Mathspace VIC 2.0 Year 10 Pre-Methods by Mathspace

© 2023 Mathspace Pty 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 Mathspace Pty Ltd Level 11, 477 Pitt St Sydney 2000 The Victorian Curriculum F-10 content elements are © VCAA, reproduced by permission. The VCAA does not endorse or make any warranties regarding this resource. The Victorian Curriculum F-10 and related content can be accessed directly at the VCAA website. Title: Mathspace Victorian Curriculum 2.0 - Year 10 (Pre-Methods): 2024 Edition Print Run: 1 ISBN: 978-1-963022-46-9 Editors and lead authors: Erin Gallagher, Elisha Templin, Sarah Butterworth, Julia Reid, Anjelica Sieger, Jacqueline Cichello, Stephen George, Grace Peake, Jonathan Kim Sing, Toni Smythe, Dot Yung Further reviewers: Melissa Archey, Warren Bardsley, Lianna Beeching, Kathryn Bennett, Benjamin Bui, Eric Chan, Cory Duker, Bahija Elmir, Wayne Gillie, Jeremy Hunter, Shahad Odeh, Helen Soar, Patricia Toutzaris, Paul Unkles, Jie Yang Writing and development team: Adam Humphries, Neil Lopez, Ma. Lourdes Ona, Bernadette Barillo, Sheramie Castillo, Abegail Fontanilla, Shaira Llanita, Lesley Lopez, Kerby Mendoza, Regine Jara, Albert Jara, Rolly Apdo, Michaela Belen, Valerie Baja, Marielle Belen, Allen Calabia, Adrian Doctolero, Dhave Fernandez, Janine Francisco, Rose Francisco, Rona Josue, Crystel Lontoc Images and Design team: Alex Duckmanton, Chastine Marquez, Keith Gimeno

Contents 1

Indices and surds

1.01

Index laws

1.02

Negative indices

1.03

Optional: Simplify surds

1.04

Optional: Add and subtract surds

1.05

Optional: Multiply and divide surds

1.06

Optional: Binomial expansions with surds

1.07

Optional: Rationalise the denominator

1.08

Optional: Fractional indices

Chapter 1 review

Algebraic fractions

2.01

Add and subtract fractions with numeric denominators

2.02

Add and subtract fractions with algebraic denominators

2.03

Optional: Add and subtract further algebraic fractions

2.04

ultiply and divide algebraic fractions with M numeric denominators

2.05

ptional: Multiply and divide algebraic fractions with O algebraic denominators

Chapter 2 review

Expand and factorise

3.01

Factorise algebraic expressions

100

3.02

Expand binomial products

105

3.03

Factorise binomial products

111

3.04

Factorise monic quadratic trinomials

116

3.05

Optional: Factorise non-monic quadratic trinomials

121

3.06

Complete the square

127

Chapter 3 review

134

Contents mathspace.co

iii

Linear equations

136

4.A

Linear graphs

4.B

Equations of more than 3 steps

4.01

Equations involving fractions

138

4.02

Rearrange equations

146

4.03

Parallel and perpendicular lines

154

4.04

Linear models

163

4.05

Graph linear inequalities

176

Chapter 4 review

191

Quadratics 194 5.01

Quadratic equations

196

5.02

Quadratic graphs

204

5.03

Solve by factorisation

228

5.04

Solve by completing the square

238

5.05

The quadratic formula

244

5.06

The discriminant

251

5.07

Optional: Further quadratic equations

260

Investigation: Derive the quadratic formula Investigation: Use bisection to solve quadratics Chapter 5 review

272

Intersections and simultaneous equations 276 6.01

Solve graphically

278

6.02

Simultaneous inequalities

286

6.03

Solve using substitution

296

6.04

Solve using elimination

304

6.05

Applications of simultaneous equations

311

Investigation: Intersections of curves 6.06

Optional: Intersections of curves Chapter 6 review

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

324

Introduction to functions (Optional) 7.01

Optional: Define relations and functions

7.02

328 330

Optional: Domain and range

341

7.03

Optional: Graphs of xn

352

7.04

Optional: Transformation of functions

363

Chapter 7 review

375

Exponentials 378 8.01

Exponential relationships

380

8.02

Characteristics of exponential graphs

390

8.03

Optional: Transformations of exponentials

403

8.04

Exponential equations

426

8.05

Compound interest

434

Investigation: Half-life 8.06

Exponential models

449

Chapter 8 review

464

Logarithms 468 9.01

Logarithms

470

9.02

Order of magnitude and logarithmic scales

480

9.03

Optional: Other bases

489

Investigation: Logarithms 9.04

Optional: Logarithm laws

493

9.05

Optional: Connect exponentials and logarithms

500

9.06

Optional: Solve exponential equations with logarithms

514

Chapter 9 review

520

Contents mathspace.co

Circles 524 Investigation: Circles 10.01

Circles at origin

526

10.02

Translations of circles

533

Investigation: Equation of any circle 10.03

Optional: Transform circle equations Chapter 10 review

544

Hyperbolas 546 Investigation: Hyperbolas

11.01

Direct and indirect proportion

548

11.02

Hyperbolas

559

11.03

Optional Transformations of hyperbolas

570

Chapter 11 review

582

Trigonometry 586 12.01

Pythagoras’ theorem

588

12.02

Trigonometric ratios

597

12.03

Angles of elevation and depression

611

12.04

Bearings

620

12.05

Optional: Sine and cosine rules

636

12.06

Optional: Pythagoras in 3D

12.07

Optional: Trigonometry in 3D Chapter 12 review

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

656

Measurement 662 13.01

Surface area of prisms and cylinders

664

13.02

Volume of prisms and cylinders

674

13.03

Surface area and volume of composite solids

684

13.04

Estimation and accuracy

695

Investigation: Rainfall collection Investigation: Instruments and methods Investigation: Scale models 13.05

Optional: Spheres

13.06

Optional: Cones

13.07

Optional: Pyramids Chapter 13 review

704

Probability 708 14.A

Review Venn diagrams and two-way tables

14.01

Independent and dependent events

710

14.02

Conditional probability

724

14.03

Mixed probability concepts

736

14.04

Optional: Geometric probability Investigation: Simulation Chapter 14 review

Univariate statistics

751

754

15.A

Representations of data

15.01

Quartiles

756

15.02

Boxplots

769

15.03

Cumulative frequency

783

15.04

Compare data sets

796

15.05

Samples and populations

811

15.06

Optional: Standard deviation

15.07

Optional: Comparisons using standard deviation Chapter 15 review

821

Contents mathspace.co

vii

Bivariate statistics

826

16.01

Two-way tables

828

16.02

Associations between categorical variables

840

16.03

Bivariate numerical data

859

16.04

Associations between numerical variables

869

16.05

Line of good fit

883

16.06

Interpolation and extrapolation

893

16.07

Misrepresentations of data

907

Investigation: Misrepresentations in the media Investigation: Conduct a statistical investigation 16.08

Optional: Correlation coefficient Chapter 16 review

Networks 17.01

Introduction to networks

17.02

Network representations

17.03

Planar graphs

17.04

Walks, paths, and trails

17.05

Optional: Traversable networks Chapter 17 review

Geometry and proofs 18.01

Deductive reasoning

18.02

Proofs with parallel lines

18.03

Congruent triangles

18.04

Similar triangles

18.05

Further geometric proofs Investigation: Shortest path

18.06

Optional: Circles and chords

18.07

Optional: Angles in circles

18.08

Optional: Tangents Investigation: Algorithms in geometry

18.09

Optional: Transformations on the plane using algorithms Chapter 18 review

viii

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

922

Further trigonometry (Optional) 19.01

Optional: Angles with exact values

19.02

Optional: The unit circle

19.03

Optional: Graphs of sine and cosine

19.04

Optional: Trigonometric equations Chapter 19 review

Functions and polynomials (Optional) 20.01

Optional: Operate with polynomials

20.02

Optional: Polynomial division

20.03

Optional: Factor and remainder theorem

20.04

Optional: Graph polynomials in factored forms

20.05

Optional: Rates of change Chapter 20 review

Counting Methods (Optional) 21.01

Optional: Counting principle

21.02

Optional: Factorial notation

21.03

Optional: Permutations

21.04

Optional: Combinations

21.05

Optional: Applications of counting methods to probability Chapter 21 review

Answers

928

Contents mathspace.co

Big ideas • Understanding and applying index laws enable us to manipulate and simplify expressions involving powers, facilitating problem-solving and exploring mathematical concepts. • Exploring irrational numbers, surds and their operations reveals the underlying principles of manipulating expressions involving roots. • Mathematical expressions can be transformed and represented in different forms using principles of equivalence and algebraic properties.

Curriculum Outcomes VC2M10A01 factorise algebraic expressions by taking out a common algebraic factor VC2M10A02 simplify algebraic products and quotients using exponent laws VC2M10AN01 define rational and irrational numbers and perform operations with surds and fractional indices VC2M10AN02 perform operations on numbers involving fractional exponents and surds

Indices and surds Chapter outline 1.01 Index laws 1.02 Negative indices 1.03 Optional: Simplify surds 1.04 Optional: Add and subtract surds 1.05 Optional: Multiply and divide surds 1.06 Optional: Binomial expansions with surds 1.07 Optional: Rationalise the denominator 1.08 Optional: Fractional indices Chapter 1 review

4 13 20 27 33 41 45 54 65

A “surd” is a term used to describe an irrational number. The name originally meant “deaf” or “unheard” because they cannot be written exactly using fractions and decimals.

1.01

Index laws After this lesson, you will be able to… • apply the multiplication law to simplify expressions. • apply the division law to simplify expressions. • apply the power of a power law to simplify expressions. • apply the zero power law to simplify expressions. • solve problems using a combination of index laws.

Multiply terms with indices

For the expression a5 × a3, since the factors have like bases we can use the multiplication law. a5 × a3 = a5 + 3 = a8 We can remind ourselves of the multiplication law by writing out the original expression in expanded form.

a´ a´ a´ a´ a ´ a´ a´ a For any base number a and any indices, also called exponents, m and n the multiplication law is: am × an = am+n When multiplying terms with a common base: • Keep the same base. • Find the sum of the indices.

Example 1 Simplify 6x2 × 4x6 using index laws.

Create a strategy Use the multiplication law.

Apply the idea 6x2 × 4x6 = 6 × 4 × x2+6 = 24x

Multiply the coeﬃcients and add the indices of x Evaluate and simplify

Reﬂect and check Notice that the coeﬃcients are multiplied separately when we use the multiplication law on the powers.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary The multiplication law: am × an = am+n

Divide terms with indices Similarly, we can avoid having to write each expression in expanded form by using the division law, which is also known as the quotient law.

a ´ a ´a´ a´ a a´a

To simplify

we can use the fact that

to get:

For any non-zero base number a, and any indices m and n we get,

When dividing terms with a common base: • Keep the same base. • Find the difference between the powers. We can also write the division law in the form: am ÷ an = am−n

Example 2 Simplify

Create a strategy Use the division law.

Apply the idea Subtract the indices of x Evaluate the subtraction

Idea summary The division law:

where a ≠ 0.

1.01 Index laws mathspace.co

Power of a power law For any base a, and indices m and n, we also have the power of a power law: (am)n = am×n To see how this law works, let’s work through an example of (52)3. (52)3 = (5 × 5)3 = (5 × 5) × (5 × 5) × (5 × 5) = 56 This gets generalised to (am)n = am×n When simplifying a term with a power that itself has a power: • Keep the same base. • Find the product of the powers. Brackets are used to show which powers are applied to any coefficient. Be careful when raising negative numbers to a power. A negative number raised to an odd power gives a negative result, but raised to an even power gives a positive result. For example: (−2x2)3 = −8x6

Example 3 Simplify (−2x2)2

Create a strategy Use the power of a power law.

Apply the idea (−2x2)2 = (−2)2 x2×2 = 4x

Multiply the powers of the variable Evaluate the multiplication and coeﬃcient

Idea summary The power of a power law: (am)n = am×n

Zero power law We also have the zero power law which says that for any non-zero base a, a0 = 1 We know using the division law that

= a m – n.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

= an – n = a0.

If m = n, then Also

= 1.

This means that a0 = 1. That is, taking the zeroth power of any number will always result in 1.

Example 4 Simplify 18a0

Create a strategy Use the zero power law.

Apply the idea Notice that only a is raised to the power of 0, but not the coeﬃcient 18. 18a0 = 18 × a0

Separate the two parts of the term

= 18 × 1

Evaluate a0

= 18

Evaluate

Idea summary The zero power law: a0 = 1 where a ≠ 0.

Mixed operations Now we are going to look at questions that can be solved using a combination of these rules. It’s important to remember the order of operations when solving such questions. • The multiplication law: am × an = am+n • The division law: am ÷ an = am−n, a ≠ 0 • The power of a power law: (am)n = am×n • The zero power law: a0 = 1, a ≠ 0 We may also come across expressions such as (am × bn)p, and we can use the power of a power law on each variable to see that: (am × bn)p = am×p × bn×p

1.01 Index laws mathspace.co

Example 5 Simplify (x5 y4)4

Create a strategy Use the power of a power law.

Apply the idea (x5 y4)4 = (x5)4 × ( y4)4

Raise each variable to the power of 4

× y Multiply the powers

y16 Simplify

=x =x

Example 6 Simplify a

Create a strategy We can use the zero power law, followed by the power of a power law, on the numerator. Then we can use the division law to simplify the expression, since there will be a power of x in both the numerator and denominator.

Apply the idea Use the zero power law

Use the power of a power law

Simplify the numerator

Use the division law

Simplify the power

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Create a strategy Use the division law then the power of a power law.

Apply the idea Use the division law Simplify Use the power of a power law Simplify

Idea summary Some expressions require a combination of the power laws to simplify them. It’s also important to remember the order of operations when simplifying these expressions. The multiplication law: am × an = am+n The division law: am ÷ an = am−n, a ≠ 0 The power of a power law: (am)n = am×n The zero power law: a0 = 1, a ≠ 0

1.01 Practice questions Understanding

3, 4

Explorer

1–4

Adventurer

1–4

Trailblazer

When simplifying a division of terms with powers of the same base, what operation is performed on the indices?

Rewrite this expression in expanded form and demonstrate how the division law for indices works.

1.01 Index laws mathspace.co

Consider the expression ( p3)2. a

Are these expressions equivalent to ( p3)2:

( p × p × p) × ( p × p × p)

iii

( p × p × p)2

( p × p × p) × ( p × p)

p3 × p3

p3 × p2

Determine whether these equations are true or false: ( p3)2 = p3+2

i c

( p3)2 = p3×2

Complete the equation: ( p3)2 = p⬚

Complete each statement: a c

x⬚ ÷ x3 = x7 (b8 × b2) ⬚ = b20

Ex 1

5a–f, 7a–g, 8a–c, 9a–g, 10a–c, 11a–d, 12a–h, 13a–c

Simplify: a

a3 × a5

6y × y

i 6

Ex 2

u2 × u6 × u3

( y⬚ × y17)1 = y17

(3m2 × ⬚)4 = 81m24

5c–j, 7c–i, 8b–e, 9d–j, 10b–f, 11b–e, 12e–n, 13d–f

a2 × a8

b × b4

3y × 4y

a × −2a

2w4 × w10 × w7

5z5 × 3z8 × z9 6bc4 × 8b6a × 0

5g4h2 × 8g3h4

8m2n4 × 7m6n

5g–l, 6, 7e–l, 8c–f, 9g–l, 10d–h, 11d–h, 12g–t, 13e–h

d h l

y3 × y7 a ma n

4y8 × 6y6 × 3y3

Simplify:

b a10 ÷ a5

c m5 ÷ m2

4y4 × (−2y6) × 4y5

Simplify: a c e

Simplify:

Fluency

2s2 t × (−5r2 s3) × 3r3 t4

(−20b4) ÷ 2b4

d f

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

d j11 ÷ j4

h l

(−4y7) × (−6y2) × (−4y2) 9y9 × 8 (−y)8 × 7y7 (−9y9) ÷ (−3y5)

9r8 ÷ (4r2)

Ex 3

Simplify: a

(j4)6

(w2)11

(5h7)2

(xy2)5

(−x9)4

(−5x4)3

i 10

Ex 4

Ex 5

Simplify:

(3a2)4

(w9 v6)4

(k5)3 × (k2)8

6 (r6)5 × 4 (r4)7

(x9 y)7 × (xy2)4

10 (r7)5 × (2rs)4

(a5)9 × (b6)10

( p10)6 × ( p4)3

((x3)4)5

(g0)7

(j8)0

(7st)0

6k0 ÷ (4x2)0

Simplify: a

24a0

10r0 + (2q3)0

Simplify: a

(x7 y2)4

(11x ) − (12x )

(4u5 v2)2

m8 ÷ m2 × m5

265

m ((x ) ) o

p5 × ( p4)3

n p

(−2x )

p9 ÷ p5 ÷ p2 (3a2 b5 c)4

(15x3)0 + 150 − 15y0

(4h0)2 + 18 ÷ (3g0)

(10x2 × 10x5)0 − 10x0 (4y4)3

(3y5)2 × (5y2)3

(c10)11 ÷ (c8)3

(−240u32) ÷ (−8u9) ÷ (−5u12)

s Ex 6

( y4)3

Simplify:

1.01 Index laws mathspace.co

Reasoning 14

15, 16ace

When simplifying the expression

15, 16

14, 16

, Clarice’s ﬁrst line of work was a3 + a2.

Is Clarice’s work correct? Explain. 15

When simplifying the expression

• Indira’s first line of work was a3 × a4 • Chad’s first line of work was a3 × a2 • Ernestino’s first line of work was Who is correct? Explain. 16

Explain why each statement is incorrect, then give the correct solution: a

3s6 × 3s2 = 3s12

c e

(2x3)5 = 2x15

(4a)0 = 4a

Problem-solving 17–19

18, 19, 21

18–20, 22

A scientist is modelling the growth of a bacteria colony. The colony doubles in size every hour. If she starts with a single bacterium, how many bacteria will there be after 6 hours? Express your answer using indices.

Will any values of x make 3x be equal to x3? Use examples to justify.

Find the value of a and b in the equation:

Determine whether each statement is true or false. a c

−xn = (−x)n for all x, only if n is odd −xn = (−x)n for all x, only if n is even

Rewrite 16p with a prime base.

Simplify: a

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

−xn = (−x)n for all x always

1.02

Negative indices After this lesson, you will be able to… • apply the negative index law to simplify expressions. • extend the concept of negative indices to include fractional bases.

Negative indices By simplifying

using the division law, we subtract the powers and get a−2. By using expansion

we see this is the same as

a´a´a 1 = a ´ a ´ a ´ a´ a a´ a This gives us the negative index law: for any non-zero base a, base to a negative power: • Take the reciprocal of the expression. • The power becomes positive.

. That is, when raising a

This also means that:

Example 1 Fill in the box to make the statement true:

Create a strategy Use the negative index law.

Apply the idea

Write the denominator as a power of 5 Apply the negative index law. −2 should be placed in the box.

1.02 Negative indices mathspace.co

Example 2 Simplify

Create a strategy Use the division law.

Apply the idea Subtract the powers and divide the coeﬃcients Simplify the power Apply the negative index law

Example 3 Simplify 20m6 ÷ 5m13 × 9m2

Create a strategy Use the division law then the multiplication law.

Apply the idea Use the division law Multiply the coefficients then use the multiplication law Use the negative index law

Idea summary The negative index law:

where a ≠ 0.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Fractional bases with negative indices When raising a fractional base to a negative power we can combine the individual rules we have seen so far.

Exploration We may see expressions in the form

We can simplify these expressions by starting with

Power law

Rewrite as division

Use negative index law

Multiply and take the reciprocal

Simplify the product

1. How could we rewrite

using positive indices?

2. Are there any values that a or b cannot have?

Example 4 Simplify: a

Create a strategy Find the reciprocal of the fraction inside the bracket then change the power to positive.

Apply the idea Find the reciprocal of the fraction then change the power to positive

Apply the power to the fraction

Evaluate the powers

1.02 Negative indices mathspace.co

Create a strategy Use the negative index law, remembering we take the reciprocal of the fraction.

Apply the idea Take the reciprocal of the fraction and change the power to positive Use Use (a × b)n = an × bn Simplify using index laws Use the negative index law again to end up with positive indices

Idea summary Negative index law with fractions:

where a, b ≠ 0.

1.02 Practice questions Understanding 1

1–3

Explorer

1–3

True or false: a

Is 4−2 less than or greater than 1?

Write

1–3

Adventurer

using a negative index?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

a−4 = −a4

Trailblazer

4b, 5a–h, 6a–h, 7ac, 8ab, 9a–c, 11a–c, 12a–f, 13ab

Fluency 4

4a, 5c–k, 6g–n, 7bd, 8bc, 9d–g, 10a–d, 11c–e, 12f–k, 13b–d

4c, 5d–l, 6i–p, 7cd, 8cd, 9gh, 10e–h, 11d–h, 12k–p, 13d–f

Complete each table: a

2−4

2−3

10−4

10−3

2−2

2−1

10−2

10−1

100

101

102

103

104

100 c

3−4

3−3

3−2

3−1

27 Ex 1

Fill in each box. Evaluate if possible: a

Rewrite with positive indices: a

a−9

p−2

3x−4

8p−3

2x−8 y3

m 3a4 b−5 c−7

4−2 k−3 l7

m−5 n−4 p4

Simplify: a

p−2 q3

Rewrite as terms with negative indices: a

Ex 2

7x−9

Simplify: a

5y9 × 4y−3

7a4 × 4a−2

5x4 × (−3x−8)

−4y × (−4y )

−4

× 4h

−2

× 2y

−5

3y−2 × 4y−3

(5mp)2 × mp−2

1.02 Negative indices mathspace.co

Evaluate: a

2−2 × 24

2 × 3−2

10 ÷ 2 + 3

20 × 2−2 + 6

3−1 + 4−1

3 × 4−2 + 5 × 2−4

g Ex 3

2x−2 × 5x5 × 4x−5

4y × 3y ÷ 2y

(7 + 32) × 2−2

6y7 × 2y−5 × 5y3

d f

−1

40x−2 ÷ 5x8 ÷ 4x−9

(2m)−3

i m

12x−9 × 2x4 × 3x−2 8y10 ÷ 2y−4 ÷ y3

14x−8 ÷ 2x × 3x6

(4m−6)4

(3p−4)−2

Simplify: a

(10x3 y−4)2

(3x−4 × 2x2)2

(4x3)−1 × ( y−2)−3

(10x3)3 × (2x4 y3)−2

Reasoning 14

Simplify: a

2y6 × 4y7 × 4y−5

c g 12

102 × 5−2

Simplify: a

Ex 4

−1

14, 16

14, 16, 18

Oreste and Donovan are writing an equivalent expression for Oreste writes down

(3y2)−2

15–18

as a single term.

and Donovan writes down x−10.

Who is correct and why? 15

We can evaluate 03 = 0, but not 0−3. Explain why 0−3 is undeﬁned.

A student wanted to express 5a−1 without negative indices and wrote their answer is incorrect, and write the correct answer.

For x > 1, what happens when we raise x to a very large negative index? That is, what happens to x−a as a gets closer to 0? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

. Explain why

For variable p, calculate the following expressions for p = −2, 0, 4, 10: • 2p • 4p • 2−p + 4p • 2−p × 4p

What do you predict will happen to the value of each expression as p increases signiﬁcantly?

Problem-solving 21 19

Write

Complete:

Solve for n:

19, 21, 23

19–24

at least three different ways.

(x3 y−5)n = x−12y20

(a−5 b3)n = a15 b−9

Why is 2−2 < 2 but

It is estimated there were 7 × 108 jellyﬁsh in the largest bloom ever observed. If each jellyﬁsh can consume 5 grams of plankton in a day, how long would it take the bloom to consume 10 tonnes (1 tonne = 1 000 000 g) of plankton?

In electrical engineering, the formula used for calculating the capacitance of a capacitor is

? Give another example where this pattern applies.

, where C is the capacitance measured in farads, Q is the charge on the capacitor measured in coulombs and V is the voltage across the capacitor measured in volts. Use this formula to answer the following questions. a

A capacitor has a charge (Q) of 15 × 10−6 C (coulombs) and a voltage (V ) of 5 volts. What is the capacitance (C) of the capacitor?

1 microcoulomb is equivalent to 1 × 10−6 coulombs. A capacitor has a charge of 20 microcoulombs, and a voltage of 4 volts, what is the capacitance of the capacitor? Remember to ﬁrst convert the charge to coulombs.

A capacitor has a capacitance (C) of 10 × 10−4 F ( farads), and a charge of 30 microcoulombs. What voltage was applied across the capacitor?

1.02 Negative indices mathspace.co

1.03

Optional: Simplify surds After this lesson, you will be able to… • understand the difference between rational and irrational numbers. • define and recognise surds. • distinguish between exact values and approximations. • simplify surds.

Rational and irrational numbers Exploration x

1. Fill in the table of values. 2. Plot the points on a Cartesian plane using happens for values of x less than 0? 3. Use a calculator to evaluate

as the y-value. What do you think

for the following values of x: 0, 0.01, 0.0001

4. What does the calculator display for

5. Is it possible to take the square root of a negative number? Why? Why not?

A rational number is a number which can be written as a fraction where both the numerator and denominator are integers. This might also look like a terminating or recurring decimal. An irrational number is a number which cannot be written as a fraction of two integers. These also cannot be written as terminating or recurring decimals. Real Numbers Rational numbers Integers 4

Whole numbers 1 0 1 2 3 4

Irrational numbers

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A particular type of irrational number is a surd. A surd is a root which cannot be simpliﬁed to a rational number. A surd is an irrational expression of the form , where x is a rational number and n is an integer such that n ≥ 2, and x > 0 when n is even. root symbol

index

radicand

surd

We would read

as the nth root of x.

Radicand The radicand is the number or expression inside the radical symbol. It is the value from which we are extracting the root. Example: In

, the radicands are 16 and 27, respectively.

Radical The radical symbol (root symbol) is used to denote roots. The entire expression, including the radical symbol and the radicand, is often referred to as the radical. Example:

and

are radicals.

Surd A surd is an irrational number that can be expressed in root form but cannot be simplified to remove the root. Example: and are surds because they cannot be simplified to a rational number. A surd is an exact value that can be approximated by a rounded decimal. We find the approximations using a calculator.

A number can have an exact value, or it could have an approximation. An approximation is a value that is close to, but not exactly the same as the exact value. Approximations are typically found by rounding the decimal form of a number. For instance, if the exact value is , one possible approximation would be 0.667. When dealing with surds, we could have exact values like = 2.924, rounded to 3 decimal places.

= 5, or approximate values like

1.03 Optional: Simplify surds mathspace.co

Example 1 Is

rational or irrational?

Create a strategy Put the number in your calculator.

Apply the idea

Reﬂect and check

When we get 3.608 826 080 139. This value goes on forever with no pattern, so it is irrational.

For this to be rational, the radicand would need to be a cube number. The ﬁrst four perfect cubes are: • 13 = 1 • 23 = 8 • 33 = 27 • 43 = 64

Also 47 is not a cube number - there is no integer which gives 47 when cubed, since 33 = 27 and 43 = 64. So this further conﬁrms is irrational. that

27 < 47 < 64, so 47 isn’t a perfect cube, and is irrational.

Idea summary A rational number is a number which can be written as a fraction where both the numerator and denominator are integers. An irrational number is a number which cannot be written as a fraction of two integers. A surd is an irrational expression of the form , where x is a rational number and n is an integer such that n ≥ 2, and x > 0 when n is even. A surd is an exact value that can be approximated by a rounded decimal. We find the approximations using a calculator.

Simplify surds We simplify surds to make them easier to understand, compare, and work with. It also helps to have a common way to express surds. Square root and cube root expressions can sometimes be written in simpler forms by simplifying any factors which are perfect squares, cubes, etc. For simplifying square roots, we can use the properties:

Combining these two, we can see that:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The same operations that can be applied when simplifying numerical radicals can be applied when simplifying radicals with algebraic expressions.

Exploration We can use prime factors to help us split a number into a product of a perfect square and a remainder. If we wanted to simplify squares.

Using the factor tree, we can see that 60 = 22 × 3 × 5, so there is a perfect square factor of 22 = 4.

60 2

, we want to see if 60 has any factors which are perfect

30 10

3 2

1. How can you identify if a number has a perfect square factor after drawing a factor tree? 2. Is it possible to get two different factor trees for the same number? 3. Is it possible to get two different prime factorisations? 4. Can every surd be simpliﬁed? 5. How would you rewrite 180 as a product of a perfect square and a remainder? 6. What would happen if we didn’t use the largest perfect square?

In general, we can use the following facts to simplify surds, for a, b ≥ 0 and m, n positive integers,

Example 2 Simplify

Create a strategy Find the highest perfect square that is a factor.

1.03 Optional: Simplify surds mathspace.co

Apply the idea We can ﬁnd that 12 = 4 × 3 where 4 is a perfect square. Write the surd as a product of its factors Evaluate Simplify

Reﬂect and check We can double check that we found the largest perfect square to factor out by looking at the radicand after simplifying. Since 3 has no factors other than 1 that are perfect squares we know this is fully simpliﬁed.

Example 3 Simplify

Create a strategy Use

to split

into two factors.

Apply the idea Write the surd as a product of its factors Rewrite 8 as 23 Simplify Evaluate the product

Idea summary We can simplify surds by writing the radicand as a product of the largest possible perfect square or cube and a remainder. These identities are useful when simplifying expressions involving surds:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1.03 Practice questions Understanding 1

Which statements are true and which are false? a

All surds are irrational.

All square roots are surds.

The cube root of a perfect square is always rational.

Some square roots are rational and some are irrational.

The cube root of a perfect cube is a surd.

4, 6, 7a–h, 8bef, 9ac

4, 5abd, 6cd, 7e–l, 8acfg, 9bc

4, 5bc, 6d, 7i–p, 8dgh, 9cd

Is each number rational, irrational, or neither? a

i m

∞

Rewrite each radicand as the product of the highest perfect square or cube factor and a whole number: b

Simplify: a

Ex 2

Trailblazer

Fill in the boxes:

a 6

1–3

Fluency Ex 1

1, 2

Adventurer

Are these surds? a

1, 2

Explorer

Simplify, and then determine if the value is rational or irrational: a

1.03 Optional: Simplify surds mathspace.co

Ex 3

Simplify: a

Reasoning 10

10, 12

11–13

A calculator states that

11, 13, 14

an exact value? is 17.029 386 366. Is this still exact? Explain.

During a physics assignment, Jordan tried to evaluate the cube root of 27 on their scientiﬁc calculator. They ended up with an answer of 19 683. Identify and correct Jordan’s error.

Are these expressions written in their simplest surd form? Explain your reasoning. b

a 13

Heiko is asked to simplify

. He notes that 80 = 4 × 20, and simpliﬁes to

Has he fully simpliﬁed the surd? If yes, explain how you know. If not, further simplify Heiko’s answer.

Use another pair of factors for simplifying

. Explain your choice.

If p is a rational number and p < 0, which statements are true and which are false? Explain your answer. is negative

is rational

Problem-solving 16, 17

is positive

15, 16, 18a–d

d 15–17, 18e–h

The volume V of a regular tetrahedron with edge length a is given by:

Find the volume of a tetrahedron with a side length of 2 cm.

Rearrange the formula to make a the subject.

Now, ﬁnd the side length of a tetrahedron with a volume of 72 cm3.

Write the expression

Simplify

Simplify:

as a single surd.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

is positive

1.04

ptional: Add and subtract O surds After this lesson, you will be able to… • understand what “like surds” are and how to identify them. • add surds. • subtract surds. • solve problems involving the addition and subtraction of surds.

Like surds We know that we can simplify 5y + 2y to give 7y as 5y and 2y are like terms, so we can collect them together. Any surd with the same index and radicand is a like surd. For example,

and

are like surds.

Exploration Sometimes terms might not appear to be like terms, but they can be simpliﬁed to be added or subtracted. and

appear to be unlike terms, but if we simplify:

So these are like terms and can be rewritten as simplify to , just as 2p + 3p = 5p.

. This would

1. Can all surds be simplified? 2. What are two surds that don’t appear to be like, but can be simplified to be like? 3. If two surds are fully simplified and have different radicands, can we say that they are not like surds or could they be simplified differently to be like surds?

1.04 Optional: Add and subtract surds mathspace.co

Example 1 Are each pair “like surds”? a

and

Create a strategy Simplify each surd and compare their radicands.

Apply the idea Express 80 using the largest possible square factor

Simplify Express 45 using the largest possible square factor

Simplify

These are like surds as they are both of the form

Reﬂect and check When simplifying

, if 4 was identiﬁed as a perfect square factor to give

it would not be fully simpliﬁed as there would still be a factor of 4 in the radicand.

and

Create a strategy Simplify each surd and compare the radicands.

Apply the idea Express 18 using the largest possible square factor

Simplify

Express 27 using the largest possible square factor

Simplify

These are unlike surds as they have different radicands.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary We can determine if two surds are like surds by simplifying them and comparing their radicands.

Add and subtract surds Similar to working with algebraic terms, we need like surds in order to add or subtract. In particular,

. For example: Evaluate the roots Evaluate the addition

A common mistake that is made when adding or subtracting surds is to combine them in the radicand. Incorrectly combining like terms Evaluate the addition Evaluate the root See how we arrived at a different, incorrect answer. If a and b are different values, we can not simplify the expression

. This can be extended to adding or

But if they have the same value, then subtracting multiples of surds, such as

We can add and subtract like surds by adding or subtracting their coeﬃcients:

The same ideas can be used with cube roots, fourth roots, etc.

Example 2 Simplify

Create a strategy Add the coeﬃcient of like surds.

Apply the idea Since both terms have irrational part

, they are like surds and we can add 10 and 14 together.

1.04 Optional: Add and subtract surds mathspace.co

Example 3 Simplify

Create a strategy Group the like surds and then add the coeﬃcients of like surds.

Apply the idea Group like surds

Simplify like surds

Example 4 Simplify

Create a strategy Simplify each surd before adding.

Apply the idea We can see that 45 = 9 × 5 where 9 is a perfect square, and 80 = 16 × 5 where 16 is a perfect square. Write the surds as products of their factors

Evaluate

and

Add like surds

Reﬂect and check To ﬁnd the largest perfect square factor, we can use a factor tree or trial and error. To use trial and error, we can divide 45 and 80 by perfect squares such as 4, 9, 16, 25, 36, … on the calculator and use the largest number with an integer remainder.

Example 5 Simplify

Create a strategy Simplify the surd and then add the like surds.

Apply the idea Write the surds as products of their factors

Simplify the cube root using 83 = 512

Subtract the like surds

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary We can add and subtract like surds:

The radicand, a, may be numerical or involve variables. Before we add and subtract surds, we make sure that all terms are already in their simplest form.

1.04 Practice questions Understanding 1

b 2

Ex 3

1–3

Adventurer

1–3

Trailblazer

To add or subtract two surds they must be ⬚.

For a surd to be in its simplest form, the radicand must not have any factors that are a ⬚.

Are each pair “like surds” or not? a

and

and and

and

Are these equations true or false? a

Fluency Ex 2

Explorer

Fill in the blanks: a

Ex 1

1–3

4a–eh, 5a–c, 6a–e, 7a–d

4d–i, 5b–e, 6d–h, 7d–i

4g–l, 5c–f, 6f–l, 7df–j

Simplify: a

1.04 Optional: Add and subtract surds mathspace.co

Ex 4

Ex 5

Simplify: a

Simplify:

Reasoning 8

8, 9

8–10

For

Simplify 4x + 5y + x − 2y.

How does 4x + 5y + x − 2y relate to

Simplify

For

. :

Choose some values of k and m to show that the equation is not generally true.

Are there values of k and m that would make the equation true?

Apply the same reasoning to

. What are your conclusions?

Identify and explain the error in each equation:

Explain why we simplify surds before adding or subtracting.

Problem-solving 12, 13 12

9–11

a 11

13, 14

13–15

Find the perimeter of the triangle in simpliﬁed surd form. 80

125

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

An athlete is participating in a race around the track shown in the diagram, which has dimensions in kilometres. a

Determine the exact distance of one lap around this track.

If the track distance is rounded to the nearest kilometre, determine by how much this approximation would overestimate the distance. Give your answer to the nearest metre.

and width A rectangle has length each surd to the nearest whole number.

4− 7

4+ 7

. Katie estimates the perimeter by rounding

What was Katie’s answer?

Is Katie’s answer an overestimate or an underestimate? Explain.

Write an expression for the exact perimeter.

What is the percentage difference between Katie’s estimate and the true value? Round your answer to two decimal places.

Emma found a piece of paper on which another student had written the ﬁrst four terms of a sequence. It read . a

Emma wanted to continue the sequence. What would the next three terms be?

The sum of the ﬁrst n integers is sequence as an exact value.

If the last term of the sequence that Emma wrote was terms in the sequence as an exact value.

1.05

. Find the sum of the ﬁrst seven terms of this , ﬁnd the sum of all the

ptional: Multiply and O divide surds After this lesson, you will be able to… • multiply surds. • divide surds. • solve problems involving multiplication and division of surds.

Multiply surds We can simplify surds using the property

If we wanted to multiply two surds together, we could combine them back in the same way. That is

1.05 Optional: Multiply and divide surds mathspace.co

This is sometimes known as the multiplication law of surds. Unlike adding and subtracting surds, these do not have to be like terms before multiplying. If we want to ﬁnd the product of surds that have a coeﬃcient, we can use the properties of multiplication to rearrange the expression.

Example 1 Simplify

Create a strategy Use the rule

Apply the idea

Apply the rule Evaluate 19 × 17

Example 2 Simplify

Create a strategy Multiply the integer terms together then the surds together.

Apply the idea Rearrange the terms

Perform the multiplication

We can simplify our answer since 363 = 121 × 3, where 121 is a perfect square.

Write the surd as a product of its factors

Evaluate

Simplify

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 3 Simplify

Create a strategy Use the distributive law: A (B − C) = AB − AC

Apply the idea Apply the distributive law Multiply the surds and integers Evaluate the products

Idea summary To multiply terms involving surds, use the rule:

Divide surds Similar to multiplying surds, we can divide surds by writing the values underneath the same symbol. That is

This is known as the division law of surds. If we want to ﬁnd the quotient of surds that have coeﬃcients, we can divide the integer terms and the surd terms separately.

Example 4 Simplify

Create a strategy Use the rule:

Apply the idea Apply the rule Perform the division

1.05 Optional: Multiply and divide surds mathspace.co

Example 5 Simplify

Create a strategy Divide the surd parts and integer parts separately.

Apply the idea Separate the integer and surd parts Write the surds as one surd Evaluate the divisions Evaluate the square root Evaluate

Reﬂect and check Similar to the multiplication example above, we can approach this question by ﬁrst simplifying . We can rewrite the original expression: Simplify Separate the integer and surd parts Simplify

Idea summary To divide terms involving surds, use the rule:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1.05 Practice questions Understanding 1

Ex 1

Ex 3

Trailblazer 1e–j, 2

Complete these statements using this example:

d 3a–df, 4ab, 5a–d, 6a–d, 7ab, 9a–e, 10ab

3c–g, 4bc, 5c–f, 6d–h, 7a–d, 9d–h, 10b–f, 11a–d

3d–h, 4cd, 5d–h, 6dh–l, 7d–h, 8, 9dh–l, 10e–h, 11

Simplify: a

Ex 2

Adventurer 1c–h, 2

Which of these statements are true and which are false? Assume all variables are positive.

Fluency 3

Explorer 1a–f, 2

Simplify:

Expand:

1.05 Optional: Multiply and divide surds mathspace.co

Simplify: a

Ex 4

Ex 5

Simplify:

Reasoning 12

13, 15

12, 14, 16, 17

Identify whether the resulting values are always a surd, sometimes a surd, or never a surd. Justify your answers with examples. a

The quotient of two surds.

The quotient of a surd and an integer.

The quotient of two integers.

Eliora claimed that if a rectangle has sides whose lengths are surds, that the area must also be a surd. Is Eliora correct? Provide an example to justify your conclusion.

The cross-sectional area of a prism can be found by dividing the volume by the height. If both the volume and the height are surds, is it possible for the height to not be a surd? Provide an example to justify your conclusion.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Give an example where a and b are not surds, but c is a surd.

Give an example where a and b are surds, but c is not a surd.

Give an example where a and b are surds, and c is also a surd.

Suppose that a and b are different prime numbers. a

Explain why

is a surd, but

Explain why

is a surd?

is not.

ISO paper, for example A4 paper, always has a length that is times the width.

A3 A0

A4 A4

Pythagoras’ theorem states that c2 = a2 + b2, where c is the length of the hypotenuse of a right-angled triangle, and a and b are the lengths of the shorter sides.

Scarlett is starting a new standard series of paper where the length is times the width. The smallest size in the . Each subsequent member series has a width of of the series has lengths that are times the lengths of the previous size. Explain why all of the paper in this series would have an area which is a surd.

Problem-solving 18, 20, 22, 24a 18

19, 20, 22–24

19, 21, 23–25

Find the exact area of these rectangles. Give your answers as simpliﬁed surds. a

125 cm

10 m

2 20 cm

Find the area of the trapezium in simpliﬁed surd form:

8 1.05 Optional: Multiply and divide surds mathspace.co

The surface area of a human body in square metres can be approximately modelled by , where A is the surface area, h is the height of the person in cm, and w is the weight of the person in kg. a

Use the model to ﬁnd the surface area of a person who is 164 cm tall and weighs 63 kg. Leave your answer in exact form.

Find the approximate surface area of the person, to the nearest hundredth of a square metre.

What are the possible height and weight ranges for a person with a surface area of 2 m2?

Find the exact perpendicular height of a triangle whose area is measures .

A rectangle has a side length of

and a width of

and whose base .

Find the exact area of the rectangle.

Round the area of the rectangle to the nearest metre.

Find an approximation for the area of the rectangle by rounding the length and width to the nearest metre before multiplying.

Is there a signiﬁcant difference between the estimated area found by rounding before and after the area calculation? Calculate the difference as a percentage of the exact area.

A big rectangular carpet is metres long and metres wide. A small rectangle, 4 metres long and metres wide, is cut out of the carpet. Find the area of the carpet that is remaining in simpliﬁed surd form.

Ciera needs to perform each of these calculations, where uses x rounded to one decimal place as an approximation.

. In her calculations Ciera

Determine how much each of her calculations differ from the exact value. Give your answer rounded to 2 decimal places. a 25

10x

x2 + 5

Vaughan needs 40 lengths of rope, each

50x − 2x2

long. Vaughan rounds the individual

rope length required up to the nearest metre and then multiplies by 40 to obtain an approximate amount to purchase from the hardware store. Determine how much less Vaughan’s estimate would be if he hadn’t used rounding for each rope length.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1.06

Optional: Binomial expansions with surds After this lesson, you will be able to… • understand how to apply the distributive law to multiply binomials. • apply the method of multiplying surds to binomial expressions. • expand and simplify binomial expressions involving surds. • identify and apply the special cases of expanding binomials, such as perfect squares and the difference of two squares.

Binomial expansions with surds We can multiply surds using the property , and we can multiply two binomial expressions using the distributive law (A + B) (C + D) = AC + AD + BC + BD. A, B, C and D can be any type of terms - they don’t need to be only variables or integers. We can now combine the two concepts together allowing us to expand and simplify binomial expressions involving surds. For example:

Interactive exploration

Discover this concept in action online

mathspace.co

Example 1 Expand and simplify

Create a strategy Expand using the distributive law: (A − B) (C − D) = A (C − D) − B (C − D)

Apply the idea Apply the distributive law

Expand both brackets

1.06 Optional: Binomial expansions with surds mathspace.co

Example 2 Expand and simplify

Create a strategy Expand using the rule for the difference of two squares: (A + B) (A − B) = A2 − B2

Apply the idea Apply the rule

Evaluate the indices

Evaluate

Reﬂect and check If we hadn’t noticed that this was a difference of two squares expression, we could have performed the multiplication as usual and then simpliﬁed to get the same result:

Example 3 Expand and simplify

Create a strategy Use the perfect square expansion (A − B)2 = A2 − 2AB + B2 to expand the brackets.

Apply the idea

Apply the rule

Evaluate the indices and product

Perform the multiplication

Add the integer terms

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary We can expand the product of two binomial expressions using the rule: (A + B) (C + D) = AC + AD + BC + BD There are three special cases of expanding binomials: • (A + B)2 = A2 + 2AB + B2 (called a perfect square) • (A − B)2 = A2 − 2AB + B2 (called a perfect square) • (A + B) (A − B) = A2 − B2 (called a difference of two squares)

1.06 Practice questions Understanding 1

Explorer 1ac, 2bc, 3a–c

(a + b) (c + d)

(x + 3) (x − 2)

(x − 5) (x + 5)

(x + 2)2

Simplify: a

When expanded and fully simpliﬁed, will these expressions have all rational terms, all surd terms, or a mix of rational and surds? a

Fluency Ex 1

Expand and simplify: a

Trailblazer 1d, 2ad, 3cef, 4cd

Expand and simplify: a

Adventurer 1bd, 2bd, 3b–d, 4ab

5a–c, 6a–c, 7a–c

5b–e, 6c–e, 7c–e, 8bc

5e–h, 6e–h, 7e–h, 8ad

Expand and simplify: a

1.06 Optional: Binomial expansions with surds mathspace.co

Ex 2

Ex 3

Expand and simplify: a

Fill in the boxes for each equation. a

Reasoning

9, 10

Explain why when expanding and simplifying there would be two terms and for there would be four terms, provided c ≠ a and d ≠ b.

For:

Find values of a, b, c, and d, so that the expansion has four terms, none of which are like terms.

Find values of a, b, c, and d, so that the expansion simpliﬁes to two terms.

Can a fully simpliﬁed product of two binomials involving surds ever have three terms? If so, give an example. If not, explain why not.

Problem-solving 11, 12 11

11–13

12–14

For the given right-angled triangle: a

Find c, the length of the hypotenuse of the triangle.

Find the exact perimeter of the triangle.

Find the exact area of the triangle.

10 + 7

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

10 - 7

For the given trapezium:

2 5

Find the exact area of the trapezium.

Find the exact perimeter of the trapezium.

17 + 5 2 17

For the given isosceles triangle: a

Find h, the height of the isosceles triangle.

Find the exact perimeter of the isosceles triangle.

Find the exact area of the isosceles triangle.

5 2

5 −1

Expand and simplify

What value of m can be substituted into rational?

1.07

. so that the expression is

ptional: Rationalise the O denominator After this lesson, you will be able to… • understand the process and importance of rationalising the denominator. • apply the technique of squaring a surd to rationalise denominators. • use conjugates to rationalise binomial denominators. • apply these techniques to rationalise denominators in solving problems.

Rationalise the denominator with perfect squares Fractions with surds in the denominator are harder to calculate with. For these fractions, we can rationalise the denominator, which is a method used to rewrite the expression without surds in the denominator.

1.07 Optional: Rationalise the denominator mathspace.co

Exploration For the expression

1. What do you think will happen if we multiply the denominator by

2. Multiply the denominator by . Does the result have the same value as the original expression? Use your calculator to verify your answer. 3. What number can we multiply by that does not change the value of the original number? 4. Using your answer, what fraction should we multiply

that will not change the

value of the expression, but will eliminate the surd from the denominator? 5. Multiply

by your answer to the previous question. Does the result have the

same value as the original expression? Use your calculator to verify your answer.

When an expression has only one term in the denominator, we can rationalise the denominator by multiplying the numerator and denominator by the surd in the denominator. To rationalise the denominator of the form

we want to multiply it by the fraction

Since we are multiplying the numerator and denominator by the same number, it is the same as multiplying by 1, which does not change the value of the expression.

Example 1 Express the fraction in simplest surd form with a rational denominator:

Create a strategy Multiply the numerator and denominator by

Apply the idea Multiply by

Evaluate the multiplication

Simplify

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 2 Simplify

Create a strategy Check if we can simplify the fraction ﬁrst before rationalising.

Apply the idea Notice that 3 is a factor of both 39 and 6 Separate into two fractions Rewrite the surds Evaluate the divisions

Reﬂect and check Rationalising the denominator would have given us the same result, but it requires more steps: Multiply by Evaluate the multiplication Simplify the surds Evaluate the division

Idea summary To rationalise the denominator of the form fraction

, we need to multiply it by the

If possible, we should simplify the surds before rationalising.

1.07 Optional: Rationalise the denominator mathspace.co

Binomial denominators Exploration For these surds: •

•

1. Calculate the product of every possible combination of two surds. That is, , and so on. 2. When multiplied together, which pairs result in a rational number? 3. How are the two expressions related? 4. For the expression

, what should we multiply by to eliminate the surd from

the denominator? Justify your answer.

When there is a binomial expression in the denominator, multiplying by

will not eliminate

the surd. Instead, we can rationalise the denominator by multiplying top and bottom by the conjugate of the denominator. For any binomial expression A + B, we can ﬁnd a conjugate, A − B, by changing the sign of the second term. For example, the conjugate of

A binomial and its conjugate are called a conjugate pair. The product of a conjugate pair takes the familiar form of the difference of two squares: (A + B) (A − B) = A2 − B2 The expression A2 − B2 will be rational even if either term A or B are surds. For example:

Example 3 Express the fraction in simplest surd form with a rational denominator:

Create a strategy Multiply the numerator and denominator by the conjugate of

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea The conjugate of

is Multiply the top and bottom by the conjugate

Use difference of two squares in the denominator

Evaluate the indices

Evaluate the denominator

Factorise the numerator

Reflect and check Factorising the numerator is not necessary to answer this question, but doing so helps us check if the fraction can be simplified further.

Example 4 Evaluate, expressing your answer in simplest surd form with a rational denominator:

Create a strategy To subtract fractions, the denominators must be the same. We will rationalise the denominators before subtracting.

Apply the idea Rationalise the denominators

Evaluate the multiplication

Find a common denominator

Evaluate the multiplication

Evaluate the subtraction

1.07 Optional: Rationalise the denominator mathspace.co

Reﬂect and check Rationalising the denominator after subtracting will result in the same answer, but may require more steps. Find a common denominator Evaluate the multiplication Evaluate the subtraction Rationalise the denominator Evaluate the multiplication Simplify the surds Evaluate the multiplication

Idea summary For any binomial expression A + B, we can ﬁnd a conjugate A − B by changing the sign of the second term. A binomial and its conjugate are called a conjugate pair. To rationalise the denominator of the form , we multiply the numerator and denominator by the conjugate of the denominator. That is, we multiply it by the fraction

Where possible we should simplify the expression before rationalising.

1.07 Practice questions Understanding 1

Simplify

Evaluate

Explorer 1, 3ab, 4a–c, 5, 6a–d

. .

Adventurer 1, 3bc, 4c–f, 5, 6d–g

Trailblazer 1, 2, 3cd, 4e–h, 5, 6d–h

Simplify

Why is

equal to

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Determine whether each statement is true or false: a

The product of a conjugate pair will result in a rational number.

A conjugate pair is a fraction with the same numerator and denominator.

A fraction with the same numerator and denominator simplifies to 1.

A product of conjugate pairs is similar to a difference of squares.

Simplify: a

For each surd, what is the simplest surd we could multiply it by to produce an integer result? a

Simplify

Evaluate

. .

Does each product show a conjugate pair or not? a

Fluency 7

7a–h, 8a–f, 9a–e, 10a–e, 11a–d, 12a–e

7d–j, 8e–j, 9d–h, 10e–j, 11c–g, 12d–h

7e–l, 8g–l, 9g–l, 10g–l, 11f–l, 12h–l

Fi nd the conjugate: a

Rationalise the denominator and simplify: a

l 1.07 Optional: Rationalise the denominator mathspace.co

Ex 1

Ex 2

Ex 3

Ex 4

Rationalise the denominator and simplify: a

Evaluate in simplest surd form with a rational denominator: a

Reasoning 13

13, 15

14, 15

For: x2 − y2 a

Explain why this expression is known as a difference of perfect squares.

Factorise the expression.

Explain how this relationship is useful in rationalising the denominator of an expression like

, if x and y are surds.

Explain why the rationalised form of

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

will contain a term of

Xander says that to rationalise Justify your answer.

, we can multiply it by

Problem-solving 16, 17, 19

. Is he correct or incorrect?

16–18, 20

18, 20–22

A rectangle has a length of

Pythagoras’ theorem states that in a right-angled triangle, c2 = a2 + b2, where c is the length of the hypotenuse and a and b are the lengths of the two shorter sides.

and width of

The shorter sides of a right-angled triangle are hypotenuse of this right-angled triangle. 18

The ﬁrst term of a sequence is

. Find the area.

and

. Find the length of the

. Subsequent terms of the sequence are found by

multiplying the current term by so on.

, so the second term would be

Find a term of the sequence that is rational.

Is the term found in part (a), the only term that would be rational?

A rectangular garden has an area of 18 m2 and width of garden in simplest form with a rational denominator.

Rewrite

Find the mean of

in the form of and

, and

. Find the length of the

where x, y, and z are real numbers. , giving the answer in simpliﬁed rational form. Assume that

x is positive. 22

Evaluate

, giving the answer in simpliﬁed rational form.

1.07 Optional: Rationalise the denominator mathspace.co

1.08

Optional: Fractional indices After this lesson, you will be able to… • rewrite square and other roots in index form. • evaluate expressions with fractional indices. • apply understanding of fractional indices to simplify fractions using negative indices.

Fractional indices with numerator of 1 We want to look at expressions of the form

Exploration Let’s explore different types of indices, starting by looking at the equation: a+b=1 • Find a combination where a and b are integers. • Find a combination where one of a or b is zero. • Find a combination where both a and b are positive. • Find a combination where both a and b are fractions. • Find a combination where a and b are equal. Next let’s look at 16a × 16b = 161 1. What must be true about a and b? 2. Use your calculator to evaluate 16a and 16b for each combination of a and b found previously. 3. What do you notice when a and b are fractions? 4. What do you notice when a and b are equal?

Let’s formalise our exploration to help us work with fractional indices, starting by writing different form. First, using the multiplication index law, we have:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

in a

The definition of a square root is a value which multiplies by itself to give the original number. We know that Comparing these two statements we see that

We can confirm this by using the power of a power rule:

In a similar way we can look at the value

. Using the multiplication law:

From the definition of a cube root: Once again, by comparison we see that

These results can be generalised for any positive denominator. That is,

Example 1 For a Rewrite in surd form.

Create a strategy Use the rule

Apply the idea

Reﬂect and check We can also write this as , but writing the root without an index is usual for a square root.

1.08 Optional: Fractional indices mathspace.co

b Evaluate

Create a strategy

Apply the idea

Take the square root of 36

From part (a) Use that 36 = 62 Simplify

Example 2 Write

with a fractional index.

Create a strategy Use the rule

Apply the idea .

Idea summary Fractional index law:

Fractional indices with other numerators Next we look at expressions of the form

We know the power of a power law: Applying this index law, and the fractional index law that we’ve just seen:

In a similar way, we can see:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For example:

or:

More generally, the fractional index law states:

When solving problems with fractional indices, it doesn’t matter whether you start with the powers or the roots (although it might be easier to do it one way for any particular expression). Let’s simplify

in two ways:

Starting with the root:

Now let’s start with the powers:

You can see that we get the same answer both ways, but starting with the powers led to working with much larger numbers.

Example 3 Rewrite each expression in index form: a

Create a strategy Use the rule

Apply the idea .

1.08 Optional: Fractional indices mathspace.co

Create a strategy Use the rule

Apply the idea .

Example 4 Evaluate: a

Create a strategy Use the rule

Apply the idea Apply the rule 33 = 27

Create a strategy Use the rule

Apply the idea Apply the rule

Solve inside the brackets, note that

22 = 4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 5 Write as a simpliﬁed fraction: a

Create a strategy Use the rules

and

Apply the idea Apply the rules

Evaluate the cube roots

Evaluate the squares

Reﬂect and check If we had evaluated the squares before the cube roots, the numbers would have become very large and would likely require a calculator.

Create a strategy Use the rules

and

Apply the idea Use the negative index law

Write as roots

Evaluate the square roots

1.08 Optional: Fractional indices mathspace.co

Idea summary When simplifying expressions with fractional indices, we may need to use many index laws to fully simplify. Fractional index law: The multiplication law: am × an = am+n The division law: am ÷ an = am−n, an ≠ 0 The power of a power law: (am)n = am×n Powers of fractions: The zero power law: a0 = 1, a ≠ 0 Negative index law: Negative index law with fractions

1.08 Practice questions Understanding

1–4

Explorer

Complete the statement:

Complete the expressions: a

e 4

0.75

Adventurer

1–4

Trailblazer

Evaluate on your calculator, rounding to two decimal places, if necessary. a

Which expressions are equivalent to a

1–4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

80.5

320.2

Fluency

Ex 1

Ex 2

Ex 3

5b–h, 6b–f, 7b–f, 8b–f, 9bd, 10e–h, 11e–h, 12a–f, 13a–d, 14ab, 15a–d, 16a–d, 17a–d

5e–l, 6c–h, 7c–h, 8c–h, 9bd, 10, 11, 12c–h, 13d–h, 14, 15c–h, 16d–h, 17

For each expression: i

Write in surd form

Evaluate

Evaluate:

Write each expression with a fractional index: a

Express with a fractional index: a

Express in surd form: a

Ex 4

5a–e, 6a–d, 7a–d, 8a–d, 9ac, 10a–d, 11a–d, 12ab, 13ab, 15ab, 16ab

Evaluate:

1.08 Optional: Fractional indices mathspace.co

Express with a fractional index: a

Simplify:

Evaluate: a

Ex 5

Evaluate:

Express in the form xk, where k is rational: a

Reasoning

Given

, explain how this expression was simpliﬁed.

Without calculating, which is greater:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

18, 19

20–22

? Explain your reasoning.

Explain circumstances where it is easier to evaluate

by writing it as

, rather than

. Use a numeric example in your explanation. 21

Blair attempted to fully simplify an expression: 1

Rewrite the cube root in index form

Rewrite the indices with a common denominator

Add the indices

Multiply the indices using power of a power law

Rewrite from index form into surd form

Identify any errors that Blair has made and explain what he has done wrong.

Correct any errors.

Phoebe attempted to fully simplify an expression with positive variables: 1 2 3

Use the power of a power law Simplify numeric coeﬃcients

Rewrite from index form into radical form

Identify any errors that Phoebe has made and explain what she has done wrong.

Correct any errors.

Problem-solving 23, 24, 27a–e 23

Multiply numeric and algebraic parts

A rectangle has a length of a

23, 24, 27b–j

and a width of

25, 26, 27e–p, 28, 29

. Find the area of the rectangle in:

Fractional index form

Surd form

A cube has sides of length

. Find the volume of the cube.

Using six different digits from 1 to 9, ﬁll in the boxes to make a true statement.

1.08 Optional: Fractional indices mathspace.co

Using each digit from 1 to 6 once, ﬁll in the boxes to make the largest possible product.

Simplify each expression in index form. Assume all variables represent positive numbers.

Simplify: a

Simplify:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1 Chapter review 1

Simplify: a

x3 × 6x2

a4 × −3a5

Find the value of a and b in the equation:

Simplify: a

4y10 × 5y−4

(4a)−3

5x4 × (−3x−8)

3y−2 × 4y−3

(5mp)2 × mp−2

(6x−2 × 2x3)4

(Optional) Simplify fully, then determine if the value is rational or irrational: a

Fill in the blanks: a

Simplify, giving your answers with positive exponents: a

(−2x3)4

Simplify, giving your answers with positive exponents: a

6m3 n7 × 9m6 n

Fill in each box. Evaluate if possible. a

−2y7 × 2y2 × −2y6 d

Simplify: a

(Optional) Simplify fully: a

(Optional) Simplify fully, given the variable represents a positive value:

(Optional) Simplify fully: a

Chapter 1 review mathspace.co

(Optional) Assuming all variables represent positive values, simplify fully: a

(Optional) Find the perimeter of the triangle in simpliﬁed surd form. 27

(Optional) Simplify fully: a

(Optional) Expand and fully simplify: a

(Optional) For this shape: a

Find the exact perimeter of the ﬁgure.

Find the exact area of the ﬁgure.

( 5 + 7 ) cm ( 5 + 7 ) cm

(Optional) Rationalise the denominator of the given expressions. Express your answer in simplest surd form: a

(Optional) A rectangular garden has an area of 21 m2 and width of of the garden in simplest form with a rational denominator.

(Optional) The initial term of a sequence is are generated by multiplying the current term by

. Consecutive terms of the sequence , thus the second term would be

, and so forth.

Identify a term of the sequence that is rational.

Is the term found in part (a) the sole term that would be rational?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

. Find the length

(Optional) Write with a fractional exponent: a

(Optional) For each expression: i

Write in surd form.

Evaluate.

(Optional) Evaluate: a

Did you know?

The energy in one lightning strike is approximately 1 × 109 joules, enough to power a 60-watt light bulb for six months! Each second, about 100 lightning strikes occur around the world, meaning the Earth experiences roughly 8 million strikes per day. If we could capture even a small fraction of this energy, it could contribute significantly to global energy needs. Index notation helps meteorologists calculate this incredible energy. However, when studying thousands or millions of strikes occurring globally, meteorologists often deal with cumulative energy levels like 1015 (a quadrillion joules) or higher. Writing and calculating these enormous values is far more efficient with index notation. Chapter 1 review mathspace.co

Big ideas Mastering the basics of algebraic operations and understanding common denominators are essential for simplifying and solving algebraic fractions.

Curriculum Outcomes VC2M10A03 apply the 4 operations to simple algebraic fractions with numerical or single variable denominators VC2M10AA03 simplify combinations of linear expressions with rational coefficients and the solution of related equations

Algebraic fractions Chapter outline 2.01 2.02 2.03 2.04 2.05

Add and subtract fractions with numeric denominators Add and subtract fractions with algebraic denominators Optional: Add and subtract further algebraic fractions Multiply and divide algebraic fractions with numeric denominators Optional: Multiply and divide algebraic fractions with algebraic denominators Chapter 2 review

70 75 80 86 91 96

Did you know that the word “fraction” comes from the Latin word “fractus,” meaning “broken”?

2.01

dd and subtract A fractions with numeric denominators After this lesson, you will be able to… • add and subtract fractions with algebraic numerators by finding a common denominator.

Add and subtract fractions with numeric denominators An algebraic fraction is a fraction with a variable in the numerator or denominator. Adding and subtracting algebraic fractions with numeric denominators is very similar to adding and subtracting purely numeric fractions. We ﬁrst need to ﬁnd equivalent fractions with a common denominator, then we add or subtract the numerators. We may be able to simplify our answer by collecting like terms, and/or dividing the numerator and denominator by common factors.

Exploration Notice that

Write another equivalent fraction with a denominator of 12.

Write a fraction equivalent to this.

What would be the lowest common denominator for

with a denominator of 15. Explain how you got and

Example 1 Simplify:

Create a strategy The fractions already have a common denominator, so we can subtract the numerators. Simplify by collecting like terms and dividing the numerator and denominator by any common factors.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea

Subtract the numerators

Collect like terms

Divide numerator and denominator by 3

Reﬂect and check To check our answer, substitute x = 1 into both the original expression and the simpliﬁed expression. We could use any value of x, but 1 is the simplest. Substitute x = 1 into the original expression

Multiply and subtract

Evaluate Substitute x = 1 into the simpliﬁed expression Both the original expression and the simpliﬁed expression gave the same result, suggesting that the answer is correct.

Example 2 Simplify:

Create a strategy Rewrite the fractions with a common denominator, then add the numerators.

Apply the idea The lowest common multiple (LCM) of 3 and 5 is 15. Rewrite the fractions to have a common denominator

Evaluate the products

Add the numerators

Collect like terms

Reﬂect and check In this case, the lowest common denominator is the product of the two denominators. The product will always be a common denominator, but may not be the lowest common denominator. For example with

, the LCM is 30, but the product of the denominators is 90. If we used 90

as the denominator, we would have to work with larger numbers than if we used 30.

2.01 Add and subtract fractions with numeric denominators mathspace.co

Example 3 Simplify: 8x −

Create a strategy Rewrite the ﬁrst term as a fraction with the same denominator as the second term, then subtract the numerators.

Apply the idea Write 8x as a fraction Rewrite the ﬁrst fraction to have a denominator of 3 Evaluate the products Subtract the numerators Collect like terms

Reflect and check To check our answer, substitute x = 1 into both the original expression and the simplified expression. Substitute x = 1 into the original expression Evaluate the products and write the fraction as a mixed number Evaluate the difference Substitute x = 1 into the simplified expression Evaluate the product and write the fraction as a mixed number Both the original expression and the simplified expression gave the same result, suggesting that the answer is correct.

Idea summary To add or subtract fractions where the numerators are algebraic terms: 1. Rewrite the fractions with a common denominator. 2. Add or subtract the numerators. 3. Simplify by collecting like terms and dividing the numerator and denominator by any common factors.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

2.01 Practice questions Explorer 1–3

Understanding 1

and

For a

Ex 2

Ex 3

and

simpliﬁes to The expression

and

is equivalent to 2z

4, 5a–e, 6a–e, 7a–c, 8ad

Fluency

Ex 1

and

Are these statements true or false? a

Find the lowest common denominator for these pairs of algebraic fractions: a

Trailblazer 1–3

Evaluate and simplify: a

Adventurer 1–3

and

simpliﬁes to The expression

4, 5e–i, 6e–i, 7c–g, 8b–g

is equivalent to 3a 4, 5g–l, 6h–l, 7e–h, 8c–h

Find the lowest common denominator.

Simplify

Simplify: a

Simplify:

Simplify each expression, by ﬁrst simplifying the fractions (where possible). a

2.01 Add and subtract fractions with numeric denominators mathspace.co

Reasoning

9, 12

9–12

10–12

Using mathematical vocabulary, explain how you can complete the equation.

Complete the equation:

Why is it important to simplify fractions before adding or subtracting them in algebraic equations?

Can we add or subtract fractions with different denominators directly? Explain your answer.

Problem-solving 13ac, 14a, 15, 17 13

Fully simplify these expressions: a

13acd, 14–17

Fully simplify these expressions: a

13a–c, 14ab, 15, 16a, 17

Write an expression for the perimeter of the triangle shown, and simplify this expression.

x 5

3x 10

Given y1 = a b

and y2 = − ,

Find y3 if y3 = y1 + y2.

Sketch the graph of y3 on the Cartesian plane.

The fuel tank in Dario’s car holds p litres of petrol. After ﬁlling the tank, Dario uses tank driving to his friend’s house, then another

of the

of the tank driving to the beach. Write an

expression for number of litres of fuel left in the tank when Dario arrives at the beach, and simplify this expression.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

2.02

dd and subtract A fractions with algebraic denominators

After this lesson, you will be able to… • determine the lowest common multiple (LCM) of two algebraic terms by considering the coefficients and variables. • apply the knowledge of finding the LCM to add or subtract fractions that have algebraic denominators. • simplify fractions with algebraic denominators.

Add and subtract fractions with algebraic denominators Just like with other types of fractions, we need to ﬁnd a common denominator to add or subtract fractions with algebraic denominators. Once we have a common denominator, we can add or subtract the numerators of the fractions and simplify the result if possible. When at least one of the denominators is algebraic, then we will have an algebraic common denominator. Remember that the best common denominator to use is the lowest common denominator, which is the lowest common multiple (LCM) of the denominators. The LCM of two algebraic terms will be the LCM of the coeﬃcients, multiplied by the different variables. For example, the LCM of 12 and 8 is 24, so the LCM of 12x and 8y is 24xy.

Exploration For 60x and 24y, let’s look at how the prime factors of the coeﬃcients can help us ﬁnd the lowest common multiple. 24

60 6 2

10 3

12 2

60 = 22 ⋅ 3 ⋅ 5

24 = 23 ⋅ 3

6 2

Between 60 and 24, what is the highest power of each prime factor (2, 3, and 5)?

What is the lowest common multiple of 60 and 24?

What is the lowest common multiple of 60x and 24y?

What is the product of 60x and 24y?

If simplifying

, which would be the easiest common denominator to use,

the lowest common multiple of 60x and 24y, or the product of 60x and 24y? Explain.

2.02 Add and subtract fractions with algebraic denominators mathspace.co

Consider this expression, where a and b are numbers: To add these fractions, we need to find a common denominator. The most efficient way to do this is to use lowest common multiple (LCM) of the denominators. In this case, a common multiple of a and b is ab. We need to rewrite the fractions with a common denominator of ab, then add the numerators. The first fraction needs to have both numerator and denominator multiplied by b, and the second fraction must have both multiplied by a: Now, we can add the numerators to get the final result:

Example 1 Simplify

Create a strategy The fractions already have a common denominator, so we can add the numerators. Simplify by dividing the numerator and denominator by any common factors.

Apply the idea Add the numerators

Evaluate the sum

Divide numerator and denominator by 7

Reﬂect and check When working with fractions (algebraic or numeric), it is always best to check our solution for any common factors. If there are common factors, we should simplify the fraction further as we did in this example.

Example 2 Simplify

Create a strategy Rewrite the fractions with a common denominator, then subtract the numerators.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea The lowest common multiple of 8y and 12y is 24y. Multiply to get a common denominator Simplify the products Subtract the numerators Collect like terms

Example 3 Write

as a single fraction.

Create a strategy Find a common denominator and combine the numerators of the fractions.

Apply the idea The lowest common multiple of 3y and 6x is 6xy. Multiply to get a common denominator Simplify the products Add the numerators

Reﬂect and check Notice in this case that the terms in the numerators are not like terms, so we cannot add them and our answer is fully simpliﬁed.

Idea summary To add or subtract fractions where the denominators are algebraic terms, we need to ﬁnd a common denominator. We can then add or subtract the numerators and simplify the result if possible. When the denominators are algebraic terms, the lowest common denominator will be the lowest common multiple of the coeﬃcients, multiplied by the different variables.

2.02 Add and subtract fractions with algebraic denominators mathspace.co

2.02 Practice questions Understanding 1

Explorer 1–3

To add or subtract algebraic fractions, we need a common denominator.

When adding or subtracting algebraic fractions, we add the numerators and add the denominators.

If at least one of the denominators is an algebraic term, then the lowest common denominator of two fractions will be an algebraic term.

Evaluate and simplify: b

15a and 24a

Ex 2

12x and 15x

simpliﬁes to The expression

Fluency 5

5c and 25c

6y and 14x

Are these expressions true or false? a

Ex 1

Find the lowest common multiple of each pair of algebraic terms. a

Trailblazer 1–4

Are these statements true or false?

a 3

Adventurer 1–4

is equivalent to

5a–d, 6a–cg, 7abh, 8a–d, 9acg, 10ac

b d

simpliﬁes to The expression

is equivalent to

5abe, 6dfh, 7b–g, 8b–f, 9b–e, 10bd

5e–h, 6ei–l, 7d–g, 8f–h, 9befh, 10

Simplify: a

Simplify:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Ex 3

Write as a single fraction: a

11, 13

Write as a single fraction: a

Reasoning

If x does not equal 0, is

Explain why it is important to ﬁnd a common denominator when adding or subtracting algebraic fractions with different denominators.

This expression was simpliﬁed incorrectly:

the same as

4y a

Explain the mistake.

Problem-solving 15a, 16 14

Complete the equation:

Simplify:

? Explain your reasoning.

7y 8x

21xy 21 32xy 32 b

Give the correct solution.

14, 15bc

14, 15d, 16, 17

A rectangle has a length of rectangle.

To get her driver’s licence, Attia took a theory test and a practical test. The theory test had t points in total and the practical test had twice as many points as the theory test. Attia scored x points in the theory test and y points in the practical. a

Write an expression for Attia’s percentage score in the: i

and a width of . Find an expression for the perimeter of the

Theory test

Practical test

Attia will get her licence if the average of her two scores is greater than 70%. Use your answers to part (a) to write this as an inequality, simplifying the algebraic expression to a single fraction.

2.02 Add and subtract fractions with algebraic denominators mathspace.co

2.03

ptional: Add and O subtract further algebraic fractions

After this lesson, you will be able to… • add or subtract the binomial numerators while keeping the denominator constant. • simplify algebraic fractions to a single expression.

Add and subtract algebraic fractions Adding and subtracting algebraic fractions with binomial numerators and numerical denominators is similar to adding and subtracting numerical fractions. The key is to have the same denominator for the fractions. If the denominators are different, we need to ﬁnd the common multiple and adjust the fractions accordingly. Once the denominators are the same, we add or subtract the numerators, and the denominator remains the same.

Example 1 Simplify

as a single fraction.

Create a strategy Since the denominators are the same, add the numerators.

Apply the idea Add the numerators

Simplify the numerator

Example 2 Simplify

as a single fraction.

Create a strategy Since the denominators are the same, subtract the numerators.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea Subtract the numerators

Distribute negative sign in (3x + 2)

Simplify the numerator

Example 3 Simplify

as a single fraction.

Create a strategy Find a common multiple of 3 and 4. Then, adjust the fractions so they have the same denominator and add the numerators.

Apply the idea Adjust the fractions to have the same denominator

Distribute 4 in (x − 3) and 3 in (2x + 5) in the numerator

Simplify the numerator

Example 4 Simplify

as a single fraction.

Create a strategy Find the common multiple of 8 and 6. Then, adjust the fractions so they have the same denominator and subtract the numerators.

Apply the idea Adjust the fractions to have the same denominator

Distribute 3 in (3x − 6) and −4 in (2x − 4) in the numerator

Simplify the numerator

2.03 Optional: Add and subtract further algebraic fractions mathspace.co

Idea summary To add and subtract algebraic fractions with binomial numerators and numerical denominators, the denominators of the fractions must be the same. We then add or subtract the numerators, and the denominator remains the same.

2.03 Practice questions Understanding 1

Explorer 1–3

Adventurer 1, 2, 4

Trailblazer 1, 3–5

Which statements are true and which are false? a

Adding algebraic fractions is similar to adding numerical fractions.

The denominators of algebraic fractions need to be different before you can add or subtract them.

When adding or subtracting algebraic fractions, the numerator is always added or subtracted, not the denominator.

When adding or subtracting algebraic fractions, we need to factorise the numerator.

When the denominators of algebraic fractions are different, we need to ﬁnd the common multiple of the denominator.

Tom needs to change this fraction to have a denominator of 12.

He is aware that the denominator needs to be multiplied by 3. Which is the correct numerator of the new fraction? A 3

3x + 2

3x + 6

x+6

x+2

Jenny is trying to add these algebraic fractions:

She has to ﬁnd the common multiple of 3 and 6. Then, adjust the fractions so they have the same denominator. Which is the correct numerator of the ﬁrst fraction after the adjustment? A 4

x + 10

x+4

2x + 10

2x + 8

Mike needs to solve this algebraic fractions:

He has to ﬁnd the common multiple of 5 and 7. Then, adjust the fractions so they have the same denominator. Which is the correct numerator of the second fraction after the adjustment? A

7x + 49

5x + 35

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

7x + 42

5x + 30

Match the 3 pairs of equivalent fractions. a

iii

Fluency 6

Ex 1

9 Ex 2

12 Ex 3

6b–e, 7, 8c–f, 9b–f, 10b–f, 11b–f, 12b–e, 13bde, 14ac

6e–h, 7, 8e–h, 9d–h, 10e–h, 11d–h, 12e–h, 13d–h, 14b–d

Simplify: a

Are these expressions equal to a

6a–d, 8a–d, 9a–c, 10a–c, 11ac, 12a–c, 13a–c

Simplify:

2.03 Optional: Add and subtract further algebraic fractions mathspace.co

13 Ex 4

Simplify: a

Reasoning

15, 16ab

15, 16a–c, 17

If a =

Are these calculations correct? Explain how you know.

and b =

, explain how you would ﬁnd the value of a + b.

c 17

Two students, Joanne and Edmon tried to solve the problem:

Joanne’s working:

Edmon’s working:

15–18

Who is correct? Explain why.

