angle of elevation

Horizontal

Angle of depression

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Compass bearings Bearing A direction from one point on the Earth’s surface to another. Bearings identify directions in navigation and surveying, using cardinal directions (north, east, south, west) and intermediate directions (northeast, southeast, southwest, northwest).

N NW NNW

NNE

WNW

NE ENE

E WSW

ESE SSW

SSE

Cardinal directions are north (N), east (E), south (S), and west (W). Intermediate directions are at 45° between cardinals.

S Compass bearing Angles either side of north or south. For example, a compass bearing of N50°E is found by facing north and moving through an angle of 50° to the east. A compass bearing speciﬁes: • Starting direction (north or south). • Acute angle (≤ 90°) of rotation. • Rotation direction (east or west). North

West

East

The bearing of point B from A is S53°E, starting at south and rotating 53° east.

53° B South

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Example 4 N A compass shows point A relative to point O:

35° O

Determine the compass bearing of A from O.

Create a strategy

Apply the idea

Identify the starting direction (north or south), measure the acute angle, and determine the rotation direction (east or west).

Point A is 35° clockwise from north. The bearing starts at north, rotates 35° east, written as N35°E.

Idea summary A compass bearing speciﬁes direction using north or south, an acute angle, and east or west (e.g., N45°E).

True bearings True bearing Measured in degrees clockwise from true north and written with 3 digits to specify the direction. For example, the direction of north is specified as 000°T, east is specified as 090°T, south is specified as 180°T and north-west is specified as 315°T. A true bearing is defined by: • Starting from north (0°) • Measured clockwise • Written with three digits (e.g. 060°) North

West

127° East

The true bearing of B from A is 127°T.

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Example 5 Determine the true bearing of point A from point O: a

35° O

Create a strategy Measure the bearing clockwise from north, and write it using three digits.

Apply the idea The angle between the north direction and the line OA measured clockwise is 35°. So, the true bearing is 035°T.

b A

42° O

Create a strategy To measure the angle from north to the given angle clockwise, subtract the given angle from 360°.

Apply the idea The angle between the north direction and the line OA measured clockwise is 360 − 42 = 318°. So, the true bearing is 318°T.

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Idea summary A true bearing is measured clockwise from north, written with three digits (e.g., 030°T).

Applications Bearings are applied in navigation and surveying to calculate distances and positions using trigonometry, often with exact ratios from right-angled triangles. To solve bearing problems: 1. Sketch a diagram with the reference point at the compass centre. 2. Label angles and lengths, aligning north upwards for true bearings. 3. Apply the right-angled trigonometry rule (SOH-CAH-TOA). To reverse direction, add or subtract 180°. This gives the angle to return to your starting point.

Example 6 A tower is 18 m from point P on the ground, making an angle of elevation of 30° to its top. Find the tower’s exact height.

Create a strategy Use the tangent ratio with θ = 30°, opposite side as height h and adjacent side of 18.

Apply the idea Write the tangent ratio

Substitute θ = 13° and

Multiply both sides by 18

Use the exact value tan 30° =

Simplify

The tower’s height is

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Example 7 A ship S is 15 km from point P on a true bearing of 060°T. Find the ship’s northward distance from P, rounded to one decimal place.

N x y 060° T

15 km

Create a strategy Use the cosine ratio with θ = 60° adjacent side as y and hypotenuse side of 15 to find the adjacent side (northward distance).

Apply the idea Write the cosine ratio

Substitute θ = 60° and

Multiply both sides by 15

Use the exact value cos 60° =

Simplify

Evaluate and round

The northward distance is 7.5 km.

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Example 8 If a hiker walks from point A to point B on a bearing of 120°, determine the bearing to return to A from B.

North

West

120°

East

B South

Create a strategy Add 180° to the original bearing to ﬁnd the return bearing.

Apply the idea Return bearing = 120° + 180° = 300°

Add 180° to reverse direction Evaluate

The return journey bearing is 300°.

Idea summary Bearing problems are solved by sketching diagrams, labelling angles and lengths, and applying trigonometry (SOH-CAH-TOA) with exact ratios.

7.02 Practice questions What do you remember? 1

Deﬁne the angle of elevation and the angle of depression, and explain their relationship in a right-angled triangle scenario: a

Angle of elevation

Identify the angle using the diagram: a

Elevation from P to Q

Depression from Q to P

Angle of depression

Q β

θ R

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Write the rules for writing a compass bearing and a true bearing: a

True bearing

Convert these compass bearings to true bearings: a

Compass bearing

N 42° E

S 75° W

S 20° E

N 65° W

Identify the compass bearing for each cardinal or intermediate direction: a

Southeast

Northwest

South

Northeast

Practice

Ex 1

A ﬂagpole is 15 m from a point R on the ground, with an angle of elevation of 30° to its top. Calculate the exact height of the ﬂagpole.

From a lookout tower 100 m high, the angle of depression to a vehicle on the ground, is 45°. Find the exact horizontal distance to the vehicle.

A kite at point K is 50 m above point G on the ground. The angle of elevation from G to K is 60°. Calculate the exact length of the kite string (direct distance).

Let x be the angle of elevation from point C to point A: a

Determine x, rounded to two decimal places.

Determine y, rounded to two decimal places.

19 y x C 10

Ex 2

Fred is on a ship and observes a lighthouse on top of a cliff. The base of the cliff is 651 m away from the ship, and the angle of elevation from Fred to the top of the lighthouse is 35°: a

If the top of the lighthouse is x metres above sea level, determine x, rounded to two decimal places.

If the lighthouse is 30 metres tall, determine the height of the cliff that the lighthouse stands on, rounded to two decimal places.

A person at point S on the ground measures the angle of elevation to a drone at D as 40°. If SD = 200 m, calculate the drone’s height above S, rounded to two decimal places.

From a cliff 80 m high, the angle of depression to a buoy is 25°. Calculate the horizontal distance to the buoy from the base of the cliff, rounded to the nearest metre.

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Ex 3

A man stands at point A looking up at the top of two poles. Pole 1 has a height of 8 m and an angle of elevation of 34° from point A. Pole 2 has a height 25 m and an angle of elevation of 57° from point A: a

Determine the distance from A to B, rounded to two decimal places.

Determine the distance from A to C, rounded to two decimal places.

Determine BC, the distance between the two poles, rounded to one decimal place.

Pole 2

25 m

Pole 1 8m A

Ex 4

Determine the compass bearing of point P from point O: a

N P 30°

O P

65°

P 30° O

26° P

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Ex 5

Determine the true bearing of point A from point O: a

30°

40°

34° O

O 25° A

Convert these true bearings to compass bearings: a

Ex 6

048°T

215°T

140°T

310°T

A tree’s shadow is 12 m long when the sun’s angle of elevation is 60°. Calculate the exact height of the tree.

A person stands 400 m from the base of a building and measures the angle of elevation to the top as 32°. Calculate the building’s height, rounded to one decimal place.

A hot air balloon is 500 m above ground, with an angle of depression to a car on the ground, 20°. Calculate the direct distance to the car, rounded to the nearest metre.

A pole is 10 m tall, and the angle of elevation from a point 15 m away from the foot of the pole, is θ. Calculate θ, rounded to two decimal places.

A ship travels on a true bearing of 070°T and then turns around to return to its starting point. What is the true bearing for the return journey?

The angle of elevation from a ship at S to the top of a 40 m lighthouse is 30°. Calculate the exact horizontal distance from S to the base of the lighthouse.

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Ex 7

A boat sails 120 km on a bearing of 235°T: a

Determine how far due west is the ship from the starting point, rounded to one decimal place.

Determine how far due south is the ship from the starting point, rounded to one decimal place.

w 235° T 120 km

Ex 8

For each true bearing from A to B, calculate the true bearing required to return to the starting point: a

072°T

127°T

198°T

253°T

304°T

346°T

053°T

161°T

For each compass bearing, calculate the compass bearing required to return to the starting point: a

N 23° E

S 67° W

N 42° W

S 78° E

S 37° W

N 55° E

S 29° E

N 71° W

A hiker walks 600 m due north from point A to point B, then due east to point C. The true bearing of C from A is 065°T.

N C

Determine the direct distance from C to A, rounded to the nearest metre.

600 m

065° T

A plane ﬂies 50 km due south of the airport at O and 40 km due west of its starting airport: a

Determine the true bearing of the plane from the airport, rounded to one decimal place.

Determine the direct distance from the plane to the airport, rounded to one decimal place.

O b°

Plane

Sydney

50 km

40 km

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A surveyor moves 10 km due west from point X to point Y, then 15 km due north to point Z. Point Z is on a true bearing of 326°T from X: a

Determine the angle ∠Y XZ.

Determine the direct distance from Z to X, rounded to one decimal place.

N d km

15 km

Y 10 km

326° T

A ship is 30 km from a port on a true bearing of 045°T: a

Calculate the ship’s x-coordinate (east-west), rounded to one decimal place:

Calculate the ship’s y-coordinate (north-south), rounded to one decimal place.

N x y P

30 km 045° T

Extend your thinking 30

For angles p, q between 0° and 90°, express the true bearing for each compass bearing: a

N p° E

S q° E

N p° W

S q° W

Two hot air balloons are anchored to the ground below each at a different location. An observer at each location measures the angle of elevation to the opposite balloon. The observers are 1600 m apart as shown in this diagram: a

Calculate the difference in height between the two balloons, rounded to the nearest metre.

Determine the direct distance between the two balloons and their respective observers, rounded to the nearest metre.

3.38°

1.22° 1600 m

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A surveyor at point A on the ground, measures the angles of elevation to the two towers, B and C. Tower B is 20 m tall with an angle of 30°; tower C is 35 m tall with an angle of 45°. Both bases are on horizontal ground, and ∠BAC = 90°. a

Calculate the exact distance from A to B.

Calculate the exact distance from A to C.

Calculate the distance between the towers’ bases, rounded to one decimal place.

A ship at S in the ocean is 2 km from a lighthouse at L on a true bearing of 040°T. The lighthouse is 50 m tall, and the angle of elevation from S to the top is θ. A buoy at B is 1.5 km from L on a bearing of 130°T: a

Calculate θ, rounded to two decimal places.

Calculate the distance from S to B, rounded to two decimal places.

A room is 8 m long, 6 m wide, and h m high:

The angle of elevation from the bottom corner P to the opposite top corner Q is 45°.

Calculate the exact height h of the room.

Calculate the angle of depression from Q to R on the 6 m wall, rounded to two decimal places.

45° P

Determine the direct distance from A to C, rounded to two decimal places.

Points A, B, and C form a triangle. AB is 800 m on a bearing of 055°T from A. BC is 1000 m on a bearing of 145°T from B: a

B A

055° T

800 m

145° T

1000 m

Determine the true bearing of C from A, rounded to one decimal place. C

A yacht sails 35 km from port A on a bearing of 115°T to point B, then 45 km on a bearing of 155°T to point C: a

Determine the angle at which the yacht changed direction at B, rounded to one decimal place.

Determine the direct distance from A to C, rounded to two decimal places.

A surveyor at point A observes a landmark B at 12 km away on a bearing of 065°T and a lighthouse C at 15 km away on a bearing of 155°T: a

Determine the distance between B and C, rounded to two decimal places.

Determine the true bearing of C from B, rounded to one decimal place.

Determine the true bearing of B from C, rounded to the nearest degree.

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7.03 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 =

332

and y =

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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.

Interactive exploration Discover this concept in action online

mathspace.co

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 cos θ

(cos θ , sin θ ) −1

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Third quadrant: • y is negative: sin θ is negative • x is negative: cos θ is negative • Gradient is positive: tan θ is positive

−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

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−1

C cos positive

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• In Quadrant 4, only cos θ is positive.

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.

7.03 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 

340

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

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−1

4 3 P − ,− 5 5

−1

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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7.04 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.

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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°

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Apply the quadrant rule

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.

7.04 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

348

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°

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tan 210°

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°). 7.04 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.

7.05 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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7.05 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

7.06 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. 7.06 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

7.06 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

368

−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

Mathspace New South Wales – Year 11 Advanced mathspace.co

cos s

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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7.07 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. 370

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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 =

372

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

7.07 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.

θ 100 m 7.07 Arc length and sector area mathspace.co

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7.08 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

378

π 1

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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

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1 π 2

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

2π

−2 −3 −4

Period

, 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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7.08 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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391

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 −2 −3 −4

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x 1 π 2

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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395

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

A right-angled triangle has an angle of 60° and a hypotenuse of 10 cm. What is the exact length of the side opposite the 60° angle? A

80 m

If cos θ =

40 m

sin 60° cos 30°

tan 30° sin 60°

cos 45° + sin 45°

A slide of length 8 m leans against a platform, making an angle of 30° with the ground. a

Find the exact height the slide reaches on the platform.

Find the exact horizontal distance from the base of the slide to the platform.

A right-angled triangle has an angle θ where sin θ = a

Evaluate each expression, leaving answers in exact rationalised form. a

and 180° < θ < 270°, what is the exact value of sin θ?

A 4

From the top of a cliff 80 m high, the angle of depression to a boat is 30°. What is the exact horizontal distance from the base of the cliff to the boat? A

5 cm

Find the exact value of cos θ.

Find the exact value of tan θ.

An observer on a boat sees a cliff in the distance. The base of the cliff is 500 m away from the boat, and the angle of elevation from the observer to the top of a tower on the cliff is 25°. a

If the top of the tower is h metres above sea level, determine h, rounded to two decimal places.

If the tower is 25 m tall, determine the height of the cliff that the tower stands on, rounded to two decimal places.

A car travels 150 km on a bearing of 210°T. a

Determine the distance travelled west, rounded to one decimal place.

Determine the distance travelled south, rounded to one decimal place.

w 210° 150 km

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A hiker is 60 km due south and 25 km due west of their base camp. a

Determine the true bearing of the hiker from the camp, rounded to one decimal place.

Determine the direct distance from the hiker to the camp, rounded to one decimal place.

O Hiker b° Camp

25 km

A surveyor at point P measures angles of elevation to two buildings, A and B. Building A is 40 m tall with an angle of elevation of 60°; building B is 50 m tall with an angle of elevation of 30°. Both bases of the buildings are on horizontal ground, and ∠APB = 90°. a

Calculate the exact distance from P to A.

Calculate the exact distance from P to B.

Calculate the distance between the buildings’ bases, rounded to one decimal place.

For each diagram, determine the value of: i

sin θ

cos θ

iii

tan θ

P θ

−1

4 3 , 5 5

P −

12 5 , 13 13

−1

60 km

−1

Determine the coordinates of P (x, y) on the unit circle, rounded to two decimal places: a

y 1

200°

x 1

−1

P 40°

−1

x 1

P −1

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Given tan θ = −2.4, 90° < θ < 180°, calculate the exact value of: a

sin θ

sin 40°

cos 220°

tan 320°

Find the exact value of each ratio without technology: a

For a unit circle with point P (c, d) at 40°, express each in terms of c, d: a

cos θ

sin 135°

cos 240°

tan 330°

Rewrite each ratio as an equivalent trigonometric ratio of a positive reference angle α: a

sin (180° − α)

cos (180° + α)

tan (360° − α)

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.

Two hikers leave a camp. Hiker A travels 5 km on a bearing of N30°E. Hiker B travels 7 km on a bearing of N70°E. Calculate the distance between them, rounded to two decimal places.

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

210°

Find the exact value of each trigonometric ratio: a

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.

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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

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.

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Big ideas • The family of trigonometric functions is expanded to include secant, cosecant, and cotangent, which are defined as the reciprocals of the primary functions; their properties are generalised from right-angled triangles to angles of any magnitude using the unit circle. • Trigonometric functions are interconnected through a system of identities, including reciprocal, complementary, and Pythagorean relationships, which provide a powerful algebraic framework for simplifying expressions and proving more complex statements. • Solving trigonometric equations requires a systematic process of first simplifying the equation, often using identities, then finding a principal value, and finally leveraging the periodic nature of the functions to determine all solutions within a specified domain.

8 Trigonometric identities and equations Chapter outline 8.01 8.02 8.03 8.04 8.05 8.06 8.07

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 Chapter 8 review

404 413 423 428 436 442 451 458

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 =

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, where 0 < θ < 90°.

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 units long.

Pythagoras’ theorem, the hypotenuse is

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

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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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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:

410

sec

cosec

cot

0°

Undefined

90°

Undefined

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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°

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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

420

sec θ

, ﬁnd:

cosec (0)

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cot (π )

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. 424

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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

426

= 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

Mathspace New South Wales – Year 11 Advanced mathspace.co

cosec

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

428

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P (cos θ , sinθ ) b −b

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.

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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°)

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tan (90° − 45°) × sin (270° − 45°)

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.

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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

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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Write the left-hand side

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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

456

on [0°, 360°] and ﬁnds θ = 150°. Correct the error.

sin θ = sin2θ

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2 sin2θ − cos θ = 1

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 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

458

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°)

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cosec2

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

tan (90° − θ ) × tan θ

cos (90° − θ ) × cosec θ

sin 30°

cot 45°

sec 60°

tan 30°

sec (90° − θ )

tan (90° − θ )

cosec (90° − θ )

cot (90° − A)

sin 135°

, evaluate:

cos (90° − θ )

If sec A = 4, ﬁnd the exact value of: a

cosec (90° − A)

Evaluate using reference angles and quadrants: a

cot A

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°:

cosec2θ

sin θ =

tan θ = −1

cos θ = 0

tan x =

sin x = 0

Solve for x, where 0 ≤ x ≤ 2π : a

cos θ = −1

= cot x.

Prove:

. Find the exact value of:

(cosec2x − 1) sin2x

sin x = 1

cos x =

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

cos2x = 1

sin2x = = 2 sec θ.

Prove the identity

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2 cos2x = 1

“Infinity is merely the absence of a bound.” Georg Cantor

Big ideas • Set theory provides a formal language and visual tools, such as Venn diagrams, to define, classify, and operate on collections of objects, forming the essential foundation for quantifying probability. • The probability of an event is a numerical measure of its likelihood, calculated by applying the rules of set theory, such as the addition and complement principles, to a defined sample space of all possible outcomes. • The probability of an event can be influenced by the occurrence of another, a concept formalised as conditional probability; when there is no influence, the events are deemed independent, which simplifies calculations for multistage experiments.

9 Probability Chapter outline 9.01 9.02 9.03 9.04 9.05 9.06 9.07 9.08

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 9 review

464 470 476 484 491 499 508 515 522

Mutually exclusive events can’t both happen — like turning left and right at once.

9.01 Sets and notation After this lesson, you will be able to… • define a set and represent it using correct notation. • determine the cardinality of a finite set using the notation n(A) or A. • identify the empty set, denoted by ∅. • define a subset and determine if one set is a subset of another. • find the complement of a set with respect to a universal set.

Sets and notation Element A member of a set. For example, 3 is a member of the set of natural numbers N = {0, 1, 2, 3, 4, ...}. 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’. Finite A value between ∞ and −∞.

A set is a collection of distinct objects, called elements, grouped together based on a common property. Sets are represented using curly brackets {} with elements listed inside, separated by commas. For example, the set of even numbers less than 10 is written as {2, 4, 6, 8}.

A = {1, 2, 3, 4} A

is the set of positive integers less than 5

The number of elements in a ﬁnite set A is called its cardinality, denoted by n(A) or A. For example, if A = {1, 2, 3}, then n(A) = 3.

n(A) = A n(A) or A represents the number of elements in set A

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The empty set is the set with no elements, denoted by ∅. For example, the set of odd numbers divisible by 2 is ∅.

∅ = {} ∅

represents the empty set, containing no elements

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Example 1 Given set D = {cat, dog, bird}: a Find the cardinality of D.

Create a strategy Count the elements in D.

Apply the idea n(D) = 3 Write the number of elements

b Is D the empty set?

Create a strategy Check if D has no elements.

Apply the idea D has 3 elements, so it is not ∅.

Reflect and check Verify set-builder notation defines all elements. Confirm cardinality by counting distinct elements. Empty set has n(∅) = 0.

Idea summary A set is a collection of distinct elements, represented using curly braces, e.g., {1, 2, 3}. The cardinality of a set A, denoted n(A) or A, is the number of elements in the set. The empty set, denoted ∅, contains no elements.

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Subsets and complements Subset B is a subset of set A, if every element of set B, is an element of set A. This is written as B ⊆ A, i.e. ‘B is a subset of A’. If the cardinality of set A is greater than the cardinality of set B, then we call set B a proper subset, written as B ⊂ A. A set A is a subset of set B, denoted by A ⊆ B, if every element of A is also an element of B.

A⊆B A⊆B

means all elements of A are in B

For example, if B = {1, 2, 3, 4} and A = {1, 3}, then A ⊂ B because 1 and 3 are in B. A⊆B

A⊂B

B 4

A 4

1 2

Complement All of the outcomes that are not in a given event. All the elements of a set that are not in a given subset. The complement of a set A with respect to a universal set U, denoted by all elements in U that are not in A.

, A′, or Ac, is the set of

= {3, 4, 5} is the set of all elements in the universal set U = {1, 2, 3, 4, 5} that are not in A = {1, 2} For example, if U = {1, 2, 3, 4, 5} and A = {1, 2}, then

= {3, 4, 5}.

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Example 2 Given U = {1, 2, 3, 4, 5}, A = {1, 3, 5}, and B = {1, 3}: a Is B ⊂ A?

Create a strategy Check if all elements of B are in A.

Apply the idea Set B has {1, 3} which are in A = {1, 3, 5}. Thus, B ⊂ A.

b Find

Create a strategy List the elements in U that are not in A.

Apply the idea Set A = {1, 3, 5}, so the elements that are not in A are 2, 4. = {2, 4} List elements that are not in A

c Find

Create a strategy List the elements in U that are not in B.

Apply the idea Set B = {1, 3}, so the elements in U that are not in B are 2, 4, 5. = {2, 4, 5} List elements that are not in B

Reﬂect and check Conﬁrm all elements of B are in A for subsets. Verify complements include all non-A or non-B elements in U.

Idea summary A subset, A ⊆ B, means all elements of A are in B. The complement, all elements in U that are not in A.

, includes

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9.01 Practice questions What do you remember? 1

Deﬁne these terms related to sets: a

Set

Empty set

Cardinality of a set

Write these sets using roster notation: a

The set of positive odd numbers less than 10.

The set of vowels in the English alphabet.

Determine whether each statement is true or false: a

The set {2, 4, 6} has cardinality 3.

The set of even numbers divisible by 3 is the empty set.

If A = {1, 2} and U = {1, 2, 3}, then

The set {1, 3} is a subset of {1, 2, 4}.

= {3}.

Write the correct notation for each description: a

Number of elements in set A

Set with no elements

Complement of set A

Set A is a subset of B

Practice Ex 1a

Ex 1b

Ex 2

Determine the cardinality of each set: a

A = {biology, chemistry, physics}

B = {1, 2, 3}

C = {a, b, c, d, e}

D=∅

E = {2, 3, 5, 7}

F = {mon, tue, wed, thu}

G = {1, 2, 3, 4, 5}

H = {piano, guitar}

Identify if these sets are empty or not: a

{square, circle, triangle}

The set of even numbers greater than 10.

The set of odd numbers divisible by 2.

∅

The set of negative even numbers greater than −5.

The set of square numbers less than 0.

Given U = {p, q, r, s}, A = {p, q, r}, and B = {p, r}: a

468

Is B ⊂ A?

Find

Find B′.

Given the universal set U = {1, 2, 3, 4, 5}, determine the complement of each set: a

A = {1, 3, 5}

B = {2, 4}

C=∅

D = {1, 2, 3, 4, 5}

E = {1, 2}

F = {3}

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Determine if the ﬁrst set is a subset of the second set: a

{1, 3}, {1, 2, 3, 4}

{a, b}, {a, c, d}

∅, {1, 2}

{2, 4, 6}, {2, 4, 6}

{red}, {red, blue}

{1, 2, 3}, {1, 2}

A music club has these members: {Alice, Bob, Clara, Dan}. The set of members who play the guitar is {Alice, Clara}: a

List the members who play the guitar.

Determine the cardinality of the guitar players set.

Determine the complement of the guitar players set with respect to the music club.

Given the universal set U = {a, b, c, d, e, f}, determine the set and its cardinality: a

Set of letters in {a, c, e}

Complement of {a, c, e}

Set of letters in {b, d}

Complement of {b, d}

Determine if these statements are true or false: a

{2, 4} ⊆ {2, 4, 6, 8}

{x, y} ⊆ {x, z}

∅ ⊆ ∅

{1, 2, 3} ⊆ {1, 2, 3}

Given the universal set U = {1, 2, 3, 4, 5, 6, 7}, write these sets in roster notation: a

Complement of {1, 3, 5, 7}

Complement of ∅

Set of even numbers

Complement of {2, 4, 6}

A dance class has members: {Emma, Finn, Grace, Henry}. The set of members who know salsa is {Emma, Grace}: a

List the salsa dancers.

Find the cardinality of the salsa dancers set.

Find the complement of the salsa dancers set with respect to the dance class.

Given the universal set U = {p, q, r, s, t}, determine if the ﬁrst set is a subset of the second set: a

{p, r}, {p, q, r, s}

{s, t}, {p, q, t}

{q}, {q, r, s, t}

{p, q, r}, {p, q}

Extend your thinking 16

A survey of students’ favourite sports gives the set S = {soccer, tennis, basketball}: a

How many subsets does S have? Explain your reasoning.

List all subsets of S.

Given U = {1, 2, 3, 4, 5, 6} and A = {1, 2, 3}, explain why A′ ⊆ U is always true.

A store sells 4 types of fruit: {apple, banana, orange, mango}. A customer can choose any combination of fruits (including none): a

Write the set of all possible combinations the customer can choose.

What is the cardinality of this set?

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Consider the universal set U = {a, b, c, d, e} and two sets A = {a, b} and B = {b, c, d}:

Determine Ac and Bc.

a 20

Is Ac ⊆ Bc? Explain.

A library has a set of 5 books: {A, B, C, D, E}. A student can borrow any combination of these books: a

How many possible borrowing combinations are there?

If the student borrows at least one book, how many combinations are possible?

Explain why the empty set ∅ is a subset of every set.

Given a universal set U, set A exists such that A ⊆ U and U ⊆ A. Given this information, explain why AC = ∅.

9.02 Set operations and complements After this lesson, you will be able to… • find the intersection or union of two or more sets. • define disjoint sets as having an empty intersection. • identify whether given sets are disjoint. • apply multiple set operations, including complements, to solve problems.

Intersection and union of sets Intersection For sets A and B, it is the set of elements that are in both A and in B. The intersection of A and B is denoted A ∩ B. Union For sets A and B, the union is the set of elements which are in A or B or both and is written as A ∪ B, i.e. ‘A union B’.

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The intersection of two sets A and B, denoted A ∩ B, is the set of elements that are in both A and B.

A ∩ B = {2, 3} A ∩ B is the set of elements common to A = {1, 2, 3} and B = {2, 3, 4} The union of two sets A and B, denoted A ∪ B, is the set of elements that are in A, in B, or in both.

A ∪ B = {1, 2, 3, 4} A ∪ B is the set of all elements in A = {1, 2, 3} or B = {2, 3, 4} or both A

A∪B

A∩B

Example 1 Let the universal set be U = {1, 2, 3, 4, 5, 6}, with sets A = {1, 3, 5} and B = {2, 3, 4}. Find: a A∩B

Create a strategy Identify elements common to both A and B.

Apply the idea A ∩ B = {1, 3, 5} ∩ {2, 3, 4} = {3}

List elements in both sets Identify common element

b A∪B

Create a strategy Combine all elements from A and B, removing duplicates.

Apply the idea A ∪ B = {1, 3, 5} ∪ {2, 3, 4} = {1, 2, 3, 4, 5}

List elements from both sets Combine and remove duplicates

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Reﬂect and check Check all unique elements from both sets are included without repetition.

Create a strategy List elements in U that are not in A.

Apply the idea = {1, 2, 3, 4, 5, 6} \ {1, 3, 5}

Identify elements in U not in A

= {2, 4, 6}

List remaining elements

Idea summary The intersection A ∩ B contains elements in both A and B. The union A ∪ B contains elements in A, B, or both.

Disjoint sets Disjoint (sets) Two sets which do not have any common elements. Two sets A and B are disjoint if they have no elements in common, i.e., their intersection is the empty set.

A∩B=∅ A and B are disjoint if A ∩ B = ∅ For example, if A = {1, 3, 5} and B = {2, 4, 6}, then A ∩ B = ∅, so A and B are disjoint. If A and B are not disjoint, they share at least one element, so A ∩ B ≠ ∅.

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Example 2 Let the universal set be U = {a, b, c, d, e}, with sets A = {a, b} and B = {c, d}. Determine: a If A and B are disjoint

Create a strategy Check if A ∩ B = ∅.

Apply the idea A ∩ B = {a, b} ∩ {c, d} =∅

List elements in both sets No common elements

Since A ∩ B = ∅, A and B are disjoint.

b A∪B

Create a strategy Combine elements from A and B, ensuring no duplicates.

Apply the idea A ∪ B = {a, b} ∪ {c, d}

List elements from both sets

= {a, b, c, d}

Combine without duplicates

Idea summary Sets A and B are disjoint if A ∩ B = ∅, meaning they have no elements in common. The union A ∪ B of disjoint sets includes all elements from both sets.

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9.02 Practice questions What do you remember? 1

Match each set operation with its correct notation and deﬁnition: a

Intersection of sets A and B

A∪B

Union of sets A and B

Complement of set A in universal set U

iii

A∩B

Disjoint sets A and B

A∩B=∅

Determine whether each statement is true or false: a

The intersection A ∩ B contains elements in A or B.

The union A ∪ B includes all elements from both sets without duplicates.

Sets A and B are disjoint if A ∩ B ≠ ∅.

The complement

Given the universal set U = {1, 2, 3, 4, 5} and set A = {2, 4}, determine: a

includes elements in A not in U.

A′

A∩U

Given sets A = {3, 6, 9} and B = {1, 2, 3}, determine: a

A ∩ B

A∪B

Practice Ex 1

Let the universal set be U = {1, 2, 3, 4, 5, 6, 7}, with sets A = {1, 3, 5, 7} and B = {2, 4, 6}. Determine: a

Ex 2

474

A∪B

If A and B are disjoint.

A∪B

At an NSW school, let A be the set of students in the chess club {Alice, Bob, Clara} and B the set of students in the debate club {Bob, Dave, Eve}. Determine: a

A ∩ B

If A and B are disjoint.

A∪B

Given U = {a, b, c, d, e, f}, A = {a, c, e} and B = {b, d, f}. Determine: a

Let the universal set be U = {oak, pine, maple, birch, cedar}, with sets A = {oak, pine} and B = {maple, birch}. Determine: a

A∩B

A∪B

In a survey at a Sydney festival, let A = {pizza, burger, sushi} be foods liked by Group A and B = {sushi, pasta} by Group B. Determine: a

A ∩ B

If A and B are disjoint.

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A∪B

Given U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {1, 3, 5, 7, 9}. Determine: b

a 11

A′

A = {1, 2, 3}, B = {4, 5, 6}

A ∩ B

If A and B are disjoint.

A∪B

A = {cat, dog}, B = {dog, bird}

A∪B

Given U = {1, 2, 3, 4, 5}, A = {1, 2} and B = {4, 5}. Determine: Ac

A∩B

A∪B

In an NSW library, let A = {fiction, biography, poetry} be books borrowed by Group A and B = {poetry, history} by Group B. Determine: a

A∩B

If A and B are disjoint.