What mistake did the other student make?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Jonathan and Jefferson were given algebraic fractions:

Jonathan says that b + c = d. Jefferson says that d − c = b. Who is correct? Explain your reasoning.

Problem-solving 20ab, 23a 19

Simplify the expression.

If x = 4, what is the value of the expression?

Given a =

,b=

,q=

Given a =

Simplify:

,b=

, express each expression in simplest form: c

and s =

a−d

c+b

, express each expression in simplest form. c

r−q

,c=

and d =

a−b ,r=

s−r

,c= b

a+b

Given p = a

19, 21–24

Consider the algebraic fraction:

a 21

19, 20cd, 21bc, 23ab

, and d =

s−q+r

p−q−r

, solve for x if a − b + c − d = 2.

If 3x − 2y = 7, then ﬁnd:

2.03 Optional: Add and subtract further algebraic fractions mathspace.co

2.04

ultiply and divide M algebraic fractions with numeric denominators

After this lesson, you will be able to… • apply the multiplication and division rules to algebraic fractions with numeric denominators. • use indices to simplify numerators and denominators in algebraic fractions. • identify and cancel common factors in the numerator and denominator.

Multiply fractions with numeric denominators We have multiplied numerical fractions in the past. When multiplying fractions, we multiply the numerators and multiply the denominators

For example:

When it comes to working with algebraic fractions with numeric denominators and applying the four operations, the process is exactly the same as when we worked with numeric fractions. Let’s have a look at a simple example of multiplying two algebraic fractions with numerical denominators. We may need to use indices to simplify the numerators. Remember that: a × a = a2 and in general am × an = am+n We should check if the ﬁnal answer can be simpliﬁed by cancelling out a common factor in the numerator and denominator using that

= 1, a ≠ 0.

Example 1 Simplify:

Create a strategy Multiply the numerators together and multiply the denominators together.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea Multiply the numerators and denominators

Evaluate

Example 2 Simplify:

Create a strategy Multiply the numerators together and multiply the denominators together.

Apply the idea Multiply the numerators and denominators

Evaluate

Factorise 12 out

Simplify

Reﬂect and check We could also have removed any common factors before we performed the multiplication: Factorise the terms

Remove the common factors

Simplify

2.04 Multiply and divide algebraic fractions with numeric denominators mathspace.co

Idea summary The product of two fractions is the product of their numerators divided by the product of their denominators. For B, D ≠ 0:

We may need to use indices to simplify the numerators. Remember that: a × a = a2 and in general am × an = am+n Before multiplying two fractions together, look for common factors that we can remove ﬁrst. This will make the resulting multiplication easier in most cases, as there will be less factors to deal with. We can use that

= 1.

Divide fractions with numeric denominators When we divide fractions, we take the reciprocal of the second fraction and multiply by it. Sometimes we say ﬂip or invert when we take the reciprocal. Multiply by the reciprocal Evaluate The process for dividing is the same as when we divide numerical fractions, but we may need to use indices.

Example 3 Simplify:

Create a strategy Multiply the ﬁrst fraction by the reciprocal of the second fraction.

Apply the idea Multiply by the reciprocal Remove the common factor

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reﬂect and check Notice that if common factors are not removed, we would end up with very large numbers which would then need to simplify. Multiply by the reciprocal Multiply the numerators and denominators Simplify

Idea summary To divide by a fraction, multiply by the reciprocal. For B, C, D ≠ 0:

Simplify if possible by removing common numerical or algebraic factors.

2.04 Practice questions Understanding 1

Adventurer 1, 2, 3bc

Trailblazer 1, 2, 3bd

Evaluate: a

Explorer 1, 2, 3ab

Are these statements true or false? a To multiply two algebraic fractions, we multiply the numerators and multiply the denominators. b When multiplying or dividing algebraic fractions with numeric denominators, the denominators must be the same before performing the operation. c To multiply or divide algebraic fractions, you must first find a common denominator. d The order of operations still applies when multiplying or dividing algebraic fractions with numeric denominators. Complete each simplification. a

2.04 Multiply and divide algebraic fractions with numeric denominators mathspace.co

Fluency Ex 1

5 Ex 2

6 Ex 3

4d–h, 5g–l, 6g–l

Simplify:

7, 9

8, 9

The following expression was simpliﬁed incorrectly:

a 8

4d–g, 5d–h, 6d–h

Simplify:

Reasoning 7

4a–d, 5a–d, 6a–d

Explain the mistake.

Amarion and Dean both simpliﬁed

Give the correct solution.

differently than the others in their group. Their

ﬁrst two steps are shown. Who could go on to be correct? Explain your answer. Amarion:

Dean:

1 2 9

1 2

Is each statement always true, sometimes true, or never true? Justify your answer. a

The quotient of two algebraic fractions is a numeric fraction.

The product of two numeric fractions is an algebraic fraction.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving 10a, 11ab, 12a 10

. Find an expression

Simplify: a

10d, 11cd, 12cd, 13, 14

Simplify: a

10bc, 11bc, 12b, 13

Simplify: a

and

2.05

and

. Find an expression for

Optional: Multiply and divide algebraic fractions with algebraic denominators

After this lesson, you will be able to… • apply multiplication to the numerators and the denominators of algebraic fractions. • apply division by using the reciprocal of the second fraction. • simplify resulting algebraic fractions by cancelling out common factors.

Multiply and divide algebraic fractions with algebraic denominators In the same way as when multiplying algebraic fractions with numeric denominators, we apply all the same fractional and algebraic thinking to algebraic denominators.

2.05 Optional: Multiply and divide algebraic fractions with algebraic denominators mathspace.co

Example 1 Simplify as a single fraction:

Create a strategy Remember to simplify a product of algebraic fraction we should cancel out common factors across numerators and denominators.

Apply the idea implify common numerical factors across numerators and S denominators Simplify common algebraic factors across numerators and denominators

Multiply the numerators and multiply the denominators

Simplify

Example 2 Simplify as a single fraction:

Create a strategy Dividing by a fraction is the same as multiplying by its reciprocal.

Apply the idea Take the reciprocal of the second fraction

Simplify common numerical factor across numerators and denominators

Simplify common algebraic factor across numerators and denominators

Multiply the numerators and multiply the denominators

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary To multiply algebraic fractions, multiply the numerators together and the denominators together. To divide algebraic fractions, multiply the ﬁrst fraction by the reciprocal of the second fraction. Always simplify the resulting fraction by cancelling any common factors in the numerators and denominators.

2.05 Practice questions Understanding 1

Multiply the ﬁrst fraction by the second fraction

Multiply the ﬁrst fraction by the reciprocal of the second fraction

Add the ﬁrst fraction to the second fraction

Subtract the second fraction from the ﬁrst fraction

Are these statements true or false? a

To simplify a product of algebraic fractions, you can cancel common factors in the numerators and denominators before multiplying.

To simplify a quotient of algebraic fractions, you can divide across the numerators and across the denominators.

Simplify:

a 2 × a3

Fluency Ex 1

Trailblazer 1–5

When dividing algebraic fractions, you should:

a 5

Adventurer 1–5

a 4

Explorer 1–3, 4a–c, 5

(x3 × x4)2

6, 7a–f, 8b, 9a–d, 10ab, 11ab

6, 7d–i, 8, 9c–f, 10b–e, 11abd

6, 7e–l, 8, 9e–h, 10d–f, 11

Simplify as a single fraction: a

2.05 Optional: Multiply and divide algebraic fractions with algebraic denominators mathspace.co

Simplify as a single fraction: a

9 Ex 2

Simplify as a single fraction: a

Reasoning

b 13a

12, 13a, 14

12–14

The product of two fractions is 1. If one of the fractions is

Complete the equations: a

Show that the given expression simpliﬁes to 1.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

, what is the other fraction?

Problem-solving 15a, 16 15

15ac, 16, 18

15b–d, 16–19

Simplify: a

Simplify

, what is the value of A?

where

Did you know?

Algebraic fractions are crucial for solving problems involving rates, proportions, and optimisation! For example, they help engineers calculate the efficiency of systems, like determining the flow of water through pipes of varying sizes. By simplifying algebraic fractions, they can model how changes in pipe diameter affect water pressure and flow rate, ensuring systems operate smoothly and effectively. Algebraic fractions allow us to break down complex relationships into manageable parts, making them essential in engineering, physics, and even economics!

2.05 Optional: Multiply and divide algebraic fractions with algebraic denominators mathspace.co

2 Chapter review 1

At the end of last month, a dam was

full. Over the current month,

was used up by the town. Meanwhile, rain filled dam is full at end of this month? 2

of the dam’s capacity

of the dam’s capacity. What fraction of the

Simplify: a

The fuel tank in Lannie’s car holds q litres of petrol. After filling the tank, Lannie uses the tank driving to her cousin’s house, then another

of the tank driving to the mall. Write

an expression for number of litres of fuel left in the tank when Lannie arrives at the mall, and simplify this expression. 4

(Optional) Two students, Waheed and Azize tried to solve the problem: Waheed’s working:

a 5

Who is correct? Explain why.

What mistake did the other student make?

Simplify: a

Simplify:

(Optional) Simplify: a

Azize’s working:

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Simplify: a

(Optional) Complete the equations: a

Simplify fully: a

(Optional)Simplify as a single fraction: a

(Optional) Show that the given expression simplifies to 1.

(Optional) Show that

(Optional) Simplify as a single fraction: a c

Chapter 2 review mathspace.co

Big ideas • Algebraic expressions can be manipulated and transformed through different ways allowing for a deeper understanding of their structure and relationships. • Algebraic fractions can be manipulated by ensuring common denominators, performing operations and simplifying through factorisation.

Curriculum Outcomes VC2M10A01 factorise algebraic expressions by taking out a common algebraic factor VC2M10A04 expand binomial products and factorise monic quadratic expressions using a variety of strategies VC2M10AA07 factorise monic and non-monic quadratic expressions and solve a wide range of quadratic equations derived from a variety of contexts

Expand and factorise Chapter outline 3.01 Factorise algebraic expressions 3.02 Expand binomial products 3.03 Factorise binomial products 3.04 Factorise monic quadratic trinomials 3.05 Optional: Factorise non-monic quadratic trinomials 3.06 Complete the square Chapter 3 review

100 105 111 116 121 127 134

Engineers use algebraic factorisation and expansion to simplify complex equations representing the motion of robotic arms, spacecraft, and drones.

3.01

Factorise algebraic expressions After this lesson, you will be able to… • identify the highest common factor in algebraic expressions. • factorise the highest common factor out from algebraic expressions. • apply the distributive law to factorise algebraic expressions with common factors.

Highest common factor Any factorising problem should start with an inspection of the terms to see if there is a common factor. If there are common factors, other than 1, these should be factored out before any other work is done.

Example 1 Find the highest common factor for the following pairs of terms: a 9ab and 12ac

Create a strategy Break down the terms into factors. Note, the HCF of 9 and 12 is 3.

Apply the idea 9ab = 3 × 3 × a × b

Factorise 9ab

12ac = 3 × 4 × a × c

Factorise 12ac

HCF = 3 × a Select the common factors = 3a Simplify b 16c3d2 and 20c2d3

Create a strategy Break down the terms into factors. Note, the HCF of 16 and 20 is 4.

Apply the idea 16c3d2 = 4 × 4 × c × c × c × d × d

Factorise 16c3d2

20c2d3 = 4 × 5 × c × c × d × d × d

Factorise 20c2d3

HCF = 4 × c × c × d × d 2 2

= 4c d

100

Select the common factors Simplify

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c 4(x − 3) and y(x − 3)

Create a strategy Note that a common factor can be an expression in brackets.

Apply the idea 4(x − 3) = 4 × (x − 3)

Factorise 4(x − 3)

y(x − 3) = y × (x − 3)

Factorise y(x − 3)

HCF = (x − 3)

Select the common factor

Idea summary The ﬁrst step in any factorising problem is to identify and factor out the highest common factor. It can help to break each term down into factors and compare.

Factorise algebraic expressions Consider 9ab + 12ac We can write each term in factors: 3×3×a×b+3×4×a×c We can see that the common factors are 3 and a, so we factor 3a out, and what is left goes into a bracket. 9ab + 12ac = 3a(3b + 4c) We can view this as factorising using the law, AB + AC = A(B + C) A is the highest common factor, which is taken out. This extends to three or more terms. AB + AC + AD = A(B + C + D)

Example 2 Factorise the expression by taking out the highest common factor: 8 + 16x − 4x2

Create a strategy The terms have a highest common factor of 4, which we can factor out using AB + AC + AD = A (B + C + D)

3.01 Factorise algebraic expressions mathspace.co

101

Apply the idea 8 + 16x − 4x2 = 4 × 2 + 4 × 4x − 4 × x2 2

= 4(2 + 4x − x )

Write the terms as products of 4 Factorise

Example 3 Factorise the expression 4(x − 3) − y(x − 3).

Create a strategy Factorise using the law, AB + AC = A (B + C). Here the common factor is the binomial expression (x − 3).

Apply the idea 4(x − 3) − y(x − 3) = 4 × (x − 3) − y × (x − 3) Write the terms as products of (x − 3) = (x − 3) (4 − y)

Factorise

Example 4 Factorise the expression 2t2k7 + 18t9k9.

Create a strategy Factorise using the law, AB + AC = A (B + C).

Apply the idea The coeﬃcients have a common factor of 2, and both terms have factors of t2 and k7, so we can factorise 2t2k7 out. 2t2k7 + 18t9k9 = 2t2k7 × 1 + 2t2k7 (9t7k2) = 2t2k7 (1 + 9t7k2)

Write the terms as products of 2t2 k7 Factorise

Idea summary We can use the distributive law to factorise an algebraic expression like AB + AC. This means writing the expression with any common factors between the terms taken outside of the brackets: AB + AC = A(B + C) Factorising is the reverse of expanding.

102

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3.01 Practice questions Understanding Ex 1

15n2m and 8nm2

16abnx and 12abmy

Ex 4

20x and 15x2n

6abmn and 8bnxy

12a7bm3y2 and 9ab5n2x3

x(x + 1) + 3(x + 1)

2x(4 − x) − (4 − x)

5y3 + 15y2 + 10y = 5y( y2 + ⬚ + ⬚)

Fill in the boxes to complete the factorisation: 2x3 + 4x2 + 2x = ⬚(x2 + 2x + 1)

ax2 − 3ax + 6a = ⬚(x2 − 3x + 6)

4kx2 + 12kx + 16k = 4k(⬚ + ⬚ + 4)

Find the highest common factor of the terms: 12p2r2t3, 9prt3, and 18p3r3. 4a–f, 5a–d, 6, 7

4c–h, 5c–h, 6, 7

Factorise: a

3ax2 + 6ax − 6a

2ax + 2ax + 4ax

4b3 − 16b2 + 20b

3xyz + 21xyz + 9xz

i Ex 3

16a2bx3y6 and 12b4m2n6x

Fluency 4

1–3

Trailblazer

Ex 2

1–3

Adventurer

Find the highest common algebraic factor of each pair of terms:

1–3

Explorer

−6efgh2 − 30efgh − 12ehg

f h

4m (n − 3) + 6(n − 3)

( y + 4) ( y + 7) + x( y + 7)

6qr2 − 36qr − 12q

−4abc2 + 8abc + 4ac

3m2(n + 2) + 5(n + 2)

2y2(x + 2) − 3(x + 2) 2p3q(r + 4) + 5q(r + 4)

5xy2 + 10xy + 15x

10rstu2 + 70rstu − 40rtu

a e

2x3 − 16x2 + 2x

Factorise: 2

4e–j, 5g–l, 6, 7

f h

5y3(z + 5) + 7(z + 5)

6a2b(c − 2) − 4b(c − 2) 2f(g + h) + (g + h)2

4(7 − x2) − 2y(7 − x2) + 5(x2 − 7)

3(8 − c2) − 5d(8 − c2) − 4(c2 − 8)

(3m − 2) (4m + 7) − (2m + 3) (4m + 7)

(6z − 5) (2z + 4) + (4z + 2) (2z + 4)

A farmer wants to fence a rectangular field with an area of 48xy − 24y. If the length of the field is 24y, find an expression for the dimension of the width of the field.

Factorise: a

20j − 15jk − 55j2k

2yz − 10xy + 12xy2z

−hf − hj + hg

3p2q2 + 4p4q4 + 5p6q6

pqr + p2q2r + p3q3r

25m − 15mn − 10m2n

3.01 Factorise algebraic expressions mathspace.co

103

Reasoning

8, 9, 11

9–11

12x4 + 18x3 − 24x2 can be factorised into the form 6x(2x3 + 3x2 − 4x). Has the expression been fully factorised? Explain your answer.

Xander was asked to factorise the expression 35x2y + 10xy2 − 5xy. He stated that the highest common factor was 5xy so the factored form of the expression is 5xy(7x + 2y). Identify and correct Xander’s error.

Explain why the highest common factor of the variables in any expression has the least possible index of any of the terms.

When factorising an algebraic expression, explain how you can check your answer.

Problem-solving 12, 14, 15 12

13–15

A rectangle’s length is three times its width. Its length can be represented by the expression: 3x2 + 6x + 3

Factorise this expression by taking out a common factor.

Interpret what the factors represent in the context of this problem.

A farmer wants to create a set of adjacent fields which all have the same width. He plans to create the smallest field in the shape of a square, with the largest field 9 times the size of the smallest field, and the middle field to have a length that is 5 units more than the smallest field. a

Write expressions for the area of each ﬁeld.

Use factorising to determine the dimensions of the entire ﬁeld.

A rectangle has area 4x2 + 8x cm2 and length 4x cm. For what values of x is it a square? For what values is it just a rectangle?

Using the digits 0 to 9, at most once each, place a digit in each box to create a true statement. ⬚x + ⬚⬚ = ⬚(⬚x + ⬚⬚)

104

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3.02

Expand binomial products

After this lesson, you will be able to… • understand and apply the rule for expanding binomial products. • visualise the expansion of binomial products using area models. • expand perfect squares. • understand and apply the difference of two squares identity. • solve problems requiring the expansion of binomial products.

Expansion of binomial products A binomial is a type of algebraic expression that consists of two terms. A binomial product refers to the result obtained when multiplying two binomials. A C

+ D

Consider the expression (A + B) (C + D). We can represent it visually using this area model. This diagram shows that (A + B) (C + D) = AC + AD + BC + BD. In other words, the area of the whole shape is the sum of the smaller areas.

Let’s look at using this model to expand (x + 3) (x + 5). Here we can see that:

x ´ x = x2

(x + 3) (x + 5) = x2 + 3x + 5x + 15

5 ´ x = 5x

We can simplify this to: x2 + 8x + 15

+ 3

3 ´ x = 3x

5 ´ 3 = 15

Example 1 Expand and simplify: (x + 10) (x + 5)

Apply the idea Use the rule (A + B) (C + D) = AC + AD + BC + BD. (x + 10) (x + 5) = x × x + x × 10 + 5 × x + 10 × 5 Apply the rule = x2 + 10x + 5x + 50 2

= x + 15x + 50

Simplify the products Combine like terms

3.02 Expand binomial products mathspace.co

105

Idea summary We can expand the product of two binomial expressions using the rule: (A + B) (C + D) = AC + AD + BC + BD

Expand perfect squares Exploration Consider the expansions of the binomials of the form (a + b) (a + b) = (a + b)2: • (x + 3) (x + 3) = x² + 6x + 9 • (x + 5) (x + 5) = x2 + 10x + 25 • (x + 6)2 = x² + 12x + 36 • (2r + 3s)2 = 4r2 + 12rs + 9s2 • (5s − 3)2 = 25s² − 30s + 9 1. What do you notice about the linear coefficient of the product? 2. What do you notice about the constant of the product? 3. Is there a general rule for this type of product?

A perfect square is something of the form (A + B)2.

We can use the same area model to determine the expansion. A

+ B

Here we can see that the expansion is: (A + B)2 = A2 + 2AB + B2 In a similar way: (A − B)2 = A2 − 2AB + B2 We can use perfect square expansions whenever we have to expand a perfect square.

106

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 2 Expand and simplify: (x + 2)2

Apply the idea Use the perfect square expansion: (A + B)2 = A2 + 2AB + B2 (x + 2)2 = x2 + 2 × 2x + 22

Use perfect squares

= x + 4x + 4

Evaluate

Idea summary A special case of expanding binomials is for perfect squares: • (A + B)2 = A2 + 2AB + B2 • (A − B)2 = A2 − 2AB + B2

Difference of two squares Exploration Consider the expansions of the binomials of the form (a + b) (a − b): • (x + 3) (x − 3) = x² − 9 • (x − 5) (x + 5) = x5 − 25 • (x + 6) (x − 6) = x² − 36 • (2r + 3s) (2r − 3s) = 4r2 − 9s2 • (5s − 3) (5s + 3) = 25s² − 9 1. What do you notice about the linear coefficient of the product? 2. What do you notice about the constant of the product? 3. Is there a general rule for this type of product?

The difference of two squares identity is very useful in many contexts in mathematics. Consider expanding (A − B) (A + B), using the distributive law: (A − B) (A + B) = A2 + AB − AB − B2 As the middle terms, AB − AB = 0, the right hand simplifies and we have (A − B) (A + B) = A2 − B2 The right hand side, A2 − B2, is a difference of two squares, which is where the identity gets its name.

3.02 Expand binomial products mathspace.co

107

Example 3 Expand and simplify: (m + 3) (m − 3)

Create a strategy Expand using the identity for difference of two squares: (A − B) (A + B) = A2 − B2

Apply the idea We are given: A = m and B = 3. (m + 3) (m − 3) = m2 − 32 2

=m –9

Apply the identity Evaluate the square

Idea summary A special case of expanding binomials is the difference of two squares: (A + B) (A − B) = A2 − B2

3.02 Practice questions Understanding 1

(2x) (3x)

(4y) (5y)

2–4

Trailblazer

(−3z) (7z)

(x + 2) (x + 1) = x2 + ⬚ + 2

(x − 3) (x + 2) = x2 − ⬚x − 6

(x + 2) (x − 5) = x2 − 3x − ⬚

(x + 3)2 = x2 + ⬚ + 9

(5p) (−2p)

(x − 3) (x − 4) = x2 − ⬚x + ⬚ (x − 3)2 = x2 − ⬚x + ⬚

(x + ⬚)2 = x2 + ⬚x + 36

(3x − 8)2 = 9x2 − ⬚ + 64

(x + 5) (x − 5) = x2 − ⬚

(x + 2) (x − 2) = ⬚2 − 4

Fill in the missing boxes. a

108

Adventurer

Fill in the missing boxes. a

2–4

Fill in the missing boxes. a

Explorer

Multiply the terms. a

1–3

(x − ⬚) (x + ⬚) = x2⬚64

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

(x + 14) (x − ⬚) = x2 − ⬚

5a–d, 6a–ce, 8a–d, 9a–d, 11ab, 12a

Fluency Ex 1

(x − 1) (x − 8)

(x + 1) (x + 4)

(x − 3) (x − 6)

(x − 4) (x + 3)

(x − 7) (x − 5)

(x − 5) (x − 1)

(x + 5) (x − 6)

(x − 9) (x + 2)

Expand and simplify: a

(2x + 3) (3x − 4)

(3x − 2) (4x + 5)

(4x + 1) (x − 2)

(3x − 4) (4x + 5)

2(x + 3) (x−2)

3(2x−5) (x + 4)

5(2x + 1) (x − 2)

−3(3x + 2) (2x − 1)

Expand and simplify: b

Expand and simplify: (x + 7)2

(c + 6)2

(7 − z)

Expand and simplify: (3x + 7)2

5(m − 5)2

(e + 5)2

(−6 − a)

(a + b)

(a − b)2

(2y − 5)2

(2y − 3)2

(x2 + 1)2

6(n + 6)2

( y − 5)2

8(q + 3)2

3x(7x − 5y)2

Expand and simplify: a

(x − 7) (x + 7)

(8 − m) (m + 8)

(7 − 6p) (7 + 6p)

(−9y − 1) (−9y + 1)

(9x − 2y) (9x + 2y)

(0.2x + 3) (0.2x − 3)

Expand and simplify: a

3x(5x − 6y) (5x + 6y)

3(9x − 8y) (9x + 8y)

a + (9 − a) (9 + a) − 10

[(5t + 9) − 6r] [(5t + 9) + 6r]

6(8x − 9y) (8x + 9y)

5x(2x − 9y) (2x + 9y)

Reasoning 14

Expand and simplify: a

Ex 3

5f–h, 6df–h, 7, 8c–h, 9d–h, 10, 11c–f, 12, 13

Expand and simplify:

a Ex 2

5d–f, 6d–f, 7, 8c–g, 9b–f, 10, 11b–d, 12ab, 13ab

14, 17

16, 17

15–18

Is (3x + 4)2 equal to 9x2 + 16? Explain your answer. 3.02 Expand binomial products mathspace.co

109

Show and explain how 36 × 24 may be multiplied using the structure of a difference of two squares.

Given two expressions (x + 3) (x + 3) and 13 × 13, perform the multiplication and describe the similarities between them.

A student wrote (a + b)2 = a2 + b2. Explain why they are incorrect and write the correct answer.

Consider the expression a

Explain how to expand and simplify the expression.

Is the result a trinomial?

Problem-solving 19, 20 19

19–22

21–24

The length of a rectangular ﬁeld is (x + 10) metres and the width of the ﬁeld is (x − 2) metres. a

Find the simpliﬁed expression for the perimeter of the ﬁeld.

Find the simpliﬁed expression for the area of the ﬁeld.

If x = 6, ﬁnd the length, width, perimeter, and area of the ﬁeld.

A cube has side length x + 3: a

Find the surface area of the cube in expanded form.

Find the volume of the cube in expanded form.

x+3

A square of side lengths measuring (x − 1) centimetres has each side enlarged by a factor of 2. Write an expression, in expanded form, for the area of the new square.

Form an expression in expanded form for the shaded area in the given ﬁgures. a

5x − 7

x 8x − 5y

8x + 5y 5x + 7

If the smallest of three consecutive integers is n, ﬁnd the product of the three integers in terms of n. Write your answer in expanded form.

110

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A rectangular prism has side lengths y, 12 − y, and y + 8. a

Write an expression for the volume of the prism in expanded form.

Determine which of these are viable options for the length y and explain why or why not in relation to both the side lengths and volume.

y = −10

y = 20

y=0

iii

y=2

y y+8

12 − y

y = 12

By trial and error or by using technology, determine the maximum volume of the prism.

3.03

Factorise binomial products

After this lesson, you will be able to… • understand the process of factorising binomial products. • factorise expressions by identifying and grouping common factors. • understand and apply perfect square factorisation. • understand and apply the factorisation of the difference of two squares. • apply the knowledge of factorising binomial products to solve problems.

Factorise by grouping We’ve seen the distributive law: A(B + C) = AB + AC We have also seen how the distributive law applies when the ﬁrst term is a binomial expression: (A + B) (C + D) = A(C + D) + B(C + D) This means that if we have an expression such as: y(x + 4) + 5(x + 4) we can write it as a product of two brackets: ( y + 5) (x + 4)

Example 1 Fully factorise 5(a + b) + v(a + b)

Create a strategy Identify the common factor between the two terms, then factorise using the rule: A (B + C) + D (B + C) = (A + D) (B + C)

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Apply the idea We can see that (a + b) is a common factor of both terms. 5 (a + b) + v (a + b) = (5 + v) (a + b)

Use the rule for factoring

Example 2 Factorise this expression by grouping in pairs: 2x + xz − 40y − 20yz

Create a strategy To factorise an expression with four terms, group the expression in pairs so that each pair has a common factor.

Apply the idea We can then factorise both pairs, since x is a common factor of (2x + xz) and 20y is a common factor of (40y − 20yz). 2x + xz − 40y − 20yz = x (2 + z) − 20y (2 + z) = (2 + z) (x − 20y)

Factorise by x and 20y Simplify

Idea summary We can use the rule (A + B) (C + D) = A(C + D) + B(C + D) = AC + AD + BC + BD. to factorise expressions in terms of the product of two binomial expressions.

Factorise using perfect squares We’ve seen how to expand perfect squares and use a model to visualise it. (A + B)2 = A2 + 2AB + B2 A

+ B

When factorising an expression of the type A2 + 2AB + B2: • Look for a common factor to both terms first as this can simplify the expression for the remaining steps. • Identify values for A and B that satisfy this form. • Write in factorised form using the rule (A + B)2 = A2 + 2AB + B2.

In our work in quadratics, the first term A, is often the variable x, so for the expression x2 + 4x + 4, we would identify A = x and B = 2, because 22 = 4. The middle term 2AB holds true as 2 × x × 2 = 4x.

112

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 3 Factorise x2 − 12x + 36

Create a strategy Using the perfect squares rule (A + B)2 = A2 + 2AB + B2.

Apply the idea We can use the perfect squares rule because half of −12 squared is (−6)2 = 36. Using the perfect squares rule, A = x and B = −6. x2 − 12x + 36 = (x − 6)2 Use the rule

Idea summary We can factorise the product of two binomial expressions using the rule (A + B)2 = A2 + 2AB + B2 (called a perfect square)

Factorise using the difference of two squares We know that the difference of two squares looks like this: A2 − B2 = (A − B) (A + B) When factorising an expression of the type A2 − B2: • Look for a common factor to both terms first as this can simplify the expression for the remaining steps. • Identify the terms that are the squares A and B. • Write in factorised form using the rule A2 − B2 = (A − B) (A + B).

Example 4 Factorise: 121m2 – 64

Create a strategy Factorise using the difference of two squares, A2 − B2 = (A + B) (A − B)

Apply the idea We can tell that this is a difference of two squares expression because there are 2 terms that are both perfect squares, and the sign between them is a minus (−). Since (11m)2 = 121m2 and 82 = 64, we can use A2 − B2 = (A + B) (A − B), where A = 11m and B = 8.

121m2 − 64 = (11m + 8) (11m − 8)

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Idea summary We can factorise expressions of the form A2 − B2 using the rule: A2 − B2 = (A + B) (A − B) (called the difference of two squares)

3.03 Practice questions Understanding 1

Identify the highest common factor: b

32(c + 5) − x(c + 5)

6d4 + 8d2

⬚ = 42

Fill in the box to make the equation true: 81 =⬚2

100 =⬚2

One linear factor of x2 − 16 is x − 4. What is the other factor?

One linear factor of n2 + 12n + 36 is n + 6. What is the other factor?

Are these expressions a difference of two squares? a2 – 16

6a–d, 7a–d, 8a–d, 9a–d

Fluency 6

4w2 – 25a2

6b–e, 7c–g, 8a–f, 9c–g, 10a–d

25 = ⬚2

100 + 25y2

6e–h, 7e–j, 8f–j, 9f–l, 10

Factorise: a

m(x + y) − 6(x + y)

x( y − z) − w( y − z)

5y(4w + 3x) − z(4w + 3x)

8y( y − 4) + 3(4 − y)

3f(g + h) + (g + h)

(2c − d) (c + 5d) − 3(d − 2c)

3x2(x + 4y) − 5y(x + 4y)

x2 + 2x + 5x + 10

x2 − 3x + 8x − 24

a + 8a + a + 8

8x + xz − 16y − 2yz

24 + 3y + 8x + xy

7xy + wx + 7yz + wz

2mp + 6 + 3p + 4m

4x + 24yz + 32xy + 3z

2x + 18yz + 12xy + 3z

Factorise: a

114

b2 – 9b – 81

1–5

Trailblazer

10xy + 25x2

2a + 18b

Ex 2

1–5

Adventurer

Ex 1

Explorer 1–3, 5

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

2y(2x2 + 3z) − (2x2 + 3z)

3df 2 − 11ydf + 3rzf − 11yrz

Ex 3

Factorise: a

x2 + 6x + 9

49 − 14p + p

e g

6c2 − 24c + 24 16w2 − 40w + 25

i Ex 4

Factorise: a

x2 − y2

e i 10

n2 − 25

f 25m2 − 49

81x2 − 16y2

3x2y − 6y4 + 9x2y3

121c2 − 49d2

8a b + 12bc

g i

2 4

b2 − 2br + r2

v2 − 1

121 − v2

16 − 9y2

x2y2 − 49

7x2 − 63

45t2 − 20

16x2y2 + 24xyz + 9z2

14h3p2r + 21h5p2 + 49hp4

16q4 + 24q2r + 9r2

Reasoning

11, 12

5q2 + 10qt + 5t2 64r2 + 48r + 9

u2 − 2uq + q2

Factorise: a

64 + 16e + e2

3x2y + x2z + 3y2 + yz (b + 4)2 − (3 − b)2

16w2 − 64wt2 + 6wt − 24t3 48df 2 − 3dh2

12–14

Does the sum of two squares, a2 + b2, equal (a + b) (a + b)? Explain your reasoning.

Identify and correct the error in each statement. a

x2 − 4x + 4 = (x + 2) (x + 2)

What strategy or combination of strategies would you use to factorise? Explain your choice.

Factorise the expression fully.

x2 + 2x + 1

2x − 6x + 5x − 15

(x + 16)2 − y2

n3 − 121n

9x2 − 12xy + 4y2

4x3 + 16x2 − x − 4

Explain why the trinomial 4a2−72a + 9 cannot be factorised using perfect squares.

Problem-solving 15ac, 16 15

For each expression:

e 14

x2 + 4x + 3 = (x + 2) (x + 2)

15bd, 16–18

Using the difference of two squares, evaluate: a

652 − 632

992 − 972

15d, 16–20

452 − 422

582 − 522

3.03 Factorise binomial products mathspace.co

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A square has area t2 + 14t + 49 units2:

Find the side length of the square.

Find the perimeter of the square.

Using four of the digits 1 to 9 with no repeats, ﬁll in the blanks to create a perfect square trinomial. x2 + ⬚⬚x + ⬚⬚

Factorise: a

x4 + 8x2 + 16

a4 − 18a2 + 81

Factorise x4 − 1.

The area of a quilt can be expressed as x2 + 6x + 4x + 24. a

Label the area model so that it represents the area of the quilt.

Express the area of the quilt in factorised form.

3.04

Area I

Area II

Area III

Area IV

actorise monic quadratic F trinomials

After this lesson, you will be able to… • find two numbers that satisfy the conditions of the trinomial’s coefficients. • factorise monic quadratic trinomials.

Factorise monic quadratic trinomials A monic quadratic trinomial is an expression with three terms (trinomial), whose highest power is 2 (quadratic) and whose coefficient of the x2 term is a 1 (monic). This means that quadratics of the form ax2 + bx + c, when a = 1 are monic quadratic trinomials. Let’s look at how we can approach trying to factorise an expression in the form x2 + bx + c. If we can ﬁnd numbers p and q so that b = p + q and c = pq then we can write x2 + bx + c = x2 + ( p + q)x + pq 2

116

Since b = p + q and c = pq

= x + px + qx + pq

By expanding the middle term we have four terms

= x(x + p) + q(x + p)

Factorise the ﬁrst two and the last two terms

= (x + q) (x + p)

Finish the factorisation

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

We can see that we have just used AC + AD + BC + BD = (A + B) (C + D) so once we have identiﬁed p and q, we can write (x + p) (x + q) To factorise a monic quadratic trinomial of the form x2 + bx + c, we want two numbers p and q so that when you add them you get the coefficient of the middle term (b) and when you multiply them you get the constant term (c). For example: • To factorise x2 + 8x + 12, we want two numbers that sum to 8 and multiply together to give 12. The two numbers would be 6 and 2. • To factorise x2 + 4x + 3, we want two numbers that sum to 4 and multiply together to give 3. The two numbers would be 3 and 1. • To factorise x2 − x − 6, we want two numbers that sum to −1, and multiply together to give −6. The two numbers would be −3 and 2.

Example 1 Factorise: x2 + 6x + 8

Create a strategy

Apply the idea

Find two numbers p and q that add to 6 and multiply to 8, then factorise using (x + p) (x + q).

Notice that 6 and 8 are positive, then both p and q must be positive. We need factors of 8 that add to 6, which are 2 and 4. x2 + 6x + 8 = (x + 2) (x + 4)

Factorise

Example 2 Factorise: x2 − 17x + 60

Create a strategy

Apply the idea

Find two numbers that add to −17 and multiply to give 60, then factorise.

We need to choose factors of 60 that have sum of −17. The factors are −5 and −12. x2 − 17x + 60 = (x − 5) (x − 12)

Factorise

Example 3 Factorise: 3x2 + 6x − 45

Create a strategy There is a common factor of 3 to factor out. Factorise the remaining quadratic.

3.04 Factorise monic quadratic trinomials mathspace.co

117

Apply the idea 3x2 + 6x − 45 = 3 (x2 + 2x − 15)

Factor out the common factor

If we now consider the trinomial in the bracket, we need to choose factors of −15 that have sum of 2. The factors are −3 and 5. 3x2 + 6x − 45 = 3 (x − 3) (x + 5)

Factorise the quadratic trinomial

Idea summary An expression of the form x2 + bx + c is a monic quadratic trinomial. To factorise expressions like this, we ﬁnd a pair of numbers p and q such that p + q = b and pq = c. Then the factorisation is x2 + bx + c = (x + p) (x + q).

3.04 Practice questions Understanding 1

1–3

Explorer

Sum: 19

Product: 90

x2 + 16x + 64

(m + 4) (m + ⬚) = m2 + 10m + 24 2

( y − 14) ( y + ⬚) = y − 12y − 28

5ab, 6a–f, 7a–f, 8a–f, 9ab

x2 + 3x + 2x + 6

Sum: 3 Product: −88

x2 + x − 30

(m + 8) (⬚ + ⬚) = m2 + 18m + 80

(x − 2) (⬚ − ⬚) = x2 − 6x + 8

5bc, 6c–h, 7c–h, 8c–h, 9a–h

t2 + 7t + 8t + 56

Factorise: a

118

Product: 24

Factorise: c

Sum: −10

Find the values of a and b that will make the equation true: y2 + ay + 35 = ( y + 5) ( y + b).

Ex 1

Product: −63

x2 + 11x + 18

Fluency 5

Complete the missing values in each equation. a

Sum: −2

For each of the quadratic expressions, list all factor pairs of the constant term. a

Trailblazer 1, 3, 4

Find a pair of numbers with the given sum and product. a

1–3

Adventurer

x2 + 11x + 24

t2 + 7t + 10

x + 7x + 12

x2 + 8x + 16

t2 + 8t + 7

y + 9y + 14

x2 + 4x + 5x + 20

y2 + 6y + 12y + 72

x2 + 17x + 72

y2 + 9y + 20

40 + 13x + x

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5cd, 6g–l, 7g–l, 8g–l, 9e–l

d h l

x2 + 16x + 64 x2 + 8x + 15

36 + 12x + x2

Ex 2

Factorise: a

x2 − 4x + 3

x2 − 16x + 60

x2 − 19x + 84

x − 8x + 7

y − 10y + 16

t − 13t + 42

i 8

y2 − 18y + 77

t2 − 20t + 96

x2 − 18x + 45 x2 − 3x − 54

x2 − x − 30

x2 + x − 20

x − 3x − 70

x + 4x − 117

y − 13y − 48

y2 − 12y − 45

x2 + 10x − 56

4x2 + 24x + 32

Factorise: e i

2x2 + 14x + 20 5x2 + 5x − 30 −8 − 6x − x2

Reasoning

f j

−2x2 − 18x + 44

10–12

x2 + 10x − 24

4x2 − 4x − 288

−12 + 7x − x2

x2 − 20x + 51

x2 − 12x + 11

x2 − x − 6 2

x2 − 12x + 35

Factorise:

i Ex 3

−4x2 + 12x + 40 x3 + 7x2 + 12x

10–12, 14

t2 + 8t − 84

x2 + 3x − 28

3x2 − 21x + 30

h l

−3x2 + 12x − 12

2x3 + 16x2 + 30x

12–16

Octavia and Rajeev are factorising −3x2 − 15x − 18. • Rajeev states that the first step in factorising the trinomial is to find two numbers that multiply to −18 (the constant term) and add to −15 (the coefficient of the middle term). • Octavia states that the first step is factorising the highest common factor, which is −3. Who is correct? Explain and factorise −3x2 − 15x − 18.

Omar and Dima both factorise the expression x2 + 24x + 144. Their work is shown. Omar:

x2 + 24x + 144 = x2 + 2(x) (12) + (12)2 = (x + 12)2

Dima:

⬚ + ⬚ = 24

⬚ × ⬚ = 144

x + 24x + 144 = (x + 12) (x + 12) = (x + 12)2

Explain and compare each of their strategies. 12

Identify and correct the error in each statement. a

x2 + 7x − 18 = (x + 2) (x + 9)

x2 − 11x + 30 = (x − 5) (x + 6)

Suppose we want to write the trinomial x2 + bx + c in the form (x + p) (x + q). For each condition, explain what must be true about the signs of p and q. a

c is negative.

c and b are both positive.

c is positive and b is negative.

3.04 Factorise monic quadratic trinomials mathspace.co

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A square table of side length x has one of its dimensions decreased by 4. This can be expressed visually by the area model shown.

x 4

Determine if the following expressions are equivalent to the area x (x − 4). If so, explain why. If not, show what they are equivalent to. a b

x2− Area II

Area I

d 15

Maximises the value of b.

(x + ⬚) (x + ⬚)

Minimises the value of c. (x + ⬚) (x + ⬚)

Minimises the value of b. (x + ⬚) (x − ⬚)

A square ﬁeld with side length x will have its length extended by a and its width extended by b. Both a and b are integers. a

What values of a and b would maximise the length of the entire ﬁeld? Explain.

What values of a and b would minimise the difference in area of the bottom left and top right ﬁeld? Explain.

17–20, 21ac

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

10m

Find an expression for the total area of the rectangles in factorised form.

19, 20, 21bd, 22, 23

One expression for the area of the rectangle is m2 + 14m + 45. The rectangle is made up of four smaller rectangles. Use the diagram to express the area of the large rectangle in factorised form.

120

Area II

x2− Area I

Problem-solving 17–19 17

Area I

Find two digits (1 to 9) to ﬁll in the blanks and create a trinomial in the form x2 + bx + c that

Area I + Area II

a b

Factorise u2 − 11u + 24.

Factorise (x2 + 2x)2−11(x2 + 2x) + 24.

The area of a patchwork artwork can be expressed as x2 + 8x + 15. a

Express the area of the quilt in factored form (x + p) (x + q).

What is the difference in the side lengths?

If both side lengths are tripled, what is the new expanded expression for the area of the artwork?

Factorise: a c

x4 − x2 − 12 (a + 3)2 + 3(a + 3) − 10

b d

y4 − 5y2 + 4

(d − 1)2 − 12(d − 1) + 36

A square with side length y metres has c metres added to its length and d metres added to its width. The resulting rectangle has an area of ( y2 + 8y + 15) m2. Find the values of c and d.

Students decide to make rectangular appreciation cards for their family members. The total area of each card is equal to x2 + 2x − 3 cm2. If the shorter sides of a card are 6 cm in length and the longer sides are 10 cm, ﬁnd the value of x.

3.05

Optional: Factorise non-monic quadratic trinomials

After this lesson, you will be able to… • understand the difference between monic and non-monic quadratic trinomials. • understand and apply the method of factorising non-monic quadratic trinomials by grouping. • apply the splitting of the middle term to factorise non-monic quadratic trinomials. • factorise non-monic quadratic trinomials.

Factorise non-monic quadratic trinomials A non-monic quadratic trinomial is an expression of the form: where a ≠ 1.

ax2 + bx + c

The general approach to factorising a quadratic trinomial of the form ax2 + bx + c: 1. Factor out any HCF. (If a is negative it is advised to factor out -1 at this stage).

3.05 Optional: Factorise non-monic quadratic trinomials mathspace.co

121

2. Find two numbers, p and q that add to give b and multiply to give ac. 3. Rewrite the trinomial with four terms in the form ax2 + px + qx + c. 4. Factorise by grouping. 5. Check that the answer will not factorise further, and verify the factorised form by expanding. We will explore this approach in the following examples.

Example 1 Factorise: 2x2 + 11x + 12

Create a strategy Find numbers that multiply to give 2 × 12 = 24 and add to give 11. Use those numbers to split the middle term and factorise by grouping.

Apply the idea 8 × 3 = 24 and 8 + 3 = 11, so split the middle term into 8x and 3x. 2x2 + 11x + 12 = 2x2 + 8x + 3x + 12

Split the middle term

= 2x(x + 4) + 3(x + 4)

Factorise the ﬁrst two and the last two terms

= (2x + 3) (x + 4)

Finish the factorisation

Reﬂect and check We can check our factorisation by expanding to make sure we get the original expression: (2x + 3) (x + 4) = 2x2 + 8x + 3x + 12 2

= 2x + 11x + 12

Expand the brackets Collect like terms

Example 2 Factorise: 6x2 − 26x + 24

Create a strategy There is a common factor of 2 to factor out. Look for numbers p and q to split the middle term, then factorise by grouping.

Apply the idea 6x2 − 26x + 24 = 2 (3x2 − 13x + 12)

Take out the common factor of 2

The trinomial in the bracket is 3x2 − 13x + 12 so we need to ﬁnd numbers p and q so that pq = 36 and p + q = −13.

122

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

p = −9 and q = −4 will do this. 6x2 − 26x + 24 = 2 (3x2 − 9x − 4x + 12)

Use p = −9 and q = −4 to split the middle term

= 2 (3x(x − 3) − 4(x − 3))

Factorise the ﬁrst two and last two terms

= 2 (3x − 4) (x − 3)

Complete the factorisation by grouping

Reﬂect and check We can check our factorisation by expanding to make sure we get the original expression: 2 (3x − 4) (x − 3) = 2(3x2 − 13x + 12) 2

= 6x − 26x + 24

Complete the binomial expansion Do the remaining expansion

Example 3 Factorise the expression by ﬁrst taking out a common factor: −6x2 − 21x + 45

Create a strategy There is a common factor of −3 to factor out. Look for numbers p and q to split the middle term, then factorise by grouping.

Apply the idea −6x2 − 21x + 45 = −3(2x2 + 7x − 15)

Take out the common factor of −3

The trinomial in the bracket is 2x2 + 7x − 15 so we need to ﬁnd numbers p and q so that pq = −30 and p + q = 7. p = 10 and q = −3 will do this. −6x2 − 21x + 45 = −3(2x2 + 10x − 3x − 15)

Use p = 10 and q = −3 to split the middle term

= −3(2x(x + 5) − 3(x + 5))

Factorise the ﬁrst two and last two terms

= −3(2x − 3) (x + 5)

Complete the factorisation by grouping

Reﬂect and check We can check our factorisation by expanding to make sure we get the original expression: −3 (2x − 3) (x + 5) = −3(2x2 + 7x − 15) 2

= −6x − 21x + 45

Complete the binomial expansion Do the remaining expansion

3.05 Optional: Factorise non-monic quadratic trinomials mathspace.co

123

Idea summary The steps in factorising a quadratic trinomial of the form ax2 + bx + c 1. Factor out any HCF. (If a is negative it is advised to factor out −1 at this stage). 2. Find two numbers, p and q that add to give b and multiply to give ac. 3. Rewrite the trinomial with four terms in the form ax2 + px + qx + c. 4. Factorise by grouping. 5. Check that the answer will not factorise further, and verify the factorised form by expanding.

3.05 Practice questions Understanding 1 2

8x2 + 11x + 3 = (8x + 3) (x + ⬚)

2x2 − 19x + 45 = (⬚ − 9) (⬚ − 5)

Ex 2

4a–d, 5a–d, 6a, 7, 11

Factorise fully:

1–3

Trailblazer

b d

2x2 + 3x − 20 = (2x − 5) (x + ⬚)

10x2 + 29x + 21 = (⬚ + 3) (5x + ⬚)

4c–f, 5b–e, 6a–e, 8, 9, 11

3x2

18x

4e–h, 5e–h, 6d–h, 8–10

3y2 + 28y + 9

2x2 − 11x − 40

3x2 − 25x + 28

12t − 13t − 4

81x + 72x + 16

12x + 7x − 10

6x2 + 41x − 56

4x2 + 40x + 100

8s2 + 6s − 54

9x2 − 42x − 15

Factorise fully: a e

124

Adventurer

Find an expression for the total area of the rectangle, 3x2 + 19x + 6, in factorised form.

Fluency 4

1–3

Fill in the boxes to complete each factorisation. c

Ex 1

Explorer

To factorise −8x2 − 40x − 48, what is a possible ﬁrst step?

1–3

3x2 − 21x − 54 8x2 − 20x + 8

72x2 − 6x − 6

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

105x2 − 50x + 5

8x2 − 21x − 9

168x2 − 88x − 16

Ex 3

Factorise fully: a

−6x2 + 5x + 6

−6x2 + 25x − 14

−12x2 − 12x + 45

−4x + 11x − 6

−9x + 24x − 16

−5x − 14x − 8

−6x2 + 22x + 40 −3x2 − 10x + 8

The length of a rectangle is L = −56y + 11 and the width is W = 5y2. W

What is the perimeter of the rectangle in terms of y in expanded form?

What is the perimeter in fully factorised form?

The side length of a regular pentagon is S = 2x2 + 21x + 49. a

What is the perimeter of the pentagon in terms of x in expanded form?

What is the perimeter in fully factorised form?

A cube has a surface area of (6x2 + 36x + 54) square units, where x > 0. a

Factorise 6x2 + 36x + 54 fully.

What is an expression for the length of a side of the cube?

A rectangle has an area of 6x2 + 23x + 20. If the length and width are linear factors of 6x2 + 23x + 20, what are the dimensions of the rectangle in terms of x?

A ball is thrown from the top of a 140 m tall cliff, with an initial velocity of 50 m/s. The height of the ball after t seconds is approximated by the quadratic expression −10t2 + 50t + 140. Factorise this quadratic.

Reasoning 12

12, 13

13, 14

Consider each quadratic expression. The factorised form of the expression is (Ax + C) (Bx + D). Determine: i

the product of A and B

the signs for A and B: opposite or the same

iii

the product of C and D

the signs for C and D: opposite or the same

15x2 + 14x − 16

−45x2 + 29x − 4

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Sheldon and Gabriella each factorised 16x2 + 48x + 36. Whose work is incorrect? Identify and explain the error. Sheldon:

Gabriella:

16x2 + 48x + 36 = 16x2 + 24x + 24x + 36

16x2 + 48x + 36 = (4x + 6)2

= 8x(2x + 3) + 12(2x + 3)

= 2 (2x + 3)2

= (8x + 12) (2x + 3) = 4(2x + 3) (2x + 3) = 4(2x + 3)2 14

Consider these polynomials: 2x2 + x + 1,

3x2 − 2x − 4,

12x2 − 12x + 1

What do the polynomials have in common?

How can you show that a polynomial where a ≠ 1 cannot be factorised?

Problem-solving 15, 16, 20 15

6x2 + 10x − 5,

15–17, 20

17–21

Tara is practicing diving. She springs up off a board 10 metres high, and after t seconds, her height in metres above the water is described by the quadratic: −5t2 + 5t + 10

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Fully factorise the quadratic.

Substitute t = 2 into the factorised quadratic and ﬁnd the value of the expression.

Substitute t = 2 into the original quadratic and ﬁnd the value of the expression.

What is happening 2 seconds after Tara jumps off the board?

Consider the ﬁgure. a

Write an expression in expanded form for the area of the shaded region.

Write the expression for the area in factorised form.

The area of a rectangular ﬁeld is given by the general expression (6x2 + 17x + 12) m2 where x is a positive whole number. a

Find the length and width of the ﬁeld in terms of x.

Write an expression for the perimeter of the ﬁeld. Describe how to determine the expression for the perimeter.

If the perimeter of a particular ﬁeld is 114 metres, ﬁnd x.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A photographer wants to put a border around her photo.

Why might the photographer use a variable to represent the width of the border?

What are the dimensions, length and width, of the total artwork if the total area of the photograph including the border is 4x2 + 24x + 35 square units?

What is the area of the photograph? Explain how you reached your solution.

Using the digits 1 to 9 with no repeats, ﬁll in the blanks to create a trinomial that can be factorised, and provide the factorised form: 4x2 + ⬚⬚x + ⬚.

Find the value of k that will make 16x2 − 24x + k a perfect square trinomial.

Quadratic trinomials can be factorised using the identity:

where m + n = b and mn = ac. Find the values m and n for the quadratic 4x2 − 14x + 12.

3.06

Complete the square

After this lesson, you will be able to… • understand how to complete the square using visual representation. • understand and apply the process of completing the square in transforming quadratic expressions. • factorise expressions from the completed square form.

Complete the square We have seen the perfect square identity (A + B)2 = A2 + 2AB + B2 and we have seen how we can use this to factorise an expression such as x2 + 4x + 4 as (x + 2)2. All quadratics of the form y = ax2 + bx + c can be transformed into the form y = a(x − h)2 + k through the process of completing the square. This form is very useful for us in graphing, solving and identifying key features of quadratics but for now we will just focus on the algebraic skills associated with completing the square.

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From earlier in this chapter you will remember that Perfect squares : (x + a)2 = x2 + 2ax + a2

(x − a)2 = x2 − 2ax + a2

Difference of two squares : (x − a) (x + b) = x2 + b2 We also used a visual to help us understand the expansion of a perfect square. Let’s use the same ideas and look at how to complete the square, and how it got that name. Let’s start with the expression x2 − 4x + 23. We will start with the ﬁrst two terms and build a perfect square using our visual. x x

We know that x × x is x2, so we can put that in.

−2

−2x

This is now a square, with a side length of (x − 2).

−2 −2x

−2

−2x

−2 −2x

To show the −4x, we will have two rectangles each with an area of −2x.

To complete the square we will use −2 × −2 = 4. If the constant is 23, and we have used 4, there is 19 remaining.

So we can rewrite x2 − 4x + 23 = (x − 2)2 + 19. Algebraically the same example would look like this: x2 − 4x + 23 = x2 − 4x + (−2)2 + 23 − (−2)2 Add the square of half the coeﬃcient of x, but also subtract that same value so the expression’s value is not altered

128

= (x − 2)2 + 23 − (−2)2

= (x − 2)2 + 19

Simplify the constant term

Identify the perfect square we created by adding the square of half the coeﬃcient of x

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 1 Using the method of completing the square, rewrite x2 + 4x in the form a(x − h)2 + k.

Create a strategy Form a perfect square from the terms and see what the required constant is.

Apply the idea Use x and half of the coeﬃcient of x to make a perfect square, (x + 2)2 (x + 2)2 = x2 + 4x + 4 2

Expand the perfect square 2

It follows that x + 4x = (x + 2) − 4

Example 2 Write 2x2 + 4x + 7 in the form a(x − h)2 + k.

Create a strategy Factorise 2 from the ﬁrst two terms. Complete the square inside the bracket, re-expand and simplify.

Apply the idea 2x2 + 4x + 7 = 2(x2 + 2x) + 7 2

Factor 2 out from the ﬁrst two terms

= 2((x + 1) − 1) + 7

Complete the square on x2 + 2x

= 2(x + 1)2 − 2 + 7

Expand the bracket

= 2(x + 1) + 5

Simplify

Idea summary Completing the square is a method that allows us to write a quadratic trinomial ax2 + bx + c in the form a(x + h)2 + k where

and

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Factorise from completed square form We have met the difference of two squares identity: (A − B) (A + B) = A2 − B2 We can use this to factorise an expression such as x2 − 9 x2 − 9 = (x − 3) (x + 3) or an expression such as x2 − 5

If we have a completed square expression such as (x + 2)2 − 25, we can factorise as: (x + 2)2 − 25 = ((x + 2) − 5) ((x + 2) + 5)

Use difference of two squares

= (x + 2 − 5) (x + 2 + 5)

Remove the brackets

= (x − 3) (x + 7)

Simplify

Example 3 Factorise x2 + 4x + 3 using the method of completing the square.

Create a strategy Complete the square. Use the difference of two squares identity (A − B) (A + B) = A2 − B2

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

Complete the square on the ﬁrst two terms

= (x + 2) – 1

Simplify

= (x + 2)2 − 12

Write as difference of 2 squares

= ((x + 2) + 1) ((x + 2) − 1)

Factorise by difference of 2 squares

= (x + 3) (x + 1)

Simplify each bracket

Example 4 Factorise 5(x + 3)2 − 15.

Create a strategy Factor out the common factor of 5 Use the difference of two squares identity (A − B) (A + B) = A2 − B2

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Apply the idea Factor out the common factor Use the difference of two squares identity Remove the unnecessary brackets

Idea summary If completing the square leaves us with an expression of the form (x + h)2 − k we can use the difference of two squares identity A2 − B2 = (A − B) (A + B) to factorise the completed square form to

3.06 Practice questions Understanding 1

1, 2

Adventurer

1, 2

Trailblazer

x2 + 6x + 36

x2 + 8x + 12

x2 − 14x − 72

x2 − 10x + 24

Identify if the correct value for ‘the square of half the coeﬃcient of x’ has been applied in each expression, ensuring that the expression value is not altered. a b c d

x2 + 4x + 18 = x2 + 4x + 22 + 18 − 22 x2 − 10x − 21 = x2 − 10x + 52 − 21

x2 + 30x − 64 = x2 + 30x + 152 − 64 + 152

x2 − 24x + 80 = x2 − 24x + 122 + 80 − 122

Fluency 3

Explorer

For each quadratic expression, calculate the square of half the coefficient of x: a

1, 2

3ac, 4, 5a–d, 6ac, 7a–h, 8a–d

3bd, 4, 5c–f, 6bc, 7c–j, 8b–e, 9a–e

Complete each equation: a c

3x2 + 6x − 8 = ⬚(x2 + 2x) − 8

4x2 + 12x − 6 = ⬚(x2 + ⬚) − 6

b d

3bd, 4, 5e–h, 6cd, 7e–l, 8e–h, 9c–h

2x2 − 10x + 1 = ⬚(x2 − 5x) + 1

6x2 + 24x − 7 = ⬚(x2 + ⬚) − 7

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Complete each solution for completing the square: a

5x2 − 10x + 4 = ⬚(x2 − 2x) + 4

6x2 + 36x − 1 = ⬚(x2 + ⬚) − 1

= ⬚(x − 2x + ⬚) + 4 − ⬚

Ex 1

= ⬚(x − ⬚)2 − 1

x2 + 18x

x2 − 8x

x2 + 10x + 31

x − 10x + 30

x − 18x + 77

x + 9x + 16

x2 − 7x + 15

(x + 4)2 − 9

(x + 5)2 − 49

(x − 1)2 – 3

2x2 + 8x − 7

5x2 + 20x − 9

7x2 − 14x + 1

9x2 + 54x − 2

4x − 11x + 7

3x + 9x + 8

4x2 − 16x + 3

3x2 + 33x + 88

2x − 7x + 6

d h

x2 + 6x + 4

x2 − 11x + 30

x2 + 24x + 63

2(x + 4)2 − 50

Factorise fully: a e

3(x − 1)2 − 12

−4(x + 2)2 + 16

Reasoning

x2 − 8x + 11

9(x + 2)2 − 4

11, 12, 14a

x2 − 6x + 5

10x2 + 20x − 11 5x2 + 5x + 1

4x2 + 20x + 16

x2 + 11x + 10

x2 − 20x + 19

x2 − 28x + 115

4(x + 2)2 − 1

−2(x − 9)2 + 32

4(x + 3)2 − 5

10–12, 14ab

2(x − 1)2 − 3

13, 14

Explain why we cannot factorise x2 + 6x + 13.

Alessandro is completing the square on x2 + 6x + 12. Here is his work:

x2 + 14x + 47

Factorise fully using the method of completing the square: e

(x − 3)2 − 1

Rewrite the following in the form a(x − h)2 + k by completing the square:

Ex 4

Factorise fully:

Ex 3

= ⬚(x + ⬚)2 − ⬚

Rewrite each quadratic in the form a(x − h)2 + k using the method of completing the square:

a Ex 2

= ⬚(x2 + 6x + ⬚) − 1 − ⬚

x2 + 6x + 12 = (x2 + 6x + 9) + 9 + 12

= (x + 3)2 + 9 + 12

= (x + 3)2 + 21

What mistake has Alessandro made? 12

Beatrix is completing the square on 3x2 + 18x − 7. Here is her work:

3x2 + 18x − 7 = 3(x2 + 6x) − 7

= 3[(x + 3)2 − 9] − 7

= 3(x + 3)2 − 9 − 7

= 3(x + 3)2 − 16

What mistake has Beatrix made? 13

An equation is written as y = 3(x − 4)2 + 2. Explain why the minimum value of y is 2. For what value of x does y take the value 2?

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Factorise each quadratic expression in 2 different ways. Show your working. a

x2 − 18x + 81

4c2 − 12c + 9

Problem-solving 15, 17 15

3a2 + 14a + 8

15–17

2x2 + 5x − 3

16–18

A square has an area of x2 − 10x + 25 cm2. a

What is the length of one of the sides?

What are the restrictions on the value of x?

A cube has a surface area of 6x2 − 48x + 96. What is the length of one of the sides?

Complete the square for the equation y = −x2 − 8x + 20.

What is the maximum value of y and at what value of x does this occur?

Using four of the digits 1 to 9 with no repeats, ﬁll in the blanks to complete the expression:

Did you know?

x2 + ⬚x + ⬚ = (x + ⬚)2 − ⬚

“Completing the square” might sound super fancy, but it’s actually an ancient math trick! Way back in 2000 BCE, Babylonian mathematicians were already onto it, using geometry to solve equations by filling in squares to complete rectangles—kind of like finishing a puzzle. Fast forward to the 9th century, and the legendary Al-Khwarizmi took things up a notch, perfecting the method to solve quadratic equations and paving the way for modern algebra. So next time you complete the square, remember: you’re doing maths just like the ancient pros did. It’s a neat little piece of history that’s still making algebra easier, even today! 3.06 Complete the square mathspace.co

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3 Chapter review 1

Find the highest common factor of each pair of terms. a

20x and 15x2 n

16a2 bx3 y6 and 12b4 m2 n6 x

5y2 + 10y + 15

Factorise: a c

k2 − 8k −4bc2 + 8bc + 4c

3p2 q2 + 4p4 q4 + 5p6 q6

A student was asked to factorise the expression 35x2 y + 10xy2 − 5xy. They stated that the highest common factor is 5xy, so the factored form of the expression is 5xy (7x + 2y). Identify and correct the student’s error.

Expand and simplify: a

(x + 1) (x + 3)

(x + 5) (x − 6)

(x − 9) (x + 4)

(4y + 3)2 = 16y2 + 24y + 9

(y − 3)2 = y2 − 9y + 9

(3x − 0.7) (3x + 0.7) d

Expand and simplify: a

Are the expansions correct or not? If not, write the correct expansion. a

(x − 2) (x − 8)

(m + 4) (m − 4)

(8 − 5p) (8 + 5p)

Expand and simplify: a

−2 (3x−4) (4x + 3)

A square of side lengths measuring (x − 4) centimetres has each side enlarged by a factor of 3. Write an expression, in expanded form, for the area of the new square.

Factorise:

134

k3 + 8k2 + k + 8

35m2 − 49

5 (a + b) + x (a + b)

5x2 (x + 4y) − 5y (x + 4y)

Factorise:

5r2 + 10rs + 5s2

x2 − 10x + 25

x2 + 16x + 64

1 − 2x + x2

x + 4x − 117

y − 13y − 48

x + 10x − 24

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x2 + 17x + 72 x2 + x − 20

Factorise fully: a

12t2 − 13t − 4

6x2 + 41x − 56

81x2 − 90x + 25

−6x2 + 22x + 40

A rectangle has an area of 20x2 + 39x + 18. If the length and width are linear factors of 20x2 + 39x + 18, what are the dimensions of the rectangle in terms of x?

The side length of the regular pentagon is S = 2x2 + 15x + 25.

What is the perimeter of the pentagon in terms of x in expanded form?

What is the perimeter in fully factorised form?

Find the value of k that will make 4x2 − 12x + k a perfect square trinomial.

Rewrite each quadratic in the form a (x − h)2 + k using the method of completing the square: a e

x2 − 8x

10x2 + 20x − 11

b f

x2 − 4x + 6

2x2 + 8x − 7

c g

x2 + 10x + 21

4x2 + 20x + 28

d h

5x2 + 5x + 1

x2 − 11x + 30

Factorise fully using the method of completing the square: a

x2 + 6x + 4

Complete the square for the equation y = 20 − 8x − x2.

What is the maximum value of y and at what value of x does this occur?

x2 + 14x + 33

x2 + 42x + 185

x2 + 3x + 3

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Big ideas • Graphs and models represent relationships between variables, with linear functions exhibiting a constant rate of change and non-linear functions demonstrating a varying rate of change. • Graphical representations of inequalities provide a visual tool to depict solution regions in the Cartesian plane. • Equations can be solved more easily by manipulating algebraic expressions and applying a series of organised steps and techniques. • Parallel and perpendicular lines have special rules that make them different from other lines.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10A07 solve problems involving linear equations, including those derived from formulas VC2M10A08 solve linear inequalities and graph their solutions on a number line VC2M10A10 solve problems involving gradients of parallel and perpendicular lines VC2M10A12 solve linear equations involving simple algebraic fractions

Linear equations Chapter outline 4.A Linear graphs 4.B Equations with more than 3 steps 4.01 Equations involving fractions 4.02 Rearrange equations 4.03 Parallel and perpendicular lines 4.04 Linear models 4.05 Graph linear inequalities Chapter 4 review

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138 146 154 163 176 191

If you bought one less ice cream per week, how much would you save in a year? You can use a linear model to measure it! 137 mathspace.co

4.01 Equations involving fractions After this lesson, you will be able to… • apply the three steps in solving equations with fractions. • use the lowest common multiple to clear fractions from equations. • apply inverse operations to isolate the variable and solve equations. • recognise when an algebraic denominator makes a solution invalid. • solve equations that have algebraic expressions in the denominators.

Equations with algebraic fractions To solve equations involving algebraic fractions, we can solve as we did with equations of three or more steps, but at some point we will clear the fractions. We can clear fractions by ﬁnding the lowest common multiple of the denominators and multiplying both sides by it. Using the lowest common multiple will eliminate the fractions and make it easier to solve the equation.

Exploration Let’s look at part of the process for solving this equation. 1

Start with the equation

Multiply by the lowest common multiple

Simplify the fraction

What do you notice about the coeﬃcients outside the brackets in the last step and the denominators in the ﬁrst step?

Will this pattern always hold? For example, does it work for

What will happen to the constant, 1, when solving

? ?

Here are some things to remember: • We often start with the inside brackets and work our way out. • The distributive law can still be used if there are more than two terms in the grouping symbol. • We must multiply the whole side of an equation by the lowest common multiple, not just the fractional terms. It is essential to simplify the equation and check your solution to ensure accuracy in your ﬁnal answer.

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Example 1 Solve:

Create a strategy We can start by expanding the brackets, and then clear the fraction. Once we have done that we can solve by using inverse operations.

Apply the idea Start with the equation

Expand the brackets

Multiply both sides by 8

Combine like terms on the left-hand side

Subtract 15x from both sides

Subtract 21 from both sides

Divide both sides by −14 and simplify

Reﬂect and check An alternative method for solving this problem involves ﬁrst multiplying through by 8, and then expanding the brackets.

Keep in mind that when using this strategy, we multiply each group, not each individual term.

Example 2 Solve:

Create a strategy First, we need to ﬁnd the lowest common multiple to clear the fraction. Then we can solve using inverse operations to isolate the variable.

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Apply the idea Start with the equation

Multiply both sides by the lowest common multiple, 10

Distribute the 10

Simplify the fractions

Expand the brackets

Add 2x to both sides

Subtract 50 from both sides

Divide both sides by 27

Reﬂect and check We can substitute the value of x to check the answer. Start with the equation

Substitute

Multiply both sides by the lowest common multiple, 10

Evaluate the brackets then expand

Multiply both sides by 27

Evaluate both sides

Since both sides are equal, we can conﬁrm that our answer is correct.

Example 3 Solve:

Create a strategy First, we need to ﬁnd the lowest common multiple to clear the fractions. Then we can solve using inverse operations to isolate the variable.

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Apply the idea Start with the equation Multiply both sides by the lowest common multiple, 12 Distribute the 12 Simplify the fractions Expand the brackets Combine like terms Subtract 73 from both sides and add x to both sides Divide both sides by −7 and simplify

Reﬂect and check We can substitute the value of x to check the answer. Start with the equation

Substitute

Write each term as a fraction with a denominator of 12 Multiply both sides by 12 to clear them Evaluate the brackets Expand the brackets and simplify Multiply both sides by 7 Evaluate both sides Since both sides are equal, we can conﬁrm that our answer is correct.

Idea summary Solving equations with algebraic fractions involves simplifying the equation, clearing fractions by multiplying both sides by the lowest common multiple, and solving for the variable using inverse operations.

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Algebraic denominators If we have an algebraic expression in the denominator, then we can multiply both sides by the algebraic expression. When we simplify, we should get an expression which we can solve using previous strategies. Since we can’t divide by 0, the value which makes the denominator zero cannot be a possible solution. For example,

cannot have a solution of x = 2. We sometimes write x ≠ 2.

Example 4 Solve:

Create a strategy To solve this equation, we can multiply both sides by the lowest common multiple, which is 7(2x + 5), to clear the fractions. Then we can solve for the variable using inverse operations.

Apply the idea Start with the equation

Multiply both sides by 7(2x + 5)

Remove the common factor

Expand the brackets and evaluate

Subtract 30 from both sides

Divide both sides by 12 and simplify

Reﬂect and check We can use cross-multiplication as an alternative method for solving problems with fractions.

Start with the equation

Cross multiply to eliminate the fractions

Expand the brackets and evaluate

Subtract 30 from both sides

Divide both sides by 12 and simplify

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Idea summary If we have an algebraic denominator, the lowest common multiple will be an algebraic expression. We can multiply both sides of the equation by the lowest common multiple and then simplify and solve. Since we can’t divide by 0, the value which makes the denominator zero cannot be a possible solution.

4.01 Practice questions Understanding 1

Ex 1

1–4

, in solving for x, would this be a valid next step?

For the equation b

3x + 2 = 15

Is each statement true or false? a

x = 1 is a solution for the equation

is a solution for the equation

is a solution for the equation 4x − 3 = 3

is a solution for the equation

Fill in the blanks: a

Fluency

5a–dg, 6, 7a–d

Trailblazer

1–4

a 3

1–3

Adventurer

What is the lowest common denominator of each set of fractions? a

Explorer

5, 6, 7a–f, 8a–e

5gh, 6e–h, 7, 8

Solve: a

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Ex 2

Ex 3

Solve: a

Ex 4

Solve: a

9, 10, 12

9–12

Reasoning 9

9, 12

Dylan attempted to solve the equation 1

8 × (3x + 6) = 4 − 2 × x

24x + 48 = 4 − 2x

26x = −44

. Here is his work:

Cross multiply Expand the brackets and simplify Add 2x to both sides and subtract 48 from both sides Divide both sides by 26 and simplify

Identify where Dylan has made an error and explain what it is.

Solve the equation correctly, showing the correct steps of work.

Complete the equation so it is true for all values of x.

Give an example of a linear equation with algebraic fractions that has no solutions. Justify your answer.

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Faith looked at the equation

and noticed that the fraction could be simpliﬁed

since it had the same numerator and denominator. She then wrote down 1 = 1 as her next line of working and said that x = 1 was the solution to the equation. a

Does x = 3 satisfy the equation?

Does the equation have any other solutions?

What value is not a possible solution?

Problem-solving 13, 16 13

13–16

14–18

A construction company has spent $15 000 000 to develop new cranes, and wants to limit the cost of development and production of a single crane to $4000. Given that the production cost of each crane is $2000, the cost for development and production of x cranes is given by 2000x + 15 000 000 dollars. a

Write an expression for the cost of one crane.

If the cost of one crane must equal $4000, ﬁnd x, the number of cranes that must be sold.

Fiona, a tennis player, has won 40 out of 55 matches in her career. Find x, the number of matches she must win in a row to raise her win percentage to 80%.

Solve for u:

If x = 2 and

At an annual charity event, the money raised was distributed in several different ways. One-third of the funds came from the silent auction, art sales comprised one-fourth of the total, and direct in-person donations made up one-ﬁfth of the revenue. The food booth raised $250. How much money was raised at the charity event? Round your answer to two decimal places.

, ﬁnd the value of k.

, ﬁnd the value of r.

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4.02 Rearrange equations After this lesson, you will be able to… • apply correct algebraic operations such as addition, subtraction, multiplication, and division to rearrange equations effectively. • solve real-world problems that require making different variables the subject.

Rearrange equations In mathematics and science, there are many formulas that are written in a specific way.

Exploration Let’s look at the formula for speed:

s d t 1.

speed distance time

Use the formula to ﬁnd each of the following: • The speed at which a person travels if they drive 100 km in 75 min • The distance travelled when walking 5 kph for 30 min • The distance travelled by a car driving 85 kph for 5 hr • The time it will take to travel 1 000 km if you walk 80 m/min • How fast a plane is travelling if it can ﬂy 2 789 km in 6 hr • How long it will take to walk to the store 2 km away if you walk 100 m/min

Which ones took the most effort to solve?

If we always wanted to ﬁnd the distance, given time and speed, how could we write the formula to make these calculations more eﬃcient?

By rearranging a formula for a variable of interest, we can make these calculations more efficient. To rearrange an equation and make a speciﬁc variable the subject, we need to perform various algebraic operations such as addition, subtraction, multiplication, and division. The key is to isolate the desired variable on one side of the equation while maintaining the balance of the equation. Here are some general steps to follow: • Identify the variable that you want to make the subject. • Perform algebraic operations on both sides of the equation to isolate the desired variable, remembering to use the correct order of operations. • Simplify the equation as much as possible, and always double-check your steps to ensure accuracy.

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Example 1 Rearrange kx = m + nx to make x the subject.

Create a strategy Notice the like terms. Rearrange so that all the terms in x are on one side of the equation.

Apply the idea Write the equation

Subtract nx from both sides

Take out the common factor of x

Divide both sides by (k − n) to make x the subject

Reﬂect and check When the variable we want to make the subject appears more than once in the equation, we should get them all on the same side and factorise it out.

Example 2 Make M the subject of

Create a strategy We need to have one instance of M in the equation, so we ﬁrst need to clear the fraction and then factorise.

Apply the idea Start with the equation

Multiply both sides by (M + A)

Simplify

Expand the brackets

Subtract M from both sides

Subtract PA from both sides

Factorise each side of the equation

Divide both sides by (P − 1)

Simplify

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Reﬂect and check We can substitute values for A and P, to see if we get the same value solving for M in the original equation and using the rearranged equation. However, this does not guarantee we haven’t made any errors. It is worth checking every step for errors.

Example 3 Make R the subject of V = IR − E.

Create a strategy I is multiplied by R, then E is taken away to give V. Reverse these operations.

Apply the idea Write the equation

Add E to both sides

Divide both sides by I

Write R as the subject

Example 4 Find the gradient and use the gradient and intercept method to sketch the graph of 6x + 2y − 12 = 0.

Create a strategy Make y the subject of 6x + 2y − 12 = 0, which will put the equation in the form y = mx + c. From this form, we can identify the gradient (m) and the y-intercept (c).

Apply the idea Make y the subject of 6x + 2y − 12 = 0. 6x + 2y − 12 = 0 2y − 12 = −6x Subtract 6x from both sides 2y = −6x + 12 Add 12 to both sides y=

−6 x 12 Divide both sides by 2 + 2 2

y = −3x + 6 Simplify Identify the gradient and y-intercept. Gradient is −3 and the y-intercept is 6.

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Sketch the graph. Plot the y-intercept (0, 6) and then apply the gradient of −3. Step across 1 unit and down 3 units. 7 6 5 4 3 2 1

x 1

−1

Draw the line. 7 6 5 4 3 2 1

x 1

−1

Idea summary We can rearrange equations and formulas to make different variables the subject. This can make it easier to do repeated calculations and for graphing equations that are not in the form y = mx + c. Carefully using inverse operations, we can rearrange equations in a similar way to solving equations. It is very important to keep the equation balanced by doing operations on both sides. However, instead of operating on numbers, we are operating on variables.

4.02 Practice questions Understanding

Explorer 1–4

Why do we rearrange equations?

For the formula, Q = m × c × ∆T:

Adventurer 1–4

Trailblazer 1–4

• Q is the amount of heat absorbed in Joules, (J) • m is the mass in grams (g) • c is the heat capacity (J/g °C) • ∆T is the temperature change, (°C)

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149

We are given the amount of heat absorbed, the mass, and temperature change for ten different substances. Which variable should we make the subject to ﬁnd the unknown values more eﬃciently? 3

To make p the subject of R = pq − 4pt2, which option could be a ﬁrst step? A

Divide pq − 4pt2 by p

Multiply both sides of the equation by 4pt2

Factorise p from pq − 4pt2

Add p to one side of the equation

Is each statement true or false? a

When we rearrange a formula, we will get a numerical value.

When rearranging a formula, the formula may involve numerical and algebraic terms.

We use inverse operations to make a variable the subject.

When we rearrange a formula, we will eliminate all of the other variables.

Fluency Ex 1

5a–d, 6a–e, 7a–c, 8ab, 10, 11a–f, 12ab, 14–17

y=x+z

y = 5 (8 + x)

3y = 2x + z

y = 5 (3 − x)

Make m the subject of these equations:

y = 3mx − 5

x = 8k (n + m)

x = −7m (p + q)

y = −5mx − 14

x = −7k (l − m)

8x3 = 4y − 12

9x2 + 3y = 12

Make k the subject of these equations: a

Ex 3

b = −2am + 10

Make y the subject of these equations: a

Ex 2

Make p the subject of the equation M = pq − 7pr2.

Make v2 the subject of the formula for kinetic energy,

Make y the subject in each equation:

150

5–9, 11e–j, 12cd, 13–15, 18, 19

Make x the subject of these equations:

5–7, 8a–c, 9, 10, 11d–h, 12bc, 14–16, 18

y − x + 1 = 0

−4y − 8x + 12 = 0

7y + 14x − 21 = 0

−6y + 24x − 30 = 0

5y − 15x + 25 = 0

3y + 6x − 9 = 0

−2y + 4x + 8 = 0

10y − 20x − 30 = 0

6x + 4y − 18 = 0

4x − 3y − 12 = 0

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Ex 4

Identify the gradient and y-intercept and graph each equation. a

5x + 2y − 10 = 0

−4x + y + 16 = 0

6x + 3y − 12 = 0

−12x + 4y − 24 = 0

The bend allowance for sheet metal is given by the formula

Make T, the thickness of the sheet, the subject of the formula. 14

Make r2 the subject of these formulas: a

The volume of a cone is given by the formula

The area of a circle is given by the formula A = π r2.

The area of a sector of a circle is given by the formula

Make the mentioned variables the subject in the following equations: a

E = mc2, make m the subject 2

A = P (1 + r)t, make P the subject

P = I R, make R the subject

V = π r2 h, make h the subject

, make h the subject

D = b2 − 4ac, make c the subject

, make z the subject

g = hx2 k, make h the subject

, make c the subject

d = uv4 − t, make u the subject

The formula for the force of an object is F = ma, where F is the force, m is the mass of the object, and a is the object’s acceleration. Write an equation that could be used to solve for the mass of the object.

The formula for the perimeter of a rectangle is P = 2l + 2w, where P is the perimeter, l is the length, and w is the width. Write an equation that could be used to ﬁnd the length of the rectangle.

Betsy knows the formula to convert Celsius into Fahrenheit is

. She wants to

convert a temperature from Fahrenheit to Celsius. Find an equation that she could use to do this conversion. 19

The displacement of an object is given by used to ﬁnd v0.

Reasoning 20

20, 22

. Write an equation that could be

20–23

Describe the similarities and differences of solving 3x + y = z and 3x + 8 = 2. Explain your thinking.

4.02 Rearrange equations mathspace.co

151

Two students, Alice and Bob, are given the Pythagora’s theorem equation to rearrange: c2 = a2 + b2 Alice rearranges the equation to make a2 the subject, while Bob rearranges the equation to make b2 the subject. Their results are as follows: Alice: a2 = c2 − b2 Bob: b2 = c2 + a2 Determine which student, if any, is correct in their rearrangement of the equation and explain your reasoning.

Pamela attempted to make r the subject of the equation

and followed these steps:

Multiplied both sides by R + r

Divided both sides by L

Subtract R from both sides of the equation

What was Pamela’s mistake? Explain and correct her error. 23

Ruﬁno has submitted the following work to write the equation for the mass of an object, m, given the kinetic energy, KE, and velocity, v:

Was Ruﬁno correct? If so, explain each step of his working. If not, ﬁx his error(s).

Problem-solving 24, 25 24

152

24, 25, 27

The area, A, of a triangle is given by the formula and b is the length of its base.

24–27

, where h is the height of the triangle

Write an equation that could be used to solve for h.

Find the height of a triangle with a base of 12 cm and an area of 18 cm2.

The volume of a sphere is given by the formula

, where r is the radius of the sphere.

Rearrange the formula to make, r the subject of the formula.

Calculate the radius of a sphere with a volume of 91 cm3 to two decimal places.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

When a certain amount of money, P, is invested, its future value, A, is given by the formula , where r is the rate per time period and n is the number of time periods. a

Make P, the principal, the subject of the equation.

How much must be invested now, if it is to grow to $3000 in 10 years with interest of 4%

compounded annually? Round to the nearest whole dollar. 27

The volume of the cylinder that ﬁts exactly into a sphere is given by the formula:

Rearrange the formula so that r is the subject.

What is the radius of the sphere if the volume of the cylinder is 13.75π mm3 and the height of the cylinder is 5 mm? Round your answer to the nearest millimetre.

Did you know?

Some of history’s biggest scientific breakthroughs happened just by rearranging formulas! Take Johannes Kepler, for example—by reshuffling astronomical equations, he uncovered the laws of planetary motion, revealing how planets orbit the sun in ellipses. This simple twist on formulas changed our entire view of the universe and set the stage for modern astronomy. Kepler’s discovery shows that sometimes, all it takes is a new perspective on old formulas to unlock the secrets of the cosmos!

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4.03 Parallel and perpendicular lines After this lesson, you will be able to… • understand the characteristics of parallel and perpendicular lines. • apply the understanding of parallel and perpendicular lines to find equations of such lines. • interpret and solve problems involving parallel and perpendicular lines.

Parallel lines Lines that never intersect are called parallel lines.

Interactive exploration

Discover this concept in action online

Parallel lines are lines that have the same gradient.

When two lines are parallel, they never cross and have no points in common.

3 2 1 −4 −3 −2 −1

−1

mathspace.co

x 1

−2 −3 −4

Let’s look at how we can identify parallel lines given their equations. Equation form

Characteristic of parallel lines

y = mx + c

Parallel lines have the same m value.

For every straight line, there exists an inﬁnite number of lines parallel to it.

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Examples y = 2x – 1 y = 4 + 2x

Here is the line y = x.

y=x+3

Two more lines in the same family of parallel lines are also graphed:

3 y=x

y=x+3

1 −4 −3 −2 −1

−1

x 1

−2

y=x-4 Notice that they have the same gradient (m-value) but different x and y-intercepts (c-values).

−3 −4 y = x − 4

The gradient of any vertical line is undefined. But by definition, any two vertical lines are parallel.

Example 1 Consider these questions about parallel lines: a Are the lines y = 3x + 2 and y = 3x – 5 parallel?

Create a strategy Lines are parallel when the gradients are the same. Identify the gradients for each equation. The equations are in the form y = mx + c. In this form, m represents the gradient.

Apply the idea y = 3x + 2, will have a gradient of 3. y = 3x − 5, will have a gradient of 3. So these lines are parallel.

b Which of these three lines are parallel? y = –3x + 2 −3y – x = 5 y + 3x = 7

Create a strategy Lines are parallel when the gradients are the same. The equations will need to be rearranged into the format of y = mx + c to compare the gradients.

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Apply the idea y = −3x + 2 is in the form y = mx + c. It has a gradient of −3. −3y – x = 5 This needs to be rearranged. the equation −3 Write y− x =5

−3 y = x + 5

−3 y x + 5 = −3 −3 1 5 y = − x − 3 3

Add x to both sides Divide both sides by −3 Write in gradient-intercept form

y + 3x = 7 This needs to be rearranged. y + 3x = 7 y = −3x + 7

Write the equation Evaluate

The lines that are parallel are y = −3x + 2 and y + 3x = 7 as both have a gradient of −3.

Example 2 Find the equation of the line that is parallel to

and passes through the point (9, −1).

Create a strategy First, ﬁnd the gradient of the given line. Next, substitute the gradient and the given coordinates into the equation y = mx + c, and solve for c.

Apply the idea The gradient of

. Since the new line is parallel to the given line, the gradient of

the new line is also . Equation of a line in gradient-intercept form −1 =

1 ×9+c 3

Substitute Substitute x = 9 and y = −1 Evaluate the multiplication

Subtract 3 from both sides

Substituting the values of m and c into the gradient-intercept form of a line, the equation is:

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Reﬂect and check y −4 −3 −2 −1

−1

We can graph the given line with the line we found to see that they are parallel.

x 1

−2 −3 −4 −5 −6 −7 −8

Idea summary If two lines are parallel, they have the same gradient. If two lines have the same gradient, they are parallel. All vertical lines are parallel to each other.

Perpendicular lines Lines that meet at right angles (90°) are called perpendicular lines.

Interactive exploration

Discover this concept in action online

mathspace.co

Perpendicular lines have gradients whose product is equal to −1. For two perpendicular lines, y = m1 x + c1 and y = m2 x + c2: m1 × m2 = −1 We say that m1 is the negative reciprocal of m2. This means the two gradients have opposite signs and are reciprocals of each other. Here are some examples of negative reciprocals:

If the gradient of the ﬁrst line is m1, then we can ﬁnd the gradient m2 of a perpendicular line using the formula:

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157

Special case: a vertical line and a horizontal line are perpendicular but do not have gradients which have a product of −1. This is because the gradient of a vertical line is undeﬁned.

Example 3 Find the equation of a line that is perpendicular to y = 6x + 10, and has the same y-intercept. Give your answer in the form y = mx + c.

Create a strategy Find the gradient of the perpendicular line and the y-intercept of the given line, then substitute both values into the equation y = mx + c.

Apply the idea The gradient of y = 6x + 10 is 6, and the negative reciprocal of 6 is

, so the gradient of the

. The y-intercept is c = 10. So the equation of the line is:

perpendicular line is

Reﬂect and check y 15 12 9

We can check that the product of the gradients is equal to −1, and we can graph both lines to see they are perpendicular.

y = 6x + 10

y=−

1 x + 10 6

6 3 x −6

−3

Idea summary Two lines are perpendicular if their gradients are negative reciprocals of each other. To test if lines are perpendicular, multiply the gradients together. If the result is −1, then the lines are perpendicular. Vertical and horizontal lines are perpendicular to each other, even though their gradients do not have a product of −1.

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4.03 Practice questions Understanding 1

Explorer 1–4

Adventurer

Trailblazer 2–6

1–5

The equations y = 2x, y = 2x + 5 and y = 2x − 7 have been graphed on the same Cartesian plane: a

What do all of the equations have in common?

What do all the lines have in common?

10 8 6 4 y = 2x + 5 2 −4

−2

y y = 2x

x 2

−2 −4 −6

y = 2x − 7

If two non-vertical lines are perpendicular, what is the product of their gradients?

For the given graph: a

Which two lines are parallel?

Find two pairs of perpendicular lines.

y = 3x + 15

10 8 6 4 2

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

Complete the statement:

The x-axis and the y-axis are a pair of ⬚ lines.

y y = 1.9x

x 2 4 6 8 10

1 y=− x 3 1 y = − x−9 3

Which of these lines is parallel to the y-axis? A

y = −1

y = 2x − 1

y=x

x=5

If the equation of a line is y = mx + c, and m > 0, which of the statements is true? A

Any line parallel to this line will have a negative gradient.

Any line perpendicular to this line will have a positive gradient.

Any line perpendicular to this line will have a gradient of zero.

Any line perpendicular to this line will have a negative gradient.

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7–9, 10ab, 11, 12, 14a–d, 15, 16, 18a

Fluency Ex 1

y = 7x − 3

y = 6x + 3

y = 7x + 4

y = 7x

y = 6x + 4

y = −7x + 3

y = 3x + 7

y = −3x − 7

Are these pairs of lines parallel? a

y = −3x − 2 and y = −3x + 9

y = −2x − 5 and y = −2x − 8

y = 7x + 8 and y = −5x + 8

y = 4x − 1 and y = 4x − 6

x = 4 and y = 5

y = 7x − 5 and y = −7x + 6

y = 3x − 5 and

y = −8x − 2 and 9x + 7

The line L passes through the point (9, −5) and is parallel to the line y = −5x + 2. Find the gradient of line L.

Parallel to the line y = −2x + 9 and passes through the point (−3, 1).

Parallel to the line y = 8x − 3 and crosses the y-axis at 5.

Parallel to the line y = −3x − 8 and crosses the y-axis at −4.

Parallel to the x-axis and passes through (−10, 2).

Parallel to the y-axis and passes through (−7, 2).

Find the gradient of the line perpendicular to: A line with gradient 6

y = −3x

y=8−x

3y − x = 5

x + 3y = −10

y = 9x

y = −9x + 5

9y = −x + 2

Are these pairs of lines perpendicular? a

y = x + 1 and y = x − 1

y = 3x − 7 and y = −3x + 6

y = −3x + 6 and

and

and y = −(3x − 27)

y = 3x − 5 and

and

The line L is perpendicular to y = 5x − 4 and crosses the y-axis at 1. a

160

A line with gradient

Are these lines perpendicular to y = 9x + 2? a

Are these lines perpendicular to y = 3x + 2? a

Find the equation of line L.

Find the equation of the line described:

8–11, 13–18

Which of these lines are parallel to y = 7x + 3?

a Ex 2

8–12, 14a–d, 15–18

Find the gradient of line L.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Find the equation of line L.

Ex 3

Given the lines L1: 8x = −7y + 3 and L2: 7y − 8x − 7 = 0 a

Find the gradient of line L1.

Find the gradient of line L2.

Are the two lines perpendicular?

Given the lines L1 : 5x + 8y − 10 = 0 and L2 : a

Find the gradient of line L1.

Are the two lines parallel?

Find the gradient of line L2.

Find the equation of the line described: a

Perpendicular to

Perpendicular to y = −6x + 10 and has the same y-intercept.

Perpendicular to

Perpendicular to the x-axis and passes through (−8, −1).

Perpendicular to the y-axis and passes through (−8, −8).

Reasoning 19

and passes through (−8, 5).

and passes through the point (0, 6).

19, 20

19, 21

19–21

A farmer wants to create an enclosure for his sheep. He puts fencing along these lines:

What is the shape of the enclosure? Justify your answer. 20

If every point on the line y = 2x + 2 is shifted 2 units up, ﬁnd the equation of the new line.

Every point on a particular line is shifted 6 units down. The resulting line has equation y = 2x − 2. Find the equation of the original line.

Problem-solving 22a, 23–25 22

22, 23, 25, 28, 30

22, 25–33

Each pair of lines are parallel. Find the value of b. a

0 = 5x − 9y + 6

y = bx + 4

0 = bx + 3y − 2 23

The line L passes through (−9, 10) and (2, −1). a

Find the gradient of line L.

Find the equation of the line that passes through (10, 8) and is parallel to line L.

The line L passes through (1, 1) and (3, 13). a

Find the gradient of line L.

Find the equation of the line that has a y-intercept of (0, −4) and is perpendicular to line L.

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161

A construction company is building a new road that runs parallel to an existing road. If the equation of the existing road is y = 0.5x + 3, ﬁnd the equation of the new road that passes through the point (2, 6).

The line L passes through (3, 2) and (−2, 4). Find the equation of the line that has a y-intercept of 1 and is parallel to line L.

Each pair of lines are perpendicular. Find the value of k. a

5x + 4y + 8 = 0

kx + 9y − 8 = 0 28

−6x + 6y − 8 = 0 kx + 10y + 8 = 0

The line L is perpendicular to y = 10x + 8 and passes through the point of intersection of the lines y = x + 5 and 8x − 10y + 60 = 0. a

Find the gradient of line L.

Find the point of intersection of y = x + 5 and 8x − 10y + 60 = 0.

Find the equation of the perpendicular line L.

The line y = 2x + 4 is perpendicular to the line cx − 4y + 7 = 0. Find the value of c.

The 4 vertices of square ABCD have been plotted on a Cartesian plane. a

Find the gradient of side AB.

Find the gradient of side BC.

Find the product of the gradients in parts (a) and (b).

Are sides AB and BC perpendicular?

y 7 6 A

5 4 D

2 C

1 1

The points P(0, −2), Q(−3, −2), R(0, 4) and S(2, 2) are graphed on the Cartesian plane:

Find the gradient of PQ.

Find the gradient of RS.

Are PQ and RS parallel?

Are QR and PS parallel?

What type of quadrilateral is PQRS?

−4 −3 −2 −1

−1

−2 P −3 −4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

y 4 R

162

S x 1

The vertices of a quadrilateral are A(2, 2), B(6, 4), C(8, 10), and D(4, 8). a

Find the gradient of each side.

Is quadrilateral ABCD a parallelogram?

Is quadrilateral ABCD a rectangle?

A quadrilateral ABCD has been graphed on the Cartesian plane. a

Find the gradient of each side: i

iii

What type of quadrilateral is ABCD? Explain your answer.

4.04

y x

−1 1 2 3 4 5 6 7 8 9 −1 D −2 −3 −4 A −5 C −6 −7 −8 B −9

Linear models

After this lesson, you will be able to… • understand the roles of independent and dependent variables in a linear model. • interpret linear models in real-world contexts, including understanding the significance of the gradient and vertical intercept. • explore the concept of negative gradients in the context of decreasing quantities. • understand the characteristics of linear graphs in real-world scenarios. • interpret linear relationships from a table of values. • solve real-world problems involving linear models.

Linear models In linear relationships we have two variables, independent and dependent variables which are x and y respectively. These will often represent physical quantities such as time, distance, cost, mass or temperature. Instead of being labelled x and y, they may be labelled with letters or names that better represent those quantities. For example, if we were modelling the rate of fuel being used in a car, the independent variable may represent the distance travelled, in kilometres, and the dependent variable may represent the volume of fuel, in litres, in the car’s fuel tank. Instead of x and y, we might use d and V as our variables. Each of these variables has units of measurement associated with them. 4.04 Linear models mathspace.co

163

Fuel consumption of a mid-sized Sports Utility Vehicle V (L) 70 60 50 40 30 20 10

d (km) 100 200 300 400 500 600 700 800 900

As we know already, the two key features of a linear equation, or straight-line graph, are the gradient and the y-intercept. In linear modelling situations, the y-intercept is often referred to as the vertical intercept, because the vertical axis may be labelled with a variable other than y. The vertical intercept represents an initial value. In our example above, the vertical intercept is 63 litres. It represents the volume of fuel in a full tank, before the car began its journey. In a real-world context the gradient represents a rate of change. Using our example above, the gradient would represent the volume of fuel used per distance travelled. In other words, the gradient is a measure of the car’s fuel consumption. The graph above is decreasing as the fuel is being used, so it has a negative gradient. If we divide the ‘rise’ of −63 by the ‘run’ of 900 we get a gradient of −0.07. This means fuel is being consumed at a rate of 0.07 litres per kilometre (or 7 litres per 100 kilometres). Because most physical quantities like distance, volume or time do not contain negative values, the graphs of most linear models tend to exist only in the ﬁrst quadrant of the Cartesian plane (like the example above). This is not always the case though - for example, temperature is a physical quantity that can have negative values.

Idea summary In linear modelling situations: • • • •

The vertical intercept represents the initial value. The gradient represents the rate of change. The independent variable goes on the horizontal axis. The dependent variable goes on the vertical axis.

Linear relationship from a table of values Real world situations often involve the collection of data, which is commonly displayed in a table of values. The independent variable is always displayed in the ﬁrst row of the table, followed by the dependent variable, in the second row.

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We may ﬁrst want to decide whether the values in the table represent a linear relationship. To do this, we could plot the values on a Cartesian plane. If we can draw a straight line through all of the points, then we have a linear relationship. Once we know that the relationship between the variables is linear, we can work out the gradient and vertical intercept, and express the relationship as a linear equation.

Example 1 Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol ( y)

0 0

10 16.40

20 32.80

30 49.20

40 65.60

a Graph the linear relationship represented in the table.

Create a strategy Write the values in the table as points, plot the points on a Cartesian plane, and draw the line that passess through the points.

Apply the idea y 70 60 50 40 30 20 10

x 10

20 30 40 50 60

b Write an equation linking the number of litres of petrol pumped (x) and the cost of the petrol ( y).

Create a strategy Find the gradient of the line and the y-value of the y-intercept, then write the equation of the line in the form y = mx + b.

4.04 Linear models mathspace.co

165

Apply the idea Find the gradient of the line: Use the coordinates (0, 0) as (x1, y1) and (10, 16.40) as (x2, y2). We can find the gradient of the line: Write the gradient formula

Substitute the values

Divide the change in y by the change in x

Evaluate Since the y-intercept is at (0, 0), b = 0. The equation of the line is y = 1.64x

c How much does petrol cost per litre?

Create a strategy The gradient of the line represents the unit rate of the cost of petrol per litre.

Apply the idea The gradient of the line is

and represents the increase in the price of petrol per 1 litre.

In other words, the cost of petrol per litre is $1.64

Reﬂect and check We can also ﬁnd the cost of petrol per litre by dividing the cost of the petrol ( y) by the number of litres (x). Divide 16.40 by 10

Evaluate

d How much would 47 litres of petrol cost at this unit price?

Create a strategy Using the equation from part (b), substitute x = 47.

Apply the idea y = 1.64x

166

Write the equation

= 1.64 × 47

Substitute x = 47

= $77.08

Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary The independent variable is often displayed in the first row of the table, followed by the dependent variable. From a table of values we can plot the graph and determine the equation for the relationship. We can use the relationship to determine the value of one of the variables given the value of the other.

4.04 Practice questions Understanding 1

1–4

y = 2x + 3

y = x2 + 2x + 1

y = 3x3 + 2x + 1

f (x) = 2x+1

The rate of change is variable.

The rate of change is constant.

The rate of change is exponential.

The rate of change is never zero.

Complete the sentences with ‘horizontal’ or ‘vertical’: b c d

The independent variable is plotted on the ⬚ axis. The dependent variable is plotted on the ⬚ axis. The line y = 5 is a ⬚ line.

The line x = −2 is a ⬚ line.

The initial value is the starting value or the y−intercept. Find the initial value for these linear models: a

C = 250 + 7.55x

Fluency 5

Trailblazer

Which of the statements is a characteristic of a linear relationship?

1–4

Adventurer

Which of the equations represents a linear model? A

Explorer

T = 12.5 + 1.5m

5–11, 16, 20

What is the gradient of the line?

What does the gradient represent? A

The initial value

The speed

Cost of petrol

Time travelled

5, 7, 12–15, 17–19

The horizontal axis represents the time in hours and the vertical axis represents the distance travelled in kilometres. a

4x + 5y = 40

F = 0.6k − 5

5, 7, 12–15, 17–19

Distance (km) 575 460 345 230 115 t (hrs) 1

4.04 Linear models mathspace.co

167

The following graphs represent the cost and revenue in dollars for x items produced and sold, respectively.

$ 600 500

C = 15x is the cost equation, indicating each item costs $15 to make.

400

R = 0.5x2 + 50 is the revenue equation.

300

Identify each equation on the graph.

Explain the meaning of the intersection points of the graphs.

200 100 Items (x) 10

Ross deposits $100 into a savings account each week.

Tammy’s parents record her height every year for the ﬁrst 12 years of her life.

Julia collected lollies from a pinata at a party and eats two lollies per day.

Janet starts running and doubles the distance she runs each day.

A dam used to supply water to the neighbouring town had this data recorded for its volume over a number of months:

Plot the points on the Cartesian plane.

Is this relationship linear?

1 112

2 106

3 110

4 80

After Sally starts running, her heartbeat increases at a constant rate as shown in the table: 0 75

2 81

4 87

Complete the table.

Graph the linear relationship represented in the table.

By how much is her heartbeat increasing each minute?

6 93

8 99

10 105

A diver starts at the surface of the water and begins to descend below the surface at a constant rate given by y = 0.8x a

Complete the table to show the depth of the diver over 4 minutes: Number of minutes passed (x) Depth of diver in metres ( y)

168

Number of minutes passed (x) Heart rate ( y)

Identify whether the situation is represented by a linear or non-linear equation:

Month (M ) Volume in billions of litres (V )

0 0

Graph the linear relationship represented in the table.

Calculate the increase in depth each minute.

Calculate the depth of the diver after 24 minutes.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

11 Ex 1

Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol (C)

20 24

30 36

40 48

Graph the linear relationship represented in the table.

Write an equation linking the number of litres of petrol pumped, x, and the cost of the petrol C.

Find the cost of petrol per litre.

Calculate the cost of 75 L petrol.

The time in minutes and the temperature in degrees Celsius (°C) are given in the table: 1 8

2 13

3 18

4 23

5 28

By how much is the temperature increasing each minute?

Find the initial temperature.

Form an equation relating x and y.

Graph the linear relationship represented in the table.

The table shows the temperature of a metal plate, in °C, after an amount of time in minutes: Time (x) Temperature (T)

10 12

Time (x) Temperature ( y °C)

0 0

1 10

2 15

3 20

4 25

5 30

Graph the linear relationship represented in the table.

By how much is the temperature increasing each minute?

Find the initial temperature.

Form an equation relating x and T.

Chirping crickets can be used as an indication of how hot or cool it is outside. For a particular species this data was recorded: Number of chirps per minute (x) Temperature (T °C)

80 16

110 19

140 22

160 24

Graph the linear relationship represented in the table.

Determine the temperature when the crickets make 130 chirps each minute.

How many chirps per minute will the crickets make if the temperature is 27 °C?

How many chirps per minute will the crickets make if the temperature is 20 °C?

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Emma is gathering data on the growth of her plant as shown in the table: Day (d) Height of plant (h cm)

8 18

Graph the linear relationship represented in the table. How tall was the plant when Emma started tracking the growth?

Determine the height of the plant on day 12.

On which day will the plant be 24 cm tall?

Write a linear equation to represent the height, h, as an equation of the day, d.

When a plane is coming in to land, the pilot requires the plane to lose altitude at a rate of 5 m/s. At the beginning of a particular descent the plane starts at an altitude of 8500 m. The altitude of the plane at various times after starting the descent are shown in the table: 0 8500

200 7500

400 6500

1600 500

Graph the linear relationship represented in the table.

Determine the altitude of the plane 500 seconds after starting the descent.

How many seconds did the plane take to descend to the ground?

Write a linear equation to represent the altitude of the plane, A, as an equation of the number of seconds after commencing descent, t.

Katrina is going to catch a taxi into the city. She knows that the taxi has a ﬂat fee of $0.30 and that the fare rises by $0.10 per kilometre travelled. The fares for particular distances are shown in the table: 6 0.90

7 1.00

8 1.10

9 1.20

Graph the linear relationship represented in the table.

Write a linear equation to represent the fare, F, as an equation of the journey’s length, d.

Calculate the fare for a 14 km journey.

Find the distance travelled for a trip that costs $0.60

Justin is looking into the details of his mobile phone plan. He knows the costs for several call lengths as shown in the table: Length of call in minutes (t) Cost (C)

170

6 14

Distance travelled, d (km) Fare, F ($)

4 10

Descent (t seconds) Altitude (A)

2 6

2 $1.00

6 $2.20

10 $3.40

14 $4.60

Graph the data in the table.

How much will it cost to make an 8 minute call?

Find the length of a call that costs $1.60

Determine the rule that connects the cost of a call, C, to the length of the call, t.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Beth’s income is based solely on the number of hours she works, and she is paid a ﬁxed hourly wage. She earns $750 for working 30 hours. a

Identify the two variables in the question and assign a variable name to each.

Sketch a graph that displays her income against her hours worked.

Form an equation relating the variables.

Calculate Beth’s income when she works 25 hours.

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

The cost for a business to operate can be expressed in terms of the total number of hours it has operated for. The cost is $150 an hour. a

Identify the two variables in the question and assign a variable name to each.

Form an equation relating the two variables.

Find the total cost for the business to operate for 22 hours.

Find the number of hours that the business needs to operate to incur a total cost of $5400.

Reasoning 21

22, 24

21, 23, 25

21, 25, 26

If a line were plotted on each of these graphs, what would the gradient represent? a

Petrol used (L)

Amount earned ($)

Hours worked (h)

Distance (km)

Heartbeat (b)

Minutes running (m)

Altitude in metres (a)

Number of seconds flying (f)

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The graph shows the conversion between Country A and Country B’s currency: a b c

Use the graph to convert 8 of currency A to currency B.

14 12

Use the graph to convert 2 of currency B to currency A.

Discuss the usefulness and limitations of this model.

8 4 2

B 1

A baseball is thrown vertically upward, and the velocity of the baseball, V in metres per second after T seconds is given by V = 120 − 32T. a

Complete the table of values: Time (T ) Velocity (V )

What does the negative value of V, when T = 4, mean in context?

Discuss the usefulness and limitations of this model.

A racing car starts the race with 250 litres of fuel. From there, it uses fuel at a rate of 5 litres per minute. Complete the table of values: Number of minutes passed, x Amount of fuel left in tank, y

Determine an algebraic rule linking 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.

Discuss the usefulness and limitations of this model.

A ball is rolled down a hill. The table shows the velocity, in m/s of the ball after a given number of seconds: Time in seconds, t Velocity, V

172

2 3 4 5 6 7 8 9 10

0 18

1 19.7

2 21.4

3 23.1

4 24.8

5 26.5

Find and interpret the gradient of the line.

Find and interpret the vertical intercept of the line.

Determine the rule that connects the velocity, V, to the time in seconds, t.

Find the velocity of the ball after 17 seconds, rounded to one decimal place.

Discuss the usefulness and limitations of this model.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

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

Find and interpret the gradient of the line.

Find and interpret the y-intercept.

Write an equation to represent the temperature of the room, y, in terms of time, t.

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

Discuss the usefulness and limitations of this model.

d e

Temp (°C)

5 3 1

t (mins) 1

Problem-solving 27, 28, 32, 33 27

29–31, 33

The graph shows the conversion between miles and kilometres:

29–31, 34, 35

Convert 10 miles to kilometres.

How many kilometres is 1 mile equal to?

14 12 10 8 6 4 2

mi 1

2 3 4 5 6 7

8 9 10

The graph shows the conversion between temperatures in Celsius and Fahrenheit: 50

°C

40 30 20 10 −20

−10

°F 10

100

−20 −30 −40 −50

Use the graph to convert as indicated: a

40 °C to Fahrenheit

50 °F to Celsius

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Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol ( y)

0 0

10 16.40

20 32.80

30 49.20

40 65.60

Calculate the cost of 47 L petrol. 30

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

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?

80 70 60

Would 80 °F be above or below normal body temperature? Normal body temperature is approximately 37 °C.

50 40 30

Write the rule for conversion between Celsius (C) and Fahrenheit (F).

Convert 35 °C into Fahrenheit.

20 10

°C 10 15 20 25 30 35 40

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 )

6000

5700

5400

5100

4800

4500

Find and interpret the gradient of the line.

What is the value of F when the line crosses the vertical axis?

Write an algebraic equation for the line relating t and F.

How many ﬁsh remain in the river after 9 years?

Determine the number of years it takes for there to be 1800 ﬁsh remaining in the river.

The table shows Peter’s earnings from ironing shirts: Shirts ironed Earnings ($)

174

°F

100

0 4

2 8

4 12

6 16

8 20

Find and interpret the gradient of the line.

Find and interpret the y-intercept of the line.

What will be his total earnings if he irons 14 shirts?

How many shirts will he have to iron in order to earn $28?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The graph shows the conversion between miles and kilometres: a

If a car is driving 32 km/hr in a school zone, at what speed is it travelling in miles per hour?

A road sign says the speed limit is 128 km/hr. Calculate this speed limit in miles per hour.

22 20 18 16 14 12 10 8 6 4 2

km 4

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

Find and interpret the gradient of the line.

Find and interpret the y-intercept.

Write an equation to represent the amount of water remaining in the bucket, y, in terms of time, x.

Find the amount of water remaining in the bucket after 54 minutes.

32 Water remaining (L) 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 5

8 12 16 20 24 28 32 36

Time (mins)

20 25 30 35

The table shows the linear relationship between the number of plastic chairs manufactured and the total manufacturing cost: Number of plastic chairs Cost (dollars)

5 135

10 185

15 235

Find and interpret the gradient of the line.

Find and interpret the y-intercept.

Form an equation relating the number of plastic chairs and the cost.

Find the total cost of manufacturing 25 plastic chairs.

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4.05

Graph linear inequalities

After this lesson, you will be able to… • graph linear inequalities on the Cartesian plane. • apply test points to verify if they satisfy the given linear inequality, and shade appropriate regions.

Graph linear inequalities We can represent a linear relationship using a line on the Cartesian plane, for example, y = −2x − 5.

2 1 −5 −4 −3 −2 −1

−1

x 1

−2 −3 −4

The line indicates the set of all points on the Cartesian plane that make the equation true.

−5 −6

Interactive exploration

Discover this concept in action online

mathspace.co

There are two key considerations when plotting inequalities. 1. Is the region shaded above or below the line? 2. Is the line included (solid) or excluded (dotted)?

The inequality y > 2x + 3 is satisfied for y-values greater than the expression 2x + 3. This corresponds to y-values ABOVE the line, which is the shaded region.

5 4 3 2 1 −4 − 4 − −3 3 − −2 2 −1

−1

x 1

The line corresponds to the equation y = 2x + 3. Since this is not included in the strict inequality y > 2x + 3, the points on the line are not included in the solution. We indicate this by using a dotted or dashed line.

−2

176

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Compare this to the inequality y ≤ 2x + 3. This represents all of the y-values less than or equal to 2x + 3. This includes all of the points BELOW the line, but also the points ON the line itself. We show that the line is included in the inequality by using a solid line.

5 4 3 2 1 −4 −3 −2 2 − −11

−11

x 1

−2 2

Example 1 Determine whether or not each of the following points satisfies the inequality y < 2x − 5: a (1, −4)

Create a strategy Substitute x = 1 into the right side of the inequality, and compare the result to the value of y, which is −4.

Apply the idea y < 2x − 5

Start with the inequality

y<2×1−5

Substitute the value of x

y < −3

Evaluate

−4 < −3

Compare to −4

This result matches the inequality, and so this point satisﬁes the inequality. b (−3, −9)

Create a strategy Substitute x = −3 into the right side of the inequality, and compare the result to the value of y, which is −9.

Apply the idea y < 2x − 5

Start with the inequality

y < 2 × (−3) − 5

Substitute the value of x

y < −11

Evaluate

−9 > −11

Compare to −9

This result does not match the inequality, and so this point does not satisfy the inequality.

4.05 Graph linear inequalities mathspace.co

177

c (3, 1)

Create a strategy Substitute x = 3 into the right side of the inequality, and compare the result to the value of y, which is 1.

Apply the idea y < 2x - 5

Start with the inequality

y<2×3-5

Substitute the value of x

y<1

Evaluate

1=1

Compare to 1

This result does not match the inequality, and so this point does not satisfy the inequality. The fact that the two values are the same tells us that this point lies on the boundary line y = 2x − 5, but due to the inequality symbol < the boundary is not included in the inequality.

Example 2 Consider the inequality 3x − 4y ≥ 24. a Find the x-intercept of the boundary line 3x − 4y = 24

Create a strategy The x-intercept occurs when y = 0.

Apply the idea 3x − 4y = 24 3x − 4 × 0 = 24

Start with the equation Substitute y = 0 into the equation

3x = 24

Simplify

x=8

Solve for x

The x-intercept for the boundary line is (8, 0).

b Find the y-intercept of the boundary line 3x − 4y = 24.

Apply the idea 3x − 4y = 24 3 × 0 − 4y = 24

Start with the equation Substitute x = 0 into the equation

−4y = 24

Simplify

y = −6

Solve for y

The y-intercept for the boundary line is (0, −6).

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c Does the origin (0, 0) satisfy the inequality?

Create a strategy Substitute x = 0 and y = 0 into the inequality and see if it is true.

Apply the idea 3x − 4y ≥ 24 Start with the inequality 3 × 0 − 4 × 0 ≥ 24 Substitute the values into inequality 0 ≥ 24 Evaluate the left-hand side The statement is false, so the origin does not satisfy the inequality.

d Graph the inequality and shade the appropriate region.

Create a strategy Draw the boundary line using the intercepts found in parts (a) and (b). Decide which side to shade based on the answer to part (c) which states that the origin is not in the region.

Apply the idea 1 −1 −1 −2

y x 1

2 3 4 5 6 7 8 9

−3 −4

Since the inequality is ≥ we know the boundary line is included so it is solid. The boundary line goes through (8, 0) and (0, −6). We know that (0, 0) is not in the required region, so we shade the opposite side.

−5 −6 −7 − 7 −8 − 8 −9 − 9

e Do points on the line 3x − 4y = 24 satisfy the inequality?

Create a strategy Is the boundary included or excluded?

Apply the idea The inequality is 3x − 4y ≥ 24. The inequality symbol ≥ means “greater than or equal to”, so points that satisfy 3x − 4y = 24 are included. Points on the boundary line satisfy the inequality.

4.05 Graph linear inequalities mathspace.co

179

Example 3 A linear inequality has points (1, 4) and (−1, 10) on its boundary line. The inequality is satisﬁed by (5, 5). Points on the boundary line itself do not satisfy the inequality. a Find the gradient of the boundary line.

Create a strategy Use m =

Apply the idea Let (1, 4) = (x1, y1) and (−1, 10) = (x2, y2) Substitute the values into the gradient formula

Simplify numerator and denominator

Calculate

The gradient of the boundary line is −3

b Find the equation of the boundary line in the form y = mx + c.

Create a strategy Use the gradient found in part (a) and one of the points that lie on the boundary and substitute into y = mx + c.

Apply the idea y = mx + c

Start with y = mx + c

4 = −3 × 1 + c

Substitute m = −3 and (1, 4)

c=7

Evaluate c

The equation of the boundary line is y = −3x + 7, which may be expressed as y = 7 − 3x

c State the inequality that satisﬁes the given conditions.

Create a strategy We are told that the point (5, 5) satisﬁes the inequality. Substitute x = 5 into the expression for the line and then compare this to the value y = 5 to see which way the inequality should go. We are also told that points on the boundary line do not satisfy the inequality, so the relationship will either be < or > and not ≤ or ≥.

180

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea 7 − 3x = 7 − 3 × 5

Substitute 5 for x

= −8

Evaluate

5 > −8

Compare to the y-value of 5

So the inequality is y > 7 − 3x.

d Graph the inequality.

Create a strategy Draw the boundary line. The given points (1, 4) and (−1, 10) can be used to draw the boundary. The boundary line should be dashed. Shade the side that contains (5, 5).

Apply the idea 10 9 8 7 6 5 4 3 2 1 −1 −1

Since the inequality is > we know the boundary line will be dashed.

The boundary line goes through (−1, 10) and (1, 4). We know that (5, 5) is in the required region, so we shade that side.

x 1

2 3 4 5 6 7 8 9

Example 4 Bernie sells his homemade pizza sauce in two sizes of containers: 200 mL and 500 mL. One week he has 4 L of sauce to sell. a Let x be the number of 200 mL bottles he ﬁlls, and y be the number of 500 mL bottles he ﬁlls. Write an inequality, with y as the subject, to represent this situation.

Create a strategy The amount of sauce in 200 mL bottles, added to the amount of sauce in 500 mL bottles, must be less than, or equal to, the amount of sauce available, 4000 mL.

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181

Apply the idea x of the 200 mL bottles contains 200x mL. y of the 500 mL bottles contains 500y mL. The amount in bottles must be less than or equal to the amount available Divide each term by the common factor of 100 Subtract 2x from each side of the inequality Divide each term of the inequality by 5

b Graph the solutions.

Create a strategy The boundary line may be expressed as 2x + 5y = 40. We can find the intercepts for this line by setting x = 0 to find the y-intercept and y = 0 to find the x-intercept. We can then decide which side of the boundary line to shade.

Apply the idea To ﬁnd the y-intercept set x = 0 2 × 0 + 5y = 40

Substitute x = 0

5y = 40

Simplify

y=8

Divide both sides by 5

To ﬁnd the x-intercept set y = 0 2x + 5 × 0 = 40

Substitute y = 0

2x = 40

Simplify

x = 20

Divide both sides by 2 The boundary line is drawn between the two intercepts.

The boundary is solid as the relationship is ≤.

A possible solution to the inequality is (0, 0) (which equates to ﬁlling no bottles), so we need to shade the side of the line that contains the origin.

6 5 4 3 2 1

x 2 4 6 8 10 12 14 16 18 20

182

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c Which combinations of bottles of sauce could Bernie ﬁll? Select all that apply. A

No 200 mL bottles and eight 500 mL bottles

Four 200 mL bottles and ten 500 mL bottles

Ten 200 mL bottles and four 500 mL bottles

Twenty 200 mL bottles and no 500 mL bottles

Create a strategy As we have already graphed the line and shaded the region we can plot the points on the graph to determine if the points are in or out of the region.

Apply the idea The four points we need to test are: y

A: (0,8)

B: (4, 10)

C: (10, 4)

8 A

D: (20, 0)

7 6 5

4 3 2 1

D x 2 4 6 8 10 12 14 16 18 20

Points A, C and D are in the acceptable region.

Reflect and check If the region had not already been plotted we could use the equation to test each set of coordinates We can express the inequality as 2x + 5y ≤ 40 A: x = 0, y = 8. 2 × 0 + 5 × 8 = 40

Substitute x = 0, y = 8

This satisﬁes the inequality, so A is a possible solution. B: x = 4, y = 10. 2 × 4 + 5 × 10 = 58

Substitute x = 4, y = 10

This does not satisfy the inequality, so B is not a possible solution. C: x = 10, y = 4. 2 × 10 + 5 × 4 = 40

Substitute x = 10, y = 4

This satisﬁes the inequality, so C is a possible solution. D: x = 20, y = 0. 2 × 20 + 5 × 0 = 40

Substitute x = 20, y = 0

This satisﬁes the inequality, so D is a possible solution.

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183

Idea summary The following are features of two variable inequalities: • • • •

A line is solid if the symbols are ≥ or ≤. A line is dotted if the symbols are > or <. If the symbols are ≥ and >, then the shaded region must be above the line. If the symbols are ≤ and <, then the shaded region must be below the line.

4.05 Practice questions Understanding 1

Explorer 1, 3–5

Adventurer 1–5

Trailblazer 1–5

The two entertainment options for the school holidays are movies and ice-skating. • Movies, m, cost $12 each • Ice-skating, s, costs $21 per session The inequality 12m + 21s ≤ 150 represents the total amount of money you can spend. What does the point (0, 0) on this graph signify?

184

The point (0, 0) signiﬁes that there is no money spent on movies or ice-skating sessions.

The point (0, 0) signiﬁes that you spent all your budget on movies and ice-skating sessions.

The point (0, 0) signiﬁes that you attended only movies.

The point (0, 0) signiﬁes that you attended only ice-skating sessions.

Are these statements true or false? a

If a point lies in the shaded region of the graph of an inequality, it is not a solution to the inequality.

The point (0, 0) lies in the solution region of the inequality x + y > 0.

If an inequality has < or > symbol, the graph of the inequality will have a dashed line.

In the context of budgeting, the inequality 10m + 20s ≤ 100 could represent spending on movies (m) and snacks (s) with a total budget not exceeding $100.

The inequalities 3x − 2y > 6 and 3x − 2y ≥ 6 will have exactly the same graph.

Changing 2x + 3y ≤ 20 to 2x + 3y ≤ 30 will shift the line upwards on the graph.

The inequality x ≥ 0 represents all points to the right of the y-axis, including the y-axis itself.

When graphing an inequality, we shade the region that contains the solutions to the inequality.

What does this shaded region mean in the context of a problem? A

The shaded region represents the line of the graph.

The shaded region represents all possible combinations of x and y-values that make the inequality false.

The shaded region represents all possible combinations of x and y-values that make the inequality true.

The shaded region looks nice and has no mathematical significance.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Is the inequality represented using a solid or dashed line on the graph? a

y ≤ 6x − 3

y ≤ 5x − 3

y ≥ 3x − 2

y>8

y ≥ 6x − 2

y < −9

6bc, 7b, 8, 10–13, 14c-h, 15, 16a, 17bc, 18

6, 7c, 9, 11, 12, 14c-l, 16d, 17cd

Is the coordinate pair a solution to the inequality? a

(6, 20); y ≥ 4x + 5 b

(3, 2); 3x + 2y ≥ 12

(3, 5); 2y ≤ x2

(−1, 7); y ≥ 2x2 − 4x

For each inequality, identify if the points satisfy it or not. a

x − y > 10 i

(−2, −9)

(15, 20)

iii

(16, 7)

(10, −6)

(−2, −3)

iii

(0, 0)

(2, 7)

(−2, 4)

iii

(−4, 29)

(2, −13)

y > 3x i

(12, 20)

y ≥ −8x − 3 i

y < −2x − 7

6ab, 7a, 8, 10–13, 14a-f, 15, 16ab, 17ab, 19

Fluency 6

Is the shaded region above or below the line? a

Ex 1

y < 5x − 7

(6, −50)

For each graph: i

Find the equation of the boundary line.

Write the inequality that represents the shaded region.

7 y 6 5 4 3 2 (0, 2) 1 −4 − 4 − −3 3 − −2 2 − −11

−11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6 −7 − 7

3 (−3, 2)

(3, 2)

x 2

3 − −4 −3 −11 −2 2 −

−11 −

x 1

−2 − 2 −3 − 3 (−3, −4) − −4 4

4.05 Graph linear inequalities mathspace.co

185

7 y 6 5 4 3 (−2, 1) 2 1 −4 − 4 − −3 3 − −2 2 −1 −1

10 8 6

x 1

−4 −3 −2 2 − −11

(−2, 4)

− −2 2 − −4 4 −6 − 6 − −8 8 − −10 10

−4 −3 −2 −1

−1

(0, 0) 1 2

−4 − 4 − −3 3 − −2 2 − −11

−11

−2

−2 2

−3

−3 3

−4

−4 4

(3, 0) x 1

Write the inequality that describes the shaded region: a

6 5 4 3 2 1 −6 − 6 − −5 5 − −4 4 − −3 3 − −2 2 − −11 −11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6

186

x 3

x 1

y 4 (0, 4)

3 1

(0, 7)

4 2

(−1, 3)

−2 −3 − −4 − 4 −5 − 5 (0, −5) −6 − 6 −7 − 7

x 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

6 5 4 3 2 1 −6 − 6 − −5 5 − −4 4 − −3 3 −2 −1 −1 −2 −3 −4 −5 − 5 − −6 6

x 1

6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 3 −4 − 4 −5 − 5 −6 − 6

7 y 6 5 4 3 2 1

x 1

−3 −2 −1−1

2 3

x 1

2 3 4 5 6 7

−2 −3 −4 −5 −6 −7

The line y = x − 5 is graphed on the Cartesian plane. a

Graph the solution to y > x − 5.

Graph the solution to y ≤ x − 5.

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Ex 2

x 1

2 3 4 5

For 3x + 4y = 12: a

Find the x-intercept.

Find the y-intercept.

Graph the line on a cartesian plane.

Write a related inequality that describes the region where (−18, 15) is located.

For y = −3x + 4: a

Find the x-intercept.

Find the y-intercept.

Graph the inequality y ≤ −3x + 4.

Do these points satisfy the inequality y ≤ −3x + 4? i

e 13

(3, 4)

(4, −8)

iii

(5, −5)

(2, 2)

6x + 3y > 12

Do the points on the line satisfy the inequality y ≤ −3x + 4?

For each inequality: i

Graph the boundary line.

Does the point (0, 0) satisfy the inequality?

iii

Graph the inequality.

y≥x+5

y > 2x − 4

2x + 3y < 12

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187

Graph the inequality: a

y ≤ 4x − 3

e i Ex 3

Ex 4

f −4x + y < 5

y < 4x − 7

g −3x + y < 7

y ≥ −3x + 5

h −2x + y < 9

3x + y < 11

For each set of conditions: i

Find the gradient of the boundary line.

Write the equation of the boundary line in the form y = mx + c.

iii

Write the inequality that satisﬁes the given conditions.

A linear inequality has solutions of (0, −3) and (−2, −7) on its boundary line and is not satisﬁed by (1, 4).

A linear inequality has solutions of (0, −4) and (−4, −20) on its boundary line and is satisﬁed by (−2, −8).

By using a graph of the inequality, find the maximum value of y with the given conditions. a

2x + y ≤ 10, when x = 3

3x + y ≤ 9, when x = 2

, when x = 3

, when x = 4

By using a graph of the inequality, find the minimum value of y with the given conditions. a

x − 2y ≤ 6, when x = 10

2x − 3y ≤ 12, when x = 8

, when x = 5

, when x = 4

Roxanne has set aside $14.00 in her shopping budget for fruit this week. Currently, oranges are on sale for $1.40 each, while apples are on sale for $1.75 each. a

Write an inequality relating the number of oranges, x, and the number of apples, y, that Roxanne buys.

Graph the solution to the inequality.

Can Roxanne buy these combinations of fruit? i

6 oranges and 7 apples

3 oranges and 4 apples

iii

7 oranges and 4 apples

1.5 oranges and 2.1 apples

A book seller makes a proﬁt of $8 on every book sold online and $5 on every book sold in-store. The company wants to make a proﬁt of at least $240 a day selling books through online and in-store sales. a

Write an inequality relating the number of books sold online, x, the number of books sold in-store, y, and the target proﬁt.

Graph the solution to the inequality.

Will the combination of books sold in-store and online be enough for the book seller to meet the target proﬁt?

d 188

y > 3x − 5

15.1 online and 29.1 in-store

25 online and 5 in-store

iii

13 online and 23 in-store

20 online and 20 in-store

Using the graph, ﬁnd the minimum number of total books ( from both online and in-store) that need to be sold to reach the target proﬁt.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning 20

21, 22

Explain the differences between the graphs of: a

4x + 5y > 20 and 4x + 5y ≥ 20

Can the inequality x + y > 10 have integer solutions where both x and y are less than 3?

4x + 5y ≤ 20 and 4x + 5y ≥ 20

Explain your reasoning. b

Can the inequality x − y > 10 have integer solutions where both x and y are less than 3? Explain your reasoning.

A gardening club wants to invest in new tools and plants. They can buy a set of tools for $12 and a bundle of plants for $21. They have a budget of $150. a

Given the inequality 12t + 21p ≤ 150 (where t represents the number of tool sets and p the number of plant bundles), explain why the club can’t buy more than 12 tool sets or more than 7 plant bundles.

Suppose the club decides to buy 5 sets of tools. Explain why the maximum number of plant bundles they can afford is 4.

Suppose the club decides to buy an equal number of tool sets and plant bundles. What is the maximum number of each they can afford? Explain your reasoning.

Suppose the budget for the gardening club was raised to $180. They decide to buy an equal number of tool sets and plant bundles. What is the maximum number of each they can afford?

Problem-solving 23, 24, 26, 27 23

23, 25–28

23, 25–30

Applicants for a particular job are asked to sit a numeracy test and verbal reasoning test. Successful applicants must obtain a minimum combined score of 43 for both tests. Write an inequality relating the applicant’s score on the numeracy test, x, and the verbal reasoning test, y.

Athletes applying for a speciﬁc sports scholarship must pass both a physical and a written test. The minimum combined score for both tests is 50. Write and graph an inequality relating the athlete’s score on the physical test, x, and the written test, y.

Rochelle is being careful with her spending so she can later purchase a house. She has allocated no more than $855 each month for both travel expenses and eating out. On average, every time she eats out it costs $19, and every travel journey costs $9. Write an inequality relating the number of times she eats out, x, and the number of her travel journeys each month, y.

Edward is setting up a monthly budget to save for a new car. He plans to spend no more than $720 each month on groceries and utility bills. On average, each grocery bill is $30, and each utility bill is $60. Write and graph an inequality relating the number of utility bills, x, and the number of grocery bills each month, y. 4.05 Graph linear inequalities mathspace.co

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In the new basketball season, William is looking to beat his personal best of scoring 102 points in total for the whole season (this does not include points from fouls). a

If he is to score x ‘two pointers’ and y ‘three pointers’ throughout the new season, write an inequality relating x, y and the number of points he hopes to score.

Graph the solution to the inequality.

In which given scenarios will William beat his personal best for the season? A

He scores 17.6 ‘two pointers’ and 27.9 ‘three pointers’

He scores 23 ‘two pointers’ and 24 ‘three pointers’

He scores 4 ‘two pointers’ and 27 ‘three pointers’

He scores 31 ‘two pointers’ and 8 ‘three pointers’

Sean is thinking about how to use his remaining spending money for snacks. A pack of dried fruit costs $6 and a bag of mixed nuts costs $4. He has $48 remaining. a

Write an inequality relating the number of packs of dried fruit, x and the number of bags of mixed nuts, y that Sean buys.

Sketch the region containing the points corresponding to all the different ways Sean could spend his remaining spending money on these snacks.

Supposing that Sean does not need to spend all of it, how many different ways can Sean spend his remaining money on some combination of dried fruit and mixed nuts?

Oprah is thinking about how to use her last arcade tokens. A game of Table Tennis costs 2 tokens and a game of Frog Days costs 5 tokens. She has a total of 30 tokens left. a

Write an inequality relating the number of times Oprah plays Table Tennis, x, and the number of times she plays Frog Days, y.

Construct a graph of the region containing the points corresponding to all the different ways Oprah could spend her remaining tokens across the two games.

Interpret the value of the x-intercept of the boundary line in the given context.

Interpret the value of the y-intercept of the boundary line in the given context.

Monica works part-time as a tutor and also as a sales associate. The number of hours she works in each role varies from week to week, and she needs to earn at least a certain amount weekly to cover her expenses. The graph of the inequality relating the number of hours she works as a tutor, x, and as a sales associate, y, is shown.

190

45 40 35 30 25 20 15

Write the rule for the inequality.

Find Monica’s hourly rate of pay for each job.

Find the amount of money she needs weekly to cover her expenses.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 5 10 15 20 25 30 35 40 45

4 Chapter review 1

Solve: a

g 2

Make the mentioned variables the subject in the following equations: a

W = Fd, make F the subject

Q = n2L, make n the subject

, make H the subject

Rearrange the formula to make, r the subject of the formula.

Calculate the radius of a cylinder with a volume of 200 cm3 and a height of 10 cm. Round your answer to two decimal places.

Are these lines parallel to y = 3x + 5? y = −3x + 5

y = 3x + 2

3x + y = −5

y = 3x

x−2

4y = x

Are these lines perpendicular to y = −4x + 1? a

S = I (1 + rt), make I the subject

The volume of a cylinder is given by the formula V = π r2h, where r is the radius of the base and h is the height of the cylinder.

a 5

x+5

y = −4x + 5

The graph shows the conversion between metres and feet: a

If a plank of wood is 10 ft in length, how long is it in metres?

In softball, the ﬁrst base is 60 ft from home plate. Calculate this length in metres.

m 8 7 6 5 4 3 2 1

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

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Javier tracked the number of apps on his phone over a number of months. Month (M ) Apps (A)

1 118

3 104

Plot the points on a Cartesian plane.

Is this relationship linear?

The time in minutes and the temperature in degrees Celsius (°C) are given in the table: 1 8

2 11

5 20

Find the initial temperature.

Form an equation relating x and y.

Graph the linear relationship represented in the table.

Rice costs a certain amount per kilogram. The table shows the cost of various amounts of rice in dollars: 0 0

10 32

20 64

30 96

Complete the table.

Graph the linear relationship represented in the table.

Find the cost of rice per kilogram.

Calculate the cost of 24 kg of rice.

Find the amount of rice that costs $44.80.

Is the coordinate pair a solution to the inequality? (2, 2) ; x + 4y > 11

(−1, 6) ; y ≤ 3x2 − 2x + 1

For each graph, write the inequality that represents the shaded region. a 4

1 −4 − 4 −3 −3 −2 2 −1 −1

192

4 17

By how much is the temperature increasing each minute?

a 11

3 14

Number of kilograms (x) Cost of rice (C)

4 96

Time (x) Temperature (y)

2 106

x 1

1 −4 − 4 −3 −2 −1 −1

−2

−3

− −3 3

−4

−4 − 4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1

For 3x + 5y = 15: a

Find the x-intercept.

Find the y-intercept.

Graph the line on a Cartesian plane.

Write a related inequality that describes the region where (−12, 9) is located.

For y = −2x + 2: a b

Graph the inequality y ≤ −2x + 2. Do these points satisfy the inequality y ≤ −2x + 2? i

c 14

(3, −6)

iii

(4, −2)

(1, 2)

Do the points on the line satisfy the inequality y ≤ −2x + 2?

Graph the inequality: a

(2, 3)

2x − 3y < 15

A linear inequality has solutions of (0, −3) and (7, −31) on its boundary line and is satisﬁed by (3, −11). a

Find the gradient of the boundary line.

Write the equation of the boundary line in the form y = mx + b.

Write the inequality that satisﬁes the given conditions.

Throughout university, Ben works as a waiter, getting paid $14 per hour, and as a valet, getting paid $18 per hour. The number of hours he works in each job can vary from week to week, and he needs to be able to at least cover his weekly expenses of $210. Write an inequality relating the number of hours he works as a waiter, x, and valet, y.

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Big ideas Quadratic equations can be represented graphically and solved through various methods, and their properties can be determined using the discriminant.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10A13 solve simple quadratic equations using a range of strategies, including null factor law VC2M10AA02 devise and use algorithms and simulations to solve mathematical problems VC2M10AA07 factorise monic and non-monic quadratic expressions and solve a wide range of quadratic equations derived from a variety of contexts VC2M10AA10 experiment with functions and relations using digital tools, making and testing conjectures and generalising emerging patterns

Quadratics Chapter outline 5.01 Quadratic equations 5.02 Quadratic graphs 5.03 Solve by factorisation 5.04 Solve by completing the square 5.05 The quadratic formula 5.06 The discriminant 5.07 Optional: Further quadratic equations Investigation: Derive the quadratic formula Investigation: Use bisection to solve quadratics Chapter 5 review

196 204 228 238 244 251 260

272

Developers use quadratics to control the speed and physics of race cars in video games!

5.01 Quadratic equations After this lesson, you will be able to… • identify the key features of quadratic graphs and how they relate to the solutions of the equation. • recognise that quadratic equations can have zero, one, or two solutions. • apply the null factor law to solve quadratic equations.

Quadratic graphs The standard form of a quadratic equation is y = ax2 + bx + c, where a ≠ 0. For example, the quadratic equation y = 2x2 − 3x + 5 has the values a = 2, b = −3, c = 5. The simplest quadratic equation is y = x2 and has the values of a = 1, b = 0, c = 0. The graph of a quadratic equation is called a parabola. This parabola is concave up as it opens upwards. A parabola will be concave up when the value of a is positive, such as y = x2.

y 7 6 5 y = x2

The point (0, 0) is the minimum point on the curve and is called the turning point or vertex.

4 3 2 1

−2

−1

x 1

−1

y 1 −2

−1

x 1

If the value of a is negative, such as y = −x2, the graph opens downwards. It is called concave down and the turning point is a maximum.

−2 y = −x2 −3 −4 −5 −6 −7

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There are many other key features including the intercepts and axis of symmetry. These key features will depend on the values of a, b, and c in the quadratic equation. axis of symmetry

y x x-intercepts Maximum x

y-intercept

y-intercept Minimum

Concave up with two x-intercepts

axis of symmetry

Concave down with no x-intercepts

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Example 1 For the parabola deﬁned by the equation y = −x2 + 3: a Is the parabola concave up or concave down?

Create a strategy For a parabola in the form y = ax2 + bx + c, concavity depends on the x2 coeﬃcient a: positive for concave up, negative for concave down.

Apply the idea Looking at the equation, the coeﬃcient of x2 is −1. So the parabola is concave down.

b What is the y-value of the y-intercept of the parabola?

Create a strategy To ﬁnd the value of the y-intercept, set the x-value equal to 0.

Apply the idea y = −x2 + 3 =3

Rewrite the equation Substitute x = 0

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Idea summary A quadratic equation is of the form y = ax2 + bx + c, where a ≠ 0. The graph is called a parabola. If a > 0, the graph is concave up as it opens upwards. If a < 0, the graph is concave down as it opens downwards. The minimum or maximum point on the curve is called the turning point or vertex. The turning point and whether it is concave up or down can tell us about the number of solutions.

Quadratic equations Quadratic equations can be written in the form ax2 + bx + c = 0. Quadratic equations can come in many forms such as: x2 = 16, (x − 5)2 = 25, 2x2 + 5x = 17 Quadratic equations can have zero, one or two solutions. This is related to the number of x-intercepts on the graph. For example: x2 = −4 has no solutions as a squared number is always positive. The related quadratic graph will have no x-intercepts.

x2 = 0 has one solution only, x = 0. The related quadratic graph will have one x-intercept.

x2 = 9, has two solutions, x = 3 and x = −3. We can also write this as x = ±3. The related quadratic graph will have two x-intercepts.

The symbol ± means “plus or minus”. We can use this as a shorthand for both the positive and negative of a number.

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Example 2 Solve for x: a x2 = 16

Create a strategy To ﬁnd x, we need to reverse the square operation.

Apply the idea x = ±4

Take the square root of both sides

Reﬂect and check Remember to always use the plus or minus sign (±) when dealing with perfect square answers.

b 81x2 − 64 = 0

Create a strategy Isolate the term with x and reverse the square operation.

Apply the idea Add 64 to both sides

Divide both sides by 81

Take the square root of both sides

Create a strategy Isolate x using inverse operations.

Apply the idea

Add 3 to both sides

Multiply both sides by 16

Take the square root of both sides

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199

Example 3 Solve (7 − x)2 = 81.

Create a strategy We can directly reverse the square operation on both sides, without expanding the binomial square on the right-hand side.

Apply the idea 7 − x = ±9

Take the square root of both sides

We need to further simplify both equations 7 − x = 9 and 7 − x = −9. 7−x=9

Write the first equation

−x = 2

Subtract 7 from both sides

x = −2

Divide both sides by −1

7 − x = −9

Write the second equation

−x = −16

Subtract 7 from both sides

x = 16

Divide both sides by −1

Idea summary Equations where the highest power is two are quadratic equations. They can be written in the form ax2 + bx + c = 0 and can have zero, one or two solutions. If we can rearrange a quadratic equation into the form x2 = k, then we can solve the equation by taking the positive and negative square roots. That is, x = ± . The symbol ± means “plus or minus”.

The null factor law The null factor law states that if a product of two or more factors is equal to 0, then at least one of those factors must be equal to 0. For example, if ab = 0 then either a = 0 or b = 0 We can use this to solve quadratic equations that are not in the form x2 = k. If we can rearrange a quadratic equation into the form (x − p) (x − q) = 0, then we know that either x − p = 0 or x − q = 0. We can solve the quadratic equation by solving each of these linear equations.

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Example 4 Solve: a (x − 6) (x + 7) = 0

Create a strategy Using the null factor law, set both factors to 0 and solve the resulting linear equations.

Apply the idea x−6=0 x=6

Set the ﬁrst factor to 0 Add 6 to both sides

x+7=0

Set the second factor to 0

x = −7

Subtract 7 from both sides

So the solutions are x = 6 and x = −7.

b 4m (m + 5) = 0

Create a strategy Using the null factor law, set both factors to 0 and solve the resulting linear equations.

Apply the idea 4m = 0

Set the ﬁrst factor to 0

m=0

Divide both sides by 4

m+5=0

Set the second factor to 0

m = −5

Subtract 5 from both sides

So the solutions are m = 0 and m = −5.

Idea summary The null factor law states that if a product of two or more factors is equal to 0, then at least one of those factors must be equal to 0. If we can rearrange a quadratic equation into the form (x − p) (x − q) = 0, then we know that either x − p = 0 or x − q = 0. We can solve the quadratic equation by solving each of these linear equations.

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5.01 Practice questions Understanding

How many solutions can a quadratic equation have?

A quadratic equation can be written in the form ax2 + bx + c = 0. Write the values of a, b and c for these quadratic equations:

2x2 − 7x + 5 = 0

−3x2 + 8x = 7

x2 = 16

−x2 = 4x − 3

A quadratic equation can be written in its factorised form (x − a) (x − b) = 0. Write the values for a and b for these quadratic equations. (x − 3) (x − 5)

Ex 4

202

(x − 6) (x + 3)

x(x − 4)

The equation x2 − 144 = 0 has a positive integer solution of x = 12. Find its other solution. 6, 7a–f, 8a–c, 9a–i

6, 7a–g, 8c–f, 9a–i

6, 7e–l, 8c–h, 9c–l

Are the graphs of these quadratic equations concave up or concave down? a

Fluency

Ex 3

Trailblazer 2–5

Given that we know 5 × 5 = 25. What is

Ex 2

1–5

Adventurer 2–5

Ex 1

Explorer

y = −x2 + 4x − 2

y = 3x2 + 8x

y = −x (x + 2) + 5

Solve: a

a2 = 2

k2 = 25

x2 = 121

t2 = 294

x2 − 121 = 0

4y2 = 100

15 = x2 − 10

36 = 25y2

−3k2 = −12

81k2 + 8 = 24

−16 = −81x2

Solve: a

(z + 3)2 = 49

(m − 7)2 = 81

(x − 6)2 = 2

(−3 + p)2 = 64

81 = (2 − n)2

(7 − x)2 = 81

(8l + 9)2 = 256

10 ( p2 − 7) = 930

x (2x − 9) = 0

Solve: a

j (j + 7) = 0

−5m (m + 5) = 0

(10x − 9) = 0

(4k − 9)2 = 0

(−3 + 7x)2 = 0

(e − 4) (e − 2) = 0

(s + 6) (s + 7) = 0

(8x − 5) (3x − 7) = 0

(3w + 8) (5w − 7) = 0

(3x − 17) (2x − d) = 0

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning

10, 11, 13

12–14

If a quadratic equation is given in the form x2 = k, how can we ﬁnd the solutions for x?

Explain why the quadratic equation x2 − 9 = 0 has two solutions.

A quadratic equation has two solutions, x = 4 and x = −3. Explain how to write the equation in the form (x − a) (x − b) = 0.

Jade used the null factor law to solve the equation (x − 4) (x + 6) = 0 and found the solutions x = −4 and x = 6. Is Jade’s answer correct? Explain why.

Explain and justify why the equation x2 + 4 = 0 has no solutions.

Problem-solving 15–17 15

15, 17, 18

17–20

The Widget and Trinket Emporium has released the forecast of its revenue over the next year. The revenue R (in dollars) at any point in time t (in months) is described by the equation: R = −(t − 12)2 + 4 a

Is the graph of R concave up or concave down?

Does the revenue reach a maximum or minimum point?

When will the revenue be zero?

The equation 4x2 + kx + 16 = 0 has one solution: x = 2. Find the value of the coeﬃcient k.

One solution to the equation ax2 − 32x − 80 = 0 is x = 4. Find the value of the coeﬃcient a.

Neville needs a sheet of paper x cm by 13 cm for an origami giraffe. The local origami supply store only sells square sheets of paper. The lower portion of the image below shows the excess area, A, of paper that will remain after Neville cuts out the x cm by 13 cm piece. The excess area, in square centimetres, is given by the equation: A = x (x − 13) a

At what lengths of x will the excess area be zero?

Does the graph of A have a minimum or maximum turning point?

For what value of x will Neville be able to make an origami giraffe with the least amount of excess paper?

13 x A

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The equation px2 + qx − 20 = 0 has solutions coeﬃcients p and q.

Given ax2 − c = 0: a b

and x = 4. Find the value of the

Solve for x, in terms of a and c. For what values of a and c, would there be: i

No solutions

One solution

iii

Two solutions

5.02 Quadratic graphs After this lesson, you will be able to… • understand the different forms of quadratic equations and how they relate to the graph of the equation. • identify the turning point and interpret its significance in the graph of a quadratic. • understand the transformations of a parabola, including translations, dilations and reflections. • calculate the y-intercept and x-intercepts of a parabola and understand their significance in sketching the graph. • apply knowledge of quadratic graphs to solve problems.

Transformations of parabolas We can graph a quadratic using transformations that can be identified from the equation. Let’s explore the possible transformations and how they are represented in the equation and graph.

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mathspace.co

A parabola can be vertically translated by increasing or decreasing the y-values by a constant number. A translation of y = x2 up by k units gives the equation y = x2 + k. y 4 y = x2 + 2

3 2

y = x2

x −4 −3 −2 −1

This graph shows y = x2 translated vertically up by 2 to get y = x2 + 2, and down by 2 to get y = x2 − 2.

y = x2 − 2

−1 −2

Similarly, a parabola can be horizontally translated by increasing or decreasing the x-values by a constant number. However, the x-value and the translation must be squared together. That is, to translate y = x2 to the left by h units we get y = (x + h)2. y 6 5 y = x2

4 3 y = (x + 2)2 2

y = (x − 2)2

1 −4 −3 −2 −1

x 1

−1

This graph shows y = x2 translated horizontally left by 2 to get y = (x + 2)2 and right by 2 to get y = (x − 2)2.

−2

A parabola can be dilated by multiplying every y-value by a constant number greater than 1. So to enlarge the parabola y = x2 by a scale factor of a, we get y = ax2. We can compress a parabola by using a scale factor between 0 and 1. y 6 5 y = 2x2 4

y = x2

This graph shows y = x2 vertically enlarged by a scale factor of 2 to get y = 2x2 and compressed by

3 2

1 −4 −3 −2 −1

−1

x2 2

x 1

a scale factor of 2 to get

−2

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Finally, we can reflect a parabola across the x-axis by multiplying by −1. So to reflect y = x2 across the x-axis, we get y = −x2. Notice that reflecting will change the concavity (in this case from concave up to concave down).

y 4 3 2 y = x2

1 −4 −3 −2 −1

−1

x 3

−2 −3 −4

y = −x2

All of these transformations can be performed on a quadratic and are represented by the equation y = a (x − h)2 + k.

Example 1 For the parabola y = −(x + 2)2 − 5: a Which successive transformations turn y = x2 into the equation y = −(x + 2)2 − 5? A

Reﬂection across the x-axis, horizontal shift 2 units to the left and vertical shift 5 units down.

Horizontal shift 2 units to the right, vertical shift 5 units down, reﬂection across the x-axis.

Reﬂection across the x-axis, vertical shift 2 units down and horizontal shift 5 units to the left.

Vertical shift 2 units up, horizontal shift 5 units to the left, reﬂection across the x-axis.

Create a strategy Identify the values of a, h, and k in turning point form where: • a represents a reflection and/or dilation • h represents a horizontal translation • k represents a vertical translation

Apply the idea y = (x + 2)2 shifts y = x2, 2 units to the left. y = −(x + 2)2 reﬂects y = (x + 2)2 about the x-axis. y = −(x + 2)2 − 5 shifts y = −(x + 2)2 down 5 units. So the answer is option A.

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b State the coordinates of the turning point of the curve.

Create a strategy Apply the transformations from part (a) to the turning point.

Apply the idea y 2 1 −4 −3 −2 −1

x 1

−1

−2 −3

The turning point of y = −(x + 2)2 − 5 is the turning point of y = x2 reﬂected across the x-axis, horizontally shifted 2 units to the left and shifted vertically 5 units down. As we can see in the graph, the coordinates of the turning point of the curve are: (0 − 2, 0 − 5) = (−2, −5).

−4 −5 −6

Reﬂect and check Because the graph has been reflected across the x-axis, the parabola is concave down. This means the turning point will be a maximum.

c Plot the graph of the equation.

Create a strategy Use the transformations and turning point we found in the previous parts.

Apply the idea 3 2 1

−4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 −5 −6 −7 −8 y = −(x + 2)2 − 5

Starting with the top dashed curve, y = x2, we need to reﬂect the curve across the x-axis, shift the curve 2 units to the left, then shift this curve down 5 units to get the ﬁnal parabola.

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Idea summary The parabola y = x2 can be transformed by: • Vertical translation by k units: y = x2 + k • If k > 0, like y = x2 + 3, then it is translated upwards. • If k < 0, like y = x2 − 1, then it is translated downwards. • Horizontal translation by h units: y = (x − h)2 • If h > 0, like y = (x − 2)2, then it is translated to the right. • If h < 0, like y = (x + 3), then it is translated to the left. • Dilated by a scale factor of a: y = ax2 • If a > 1, or a < −1, like y = 2x2, then it is vertically enlarged (stretched). • If −1 < a < 1 (with a ≠ 0), like y = 0.5x2, then it is compressed. • Vertically reﬂected across the x-axis: y = −x2 All of these transformations can be performed on a quadratic and are represented by the equation y = a (x − h)2 + k.

Quadratics in turning point form A quadratic equation in turning point form is written as y = a(x − h)2 + k, where a, h, k are numbers. We can find the key features and then use these to graph the parabola. The turning point is (h, k). The axis of symmetry is x = h. To find the y-intercept, substitute x = 0 and evaluate. To find the x-intercepts, if required, substitute y = 0 and solve the equation using inverse operations. Going the other way, we can also find the equation of a quadratic from the graph using these steps. 1. Identify the turning point from the graph and fill in the values of h and k in y = a(x − h)2 + k. 2. Substitute in any other point on the curve in for x and y to solve for the value of a. 3. Write the equation.

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Example 2 For the quadratic equation y = −2 (x − 3)2 + 8: a Is the parabola concave up or concave down?

Create a strategy Concavity depends on the sign of a: positive for concave up, negative for concave down.

Apply the idea Since the quadratic equation is y = −2 (x − 3)2 + 8 and a = −2 < 0, the parabola is concave down.

b Is the parabola more or less steep than y = x2?

Create a strategy Look at the value of a. Use the fact that as the value of a gets further from 0, the parabola becomes more steep and narrow.

Apply the idea Since we have that a = −2, and −2 is further away from zero than 1 is, the parabola is more steep than y = x2.

c Find the coordinates of the turning point.

Create a strategy A quadratic of the form y = a(x − h)2 + k has a turning point at (h, k).

Apply the idea The turning point of the quadratic equation y = −2 (x − 3)2 + 8 is at (3, 8).

d Find the y-intercept.

Create a strategy Substitute x = 0 into the equation and solve for y.

Apply the idea y = −2(x − 3)2 + 8 Write the equation = −2(0 − 3)2 + 8 Substitute x = 0 = −2(9) + 8

Evaluate the power

= −10

Evaluate

The y-intercept is at (0, −10).

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e Sketch the graph of this quadratic. Include the axis of symmetry.

Create a strategy Plot the y-intercept and turning point and draw a curve through the points.