A∪B

Given U = {x, y, z, w}, A = {x, y} and B = {z, w}. Determine: a

A∩B

At a school, A is the set of students playing soccer {Emma, Finn, Grace} and B is the set playing netball {Grace, Harry, Ivy}. Determine:

a 15

A∪B

Determine if these pairs of sets are disjoint: a

Let U = {red, blue, green, yellow}, A = {red, blue} and B = {blue, green}. Determine: a

A∩B

Ac ∪ A

A ∩ Bc

(A ∪ B)c ∩ U

A = {1, 2, 3}, B = {2, 4, 6}

Determine if these pairs of sets are disjoint: a

A = {apple, banana}, B = {cherry, date}

Extend your thinking 18

A school offers extracurricular activities. Let the universal set U be all students, with A = {students in drama club} and B = {students in music club}. Suppose A ∩ B = {Jack, Lily}, A ∪ B = {Jack, Lily, Mia, Noah, Olivia} and A′ = {Noah, Olivia, Peter}. Determine: a

A and B

Are A and B disjoint? Explain.

At an NSW community event, a survey asks attendees about their hobbies. Let U be all attendees, A those who enjoy gardening, and B those who enjoy hiking. Explain why (A ∩ B)c represents attendees who do not enjoy both gardening and hiking. Then, if U = {James, Jane, Jack, Jill, John, Jenny}, A = {James, Jane, Jack} and B = {Jane, Jack, Jill}, determine .

A school has two teams for a trivia competition. Team A’s members are A = {Amy, Ben, Cara} and Team B’s members are B = {Cara, Dan, Ella}. The universal set U is all students in the school. If a student can only be on one team or neither, explain why A and B are not disjoint. Then, propose a change to make them disjoint and determine the resulting A ∪ B if A becomes {Amy, Ben}.

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In an NSW book club, let U be all members, A those who read sci-ﬁ, and B those who read fantasy. If A = {Eva, Finn, Gina}, B = {Finn, Hugh, Iris}, and = {Jack}, determine:

a 22

A ∪ B

An NSW school tracks student participation in two events. Let A be students in the art show and B be students in the science fair. If A ∩ B = {Kate, Leo} and A ∪ B = {Kate, Leo, Mia, Nick, Owen}, explain how to ﬁnd elements of A and B. Then, if Ac = {Nick, Owen, Paul}, determine A and B.

9.03 Venn diagrams After this lesson, you will be able to… • represent sets and their relationships using a Venn diagram. • identify regions in a Venn diagram corresponding to intersection, union, and complement. • solve problems by populating a Venn diagram with given information. • establish and use the cardinality rule for the union of two sets, A ∪ B  = A  + B  − A ∩ B . • calculate the number of elements in any region of a two-set Venn diagram.

Represent sets with Venn diagrams Venn diagram Graphical representations, using several typically overlapping circles, showing elements of sets in relation to properties or attributes. They are drawn for some speciﬁed universal set. A Venn diagram is a visual representation of sets, where each set is depicted as a circle within a universal set U, shown as a rectangle. Overlapping circles indicate elements common to multiple sets. For two sets A and B: • The region inside the circle for A represents elements in A. • The overlap of the circles represents the intersection A ∩ B. • The combined area of both circles represents the union A ∪ B. • The area outside A but within U represents the complement . A

A∪B A∩B A∪B

476

This Venn diagram shows sets A and B within the universal set U (depicted as the rectangle). The shaded overlap represents A ∩ B, and the area outside A represents .

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Venn diagrams help interpret set operations in practical situations, such as identifying common or unique elements between groups. It also illustrate relationships between events in a sample space. S A

B The rectangle represents the sample space S, containing all possible outcomes of an experiment.

S A

B Circles labelled A and B represent events, subsets of S.

S A

B The overlap, A ∩ B, shows outcomes where both A and B occur.

S A

B The combined area, A ∪ B, represents outcomes where A or B occurs.

S A The area outside A but inside S, , represents outcomes where A does not occur.

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Interactive exploration Discover this concept in action online

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Example 1 In a school of 50 students, 30 take art, 25 take music and 10 take both. a Draw a Venn diagram to represent this.

Create a strategy

Apply the idea

Use two circles for art and music within universal set U. Place 10 in the overlap and distribute remaining students.

Let A = Art, B = Music. The overlap is A ∩ B = 10. Students who take art only is 30 − 10 = 20. Students who take music only is 25 − 10 = 15. Students who take neither is 50 − (20 + 10 + 15) = 50 − 45 = 5. Art

Music

15 5

b Calculate the number of students taking art or music.

Create a strategy Sum the students in the art only, music only and overlap region of the Venn diagram from part (a).

Apply the idea From the Venn Diagram, art only has 20, music only has 15 and the overlap region has 10. Total = 20 + 15 + 10 = 45 students

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Add the values Evaluate

c How many students take neither art nor music?

Create a strategy Use the number of students outside both circles in the Venn diagram.

Apply the idea From the Venn diagram, outside both circles has 5. So, 5 students take neither.

Reﬂect and check To verify, add all the values: 20 + 15 + 10 + 5 = 50, matching U .

Idea summary A Venn diagram represents sets as circles within a universal set U, with overlaps showing A ∩ B and combined areas showing A ∪ B. Venn diagrams are used to visualise and interpret set operations in practical contexts.

Cardinality rule for union of sets The number of elements in the union of two sets A and B, denoted by A ∪ B, can be calculated using the formula:

A ∪ B = A + B − A ∩ B A ∪ B is the number of elements in A ∪ B are the number of elements in A and B, A, B respectively is the number of elements in A ∩ B, A ∩ B subtracted to avoid double-counting This rule accounts for elements in A ∩ B being counted twice when adding A and B.

Example 2 In a group of 40 students, 22 study biology, 18 study chemistry and 8 study both. a Draw a Venn diagram to represent this.

Create a strategy Use two circles for biology and chemistry within universal set U. Place 8 in the overlap and distribute remaining students.

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Apply the idea Let A = Biology, B = Chemistry. The overlap is A ∩ B = 8. Students who take biology only is 22 − 8 = 14. Students who take chemistry only is 18 − 8 = 10. So, outside both circles is 40 − (14 + 10 + 8) = 8. Biology

Chemistry

10 8

b How many students study biology or chemistry?

Create a strategy Use the formula A ∪ B = A + B − A ∩ B.

Apply the idea The students study biology is A = 22. The students study chemistry is B = 18. The students study both is A ∩ B = 8.

A ∪ B  = A  + B  − A ∩ B 

Write the formula

= 22 + 18 − 8

Substitute the values

= 32

Evaluate

So, 32 students study biology or chemistry.

c How many students study only chemistry?

Create a strategy Subtract the overlap, A ∩ B, from the students who study chemistry, B.

Apply the idea Only chemistry = A ∩ B − B = 18 − 8

Substitute the values

= 10

Evaluate

So, 10 students study only chemistry.

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Subtract A ∩ B from B

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Reflect and check To verify, add the values: 14 + 10 + 8 + 8 = 40, matching U.

Idea summary The cardinality of the union of two sets is given by A ∪ B = A + B − A ∩ B, accounting for elements counted twice in A ∩ B. This rule is useful for calculating the total number of elements in A ∪ B in practical problems.

9.03 Practice questions What do you remember? 1

Write the proper notation for the regions of a Venn diagram of two sets A and B within the universal set U : a

Overlap of the circles

Combined area of both circles

Area outside A but within U

Rectangle surrounding the circles

Write the formula for the number of elements in the union of two sets A and B.

Determine whether each statement is true or false: a

The intersection A ∩ B contains elements common to both A and B.

If A ∩ B = ∅, then A ∪ B = A + B.

The complement

includes elements in A.

In a group of 25 people, 15 like to hike, and 10 like to swim. If 5 like both, how many people like hiking or swimming (or both)?

Practice Ex 1

Ex 2

In a survey of 70 people, 40 enjoy reading, 35 enjoy hiking, and 15 enjoy both: a

Represent this information in a Venn diagram.

Calculate the number of people enjoying reading or hiking.

How many people enjoy neither reading nor hiking?

In a survey of 75 people, 45 use a smartphone, 35 use a tablet, and 20 use both: a

Draw a Venn diagram to represent this.

How many people use a smartphone or a tablet?

How many people use neither?

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In a class of 40 students, 22 study biology and 18 study chemistry, 10 study both subjects: a

Represent this information in a Venn diagram.

How many students study only Biology?

How many students study only Chemistry?

How many students study neither subject?

The Venn diagram shows the results of a survey of 50 people about owning pets: a

How many people own only a dog?

How many people own only a cat?

How many people own both a dog and a cat?

How many people own neither?

Dog

Cat

25 10

The Venn diagram shows a club with 35 members and their sports activities: a

How many members play tennis only?

How many members play badminton only?

How many members play both tennis and badminton?

Tennis

Badminton

15 8

In a school of 100 students, 60 take mathematics, 50 take science, and 30 take both. A student is chosen at random: a

How many students take only mathematics?

What is the number of students that take at least one subject?

Mathematics

Science

U  = 45, A = 25, B = 20, A ∩ B = 0, find: b a A ∪ B

(A ∪ B) 

U  = 80, A = 50, B = 40, A ∩ B = 25, find: b (A ∪ B)c a B ∩ Ac

The Venn diagram shows a set U  = 30 and subsets where A is French and B is Spanish. Find: a c

A  A ∩ B 

French

Spanish

B  8

5 7

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U  = 70, A = 45, B = 35, A ∩ B = 20, find: b (A ∪ B)c a B ∩ Ac

U  = 55, A = 30, B = 25, A ∩ B = 15, find: b A ∪ B  a A ∩ Bc

U  = 65, A = 40, B = 30, A ∩ B = 20, find: b (A ∪ B)c a A ∩ Bc

The Venn diagram shows a set U  = 90 and subsets where A is movies and B is music. Find: a b c

A  B  A ∩ B 

Movies

Music

Extend your thinking 18

The Venn diagram shows a group of 60 students and their food preferences: a

How many students like pizza only?

How many students like burgers only?

How many students like sushi only?

How many students like none of these foods?

Pizza

Burger 10

30 8

25 7

20 15 Sushi 19

U  = 100, A = 60, B = 50, C = 40, A ∩ B = 25, A ∩ C = 20, B ∩ C = 15, A ∩ B ∩ C = 10 a

Represent this information in a Venn diagram.

Find A ∩ Bc ∩ C c.

Find B ∩ Ac ∩ C c.

Find A ∪ B ∪ C.

A company has 80 employees. 50 speak English, 40 speak Spanish, and 25 speak both. Explain why the formula A ∪ B = A + B − A ∩ B is necessary when calculating the number of employees who speak at least one language.

In a survey, 70 people were asked about their hobbies. 40 enjoy painting, 35 enjoy gardening, and 20 enjoy both. A student incorrectly calculates the number of people who enjoy painting or gardening as 40 + 35 = 75: a

Identify the error in the student’s calculation.

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A library has 120 members. 80 borrow ﬁction books, and 60 borrow non-ﬁction books. 40 borrow both. The library wants to offer a discount to members who borrow both types of book: a

How many members borrow only ﬁction books?

How many members borrow only non-ﬁction books?

How many members are eligible for the discount?

9.04 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).

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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

is an event, a subset of the sample space S

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

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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

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 is the number of outcomes in the S 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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Interactive exploration Discover this concept in action online

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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”.

9.04 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

490

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.

Mathspace New South Wales – Year 11 Advanced mathspace.co

Calculate P (A ∩ B).

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.

9.05 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.

9.05 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.

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Find P ( ).

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)

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.

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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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9.06 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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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.

9.06 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.

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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 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.

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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.

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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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9.07 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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9.07 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.

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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

514

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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9.08 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).

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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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9.08 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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9 Chapter review 1

Let the universal set be U = {1, 2, 3, 4, 5, 6, 7, 8}, with sets A = {1, 3, 5, 7} and B = {1, 2, 3, 4}. Which of these is the set A′ ∩ B? A

{1, 3}

{5, 7}

{2, 4}

{6, 8}

The Venn diagram shows the results of a survey of 80 people about device ownership. How many people own a laptop but NOT a tablet? Laptops

Tablets

20 15

A 3

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0.1

0.3

0.4

0.7

A Python coding club has these members: {Leo, Mia, Noah, Ava}. The members who know Python are Leo and Noah: a

List the set of members who know Python.

Determine the cardinality of the set of Python programmers.

Determine the complement of the Python programmers set with respect to the coding club.

A survey of students’ favourite subjects gives the set S = {Mathematics, Art, History}: a

How many subsets does S have? Explain your reasoning.

List all subsets of S.

Consider the universal set U = { p, q, r, s, t} and two sets A = {p, q} and B = {q, r, s}: a

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

Determine A′ and B′.

Is B′ ⊆ A′? Explain.

At a university, let S be the set of students on the soccer team {Alex, Ben, Chloe} and B be the set of students on the basketball team {Chloe, David, Eva}. Determine: a

S ∩ B

If S and B are disjoint.

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S∪B

At a school, S is the set of students playing soccer {Liam, Olivia, Noah} and T is the set playing tennis {Noah, Emma, Sophia}. Determine: a

S ∩ T

If S and T are disjoint.

S∪T

A company has volunteer programs. Let the universal set U be all employees, with A = {employees at the animal shelter} and B = {employees at the food bank}. Suppose A ∩ B = {Ken, Laura}, A ∪ B = {Ken, Laura, Mike, Nora, Oscar} and A′ = {Nora, Oscar, Paul}. Determine: a

A and B

If A and B are disjoint. Explain.

In a class of 35 students, 20 study Physics, 17 study Art and 8 study both subjects: a

Represent this information in a Venn diagram.

How many students study only Physics?

How many students study only Art?

How many students study neither subject?

Consider the following information:

U  = 120, A = 70, B = 60, C = 50, A ∩ B = 30, A ∩ C = 25, B ∩ C = 20, A ∩ B ∩ C = 10.

Represent this information in a Venn diagram.

Find A ∩ B′ ∩ C′.

Find C ∩ A′ ∩ B′.

Find (A ∪ B ∪ C)′.

The Venn diagram shows a set U  = 50 and subsets where A is German and B is Japanese. Find: Japanese

German

12 15

a 13

A 

B 

A ∪ B 

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

How many customers buy only digital games?

How many customers buy only physical games?

How many customers are eligible for the discount?

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A die is rolled. Deﬁne A as rolling a number greater than 2, B as rolling an even number. Determine: a

A ∩ B

P (A ∩ B)

A box has 5 apple and 7 orange candies. A candy is picked. Calculate: a

P (apple)

P (not apple)

A die is rolled. Deﬁne A as rolling a multiple of 2, B as rolling a number less than 6. Sketch a Venn diagram showing S, A, B, A ∩ B, A′. List outcomes in each region.

A card is drawn from a deck of 52 cards. Deﬁne event A as drawing a heart, event B as drawing a jack. Find: a

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: a

P (A ∩ B)

P (A ∪ B)

A card is drawn from 52 cards. Event A is drawing a jack, event B is drawing a red card. A student claims P (A ∪ B) = P (A) + P (B). Correct the error using a Venn diagram description.

A bag contains 4 blue balls and 3 yellow balls. Two balls are drawn without replacement:

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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 survey of 150 employees found 90 use a laptop, 50 use a desktop, and 30 use both: a

Construct a Venn diagram to represent this data.

Find the probability an employee uses a laptop.

Find the probability an employee uses a desktop given they use a laptop.

Construct a two-way table for the data.

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 program: 180 use the gym, 140 use the pool, and 70 use both. A member is chosen at random: a

Find the probability they use the gym given they use the pool.

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)?

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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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Big ideas • A random variable provides a numerical value for an experimental outcome, and its classification as discrete or continuous dictates how its data is organised and visualised using tools like frequency distributions and histograms to reveal key features of the distribution. • The relative frequency of an outcome observed in data provides a practical, experimental estimate of its theoretical probability, with the reliability of this estimate increasing as the number of trials grows.

10 Data Chapter outline 10.01 Random variables 10.02 Organise and graph datasets 10.03 Analyse data Chapter 10 review

528 532 541 546

Your chance of seeing a shooting star on a clear night? Around 1 every 10 minutes.

10.01 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.

10.01 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.

10.02 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.

10.02 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?

Retail stores use frequency tables to make smarter decisions about restocking! By tracking how often each product is purchased, they can identify customer favourites, reduce waste, save money, and ensure shelves are always stocked with what shoppers want most, creating a better shopping experience. 540

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10.03 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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10.03 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

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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.

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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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10 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

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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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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 a

Create cumulative frequency polygons for both datasets on the same axes.

Compare the medians of the two classes using the polygons.

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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:

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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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“You can have data without information, but you cannot have information without data.” Daniel Keys Moran

Big ideas • The average rate of change of a function over an interval measures the overall change between two points and is represented geometrically by the gradient of the secant line connecting them. • The instantaneous rate of change at a specific point is the limiting value of the average rate of change over an ever-shrinking interval, and is defined precisely as the gradient of the tangent line to the function’s graph at that point.

11 Introduction to rates of change Chapter outline 11.01 11.02 11.03 11.04 11.05

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 11 review

552 557 563 568 575 584

The fastest recorded tennis serve hit 263 km/h. That’s over 70 metres in one second!

11.01 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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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

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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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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)).

11.01 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?

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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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11.02 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).

Interactive exploration Discover this concept in action online

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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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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

11.02 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?

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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

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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

562

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?

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11.03 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

[t1, t2]. In contrast, instantaneous speed is the speed of an

d 40

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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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].

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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.

11.03 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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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.

11.04 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.

11.04 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?

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.

0.01

(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 572

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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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t 2s

11.05 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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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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11.05 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)

580

4.01

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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

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0.998

0.999

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1.001

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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11 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

584

d(t) = 40t + 5t2, t = 1 to t = 3 3

d(t) = 4t − 2t, t = 0 to t = 2

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d(t) = 2t2 + 3t, t = 2 to t = 4

d(t) = 80t + 10t2, t = 2 to t = 4

= 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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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.

12 The derivative Chapter outline 12.01 12.02 12.03 12.04 12.05 12.06 12.07 12.08 12.09 12.10 12.11

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 12 review

590 602 606 613 623 636 643 648 655 662 675 684

The chain rule is like peeling an onion — one function inside another.

12.01 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 ≠ 0 592

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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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12.01 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).

12.01 Gradient of a curve mathspace.co

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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

600

−2

f ′(x) = 0

⬚

What value does the gradient of the secant appear to approach?

Mathspace New South Wales – Year 11 Advanced 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. 12.01 Gradient of a curve mathspace.co

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12.02 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.

602

Function

Derivative

f (x) = mx + c

f ′(x) = m

Mathspace New South Wales – Year 11 Advanced 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).

12.02 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

604

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

Mathspace New South Wales – Year 11 Advanced mathspace.co

f (x) = x + 3

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)?

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12.03 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 606

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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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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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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.

12.03 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)).

12.04 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.

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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

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= 4x3.

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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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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

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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.

12.04 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) =

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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(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.

12.05 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

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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

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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°).

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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.

12.05 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:

632

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 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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12.06 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 12.06 Chain rule mathspace.co

637

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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12.06 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

640

iii

y = (2x4 + 6)5

y = (4x + 3) − 1

e 8

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 mathspace.co

y = 3(x + 5)5

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

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Did you know?

When you press the accelerator in a car, you’re applying the chain rule in motion! Your car’s speed depends on how far you press the pedal — the deeper you press, the more fuel is released into the engine. That fuel changes how much force the engine produces, which in turn affects how quickly your position changes on the road. The relationship between pedal pressure, speed, and distance is a real-world example of how rates of change are linked through composite functions.

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12.07 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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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.”

12.07 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

646

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 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).

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Extension: Proof of product rule Extension online

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12.08 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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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

12.08 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 = :

652

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 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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6 8 10 12 14 16 18

Extension: Proof of quotient rule Extension online

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12.09 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.

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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

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.

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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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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.

12.09 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

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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

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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.

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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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12.10 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

The function y = x + 1 is always increasing. The gradient f ′(x) = 1 is always positive.

x −2

−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

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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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12.10 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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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 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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12.11 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

Set

3(t − 4t + 3) = 0

Factorise 3

3(t − 1)(t − 3) = 0

Factorise the quadratic

3t − 12t + 9 = 0

1, t = 3

The particle is at rest at t = 1 s and t = 3 s.

678

Write the velocity function

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Use null factor law

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.

12.11 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?

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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

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?

−1 −2

a 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 =

682

, 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 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.

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12 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.

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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 12 review mathspace.co

685

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

686

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 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.

13 Exponential functions Chapter outline 13.01 Exponential graphs 13.02 Tangent gradient at y-intercept 13.03 Euler’s number and derivatives Investigation: Differentiation of exponential functions 13.04 Applications Chapter 13 review

690 702 706 712 714 720

In just 10 steps, you can walk 10 metres. But 10 exponential steps? You’d pass Saturn.

13.01 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. 690

Mathspace New South Wales – Year 11 Advanced 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.

13.01 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

692

−1

Mathspace New South Wales – Year 11 Advanced mathspace.co

x 1

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

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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.

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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.

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13.01 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

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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, ⬚)

13.01 Exponential graphs mathspace.co

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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

698

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 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.

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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

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Q(0, 1) 1 2 3 4 5

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.

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13.02 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).

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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.

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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.

13.02 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: i

h = 0.1

h = 0.01

iii

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, rounded to three decimal places: a

y = 1.8x

y = 3.2x

y = 0.8x

y = 5x

y = 1.3x

y = 4x

y = 2.2x

y = 0.9x

y = 3.8x

y = 1.7x

Estimate the gradient of the tangent at x = 0 for y = 2.5x using the secant line with: h = 0.1

h = 0.01

iii

h = 0.001

Compare with the gradient for y = 2x using h = 0.01.

Estimate the gradient of the tangent at x = 0 for y = 3.5x using the secant line with:

h = 0.1

h = 0.01

iii

h = 0.001

Compare with the gradient for y = 3x using h = 0.001.

Calculate the gradient of the secant at x = 0 with h = 0.001 for the following functions, rounded to three decimal places: a

y = 2.3x

y = 4.2x

y = 1.4x

y = 0.7x

y = 1.6x

y = 2.4x

y = 3.3x

y = 0.6x

y = 0.5x

i 11

h = 0.001

Estimate the gradient of the tangent at x = 0 for y = 4.5x using the secant line with:

iii

h = 0.01

Compare the results for different h values.

h = 0.1

y = 4.8

y = 1.9

y = 2.9

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 • y = 3x • y = 3.5x • y = 2x • y = 4x

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Extend your thinking 12

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): a

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 = 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

13.03 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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707

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) 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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Write the point-slope form

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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.

13.03 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

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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.

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“No human investigation can be called true science without passing through mathematical tests.” Leonardo da Vinci

INVESTIGATION

Differentiation of exponential functions Investigate Euler’s number, e, and conclude that it is the unique number such that

(ex) = ex,

meaning e is its own derivative. Use numerical and graphing methods to explore this property.

Objectives • To estimate the limit

for various a > 0 using technology, and

recognise that it equals 1 when a = e. • To derive and verify that property.

(ex) = ex, and conclude that e is unique in this

• To use graphing software to visualise that ex is its own derivative, and compare with other exponential functions.

Exploring the limit Euler’s number, e, is special because the derivative of ax involves the limit limh → 0 spreadsheet to estimate this limit for different a > 0.

. Use a

This limit is the gradient of ax at x = 0. Test values like a = 2, 2.718, 3 to ﬁnd when the limit equals 1. Why is this limit signiﬁcant for the derivative of exponential functions? Step 1: Open a spreadsheet. In row 1, enter a values: 2, 2.71828(≈ e), 3(e.g., cells B1, C1, D1). Step 2: In column A (from A2), list decreasing h values: 0.1, 0.01, 0.001, 0.0001 Step 3: In cell B2, enter

, using a from B1 and h from A2

. Drag to ﬁll the table.

Step 4: Record the approximate limit for each a as h approaches 0. Step 5: For a ≈ 2.71828, check if the limit is 1. Hypothesise why this makes e unique.

Investigate

712

What are the approximate limits for a = 2, 3, e?

Why does a = e yield a limit of 1? How does this relate to the derivative?

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Deriving the derivative The derivative of f (x) = ax can be found using ﬁrst principles:

(ex) = ex.

For a = e, the limit is 1, so

This shows ex is its own derivative, a unique property. Later, you will learn that for general a, the constant is ln a. Numerically verify this for ex and compare with other bases.

Step 1: In a spreadsheet, for x = 1, compute e1 ≈ 2.71828. Step 2: For a = 2, compute previous section times 21.

for h = 0.1, 0.01, 0.001. Compare to

for the same h values. Compare to your limit from the

Step 3: Use a graphing calculator to plot ex and its derivative. Verify they coincide. Plot 2x and k × 2x (using your limit for k) to compare.

Investigate approach e1, as h approaches 0? Why is this signiﬁcant?

Does

How does the derivative of 2x differ from ex graphically and numerically?

Why do the graphs of ex and its derivative coincide?

Discussion (ex) = ex a unique property compared to other bases like 2x?

Why is

How did graphing ex and its derivative clarify its self-derivative property?

How does the limit limh → 0

= 1 explain why ex is its own derivative?

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13.04 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.

13.04 Practice questions What do you remember? 1

716

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)

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.

Mathspace New South Wales – Year 11 Advanced mathspace.co

y = 4(1.2x)

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

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.

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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

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

718

Complete the table of values:

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

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.

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13 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 mathspace.co

Population (P)

Time (years)

720

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

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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

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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.

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.

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Consider the function y =

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.

from the graph of

on the same set of axes.

A radioactive isotope decays according to M = 300e− 0.02t, 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 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.

Did you know?

Epidemiologists use exponential functions to track how diseases spread through populations! For instance, if one infected person passes the illness to two others, and each of them does the same, the numbers grow like 1, 2, 4, 8, 16… By studying this exponential growth, scientists can forecast infection surges, estimate hospital bed demand, and plan strategies such as vaccination drives or lockdowns to slow the spread. Exponential functions make it possible to predict and manage the impact of contagious diseases in the real world!

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Big ideas • A logarithm is an exponent, and this fundamental inverse relationship with exponentiation is the foundation for the logarithm laws, which provide a complete algebraic toolkit for manipulating and simplifying logarithmic expressions. • The logarithm laws are the primary tool for solving exponential equations where bases cannot be matched, and for simplifying logarithmic equations into a solvable form, with the change of base formula allowing for numerical evaluation. • The logarithmic function is the inverse of the exponential function, a relationship that is visually represented by their graphs being reflections of each other across the line y = x and by their key features, such as asymptotes and intercepts, being interchanged.

14 Logarithmic functions Chapter outline 14.01 14.02 14.03 14.04 14.05

Logarithms Exponential-logarithmic equivalence Logarithm laws and properties Logarithm expressions and equations Logarithmic graphs Investigation: Logarithms Chapter 14 review

726 735 742 748 755 763 767

The pH scale measures acidity using logarithms. Lemon juice is about 100 000 times more acidic than water!

14.01 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:

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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

Write the exponential form Substitute a = 3, x = 81 and y = 4

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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

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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

as 9−2

= −2.

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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

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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: ey = x y

e = 10

Write the exponential form Substitute x = 10

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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

14.01 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

ex = 5

10−1 = 0.1

ay = x

log (1) = 0

log−2 (4) = 2

State whether each statement is true or false: a

732

24 = 16

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)

ln (10)

log (99)

log (0.05)

ln (1) = 0

loga (b) = c

ln (e5) = 5

log (0.001) = −3

log (x) = 10

ln (x5) = 5

log (x) = −3

Solve for x: ln (x) = 0

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: a

Rewrite in exponential form:

a 12

= −2

log2 (16)

a 11

log (100) = 2

a 10

Evaluate without technology:

a 9

log6 (36) = 2

log3 (x) = 2

Verify whether each statement is true by converting to exponential form: a

ln (e3) = 3

log2 (32) = 5

log (100) = 3

log7

= −2

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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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14.02 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

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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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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

14.02 Practice questions What do you remember? 1

Determine whether each statement is true or false: a b c

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 =

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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.

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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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14.03 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

742

= −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

= x, loga (a) = 1,

= − loga (x), derived from the logarithm deﬁnition.

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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

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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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14.03 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

746

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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14.04 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

748

= 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

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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.

14.04 Practice questions What do you remember? 1

Determine whether each statement is true or false: a c

loga (s ) = m loga (s)

b d

loga (an) = n

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

Complete each statement for solving equations: a

loga (st) = loga (s) + loga (t)

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

log13 (w − 4) = 2

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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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14.05 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.

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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).

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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

−2

−1

1 −1

x=0 y=x

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y=0

(1, 0)

−2

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y = log2 (x)

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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14.05 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.

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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 = ex, y = ln(x)

y = 4x, y = log4 (x)

y = 6x, y = log6 (x)

y = 5x, y = log5 (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

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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.

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

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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.

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INVESTIGATION

Logarithms Investigate logarithms and understand the heating and cooling formula.

Objectives • To apply logarithmic functions to solve a real-world problem • To understand the heating and cooling formula • To practise identifying logarithmic data • To investigate the relationship between exponents and logarithms

Activity title Murder Mystery For this investigation we are going to be solving a murder mystery! You and your class have rented out a mansion for an end-of-year celebration. Everyone is having a lot of fun hanging out, playing games, and eating food. Who did it? Suddenly, you all hear a loud scream coming from the upstairs of the mansion. Everyone rushes upstairs to ﬁnd one of your classmates, who was on his way to the bathroom, has discovered a dead body in the hallway, outside the library. The man who has died is the owner of the mansion and no one seems to know who did it. Among us The only thing everyone agrees on is that it must be someone in the class because there was no one else in the mansion. At the time the body was found the entire class is present. Time of death One of your friends comes up with the suggestion that if you knew the time of death, you might be able to determine who was with the mansion owner at that moment and therefore, most likely, who the murderer is. Measurements Another one of your friends has the idea that knowing the temperature of the body would help to ﬁgure out the time of his death, so immediately a thermometer is found and body temperature of the corpse is taken. Temperature Based on the thermometer reading, the body was 33.89°C at 6:00 p.m. when the body was found. Two hours later the body’s temperature is taken again and is now found to be 30.33°C.

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Pre-Investigation Work together in small groups to answer the following questions. We want to come up with a method to determine what time the man was killed. Let’s look at a smaller example ﬁrst to get the idea. • Do you think that a corpse will cool off similarly to things like coffee or tea? • How would it be similar? How would it be different? • What is a good estimate for the starting temperature of coffee or tea? • After what amount of time do you think a cup of coffee or tea would get cold? • At what temperature would you consider a cup of coffee or tea to be cold? Why? Next: • Would you consider the same temperature as being cold for a body? What factors do you think might affect how fast something cools off? • On your piece of paper graph your estimated change in temperature over time, plotting temperature on the y-axis and time on the x-axis. Be sure to label your axes and title your graph. • Do you think a linear representation would be useful in modelling the change in temperature over time? If not, what type of graph would make the most sense for this situation? Explain. Optional: If you have a cooking thermometer available complete the following activity to test your ﬁndings. Other thermometers may break during this experiment so make sure it is a cooking thermometer. Take note of the thermostat in the room you are in. If there is no thermostat be sure to get a reading of the room’s temperature. Make a cup of coffee or a cup of tea. Measure and record the temperature of the drink immediately after it is made, then continue measuring the temperature: • 15 minutes after it has been made • 2 minutes after it has been made • 20 minutes after it has been made • 4 minutes after it has been made • 30 minutes after it has been made • 5 minutes after it has been made • 60 minutes after it has been made • 10 minutes after it has been made

Investigate

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Was your guess of the initial temperature close to the real initial temperature?

Was your guess of how long it would take the drink to reach the temperature you considered to be cool close?

Make a graph to represent the data you are using. Be sure to label the axes and title the graph. Like your ﬁrst graph, time should be on the x-axis and temperature should be on the y-axis.

Did the graph look as you expected it to? Why or why not?

Make some observations about the graph in the context of the situation.

What temperature do you think the drink would be after 2 hours? What about after 3 hours? How do you know?

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Investigation Now let’s apply this knowledge about heating and cooling to the murder mystery in the mansion. Use the facts below about the crime scene to help you in your investigation. Work in groups to solve the murder mystery before the killer strikes again. The facts The mansion owner was found in the hallway near the library. There is only one pathway in front of the library, and it leads to the bathroom. The temperature of the body was 33.89°C at 6:00 p.m. The temperature of the body was 30.33°C at 8:00 p.m. The thermostat was found to be set at 21.11°C. After much discussion, everyone was able to remember the times that they were up in the hall passing by the library to walk to the bathroom. Cut out the time cards provided based on the number of students in your group. If you have more participants than cards, some people should pair up. In this case, they will be assumed to have been going to the bathroom together. Place the cards with times on them into an opaque bag. Each member (or pair) of the group should pick a piece of paper from the bag. This will indicate what time you were passing through the hall to go to the bathroom. Time Cards • 4:00 p.m. • 4:10 p.m. • 4:20 p.m. • 4:30 p.m. • 4:40 p.m.

• 4:50 p.m. • 5:00 p.m. • 5:10 p.m. • 5:20 p.m.

Did you know? An exponential equation such as ex = 3 can be rewritten as the equivalent logarithmic equation as x = loge 3. We often use the special notation ln for the natural logarithm, instead of loge. Newton’s formula for cooling: T (t) = Tenv + (T0 − Tenv) e−kt T (t) is the temperature of the body at time t, Tenv is the temperature of the surrounding environment, T0 is the initial temperature of the body (look up average body temperature for this), t is the time after death in hours, and k is a constant that can be calculated using the formula:

Investigation: Logarithms mathspace.co

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Let t represent the time (in hours) after death that the body was found. Using this fact, at time t when the body was found, the temperature of the body was 33.89°C. Two hours later, at time t + 2, the temperature of the body was 30.33°C. Substitute the values you know and solve for t to ﬁnd the time that elapsed before the body was found. Assume the average living body temperature is 37°C.

Investigate 1.

What time was the mansion owner killed?

Based on everything you have found, who was the murderer?

Discussion 1.

Summarise what you have learnt about the heating and cooling formula.

Do all things heat and cool at the same rate?

What considerations need to be made when using the formula?

Did you know?

Logarithms don’t just live in your maths textbook — they also help solve real murder cases! When a person dies, their body begins to cool down to match the surrounding air temperature. This cooling process doesn’t happen at a constant rate — it’s exponential, which means the temperature change slows down over time. So when your class investigates the mansion murder mystery, you’re stepping right into the shoes of a forensic scientist! 766

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14 Chapter review 1

What is the value of log3 (81)? A

log(9)

log(1.25)

log(20)

x=2

x=0

x=3

y=3

43 = 64

2− 4 =

e2 = e2

91.5 = 27

log7 (49) = 2

log5

log2 (64) = 6

log(100)

log2

log(1000) = 3

= −3

Evaluate without technology: a

Rewrite in exponential form: a

Rewrite in logarithmic form: a

What is the equation of the vertical asymptote of the graph of y = log2 (x − 3)? A

Which of the following is equivalent to log(5) + log(4)? A

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

5x = 50

6x = 11

2x = 0.8

Solve the following equations, rounded to four decimal places: a

3x = 25

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:

log2 (14)

a e

log11 (11)

log5

log15 (1)

log7 (74)

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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. 18

Simplify each logarithmic expression and evaluate it. Then, state whether the result is rational or irrational. a

log3 (54) − log3 (2)

log5 (w − 2) = 3

log3 (w + 1) = 4

log4 (w − 2) + log4 (w + 2) = 3

log3 (log2 (x − 1)) = 2

Solve for w: a

Solve for w and state if the result is rational or irrational: a

log2 (16) + log2 (5)

log2 (w + 1) + log2 (w − 1) = 3

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

Domain

Range

, where t is time in hours. Find the time

Asymptote

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:

768

y = 10x, y = log(x)

y = 6x, y = log6 (x)

Mathspace New South Wales – Year 11 Advanced mathspace.co

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?

Earthquakes are measured using a logarithmic scale! The Richter scale, which measures the energy released by an earthquake, is logarithmic — meaning each step up represents ten times stronger ground shaking and about 32 times more energy released. By understanding this logarithmic relationship, engineers can design buildings that better withstand tremors, while emergency services can predict which areas are likely to experience the most severe impact.

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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.

15 Transformations Chapter outline 15.01 Reflections in axes 15.02 Horizontal and vertical translations 15.03 Dilations Chapter 15 review

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Shifting furniture in your room is literally translating it across the floor.

15.01 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?

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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)

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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.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).

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15.01 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

776

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 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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15.02 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

778

(2, 0) 1

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Interactive exploration Discover this concept in action online

mathspace.co

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

15.02 Horizontal and vertical translations mathspace.co

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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.

15.02 Practice questions What do you remember? 1

782

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 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.

15.03 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).

Interactive exploration Discover this concept in action online

mathspace.co

15.03 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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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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15.03 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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−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.

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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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15 Chapter review 1

Which equation represents the graph of y = 2x + 5 after a reﬂection in the y-axis? A

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.

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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)

−3 −2

794

(1, 1)

2 y = x2 9

x −1

Determine the type and factors of the dilation applied.

Find the equation of the transformed graph.

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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, this transformation reduces the graph, making it a smaller scale model.

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.

Did you know?

Mathematical transformations can turn simple shapes into complex masterpieces! By combining rotations, reflections, and dilations, you can generate stunning patterns — like those seen in tessellations and cosmic designs. These transformations are the foundation of digital art, animation, and even architectural design, helping creators bring symmetry and movement to life. They reveal how simple rules can produce extraordinary complexity, much like the universe itself. Through transformations, maths becomes not just numbers and equations, but a language of beauty, movement, and creativity.

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Big ideas • Complex functions can be constructed by applying a systematic sequence of transformations to a basic parent function, where the non-commutative nature of these operations means the order, typically dilations and reflections followed by translations, is critical to determining the correct final graph and equation. • While the algebraic rules for transformations are universal, their geometric impact is specific to the parent function, systematically shifting key features such as the asymptotes of exponential, logarithmic, and reciprocal functions, the vertex of an absolute value function, or the centre of a circle.

16 Advanced applications of transformations Chapter outline 16.01 16.02 16.03 16.04 16.05 16.06

Linear, quadratic and cubic functions Exponential and logarithmic functions Reciprocal and absolute value functions Circles and translations Order of transformations Multiple transformations Chapter 16 review

798 811 824 832 839 850 858

Roller doors roll up in a spiral — another sneaky logarithmic pattern.

16.01 Linear, quadratic and cubic functions After this lesson, you will be able to… • apply translations to linear functions, and reflections to quadratic functions. • apply dilations to cubic functions. • determine the equation of a function after one or more transformations have been applied. • identify the key features of a transformed graph, including its vertex, intercepts, domain and range. • describe a sequence of transformations that maps one function to another.

Translations of linear functions Domain For a function or relation, it is the set of real numbers on which the function or relation is deﬁned. 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-intercept s are sometimes taken to mean the signed distance from the point at which the curve crosses the axis to the origin, for example, for the line y = mx + c, the y-intercept is c. Range The set of values of the dependent variable for which a function is deﬁned. 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. Function A function f assigns to each element of one set S precisely one element of a second set T.

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A translation shifts a linear function’s graph horizontally or vertically. For a linear function f (x) = mx + c, translations are applied as:

g(x) = f (x − a) + b = m(x − a) + c + b g(x)

is the transformed function

a is the horizontal shift: right if a > 0, left if a < 0 b is the vertical shift: up if b > 0, down if b < 0 The domain of a translated linear function remains (−∞, ∞). The range also remains (−∞, ∞) unless m = 0, in which case the function is a horizontal line with range y = c + b. Translations do not restrict input values but in speciﬁc cases may affect the range.

Exploration Consider the function f (x) = 3x − 4. Explore the effect of translating it right by 3 units and up by 2 units: 1. Determine the equation of the transformed function. 2. Calculate the new y-intercept and compare it to the original. 3. Sketch the original and transformed graphs, labelling the y-intercept. 4. Discuss how the translations affect the position of the graph without changing its slope.

Example 1 The function f (x) = 3x − 2 is translated right by 1 unit and up by 4 units: a Determine the equation of the transformed function.

Create a strategy Use the translation form g(x) = m(x − a) + c + b with a = 1 and b = 4 to ﬁnd the transformed equation.

Apply the idea f (x) = 3x – 2 f (x − 1) + 4 = 3(x − 1) − 2 + 4

Write the function Apply translations with a = 1 and b = 4

= 3x − 3 + 2

Evaluate each term

= 3x − 1

Simplify

The transformed function is g(x) = 3x − 1.

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b Determine the y-intercept.

Create a strategy Substitute x = 0 into the transformed equation to ﬁnd the y-intercept.

Apply the idea g(x) = 3x – 1

Write the transformed function

g(0) = 3 × 0 − 1

Substitute x = 0

= −1

Evaluate

The y-intercept is at (0, −1).

c Sketch the graph of the transformed function.

Create a strategy Plot the y-intercept and use the slope to ﬁnd another point, then draw the line.

Apply the idea y 3

(1, 2)

2 1 −1 (0, −1) −1

g(x) = 3x − 1

First, plot the y-intercept at (0, −1). Starting from this point, move 1 unit to the right and 3 units up (because the gradient is 3) to plot a second point.

x 1

Then, draw a straight line through these two points to complete the graph.

−2 −3

Reﬂect and check The original y-intercept was (0, −2). The vertical shift up by 4 units and horizontal shift right by 1 unit changed this point to (1, 2). y 3 2

(1, 2)

−1

1 −1

g(x) = 3x − 1 −2 −3

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f (x) = 3x − 2 (0, −2)

Idea summary A translation of a linear function y = mx + c by a units horizontally and b units vertically results in g(x) = m(x − a) + c + b. The domain and range remain (−∞, ∞) unless m = 0 where the transformed range is y = c + b. Sketching shows the translated line with updated intercepts.

Reflections of quadratic functions Reﬂection A transformation of a shape formed by creating a mirror image on the other side of a given line. Vertex A vertex is a point in the plane where lines meet and do not extend beyond, or a point in space where several edges meet. A vertex can also refer to a node in a network. It is also the turning point of a parabola.

A reflection flips a quadratic function’s graph over an axis. For a quadratic f (x) = ax2 + bx + c, reflections are:

g(x) = −f (x) g(x)

is the transformed function

−f (x) − is the reﬂection over the x-axis, ﬂipping the graph vertically

g(x) = f (−x) f (− −x) is the reflection over the y-axis, flipping the graph horizontally The vertex changes position based on the reflection, but the domain remains (−∞, ∞). The range depends on the concavity and reflection.

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Example 2 Reﬂect the function f (x) = (x − 1)2 + 2 over the x-axis: a Determine the equation of the transformed function.

Create a strategy Apply the reﬂection g(x) = −f (x) to ﬁnd the new equation.

Apply the idea f (x) = (x − 1)2 + 2

Write the function

Reﬂect over the x-axis

Apply distributive property

−f (x) = −((x − 1) + 2) = −(x − 1) − 2 The transformed function is g(x) = −(x − 1)2 − 2.

b Determine the vertex of the transformed function.

Create a strategy Compare the resulting equation to vertex form y = a(x − h)2 + k where (h, k) is the vertex.

Apply the idea The transformed function g(x) = −(x − 1)2 − 2 has h = 1 and k = −2. The vertex is (1, −2).

c Determine the y-intercept of the transformed function.

Create a strategy Substitute x = 0 into the transformed function.

Apply the idea g(x) = −(x − 1)2 – 2 2

g(0) = −(0 − 1) − 2 2

Write the transformed function Substitute x = 0

= −(−1) − 2

Evaluate the expression inside the brackets

= −1 − 2

Evaluate the power

= −3

Evaluate

The y-intercept is at (0, −3).

Reﬂect and check The transformed function is not in the general form g(x) = ax2 + bx + c, so −2 in g(x) = −(x − 1)2 − 2 cannot be considered as the y-intercept. For this reason, the y-intercept has to be calculated.

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d Determine the range of the transformed function.

Create a strategy Check if the parabola opens upwards or downwards. Then, use the y-value of the vertex as the starting or ending value for the range.

Apply the idea Since a = −1 < 0, the parabola is concave down, with a maximum at the vertex y = −2. The range is (−∞, −2].

e Sketch the graph of the transformed function.

Create a strategy Plot the vertex and the y-intercept. Since the parabola is concave down, sketch a smooth curve through these points that opens downwards.

Apply the idea y 1 x −1

−1 −2

(1, −2)

The graph of g(x) = −(x − 1)2 − 2 is a parabola opening downwards, with vertex at (1, −2) and y-intercept at (0, −3).

−3 (0, −3)

Idea summary A reflection of a quadratic over the x-axis ( g(x) = −f (x)) flips it vertically, altering the vertex and range. Reflection over the y-axis ( g(x) = f (−x)) flips it horizontally, and the domain remains the same. Sketching shows the flipped parabola.

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Dilations of cubic functions Dilation A process of stretching or compressing the graph of a function. This could happen either in the x or y direction or both. A dilation stretches or compresses a cubic function. For a cubic f (x) = x3, a vertical dilation is:

g(x) = lf (x) l is the dilation factor: stretches if ∣l∣ > 1, compresses if 0 < ∣l∣ < 1 The stationary point at (0, 0) remains at the origin. The domain and range are (−∞, ∞).

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Example 3 Dilate the function f (x) = x3 vertically by a factor of : a Determine the equation of the transformed function.

Create a strategy Apply the dilation g(x) = ax3 with a =

to ﬁnd the new equation.

Apply the idea Write the function The transformed function is g(x) =

Apply vertical dilation by factor x3.

b Determine the stationary point.

Create a strategy Identify the stationary point, which remains at the origin for vertical dilations of a cubic function.

Apply the idea The stationary point remains at (0, 0).

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c Sketch the graph of the transformed function.

Create a strategy Plot the stationary point and key points, connect with a smooth curve, and compare to the original function f (x) = x3.

Apply the idea For x = 2: Write the function Substitute x = 2 Evaluate For x = −2: Write the function Substitute x = −2 Evaluate y 6 4

g( x ) = −2

(2, 4)

1 32 x (0, 0) 2

−1

x 2

are connected with a smooth curve.

−2

(−2, −4)

x3 is less steep than y = x3. Points

The graph of y =

−4 −6

Reﬂect and check The graph of g(x) = vertically.

x3 is less steep than f (x) = x3 , as the dilation factor

compresses the curve

y 6

f (x) = x3

g( x ) = −2

1 3 x 2

−1

−2

(−2, −4)

(2, 4) (0, 0) 1

x 2

−4 −6

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Idea summary A vertical dilation of a cubic function f (x) = x3 by factor l results in g(x) = lx3, stretching or compressing the graph. The stationary point stays at (0, 0). Sketching shows the adjusted steepness.

Combined transformations Transformation 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, enlargements are all examples of transformations. Combining translation, reflection, and dilation transforms linear, quadratic, or cubic functions. The order of transformation matters, typically applied as dilation/reflection first, then translations.

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Example 4 Transform f (x) = x2 by reﬂecting over the x-axis, dilating vertically by 2, and translating right by 1 unit and up by 3 units: a Determine the equation of the transformed function.

Create a strategy Apply the transformations in the correct order: 1. Reﬂection 2. Dilation (determined by l) 3. Translations (determined by a and b)

Apply the idea f (x) = x2

Write the function 2

−f (x) = −x

Reﬂect over the x-axis 2

−2f (x) = −2x

Dilate vertically by a factor of 2 2

−2f (x − 1) = −2(x − 1)

Translate right by 1

−2f (x − 1) + 3 = −2(x − 1) + 3 2

The transformed function is g(x) = −2(x − 1) + 3.

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Translate up by 3

b Determine the vertex of the transformed function.

Create a strategy

Apply the idea

Compare the transformed function to vertex form y = a(x − h)2 + k where (h, k) is the vertex.

The equation g(x) = −2(x − 1)2 + 3 has h = 1 and k = 3. The vertex is (1, 3).

c Determine the range of the transformed function.

Create a strategy

Apply the idea

Check if the parabola opens upwards or downwards. Then, use the y-value of the vertex as the starting or ending value for the range.

Since l = −2 < 0, the parabola is concave down, with a maximum at y = 3. The range is (−∞, 3].

d Sketch the graph of the transformed function.

Create a strategy Set x = 0 to determine the y-intercept, then plot it and the vertex. Draw the parabola with a smooth curve, noting its downward concavity.

Apply the idea Set x = 0 to determine the y-intercept: g(x) = −2(x − 1)2 + 3

Write the function

Substitute x = 0

= −2 × (− 1) + 3

Evaluate the expression inside the brackets

= −2 × 1 + 3

Evaluate the power

Simplify

g(0) = −2(0 − 1) + 3

The y-intercept is at (0, 1). y 3

(1, 3)

2 1 −1

(0, 1) x 1

−1 −2 −3

The graph of g(x) = −2(x − 1)2 + 3 is a parabola opening downwards, with vertex (1, 3) and y-intercept at (0, 1). Points are connected with a smooth curve.

g(x) = −2(x − 1)2 + 3

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Reﬂect and check The original function f (x) = x2 has a vertex at (0, 0) and is concave up. The transformed function g(x) = −2(x − 1)2 + 3 is reﬂected over the x-axis, stretched vertically by a factor of 2, and shifted right by 1 unit and up by 3 units, resulting in a concave down parabola with vertex at (1, 3). f (x) = x2

(1, 3)

2 1 (0, 1) −1

−1 −2 −3

g(x) = −2(x − 1)2 + 3

Idea summary Combining translation, reﬂection, and dilation transforms functions. The order of transformations affects the result, impacting the vertex and range. Sketching visualises the combined effects.

16.01 Practice questions What do you remember? 1

Which is the correct vertex form for a quadratic function that has been transformed by a vertical dilation by a factor of l and a translation of (a, b)? A

y = ax2 + bx + c

y = l(x − a)2 + b

y = l(x − a)(x − b)

Describe the effect of the transformations on y = f (x): a

808

y = f (x + 2)

y = −2f (x)

y = f (x) − 3

Identify the effect of the transformations on the stationary point of a cubic function y = x3: a

Vertical dilation by a factor of l

Horizontal translation by a units

Identify the domain and range for these functions formed after any transformation: a

Linear function, y = mx + c

Quadratic function, y = a(x − h)2 + k with a < 0

Mathspace New South Wales – Year 11 Advanced mathspace.co

Practice Ex 1

Apply these transformations to the linear function f (x) = 2x + 1. For each transformation: i

Determine the equation of the transformed function.

Determine the y-intercept.

iii

Sketch the graph of the transformed function.

Translate right by 3 units

Translate up by 2 units

Translate left by 1 unit

Translate down by 4 units

Apply transformations to f (x) = x3, ﬁnd the equation, and sketch the graph, labelling the stationary point: a

Ex 3

Ex 4

Dilate by , translate down by 3 units

Determine the stationary point for cubic functions formed after transformation and sketch the graph, labelling the stationary point: a

Ex 2

Translate right by 2 units, up by 1 unit

y = 3(x + 2)3 − 1

x3 + 5

Describe in words the transformations from f (x) = x2 to g(x) = 2(x + 1)2 − 3 and sketch the graph, labelling the vertex.

Reﬂect the function f (x) = (x + 1)2 + 4 over the x-axis:

Determine the equation of the transformed function.

Determine the vertex.

Determine the range.

Sketch the graph of the transformed function, labelling the vertex.

Dilate the function f (x) = x3 vertically by a factor of : a

Determine the equation of the transformed function.

Determine the stationary point.

Sketch the graph of the transformed function.

Transform f (x) = x2 by reﬂecting over the x-axis, dilating vertically by , and 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 graph of the transformed function, labelling the vertex.

Determine the equation of the quadratic in vertex form with parameters and sketch the graph, labelling the vertex: a

a = 2, h = 3, k = −1

a = −1, h = −2, k = 5

a = , h = 0, k = −2

a = −3, h = 2, k = −4

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Describe the direction of translation from y = x2 and sketch the graph: a

y = x2 + 3

y = (x + 4)2

y = x2 − 2

y = 2(x + 1)2 − 3

x2 + 4

Describe the shape of each quadratic function compared to y = x2 in terms of concavity and width, and sketch the graph, labelling the vertex: a

Find the parameters a, h, k for quadratics in vertex form: a

y = (x − 5)2

y = 3x2

Given y = 2(x − 1)2 + 3, describe the transformations to obtain y = −2(x + 2)2 and sketch the graph.

Extend your thinking 17

810

A quadratic function y = x2 is transformed to have a vertex at (2, −4) and pass through (0, 4): a

Determine the equation of the transformed quadratic function.

Sketch the graph, labelling the vertex.

Determine the range.

A linear function f (x) = 3x − 2 is translated to pass through (1, 5) while maintaining the same slope: a

Determine the equation and sketch the graph, labelling the y-intercept.

Describe the transformation.

A cubic function f (x) = x3 is transformed by a vertical dilation by a factor of 2. It is then translated so that its stationary point of inﬂection is at (−3, 4): a

Determine the equation of the cubic function.

Sketch the graph, labelling the stationary point.

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16.02 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.

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16.02 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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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.

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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 x

The transformed function is g(x) = −2 .

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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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.

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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.

16.02 Exponential and logarithmic functions mathspace.co

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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

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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 The transformed function is g(x) = 3 × 2x.

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Apply dilation

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.

16.02 Exponential and logarithmic functions mathspace.co

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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.

16.02 Exponential and logarithmic functions mathspace.co

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16.02 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

822

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 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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16.03 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.

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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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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

16.03 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

Solving for 1 = x + 1: 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:

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.

16.03 Reciprocal and absolute value functions mathspace.co

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16.03 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

830

by translating right by 3 units and up by 4 units, then reﬂecting over the

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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

16.04 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

832

is the radius of the circle, where r > 0

Mathspace New South Wales – Year 11 Advanced 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

mathspace.co

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.

16.04 Circles and translations mathspace.co

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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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16.04 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

16.04 Circles and translations mathspace.co

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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:

838

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 mathspace.co

16.05 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.

16.05 Order of transformations mathspace.co

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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 2

2f (x) = 2x

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 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).

16.05 Order of transformations mathspace.co

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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 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

16.05 Order of transformations mathspace.co

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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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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.

16.05 Order of transformations mathspace.co

845

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

−1

x=0

x 1

−1 −2

y = −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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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.

16.05 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. 16.05 Order of transformations mathspace.co

847

Ex 2

848

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

Mathspace New South Wales – Year 11 Advanced mathspace.co

and translating right by 1 unit.

Translation then dilation

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).

16.05 Order of transformations mathspace.co

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16.06 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) =

850

+ 1.

Mathspace New South Wales – Year 11 Advanced 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.

16.06 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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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.

854

Mathspace New South Wales – Year 11 Advanced 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.

16.06 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.

16.06 Multiple transformations mathspace.co

855

856

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 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!

16.06 Multiple transformations mathspace.co

857

16 Chapter review 1

Which of the following is the vertex of the parabola y = −3(x + 2)2 − 4? A

(−2, −4)

(−2, 4)

(2, 4)

y=1

What is the horizontal asymptote of the function y = 5x − 1 − 3? A

(2, −4)

y=5

y = −1

y = −3

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

Apply a translation to the linear function f (x) = 3x − 2 by shifting it up by 5 units: a

Determine the equation of the transformed function.

Determine the new y-intercept.

Determine the vertex of the quadratic function y = 2(x − 1)2 + 1 and sketch its graph.

Reﬂect the function f (x) = (x + 2)2 + 3 over the x-axis: a

Determine the equation of the transformed function.

Determine the vertex.

Determine the range.

Transform f (x) = x3 by dilating vertically by a factor of 2, translating right by 1 unit, and up by 4 units. Find the equation and the stationary point.

Find the x-intercepts and the y-intercept for the quadratic function y = 16 − (x − 2)2.

Describe in words the transformations that have been applied to f (x) = x2 to obtain the graph of g(x) = 3(x − 1)2 − 4.

A quadratic function has its vertex at (−1, 5) and passes through the point (1, −3): a

Find the equation of the function.

Determine the range of the function.

The linear function f (x) = −x + 3 is translated so that it passes through the point (2, 4) while maintaining the same slope. Determine the transformation that was applied.

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:

858

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.

Mathspace New South Wales – Year 11 Advanced mathspace.co

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) =

− 2.

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.

Chapter 16 review mathspace.co

859

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.

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?

Advanced transformations are vital in fields like computer graphics and animation! By combining multiple transformations — such as rotations, reflections, translations, and dilations — designers can create lifelike 3D models and realistic motion on screen. These mathematical techniques allow digital artists to manipulate shapes with precision, ensuring objects move smoothly and appear natural from every angle. Transformations make it possible to visualise, simulate, and enhance complex designs in technology, engineering, and even virtual reality!

860

Mathspace New South Wales – Year 11 Advanced mathspace.co

“The essence of mathematics is not to make simple things complicated, but to make complicated things simple.” Stan Gudder

Answers

13 a x = −3, y = −4

b a = 11, b = 6

c n = 5, m = −3

d y = −5, z = 3

e x = −8, y = −3

f p = 6, q = 4

g x = −6, y = −3

h x = −3, y = 5

1.01 Systems of linear equations What do you remember? 1 a Subtraction c Addition

14 First, solve the system for x and y:

b Division d Multiplication

2 a Subtraction property b Multiplication property 3 Substitute x = 3: 2 × 3 − 1 = 6 − 1 = 5, which equals 5, so it is a solution. 4 The substitution method involves isolating one variable in one equation and substituting the resulting expression into the other equation. This creates an equation with a single variable that can be solved. 5 The elimination method involves adding or subtracting two equations to eliminate one variable, allowing the other to be solved.

x + 11 = 3x + 19

et the two expressions for S y equal to each other

11 – 19 = 3x – x

Group the x terms and constant terms

−8 = 2x

Simplify both sides

−4 = x

Divide by 2

Now substitute x = −4 into the ﬁrst equation:

b x=3

c x = 18

d x=3

7 a x=3

b x=4

c x=8

8 a x=2

b x=3

c x=3

d x = −1

9 a x=7 c No solution 10 a x = −5 c x=1

y=7

Solve for y

Next, verify the solution in both equations. Equation 1 : y = x + 11 7 = (−4) + 11 =7

Substitute x = −4 and y = 7 The solution is correct

Equation 2 : y = 3x + 19 7 = 3(−4) + 19

Substitute x = −4 and y = 7

= −12 + 19

Simplify the right-hand side

The solution is correct

15 a x = 6, y = 7

b a = −9, b = 6

c j = 2, k = 8

d d = 8, e = 4

e s = 9, t = 6

f p = 0, q = 5

d x=1

i w = −3, x = 6

j x = −5, y = −3

b d x = −7

11 a x, as it has a coefficient of 1. b Multiply the second equation by 5 and add it to the first equation. c The point of intersection of the graphs of the two linear equations. 12 a Multiply the first equation by 5 and the second equation by 7. b 15x + 35y = −120 28x − 35y = 77 c x = −1, y = −3

862

Substitute the value of x

The solution is (−4, 7).

Practice 6 a x=4

y = (−4) + 11

Mathspace New South Wales – Year 11 Advanced mathspace.co

16 The magazine costs $23 and the novel costs $36. Extend your thinking 17 2 hours 18 19 a Equation: x + (3x − 200) = 1200, Solution: x = 350 b Second investment = 3(350) − 200 = 850 dollars

20 Equation:

Practice y = 2x + 2

7 a x = 7, x = −2

x+y=5 x+y=5

Write equation 2

x + (2x + 2) = 5 Substitute the expression

for y from equation 1 into 2

3x + 2 = 5

b f = −11, f = 5

c h = −8, h = −11

d x = 10

e a = −7, a = −8

f x = 9, x = −8

8 a

Simplify the left side

3x = 3

Subtract 2 from both sides

x=1

Divide by 3 to solve for x

Substitute x = 1 into equation 2 to solve for y: x+y=5

Write equation 2

1+y=5

Substitute the value of x

y=4

ubtract 1 from both sides to S solve for y

The solution is (1, 4).

9 a x = 0, x = −9 c x = 0, x = 9

d y = 2, y = −3

e a = 3, a = 5

f x = 0, x = 2, x = 5

10 a m = 5, m = −2

b x = 2, x = 10 d g = 2, g = 6 f x = 3, x = −6

21 a

a + o = 10

c y = −2, y = −4

2a + 1.5o = 17.5

e x = 6, x = −7

b 5 apples, 5 oranges 22 a C = 500 + 10x, R = 20x b x = 50 units; costs equal revenue at $1000. 23 a One solution: x = 3.4, y = −1.2

11 a

b No solution 12 a x = 4, x = 6

c Infinitely many solutions d Infinitely many solutions 24 a Multiply the first equation by 10 and the second by 100: 20x − 18y = 56 65x + 70y = −75 b (1, −2)

1.02 Solve quadratic equations

f x = 0.39, x = −3.28

13 a

e x = 1, x = 6

14 a x = 9, x = −9

What do you remember? 1 a x = 0, x = 7

b k = −4, k = 2 d m = −6, m = 6

c 2 a a = 2, b = −5, c = 3

b a = 1, b = −7, c = 4

c a = −3, b = 0, c = 9

d a = 5, b = 1, c = 0

c x = 5, x = −3

d x = 5, x = −9

f x = ±3

15 12 cm × 5 cm 16 5 cm × 12 cm

3 4 a 16

17 5.81 seconds b 36

b True

c True

d False

5 C 6 a True

Answers mathspace.co

863

Extend your thinking 18 a x = ±2, x = ±3

Practice b x = 0, x = 3

6 a Δ = 0, one real root (repeated) b Δ = 64, two real rational roots

c x = 1, x = −2

e x = 16

f No real solutions.

c Δ = −8, no real roots d Δ = 0, one real root (repeated)

19 10 metres

e Δ = 0, one real root

20 11

f Δ = 16, two real rational roots

21 The other root is the conjugate, x = 4 + The value of k is k = 11.

g Δ = −7, no real roots

7 a

22 x = 5 The side lengths are 5, 12, and 13.

23 Eddie made a mistake in the ﬁrst line when substituting into the quadratic formula. , not . The value of c is

9 Yes, it reaches the height at t ≈ 1.66 seconds and t ≈ 3.44 seconds. 10 Yes, it is possible at 21 km/h. 11 a Δ = 5, two real irrational roots b Δ = 0, one real root (repeated) 12 x = 5 (Factorising) (Quadratic formula)

13 The correct solutions are

and

Extend your thinking 14 m = ±8, ±10, ±17, …

24 6 m × 4 m

1.03 The discriminant

16 The other root is

What do you remember?

1 • Δ > 0 for two distinct real roots

18 a x2 + 2(p − q) x + p2 − q2 = 0

• Δ = 0 for one real root

b x = 3p

• Δ < 0 for no real roots

19 b = −(r + s), c = rs

2 a = 3, b = −6, c = 2; Δ = 12 3 a True

b True

c False

20 k < −4 d True

4 It indicates the number and nature of roots, guiding the choice of solution method. 5 Yes, it solves any quadratic (a ≠ 0), but solutions may be complex. No, not all have real solutions if the discriminant is less than 0.