Apply the idea Since the axis of symmetry occurs at x = h, then the axis of symmetry for y = −2(x − 3)2 + 8 is x = 3. 10 8 6 4 2

y (3, 8)

1 2 3 4 5 6 7 8 9 −2 −1 −2 −4 −6 −8 −10 (0, −10)

Reﬂect and check Graphing using as many key features as possible helps us to identify any errors we may have made when ﬁnding the key features.

Example 3 A parabola has the equation y = x2 + 4x − 1: a Express the equation of the parabola in the form y = (x − h)2 + k by completing the square.

Create a strategy Add and subtract

into the expression and factorise the perfect square part.

Apply the idea For x2 + 4x – 1 we need to add and subtract x2 + 4x − 1 = x2 + 4x + 4 − 4 – 1

Factorise the perfect square part

Simplify the constant

= (x + 2) − 4 – 1 = (x + 2) – 5

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Adding and subtracting 4

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b Find the y-intercept of the curve.

Create a strategy Substitute x = 0 into the equation of the parabola.

Apply the idea y = (0)2 + 4(0) – 1

Substitute 0 to x

y = −1

Evaluate

Reﬂect and check We can check that we have correctly completed the square by ﬁnding the y-intercept using that form as well. y = (x + 2)2 – 5

Write the equation

y = (0 + 2) – 5

Substitute x = 0

y = 4 – 5

Evaluate the brackets and indices

y = −1

Evaluate the subtraction

This is the same value we found when using the standard form, which tells us we correctly completed the square.

c Find the turning point of the parabola.

Create a strategy We can find the turning point using the equation found in part (a).

Apply the idea The equation in turning point form is: y = (x + 2)2 − 5. We can write this as y = (x − (−2))2 − 5. The coordinates of the turning point are (−2, −5).

Reﬂect and check If we hadn’t found the equation in turning point form, we could have used the idea that the axis of symmetry is at

From the equation, y = x2 + 4x − 1, a = 1, b = 4.

Substitute a and b Evaluate

The equation of the axis of symmetry is x = −2, which is also the x-coordinate of the turning point. y = x2 + 4x – 1

Write the equation

y = (−2) + 4(−2) – 1

Substitute x = −2

y = −5

Evaluate

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d Is the parabola concave up or down? A

Concave up

Concave down

Create a strategy Look at the coefficient of x2. A positive coefficient of x2 means the parabola is concave up while a negative coefficient of x2 means the parabola is concave down.

Apply the idea Since there is a positive coeﬃcient of x2 in the equation of the parabola y = x2 + 4x − 1, the parabola is concave up. So, the correct answer is Option A.

e Plot the curve y = x2 + 4x − 1.

Create a strategy Plot the intercepts and turning point and draw a curve through the points.

Apply the idea 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

x 1 2 3 4

Idea summary If an equation of a parabola is in the form y = a (x − h)2 + k, then: • The turning point form will be (h, k). • The axis of symmetry will be x = h. • The y-intercept, can be found by setting x = 0 and evaluating to get (0, a(h)2 + k). These key features can be used to plot the parabola.

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Quadratics in factorised form A parabola will always have one y-intercept and can have zero, one or two x-intercepts. A quadratic equation in factorised form is written as y = a (x − p) (x − q) where p and q are the x-values of the x-intercepts. y 7

−4 −3 −2 −1

2 1 −1

−1

x 1

−1

−2

x 1

−3

−4

y = (x − 4)2 We can graph a quadratic in factorised form using these steps: 1. Find the y-intercept by substituting x = 0, this gives us y = apq. 2. Find the x-intercepts by setting y = 0 and using the null factor law. 0 = a (x − p) (x − q), so that x = p or x = q, and the two x-intercepts are ( p, 0) and (q, 0). 3. Find the axis of symmetry by ﬁnding the average of the two intercepts, 4. The x-value of the turning point is the axis of symmetry, so to ﬁnd the y-value, we can substitute the x-value into the original equation and evaluate for y. If the equation of a parabola factorises into the form y = a (x − p) (x − q), then it is possible to use these key features to plot the parabola.

Example 4 Consider the graph shown: a Find the coordinates of the turning point.

Create a strategy Find the point on the parabola where the direction changes from downward to upward.

Apply the idea Based on the given graph, the coordinates of the turning point of the parabola are (−3,−6).

10 8 6 4 2 −7 −6 −5 −4 −3 −2 −1 −2

x 1

−4 −6

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b Find the equation in the form y = a(x − h)2 + k.

Create a strategy Use the coordinates of the turning point found in part (a) to fill in h and k in the quadratic equation of the form y = a(x − h)2 + k. We can then use another point on the curve to find the value of a.

Apply the idea Since the coordinates of the turning point of the parabola are (−3, −6), then the equation is y = a(x − (−3))2 − 6 which simplifies to y = a(x + 3)2 − 6 To determine the value of a, we can use one of the points that passes through the curve of the parabola, such as (0, 12).

Write the equation

Substitute the values of x and y

Add 6 to both sides

Evaluate the addition

Evaluate the power

Divide both sides by 9

Evaluate So, the quadratic equation of the given graph is: y = 2(x + 3)2 − 6

Reﬂect and check We could have used any point we can identify from the graph, it does not need to be the y-intercept at (0, 12). For example, we could also have used (−5, 2), (−1, 2), or (−6, 12).

Example 5 For the equation y = (2 − x) (x + 4): a State the y-value of the y-intercept.

Create a strategy Substitute x = 0 into the equation and solve for y.

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Apply the idea y = (2 − x) (x + 4) Write the equation y = (2 − 0) (0 + 4) Substitute x = 0 =8

Evaluate the product

b Determine the x-values of the x-intercepts.

Create a strategy Substitute y = 0 into the equation and solve for x using the null factor law.

Apply the idea y = (2 − x) (x + 4)

Write the equation

0 = (2 − x) (x + 4)

Substitute y = 0

2 − x = 0 Equate the ﬁrst factor to 0 x = 2 Add x to both sides x+4=0 x = −4

Equate the second factor to 0 Subtract 4 from both sides

So the x-values of the x-intercepts are x = 2, x = −4.

c Determine the coordinates of the turning point of the parabola.

Create a strategy The x-coordinate of the turning point is halfway between the x-intercepts.

Apply the idea Find the average of the x-intercepts Evaluate Substitute x = −1 into the equation Evaluate The coordinates of the turning point are (−1, 9).

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d Plot the graph of the parabola.

Create a strategy Plot the intercepts and the turning point and draw a curve through the points.

Apply the idea 9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1

−1 −2 −3 −4

x 1

Idea summary If a quadratic equation can be factorised into the form y = a (x − p) (x − q), then: • The y-intercept will be (0, apq). • The x-intercepts will be ( p, 0) and (q, 0). • The x-value of the turning point is

. The y-value of the turning point will

be found by substituting in the x-value turning point into the equation. These points can be used to plot the parabola.

Quadratics in standard form A quadratic equation in standard form is written as ax2 + bx + c = 0, where a ≠ 0. If a > 0, the graph is concave up and the turning point is a minimum. If a < 0, the graph is concave down and the turning point is a maximum. To graph the parabola, we can: 1. Find the axis of symmetry using

2. Find the turning point by substituting in the axis of symmetry for x and evaluating for y. 3. Find the y-intercept by setting x = 0 and evaluating to get y = c. If it is able to be factorised, we can use the null factor law to ﬁnd the x-intercepts. The turning point and y-intercept can provide a sketch of the graph. If other points are required, they can be found by substituting x-values into the equation. 216

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Example 6 For the equation y = 4x + x2: a Factorise the expression 4x + x2.

Create a strategy Take out the common factor of the two terms.

Apply the idea The common factor of the two terms is x. 4x + x2 = x(4 + x)

b Solve for the x-values of the x-intercepts of y = 4x + x2.

Create a strategy Substitute y = 0 into the equation and solve for x using the null factor law.

Apply the idea y = 4x + x2 Write the equation y = x (4 + x) Factorise 0 = x (4 + x) Substitute y = 0 x = 0 Equate the ﬁrst factor to 0 4+x=0

Equate the second factor to 0

x = −4 Subtract 4 from both sides So the x-values of the x-intercepts are x = 0, x = −4.

c Determine the y-value of the y-intercept of the graph.

Create a strategy Substitute x = 0 into the equation then solve for y.

Apply the idea y = 4x + x2

Write the equation

= 4 × 0 + 02

Substitute x = 0

Evaluate

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d Find the coordinates of the turning point.

Create a strategy Use the equation

and substitute the value of x into the equation of the parabola.

Apply the idea From the equation, y = 4x + x2, a = 1, b = 4. Substitute a and b

Evaluate The equation of the axis of symmetry is x = −2, which is also the x-coordinate of the turning point. y = 4x + x2 Write the equation = 4(−2) + (−2)2 Substitute x = −2 = −4

Evaluate

The coordinates of the turning point are (−2, −4).

Reﬂect and check We could also have used the average of the x-intercepts to ﬁnd that

e Plot the graph of the parabola.

Create a strategy Plot the intercepts and turning point and draw a curve through the points.

Apply the idea y 4 3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

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Example 7 For the quadratic y = 3x2 − 6x + 8: a Find the axis of symmetry.

Create a strategy Use the formula

Apply the idea From the equation y = 3x2 − 6x + 8, we have a = 3 and b = −6.

Substitute the values of a and b

Evaluate the numerator and denominator

Simplify the fraction

Reﬂect and check The turning point lies on the axis of symmetry, which means x = 1 is the x-value of the turning point.

b Find the turning point of the parabola. Give your answer in the form (a, b).

Create a strategy Substitute x = 1 into y = 3x2 − 6x + 8 to ﬁnd the y-coordinate of the turning point.

Apply the idea y = 3(1)2 − 6(1) + 8 =5

Substitute 1 to x Evaluate

So, the turning point of the parabola is (1, 5).

c What is the y-intercept of this quadratic curve?

Create a strategy Substitute x = 0 into y = 3x2 − 6x + 8 to ﬁnd the y-intercept.

Apply the idea y = 3(0)2 − 6(0) + 8 =8

Substitute 0 to x Evaluate

So, the y-intercept is (0, 8).

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d Plot the graph of the equation.

Create a strategy Plot the y-intercept and the turning point and draw a curve through the points.

Apply the idea y 11 10 9 8 (0, 8) 7 6 5 (1, 5) 4 3 2 1 −4 −3 −2 −1

−1

x 3

Idea summary A quadratic equation in standard form is written as y = ax2 + bx + c, where a ≠ 0. To graph the parabola, we can: 1. Find the axis of symmetry using

2. Find the turning point by substituting in the axis of symmetry for x and evaluating for y. 3. Find the y-intercept by setting x = 0 and evaluate to get y = c. If a > 0, the graph is concave up and the turning point is a minimum. If a < 0, the graph is concave down and the turning point is a maximum. If it is able to be factorised, we can use the null factor law to ﬁnd the x-intercepts.

5.02 Practice questions Understanding 1

220

Explorer 1–4

Adventurer 1–4

Describe the basic shape of a parabola.

Given that a general quadratic is y = ax2 + bx + c: i

What does the sign of a tell us?

What does changing the value of c do?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Trailblazer 1–4

Complete the following statement and ﬁll in the key features of a parabolic curve. The graph is concave ⬚.

y 10 8 6 4 2 –2

Ex 2

Ex 3

What feature is at the point (h, k) on the parabola deﬁned by y = a (x − h)2 + k?

What features are at the points ( p, 0) and (q, 0) on the parabola defined by y = a (x − p) (x − q)? 5ab, 6ac, 7a, 8a–c, 9, 10ad, 11ab, 12, 13ab, 14a, 15a–e, 16a–c

Fluency

Ex 1

5ac, 6ac, 7ab, 8ace, 9, 10b–d, 11bc, 12, 13bc, 14a, 15c–g, 16c–f

5ad, 6bd, 7cd, 8d–f, 9, 10d–f, 11cd, 12, 13cd, 14, 15h–l, 16d–h

For each equation: i

Describe the transformations from y = x2.

What is the turning point?

iii

Sketch the graph of the parabola.

y = (x − 3)2 − 4

y = −(x + 3)2 − 6

y = 3 (x + 1)2 + 4

y = −0.5 (x − 6)2 + 1

For each quadratic: i

Is the parabola concave up or down?

Is the parabola more or less steep than the parabola y = x2?

iii

Find the coordinates of the turning point.

Find the y-intercept.

Sketch the graph of this quadratic. Include the axis of symmetry.

y = 2 (x − 1)2 − 3

y = 3 (x + 1)2 + 2

y = −3 (x + 5) − 4

For each quadratic equation: i

Write in the form y = a (x − h)2 + k.

Find the coordinates of the turning point.

iii

Sketch the graph of the parabola.

y = x2 + 6x

y = −x2 + 4x − 3

y = −x2 − 8x − 18

y = x2 + x + 3

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Ex 4

Each of the graphs has an equation of the form, y = a(x − h)2 + k. Find the values of a, h and k for each graph and write the equation. a

2 −3

−2

−1

x 1

−2

−1

8 6

x 1

−2

−6

−5 −4 −3 −2 −1 −2

−8

−4

−5 −4 −3 −2 −1

f 5

−1

x 1

3 −3 −2 −1

−2

−1

−3

−2

−4

−3

x 1

The parabola y = x2 has been translated 2 units to the right and 6 units down. a

222

y 12

−2

−4

x 1

2 −3

−1

−4

Find the equation of the parabola.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Sketch the graph of the parabola.

Ex 5

For each quadratic: i

Find the y-intercept.

Find the x-intercepts.

iii

Find the axis of symmetry.

Find the coordinates of the turning point.

Sketch the graph of the parabola.

y = x (x + 4)

y = x (4 − x)

y = (x − 2) (x − 6)

y = (x − 3) (x + 1)

y = (5 − x) (x + 1)

y = (1 − x) (3 − x)

A parabola has x-intercepts at x = 1 and x = −7. State the equation of the parabola in the form y = a (x − p) (x − q) for each of these cases, and make a sketch of the curve. a

The turning point is a minimum turning point and the y-intercept is at (0, −7)

The turning point is a maximum turning point and the y-intercept is at (0, 14)

The turning point is a minimum turning point and the y-intercept is at (0, −21)

The turning point is a maximum turning point and the y-intercept is at (0, 3.5)

a Find the equation of the parabola in factorised form. b

y 14 12

What is the turning point?

10 8 6 4 2

−2 −1 −2

−4

These parabolas each have x-intercepts at x = ± . What is the equation of each parabola in the form y = a (x − p) (x − q) and y = a (x − h)2 + k? a

7 6 5 4 3 2 1

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

x 1

7 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

x 1

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−4 −3 −2 −1

−1 −2 −3 −4 −5 −6 −7

Ex 6

Ex 7

9 8 7 6 5 4 3 2 1

x 1

−4 −3 −2 −1−1

x 1

−2 −3 −4 −5

For each quadratic: i

Factorise to ﬁnd the x-intercepts.

Express the equation in the form y = a (x − h)2 + k by completing the square.

iii

Find the coordinates of the turning point.

Sketch the graph of the parabola.

y = x2 − 12x + 32

y = −2x2 − 8x + 10

For each quadratic: i

Describe the concavity of the parabola.

Factorise, and ﬁnd the x-intercepts.

iii

Find the y-intercept.

Find the axis of symmetry.

Find the coordinates of the turning vi point.

Sketch the graph and identify these key features on it.

y = x2 − 4x + 3

y = x2 − 36

y = 4 − x2

y = x2 + 4x + 3

y = x + 6x + 5

y = x2 − 8x + 15

y = −x2 + 2x + 24

y = −x2 − 2x + 8

y = x − 2x − 8

y = 8 + 2x − x2

y = x2 + 2x − 3

y = 2x2 + 9x + 9

For each quadratic: i

What is the axis of symmetry?

What are the coordinates of the turning point?

iii

What is the y-intercept?

Sketch a graph of the parabola.

y = x2 − 4x + 1 2

y = x2 + 6x + 7

y = −4x − 4x + 1

y = −(x2 + 8x + 11)

y = x2 + 6x − 2

y = −x2 + 10x − 23

y = x2 + 2x − 11

224

7 6 5 4 3 2 1

y = −x + 8x − 9

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning 17

17, 18, 23

17, 18, 20, 23

19, 21, 22, 24

Three different ways of expressing a quadratic equation are: y = a (x − p) (x − q), y = a (x − h)2 + k, and y = ax2 + bx + c. How can you ﬁnd the y-intercept for each form? Explain which form you would use to identify the y-intercept if you had all three available.

Determine whether each statement is true or false. Justify your answer. a

The turning point of a quadratic graph is always at the origin (0, 0).

If the coeﬃcient of the x2 term in a quadratic equation is negative, the graph will open upwards.

All quadratic graphs have a line of symmetry.

The y-intercept of a quadratic graph is always equal to the constant term in the quadratic equation.

The solutions of a quadratic equation set equal to zero can be of a quadratic equation can be determined by ﬁnding the x-intercepts of the corresponding quadratic graph.

Given a quadratic of the form y = a (x − h)2 + k, what are the conditions on a, h, and k so that: a

The turning point is a minimum turning point and the parabola has two x-intercepts?

The turning point is a maximum turning point and the parabola has one x-intercept?

The turning point is a minimum turning point and the parabola has no x-intercepts?

The turning point of a parabola is located at (2, 5). a

Determine the equation of the parabola if the y-intercept is at (0, 9).

The parabola is translated 2 units to the right and 7 units down. Determine the equation of the transformed parabola.

Explain the change that would be made to the equation in part (a) if the graph is translated a units to the right and b units upwards.

For the equations: A = (x − 2) (x − 4) and B = 2 (x − 2) (x − 4)

Which key features are the same for A and B?

Which key features are different for A and B?

Marvin wants to understand how turning point form and standard form of a quadratic function are related. Consider the turning point form: y = a(x − h)2 + k a

Expand a(x − h)2 + k and write the equivalent equation to y = a(x − h)2 + k in standard form.

Determine the value of b in the equation from part (a).

Determine the value of c in the equation from part (a).

Explain how the equations are related.

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Describe the relationship between the turning point of a quadratic graph and the coeﬃcients a and b of y = ax2 + bx + c.

Explain what possible differences there can be, if: a

Two different quadratic curves have the same intercepts.

Three different quadratic curves have the same turning point and share one common intercept.

Problem-solving 25–27 25

25–28

26–30

The height h, in metres, reached by a ball thrown in the air after t seconds is given by the equation: h = 10t − t2 a

Complete the table of values for h = 10t − t2: t

Graph the relationship h = 10t − t2.

Find the height of the ball after 5.5 seconds have passed.

Find the maximum height reached by the ball.

When an object is thrown into the air, its height above the ground is given by the equation: h = 163 + 38s − s2 where s is its horizontal distance from where it was thrown.

Graph the relationship h = 163 + 38s − s2.

Find s, how far horizontally the object is from where it was thrown at the point when it reaches its greatest height above the ground.

Find the maximum height reached by the object.

Over the summer, Susana and her friends build a bike ramp to launch themselves into the local lake. Susana decides that the shape of the ramp will be parabolic, and thinks that the parabola is given by the equation:

1 ( x + 2)2 + 2 4 x

where y is the height in metres above the ground, and x is −6 the horizontal distance in metres from the edge of the lake.

226

If the ramp starts 6 m back from the edge of the lake, how high is the start of the ramp?

At what height will the rider leave the ramp?

At what other distance x is the rider also at a height of 3 m?

Graph the shape of the ramp on a Cartesian plane.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The main cable on a suspension bridge is approximately parabolic in shape. A point (x, y) on the parabola represents the height, y m, above the roadway x m from the leftmost supporting pole. The lowest point of the main cable is 10 m above the roadway and occurs 150 m from the leftmost post. The main cable meets the supporting poles 90 m above the roadway. Determine an equation for the parabola formed by the main cable. y

(150, 10) x

Elise is the owner of a restaurant. Elise wants to install new wooden ﬂoors in several rooms. The rooms in the restaurant are square. The wood costs $6.25 per square metre. The cost of the ﬂooring in terms of its side length is shown by the quadratic function C = 6.25x2. a

Determine how much Elise should expect to spend on ﬂooring if the room has side lengths of 12 m.

Determine how much the price would change if the side lengths decreased by 3 m.

Graph the quadratic model from part (a).

Describe what changes and what stays the same about the graph of the quadratic model if the cost per square metre decreases.

A quadratic equation in factorised form is y = a (x − m) (x − 12). If the equation has a turning point at the point (8, 15), show how we can ﬁnd the values of a and m.

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5.03 Solve by factorisation After this lesson, you will be able to… • solve quadratic equations by factorisation. • identify x-intercepts on a parabola and use them to solve a quadratic equation. • find the equation of a graph by identifying the intercepts and using them to write an equation.

Solve by factorisation We have seen how to solve quadratic equations using the null factor law. We can use these steps to solve factorisable quadratic equations: 1. Gather together all terms to one side of the equation and make it equal to 0, in the form ax2 + bx + c = 0. 2. Factorise the quadratic using any appropriate techniques, such as: • Finding the highest common factor • Using patterns like a2 − b2 = (a − b) (a + b) and a2 + 2ab + b2 = (a + b)2 • Splitting the middle term and using grouping • For x2 + bx + c, ﬁnding two numbers whose sum is b and whose product is c • Using the cross product method 3. Use the null factor law to split the quadratic equation into two linear equations and make each linear equation equal to 0. 4. Solve the linear equations to get the solutions to the original quadratic equation. 5. Check the solutions by substituting back into the original equation.

Interactive exploration

Discover this concept in action online

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Example 1 Solve the quadratic equations: a 3y − 15y2 = 0, which has two solutions. Solve for both values of y.

Create a strategy Factorise 3y − 15y2 and use the null factor law to solve.

Apply the idea 3y − 15y2 = 0 Write the equation 3y(1 − 5y) = 0 Factorise the quadratic expression 3y = 0 Set the ﬁrst factor equal to 0 y = 0 Divide both sides by 3 1 − 5y = 0 Set the second factor equal to 0 5y = 1 Add 1 to both sides Divide both sides by 5 So the solutions are y = 0 and

Reﬂect and check We can substitute y = 0 and

to check that these are the solutions of 3y − 15y2 = 0.

For y = 0: LHS = 3(0) − 15(0)2 =0

Substitute 0 for y Evaluate

= RHS For

: Substitute

for y

Evaluate the power Simplify

15 25

Evaluate the subtraction

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b x2 − 22x + 121 = 0

Create a strategy Factorise x2 − 22x + 121 and use the null factor law to solve.

Apply the idea x2 − 22x + 121 = 0

Write the equation

(x − 11) (x − 11) = 0

Factorise the quadratic expression

x − 11 = 0 x = 11

Set one of the factors equal to 0 Add 11 to both sides

Reﬂect and check Notice that x2 − 22x + 121 is in the form a2 + 2ab + b2, where a = x and b = −11. x2 − 22x + 121 = 0 2

Write the equation

Use the pattern

Factorise

(x) + 2 (x) (−11) + (−11) = 0 (x − 11) = 0 x − 11 = 0 x = 11

Set one of the factors equal to 0 Add 11 to both sides

Example 2 Solve x2 = 17x + 60 for x.

Create a strategy Move all terms to one side, then factorise and use the null factor law to solve.

Apply the idea x2 = 17x + 60 2

Write the equation

x − 17x − 60 = 0

Move all the terms to one side

(x − 20) (x + 3) = 0

Factorise the quadratic expression

x − 20 = 0 x = 20 x + 3 = 0 x = −3

Set the ﬁrst factor equal to 0 Add 20 to both sides Set the second factor equal to 0 Subtract 3 from both sides

So the solutions are x = 20 and x = −3.

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Idea summary We can solve most quadratic equations using these steps: 1. Gather together all terms to one side of the equation and make it equal to 0, in the form ax2 + bx + c = 0. 2. Factorise the quadratic expression using any appropriate techniques. 3. Use the null factor law to split the quadratic equation into two linear equations and make each linear equation equal to 0. 4. Solve the linear equations to get the solutions to the original quadratic equation. 5. Check the solutions by substituting back into the original equation.

Equations from graphs As well as drawing graphs from quadratic equations, we can also find the equation from the graph of a parabola. We have previously done this using the turning point, but it is not always easy to identify the coordinates of the turning point from the graph. Sometimes we will know other features, such as the x-intercepts, but not the turning point. To find the equation of a quadratic, we can follow these steps: 1. Identify the x-values of the x-intercepts. These occur at the point where y = 0. 2. Substitute the x-values found in Step 1 for p and q in the equation y = a (x − p) (x − q). 3. Find the value of a by using another known point or by using second differences. The first difference is the difference between consecutive y-values. The second difference is the difference between consecutive first differences. The second difference of any quadratic relationship is a non-zero constant. We can identify the first and second differences from a table of values or from a graph. 11 10 9 8 7 6 5 4 3 2 1 −4 −3 −2 −1

For our basic parabola, y = x2, where a = 1, the first differences are:

1−0=1 4−1=3 9−4=5

The second differences are: 3−1=2 3

5−3=2 x

1 1

In general,

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Example 3 Write an equation for the quadratic relation shown by the graph.

y 4 3 2 1 x −4 −3 −2 −1

−1 −2

Create a strategy For this graph, we can easily read the coordinates of the x-intercepts, so we can write an equation using the factorised form of a quadratic given by: y = a (x − p) (x − q) where p and q are the x-values of the x-intercepts.

Apply the idea The coordinates of the x-intercepts are (−1, 0) and (2, 0), so p = −1 and q = 2. To ﬁnd the value of a, we can use one of the points that passes through the curve of the parabola, such as the y-intercept, (0, −1). Write the factorised form of a quadratic equation Substitute p = −1 and q = 2 Substitute x = 0 and y = 1

Evaluate the brackets

Divide both sides by −2

Simplify The quadratic equation of the graph is:

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Reflect and check Instead of substituting a point to find the value of a, we can also look at the second differences. For this graph, the first differences are 1, 2, 3, …, so the second differences are all 1.

y 4

3 2 2

x −4 −3 −2 −1

−1

−2

Alternatively, we could have use the coordinates of points on the graph to create a table of values, then calculate the ﬁrst and second differences. x

−1

First Difference −1 − (−1) = 0 0 − (−1) = 1 2−0=2 5−2=3

Second Difference 1−0=1 2−1=1 3−2=1

Idea summary To ﬁnd the equation of a quadratic, we can use these steps: 1. Identify the x-intercepts from the graph or otherwise. 2. Substitute the values for p and q found in Step 1 into y = a (x − p) (x − q). 3. Find the value of a by using another known point or by using second differences.

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5.03 Practice questions Understanding 1

Adventurer 1, 2c–f, 4, 5cd

Trailblazer 2e–h, 4, 5ad

Which two options are solutions to the equation x (x − 7) = −10? A

Explorer 1, 2a–d, 3, 4, 5ab

x=5

x = −5

x=1

x=2

Find possible values of a and b: a

a + b = 5, ab = 6

a + b = −11, ab = 18

a + b = 8, ab = 16

a + b = −6, ab = 9

a + b = −1, ab = −2

a + b = −7, ab = 12

a + b = 0, ab = −9

a + b = 2, ab = −15

To factorise x2 + 11x + 24, we need to find two numbers: a

What should their product be?

What should their sum be?

What are the two numbers?

Factorise x2 + 11x + 24

One expression for the area of the rectangle below is m2 + 14m + 45. The rectangle is made up of four smaller rectangles. Use the diagram to express the area of the large rectangle in factorised form. m

For the given graphs, ﬁnd the solution(s) to the equation. a

(x + 2) (x − 3) = 0

(x − 1) (x + 4) = 0

2 x

−6

234

−4

−2

x −6

−4

−2

−4

−6

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

(x − 3) (x + 3) = 0

(x + 1) (x − 2) = 0

y 6

2 −6

−4

−2

x 2

−6

−2

−4

−2

−6

−8

Ex 1

6ef, 7d–h, 8c–f, 9d–g, 10a–f, 11ab

6e–h, 7g–l, 8e–h, 9g–l, 10d–l, 11cd

Solve: a

x (x − 9) = 0

2m (m − 8) = 0

c (5c − 12) = 0

( y − 6) ( y + 11) = 0

(k − 3) (k − 5) = 0

(3x − 9) (2x − 5) = 0

5 (x + 5) (x − 5) = 0

(7a − 2)2 = 0

Solve: a

x2 − 64 = 0

x2 − 9 = 0

x2 + 12x = 0

x2 − 8x = 0

x2 + 5x + 4 = 0

x2 + 7x + 10 = 0

x − 8x + 15 = 0

x2 − 2x − 24 = 0

x2 − x − 20 = 0

x2 + 3x − 10 = 0

x2 − 14x + 49 = 0

10x2 − 10x − 120 = 0

x − 14x + 40 = 0

Solve: a

Ex 2

−4

−6

6a–d, 7a–h, 8a–d, 9a–d, 10a–c

−2

−4

Fluency

2x2 − 32x = 0 2

4x + 24x + 20 = 0

3x2 − 30x + 75 = 0

3x2 − 21x − 54 = 0

−2x2 + 30x − 112 = 0

−5x2 + 10x + 40 = 0

−3x2 − 3x + 396 = 0

Solve: a

m2 = 14m

m2 = m + 20

x2 = 81

−3m (m + 2) = 0

m − 27m = −182

3 (n2 + 8) = −18n

2y − 6y2 = 0

2y − 10y2 = 0

15 − 11b − 12b2 = 0

6k2 = −7 + 13k

x2 = 24x − 144

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Solve: a

Ex 3

x(x + 2) = 48 2

x2 + 8x + 16 = 36

4x = 400

3x2 − 7x − 20 = 0

5x2 = −320 + 80x

−5x2 = −53x + 72

−10x + x2 = 2 − 7x − x2

−8x + x2 = −6 − x − x2

( y − 3) = 2y + 2

( y + 1)2 = 4y + 4

6x2 − 27x + 28 = −3x2 + 3x + 3

(5x2 − 13x + 6) (2x2 − 13x + 20) = 0

For each graph, complete the following: i

What are the solutions of the related quadratic equation?

Determine the y-intercept(s).

iii

Write the equation in factored form.

18 16 14 12 10 8 6 4 2

12 10 8 6 4 2

−2 −1 −2

−2

−4

−2

2 −8 −6 −4 −2 −2

−8 −10 −12

−4

−14

−6

−16

−8

Reasoning

−6

236

−4

x 4

2 2

−2 −4

12, 15

x 2

13–16

For x2 + kx + 24, where k is an integer: a

Show that k = 10 is a possible value of k so that the expression can be factorised.

Find all of the integer values of k that allow the expression to be factorised over the integers.

For each integer value of k in part (b), state the solutions to x2 + kx + 24 = 0.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For x2 + kx + 28: a

Find the largest integer value of k that makes the expression factorisable. Explain.

Find the integer solutions to x2 + kx + 28 = 0 so that the difference between the two solutions is as large as possible.

For x2 + 4x + c: a

Find all positive integer values of c that allow x2 + 4x + c to be factorised as a product of rational linear factors. Explain your process.

Find all possible integer solutions to x2 + 4x + c = 0 for positive integer values of c.

Explain why it is not possible to list all negative integer values of c that allow x2 + 4x + c to be factorised into the product of rational linear factors.

Explain why there are always solutions to x2 + 4x + c = 0 for c < 0.

In general, a quadratic can have 0, 1, or 2 x-intercepts. If its equation can be factorised using rational numbers, how many x-intercepts will it have? Explain.

Explain why we can say that the solutions to (ax + b) (cx + d) = 0 are

and

but we cannot say anything about the solutions for (ax + b) (cx + d) = e, where e ≠ 0 without further work.

Problem-solving 17a–c, 18, 19, 21 17

17d–h, 18, 19, 21

17e–j, 18–22

Solve: a

On the graph of y = x2 − 4, there are two points where y = 12. Find the x-coordinates of these two points.

Find the values of m and n that make both sides of the following equation equal: y2 + my + 60 = ( y + 10) ( y + n)

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Software engineers are designing a self-serve checkout system for a supermarket. They notice that the traﬃc through the store during the day is described by the equation: C = −t (t − 12) where C is the number of customers and t is the number of hours after the store opens. To meet the peak demand, the engineers allow for an extra checkout machine to automatically turn on when the number of customers ﬁrst reaches 27 people, and to automatically turn off when it next falls below 27 people. a

Find the times t when the number of customers is equal to 27 people.

After how many hours will the extra checkout machine turn on after opening?

How many hours will the extra machine be on for?

The height of a projectile above the ground is given by y = 10 + 5t − 5t2. Height is measured in metres and time in seconds. Find the time, t, at which the projectile hits the ground.

Write the equation of a quadratic in factorised form that has the following key features and draw a graph of the curve: a

Two x-intercepts that have the same values with opposite signs

One x-intercept that is a fraction and the other is a prime number

One unique x-intercept that is less than 1 but greater than 0

x-intercepts of x = 7, x = −2 and has a y-intercept at (0, 14)

x-intercepts of x = −1, x = −10 and has a y-intercept at (0, 30)

5.04 Solve by completing the square After this lesson, you will be able to… • solve quadratic equations by completing the square.

Algebraic method The process for completing the square: 1. Factor out the coeﬃcient of the x2 term, unless it is 1. 2. Halve the coeﬃcient of x and square it, add and subtract this value. 3. Factorise the terms that now form a perfect square. 4. Combine the constant and term left over from completing the square.

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Interactive exploration

Discover this concept in action online

mathspace.co

Example 1 Using the completing the square method, rewrite into the form (x + a)2 + b: x2 − 14x + 59

Apply the idea The coefficient of the x term is −14, and half of this is −7. x2 − 14x + 59 = x2 − 14x + 72 − 72 + 59 = (x − 7)2 − 72 + 59 2

= (x − 7) + 10

Add 72 − 72 Factorise the perfect square part Evaluate the constant

Example 2 Solve x2 − 7x + 8 = 0.

Create a strategy Complete the square and then solve.

Apply the idea Complete the square: The coefficient of the x term is −7, and half of this is

. Add

Factorise the perfect square part

Evaluate the exponent

Evaluate the constant

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Solve: Collect like terms

Take the square root of both sides

Simplify the square root

Add

to both sides

Evaluate So the solutions are

Simplify using a common denominator and

Example 3 Solve 4x2 + 11x + 7 = 0.

Create a strategy Divide the equation by the coeﬃcient of x2, then complete the square and then solve.

Apply the idea Divide the equation by the coefficient of x2: Write the equation

Divide both sides by 4

Evaluate

Complete the square: The coefficient of the x term is

, and so

Add

Factorise the perfect square part

Evaluate the exponent

Evaluate the constant

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Solve: Collect like terms Take the square roots of both sides Evaluate Subtract

from both sides

Evaluate Simplify using a common denominator We can see that we can still simplify each x-value. Write the positive x-value Evaluate the subtraction Evaluate Write the negative x-value Evaluate the subtraction Simplify the fraction

Idea summary The following are the steps of completing the square algebraically. 1. Factor out the coeﬃcient of the x2 term, unless it is 1. 2. Halve the coeﬃcient of x and square it, add and subtract this value. 3. Factorise the terms that now form a perfect square. 4. Combine the constant and term left over from completing the square. We can then solve using inverse operations.

5.04 Practice questions Understanding 1

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

When solving x2 + 24x = 10, four different students showed their ﬁrst step of working. Identify if there is an error or not. a

Asad: x2 + 24x + 144 = 10

Blanche: x2 + 24x + 144 − 144 = 10

Cordelia: x2 + 24x + 144 = 10 + 144

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What should be added to x2 + 10x in order to make it a perfect square trinomial?

What should be added to x2 − 8x in order to make it a perfect square trinomial?

Which value would complete the square for x2 + 6x?

Ex 2

5a–d, 6a–c, 7ab, 8a, 9a–d

x2 − 2x + 46

5c–f, 6b–e, 7ac, 8b, 9c–f

5e–h, 6c–f, 7cd, 8b, 9e–h

x + 3x + 6

x2 + 2x + 2

x2 − 14x + 56

x2 − 6x + 5

x − 7x + 14

x + 5x − 1

x2 − 2x + 4

(4x − 4)2 = 4

2 (x − 3)2 = 8

(x + 2)2 = 20

x2 − 6x + 8 = 0

x2 − 9x + 8 = 0

x2 − 2x − 32 = 0

6x2 + 48x + 24 = 0

x2 + 11x + 5 = 0

Solve: a

(x + 6)2 = 4

(x + 5)2 − 2 = 15

Solve: x2 + 18x + 32 = 0

Solve: a

Rewrite in the form (x − h)2 + k:

a Ex 3

For x2 − 3x − 6 = 0, what should be added and subtracted to the left side of the equation in order to complete the square?

Fluency Ex 1

2x2 − 12x − 32 = 0

Solve, leaving your answers in exact form: a

x2 + 24x + 5 = 0

x − 7x + 8 = 0

2x2 − 5x − 4 = 0

2x2 − 8x − 28 = 0

3x2 − 5x + 1 = 0

−4x2 − 20x + 8 = 0

Reasoning

10, 13

10–13

Explain how to ﬁnd the number to add to x2 + bx to make it a perfect square trinomial.

Explain the relationship between solving equations by completing the square and solving equations in the form x2 = k where k is a positive number.

Explain why we might use the method of completing the square to ﬁnd the exact solutions to some quadratic equations rather than by sketching a graph or factoring.

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Tyler attempted to solve the quadratic equation x2 − 4x + 3 = 0 using the completing the square method. Tyler’s solution is shown below. Identify the errors in the solution and explain what Tyler should have done instead. Tyler’s solution:

Complete the square

Simplify fractions and exponents

Factorise the perfect square part

Evaluate and simplify

Collect like terms

Take the square root of both sides

Add 4 to both sides

Evaluate 4 + 1 and 4 − 1

Problem-solving 14, 15 14

Given

14, 15

15–17

The revenue y (in millions of dollars) of a company x years after it ﬁrst started is modelled by: y = 12.5x2 − 64x + 135 By completing the square, use this equation to predict the number of years x after which the revenue of the company will reach $1167 million.

A model submarine performs a parabolic dive so that its depth, y, below the surface after covering a horizontal distance x was given by: y = x2 − 20x All distances are in metres. By completing the square:

Find the lowest depth of the submarine beneath the surface.

Find the horizontal distance(s) travelled when the submarine is 10 m beneath the surface.

The headlights of a car are shaped like a parabola, to direct all light from the bulb in the same direction. For a particular model of car, the bottom of the headlight can be modelled by the equation:

y (0, 54)

where x and y are measured in centimetres, and the y-intercept represents the top left point of the headlight. Determine the width and height of the headlight by ﬁrst rewriting the equation in turning point form. Explain your reasoning. 17

Consider the quadratic equation ax2 − c = 0. Given the following conditions, state whether the equation has zero, one, or two solutions. a

a and c are positive

a and c are negative

a and c have opposite signs

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5.05 The quadratic formula After this lesson, you will be able to… • understand what the quadratic formula is and how it is derived. • solve quadratic equations using the quadratic formula.

The quadratic formula The quadratic formula gives the solutions to any quadratic equation. It can be used to solve quadratic equations which cannot be solved by factorisation using rational numbers. Before we can use the quadratic formula, we have to rearrange the quadratic equation into the form ax2 + bx + c = 0, where a, b, and c are any number and a ≠ 0. Once the equation is in this form, the solutions are given by the quadratic formula:

a b c

The coeﬃcent of x2 The coeﬃcent of x The constant

Exploration We have seen that for an equation in the form ax2 + bx + c = 0, we can complete the square and then solve for x using inverse operations. What if instead of working with numbers, we solved for x in terms of variables? Here is the start of the process:

1. Explain what is happening in each step by ﬁlling in the blanks. 2. Complete the solution and solve for x in terms of a, b, and c. The result is the quadratic formula. 3. What would happen if we forgot to include both the positive and negative square root? 4. Recall that

is the axis of symmetry. Where do you see this in the quadratic

formula? Why do you think this is the axis of symmetry?

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Example 1 Solve 4x2 + 7x + 3 = 0 for x by using the quadratic formula or otherwise.

Create a strategy Identify the values of a, b, and c, then substitute them into the quadratic formula.

Apply the idea Write the quadratic formula

Substitute a = 4, b = 7, c = 3

Simplify the expression inside the square root

Take the positive square root

Evaluate the numerator

Simplify

Take the negative square root

Evaluate the numerator

Simplify

So the solutions are

and x = −1.

Reflect and check We can also get the same solutions by factorising the equation. To do this, multiply the leading coefficient (4) by the constant term (3): 4 × 3 = 12. Then, two numbers whose product is 12 and sum is the middle coeﬃcient (7). These numbers are 3 and 4. So we can: 4x2 + 7x + 3 = 0 2

Write the equation

4x + 4x + 3x + 3 = 0

Split 7x

4x(x + 1) + 3(x + 1) = 0

Factor each pair of terms

(4x + 3) (x + 1) = 0

Factor out x + 1 Equate the first factor to 0

Subtract 3 from both sides

Divide both sides by 4 x + 1 = 0 x = −1

So the solutions are

Equate the second factor to 0 Subtract 1 from both sides

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Example 2 Solve, leaving your answer in exact form. a 5x2 − 15x + 2 = 0

Create a strategy Identify the values of a, b, and c, then substitute them into the quadratic formula.

Apply the idea Write the equation

Write the quadratic formula

Substitute a = 5, b = −15, c = 2

Simplify the expression inside the square root

So the solutions are

and

b 10 − 6m + 2m2 = m2 + 8m + 9

Create a strategy Rearrange the equation into the form ax2 + bx + c = 0, and use the quadratic formula.

Apply the idea Write the equation

Subtract m2, 8m, 9 from both sides

Write the quadratic formula

Substitute a = 1, b = −14, c = 1

Simplify the expression inside the square root

Simplify the surd

Simplify the fraction

So the solutions are

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Idea summary For a quadratic equation of the form ax2 + bx + c = 0, the solutions are given by the quadratic formula:

where a, b, and c are any number and a ≠ 0.

5.05 Practice questions Understanding 1

Explorer 1, 2a–d, 4

Adventurer 1, 2b–e, 3, 4

Trailblazer 1, 2e–h, 3–5

Is each statement true or false? a

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

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

The equation 3x2 + 3x − 7 = 0 can be solved using the quadratic formula.

The quadratic formula will always give 2 unique solutions.

For the equation 5x2 − x − 2 = 0, we would set a = 5, b = 1, and c = −2. , ﬁnd the values of a, b and c for each quadratic equation:

For a

x2 + 7x + 10 = 0 2

x2 − 3x − 4 = 0

2x + 9x = 0

8x2 − 3x = 0

−8x2 + 3x = 0

−2x2 + 9x + 5 = 0

4x2 + 3x = 5

3x2 − 8x + 2 = 9x − 7

Jeremy is using the quadratic formula for the equation 9y2 = 8y. He has correctly identified that b = 8, what are the values of a and c?

To solve x2 − x = 2, Emma substitutes b = −1 into the quadratic formula. Find the values of a and c that she must substitute into the quadratic formula.

Given the quadratic equation x2 − 4x + k = 0, where k is a constant. If one of the solutions to a quadratic equation is , what is the other solution?

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Fluency Ex 1

6a–f, 7a–e, 8a–d

x2 + 5x + 6 = 0 2

4x2 − 10x + 4 = 0

2x2 + 7x + 3 = 0

4x2 − 17x − 15 = 0

5x2 + 7x − 6 = 0

5x2 − 13x − 6 = 0

5x2 + 21x − 20 = 0

3x − 11x − 20 = 0

Solve, rounding your answers to three decimal places: a

x2 + 3x − 6 = 0

x2 + 7x − 3 = 0

4x2 + 7x + 2 = 0

5x2 + 9x + 2 = 0

7.1x2 + 5.3x − 1.5 = 0

1.8x2 + 5.2x − 2.3 = 0

2x = 14

3x2 + x − 30 = 0

2x (x − 4) = 3x + 1

3x (x + 4) = −3x + 4

Solve, leaving your answers in exact form: a

x2 − 7x + 9 = 0

x2 − 5x − 2 = 0

−2x2 − 15x − 4 = 0

3x2 + 9x − 4 = 0

7x2 − 19x + 3 = 0

−5x2 − 15x + 3 = 0

5x = (x − 5) (3x + 3)

3n2 = 2n2 − 2n + 7

Reasoning 9

x2 − 5x + 6 = 0

2x + 6x − 8 = 0

Ex 2

6g–l, 7e–j, 8g–j

Solve using the quadratic formula: a

6e–h, 7c–h, 8c–g

9, 10a–c, 13

12 − 8m + 2m2 = m2 + 12m + 15

9, 10c–f, 13

10e–h, 11, 12

For x (x + 9) = −20: a

Solve it by the method of factorisation.

Check your solution by solving it using the quadratic formula.

Solve each quadratic equation using two different methods and comment on the eﬃciency of both methods: a

x2 − 10 = 15

x2 + 24x + 63 = 0 f

x2 − 7x + 6 = 0

25y2 = 36

4x2 + 5x + 1 = 0

x2 − 7x = 0

5k2 − 17k + 13 = 0 h

x2 + 9x + 20 = 0

For the equation in terms of x: mx2 − 3x − 5 = 0, m ≠ 0

248

Given that it has two unique solutions, determine the possible values of m.

There is one value of m that must be eliminated from the range of solutions found in the previous part. Determine the solution and explain why it must be eliminated.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For b2 − 4ac, the part of the quadratic formula under the square root:

What type of number can we not take the square root of?

The value of b2 − 4ac determines the number of solutions that a quadratic equation has. i

How many solutions does a quadratic equation have if b2 − 4ac > 0? Explain.

How many solutions does a quadratic equation have if b2 − 4ac = 0? Explain.

iii

How many solutions does a quadratic equation have if b2 − 4ac < 0? Explain.

Harriet solved a quadratic using the quadratic formula and got the following result:

What can be concluded about the solutions of the equation?

Problem-solving 15a, 16, 18

15a, 17, 19, 20

The solutions to a quadratic equation are x =

For each of the given solutions to a quadratic equation:

. Find the quadratic equation.

Find the values of a, b and c.

Write down the quadratic equation that has these solutions.

a 16

14, 15b, 18, 20, 21

An object is launched from a height of 80 m with an initial velocity of 107 m/s. After x seconds, its height, h, is given by: h = −10x2 + 107x + 80 Find the number of seconds, x, after which the object is 30 m above the ground. Round your answer to one decimal place.

An object is launched from a height of 8 m with an initial velocity of k m/s. After t seconds, its height, h, is given by: h = −10t2 + kt + 8 The object was in ﬂight for 4 seconds before reaching the ground. What was its initial velocity k?

5.05 The quadratic formula mathspace.co

249

Yuri is playing cricket and hits a six with an initial velocity of 101 metres per second, from a height of 3 metres. After x seconds, its height (in metres) is given by y = −10x2 + 101x + 3

Determine the maximum height the ball reaches.

LaDeana is in the crowd and caught the ball in the stands from a height of 15 metres above the ground. Determine the number of seconds after Yuri hits the ball that LaDeana caught it. Round your answer to two decimal places.

The stopping distance of a car when the brakes are applied can be modelled by the equation in m/s.

where y is the stopping distance in metres and x is the initial speed

Determine the speed the car was travelling if its stopping distance was 100 m, rounding your answer to the nearest hundredth.

It takes 399 m to stop when travelling at the speed limit of 70 m/s. If it takes Ray 496 m to stop, determine how fast over the speed limit Ray was driving.

A rectangular park with dimensions 3 metres by 10 metres has a uniform pathway that is x metres wide running around it. a

If the total area including the pathway is 120 m2, show that the area of the pathway is given by the equation 4x2 + 26x − 90 = 0.

Find the width of the pathway.

The diagram shows a line divided in two parts, one of length a and the other of length b. in the golden ratio if

a+b

250

. Show that if

, then x2 − x − 1 = 0.

Let

Find the solutions to this quadratic equation correct to three decimal places.

Which of the solutions is the golden ratio? Justify your answer.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5.06 The discriminant After this lesson, you will be able to… • define what the discriminant is. • identify how the discriminant determines the number and type of solutions. • analyse different cases of the discriminant: less than, equal to and greater than zero. • predict the nature of the roots of a quadratic equation based on the discriminant. • apply the knowledge of the discriminant to solve problems.

The discriminant We have many ways to solve quadratic equations. Before we try to solve, we may want to know how many solutions or roots a quadratic equation will have. For ax2 + bx + c = 0, and a ≠ 0, the quadratic formula is shown. We will look at what happens if the value under the square root sign, the discriminant, takes on different values.

b2 − 4ac

discriminant

The discriminant determines the number and type of real solutions a quadratic function will have. The symbol Δ or D are often used to represent the discriminant b2 − 4ac.

Interactive exploration

Discover this concept in action online

mathspace.co

For quadratic equations in the form ax2 + bx + c = 0, we can have an idea of what the parabola will look like by observing the sign of a and the value of the discriminant. y 4 3 2

a>0

−4 −3 −2 −1

−1

−2 −3

x 1

a<0

If b2 − 4ac < 0, then the value under the square root sign is negative. It is not possible to get real solutions if we square root a negative value. In this case, there are no real roots. If there are no real roots, the parabola will have no x-intercepts.

−4

5.06 The discriminant mathspace.co

251

y 4 3

If b2 − 4ac = 0, then the square root is 0. The

2 1 −4 −3 −2 −1

x 1

−1

−2

quadratic equation becomes

, which is the

equation for the axis of symmetry. In this case, the two roots are identical and the equation has only one real solution

−3 −4 y 4 3 2 1 −4 −3 −2 −1 −1

x 1

If b2 − 4ac > 0, then the square root will have two values, one for and one for In this case there are two distinct real roots.

−2 −3 −4

When the discriminant is positive, we can also determine if the roots will be rational or irrational. We sometimes call this the nature of the roots. If the discriminant is a square number, for example: 1, 4, 9, 16, 25 …, and the coefficients of the quadratic are rational, we will have two rational roots. The square root of a square number is rational. If the discriminant is any other positive number, we will end up with surds, so two irrational roots.

Example 1 For 4x2 − 6x + 7 = 0: a Find the value of the discriminant.

Create a strategy Substitute a = 4, b = −6, and c = 7 into Δ = b2 − 4ac.

Apply the idea Δ = b2 − 4ac 2

252

Write the formula

= (−6) − 4 × 4 × 7

Substitute a = 4, b = −6, c = 7

= 36 – 112

Evaluate the index and multiplication

= −76

Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b Using your answer from the previous part, determine the number of real solutions the equation has. A

Create a strategy If Δ is negative, the number of real solutions is 0.

Apply the idea Since Δ = −76, the number of real solutions is 0. The answer is option B.

Example 2 For x2 + 38x + 361 = 0 a Find the value of the discriminant.

Create a strategy Substitute a = 1, b = 38, and c = 361 into Δ = b2 − 4ac.

Apply the idea Δ = b2 − 4ac 2

Write the formula

= 38 − 4 × 1 × 361

Substitute a = 1, b = 38, c = 361

= 1444 – 1444

Evaluate the index and multiplication

Evaluate

b Using your answer from the previous part, detemine whether the solutions to the equation are rational or irrational.

Create a strategy The solutions are rational if Δ is a square number, while irrational if Δ is not a square number.

Apply the idea = 0, so there will be only one solution to x2 + 38x + 361 = 0 and it will be rational.

5.06 The discriminant mathspace.co

253

Example 3 Identify the graph of the quadratic y = ax2 + bx + c, where a > 0 and b2 − 4ac < 0. A y

x –10

–5

–10

–5

–10

C y

x –10

–5

x –10

–5

–10

Create a strategy Consider the signs of a and the discriminant b2 − 4ac.

Apply the idea Notice that a > 0, so it means that the parabola is concave up. Since b2 − 4ac < 0, then the parabola has no x-intercepts. This means that the graph of the quadratic is a concave up parabola and has no x-intercepts. So, the correct answer is Option A.

254

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 4 Find an expression for the discriminant of x2 + 5x + p − 5 = 0.

Create a strategy Substitute a = 1, b = 5, and c = ( p − 5) into Δ = b2 − 4ac.

Apply the idea Δ = b2 − 4ac

Write the formula

= 5 − 4 × 1 × ( p − 5) Substitute a = 1, b = 5, c = ( p − 5) = 25 − 4 ( p − 5)

Evaluate the index and multiplication

= 25 − 4p + 20

Expand the brackets

= 45 − 4p

Simplify

Reﬂect and check Without knowing the value of p, we cannot make any conclusions about the number or nature of the roots. We may be asked to solve for a value or range of values of p, which gives a certain number of solutions.

Idea summary Discriminant of a quadratic Δ: • b2 − 4ac < 0, 0 real solutions, the parabola has no x-intercepts • b2 − 4ac = 0, 1 real solution, 2 equal real roots, the parabola touches the x-axis once • b2 − 4ac > 0, 2 real solutions, 2 distinct real roots, the parabola passes through two different points on the x-axis. For a quadratic with rational coeﬃcients, if the discriminant is a square number like 0, 1, 4, 9, 16 then the roots will be rational. Otherwise, they will be irrational.

5.06 Practice questions Understanding

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

When graphing a particular parabola, Katrina used the quadratic formula and found that b2 − 4ac = −5. How many x-intercepts does the parabola have?

When graphing a particular parabola, Tony used the quadratic formula and found that b2 − 4ac = 0. How many x-intercepts does the parabola have? 5.06 The discriminant mathspace.co

255

The solutions of a quadratic equation are 9 and −9. What can be said about the value of b2 − 4ac?

Would the discriminant be positive, negative, or zero for these scenarios? a

No real solutions

One real solution

Two distinct solutions

Two distinct real roots

Two equal real roots

Non-real roots

No x-intercept

One x-intercept

Two x-intercepts 5a–d, 6a–e, 7a–d, 8a–d, 9a–c

Fluency 5

6 Ex 1

Ex 2

256

5e–h, 6g–l, 7e–h, 8e–h, 9d–f

For each equation in the form ax2 − bx + c = 0: i

State the values of a, b and c.

Calculate the discriminant.

x2 + 11x + 28 = 0 2

x2 − 5x + 6 = 0

4x − 7x − 15 = 0

−x2 − 5x − 8 = 0

16x2 − 24x + 9 = 0

−2x2 + 3x − 2 = 0

x2 + 5x +

= 0

For each quadratic equation: i

Calculate the discriminant.

State the number of real solutions for the equation.

x2 + 6x + 9 = 0

3x2 − 5x + 7 = 0

x2 − 4 = 0

−x2 − 8x − 16 = 0

−2x2 + 5x − 6 = 0

−x2 + 6x + 1 = 0

−x2 + 25 = 0

2x2 − 2x = x − 1

3x2 − 5x = 3x + 1

−3x2 − 5x = −4x2 − 2

k 7

5b–e, 6d–h, 7d–g, 8c–f, 9c–e

4x − x = x − 5

For each equation: i

Find the value of the discriminant.

Are the roots to the equation rational or irrational?

x2 + 22x + 121 = 0

x2 − 8x − 48 = 0

4x2 + 4x − 6 = 0

4x2 − 4x + 1 = 0

x2 − 16 = 0

9x2 − 42x = −49

2x2 − 9x = −2

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Ex 3

For each graph of a quadratic y = ax2 + bx + c, with discriminant Δ = b2 − 4ac: i

Is a > 0 or a < 0?

b 8

−8 −6 −4 −2 −2

x 2

−4

−6

−8

d 8

−4

−6

−8

f 8

−8 −6 −4 −2 −2

x 2

−8 −6 −4 −2 −2

x 2

−8 −6 −4 −2 −2

Is Δ > 0, Δ < 0 or Δ = 0?

2 −8 −6 −4 −2 −2

−4

−6

−8

5.06 The discriminant mathspace.co

257

h 8

2 −8 −6 −4 −2 −2

Ex 4

x 2

−8 −6 −4 −2 −2

−4

−6

−8

x 2

Find an expression for the discriminant: a

mx2 + 3x − 2 = 0

x2 − 4nx − 2 = 0

x + 2x + p − 6 = 0

−px2 + 4x + p = 0

mx2 − 2mx + m − 3 = 0

Reasoning

10, 12

11, 12, 14

11, 13, 14

Explain how the discriminant determines the nature of the roots to a quadratic equation and the three possibilities.

Determine the range of values of the constant k such that the equation 3x2 + kx + 12 = 0 has no real solutions. Justify your answer.

Use the discriminant and your knowledge of the graphs of y = 3x2 − x + 8 and y = −x2 + 2x − 4 to explain why the graphs do not intersect.

A quadratic equation with rational coeﬃcients has an axis of symmetry at x = 3 and an units from the axis of symmetry. x-intercept which is

Is this the only x-intercept? If so, explain why. If not, explain why not and state the other intercept.

What does this tell us about the value of the discriminant?

Is it possible for a quadratic equation to have one rational root and one irrational root? If so, give an example. If not, explain why not.

Problem-solving 15b, 16, 17, 19, 20a, 21 15

258

15a, 17, 19, 20b, 22

15c, 18, 20c, 22–24

Find the values of the variables that result in one real solution. a

n for x2 − 8nx + 1296 = 0

m for (m + 4) x2 + 2mx + 2 = 0

k for 16x2 + 8x + k = 0

For x2 + 12x + k + 10 = 0: a

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

What is the smallest possible integer value of k?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For what value(s) of k will the graph of y = 4x2 − 4x + k − 15 just touch the x-axis?

For what range of values of a will the quadratic equation ax2 − bx + c = 0, where b = 6 and c = 10, have two distinct real solutions?

Given that mx2 + 2x − 1 = 0 has two distinct solutions, determine the possible values of m.

For each equation: a

2x2 + 8x + k = 0

kx2 + 20x + 2 = 0

(k + 4) x + 10x + 3 = 0

Find the discriminant in terms of k.

Find the value of k so that the equation has one real solution.

iii

Find the values of k so that the equation has real solutions.

Find the values of k so that the equation has no real solutions.

Find the values of k so that the equation has real and distinct solutions.

For the graph of the quadratic, y = m − 9x − 3x2: a

Find the possible values of m, if the graph has no x-intercepts.

State the largest possible integer value of m.

A right-angled triangle with side lengths x units, x + p units and x + q units, ordered from shortest to longest is shown where no two sides of the triangle have the same length. a

Complete the statement:

p and q have lengths such that 0 < ⬚ < ⬚.

Write a quadratic equation that describes the relationship between the sides of the triangle in terms of x.

Find the discriminant of this quadratic equation.

State the number of real solutions to the quadratic equation. Explain your answer.

Find the value of x in terms of p when q = 2p.

x+q x+p

A farmer wants to build a rectangular fence around his field. The length of the fence is x metres and the width is x − 5 metres. The area of the field can be represented by the quadratic equation A = x2 − 5x. The farmer wants to know the possible dimensions of the field that will give him an area of at least 50 square metres. a

Write the quadratic equation representing the area of the ﬁeld when it is at least 50 square metres.

Find the discriminant of the quadratic equation.

Determine the number of real solutions to the quadratic equation and explain your answer.

What is the possible length of the field that will provide an area of at least 50 m2?

Considering your answer to part (c), explain why the real-world context is important in determining the number of solutions to this problem.

5.06 The discriminant mathspace.co

259

A water tank has the shape of a parabolic container with the equation y = −ax2 + h, where y represents the height of the water in the tank, x represents the horizontal distance from the centre of the tank, a is a positive constant and h is the maximum height of the tank. The tank is ﬁlled with water up to a height of k, where 0 < k < h. a

Write the quadratic equation that represents the height of the water’s surface in the tank in terms of x, a and k.

Find the discriminant of the quadratic equation representing the water surface in terms of a and k.

Determine the number of real solutions to the quadratic equation when the discriminant is positive. Explain what the solutions represent in context.

If the width of the water surface is w, ﬁnd the relationship between a, k, and w.

5.07 Optional: Further quadratic equations After this lesson, you will be able to… • use the structure of a quadratic equation to identify ways to rearrange it. • identify the advantages and disadvantages of each method for solving quadratic equations. • choose appropriate methods for solving quadratic equations based on the structure of the quadratic expression. • create and solve quadratic equations for real-world contexts.

Further quadratic equations We have looked at a number of ways to solve quadratic equations. These include: • Factorising the quadratic expression and then using the null factor law. In particular, when the quadratic expression is non-monic, we can use techniques for non-monic quadratic trinomials. • Completing the square and then taking the square root of both sides of the equation. • Using the quadratic formula. Is the equation in standard form?

Yes

Is it easily factorisable? No

No Can we get to Yes a(x − h)2 = k easily?

Complete the square

Yes

Do I want to know the turning point as well?

No No Rearrange to standard form 260

Quadratic formula

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Yes

Factorise

Example 1 Answer these questions: a Solve 5x2 + 22x + 8 = 0

Create a strategy We can try to factorise the equation and use the null factor law, but if it takes too long or doesn’t work, then we can use the quadratic formula.

Apply the idea 5x2 + 22x + 8 = 0 2

Write the equation

5x + 2x + 20x + 8 = 0

Split 22x

x(5x + 2) + 4(5x + 2) = 0

Factor each pair of terms

(5x + 2) (x + 4) = 0

Factor out 5x + 2 Equate the ﬁrst factor to 0

Subtract 2 from both sides

Divide both sides by 5

x+4=0 x = −4 So the solutions are

Equate the second factor to 0 Subtract 4 from both sides

and x = −4.

Reﬂect and check As an alternative approach to ﬁnding the solutions, we can use the quadratic formula. Write the quadratic formula

Substitute a = 5, b = 22, c = 8

Simplify the expression inside the square root Take the positive square root

Evaluate and simplify

Take the negative square root

Evaluate and simplify

So we get the same answers:

and x = −4

This involves more numerical operations, but will always work.

5.07 Optional: Further quadratic equations mathspace.co

261

b Find the x-intercepts and the turning point for the quadratic y = 3x2 + 5x + 2.

Create a strategy Use the flow chart to determine a good method to solve. Because we also need to identify the turning point, completing the square is a good option.

Apply the idea To complete the square on the equation y = 3x2 + 5x + 2, we follow these steps: Write the equation

Subtract 2 from both sides

Factor out the coefficient of x2

Take half the coefficient of x, square it and add both inside and outside the brackets. Half of

is . Add

Complete the square

Distribute 3

Square

Remove common factor of 3

Add 2 to both sides

Simplify

This means that the turning point is

To find the x-intercepts, this is the same as finding the solutions to 3x2 + 5x + 2 = 0. Which is the same as solving

Write the equation

Add

Add to both sides

Divide both sides by 3

Take the positive and negative square root

Subtract 262

from both sides

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

So x can equal either

So the x-intercepts occur at x = −1 and

Reﬂect and check As an alternative approach to ﬁnding the solutions, we can factorise. 3x2 + 5x + 2 = 0

Write the equation

(3x + 2) (x + 1) = 0

Factorise Set the ﬁrst factor to 0 Solve for x

x+1=0 x = −1

Set the second factor to 0 Solve for x and x = −1

So we get the same answers:

c Solve x2 − 7x + 9 = 0

Create a strategy We cannot factorise because no two numbers add to −7 and multiply to 9. We can either use the quadratic formula or complete the square. Let’s use the quadratic formula.

Apply the idea Write the quadratic formula

Substitute a = 1, b = −7, c = 9

Evaluate

So the solutions are

and

5.07 Optional: Further quadratic equations mathspace.co

263

Example 2 Solve: a

Create a strategy Rearrange the equation into the form ax2 + bx + c = 0, then use the quadratic formula.

Apply the idea

Rewrite the equation

Multiply both sides by 5x

Subtract 5x2 from both sides

Multiply both sides by −1 and rearrange Write the quadratic formula

Substitute a = 5, b = −11, c = −12

Simplify the expression inside the square root

Take the positive square root

Evaluate and simplify

Take the negative square root

Evaluate and simplify

So the solutions are x = 3 and

Create a strategy Rearrange the equation into the form ax2 + bx + c = 0, then factorise the equation and use the null factor law. Rewrite the equation

Multiply both sides by x (x + 1)

Simplify 7x (x + 1)

Simplify the right-hand side

Combine like terms and rearrange

Factorise

264

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Equate the ﬁrst factor to 0

Subtract 1 from both sides

Divide both sides by 3 Equate the ﬁrst factor to 0

Add 1 to both sides

Divide both sides by 2

So the solutions are

and

Example 3 A parabola is described by the function y = x2 + 4x − 5. a Find the x-values of the x-intercepts of this parabola.

Create a strategy Substitute y = 0 into the equation and then solve for x.

Apply the idea 0 = x2 + 4x − 5

Substitute y = 0

0 = (x + 5) (x − 1)

Factorise

x+5=0

Equate the first factor to 0

x = −5

Subtract 5 from both sides

x–1=0 x=1

Equate the second factor to 0 Add 1 to both sides

So the x-intercepts are x = −5 and x = 1.

b What is the y-value of the y-intercept for this parabola?

Create a strategy Substitute x = 0 into the equation and then solve for y.

Apply the idea y = 02 + 4 × 0 – 5 = −5

Substitute x = 0 Evaluate

5.07 Optional: Further quadratic equations mathspace.co

265

c What is the equation of the vertical axis of symmetry for this parabola?

Create a strategy To ﬁnd the x-value of the vertex or turning point, use the formula

Apply the idea

Substitute a = 1, b = 4 Evaluate

Reﬂect and check We can ﬁnd the x-value of the vertex by calculating the average of the two x-intercepts in part (a).

Get the average Evaluate

d Find the y-coordinate of the vertex for this parabola.

Create a strategy Substitute the value x = −2 found in part (c) into the function.

Apply the idea y = (−2)2 + 4 × (−2) – 5 Substitute x = −2 = −9 Evaluate e Plot the graph of the equation.

Create a strategy Plot the vertex or turning point and intercepts found in the previous parts, then sketch the parabola through these points.

Apply the idea 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

266

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

y x 1

2 3

Idea summary We can solve quadratic equations in many ways. There are beneﬁts and drawbacks to each strategy. • Factorising the quadratic expression and then using the null factor law. • Completing the square then using inverse operations. • Using the quadratic formula. Once we have the solutions, we can plot the x-intercepts on the graph.

5.07 Practice questions Explorer

Understanding 1

1–4

Trailblazer 1–4

Are the following statements true or false? a

Completing the square is a method used to solve linear equations.

Factorising is a method used to ﬁnd the roots of a quadratic equation by expressing it as a product of two binomials.

The quadratic formula can be used to ﬁnd real solutions of any quadratic equation.

A quadratic equation always has exactly two real solutions.

When completing the square for the quadratic equation x2 + 6x + 11 = 0, what value should be added and subtracted to the equation?

Given that one factor of the quadratic equation x2 − 5x + 6 = 0 is (x − 2), what is the other factor? A

(x + 2)

5a–g, 6a–d, 7, 8, 9ab, 10a

Fluency 5

(x + 3)

(x − 4)

(x − 3)

Which of the following is the correct quadratic formula for solving a quadratic equation of the form ax2 + bx + c = 0? A

Ex 1

Adventurer 1–4

5b–i, 6c–g, 9bc, 10cd

5e–l, 6f–j, 9cd, 10bd, 11

Solve: a

5x2 + 45x = 0

17x2 = 3x

81x2 − 49 = 0

4x2 − 32x + 60 = 0

−2x2 − 3x + 5 = 0

5x2 − 14x + 8 = 0

12 − 7b − 12b = 0

4x2 − 3x − 10 = 0

4x2 − 13x + 2 = 0

4k2 = −15 + 17k

3x2 = 2 − 6x

4x = 2 + 8x

5.07 Optional: Further quadratic equations mathspace.co

267

Ex 2

Ex 3

Solve: a

(3x2 + 11x + 10) (2x2 + 11x + 12) = 0

The equation 4x2 + kx + 36 = 0 has one solution x = −3. Find the value of the coeﬃcient k.

If you substitute x = 6 into ax2 − 6x + 144, what must be the value of a for the expression to be equal to 0?

For each equation:

What is the y-intercept?

What are the x-intercept(s)?

iii

What is the equation of the axis of symmetry?

What are the coordinates of the turning point?

Sketch the graph of the parabola.

y = x2 + 2x

y = 2x2 − 8

y = −2x2 + 12x − 10

y = 9x2 − 54x + 72

For each equation: i

What is the y-intercept?

What are the coordinates of the turning point?

iii

What are the x-intercepts?

Sketch the graph of the parabola.

y = x2 + 3x − 2

c 11

268

y = 2x − 9x + 2

y = x2 − 62x + 945

y = 3x2 − 25x + 29

A parabola is of the form y = (x − h)2 + k. It has x-intercepts at (9, 0) and (10, 0). a

Determine the axis of symmetry of the curve.

Find the equation of the curve in the form y = (x − h)2 + k.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning 12

12–14, 16

12–15

13, 16, 17

Jeric is solving the quadratic equation x2 − 6x + 8 = 0 using factorising. Is there an error in his solution? If so, identify and correct it. If not, ﬁll in the reasoning for his steps. x2 − 6x + 8 = (x − 2) (x − 4) x = 2, 4

⬚ ⬚

Solve the quadratic equation x2 + 2x − 8 = 0 and verify the solution by substitution. Explain your process.

Three students, Alice, Grace and Bob, are solving the quadratic equation x2 − 7x + 10 = 0. Alice uses factorising, Grace completes the square and Bob uses the quadratic formula. Will they arrive at the same solutions? Explain your reasoning.

Sarah attempted to solve the quadratic equation x2 + 15x + 54 = 0 using the factorising method. Identify the errors in Sarah’s solution and explain what Sarah should have done instead. Sarah’s solution:

x2 + 15x + 54 = 0

Given

x + 15x + 54 = (x + 9) (x + 6) Factoring the quadratic equation 16

x = 9, 6

Finding the solutions

Tom attempted to solve the quadratic equation x2 − 2x − 3 = 0 using the quadratic formula. Identify the errors in Tom’s solution and explain what Tom should have done instead. Tom’s solution:

Applying the quadratic formula

Substituting the values

Simplifying the expression

Simplifying the surd

Finding the solutions

5.07 Optional: Further quadratic equations mathspace.co

269

We can solve some quartic equations using substitution. a

The equation x4 − 13x2 + 36 = 0 is reducible to a quadratic by substituting u for x2.

x4 − 13x2 + 36 = 0 2

u − 13u + 36 = 0

Original quadratic Let u = x2

Continue to solve for x, showing all your work. How many solutions do you get? b

Apply the same technique to solve x4 − x2 − 20 = 0. How many solutions do you get? Explain.

Problem-solving 18, 19, 22

19, 20, 23

19, 21, 23, 24

A rectangle has one side 3 cm longer than the other side, and its area is 108 cm2. Determine the length of the shorter side of the rectangle.

The length of a rectangular mat is twice its width and the diagonal length is 55 cm. Find the width of the mat, rounded to the nearest centimetre.

Mae throws a stick vertically upwards. After t seconds, its height h metres above the ground is given by the formula h = 25t − 5t2.

At what time(s) will the stick be 30 m above the ground?

How long does the stick take to hit the ground?

Can the stick ever reach a height of 36 m? Explain.

A t-shirt is ﬁred straight up from a t-shirt cannon at ground level. After t seconds, its height above the ground is h metres, where h = −16t2 + 48t. a

For what values of t is the t-shirt 11 m above the ground?

How long was the t-shirt at least 11 m above the ground?

How long was the t-shirt at least 35 m above the ground?

Let m and n be the solutions to the quadratic equation value of m2 + n2, rounded to the nearest whole number.

270

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. Determine the

An interplanetary freight transport company has won a contract to supply the space station orbiting Mars. They will be shipping stackable containers, each carrying a fuel module and a water module, that must meet certain dimension restrictions. The design engineers have produced a sketch for the modules and container. Each component has a square cross-sectional area of x2 cm2, and the heights of both modules sum to the height of the container.

17 cm

Water Module

663 cm x

6426 cm x2

Fuel Module

Container

Write an equation that equates the height of the container and the sum of the heights of the modules.

Find the possible values of x.

Find the tallest possible height of the container. Give your answer to two decimal places.

Investigation: Derive the quadratic formula

Investigate online

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Investigation: Use bisection to solve quadratics

Investigate online

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5.07 Optional: Further quadratic equations mathspace.co

271

5 Chapter review 1

Are the graphs of these quadratic equations concave up or concave down? a

y = −x (x + 2) + 5

y = 9x2 + 4x

55 = x2 − 45

36 = (n − 2)2

Solve: a

y = x2 + 6x − 9

x2 = 81

A company’s revenue, R (in dollars), after t months is described by the equation: R = − (t − 12)2 + 36

272

x (2x − 16) = 0

Does the revenue reach a maximum or minimum point?

When will the revenue be zero?

For each equation: i

Describe the transformations from y = x2.

What are the coordinates of the turning point?

iii

What are the coordinates of the y-intercept?

Sketch the graph of the parabola.

y = 2 (x − 4)2

y = − (x + 1)2 + 7

For the quadratic equation y = (x − 1) (x + 3): a

Find the y-intercept.

Find the x-intercepts.

Write the equation of the axis of symmetry.

Find the coordinates of the turning point.

Sketch the graph of the parabola.

For the quadratic equation y = x2 + 4x − 1: a

Write in the form y = a (x−h)2 + k.

Find the coordinates of the turning point.

Sketch the graph of the parabola.

Solve by factorisation: a

x2 − 12x = 0

x2 + 9x − 10 = 0

−x2 + 24x − 144 = 0

3x2 − 34x + 75 = 0

For each graph: i

What are the solutions of the related quadratic equation?

Write the equation in factored form.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

−4 −3 −2 −1 −2 −4

4 2 −6 −4 −2 −2

x 2 4 6 8 10 12

−4 −6 x 1

−8

−10 −12

Solve for y = 0 by completing the square: a

18 16 14 12 10 8 6 4 2

y = x2 − 14x + 40

y = −x2 + 6x − 9

y = 2x2 + 9x + 10

A rocket is launched into the sky and its altitude in metres above the ground after travelling a horizontal distance x metres is given by y = −x2 + 24x By completing the square:

Find the maximum altitude of the rocket above the ground.

Find the horizontal distance(s) travelled when the rocket is 85 m above the ground.

Solve by using the quadratic formula, leaving your answer in exact form: a

3x2 − 11x − 20 = 0

12 − 8m + 2m2 = m2 + 12m + 15

The solutions to a quadratic equation are:

Write down the quadratic equation that has these solutions.

What can be concluded about the solutions of the equation?

For each quadratic equation: i

Calculate the discriminant.

State the number of real solutions for the equation.

x2 − 4 = 0

c 14

x2 + 6x + 9 = 0

−3x2 − 5x = −4x2 − 2

For each equation: i

Find the value of the discriminant.

Are the roots to the equation rational, irrational or are there no real roots?

x2 + 22x + 121 = 0

2x2 + 2x − 9 = 0

5x2 + 4x + 8 = 0

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For each graph of a quadratic y = ax2 + bx + c, with discriminant Δ = b2 − 4ac: i

Is a > 0 or a < 0?

b 8

−8 −6 −4 −2 −2

x 2

−4

−6

−8

d 8

x 2

−8 −6 −4 −2 −2

Is Δ > 0, Δ < 0 or Δ = 0?

−8 −6 −4 −2 −2

−4

−6

−8

For the equation in terms of x: mx2 − 3x + 5 = 0 Given that it has two distinct solutions, determine the possible values of m.

(Optional) For the equation: y = 3x2 + 18x + 24 a

What are the coordinates of the y-intercept?

What are the coordinates of the x-intercept(s)?

What is the equation of the axis of symmetry?

What are the coordinates of the turning point?

Sketch the graph of the parabola.

(Optional) Let a and b be the solutions to the quadratic equation value of a2 + b2.

(Optional) A rock is thrown vertically upwards. After t seconds, its height h metres above the ground is given by the formula h = 3 + 14t − 5t2.

274

How long does the rock takes to hit the ground?