864

and

Mathspace New South Wales – Year 11 Advanced mathspace.co

Chapter 1 review 1 C 2 C 3 C 4 a x = −3, y = −1

b n = 3, m = 6

c q = −3, p = 17

d x = −2, y = 5

5 a x = 4, y = 2 c p = 5, q = 3 6 a

b a = 4, b = 2 d b

c x = 11

d x=6

7 The mouse costs $300 and the laptop costs $950. 8 a

20 The discriminant is Δ = (−10)2 − 4(1)(25) = 0. This indicates one real root. The most eﬃcient method is to factorise the perfect square: (x − 5)2 = 0, so x = 5. 21 The discriminant is Δ = (2)2 − 4(3)(−4) = 52. Since Δ > 0 and is not a perfect square, there are two real, irrational roots. The quadratic formula is the most eﬃcient method.

c + t = 12

22 m < 4, m ≠ 0

b 7 coffees, 5 teas 9 a C = 800 + 15x, R = 35x b x = 40 units; the company breaks even when it sells 40 items, with both costs and revenue equal to $1400. 10 a x = 7, x = −4 c p = −6, p = −8

Solution:

4.5c + 3.5t = 49

b a = 3, a = −7 d m=8

11 a

23 24 a Using Pythagoras’ theorem: x2 + (x + a)2 = (x + 2a)2. Expanding gives x2 + x2 + 2ax + a2 = x2 + 4ax + 4a2. Collecting terms on one side gives x2 − 2ax − 3a2 = 0. b Factorising gives (x − 3a)(x + a) = 0. The solutions are x = 3a and x = −a. Since side length x must be positive, x = 3a.

2.01 Functions and relations 12 a x = 7, x = −5

b a = 6, a = −4

c p = 3, p = −5

d x = 6, x = −2

What do you remember? 1 Relation

13 a

14 a

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

15 x = 9. The side lengths are 9, 12, and 15. 16 Approximately 7.42 metres 17 a x = ±1, x = ±3 c x = 4, x = 9

b x = 7, x = −1 d

18 a Δ = 0, one real rational root (repeated) b Δ = 81, two real rational roots c Δ = −11, no real roots d Δ = 37, two real irrational roots 19

b False, because a function assigns at most one y-value per x-value. c False, only functions pass the vertical line test. d True 3 a Correct

b Correct

c Incorrect

d Incorrect

Practice 4 No, because x = 4 corresponds to y = 14 and y = 11. 5 a Both relation and function b Relation

Answers mathspace.co

865

6 a Function

b Relation

c Function

d Function

7 a y

12 a Total GB Total charge used (dollars) b Yes

4 2 x −4

−2

−2 −4

13 a

b Yes

8 a

x −4 −3 −2 −1 y

−1 −1 −1 −1 −1

2 1

b Yes

2 3 4

−2

9 a

−3

−4

8 6 4 2 −8 −6 −4 −2 −2 −4 −6 −8

No, because a vertical line can be drawn that intersects the graph at more than one point.

2 4 6 8

9 8 7 6 5 4 3 2 1

b (−5, −10) or (−5, −4) 10 a Call length International call (minutes) cost (dollars) 1

1.80

2.90

4.00

5.10

6.20

11 a Number of Total cost t-shirts (dollars)

866

−4 −3 −2 −1 −1

b Yes −4 −3 −2 −1

x 1

2 3 4

Yes, because no vertical line intersects the graph more than once. Extend your thinking 14 a

b Yes

Mathspace New South Wales – Year 11 Advanced mathspace.co

y (Machine 1)

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.

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

b 2.67

14 a c 15 a t = −3, 3

c Change (1, 1) to (1, 2) and (2, 4) to (2, 3):

c 80.99

d 2187.00

b 12 d b t = −2, 2 d

{(−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.

16 a x = −3, 2 c x = −2, 1

b x = −4, 3 d

17 −9

2.02 Variables and substitution

18 −49 What do you remember?

19 −46

1 The independent variable is x, which is the input. The dependent variable is y, which is the output and depends on x. 2 Independent variable: t. Dependent variable: f (t).

20 a b −1 21 32 22 a 18

3 Independent variable: t (time in hours). Dependent variable: d (distance in km). Practice

Extend your thinking 2

4 a g(x) = x − 10

b g(−3)

23 f (7) = −6

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 It represents the total cost of buying 6 bottles.

b 966

c −3.21

24 6 25 2a + h + 5 26 27 x = 13 or x = 127 28 T (x) = 4x2 − 394x + 26 349

d 46

Answers mathspace.co

867

2.03 Characteristics of functions

What do you remember? 1 a True

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}

b {2, 3, 5}

c Yes, it is a function.

−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

b All real numbers.

4 a Domain: {1, 2, 3, 4}; Range: {5, 6}; Function: Yes b Domain: {2, 3, 4}; Range: {1, 2}; Function: No 5 a Domain: All real x; Range: All real y b Domain: All real x; Range: y ≥ 2

c x=0

Chapter 2 review 1 C 2 A

c Domain: All real x; Range: y ≥ −2

3 C

d Domain: All real x; Range: y = 3

4 No, because x = 5 maps to both y = 12 and y = 9.

6 a x = −3, 3

b −9

7 a x=3

b x = −3, 3 d x = −4

c No real zeroes

b No x-intercept

8 a 1

d y>0

c All real x b y ≤ 16

9 a 16

c All real x

10 a i Function

5 a Function

b Relation

c Function

d Relation

6 a

10 8 6 4 2 −8 −6 −4 −2 −2 −4 −6 −8 −10

ii Domain: −7 ≤ x ≤ 7 Range: 0 ≤ y ≤ 7 b i Not a function ii Domain: −9 ≤ x ≤ 9 Range: −9 ≤ y ≤ 9

x 2 4 6 8

b Either (−6, −11) or (−6, −5)

11 a x ≥ −1

b y≥0

c x = −1

12 a 5

b All real x

c y≥5

−2

−1

−9

−2

−1

Range: All real y

868

−5

19 a All real x except x = −2 and x = 2

Practice

Mathspace New South Wales – Year 11 Advanced 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)}.

3.01 Characteristics of linear graphs

8 a −4 b 68

What do you remember?

c −5.92 d 2x2 + 4xa + 2a2 + 5x + 5a − 7 9 a

1 a True d −3

c 12

10 a 7

b 4

c False

d True

2 a Gradient-intercept form b General form c Gradient-intercept form

c 2

b (0, −6)

3 a 3

d 11 a

b False

c 3x − y − 6 = 0 x

Gradient

y-intercept

b Yes, because each number of muﬃns x maps to exactly one cost y. 12 b2 + 2bk + k2 + 3b + 3k

5 There is a gradient, m = 1.

13 14 a Domain: {x ∈  : x ≠ 3}; Range: {y ∈  : y ≠ 0} b Domain: ; Range: {y ∈  : y ≥ −4} c Domain: {x ∈  : x ≥ −5}; Range: {y ∈  : y ≥ 0} d Domain: ; Range:  15 a

Viewing hours (H)

Loyalty points (P )

22.5

10 20

Practice b y = −2x − 5

6 a c

7 a y = 2x − 7

b y = −x + 4

8 a

d y = −x − 3

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 b Domain: {1, 5, 7}; Range: {2, 4}; Function: No, because x = 1 maps to both y = 2 and y = 4.

9 a

3 2 1

17 a x = −5, 5

b −25

18 a x = 3

b x = −7, 7

−2

c No real zeroes

d x = −5

−3

19 a No, because l > 0.

b P (l) > 6

−4 −3 −2 −1 −1

x 1

2 3 4

−4

Answers mathspace.co

869

b 2

−3 −2 −1 −1

1 2 3 4 5 6

−4 −3 −2 −1 −1

−2 −3

2 3 4

−2

−4

−3

−5

−4

10 a

3 2 1 −4 −3 −2 −1 −1

x 1

2 3 4

−2

11 a 3x − y − 7 = 0

−3

c x − 2y − 6 = 0

d x+y−1=0

−4

e 3x + 5y − 10 = 0

f 2x − 3y − 12 = 0

12 a i x-intercept: (2, 0), y-intercept: (0, 5)

2 1 −1 −1

b 2x + y − 4 = 0

1 2 3 4 5 6 7 8 9 −1

−2

e 4

−4 −3−2 −1 −1

b i x-intercept: (3, 0), y-intercept: (0, 2) y

2 1

−1

x 1

3 x 1 2 3 4 5 6 7 8 9

2 1

−2 −3 −4

870

Mathspace New South Wales – Year 11 Advanced mathspace.co

x −1

1 −1

c i x-intercept: (3, 0), y-intercept: (0, −4) ii

Extend your thinking 13 a y = 3x + 5

y 1

−1

b ( 0, 5); the ﬁxed charge of $5 when no time is parked. 14 a The y-intercept is (0, 5), meaning the ﬁxed fee is $5 when 0 minutes are used.

−2 −3

b The x-intercept is (−2.5, 0). This is not meaningful because the number of minutes x cannot be negative in a phone plan.

−4 −5

c ii

y 12

d i x-intercept: (8, 0), y-intercept: (0, 2)

y = 2x + 5

(1, 7)

(0, 5)

x −1

1 2 3 4 5 6 7 8 9

−1

15 a 5x + 3y − 15 = 0

e i x-intercept: (7, 0), y-intercept: (0, 2) ii

y 3

b The x-intercept is (3, 0), meaning 3 muﬃns and 0 cookies weigh 15 kg. The y-intercept is (0, 5), meaning 0 muﬃns and 5 cookies weigh 15 kg. c

2 1

x −1

2 3 4 5 6 7

5x + 3y = 15

−1

2 1

f i x-intercept: (1, 0), y-intercept: (0, −5) ii 1 −1

y 5 (0, 5)

−1 −2 −3

(3, 0) x 0

y x 1

16 a , y-intercept: (0, −1)

b x-intercept:

. Substituting into

c Midpoint:

−4 −5

x − 1:

⋅1–1=

which is true.

Answers mathspace.co

871

3.02 Equations of lines

y 2

What do you remember?

1 a

x 1

b c = y-intercept −1

2 a By setting y = 0 and solving the equation for x. b By setting x = 0 and solving the equation for y. 3 a Perpendicular

−2

b Parallel

c Neither

d Neither

4 a i m1 = m2

ii m1 × m2 = −1

b m1 = m2 and c1 = c2 5 a True

b True

c False

−1

6 Answers may vary: • Railway tracks are an example of parallel lines.

• The corner of a piece of paper shows two perpendicular edges.

1 x

Practice

−1

7 a y = 2x + 1

b y = −x + 7

8 a 8x − 6y + 7 = 0

−1 −2

b 2x + y + 2 = 0

c 2x − 4y − 7 = 0

d 9x + 15y − 50 = 0

e 3x − 3y + 10 = 0

f 5x − 2y + 7 = 0

10 a y = 2x + 2

11 a

9 a

y 4

1 x −1

−1

872

Mathspace New South Wales – Year 11 Advanced mathspace.co

12 a x − y + 2 = 0

b y−3=0

13 a x − 2y − 8 = 0

b 3x + 2y + 13 = 0

14 a (2, 0)

b y=x−2

15 a • For

: use rise/run from

4 a 3p + 9 < 24

y-intercept.

• For 4x + y − 12 = 0: Already in general form, ﬁnd intercepts.

• For y − 4 = 3(x + 2): Not a standard form; easiest is to convert to gradient-intercept form, then use rise/run.

c x+6≥7

e −4x ≤ 3 Practice b m < −11

c x < −5

d x < −26

e x>2

f x<6

6 a x ≥ 18

b x ≥ −2

c x ≤ −3

d k ≤ 72

e y≥2

f x ≥ −5.25

y 10

3 y = − x+5 8 2

y = −4x + 12

y − 4 = 3(x + 2) −2

b 5x + 3 ≥ 23

5 a x<5

−4

3 B

x 2

−2

7 a −20−18 −16 −14 −12−10 −8 −6 −4 −2 0 2

8 10 12 14 16 18 20

b −20−18 −16 −14 −12−10 −8 −6 −4 −2 0 2

Extend your thinking

c 16 a

−20−18 −16 −14 −12−10 −8 −6 −4 −2 0 2

b 2x + 4y − 3 = 0

17 a

d −5 −4

−3 −2

−1

c Yes, it lies on the path. Substituting into the general form gives 2×1+

− 3 = 2 + 1 – 3 = 0.

8 C 9 a i x>1 ii −4 −3 −2 −1 0 1 2 3 4 5

b 6x + 4y + 61 = 0

18 a c Gradients are

. The product is −1, so

and

they are perpendicular. b x + 2y − 13 = 0

19 a c The gradient of decreasing by

means the crop height is cm per day.

b i x>3 ii −1 0 1 2 3 4 5 6 7 8

c i x ≤ −4 ii −8 −7 −6 −5 −4 −3 −2 −1 0 1 2

d i x ≥ 7.5 ii −1 0 1 2 3 4 5 6 7 8 9 10

3.03 Linear inequalities

10 a 3p − 3 ≤ 24 What do you remember? 1 3 ≤ x ≤ 7 includes both end points, whereas 0 ≤ x < 4 only includes the left hand end point. 0 ≤ x < 4 includes the number 0 and every number up to 4 but not including 4. 3 ≤ x ≤ 7 includes the number 3 and every number up to 7 including 7.

b p≤9 c p=9 11 p + 21.14 ≤ 43, p ≤ 21.86 12 a x + 410 ≥ 550 b 0

100 200 300 400 500

2 D

Answers mathspace.co

873

Extend your thinking

6 a

13 a 2w + 22.10 ≥ 40

b y=3

b w ≥ 8.95 c i N o, washing exactly 9 windows would generate $18, which, with the $22.10 he already has, gives $40.10, enough to afford the equipment. ii No, washing 8 windows would generate $16, giving a total of $38.10, which is not enough to afford the equipment. iii No, he gets paid for washing whole windows, not parts. iv Yes, washing 9 windows gives $40.10, suﬃcient to afford the equipment. More windows would provide excess funds. 14 Step 2 is incorrect. Dividing both sides by −2 requires reversing the inequality symbol, so x < −7.

c d The amount by which the temperature is rising every minute. e The temperature of the room before the heater has been turned on. f 13°C g Example answer: The model is useful to track the temperature for a couple of hours, but eventually it can be assumed that the heater will be turned off, or a window or door will be opened, which will affect the temperature and the model will no longer be accurate. For instance after two hours the temperature according to the model will be 43°C, so the heater will most likely be turned off.

15 The minimum should be at −9 instead of −8, and the dot at that point should be ﬁlled as −9 is included in the solution.

7 a

3.04 Linear models

Cost (C) 18.00 20.50 23.00 25.50 30.50 43.00 in dollars

Time (t) in minutes

What do you remember? b 1 a (0, 6) 2 a

b (8, 0)

Number of columns (c)

Number of blue boxes (b)

b b = 2c − 1 3 a 85 beats/minute

8 a b 70 hours

5 a b y = 28 c The amount of water that is ﬂowing out of the hole every minute. d The capacity of the bucket. e f 18 L

874

t 2 4 6 8 10 12 14 16

c t≥0

Practice 4 a 135 mg

b 3 beats/minute

c H = 3t + 49

50 45 40 35 30 25 20 15 10 5

Mathspace New South Wales – Year 11 Advanced mathspace.co

Number of minutes passed, x

20 50

Amount of fuel left in 250 225 200 175 150 tank, y

b y = 250 − 5x c The amount of fuel in the car is decreasing at a constant rate of 5 litres per minute. d 0 ≤ x ≤ 50

c 8.4 metres 10 a

d $3360

b y = 1.4x

9 a 1.4 m/min

d 9 min

4500 4000 3500 3000 2500 2000 1500 1000 500

b $545 b 77.5° C

20 a x ≤ 4 b i Viable t 2

8 10 12 14

iii Viable

iii Non-viable

f 2200 ﬁsh g t = 14

ii Non-viable iv Viable

22 a f > 2m b i Non-viable

e F = −200t + 4800

ii Non-viable iv Viable

21 a b i Viable

d F = 4800

ii Viable iv Non-viable

23 a

11 a I = 25h

b $625

12 a C = 60h + 75

c 5 hours

b $375

c 4 hours b h ≤ 2 hours b 1.8°F d F = 1.8C + 32

c Below

c To get an overall grade of C Uther must score at least 86 on the last exam.

24 a G = −1.2t + 56

c Not viable, as hours cannot be negative. 14 a 50°F

b x ≥ 86

Extend your thinking

13 a 85.00 + 45.50h ≤ 220.00

e 149°F

b 21.67 s c Max amount of gas permitted at 17 seconds is 35.6 g. Therefore, Peter is exposed to more than this amount (40 g). d 12.8 g

15 a 12 Euros

b 10 AUD

c 0.60 Euros

e 279 Euros

b 120

18 a P = 55t + 160

c C = −2t + 305

c The rate of change of the ﬁsh population each year.

b $700

19 a 210 cups of coffee

b −200

4500 4000 3500 3000 2500 2000 1500 1000 500

17 a E = 10n + 340 c Company A

16 a

e 32 hours

Cost

25 In capsule form. 26 a 25 L

b 5L

27 a 40 years

b In the year 2067

28 a 100 customers

b 20 customers

c 9:00 p.m. 29 a 40 months c

d 5:00 p.m. b After 35 months

months

30 a 30c + 5k ≤ 300 Hours 5

10 15 20 25 30 35

c C = 120h

b k ≤ 24 c Non-viable, as 30 × 8 + 5 × 15 = 315 exceeds the budget of $300.

Answers mathspace.co

875

3.05 Simultaneous equations

b i

7 y 6 5 4 y = −4x + 5 3 2 1

What do you remember? 1 Simultaneous linear equations are two or more linear equations with the same variables, solved to find values satisfying all equations. They model real-world scenarios with multiple unknowns, like sales quantities or costs. 2 Define variables for unknowns (e.g., x for one item, y for another). Use total quantities for one equation and total revenue or cost for another, based on given prices or conditions. 3 The break-even point is where revenue equals cost, yielding no profit or loss. Write cost and revenue equations (e.g., C = fixed + variable × x, R = price × x), equate (C = R), solve for units x, and find cost/revenue.

−7−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6 −7

y = −4x − 4

ii No solution c i

4 Graphical: Plots lines to visualise intersection, intuitive but less precise. Substitution: Isolates one variable, effective for simple coefficients. Elimination: Aligns coefficients to eliminate a variable, efficient for aligned equations. All ensure viable solutions (e.g., integers) in contexts like sales.

6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

x 1 2 3 4 5 6

y=x−2

y = 2x − 1

ii (−1, −3)

5 B

d i

6 E 7 a (4, −6)

y = −x − 2

5 4 3 2 1

d (−4.5, −2)

Practice 8 a i

y = −2x + 1 6 y

b (−1, 3)

c (6, −3)

4 y 3 2 1

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10

x 1

2 3 4

y = 2x − 5

ii (1, −3)

876

1 2 3 4 5 6 7

Mathspace New South Wales – Year 11 Advanced mathspace.co

−6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

ii (1, −1)

x 1 2 3 4 5 6

y = −x

Baker purchases

9 a 7

Cost per kg of sugar (dollars)

6 5

12 a Let x = the number of parts Elvia completes and y = the number of parts Kang completes.

4 3 2

Cost per kg of flour (dollars)

1 0

b The solution (2.5, 1.75) means the cost per kilogram of ﬂour is $2.50 and per kilogram of sugar is $1.75, which is viable as both costs are positive. 10 a b

y+x=4

y − x = −2

6 y 5 4 y+x=4 3 2 1

0 1 2 3 4 5 6 7 8 9 10

c Kang did 3 parts of the assignment and Elvia did 6 parts. 13 a

5x + 3y = 14

2x + 6y = 13

x 4

1 2

Cost per orange (dollars)

4 3 2

Cost per apple (dollars)

11 a Let x = Mohamad’s age and y = Rochelle’s age

Shop purchases 5

c x = 3, y = 1

y = 2x − 9

10 Number of parts 9 Kang completes 8 x+y=9 7 6 5 y = 2x − 9 4 3 2 Number of parts 1 Elvia completes

1 2 3 −1 −2 y − x = −2 −3 −4 −5 −6

x+y=9

y + x = 21 y = 4x + 11

22 Rochelle’s age 20 18 y + x = 21 16 14 12 y = 4x + 11 10 8 6 4 2 Mohamad’s age 0 2 4 6 8 10 12 14 16 18 20 22

c Rochelle is 19 years old and Mohamad is 2 years old.

c The solution (1.88, 1.54) means the cost per apple is $1.88 and per orange is $1.54, which is viable as both costs are positive. 14 a

y = 200 + 75x

y = 650

Miji’s laptop savings

b 900 800 700 600 500 400 300 200 100 0

Amount of money

Time (months) 1

2 3 4 5 6 7 8

Answers mathspace.co

877

c The intersection at (6, 650) means the student saves $650 in 6 months, enough to afford the laptop on release. 15 a = $0.90 x + y = 2000

16 a

2 0.08x + 0.09y = 177

b x = $300, y = $1700 c Valentina’s choice will not yield the highest return. It would be better to invest all $2000 in Account B since it earns more interest than Account A. 17 a 2017 b Yes. It is possible that in 2017 people consumed over 12 billion gallons each of bottled water and soda. 18 (80, 70), meaning 80 students and 70 adults attended. 19 a

x + y = 137

x − y = 29

b Judy scored 83 on Geography. c Judy scored 54 on Maths. 20 a

12x + 5y = 70

3x + 25y = 65

b x = $5 c y = $2 21 a

C = 150 + 2.50x

R = 5x

b (60, 300); $300 for 60 cups c Selling 60 cups results in both total cost and total revenue of $300, no proﬁt or loss. 22 a

R+C=9 6.45R + 5.75C = 55.25

b Fred purchased 5 rhododendrons and 4 chrysanthemums. Extend your thinking 23 The point of intersection on the graph has negative x and y values. In terms of the height of a plant over time, only positive values make sense. Since the two lines do not intersect in the ﬁrst quadrant, they will never reach the same height on the same day.

878

Mathspace New South Wales – Year 11 Advanced mathspace.co

24 If you need an exact solution to the system and the solution contains irrational numbers or fractions not easily readable on the scale of your graph, then solving the system by graphing is not feasible. 25 Answers will vary. For example: Maria has 15 kg of rice and uses

kg each day.

Arjun has 12 kg of rice and uses 1 kg each day. The solution to the system (14, 5) means that after 14 days, Maria and Arjun both have 5 kg of rice remaining. 26 Let x = the cost of an order of oxtail stew and y = the cost of a slice of sweet potato pie. 1

2x + y = 38.49

2 5x + 3y = 99.22 Since oxtail stew costs $16.25 and sweet potato pie costs $5.99, an order of four oxtail stews and two slices of sweet potato pie costs $76.98. 27 a Let x = the cost per orange and let y = the cost per banana. 1

2x + 3y = 18.30

5x + 7y = 44.03

b One way to use the elimination method in finding the cost per orange and banana would be multiplying the first equation by −5 and the second equation by 2. That way, the x-terms would be eliminated after adding the equations and we would begin solving for the cost per banana. Another way to use the elimination method in finding the cost per orange and banana would be multiplying the first equation by 7 and the second equation by −3. That way, the y-terms would be eliminated after adding the equations and we would begin solving for the cost per orange. The system could be converted to gradient-intercept form for graphing as well, and the point of intersection would be where we find the cost per orange and the cost per banana. 28 a

x + y = 80 0.10x + 0.05y = 5.50

Omeida has 30 pieces of 10 cent and 50 pieces of 5 cent.

x + y = 80

5 4 3 2 1

0.05x + 0.20y = 5.50

The solution would be 70 pieces of 5 cent and 10 pieces of 20 cent. This changes the solution by increasing the number of lower-value coins (from 50 to 70) and decreasing the number of higher-value coins (from 30 ten-cent coins to 10 twenty-cent coins) to maintain the total coin count and value. The solution is still viable.

−3 −2 −1

(2, 0)

−1 −2 −3 −4 −5

y 3

Chapter 3 review

1 B (−3, 0)

2 D

−3 −2 −1

3 A 4 a y = −2x + 4

−1 −2

(0, −2)

−3

f y = −3x − 5

5 a

y 3 2 (0, 2)

4 − ,0 3

−3 −2 −1

3 2 1 −3 −2 −1

−1

f (−2, 0)

(0, 4) y 3 2 1

−2 −3

−3

−3 −2 −1

−1

−3 (0, −3)

−3 −2 −1

x 1

−2

(2, 0)

−2

(2, 0) x 1

3 2 1

y x

−1 −2 −3 −4 −5 −6 (0, −6) −7

6 a y = 5x + 10 b The y-intercept is (0, 10); this represents the ﬁxed charge of $10 when no classes are attended.

Answers mathspace.co

879

7 a The y-intercept is (0, 8); this represents the ﬁxed monthly fee of $8 when no movies are rented. . This is not b The x-intercept is meaningful because the number of movies rented x cannot be negative. c

y 20 18 16 14 12 (1, 11) 10 8 (0, 8) 6 4 2 0

b 56x + 42y − 159 = 0 c Ask your teacher for worked solutions. 14 a x ≤ 3

b x<4

e x≥4

f x>0

−2 −1 0

−6 −5 −4 −3 −2 −1 0

ii −2 −1 0

c i x < −2 x 3

d i x≤2 ii −2 −1 0

e i x<9 ii 0 1 2 3 4 5 6 7 8 9 10 11 12

f i

−2 −1 0

16 p + 18.50 ≤ 50, so p ≤ 31.50

17 a 5w + 15 ≥ 50

30 20

Number of(100, apples(x) 0)

0 10 20 30 40 50 60 70 80 90 100

9 a 4x − y − 5 = 0

b 2x + y − 3 = 0

c x−y+4=0

d 3x + y + 1 = 0

e x − 2y + 4 = 0

f 2x + 3y − 15 = 0

g 3x − 4y − 24 = 0

h 5x + 2y − 7 = 0 b y = 2x + 3

c y = −x + 1

11 y = 3x − 1 12 y = −3x + 4

Mathspace New South Wales – Year 11 Advanced mathspace.co

b w≥7

c The solution w ≥ 7 means Ben must wash at least 7 cars to afford the video game, as the number of cars must be a whole number. 18 Step 2 is incorrect. When dividing both sides of the inequality −3x > 18 by −3, a negative number, the inequality symbol must be reversed. The correct step is x < −6. 19 a b (0, 30) c The gradient

represents the rate at which

water is ﬂowing out of the tank, losing per minute.

litre

d The y-intercept (0, 30) represents the initial amount of water in the tank, which is 30 litres before the tap is opened. e

880

b i x≤2

Number of oranges(y) 80 (0, 80)

10 a y = −3x + 10

d x<3

b The x-intercept is (100, 0); this means 100 apples and no oranges weigh 20 kg. The y-intercept is (0, 80); this means no apples and 80 oranges weigh 20 kg.

c x≤7

15 a i x > 2

8 a 0.2x + 0.25y − 20 = 0

13 a

20 a −20 m/min

b y = −20x + 500

d i

y 5

c 360 m 21 a C = 75h + 90

4 y = −x + 4

b $352.50

c 5 hours 22 a 150 + 35.50p ≤ 1200

≈ 29.58, so the maximum number

b p ≤

−1 −1

of people is 29. 23 a E = 20n + 400

b $1040

24 a C = −2.5t + 80

b 12 hours

c The maximum safe amount for 18 hours is C = −2.5(18) + 80 = 35 mg. Since the scientist was exposed to 40 mg, they are at risk. 25 a i

x 3

ii 26 140 children and 60 adults attended. 27 a p + c = 150

28 2s + c = 25 (3, 0) 1

2 3 4 5 6 7

The cost of 4 sandwiches and 2 coffees $50.00. 29 a Let t be the number of 10c coins and w the number of 20c coins.

y=x−3

t + w = 50 0.10t + 0.20w = 7.50 7 6 5 4 3 2 1

There are 25 10c coins and 25 20c coins.

b Revised system: t + w = 50

(2, 4)

0.10t + 0.20w = 12.00 Using the second equation:

y = −x + 6

1 2 3 −1 −2 y = 3x − 2 −3

x 5

0.10t + 0.20w = 12.00 0.10t + 0.20(50 − t) = 12.00 0.10t + 10 − 0.20t = 12.00 −0.10t + 10 = 12.00 −0.10t = 2.00

ii (2, 4) 5 4 3

t = −20

Using the ﬁrst equation:

y = −2x + 5

t + w = 50 −20 + w = 50

2 1 −1

5s + 3c = 65

ii (3, 0)

c i

b The student scored 84 on the physics test and 66 on the chemistry test.

y = −2x + 6

−1 −1 −2 −3

−1

2 x+1 3

p − c = 18

7 6 5 4 3 2 1

b i

9 11 , 5 5

1 2 −1 y = x − 1 −2 −3

(2, 1) 3

x 4

w = 70 Therefore, there are −20 10c coins and 70 20c coins. However, this is impossible because the number of coins cannot be negative.

ii (2, 1)

Answers mathspace.co

881

4.01 Characteristics of quadratics

13 a For y = x2 − 2x − 3: Δ = 16, two rational roots; For y = x2 + 2x + 2: Δ = −4, no real roots

What do you remember?

1 a Up, minimum

b Down, maximum

2 a iii

c i

b ii

y = x2 − 2x − 3

d iv

3 a The value Δ = b − 4ac in a quadratic equation

y = x2 + 2x + 2

x −8

to ﬁnd roots

b The formula

−6

−4

−2

4 a a = 3, b = −5, c = 2

b a = 1, b = −2, c = 4

5 a iii

c i

b ii

−4

x = −1

x=1

6 For a quadratic equation ax + bx + c = 0 to have rational solutions, two conditions must be met: the coeﬃcients a, b, and c must be rational numbers, and the discriminant, Δ = b2 − 4ac, must be a perfect square.

14 k > 4 15 a No real roots (Δ = −7) b One rational root (Δ = 0) 16 a k = 16

Practice 7 a (0, 8)

b x=3

8 a x = 1, x = 5

b (0, 10)

9 a x = −2, x = 3

b (0, 6)

c x = 0.5

b k < 16

17 a Approximately 3.13 seconds. b Approximately 12.48 metres. c 12

(1.53, 12.48)

10 8

10 a Δ = 4

b Two distinct rational roots c Concave up, vertex at (2, −1) below the x-axis, crosses at two points due to Δ > 0.

4 2

11 a Δ = −24

(0, 1)

(3.13, 0) t 1

b No real roots c Domain: all real numbers. Range: [2, ∞). d

18 Zero x-intercepts. A concave up parabola with its vertex above the x-axis will never cross the x-axis.

5 (0, 5) 4

19 Function A: y = −(x − 1)(x − 5), Vertex A: (3, 4). Function B: y = 2(x − 1)(x − 5), Vertex B: (3, −8). Vertical distance is 4 − (−8) = 12 units.

3 2

(1, 2)

20 a Factored form: (2x + 3)(x − 1)

x=1 −1

12 a i 0

882

Extend your thinking

x 3

ii One rational root

Expanded form: 2x2 + x – 3 b 7 m2 c x > 1 ensures the width (x − 1) is positive.

b i −56

ii No real roots

21 50 units, $1000

c i 1

ii Two rational roots

22 a x2 − (4 + m)x + 3 = 0

Mathspace New South Wales – Year 11 Advanced mathspace.co

b m < −4 − 2

c Domain is .

or m > −4 + 2

Range is [−7, ∞).

y 6

y = x − 4x + 5

−1

−2

5 −2

x 1

−4

y = −8x + 2 −3

x=2

(0, 5)

−1

−6

(2, −7)

23 Δ = (k + 3)2 − 4(1)(−k) = k2 + 10k + 9 > 0. Factoring gives (k + 9)(k + 1) > 0. The solution is k < −9 or k > −1. 24 For two real solutions, the discriminant − 4π k = 16π − 4π k ≥ 0 gives k ≤ 4. Δ= Since coeﬃcients a = π and b = are irrational, no value of k produces two rational roots.

6 a y = (x − 2)2 + 3; Vertex: (2, 3) b y = 3(x + 2)2 − 17; Vertex: (−2, −17) 7 a a=1 b y = (x − 1)2 + 2 8 a y = −(x + 3)2 + 11 b Vertex: (−3, 11); Axis of symmetry: x = −3; Range: (−∞, 11] c

4.02 Completed square form 9 a

What do you remember? 1 a True

b False

c False

d True

b x=3

2 a (3, 5)

b Vertex: c

d [5, ∞)

c Concave up

; Concave up 2

−5 −4 −3 −2 −1

3 16 Practice 4 a y = (x + 2)2 − 1 b Vertex: (−2, −1); Axis of symmetry: x = −2

5 31 − ,− 2 2

c (0, 3)

x 1

−2 (0, −3) −4 −6 −8 −10 −12 −14 −16 −18

5 x=− 2

x = −2

10 Vertex: (1, 1); y-intercept: (0, 4) 4

11 a y = −(x + 2)2 + 5

(0, 3)

(−2, 5)

x −4 −3 −2 −1

(−2, −1) 5 a y = 3(x − 2)2 – 7 b Vertex: (2, −7); Axis of symmetry: x = 2

y 5

x = −2

1 (0, 1)

−5 −4 −3 −2 −1 −1

Answers mathspace.co

883

12 a y = 3(x + 1)2 + 4

b Axis of symmetry: x = −1; Range: [4, ∞)

x=1

13 y = (x + 5)2 − 12 2 (0, 2)

Extend your thinking 1

14 a A = −(x − 5)2 + 25; Vertex: (5, 25) 2

b Maximum area: 25 m ; Dimensions: 5 m by 5 m

3 ,0 3

1−

−1

15 k =

3 ,0 3

(1, −1)

16 a h = −5(t − 2)2 + 21; Vertex: (2, 21) b 21 m at t = 2 s

6 a y-intercept: (0, −2); Axis: x = −3 b y-intercept: (0, 5); Axis: x = 2

17 y = (x − 2) − 3

4.03 Graph parabolas

7 a

What do you remember? 1 a True

b Vertex: (2, 7); Range: (−∞, 7]

b True

c False

2 a c Concave down

d True

b (0, c)

(2, 7)

x=2

Practice 4 a (1, 0), (5, 0)

4 + 14 ,0 2

4 − 14 ,0 2 (0, −1)

b (0, 5) c Axis of symmetry: x = 3; Vertex: (3, −4) d Domain: all real numbers; Range: [−4, ∞) e

9 a Axis of symmetry: x =

y 5 (0, 5) 4 3 2 1 1

(5, 0) 3

c 5

4 (0, 4) 3

(3, −4)

3 2 x

5 a

−1

b (0, 2)

−2

c Axis of symmetry: x = 1; Vertex: (1, −1)

−3

d Domain: all real numbers; Range: [−1, ∞)

884

; Vertex:

b Concave up; Range:

x=3

(1, 0)

−1 −2 −3 −4

8 Vertex: (1, 1); No x-intercepts

Mathspace New South Wales – Year 11 Advanced mathspace.co

3 11 ,− 2 4

4.04 Equations of parabolas

10 b y-intercept: (0, −2); Axis: x = 2 c

(2, 2)

2 1

(2 − 2, 0) (2 + 2, 0) x −1

1 a True

b True

c False

2 a iii

b ii

c i

3 Equate coeﬃcients: a = p, b = q, c = r

4 a y = (x − 3)2 – 2

x=2 (0, −2)

b Domain: ; Range: [−2, ∞) c

(1, 2)

(1, 4)

x=3

3 (0, 3) 1

x=1 (3, 0)

−1

(−1, 0)

d False

Practice

−1 −2

What do you remember?

−2

(3, −2)

−1

5 a y = x2 − 2x − 8

12 a x-intercepts: (−2, 0), (4, 0); y-intercept: (0, −8) b Vertex: (1, −9); Axis: x = 1 c

y 2

(4, 0)

(−2, 0) −2

2 −2

x=1

−4

c a = 1, b = −4, c = 2 or a = 1, b = 2, c = −4 6 a y = (x − 2)(x − 4)

b y = x2 − 6x + 8

7 a y = x2 + 2

b [2, ∞) 2

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

−6 −8

b Domain: ; Range: [−9, ∞)

(0, −8) (1, −9)

b (−∞, 3]

11 a y = 2x – 2 b

y 1

Extend your thinking

x=0 (1, 0)

(−1, 0) −2 −1

13 a A = −x + 15x

−1

b Vertex: (7.5, 56.25) 2

Area: 56.25 m

−2

c 7.5 m ×7.5 m

(0, −2)

14 a Vertex: (2, 21); Maximum height: 21 m b t ≈ 4.05 s

Extend your thinking

15 a = 16 y = x2 − 4x

12 a y =

(x + 2)(x – 6)

b x=2

Answers mathspace.co

885

13 a y = −(x − 2)2 + 4

7 a (4, 5), (2, 1)

b (4, 0)

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

4.05 Solve quadratic systems

f (x) (2, 1)

−1

What do you remember? 1 a True

(4, 5)

b False

c False

d True

2 x2 + x − 2 = 0

x 4

8 a x = −1 and x = 2 b

3 B

5 4

g(x)

Practice

(2, 4)

4 a (1, 0), (6, 5) b

7 6 5 4 3 2 1

f (x)

2 1

−3 −2 −1 −1

(6, 5)

(1, 0) 1

−1 −2 −3 −4

(−1, −2) −2

g(x)

f (x)

x 1

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

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.

(0, 3), (3, 0) 6 a

11 (2, 1), (4, 5) Extend your thinking

b Δ = 5 > 0; There are two intersection points.

12 There are two possible values: m = 2 or m = −6. 13

886

Mathspace New South Wales – Year 11 Advanced 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

4.06 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 mathspace.co

887

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

4.07 Quadratic models

200 100

What do you remember? 1 a True

b True

c False

d 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

888

(20, 400)

Mathspace New South Wales – Year 11 Advanced mathspace.co

14 4 –

<x<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]

4.08 Cubic functions What do you remember? 1 a True

b False

c True

d False

2 If k > 0, the graph rises to the right and falls to the left. If k < 0, the graph falls to the right and rises to the left. 3 (a, 0), (b, 0), (c, 0)

Practice

b 8

4 a Domain: ; Range:  b

6 4

(2, 4)

(−3, 0) (−2, 0) 2

1 3 2 f ( x) = 2 x

−3

(0, 0) 1 −2

−1

−2

−1

−3

8 a x-intercepts: (−3, 0), (1, 0), (2, 0); y-intercept: (0, −6)

−4

(−3, 0) −4 −3 −2 −1

c Domain is . Range is . y 10 (0, 8) 8 6 4 2 1

9 a (0, 5)

6 f (x) = −x3 6

−1

(2, 0) x 2

f (x) = (x + 1)(x − 2)(x − 4)

(−1, 1)

−2 −4 −6 (0, −6) −8 −10 −12 −14

(4, 0) x

(2, 0)

−2 −1 −2 −4 −6 −8

y 8 6 4 2 (1, 0)

f (x) = −(x + 3)(x − 1)(x − 2)

b The graph rises to the right and falls to the left.

(−1, 0)

−8

5 a x-intercepts: (−1, 0), (2, 0), (4, 0); y-intercept: (0, 8)

−2

−6 (0, −6)

−2

(−2, −4)

(1, 0) x 1

−4

−1

2 −2

b x-intercepts: (0, 0), (5, 0), (6, 0); y-intercept: (0, 0). However, within the practical domain of (0, 5), there are no intercepts. The points (0, 0) and (5, 0) are the boundary points where the volume is zero. c

(0, 0)

x 1 (1, −1)

−4 −6

7 a x-intercepts: (−3, 0), (−2, 0), (1, 0); y-intercept: (0, −6)

V(x) = 4x(6 − x)(5 − x) 100 V 90 80 (1, 80) 70 (3, 72) 60 50 40 30 20 10 (0, 0) (5, 0) x 0

Answers mathspace.co

889

10 a

Chapter 4 review

1.4 m 1.3 1.2 1.1 1 0.9 0.8 0.7 m(x) = 0.002x3 0.6 0.5 0.4 0.3 0.2 0.1 0

1 B 2 A 3 B 4 a i (−1, 0), (5, 0) iii x = 2 b i (3, 0), (−2, 0) x

c i (1, 0), (4, 0) d i (−3, 0), (2, 0)

Extend your thinking

iii x = −0.5

11 a f (x) = k(x + 1)(x − 2)(x − 4). Substituting (0, 8) gives k = 1, so f (x) = (x + 1)(x − 2)(x − 4).

5 a i (0, 6)

ii (0, 6) ii x = 2 ii x = −2

iii (−2, −9) c i (0, 9)

10 (0, 8)

(4, 0) x

(2, 0) 1

ii x = 1

iii (1, 12)

5 −2 −1

ii (0, 8)

iii (2, −2) b i (0, −5)

y 15

(−1, 0)

ii (0, −6)

iii x = 0.5 iii x = 2.5

b x ≈ 15.87 cm

ii (0, 5)

−5

d i (0, −3)

ii x = 1

iii (1, −4) 6 a Factored: (3x + 2)(x − 2)

−10

f (x) = (x + 1)(x − 2)(x − 4) −15

Expanded: 3x2 − 4x − 4 b 28 m2 c To ensure the width (x − 2) is a positive length, x must be greater than 2.

12 The deflection is positive for 0 < x < 4 and x > 6. 13 The company makes a profit when selling between 0 and 500 items (0 < x < 5) or more than 1000 items (x > 10). 14 The population increases for the first two years (0 < t < 2) and after eight years (t > 8). 15

c The parabola is concave up with its vertex at (4, −4), which is below the x-axis. Since the discriminant is positive, it crosses the x-axis at two distinct points. b k > 25

9 a x − (6 + m)x + 9 = 0 b m < −12 or m > 0 10 a y = (x + 3)2 − 4 b Vertex: (−3, −4); Axis of symmetry: x = −3 c (0, 5)

r 8

The maximum volume is approximately 465 m3 when r ≈ 6.67 m.

890

b Two distinct real roots

8 a k = 25

500 V 450 2 400 V ( r ) = π r (10 − r ) 350 300 250 200 150 100 50 0

7 a Δ = 16

Mathspace New South Wales – Year 11 Advanced mathspace.co

(−5, 0)

(−1, 0)

y 5 (0, 5) 4 3 2 1 x

−5 −4 −3 −2 −1

−1 −2 −3 (−3, −4) −4

15 a x-intercepts: (−1, 0), (5, 0); y-intercept: (0, −5) b Vertex: (2, −9); Axis of symmetry: x = 2 c

y 4 2

(5, 0) x

(−1, 0)

−1

−2 −4

−8

17 a y = 2(x + 1)(x − 3)

b y = 2x2 − 4x – 6

18 a y = 2x2 − 1

b Range: [−1, ∞)

19 a = 3, b = 13 x 2

20 a (1, 3), (5, 7) b

(5, 7)

7 (3, −7)

12 a y = 2(x − 2) − 5

b Axis of symmetry: x = 2; Range: [−5, ∞)

13 a A = −(x − 10)2 + 100; Vertex: (10, 100)

(1, 3)

b Maximum area: 100 m ; Dimensions: 10 m by 10 m

x 0

14 a (−1, 0), (5, 0) b (0, −5) c Axis of symmetry: x = 2; Vertex: (2, −9)

−6 −8

(3, 0) x (5, 0) x

(−1, 0)

−4

−2

−1

21 a (0, −3), (3, 0) b

d Domain: ; Range: [−9, ∞) e

16 y = x2 − 2x – 8

y 10 (0, 11) 8 6 4 2 −2 −4 −6

(2, −9)

b Vertex: (3, −7); Axis of symmetry: x = 3 c Domain: ; Range: [−7, ∞)

(0, −5)

−6

11 a y = 2(x − 3)2 – 7

−1 −2 −3 (0, −3)

(0, −5)

−4

(2, −9)

22 m = 2 or m = −10 23 k < −2 or k > 6

Answers mathspace.co

891

24 a −3 ≤ x ≤ 4 b

32 a x-intercepts: (−2, 0), (1, 0), (4, 0); y-intercept: (0, 8)

(4, 0) x

(−3, 0) −3 −2 −1

y 10

(0, 8)

(1, 0) (4, 0)

(−2, 0)

−5

−2 −1

−5

−10

−10 −15

25 a x < −5 or x > 2 b

33 a f (x) = 2(x + 2)(x − 1)(x − 3)

(2, 0)

(−5, 0) −5 −4 −3 −2 −1

y 15

2 3

(0, 12)

−5

(1, 0) (3, 0) x

(−2, 0) −2

−1

−10

−5

−10 −15

26 x ≤ −2 or x ≥ 4 −3 −2 −1 0

5.01 Further domain and range 1

What do you remember?

27 k > 16 28 a x = 10, 30 (1000 and 3000 units) b Maximum proﬁt: $100 000 at x = 20 (2000 units) c Domain: (10, 30); Range: (0, 100] 29 a A(x) = −x2 + 80x

b Relation

c Function

d Function

2 Square brackets [ ] include the endpoints, while parentheses ( ) exclude the endpoints.

b Maximum area: 1600 m ; Dimensions: 40 m by 40 m

3 The domain of a relation is the set of all possible x-values (inputs) for which the relation is deﬁned.

c (0, 80)

4 a False

30 4 < t < 6 8

c False

d True

5 a i 0≤x<4 ii All real numbers greater than or equal to 0 and less than 4

6 4

(−1, 2) −2

b True

Practice

−1

2 −2 −4

(0, 0) 1

(1, −2)

−6 −8

892

1 a Function

Mathspace New South Wales – Year 11 Advanced mathspace.co

b i −3 < x ≤ 1 ii All real numbers greater than −3 and less than or equal to 1 c i −2 ≤ x ≤ 5 ii All real numbers greater than or equal to −2 and less than or equal to 5

d i 1<x<7

b i {−2, 0, 2, 4}

ii All real numbers greater than 1 and less than 7 e i x≤0

c i {0, 1, 2, 3}

f i x≥3

d i {−1, 0, 1}

ii {−1, 0, 1}

iii Relation

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, ∞)

14 Domain:

16 Domain: Extend your thinking

c (−∞, 5]

17 a (−∞, 2) ∪ (2, ∞)

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

10 a i (−∞, ∞)

ii [3, ∞)

b i (−∞, ∞)

ii (−∞, 2]

c i [0, ∞)

ii [0, ∞)

c x = 2 makes the denominator zero, so the function is undeﬁned. d As x approaches inﬁnity, y approaches 0 but never equals it, since b (−∞, ∞) 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.

ii (−∞, 2] ii [0, ∞)

f i (−∞, ∞)

ii (−∞, ∞)

g i (−∞, ∞)

ii (−∞, ∞)

h i (−∞, ∞)

ii (−∞, ∞)

b [−1, 3]

11 a i [−5, 5]

ii [−5, 5]

c [−3, 1]

ii [−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]

19 a (x − 1)2 + (y + 1)2 = 4

20 a The student is incorrect as y = −4 is a valid solution, and 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.

d Domain: (−∞, ∞); Range: (−∞, 4]

21 a (−∞, ∞)

e Domain: [−1, 3]; Range: [2, 4]

b (0, 1]

f Domain: [−1, 1]; Range: [−1, 1] ii {2, 3, 4, 5}

= 0 has no solution.

18 a [0, ∞)

d i (−∞, ∞)

b i [−4, 4]

; Range: [0, 3]

b (−∞, 0) ∪ (0, ∞)

e i [−2, ∞)

c i [−4, 4]

; Range: [−2, 2]

15 Domain: [−3, 3]; Range: [−3, 3]

b (−∞, −3) ∪ (−3, ∞)

13 a i {1, 3, 5}

ii {1}

iii Function

ii All real numbers less than or equal to 0

iii Relation

ii {0, 1, 2, 3}

iii Function

c The denominator x2 + 1 is always positive, so y is always positive, and y = 0 would require x2 + 1 = ∞, which is impossible. d The maximum value is y = 1 when x = 0, since x2 + 1 is minimised at 1.

Answers mathspace.co

893

5.02 Even and odd functions What do you remember? 1 a f (−x) = f (x)

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 4 a Even

b Neither

c Neither

d Even

e Even

f Odd

g Neither

h Even

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

7 a i 2x2

ii Even

b i 0

ii Both even and odd

c i x4

ii Even

d i x +x −x

ii Neither

8 a Even

b Odd

c Even

d Neither b (−2, −6)

9 a (−3, 82) 4

10 a f (x) = x − 2x

c f (x) = x2 + x + 1 8

b f (x) = x3 + x 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 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

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 the trivial function f (x) = 0 satisﬁes both conditions. 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.

5.03 Composite functions What do you remember? 1 A composite function combines two functions where the output of one is the input of another. For example, if f (x) = x + 2 and g(x) = x2, then g(f (x)) = (x + 2)2. 2 f (g(x)) applies g ﬁrst, then f, giving 2(x − 1). g(f (x)) applies f ﬁrst, then g, giving 2x − 1. 3 The domain of g(f (x)) is all x in the domain of f such that f (x) is in the domain of g. 4 a False

894

Mathspace New South Wales – Year 11 Advanced mathspace.co

b Odd

b True

c True

d False

Practice

22 g(f (x)) =

5 a 4x2 + 4x − 2

b 2x2 − 5

c 22

d 13

6 a 11

c −10

b 3

7 a (x + 5)2

d −18

b x2 + 5

c 25

d 5

8 a 28

c 6

d 110

ii y ≥ 0

9 a i b i x≥0

ii y ≥ 5

10 a i (2, ∞)

ii (0, ∞)

b i (0, ∞)

ii (−2, ∞)

11 a

−1

(x − 32) + 273.15 represents

Fahrenheit to Kelvin conversion. 23 The domain of f (g(x)) =

requires

≥ 0, so

the domain is x > 0. The domain of g(f (x)) = requires x > 0 to ensure the value inside the square root is non-negative and the denominator is not zero. Although the domains are the same, (0, ∞), the constraints that determine them arise from different parts of the composite functions.

5.04 Piecewise functions What do you remember? 1 a A function deﬁned by different rules over distinct intervals of its domain.

b x≠0

b The graph has no gaps or breaks, and the function values meet at the boundaries of the intervals.

12 [0, ∞) 13 x ≠ 2 14 [1, ∞)

c Check if f (−x) = f (x) for all x in the domain. d Check if f (−x) = −f (x) for all x in the domain.

15 a 1

b 2

16 a x =

b x=1

17 a

b x ≠ ±1

18 a

b x ≥ −1

2 Evaluate the function’s value at the boundary using the rule for the left interval and the limit from the right interval. If they are equal, the function is continuous at that point. 3 Key features include turning points, intercepts, and the coordinates of endpoints for each interval. It is also important to check for continuity at the boundaries.

19 a Domain: (−∞, ∞), Range: (−∞, ∞) b Domain: (−∞, ∞), Range: [1, ∞) c Domain: (−∞, ∞), Range: [2, ∞)

4 a False

d Domain: (−∞, ∞), Range: [1, ∞) Extend your thinking

Practice

20 x = 0, −2

5 a 1

21 Sketch transforms f (x)’s output through g(x)’s function.

6 a

b False

c True

d True

b 3

c 3

d −1

6 5 4

2 2

y=x

y = (x + 1)

−2

3 (0, 3)

2 −2

1 −4 −3 −2 −1

x 1

2 3 4

y=x+1

Answers mathspace.co

895

14 y 13 12 11 10 9 8 7 6 5 4 3 2 1 −8 −6 −4 −2

22 y (8, 22) 20 18 16 (6, 14) 14 12 10 8 6 4 (2, 2) (5, 2) 2

(2, 6)

(0, 0)

−2−2

x 2

7 a c 2

x 1

−1

(1, −1)

−2

−3 −4 (0, −4) −5

(−5, 5)

22 y (3, 20) 20 18 16 (3, 15) 14 12 10 8 6 4 2 (0, 0) x

−8 −6 −4 −2

8 a 4

3 2

(−2, 0) −4

−3

−2

−1

2 4 6 8

7 ,0 3

1 1

−1

−2

y 18 16 14 12 10 (4, 8) 8 (9, 9) 6 (7, 7) (0, 6) 4 2 x −2

−3 −4

b 4

(−1, 1) (−1.5, 0) −4

−3

−2

−1

2 1 −1 −2 −3 −4

Mathspace New South Wales – Year 11 Advanced mathspace.co

(−1, −3)

896

(1, −4)

(0, −4)

(2, 0) 1

(2, −1)

13 a 6.70

c 4

b 13.60 c

16$

2 1

(−3, 0) −4

−3

−2

−1

(5, 13.30)

14$

(1, 0)

F (d)

12$

10$

−2

(5, 10.10)

−3 (0, −3)

−4

(−1, −4)

4$ 2$

d (−4, 16)

−10

−8

(−5, −1) −6

(−4.5, −2)

−4

16 y 14 12 10 8 6 4 2 (0, 0) −2 −2 −4 −6 −8 −10 −12 −14 −16

0 km 1 km 2 km 3 km 4 km 5 km 6 km

d [4.50, 13.30] ∪ (10.10, 17.10] (9, 3)

(4, 2) 4

x 6

14 a f (−1) = −1, f (4) = 1 b Domain = , Range = (−∞, 0) ∪ (0, ∞) ∪ {2} c

4 f (x) 3 2

(4, −16)

1 −2 −1 −1

x 1

−2 −3 −4

10 x =

11 a Odd

15 a Removable discontinuity at x = 1.

b Neither

c Neither

d Even

y 4

12 a 30

b 65 c

C(x)

70$ −2

60$ 50$

−1

x 1

−2

40$ 30$

Extend your thinking

20$

16 a a = −0.5

10$ x 0GB 1GB 2GB 3GB 4GB 5GB 6GB

d [20, 80]

b y=5

17 There is an overlap between the domains of the second rule and the third rule. For the input values within x ∈ [2, 4], there are two output y-values. Therefore, the relation is not a function.

Answers mathspace.co

897

18 a 100 L/h

d The numerator is a constant 1, so y cannot be 0. The function approaches y = 0 as a horizontal asymptote.

R(t)

80 L/h

8 a (x − 2)2 + (y + 1)2 = 9

60 L/h

b [−1, 5]

40 L/h

c [−4, 2]

20 L/h 0h

t 1h

The function is continuous because the pieces meet at the boundary points. At t = 1, the ﬁrst piece is 100 and the second is 100 − 20(1 − 1) = 100. At t = 3, the second piece is 100 − 20(3 − 1) = 60 and the third is 60 − 10(3 − 3) = 60. Since the values match at the boundaries, the graph is unbroken. b The total volume drained is equal to the area under the graph of R(t) from t = 0 to t = 4. This area can be calculated by splitting it into a rectangle and two trapeziums.

9 a Even

b Odd

10 (−2, −28) 11 Odd function: f (−x) = −f (x)

Write the formula

f (−0) = −f (0)

Substitute x = 0

f (0) = −f (0)

(100 + 60) × 2 = 160

Area3 − 4 =

(60 + 50) × 1 = 55

Total Volume = 100 + 160 + 55 = 315 litres

Add f (0) to both sides

f (0) = 0

Divide both sides by 2

Thus, the function passes through (0, 0). 12 The student’s claim is incorrect. An odd function requires f (−x) = −f (x). f (−x) = (−x)5 + 3 = −x5 + 3 −f (x) = −(x5 + 3) = −x5 − 3 −x5 + 3 ≠ −x5 − 3 The constant term +3 (i.e., 3x0) has an even power, so the function is not odd. 13 a 9x2 + 12x + 3

Chapter 5 review 1 A

Evaluate

2f (0) = 0

Area0 − 1 = 1 × 100 = 100 Area1 − 3 =

c Neither d Odd

b 3x2 − 1

c 24

d 11

14 a x = 3

b x=

15 a i [5, ∞)

ii [0, ∞)

b i [0, ∞)

ii [−5, ∞)

2 B 3 A 4 a [4, ∞)

16 x = 0, −1

b (−∞, −5) ∪ (−5, ∞) c (−∞, 7]

17 E(U (a)) = 0.8624a

d (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)

$200 AUD = 172.48 Euros

5 a i (−∞, ∞)

ii (−∞, 5]

18 a 2 19

b 5

b i [−6, 6]

ii [−6, 6]

c i [−4, ∞)

ii [0, ∞)

d i [−5, 5]

ii [−2, 2]

d −7

c 5 y

6 Domain:

; Range: [0, 5]

7 a (−∞, 3) ∪ (3, ∞) b (−∞, 0) ∪ (0, ∞) c When x = 3, the denominator becomes 0, making the expression undeﬁned.

898

Mathspace New South Wales – Year 11 Advanced mathspace.co

(−1, 0) −3 −2 −1

−1

(1, 0) 1

x 3

20 a 35

Practice

b 85 c

4 a

80 C (h) ($) 70

−2 −1

f (x)

−1 −2

−4

−20 20

60 50

20 10

h (hours)

2 3 4

−2 −3

21 a a = −2

−4

b 12

90 P (t) (units/hr) 80 70 60 50 40 30 20 10 0

−4 −3 −2 −1 −1

d [15, 85]

22 a

5 a

−2 −1

f (x)

1 10

−10

−5

b Quadrants 2 and 4 6 a f (x) approaches 0 from the positive side

t (hours) 3

b f (x) → ∞

7 a Quadrants 2 and 4 The function is continuous because the pieces connect at the boundaries. At t = 2, the ﬁrst piece is 80 and the second is 80 − 15(2 − 2) = 80. At t = 4, the second piece is 80 − 15(4 − 2) = 50, and the third is 50 − 5(4 − 4) = 50. Since the function values match at the boundaries, the graph is unbroken. b From the graph, the maximum value is 80 units/hr and the minimum value in this interval is P (4) = 50 units/hr. The range is [50, 80].

b f (x) = 3 c d x = 0, f (x) = 0 8 a 0 9

3 1 −4 −3 −2 −1 −1

x 1

2 3 4

−2 −3 −4

What do you remember? 1 a Rectangular hyperbola

10 a No, because x = 0 makes the function undeﬁned, and f (x) = 0 has no solution

b x = 0, f (x) = 0 b True

d −∞

6.01 Graphs of reciprocal functions

2 a False

c ∞

b 0

c True

d True

3 As x → −∞, f (x) → 0. Speciﬁcally, if k > 0, f (x) → 0−; if k < 0, f (x) → 0+.

since

≠ 0.

b Multiplying both sides of f (x) =

by x gives

x × f (x) = 8, which is equivalent.

Answers mathspace.co

899

11 As x → ∞, f (x) =

→ 0+ and f (x) =

→ 0−.

6 a D = 20T, k = 20

b D = 160 metres

As x → −∞, f (x) =

→ 0− and f (x) =

→ 0+.

7 a W = 9.8M, k = 9.8

b W = 735 Newtons

As x → 0+, f (x) =

→ ∞ and f (x) =

→ −∞.

8 a P = 0.2A, k = 0.2

b P=7L

As x → 0−, f (x) =

→ −∞ and f (x) =

→ ∞.

9 a F = 4M, k = 4

b F = 20 Newtons

10 a E = 2S2, k = 2

b E = 450 Joules

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.

11 a I = 2000A , k = 2000 b I = 3.2 Watts per square metre 12 a k = 1

b C = 81 3

Extend your thinking b f (x) =

12 a f (x) =

13 A : f (x) = , B : f (x) = , C : f (x) = , where c = 4, b = 2, a = 1 b t = 3 hours

14 a t = c n = 8 workers

13 a V = S , k = 1

b V = 125 cubic metres

14 a F = 4X, k = 4

b F = 12 Newtons

15 a C = 0.15D, k = 0.15

b D = 400 kilometres

16 a H = 2I2, k = 2

b I = 8 Amperes

Extend your thinking 17 a C = P 3, k = 1 b P=5 c Since C = kP 3, doubling P results in C = k(2P )3 = 8kP 3, which is 8 times the original cost.

15 a c = b

c 80

18 a Brightness: B = 50I2, k = 50, Power: P = 12.5I3, k = 12.5

b Brightness: B = 450 Lumens, Power: P = 337.5 Watts

B or P 50

n 0

B = 50I2

6.02 Direct variation models

1 One quantity is a constant multiple of a power of another, expressed as y = kxn. 2 y = kx

3 The constant of variation, which scales the relationship between y and xn. 4 a Yes

b No

c No

d No

Practice 5 a k = 20

900

P = 12.5I3

What do you remember?

b V = 180

Mathspace New South Wales – Year 11 Advanced mathspace.co

0.2

0.4

0.6

0.8

The graph of B = 50I2 is a parabola opening upward, starting at (0, 0). The graph of P = 12.5I3 is a cubic curve, also starting at (0, 0), increasing less steeply than B for 0 ≤ I ≤ 1. 19 a D = 5T 2, k = 5 b The graph is a parabola opening upward, starting at the origin, since D = 5T 2 is a quadratic function.

6.03 Inverse variation models

18 a T = b C = 30M 2 T

What do you remember? 1 When one quantity increases, the other quantity decreases proportionally, or vice versa, such that their product or a related expression remains constant.

6.04 Introduction to absolute value functions What do you remember?

2 y= 3 a Yes

c Cost: C = 10 800 dollars; Machines: M = 4

b No

c No

d Yes

4 The constant k is the constant of proportionality. For an inverse variation, it represents the constant product of the two variables, that is, k = xy.

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.

5 The graph is a hyperbola.

b For b = , the distance from zero is . Thus, b =

Practice 6 a V=

b V = 20

7 a k = 20

b F=5

8 a k = 16

b I=1

9 a T=

b T=6

10 a k = 220

b F = 550

11 a R =

b R = 2.5

12 a C =

b D=6

13 a T =

b T=3

or b =

c For c = −4, there is no solution because absolute value cannot be negative, as distance is non-negative. d For d = 0, the distance from zero is 0. Thus, d = 0. 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). 4 5 No, = a for all real a, as both yield the non-negative distance from 0.

Extend your thinking

Practice

14 D = 5

6 a x = 7 c x − 4 = 2.5

15 C = $960

If speed doubles, time halves, maintaining a constant product.

b x + 1 = 3

7 • f (−6) = 6

16 For positive k and n, as x increases, y decreases because xn is in the denominator, and their product y × xn = k remains constant. For example, the time T to travel a ﬁxed distance varies inversely with speed

• f (0) = 0 • f (3) = 3 8 l − 50 = 0.5 .

9 t − 20 = 1.5

17 D = 4

Answers mathspace.co

901

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

4 a The graph becomes steeper. b The graph becomes wider. c The graph becomes a horizontal line at y = b. This is a degenerate case, and for the graph to have a V-shape, we require a ≠ 0. 5 a Two solutions

b One solution

c No solutions

c x  = Practice Extend your thinking

6 a x = 5

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.