Can the rock ever reach a height of 13 m?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

. Find the

“Mathematics, rightly viewed, possesses not only truth but supreme beauty—a beauty cold and austere, like that of sculpture.” –Bertrand Russell

Big ideas Solutions to simultaneous equations and inequalities can be graphically and algebraically determined through various methods. There is often a more efficient method that can be used based on the structure of the equation. Curriculum Outcomes VC2M10A08 solve linear inequalities and graph their solutions on a number line VC2M10A09 solve simultaneous linear equations, using algebraic and graphical techniques including using digital tools VC2M10A16 solve equations graphically or using systematic numerical guess-check-and-refine with digital tools, with consideration of whether all solutions have been found VC2M10AA09 solve linear and non-linear simultaneous equations using graphing or systematic guess-check-and-refine with digital tools VC2M10AA10 experiment with functions and relations using digital tools, making and testing conjectures and generalising emerging patterns

6 Intersections and simultaneous equations Chapter outline 6.01 6.02 6.03 6.04 6.05 6.06

Solve graphically Simultaneous inequalities Solve using substitution Solve using elimination Applications of simultaneous equations Investigation: Intersections of curves Optional: Intersections of curves Chapter 6 review

278 286 296 304 311

324

Businesses use optimisation to improve efficiencies. In 1987 a large airline removed a single olive from each of its inflight salads, reducing costs by $ 55,000 a year!

6.01

Solve graphically After this lesson, you will be able to… • plot pairs of linear equations on a Cartesian plane. • identify the point of intersection of two graphs as the solution to the simultaneous equations. • recognise the significance of parallel lines and their impact on solutions. • apply graphical methods to real-life equations to find their solutions. • draw conclusions about the presence or absence of simultaneous solutions based on the properties of the graphed lines.

Solve graphically To ﬁnd the solution to a pair of linear equations, which we call simultaneous equations, we can graph both lines on a single Cartesian plane and find any points of intersection.

Interactive exploration Discover this concept in action online

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If two lines are parallel and are not the same line, they will never intersect.

Example 1 Consider the following linear equations: y = 5x − 7 y = −x + 5 a Sketch the two lines representing these equations on a Cartesian plane.

Create a strategy Use the gradient and y-intercept from the equations to graph the lines.

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Apply the idea 8

For the line y = 5x − 7, m = 5 and c = −7. Plot the y-intercept of (0, −7).

6 4 2

−5 −4 −3 −2 −1 −2

Use the gradient of 5 to find another point, by moving 1 unit to the right and 5 units upwards, which is the point (1, −2). Draw a line through these two points.

−4 −6 −8

y = 5x − 7

For the line y = −x + 5, m = −1 and c = 5. Plot the y-intercept of (0, 5).

8 6

y = −x +5

4 2

−5 −4 −3 −2 −1 −2

1 2 3 4 5

−4

Use the gradient of −1 to find another point, by moving 1 unit to the right and 1 unit downwards, which is the point (1, 4). Draw a line through these two points.

y = 5x −7

−6 −8

b What is the point of intersection for the two graphs?

Create a strategy Use the graphs to identify the point of intersection.

Apply the idea y

The point where the two lines meet is at (2, 3). This is the point of intersection.

8 y = -x +5

The intersection point (2, 3) satisfies both equations, y = 5x – 7 and y = –x + 5.

4 2 -5 -4 -3 -2 -1 -2 -4

x 1 2 3 4 5

When x = 2: • The first equation is y = 5 x 2 – 7 = 3. • The second equation is y = –2 + 5 = 3.

y = 5x -7

-6 -8

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Idea summary We can solve any pair of simultaneous equations by plotting their graphs and ﬁnding the point of intersection. If two different lines have the same gradient, then they will be parallel and will have no point of intersection. So their equations will have no simultaneous solution.

6.01 Practice questions Understanding

Explorer 1–3

Adventurer

Trailblazer

1–3

What does the point of intersection of two straight lines on a graph represent?

If a pair of linear equations has no solutions, what would this look like graphically?

State the number of solutions for the simultaneous equations: a

6 5 4 3 2 1

x 1 2 3 4 5 6

−6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

6 5 4 3 2 1

6 5 4 3 2 1 −6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

280

6 5 4 3 2 1

x 1 2 3 4 5 6

5 4 3 2 1

x 1 2 3 4 5 6

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

1 2 3 4 5

3 2 x 1 2 3 4 5 6

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1 −4 −3 −2 −1

−1

−2 −3 −4

x 1

4b–d, 5ab, 6a–d, 8, 9, 11, 12ab, 13ab

Fluency 4

4ace, 5bc, 6e–h, 7, 9, 11, 12ab, 13ab

4aef, 5bc, 6i–l, 7, 10, 11, 12cd, 13cd

Identify the solution to the simultaneous equations: a

y 10 8 6 4 2

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

10 8 6 4 2

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

5 4 3 2 1

4 3 2 1 −3 −2 −1−1

x 2 4 6 8 10

x 1 2 3 4 5 6 7

−2 −3 −4 −5 −6 −7 −8 −9

2 4 6 8 10

−1−1 −2 −3 −4 −5 −6 −7

10 8 6 4 2

x 2 4 6 8 10

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

f 4

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

1 −9−8−7−6−5−4−3−2−1 −1

x 1 2 3 4

−2 −3 −4

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Use the graph to solve the simultaneous equations: y

5 4 3 2 1 −2

−1

x − 2y = 4

y=x−2 1

−1 −2 −3 −4 −5

x 8

y = −8 + 2x

y=x−2

y = −8 + 2x

y = −8 + 2x x − 2y = 4 x − 2y = 4 Ex 1

For each pair of linear equations: i

Sketch the lines on the same Cartesian plane.

Find the solution.

y = 4x − 4

y=x+9

y = 8 − 2x y = −x – 9 e

y = 3x + 6

y = −x + 2 i

y = 1.5x

y = −x − 3

y = −1 j

y = 2.5x − 4

y=x+3 y = 4x − 3

2x − 4y = −8

y = 3x + 2

y = 4 − 3x y = 7 − 2x k

y = 4 − 0.5x y = 12 − 1.5x 7

y = −x + 3 −4x + 2y = −8

A band plans to record a demo at a local studio. The cost of renting studio A is $250 plus $50 per hour. The cost of renting studio B is $50 plus $100 per hour. The cost, y, in dollars of renting the studios for x hours can be modelled by the simultaneous equations: • Studio A: y = 50x + 250 • Studio B: y = 100x + 50

Sketch the two lines on the same Cartesian plane.

What coordinates satisfy both equations?

What do the coordinates from part (b) mean in this context?

Michael plans to start taking an aerobics class. Non-members pay $4 per class. Members pay a $10 one-time fee, but only have to pay $2 per class. The cost, y, of taking x classes can be modelled by the simultaneous equations: • Non-members: y = 4x • Members: y = 2x + 10

282

Sketch the two lines on the same Cartesian plane.

What coordinates satisfy both equations?

What do the coordinates from part (b) mean in this context?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A company’s cost is represented by C = 0.4x + 2015, and its revenue is represented by R = 3x, where x represents the number of items produced. y 2790 2325 1860

1395 930 465

x 155 310 465 620 775 930 1085 1240

What coordinates satisfy both equations?

What do the coordinates from part (a) mean in this context?

The cost of manufacturing toys, C, is related to the number of toys produced, n, by the formula C = 400 + 2n. The revenue, R, made from selling n toys is given by R = 4n. a

Sketch the graphs of cost and revenue on the same Cartesian plane.

What coordinates satisfy both equations?

What do the coordinates from part (a) mean in this context?

Laura is going on maternity leave. Her living expenses are represented by y = 1200x. Her bank account balance is represented by y = 10 000 + 772.60x. For both equations, x is time in weeks. a

Sketch both equations on the same Cartesian plane using technology.

Estimate the solution to one decimal place.

What does the point of intersection of the two lines mean in this context?

A pair of linear equations has no solutions. One of the equations is y = −3x − 2. Determine if the following can be the other equation: a

y = −3x − 3

y = −3x + 2

A pair of simultaneous linear equations has inﬁnite solutions. One of the equations is y = 4x − 5. Determine if the following can be the other equation: a

y = 8x − 10

Reasoning

b 14, 15, 17, 18

y = 4x + 5

c 15–18

3y = 12x − 15

2y = 8x − 10

16–19

Is it possible to have a pair of linear equations with exactly two solutions? Explain why or why not.

In a pair of linear equations, one equation is a scalar multiple of the other. For example, 2x + 4y = 6 and x + 2y = 3. What can you say about the solutions of the simultaneous equations?

6.01 Solve graphically mathspace.co

283

The graph of y = 3x + 2 is shown.

10 8 6 4 2

If a second line intersects this line at the point (0, 2), can we determine the equation of the second line? Explain your answer.

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

The graph of y = 5x + 3 is shown.

2 4 6 8 10

A second line, y = mx + c, intersects this line at the one point (0, 3).

What is the value of c? Explain your answer.

What value of m is not possible? Explain your answer.

1 −2

−1

x 1

−2

Consider the simultaneous equations: y = −2x − 5 y = −2x + 4 Without graphing, is there a solution? Explain your answer.

A company needs to ﬁnd the best way to distribute goods from multiple suppliers to multiple customers. The distribution amounts must satisfy the simultaneous linear equations: y = 3x + 15 y = 9x − 10 Explain the implications of having non-integer solutions in this context and discuss how the company might handle such solutions.

Problem-solving 20, 21 20

284

20, 21, 23

21–24

A rectangle is to be 3 m longer than it is wide, with a total perimeter of 18 m. a

Let y represent the length of the rectangle and x represent the width. Create two equations that represent this information.

Sketch the two lines on the same Cartesian plane.

Find the length and width of the rectangle.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

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

Write a pair of linear equations to represent the situation.

Graph the simultaneous equations.

How old are Rochelle and Mohamad?

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 6 pieces of work to do for another subject. a

Write a pair of linear equations to represent the situation.

Graph the simultaneous equations.

Solve the simultaneous equations and interpret the result in terms of this context.

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

20 18 16 14 12 10 8 6 4 2

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

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. 25 20 15 10 5 −40 −30 −20 −10 −5 −10 y1 −15 −20 y2 −25

x 10 20 30

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6.02

Simultaneous inequalities

After this lesson, you will be able to… • define simultaneous inequalities. • find solutions to simultaneous inequalities and represent them graphically. • interpret and solve problems involving simultaneous inequalities in given contexts. • apply the concepts of simultaneous inequalities to real-world scenarios.

Simultaneous inequalities Simultaneous inequalities is a set of inequalities which have the same variables. The solution to simultaneous inequalities is the set containing any ordered pair that makes all the inequalities true. A solution can also be represented graphically as the region of the plane that satisﬁes all inequalities. 4

1 −4 −3 −2 −1

−11

x 1

−2 2

−4 −3 −2 −1

−1

x 1

−2

−3 3

−3

−4 4

−4

The solutions to simultaneous inequalities in a given context are viable if the solutions make sense in the context, and are non-viable if they do not make sense.

286

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Example 1 y

Select the inequalities that describe the shaded region. 12

A x + y ≤ 9, x ≥ 3 B x + y > 9, x < 3

C x + y > 9, x > 3 D x + y ≤ 9, x ≤ 3

6 3 x −3

−3

Create a strategy If the lines are dotted and the shaded region is above or on the right of each line, then the inequality symbol is >.

Apply the idea The answer is option C.

Example 2 Consider the lines y = 4x − 2 and y = 3x − 3. a Find the x-coordinate of the point at which the two lines intersect.

Create a strategy

Apply the idea 4x − 2 = 3x – 3 Form an equation by substitution of y

Form an equation by substitution of y and solve.

x − 2 = −3

Subtract −3x from both sides

x = −1

Add 2 to both sides

b Find the y-coordinate of the point of intersection.

Create a strategy

Apply the idea

Substitute x = −1 into one of the equations.

We can use y = 4x − 2. y = 4x – 2 = 4 × (−1) – 2

Substitute x = −1

= −6

Evaluate

6.02 Simultaneous inequalities mathspace.co

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c Graph the region that satisfies both y < 4x − 2 and y < 3x − 3.

Create a strategy • Plot the intersection point and the y-intercepts of each inequality. • The inequality symbols are both < so the lines must be dotted. • Since it is an “and” inequality, shade the region which satisfies both inequalities.

Apply the idea We found from parts (a) and (b) that the intersection point is at (−1, −6).

y 1 x −2

−1

1 −1

y < 4x − 2 has a y-intercept at (0, −2), while y < 3x − 3 has a y-intercept at (0, −3). The shaded region must be below both lines.

−2 (0, −2) −3 (0, −3) −4 4 −5 − 5 −6 − 6 (−1, −6) −7 − 7

Reﬂect and check We can perform a region test. This is where we take a point such as (0,0) and test if it satisfies the inequality or not by plotting it on the Cartesian plane and checking to see if it falls in the shaded region. We can see that the origin, point (0, 0), does not fall in the region, so (0,0) does not satisfy the inequality.

Example 3 Applicants for a particular university are asked to sit a quantitative reasoning test and verbal reasoning test. Successful applicants must obtain a minimum score of 14 on a quantitative reasoning test and a minimum combined score of 29 for both tests. a Write simultaneous inequalities for this scenario, where x represents the quantitative reasoning test score and y represents the verbal reasoning test score.

Create a strategy

Apply the idea

The quantitative reasoning will show an inequality with 14 as the lowest possible solution. The total of the two scores must sum to 29 or more. 288

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b Sketch the graph of the simultaneous inequalities.

Create a strategy

Apply the idea

Graph each inequality on the same Cartesian plane and shade the region that satisfies both.

y 30 25 20 15 10 5 x 5

c Does the solution

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

Apply the idea No. This point satisfies both inequalities, but a test score is likely a positive integer or a simple fraction such as

or .

Idea summary A set of simultaneous inequalities is a set of two or more inequalities with the same variables. The solutions of simultaneous inequalities are the values of the variables that make all the inequalities true. It is shown as a shaded region on the Cartesian plane.

6.02 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

The graph shows the region for the simultaneous inequalities: y > − 3x + 6 y>4 Determine whether each point is located in the region. a

(−3, 2)

(2, 5)

(−1, −6)

(3, 2)

(4, 4)

Trailblazer 1–4 7 6 5 4 3 2 1 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

x 1 2 3 4 5 6 7

6.02 Simultaneous inequalities mathspace.co

289

The graph shows the region for the simultaneous inequalities: x+y≤4

7 6 5 4 3 2 1

x≥1 y≥0 What are the coordinates of the three vertices of the region?

Which ordered pair satisﬁes the simultaneous inequalities?

−3 −2 −1−1 −2 −3 −4 −5 −6 −7

1 2 3 4 5 6 7

2x − y > −6 2x + 2y < 0 A 4

(2, 6)

(1, −3)

(−6, 1)

The region can only include individual points and cannot include lines.

The solution set to a linear inequality is a region of the graph and can include an infinite number of points.

All the points on the boundary line of a linear inequality are included in the solution if the inequality is “less than or equal to” or “greater than or equal to”. 5ab, 6, 7ab, 8, 10, 11, 13, 14 5, 7ac, 8, 10, 11, 13, 15

5, 7ad, 9, 10, 12, 15

Write the simultaneous inequalities that describe the shaded region: a

5 4 3 2 1 −5 5 −4 4 −3 3 −2 2− −11 −11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5

3 2 1

x −4 − 4 − −3 3 −2 2 −1 −1

1 2 3 4 5

−11 −

−2 − 2 −3 −3 −4 − 4

3 2 1 −7 −6 −5 5− −4 4− −3 3− −2 2− −11 −11 −2 − 2 −3 3 −4 4 −5 5 −6 −7

290

Is the statement true or false?

Fluency Ex 1

(−3, 6)

4 3

1 2 3

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1 −4 −3 −2 −1

−1

−2 −3 −4

Ex 2

Ex 3

For 2x − y < 4 and 6x + 3y > 0: a

Sketch the boundary lines, 2x − y = 4 and 6x + 3y = 0, on the same set of axes.

Does the point (2, 2) satisfy the simultaneous inequalities?

Graph the solution set to the simultaneous inequalities 2x − y < 4 and 6x + 3y > 0.

For each pair of simultaneous inequalities: i

Find the x-coordinate of the point at which the two lines intersect.

Find the y-coordinate of the point of intersection.

iii

Graph the region.

x ≤ 5 and y < 3

y ≤ 3x − 4 and y > −4x + 1

3x + y > 5 and 3x + y < 7

y < 2x − 4 and y < −3x + 6

For y > x − 5, y > −x − 5, and y < 0: a

Graph the lines y = x − 5, y = −x − 5, and y = 0 on the same set of axes.

Does the point (1, −2) satisfy the simultaneous inequalities?

Shade the region described by the inequalities y > x − 5, y > −x − 5, and y < 0.

Calculate the area of the region.

For x > −2, y > 1 and y ≤ −0.75x + 2.5: a

Graph the region which satisfies the inequalities.

Calculate the area of the region.

Applicants for a particular university are asked to sit a numeracy test and verbal reasoning test. Successful applicants must obtain a minimum score of 17 on the numeracy test and a minimum combined score of 37 for both tests. Let x and y represent an applicant’s score on the numeracy and verbal reasoning test respectively. a

It’s not possible to get a negative score therefore y ≥ 0 is one inequality. Write two more inequalities from the information in terms of x and y.

Graph the solution to the simultaneous inequalities.

Determine whether each point represents a score that would make the applicant successful. i

(14, 16)

(33, 18)

iii

(20, 13)

(17, 22)

If an applicant obtains a score of 24 in the numeracy test, ﬁnd the minimum integer score they need to obtain in the verbal reasoning test to be successful.

An airline is checking passengers into two flights, A and B, simultaneously. Due to passenger numbers, there must be at least 10 staff at check-in for flight A and at least 7 staff at check-in for flight B. Since there must be staff on hand for other services, the airline can only allocate at most 23 staff for check-in of both flights. Let x and y represent the number of staff attending check-in of flights A and B respectively. a

Write three inequalities representing the possible values of x and y.

Graph the solution to the simultaneous inequalities.

If 14 staff are allocated to checking in passengers of ﬂight A, what is the maximum number of staff that can be allocated to checking in passengers of ﬂight B?

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An old Air Force plane used for skydiving can ﬁt no more than 32 divers at a time, and has a maximum carrying capacity of 3900 kilograms. When carrying a parachute and equipment an average female’s weight is 100 kilograms and an average male’s weight is 150 kilograms. Let x represent the number of women. Let y represent the number of men. a

Write four inequalities in x and y that represent the number of men and women who can be on the plane at one time.

Graph the inequalities on the same set of axes.

Which combinations of men and women can be on the plane at the same time? A

25 women and 11 men

19 women and 12 men

25 women and 9 men

7 women and 10 men

A bicycle manufacturer employs a mechanic and a painter to construct two types of bikes: racing bikes and mountain bikes. The time each worker spends on each bike is shown (assume both workers can work on the same bike at the same time). Let x represent the number of racing bikes built and y represent the number of mountain bikes built. Each worker can work at most 40 hours in a week.

Racing Bike Mountain Bike

Mechanic 5 hours 8 hours

Painter 10 hours 4 hours

Write an inequality relating x and y to the total time spent working by the mechanic.

Write an inequality relating x and y to the total time spent working by the painter.

Lamborgotti Motors produces two types of car: a sedan and a coupe. Each type of car requires assembly in two different factories. The time that each car is required to be in each factory is given in the table. Factory A Factory B Each factory can operate for 18 hours a day. Sedan 3 hours 4.5 hours Let x represent the number of sedans Coupe 2 hours 1.5 hours produced each day, and y represent the number of coupes produced each day.

Write an inequality for Factory A.

Write inequalities for x and y given the number of cars cannot be a negative number.

Graph the region deﬁned by the inequalities for Factory A and Factory B

Write an inequality for Factory B.

Mahindra produces two types of tractors: an All Purpose and a Harvester. Each type of tractor requires production in two different factories. The time that each tractor is required to be in each factory is given in the table. Each factory can operate for 18 hours a day. Let x represent the number of All Purpose tractors produced each day, and y represent the number of Harvesters produced each day.

292

All Purpose Harvesters

Factory A 4 hours 2 hours

Factory B 6 hours 1.5 hours

Write an inequality for Factory A.

Write inequalities for x and y given the number of tractors cannot be a negative number.

Graph the region deﬁned by the inequalities for Factory A and Factory B.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Write an inequality for Factory B.

Reasoning 16

16, 18ab, 20

16, 18ac, 20

17, 18cd, 19

Without graphing, determine whether the simultaneous inequalities have no solutions or infinitely many solutions. Explain your answer. a

y > 4 and y < 4

x ≤ −2 and x ≥ −2

9x − y < 7 and 9x − y > 7

2x − y ≤ 7 and 2x − y ≥ 7

Tracy has experimented with growing two different types of banana trees on her land in different proportions. Let x and y represent the decimal proportion of area Type A and Type B banana trees respectively. a

She has found that bananas grow best if Type A takes up at most 35% of the plantation area. Write two inequalities representing the possible values of x.

She has found that bananas grow best if Type B takes up at least 20% of the plantation area. Write one inequality representing the possible values of y.

The whole area must be planted with a combination of Type A and Type B. Write one inequality representing the possible values of x and y.

Graph the solution to the simultaneous inequalties.

Do these points correspond to appropriate areas for Type A and Type B banana trees? i

(0.28, 0.68)

(0.58, 0.68)

iii

(0.13, 0.57)

(0.09, 0.11)

Is it possible for Type B trees to grow optimally if they take up 100% of the plantation area? Explain your answer.

A company produces cans of cat food and dog food. Each product passes through two machines in their production, and the amount of time each machine takes to process one pallet of pet food is shown in the table. Each machine can operate for 12 hours in a day.

Let x represent the number of pallets of cat food processed each day, and y represent the number of pallets of dog food processed each day.

Cat Food Dog Food

Write an inequality for Machine A.

Can the values of x and y be negative? Explain your answer. Write any additional inequalities if necessary.

Graph the region deﬁned by the inequalities for Machine A and Machine B on a Cartesian plane.

Machine A 2 hours 4 hours

Machine B 3 hours 1.5 hours

Write an inequality for Machine B.

Is it possible for a coordinate point to solve one part of the linear inequality pair but not the other? Explain with a diagram what this would look like.

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A light bulb manufacturer produces two types of light bulbs: incandescent and ﬂuorescent. Let x represent the number of incandescent bulbs produced in a day, and y represent the number of ﬂuorescent bulbs produced in a day. a

If it’s not possible for a number of lightbulbs to be less than zero, write an inequality for x and y to represent this fact.

Up to 200 light bulbs can be produced in one day. Write an inequality to represent this.

The ﬁnal inequality representing the limits on the number of light bulbs that can be produced each day is:

Explain the meaning of this ﬁnal inequality in the context of the question.

Problem-solving 22, 23, 25 21

21, 23–26

There are x surfers competing for wildcard entry into a pro-surﬁng competition, of which Joanne is one. She needs her total score, y, from all judges to be at least 35 in order to be chosen. The total score possible from all judges is at most 50. a

x > 1 is one of the inequalities that can be formed for the situation. Write down the two other inequalities related to y to complete the system.

Graph the solution to the simultaneous inequalities.

If Joanne receives scores of 8.4, 7.3, 7.3, 7.8, 8.9 from the judges, will she be chosen as a wildcard entry?

Whether Joanne is chosen as a wildcard entry is dependent on the number of surfers competing for the spot. True or false?

Roald has 24 centimetres of leftover wood that he is trying to make a rectangular photo frame out of. Let x represent the length of the photo frame and let y represent the width. a

Write three inequalities to represent the situation.

Graph the solution to the simultaneous inequalities.

Are these possible dimensions of the photo frame?

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21, 23, 25

A length of 6 cm and a width of 4 cm

A length of 7 cm and a width of 6 cm

iii

A length of 9 cm and a width of 8 cm

A length of 7 cm and a width of 4 cm

If Roald wanted to create a square frame, what are the dimensions of the largest possible frame?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Throughout university, Tom works as a mentor, earning $15 per hour, and as a ﬁtness instructor earning $18 per hour. The number of hours he works in each job can vary from week to week, but he never works more than 29 hours in total each week, and he needs to be able to at least cover his weekly expenses of $270. Let x represent the number of hours he works as a mentor and y represent the number of hours as ﬁtness instructor.

Write four inequalities representing the possible values of x and y.

Graph the solution to the simultaneous inequalities.

If Tom works 6 hours as a mentor in one week, ﬁnd the minimum number of hours y he can work as a ﬁtness instructor so that he can cover his expenses.

Is it possible for him to work the same number of hours in both jobs and still be able to cover his expenses?

A warehouse is stocked with two types of storage containers: square boxes and rectangular boxes. The square boxes cost $55 each and have a volume of 2 m3 and the rectangular boxes cost $65 each and have a volume of 7 m3. The warehouse has a total storage space of 200 m3 and there is a budget of $800 to purchase the containers. Let x represent the number of square boxes purchased, and y represent the number of rectangular boxes purchased.

If it’s not possible for a number of boxes to be less than zero, write two inequalities for x and y to represent this fact.

Write an inequality relating x and y for the total budget for the warehouse.

Write an inequality relating x and y for the total storage space of the warehouse.

Eileen is spending the week at a sports camp. She is able to attend coaching sessions for her two favourite sports: basketball and football. She can attend a maximum of 26 sessions and a minimum of 20 sessions per week. Eileen’s basketball coach has told her that she must attend at least 5 basketball sessions. Her football coach has told her that she must go to at least 8 more football sessions than basketball sessions. Let x represent the number of basketball sessions she attends, and y represent the number of football sessions she attends. a

From this information, write four inequalities.

Graph the region deﬁned by the four inequalities.

List the four vertices of the region.

The vertices of the region represent the maximum and minimum number of session Eileen can attend. What is the maximum number of basketball sessions that she can attend?

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Edward is going to a music camp for one week. Edward plays drums and guitar and can attend lessons for both instruments. He can attend a maximum of 18 lessons and a minimum of 13 lessons per week. Edward’s guitar teacher has told him that he must not go to more than two times as many guitar lessons as drum lessons. His drum teacher has told him he must attend at least ﬁve drum lessons. Let x represent the number of drum lessons he attends, and y represent the number of guitar lessons he attends. a

From this information, construct a set of ﬁve inequalities.

Graph the region deﬁned by the ﬁve inequalities.

List the ﬁve vertices of the region.

Using the ﬁve vertices determine the maximum number of guitar lessons that Edward can attend.

6.03 Solve using substitution After this lesson, you will be able to… • utilise the substitution method to solve for both variables in two linear simultaneous equations. • demonstrate how to rearrange equations to isolate a variable when equations are not ready for direct substitution and then apply the substitution method.

Substitute variables When solving simultaneous equations, we are looking for values for the variables which make both equations true. This will be the point of intersection of the graphs of the equations. This section focuses on solving for the point of intersection of two linear equations, but the same process can be applied to any two equations with two unknowns. We can rearrange the equations to suit our purpose as long as we always maintain the equivalence. Let’s look at some methods in the following examples.

Example 1 We want to solve these simultaneous equations using the substitution method.

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y = x + 11

y = 3x + 19

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a First solve for x.

Create a strategy Substitute the expression for y from equation 1, into the variable y in equation 2 and then solve for x.

Apply the idea x + 11 = 3x + 19

Substitute 1 into 2

11 = 2x + 19

Subtract x from both sides

−8 = 2x

Subtract 19 from both sides

x = −4

Divide both sides by 2

Using substitution, we have found that x = −4.

b Solve for y.

Create a strategy Substitute the value of x into one of the equations and solve for y.

Apply the idea y = x + 11

y = −4 + 11

Substitute x = −4 into 1

y=7

Evaluate

Reﬂect and check To check that x = −4, y = 7 is also a solution to 2 , substitute the value of x into 2 and see if we get the same value for y: y = 3x + 19

y = 3(−4) + 19

Substitute x = −4 into 2

y=7

Evaluate

y = 7 is the same value, this confirms that x = −4, y = 7 is the solution. We can also write this as (−4, 7), the intersection point of the two lines.

Idea summary To solve simultaneous equations using the substitution method: • Substitute the expression for one variable, usually y, into the other equation. • Solve the new equation for the other variable, usually x. • Substitute the result back into one of the equations to find the value of the other variable.

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Rearrange and substitute In the previous example, both equations were already written with y as the subject. As such, it was easy to substitute the expression for y from one equation into the other. When the equations are not written with one of the variables as the subject, we can choose any of the variables to isolate in one equation in order to substitute it into the other.

Example 2 Solve these simultaneous equations using the substitution method. p + 7q = –4

–6p + 2q = –20

Create a strategy Equation 1 looks to be the easiest to rearrange as the coefficient of p is already 1. Use 1 to get an expression for p in terms of q. Then we will substitute that expression into 2 and solve.

Apply the idea p + 7q = –4

p = –4 – 7q Subtract 7q from both sides

We now have an equation for p, we will call this equation 3 . p = –4 – 7q

Substitute the value of p into 2 and solve for q. –6p + 2q = –20

–6(–4 – 7q) + 2q = –20

24 + 42q + 2q = – 20

24 + 44q = –20

2 Substitute p Expand the brackets Collect like terms

44q = –44

Subtract 24 from both sides

q = –1

Divide both sides by –44

Finally, we now find the value for p. p = –4 – 7q

p = –4 – 7 × –1

Substitute q = –1

p = –4 + 7

Evaluate the multiplication

p = 3

Evaluate

So the solution is p = 3 and q = –1.

Reﬂect and check If p = 3 and q = –1 then these values should satisfy both equations. We can do a quick check at 1 p + 7q = –4. If p = 3 and q = –1, then we have 3 – 7 = –4 so this is true. At 2 –6p + 2q = –20, if p = 3 and q = –1 then we have –18 – 2 = –20 which is also true.

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Example 3 A man is ﬁve times as old as his son. Four years ago, the man was nine times as old as his son. Find their present ages. a Use the statement “a man is five times as old as his son” to set up equation 1.

Create a strategy Let x represent the present age of the man, and y represent the present age of the son.

Apply the idea The man is ﬁve times as old as his son, so x is equal to 5 times y. x = 5y

b Use the statement “4 years ago the man was 9 times as old as his son” to set up equation 2. Write this with y the subject.

Create a strategy To ﬁnd their ages 4 years ago, we can subtract 4 from x and y.

Apply the idea We can express the age of the man from four years ago as x − 4 and we can express the age of the son as y − 4. At this time, the age of the man is 9 times the age of the son, this will give us equation 2. 4 years ago the man is 9 times as old as the son Expand the brackets Add 36 to both sides Divide both sides by 9 2 2 is

c Solve for y to ﬁnd the age of the son.

Create a strategy Substitute equation 1 into equation 2.

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Apply the idea 2 Substitute 1 to 2 Multiply both sides by 9 Subtract 5y from both sides Divide both sides by 4 The son is 8 years old.

d Solve for x to ﬁnd the age of the man.

Create a strategy Substitute the value of y into equation 1.

Apply the idea x = 5y

x=5×8

Substitute y = 8

x = 40

Evaluate

The man is 40 years old.

Reflect and check Check the ages for the son (8) and man (40) against the initial statement. A man is 5 times as old as his son, this is true as 5 × 8 = 40. Four years ago, the man was 9 times as old as his son, this is true as 4 years ago the man would be 36, and the son would be 4 and 4 × 9 = 36.

Idea summary To solve simultaneous equations using the substitution method, we may need to rearrange one of the equations so that one variable is the subject before we can perform the substitution.

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6.03 Practice questions Understanding 1

Explorer

Adventurer

1, 2

Trailblazer 1, 3, 4

1–3

A group of students were solving the equations: x+y=6 y = 2x − 3 a

Tomi substituted x from the ﬁrst equation into the second.

Carla substituted y from the second equation into the ﬁrst.

Regie substituted y from the ﬁrst equation into the second.

Junard substituted x from the second equation into the ﬁrst.

Write the equation that each student would then have to solve and comment on the eﬃciency of each. 2

Is (5, 2) a solution of the simultaneous equations? x+y=7 x−y=3

State whether these statements are true or false. a

The method of substitution involves isolating one variable in one equation and substituting this expression into the other equation.

The substitution method can be used to solve any simultaneous linear equations.

Once you solve one of the equations for one variable, you must substitute it back into the ﬁrst equation.

Solving simultaneous equations by substitution will always result in the same solution, regardless of which variable you solve for ﬁrst.

Simultaneous linear equations can have exactly two solutions, no solutions, or inﬁnitely many solutions.

Is (4, 17) a solution of the simultaneous equations? y = 6x – 7 4x + 2y = 67

Fluency Ex 1

5a–e, 6, 7a–d, 8, 10

5c–g, 6, 7c–f, 8, 10

5g–l, 6, 7e–h, 8, 9, 11

Solve the simultaneous linear equations using substitution: a

y = 6x + 56

y = 2x + 24 e

y = 3x + 14

x+y=2 i

y = −3x − 23

y = −5x − 22

y = 5x + 37

y = −4x – 27

y = 7x + 38 5y = 35x + 265 2y = 10x + 54 y = −3x + 22

y = −3x + 9

x + y = 10 6x − y = 0 j

y = 2x – 15

x + 4y = −37 3x − 4y = 20

y = 4x – 28 3x + 5y = 44

5x − 2y = −10

2x + 3y = 48

y = −5x + 3 4x − 5y = 8

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For these simultaneous equations: 3x − 7y = 4 −12x + 28y = −16

Ex 2

Rearrange 3x − 7y = 4 to make x the subject in terms of y.

Substitute your expression for x from part (a) into −12x + 28y = −16 and solve for y.

Solve the simultaneous equations using the substitution method: a

2x + 3y = 26

2x + 7y = 23

5x + 4y = 10

4x + 3y = 17

8x − 5y = 36 5x + 6y = 0 3x + 7y = −17 9x − 4y = 6 e

−7f + 2g = −0.4

−4a + 7b = 11.5

−2.1f + g = 0.36 −2a + 2.1b = 2.25 Ex 3

A mother is currently 10 times older than her son. In two years time, she will be 7 times older than her son. Let x and y be the present ages of the son and mother respectively. This can be modelled by the simultaneous equations: y = 10x y + 2 = 7(x + 2) What are the current ages of the son and the mother?

Farzad bought some fresh produce. He picked up 2 apples and 3 kiwi fruit. The cost of Farzad’s shopping was $10.08. Eileen went to the same shop and bought 5 apples and 7 kiwi fruit. The cost of Eileen’s shopping was $24.08. Let a and k be the price of apples and kiwi fruit respectively. This can be modelled by the simultaneous equations: 2a + 3k = 10.08 5a + 7k = 24.08

Solve the simultaneous equations using the substitution method.

What is the cost of one apple and one kiwi fruit?

The number of new jobs created in Venhurst can be modelled by the simultaneous equations y = 4x − 190 000 y = x + 560 000 where x is the number of jobs created in 2008 and y is the number of jobs created in 2013.

Solve the simultaneous equations using the substitution method.

What does the solution to the simultaneous equations mean in this context?

The members in a group can be modelled by the simultaneous equations y=x+6 x + y = 22 where x and y are the number of adults and children in the group respectively.

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Solve for the values of x and y.

What does the solution to the simultaneous equations mean in this context?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning

12–14

12, 13, 15

12, 13, 15, 16

If you solve simultaneous equations by substitution and ﬁnd that the result is a false statement such as 0 = 5, what does that tell you about the equations?

What does it mean if you solve simultaneous equations by substitution and get a true statement, such as 0 = 0 or 5 = 5?

For these simultaneous equations: 3x + y = 7 2x − y = 4 Explain the steps you would take to solve the simultaneous equations and why you would do it in that way.

Do the equations y = 3x + 5 and y = 3(x + 5) have inﬁnitely many solutions? Explain your answer.

For these simultaneous linear equations: −6x − 2y = −28 2x + 16y = 40 4x − 2y = 12 a

Find the values of x and y that satisfy the ﬁrst two equations.

Does this solution satisfy the third equation?

Is this the only solution to the simultaneous equations? Justify your answer.

Conﬁrm the results by graphing the three linear equations on the same Cartesian plane.

Problem-solving 17, 18, 20a–d 17

17, 19, 20b–e

19, 20c–h, 21

A taxi driver charges a fee for the initial pick up and an additional amount per km travelled. A journey of 10 km costs $105, while a journey of 15 km costs $155. a

Find the amount of the ﬁxed charge and the charge per km.

How much does a person have to pay for travelling a distance of 25 km?

The coach of a tennis team buys 7 rackets and 6 balls for $380. Later, he buys 3 rackets and 5 balls for $175. Find the price of rackets and balls respectively.

Mikhail sells sausages and soft drinks. Each sausage costs $1.50 and each soft drink costs $0.50. At the end of the day, he made a total of $78.50 for a total of 87 sausages and soft drinks combined. How many sausages and soft drinks were sold?

Use the substitution method to ﬁnd the solution(s) for each pair of simultaneous equations. a

y = x2

y = −5x d

(x + 3) + ( y − 3) = 4

(x + 1)2 + ( y + 4)2 = 25 y = −3x − 2

y = (x − 5) (x − 2)

y = 5x + 4 y = x2 + x + 8

y = 24 2

x = −5 g

y = x2 − 5x

x2 + ( y + 7)2 = 7

y = −x + 1 y = −7 h

y = (x − 1)2 x − 3y = 1

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The plan for a proposed tunnel through a mountain side is shown in the diagram:

y nnel

m the tu

Exit fro

ed tunn

Entrance to the tunnel

Propos

Sea level

The height of the mountain above sea level is given by the equation height of the tunnel above sea level is given by the equation the height above sea level in kilometres.

, and the , where y represents

Use the substitution method to solve for y, the heights above sea level at which cars will enter and exit the tunnel.

6.04

Solve using elimination

After this lesson, you will be able to… • apply the elimination method to solve equations simultaneously.

Elimination method As long as equivalence is maintained in both the left and right-hand sides of any given equation, we can combine them in specific ways to make other equations that are still true. For example, if we know that:

a=b

c=d

Then we can combine these two equations to create others. We label these new equations in a similar way, so that we can reference them in our working later.

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1 + 2

a+c=b+d

1 − 2

a−c=b−d

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

We can also take multiples of equations before combining them with addition or subtraction, for example: 3× 1 −4× 2

3a − 4c = 3b − 4d

× 1 +3× 2

5 6

Equations 3, 4, 5 and 6 are all different, but they all maintain the equivalence that was present in equations 1 and 2. To solve equations simultaneously, we can use specific combinations of equations to create situations where we can eliminate one of the variables. This is the elimination method.

Example 1 Use the elimination method to solve the pair of simultaneous equations: 3x + y = 10

−4x + y = −11

Create a strategy We can remove the variable y, by subtracting the equations.

Apply the idea 3x + y = −

(−4x + y = −11)

7x =

x =

Now that we have a value for x, substitute this into equation 1 and solve for y. 3x + y = 10

3(3) + y = 10

Substitute x = 3

9 + y = 10 y=1

Evaluate Subtract 9 from both sides

The solution to simultaneous equations is (3, 1).

Reflect and check Substitute the values for x and y into equation 2 to confirm. −4x + y = −11

−4(3) + 1 = −11

Substitute

−12 + 1 = −11

Evaluate

−11 = −11

True

Confirmed that (3, 1) also solves equation 2.

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Example 2 Use the elimination method to solve the pair of simultaneous equations: 2x + y = 4

x = 4y − 34 2

Create a strategy Rearrange 2 to have the same form as 1 , then identify a multiple to eliminate one of the variables. We can remove the variable x, by doing 1 − 2 × 2 .

Apply the idea Start with rearranging 2 by subtracting 4y from both sides, then call it 3 . x = 4y – 34 2

x − 4y = −34

To eliminate x, multiply 3 by 2.

x − 4y = −34

2x − 8y = −68

3 ×2

Then, subtract 3 × 2 from 1 . 2x + y = 4 −(2x – 8y = −68) 9y = 72 y=8 Now that we have a value for y, substitute this into equation 1 and solve for x. 2x + y = 4

2x + 8 = 4

Substitute y = 8

2x = −4

Subtract 8 from both sides

x = −2

Divide both sides by 2

The solution to simultaneous equations is (−2, 8).

Reflect and check Substitute the values for x and y into equation 2 to confirm. x − 4y = −34

−2 − 4(8) = −34

Substitute

−2 − 32 = −34

Evaluate

−34 = −34

True

Confirmed that (−2, 8) also solves equation 2.

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Idea summary To solve equations simultaneously, using the elimination method we want to combine equations so the new equation has only one variable. If the coeﬃcient of the variable we want to eliminate in both equations has the same value but different sign, we can add the equations to eliminate the variable. If the coeﬃcient of the variable we want to eliminate in both equations has the same value and the same sign, we can subtract the equations to eliminate the variable.

6.04 Practice questions Understanding 1

Explorer 1, 2

Adventurer

Trailblazer

1–3

The method of elimination involves: A

Substituting one variable into another equation

Graphing the equations

Adding or subtracting equations to eliminate a variable

Isolating a variable

Explain how you would begin the process to eliminate x from the following set of equations. 5x − 9y = 2 7x − 8y = 1

Comment on whether each statement is true of false. a

The elimination method is best used when the coeﬃcients of one variable are the same in both equations.

When using the elimination method, the solution represents where the lines of the equations intersect.

There are situations when the elimination method will not work.

Fluency 4

4, 5, 6ab, 7a–c, 8a–d, 9

4, 5, 6bc, 7b–d, 8b–f, 10

4, 6b–d, 7d–f, 8d–h, 11

For the simultaneous linear equations:

What number should the second equation be multiplied by to eliminate y?

Eliminate y and ﬁnd the value of x that satisﬁes both equations.

Substitute the x-value to ﬁnd the value of y that satisﬁes both equations.

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For the simultaneous equations: 3x + 7y = −24 4x − 5y = 11

What value can we multiply each equation by to eliminate the variable y?

Solve for x and y.

For each pair of simultaneous equations: i

Rewrite the equations as equivalent equations with lowest common multiple integer coefficients.

Solve for x and y.

0.2x + 0.3y = 0.5

0.5x + 0.4y = 0.2

Ex 1

Use the elimination method to solve each pair of linear equations: a

3x + 6y = 2

2x + 5y = 44

3x + 3y = −7 6x − 5y = −28 c

8x + 3y = −11

2x − 5y = 1

−8x − 5y = 29 −3x − 5y = −39

f =1

Ex 2

Use the elimination method to solve each pair of linear equations: a

3x + 2y = 22

x = 4y − 16

2 2x = 3y + 9

2y = 2x − 4

6x = y + 15

4y − x = 1

3x − 2y = −15

7x + 13 = 3y − 11

2.5x + 3y = 11

x − 2y = −5

2 1.5x = 2.5y − 2

4.2x − 1.8y = 13.8 3.4x = 2.2y + 8.2

3 1 13 x+ y = 4 2 2 2 1 2 2 x− y= − 3 3 3 1

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4x − y = 13

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1 2 1 2

The train in an amusement park has 15 cabs that can hold a total of 56 people. Some cabs hold 4 people while some hold 6 people. Let x be the number of four-passenger cabs and y be the number of six-passenger cabs. This situation can be represented by the simultaneous equations: x + y = 15 4x + 6y = 56

Find the number of four-passenger cabs and six-passenger cabs using the elimination method.

In terms of this context, is the solution valid?

Nine pens and 2 rulers cost $26, while 3 pens and 4 rulers cost $22. Let x and y be the price of the pen and ruler respectively. This can be modelled by the simultaneous equations: 9x + 2y = 26 3x + 4y = 22

Solve the simultaneous equations using elimination.

Explain what the solution means in this context.

Marjan spent $55.25 to purchase 9 ﬂowers. She bought rhododendrons which cost $6.45 each and chrysanthemums which cost $5.75 each. This can be modelled by the simultaneous equations R+C=9 6.45R + 5.75C = 55.25 where R is the number of rhododendrons and C is the number of chrysanthemums that Marjan bought. a

Solve the simultaneous equations using elimination.

Explain what the solution means in this context.

Reasoning 12

12, 14

13, 14

13, 15

Explain how you could use the elimination method to solve the simultaneous equations: 2x − 3y = 8 4x + 3y = 12

Without solving, how can you determine the number of solutions to these simultaneous equations using the elimination method? 3x − 2y = 6 −3x + 2y = −6

Describe a situation where it would be more eﬃcient to use the elimination method rather than the substitution method to solve simultaneous equations and explain why.

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Elias solved nonlinear simultaneous equations using elimination. His work is shown. Aria chose to solve by graphing, and she found two solutions. a

Explain the mistake Elias made that caused him to only ﬁnd one solution.

Find the second solution.

y = 3x2 − 6 y = −x2 – 2 y = 3x2 – 6 3y = −3x2 – 6

Original equations

Multiply equation 2 by 3

4y = −12 Add the equations, solve for y y = −3 −3 = 3x2 − 6 Substitute y, solve for x 2 3 = 3x 1 = x2 1=x The solution is (1, −3).

Problem-solving 17, 20ab 16

6, 18, 20b–d

16, 19, 20d–f

In a game show, Nigel has the opportunity to answer 30 questions. For every question he answers correctly, he earns 6 points, but for every question he answers incorrectly, he loses 3 points. Let x represent the number of correct answers and y represent the number of incorrect answers. a

If Nigel’s total score was 90 points, write a pair of simultaneous equations that represent this problem.

Solve these equations to ﬁnd out the number of correct and incorrect answers Nigel provided.

For two numbers, x and y: • The sum of y and two times x is 27. • The difference between ﬁve times x and y is 43. Find the two numbers.

For two numbers, x and y: • The sum of the x and y is 10. • The difference between six times x and ﬁve times y is 5. Find the two numbers.

310

The proportion of the workforce of a particular country that are teenagers is modelled by 3.3x + y = 70.8, where x is the number of years since 2015 and y is the proportion as a percentage. The proportion of the workforce that are pensioners is modelled by 3.2x − y = −64.5. a

Estimate the year in which the proportion of the workforce that are teenagers and the proportion of the workforce that are pensioners are the same.

Estimate the percentage of the workforce that are teenagers or pensioners during the estimated year in part (a).

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Use the elimination method to solve the simultaneous equations. a

x2 − y = −2

−x + y = 4 c

y = 3x2 − 2

x2 + y = 5 −x + y = −1

x2 + y2 = 100

y = x2 2x2 + 10y = 68 e

x2 + y2 = 25

(x + 3)2 + ( y + 1)2 = 25

(x + 3)2 − y = 6

6.05 Applications of simultaneous equations After this lesson, you will be able to… • define variables that represent unknown quantities to translate information into at least two equations. • apply simultaneous equations to real-life scenarios. • define and understand the concept of the break-even point. • perform break-even analysis using graphical representations.

Applications of simultaneous equations Our ﬁrst step is to deﬁne the variables that represent the quantities in the problem. Then translate the pieces of information about the quantities into equations. Lastly, by using either the substitution method, elimination method, or a graphical method, we solve the equations simultaneously.

Example 1 When comparing some test results, Christa noticed that the sum of her geography and science test score was 172, and that their difference was 18. Given that her geography score is x and her science score is y, and that she scored higher in the geography test: a Use the sum of the test scores to form an equation. We will refer to this as equation 1.

Create a strategy Write an equation in terms of x and y.

Apply the idea The sum of her two test scores is 172, where x and y are the scores. Equation 1: x + y = 172

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b Use the difference of the test scores to form an equation. We will refer to this as equation 2.

Create a strategy

Apply the idea

Write an equation in terms of x and y.

The difference between the scores is 18, where x and y are the scores. Since x > y we can subtract y from x. Equation 2: x − y = 18

c Use these two equations to ﬁnd her geography score.

Create a strategy We can use the elimination method to solve for x.

Apply the idea Notice that the coeﬃcient of y in both equations is the same, but with opposite sign. So we can eliminate the variable y by adding the two equations together. +

x + y = 172 x - y = 18 = 190 2x x = 95

So her geography score is 95.

d Find her science score.

Create a strategy Substitute the x-value into one of the equations to solve for y.

Apply the idea x + y = 172

Write equation 1

95 + y = 172

Substitute the value of x

95 + y − 95 = 172 – 95 Subtract 95 from both sides y = 77

Simplify

So her science score is 77.

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Example 2 Consider the following phone plans: • GO SMALL plan: This plan has a $20 monthly base charge and charges 90c per minute for all calls. • GO MEDIUM plan: This plan has a $26 monthly base charge and then charges 70c per minute for all calls. a Complete the table of values for various total monthly call times for the two plans: Call time (in minutes) 20 30 40 50

Total cost for GO SMALL plan (S)

Total cost for GO MEDIUM plan (M )

Create a strategy Add the monthly base charge to the number of minutes times the rate per minute.

Apply the idea For the call time of 20 minutes we can ﬁnd the cost for the GO SMALL plan by: GO SMALL cost = 20 + 0.9 × 20 = $38.00

Add the base charge to the rate times the minutes Evaluate

We can ﬁnd the cost for the GO MEDIUM plan by: GO MEDIUM cost = 26 + 0.7 × 20 = $40.00

Add the base charge to the rate times the minutes Evaluate

Continuing in this way for the rest of the call times gives us this table: Call time (in minutes) 20 30 40 50

Total cost for GO SMALL plan (S) 38.00 47.00 56.00 65.00

Total cost for GO MEDIUM plan (M ) 40.00 47.00 54.00 61.00

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b Sketch a graph for the two plans.

Create a strategy Use the points from the table to graph both plans.

Apply the idea Cost 65 60 55

To include the points from the table, the vertical or cost axis needs to go over 60 and the horizontal or call time axis needs to go over 50. We will go a bit further than this along both axes, in order to see more detail in the graph. We will also label the lines S and M.

50 45 40 35 30 M 25

20 15

To graph the lines we can use two points for each plan from the table. For the GO SMALL plan we can use (20, 38) and (40, 56). For the GO MEDIUM plan we can use (30, 47) and (50, 61).

10 5

Call time 5 10 15 20 25 30 35 40 45 50

c Using the graph, determine how many minutes of calls would need to be made so that the monthly bill costs the same on both plans.

Create a strategy Use the point of intersection.

Apply the idea Cost 65 60

The point of intersection is at (30, 47). This means that for a call time of 30 minutes, both plans will cost 47.

55 50 45 40 35 30 25 20

M S

15 10 5

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Call time 5 10 15 20 25 30 35 40 45 50

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary To solve a problem with two unknown quantities: • Deﬁne variables to represent the quantities. • Translate the information into two equations. • Use the substitution method, elimination method or graphical method to solve the equations simultaneously.

Break-even point It’s important for businesses to make a profit. The point at which a business makes a profit is known as the break even point. To find this point, we use break-even analysis. Break-even analysis looks at the graphs of cost and revenue together to determine where they intersect. This point of intersection is the break-even point where revenue equals expenses. Here is an example of a break-even analysis for a single day of operation for the company Lovely Lemonade. Lovely Lemonade’s break even analysis 1200 Profit

Money (dollars)

1000

Break even point

800 600 400

Loss

200 1000

2000

3000

4000

5000

Number of drinks (units) Revenue

Expenses

The expenses line (red) starts off higher than the revenue line (green). For a low number of units sold, Lovely Lemonade will lose money. For example, if they only sell 1000 units, the green line tells us they earn $200, the red line tells us they spend $600, and so overall they make $200 − $600 = −$400. The negative sign means overall they lose $400. If they sell 5000 units the graph tells us that they will earn $1100 but only spend $1000. Overall they make $1100 − $1000 = $100, and the positive sign means a profit of $100. The important point is when the two lines intersect - this is the break-even point. In this example we can see that selling 4000 units means they don’t make any profit, but they don’t lose any money either - both their revenue and their expenses are $900. We can also use break-even analysis to compare two different pricing plans for the same product or service to see which one offers the best deal for a particular situation.

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Here’s an example comparison between two energy plans provided by rival companies, Thorgate and Callisto: Thorgate vs Callisto pricing comparison 18 Callisto cheaper

Price (dollars)

15 12

Break even point

9 6 Thorgate cheaper

3 10

Energy per day (kWh) Thorgate

Callisto

In this scenario, the break-even point at (40, 10) tells us that both plans charge $10 for 40kWh. For energy amounts less than the break-even point, we can see that Thorgate is cheaper, and for energy amounts more than the break-even point, Callisto is cheaper. Knowing how much energy someone plans on using can then determine the best plan for them.

Example 3 The cost for a furniture manufacturer to make an armchair is $500 per armchair plus a ﬁxed setup cost of $3500. The armchairs will sell for $650 each. a Write an expression to represent the cost of manufacturing x armchairs.

Create a strategy Add the ﬁxed cost and the variable cost.

Apply the idea The ﬁxed cost is $3500. The variable cost is $500 per armchair, which is $500 × x. Cost = 3500 + 500x

Add the ﬁxed and variable costs

b Write an expression to represent the revenue generated from the sale of x armchairs.

Create a strategy The amount of revenue generated is equal to the number of pieces sold multiplied by the price at which they sell.

Apply the idea Revenue = 650x 316

Multiply the selling price and the number of armchairs sold

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c Find the break-even point.

Create a strategy Equate the expressions for the cost and the revenue and solve for x.

Apply the idea At the break-even point, the cost is equal to the revenue, so we equate their expressions. Equate the revenue and cost

Subtract 500x from both sides

Simplify

Divide both sides by 150

Evaluate

So the break-even point is when x = 24 since we can only sell whole numbers of armchairs.

Example 4 David decides to start his own yoga class. The cost and revenue from running the class have been graphed: 54

Money (dollars, $)

48 Cost

42 36 30 24 18 12

Revenue

x 1 2 3 4 5 6 7 8 9 10 11 Number of students

a Find the revenue that David makes for each student.

Create a strategy Use the revenue line on the graph.

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Apply the idea 54

For 1 student, x = 1. We move from x = 1 on the x-axis up to the revenue line, then across to the y-axis to get y = 6.

Money (dollars, $)

48 42

So the revenue per student is $6.

36 30 24 18 Revenue

12 6

x 1 2 3 4 5 6 7 8 9 10 11 Number of students

b How many students must attend his class so that David can cover the costs of running the class?

Create a strategy

Apply the idea

The number of students is given by the intersection point of the revenue and cost curves.

The point of intersection is (6, 36). So if David get 6 students, he can cover the cost of running the class.

c How much proﬁt does David make if there are 9 students in his class?

Create a strategy Find the difference between the revenue and the cost.

Apply the idea 54 48 Money (dollars, $)

To ﬁnd the revenue and cost for 9 students, we ﬁnd x = 9 on the x-axis and move up to the two lines and then across to the y-axis from each.

Cost

We can see that cost is $45 and the revenue is $54.

36 30 24 18

Revenue

12 6

x 1 2 3 4 5 6 7 8 9 10 11 Number of students

Profit = 54 – 45 = $9

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Subtract the cost from the revenue Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary The break-even point is where the cost and revenue are equal. To ﬁnd the break-even point, we can either: • Look at the graphs of cost and revenue together to determine where they intersect. • Equate the expressions for revenue and cost and solve for the independent variable.

6.05 Practice questions Understanding 1

Explorer 1–3

y = 3x – 1

2x + y = 7

y = 4x + 2

y = −x + 5

3x − 2y = 6

2x + 3y = 5 x−y=1

State whether the substitution method or the elimination method would be more eﬃcient for solving.

Explain why you chose the method in part (i). y = 6x + 56 y = 2x + 24

7x − 4y = 15

7x + 5y = 60

−5x + 16y = 82 25x − 4y = 122

y = −3x − 23 x + 4y = −37

Identify whether each statement is true or false. a

All simultaneous equations can be solved using any method.

Only some simultaneous equations can be solved by graphing.

Graphing is only useful to be used if the exact point of intersection can be determined from the graph.

Fluency 4

y = 2x + 3

For each of the simultaneous equations:

Trailblazer 1–3

For the simultaneous equations, explain which strategy you would use and why: a

Adventurer 1–3

4–9, 14, 16ac

5–8, 10–13, 16bd

5–8, 11–13, 15, 16ab

A family owns two businesses that made a combined profit of $6 million in the previous financial year, with Business B making 2 times as much profit as Business A. Let x and y be the profits (in millions) of Business A and Business B respectively. a

Write a pair of linear equations to represent this situation.

Find Business A’s proﬁt.

Find Business B’s proﬁt.

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Ex 1

When comparing some test results, Akira noticed that the sum of his Geography score and Science score was 184, and that their difference was 22. He also noticed that he scored higher for the Geography test. Let x and y be his Geography score and his Science score respectively.

Ex 2

Write two equations in terms of x and y from the information given.

Find his Geography score.

Find his Science score.

For a phone company, customers can choose between two phone plans: • GO SMALL plan: $24 monthly base charge, and 80c per minute for all calls • GO MEDIUM plan: $32 monthly base charge, and 60c per minute for all calls a

Complete the table of values for various total monthly call times for the two plans: Call time (in minutes) 25 30 35 40

b 7

Ex 3

Total cost for GO SMALL plan

Total cost for GO MEDIUM plan

Find how many minutes of calls would need to be made so that the monthly bill costs the same on both plans.

Toby’s piggy bank contains only 5c and 10c coins. It contains 70 coins with a total value of $3.85. a

Set up two equations by letting x and y be the number of 5c and 10c coins respectively.

Solve the simultaneous equations.

Kerry currently pays $50 a month for her internet service. She is planning to switch to a ﬁbre optic cable service. a

Write an equation for the total cost T of Kerry’s current internet service over a period of n months.

For the ﬁbre optic cable service, Kerry pays a one-off amount of $1200 for the installation costs and then a monthly fee of $25. Write an equation of the total cost T of Kerry’s new internet service over n months.

Determine how many months it will take for Kerry to break-even on her new internet service.

There are 36 employees in a company, and the full-time employees outnumber the part-time employees by 16. Find the number of part-time employees and full-time employees in the company.

Two twin soccer players, Arden and Benton, are having a contest over who will score the most goals in the season. The twins have scored a combined total of 34 goals so far over the season and Arden has scored 6 more goals than Benton. Let x and y be the number of goals scored by Arden and Benton respectively.

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Write a pair of linear equations to model this situation.

Find the number of goals Arden has scored this season.

Find the number of goals Benton has scored this season.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Christa has $4000 to invest, and wants to split it up between two accounts: • Account A earns 7% annual interest. • Account B earns 8% annual interest. Her target is to earn $303 total interest from the two accounts in one year. Let x and y be the amounts, in dollars, that she invests in accounts A and B respectively.

Ex 4

Write a pair of linear equations to represent this situation.

Solve for y, the amount Christa invests in account B.

Solve for x, the amount Christa invests in account A.

The length of a rectangle is 12 units more than the width, and the perimeter of the rectangle is 56 units. Let y be the width and x be the length of the rectangle. a

Write a pair of linear equations to represent this situation.

Solve the simultaneous equations.

Consider the diagram of the rectangle: a

Solve for x and y.

Find the length of the rectangle.

Find the width of the rectangle.

3y + 384

8x + 60 D

B 15y + 48

64x + 12

Maria and Buzz both walk from their houses to the bus stop every morning. Maria walks 1.5 kilometres further, and together they walk 3.3 kilometres. Find the distance Buzz walks.

Qin invested $16 000 in total in two stocks, A and B. In one year, the investment in stock A made a 14% return, while the investment in stock B fell by 6%. The total annual interest from both stocks was $700.

Set up two equations by letting x and y be the amounts that she invested in stocks A and B respectively.

Solve the simultaneous equations.

Explain what the solution means in this context.

A clothing manufacturer is deciding whether to employ people or to purchase machinery to manufacture their line of t-shirts. The cost of employing people to make the clothing is modelled by y = 800 + 60x and the cost of using machinery is modelled by y = 3200 + 20x, where y is the cost and x is the number of t-shirts to be made. a

Sketch a graph that depicts the two cost functions.

For what value of x will the cost of making t-shirts be the same, whether they are made by people or by machines?

For what values of x will it be more cost eﬃcient to use machines to manufacture the t-shirts?

For what values of x will it be more cost eﬃcient to employ people to manufacture the t-shirts?

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Reasoning 17

17, 18

17, 19

Barry is considering opening a restaurant. He has calculated his ﬁxed costs at $10 000 per month. He expects to generate a revenue of $12 per meal, but the variable cost for each meal is $15. a

Write equations for the total cost C and total revenue R for selling x meals.

Can Barry reach the break-even point? Explain.

How does this impact Barry’s restaurant idea and what changes can he make for it to succeed?

Pensacola high school hosts an annual badminton tournament. Student tickets cost $3.00 and general admission tickets cost $5.50. a

If 350 tickets were sold for a total of $1450, how many of each type of ticket were sold?

Explain which method you used to solve and why.

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. a

How much is the cost of an order of oxtail stew and a slice of sweet potato pie?

Explain which method you used to solve and why.

Problem-solving 20

20, 21

22–24

A bacteria population doubles every day, following the equation y = 2x. A scientist predicts that the population will increase by 3 times the number of days, following the linear equation y = 3x. At what point will the scientist’s prediction first match the actual population?

A circular garden has a radius of 1 unit and its centre is at the point (2, 1). A straight path is built through the garden, following the equation y = x − 1. Find the points where the path intersects the garden’s boundary.

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For the simultaneous equations: y = −4x2 + 8x

y = 2x + c

Determine the value of c that will result in one unique solution to the simultaneous equations.

Determine the values of c that will result in two real solutions to the simultaneous equations.

A chef is experimenting with a new recipe that requires a speciﬁc ratio of ingredients. If 2 is added to the numerator and denominator it becomes

and if 3 is subtracted from the

numerator and denominator it becomes . Find the original ratio of the ingredients.

An engineer is designing a playground that will be placed at the centre of a hyperbolicshaped park. The engineer wants to place a fence along a straight line that intersects the park at two points. The situation can be modelled by the simultaneous equations: • Park: • Fence: Find the points where the fence intersects the park. Round your answers to two decimal places.

Investigation: Intersections of curves Investigate online

mathspace.co

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6 Chapter review 1

For each pair of linear equations: i

Sketch the lines on the same Cartesian plane.

Find the solution.

x = −4

2y = 5x − 12

y = −3x − 9 y = x − 3 2

Use the graph to solve the simultaneous equations: y = −3x −13

y = −x − 3

−8

−7

5 4 3 2 1

−6

−5

−4

−3

−2

−1

x + 2y = −6

y = −x − 3

y = −3x − 13 x + 2y = −6 3

−1 −2 −3 −4 −5

x 1

y = −3x − 13 x + 2y = −6

A boutique’s expenses are represented by E = x + 1220, and its sales revenue by S = 1.95x, where x is the number of dresses sold. a

What coordinates satisfy both equations?

What do the coordinates from part (a) mean in this context?

y 3500 3000 2500 2000 1500

E S

1000 500 x 256 512 768 1024 1280 1536

A pair of linear equations has inﬁnitely many solutions. One of the equations is y = 2x + 5. Which of these is the other equation? A

y = 2x + 7

Consider the simultaneous equations:

y = −3x − 4

y = −3x + 7

Without graphing, is there a solution? Explain your answer.

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−y = −2x − 5

A triangular garden has a base and two equal-length sides. One of the equal sides is 2 m longer than the base. The garden has a total perimeter of 22 m. a

Let y represent the length of one of the equal sides and x represent the base. Create two equations that represent this information.

Sketch the two lines on the same coordinate plane.

Find the length of one of the equal sides and the base of the triangle.

For the simultaneous inequalities y ≤ 2x − 1 and y > x − 4: a

Find the x-coordinate of the point at which the two lines intersect.

Find the y-coordinate of the point of intersection.

Graph the solution region.

Write the simultaneous inequalities that describe the shaded region. 8 7 6 5 4 3 2 1

-4 4-3 3-2 2 -11-11 -2 2 -3 3 -4 4 -5 5 -6 6

x 1

2 3 4 5

For y > x − 4, y ≥ −x − 4, and y < 1: a

Graph the lines y = x − 4, y = −x − 4, and y = 1 on the same set of axes.

Does the point (2, 0) satisfy the simultaneous inequalities?

Shade the region described by the inequalities y > x − 4, y ≥ −x − 4, and y < 1.

Calculate the area of the shaded region.

ChocDelight produces two types of chocolates: milk chocolate and dark chocolate. Each type of chocolate is made in different machines. The time that each chocolate requires in each machine is detailed in the table. Each machine can operate for 24 hours a day. Let x represent the number of milk chocolates produced each day, and y represent the number of dark chocolates produced each day. a

Write an inequality for Machine A.

Write an inequality for Machine B.

Write inequalities for x and y, given the number of chocolates cannot be a negative number.

Graph the region deﬁned by the inequalities for Machine A and Machine B.

Milk Chocolate Dark Chocolate

Machine A 5 hours 3 hours

Machine B 4 hours 6 hours

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Mirela has 16 centimetres of leftover cardstock paper that she wants to use to create a rectangular bookmark. Let x and y represent the length and width of the bookmark respectively. a

Write three inequalities to represent the situation.

Graph the solution to the simultaneous inequalities.

Are these possible dimensions of the bookmark?

d 12

A length of 4 cm and a width of 3 cm

A length of 5 cm and a width of 4 cm

iii

A length of 6 cm and a width of 5 cm

A length of 5 cm and a width of 2 cm

If Mirela wanted to create a square bookmark, what are the dimensions of the largest possible bookmark?

Solve the simultaneous linear equations using substitution: a

y = 4x + 40 y = 3x + 30

y = −2x – 11

c 2a + b = 23

4x + 2y = −22

d −3f + 2g = 6

b = −2a + 35

2f + 5g = 53

Use the substitution method to ﬁnd the solution(s) for each pair of simultaneous equations. a

(x + 2)2 + (y − 4)2 = 10

y = 4x + 2

x = −1 y = −x2 + x + 6 14

Solve the simultaneous linear equations using elimination: a

2x + 3y = 8

3x − 6y = 48

x − 3y = −14

x + 2y = 10

0.1x + 0.2y = 0.3

x − y = −22.5

Use the elimination method to ﬁnd the solution(s) for each pair of simultaneous equations. a

x2 − y = 9

x2 + y2 = 80

−x + y = 3 16

A couple runs two restaurants that made a combined profit of $10 000 in the previous financial year, with Fine Diners making 3 times as much profit as Casual Eats. Let x and y be the profits of Casual Eats and Fine Diners respectively.

Write a pair of linear equations to represent this situation.

Find Casual Eats’ proﬁt.

Find Fine Diners’ proﬁt.

Alfred invested $20 000 in total in two bonds, C and D. In one year, the investment in bond C had a 10% return, while the investment in bond D rose by only 4%. The total annual interest from both bonds was $1340. a

Set up two equations by letting x and y be the amounts that he invested in bonds C and D respectively.

Solve the simultaneous equations.

Explain what the solution means in this context.

Juma and Jian both cycle from their homes to the gym every evening. Juma cycles 2.2 kilometres further, and together they cycle 5.8 kilometres. Find the distance Jian cycles.

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“Everything connects to everything else.” –Leonardo da Vinci

Big ideas Functions and their properties play a crucial role in understanding and analysing relationships between variables.

Curriculum Outcomes VC2M10AA06 apply understanding of polynomials to sketch a range of curves and describe the features of these curves from their equation VC2M10AA08 use function notation to describe the relationship between dependent and independent variables in modelling contexts

Chapter outline

Introduction to functions (Optional) Chapter outline 7.01 7.02 7.03 7.04

Optional: Define relations and functions Optional: Domain and range Optional: Graphs of xn Optional: Transformation of functions Chapter 7 review

330 341 352 363 375

Theme parks use functions to analyse visitor patterns and improve experiences!

7.01

ptional: Define relations O and functions After this lesson, you will be able to… • understand what a relation is and how it can be represented. • classify relations. • understand what a function is. • write a function equation using function notation. • apply the vertical line test to determine if a graph is a function.

Functions and relations A relation is a relationship between sets of information. For example, consider the names of students in a class and their heights. If we choose a specific height, such as 162 cm, multiple students might have that height. This data can be expressed as a relation. Relations can be represented using tables, rules, graphs, ordered pairs, or diagrams mapping elements of two different sets. Relations can be classiﬁed by how sets of inputs and outputs are related. The following diagrams show the four types of relations for sets of inputs (x-values) and outputs (y-values). one-to-one x

1 2 3

2 4 6

One input relates to one output. one-to-many y x 0 4 9

−3 −2 0 2 3

One input relates to two or more outputs.

many-to-one x y −1 0 1 2

1 0 4

Two or more inputs related to one output. many-to-many x

1 2 3

5 7 9

Two or more inputs are related to two or more outputs.

A function is a special type of relation where each input corresponds to only one output. This means functions can be one-to-one or many-to-one mappings. Functions are often described using dependent and independent variables. The independent variable is the input of the function, usually x. The dependent variable is the output of the function, usually y.

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Functions can be written in function notation. This notation highlights the independent variable and names the function, which is helpful when working with multiple functions. To use function notation, the equation must represent a function and be of the form y =. When we are writing in function notation, instead of writing “y =”, we write “f (x) =”. This can be interpreted as “f is a function of the variable x” and read as “f of x”. Common letters to use for general functions are lower case f, g and h.

f (x) = 3x2 f is the function name f (x) is the function output, or y-value 3x2 is what the function does to the input (x) We can show substitution by replacing the input with a particular value. For example, instead of saying “substitute x = 3 into y = 2x + 1”, we can deﬁne f (x) = 2x + 1, then say “find f (3).” This means to substitute x = 3 to find the output. We read f (3) as “f of 3”. The result of the substitution is the output. For f (x) = 2x + 1, f (3) = 7. This shows that when x = 3, y = 7. We can write this as a point, (3, 7).

Example 1 Does the set of points describe a function? a {(2, 5), (7, −3), (5, 2), (−4, −9)}

Create a strategy Identify if the sets of points represent a many-to-one, one-to-many, many-to-many or one-to-one relationship. The relation is a function if it is a many-to-one or one-to-one relationship.

Apply the idea Let’s analyse the given set of points: {(2, 5), (7, −3), (5, 2), (−4, −9)} For a set of points to be a function, each x-value must have a unique corresponding y-value. • For x = 2, there is only one y-value which is (5). • For x = 7, there is only one y-value which is (−3). • For x = 5, there is only one y-value which is (2). • For x = −4, there is only one y-value which is (−9). Since this demonstrates a one-to-one mapping, it is a function.

b {(2, 5), (2, 7), (−3, −4), (−9, 13)}

Create a strategy If the same input (x-value) has different outputs (y-values), then it is not a function. If different inputs (x-values) have the same outputs (y-values), it can still be a function.

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Apply the idea Let’s analyse the given set of points: {(2, 5), (2, 7), (−3, −4), (−9, 13)}. For a set of points to be a function, each x-value must have a unique corresponding y-value. Let’s check: • For x = 2, there is more than one y-value, which are (5) and (7). • For x = −3, there is only one y-value, which is (−4). • For x = −9, there is only one y-value, which is (13). Since there are x-values with multiple y-values, it is a one-to-many relation. This set of points is not a function.

Example 2 The pairs of values in the table represent a relation between x and y. Do they represent a function? x y

−4 −4

−3 −3

−2 −2

−1 −1

0 0

1 −1

2 −2

3 −3

4 −4

Create a strategy Determine whether each input (x-value) has a unique output (y-value).

Apply the idea The table represents a many-to-one relation because each x-value only appears once, but multiple x-values share the same y-value. For example, x = −3 and x = 3 both have a single y-value, which is −3. Given the many-to-one relation between x and y-values, the pairs of values in the table represent a function.

Example 3 Consider the function f (x) = 4x + 4. a Evaluate f (2).

Create a strategy Substitute the x-value into the expression.

Apply the idea f (2) = 4(2) + 4

332

Substitute x = 2

=8+4

Evaluate the multiplication

= 12

Evaluate the sum

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b Evaluate f (−5).

Apply the idea f (−5) = 4(−5) + 4

Substitute x = −5

= −20 + 4

Evaluate the multiplication

= 16

Evaluate the sum

Reﬂect and check f (−5) = 16 means that (−5, 16) is a point on the graph of f (x) = 4x + 4.

Idea summary A relation is an association between elements of two sets. It can be represented in various ways, including tables of values and graphs. We can deﬁne relations as being: • • • •

One-to-one: one x-value maps to one y-value. Many-to-one: many x-values correspond to one unique y-value. Many-to-many: multiple x-values map to multiple y-values. One-to-many: one x-value may map to multiple y-values.

A function is a special type of relation where each input has only one output. This means functions can be one-to-one or many-to-one mappings.

Vertical line test To determine if a graph represents a function, we use the vertical line test. If a vertical line intersects a graph at more than one point, then the graph does not represent a function. This shows that each x-value does not have a unique y-value, making the graph a one-to-many or a many-to-many relation.

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Relations

Many-to-one and one-to-one mappings are functions as they would pass the vertical line test. Many-tomany and one-to-many are only relations as they would not pass the vertical line test.

y x

many-to-many Functions y

one-to-many y x

one-to-one

many-to-one

Example 4 Do the graphs represent a function? y

1 −4 −3 −2 −1

−1

−4 −3 −2 −1 −1

−3

−4

x 1

−2

x 1

−6

−4

−2

−2 −4 −6

Create a strategy Apply the vertical line test. The vertical line test states that for a graph to represent a function, any vertical line drawn on the graph should intersect the graph only once. 4

Apply the idea

3 2 1 −4 −3 −2 −1 −1

x 1

No matter where we draw a vertical line on the graph, each vertical line intersects the graph at exactly one point. This means that this graph is a function.

−2 −3 −4

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When we draw vertical lines in various places on the graph, each line intersects the graph at exactly one point. Since the vertical line test is satisfied, this graph represents a function.

3 2 1

−4 −3 −2 −1 −1

−2 −3 −4

When we draw vertical lines in various places on the graph, the line intersects the graph at more than one point. This graph fails the vertical line test, so it does not represent a function.

7 y 6 5 4 3 2 1 −7−6−5−4−3−2−1 −1 −2 −3 −4 −5 −6 −7

1 2 3 4 5 6 7

Idea summary The vertical line test is a graphical method to determine if a graph represents a function. If any vertical line intersects the graph at more than one point, then it’s not a function.

7.01 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–4

Trailblazer 1–5

Are the statements true or false? a

When working with a function, substituting a certain value of x into the equation gives only one value of y for that value of x.

All relations are functions.

Some relations are functions.

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All functions are relations.

A relation always passes the vertical line test.

The graph of every straight line is a function.

The graph of every straight line is a relation.

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

Imagine you have a machine that represents a function. If you feed an item into the machine, it gives you an output. Can you get different outputs for the same input? Why or why not?

For a function f, what does f (3) represent?

Suppose that (7, −6) is an ordered pair that satisﬁes the function g. Write this situation using function notation.

Suppose that the value of the function f is 22 when x = 7. Write this situation using function notation. 6, 7ad, 8, 13ab, 14, 17a–e, 18, 20

Fluency Ex 1

Ex 2

{(1, 6), (2, 12), (3, 18), (4, 24)}

{(10, 2), (2, 4), (3, 5), (10, 3)}

{(−2, 4), (−1, 4), (0, 5), (1, 3)}

{(−2, −3), (−1, 0), (0, 2), (1, 4), (1, −1)}

Is each table a function?

x y

−4 −4

−3 −3

−2 −2

−1 −1

0 0

1 1

2 2

x y

−2 1

−2 2

0 3

1 4

2 3

2 4

3 3

4 4

5 5

x y

−9 10

−7 10

−6 10

−5 10

−3 10

−2 10

3 10

5 10

10 10

4 4

3 3 16

4 4 18

Given the points in the table. i

Plot the points on the Cartesian plane.

Is this relation a function?

336

6, 10, 12, 13, 15, 16, 17c–j, 19, 20

Does each set of points describe a function?