7 a

b x + 2 = 4 y 6 5 4

14 a f (v) = v − 500

b Piecewise function:

2 1 −6 −5 −4 −3 −2 −1

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. c g(x) = −2x − 1 − 1

x 1

Symmetry: x = −3 Domain: , Range: y ≥ 0 b

y 6 5 4 3 2 1

x 1

2 3 4 5 6 7 8

Symmetry: x = 4 Domain: , Range: y ≥ 0 c

6.05 Absolute value functions

6 5

What do you remember?

1 Absolute value represents the distance of a number from 0 on a number line, always non-negative. 2 x=

,y=0

3 y≥0

3 2 1 −2 −1

x 1

Symmetry: x = 1 Domain: , Range: y ≥ 0

902

Mathspace New South Wales – Year 11 Advanced mathspace.co

y 10

x −2 −1

2 3 4 5 6

2 x

Symmetry: x = 2

−16 −12 −8 −4

Domain: , Range: y ≥ 0 8 a

9 a

2 3 4

y 4

2 1

x 1

2 3 4

−2

−3 −4

−4 −3 −2 −1 −1

x = 1, 3 b

8 12 16 20 24

x = −16, 24

−4 −3 −2 −1 −1

Vertex:

Line of symmetry: x =

x-intercept:

y-intercept: (0, 2)

x −4 −2

y 10

2 4 6 8 10

x = −6, 12 c

y 14

2 x

−1

Vertex: (3, 0)

Line of symmetry: x = 3 y-intercept: (0, 9)

2 x

x = −15, 13

2 3 4 5 6 7

x-intercept: (3, 0)

−14−12−10 −8 −6 −4 −2

2 4 6 8 10 12

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

e x=

Answers mathspace.co

903

g x=

,x=

h m = −4, m = 4

i u = −7, u = 3

j v = −3, v = 5

x =

16 Function: f (t) = t − 25

11 a  x − 50 = 2 gives the solutions x = 48 mm and x = 52 mm.

At t = 22, 3°C

b  s − 60 = 5 gives the solutions s = 55 km/h and s = 65 km/h.

At t = 28, 3°C

c x − 2 = 4; x = −2, x = 6 12 a x = −1, x = 7

b x = −5, x = 1

c x = −4, x = 2

and

13 f (d) = d − 25, d = 22, 28 mm 14 a

At t = 25, 0°C 17 a Two intersections with y = k above x-axis. b One intersection at vertex. 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.

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.

1 x 1

6.06 Circles and semicircles

What do you remember? x = 1, 5

1 a x2 + y2 = r2

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.

2 1 x −2

−1

c y= d Circle 2 r = 10.

−1

3 a Centre: (0, 0), Radius: b (±4, 0)

x = −1, 0

Practice

Extend your thinking 15 a x =

4 For a point (x, y) on the circle, the distance from the origin (0, 0) is 6. By Pythagoras’ theorem,

,x=3

b 7

= 6. Squaring both sides gives x2 + y2 = 36.

6 5

(3, 5)

4 3

1 5 2 − , 3 3 1

−4 −3 −2 −1 −1

904

x 1

2 3 4

Mathspace New South Wales – Year 11 Advanced mathspace.co

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.

6 5 4 3 2 1

3 2 1

r=5

−6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

x 2 3 4 5 6

−4 −3 −2 −1 −1

2 3 4

−2 −3 −4

10 a

7 a x2 + y2 = 9

b x2 + y2 = 72 = 49

8 a Upper semicircle (positive y-values)

b Left semicircle (negative x-values) 9 a

−1

1 −1

2 1 x −2

−1

b 8

−1

−2

−4

−6

1 1

−8

−1

−8 −6 −4 −2 −2

−3 −2 −1

−2

−3

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−8 −6 −4 −2 −2

−4 x 1 2 3 4 5

−6 −8

Answers mathspace.co

905

4 a Quadrants II and IV

10 8 6 4 2

b f (x) = 2 c x= d x = 0, y = 0

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

2 4 6 8 10

5 a 0

b y=

11 a y =

b f (x) =

c f (x) =

d f (x) =

e f (x) =

f f (x) =

x2 + y2 = 81

6 4

b t = 4 hours

8 a D = 15T

b D = 150 m

9 a k = 25

b H = 400 J 3

10 a C = 2S

x 2 4 6 8 10

b C = $250

−6

c Since the cost varies directly with the cube of the side length, the equation is C = kS3. When the side length is doubled from S to 2S:

−8

Cnew = k(2S)3

−4

−10

13 a x2 + y2 = 25

Cnew = k × 8S3 Cnew = 8(kS3)

b y=

Cnew = 8Cold Thus, the cost increases by a factor of 8.

Extend your thinking 14 From x2 + y2 = r2, solve for x: x2 = r2 − y2, so . Right semicircle:

(positive x).

Left semicircle:

(negative x).

15 Width at ground level: 8 metres Maximum height: 4 metres 16 Width at the base: 26 metres Maximum height: 13 metres b [−6, 6]

17 a

d −∞

c n = 12 drivers

−10 −8 −6 −4 −2 −2

6 a f (x) =

7 a t=

12 a y = ± b

c ∞

b 0

11 a D = 3T 2 b

45 D 42 39 36 33 30 27 24 21 18 15 12 9 6 3 0

T 1

Chapter 6 review 1 C

12 a k = 8

b F=2N

2 A

13 a k = 198

b F = 396 Hz

3 C

14 D = 4 m

906

Mathspace New South Wales – Year 11 Advanced mathspace.co

21 a

15 D = 4 m 16 a Verify the identity values.

= a for the given

6 5 (0, 5)

For a = −9:

4 3 2

5 − ,0 2

−5 −4 −3 −2 −1

Since 9 = 9, the identity holds for a = −9. For a = 5:

1 1

Vertex: Line of symmetry: x = x-intercept:

Since 5 = 5, the identity holds for a = 5.

y-intercept: (0, 5)

Thus, the identity is veriﬁed for both values.

Apply the identity

y 12 (0, 12) 10 8

Evaluate the absolute

values

Sum the terms

4 2

17 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. 18 a f (v) = v − 450 b Piecewise function: . Vertex: (450, 0).

1 2 3 4 5 6 7 8

Vertex: (4, 0) Line of symmetry: x = 4 x-intercept: (4, 0) y-intercept: (0, 12) 22

(8, 5)

19 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 20 a x = −2, 3

b x = −12, 6

c x = 4, 8

d x = −5, −3

(4, 0)

−1

(−2, 5)

4 3 2 1 x −2 −1

1 2 3 4 5 6 7 8

Solutions: x = −2, 8 23 Solutions: x = −2, 8; Ordered pairs: (−2, 7), (8, 7) 24 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.

Answers mathspace.co

907

25 Centre: (0, 0), Radius: 3

8 a Opposite = 4 cm, Adjacent = b

y 3

9 a Other side = 6 m, Hypotenuse =

b 18 m2

1 −3 −2 −1

x 1

−1

10 a

d 1

11 a θ = 60° b θ = 60° c θ = 45° d θ = 60°

−2 −3

12 a

26 a

13 1

27 Width at the base (y = 0): 14 m; Maximum height (at x = 0): 7 m

14 6 cm Extend your thinking

b y ∈ [−4, 4]

28 a

7.01 Exact trigonometric values What do you remember?

16 For the triangle, sin 45° =

1 a

c 1

2 a 54°

b 30°

c (90 − y)° d 71°

sin θ =

Practice sin

cos

tan

so sin 45° = cos 45°. Generally, in a right-angled triangle, if θ is one acute angle, the other is 90° − θ . The opposite side for θ is the adjacent side for 90° − θ , so

, cos 45° =

= cos (90° − θ ).

17 a

18 a

19 a

30° 45°

60°

5 a

c 1

What do you remember? 1 a Angle between the horizontal and the line of sight to an object above the observer.

6 a b sin 60° =

, cos 60° = , tan 60° =

7 a θ = 30°

b θ = 45°

c θ = 60°

d θ = 60°

908

7.02 Angles of elevation, depression and bearings

Mathspace New South Wales – Year 11 Advanced mathspace.co

b Angle between the horizontal and the line of sight to an object below the observer. They are equal to each other when measured between two points, as they are alternate angles between parallel horizontals.

2 a θ

e 124°T

f 166°T

b β

g 233°T

h 341°T

3 a Starts from north or south, measures an acute angle (0° to 90°) towards east or west, written as N/S a° E/W.

25 a S 23° W

b N 67° E

c S 42° E

d N 78° W

e N 37° E

f S 55° W

b Measured clockwise from north, written using three digits, often with T, e.g., 046°T.

g N 29° W

h S 71° E

4 a 042°T

b 255°T

c 160°T

d 295°T

5 a S 45° E

b N 45° W

c S 0° E

d N 45° E

26 1420 m 27 a 218.7°T

b 64.0 km

28 a 56°

b 18.0 km

29 a 21.2 km

b 21.2 km

Practice

Extend your thinking

30 a p°T

b (180 − q)°T

c (360 − p)°T

7 100 m 8 9 a x = 37.23°

b y = 52.77°

10 a x = 455.84 m

b 425.84 m

d (180 + q)°T

31 a 60 m

b 1601 m

32 a

b 35 m

c 49.2 m 33 a 1.43°

b 2.50 km

11 128.56 m

34 a 10 m

b 59.04°

12 172 m

35 a 1045.36 m

b 119.0°T

36 a 140.0°

b 75.25 km

37 a 19.21 km

b 193.7°T

13 a AB = 11.86 m

b AC = 16.24 m

c BC = 4.4 m

c 014°T

14 a S 65° W

b N 30° W

c E 26° S

d N 30° E

15 a 030°T

b 320°T

c 056°T

d 205°T

16 a N 48° E

b S 35° W

1 a Quadrant 4

b Quadrant 1

c S 40° E

d N 50° W

c Quadrant 2

d Quadrant 3

7.03 Unit circle What do you remember?

2 0°, 90°, 180°, 270°, 360°

18 250.0 m

3 a Quadrants 1 and 2

b Quadrants 1 and 4

c Quadrants 1 and 3

19 1461.9 m

4 a Quadrant 2

20 33.69° 21 250°T

Practice

5 a 0.67

b −0.47

b Quadrant 3

c −0.14

23 a 98.3 km

b 68.8 km

6 a True

b True

24 a 252°T

b 307°T

7 a False

b False

c 018°T

d 073°T

d 0.77

Answers mathspace.co

909

8 a i

iii

b i

iii

c i

iii

d i

iii

3 a − sin θ

b cos θ

c − tan θ Practice

9 a (0.82, 0.57)

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 −1

4 a 60°, cos 60°

b 45°, sin 45°

c 30°, tan 30°

d 30°, cos 30°

5 a

6 a −tan 15° = −0.27

b −sin 60° = −0.87

c cos 30° = 0.87 b

7 a c 8 a Quadrant 2, positive b Quadrant 3, negative c Quadrant 4, negative

c Undeﬁned

9 a 60°

b 60°

10 a sin 30°

b cos 90°

c tan 30°

d 1

16 a (−0.87, −0.50)

b (−0.87, −0.50)

c (0.50, 0.87)

d (−0.87, 0.50)

17 a

c −1

18 a 0

b 0

c 0

d −1

11 Related angle: 45°, 12

sin θ

cos θ

tan θ

150°

Extend your thinking 19 a b

b −a

13 a

c 14 a Positive angle: 60°, Related angle: 60°

20 Q(−a, b), R(−a, −b), S(a, −b) 21 a −2 sin x

b Positive angle: 240°, Related angle: 60°

b 2 cos x

15 a −sin θ

b cos θ

c tan θ

d −sin θ

16 a Negative

b Negative

22 (−0.407, 0.914) 23 a 0.26

b −0.97

c −0.26

d −0.97

c Positive

24 Ask your teacher for worked solutions.

7.04 Related angles and identities

18 a sin 45°, positive

What do you remember? 1 a θ c θ − 180° 2 a All c tan θ

910

b 180° − θ d 360° − θ b sin θ d cos θ

Mathspace New South Wales – Year 11 Advanced mathspace.co

b −cos 30°, negative

c −tan 60°, negative 19 a

b −2

sin θ

cos θ

tan θ

14 178.31 cm2 15 C = 90.0°

−120°

16 B = 38.7°, 141.3°

−225°

−1

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.

21 Related angle: 30°, cos 510° = −0.87 Extend your thinking 22

18 Ask your teacher for worked solutions. 19 Ask your teacher for worked solutions.

Extend your thinking 23 Ask your teacher for worked solutions. 24 Ask your teacher for worked solutions. 25 θ = 30°, 150°, −210°, −330° 26 At 270°, coordinates are (0, −1). Thus,

, which is undeﬁned.

20 303° 21 20.53 km 22 AC = 19.29 23 Ask your teacher for worked solutions. 24 x = 18.00 cm

7.05 Sine and cosine rules

7.06 Radians

What do you remember?

1 a b c2 = a2 + b2 − 2ab cos C c 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°

1 a True

b True

c False

2 a i

b ii

c iv

3 a

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°

d iii

Practice

8 a

8 c = 14.26 9 41.57 cm2 10 b = 15.73 11 A = 43.1° 12 Two solutions 13 q = 9.32

9 a

10 a

b −1

c 1

d 0

e 11 a

f −1 b

Answers mathspace.co

911

12 a

b 25π cm2

13 a 14 209.44 m2

Extend your thinking 13 sin (s + π ) = 14 sin ( θ ) =

15 30.00 cm

, 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 ).

16 24 + 6π cm 17 150.80 m2 18 θ = 0.75 radians Extend your thinking 19 20 2227.06 cm2 21 186.63 m2

22 θ = 0.16 radians, P = 215.71 m 16 cos ( θ ) =

and tan ( θ ) =

7.07 Arc length and sector area What do you remember? 1 a l = rθ

7.08 Graphs of trigonometric functions What do you remember? 1

2π

sin θ

cos θ

b 8π cm2 2

3 a

c P = r( θ + 2) 2 a 2π cm

4 θ =2 Practice

3 • Period: 2π

5 a 5π cm

• Amplitude: 1

2π

−1

• Midline: y = 0

• Domain: x ∈ 

6 a π cm

• Range: y ∈ [−1, 1]

7 a 8.40 cm

b 31.42 m

• Symmetry: Odd, sin (−x) = − sin x, with point symmetry about the origin

8 a 22.50 cm2

b 104.72 m2

9 a

10 a

y 1 0.5 x

11 a r = 14.32 cm

b r = 7.98 m

−0.5

12 a θ = 1.50

b θ = 1.67

−1

912

Mathspace New South Wales – Year 11 Advanced mathspace.co

1 π 2

1π

3 π 2

y 1

θ tan θ

0.5

−1

−1π

3 − π 2

1 − π 2 −0.5

−1

7 • Period: π • Domain: x ∈ , x ≠ integers

+ kπ, where k are 19

• Range: (−∞, ∞) • y-intercept: (0, 0) • x-intercepts: x = kπ, where k are integers • Symmetry: Odd, tan (−x) = −tan x, with point symmetry about the origin 8

sin ( θ )

cos θ )

−1

3 2 1

( −π , 0)

1 − π −1 2 −2

(0, 0) (π , 0) x

1 π 2

21 x = 0, π, 2π

−3

22 a Periodic

−4

c π

b Amplitude d (−∞, ∞)

e Practice 9 a y = sin x

b y=1

10 a −1

b y=0

11 a True

b False

c 1

23 tan

= 1; tan (nπ ) = 0

24 A, C, D c True

25 Increasing 26 Negative

12 x = π 13 a Increasing

b Decreasing

Extend your thinking

14 a Increasing

b Decreasing

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].

15 x = 0 16 17

28 y = cos x is y = sin x translated

units left

Answers mathspace.co

913

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.

as θ approaches

b −c

15 a

16 a sin α

b −cos α

c − tan α

d cos α

17 θ = ±120°, ±240° 18 12.99

31 On the unit circle, a point at angle θ has coordinates (cos θ , sin θ ). Since

14 a d

(where cos θ → 0),

tan θ grows without bound (positive or negative). Thus, tan x takes all real values, giving a range of (−∞, ∞). 32

19 17.29 20

C ≈ 74.6° or C ≈ 105.4°

21 4.51 km 22 a

c 144°

d 330°

23 a

33 On the unit circle, at angle θ + π, coordinates are (−cos θ , −sin θ ).

24 a

Thus, tan ( θ + π ) = periodicity of π.

25 a 2π cm

b 10π m2

26 a

b 54π cm2

= tan θ , proving

Chapter 7 review

27 Central angle is

1 B

24 sin

2 C 3 B

. Chord length is

. Difference is 5π – 24 sin

cm.

28 Central angle is 1 radian. Triangle area is b

4 a

sin (1) m2.

d 2 29 a −1

b y=0

5 a 4m

6 a

7 a 233.15 m

b 208.15 m

8 a 75.0 km

b 129.9 km

9 a 202.6°T

b 65.0 km

10 a

32 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.

30 a π

c 89.6 m 11 a i

iii

b i

iii

12 a (0.77, 0.64)

b (−0.94, −0.34)

13 a

914

Mathspace New South Wales – Year 11 Advanced mathspace.co

b (−∞, ∞)

8.01 Secant, cosecant and cotangent What do you remember? 1 a cosec θ = b sec θ =

c cot θ =

17 Use the side ratios: 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° c False

Practice 4 a

5 a

6 a

b 2

c 1

and

. Therefore,

sec θ × cos θ =

18 The student used sin 30° = reciprocal.

instead of the

Correct value: cosec 30° =

19 sec 45° =

, cosec 45° =

= 2.

, cot 45° = 1.

20 sec θ = , cosec θ =

8.02 Unit circle and sec, cosec and cot

d What do you remember?

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 =

b As θ → 90°, the adjacent side approaches 0, →

which is undeﬁned. 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

c 1

d cosec θ

14 a 2

b 1 b

2 a

3 a

b 0, π

c 0, π

4 a

b 2

5 a 2

6 a

= 1.

→

so sec θ =

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.

Practice

d 2

e −1

f −2

g Undeﬁned

i −1

c 7 a

15 If the reference angle is 0°, the opposite side length is 0, and the adjacent side is equal to the hypotenuse. Extend your thinking 16 cosec θ =

, cot θ =

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

Answers mathspace.co

915

11 a 1

b 1

c 1

12 a

b 0

b In mathematics, a value is considered undeﬁned when it does not have a meaningful or valid result. Example: The expression

13 a 1

f 2

g 0

because division by zero does not produce a meaningful result. c The quotient is the result obtained by dividing one number by another.

Extend your thinking 14 At θ =

is undeﬁned

, the point on the unit circle is (0, 1), so

x = 0. Since sec θ = , it is undeﬁned when

Example: In the division 12 ÷ 4, the quotient is 3. b

2 a

x = 0. 15 On the unit circle, sec θ = sec θ =

and cos θ = x, so

3 a

4 a 90° + 180°n

b 180°n

c 90° + 180°n

d 180°n

instead of the

16 The student used

Practice

reciprocal. Correct value:

5 a

17 sec θ =

, cosec θ = 2, cot θ =

18 sec θ =

, cosec θ = 2, cot θ =

True

b 19 On the unit circle, cot θ = , cos θ = x, and sin θ = y. Thus, cot θ = 20 Q =

21 a

,R=

. and S =

c −2

8.03 Reciprocal and quotient identities

False c

What do you remember? 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

916

Mathspace New South Wales – Year 11 Advanced mathspace.co

True

7 a

6 a

False

c d

8 a cos θ

b sin θ

9 a

10 a Undeﬁned

b Undeﬁned

c Undeﬁned

d Undeﬁned

Answers mathspace.co

917

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.

, it is undeﬁned

14 b

15 sec θ =

, cosec θ =

, cot θ =

12 a

17 sec θ =

, tan θ =

, cot θ =

18 The student is incorrect. Using cot θ =

tan 45° = 1, so cot 45° = = 1.

8.04 Complementary angle identities What do you remember? 1 a cos θ

b cot θ

c cosec θ

d sec θ

2 a θ = 90° + 180°n c θ = 180°n

3 a True

918

Mathspace New South Wales – Year 11 Advanced mathspace.co

b θ = 180°n d θ = 90° + 180°n

b False

c True

d False

Practice

5 a

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°). b

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°). 6 a 1

b 1

c 1

d 1

7 a

8 a sin θ

b tan θ

c cosec θ

9 a

Thus, the identity holds. d 10 a

b 0

d 0

11 cot (90° − θ ) = tan θ = 3

Thus, the identity holds.

Extend your thinking 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° =

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°.

Answers mathspace.co

919

14 a sin θ =

b cosec θ =

15 cos 30° = sin 60° =

11 a Using complementary identity, sin (90° − θ ) = cos θ , which is true. b tan (90° − θ ) × sin θ = cot θ × sin θ =

, while

× sin θ = cos θ , which is true.

sin 30° = cos 60° = .

c cos (90° − θ ) ÷ cos θ = sin θ ÷ cos θ = tan θ , which is true.

16 Ask your teacher for worked solutions. 17 a cos θ =

d cot(90° − θ ) × cos θ = tan θ × cos θ = × cos θ = sin θ , which is true.

b sec θ =

18 Ask your teacher for worked solutions.

12 a 1

b 1

d 1

8.05 Evaluate expressions with identities

13 a

What do you remember? 1 a cosec θ

Extend your thinking

b sec θ

14 For 120° in Quadrant 2, the reference angle is 180° − 120° = 60°. Since cos is negative in

2 a Quadrant 2, 60° b Quadrant 3, 45°

Quadrant 2, cos 120° = −cos 60° =

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

15 The statement is incorrect.

b 0

d b − cos 30° = d cot 60° =

7 a

8 a cos2θ

b sin2θ

c cot2θ

d tan2θ

9 a 0

10 a

b cos θ

c cos θ

while cos 120° = −cos 60° =

. The correct

identity is sin (90° − θ ) = cos θ , so sin (90° − 120°) = cos 120°.

cos 135° = − cos 45° =

Mathspace New South Wales – Year 11 Advanced mathspace.co

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 = =

which does not equal cos θ . The statement is false.

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

920

16 The northward component in true bearings is

5 a

c − tan 45° = −1

sin (90° − 120°) = sin (−30°) = − sin 30° =

d True

Practice

6 a sin 60° =

b θ = 180°n

3 a

7 a Ask your teacher for worked solutions.

b Ask your teacher for worked solutions. 8 a sec2θ = 5

Practice

4 a

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.

10 a 1

b 1

11 a i Ask your teacher for worked solutions. ii Ask your teacher for worked solutions. ii

b i c 12 Correctly,

+ 1 = tan2θ + 1 = sec2θ .

13 Ask your teacher for worked solutions. Extend your thinking 14 a b For θ in Quadrant

, the

reference angle is 2π − θ or 360° − θ in degrees. Since sec2θ = 5 a 1

b sec θ

c 1

d 2 + sec θ cosec θ

e tan2θ + sin2θ

f sec2θ − 2

, 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 =

6 a

in Quadrant 1, the

corresponding cos θ yields the same sec2θ in Quadrant 4. 15 cosec2θ = 16 Ask your teacher for worked solutions. 17 18

c 2

19 Ask your teacher for worked solutions. 20

Answers mathspace.co

921

21 Ask your teacher for worked solutions.

Extend your thinking

8.07 Trigonometric equations

14 For all θ , cos θ  ≤ 1, so cos θ = 1.5 is impossible.

What do you remember? 1 a True

b True

c False

d False

2 a First and second quadrants b Second and third quadrants

16 a θ = −180°, 0°, 90°, 180°

c First and third quadrants

b θ = −180°, −60°, 60°, 180°

d Third and fourth quadrants 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.

17 θ = 0.644, 5.640 18 Ask your teacher for worked solutions. 19 x = 90°, 270° 20 θ =

Practice 4 a θ = 30°, 150°

d θ = 0°, 180°, 360° b θ=

c θ = 0, π, 2π, 3π, 4π

21 θ = 0°, 60°, 180°, 300°, 360°

b θ = 150°, 210°

c θ = 120°, 300° 5 a θ=

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°.

Chapter 8 review 1 B

d θ = π, 3π

2 C

6 a θ = 60°, 120°

b θ = 120°, 240°

c θ = 135°, 315°

d θ = 30°, 150°

3 C

7 a θ = 30°, 150°, 270°

4 a

5 a

6 a

d −1

b θ = 0°, 360°, 131.81°, 228.19° c θ = 45°, 225°, 63.43°, 243.43° b x=

8 a

9 θ ≈ 22.62°

7 a i

10 a θ =

b θ =

11 a θ = 143.13° 12 a θ =

b θ = 157.38° ,

b Ask your teacher for worked solutions. 8 a

9 a

b −2

d −1

10 a 2

d 2

b θ = π, 3π, 5π 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°.

922

Mathspace New South Wales – Year 11 Advanced mathspace.co

11 B = 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. b

15 a

16 a cos θ × sec θ = 1

9.01 Sets and notation What do you remember?

b cot θ × tan θ = 1

1 a A set is a collection of distinct objects, called elements, grouped based on a speciﬁc property.

d sin θ × cosec θ = 1

17 a

b 1

c 2

b The cardinality of a set A, denoted n(A) or A, is the number of elements in the set.

18 a

c The empty set, denoted ∅, is the set with no elements.

19 a 4

20 a

21 a 0

2 a {1, 3, 5, 7, 9}

b {a, e, i, o, u}

3 a True

c True

b False

4 a n(A) or A

22 For 210° in Quadrant 3, the reference angle is 210° − 180° = 30°. Since sin is negative in Quadrant 3, sin 210° = −sin 30° =

b cos2x

23 a tan x c −tan2x 24 a 10

d cos2A b

25 a

b θ = 60°, 120°

c θ = 135°, 315°

d θ = 90°, 270°

28 a x = c x=

b x=

d x = 0, π, 2π

, A′, or Ac

b ∅ d A⊆B

Practice 5 a n(A) = 3

b n(B) = 3

c n(C) = 5

d n(D) = 0

e n(E) = 4

f n(F) = 4

g n(G) = 5

h n(H) = 2

6 a Not empty

b Not empty

c Empty

d Empty

e Not empty

f Empty

7 a Yes

26 Ask your teacher for worked solutions. 27 a θ = 180°

d False

8 a

b {s}

= {2, 4}

c {q, s} b

= {1,3,5}

= {1, 2, 3, 4, 5}

=∅

= {3, 4, 5}

= {1, 2, 4, 5}

9 a Yes

b No

e Yes

f No

10 a {Alice, Clara}

c Yes

d Yes

b n({Alice, Clara}) = 2

c {Bob, Dan}

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°

11 a {a, c, e}, n = 3

b {b, d, f}, n = 3

c {b, d}, n = 2

d {a, c, e, f}, n = 4

12 a True

30 a x = 0, π, 2π, 3π, 4π

b False

13 a {2, 4, 6}

b π=

c x=

d x=

31 Ask your teacher for worked solutions.

c True

d True

b {1, 2, 3, 4, 5, 6, 7}

c {2, 4, 6}

d {1, 3, 5, 7}

14 a {Emma, Grace} b n({Emma, Grace}) = 2 c {Finn, Henry} 15 a Yes

b No

c Yes

d No

Answers mathspace.co

923

6 a Yes, since A ∩ B = ∅.

Extend your thinking 16 a 23 = 8 subsets. Each element can either be included or excluded, giving 2 choices for each of the 3 elements. b ∅, {soccer}, {tennis}, {basketball}, {soccer, tennis}, {soccer, basketball}, {tennis, basketball}, {soccer, tennis, basketball} 17 The complement A′ contains all elements in U that are not in A. Since U includes all possible elements, A′ only contains elements from U, making it a subset of U. 18 a ∅, {apple}, {banana}, {orange}, {mango}, {apple, banana}, {apple, orange}, {apple, mango}, {banana, orange}, {banana, mango}, {orange, mango}, {apple, banana, orange}, {apple, banana, mango}, {apple, orange, mango}, {banana, orange, mango}, {apple, banana, orange, mango} b 24 = 16

b {oak, pine, maple, birch} 7 a {Bob} b {Alice, Bob, Clara, Dave, Eve} c No, since A ∩ B ≠ ∅. 8 a ∅ c {a, b, c, d, e, f} 9 a {sushi} b {pizza, burger, sushi, pasta} c No, since A ∩ B ≠ ∅. 10 a {1, 3, 5, 7, 9} b ∅ c {1, 2, 3, 4, 5, 6, 7, 8, 9} 11 a {blue} c {red, blue, green} 12 a Yes, A ∩ B = ∅.

b 2 − 1 = 31

21 The empty set has no elements, so it satisﬁes the condition that all its elements (none) are in any set, making it a subset of every set. 22 If A is a subset of the universal set and the universal set is also a subset of A, it means that every element exists in both sets. As such, there are no elements in U that are not in set A, and hence the complement must be the empty set.

9.02 Set operations and complements What do you remember? b i

c ii

d iv

2 a False

b True

c False

d False

3 a {1, 3, 5}

b {2, 4}

4 a {3}

b {1, 2, 3, 6, 9}

14 a {3, 4, 5}

b ∅

c {1, 2, 4, 5}

15 a {poetry} b {fiction, biography, poetry, history} c No, since A ∩ B ≠ ∅. 16 a {x, y, z, w}

b {x, y}

17 a Yes, A ∩ B = ∅.

c ∅

b No, A ∩ B = {2}.

Extend your thinking 18 a A = {Jack, Lily, Mia}, B = {Jack, Lily, Noah, Olivia} b No, since A ∩ B = {Jack, Lily} ≠ ∅. includes attendees not in both A and B. = {James, Since A ∩ B = {Jane, Jack}, Jill, John, Jenny}.

20 A ∩ B = {Cara} ≠ ∅, so not disjoint. Changing A to {Amy, Ben} makes A ∩ B = ∅. Then, A ∪ B = {Amy, Ben, Cara, Dan, Ella}. 21 a {Eva, Finn, Gina, Hugh, Iris}

Practice b {1, 2, 3, 4, 5, 6, 7}

c {2, 4, 6}

924

c No, since A ∩ B ≠ ∅.

1 a iii

5 a ∅

b No, A ∩ B = {dog}.

b {Emma, Finn, Grace, Harry, Ivy}

b No, because c and d are in Ac but not in Bc. 20 a 2 = 32

b {green, yellow}

13 a {Grace}

19 a Ac = {c, d, e}, Bc = {a, e} 5

b {a, c, e}

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b {Hugh, Iris, Jack} 22 Use A ∪ B and to ﬁnd A. Since = {Nick, Owen, Paul}, A = {Kate, Leo, Mia}. Then, B = {Kate, Leo, Nick, Owen} using A ∪ B and A ∩ B.

9.03 Venn diagrams What do you remember? 1 a A∩B

b A∪B

b 10

15 a 15

b 40

16 a 20

b 15

17 a 55

d U

14 a 15

b 45

2 A ∪ B = A + B − A ∩ B

Extend your thinking

3 a True

18 a 30

b True

c False

4 20

19 a

c 30

b 25

c 20

Practice 5 a

25 Reading

Hiking 10

6 a

C b 25

Tablet

15 15

b 60 7 a

c 15 Biology

Chemistry

c 20

d 100

20 Without subtracting A ∩ B, the employees who speak both languages are counted twice (once in A and once in B). The formula ensures these employees are counted only once, giving the correct total of 50 + 40 − 25 = 65 employees who speak at least one language. 21 a The student added A and B without subtracting A ∩ B, double-counting the 20 people who enjoy both hobbies. b 55 22 a 40

c 10 Smartphone

20 10

b 60

d 15

b 20

c 40

9.04 Probability and events

What do you remember? 10 b 12

c 8

d 10

8 a 30

b 25

c 15

9 a 20

b 15

c 8

10 a 30

b 80

11 a 45

b 0

12 a 15

b 15

13 a 18

b 15

1 a S = {1, 2, 3, 4, 5, 6} c A = {2, 4, 6}

d 10

2 a i

b ii

b S = 6 d c iii

d iv

Practice 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.

c 10

Answers mathspace.co

925

4 a

9.05 Mutually exclusivity

What do you remember?

c 5 a P (A ∩ B) = 0

1 Two events that cannot have simultaneous outcomes in the same chance experiment.

b P (A ∪ B) = 1

6 a S = {R1, R2, R3, B1, B2, G1}

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.

c A ∪ B = {R1, R2, R3, B1, B2} d

4 P(A ∪ B) = P(A) + P(B)

7 a A ∩ B = {5}

5 It indicates that events A and B are mutually exclusive, meaning they cannot occur simultaneously.

c A ∪ B = {1, 3, 4, 5, 6} d 8 a

9 a

Practice 6 a A = {13 hearts} b B = {13 spades}, A ∩ B = {13 hearts} ∩ {13 spades} = ∅, so A and B are mutually exclusive.

c P (A ∪ B) = 1 10 a A ∩ B = {HH}

b c

Extend your thinking 11

7 a C = {blue1, blue2}

8 a

, P (A ∩ B) = 0

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 ≠ ∅.

14 15

b , it means picking a number that is

14 a

both a multiple of 3 and even. b

926

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9.06 Multistage events and conditional probability

15 a b

What do you remember? 16 a A = {HH, HT, TH}, B = {HH}

1 a A multistage event involves multiple steps, each with its own set of outcomes.

b 17 a

b Conditional probability is the probability of an event A occurring given that another event B has occurred, denoted P (AB).

18 a

c It represents the probability of event A occurring given that event B has already occurred.

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. , P (A ∩ B) = 0 (mutually

exclusive). Then 21 The complement rule applies as

. 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):

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

c False

3 a It arranges outcomes in rows and columns to show all possible combinations, making it easier to count favorable outcomes. b Each branch represents an independent or conditional probability, and multiplying them gives the probability of the combined outcome. 4 a iii

b i

c ii

Practice 5 a C 4 7

23 Addition rule:

b True

, P (red ∩ blue) = 0, so

3 7

R By outcomes:

C CC

1 2 1 2

R CR C RC

4 6 2 6

R RR

red ∪ blue = {r1, r2, r3, r4, r5, b1, b2, b3, b4},

A ∪ B = 9, so

b 6 a

M S

Answers mathspace.co

927

7 a

c 1

12 a

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.

2 5 3 5

R 3 5 2 5

8 a 4 b {RR, RB, BR, BB}

2 5 3 5

c d A tree diagram 9 a

1 2 1 2

H 1 2 1 2

1 2 1 2

10 a

4 7 3 7

4 6 2 6

11 a

928

2nd Dark

2nd Milk

1st Dark

1st Milk

15 a

R BR

B RB

13 a

R RR

B BB

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N NN d

14 a 36 TT

R NR

H TH

N RN

1 2 1 2

H HH

R RR

d H 1 4 3 4

1 4 3 4

NH HNH

NHH

NH NHNH

16 a

Coffee

Tea

20 a R

4 5

1 5

N RN R NR

3 5 2 5

R RR

3 5 2 5

N NN

c c

T C

Extend your thinking

21 a

3 5 2 5

R 3 5 2 5

3 5 2 5

18 a

1 3

R RR

2 3

. G GG c

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.

1st Dark

22 a

1st Milk

6′ E 6′

Probability of winning the prize and rolling a 6:

2nd Milk

6 E6

R GR

2nd Dark

19 a 36

6′ P 6′

1 6 5 6

G RG

6 P6

1 6 5 6

P 17 a

Mathematics

Science

c 70

30 50

Answers mathspace.co

929

So, different restricted sample spaces cause the inequality.

From the table: P (MS) = P (SM) =

and

c New total would be 201, math has 121, while both has 50. P (SM) =

, 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

6 6

3 outcomes for E and H out of 12:

B RB R BR

1 2 1 2

B BB

B B R

E 1 2

1 P (AB) =

, provided B ≠ 0 , provided P (B) ≠ 0

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.

24 a

What do you remember?

2 P (AB) =

Probability:

1 2 1 2

H EH

4 a

5 a 2

b 3

Practice T ET 6 a

1 2

1 2 1 2

Probability of E and H:

H OH

S A

T OT .

930

T H

9.07 Conditional probability formulas

3 5

R RR

1 4 3 4

Mathspace New South Wales – Year 11 Advanced mathspace.co

25 10

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

9.08 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

= P (A), so independent.

b 0.55

Answers mathspace.co

931

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 9 review 1 C

13 a P (A ∩ B) =

, P (A) × P (B) =

, so

independent.

3 C

b P (A ∪ B) =

4 a {Leo, Noah}

c P (AB) =

b 2 c {Mia, Ava}

d P (BA) = 14 a 0.001

b 0.729

15 a b P (AB) =

2 C

= P (A), so independent.

5 a 23 = 8 subsets. Each element can be included or excluded, giving 2 choices for each of the 3 elements. b {∅, {Math}, {Art}, {History}, {Math, Art}, {Math, History}, {Art, History}, {Math, Art, History}} 6 a A′ = {r, s, t}, B′ = {p, t}

16 a 0.12

b 0.68

17 a 0.42

b 0.46

18 a 0.0025

b 0.0975

19 a

b No, because p ∈ B′ but p ∙ A′. 7 a {Chloe} b {Alex, Ben, Chloe, David, Eva} c No, since S ∩ B ≠ ∅. 8 a {Noah}

b P (AB) =

= P (A), so not

independent.

b {Liam, Olivia, Noah, Emma, Sophia} c No, since S ∩ T ≠ ∅. 9 a A = {Ken, Laura, Mike}, B = {Ken, Laura, Nora, Oscar} b No, since A ∩ B = {Ken, Laura} ≠ ∅.

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Mathspace New South Wales – Year 11 Advanced mathspace.co

10 a

21 a

Arts

Physics

B 12

9 6

b 12

c 9

11 a

4 7

3 6

3 7

4 6

22 a

Laptop

b 25

c 15

d 5

12 a 23

b 20

c 35

13 a 60

b 40

c 100

14 a {4, 6}

15 a

b A

D′

L′

23 a c

Roll 1

Roll 2

b 24 a

18 a No, because black queens exist, so A ∩ B ≠ ∅.

b 19 a 0

Desktop

B 2

17 a

Y YY

B YB

B b

Y BY

2 6

d 6

B BB

3 6

b 1

20 The claim is incorrect because A ∩ B = ∅ (red jacks exist). Venn diagram: circles for A and B overlap. Correct formula: P (A ∪ B) = P (A) + P (B) − P (A ∩ B). Thus, P (A ∪ B) =

Pool

Gym

110

70 50

Answers mathspace.co

933

P′

110

G′

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.

P (GymPool) =

P (PoolGym) =

. Different sample

spaces cause inequality. c P (PoolGym) =

, less than original

25 a

26 a

29 a

ii {0, 1, 2} b i Y is the number shown on the die. 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.

b Table shows Pop ∩ Under 30 = 30, Under 30 = 40. Thus,

28 a 0.0196

4 a i X is the number of heads obtained in two coin ﬂips.

ii {1, 2, 3, 4, 5, 6}

27 a

P (PopUnder 30) =

Practice

. b

≈ 0.286 c

ii {1, 2, 3, 4} 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}

30 a

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. 9 a The possible outcomes are red or blue.

31 a 0.0784

d b 784

10.01 Random variables What do you remember? 1 A variable whose value is the numerical outcome of a random experiment.

934

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b {0, 1} 10 a G is the genre selected, with action = 1, comedy = 2, drama = 3, sci-ﬁ = 4. b {1, 2, 3, 4} c Discrete 11 a J is the volume of juice dispensed in millilitres. b Continuous

Extend your thinking

6 a

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.

13 Possible values: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Discrete, as the values are countable. 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.

0.3

0.1

0.3

0.1

10.02 Organise and graph datasets What do you remember?

1 The number of times a particular value or group of values occurs in a dataset.

7 a

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

0.25

number of data values.

b 6.5

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.

8 a

7 6

0.2 b

Frequency

Practice 5 a

0.167 0.083

0.267 0.333 0.133 0.067

5 4 3 2 1 0

0.2

0.267

0.333

0.2

0.467

0.8

0.933

Number of hours b Mode is 2, with the highest frequency of 7.

0.133 0.067

Answers mathspace.co

935

9 a

11 a

Relative frequency

0.7

Class Interval

0.6 0.5 0.4 0.3 0.2 0.1 0.0

Frequency

Cumulative Frequency

b Mode is 6, with the highest relative frequency of 0.333. c Cumulative frequency

16 14 12 10 8

20 15 10 5 0

4 2 0

c The median is approximately 31.7; the modal class is 30–40.

12 a

Cumulative frequency

10 a 7 6 5 4 3

15 10 5 0

Number of hours

b 4.5 10

Race time (minutes)

b Mode is the interval 20–25, with the highest frequency of 8. c 20.0 17.5

15 10 5 0

15.0

Number of hours

12.5 10.0 7.5 5.0 2.5 0.0

c 20

Cumulative frequency

936

Daily water consumption

Quiz Score (out of 10)

Frequency

Cumulative frequency

Quiz Score (out of 10)

Cumulative frequency

10–20 20–30 30–40 40–50 50–60

15 20 25 Race time (minutes)

Mathspace New South Wales – Year 11 Advanced mathspace.co

Extend your thinking

Cumulative frequency

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.

20 15 10 5 0

100

Test scores b Mode is the interval 40–60, with the highest frequency of 8. c 53.75 14 a

0.333

0.267

0.067

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. 17 a 25 Cumulative frequency

13 a

Frequency

10 5 0

15 20 25 Temperature

b Approximately 20°C

Number of hours student slept Mode is 6 or 7 (frequency 5).

Cumulative frequency

14 12 10 8 6

8 6

Group A Group B

4 2 0

Hours spent on extracurricular activities

2 0

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

Cumulative frequency

Number of hours student slept

b • Group A: The median is 2, corresponding to a cumulative frequency of 5. • Group B: The median is 3, corresponding to a cumulative frequency of 5. Group B’s median is higher, indicating more hours spent on activities.

Answers mathspace.co

937

10.03 Analyse data

c The difference is small and likely due to random variation, not strong evidence of bias.

What do you remember?

d More ﬂips (600) would reduce variability, likely making the estimate closer to 0.5.

1 The proportion of times an outcome occurs,

14 a 0.24

calculated as

P (X = x) ≈

for a random

variable X. 3 a True

b True

c False

d False

Practice 4 a P (X = 4) ≈ 0.25

≈ 0.167. The

b Theoretical P (X = 2) = estimate is higher.

2 It approximates the probability as

b P (even) = 0.43

5 a P (defective) = 0.075 b P (not defective) = 0.925 6 a P (defective) ≈ 0.075

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.

Chapter 10 review 1 C

b P (not defective) ≈ 0.925

2 B

c 60 widgets

3 B

7 a P (online) ≈ 0.4 b P (not online) = 0.6 c 200 students

4 a H is the number of heads obtained in four coin tosses. b {0, 1, 2, 3, 4}

8 a 0.18 b P (X = 5) ≈ 0.18 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

5 a The random variable C represents the outcome of the draw, encoded numerically. 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. 7 a

b P (not heart) ≈ 0.767

0.25

0.35

0.25

0.1

0.05

0.25

0.35

0.25

0.1

0.05

0.25

0.6

0.85

0.95

c 117 hearts Extend your thinking 12 P (heads) ≈

= 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 b Theoretical P (heads) = 0.5. The estimate is slightly lower.

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12 th

b The median is the average of the 6 and 7 values. From cumulative frequency, the 6th value is 9 and the 7th value is 10. Median is

Cumulative frequency

8 a

30 25 20 15 10 5 0

= 9.5. x

0.25

Frequency

11 a

50 40 30 20 10 0

Plant height (cm)

1 0

b The modal class is 20 − 30, with a frequency of 18.

c The median is the 25th value. On the polygon, at cumulative frequency 25, the height is approximately 26 cm.

Number of hours

b Modes are 5, 7, 8, each with a frequency of 5. c 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.

9 a

12 a

10 5 0

Number of hours

Frequency

d 7 10 a 10

Frequency

Commute times (minutes)

Cumulative frequency

3 2

Number of hours

The modes are 7, 8, each with a frequency of 5.

2 0

Commute times (minutes)

b The modal class is 20 − 30, with a frequency of 9.

Answers mathspace.co

939

11.01 Average rate of change

Cumulative frequency

c 16 14 12

What do you remember?

10 8 6 4 2

change in y divided by the change in x.

2 It represents the gradient of the secant line connecting (a, f (a)) and (b, f (b)).

Number of hours

3 The average rate of change is equal to the gradient m.

d 8 13 a Cumulative frequency

, the

1 The average rate of change is

4 A secant line is a straight line passing through two points on the graph of a function.

16 14 12 10

Class B

Class A

6 4

5 a 2 e −1

2 0

Quiz scores

b The median is the 8 score. For Class A, the polygon shows the 8th score is 8. For Class B, it’s 7. Class A’s median is higher. 14 a 0.14

b 0.385

c 70 times 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. 19 a 0.4 b The theoretical probability of rolling a 5 or 6 is

b 2

c 3

d 2

f 0.707

6 a 100 litres/hour

b 100 litres/hour

7 a 60 km/h b 60 km/h c It is constant, equal to the gradient 60. 8 a 2 m/year b 3 m/year

15 a 0.04

≈ 0.333. The experimental probability of

0.4 is higher. 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 theoretical probability of .

940

Practice

Mathspace New South Wales – Year 11 Advanced mathspace.co

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

12 a −1

c 0.2

b 1

13 a $173

d 3

b $233

Extend your thinking 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. 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

11.02 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.

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 m/s

b 60 m/s

c 50 m/s

d 50 m/s

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 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

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 mathspace.co

941

11.03 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.

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Mathspace New South Wales – Year 11 Advanced mathspace.co

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.

11.04 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

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.

11.05 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 mathspace.co

943

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 11 review

b 6.015 m/s

17 50 km/h

1 B

18 a 14.03 m/s

2 B 3 C 4 a 3

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

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

c Both are 80 km/h. The linear function has a constant rate of change.

944

b 40.05 m2/min

Mathspace New South Wales – Year 11 Advanced mathspace.co

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.

12.01 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 mathspace.co

945

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

12.02 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.

946

Mathspace New South Wales – Year 11 Advanced 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

12.03 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 mathspace.co

947

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.

12.04 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 the function. That is,

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.

948

Mathspace New South Wales – Year 11 Advanced mathspace.co

[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

[ 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 mathspace.co

949

12.05 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

950

ii iv −2

Mathspace New South Wales – Year 11 Advanced mathspace.co

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.

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

12.06 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 mathspace.co

951

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

= 36(4x + 3)8

9 a

= −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)

15 a y = −8x − 7

b y = 16x − 15

16 a

b f ′(−2) = −3

17 x = 1 or x = 6

18 m = = −5(x + 6)−6

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

= 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.

10 a x = 5

952

= 4(3x2 + k)3 × (6x).

b x = 2 or x = −6

Mathspace New South Wales – Year 11 Advanced mathspace.co

12.07 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 mathspace.co

953

12.08 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

. iii

954

ii No, numerator is −12.

Mathspace New South Wales – Year 11 Advanced 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) =

12.09 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 mathspace.co

955

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

e 7 a

b y = 5x − 9

d y = 21x − 33 b

8 a

with a very

small value for h, such as 0.001.

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 .

−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

12.10 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.

956

Mathspace New South Wales – Year 11 Advanced mathspace.co

−15

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 mathspace.co

957

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

x 1

−1

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

y = f ′(x) 1

x 2

−2 −3

958

. So, f ′(2) ≈ 32.

b Approximately 32.24

−3

x 0

The gradient is

−2

P (2, 16)

Mathspace New South Wales – Year 11 Advanced mathspace.co

d i The gradient is −1. ii Approximately −1.101

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.

12.11 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 mathspace.co

959

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 12 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.

960

Mathspace New South Wales – Year 11 Advanced mathspace.co

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:

12 a = 3, b = 5, c = −1

13.01 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

d iv

b False

c False

d True

3 a Growth b Decay

c Decay

d Decay

4 a Not exponential

b Exponential

2 a True

16 17

c i

= 5(2x − x3)4(2 − 3x2)

18 x = 0, x = −2, x = 2 19 a = 2, c = 1 20

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 mathspace.co

961

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

962

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

Mathspace New South Wales – Year 11 Advanced 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) =

13.02 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

c iii

5 a i m ≈ 0.184 ii m ≈ 0.182 Extend your thinking

6 a i m ≈ 1.623 ii m ≈ 1.515 20 a 5

iii m ≈ 1.505

b The gradients converge to approximately 1.505, showing a reliable estimate of the tangent gradient.

21 2409.20 22 23 a

iii m ≈ 0.182

b The gradients are consistent, indicating a stable approximation of the tangent gradient at approximately 0.182.

19 a

d ii

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

e m ≈ 0.263

f m ≈ 1.396

g m ≈ 0.792

h m ≈ −0.105

i m ≈ 1.344

j m ≈ 0.532

8 a i m ≈ 0.960 ii m ≈ 0.921

iii m ≈ 0.917

b 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.

Answers mathspace.co

963

9 a i m ≈ 1.335

= kex

ii m ≈ 1.261 iii m ≈ 1.254 b The gradient for y = 3.5 (m ≈ 1.254) is greater than for y = 3x (m ≈ 1.099), indicating a steeper tangent as the base increases. 10 a m ≈ 0.833

b m ≈ 1.436

c m ≈ 0.337

d m ≈ −0.357

e m ≈ 0.470

f m ≈ 0.876

g m ≈ 1.195

h m ≈ −0.511

i m ≈ 1.570

j m ≈ 0.642

k m ≈ 1.065

l m ≈ −0.693

11 Ordering from smallest to largest:

−0.223, 0.406, 0.696, 1.105, 1.261, 1.396

Corresponding to: y = 0.8x, 1.5x, 2x, 3x, 3.5x, 4x

derivative is kax where k = ln a. Practice 5 a e3

b e

6 a

b e2

c 1

d 2

7 a y=x+1 4

b 4

c y = e x − 3e

d y = 3x + 3 − 3 ln 3

8 a First: 4e , Second: 4e b First: ex, Second: ex

c First: 2ex, Second: 2ex d First: ex + 1, Second: ex e First: 5ex + 2, Second: 5ex

Extend your thinking 12 As h → 0, the secant line between (0, ka0) 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). 13 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

b m ≈ 0.696

c m ≈ 0.921

d m ≈ 1.105

13.03 Euler’s number and derivatives What do you remember? 1 Euler’s number e ≈ 2.71828182845... is an irrational constant where

f First: ex − 2x, Second: ex − 2 9 a y = −x + 1 2

(ex) = ex.

2 The derivative of y = ex is ex, meaning the function equals its own derivative.

Mathspace New South Wales – Year 11 Advanced mathspace.co

d y = − e3(x + 3) + e−3

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

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

13 a (0.69, 2)

14 a m ≈ 0.406

964

(ax) = ax; for other a, the

4 Only for a = e does

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 b c

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

13.04 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 mathspace.co

965

9 a y-intercept: (0, 2000). Asymptote: R = 0 b

12 a

R 2500

R = 2000(1.1t)

2000 (0, 2000)

Time (months) (t)

Efficiency (%) (E)

80 68

57.8

49.13

1500

E 80 (0, 80)

1000

E = 80(0.85t)

500 t

R=0 0

c R→∞

d $2928 10 a

Time (hours) (t)

Concentration (mg) (C)

500 475

c E→0 d 35 percent

451.25 428.6875

13 a y-intercept: (0, 150). Asymptote: N = 0 b

C 500 (0, 500)

N 800

C = 500(0.95 )

400

600

N = 150(2.5t)

400

300

200

(0, 150) t

N=0

100 t

C=0 0

Extend your thinking

11 a Domain: t ≥ 0

14 a Domain: t ≥ 0

Range: F ≥ 300

Range: y ≥ 1000

b y→∞

600

c 1221

F = 300(1.4t)

d k ≈ 0.139

(1, 420)

400

e Approximately 259

(0, 300) 200 t 0

d 5859 cells

d 5 hours

c N→∞

c C→0

c F→∞ d 2259 ﬁshes

966

E=0 1

Mathspace New South Wales – Year 11 Advanced mathspace.co

15 a y-intercept: (0, 200). Asymptote: M = 0 b

5 a

−3

−2

−1

150

M = 200(e−0.05t)

−1 −5 −25 −125

200 (0, 200)

b As x → ∞, y → −∞.

100

As x → −∞, y → 0−.

Horizontal asymptote: y = 0 approached from below as x → −∞. t

M=0 0

c Domain: x ∈ . Range: y < 0.

c 121 grams

6 a (0, 1)

d 14 days

b No, since 3x > 0 for all x ∈ .

e 22.1%

c x∈

16 a Domain: t ≥ 0

d y>0

Range: B ≥ 4000

e 243

f m = 81

B = 4000(1.08t) (1, 4320) 4000 (0, 4000)

5000

7 a

3000 2000 1000 t 0

c $5039

d 3 years

f (n)

g(n)

162

486

b f (n) increases arithmetically by 4. g(n) increases geometrically by a factor of 3. 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 13 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)

y=0 −2

−1

b c m ≈ 4.176

Answers mathspace.co

967

d The tangent gradient at x = 0 is k ln a. Increasing k proportionally increases the gradient; a negative k reverses its direction. x

12 For f (x) = ka , 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.

b A→∞ c $9733 d 6 19 a y = 4−x b Reﬂection across the y-axis. c 7

d e2

14 a y = x + 1

b y = e−2 x + 2e−2

c y = ex

d y = (e3 )x − 2(e3 )

13 a e−3

b e3

c 1

5 3 2

15 a y = e2(x − 1)

y = 4x

−3 −2 −1

x 1

−1

d (0, 1) 0.05t

16 No solution for t ≥ 0, as P ′(t) = 100e 17 a

Time ( years) (t)

≥ 100.

, as

20 a y-intercept: (0, 300). Horizontal asymptote: M = 0.

Value 25 000 21 250 18 062.5 15 353.125 ($)(V ) b

e For y = 4x, as x → ∞, y → ∞. For y = x → ∞ y → 0+.

M 300 (0, 300) 250

200

25 000 (0, 25 000)

150

20 000

100

15 000

50 t

M=0

10 000 0

5000

V=0 1

c 201 d 35

c V→0

14.01 Logarithms

d $11 093 18 a

What do you remember?

A 8000

1 a True

(2, 8652.8)

(0, 8000)

2 a log2 (16) = 4

6000

c log (0.1) = −1 3 a True

4000 2000

A=0 0

968

b True

t 3

Mathspace New South Wales – Year 11 Advanced mathspace.co

b True

c True b ln (5) = x d loga (x) = y c True

d False

14.02 Exponential-logarithmic equivalence

Practice 4 a log3 (9) = 2

c ln (e) = 1

f logb (m) = n

What do you remember?

5 a 62 = 36

1 a True

b 102 = 100

b True

2 a log3 (9) = 2

c True b log(1000) = 3

c ln(e) = 1

e log4 (64) = 3

g ln(1) = 0

i log6 (36) = 2

j logb (m) = n

3 a

d 3 = 27

c 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

10 a 2.8751

b ac = b

d 10−3 = 0.001

Practice

c e

4 a x = 1.1761

b x = 2.3026

c x = 3.9120

d x = −1.3010

d 0.001

b 3.9120 d −2.0969

c 2.9957 11 a log7 (343) = 3

b log11 (121) = 2 d ln(x) =6

12 a 9

b 1

c 0.01

d 64

13 a True

b True

c False

d True

b x=2

c x = 12

d x=3

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

5 a e x=4

8 a 52 = 25 −2

c 10

Extend your thinking

b e3 = e3

= 0.01

d 23 = 8

14 a 13.86 years

b 13.52 years

e ac = b

f 70 = 1

c 10.14 years

d 20.27 year

h 104 = 10 000

15 0.0001 mol/L y

16 For a ≤ 0, a 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. 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.

9 a x = 3.9069

b x = 2.7268

c x = 1.2091

d x = 1.6610 b x = ln(7)

10 a c 11 a 1.771

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

Answers mathspace.co

969

9 a LHS:

Extend your thinking 13 60 dB

RHS:

15 x = ±2

−log4 (16) = −log4 (42) = −2

16 17 loga (b) is undeﬁned for b ≤ 0 or a = 1.

Thus, the identity holds. b LHS:

14.03 Logarithm laws and properties What do you remember? 1 a Product law c Quotient law 2 a 1

b 0

b Quotient law

RHS: −log4 (25) = −log5 (52)

d Special property c −1

d 5

3 a • Product law: loga (xy) = loga (x) + loga ( y)

= −2 Thus, the identity holds. c LHS:

= loga (x) − loga ( y)

• Quotient law: loga

• Power law: loga (xn) = n loga (x)

• Inverse property 1: loga (ax) = x

RHS: −log7 (49) = −log7 (72)

• Inverse property 2:

= −2

• Identity property: loga (a) = 1

Thus, the identity holds.

• Zero property: loga (1) = 0 • Reciprocal property: loga

= −loga (x)

d LHS:

Practice 4 a 1

b 0

c −1

d 3

e 11

f −2

h 4

RHS: −log2 (8) = −log2 (23) = −3 Thus, the identity holds.

5 a log10 (21)

b log4 (4)

c log6 (125)

d log3 (3)

e log10 (10)

f log7 (81)

c log5 (12.5)

d log3 (9)

b log8 (z) − log8 (64)

e log2 (81)

f log7 (343)

6 a log5 (6) + log5 ( y) c 3 log4 (x) + log4 (z)

d 4 log10 ( p) − 3 log10 (q)

e log3 (2) + log3 (x)

f log6 ( y) − log6 (36)

g 4 log2 (a) + log2 (b) 7 a log5 (50)

b log4 (8)

11 a

h 2 log5 (m) − 5 log5 (n)

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)

970

10 a log6 (18)

Mathspace New South Wales – Year 11 Advanced mathspace.co

12 a −3

b 5

e a2

f −r

c k

d −2

13 a loga(b2c)

7 a w = 173

8 a

9 a w ≈ 1.7075

b w ≈ 2.1414

c w ≈ 0.8416

d w ≈ −0.8239

10 a x = 2

14 Ask your teacher for worked solutions

11 a 1.9842

b 0.1878

15 Ask your teacher for worked solutions.

12 a 2, 3

b −3, −2

Extend your thinking

13 30

14.04 Logarithm expressions and equations

What do you remember? 1 a True 2 a

b True

16 x = 8, y = 8 c True b w=a

d True m

3 a A logarithm that evaluates to a rational number (an integer or fraction). 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. d An expression that contains a logarithm, such as loga (x). 4 a Rational

b Rational

c Irrational

d Irrational

5 a w = 121

15 x = 8, x = 1

b w=2

Practice 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. d log10 (1000) = 3.0000. The result is rational. e log17 (2) ≈ 0.2447. The result is irrational.

17 t = 4.00

14.05 Logarithmic graphs What do you remember? 1 a (1, 0)

b x=0

2 a y = loga (x)

c x>0

d y∈

b (1, 0)

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. Now, we can test for the y-intercept by substituting x = 0 into this exponential form. This gives us the equation: 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.

g log13 (676) ≈ 2.5404. The result is irrational.

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.

h log15 (90/7) ≈ 0.9431. The result is irrational

4 a x=0

f log10 (9) − log10 (9) = 0. The result is rational.

b (1, 0)

Answers mathspace.co

971

11 Range: [0, 2]

Practice 5 a True

b False

c True

6 Domain: x > 0, Range: y ∈ 

(169, 2)

2 x=0

y 3

x=0

2 1

(1, 0) 2

−1

(1, 0)

100

150

−2

12 a iv

−3

b iii

c i

13 • Graph A: y = log4 (x) 7 y = log13 (x) is closer to the x-axis because higher bases result in ﬂatter graphs. 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

b y∈

10 a

f (x)

0.07

0.13

c x=0

d (1, 0)

• Graph B: y = log9 (x) • Graph C: y = log20 (x) 14 h < 0 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).

maps to .

d (2, 25) on y = 5x maps to (25, 2) on y = log5 (x) 16 a

4 3

x = −3

x=0

2 1

y = 2x

−4 −3 −2 −1 −1 x

(4, 0) −2

y=x

−2 f (x) = log7 (x + 3) − 1 −4

c Domain: x > −3, Range: y ∈ 

(0, 1) (1, 0) y = 0 x −2 −3

2 3 4

y = log2 (x)

−4

They are reﬂection across the y = x. The point (0, 1) on y = 2x maps to (1, 0) on y = log2 (x). b

3 2

y=0

1 (0, 1)

−4 −3 −2 −1 −1

y=x

−2 −3 −4

x=0

972

Mathspace New South Wales – Year 11 Advanced mathspace.co

(1, 0) 1

2 3 4

y = log 1 ( x) 2

They are reﬂection across the y = x.

(−1, 4) 4

maps to (1, 0) on

The point (0, 1) on .

y = log 1 ( x)

2 1

x=0

(1, e)

y=e

y=0 x (1, 1 0) 2 3 4

−4 −3 −2 −1 −1 −2

y=x

−3

−2

y = log5 (x)

y=4

−4 −3 −2 −1 −1

y=x

−2 −3

(−1, 6) 6 y=x

4 2

(4, 1)

x 1

They are reﬂection across the y = x. The point (1, 6) on y = 6x maps to (6, 1) on y = log6 (x).

(1, 4)

(6, 1)

−2

3 2

y=x

y = log6 (x)

They are reﬂections across y = x. The point (0, 1) on y = 5x maps to (1, 0) on y = log5 (x). 4

−2

−3

(1, 6)

y = 6x

1 (0, 1) (1, 0) y = 0 x −1

maps to (4, −1)

y 6

y=5

−3

The point (−1, 4) on . on

x=0 y 3

−3 −2 −1

(4, −1)

They are reﬂection across the y = x.

y = In(x)

They are reﬂections across y = x. The point (1, e) on y = ex maps to (e, 1) on y = ln(x).

2 3 4

−4

−2

y=x

(e, 1)

1 (0, 1)

−4 −3 −2 −1 −1

−2

2 3 4

2 −2

(6, −1)

y = log4 (x)

−4

They are reﬂection across the y = x. The point (1, 4) on y = 4x maps to (4, 1) on y = log4 (x).

They are reﬂection across the y = x. The point (−1, 6) on on

maps to (6, −1)

Answers mathspace.co

973

17 a

−2

y = 3x −1

y=x

x 1

y = log3 (x)

−2 −2

−1 −2

y = 8x

y=x 18 a

y = log8 (x) 2

−2

grows

(1, 0) −2

−1

−2

decreases,

6 4

y 2

x 2

also

decreases. They are inverses, so their graphs are reﬂections across y = x (e.g., (0, 1) maps to (1, 0)).

y=x

x 2

−1

y=x

(0, 1)

With a larger base, the exponential function y = 8x grows even more rapidly, and the logarithmic function y = log8 (x) grows more slowly.

−2

rapidly as x increases, while its inverse, , grows slowly

−4

−2

The exponential function

−2

−4

The exponential function y = 3x grows rapidly as x increases, while its reﬂection, the logarithmic function y = log3 (x), grows slowly. 8

x 2

y=x

−3

(1, 0)

−4

The exponential function

(0, 1)

−2

y=x

grows

−1

−1 −2

rapidly, and the logarithmic function grows slowly, with their graphs being reﬂections across y = x.