6, 7, 9, 11, 13, 15, 16ab, 17b–h, 20

x y

−4 4

−3 3

0 2

−1 1

0 0

1 1

2 2

3 3

x y

−3 0

−2 1

1 2

2 3

3 4

4 3

5 4

6 5

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A relation is deﬁned as y = 1 if x is positive and y = −1 if x is 0 or negative. a

Complete the table. x y

−4

−3

−2

−1

Plot the points on the Cartesian plane.

Is this relation a function?

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

Plot the ordered pairs on the Cartesian plane.

Is this relation a function?

Which ordered pair would need to be removed from the set so that the remaining ordered pairs represent a function?

For which of these values of k does the relation {(2, 1), (7, k), (5, 8), (3, 9)} not represent a function? A

Ex 3

k=7

k=8

k=9

o such value of N k exists

Find a value of k such that the relation {(6, 2), (8, 5), (1, 7), (k, 4)} does not represent a function.

Do the equations describe a function?

y = 9x

y = x2 + 2

y2 = x

y = −x3

3x + 5y = 12

f (−4)

y2 + x2 = 25

5x − 5y + 1 = 0

If f (x) = 3x − 1, ﬁnd: a

f (3)

Consider the function f (x) = 2x2 − 2x + 5. Evaluate

For each function, evaluate the function outputs when:

x = 2

x = −3

f (x) = 4x + 1

g(x) = (x + 1) (x + 2)

h(x) = x2 − 3x − 2

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Ex 4

Is this graph a function? a

y 4

4 3

x 1

-2 -3

-3

-4

y 4

1 -4 -3 -2 -1 -1

-2 -3

-3

-4

-4 -3 -2 -1 -1

-2 -3 -4

7 y 6 5 4 3 2 1 -7-6-5-4-3-2-1 -1 -2 -3 -4 -5 -6 -7

-4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11

-2

4 y 3 2 1

-4 -3 -2 -1 -1

-3

-2

-4 -3 -2 -1 -1

-3

x 1

4 2

1 -4 -3 -2 -1 -1

-2 -3 -4

1 2 3 4 5 6 7

CheapCalls Mobile charges $1.10 a minute plus a connection fee of 70c for an international call. a

Complete the table. Call length (minutes) International call cost (dollars)

338

-2

-4 -3 -2 -1 -1

3 1

-4 -3 -2 -1 -1

Is this relation a function?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A particular store is offering one free T-shirt for every two T-shirts purchased, where each T-shirt costs $22. a

Complete the table. Number of T-shirts (x) Total cost

b 20

Is this relation a function?

The model P (t) = 3500 × 2t describes the population of fish in a lake after t years, where t = 0 is the initial time when the population of fish was first recorded. Find and interpret the value of P (2).

Reasoning

21, 22, 25

21, 23, 27

21, 24, 26, 28

Given the ordered pairs (a, b), (c, d), (e, f ), if this set represents a function, what can you infer about the values of a, c, and e?

Sharon works at a pizza shop and makes $15 per hour. Her wages can be represented by the function W (t) = 15t, where t represents the number of hours that she works.

Explain what the expression W (19) represents.

Explain what the equation W (t) = 165 represents.

Find her wages after working 26 hours.

A bookstore offers a yearly membership for $120 which includes unlimited book rentals. Additionally, the bookstore provides an option for non-members to rent a book for $5 per book. Is the relation between the number of books rented and the total rental cost a function? Explain why or why not.

Imagine a cinema where each seat corresponds to a unique ticket number. If the seating chart represents a function, what can you infer about any two people having tickets with the same number?

John lists his pizza topping preferences: • For Margherita, he likes olives. • For Pepperoni, he likes jalapeños. • For Vegetarian, he likes bell peppers. Does this represent a function? Justify your answer.

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You have a coffee machine that gives speciﬁc types of coffee for each button pressed. If pressing the “Espresso” button always gives an espresso shot and pressing the “Latte” button always gives a latte, can this machine’s operation be described by a function? Why or why not?

In a small town, every family owns at least one car, but multiple families can own the same type of car. Is the relation between families and the type of car they own a function? Explain.

The local bakery uses a set of ingredients (flour, sugar, eggs) to make a set of pastries (cakes, cookies, muffins). If each pastry type requires all three ingredients but in different quantities, explain whether the relation of ingredients to pastries is a function or not.

Problem-solving 29, 32 29

30–33

Consider the functions f (x) = 7x + 2 and g(x) = 15 − 2x. Find the value of the following: a

30, 32

f (1) + g(2)

f (−2) − g(7)

f (0) × g(0)

For each of the following functions, evaluate the function outputs when x = a + 1. a

f (x) = 4x + 1

h(x) = x2 − 3x − 2

g(x) = (x + 1) (x + 2)

Consider the function f (x) = x2 + 2x + k. a

Simplify the expression for f (2).

If f (2) = 15, ﬁnd the value of k.

You have two sets of numbers. Set A: 1, 2, 3 and Set B: 4, 5, 6. Design a relation between these sets such that it is a function. Then add one more ordered pair to make this relation no longer a function.

Given the relation (2, 4), (3, 9), (4, 16), (5, 25), determine if this represents a function. If yes, brieﬂy explain the real-world scenario it could depict.

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7.02

Optional: Domain and range After this lesson, you will be able to… • use algebraic methods to determine the domain and range. • utilise graphs to identify domain and range. • distinguish between the domain and range in different types of functions.

Domain and range of a function The domain is the set of all allowable input values (usually x-values) for which the function is deﬁned. We can find the domain either algebraically or from the graph of the function. A function is deﬁned for all values of x except where: • The denominator of a fraction is zero, the position of a vertical asymptote on the graph. • The number under an even root is negative. • The argument of a logarithm is negative or zero. The domain may also be found from the graph. The domain is all x-values for which the function has a corresponding y-value. 4 3 2

The domain is the set of allowable x-values which, for this example, is all x greater than 0, written as x > 0. The vertical asymptote at x = 0 means the graph will get infinitely close to, but will never actually reach, x = 0.

y Vertical asymptote

−3 −2 −1 −1

This is confirmed by looking at the graph, we can see that if x = 0, f(0) is undefined.

−2 −3 −4

The range is the set of all possible output values (usually y-values) as x varies over the domain of the function. Just like the domain, the range can be determined either algebraically or from the graph of the function. When considering the range of a function, we must remember that: • The minimum and maximum values of a function will limit its range. • A horizontal asymptote will limit a function’s range. 4

The range of the function is the set of all possible y-values, which in this case excludes y = 0.

3 Horizontal asymptote

The graph approaches the horizontal asymptote y = 0, but it will never intersect the x-axis.

2 1

−8 −6 −4 −2 −1 −2 −3

x 2 4

From the graph, we can see that as x moves away from 2 (vertical asymptote), the function values either approach infinity or negative infinity, confirming that the range is all real numbers except 0.

−4

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The domain and range of a function are dependent on the type of function. Function type Linear Quadratic

Equation y = mx + c y = ax2 + bx + c

Exponential

y = ax where a > 0 All real numbers and a ≠ 1 y = loga(x) where x>0 a > 0 and a ≠ 1 All real numbers except where p (x) = 0

Logarithmic Rational

Root

Domain All real numbers All real numbers

If n is even, the domain is x ≥ 0. If n is odd, the domain is all real numbers.

Range All real numbers If a is positive, the range is y ≥ (turning point’s y-value) If a is negative, the range is y ≤ (turning point’s y-value) y>0 All real numbers Typically needs more in-depth analysis. Watch for horizontal asymptotes for clues. If n is even, the range is y ≥ 0. If n is odd, the range is all real numbers.

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Example 1 State the domain of y =

Create a strategy Identify the value(s) of x for which the function is undefined algebraically, and write an inequality that excludes the value(s).

Apply the idea The function is undeﬁned when the denominator is equal to zero. x−3≠0 x≠3

Write an inequality for where the function is deﬁned Add three to both sides to solve for x

So the domain is x ≠ 3.

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Reﬂect and check The domain of x ≠ 3 means that a y-value can be calculated for any real x-value except x = 3. We can check our answer by graphing the function using technology:

y 4 3 2

The graph shows that there is a vertical asymptote at x = 3. This means that the function is undeﬁned at x = 3.

1 -1

x 1

-1

-2

1 x-3

-3 -4

Example 2 The graph of f (x) =

is shown.

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

a What is the domain of the function? Express as an inequality.

Create a strategy

Apply the idea

Use the graph to identify the values of x for which the function is deﬁned.

We can see that the graph starts at x = −1 and extends indeﬁnitely to the right. 4

3 2 1 −4 −3 −2 −1

−1

−2 −3

x 1

Domain

−4

So, the domain is x ≥ −1.

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b Is there a value in the domain that can produce a y-value of −2?

Create a strategy Determine if there is an x-value that, when substituted into the function, produces a value of −2. Remember, a square root always produces non-negative values.

Apply the idea The square root function always produces non-negative outputs. It’s impossible for the function to produce a value of −2 since that’s a negative value. So, there’s no x-value in the domain that will produce an output of −2 for the function

Reﬂect and check Looking at the graph, we can see that it does not extend down to y = −2. This verifies our answer.

Example 3 State the range of y = x2.

Create a strategy Draw a graph or use technology to graph the function, then identify the range.

Apply the idea

Reflect and check

The graph of the function is an upward-facing parabola with its turning point (vertex) at the origin (0, 0). The minimum value of y is 0, and y increases to positive inﬁnity.

Alternatively, we can use the fact that even powers always result in non-negative outputs: x2 ≥ 0 which means that y ≥ 0.

y = x2

3 Range

2 1

−4 −3 −2 −1

−1

x 1

−2 −3 −4

So the range is y ≥ 0.

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Example 4 State the domain and range of y = 3x.

Create a strategy Use technology to graph the function, then use the graph to identify the domain and range.

Apply the idea y 75

25 x −4 −3 −2 −1

On the x-axis, the graph extends from negative infinity to positive infinity. So, the domain is all real numbers. The y-value is always positive. As x approaches negative infinity, y approaches 0 (but never equals 0). This is because there is an invisible, horizontal line (called an asymptote) on the x-axis that the graph cannot cross. As x approaches positive infinity, the function approaches infinity so, the range is y > 0.

Reﬂect and check As shown in the table given earlier in the lesson, all exponential functions have a domain of all real numbers. The range is restricted by the horizontal asymptote.

Idea summary The domain is all acceptable input values (usually x-values) for a function. Certain conditions limit the domain of functions, such as: • zero denominators • negative values under even roots • negative or zero values in the argument of logarithms The range is the set of possible output values (usually y-values) as x varies over the domain of the function.

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7.02 Practice questions Understanding

Trailblazer 1–3

If a function’s equation does not imply any restrictions on x, what can we say about its domain?

What are three conditions that can limit the domain of a function?

True or false: a

The domain of a rational function is the set of all real numbers except for those that make the denominator 1.

If the graph of a function lies entirely above the x-axis, then its range is only positive y-values.

The range of a quadratic function is all real numbers.

4ab, 5, 6bc, 7a–d, 8ab, 4–6, 7c–g, 8bc, 10, 11, 9, 11, 12, 14a–d, 15a–d, 16 13, 14b–e, 15c–f, 16, 17

The function

is shown. 4

3 2 1 −1 −1

x 1

−2 −3 −4

346

4, 5, 7d–h, 8cd, 11, 14e–h, 15e–h, 18

State the domain of each function. a

Ex 2

Adventurer 1–3

Fluency Ex 1

Explorer 1–3

What is the domain of the function?

Is there a value in the domain that can produce a y-value of −2?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For each of the given graphs, state the domain. a

4 3

1 −4 −3 −2 −1

−7 −6 −5 −4 −3 −2 −1 −1

−2

−3

−4

10 8 6 4 2

−1

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

3 2 1 x

−4 −3 −2 −1 −1

2 4 6 8 10

x 1

−2 −3 −4 −5

For the given graphs: i

State the domain.

1 −4 −3 −2 −1

State the range.

−1

x 1

1 −7 −6 −5 −4 −3 −2 −1 −1

−2

−3

−4

x 1

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4 3

−4 −3 −2 −1

1 2 3

−2

−3

−4

−4 −3 −2 −1

−1

−4 −3 −2 −1

348

x 1

−1

−2

−3

−4

10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

−2

3 1

x 1

−1

−2

−7 −6 −5 −4 −3 −2 −1 −1

6 4 2

−12−10−8−6−4−2 −2

2 4 6 8 10

x 2 4 6 8 10 12

−4 −6 −8

For the given functions: i

Plot the graph of the function.

What is the domain of f (x)?

iii

What is the range of f (x)?

f (x) = −x + 1

f (x) = 3x − 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

f (x) = −4

Given the graph of the function y = f (x): a

What is the absolute maximum of the graph?

What is the range of the function?

What is the domain of the function?

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

Given the graph of the function y = f (x): a

What is the absolute minimum of the graph?

What is the range of the function?

What is the domain of the function?

x 1 2 3

80 70 60 50 40 30 20 10 −8−6−4−2

For the quadratic function that passes through the points shown in each table of values: i

Does the function have a maximum or minimum point?

What is the minimum value of the function?

iii

What is the range of the function?

x 2 4 6 8 10 12 14 16 18

x y

1 17

x y

−5.5 −12

2 7

3 1 −5 −7

4 −1

5 1

6 7

−4.5 −4

−4 −3

−3.5 −4

7 17 −3 −7

−2.5 −12

Given the function f (x) = (x − 2)2, for all real x. a

Complete the table of values. x y

0 ⬚

1 ⬚

2 ⬚

3 ⬚

4 ⬚

Graph the function.

Is there a value in the domain that can produce a y-value of −9?

7.02 Optional: Domain and range mathspace.co

349

Ex 3

Given the parabola deﬁned by the equation y = x2 + 5: a

Is the parabola concave up or concave down?

What is the y-intercept of the parabola?

What is the minimum y-value of the parabola?

What is the range of the parabola?

State the range of the functions: a e

15 Ex 3

y = x2 + 1

y = 2x2

y = −x2 − 5

c g

y = x2 + 9

y = 4 − x2

y = −x2

y = (x + 1)2

For each function: i.

Use technology to graph the function.

ii.

State the domain and range of the function.

y = 2x

y = log2 x

y = log(x − 1)

e 16

y = x2 − 7

y=3 +4

The function

has been graphed.

What is the domain of the function?

What is the range of the function?

4 2 x −6 −4 −2

−2 −4 −6

Given the function y = x − 1. a

Complete the table of values. x y

−2 ⬚

−1 ⬚

0 ⬚

1 ⬚

2 ⬚

Sketch the graph of the function.

Is there a value in the domain that can produce a y-value of

State the domain and range for: a

y = x3 + 2x2 − 1

Reasoning

19–21

19, 21, 22

19–22

If the range of a function is given as all non-negative values, does that imply that the function’s graph is always above the x-axis? Justify your answer.

Explain why the domain of the function

A factory produces a number of items, y, as a function of hours worked, x, given by y = 100 − x2. Why can’t the factory work more than 10 hours?

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is all real numbers.

A theme park’s popularity y is modelled by the function y = 2x, where x is the number of years since its opening. What can we say about the popularity of the park over time?

Problem-solving 23, 26, 28, 29 23

24–28

24–29

The function f is used to determine the area of a square given its side length. a

Which of the following values is not part of the domain of the function? A

−8

6.5

For n ≥ 0, what is the area function for a side length of n?

Plot the graph of the function f.

A whale jumps above the surface of the ocean and then dives back under. Its height above the surface can be sketched on a set of axes where x represents time and y represents distance above or below the surface. State the domain of the graph.

A function f (n) describes the maximum height of a ball dropped from a certain height after the nth bounce. What is the domain of the function?

The graph of a function is shown.

What is the range of the function?

150

A ball is thrown from an apartment window in a high-rise building. The height of the ball above the ground, y, can be modelled using the function shown in the graph, where x is the time in seconds. If the ball is thrown at x = 0, what is the range of the function in this context?

125 100 75 50 25 −8 −6 −4 −2 −25

x 2 4 6 8 10

A comet is travelling through the solar system. It passes by the Earth before curving around the sun and heading back out into deep space. The distance of the comet from Earth, y, is tracked by satellite telescopes and expressed as a function of time, x, since the comet was ﬁrst spotted.

What is the domain of this function?

The satellites detect that, at its closest point, the comet was a distance of 0.14 AU (Astronomical Units) from Earth. What is the range of the function?

The number of bees in a colony is initially measured and left to form a hive. The number of bees in the hive is measured each day over the course of one week. The function H (x) = 30 (3)x is found to model the number of bees, H, after x days. a

What is the domain of the function?

What is the initial number of bees in the colony?

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An industrial process involves heating a material from 25 °C to 80 °C in order to remove impurities. Using the standard approach, the temperature of the material over time is given by the function T = −t (t − 16) + 25. A graph of the temperature proﬁle is shown, where temperature T is given in degrees Celsius (°C) and time t is given in minutes. 90

T°C

80 70 60 50 40 30 20 10

t (minutes) 1

What is the domain of this function?

What is the range of this function?

A new heating process is being developed that uses more energy but takes less time.

This regime is based on a different quadratic function, and the table shows the theoretical value of the temperature at different points in time using the new function. t T

0 25

1 45

3 80

4 96

30 96

31 80

32 63

33 45

34 25

35 4

Over what domain does this new function have the same range as the standard process? d

Which of the two domains found in part (c) should be used for the new heating process?

7.03

Optional: Graphs of xn After this lesson, you will be able to… • Identify cubic functions and explain their characteristics. • complete tables of values for given cubic functions to understand their behaviour. • sketch graphs of cubic functions and polynomial curves using data points and equations. • understand and describe the end behaviour of polynomial functions with even and odd degrees. • explore the effects of the degree and leading coefficient on a polynomial curve’s shape.

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Cubic function

We have seen linear and quadratic functions that had an x and x2 term respectively. These functions are part of a larger group of functions called polynomial functions. Polynomials are classified by their degree, which is the highest power in the function. For example: Equation

Degree

y=2 y = 3x y = −9x2 y = 6x3

0 1 2 3

Function type Constant Linear Quadratic Cubic

The cubic function, is a type of polynomial function where the highest power is 3. The graph of a cubic function of the form y = ax3 + c can have different shapes based on the values of a and c.

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y = x3: • has a point of inﬂection at (0, 0) where the steepness of the curve ﬂattens out before beginning to get steeper again. • is an increasing function because as x increases, y also increases.

Example 1 Sketch the graph of y = 3x3 − 5.

Create a strategy Create a table of values and plot the points to sketch the graph.

Apply the idea We will substitute several x-values into the function to find the corresponding y-values. x −2 −1 0 1 2

3(−2)3 − 5 3(−1)3 − 5 3(0)3 − 5 3(1)3 − 5 3(2)3 − 5

y −29 −8 −5 −2 19

Coordinates (−2, −29) (−1, −8) (0, −5) (1, −2) (2, 19)

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353

Now, we can plot the points on a Cartesian plane. y 16 15 12 9 6 3 −2

−1 −3

2 y = 3x3 – 5

−6 −9 −12 −15 −18 −21 −24 −27

Reﬂect and check This graph is narrower and shifted down from the basic graph of y = x3. 8

y = x3

6 4 2 –4 –3 –2 –1

–2 –4

x 1

y = 3x3 – 5

–6 –8

Idea summary The cubic function, y = x3, has a point of inﬂection at (0, 0) where the steepness of the curve ﬂattens out before beginning to get steeper again. The shape of the basic curve y = x3 is always increasing.

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Graphs of xn The shape of the general polynomial y = kxn changes based on the degree, n. We can identify similarities and differences based on the degree.

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For a function in the form y = kxn, where n is a positive integer, we call k the leading coeﬃcient and n the degree. We can identify trends in the end behavior of a function based on its degree and leading coefficient. For polynomials with an even degree, the end behaviour will tend toward the same direction. y

Positive leading coeﬃcient, even degree

Example of an even degree: y = 3x6

Negative leading coefficient, even degree

Example of an even degree: y = −2x8

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355

For polynomials with an odd degree, the end behaviour will tend toward opposite directions. y

Negative leading coefficient, odd degree

Example of odd degree: y = 8x5

Negative leading coefficient, odd degree

Example of odd degree: y = −7x9

We can describe the end behaviour of a polynomial using mathematical notation. For example, we can describe the end behaviour of y = −7x9 by writing: • As x → −∞, y → ∞. • As x → ∞, y → −∞. We read this as, “As x approaches negative infinity, y approaches infinity.” Between these ends, y = kxn will have a turning point if n is even and a point of inflection if n is odd. The higher the power of n, the flatter the point of inflection or turning point will be. y

Odd degree and n = 3 356

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Odd degree and n = 9

Even degree and n = 2

Even degree and n = 10

Example 2 Consider the function y = x7. a Describe the end behaviour of the function.

Create a strategy To describe the end behaviour of a polynomial function, we look at the degree of the polynomial and the sign of the leading coeﬃcient.

Apply the idea y

The given function is y = x7, which has an odd degree (7) and a positive leading coeﬃcient (1). The end behaviour is: • As x → −∞, y → −∞ • As x → +∞, y → +∞

b Sketch the graph of the function.

Create a strategy We can sketch the graph of the function by sketching a curve that reflects the described end behaviour and the properties of an odd-degree polynomial with a positive leading coefficient. Since n = 7, the point of inflection will be fairly flat.

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Apply the idea

Reﬂect and check 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

We can also graph using technology.

x 1 2 3 4 5

Idea summary The degree of a polynomial and the leading coefficient determine the end behaviour. • If the degree is even, the end behaviour tends toward the same direction • Positive leading coefficient: both ends tend toward +∞. • Negative leading coefficient: both ends tend toward −∞. • If the degree is odd, the end behaviour tends toward opposite directions • Positive leading coefficient: the left side falls toward −∞ and the right side rises toward +∞. • Negative leading coefficient: the left side rises toward +∞ and the right side falls toward −∞.

7.03 Practice questions Understanding

Explorer 1–3

Adventurer 1–3

For the general equation y = kxn what does n represent?

For y = x5, is the function increasing or decreasing?

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Trailblazer 1–3

Is each statement true or false for functions of the form y = kxn? a

This graph could have a positive leading coeﬃcient and an odd degree.

This graph could have a negative leading coeﬃcient and an even degree.

If k = −3, then an even degreed function would open upwards.

If k = 2 and n = 9, then x → ∞ as y → −∞.

If n = 9, then there would be a point of inflection at (0, 0).

The graph of y = 3x6 would have a ﬂatter turning point than y = 3x2. 5, 6a–d, 7a–d, 8–10, 11a–d, 12a–d, 13a–d, 14, 15

Fluency Ex 1

Sketch the graph of y = 2x3 + 4.

For each of the given graphs:

5, 6aefg, 7a–d, 9, 10, 11c–f, 12a–d, 13c–f, 15

4, 5, 6e–h, 7adef, 9, 10, 11e–h, 12e–h, 13e–h, 15

Is the degree odd or even?

What are the intervals where the function is increasing and decreasing?

iii

Describe the end behaviour of the function.

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

f (x) = x6

f (x) = −3x4

3 2 1

x −4 −3 −2 −1

1 2 3 4 5

−1

x 1

f (x) = 2x5

−2 −3 −4

What is the end behaviour of the graph of f (x)? a

b f

f (x) = −x7

f (x) = 4x3

c g

f (x) = −x8

f (x) = −5x3

f (x) = 3x2

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359

For each of these graphs of the function y = f (x): i

Is the degree odd or even?

Determine the behaviour of the function as x → ∞.

iii

Determine the behaviour of the function as x → −∞.

−5 −4 −3 −2 −1 −2 −4 −6 −8 −10

8 6

x 1 2 3 4 5

4 2 x −4

1 2 3 4 5

−4 −6 −8

3 2 1 x 1

−1

360

1 2 3 4 5

−2

−5 −4 −3 −2 −1

Ex 2

−1

−2

10 8 6 4 2

−3 −2

y 10

−5 −4 −3 −2 −1 −2 −4 −6 −8 −10

10 8 6 4 2

Consider the function y = x5. a

Describe the end behaviour of the function.

Sketch the graph of the function.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1

Consider the functions y = x2 and y = x4. a b

For even powered functions in the form y = xn, desrcibe what you notice as n increases.

The function f (x) = x3 is transformed into the function g (x) = kx3. a

Determine the values of k that will make the graph of g (x) wider than the graph of f (x) and reflected across the x-axis.

Choose an appropriate value for k based on your previous answer, and sketch the graph of the functions f (x) = x3 and g (x) = kx3 on the same Cartesian plane.

Sketch each graph using technology. a

y = 2x2

y = −4x3

y = −0.5x7

y = 8x8

Sketch each graph using technology. a

y = 2x2 + 5

y = 5x4 − 4

y = −0.25x7 + 1

y = 12x8 − 5

Sketch each graph using technology. a e

Sketch the graphs of the functions on the same Cartesian plane.

y = −3(x + 1)2

y = 4(x − 3)3

y = −8(x − 7)8

y = −9x4

y = −7x3 − 3

y = 9(x + 9)4

y = 7x5

y = −14x10

y = −2x5 + 6

y = −4x10 + 1

y = −7(x − 6)5

y = 14(x + 2)10

Sketch the graph of these functions on the same Cartesian plane: y = x3 and y = x3 − 4

The given graph models the revenue earned over the ﬁrst year of a small business launching. 40 000

Revenue in $ ( y)

35 000 30 000 25 000 20 000 15 000 10 000 5000

Time in months (x) 1 2 3 4 5 6 7 8 9 10 11 12

Identify when the revenue is increasing.

Identify when the revenue is negative.

Predict the revenue after one year.

Predict when the revenue reaches $25 000.

7.03 Optional: Graphs of xn mathspace.co

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Reasoning 16

16, 17

16–18

The given graph shows the population of crows at a local park over 15 weeks.

Estimate when the number of crows is decreasing. 80

Predict the number of crows after 10 weeks.

Predict when the number of crows will reach 70.

Describe what happens as the number of weeks increases indeﬁnitely, as x → ∞. Explain if this is reasonable.

Number of crows ( y)

60 50 40 30 20 10

Time in weeks (x) 2

Explain how the end behaviour of a function is determined by the leading term.

Florin is using the following graph to describe information about the equation of the polynomial function. Florin states that there is not enough information to determine anything about the equation of the function. Determine whether Florin is correct and explain your reasoning.

y 8 6 4 2

−8 −6 −4 −2 −2

2 4 6 8

−4 −6 −8

Problem-solving 19, 20a–c 19

19, 20a–c, 21

The volume of a sphere has the formula

The graph relating r and V is shown. a

Complete the table of values, rounded to two decimal places. r V

362

A sphere has radius measuring 3.5 m. Determine the exact interval that the volume of the sphere falls within. According to the graph, what is the radius of a sphere of volume 288π m3?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

19–21

324π

V (cubic metres)

288π 252π 216π 180π 144π 108π 72π 36π

r (metres) 1

A cube has side length 5 cm and a mass of 375 g. The mass of the cube is directly proportional to the cube of its side length. a

Let k be the constant of proportionality for the relationship between the side length, x, and the mass, m, of a cube, such that m = kx3. Solve for k.

Now state the equation relating the mass and side length of the cube.

Sketch a graph of this equation.

Find the mass of a cube with side length 6.5 cm, to the nearest gram.

A cube has a mass of 576 g. Determine what whole number value its side length is closest to.

The mass in grams, M, of a cube of cork is given by the formula M = 0.28l3, where l is the side length of the cube in centimetres. a

What is the mass of a cubic centimetre of cork to the nearest 0.01 of a gram?

Complete the table of values, to the nearest 0.01 of a gram if necessary. Side length(cm) Mass (g)

Sketch the graph of the function.

7.04

Optional: Transformation of functions After this lesson, you will be able to… • define what a transformation of a function is and identify the different types of transformations. • understand and explain the concepts of shifts, stretches, compressions, and reflections. • analyse the effects of transformations on the graph of a function. • apply the knowledge of transformations to change the position or shape of a function’s graph. • identify and describe the transformations applied to a given function.

Transformation of functions A transformation of a function is a change in the position or shape of its graph. Transformations can be categorised as translations (shifting the graph up, down, left, or right), dilations (stretching or compressing the graph), or reﬂections (creating a mirror image of the graph across a line). In this lesson, we will explore different types of transformations and how they affect the graphs of various functions. A translation is a transformation in which every point in a ﬁgure is moved in the same direction and by the same distance.

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A dilation is a transformation in which every point in a figure is scaled by the same factor. Dilations can be described as a compression, where all points are compressed towards the x-axis, or as an enlargement where all points are pulled away from the x-axis. A reflection is a transformation that produces the mirror image of a figure across a line. Let’s consider the quadratic function y = a(x − h)2 + k. This function represents a parabola with turning point (h, k) and a vertical stretch or compression factor of a.

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As we have experimented with applets and summarised the effects of a, b, h and k for each type of function, did you notice that each variable always had the same effect? We found that: • a dilates the graph by a factor of a from the x-axis. • When a < 0 the graph was reﬂected across the x-axis. • h translates the graph h units horizontally. • k translates the graph k units vertically. 1 • b dilates the graph by a factor of from the y-axis. b • When b < 0, the graph was reflected across the y-axis. These transformations work in the same way for a general function f (x). Let’s review and summarise them. Transformation −f (x) af (x) f (x − h) f (x) + k f (−x) f (bx)

Effect on graph reflects f (x) across the x-axis dilates f (x) by a factor of a from the x-axis translates f (x) by h units horizontally translates f (x) by k units vertically reflects f (x) across the y-axis 1 dilates f (x) by a factor of from the y-axis b

Effect on coordinates multiplies y-coordinate by −1 multiplies y-coordinate by a adds h to each x-coordinate adds k to each y-coordinate multiplies x-coordinate by −1 divides x-coordinate by b

Example 1 If the graph of y = −x2 is translated horizontally 5 units to the right and translated vertically 6 units upwards, what is its new equation?

Create a strategy A horizontal translation involves moving to the left or right, where right is the positive direction and left is the negative direction. A vertical translation involves moving to up or down, where up is the positive direction and down is the negative direction.

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Apply the idea When a graph of y = f (x) is translated horizontally by h units to the right, its new equation is y = f (x − h). When a graph of y = f (x) is translated vertically by k units upwards, its new equation is y = f (x) + k. The new equation is y = −(x − 5)2 + 6.

Example 2 Determine the coordinates of the corresponding point after the given transformations: a Point A is (5, 2). The transformation is y = f (x + 4) − 2.

Create a strategy Start inside the bracket and work outwards and for each transformation consider if it is affecting the x-values or y-values.

Apply the idea First, x + 4 means the point will be shifted 4 units to the left. When moving to the left, the x-values are decreasing. This means we need to subtract 4 from the x-values. x=5−4=1 Then, the −2 means the point will be shifted 2 units down. When moving downward, the y-values are decreasing. This means we need to subtract 2 from the y-values. y=2−2=0 The new coordinates are (1, 0). b Point A is (−3, 2). The transformation is y =

f(x−2).

Apply the idea First, the

is a vertical dilation, which means we need to multiply the y-value by . y=2×

2 3

Then x − 2 means the point is translated 2 units to the right. When moving to the right, the x-values are increasing. This means we need to add 2 to the x-value. x = −3 + 2 = −1 The new coordinates are

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Idea summary Here is a summary table demonstrating how transformations apply to a general function f (x). Transformation Effect on graph −f (x) reflects f (x) across the x-axis dilates f (x) by a factor of a from af (x) the x-axis translates f (x) by h units f (x − h) horizontally translates f (x) by k units f (x) + k vertically f (−x) reflects f (x) across the y-axis 1 dilates f (x) by a factor of f (bx) b from the y-axis

Effect on coordinates multiplies y-coordinate by −1 multiplies y-coordinate by a adds h to each x-coordinate adds k to each y-coordinate multiplies x-coordinate by −1 divides x-coordinate by b

7.04 Practice questions Understanding 1

366

y = f (−x)

y = −f (x)

y = f (x) + 2

y = 0.5f (x)

y = f (0.5x)

y = f (x) + 0.5

y = 2f (x)

y = x3

Which function remains unchanged after reﬂecting across the y-axis? A

Trailblazer 1–4

Which transformation corresponds to a vertical compression by a factor of 0.5? A

Adventurer 1, 2, 4

Which transformation represents a reﬂection across the y-axis? A

Explorer 1, 2, 4b–d

y = x2

y=x

y = −x

State whether the following statements are true or false. a

If y = f (x) is transformed to y = f (x) − 3, the resulting graph will move upwards by 3 units.

The transformation y = 3 + f (x) will shift the graph of y = f (x) three units downwards.

For the transformation y = f (x) − 4, the maximum and minimum values of the function (if they exist) will also decrease by 4.

y = −f (−x) will reﬂect the graph of y = f (x) ﬁrst across the y-axis and then across the x-axis.

A transformation in the form y = f (bx) where b > 1 will horizontally compress the graph.

A transformation of the form of the graph.

where b > 1 will result in a horizontal compression

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5, 7a, 9a, 10, 11a, 12, 13a, 5, 6ac, 7ab, 9, 11ad, 12, 14a–d, 15c, 18, 19 13b, 14b–f, 15b, 16, 18, 19bd

Fluency

For each equation, describe the type of transformation from f (x) to g (x): a

g (x) = f (x + 4)

g (x) = 6f (x)

g (x) = f (x) − 2

g (x) = −f (x)

If the x-intercepts of the graph of y = f (x) are (−5, 0) and (6, 0), determine the x-intercepts of: a

5, 6, 7cd, 8, 9, 11bc, 13cd, 14e–g, 15a, 17, 18, 19ac

y = f (x + 4)

y = f (x − 4)

y = 3f (x)

y = f (−x)

For each table: i

Describe the type of transformation from f (x) to g (x).

Write an equation for g (x) in terms of f (x) in the form g (x) = f (x) + k or g (x) = af (x) depending on the transformation.

f (x) 2 5 8 11 g (x) 4 7 10 13

14 16

f (x) g (x)

1 3

2 6

3 9

4 12

5 15

f (x) −2 −1 0 1 g (x) −4 −3 −2 −1

2 0

f (x) g (x)

4 −2

6 −3

8 −4

10 −5

12 −6

Ex 1

The graph of is translated horizontally 6 units to the right and vertically 5 units upwards. State the new equation after the transformations are applied.

Ex 2

Point A lies on the graph of y = f (x). Determine the coordinates of the corresponding point after the given transformation:

Point A is (−3, −1). The transformation is y = f (x − 5) − 2.

Point A is (0, 3). The transformation is

Point A is (1, −2). The transformation is y = f (3x) − 2.

Point A is (12, 3). The transformation is y = 6f (6x).

Consider the graph of the function f (x). Select the transformed function g (x) deﬁned by adding 2 to the x-values. A

g (x) = (x + 2)3

g (x) = (x − 2)3

g (x) = x + 2

g (x) = x3 − 2

5 4 3 2 1 −5 −4 −3 −2 −1 −1 y = f (x) −2 −3 −4 −5

x 1 2 3 4 5

7.04 Optional: Transformation of functions mathspace.co

367

For each function, describe the transformation applied to f (x) that is required to produce the graph shown: a

f (x) = x

f (x) = x2

1 −4 −3 −2 −1

−1

x 1

4 −5 −4 −3 −2 −1

−2

−1

−3

−2

−4

−3

f (x) = x – 1 4

1 −1

−4 −3 −2 −1

x 1

−2

−3

−4 −3 −2 −1

−4

x 1

−1

For what values of a will the transformation from f (x) to g (x) = af (x) involve a reﬂection across the x-axis?

Consider these graphs: i

Describe the transformation applied to f (x) to produce g (x).

Write an equation for g (x) in terms of f (x).

4 3

−1

−2

368

g (x)

f (x)

−4 −3 −2 −1

−1

−2

−3

−4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

f (x)

x 1

g (x)

1 −4 −3 −2 −1

x 1

6 g (x) 4

2 f (x)

−8 −6 −4 −2 −2

2 x −8 −6 −4 −2 −2

g (x)

f (x) 2

x 8

−4

−6

Find the resulting equation when y = x2 undergoes these transformations. a

A dilation by a factor of 10

A dilation by a factor of 5 and a reﬂection across the x-axis

A vertical translation of 10 units up

A horizontal translation of 2 units to the right

A reﬂection about the x-axis and a vertical translation of 7 units down

A horizontal translation of 4 units to the left and dilation by a factor of 8

A horizontal translation of 6 units to the right and a vertical translation of 5 units upwards

Consider the graph of f (x) and the table of values for g (x). i

Describe the transformation applied to f (x) to produce g (x).

Write an equation for g (x) in terms of f (x).

y 4 3

g (x)

−1 3

0 0

1 −1

2 0

−3

−2

−1

3 3

2 f (x)

1 −4 −3 −2 −1

−1

x 3

−2 −3 −4

f (x) −4 −3 −2 −1

9 8 7 6 5 4 3 2 1

−1

g (x)

x 1

7.04 Optional: Transformation of functions mathspace.co

369

y 4

x g (x)

−4 12

−2 8

0 4

2 0

4 −4

f (x)

−4 −3 −2 −1

−1

x 1

−2 −3 −4

Consider the graph of the function y = f (x):

Graph y = g (x) where g (x) = f (−x).

Describe the transformation from f (x) to g (x).

8 6 4 2 x −10 −8 −6 −4 −2 −2

Consider the graph of the function y = f (x): 6

Graph y = g (x) where

Describe the transformation from f (x) to g (x):

4 2 −6 −4 −2 −2

−4 −6

Consider the table of values of f (x) and the graph of g (x) = af (x): x f (x) a b

−3 9

−2 4

−1 1

0 0

1 1

2 4

3 9

Identify the type of transformation needed to get from f (x) to g (x): Determine the value of a.

y 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5

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x 1

y = g (x)

For each graph that has been transformed from y = x2: i

Determine the vertical translation.

Determine the horizontal translation.

iii

Has there been a reﬂection across the x-axis?

Has the quadratic had a change in dilation from the standard y = x2 curve?

2 x

−6 −4 −2

x −6 −4 −2

−2

−4

−6

4 2

x −6 −4 −2

Reasoning 20

−6 −4 −2

−2

−4

−6

20–22, 23a, 26

20, 21, 23, 24, 26

20, 21, 23–27

For each type of transformation, explain whether or not these key features change. If they change, explain how. • Intervals where the function is increasing or decreasing • Intervals where the function values are positive, negative, or zero • End behaviour a

Horizontal translations

Vertical stretches and compressions

Reﬂections across the x-axis

7.04 Optional: Transformation of functions mathspace.co

371

Stavros has a bank account with $100. Every week, Stavros deposits an additional $50 into the account. A graph of Stavros’ bank balance reveals that the growth of his account is linear. Suppose that Stavros was instead depositing an additional $100 every week. Explain whether or not this could be represented by a vertical stretch of the current graph.

Balance 500 400 300 200 100 1

The functions f (x) and g (x) = f (x + k) have been graphed on the Cartesian plane.

Determine the value of k.

Explain what k represents in this context.

Weeks 7

f (x)

g (x)

2 x −4 −2

−2 −4

Oreste draws the graph of g (x) = −f (x), as shown. a

Describe the error Oreste made.

Describe the type of graphs that f (x) could represent for Oreste’s error to still give him the correct answer.

3 2

g (x)

f (x)

−4 −3 −2 −1

−1

x 1

−2 −3 −4

372

Maddox and Paulina are going to a lake 2 kilometres from their home to go ice skating. Their mother told them that if the ice is not thick enough they must turn around and come straight home. It took them 20 minutes to walk to the lake at a constant rate and when they arrived Maddox could tell the ice was not thick enough so they immediately turned around and went home which took them an additional 20 minutes. a

Graph Paulina and Maddox’s distance from home over their entire journey.

If Paulina and Maddox had walked twice as fast, describe how your graph would change.

Describe the transformations if instead you were graphing Paulina and Maddox’s distance from the lake.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Consider the graphs of y = f (x) and y = g (x) shown. a

Describe a series of transformations to get from f (x) to g (x).

Phoebe thinks she has two different sets of transformations to get from f (x) to g (x), but she isn’t sure which one is correct.

Can you ﬁnd a second series of transformations that is different to your answer in part (a)?

6 4 2 −8 −6 −4 −2 −2 −4

x 2

g (x)

f (x) −6 −8

Here are the graphs of f (x) = x3 and g (x) =

(x − 3)3 + 5:

Describe how the graphs are different and how they are similar.

Describe the transformations from f (x) to g(x).

10 9 8 7 6 5 4 3 2 1

−3 −2 −1 −1 −2

f (x)

g (x) x

1 2 3 4 5 6 7 8

Yasushi and Peter are looking at the graphs of a few types of functions. a

Yasushi claims that a translation of a units to the right and a translation of b units upwards can be performed in either order and still get the same result.

Determine whether or not Yasushi is correct. Support your answer with an example or counter-example. b

Peter thinks that the same might be true for other types of transformations, and that the order in which transformations is applied doesn’t matter.

Determine whether or not Peter is correct. Support your answer with an example or counter-example.

Problem-solving 28, 29

28, 30

28, 31, 32

A classroom’s noise level is modelled by y = f (x), where x is the number of students. With a new seating arrangement, the noise is predicted to be y = 2f (x − 5). If 20 students originally resulted in a noise level of 50 dB, what will be the noise level with 25 students in the new seating arrangement?

A café owner models his daily coffee sales with the function y = j(x). After a marketing campaign, the sales increase by a factor of 1.5. If he originally sold 100 cups at noon, how many cups will he sell at noon after the campaign?

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A farmer uses the function y = n(x) to model the amount of sunlight his crops receive where x is the number of hours after 12 noon. Due to a change in the season, the sunlight pattern is compressed horizontally by a factor of 2. If his crops originally received maximum sunlight at 6 p.m., at what time will they now receive maximum sunlight?

A scientist is modelling the light intensity of a star over time using the function y = p(t). She notices that after a certain event, the light intensity behaves as y = 2p(0.5t + 3) − 4. If the original peak intensity was at time t = 10 hours, at what time t will the new peak intensity occur?

The effectiveness of a drug over time is modelled by y = v(x). After a reformulation, its effectiveness is modelled by y = v(1.5x − 2) − 3. If the original drug reached its peak effectiveness in 6 hours, how long will it take for the reformulated drug to reach its peak effectiveness?

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7 Chapter review 1

(Optional) Does each set of points describe a function? a

{(2, 5), (7, −3), (5, 2), (−4, −9)}

{(2, 5), (2, 7), (−3, −4), (−9, 13)}

{(1, 5), (1, 1), (7, −2), (−5, −10)}

{(1, 5), (7, −2), (−5, −10), (13, −13)}

{(1, 5), (1, 7), (−2, −5), (−5, −10)}

{(−1, −9), (0, 0), (1, 9), (2, 18)}

{(−2, 4), (−1, 1), (2, 4), (6, 36)}

(Optional) If f (x) = −6x + 4, ﬁnd: a

x −2 −1 0 1 y 0 5 10 15

2 20

x y

−4 −3 0 4 3 2

−1 1

0 0

1 1

2 2

3 3

4 4

(Optional) Is each graph a function? a

−4 −3 −2 −1

−1

−4 −3 −2 −1

−2

−3

−3 −4

1 −1

x 1

−1

−4

−4 −3 −2 −1

−2

3 1

x −4 −3 −2 −1

−1

x 1

−2

f (0)

(Optional) Is each table a function? a

f (4)

−3

−4

(Optional) State the domain of each function. a

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(Optional) Consider the function shown.

What is the domain of the function?

At what x-values does the function reach a minimum?

10 8 6 4 2

−5 −4 −3 −2 −1 −2

2 3 4 5

−4 −6

(Optional) Determine the domain and range of the following functions: a

x −6 −4 −2

−8 −6 −4 −2 −2

−2

x 2

−6

−8

x −6 −4 −2

−2

−4

−6 −5 −4 −3 −2 −1 −1

−6

−2

(Optional) State the range of the functions: a

y = x2

y = x2 + 4

(Optional) For the function y = 3x3: a

Complete the table of values: x

−1

−0.5

0.5

y = 3x b

376

−4

Sketch the graph of y = 3x3

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

y = −x2

y = (x − 1)2 − 2

(Optional) Consider the function y = x6. a

Describe the end behaviour of the function.

Draw the graph of the function.

(Optional) Velin is using the following graph to describe information about the equation of the polynomial function. Velin states that there is not enough information to determine anything about the equation of the function. Determine whether Velin is correct and explain your reasoning.

6 4 2

−8 −6 −4 −2 −2

−4 −6 −8

(Optional) Describe the transformations applied to f (x) to derive the equation y = 2f (x) + 4.

(Optional) Consider the equation y = 4 − x2.

Describe of the transformation(s) applied to y = x2 to achieve the given function.

Draw a graph of y = x2 and the transformed function on the same coordinate plane.

b 14

(Optional) Consider the equation y = 53−x − 1. Describe the transformation(s) applied to y = 5x to achieve the given function.

Draw a graph of y = 5x and the transformed function on the same coordinate plane.

b 15

(Optional) Consider the table of values of f (x) and the graph of g (x) = af (x): x f (x)

−3 9

−2 4

−1 1

0 0

1 1

2 4

3 9

Determine the type(s) of transformation needed to get from f (x) to g (x).

Determine the value of a.

y 2 1 x −6

−3

−1 −2

y = g (x)

−3 −4

(Optional) Write an equation for each of the following descriptions and describe how the domain and range changed from the parent function: a b

The parent function y = −3x is horizontally translated by 4 units left and vertically translated by 1 unit up.

The parent function y = x2 is vertically translated by 6 units up and horizontally translated by 3 units right.

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Big ideas • Exponentials demonstrate consistent growth or decay through multiplication by a constant factor. • Mathematical models provide powerful tools for analysing and predicting phenomena characterised by continuous change and accumulated effects over time.

Curriculum Outcomes VC2M10A11 explore the connection between algebraic and graphical representations of relations such as simple quadratic, reciprocal, circle and exponential, using digital tools as appropriate VC2M10A14 solve simple exponential equations VC2M10A15 use mathematical modelling to solve applied problems involving inverse proportion, growth and decay, including in financial contexts to establish the compound interest formula as repeated applications of simple interest; formulate problems, choosing to apply linear, quadratic or exponential models; interpret solutions in terms of the situation; evaluate and modify models as necessary and report assumptions, methods and findings VC2M10A16 solve equations graphically or using systematic numerical guess-check-and-refine with digital tools, with consideration of whether all solutions have been found VC2M10N01 recognise the effect of using approximations of real numbers in repeated calculations and compare the results when using exact representations VC2M10AA05 describe, interpret, and sketch parabolas, hyperbolas, circles and exponential functions and their transformations

8 Exponentials Chapter outline 8.01 Exponential relationships 8.02 Characteristics of exponential graphs 8.03 Optional: Transformations of exponentials 8.04 Exponential equations 8.05 Compound interest Investigation: Half-life 8.06 Exponential models Chapter 8 review

380 390 403 426 434 449 464

Carbon-14 dating uses exponentials to help archaeologists date ancient artefacts that are up to 50 000 years old!

8.01 Exponential relationships After this lesson, you will be able to… • differentiate between linear and exponential relationships. • identify the characteristics of exponential growth and decay. • recognise the growth or decay factor in exponential relationships. • interpret tables, equations, and graphs that represent exponential relationships. • formulate equations for exponential relationships based on provided data or patterns. • describe real-world scenarios that can be modelled using exponential relationships.

Exponential relationships We have seen that for a linear relationship, growth or decay are constant. x

−5

The table shows an example of linear growth, an increase of one unit in x has an increase of three units in y.

−5

The table shows an example of linear decay, an increase of one unit in x has an decrease of ﬁve units in y.

−5

Exponential growth is not linear. Each time x increases by one, we multiply by the same amount each time. The multiplication factor will be a number greater than 1. This is sometimes called the constant factor, common ratio, or growth factor. x

´3 3 =3 1

380

The table shows an example of exponential growth, an increase of one unit in x means that we triple the y-value.

´3 ´3 9 27 =3 =3 3 9

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

If we plot the table of values on a graph, we get a curve that is increasing at an increasing rate.

Every one unit moved in the x-direction causes the y-coordinate to triple.

21 18

y = 3x

We are multiplying by 3.

12 9 6 3

x 1

We may also have exponential decay. Each time x increases by one, we multiply by the same amount each time, but this time we multiply by a number between 0 and 1. We sometimes call this the decay factor. x

100

0.1

The table shows an example of exponential decay, an increase of one unit in x means that the y-value is one-tenth of the previous value, so the constant factor is 0.1

´0.1 ´0.1 ´0.1 10 1 0.1 = 0.1 = 0.1 = 0.1 1 100 10 100

If we plot the table of values on a graph, we get a curve that is decreasing at an decreasing rate.

Every one unit moved in the x-direction means that the y-coordinate is divided by 10.

80 70 60 50

We multiply by 0.1 each time.

y = 100(0.1x)

40 30 20 10

x 1

We need to be able to recognise exponential relationships from equations, tables of values, and graphs.

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Example 1 For y = 2x: a Complete this table of values. x y

−3

−2

−1

Create a strategy Substitute in each x-value and then simplify.

Apply the idea For each x-value, we can evaluate 2x. Let’s show our work for the ﬁrst one. y = 2x

Write the equation

−3

Substitute the ﬁrst value of x from the table

= 0.125

Express as a decimal

−3 2−3 0.125

x 2x y

−2 2−2 0.25

−1 2−1 0.5

0 20 1

1 21 2

2 22 4

3 23 8

b Is y = 2x a linear or an exponential relationship?

Create a strategy We will check if the y-values in the table are increasing by a constant amount or have a factor they are being multiplied by.

Apply the idea x y = 2x

−3 0.125

−2 0.25

−1 0.5

0 1

1 2

2 4

3 8

We can ﬁnd the ratio between each pair of numbers, if they are all the same, then it is exponential. We’ll show our work for the ﬁrst three pairs.

Since these are all the same this is an exponential relationship. The multiplying factor is greater than 1, so this is exponential growth to be more speciﬁc.

Reﬂect and check Without the use of a table, we can tell this is an exponential relationship as there is a variable in the exponent.

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Example 2 Which graphs show exponential relationships? A y

2 −4 −3 −2 −1 −2

x 1

−4 −3 −2 −1 −2

−4

−6

−8

C y

D 8

−4 −3 −2 −1 −2

x 1

2 −4 −3 −2 −1 −2

−4

−6

−8

x 1

Create a strategy Look at the shape of the graph. An exponential relationship will be increasing at an increasing rate or decreasing at a decreasing rate.

Apply the idea Graph B is a straight line, indicating a linear graph. Graph C has a U-shape, representing a quadratic graph. Graphs A and D are exponential curves, with A showing exponential decay and D showing exponential growth. So, the correct answers are options A and D.

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Example 3 For each table of values, decide whether the relationship is exponential or not. If it is exponential, ﬁnd what we multiply by each time. a

x y

−3 −2

−2 1

−1 4

0 7

1 10

2 13

3 16

Create a strategy

Apply the idea

Look at the pattern in the y-values.

As x increases by 1, y is increasing by 3. This is a linear relationship, so it is not exponential.

x y

−3 0.015 625

−2 0.0625

−1 0.25

0 1

1 4

2 16

3 64

Create a strategy Look at the pattern in the y-values.

Apply the idea We can see from the right-hand side of the table the y-values: 1, 4, 16, 64 This suggests an exponential relationship where we multiply by 4. To check: 0.015 625 × 4 = 0.0625 0.0625 × 4 = 0.25 0.25 × 4 = 1 This is an exponential relationship with a growth factor of 4.

x y

−3 8

−2 4

−1 2

0 1

1 0.5

2 0.25

3 0.125

Create a strategy Look at the pattern in the y-values.

Apply the idea If we look at the y-values, we can see that each value is half of the previous value. To check:

This is an exponential relationship with a decay factor of 0.5

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x y

−3 1

−2 2

−1 6

0 24

1 120

2 720

3 5040

Create a strategy

Apply the idea

Look at the pattern in the y-values.

Going from 1 to 2 means multiplying by 2, but going from 2 to 6 means multiplying by 3 The ratio is not the same, so this is not an exponential relationship.

Example 4 This pattern shows an exponential relationship.

Step 1

Step 2

Step 3

Step 4

a Find the growth or decay factor.

Create a strategy We can write the number of squares in each step in a table of values.

Apply the idea Step Number of squares

1 2

2 4

3 8

4 16

Looking at the number of squares, to get from one step to the next, we multiply by 2. The growth factor is 2.

Reﬂect and check We may do this without a table of values by observation.

b Find the number of squares for Step 5.

Create a strategy

Apply the idea

We need to use the growth factor to go from Step 4 to Step 5.

To go from one step to the next, we multiply by the constant factor. Step 5 = 2 × Step 4 Step 5 = 2 × 16 Step 5 = 32

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Idea summary Exponential relationships are relationships in which y-values change by multiplying by a ﬁxed amount when the x-value increases by 1. Exponential growth relationships look like curves increasing at an increasing rate. Exponential relationships have an equation which has a variable in the exponent like y = 2x.

18 y 16 14 12 10 8 6 4 2

x 1

Exponential decay relationships look like curves decreasing at a decreasing rate and have an equation which has a variable in the exponent like y = 0.5x.

18 y 16 14 12 10 8 6 4 2

x 1

8.01 Practice questions Understanding 1

Adventurer 1ac, 2–3

Trailblazer 1cd, 2–3

For each table of values: i

Explorer 1ab, 2–3

Describe the pattern in words.

Find the next two terms.

x y

0 5

3 40

x y

0 5

x y

0 1 2 3 2.5 7.5 22.5 67.5

x y

0 40 000

1 10

2 20

Describe the pattern in words.

Find the number of circles for Step 4.

1 10

2 15

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1 2 3 4000 400 40

Step 1

386

3 20

Step 2

Step 3

Describe the pattern in words.

Find the number of hexagons for Step 4.

Step 1 Step 2

Fluency 4 Ex 1

4–5, 6ab, 7–8

Complete the table of values. 0 50 1

5 y b 5

4–8

For y = 5x: x

Ex 2

4–5, 6bc, 7–8

1 51 5

5 55 3125

Is y = 5x a linear or an exponential relationship?

Are these relationships linear, exponential or quadratic? a

y = 4x 2

y = 4x + 7 y = 4 × 7x

y = 4x + 7

−5 −4 −3 −2 −1 −2

x 1

2 3 4 5

50 40

−4

−6

−8

−10

6 Ex 3

Step 3

−4 −3 −2 −1

x 2 3 1 y 0.8 1.6 2.4

4 3.2

x y

1 0.125

x 1

2 0.5

3 2.0

4 8.0

For these exponential relationships: Find y when x = 5.

Find the growth or decay factor.

x 1 2 3 y 10 35 122.5

x y

1 2

2 30

3 450

4 6750

x 1 2 3 y 0.123 0.246 0.492

x y

1 4096

2 1024

3 256

4 64

4 428.75 4 0.984

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Ex 4

Consider the pattern: a

Find the growth or decay factor.

Find the number of triangles for Step 4. Step 1 Step 2

The total volume of water that has dripped from a tap is measured each minute: a

Describe the relationship between time and volume.

Complete the values in the table.

Find the volume after 8 minutes.

Reasoning 9

9–10

The cumulative rainfall over a series of 4 days is represented on the graph. a

Time (minutes) 1 2 3 4 5 6

Describe the relationship between time and volume.

Determine if the points represent a linear or exponential relationship. Explain your reasoning.

Assuming the pattern continues over the next few days, ﬁnd the cumulative volume on day seven.

Step 3

Volume (mL) 3 9 27 81

10–11

Volume (mL)

32 28 24 20 16 12 8 4

Time (days) 1

Consider the graphs: a

Find the rule for graph B.

Find the rule for graph A.

Determine which is an exponential relationship and which is a linear relationship. Explain your reasoning.

16 14

12 10 8 6 4 2

x 1

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Consider the table of values for x ≥ 1 where y1 = (1.05)x and y2 = 5x. Is it correct to say that y2 is always greater than y1 for all values of x ≥ 1?

x y1 y2

1 1.05 5

2 1.10 10

3 1.16 15

4 1.22 20

5 1.28 25

Explain your reasoning.

Problem-solving 13, 15 12

12–13, 15

12–15

Consider the 7 plotted points:

Find three points that form a linear relationship.

For each 1 unit increase in x, determine how much the linear relationship increases by.

Find the three points that form an exponential relationship.

For each 1 unit increase in x, determine the constant ratio for the exponential relationship.

F E

20 C

15 10

5 A

B 1

x 2

A population of bacteria doubles every hour. Initially (at time = 0), there are 10 bacteria. Create a table showing the population of bacteria for the next 5 hours and determine if the relationship between time and population is linear or exponential.

A radioactive substance decays by 20% every hour. Initially there are 100 grams of the substance. Create a table showing the mass of the substance for the next 6 hours and determine if the relationship between time and mass is linear, quadratic or exponential.

A pile of bricks is decreased by 500 bricks every pallet. Initially there are 20 000 bricks in the pile. Create a table showing the number of bricks remaining after the ﬁrst 5 pallets are loaded and determine if the relationship between number of bricks in the pile and number of pallets is linear, quadratic or exponential.

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8.02 Characteristics of exponential graphs After this lesson, you will be able to… • recognise the standard form of exponential relationships as y = bx. • identify the y-intercept and horizontal asymptote of exponential curves. • determine the behaviour of exponential curves based on the value of b. • use tables of values, technology, or equations to sketch graphs of exponential relationships. • differentiate between exponential growth and exponential decay based on the value of b.

Characteristics of exponential graphs Exponential relationships are where we multiply to get from one y-value to the next for each increase in 1 in the x-values. An exponential relationship can be modelled by y = bx where b is the growth or decay factor. 8

All equations of the form y = bx have these features in common: • The growth or decay factor is b. • The y-intercept is at (0, 1). • There is a horizontal asymptote at y = 0.

7 6 5 0<b<1

The value of b determines whether the curve increases or decreases. • If 0 < b < 1, the curve decreases. • If b > 1, the curve increases.

b>1

4 3 2 1

−4 −3 −2 −1

x 1

Notice the dashed line at y = 0. This line is called an asymptote. An exponential curve can get inﬁnitely close to an asymptote, but it can never cross it. This means that an exponential relationship of this form will not have an x-intercept.

Interactive exploration

Discover this concept in action online

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mathspace.co

As x increases in exponential growth relationships, the graph increases faster for higher values of b. As x increases in exponential decay relationships, the graph decreases faster for smaller values of b. To draw the graph of an exponential relationship, we can use a variety of strategies, including: • Completing a table of values and drawing the curve through the points • Using technology, such as a graphing calculator • Identifying key features from the equation

Example 1 Consider the equation y = 3x. a Complete the table of values. −5

−4

−3

−2

−1

Create a strategy Substitute x = −2, −1, 0, 10 into y = 3x.

Apply the idea Substitute x = −2

Apply the negative exponent law

Evaluate the exponent

Substitute x = −1

Apply the negative exponent law

Substitute x = 0

Apply the zero exponent law

, to save time for x = 1, 2, 3, 4, 5, take the reciprocals of the corresponding y-values

Since

for x = −1, −2, −3, −4, −5. y = 310

Substitute x = 10

= 59 049 x y

−5

−4

Evaluate the exponent −3

−2

−1

243 59 049

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b Graph the exponential curve.

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

Apply the idea Because the graph increases so quickly, we need to choose an appropriate scale. To see most of the points, we can plot the plots from x = −3 to x = 3. y 27 24 21 18 15 12 9 6 3 −4 −3 −2 −1

x 1 2 3 4

c Is y = 3x an increasing curve or a decreasing curve?

Create a strategy Consider the y-values in the table from part (a) from left to right.

Apply the idea The y-values in the table are in increasing order, so y = 3x is an increasing curve.

Example 2 Consider the equation

a Determine whether the exponential curve is increasing or decreasing.

Create a strategy

Apply the idea x

For equations in the form y = b , the curve is increasing if b > 1 and the curve is decreasing if 0 < b < 1.

392

Since

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

, the curve is decreasing.

b Identify the equation of the horizontal asymptote.

Create a strategy For any relationship in the form y = bx, the horizontal asymptote is on the x-axis.

Apply the idea

Reﬂect and check

The equation of the horizontal asymptote is y = 0.

Because an asymptote is a line, it must be represented by an equation. Since exponential curves have horizontal asymptotes, the equation of the line will be in the form y = c where c is any real number.

c State the y-intercept.

Create a strategy Substitute x = 0 to ﬁnd the y-value of the y-intercept.

Apply the idea Write the equation

Substitute x = 0

Apply zero exponent law

The y-intercept is at (0, 1).

Reﬂect and check For any exponential equation of the form y = bx, the y-intercept will be at (0, 1).

d Find one other point on the curve.

Create a strategy Choose any value of x and substitute it into the equation to ﬁnd y.

Apply the idea Let x = −1. Write the equation

Substitute x = −1

Apply negative exponent law

Another point on the curve is (−1, 4).

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e Sketch the graph.

Create a strategy

Apply the idea

Plot the key points and key features found in parts (a) through (d).

7 6 5 4 3 2 1 −4 −3 −2 −1

x 1

Example 3 Consider the equation y = 8x. a Can the value of y ever be negative or zero?

Create a strategy Recall the location of the asymptote of the exponential relationship y = bx.

Apply the idea Since y = 8x is an exponential relationship of the form y = bx, there is an asymptote at y = 0. This means its value of y can never be equal to or lower than 0. So it cannot be negative or zero.

b As the value of x increases towards ∞, what can we say about the value of y? A

y tends towards 8.

y tends towards ∞.

Create a strategy Substitute some positive values of x into y = 8x.

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y tends towards 0.

Apply the idea We can use x = 5, 10. y = 85

Substitute x = 5

= 32 768

Evaluate

y=8

Substitute x = 10

= 1 073 741 824

Evaluate

We can see the values are increasing into inﬁnite values or ∞. The answer is option B. c As the value of x decreases towards −∞, what value does y approach? A

∞

Create a strategy

Apply the idea

Recall the property that as x decreases in exponential growth curves, y approaches the horizontal asymptote.

The answer is option A.

d Determine the y-value of the y-intercept of the curve.

Create a strategy Substitute x = 0 into y = 8x.

Apply the idea y = 80 =1

Substitute x = 0 Apply the zero exponent law

So, the y-intercept is at (0, 1). e How many x-intercepts does the curve have?

Create a strategy Use the answer from part (a).

Apply the idea We knew from part (a) that y = 8x can never have y = 0 because of the horizontal asymptote. However, x-intercept can only exist when y = 0. So the curve has 0 x-intercepts.

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Which of these could be the graph of y = 8x? A

y 8

2 −4 −3 −2 −1 −2

x 1

−4

−6

−8

−4 −3 −2 −1 −2

x 1

−4 −3 −2 −1 −2

−4 −3 −2 −1 −2 −4

−6

−8

x 1

Use the features found from previous parts.

Apply the idea The key features of y = 8x are: • The y-values can never be negative. • As x approaches ∞, y approaches ∞. As x approaches −∞, y approaches 0. • The previous feature tells that the curve is increasing from left to right. • The y-intercept is at (0, 1).

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Create a strategy

396

−4

All of these features can be found in option D.

Example 4 Consider the graphs of

and y = 2x.

a Sketch a graph of y = 2x on the Cartesian plane.

Create a strategy Plot the y-intercept and substitute x = 1, 2 into y = 2x for more points.

Apply the idea Recall that the y-intercept for curves of the form y = bx is at (0, 1). For more points, substitute x = 1, 2 into y = 2x: y = 21 =2 y = 22 =4

Substitute x = 1

Evaluate the exponent

Substitute x = 2

Evaluate the exponent

So we can plot the points (0, 1), (1, 2), and (2, 4).

−4 −3 −2 −1

−1

−2 −3 −4

b Sketch

on the same Cartesian plane.

Create a strategy Plot the y-intercept and substitute x = −1, 1 into

for more points.

Apply the idea Recall that the y-intercept for curves of the form y = bx is at (0, 1). For more points, substitute x = 1, 2 into

4 3

Substitute x = −1

Apply the negative exponent law

1 −4 −3 −2 −1

Substitute x = 1

−2

Evaluate the exponent

So we can plot the points (−1, 2), (0, 1), and

−1

−3

−4

8.02 Characteristics of exponential graphs mathspace.co

397

c Compare the graphs of y = 2x and

Create a strategy Use the graph from part (b) to compare the key features of each curve.

Apply the idea The two graphs are reﬂections across the y-axis. The graph of y = 2x is increasing while the graph of

is decreasing.

Both graphs have the same horizontal asymptote and y-intercept.

Idea summary An exponential relationship can be modelled by y = bx

All equations of the form y = bx have these features in common:

• The y-intercept is at (0, 1) • There is a horizontal asymptote at y = 0

6 5 0<b<1

b>1

4 3 2 1

−4 −3 −2 −1

x 1

The value of b determines whether the curve increases or decreases. • If 0 < b < 1, the curve decreases • If b > 1, the curve increases As x increases in exponential growth relationships, the graph increases faster for higher values of b. As x increases in exponential decay relationships, the graph decreases faster for smaller values of b.

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8.02 Practice questions Understanding

Explorer 1–4, 5a–d

Adventurer 1–4, 5b–f

Trailblazer 1–4, 5e–h

What part of the exponential equation y = bx determines whether it increases or decreases?

What is the name of the line that the graph of an exponential curve approaches but never crosses?

Is each statement true or false for y = bx? If it is false, correct the statement.

If b > 1, then it is exponential growth.

If b < 0, then it is exponential decay.

If b is a fraction, then it is exponential decay.

If b > 0, then it is exponential growth.

The y-intercept is always (0, 1).

Describe what happens to y as the value of x increases. a

5 y = 4x

1x 4

-5 -4 -3 -2 -1 -1

2 3 4 5

−5 −4 −3 −2 −1 −1

2 3 4 5

-2

−2

Determine whether these are increasing or decreasing exponential curves: a e

y = 3x

b f

y = 0.97x

y = 0.5x

y = 1.5−x

y = 1.05x

8.02 Characteristics of exponential graphs mathspace.co

399

Fluency 6 Ex 1

7 Ex 2

6a–d, 7abf, 8, 10–11

400

6c–f, 7d–h, 8, 11

For each exponential relationship: i

Complete the table of values.

iii

Determine whether the curve is increasing or decreasing.

x −1 0 1 y = 4x

x −1 0 1 y = 10x

−1

x −1 0

−1

Graph the exponential curve.

−1

y = 5x

For each exponential relationship: i

Determine whether the curve is increasing or decreasing.

Identify the equation of the horizontal asymptote.

iii

Find the y-intercept.

Find one other point on the curve.

Sketch the graph.

y = 2x

e Ex 3

6b–e, 7c–f, 8–9, 11

y = 4x

y = 7x

y = 8x

For the equation y = 6x: a

Can the value of y ever be zero or negative?

As the values of x get larger and larger, what can we say about the value of y?

As the values of x get smaller and smaller, what value does y approach?

Find the y-value of the y-intercept of the curve.

How many x-intercepts does the curve have?

Sketch the graph of y = 6x.

Consider the expression 2x. a

Evaluate the expression when x = −4.

Evaluate the expression when x = 0.

Evaluate the expression when x = 4.

For what values of x will 2x have a negative value?

Describe what happens to the value of 2x as x gets larger.

Describe what happens to the value of 2x as x gets smaller.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The graph of y = 4x is shown.

Is each y-value of the curve positive or negative?

What is the equation and name of the horizontal line, which y = 4x gets closer and closer to but never intersects?

5 4 3 2 1 x −3

−2

−1

11 Ex 4

Consider the graphs of y = 5x and a

Sketch the graph of y = 5x.

Compare the graphs of y = 5x and

Reasoning 12

12–13

Sketch the graph of

12–14

12–15

The graphs of two exponential relationships, R and S, are shown. a

One of the graphs is of y = 3x and the other graph is of y = 5x. Which is the graph of y = 5x?

For negative values of x, which curve will have smaller values of y? Explain your reasoning.

y R S

Do either of the exponential curves y = 9x or

have x-intercepts? Explain your reasoning.

For the exponential relationships

and

Is each statement true or false? i

All of the curves have a maximum value.

All of the curves pass through the point (1, 2).

iii

All of the curves have the same y-intercept.

None of the curves cross the x-axis.

What is the y-intercept of each curve?

Describe what happens to the values of y as x gets increasingly larger.

For

and

, which is decreasing more rapidly for x > 0? Justify your answer.

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401

Problem-solving 16, 18–19 16

16cd, 17, 19–21

Find the missing coordinate in each ordered pair so that the pair is a solution to y = 5x: a

16cd, 17–19

(3, ⬚)

(⬚, 5)

(−1, ⬚)

Find the missing coordinate in each ordered pair so that the pair is a solution of a

(0, ⬚)

(⬚, 9)

(−1, ⬚)

If the point (3, m) lies on the curve y = 2x, solve for m.

The graph of y = 2x is shown: a b

128 120 112 104 96 88 80 72 64 56 48 40 32 24 16 8

Find the value of y when x = 7. Find the value of x when y = 128.

−1

x 1

2 3 4 5 6 7 8

The graph of y = 3x is shown. Find the length of the interval PQ.

y P

Q (0,1) −5 −4 −3 −2 −1

402

For the exponential equation y = 3−x: a

Rewrite 3−x with a positive index using the index laws.

Sketch the graph of y = 3−x.

Compare the graphs of y = 3x and y = 3−x.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

2 3 4 5

8.03 Optional: Transformations of exponentials After this lesson, you will be able to… • describe the effect of vertical translations on exponential graphs. • analyse vertical dilations of exponential graphs and distinguish between vertical stretches, compressions and reflections. • determine the equation of an exponential function from a given graph by identifying the y-intercept value and the growth or decay factor. • apply combined transformations to exponential graphs, understanding the sequence of dilation, reflection and translation.

Vertical translation of an exponential graph The graph and key features of y = bx depend on the value of b. The key features of y = bx for b > 1 are: • Increasing at an increasing rate • y-intercept is at (0, 1) • The x-axis, y = 0, is the horizontal asymptote • A point on the graph is (1, b)

y 12 10 8 6 4 2

−4 −3 −2 −1 −2

−4

The key features of y = bx for 0 < b < 1 are: • Decreasing at an decreasing rate • y-intercept is at (0, 1) • The x-axis, y = 0, is the horizontal asymptote • A point on the graph is (−1, b)

y 12 10 8 6 4 2 −4 −3 −2 −1 −2

x 1

−4

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Any exponential relationship can be described as a function, which is a special type of relation where each input corresponds to only one output. Functions can be written in function notation. This notation highlights the independent variable and names the function, which is helpful when working with multiple functions. When we are writing in function notation, instead of writing “y =”, we write “f (x) =”. For example, we can write y = 3x as f (x) = 3x, which means “f is a function of x”. Common letters to use for general functions are lower case f, g, and h.

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The general rule for a vertical translation is: y = bx + k y

4 2

−4 −3 −2 −1 −2

−4 −3 −2 −1

x 1

Vertical translation of 3 units upwards: f (x) = 2x + 3

Vertical translation of 3 units downwards: f (x) = 2x − 3

The y-intercept and the horizontal asymptote will move k units vertically with the curve, for example: y

The graph shown is

Notice the horizontal asymptote has moved downwards from y = 0 to y = −4.

10 8

The y-intercept has moved downwards from (0, 1) to (0, −3).

6 4 2 −4 −3 −2 −1

−2

x 1

The curve has the same shape as

−4

404

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 1 This is a graph of y = 3x. a How do we shift the graph of y = 3x to get the graph of y = 3x − 4?

Move the graph 4 units to the right.

Move the graph downwards by 4 units.

Move the graph 4 units to the left.

Move the graph upwards by 4 units.

−4 −3 −2 −1

−1

x 1

−2 −3 −4

Create a strategy Use the fact that when a graph of y = f (x) is moved downwards by k units, its new equation is y = f (x) − k.

Apply the idea Since the new equation is y = 3x − 4 and k = −4 < 0, then it means that we need move the graph downwards by 4 units to get the graph of y = 3x − 4. So, the correct answer is option B.

b Plot y = 3x − 4 on the same graph as y = 3x.

Create a strategy Plot the graph of y = 3x, then shift each point down 4 units to get the graph of y = 3x − 4.

Apply the idea 5

4 3 2 1 −5 −4 −3 −2 −1 −1

Substitute x = 0 and x = 1 to ﬁnd two points on the curve y = 3x.

y = 3x

Then, shift each point down 4 units. The horizontal asymptote for y = 3x − 4 is y = −4. x 1

2 3 4 5

−2 −3 −4 −5

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Connect the points to draw the curve y = 3x − 4.

4 3 2 y = 3x

−5 −4 −3 −2 −1 −1

2 3 4 5 y = 3x − 4

−2 −3 −4 −5

Example 2 The graph of

is shown. 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

a Describe a transformation of the graph of

that would obtain

Create a strategy Use the fact that when a graph of y = f (x) is moved upwards by k units, its new equation is y = f (x) + k.

Apply the idea Since the new equation is

and k = 3 > 0, then it means that the transformation of the

graph of g (x) is a translation up 3 units from

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

on the same set of axes as

b Sketch the graph of

Create a strategy Plot the graph of

. Use the transformation found from part (a) to plot the graph of

Apply the idea From the graph of

, we move this graph upwards by 3 units.

Substitute x = −1 and x = 0 to ﬁnd two points on the

curve

4 3 2 1

f ( x) =

-5 -4 -3 -2 -1 -1

1 2

Then, shift each point up 3 units.

The horizontal asymptote shifts up 3 units.

2 3 4 5

-2 -3 -4 -5

8 7

Connect the points to draw the curve

6 5 x 1 4 g( x) = 2 + 3 3 x 2 1 1 f ( x) = 2

-5 -4 -3 -2 -1 -1

2 3 4 5

-2

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c Write the equation of the horizontal asymptote of

Create a strategy

Apply the idea x

For any function in the form y = b ± k, the horizontal asymptote is at the y = ±k.

The equation of the horizontal asymptote is y = 3.

Idea summary If the function bx is translated vertically, the equation will be of the form y = bx + k If k > 0, it is translated upwards. If k < 0, it is translated downwards. The key features become: • A y-intercept at (0, 1 + k) • Horizontal asymptote at y = k

Vertical dilation of an exponential graph We will now look at exponential functions of the form y = a (bx).

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The graph of y = a (bx) is a vertical dilation of y = bx. We can compare y = 3 (2x) to y = 2x. This will be a vertical enlargement (stretch). Every point on the graph of y = 3(2x) is three times as far away from the x-axis as the point on y = 2x at the corresponding x-coordinate.

y 12 10

For example, we can see (2, 4) → (2, 12).

8 6 4 x

y = 3 (2 )

−6 −5 −4 −3 −2 −1 −2

408

The y-intercept for y = 2x is at (0, 1), and the y-intercept for y = 3 (2x) is at (0, 3).

y = 2x x

Both curves have y = 0 as the horizontal asymptote.

1 2 3 4 5 6

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

We can compare y = 0.5 (2x) to y = 2x. This will be a vertical compression. Every point on the graph of y = 0.5 (2x) is half as far away from the x-axis as the point on y = 2x at the corresponding x-coordinate.

y 12 10

For example, we can see (3, 8) → (3, 4).

The y-intercept for y = 2x is at (0, 1), and the y-intercept for y = 0.5 (2x) is at (0, 0.5).

6 y = 2x

4 2

y = 0.5(2 )

−6 −5 −4 −3 −2 −1 −2

Both curves have y = 0 as the horizontal asymptote.

1 2 3 4 5 6

We can compare y = −3 (2x) to y = 2x. This will be a vertical stretch and a reﬂection across the x-axis. There is still a dilation by a scale factor of 3, but the negative also produces a reﬂection in the x-axis.

y 8

y = 2x

4 x -6 -5 -4 -3 -2 -1 -4

1 2 3 4 5 6 y = -3(2x )

For example, we can see (2, 4) → (2, −12). The y-intercept for y = −3 (2x) is at (0, −3). Both curves have y = 0 as the horizontal asymptote.

-8 -12

Example 3 Of the two functions y = 2x and y = 3 × 2x, which is increasing more rapidly for x > 0? A y = 2x

y = 3 × 2x

Create a strategy Create a table of values to compare the values of y between the two functions y = 2x and y = 3 × 2x.

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409

Apply the idea For x = 1: y = 2x

= 21

Substitute 1 to x Evaluate x

y = 3 × 2

= 3 × 21

Substitute 1 to x

Evaluate

Similarly, by substituting the remaining x-values into y = 2x and y = 3 × 2x, we get: x x

y=2 y = 3 × 2x

1 2 6

2 4 12

3 8 24

4 16 48

5 32 96

Based on the table, the function y = 2x is increasing more rapidly for x > 0. So, the correct answer is option B.

Example 4 Consider the function

a Find the coordinates of the y-intercept.

Create a strategy Substitue x = 0 to ﬁnd the y-value of the y-intercept.

Apply the idea Write the function

Substitute x = 0

Use the zero power law

Evaluate

The y-intercept is at (0, 2).

b What is the equation of the horizontal asymptote?

Create a strategy For any function in the form y = a (bx), the horizontal asymptote is on the x-axis.

Apply the idea The equation of the horizontal asymptote is y = 0. 410

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c Is the function increasing or decreasing?

Create a strategy For functions in the form y = bx the function is increasing if b > 1 and the function is decreasing if 0 < b < 1.

Apply the idea The constant factor of this function is . Since

, the function is decreasing.

d Graph the function.

Create a strategy Plot the key points and key features found in parts (a) through (c). Use the fact that a function of the form y = a (bx) is dilated by a factor of a.

Apply the idea , we multiply the y-values by 2.

From the graph of 11 10 9 8 7 6 5 4 3 2 1

graph of

−4 −3 −2 −1

Then, multiply the y-values of the points by 2.

f ( x) =

1 5

-4 -3 -2 -1

11 10 9 8 7 6 5 4 3 2 1

Substitute x = −1 and x = 0 to ﬁnd two points on the

x 3

Connect the points to draw the curve of

The horizontal asymptote stays at y = 0.

x 1

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411

Idea summary If the function y = bx is dilated or reﬂected vertically, the equation will be of the form y = a (bx) If ∣a∣ > 1 then it is stretched and if 0 < ∣a∣ < 1 then it is compressed. Negative values of a result in a reﬂection across the x-axis. The key features become: • A y-intercept at (0, a) • Horizontal asymptote at y = 0

Find the equation In the previous section, we saw that functions of the form y = a (bx) always have a y-intercept at (0, a). Recall that the value of b is the ratio between successive terms in an exponential function. These two facts can help us identify the equation of any exponential function that has been dilated or reﬂected.

Example 5 The equation of the graph is of the form y = a (b)x.

3 2 1 −4 −3 −2 −1

x 1

2 3 (1, −1)

−1 −2 −3 −4 −5 (0, −5) −6 −7 −8 −9

a What is the y-intercept? What does this tell us about the equation?

Create a strategy

Apply the idea

Check the point where the curve crosses the y-axis.

Based on the graph, the y-intercept is at (0, −5). This tells us that a = −5.

b What is the decay factor? What does this tell us about the equation?

Create a strategy Use the coordinates of two successive points on the graph of the exponential function, and ﬁnd the ratio of their y-values.

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea We are given the points (0, −5) and (1, −1), which are successive points since x = 0 and x = 1. Dividing the y-value of the point on the right by the y-value of the previous point gives us

This tells us

Reﬂect and check A different method would be to use the fact that a = −5 to rewrite the equation as y = −5 (bx). Then, substitute one of the points and solve for b. Substituting one of the points, let’s say (1, −1), to y = a(b)x where a = −5, we will get the value of b. Substitute the values Evaluate the power Divide both sides by −5 Evaluate Since

then this means that the decay factor is .

c What is the equation?

Create a strategy Use the y-intercept and the decay factor found in parts (a) and (b).

Apply the idea Since a = −5 and

, the equation is given by:

Idea summary For any function of the form y = a (bx), we can ﬁnd the equation from a graph by identifying the y-value of the y-intercept and the growth or decay factor. The y-value of the y-intercept is a in the equation, and the growth or decay factor is b.

y = a (bx) a b

y-value of the y-intercept growth or decay factor

8.03 Optional: Transformations of exponentials mathspace.co

413

Apply both transformations We will now explore exponential functions of the form y = a (bx) + k.

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Consider graphing y = 3 (2x) − 4. We can start by graphing y = 2x with: • y-intercept at (0, 1) • Horizontal asymptote at y = 0 • Another point at (2, 4)

y 12 10 8

y = 2x

6 4 2

−4 −3 −2 −1 −2

−4

We can then apply the dilation to get y = 3 (2x) with: • y-intercept at (0, 3) • Horizontal asymptote at y = 0 • Another point at (2, 12)

y 12

y = 3(2x)

10 8 6 4

y = 2x

−4 −3 −2 −1 −2

−4

We can then apply the translation to get y = 3 (2x) − 4 with: • y-intercept at (0, −1) • Horizontal asymptote at y = −4 • Another point at (2, 8)

y 12 10 8 6 4 x

y = 3(2 )

−4 −3 −2 −1 1 2 3 −2 y = 3(2x) − 4

x 4

−4

414

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

In general, for y = a (bx) + k • The asymptote would be y = k. • The y-intercept will be at (0, k + a), so it is above the asymptote if a is positive, and below the asymptote if a is negative. If we are doing multiple transformations, we should go in order of dilation, reflection and translation.

Example 6 Sketch the graph of y = −0.5 (2x) + 1.

Create a strategy We can consider the transformations to get an idea of what the graph will look like and then identify the key features from the equation.

Apply the idea Starting the function y = 2x, first there was a compression with a scale factor of 0.5. Next, there was a reflection across the x-axis. Finally, there was a vertical translation upwards by 1 unit. Let’s graph using the key features. To find the y-intercept we can set x = 0 and evaluate. y = −0.5 (2x) + 1 y = −0.5 (20) + 1 y = −0.5 (1) + 1 y = −0.5 + 1 y = 0.5 The horizontal asymptote is y = 1. To find another point we can pick an x-value and evaluate. y = −0.5 (2x) + 1 y = −0.5 (21) + 1 y = −0.5 (2) + 1 y = −1 + 1 y=0 We now have two points and the asymptote to draw the graph.

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

8.03 Optional: Transformations of exponentials mathspace.co

415

Reﬂect and check We can see each transformation applied one at a time as well. 4

Original function of y = 2x.

3 2 1 −4 −3 −2 −1

x 1

−1

−2 −3 −4

Compression with the scale factor of 0.5.

3 2 1 −4 −3 −2 −1

x 1

−1

−2 −3 −4

Reﬂection across the x-axis.

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

416

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Translation upwards 1 unit.

3 2 1 −4 −3 −2 −1

x 1

−1

−2 −3 −4

Idea summary A function of the form y = a (bx) + k has both: • Vertical translation, moving the graph up or down. • Vertical dilation, stretching the curve in the y-direction. We can ﬁnd the values of a, b, and k for a given curve. • y = k will be the asymptote. • The y-intercept is at (0, a + k). We can ﬁnd b using any other point on the curve.

8.03 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

Trailblazer 1–4

For y = 2x and y = 2x + k: a

Are these key features the same or different? i

Rate of change

Asymptote

iii

y-intercept

Shape

Increasing or decreasing

What happens if k is positive?

What happens if k is negative?

8.03 Optional: Transformations of exponentials mathspace.co

417

For y = 2x and y = a × 2x, with a > 0: a

Are these key features the same or different? i

Rate of change

Asymptote

iii

y-intercept

Shape

Increasing or decreasing

What happens if a > 1?

What happens if 0 < a < 1?

What would happen if a was negative?

Match each graph to one of the equations: i

y = 2x

b 8

x 1

−4

−6

−8

−4 −3 −2 −1 −2

418

x 1

−4 −3 −2 −1 −2

iii

−4 −3 −2 −1 −2

y = −2x

−4 −3 −2 −1 −2

−4

−6

−8

x 1

For f (x) = 3x, will the transformation change the equation of the horizontal asymptote? a

A vertical dilation by a factor of 5

A translation down 2 units

A reflection across the x-axis

A vertical dilation by a factor of

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Fluency Ex 1

5–11, 13, 14c–f, 16, 18, 19bc

y = 2x + 4

y = 2x − 6

is shown.

The graph of a

Describe a transformation of the graph of .

that would obtain b

Sketch the graph of set of axes as

on the same .

Write the equation of the horizontal asymptote of .

Ex 3

5–11, 14cf–h, 15–16, 18, 19cd

Describe how the graph of y = 2x is transformed to get each graph: a

Ex 2

5–9, 12–13, 14abg, 16–17, 19ab

The graph of f (x) = 5x is shown. a

Describe a transformation of the graph of f (x) = 5x that would obtain g (x) = −5x.

Sketch the graph of g (x) = −5x on the same set of axes as f (x) = 5x.

As x increases, what happens to the function values of g (x)?

Of the two functions y = 4x and

The graph of f (x) = 8x is shown. a b

y = 5 (2)x

y 5 4 3 x 2 1 f ( x) = 1 3 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5

5 4 3 2 1

2 3 4 5

f ( x) = 5 x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

2 3 4 5

, which is increasing more rapidly for x > 0?

Describe transformations of the graph of f (x) = 8x that would obtain g (x) = 0.5 (8x). x

Sketch the graph of g (x) = 0.5 (8 ) on the same set of axes as f (x) = 8x.

y 8 6 4 2 -4 -3 -2 -1 -2

f ( x) = 8x x 1

-4 -6 -8

8.03 Optional: Transformations of exponentials mathspace.co

419

The graphs of A, B, and

are shown.

What transformation has been applied to to get the graph of A?

11 10 9

What transformation has been applied to to get the graph of B?

8 7 y=

1 6

5 4

3 2 1

-1

For each graph, describe the transformation applied to graph A to produce graph B. a

1 −4 −3 −2 −1

−1 B

x 1

−2

−3

−4 −3 −2 −1

420

−4 −3 −2 −1

y 11 10 9 8 7 6 5 4 3 2 1

−4

x 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

−2

12 11 10 9 8 B 7 6 5 4 3 2 1 −1

x 2

x 1

Two exponential functions, R and S, are graphed.

One of the graphs is of y = 5x and the other graph is of y = 5x − 2. Which is the graph of y = 5x − 2?

x S R

For the function g (x) = 3 (4)x: a

What transformation has been applied to f (x) = 4x to obtain the graph of g (x) = 3 (4)x?

Complete the table of values: −2

x f (x) = 4x

−1

g (x) = 3 (4)x c Ex 4

For each exponential function: i

Find the coordinates of the y-intercept.

What is the equation of the horizontal asymptote?

iii

Is the function increasing or decreasing?

Graph the function.

y = 3x − 5

e 15

Sketch the graphs of f (x) = 4x and g (x) = 3 (4)x on the same Cartesian plane.

y = 4x + 3

d y = 2 (5)x

The equation of the graph is of the form y = a (bx). a

What is the y-intercept? What does this tell us about the equation?

What is the growth factor? What does this tell us about the equation?

What is the equation?

(1, 20)

18 16 14 12 10 8 6 (0, 5) 4 2 −2

−1

x 1

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421

Ex 5

The equation of the graph is of the form y = a (b)x.

−2

What is the decay factor? What does this tell us about the equation?

What is the equation?

−1

y x 1

−1 −2 (1, −2) −3 −4 (0, −4) −5 −6 −7 −8 −9

This table of values is for an exponential function. x

Find the growth or decay factor.

What is the y-intercept? What does this tell us about the equation?

Find the y-intercept.

What is the exponential equation for the table of values?

This table of values is for an exponential function. x

−4

−3

−2

−1

112

What is the exponential equation for the table of values? Ex 6

For each exponential function: i

Find the coordinates of the y-intercept.

Find the equation of the horizontal asymptote.

iii

Describe the transformations from y = 2x.

Graph the function.

y = 3 (2x) + 1

Reasoning

21–22, 24

y = −(2x) + 4

y = 0.5 (2x) − 3

20, 22, 24

y = −3 (2x) − 1

20, 23, 25

When the only transformation applied to the function f (x) = bx is a dilation by a factor of a, what are the coordinates of the y-intercept? Explain your reasoning.

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How could we use the graph of y = 7x to draw the graph of y = 7 − x?

The graphs of two exponential functions, A and B, are shown.

One of the graphs is of y = 3x and the other graph is of A

Which function has its graph below that of the other when x < 0? Explain your reasoning.

Consider the equation y = −10x. a

Jeannine thinks she has found a set of solutions for the equation as shown in the table: x

−2

−1

−10

−100

−1000

he notices that all the y-values are negative and concludes that for any value of x, y S must always be negative. Is she correct? Explain your reasoning. b 24

Graph y = −10x.

Find the values of x for which y = 0.

The function y = 3 (2x) is shown. a

10 8

Describe a transformation of the graph of y = 3 (2x) that would obtain y = −3 (2x). x

Sketch the graph of y = −3 (2 ) on the same set of axes as y = 3 (2x).

The number of bacteria over time is to be modelled by an exponential function, with x representing time and y representing the number of bacteria. Which function should be used? Explain.

y y = 3(2x)

4 2 −10 −8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

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Emir graphed y = 2 (3x) − 1 by transforming a few points. His work is shown. −1

y = 3x

y = 3x − 1

5 4 3 2 1

y = 2 (3 ) − 1

−4 −3 −2 −1

y = 2 (3 ) − 1

y = 2 (1) − 1

y = 1

Horizontal asymptote: Translated downwards 1 unit to y = −1

1 −4 −3 −2 −1

One other point: (1, 5) a

Did Emir or Yalile draw the graph correctly?

Explain and correct the error.

x 1

−1

−2 −3

26, 28

The graph of f (x) = 2x is shown. a

Describe a transformation of the graph of f (x) = 2x that would obtain g (x) = 2− x.

Sketch the graph of g (x) = 2− x on the same set of axes as f (x) = 2x.

Is g (x) = 2−x an increasing or decreasing function?

27, 29–31

f (x) = 2x

3 2 1 −5 −4 −3 −2 −1 −1 −3 −4 −5

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−2

424

−1

−3

y-intercept: (0, 1)

Problem-solving 26–27

−2

Yalile graphed y = 2 (3x) − 1 by ﬁnding some key features. Her work is shown.

x 1

2 3 4 5

For the function f (x) = 4x: a

Sketch the graph of f (x) = 4x.

Complete the table of values for −3

−2

−1

g (x)

Sketch the graph of

Compare the graphs of f (x) = 4x and

How can the graph of

be transformed to get the graph of

The equation of the original function is form a

. The equation of the new function is of the

Complete the table. Point on original graph

(−1, 2)

(0, 1)

Point on new graph

(−1, ⬚)

(0, 3)

(2, ⬚) (2, ⬚)

Find the equation of the new function.

Sketch the graphs of the original and new function on the same set of axes.

For positive x-values, is the graph of the new function above or below the graph of

For negative x-values, is the graph of the new function above or below the graph of

What is the equation of this exponential graph? 1 −3 −2 −1

−1 −2 −3 −4 −5 −6 −7 −8 −9 −10

y x 1

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The equation of the graph is of the form y = a (bx) + k. −1

x x

y = a (b ) + k y = a (bx) y = bx a

⬚ ⬚

−1

⬚ ⬚

6 4

Complete the row of the table for y = a (bx).

What is value of a?

Complete the row of the table for y = bx.

What is the equation of the graph?

(1, 2)

What is the asymptote? What does this tell us about the equation?

(2, 8)

x −2

−1 −2

1 (0, −1)

Find the equation of the graphs below in the form y = a (bx) + k.

8 6 4 2

10 5 −3

−2

−1

(1, 0) 2

−5 −10

−2

−1

−2 −4 −6 (−1, −8) −8 −10 −12

x 1

8.04 Exponential equations After this lesson, you will be able to… • understand the definition of an exponential equation. • apply a systematic approach to solving an exponential equation. • use the power of a power law to rewrite exponential expressions with the same base. • solve exponential equations graphically.

Exponential equations An exponential expression is an expression of the form bx where the base, b, is a positive number and x is a variable. An exponential equation is an equation in which one, or both, sides are exponential expressions, for example: 3x + 2 = 81. 426

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The key property we need is that: If bx = by, then x = y To solve an exponential equation, we can: 1. Write in the form bx = c. 2. Rewrite in the form bx = b⬚.

3. Set x = ⬚, and solve if necessary.

We may need the power of a power law, (am)n = amn, to rewrite with the same base.

Example 1 Solve 4x = 48.

Create a strategy Set the exponents equal and solve for x.

Apply the idea Both sides of the equation are exponentials with the same base. So the powers must be equal. 4x = 48

Write the equation

x=8

Equate the exponents

Example 2 Solve

Create a strategy Use the negative exponent law to rewrite with the same base.

Apply the idea Write the equation

Use the negative exponent law

Equate the exponents

Example 3 Solve 25x + 1 = 1253x − 4.

Create a strategy Write both sides as exponents with the same base.

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Apply the idea 25 and 125 are both powers of 5 since 52 = 25 and 53 = 125. So we can write both sides with a base of 5. 25x + 1 = 1253x − 4

Write the equation

2 x+1

3 3x − 4

Write the bases as powers of 5

3(3x − 4)

Use the power of a power rule

9x − 12

Use distributive law

2x + 2 = 9x − 12

Equate the powers

(5 )

2(x + 1)

2x + 2

= (5 ) =5

2 = 7x − 12

Subtract 2x from both sides

14 = 7x

Add 12 to both sides

x=2

Divide both sides by 7

Reﬂect and check We can check our answer by substituting x = 2 into the original equation. 25x + 1 = 1253x − 4

Write the equation

2+1

3(2) − 4

= 125

Substitute x = 2

25 = 125

Simplify the exponents

15 625 = 15 625

Evaluate on a calculator

Idea summary An exponential equation is an equation in which one, or both, sides are exponential expressions, for example: 3x + 2 = 81. The key property we need that: If bx = by, then x = y To solve an exponential equation, we can: 1. Write in the form bx = c. 2. Rewrite in the form bx = b⬚. 3. Set x = ⬚, and solve if necessary.

We may need the power of a power law, (am)n = amn, to rewrite with the same base.

Solve exponential equations graphically When we cannot solve equations that involve exponentials by expressing both sides of the equation with the same base, one way that we can get an approximate solution is to use a graph. For example, suppose we were asked to solve 2x + 2 = 3−x We can plot the graph of y = 2x + 2 and the graph of y = 3−x on the same axes. The solution to the equation is the x-coordinate of the point of intersection of the two curves.

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Interactive exploration

Discover this concept in action online

mathspace.co

Example 4 For the equation 2x = 3x + 1: a Using technology, graph y = 2x and y = 3x + 1 on the same axes. Use a window of x from −5 to 5, and y from −5 to 20.

Apply the idea 18 16 14 12 10 8 6 4 y = 3x + 4 2

y = 2x x

−4 −3 −2 −1 −2

b How many solutions are there to the equation 2x = 3x + 1?

Create a strategy We can see solutions on a graph by looking for points of intersection.

Apply the idea 18 16 14 12 10 8 6 4 y = 3x + 4 2 −4 −3 −2 −1 −2

y = 2x x 1

We can see two points of intersection on the graph, so there would be two solutions to the equation (as far as we can tell from this graph).

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c Using technology, ﬁnd each solution rounded to one decimal place, where appropriate.

Create a strategy The solutions we can see from the original graph are close to −1, and close to 4. We need to zoom in on each solution.

Apply the idea 0.9

0.8 0.7

The curves cross between x = −1.2 and x = −1.1, closer to x = −1.2, so we could give the solution as x = −1.2 to one decimal place.

0.6 0.5 0.4 0.3 0.2 −1.4 −1.2

−1 −0.8 −0.6

0.1 x

The solution appears to be exactly at x = 4.

We can show that this solution is exactly at 4. 24 = 16 and 3(4) + 4 = 16 16

x 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4

Idea summary We can get approximate solutions to equations that involve exponentials by graphing the function on each side of the equation using technology and ﬁnding the x-coordinate of any points of intersection.

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8.04 Practice questions Understanding 1

Explorer 1, 3–4

1–4

Which statements are true? a

If bx = ay, then x = y

If ax = ay, then x = y

If ax = bx, then a = b

, then x = y

When solving an exponential equation with a variable in the exponent, why do we want the bases on both sides of the equation to be the same?

Rewrite each number using the given base. a

125 with a base of 5

32 with a base of 2

64 with a base of 4

81 with a base of 3

with a base of 6

89 = 8⬚

c g

5–8, 9a–f, 10a–c, 11a–dg, 12ace, 13, 14ab, 15

4⬚ = 64

(4)2 = (22)⬚

6–8, 9d–i, 10b–d, 11d–i, 12b–f, 13, 14bc, 15

10⬚ = 1

(32)2 = 9⬚

6, 9h–l, 10d–f, 11e–j, 12d–h, 14cd, 15–16

Solve: a

73 = (⬚)3

Fluency 5

with a base of 10

Complete each statement: a

Ex 2

Trailblazer

Ex 1

Adventurer

10x = 100

10x = 0.01

10x = 0.0001

6x = 6–3

3x = 27

2x = 64

5n = 625

Solve: a

3x = 36

7x = 1

9y = 81

For each equation: i

Rewrite both sides of the equation with a base of 2.

Solve for x.

8x = 4

Solve: a

5x = 0.2

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Ex 3

Solve: a

9y = 27 2 x+7

b 3

ax − 1 = a4 −3x − 1

= 125

9x + 3 = 27x

8x + 3 = 322x − 1

36x − 1 = 2162x + 5

= 3125

35x − 10 = 1

4x + 4 = 8

81x − 1 = 93x + 5

Solve: a

5 × 10x = 500

10 × 2x = 80

30 × 2x − 6 = 15

54 × 3x − 6 = 6

40 × 3x + 1 = 3240

18 × 52x − 1 = 450

9x + 2 − 8 = 1

4x + 32 = 288

4−2x × 4x = 64

5x × 5x − 4 = 25

2x × 8x + 1 = 16

Solve: 2x × 2x + 2 = 16

(2 )

10x + 33

i 12

25y = 125

b f

Consider the equation 2x = 8. a

Sketch the graph of y = 2x.

Using the graph, ﬁnd the x-value when y = 8.

Solve the equation 2x = 8.

For each equation: i

Identify two functions that could be graphed to solve the equation.

Using technology, graph both curves and estimate the solution to two decimal places.

2x = 5

Consider the equation

Ex 4

4 × 3x = 20

2x − 2 = 2x + 5

Using technology, ﬁnd the solution(s) to the equation, rounding the answers to one decimal place. 16

Consider the equation

−3(2x) = −3x − 4

Using technology, ﬁnd the solution(s) to the equation, rounding the answers to one decimal place.

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Reasoning

17, 19

17–19

18, 20, 21

Explain why 23x − 1 = 27x − 9 cannot be solved using index laws, while 23x − 1 = 8x − 9 can.

Explain the reasoning for each step of this solution. 1

Given equation

2 3 4 5 6 7 19

Lorena was trying to ﬁnd the solution to the equation Her working is shown.

Find and explain Lorena’s mistake.

Correct her solution.

3 4 5

Explain how to solve the equation roots.

using your knowledge of squares and square

Julian wants to ﬁnd the solution to the equation 200 (1.02)x = 220. To do this, he uses his graphing calculator to graph y = 200 (1.02)x, then uses his calculator to generate a table of values. His calculator shows this table of values for y = 200 (1.02)x. Explain how Julian can use the table to estimate the solution to the equation 200 (1.02)x = 220.

x 0 1 2 3 4 5 6 7

y 200 202 208.08 212.24 216.49 220.82 225.32 229.74

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Problem-solving 22, 24, 25acf 22

23–24, 25b–f

23–24, 25b–h, 26

A certain type of cell splits in two every hour and each cell produced also splits in two each hour. The total number of cells after t hours is given by: N = 2t Find the time when the number of cells will reach the following amounts: a

1024

Given the points (3, n), (k, 16) and value of:

a 24

4096

all lie on the curve with equation y = 2x, ﬁnd the

The frequency, F, in Hertz (Hz) of the nth key of an 88-key piano is given by a

Find the frequency of the forty-ninth key.

Find the frequency of the 40th key to the nearest whole number.

Use technology to ﬁnd the value of n that corresponds to the key with a frequency of 1760 Hz.

Solve these exponential equations: a

27 (2x) = 6x

Using the digits 1 to 9 each at most once, ﬁll in the boxes to write a valid exponential equation.

8.05 Compound interest After this lesson, you will be able to… • apply the formula for calculating simple interest. • apply the compound interest formula for varying compounding frequencies. • convert interest rates and time periods based on different compounding intervals. • solve real-world problems involving simple and compound interest.

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Repeated applications of simple interest Simple interest is calculated as a ﬂat percentage of the amount that was borrowed or invested, called the principal. We can calculate the simple interest of a loan using the formula:

I = PRT I P R T

is the total interest earned is the principal (the initial amount borrowed/invested) is the interest rate, as a decimal is the number of time periods (the duration of the loan/investment)

So the total amount in the account will be:

A=P+I A P I

future value in the account principal amount interest accumulated

Exploration Taryn and Elijah each open an account with $1 000 that interest at 10% per annum. 1. Taryn earns simple interest. Complete the table to show how the account balance grows over 5 years. Year 1 2 3 4 5

Opening balance 1000 1100 1200

Interest 100 100

Closing balance 1100 1200

2. Elijah’s interest payment changes each year as he earns interest on the closing balance in the account from the previous year instead of on the principal. Complete the table to show how Elijah’s account balance grows over 5 years now. Year 1 2 3 4 5

Opening balance 1000 1100 1210

Interest 0.1 × 1000 = 100 0.1 × 1100 = 110 0.1 × 1210 = 121

Closing balance 1000 + 100 = 1100 1100 + 110 = 1210

3. Plot the closing balances for the ﬁrst 5 years for Taryn and Elijah’s accounts on the same Cartesian plane. How does the growth of Taryn’s account compare to the growth of Elijah’s account? 4. Which investment option would earn more money in the long run?

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When an account earns simple interest, it grows by the same amount each time. The growth in the balance of a simple interest account is linear. When the interest for the next period is calculated on the principal plus the previously earned interest, this is known as compound interest. The growth in the balance of a compound interest account is exponential.

Example 1 The graph shows both an investment with simple interest, and one with compound interest, labelled Investment A and Investment B respectively.

Investment ($1000) 10 8 6

Investment A

4 Investment B

Time (years)

a Which investment has a higher principal amount?

10 12

Create a strategy Choose the graph with the higher y-intercept.

Apply the idea The graph for Investment A has a higher y-intercept than the graph for Investment B. So, Investment A has a higher principal amount.

b Which investment has a higher ﬁnal amount after 10 years?

Create a strategy Find the point on each graph that lines up with x = 10.

Apply the idea From the Cartesian plane we can see that when x = 10 the value of Investment B is higher that the value of Investment A.

Investment ($1000) 10 8

So, Investment B has a higher ﬁnal amount after 10 years.

Investment A

4 Investment B

Time (years) 2

436

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c After how many years are the investments equal in value?

Create a strategy Find the intersection point of the graphs.

Apply the idea 10

From the the Cartesian plane we can see that the two graphs intersect at a point where x = 9.

So this means that after 9 years the investments are equal in value.

Investment ($1000)

Investment A

4 Investment B

Time (years) 2

10 12

Example 2 $3200 is invested for three years at a rate of 6% p.a., compounding annually. a Complete the table to determine the ﬁnal value of the investment. Round to the nearest cent for each year. First year Second year Third year Fourth year

Create a strategy

Balance + Interest − $3200 + $192 $3392 + ⬚ ⬚

Total balance $3200 $3392 ⬚ ⬚

Interest earned $192 $203.52 ⬚ −

To get the value in the “Interest earned” column, calculate the simple interest on the amount in the “Total balance” column. The values of “Total balance” and “Interest earned” are copied down to the next row into the “Balance + Interest” column, and add them together to give the new “Total balance”.

Apply the idea We bring down the “Interest earned” of $203.52 from the row above then evaluate the sum $3392 + $203.52 = $3595.52 to ﬁll in the second box. First year Second year Third year Fourth year

Balance + Interest − $3200 + $192 $3392 + $203.52 ⬚

Total balance $3200 $3392 $3595.52 ⬚

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We calculate interest on $3595.52 and put it in the interest earned column. $3595.52 × 0.06 = $215.73 Balance + Interest −

First year Second year Third year Fourth year

$3200 + $192 $3392 + $203.52 ⬚

Total balance $3200 $3392 $3595.52 ⬚

Interest earned $192 $203.52 $215.73 −

Total balance $3200 $3392 $3595.52 $3811.25

Interest earned $192 $203.52 $215.73 −

We bring down the “Total balance” of $3595.52 and the “Interest earned” of $215.73 from the row above then evaluate the sum $3595.52 + $215.73 = $3811.25 to ﬁll in the second box. Balance + Interest −

First year Second year Third year Fourth year

$3200 + $192 $3392 + $203.52 $3595.52 + $215.73

b Calculate the total interest earned over the three years.

Create a strategy Add all the values on the “Interest earned” column from part (a).

Apply the idea Total interest earned = 192 + 203.52 + 215.73 Add the values = $611.25

Evaluate

Idea summary We can calculate simple interest with the formula:

I = PRT

I interest accumulated P principal amount R rate of interest, as a decimal T time in years Simple interest pays the same amount of interest in Investment value each period because it is based on a percentage of the 10 principal only. 8 The growth in the balance of a simple interest account is linear. 6 Compound interest pays increasing amounts of interest Simple interest 4 in each period because it is based on a percentage of the Compound interest balance in the account, which grows as interest is added. 2 The growth in the balance of a compound interest account is Time 2 4 6 8 10 12 14 exponential.

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Compound interest formula It can be time consuming to use a table of values to calculate the balance for an account, especially as the time increases. We could use a spreadsheet to do repeated calculations, but a formula would be helpful. Let’s look investing $1000 at 10% p.a. using compound interest. Using the idea that A = P + I in a table, we get: Year 1 2 3

Opening balance 1000 1100 1210

Compound interest calculation 1000 × 0.1 = 100 1100 × 0.1 = 110 1210 × 0.1 = 121

Closing balance 1000 + 100 = 1100 1100 + 110 = 1210 1210 + 121 = 1331

Note that we can go straight to the closing balance from the opening balance by multiplying by 1.1 since we are doing 1000 + 0.1(1000) = 1.1(1000) to get the closing balance. Year 1 2 3

Opening balance 1000 1100 1210

Closing balance 1000 × 1.1 = 1100 1100 × 1.1 = 1210 1210 × 1.1 = 1331

This is a quicker way of calculating, but we can go one step further. Year 1 2 3

Opening balance 1000 1100 1210

Closing balance 1000 × 1.1 = 1100 1000 × 1.1 × 1.1 = 1210 1000 × 1.1 × 1.1 × 1.1 = 1331

Note from the table that to get to the closing balance in year 3, we have multiplied the principal by 1.1 three times. $1 331 = $1 000 × 1.1 × 1.1 × 1.1 = $1 000 × 1.13 Generalising this, we can write the compound interest formula as:

A = P (1 + r)n A P r n

closing balance or future value of the investment principal amount interest rate per time period, as a decimal number of time periods

If we want to ﬁnd the amount of interest earned, we can subtract the principal amount from the future value of the investment. Interest = Future value − Principal amount

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Example 3 Valerie borrows $1250 at a rate of 7.6% p.a., compounding annually. a Which of these expressions represents the amount Valerie must repay after 15 years, assuming that she hasn’t paid anything back? A

1250 × (1 + 0.076) × 15

1250 × (1 + 0.076)15

Create a strategy Use the compound interest formula: A = P (1 + r)n

Apply the idea • P = 1250 • • n = 15 A = P (1 + r)n

Write the formula 15

= 1250 (1 + 0.076)

Substitute the known values

The correct answer is option D.

b How much interest was generated on the loan over the ﬁfteen years? Round your answer to two decimal places.

Create a strategy Subtract the principal from the future value.

Apply the idea In part (a), we have the future value, 1250 (1 + 0.076)15. So we have: Interest = 1250 (1 + 0.076)15 − 1250

Substitute the values

= 3750.54 − 1250 Evaluate the future value = $2500.54

Evaluate

The interest generated on the loan over the fifteen years was $2500.54

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Idea summary We can calculate compound interest with the formula:

A = P (1 + r)n A closing balance or future value of the investment P principal amount r interest rate per time period, as a decimal n number of time periods If we only want to ﬁnd the amount of interest earned, we can simply subtract the principal amount from the future value of the investment. Interest = Future value − Principal amount

Different periods The examples we have looked at so far have all had interest calculated annually. However, interest can be calculated over various time periods. To convert the annual interest rate to a periodic interest rate, we can divide it by the number of compounding periods in a year. For example, for an investment that earns an interest rate of 6% p.a., compounding quarterly. Since there are 4 quarters in a year:

It is also important to match the total number of periods to the duration of the investment. To do this, we multiply the number of years by the number of compounding periods per year. If interest is calculated quarterly for 5 years, there will have been: 4 quarters per year × 5 years = 20 time periods

Example 4 A $9450 investment earns interest at 2.6% p.a., compounded monthly over 14 years. What is the future value of the investment?

Create a strategy Use the compound interest formula: A = P (1 + r)n

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Apply the idea We are given that: P = 9450

n = 14 × 12 months = 168 months

Substitute the values

Evaluate

The future value is $13 593.90

Example 5 Victoria borrows $35 000 at a rate of 4.8% p.a., compounded monthly. a After 4 months, Victoria repays the loan all at once. How much money does she pay back in total?

Create a strategy Use the compound interest formula: A = P (1 + r)n

Apply the idea We are given that: P = 35 000 n = 4 months

Substitute the values

Evaluate

The total amount that Victoria repaid after four months is $35 563.37

b How much interest was generated on the loan over the four months?

Create a strategy Use the fact that the amount of interest generated is the difference between the ﬁnal amount repaid and the initial amount borrowed.

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Apply the idea Use the total amount that Victoria repaid after four months from part (a). Interest = 35 563.37 − 35 000 = 563.37

Subtract the two values Evaluate

Idea summary If interest is compounded daily, weekly, fortnightly, monthly, quarterly, or half-yearly, then we need to convert the interest rate and number of time periods to the same units. To convert the interest rate r, divide the annual rate by the number of compounding periods in a year. To convert the number of time periods n, multiply the number of years by the number of compounding periods in a year. We usually use the following values when converting the rate and periods: • • • • • •

There are 365 days in a year. There are 52 weeks in a year. There are 26 fortnights in a year. There are 12 months in a year. There are 4 quarters in a year. There are 2 half-years in a year.

8.05 Practice questions Understanding 1

Explorer 1–3, 5

Adventurer 1–3, 5

Trailblazer 1–2, 4–5

For compound interest, are these statements always true, sometimes true, or never true? a

The account balance grows exponentially.

The amount of interest accrued for the ﬁrst and second periods will be the same.

Assuming the account remains untouched, the amount of interest accrued per period increases.

The interest rate is applied to the initial principal.

The interest rate is applied to the current balance.

The ﬁnal balance of an investment is more than an investment with simple interest.

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The table shows the bank balance of two accounts. Xuan invested in a simple interest yielding account for 3 years at 7%, and Florence invested in a compound interest account, compounding annually for 3 years also at 7%. Account A Account B

Principal $5000 $5000

Year 1 $5350.00 $5350.00

Year 2 $5724.50 $5700.00

Year 3 $6125.22 $6050.00

According to the data, identify which account is Xuan’s. 3

$1000 is invested for 10 years with an interest rate of 10% p.a., compounded annually. Which graph represents the growth of the investment? Graph 1 Graph 2 Investment ($)

Investment ($)

3000

2500

2000

1500

1000

500

500 T (years) 1

T (years) 1

2 3 4 5 6 7 8 9 10

Determine if each graph can represent a savings account that earns compound interest over 50 years. a

Balance ($)

Time (years)

Balance ($)

Time (years)

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Balance ($)

Time (years)

Frasier is calculating the compound interest accumulated on his loan. He types this into his calculator: 9000

0.07

)

(

)

Identify how much Frasier borrowed in dollars.

Identify the annual interest rate as a percentage.

Identify if the interest is being compounded weekly, monthly, quarterly or annually.

Identify how many years he is calculating the interest for.

Fluency Ex 1

(

6–12, 14–15

6–7, 9–11, 13–15

6, 8, 10–11, 14–15

Investment ($1000) The given graph shows two investments, one with simple interest (Investment A) and another with compound 10 interest (Investment B):

Which investment has a higher principal amount?

Which investment has a higher ﬁnal amount after 10 years?

After how many years will the investments be equal in value?

2 Time (years) 2

Complete the table to determine the ﬁnal value of the investment. Round to the nearest cent for each year. First year Second year Third year Fourth year

Ex 2

$3200 is invested for three years at a rate of 5% p.a., compounded annually. a

Balance at beginning of year $3200 $3360

Interest earned $160 $168 −

Calculate the total interest earned over the three years.

$3500 is invested for three years at a rate of 3.8% p.a., compounded annually. a

Complete the table to determine the ﬁnal value of the investment. Round to the nearest cent for each year. First year Second year Third year Fourth year

Balance at beginning of year $3500 $3633

Interest earned $133 $138.05 −

Calculate the total interest earned over the three years.

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A $2090 investment earns interest at 4.2% p.a. compounded annually over 5 years. a

Complete the table to determine the ﬁnal value of the investment if each period is rounded to the nearest cent. Year Start of investment, 0 First year, 1 Second year, 2 Third year, 3 Fourth year, 4 Fifth year, 5

10 Ex 3

11 Ex 4

14 Ex 5

446

$2090 + $87.78

Value $2090 $2177.78

Interest earned $87.78 $91.47

−

Complete the table to determine the ﬁnal value of the investment if each period is rounded to the nearest dollar. Year Start of investment, 0 First year, 1 Second year, 2 Third year, 3 Fourth year, 4 Fifth year, 5

Balance + interest −

Balance + interest − $2090 + $88

Value $2090 $2178

Interest earned $88 $91

−

What is the difference in the two values after 5 years?

Tauren borrows $500 000 at a rate of 1.2% p.a., compounding annually. a

Write an expression that represents the amount Tauren must repay after 24 years, assuming that he hasn’t paid anything back.

How much interest was generated on the loan over the twenty four years? Round your answer to two decimal places.

Calculate the future value, A, of each investment: a

A $7150 investment earns interest at 2.5% p.a., compounded weekly over 3 years.

A $5420 investment earns interest at 5% p.a., compounded monthly over 4 years.

Tiara borrows $5000 at a rate of 4.5% p.a., compounded annually. a

After 3 years, Tiara repays the loan all at once. How much money does she pay back in total?

How much interest was generated on the loan over the three years?

Ivan borrows $3000 at a rate of 6.9% p.a., compounded annually. a

After 6 years, Ivan repays the loan all at once. How much money does he pay back in total?

How much interest was generated on the loan over the six years?

Vera borrows $25 000 at a rate of 6% p.a., compounded monthly. a

After 5 months, Vera repays the loan all at once. How much money does she pay back in total?

How much interest was generated on the loan over the ﬁve months?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Calculate the interest that each investment accumulates: a

A $5970 investment earns interest at 2.1% p.a., compounded annually over 17 years.

A $5460 investment earns interest at 2.8% p.a., compounded monthly over 10 years.

A $6240 investment earns interest at 2.1% p.a., compounded weekly over 6 years.

Reasoning 16

16–17

16, 18–19

Evangelina wants to invest $1400 at 5% p.a for 5 years. She has two investment options, compounding quarterly or compounding daily. a

Calculate the value of the investment if it is compounded quarterly.

Calculate the value of the investment if it is compounded daily.

Calculate how much extra the investment is worth if it is compounded daily rather than quarterly.

What effect does the frequency of compounding have on the total amount of interest earned?

Jadria applies for a loan of $9000 and she is offered the choice between two loan plans: • Plan 1: Simple interest at a rate of 2.9% • Plan 2: Compound interest at a rate of 2.75% compounded monthly

Determine which plan Jadria should choose if she wants to pay back the loan over 3 years. Explain your reasoning.

Jadria decides that she needs longer to pay off the loan. Determine if the plan Jadria chose in part (a) will always be the better choice. Explain your reasoning.

Hassan plans to borrow $25 000 to buy a new car. After shopping around, he ﬁnds three different offers: • 4.75% compounded annually • 4.7% compounded quarterly • 4.65% compounded monthly Which offer should Hassan choose? Explain your reasoning.

Dorian’s investment of $6000 earns interest at 2% p.a., compounded annually over 3 years. He uses this table to predict the value of his investment. Year Start of investment, 0 First year, 1 Second year, 2 Third year, 3

Balance + interest − $6000 + $120 $6120 + $122 $6242 + $125

Value $6000 $6120 $6242 $6367

Interest earned $120 $122 $125 −

Find the value of the investment after 3 years using the compound interest formula.

Explain why there is a difference between your answer to part (a) and the value of the investment after 3 years in the table.

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Problem-solving 20–24

21–25

21–23, 25–27

Rory is planning to make an investment which accumulates monthly compound interest at a rate of 4.8% p.a. If she aims to have her investment reach $90 000 after 11 years, ﬁnd the principal amount that she should invest.

Cade has a savings account which accumulates compound interest yearly at a rate of 2.9% p.a. If he aims to have $60 000 in his account after 4 years, ﬁnd the principal amount that he should invest.

Han opened a savings account at the beginning of February 2023, where the interest is compounded monthly. His statements for March, April and May show his account balance at the beginning of each month:

Find the monthly interest rate, r, for his investment.

How much did Han deposit into this savings account when he opened it?

Month March April May

Balance $3825 $3901.50 $3979.53

Meerah invests $600 into an account which accumulates interest at a rate of 5.25% p.a., compounding annually. a

Use technology to graph of the relationship between the number of years passed and the Meerah’s account balance.

How many years will it take for the account to reach a total value of $2000?

David places $800 into an investment fund where interest is compounded annually at a rate of 6% per annum. Calculate the number of years required for his investment to reach $3200.

Dietrich wants to put a deposit on a house in 5 years. In order to ﬁnance the $80 000 deposit, he decides to put some money into a high interest savings account that pays 5.5% p.a. interest compounded monthly. If P is the amount of money that he must put into his account now to accumulate enough for the deposit, ﬁnd P.

A high interest savings account is advertised as earning 5% interest compounded monthly. At the end of the year, the actual interest earned is higher than 5%. Find the actual interest rate after one year. Write your answer as a percentage, rounded to three decimal places.

Mia invests $1000 in a savings account with interest compounded quarterly at an annual rate of 5%. Determine how many years it will take for her investment to grow to $2500.

Investigation: Half life

Investigate online

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8.06 Exponential models After this lesson, you will be able to… • differentiate between exponential growth and exponential decay. • identify and define key features of exponential functions. • express exponential functions algebraically, highlighting either the growth or decay factor or rate. • apply knowledge of exponential functions to solve real-world problems. • interpret the implications of the growth or decay rate in practical scenarios.

Exponential growth 10

Exponential growth happens when a quantity increases by a certain percentage over time. We can describe this kind of growth with an exponential formula:

9 8

y = abx

When we are thinking about growth or decay, this is with respect to time. So usually we are only interested in positive x-values. • a is the starting value (or initial value), when x =0. • b is the growth factor, which shows how fast the value grows.

6 5

y = 3(1.6x)

4 3 2 1

x 1

If b > 1, we have growth. If 0 < b < 1, it’s decay.

The compound interest formula: A = P (1 + r)n Is an example of exponential growth where r is the growth rate (interest rate), of the investment. So if we connect the exponential growth formula, y = abx, with the compound interest formula we can see that b = 1 + r. This is greater than 1, also confirming that it is growing. We can use the key features of an exponential relationship to build its equation. First, we must find the y-intercept (or the initial value), then find the growth factor (b) or growth rate (r).

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Exploration Here are three exponential relationships represented in three different ways: Relationship 1: 8

Relationship 2: x y

0 1

1 1.15

2 1.3225

3 1.5209

4 1.7490

Relationship 3:

Reagan decided to get a part-time job to begin saving for a new car. He was hired by a company that pays $21 500 in the ﬁrst year, and he will receive a 5% raise each year after that.

5 4 3 2 1

x 1

1. Which of the relationships would you model using y = abx and why? 2. Which of the relationships would you model using y = a (1 + r)x and why? 3. In what type of situations would you prefer knowing the growth factor, b, over the growth rate, r?

In situations where growth is given as a percentage, it is easiest to model the equation in the form y = a (1 + r)x. When trying to determine the equation from a table or a graph, it is easiest to ﬁnd the equation in the form y = abx. We usually refer to the growth rate as percentage, just like in the compound interest formula. If you need to switch between forms, remember that b=1+r • b is the growth factor • r is the growth rate, usually a decimal or percentage

Example 1 Identify the y-intercept and the growth rate for each exponential equation: a y = 3 (1.725)x

Create a strategy This equation is in the form y = abx. We can identify the y-intercept and the growth factor, but then we need to use b = 1 + r to solve for the growth rate.

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Apply the idea The growth factor, b, is 1.725 1.725 = 1 + r r = 0.725 The y-intercept is (0, 3), and the growth rate is 72.5%.

Reﬂect and check In the formula, the growth rate is represented as a decimal. When we are discussing the rate outside of the formula, it is best to state it as a percentage.

Create a strategy This equation is in the form y = abx. We can identify the y-intercept and the growth factor, but then we need to use b = 1 + r to solve for the growth rate.

Apply the idea For y = abx, we can use that a is the y-intercept and b is the growth factor.

To ﬁnd the growth rate: b = 1 + r

Equation for growth factor

2 = 1 + r

Substitute b = 2

1 = r

Subtract 2 from both sides

The y-intercept is

and the growth rate is 1 or 100%.

Reflect and check When something grows by 100%, it is being increased by its own value. In other words, we are doubling the output each time.

Example 2 Justin purchased a piece of sports memorabilia for $2900, and it is expected to increase in value by 9% per year. a Write an equation, V, to represent the value of the piece of sports memorabilia after t years.

Create a strategy Since the memorabilia is predicted to increase in value by a percentage, we will use the growth rate form of the equation, y = a (1 + r)x with the given variables: V = a (1 + r)t. 8.06 Exponential models mathspace.co

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Apply the idea The initial value of the memorabilia is $2900. This represents the value when t = 0, so this is the y-intercept. The growth rate is 9% which is 0.09 as a decimal, so the equation is V = 2900 (1 + 0.09)t.

Reﬂect and check Notice the similarities to the compound interest formula A = P (1 + r)n.

b Evaluate the equation for t = 8 and interpret the meaning in context.

Create a strategy

Apply the idea

In this context, t represents the time in years, V = 2900 (1 + 0.09)8 ≈ 5778.43 which tells us and the output, V, represents the value of that the memorabilia will be worth approximately Justin’s sports memorabilia. We will evaluate the $5778.43 after 8 years have passed. equation and apply these units to interpret the meaning of the solution.

Idea summary An exponential growth relationship can be expressed as y = abx a is the y-intercept, or initial value b is the growth factor In cases where we want to define the specific growth rate, like an interest rate we use b = 1 + r, and end up with the compound interest formula: y = a(1 + r)x

Exponential decay We have seen that exponential relationships can be expressed in the form:

y 4 3 2

y = abx x

y = 2 (0.3)

x 1

3 4

where: • a is the starting value (or initial value), when x = 0. • b tells us how fast it is decaying if 0 < b < 1.

For exponential growth, we saw that for a growth rate r: y = a (1 + r)x Exponential relationships can be classiﬁed as exponential decay based on the value of the decay factor. We get exponential decay when 0 < b < 1. 452

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Using b = (1 − r) where b is the decay factor and r is the decay rate, we can express our exponential equation y = abx as y = a (1 − r)x Notice that the base of the exponent is (1 − r). This is the decay factor because it is less than 1.

Exploration Consider these exponential relationships: Relationship 1: A ball is dropped from a ledge. The height of the ball rebounds to 80% of its previous height after each bounce. 9

x 2

Relationship 2

Relationship 3

Relationship 4: Relationship 5: x y

0 10

1 9.6

2 9.216

3 8.847 36

1. Which ones can be classiﬁed as exponential decay? 2. Explain how you identiﬁed the relationships that decay.

An equation for exponential decay will have a decay factor that is less than 1, which will cause the y-values to decrease over time. When a percentage is used, sometimes it is the value of r, while other times it is the value of b. For example, these two scenarios are the same: • Dara has a bag of 400 cookies. Each day she leaves 80% for the next day. C = 400 (0.80)x = 400 (1 − 0.20)x • Dara has a bag of 400 cookies. Each day Dara eats 20% of the cookies in her bag. C = 400 (1 − 0.20)x = 400 (0.80)x

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We can use the key features of an exponential relationship to build its equation. 1. Find the y-intercept. 2. Find the decay factor or decay rate. 3. Choose the specific form of the equation depending on whether you want to highlight the decay factor or the decay rate.

Example 3 Identify the y-intercept and the decay rate for each of these exponential equations: a

Create a strategy This equation is in the form y = abx. We can identify the y-intercept and the decay factor, but then we need to use r = 1 − b to solve for the decay rate.

Apply the idea r = 1 – b

Equation for decay rate

r = 1 − 0.625

Substitute b = 0.625

r = 0.375

Evaluate the subtraction

As a percentage, the decay rate is r = 37.5%. Since

, the y-intercept is

Create a strategy This equation is in the form y = abx. We can identify the y-intercept and the decay factor, but then we need to use r = 1 − b to solve for the decay rate.

Apply the idea We see a = 1 and

. Next, we need to solve for the decay rate. Equation for decay rate

Substitute

Evaluate the subtraction

The y-intercept is at (0, 1), and the decay rate as a percentage is 75%.

Reﬂect and check Notice that the original equation tells us that we are keeping losing 75%, as shown with the decay rate.

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. This means that we are

Example 4 A frog population, F, has been following the model F = 480(0.952)n, where n is the number of years. a What is the initial population of frogs?

Create a strategy The initial value will be when time is 0.

Apply the idea When n = 0, F = 480, so the population was 480.

Reﬂect and check Since negative values for time do not make sense, the y-intercept will be the leftmost point on the graph. In this case, we refer to it as the initial value.

b What is the decay rate? Explain what it means in this context.

Create a strategy

Apply the idea

We can ﬁnd the decay rate using that for y = abx, b = 1 − r has decay rate of r.

Since b = 0.952: b=1−r 0.952 = 1 − r r = 1 − 0.952 r = 0.048 This means that the decay rate is 4.8%. The frog population is declining at 4.8% per year.

c Find the population of frogs after 3 years, rounding to the nearest integer.

Create a strategy We can substitute n = 3, and then evaluate using a calculator.

Apply the idea F = 480(0.952)n F = 480(0.952)3 F ≈ 414 After 3 years, there would be 414 frogs remaining.

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d Using technology, graph the frog population for 0 < n < 120.

Create a strategy We can input the equation and then adjust the view using the zoom so we can see from n = 0 to n = 120.

Apply the idea A good window setting is 0 < n < 120 and 0 < F < 480. 480 440 400 360 320 280 240 200 160 120 80 40

x 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120

e Use your graph to determine approximately how many years it will take for the population to reach half of the original population.

Create a strategy We want to ﬁnd half of 480 on the vertical axis and then slide over to ﬁnd where the graph reaches that value.

Apply the idea Half of 480 is 480 440 400 360 320 280 240 200 160 120 80 40

x 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120

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Using our original scale we can see this is between n = 10 and n = 15, so we can zoom in to get a better estimation. 275 F 270 265 260 255 250 245 240 235 230 225 220 215 210 205

x 11

From the zoomed in graph, we can see that it is just after 14 years where the population reaches half of the original value.

Reﬂect and check We can solve for the exponent using a graph, table of values, or writing both sides of the equation with the same base if possible.

Idea summary An equation for exponential decay can be expressed as y = abx a is the y-intercept, or initial value b is the decay factor In cases where we want to define the specific decay rate, like an interest rate we use b = 1 − r, and end up with the compound interest formula: y = a(1 − r)x

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8.06 Practice questions Understanding 1

Explorer 1–4

A store owner checked the sales report for the previous month and found that each week the sales were

In a laboratory, the number of bacteria in a petri dish is recorded, and the bacteria are found to triple each hour.

A sequence starts with 2, 3 and the next term is always the product of the previous two terms.

A piece of land was purchased for $80 000. The value of the land has slowly been decreasing by 2% annually.

Michael has $1300 in his checking account. He gets paid $600 every 2 weeks.

For each exponential equation, where x represents time: i

Is it exponential growth or exponential decay?

What is the initial amount?

iii

What is the growth or decay rate as a percentage?

y = 0.723 (1 − 0.05)x x

y = 5.9 (1.025)

y = 2500 (0.877)x

y = 10 000 (2)x

What is the initial size of the rat population?

At what rate is the population is decreasing?

This is a table of values for an exponential relationship: 1 5

2 7

3 9.8

4 13.72

5 19.208

What is the growth factor?

What is the constant percent rate of change?

Fluency

5–9, 11, 13–14

6–9, 11, 13–14

6–8, 10–13

The mass in kilograms of a baby orangutan at n months of age is given by the equation M = 1.8 (1.1n), for ages up to n = 6 months. a

Find the mass of a baby orangutan at 3 months of age to the nearest tenth of a kilogram.

Complete the table. Round your answers to four decimal places. Months Mass (kg)

c 458

The local rat population is changing according to the exponential equation y = 840 (1 − 0.04)t where t is the number of years that have passed.

x y

of the previous week’s sales.

c 3

Trailblazer 1–4

Are these situations exponential growth, exponential decay, or neither? a

Ex 1, 3

Adventurer

1–4

Sketch a graph of M.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

6 Ex 2

Marlow purchased a piece of sports memorabilia for $1300, and it is expected to increase in value by 8% per year. a

Write an equation in terms of y to represent the value of the piece of sports memorabilia after v years.

Find the value of the piece of sports memorabilia after 10 years.

The population of Juanita’s home town is 27 400 and has a growth rate of 6% each year. Estimate the population after 7 years.

The population of some native bees is declining at a rate of 8% per year. If there are 10 000 bees in a hive now, ﬁnd the number of bees in 5 years time.

A property was valued 9 years ago to be worth $755 000. Its value depreciated at 4.4% p.a. due to poor maintenance. Calculate its depreciated value. Round your answer correct to the nearest dollar.

The price of train tickets in 1996 was $3. If the value inﬂated at an average rate of 6.3% per annum, what would the price have been in 2005?

An item costs $14 000 today. Find its price, P, 6 years ago if the inﬂation rate was an average of 2.5% per annum. Round your answer to the nearest dollar.

Winston and Rasiah want to organise a tournament. They want a knockout tournament, where the winner of each round progresses to the next round until there are only two players left. The diagram shows the draw for the final, semi-final and quarter final rounds. a

Ex 4

Quarter-Final

Final

Winner

Complete the table of values for the total number of players, p, that the competition can accommodate given a number of rounds, r. Number of rounds (r) Number of players ( p)

Semi-Final

Winston and Rasiah want to make sure that each round of the tournament has every spot ﬁlled. Find the values of p for which a tournament can be formed.

Write an equation relating r and p.

Over two weeks, they can ﬁt in 8 rounds of play. How many players can they accept into the tournament?

The number of fungal cells, N, in a colony after t minutes is given by N = 5000 (1 + 0.023)t. a

What is the initial population of fungal cells?

What is the constant percent rate of change? Explain what it means in context.

Find the population of fungal cells after 3 hours, rounding to the nearest integer.

Graph the fungal cell population for 0 < t < 120.

Use your graph to determine approximately how many minutes it will take for the population to reach 10 times the original population.

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Djamila loves treating their dog. They realise that they are currently giving their dog 2 kg worth of treats every week, which is way too much. Their plan is to cut the amount of treats they give each week in half until they are giving less than 50 g of treats per week. a

Use a table of values to determine how many weeks it will take to get down to less than 50 g of treats per week.

Write an equation relating week number to the weight of treats eaten.

Reasoning

15–16

15–17

15, 17–18

Explain the difference between a decay rate and a decay factor.

When a heated substance such as water starts to cool, its temperature T at time t minutes after being left to cool is given by an exponential relationship. a

The equation can be written in the form T = a (1 − r)t. Explain what the variables a and r mean within context.

Substance P starts off at a temperature of 145 °C and is left to cool in a room whose temperature is a constant 0 °C. By using the table, ﬁnd the equation for the temperature of the substance. Minutes passed (t) Temperature (T )

0 145

1 87

2 52.2

3 31.32

4 18.792

Find the temperature after 10 minutes, correct to two decimal places.

Will the temperature of the substance will ever reach 0 °C? Explain your reasoning.

The temperature of another substance Q which also starts off at a temperature of 145 °C is modelled by T = 145 (1 − 0.1)t. Which substance will cool more rapidly?

An online business that sells phones online has been growing very quickly over the last year. They have asked you to model their growth over the next three quarters, and further into the future. a

Looking at the quarterly data, you start by assuming the number of customers changed in the same way each quarter over the ﬁrst year.

Complete the table under this assumption to predict the number of phones sold over the next three quarters. Quarter (N ) Online customers (C)

460

1 480

2 2880

3 17 280

4 103 680

Find an equation for the number of sales (C) in terms of the number of quarters that have passed (N ).

Predict the number of sales they will have 2.75 years after they started the business.

Using your answer to part (c), explain whether or not this is a reasonable goal for the company.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A teaspoon of sugar, weighing 4 grams, is poured into a cup of hot water and immediately begins to dissolve. Each second the amount of undissolved sugar remaining is amount present in the previous second. a

Complete the table of values: Seconds passed (t) Undissolved sugar in grams ( y)

Write an equation linking undissolved sugar, y, and time, t.

Describe the change in the amount of undissolved sugar over time.

Use the model to detemine if all the sugar will eventually dissolve. Explain your reasoning.

The water is heated more so that when the experiment is performed again, the sugar dissolves twice as quickly. Will more or less sugar dissolve between the second and third seconds? Explain your reasoning. 19–20, 22–23, 25–26

19, 22–24, 26–27

In the diagram, the number of squares at each stage increases exponentially. a

Find the number of squares in the next stage of the sequence.

The number of squares can be modelled by y = 4x − 1 where x is the number of the stage. At what stage will there be 16 384 squares?

One person in a city is infected with a virus. During the ﬁrst day, they infect ﬁve more people with the virus. a

Copy and complete the table showing the number of newly infected people for that day. Day Number of people infected

Problem-solving 19–21, 23, 25–26 19

of the

0 1

If this trend continues, write an expression for the number of people infected on day n.

There are 345 800 people in the city in total. Given that the trend continues, after how many days will the entire city have been infected?

If everyone takes a daily vitamin C supplement, they can restrict the spread so that each new person only infects an average of two new people on their ﬁrst day. How many people would be infected on day 10 if this was followed from the beginning?

The value of a painting valued at $700 appreciates in value at a rate of 9% per year. a

Sketch the relationship between the number of years passed and the painting’s value on a Cartesian plane.

How many years will it take for the painting’s value to reach a total value of $2100?

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Futsal is a modiﬁed form of football, typically played indoors on a smaller pitch with fewer players. A futsal ball has less bounce than a full size soccer ball and is expected to bounce back off the ground to

of the height from which it falls after every subsequent bounce.

Write an equation to represent the height of the nth bounce if it was dropped from an initial height of 5 metres.

After how many bounces will the ball have a height of

metres?

The population of a certain city is doubling each year. In 2012, the population was 10 000 people. In what year will the population reach 2 560 000?

To test the effectiveness of a new antibiotic, a bacteria is introduced to a body and the number of bacteria is monitored. Initially, there are 19 bacteria in the body. After four hours, the number is found to double. After how many hours will there be 1216 bacteria in the body?

Juanita and Pedro are taking part in a study, and have to eat the same food and in equal quantity. They were chosen such that Juanita incorporates lots of sugar into their diet while Pedro consumes limited amounts of sugar. After eating, their blood sugar level peaked, and was then continually measured, with the following functions modelling their blood sugar level at time t hours: • Juanita: J = 179 (0.3)t • Pedro: P = 132 (0.6)t a

Find the highest blood sugar level measured for: i

Juanita

Pedro?

Pedro

Compare the reactions to sugar between Juanita and Pedro.

A new car purchased for $38 200 depreciates at a rate r each year. The value of the car for the ﬁrst two years is shown in the table: Years passed (n) Value of car ($A)

462

Find the rate, as a percentage, that the blood sugar level decreased each hour for: i

Juanita?

0 38 200

Find the value of r.

Write the rule for A, the value of the car n years after it is purchased.

Find the value of the car after 6 years.

A new motorbike purchased for the same amount depreciates according to the model V = 38 200 (1 − 0.03) n . Which vehicle depreciates more rapidly?

1 37 818

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

2 37 439.82

In a memory study, subjects are asked to memorise some content and then asked to recall what they can remember each day after. The given graphs represent the percentage of content remembered by two participants, Maximilian (P ) and Lucy (Q), after t days. Recall percentage 100% 80% 60% 40% Q

20% P 2

Days 4

Create equations to model the percentage of the content Maximilian and Lucy still recall, respectively.

According to the model, determine if either of them will completely forget the content and explain your reasoning. Is this model realistic?

Another study, similar to the one above, was conducted by teachers to see how much of the content being taught was retained by students. The given graphs represent the percentage of content remembered by two students, Raymond (R) and Sunyi (S), after t days. Recall percentage 100% 80% 60% 40% S

20%

Days 2

Do the given equations accurately model the percentage of the content Raymond and Sunyi recall, respectively: R = 100 (1 − 0.15)t and S = 100 (1 − 0.13)t? Explain your reasoning.

Compare the amount of information retained by Maximilian, Lucy, Raymond, and Sunyi. Identify and explain various factors that could account for any differences.

8.06 Exponential models mathspace.co

463

8 Chapter review 1

Are these relationships linear, exponential or quadratic? a

y = 5x + 7

y 1 2 3 x 0.8 1.6 2.4

b d

4 3.2

y = 2x2 + 7 y x

1 0.125

x 1 2 3 4 5

50 40 30 20

−8

−10

−4 −3 −2 −1

For these exponential relationships: i

Find the growth or decay factor.

x 1 2 3 y 2000 500 125

4 31.25

Find y when x = 5. x y

1 8

2 20

3 50

4 125

Consider the pattern: a

Find the growth or decay factor.

Find the number of cubes for step 4. Step 1

464

−6

−4

4 8.0

3 2.0

−5 −4 −3 −2 −1 −2

2 0.5

Step 2

Step 3

For each exponential relationship: i

Complete the table of values.

Graph the exponential curve.

iii

Is the curve increasing or decreasing?

x −1 0 1 x y=6

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

−1

For the equation y = 5x: a

Identify the equation of the horizontal asymptote.

Find the coordinates of the y-intercept.

Find the coordinates of one other point on the curve.

Sketch the graph.

The graph of

is shown.

How many x-intercepts does the curve have?

Is each y-value positive or negative?

9 8 7 6 5 4 3 2 1 −3 −2

The graph of y = 4x is shown. a

Find the value of y when x = 4.

Find the value of x when

−1

x 1

−1

14 12 10 8 6 4 2 −5 −4 −3 −2 −1

(Optional) Describe how the graph of y = 10x is transformed to get each graph: a

x 1 2 3 4 5

y = 10x + 2

y = −10x

y = 3 (10)x

(Optional) The graph of f (x) = 7x is shown. a

Describe a transformation of the graph of f (x) = 7x that would obtain g (x) = 7x − 4.

Sketch the graph of g (x) = 7x − 4 on the same set of axes as f (x) = 7x.

Write the equation of the horiztonal asymptote of g (x) = 7x − 4.

y 7 6 5 4 3 2 1 f (x) = 7x −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

1 2 3 4 5

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(Optional) Two exponential relationships, X and Y, are graphed.

One of the graphs is of y = 3x and the other graph is of

Which is the graph of

? x

(Optional) For the exponential function a

Find the coordinates of the y-intercept.

What is the equation of the horizontal asymptote?

Is the function increasing or decreasing?

Graph the function.

Solve: a

8x = 88

6x−4 = 36

d 5−4x−1 = 1

105x+18 = 100x−3

2x × 2x+4 = 8

3x × 9x+1 = 27

all lie on the curve with equation y = 3x, ﬁnd the c

$4500 is invested for three years at a rate of 4% p.a., compounded quarterly. a

Complete the table to determine the ﬁnal value of the investment. Round to the nearest cent for each year. First quarter Second quarter Third quarter Fourth quarter

466

c g

Given the points (0, n), (k, 243) and value of: a

3x = 81

Solve: a

Balance at beginning of year $4500 $4680

Interest earned $180 $187.20

Calculate the total interest earned over the ﬁrst three quarters.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

−

Lara borrows $30 000 at a rate of 6.2% p.a., compounding annually. a

Write an expression that represents the amount Lara must repay after 25 years, assuming that she hasn’t paid anything back.

How much interest was generated on the loan over the 25 years? Round your answer to two decimal places.

Calculate the future value, A, of each investment: a

A $1380 investment earns interest at 3.9% p.a., compounded weekly over 4 years.

A $9670 investment earns interest at

p.a., compounded monthly over 2 years.

Weylan opened a savings account at the beginning of August 2019, where the interest is compounded monthly. His statements for September, Month Balance October and November show his account balance at the September $6375 beginning of each month: October $6502.50 a Find the monthly interest rate, r, for his investment. November $6632.55 b How much did Weylan deposit into this savings account when he opened it?

Yexel has a savings account which accumulates compound interest yearly at a rate of 3.8% p.a. If he aims to have $40 000 in his account after 5 years, ﬁnd the principal amount, P, that he should invest.

A piece of artwork was purchased for $7700, and it is expected to increase in value by 5.3% per year. a

Write an equation in terms of y to represent the value of the piece of artwork after n years.

Find the value of the piece of artwork after 15 years.

A car was valued 7 years ago to be worth $46 000. Its value depreciated at 17.4% p.a. Calculate its depreciated value now. Round your answer to the nearest dollar.

The population of a colony of bees is given by P = 10 300 (1 − 0.098)t, where t is the time in years. a

What is the initial population of bees?

Is the population increasing or decreasing?

What is the percent rate of change?

Find the population of bees in 4 years time, rounding to the nearest integer.

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Big ideas Understanding logarithms and their relationship with exponential functions is essential for simplifying and comparing numbers on different scales, solving exponential equations and analysing growth patterns.

Curriculum Outcomes VC2M10M02 interpret and use logarithmic scales in applied contexts involving small and large quantities and change VC2M10AA04 explore the inverse relationship between exponential functions and logarithmic functions and the solution of related equations VC2M10AN03 use the definition of a logarithm to establish and apply the laws of logarithms and investigate logarithmic scales in measurement

Chapter outline

Logarithms Chapter outline 9.01 9.02 9.03 9.04 9.05 9.06

Logarithms Order of magnitude and logarithmic scales Optional: Other bases Investigation: Logarithms Optional: Logarithm laws Optional: Connect exponentials and logarithms Optional: Solve exponential equations with logarithms Chapter 9 review

470 480 489 493 500 514 520

Logarithms help measure earthquakes, turning massive vibrations into manageable numbers!

9.01

Logarithms After this lesson, you will be able to… • represent numbers as powers of ten and find their base ten logarithms. • explain what is base two logarithms and represent numbers as powers of two. • recognise the importance and application of logarithmic scales in graphing large ranges of values. • use the appropriate mathematical notation for representing logarithms. • evaluate logarithms using technology. • rewrite equations involving logarithms in exponential form.

Logarithms If a number is a power of ten such as 10, 1000, or 0.01, the number can be written as a base ten logarithm. The base 10 logarithm is the power of 10 that makes the number. For example: log10 (1000) = 3 because 1000 = 103. So, 3 is the base 10 logarithm of the number 1000. log10 (1) = 0 because 1 = 100. log10 (0.01) = −2 because 0.01 = 10−2. In general:

log10 (10a) = a a Base 10 logarithm Here is a table with the logarithms of numbers that can be written with base ten: Basic number 1 000 000 100 000 10 000 1000 100 10 1 0.1 0.01 0.001 0.0001 0.00 001 0.000 001

470

Power of 10 106 105 104 103 102 101 100 10−1 10−2 10−3 10−4 10−5 10−6

Logarithm 6 5 4 3 2 1 0 −1 −2 −3 −4 −5 −6

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

If the number is not a power of ten, then the logarithm can be found using the log button on the calculator. For example, log10(250) = 2.3979 because 102.3979 = 250. log10(0.0023) = −2.6383 because 10−2.6383 = 0.0023.

Example 1 Consider the number 0.000 001. a Express the number as a power of 10.

Create a strategy Count the number of decimal places, and write as a power of 10 with a negative index.

Apply the idea Since the number is smaller than 1, the power is negative. There are six decimal places in the answer, so the power is −6. 0.000 001 = 10−6

Express as a power of 10

b Find the base ten logarithm of the number.

Create a strategy The base ten logarithm is the exponent when written as a power of 10.

Apply the idea

Reﬂect and check

The base ten logarithm is −6.

We can write this as an equation using proper notation: log10 (0.000 001) = log10 (10−6) = −6

Idea summary The base 10 logarithm of a number is the exponent when the number is written as a power of ten. If the number cannot be written as a power of ten, the logarithm can be found using a calculator. For example, log10 (250) = 2.3979 because 102.3979 = 250.

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Base two logarithms The base two logarithm of a number is equal to the power of 2 that makes the number. For example, log2 8 = 3 because 23 = 8. log2 32 = 5 because 25 = 32. log2 0.125 = −3 because

Some useful base two logarithms to know are shown in the table: Basic number 64 32 16 8 4 2 1 0.5 0.25 0.125

Power of 2 26 25 24 23 22 21 20 2−1 2−2 2−3

Logarithm base 2 6 5 4 3 2 1 0 −1 −2 −3

Example 2 Evaluate:

Create a strategy Find the power to which 2 must be raised to get .

Apply the idea We can write base 2 of

Reﬂect and check −2

as 2 . Therefore, the logarithm

is equal to −2.

The answer makes sense because 2 to the power of −2 equals . Therefore, the logarithm base 2 of

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is indeed −2.

Example 3 Rewrite the equation log2 0.125 = −3 in exponential form.

Create a strategy Exponential form will be 2⬚ = ⬚, since 2 is the base of the logarithm.

Apply the idea

The exponential form will be 2⬚ = ⬚, so we can use the fact that the logarithm is equal to the index on the 2. The logarithm is equal to −3, so we have 2−3 = ⬚.

We are taking the base 2 logarithm of 0.125, so we have 2−3 = 0.125.

Reﬂect and check The answer makes sense because

Idea summary The base two logarithm of a number is equal to the power of 2 that makes the number. 23 = 8 therefore log2 8 = 3 In general, the base of the logarithm and the base of the exponential form must be the same. What the logarithm is equal to will become the index of that base.

a x = y ←→ loga y = x base

a > 0, a ≠ 1, y ≠ 0

Logarithmic scales on graphs Logarithmic scales are often used when there is a large range of values involved with the variables under consideration. Consider the set of ﬁve ordered pairs: x y

1 20

2 200

3 3631

4 52 481

5 250 000

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As seen on the graph of the ﬁve points, the scale on the y-axis is so huge, that the ﬁrst 3 points look like they almost have the same y-value. y (5, 250 000)

250 000 200 000 150 000 100 000 (4, 52 481)

50 000

(1, 20)

(2, 200)

(3, 3631)

To avoid losing data because of the scale, we can use the base 10 logarithm of y instead of the actual y-values as shown in the table: x

1 1.301

log10 y

2 2.301

3 3.560

4 4.720

5 5.398

Using the logarithm of the y-values gives us this graph: Even though the y-values are far more manageable in this form, we need to remember that the actual data point y-values, rounded to 3 decimal places, are given by 101.301, 102.301, 103.560, 104.720, and 105.398.

(5, 5.4)

(4, 4.72)

4 (3, 3.56) 3 (2, 2.3)

(1, 1.3)

x 1

Example 4

Number of countries

The histogram shows the area (in km2) for 12 countries, plotted using a log scale.

474

10 9 8 7 6 5 4 3 2 1 0

4 Log of area

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

How many countries have an area of between 10 000 km2 and 100 000 km2?

Create a strategy Count the number of 0s of each number to find the log of each area. Then, find their corresponding value in the vertical axis.

Apply the idea 10 000 has 4 zeros while 100 000 has 5 zeros. This means the logs of the areas are 4 and 5, respectively. We can see from the histogram that the 4–5 column corresponds to 3 in the vertical axis. So there are 3 countries with an area between 10 000 km2 and 100 000 km2.

Idea summary We can use logs to transform either (or both) axes depending on what is needed to better understand the data.

9.01 Practice questions Understanding 1

Adventurer 1–3

Trailblazer 1–3

Evaluate: a

102

10−1

104

103

100

101

10−2

105

The base ten logarithm of a number is the power of ten that makes the number. Copy and complete: a b c d e f

Explorer 1–3

The base ten logarithm of 100 is ⬚ because 10⬚ = 100

The base ten logarithm of 1000 is ⬚ because 10⬚ = 1000 The base ten logarithm of 0.01 is ⬚ because 10⬚ = 0.01 The base ten logarithm of 1 is ⬚ because 10⬚ = 1

The base ten logarithm of 65 is 1.8129 because 10⬚ = 65

The base ten logarithm of ⬚ is 2.3502 because 102.3502 = ⬚

The base two logarithm of a number is the power of two that makes the number. Copy and complete: a b c d e

The base two logarithm of 8 is ⬚ because 2⬚ = 8

The base two logarithm of 16 is ⬚ because 2⬚ = 16

The base two logarithm of 0.5 is ⬚ because 2⬚ = 0.5 The base two logarithm of 1 is ⬚ because 2⬚ = 1

The base two logarithm of 0.125 is ⬚ because 2⬚ = 0.125 9.01 Logarithms mathspace.co

475

Fluency Ex 1

Ex 2

Express the number as a power of 10.

Find the base ten logarithm of the number.

1 000 000

0.000 1

Express the number as a power of 2.

Find the base two logarithm of the number.

0.001

0.125

256

log10 1000 = 3 can be written in exponential form as 103 = 1000. Write these logarithms in exponential form: log10 100 = 2

log10 1 = 0

log10 10 000 = 4

log10 0.01 = −2

Use technology to ﬁnd the base ten logarithm, rounding your answer to four decimal places: log10 45

log10 7692

log10 5.16

log10 0.005 79

Use technology to ﬁnd the missing number, rounding your answer to three decimal places: log10 ⬚ = 1.57

log2 4 = 2

log10 ⬚ = −1.2

log2 1 = 0

log10 ⬚ = 0.8

log2 16 = 4

log10 ⬚ = 2.31

log2 8 = 3 can be written in exponential form as 23 = 8. Write these logarithms in exponential form: a

a Ex 3

4–6, 7cd, 8cd, 9–13

For each number:

a 8

4-6, 7bc, 8bc, 9–12

For each number:

a 7

4–6, 7ab, 8ab, 9–12

The points shown are plotted using a log scale. For each point, which coordinates represent the true value, (x, y)? a

(1, 2)

L J

(1, 100)

(1, 20)

Point K: A

log10 y

Point J: A

−1

(2, −1)

log2 0.5 = −1

x 1

3 K

−2

(2, −10)

Point L: A

476

−4

(4, 36)

(4, 3.6)

(7, 0.00316)

(7, 316.228)

Point M: A

Ex 4

(4, 3981)

8 M

−3

(7, −2.5)

The table shows the area (in km2) for 12 countries, using a log10 scale: a

How many countries have an area between 10 000 km2 and 100 000 km2?