974

Mathspace New South Wales – Year 11 Advanced mathspace.co

decreases,

also

decreases. They are reﬂections across y = x (e.g., (0, 1) maps to (1, 0)).

g(x) = − log7 (3 − x). This function intersects f (x) twice. This means:

y 2

• h2 = 3

1 (0, 1) −2

−1

• k2 = 0

(1, 0)

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

−2

decreases,

also

(2.62, 0.50)

decreases. They are reﬂections across y = x (e.g., (0, 1) maps to (1, 0)). d

x 1

−1

2 1

(0, 1)

−1

−2

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)

25 79.03 dB

−1

y=x

(1, 0) −2

Extend your thinking 20 h = 0.5, k = 1 21 t ≈ 12.34 22 h = 2, k = 1 23 t ≈ 8.67 24 A possible solution requires choosing simple functions that meet the criteria. Let’s choose f (x) = log7 (x). This means: • h1 = 0 • k1 = 0 For g(x), we need a function that is reﬂected both horizontally and vertically. Let’s choose

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. 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

c The point (4, 0.5) is on the logarithmic graph. Its reflection is (0.5, 4). Check this point in the exponential function: = 4. The point satisfies the y = 160.5 = equation, confirming the inverse relationship.

Chapter 14 review 1 B 2 C 3 C 4 a log4 (64) = 3

c ln(e2) = 2

Answers mathspace.co

975

5 a 72 = 49

b 103 = 1000

• Range: y ∈ .

d 2 = 64

y 3

6 a 3

b 4

7 a 17.33 years 8 a 3.16 × 10

−6

c 2

d −5

b 8.93 years

(−2, 0)

−4 −3 −2 −1 −1

mol/L

9 a x ≈ 2.9299

b x ≈ 2.4307

c x ≈ 1.3383

d x ≈ −0.3219

10 a x ≈ −0.2041

b x = 5.0000

11 a 2.096

c 2.113

25 h = 2, k = 2

b 1.302

d 3.807

12 x = ±2 13 a 1

b −2

f 0

c 4

14 a log2 (35)

b log3 (27)

c log5 (16)

d log4 (45)

15 a log3 (5) + log3 (a) c 5 log2 (x) + log2 ( y)

d 5

1 (0, 1) 1

x 2 3 4

−2 −3

26 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).

b log7 (z) − log7 (49) d 2 log(m) − 5 log( p)

17 158 18 a log2 (80), irrational

b log3 (27) = 3, rational

19 a w = 127, rational

b w = 80, rational

20 a w = 3

21 a x = 80

b x = 513

maps to (4, −1) on

d Point (−1, 4) on .

27 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. 28 a y = log3 (x) b y = 3x

c Ask your teacher for worked solutions.

15.01 Reflections in axes

22 t = 12 hours 23 a x > 0

b y∈

c x=0

d (1, 0)

24 • Asymptote: x = −4. • x-intercept: (−2, 0). • y-intercept: (0, 1). • Domain: x > −4.

976

Mathspace New South Wales – Year 11 Advanced mathspace.co

What do you remember? 1 a Reﬂects the graph in the y-axis. b Reﬂects the graph in the x-axis. b (x, −y)

2 a (−x, y) 2

3 a The function y = x is even, so f (−x) = (−x)2 = x2 = f (x), meaning the graph is symmetric about the y-axis and remains unchanged.

d i y = −(−x) − 2 = x − 2

y 1

y = x2

−1

y = −x − 2

(0, 0) −2

−1

1 −1

y=x−2

−2

−3

4 a False

b True 6 a i y = −(x2 + 2) = −x2 − 2

Practice

5 a i y = (−x)2 − 1 = x2 − 1 ii

y 3

y = x2 + 2

y = x2 − 1

−1 x

−1

y = −x2 − 2

−2

−3 −1

b i y = −(−3x + 2) = 3x − 2 ii

b i y = 2(−x) + 1 = −2x + 1 ii

y 3

4 3 2

y = 2x + 1

1 −1

−1

x 1

−1

y = 2x + 1 x

−2

−3

−1

y = 3x − 2

c i y = −(x3 − 1) = −x3 + 1

−2

c i y = (−x)3 + 2 = −x3 + 2 ii

y = −3x + 2

y 3

y = −x + 1

y = −x3 + 2

y = x3 + 2

−1

1 x

−1

y = x3 − 1

1 −1

−1

Answers mathspace.co

977

d i y = −(x − 3) = −x + 3 ii

y 3

y = −x + 3

1 −1

−3 −2 −1

−1

12 7 a y = −3x + 2

b y=x +1

c y = −2x3

d y = x2 − 3

8 a y = −4x + 1

b y = −x2 + 2

c y = −x − 1

d y=x−1

−1

−1 −2 −3 −4

9 a (0, −3) 4

y 5 4 3 (0, 3.14) 2 y = 2π x + π 1 y = −2π x + π x

(3, −2)

−2

(−3, −2)

y=x−3

−3

1 −1

−2

(3, 2)

3 (0, 3)

y = 2x − 3 2

1 −1

y = −2x + 3

−1

y = x2

−2 −3 (0, −3)

−2

−4

−1

y = −x2

(0, 0)

−1

−3

−4

y 2

14 a y = −(x + 2)2 b (−2, 0)

1 −1

y = −x − 2

(2, 0)

(−2, 0)

1 −1

y = (x + 2)2

y=x−2

−3

−4

−2

3 2

−1

x 1

−2

11 a (−3, −2)

y = −(x + 2)2 −3

b (3, 2)

−4

15 a y ≤ −1

b −∞ < y < ∞ 2

16 a y = 3x + 2x + 1 17 (1, −4), no.

Mathspace New South Wales – Year 11 Advanced mathspace.co

(−2, 0)

−2

978

−2

10 a (−2, 0)

−2

b y = −2x3 − x + 1

Extend your thinking 2

18 a Equation: y = x − 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. b

(−2, 0) −2

−1 2

Practice 4 a y = (x− 4)2 − 2 b (4, −2)

(2, 0)

3 If a > 0, the graph shifts right by a units; if a < 0, it shifts left by a units.

y 2

−1

y=x −4

−2

y = (x − 4)2 − 2

y=x −2

−3

−1

−4

−2

(0, −2)

(4, −2)

19 Fixed point: (0, 0). For y = −x , 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.

5 a y = 3x + 7 b

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. b

y = 3x + 7

−1

y = −2x − 3

5 4

y = 3x + 2

2 (0, 2) 1

−1

3 (0, 3)

y = 2x + 3

y 7 (0, 7) 6

2 1

6 a y = (x − 5)2

b y = (x + 3)2

d y = x2 − 2

c y=x +4

−1

7 a y = 2(x − 2) + 1

−2

b y = 2(x + 4) + 1

c y = 2x + 4

−3 (0, −3)

−4

d y = 2x − 4

8 a (2, −3)

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.

b (0, −2)

9 a Horizontal translation 1 unit right and vertical translation 2 units up. b Original: y = x2 + 2, vertex (0, 2). Translated: y = (x − 1)2 + 4, vertex (1, 4). y

15.02 Horizontal and vertical translations

y = (x − 1)2 + 4 6

What do you remember? 1 a Shifts the graph 3 units to the right.

y = x2 + 2

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

−2

d True

(1, 4)

(0, 2)

x 2

Answers mathspace.co

979

10 a Translations: 2 units left and 1 unit down. New point: (−2, −1). b

b Original: y = x2, vertex (0, 0). Translated: y = (x − 2)2 + 1, vertex (2, 1).

y = (x + 2)3 − 1 −3

−2

3 2 y=x

−1

y = x2

(0, 0)

1 −1

(−2, −1)

y = (x − 2)2 + 1

(2, 1)

(0, 0)

−2

−1

x 3

11 (−1, 0) and (1, 0)

20 (−4, 3)

12 y = 2(x − 3) + 3

21 (−2, −8)

13 y = (x − 3) + 5

15.03 Dilations

14 Horizontal translation 4 units right and vertical translation 7 units down.

What do you remember? 1 a A stretch from the y-axis, scaling x-coordinates.

15 y = (x + 2)3 + 3 16 (0, 2)

b A stretch from the x-axis, scaling y-coordinates.

Extend your thinking 17 a Translations: 3 units right and 7 units up. New maximum proﬁt at vertex (5, 11) is 11. b Original: P (x) = −x2 + 4x, vertex (2, 4). Translated: P (x) = −(x − 3)2 + 4x − 5, vertex (5, 11). y

c Horizontal dilation by a factor of 2. d Vertical dilation by a factor of 3. 2 a False

b False

c True

d False

3 The graph compresses vertically, becoming ﬂatter, as y-coordinates are halved. 4 The graph enlarges, with x- and y-coordinates doubled, appearing stretched outward.

12 10

P(x) = −(x − 3)2 + 4x − 5 8

Practice

5 a y = 4x2 − 4

P(x) = −x + 4x

−4 −2

x 2

f (x) = 4x2 − 4

18 a = 3 and b = −2.

−2

19 a Equation: y = (x − h)2 + k. Domain: . Range: [k, ∞).

f (x) = x2 − 4

Mathspace New South Wales – Year 11 Advanced mathspace.co

−2 −4

980

11 a (2, 2)

b y = 2x2 − 8 y

12 a

4 2

f (x) = x2 − 4 −2

−4

y 14

−2

f (x) = 2x2 − 8

b 12

−4

−6

8 (0, 7.5)

−8

f (x) = x2 + 3

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.

x −6

2 f (x) = −x + 5

5 4 3

−1

(0, 1)

b Original function: y = x2 Transformed function:

14 a 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

8 a Combined dilation by a factor of k = 0.4.

f (x) = x2

b Original function: y = x3 − 1

Transformed function:

8 6

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). c (0.5, 0.5)

Extend your thinking

2 1

(0, 4)

7 a Horizontal dilation by a factor of k = 3.

10 a (2, 1)

f (x) = −x + 6.25

(0, 5)

−2

f (x) = 4x2 + 1

−2

y 6 (0, 6.25)

6 a y = x2 + 4

f (x) = x + 4

−4

13 a f (x) = −x + 6.25

Stretches vertically from the x-axis, steepening the parabola. Points like (0, −8) correspond to (0, −4) on f (x) = x2 − 4.

4 2 (0, 3)

Vertical dilation (l = 2):

4 2 −4

−2

x 2

b (1, 3) d (2, 2)

Answers mathspace.co

981

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.

5 a y = −x3 + 4x b (0, 0), (2, 0), (−2, 0) c

y 3

y = x − 4x 2 1

16 a b (1.5, 1.5)

−2

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. d f (x) = x2

−1

x 1

−1

y = −x + 4x −2 −3

6 a y = −(x − 3)2 b (3, 0)

c 2

3 1

x −3 −2 −1

−1 −2 −3

17 Sequential dilations: • Horizontal by k = 2 gives .

• Combined dilation by k = 2 gives

(3, 0) 1

y = − (x − 3)2

−4

, then

vertical by l = 2 gives

y = (x − 3)2

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.

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, 0) −3 −2 −1

Chapter 15 review

(3, 0) 1

−2

y = x2 − 9

1 A

−4

2 C

−6

3 B

−8

4 a (−4, −5) b (4, 5) c

5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 (−4, −5) −5

982

8 Ask your teacher for worked solutions.

(4, 5)

x 1

2 3 4

(4, −5)

Mathspace New South Wales – Year 11 Advanced mathspace.co

9 a y = (x + 3)2 − 1

P(t)

b (−3, −1)

(4, 16)

4 2 2 P(t) = −(t − 2) + 8(t − 2)

x −4 −3 −2 −1

(0, −1)

(−3, −1)

y = 2x − 1

y = 2x + 3 1

y = x2 − 1 2

x −1

1 −1

1 x

(0, −1)

−3 −2 −1

11 a Horizontal translation 1 unit right and vertical translation 2 units up. The new point is (1, 2). y

y = 3x2 − 3

3 3

y = (x − 1) + 2

y = x3 1

y = x2 − 1 −2

(0, 0) −1

(0, −1)

b i y = 3 (x2 − 1) = 3x2 − 3

3 2

1 −1

(1, 2)

14 a i

3 (0, 3)

13 The horizontal shift is from x = 1 to x = 4, so a = 3. The vertical shift is from y = 5 to y = 2, so b = 2 − 5 = −3.

10 a y = 2x − 1 b

P(t) = −(t)2 + 8t

y = x2 − 1 y = (x + 3)2 − 1

(6, 13)

−1

x 1

−1

(0, −1)

−2

−3 (0, −3)

12 a The original vertex is at t = 4, with a maximum population of P (4) = 16. The translated vertex is at (4 + 2, 16 − 3) = (6, 13). The new maximum population is 13 (in thousands, etc.), occurring in month 6.

15 a y = 3x2 + 3 b

y 8

y = 3x2 + 3 6

y = 9x2 + 1

4 2

(0, 3) (0, 1)

−1

x 1

Answers mathspace.co

983

16 a Horizontal dilation by a factor of 3 and vertical dilation by a factor of 2.

b Domain: (−∞, ∞), Range: (−∞, k] Practice

5 a i g(x) = 2x − 5 iii

17 a

ii (0, −5)

b (9, 4.5)

(3, 9)

−1

y = x2

−2

(9, 4.5)

−4

(0, −5)

2 x −8 −6 −4 −2

2 4 6 8

b i g(x) = 2x + 3 iii

ii (0, 3)

18 a y = 4x2

b (0.5, 1)

c The statement is incorrect. A combined dilation by a factor of k = 0.25 is suitable for creating a smaller scale model because when 0 < k < 1, the transformation reduces the graph by scaling both x and y-coordinates by that factor.

(0, 3)

2 x −1

c i g(x) = 2x + 3 y 6

• 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.

16.01 Linear, quadratic and cubic functions What do you remember?

(0, 3) 2 x −2

−1

d i g(x) = 2x − 3 iii

ii (0, −3)

y 2

1 B 2 a Horizontal translation left by 2 units b Reﬂection over x-axis, vertical dilation by 2 c Vertical translation down by 3 units 3 a The stationary point remains at (0, 0). b The stationary point shifts to (a, 0). 4 a Domain: (−∞, ∞), Range: (−∞, ∞)

984

ii (0, 3)

iii

19 •

Mathspace New South Wales – Year 11 Advanced mathspace.co

x −1

1 −2

(0, −3) −4

8 Dilation by 2, translation left by 1 unit, translation down by 3 units.

6 a g(x) = (x − 2)3 + 1

2 1

(2, 1)

−4 −3 −2 −1 −1

2 3 4

−2

−3

−2

−1

−3

−2

−4

(−1, −3)

9 a g(x) = −((x + 1)2 + 4) = −(x + 1)2 − 4

b (−1, −4)

c Concave down, so range is (−∞, −4]

2 1

−4 −3 −2 −1 −1

y −3

−4

−1

2 3 4

(−1, −4) (0, −3)

−4 −6 −8

7 a (−2, −1)

10 a

6 4

b (0, 0)

2 −4

x 1

−2

−2 −3

−2

−3

−2

−1

(−2, −1) −2

−4

−6

2 −2

b (0, 5)

−1 (0, 0) −2

x 1

−4

y 6

−8

−6

(0, 5)

4 2 x −2

−1

Answers mathspace.co

985

d y = −3 (x − 2)2 − 4y

11 a

b (2, −1)

y −4 −3 −2 −1 −2 −4

c Concave down, so range is (−∞, −1] d

−8

−10 −12 −14

(2, −1)

−1

(2, −4)

−6

x 1

x 2 3 4

−16 −2

13 a 5 units right

12 a y = 2 (x − 3) − 1

2 1

(5, 0)

−3 −2 −1 −1

2 3 4 5

x 6

(3, −1)

−2 −3

b 3 units up

−4

b y = −(x + 2) + 5

(−2, 5)

5 4 3 2 1

(0, 3) 2

−6 −5 −4 −3 −2 −1 −1 −2 −3 −4

x −2

−1

c 4 units left

2 1 −4 −3 −2 −1 −1 −2 −3

x 1

2 3 4

(0, −2)

−4

986

Mathspace New South Wales – Year 11 Advanced mathspace.co

(−4, 0) −6 −5 −4 −3 −2 −1

d 2 units down

Extend your thinking

17 a y = 2 (x − 2)2 − 4 b

2 x −2

−1

1 −2

(0, 4) 2

x 1

(0, −2)

−2

(2, −4) 14 a a = 2, h = −1, k = −3 b

c [−4, ∞)

15 a Concave up, narrower.

18 a g(x) = 3x + 2

(1, 5)

g (x) = 3x + 2 4

3 2 (0, 2)

1 x

−1

(0, 0)

−1

b Concave down, wider.

19 a g(x) = 2 (x + 3)3 + 4 x

−1 (0, 0)

f (x) = 3x − 2

b Vertical translation up by 4units.

−2

−1

y 10

8 −2

(−3, 4)

−4

4 2

−5 −4 −3 −2 −1 −2

16 Reﬂection over x-axis, translation left by 3 units, translation down by 3 units. 4 3 2

y = 2(x − 1)2 + 3

1 −4 −3 −2 −1 −1

x 1

2 3 4

y = −2(x + 2)2 −2 −3 −4

x 1

16.02 Exponential and logarithmic functions What do you remember? 1 A 2 a y = −2

b x=3

c y=1

d x=0

3 a Domain: (−∞, ∞), Range: (−∞, 1) b Domain: (3, ∞), Range: (−∞, ∞) c Domain: (−∞, ∞), Range: (−1, ∞) d Domain: (−2, ∞), Range: (−∞, ∞)

Answers mathspace.co

987

7 a y = 2 × 3x

Practice 4 a y = 3x + 2 + 1

b (0, 2)

b y=1

c 6

c (0, 10)

y 25

(1, 6)

(1, 28)

5 4

3 2 (0, 2)

10 (0, 10) 5 x −3

−2

−1

b (1, 0) c 2

b (0, ∞)

c y=0

y 3

2 1

(−1, 3)

−2 x

−2

−1

1 , −2 3

6 a y = log2 (x − 1) − 2 b (1, ∞)

9 a (0, 1)

b (0, −1)

d (0, 8)

10 a x = −1

b x=2

c x=4

d x = −3

11 Transformations: Horizontal shift right by 2 units, vertical shift up by 1 unit

c x=1 y

1 x 1 −2

−1

(0, 1)

1 −3

(1, 0) 1

−1

8 a y = 2 log3 x

5 a y = 3−x

−1

(3, −1) 2

(2, −2)

−3

(2, 2) (0, 1.25) x

988

Mathspace New South Wales – Year 11 Advanced mathspace.co

−1

12 Transformations: Reﬂection over x-axis, vertical dilation by 2, vertical shift down by 1 unit

15 Equation: y = 6 × 2x − 2 + 1 y

x y

−1

6 5

−2

−3 (0, −3)

−4

(1, −5)

−5

2 1

−6 −7

16 Equation: y = log3 (x − 2) + 3 − log3 (2)

13 Domain: (−2, ∞)

y y

4 3

(1, 0) −1

(−1, −1)

(4, 3)

−1

−2

16.03 Reciprocal and absolute value functions

Extend your thinking 14 P (t) = 1500 × 20.02t 1600

x 1

−2

(2, 7)

What do you remember? 1 a ii

P (5, 1608)

1400 (0, 1500)

b iv

c i

d iii

2 a Domain: (−∞, ∞), Range: [0, ∞)

1200

b (0, 0)

1000

3 a x=0

b y=0

800 600

Practice

400 200

t 1

4 a b Vertical asymptote: x = 3 Horizontal asymptote: y = −4 c Domain: (−∞, 3) ∪ (3, ∞) Range: (−∞, −4) ∪ (−4, ∞) x-intercept: y-intercept:

Answers mathspace.co

989

y −1

x 1

−1 −2

g(x)

b Reﬂection over x-axis, horizontal shift left by 1 unit, vertical shift down by 2 units

x=3

−3

c Vertical dilation by 2, horizontal shift left by 3 units, vertical shift down by 1 unit

y = −4

−4

12 a Vertical dilation by 3, horizontal shift right by 2 units, vertical shift up by 1 unit

d Reﬂection over x-axis, vertical dilation by 3, horizontal shift right by 1 unit, vertical shift up by 2 unit

−5 −6 −7

Extend your thinking

5 a g(x) = 2x − 1 − 1

b (1, −1)

13 a

14 a

c [−1, ∞) d x-intercepts:

15 a f (x) = 2x − 1 + 1

b Domain: (−∞, ∞), Range: [1, ∞)

g(x)

16 a f (x) = 3x + 2 − 1

1 x −1

1 −1

b Domain: (−∞, ∞), Range: [−1, ∞)

16.04 Circles and translations

(1, −1) What do you remember?

−2

1 B

6 a

2 Completing the square 3 Domain: [a − r, a + r], Range: [b − r, b + r]

7 a Vertical asymptote: x = −2 Horizontal asymptote: y = −3 b Vertical asymptote: x = 1 Horizontal asymptote: y = 1 c Vertical asymptote: x = 4 Horizontal asymptote: y = 2 d Vertical asymptote: x = −3 Horizontal asymptote: y = −1 8 a f (x) = x + 2 + 3

b f (x) = −2x

c f (x) = x − 3 − 1

d f (x) = 3x + 2

9 a Vertex: (3, 2)

b Vertex: (−1, −1)

Range: [2, ∞) Range: (−∞, −1] 10 (1, 0) and 11 Domain: x ≠ −1 Range: y ≠ −3

990

Mathspace New South Wales – Year 11 Advanced mathspace.co

4 Constants deﬁning the circle’s position and size Practice 5 a Centre at (3, 2), radius 4 b Domain: [−1, 7], Range: [−2, 6] c

y 6

(3, 6)

(3, 2)

(−1, 2)

x 2

(7, 2)

(3, −2)

c i Centre at (0, −5)

6 a (x + 1)2 + ( y − 4)2 = 9 b Centre at (−1, 4), radius 3

ii Radius 4

iii

(−1, 7) y

−4

(−4, 4)

y −2

(2, 4)

−4

(−1, 1)

−8

−2

d i Centre at (1, 0)

7 a (x + 1) + ( y − 3)2 = 4

ii Radius 2

iii

y 2

(−3, 3)

(−1, 3)

(1, 3)

(1, 0) −1

−2

−1

x 3

−2

8 a i Centre at (4, 1)

9 a ( x + 3)2 + ( y − 1)2 = 16, centre at (−3, 1), radius 4

ii Radius 5 iii

1 −1

(−1, 1) −3

(0, −5)

−6

(−1, 5)

2 −2

(−1, 4)

b ( x − 4)2 + ( y + 2)2 = 4, centre at (4, −2), radius 2

10 a i (x − 1)2 + ( y + 4)2 = 9

(4, 1) 2

ii No x-intercepts iii

y x

−2

2 −2

b i Centre at (−2, 3)

−4

ii Radius 3 iii

(1, −4)

−6

(−2, 3) 2 x −4

−2

Answers mathspace.co

991

b Domain: [−3, 3], Range: [−6, 0]

b i (x + 2)2 + y2 = 25

ii x-intercepts: (−7, 0), (3, 0) iii

y −2

(−7, 0) −6

−4

(0, 0)

(−2, 0)

(3, 0) x

−2

2 −2

(−3, −3)

(0, −3)

(3, −3)

−4

−2

(0, −6)

−4

12 a No x-intercepts 2

y-intercept(s) at (0, 3)

c i x + ( y − 2) = 1

ii No x-intercepts iii

(2, 5)

y 4

(0, 3)

(2, 3)

2 (0, 2)

2 1

(2, 1)

x −2

−1

x 3

13 a i Right by 1 unit, up by 1 unit

d i (x − 3)2 + ( y + 1)2 = 16

ii (x − 1)2 + ( y − 1)2 = 1

ii x-intercepts: iii

(4, 3)

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

ii (x − 1)2 + ( y − 1)2 = 1 x 2

d i Left by 3 units, up by 5 units

ii (x − 1)2 + ( y − 1)2 = 1

(3, −1)

−2

Extend your thinking −4

14 (x − 1)2 + ( y − 2)2 = 25 y 6

11 a Domain: [0, 4], Range: [−1, 3] y

4 2

(2, 3)

−4 −2 −2

2 1 (0, 1)

(2, 1)

x 1 −1

(2, −1)

992

−4

(4, 1)

Mathspace New South Wales – Year 11 Advanced mathspace.co

−6

(1, 2) 2

x 6

(3, −2)

15 There are two possible circles that satisfy the given conditions:

16.05 Order of transformations What do you remember? 1 B 2 a Dilation or reﬂection

b Horizontal translation

These correspond to centres on the line y = x:

c l

d l when l < 0 2

3 a For y = x , 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.

y 4 3

(2.82, 2.82)

(1, 2)

1 −1

b Translation then reflection moves the vertex horizontally, then reflects it vertically; the vertex remains the same as reflection then translation for horizontal shifts.

(0.18, 0.18) 4

−1

Practice

16 (x − 2)2 + ( y − 1)2 = 4

4 a g(x) = 3 (x − 2)2

b (2, 0)

3 2

(2, 1)

(4, 1)

e (2, 0)

g (6, 0)

x 1

c g(x) = 3 (x − 2)

h Vertical dilation then translation:

−1

g (x) = (x − 2)2

5 g (x) = 3(x − 2)2 4

17 a (x − 3) + ( y − 2) = 100 y

(3, 12)

f (x) = x

(−7, 2)

−4

(3, 2) 4

(13, 2) x

−2

1 1

x 3

Translation then vertical dilation: g (x) = (x − 2)2

−4

g (x) = 3(x − 2)2

6 5

(3, −8)

(2, 0)

−1

4 3

b 100π m

f (x) = x

−2

1 −1

(2, 0) 1

x 3

Answers mathspace.co

993

Horizontal dilation then translation: 2

8 f (x) = x

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

2 x

(2, 0)

−8

−6

−4

−2

Translation then horizontal dilation: y

g (x) = (x − 2)2

f (x) = x2 8

x g( x) = −2 3

(6, 0)

9 a i g(x) = −(x − 2)3

ii (2, 0)

b i g(x) = −(x − 2)3

ii (2, 0)

10 a i g(x) = (2x − 2)2

ii (1, 0)

b i g(x) = (2x − 1)2

2 −4

ii y = 0

d i

−2

5 a b Vertical asymptote: x = 0 Horizontal asymptote: y = 3 c d Reﬂection then translation 5

ii y = 3

b i

ii y = 6

c i

ii 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

Extend your thinking y=3

13 a P (4, 16) b Q(4, 14.5)

2 1

x=0 −1

11 a i

−2

−1

−2

c The answers differ because the order of transformations affects the ﬁnal position of the point. 14 a

Translation then reﬂection: 3

2 1 −2

−1

ii (0.5, 0)

x x

x=0

−1 −2 −3

y = −3

−2

−1

−1 −2

−4

6 a i g(x) = 2 × 2

x+1

b i g(x) = 2 × 2x + 1 c i

994

x+2

or 2

ii y = 0 ii y = 0 ii y = 0

Mathspace New South Wales – Year 11 Advanced mathspace.co

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 2 units to

y 6

(−4, 5)

the left, changing the function to

3. Reﬂect the function over the x-axis, which ﬂips the graph upside down, resulting in

(−2, 2)

y=1 y 2

−1

x 1

−1 −2

g(x)

5 a Equation:

x 1

b Vertex: (2, −1) c Range: (−∞, −1]

−3

d • Point at x = 1: (1, −3)

−4

• Point at x = 3: (3, −3)

3 2

−5 −4 −3 −2 −1

f (x)

1 −2

• y-intercept: (0, 2) y

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) 2

16.06 Multiple transformations

(0, −2)

(3, −1.5)

−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

b Horizontal translation d Vertical dilation

6 a b Vertical: x = 1, Horizontal: y = 0 c (−∞, 1) ∪ (1, ∞) d (−∞, 0) ∪ (0, ∞) e

Practice 4 a Equation: g(x) = 2−x−2 + 1 b None c Horizontal: y = 1

y=0 −4

d (−∞, ∞) e • Range: (1, ∞) • Point at x = −2: (−2, 2) • Point at x = −4: (−4, 5) • y-intercept:

−2

8 6 4 2

−2 −4 −6 −8 −10

x=1 x 2

7 a g(x) = 2 log (3 (x − 1)) − 3 b x=1

c Range: (−∞, ∞)

Answers mathspace.co

995

2 1

x=1 f (x) = log(3x) 2

−1 −2 −3 −4 −5 −6 g(x) = 2log (3(x − 1)) − 3 −7

8 a g(x) = 2x + 1

b (−2, 1)

11 a y

3 2

(−2, 1) −4

−2

f (x) = x3 x 2

−1 −2 −3 −4

b (0, 1)

c No x-intercept(s) d (0,1)

Extend your thinking

y 5 g(x) = 2| x | + 1 4 3

Domain: (−∞, −1) ∪ (−1, ∞)

1 (0, 1)

−4 −3 −2 −1 −1

9 a g(x) = −ex + 2

x 2 3 4

b (−∞, 2)

c y=2 d

f (x) = 2ex

3 2

y=2

−4 −3 −2 −1 −1

2 3 4

g (x) = −e + 2

−4

b (−∞, ∞)

10 a 4

right by 2

gives the equation

. 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. 15 1. Translate left by 1

Chapter 16 review 1 B

3 D

−5 −4 −3 −2 −1 −1

16 g(x) = −(x− 1)2 + 2

2 C

3 1

and then

translating the graph of

3. Translate up by 2

−3

x 1 2 3 4

−2 −3 −4

996

14 Applying a vertical dilation by

2. Vertically reﬂect and dilate by factor

−2

Range: [−3, ∞) 13 Equation:

f (x) = | x |

12 Equation: g(x) = 2 (x − 2)2 − 3

Mathspace New South Wales – Year 11 Advanced mathspace.co

4 a g(x) = 3x + 3

b (0, 3)

5 Vertex: (1, 1)

y y

−2

x 1

−2

−4 (0, −4)

(1, 1)

−1

−6 x

−8

15 k = −2, a = 3. The equation is y = 3 × 3x − 2.

6 a g(x) = −(x + 2) − 3

16 Vertex: (−3, 1), Range: (−∞, 1]

b (−2, −3)

17 Vertical asymptote: x = 5, Horizontal asymptote: y = −2

c (−∞, −3] 7 Equation: g(x) = 2 (x − 1)3 + 4. Stationary point: (1, 4).

18 a

8 x-intercepts: (−2, 0), (6, 0)

b Domain: x ≠ 1, Range: y ≠ −3

y-intercept: (0, 12)

19 y = 2x + 2 − 3.

9 Vertical dilation by a factor of 3, horizontal translation 1 unit to the right, and vertical translation 4 units down.

20 Centre at (−1, 5), radius 6 21 a (x − 2)2 + ( y + 3)2 = 9 b Centre at (2, −3), radius = 3

10 a y = −2(x + 1)2 + 5

22 Equation: (x + 2)2 + ( y − 5)2 = 9

b Range: (−∞, 5] 11 The new line is g(x) = −x + 6. The transformation was a vertical translation up by 3 units. 12 (0, 6)

Domain: [−5, 1], Range: [2, 8] 23 (x − 2)2 + ( y − 1)2 = 25 b

24 a

13 a g(x) = log3 (x − 2) + 1 b Domain: (2, ∞), Asymptote: x = 2 c

−1

25 Ask your teacher for worked solutions. b (1, 4)

26 a

c x-intercept:

(5, 2) y-intercept:

1 x 1

(1, 4)

4 3

14 Transformations: Reﬂection over the x-axis, vertical dilation by a factor of 3, and a vertical translation down by 1 unit.

2 1 x −2 −1

−1

Answers mathspace.co

997