How many countries have an area between 10 000 km2 and 1 000 000 km2?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Log of area

Number of countries

3–4 4–5 5–6 6–7

2 2 8 1

The histogram has been plotted on a log10 scale: a

What is the total frequency of scores between 1 and 100?

What is the total frequency of scores larger

6 5

than c

Frequency 7

What is the total frequency of scores smaller than

2 1 Log value −3

−1

−2

The histogram has been plotted on a log2 scale: a

What is the total frequency of scores between 1 (20) and 4 (22)?

What is the total frequency of scores larger than 0.5 (2−1)?

What is the total frequency of scores less than 2 (21)?

Frequency 7 6 5 4 3 2 1 Log value −2

Reasoning 14

14–15

The populations of 9 Australian cities are shown in the table: a

Suppose you want to create a frequency table or frequency histogram. Explain why using a regular scale would not be ideal for your model.

Convert the data to a log10 scale.

Create a frequency table or frequency histogram of the data.

−1

14–16

City Melbourne Adelaide Wollongong Geelong Hobart Toowoomba Ballarat Bunbury Warragul

Population 4 929 200 1 396 783 310 479 278 708 231 707 146 955 105 924 80 106 44 625

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In the graph shown, the horizontal scale is linear and the vertical scale is logarithmic. Which of the three plots represents exponential growth?

Which of the three plots represents growth that is less than exponential?

Which of the three plots represents growth that is greater than exponential?

Complete the table of values. 0 0

1 1

2 2

3 3

4 4

At this rate, how many bacteria will be present in the petri dish after 7 hours?

Describe the rate of increase of the actual number of bacteria.

Problem-solving 17, 18

17–19

18–20

A ﬁnancial planner has clients with a large range of annual incomes. To manage the data more easily, the income (in dollars) of each client is ﬁrst converted to a log10 scale. a

What income would result in 4.9 on a log10 scale?

What range of incomes corresponds to a range of 5 to 6 on the log10 scale?

The lifespans of various animals and insects are shown in the table: Animal/Insect Ant African elephant Bowhead whale Bumblebee Chicken Dog Emperor penguin Fly Giraffe Hamster Ladybug Little penguin Monkey Shark

478

In a laboratory, the number of bacteria in a petri dish is recorded, and the bacteria are found to follow a base 2 logarithmic scale. Number of hours passed Number of bacteria using a log2 scale Actual number of bacteria

Lifespan (days) 20 25 500 71 400 21 2920 5110 8500 18 10 120 780 365 2190 7300 10 980

Lifespan in a log10 scale 4.41 1.32 3.47 3.93

2.89 3.34 4.04

Complete the table by converting the lifespans to a log10 scale, rounding each value to two decimal places.

Create a histogram to represent this data on a log10 scale.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The Richter scale is used to measure the intensity of earthquakes. The formula for the Richter scale rating of a quake is given by R = log10 x, where x is the relative strength of the quake. The table shows how quakes are categorised according to their Richter scale rating: Richter rating

Minor 2–3.9

Light 4–4.9

Moderate 5–5.9

Strong 6–6.9

A seismograph measures the intensity of a quake to be x = 5711. Determine the Richter scale rating R of this quake, rounded to one decimal place.

In which category does the quake fall?

The number of layers, y, resulting from a rectangular piece of paper being folded in half x times, is shown in the graph. a

Plot the data on a coordinate plane where the horizontal axis is the number of times the paper is folded and the vertical axis is log2 y.

What does the point (5, 5) on your graph represent in context?

If a rectangular piece of paper is folded so it has 1024 layers, how many times it was folded.

If a rectangular piece of paper of thickness 0.02 mm is folded to a thickness of 40.96 mm, how many times it was folded?

y 10 9 8 7 6 5 4 3 2 1

x 1

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9.02 Order of magnitude and logarithmic scales After this lesson, you will be able to… • determine the order of magnitude of a given number. • describe the purpose and application of logarithmic scales. • use logarithmic scales to compare and differentiate between numbers with a broad range. • understand the concept behind the Richter scale and its logarithmic nature. • interpret sound levels using the decibel scale and recognise the logarithmic basis behind it.

Orders of magnitude The order of magnitude of a number gives an indication of the size of the number using powers of ten. It is a useful way to compare numbers. 5000 has an order of magnitude of 3 because it can be written as 5 × 1000 = 5 × 103. 0.007 has an order of magnitude of −3 because it can be written as 7 ÷ 1000 = 7 × 10−3. The table shows a comparison between the basic numbers, powers of 10, and order of magnitude. Basic number

Power of 10

Order of magnitude

1 000 000 100 000 10 000 1 000 100 10 1 0.1 0.01 0.001 0.000 1 0.000 01 0.000 001

6 5 4 3 2 1 0 −1 −2 −3 −4 −5 −6

10 105 104 103 102 101 100 10−1 10−2 10−3 10−4 10−5 10−6

Increasing order of magnitude

Decreasing order of magnitude

The order of magnitude forms a logarithmic scale of base 10.

Example 1 What is the order of magnitude of the number 6 × 109?

Create a strategy

Apply the idea

If a number is written in scientiﬁc notation, the order of magnitude is the power.

The number 6 × 109 has an order of magnitude of 9.

480

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Example 2 A bakery sells 450 cupcakes per week. What is the order of magnitude of the total number of cupcakes sold over 8 weeks?

Create a strategy

Apply the idea

Multiply 450 by 8 and write the product in scientiﬁc notation.

Total = 450 × 8

Multiply

= 3600

Evaluate 3

= 3.6 × 10

Write in scientiﬁc notation

The order of magnitude is 3.

Idea summary The order of magnitude of a number gives an indication of the size of the number using powers of ten. If a number is written in scientiﬁc notation, the order of magnitude is the power of 10.

Compare using logarithms Orders of magnitude and other logarithmic scales can be useful when comparing numbers that have a very large range. For example, we can compare the weight of a mouse (19 grams) with that of a cat (3600 g). Steps Original information Write using scientific notation Orders of magnitude

Cat 3600 g 3.6 × 103 3

Mouse 19 g 1.9 × 101 1

The two animals differ by 2 orders of magnitude, which means the mouse is roughly 100 times smaller than the cat. • An increase by 1 order of magnitude means the number is 10 times larger. • A decrease by 1 order of magnitude means the number is 10 times smaller. Similarly, we can use logarithmic scales to compare numbers with a large range, like the distance of the planets from the sun. Planet Mercury Venus Earth Mars Jupiter

Distance from the sun, d (km) 57 910 000 108 200 000 149 600 000 227 900 000 778 500 000

log10 (d) 7.76 8.03 8.17 8.36 8.91

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Consider the difference between Mercury and Jupiter’s distances from the sun. The log values differ by a little bit more than 1, so Jupiter’s distance from the sun is a little more than 10 times the distance of Mercury. To ﬁnd a more exact answer: 8.91 − 7.76 = 1.15 10

1.15

= 14.13

Find the difference between Jupiter and Mercury Find 10 to the power of 1.15 and evaluate

Therefore, Jupiter is just over 14 times the distance of Mercury from the sun. • On a log scale with base ten, every 1 unit up means increasing by a factor of 10, and every 1 unit down means decreasing by a factor of 10. • On a log scale with base two, every 1 unit up means increasing by a factor of 2, and every 1 unit down means decreasing by a factor of 2. The Richter scale for measuring earthquakes and the decibel scale for measuring sound are two common applications of logarithmic scales. Measuring the size of earthquakes Since the 1930’s, the Richter magnitude scale has been used to describe the intensity or ‘size’ of an earthquake. Due to the number of variables involved in measuring an earthquake, its value on the Richter scale is calculated using the log scale; the size of the seismic wave is recorded with the distance the recording was from the source of the earthquake. The value on the Richter scale indicates the intensity and can be compared with other earthquakes using powers of 10 to describe how much stronger it was. For example, an earthquake of size 6 on the Richter scale is 103 times more intense than an earthquake of size 3. Measuring level of sound Sound levels are a measure of the intensity of the sound waves on your eardrum. The numbers given by the decibel (dB) scale indicate an intensity 10 times stronger for every 10 dB. The table shows the difference in intensity of a sound of a particular size when compared to another sound. Decibels 50 60 70 80 90

50 0 −10 −100 −1000 −10 000

60 +10 0 −10 −100 −1000

70 +100 +10 0 −10 −100

80 +1000 +100 +10 0 −10

90 +10 000 +1000 +100 +10 0

A sound with a reading of 70 on the decibel scale is 102 times more intense than a reading of size 50. A sound with a reading of 90 on the decibel scale is 103 times more intense than a reading of size 60.

Example 3 The Richter Scale is a logarithmic scale used to measure the magnitude of an earthquake or tremor. An earthquake that registers 4.0 on the Richter scale is 10 times stronger than an earthquake that measures 3.0. How many times stronger is a quake of 7.6 than one of 5.2? Round your answer to the nearest whole number. 482

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Create a strategy Convert the values to linearly comparable measures: 10x and 10y.

Apply the idea Convert to base 10 Use the index law Evaluate the exponent Evaluate the expression and round

Example 4 The decibel scale, which is used to record the loudness of sound, is a logarithmic scale. • In the decibel scale, the lowest audible sound, with intensity 10−12 watts/m2 is assigned the value of 0. • A sound that is 10 times louder is assigned a decibel value of 10. • A sound 100 (102) times louder is assigned a decibel value of 20. • A sound 1000 (103) times louder is assigned a decibel value of 30. If the sound of a normal speaking voice is 50 decibels (dB), and the sound in a bus terminal is 80 decibels (dB), then how many times louder is the bus terminal compared to the speaking voice?

Create a strategy Subtract the dB of the normal speaking voice from the dB of the bus terminal sound.

Apply the idea Decibel value = 80 – 50

Subtract 50 dB from 80 dB

= 30 dB

Evaluate 3

On the decimal scale, a sound 1000 (10 ) times louder is assigned a decibel value of 30, so the bus terminal sound is 1000 louder than the normal speaking voice.

Idea summary Orders of magnitude and logarithmic scales are useful to compare very large and very small numbers. An increase by 1 order of magnitude means the number is 10 times larger. A decrease by 1 order of magnitude means the number is 10 times smaller. On a log scale with base b, every 1 unit up means increasing by a factor of b, and every 1 unit down means decreasing by a factor of b.

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9.02 Practice questions Understanding 1

100

1000

1 000 000

0.01

0.1

0.001

Write using scientiﬁc notation: a

600

7000

4 800 000

230

0.0076

0.012

0.00009

0.1

0.5

Find the base ten logarithm of the number. 1 000

0.000 01

Find the base two logarithm of the number. 8

Fluency

Ex 2

Trailblazer 1–4

Ex 1

Adventurer 1–4

Write as powers of 10:

a 4

Explorer 1–3

5–9, 11, 12, 14

5–9, 11–15

What is the order of magnitude in the number 6 × 10−7?

For each number: i

Write the number using scientiﬁc notation.

Identify the order of magnitude.

8 000

964 400

0.0008

6–7, 9–11, 14, 16

0.2574

A coffee shop sells 580 cups of coffee in a week. What is the order of magnitude of the total number of coffees sold over 12 weeks?

One university has 6000 students enrolled, while another university has 7000 students enrolled. What is the order of magnitude of the total number of students enrolled at both universities?

The average distance of Mercury to the Sun is 5.791 × 107 km while the average distance of Uranus to the Sun is 2.871 × 109 km. By what order of magnitude does the distance between these planets and the Sun differ?

The scale shown is logarithmic with base two:

484

Find the value of A.

Find the value of B.

Find the value of C.

How many times larger is the number represented by A than the number represented by B on the scale?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

–5 –4 –3 –2 –1 0 1 2 3 4 5

11 Ex 3

The Richter Scale is a base 10 logarithmic scale used to measure the magnitude of an earthquake, given by R = log10 x, where x is the relative strength of the quake. Some past earthquakes and their Richter scales are shown in the table: a

How many times stronger was the earthquake in Sumatra compared to the earthquake in China? Round your answer to the nearest whole number.

How many times stronger is a quake of 9.1 than one of 6.3? Round your answer to the nearest whole number. Earthquake Sumatra 2004 China 2008 Haiti 2010 Italy 2009

Richter scale 9.1 7.9 7.0 6.3

Earthquakes are often preceded by lesser tremors, known as foreshocks, and followed by lesser tremors known as aftershocks. The graph shows some readings from an area before, during and after an earthquake. a

The main earthquake occurred at 5:00 a.m. What was its magnitude?

The last aftershock was measured at 9:30 a.m. How many times larger than this aftershock was the largest foreshock?

5.5

Magnitude

5 4.5 4 3.5 3 2.5 2 1.5 1 0.5

Time (hours) 1 2 3 4 5 6 7 8 9 10

The pH scale is a base 10 logarithmic scale used to measure the acidity (concentration of hydrogen ions) of a solution. The lower the pH, the higher the concentration of hydrogen ions. For example, the concentration of hydrogen ions for a solution with a pH of 1 is 10 times greater than a solution with a pH of 2. Jasper is a chemistry student who has 3 solutions. One is marked as having a pH of 10, one has a pH of 8, and one has a pH of 5.4. a

How many times greater is the concentration of hydrogen ions in the pH 8 solution compared to the pH 10 solution?

How many times greater is the concentration of hydrogen ions in the pH 5.4 solution compared to the pH 10 solution?

How many times greater is the concentration of hydrogen ions in the pH 5.4 solution compared to the pH 8 solution?

9.02 Order of magnitude and logarithmic scales mathspace.co

485

14 Ex 4

The decibel scale, which is used to record the loudness of sound, is a logarithmic scale with base 10. In the decibel scale, the lowest audible sound is assigned the value of 0. The table shows the decibel value for various types of noise.

486

How many times louder is the sound of industrial noise than the sound of a wind turbine?

If the sound of a normal speaking voice is 50 decibels, and the sound in a bus terminal is 100 decibels, how many times louder is the bus terminal compared to a normal speaking voice?

Type of noise Jet plane Pneumatic drill Industrial Stereo music Inside a car Office Household Wind turbine Bedroom Falling leaves

Decibel value 150 120 110 100 90 70 60 50 30 20

The decibel scale for sound is a base 10 logarithmic scale. Sound Intensity (watts/m2)

Relative intensity (decibels)

102 101 100 10−1 10−2

jack hammer rock concert ambulance siren school dance

140 130 120 110 100

10−3

electric drill

10−4 10−5 10−6 10−7 10−8 10−9 10−10 10−11 10−12

lawn mower hair dryer normal conversation refrigerator library

80 70 60 50 40 30 20 10 0

whisper at 2 m normal breathing barely audible

How many times more intense is the sound of a refrigerator compared to normal breathing?

How many times more intense is the sound of a rock concert compared to an electric drill?

How many times more intense is the sound of a lawn mower compared to a whisper at 2 m?

How many times more intense is the sound of a jack hammer compared to a normal conversation?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The decibel (dB) scale for sound is based on a base 10 logarithm. Three sounds are measured for their intensity. • Sound A was measured to be 125 dB. • Sound B was measured to be 97 dB. • Sound C was measured to be 61 dB. a

How many times more intense was sound A than sound B?

How many times more intense was sound B than sound C?

How many times more intense was sound A than sound C?

Reasoning

17, 18

18, 19

The diameter of a human strand of DNA is about 2.5 × 10−9 m, while the diameter of a hydrogen nucleus is about 1.8 × 10−15 m. By how many orders of magnitude does the diameter of these objects differ? Explain your answer.

The Richter scale is a base 10 logarithmic scale. Richter magnitude 1–2 3 4 5 6 7

Possible effects Usually not felt except by instruments May be felt but rarely causes damage Like vibrations from heavy traffic Strong enough to wake people sleeping Ground cracks, houses collapse Few buildings survive, landslides

How much bigger is an earthquake that is strong enough to crack the ground compared to one that is like heavy traﬃc?

How much bigger is an earthquake of magnitude 6.3 than one of magnitude 4.8?

Would an earthquake of magnitude 6.2 be suﬃcient to crack walls in buildings? Justify your reasoning.

The time taken (t years) for A grams of a radioactive substance to decompose down to y grams, where k is a constant related to the properties of a particular substance is given by:

Use these substances to answer the questions below: • R1 has a constant of k = 0.000 43 • R2 has a constant of k = 0.000 47 a

Use technology to ﬁnd how long it takes 140 units of R1 to decompose down to 105 units. Round your answer to the nearest year.

Use technology to ﬁnd how long it takes 200 units of R2 to decompose down to 150 units. Round your answer to the nearest year.

Use technology to ﬁnd the half-life of each substance. That is, determine how long it takes a quantity of a substance to decompose down to half the original quantity. Round your answers to the nearest year.

Will it take twice the half-life for each substance to decompose completely? Explain your answer. 9.02 Order of magnitude and logarithmic scales mathspace.co

487

Problem-solving 21a, 22 20

21, 22

The table of values shows a log scale relating x and y. Form a logarithmic equation relating x and y. log scale measure ( y) 0 1 2 3 4

linear measure (x) 5 50 500 5000 50 000

= = = = =

The table shows the time it takes, in years, for a $1 investment to grow to I at different annual interest rate, using a log2 scale: a

20–23

log2 I 0 1 2 3 4 5

Find the number of years it will take for a $1 investment to double in value at the given annual interest rates: i

iii

7% 0 10.24 20.49 30.73 40.98 51.22

8% 0 9.01 18.01 27.02 36.03 45.03

9% 0 8.04 16.09 24.13 32.17 40.22

How many years will it take for a $1 investment to quadruple in value at 9% annual interest rate?

The Richter Scale is a base 10 logarithmic scale used to measure the magnitude of an earthquake, given by R = log10 x, where x is the relative strength of the quake. This means an earthquake that measures 4.0 on the Richter Scale will be 10 times stronger than one that measures 3.0. The aftershock of an earthquake measured 7.9 on the Richter Scale, and the main quake was 6 times stronger than the aftershock. Find the magnitude of the main quake on the Richter Scale, rounded to one decimal place.

The population of shrimp is tracked over time and recorded as shown: Days Shrimp

488

1 4

2 16

3 64

4 256

5 1024

Plot the data on a coordinate plane where x is the number of days and y is the population of shrimp.

Convert the data to a base 4 logarithmic scale.

Plot the data on a coordinate plane where x is the number of days and y is the number of shrimp using a base 4 logarithmic scale.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

9.03

Optional: Other bases

After this lesson, you will be able to… • extend the understanding of logarithms to bases other than ten. • rewrite expressions in exponential form and logarithmic form. • evaluate logarithms with different bases using both calculation and estimation methods. • apply understanding of logarithms to solve real-world problems.

Logarithm to bases other than ten The logarithm of a number to base 10 is the power of 10 that makes the number. 102 = 100 can be written as a logarithm: log10 100 = 2, or just log 100 = 2. If we wanted to find log 81, then we can press the “log” button on a calculator and then enter 81. This gives us 1.908, rounded to three decimal places. If there is no base specified then the base is assumed to be 10. Logarithms can also be written with other bases. For example, log5 25 = 2, since 52 = 25. The base here is 5. log3 81 = 4, since 34 = 81. The base here is 3. The definition of a logarithm is that if y = Bx then logB y = x. y = Bx is called exponential form and logB y = x is called logarithmic form.

Example 1 Evaluate

Create a strategy Use the deﬁnition: ‘If y = Bx then logB y = x.’

Apply the idea Let x be the index or the subject of the equation. So we have: Write the equation

Write in exponential form

Write the denominator as a power of 8

Use the negative index law

Equate the indices

9.03 Optional: Other bases mathspace.co

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Example 2 Consider the expression log 8892. a Within what range is the value of log 8892? A

between 0 and 1

between 4 and 5

between 2 and 3

between 3 and 4

between 1 and 2

Create a strategy Consider some powers of 10. Use the fact that 100 = 1, 101 = 10, 102 = 100, and so on.

Apply the idea Since 8892 is greater than 103 = 1000 but less than 104 = 10 000, then the range of the value of log 8892 is between 3 and 4. So, the correct answer is Option D.

b Find the value of log 8892, rounded to four decimal places.

Create a strategy

Apply the idea

Use a calculator and enter the expression into your calculator by pressing the log button.

log 8892 = 3.9490

Idea summary The deﬁnition of a logarithm is that if y = Bx then logB y = x. In logB y, B is the base of the logarithm. By convention, if the base is not specified, then B = 10. y = Bx is called exponential form and logB y = x is called logarithmic form.

9.03 Practice questions Understanding 1

1–3

Adventurer 1–3

Trailblazer 1–3

What number makes both equations true? a

2⬚ = 16

7⬚ = 7

log2 16 = ⬚ log7 7 = ⬚

490

Explorer

4⬚ = 64

log4 64 = ⬚

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3⬚ = 81

log3 81 = ⬚

3⬚ = 27

log3 27 = ⬚

2 3

To evaluate the expression log2 8, we can ask ourselves: “⬚ to what power is ⬚?” Identify the base of the expression: a

log5 125

Ex 1

52 = 25

4−2 = 0.0625

f j

3x = 81

42 = 16

61 = 6

20 = 1

9x = 81

4−1 = 0.25

30 = 1

(9.8)0 = 1

log5 5 = 1

log2 0.125 = −3

log6 36 = 2

log2 2 = 1

log8 1 = 0

log5.8 33.64 = 2

log8.4 70.56 = 2

Evaluate: a

log2 8

log4 16

log5 1

log5 125

log2 16

log3 3

log5 0.2

log2 0.25

Evaluate: log36 6

log25 5

log8 2

log27 3

log0.2 25

log0.9 0.81

Find the value of x: 8 log10 (x) = 16

4 log3 (x − 6) = 16

Without evaluating, ﬁnd the two integers that the value of each logarithm lies between. a

log2 (13)

Reasoning 10

log 100

4e–l, 5c, e–l, 6d, g–l, 7d–l, 8, 9cd

log4 16 = 2

a Ex 2

4f–l, 5c–h, 6b–f, ik, 7c–fj, 8a, 9ab

loga bc

Rewrite in exponential form:

i 7

Rewrite in logarithmic form:

i 5

log9 3

4a–f, 5a–g, 6a–f, i, 7ab, 8a, 9a

Fluency 4

log10 (539)

10, 11

log7 (23)

10, 11

d 10, 12–14

Evaluate: i

log10 (−100)

log10 (−1)

iii

log10 (0)

log10 (−10)

Explain your results from part (a).

If logx y = 2, ﬁnd the value of logy x. Justify your answer.

If log7 x = y, ﬁnd the value of log7 x2. Justify your answer.

9.03 Optional: Other bases mathspace.co

491

Consider the expression 3log3x. a

Let 3log3x = a. Rewrite this as an equation in logarithmic form.

Use your answer from part (a) to simplify the expression 3log3x. Explain your reasoning.

Describe the possible values of y in the equation log4 y = x, for all values of x.

Problem-solving 15aef, 16a, e–i, 17bef 15

(log5 625) (log5 125)

loga (a9)

For each logarithmic equation: i

Rewrite in exponential form.

Find the value of x.

log4 (64) = x

log3 (243) = x

log125 (5) = x

logx 64 = 2

logx 125 = 3

log2 x = 5

log3 x = 4

log25 125 = x

log4 32 = x

logx 64 = 1.5

logx 243 = 2.5

log4 (x + 6) = 3

log7 (10x + 2311) = 4

log2 (1 − 3x) = −4

log5 (x + 4) = 3

log2 (4x − 16) = 5

log9 (7x + 652) = 3

log3 (log3 x) = 1

Find the value of x.

a c

15, 16c–l, 17a, d–h, 18

Evaluate: a

15b, d–f, 16b, e–j, 17a, c–e, 18a

Solve for x: a

9 log8 (4x) − 18 = 0

Investigation: Logarithms

Investigate online

492

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

mathspace.co

9.04

Optional: Logarithm laws

After this lesson, you will be able to… • understand the special cases of logarithms. • describe and prove the product law of logarithms. • describe and prove the quotient law of logarithms. • describe and prove the power law of logarithms. • apply the laws of logarithms to simplify expressions and solve problems without using technology.

Special cases First, consider the deﬁnition that if y = Bx then x = logB y. It follows that, logB Bx = x Substituting x = 0 and x = 1 gives the following special cases: logB 1 = 0 logB B = 1

Example 1 Evaluate log3 3.

Create a strategy Use the logarithm law: logB B = 1.

Apply the idea log3 3 = 1

Idea summary From the deﬁnition that if y = Bx then x = logB y we can show, logB Bx = x logB 1 = 0 logB B = 1

9.04 Optional: Logarithm laws mathspace.co

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Product, quotient and power laws The product law of logarithms states that if you have the logarithm of a product, you can write it as the sum of the logarithms of the factors. logB ( pq) = logB p + logB q For example, log10 20 = log10 10 + log10 2. To prove this, let p = Bm and q = Bn so that logB p = m and logB q = n. Bm × Bn = Bm + n m

logB (B × B ) = logB B

Start with the index multiplication law m+n

logB (B × B ) = m + n

Take the logarithm of both sides Apply logB Bx = x

logB pq = logB p + logB q Substitute p = Bm and q = Bn The quotient law of logarithms states that if you have the logarithm of a quotient, you can write it as the difference of the logarithms of the numerator and denominator.

For example, log10 2 = log10 20 − log10 10. To prove this, let p = Bm and q = Bn so that logB p = m and logB q = n. Start with the index division law

Take the logarithm of both sides

Apply logB Bx = x

Substitute p = Bm and q = Bn

The power law of logarithms states that if you have a logarithm to a power, you can write it as the product of the logarithm and the power. logB pn = n logB p For example, log10 52 = 2 log10 5. (Bm)n = Bmn mn

logB (B ) = logB B

Start with the index power of a power law mn

logB (B ) = mn logB pn = n logB p

Take the logarithm of both sides Apply logB Bx = x Substitute p = Bm

Example 2 Simplify each of the following expressions without using a calculator. Leave answers in exact form. a log10 11 + log10 2 + log10 9

Create a strategy Use the logarithm law: logB p + logB q = logB ( p × q).

494

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea log10 11 + log10 2 + log10 9 = log10 (11 × 2 × 9) = log10 198

Use the law: logB p + logB q = logB ( p × q) Simplify

b log10 12 − (log10 2 + log10 3)

Create a strategy and logB p + logB q = logB ( p × q)

Use the logarithm laws:

Apply the idea Use logB p + logB q = logB ( p × q)

Simplify

Use

Simplify

Example 3 Simplify log 2x + log 50y to an expression with a single logarithmic term.

Create a strategy Use the logarithm law: logB pq = logB p + logB q

Apply the idea log 2x + log 50y = log (2x × 50y)

Use the law: logB pq = logB p + logB q

= log (100xy)

Simplify

= log 100 + log xy

Use the law: logB pq = logB p + logB q

= log 10 + log xy

Write the ﬁrst log as a power of 10

= 2 + log xy

Use the law: logB Bx = x

Example 4 Simplify

Create a strategy Use the logarithm law: logB pn = n logB p

9.04 Optional: Logarithm laws mathspace.co

495

Apply the idea Write the log as a power of 7 Use the law: logB pn = n logB p Remove the common factor: log4 7

Idea summary The deﬁnition of logarithms and the index laws give us the following results: Product law of logarithms Quotient law of logarithms Power law of logarithms

9.04 Practice questions Understanding 1

logb (xy) = logb (x) ⬚ logb ( y)

log10 25 = 5 log10 2

log10 (50) = log10 10 + log10 5

log2 7 + log2 11 = log2 77

log5 100 − log5 4 = log5 25

3, 4a–e, 5a–e, g, 6a–e, 3, 4a–e, 5b, e–j, 6c–g, 3, 4c–h, 5e–l, 6f, h–l, 7a–c, g, 8a–c, g, 9a–g, 7c–g, 8b–e, g, 9b–j, 10d–g, 7d–h, 8b–f, h, 9d, i–p, 10a–d, 11a–f, 12a–d, 13a–c 11c–h, 12b–e, 13b–d, g 10e–j, 11g–l, 12e–h, 13d–h

Identify if each expression is equivalent to log 35: a

log 30 + log 5

log 70 − log 2

log 7 + log 5

Identify if each expression is equivalent to 1: a e

496

logb (xy ) = y⬚ logb (x)

Fluency

Trailblazer 1, 2

Identify what law was used to get the expression on the right-hand side of each equation. a

Adventurer 1, 2

Complete each statement with the correct operation symbol: a

Explorer 1, 2

log 2 − log 1

b f

log7 7

log12 3 + log12 4

log5 1

4 log 10 − 3

2 log 10 − log 10

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Ex 1

Simplify: a

log7 7x

log2 2a

log5 53

e i

2 loga a

log20 20

log7 1

log78 1

logk 1

log8 (xy)

log2 (3s)

log2 (11cd)

logb (5 × 4)

log2 (39)

Rewrite the logarithm without any powers. a log2 ( y3) b log2 (r6)

log4 ( y7)

loga (B–5)

log (2x + 3)−1

logb (20 ÷ 7)

log (x4)

Rewrite the logarithm without any powers or surds. log2 (x + 7)6

log (2x + 5)6

log (4x + 3)−8

Rewrite the logarithm as the sum or difference of logarithms without any powers.

h l

log7 ( fg)

Ex 2

7 log4 48

log9 9

Rewrite the logarithm as the sum or difference of two or more logarithms.

−5 log7 72

log (u3 v5)

log (7x)6

log (7x4)

log (5x−1)

log (3x)−7

m log (7x−6)

log (2x)−1

Write the expression as a single logarithmic term. a

log 7 + log 12

log 11 + log 5 + log 3

log10 42 − log10 7

log10 8 − log10 32

log10 12 − (log10 2 + log10 3)

log10 2 + log10 3 − log10 7

5 (log10 3 + log10 6)

h 3 (log10 6 − log10 3)

3 log10 22 − 3 log10 11

7 log10 5 − 21 log10 25

Evaluate: log10 (2) + log10 (5)

log3 (108) − log3 (4)

loga (a )

log (10) + log (10)

log6 (4) + log6 (9)

log2 (15) − log2 (30)

logb (b6 + b4) − logb (b5 + b3) 9.04 Optional: Logarithm laws mathspace.co

497

Ex 3

Ex 4

Write the expression to have a single logarithmic term. a

6 log x + 5 log y

log (25x) + log (4y)

3 log (x5) − 2 log (x4)

log (x) + 1

log (x) − 1

log (x4) + log (x2)

Evaluate: a

Reasoning 14

14–17

15, 17–21

Pei has attempted to fully simplify an expression and showed their work: log3 (72) − log3 (8) + log3 (3) = log3 (72) − log3 (24)

Use logB ( pq) = logB p + logB q

= log3 (3)

Use logB

Use logarithm law logB B = 1

Identify where Pei has made an error and explain what it is.

Fully simplify the expression, showing the correct steps of work.

= logB p − logB q

Explain how 210 = 1024 can be used to make the estimation 10 log10 (2) ≈ 3.

Cinzia and Uri use different methods to simplify the logarithmic expression: log10 (x7) − log10 (x4) • Cinzia uses the logarithm law logB = logB p − logB q ﬁrst, then the logarithm law logB pn = n logB p. • Uri uses the logarithm law logB pn = n logB p ﬁrst, then combines the like terms. Determine whether either of Cinzia or Uri’s methods are invalid. If a method is valid, show the work for that method.

Using the proof shown: Let logx y = b. Then, xb = y by the deﬁnition of a logarithm. By taking the log of both sides, we maintain equality. So, xb = y is equivalent to log xb = log y. By using the power law that states: loga pn = n loga p we have b log x = log y Solving for b, we ﬁnd a b c

498

Complete the proof by ﬁlling in the blank: Therefore, logx y = ⬚ Use the proof to rewrite the expression log2 9.

Use a calculator to evaluate the expression from part (b), rounding to two decimal places.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

= logB p − logB q is shown. Justify each step in the

The proof for logarithm law logB proof.

Let p = Bm and q = Bn. Then, logB p = m and logB q = n. Then:

Therefore, logB 19

= logB p − logB q.

The product law of logarithms states that: logB ( pq) = logB p + logb q Show this product law is true.

Using the product property of logarithms that states logB ( p) + logB (q) = logB ( pq), prove that logB pn = n logB p.

Using the power property of logarithms that states logB pn = n logB p, explain why loga (1) = 0.

Problem-solving 22a–d, 23ac, 24a–e 22

22c–f, 23ab, 24c–h

22d–h, 23bd, 24g–l, 25

When rounded to three decimal places, we have: • log (2) = 0.301 • log (5) = 0.699 • log (3) = 0.477 • log (7) = 0.845 Estimate the logarithm, rounded to two decimal places:

log (6)

log (25)

log (42)

log (500)

log5 (7)

log3 (28)

log10 (21)

log8 (81)

Suppose that logk (a) = 1.64. Evaluate the expression, rounded to 2 decimal places where necessary. a

logk (ka4)

loga (k5) − logk (a)

logk (k2 a3) + logk (a2)

9.04 Optional: Logarithm laws mathspace.co

499

Solve for x: b

log5 (4x2) − log5 (x) = 3

log7 (2x) + log7 (3) = 3

log3 (8x) − log3 (40) = 2

log10 (x) + log10 (9) = log10 (45)

log10 (48) − log10 (3) = log10 (x)

log3 (x) + log3 (7) = log3 (56)

3 log (x) = log (125)

c e g

loga (ax − 2) = 10

log5 (84) − log5 (x) = log5 (7)

log4 (27) − log4 (x) = 3

log6 (x) + log6 (x − 5) = 2

log2 (x) − log4 (x) = log64 (x)

For equation 2 logb (x) = 1 + logb (6x − 8b) a

Write the equation without logarithms.

Find x in terms of b where b is a positive constant and x is positive.

9.05 Optional: Connect exponentials and logarithms After this lesson, you will be able to… • understand the relationship between exponential and logarithmic functions. • identify the general forms of exponential and logarithmic functions and their properties. • find inverse functions. • use function notation for inverse functions. • interpret tables of values to understand the relationship between a function and its inverse. • graph inverse functions and identify their line of reflection. • write the equation of an inverse function.

Exponential and logarithmic functions An exponential function with base b, has the form: y = bx A logarithmic function with base b has the form: y = logb x We use logarithmic functions when we are interested in ﬁnding out the power needed ( y) to raise base b to equal the value x. Consider the table of values for the function y = 3x x y

500

−4

−3

−2

−1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The graph of the function looks like this: 80 y 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 −3

−4

−2

−1

y = 3x

x 1

We call y = 3x an increasing function because the y-values increase as the x-values increase. If the base is less than 1, the function will be decreasing. There is a horizontal asymptote at y = 0 because any base raised to a power will always be greater than zero. Now consider the table of values for the logarithmic function with the same base, y = log3 x. x y

−4

−3

−2

−1

The graph of the function looks like this: 4

y = log3x

2 1 −1

x 5

20 25 30 35 40 45 50 55 60 65 70 75 80

−2 −3 −4

The graph of y = log3 x is also an increasing function. If the base is less than 1, then it will be a decreasing function. The graph has a vertical asymptote at x = 0 because we can only take the logarithm of a positive number with a positive base that’s not equal to one. We say the set of possible x-values for the function y = logb x is: all real values where x > 0 and b > 0, b ≠ 1

9.05 Optional: Connect exponentials and logarithms mathspace.co

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Example 1 Consider the function f(x) = log10 (3x + 9) Evaluate f (4) to two decimal places.

Create a strategy Substitute x = 4 into f (x) = log10 (3x + 9).

Apply the idea f (4) = log10 (3 × 4 + 9)

Substitute 4 for x

= log10 21

Evaluate what’s inside the brackets

≈ 1.32

Evaluate the logarithm

Example 2 Consider the function y = log3 x. a Complete the table of values: x y

⬚

Express all values as simplified fractions.

Create a strategy Substitute the values of x into y = log3 x.

Apply the idea For

, to x

Substitute

Express

as powers of 3

Use the law logB pn = n logB p

Evaluate

Substituting the other values of x into y = log3 x, the complete table of values is given by: x y

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b Consider the behaviour of log3 x as x changes. Which of the following statements is correct? A

log3 x is a decreasing function.

log3 x is increasing in some intervals of x, and decreasing in other intervals.

log3 x is an increasing function.

Create a strategy Consider the table of values from part (a).

Apply the idea Based on the table of values from part (a), as x increases the value of y also increases. x −5

−1

The correct answer is option C.

c What value does log3 x approach as x approaches 0? A

−∞

−3

∞

Create a strategy

Apply the idea

Consider the graph of log3 x. Consider what happens to the value of y as you substitute in smaller and smaller values of x based on its graph.

Based on the graph of log3 x, as x approaches 0 the value of y approaches −∞. y 4 3 2 1 −1

−1

x 1

−2 −3 −4

The correct answer is option A.

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d What happens when x = 0? A

y = −3

y = −∞

Create a strategy

y is undeﬁned

Apply the idea

Check for intervals of x where the values of y are In the expression log3 x, x is the result when 3 is deﬁned based on the graph of y = log3 x. raised to some power and x must be positive. y 4 3 2 1 −1

−1

x 1

−2 −3 −4

This means that y is only deﬁned for positive values of x, so the correct answer is option C.

Reﬂect and check There is a vertical asymptote at x = 0, which means x cannot be equal to zero.

Idea summary The general form of an exponential function is:

y = bx b

is the base

is the index

is the result of raising the power

The general form of a logarithmic function is:

y = logb x b

is the base

is the result of raising the power

is the index

An exponential function is the result of raising the base to a power, while a logarithmic function is equal to the index (or power).

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A number raised to a power is always positive, therefore y = bx has a horizontal asymptote on the x-axis. You cannot ﬁnd the logarithm of numbers less than or equal to zero, so y = logb x has a vertical asymptote on the y-axis. Both functions are increasing functions for b > 1 and decreasing functions for b < 1.

Inverse functions Notice that the x- and y-values are swapped in the table of values for the two functions y = 3x and y = log3 x. The functions are inverse functions which are like the opposites of each other, so the coordinates are in opposite order. When we have an equation in two variables, such as y = 3x, we say y is a function of x. We sometimes write functions using function notation. For example, f (x) = 3x means f is a function of x. Function notation gives us greater ﬂexibility when communicating mathematically. The inverse of function f (x) is written using the notation f −1 (x). The inverse function of f (x) = 3x is f −1 (x) = log3 x. f (4) = 34 = 81 so f (x) contains the ordered pair (4, 81). f −1 (81) = log3 81 = 4 so f −1 (x) contains the ordered pair (81, 4).

Example 3 Find the inverse function of f (x) = 2x

Create a strategy Switch the x and y terms in y = 2x and ﬁnd the equation of y.

Apply the idea y = 2x x = 2y

Switch the x and y y

log2 x = log2 2

Get the logarithm of base 2 on both sides

log2 x = y log2 2

Use the law logB pn = n logB p

y = log2 x

Use the law logB B = 1 x

The inverse function of y = 2 is y = log2 x.

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Example 4 Consider the function y = log4 x shown in the graph.

y 4 3 2 1 −2 −1 −1

x 1 2 3 4 5 6 7 8 9

−2 −3 −4

a Complete the table of values. x

⬚

Create a strategy

Substitute the values of x into y = log4 x.

Apply the idea For x = 1, y = log4 1 = log4 4

Substitute 1 to x 0

Express 1 as powers of 4

= 0 × log4 4

Use the law logB pn = n logB p

Evaluate

Substituting the other values of x into y = log4 x, the complete table of values is given by: x

b The given curve is the inverse of log4 x. Notice that the points on the curve are the inverses of the points found in part (a).

y 4 3 2 1 x −4 −3 −2 −1 −1 −2

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1 2 3 4 5 6

Find the equation of the inverse function of y = log4 x.

Create a strategy Switch the x and y terms in y = log4 x and ﬁnd the equation of y.

Apply the idea y = log4 x x = log4 y

Switch the x and y

We know that if logB p = A, then BA = p. This means that the inverse function of y = log4 x is given by: y = 4x

Idea summary Inverse functions use the notation f −1 (x), and have opposite pairs of x- and y-values. y = bx and y = logb x are inverse functions.

Graphs of inverse functions 4

When inverse functions are graphed on the same Cartesian plane it can be seen that they are reﬂections of each other about the line y = x

y = 4x y=x

3 2 1

−4 −3 −2 −1 −1

2 3 4 y = log4x

−2 −3 −4

Example 5 Consider the graph of 3x.

y 4

a Sketch the inverse function on the same coordinate plane.

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

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Create a strategy Use the fact that the points on the curve of y = 3x are the inverse of the points on the curve of the inverse function of y = 3x.

Apply the idea Based on the graph of y = 3x, the points (0, 1) and (1, 3) lie on the curve of y = 3x.

y 4 3

(1, 3)

2 1 (0, 1) −4 −3 −2 −1 −1

x 2

−2 −3 −4

The inverse points corresponding to (0, 1) and (1, 3) are (1, 0) and (3, 1), respectively. So, these points can be used to sketch the graph of the inverse function of y = 3x.

y 4 3

(1, 3)

2 1 (0, 1) (1, 0) 1 2 3

−4 −3 −2 −1 −1

(3, 1)

−2 −3 −4

b Write the equation of the inverse function.

Create a strategy Switch the x and y terms in y = 3x and ﬁnd the equation of y.

Apply the idea y = 3x x = 3y

Switch the x and y y

log3 x = log3 3

Find the logarithm of base 3 on both sides

log3 x = y log3 3

Use the law logB pn = n logB p

y = log3 x

Use the law logB B = 1 x

The inverse function of y = 3 is y = log3 x.

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Idea summary When inverse functions are graphed, they are reﬂections of each other about the line y = x.

9.05 Practice questions Understanding

Explorer 1–3, 4ab, 5ab

Adventurer 1–3, 4ac, 5ac

Trailblazer 1–3, 4bd, 5bd

Fill in the blank to complete the sentence:

The inverse of an exponential function is a ⬚ function of the same base.

For the function f (x) = loga x: a

Does the value of a make f (x) = loga x an increasing function? i

iii

a=2

Does the value of a make f (x) = loga x a decreasing function? i

a=0

iii

a=2

When a = 1, what does the graph of f (x) = loga x look like?

Determine whether the function could represent a logarithmic function. a 4

y 4

1 −4 −3 −2 −1 −1 −2

x 1

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

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d 4

g 4

Ex 2

8 3

f (x)

64 = 43

log4 (16) = 2

6ab, 7, 8ac, 9, 11a, 12a, 13

y = log3 (5x + 2)

y = log2 (x2 − 9)

y = log6 (x − 4)

Complete the table of values:

1 16

1 4

⬚

⬚ 3

⬚ 4

Sketch the graph of the function on a coordinate plane.

Is the function increasing or decreasing?

What value does log4 x approach as x approaches 0?

What happens when x = 0?

Write the inverse function. f (x) = 5x

log7 (1) = 0

6ac, 7, 8ac, 9, 11, 12b, 13, 14 6cd, 7, 8bd, 10, 12cd, 13–15

Using y = log4 x.

510

−4

25 = 52

−4

f (x) = log8 (x)

Ex 3

−3

Find the value of y when x = 5. a

−2

−3

−4 −3 −2 −1 −1

−2

Fluency 6

Rewrite in exponential form. a

Ex 1

Rewrite in logarithmic form. a

x 1 2 4 f (x)

−4 −3 −2 −1 −1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

f (x) = log7(x)

Ex 4

Using f (x) = log2 x. a

Write the inverse, g (x), of the given function.

Complete the table of values for the inverse. x

−2

−1

g (x)

Ex 5

Use the inverse to create a table of values for f (x) = log2 x.

Draw the graph of f (x) = log2 x.

Find the x-intercept for f (x) = log2 (x).

Write the equation of the vertical asymptote for f (x) = log2 (x).

Is f (x) = log2 x an increasing or decreasing function?

Using

Find the inverse of the given function.

Graph the inverse.

Use the inverse to create a table of values for

Draw the graph of

Find the function value when x = 0.

on the same coordinate plane.

an increasing or decreasing function?

For each graph: i

Sketch the inverse function on the same coordinate plane.

Write the equation of the inverse function.

1 −4 −3 −2 −1 −1

x 1

−4 −3 −2 −1 −1

−2

x 1

−2

−3

−4

For each logarithmic equation: i

Write the inverse in exponential form.

Graph the function and its inverse on the same coordinate plane.

f (x) = log10 x

f (x) = log4 (x)

f (x) = log5 (x)

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Using

Find the x-value of the x-intercept.

Complete the table of values: x

State the equation of the vertical asymptote.

Sketch the graph of the function on a coordinate plane.

an increasing or decreasing function?

The graph of f (x) = logk x is shown. a

Determine the value of the base, k.

Write the equation of f (x).

5 4 3 2 1 −1 −1 −2 −3 −4

x 1 2 3 4 5 6 7 8 9

−5

The graph of line y = x are shown.

and its reﬂection across the 4

Are the two curves inverse functions of each other?

Write the equation of the reﬂection of across y = x.

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

Reasoning

16, 17

Explain why f (x) = 2x and g (x) = log2 x are inverse functions.

For the function y = loga x, where a is a value greater than 1:

512

16–18

For what values of x will loga x be negative?

For what values of x will loga x be positive?

Is there a value that loga x will always be greater than? Use a graph to explain your answer.

Is there a value that loga x will always be less than? Use a graph to explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For the functions f (x) = log3 x and g (x) = log5 x: a

Which function has larger function values for values of x that are greater than 1? Explain your reasoning.

Justify your answer to part (a).

Problem-solving 19, 21, 23a

19, 21, 23ab, 24

20, 22, 23bcd, 24, 25

lies on the curve

Find the value of h, given the point

Given the points (8, n), (k, 4) and (m, −2) all lie on the curve with equation y = log2 x, ﬁnd the value of: a

For the equation 2x = 8: a

Sketch the graph of y = 2x.

Sketch the inverse of y = 2x.

Using the graph from part (b), ﬁnd the value of the function at x = 8.

Solve the equation 2x = 8.

Given the equation log9

Sketch the graph of y = log9 x.

Sketch the inverse of y = log9 x.

Using the graph from part (b), ﬁnd the value of the function at

Solve the equation log9

For each logarithmic equation: i

Use technology to graph each side of the equation on the same coordinate plane.

Estimate the solution to the equation.

log2 (x) = 4.2

log3 (x + 1) = −1

−log2 (x + 3) = 2.4 d

log4 (x − 3) = 0.3

For y = 2 log10 x a

Complete the table of values for y = 2 log10 x x

0.01

0.1

100

y b

Draw the graph of y = 2 log10 x

Draw the graph of y = log10 x on the same coordinate plane.

Explain how the y-values of y = 2 log10 x compare to the y-values of y = log10 x for the following values of x: i

0<x<1

x=1

iii

x>1

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For these functions: • f (x) = log2 x • g (x) = log2 x + 5 • h (x) = 5 log2 x a

Determine the y-values for each function at x = 4, x = 128 and x = 21000 f (x)

i b

g (x)

iii

h (x)

Describe the relationship between the function values of f (x), g (x) and h (x) as x increases.

9.06 Optional: Solve exponential equations with logarithms After this lesson, you will be able to… • apply the method of taking logarithms to solve exponential equations. • solve real-world problems involving exponential equations.

Solve exponential equations We previously saw that one method to solve an exponential equation is to write both sides of the equation with the same base. 2x = 128 x

Both sides can be written with same base

2 =2

Write both sides as base 2

x=7

Equate the powers

It’s not always possible to write both sides of the equation with the same base. In this situation, we use the idea that logarithms and exponentials are the inverse of each other, and we can solve the exponential equation by taking the logarithm of both sides of the equation. Both sides cannot be written with same base

Take log of both sides

Use the log power law

Isolate the variable

Evaluate with calculator

Note that the logarithm of any base can be used to solve the equation, but it makes sense to use base ten logarithms as the calculator log button can be used, evaluating the answer to a number of decimal places.

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Example 1 Consider the equation 2x = 11. a Rearrange the equation into the form

Create a strategy We make x the subject of the equation by using the laws of logarithm.

Apply the idea Start with the equation

Take the logarithm of both sides of the equation

Use the law logB pn = n logB p

Divide both sides by log 2

Simplify the expression

b Evaluate x to three decimal places.

Create a strategy

Apply the idea

Evaluate the expression found from part (a) using a calculator.

Start with the equation Evaluate and round to three decimal places

Example 2 Solve 3 (10x) = 6 for x. Round your answer to two decimal places.

Create a strategy Take the logarithm of base 10 on both sides of the equation and use the law logB pn = n logB p.

Apply the idea Start with the equation

Divide both sides by 3

Evaluate

Take the logarithm of both sides

Use the law logB pn = n logB p

Use the law logB B = 1

Evaluate and round to two decimal places

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Reﬂect and check We used the law log An = n log A which is a consequence of inverse functions. The reason we chose base 10 was to make it easier to calculate the solution using a calculator. In other cases, it might be better to use a logarithm with the same base as the exponential so that the solution is a logarithmic expression.

Example 3 Solve 5y − 4 = 6y + 3 for y. Round your answer to two decimal places.

Create a strategy Take the logarithm of base 10 on both sides of the equation and use the law logB pn = n logB p.

Apply the idea Start with the equation Take the logarithm of both sides Use the law logB pn = n logB p Use the distributive property Add 4 log 5 to both sides Subtract y log 6 from both sides Factor out the common term y in the left-hand side Divide both sides by log 5 − log 6 Evaluate and round to two decimal places

Idea summary Exponentials and logarithms of the same base are inverse functions of one another. If we cannot solve an exponential equation by writing both sides with the same base, then the method of taking the logarithm of both sides can be used to solve the equation.

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9.06 Practice questions Understanding 1

Adventurer 1, 2, 3c–f

4x = 13

8x = 4

7x = 2

5 (4)x = 100

10 (6)x − 12 = 8

4a–e, 5a–e, g, 6a–c, e, 7a–d, 8ac, 9ab, 10ab, 11ab

Fluency i

Write the equation in the form

Ex 1

Evaluate x rounded to three decimal places.

13x = 5

5x = 13

−x

= 12

9x = 27x − 1

72x + 3 = 1

4g–l, 5f, i–l, 6d–h, 7e–h, 8bd, 9cd, 10cd, 11cd, 12

29x = 5 x

5 + 4 = 3129

3x = 2

4x = 6.4

49x = 169

(0.4)x = 5

6 (2 ) = 42

27x + 4 = 1924

Solve for x, rounding to two decimal places. a

10x = 3.68

10x = 39.671

10x + 2 = 345.5

10x − 3 = 338

104x = 357

104x − 4 = 629

4 × 10x = 794.4

5 × 10x − 4 = 111

4 × 104x − 5 = 95

103x − 1 + 9 = 40

105x + 4 − 7 = 21

Solve for x, rounding your answer to two decimal places: a

12x = 23

5x = 9

3x = 100

0.8x = 3

5 × 2x = 300

200 × 3x = 520

120 × 7x = 3

50 (0.8)x = 325

12x + 5 = 6.5

20 (3x) = 4

Solve for the exact value of x. a

12x − 40 = 120

e 8

5x − 9 = 25

4b, d–h, 5b–h, 6b–e, 7b–e, 8ac, 9ab, 10ac, 11ac

Write the ﬁrst step required to solve for x.

Ex 2

2 (3)x = 18

Using a calculator, determine x rounded to 3 decimal places.

Trailblazer 1, 2, 3e–h

Determine whether logarithms are required to solve the equation without a calculator: a

Explorer 1, 2, 3a–d

7x − 3 = 8.5

3 (54x + 5) = 3

Make x the subject of the equation.

Evaluate x to three decimal places.

83x = 580

4x − 3 = 25

3x + 4 = 200

42x − 8 = 70

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Solve for x, rounding your answer to two decimal places: a

Ex 3

4x + 1 = 7

74x + 3 = 6

56a − 5 = 12

3y + 5 = 62y

25y + 2 = 34 − 3y

92y − 1 = 75 − y

Solve for the exact value of y. a

35x = 6 3y = 2y + 1

6y + 5 = 5y + 2

Solve for y, rounding your answer to two decimal places. a

6y − 3 = 11y

4y + 2 = 5y − 7

5y + 4 = 83y − 4

Akiko’s solution to the equation 311 − 4w − 7 = 6 is shown. Determine if her solution is correct by solving the problem a different way.

Reasoning

13, 14, 16a

14, 15, 16b, 17

Explain why 23x − 1 = 27 requires logarithms to solve, while 23x − 1 = 4x − 9 does not.

Explain why 3x = −2 has no solution.

Use the equation Bx = A to show

For each of these incorrect sets of working:

518

for A, B ≥ 0.

Which step was incorrect? Explain your answer.

Rearrange the original equation into the form

iii

Evaluate a to three decimal places.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Hussam uses logarithms of base 3 to solve the equation 3x − 2 = 164x. His work is shown. Hussam has found the exact solution, but he wants to estimate the solution rounded to three decimal places. He does not know how to do this because his calculator only allows him to use logarithms of base 10. Explain how Hussam can rewrite the solution using base 10 logarithms.

Problem-solving 18–20 18

18–21

18–23

The population of mice, t months after initial observation, is modelled by: P (t) = 500 (1.2t). a

Find the initial population.

By what percentage is the population increasing by each month?

Find when the population reaches 1500, for the first time.

A population of wallabies, t years after initial observation, is modelled by P (t) = 800 (0.85t). a

Find the initial population.

By what percentage is the population decreasing each year?

Find when the population reaches 200, for the first time.

The growth of a population of mites is modelled by P = 20 (3x), where P is the population after x weeks. After how many weeks will the population of mites have grown to 5000?

A microbe culture initially has a population of 900 000 and the population increases by 40% every hour. Let t be the number of hours passed. Find when the population reaches 7 200 000, for the first time.

A piece of art that costs $6000, appreciates at approximately 6.2% p.a. After how many full years, n, will the value of the piece of art be over $15 000?

Samir makes an investment of $1300, compounding monthly at a rate of 5.9% p.a. How many years will it take for this investment to reach $9000?

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9 Chapter review

For each number: i

Express the number as a power of 10.

Find the base ten logarithm of the number.

0.01

10 000

For each number: i

Express the number as a power of 2.

Find the base two logarithm of the number.

0.5

The histogram shows the population for 12 cities in Australia, using a log10 scale: How many cities have a population between 100 000 and 1 000 000?

10 9 8 7 6 5 4 3 2 1 0

Number of cities

1 2 3 4 5 6 Log of population

The weights of various animals are shown in the table: Animal Ferret Cat Whale Rhinoceros Arabian wolf Hamster Shetland pony

Weight (in kg) 0.8 3.2 32 840

Weight in log10 scale

3.44 1.40 −0.82 127

Complete the table, rounding each value to two decimal places. 5

520

A maths tutor has clients with a large range of ages. To manage the data more easily, the age of each client is ﬁrst converted to a log2 scale. a

What range of ages corresponds to a range of 4 to 5 on the log2 scale?

What age would result in 3.2 on a log2 scale? Round to the nearest whole number.

For the number 0.000 04: a

Write the number using scientiﬁc notation.

Identify the order of magnitude.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A cafe sells 450 glasses of milk tea in a week. What is the order of magnitude of the total number of milk teas sold over 12 weeks?

The decibel scale, which is used to record the loudness of sound, is a logarithmic scale. In the decibel scale, the lowest audible sound is assigned the value of 0. The table shows the decibel value for various types of noise.

How many times louder is the sound of stereo music than the sound of a hair dryer?

If a sound is measured to be 136 dB and another sound is measured to be 85 dB, how many times more intense is the ﬁrst sound than the second sound? Write your answer as a power of 10.

Type of noise Jet plane Rock concert Ambulance siren Stereo music Electric drill Hair dryer Refrigerator Bedroom Falling leaves

Decibel value 150 120 110 100 90 70 50 30 20

The Richter Scale is a base 10 logarithmic scale used to measure the magnitude of an earthquake, given by R = log10 x, where x is the Earthquake Richter scale relative strength of the quake. Some past 8.1 earthquakes and their Richter scales are shown in Mexico 2017 Turkey 2023 7.8 the table. Japan 2021 7.0 How many times stronger was the earthquake in Botswana 2017 6.5 Turkey compared to the earthquake in Botswana? Round your answer to the nearest whole number.

(Optional) Evaluate: a

log3 27

log12 1

log2 0.25

(Optional) Without evaluating, ﬁnd the two integers that the value of each logarithm lies between. a

log2 (13)

log10 (0.073)

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(Optional) For each logarithmic equation: i

Rewrite in exponential form.

Solve for x.

log3 (27) = x

log4 (x + 4) = 2

log7 (6s)

log (a12)

log7 (3 − 2x) = 1

log4 (77)

log6 (x − 10)8

(Optional) Write the expression as a single logarithmic term. a

log2 4 + log2 a − log2 13

− log3 m − log3 n

log 5 + log x − 3 log y

(Optional) Evaluate: b

logb (b7)

log4 (32) − log4 (8) d

f (x) = log6 (x)

(Optional) Write the inverse function, g (x). a

a 17

(Optional) Rewrite the logarithm without any powers or surds. a

logx (16) = 4

(Optional) Rewrite the logarithm as a sum or difference of two or more logarithms. a

f (x) = 4x

(Optional) Using f (x) = log5 x. a

Write the inverse, g (x), of the given function.

Complete the table of values for the inverse: x

−2

−1

g (x)

Use the inverse to create a table of values for f (x) = log5 x.

State the equation of the vertical asymptote.

Sketch the graph of f (x) = log5 x on a coordinate plane.

Is y = log5 x an increasing or decreasing function?

(Optional) The graph of f (x) = logk x is shown. a

Determine the value of the base, k.

Write the equation of f (x).

5 4 3 2 1 −1 −1 −2 −3 −4 −5

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x 1 2 3 4 5 6 7 8 9

(Optional) For each graph: i

Sketch the inverse function on the same coordinate plane.

Write the equation of the inverse function.

b 4

1 −4 −3 −2 −1

x 1

−1

−4 −3 −2 −1

−2

−1

x 1

5x + 2 = 600

104x+3 − 9 = −2

11x+1 = 4x

−2

−3

−4

(Optional) i

Write the equation in the form

Evaluate x, rounded to three decimal places.

17x = 3

9x = 529

(0.8)x = 5

(Optional) Solve for x, rounded to two decimal places. a

10x = 2.65

5 × 102x−5 = 11

65x = 485

8x+11 = 57

32x−7 = 102

(Optional) A plant species initially has a population of 7000 and decreases by 5% each year. How many years will it take for the population to reach 600 plants? Round your answer to the nearest whole number.

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Big ideas The equations representing circles provide a systematic way to understand their geometry and position in a Cartesian plane. Curriculum Outcomes VC2M10A11 explore the connection between algebraic and graphical representations of relations such as simple quadratic, reciprocal, circle and exponential, using digital tools as appropriate VC2M10AA05 describe, interpret, and sketch parabolas, hyperbolas, circles and exponential functions and their transformations VC2M10AA10 experiment with functions and relations using digital tools, making and testing conjectures and generalising emerging patterns

Chapter outline

Circles Chapter outline Investigation: Circles 10.01 Circles at origin 10.02 Translations of circles Investigation: Equation of any circle 10.03 Optional: Transform circle equations Chapter 10 review

526 533

544

Circular dome structures, like Rome’s Pantheon, have incredible strength and stability!

Investigation: Circles Investigate online

10.01

mathspace.co

Circles at origin

After this lesson, you will be able to… • identify the equation of a circle centred at the origin. • identify the radius of the circle given points or equation. • graph circles centred at the origin. • analyse the position of a point relative to a circle (inside, on, or outside) using the circle’s equation.

Circles centred at the origin All points on a circle are the same distance from the centre. The radius tells us the distance from the centre to any point on the circle. A circle centred at the origin is represented by the equation, x2 + y2 = r2 where r is the radius. For example, the equation that represents the circle is x2 + y2 = 4.

y 2

Its centre is at (0, 0) and its radius is 2 units. 1 x −2

−1

−1 −2

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Exploration Match each equation with the correct graph. • x2 + y2 = 1

• x2 + y2 = 25 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 − −5

x 1 4

Circle 3

10 8 6 4 2

x 2 4 6 8 10

Circle 2

1 3 4 1 2 1 4 – 5 −1 – 3 – 1 – 1 4 4 2 4 1 4 1 2 3 4 −1 5 4

• x2 + y2 = 100

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

1 2 3 4 5

Circle 1 5 4

• x2 + y2 = 64

1 2

3 4

5 4

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

Circle 4

What strategies did you use to match the equation to the graph?

If the radius of a circle centred at the origin is 3 units, what is the equation of the circle?

What is the radius of the circle with the equation x2 + y2 = 49?

To check whether a point is inside, on or outside a circle, we can compare the distance between that point and the centre of the circle to the radius. • If x2 + y2 < r2, then the point is inside the circle. • If x2 + y2 = r2, then the point is on the circle. • If x2 + y2 > r2, then the point is outside the circle. These conditions are the same as substituting the point into the equation of the circle and comparing the values on each side.

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Example 1 State the equation of the circle.

5 x −10

−5

5 −5

−10

Create a strategy Identify the radius, then substitute it into the equation x2 + y2 = r2

Apply the idea Looking at the x-intercepts and y-intercepts, we can see that the radius of the circle is 7. x2 + y2 = r2

Write the equation

x +y =7

Substitute r = 7

Evaluate 72

x + y = 49

Example 2 Consider the circle with centre (0, 0) that passes through the point (3, −4). a Find the radius of the circle.

Create a strategy The radius is the distance between the centre and any point on the circle. To calculate the distance, use Pythagoras’ theorem: a2 + b2 = c2.

Apply the idea We can form a right-angled triangle with the distance between the origin and (3, −4) as the hypotenuse. Let r be the radius. a2 + b2 = c2 2

Write the formula

Substitute a = 3, b = −4, c = r

Evaluate the squares

Add the left-hand side

3 + (−4) = r

9 + 16 = r 25 = r

r=5

Take the positive square root of both sides

The radius of the circle is 5 units.

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Reﬂect and check We only consider the positive square root because the radius represents a length, and length is always non-negative.

b Determine the equation of the circle.

Create a strategy 2

Apply the idea 2

Use the equation x + y = r .

From part (a), we know that r = 5. x2 + y2 = r2

Write the equation

x +y =5

Substitute the radius

Evaluate 52

x + y = 25

Example 3 Determine whether the point (−8, 5) lies inside, outside, or on the circle represented by the equation x2 + y2 = 117.

Create a strategy Substitute in the coordinates to determine whether the left-hand side is less than, greater than or equal to the right-hand side. Recall that: • If x2 + y2 < r2, then the point is inside the circle. • If x2 + y2 = r2, then the point is on the circle. • If x2 + y2 > r2, then the point is outside the circle.

Apply the idea x2 + y2 = (−8)2 + 52

Substitute x = −8, y = 5

= 64 + 25

Evaluate the squares

= 89

Perform the addition

< 117

Compare to the circle equation

Since 89 < 117, the point (−8, 5) lies inside the circle x2 + y2 = 117.

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Idea summary All points on a circle are the same distance from the centre. The radius tells us the distance from the centre to any point on the circle. A circle centred at the origin is represented by the equation, x2 + y2 = r2 where r is the radius. To check whether a point is inside, on or outside a circle, we can compare the distance between that point and the centre of the circle the radius. • If x2 + y2 < r2, then the point is inside the circle. • If x2 + y2 = r2, then the point is on the circle. • If x2 + y2 > r2, then the point is outside the circle

10.01 Practice questions Understanding 1

Explorer

Adventurer

1–3

State the radius of each circle: a

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

10 8 6 4 2

8 x 3

9 12

4 −16 −12 −8 −4 −4

−6

−8

−9

−12

−16

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

2 4 6 8 10

−12 −9 −6 −3 −3

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

1 2 3 4 5

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Trailblazer 2, 3

1–3

x 4

8 12 16

Use the graph of the circle to complete the statement: 8

Every point on the circle is exactly ⬚ units away from the point (⬚, ⬚).

6 4 2

−8 −6 −4 −2 −2

−4 −6 −8

If a point (x, y) lies on the circle with equation x2 + y2 = r2, what is the distance between the point and the centre of the circle? 4–6, 7a–d, 8, 9ab, 10, 11

Fluency

5, 6, 7c–f, 8, 9bc, 10, 11, 13

What is the equation of the circle with centre at the origin and a radius of 5 units?

What is the radius of the circle with equation x2 + y2 = 81?

A circle with centre (0, 0) has a y-intercept at (0, −8). Write the equation of the circle.

For the equations of circles given: i

Identify the radius of the circle.

Sketch the graph of the circle.

x2 + y2 = 36

x2 + y2 = 9

e Ex 1

4–6, 7c–f, 8, 9ab, 10–12

x2 + y2 = 16

x2 + y2 = 0.64

x2 + y2 = 169

Write the equation of each circle shown: a

4 3

6 4

2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

y 8

x 1

2 3 4 5

2 −8 −6 −4 −2 −2

x 2

−4 −6 −8

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−12−10−8 −6 −4 −2 −2 −4 −6 −8 −10 −12

Ex 2

Ex 3

12 10 8 6 4 2

−12 −10 −8 −6 −4 −2 −2 −4 −6

2 4 6 8 10 12

x 2 4 6 8 10 12

−8 −10 −12

For circle with centre (0, 0) that passes through the points given: i

Find the exact radius of the circle.

(−3, 4)

(8, −6)

Find the equation of the circle.

(−5, −10)

If a circle has its centre at the origin and passes through the point (6, 8), what is the equation of the circle?

Determine whether the following points are inside, outside, or on the circle with equation x2 + y2 = 25. a

(−3, 2)

(4, 3)

(1, 6)

(−5, 0)

Is the point (−4, 2) inside, outside, or on the circle with equation x2 + y2 = 21?

Is the point (7, 1) inside, outside, or on the circle with equation x2 + y2 = 50?

Reasoning

14, 15

15, 16

15–17

How can you determine if a point (a, b) lies inside, outside, or on the circle with equation x2 + y2 = r2?

A circle with its centre at (0, 0) passes through the points (7, 24) and (20, 15). Prove that the points are on the same circle.

Explain why the equation of a circle with its centre at the origin and a radius of r is given by x2 + y2 = r2.

A circle has its centre at the origin and passes through the point (5, 6). Determine the equation of the circle and justify your answer.

Problem-solving 18, 19, 21, 23 18

18, 19, 21–23

19, 20, 22–24

A circle with centre (0, 0) has an x-intercept at (6, 0). a

Find the equation of the circle.

Find the exact area enclosed by the circle.

A circle with its centre at the origin has an area of 64π. Find the equation of the circle.

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A circle with its centre at the origin has a circumference of 12π. What is the equation of the circle?

For circle with centre (0, 0) that passes through the point (−4, 4), ﬁnd the equation of the circle.

A circle with its centre at the origin has an equation of x2 + y2 = 9. Find the area of the part of the circle that is in Quadrant 1.

A circle with its centre at the origin has an equation of x2 + y2 = 81. Use technology to estimate the coordinates of the points where the circle intersects the line y = 3x.

A circle with its centre at the origin has an equation of x2 + y2 = 64. Find the length of the horizontal line that is within the circle, parallel to the x-axis and has a y-intercept of 4.

y 8 6 4 2 −8 −6 −4 −2 −2

x 2

−4 −6 −8

10.02

Translations of circles

After this lesson, you will be able to… • identify the changes in the equation of a circle due to its translation. • understand the general equation of a circle and how it represents the centre of the circle. • write the equation of a circle that has been translated.

Translations of circles Similar to other shapes and functions, a circle can be transformed with horizontal and vertical translations.

Interactive exploration Discover this concept in action online

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A circle can be vertically translated by increasing or decreasing the y-values by a constant number, k. When k is added to y, the circle is translated down by k. When k is subtracted from y, the circle is translated up by k. y

y x2 + ( y − 2)2 = 1

3 2

2 2

x + y =1

x2 + y2 = 1

1 x

−2

−1

x −2

−1

−2

−3

The circle with equation x2 + y2 = 1 is translated up by 2.

x2 + ( y + 2)2 = 1

The circle with equation x2 + y2 = 1 is translated down by 2.

A circle can be horizontally translated by increasing or decreasing the x-values by a constant number, h. When h is added to y, the circle is translated left by h. When h is subtracted from y, the circle is translated right by h. y

y 3

2 x2 + y2 = 1 1

2 x2 + y2 = 1 1 x

−3 −2

−1

1 −1 −2

(x − 2)2 + y2 = 1

x −3 −2

−1

(x + 2)2 + y2 = 1

−3

−1 −2 −3

The circle with equation x2 + y2 = 1 is translated right by 2.

The circle with equation x2 + y2 = 1 is translated left by 2.

In general, the equation of a circle is given by:

(x − h)2 + ( y − k)2 = r2 (h, k) coordinates of the centre of the circle (x, y) coordinates of any point on the circle r radius of the circle

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Example 1 The equation of a circle is given by (x − 4)2 + ( y + 1)2 = 36. a Find the coordinates of the centre of this circle.

Create a strategy Equations of the form (x − h)2 + ( y − k)2 = r2 represent circles with centre at (h, k).

Apply the idea The equation (x − 4)2 + ( y + 1)2 = 36 can be written as (x − 4)2 + ( y − (−1))2 = 62 and it is now of the form (x − h)2 + ( y − k)2 = r2 where h = 4 and k = −1. The centre of the circle is (4, −1). b Find the radius of the circle.

Create a strategy

Apply the idea

Equations of the form (x − h)2 + ( y − k)2 = r2 represent circles with radius of r.

In part (a) we rewrote the equation (x − 4)2 + ( y + 1)2 = 36 in the form: (x − 4)2 + ( y − (−1))2 = 62 So the radius of the circle is 6 units.

c Plot the graph of the circle.

Create a strategy Plot the centre of the circle then count the units equal to the radius in each direction (left, right, up, down), and draw a circular curve connecting them.

Apply the idea 5 4 3 2 1 −2 −1 −1

−2 −3 −4 −5 −6 −7

We can plot the point that represents the centre of the circle at (4, −1).

x 1 2 3 4 5 6 7 8 9 10

Now we count 6 units in all four directions to get the points (4, 5), (4, −7), (10, −1) and (−2, −1). Connect the points using a round curve passing through the four points to graph the circle.

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Example 2 A circle has its centre at (3, 3) and a radius of 6 units. a Plot the graph of the circle.

Create a strategy Plot the centre of the circle then count the units equal to the radius in each direction (left, right, up, down), and draw a circular curve connecting them.

Apply the idea 9 8 7 6 5 4 3 2 1 −3−2 −1 −1 −2 −3

We can plot the point that represents the centre of the circle at (3, 3).

Now we count 6 units in all four directions to get the points (3, 9), (3, −3), (9, 3) and (−3, 3). Connect the points using a round curve passing through the four points to graph the circle. x 1 2 3 4 5 6 7 8 9

b Write an equation of the circle, in the form (x − h)2 + ( y − k)2 = r2.

Create a strategy Remember the equation (x − h)2 + ( y − k)2 = r2 represents a circle, where (h, k) are the coordinates of its centre and r is its radius.

Apply the idea We have the centre at (3, 3) and the radius is 6 units. (x − h)2 + ( y − k)2 = r2 2

(x − 3) + ( y − 3) = 6

Rewrite the equation Substitute h = 3, k = 3, r = 6

(x − 3) + ( y − 3) = 36 Evaluate 62 The equation of the circle is (x − 3)2 + ( y − 3)2 = 36.

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Example 3 The graph of x2 + y2 = 4 is translated 3 units left and 6 units down. Find the new equation.

Create a strategy By using the equation of circle (x − h)2 + ( y − k)2 = r2, substitute the translated coordinates into the equation.

Apply the idea A translation of 3 units left means h = −3, and 6 units down means k = −6. So by substituting these values, we have the new equation: (x + 3)2 + ( y + 6)2 = 4

Reﬂect and check 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

Here is the graph of x2 + y2 = 4 before and after the translations:

y x 1 2 3

Idea summary The general equation of a circle is given by:

(x − h)2 + ( y − k)2 = r2 (h, k) (x, y) r

Coordinates of the centre of the circle Coordinates of any point on the circle Radius of the circle

Circles with equation x2 + y2 = r2 can be translated vertically or horizontally. • • • •

When k is subtracted from y, the circle is translated up by k units. When k is added to y, the circle is translated down by k units. When h is subtracted from x, the circle is translated right by h units. When h is added to x, the circle is translated left by h units.

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10.02 Practice questions Understanding 1

Explorer

Adventurer

1–3

Determine if each circle has been translated away from the origin: a

x2 + y2 = 25

x2 + ( y − 2)2 = 4

(x − 5)2 + ( y + 1)2 = 1

(x + 1)2 + y2 = 16

Find the centre and radius of each circle: a

1 −6 −5 −4 −3 −2 −1 −1

3 2 1

−3

−4

−4 −3 −2 −1 −1

−5

−2

−6

−3

2 1 x −3 −2

−1

−2

−1 −2 −3 −4

Trailblazer

1–3

Using the graph of the circle: a

What are the coordinates of the centre?

What is the length of the diameter?

8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2

x 1

x 1 2 3 4 5

6 5 4 3 2 1 −6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6

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x 1 2 3 4 5 6

4ab, 5a–d, 6ab, 7, 8, 9a–c, 10ab

Fluency 4

Ex 1

4bc, 5b–e, 6bc, 7, 8, 9b–d, 10bc

For each equation, identify the: i

Centre of the circle

Exact radius of the circle

(x − 5)2 + ( y + 6)2 = 17

(x − 5)2 + ( y − 5)2 = 162

(x − 0.4)2 + ( y + 3.8)2 = 2

For each equation: i

State the centre of the circle.

State the radius of the circle.

iii

Sketch the graph of the circle.

(x − 2)2 + ( y + 2)2 = 36

(x − 2)2 + ( y + 2)2 = 16

(x + 1)2 + ( y − 3)2 = 16

(x + 3)2 + ( y + 4)2 = 25

(x − 6)2 + y2 = 49

e 6

Ex 2

4cd, 5ef, 6cd, 7–9, 10cd

Write the equation of the circle given its centre and radius: a

Centre (7, −1); r = 3

Centre (−2, −3); r = 5

Centre (−4.3, 0); r = 6

Centre (−10, 2); r = 12

Write the equation.

For each circle: i

Sketch the graph.

A circle whose centre is at (1, 2) and has a radius of 4 units.

A circle whose centre is at (3, −2) and has a radius of 4 units.

Find the equation of each circle: a

y 5 4 3 2 1

−3 −2 −1 −1 −2 −3 −4 −5

x 1 2 3 4 5 6 7

9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −−1

x 1 2 3 4 5

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−3 −2 −1 −1 −2 −3 −4 −5 −6 −7

Ex 3

3 2 1

1 2 3 4 5 6 7

11 10 9 8 7 6 5 4 3 2 1

−1 −1

x 1 2 3 4 5 6 7 8 9 10 11

Write the new equation after the circle with equation x2 + y2 = 25 is translated: a

10 units upwards.

10 units downwards.

10 units to the right.

10 units to the left and 4 units upwards.

Describe how each of the circles has been translated from the graph of x2 + y2 = r2 where r is the radius of the translated circle. a

12 10

2 x

6 −4

4 2 −12 −10 −8 −6 −4 −2 −2

1 −1

x 1

−2 −3 −4 −5 −6 −7

540

−6

y −3 −2 −1

−4

2 −2

x 2

−2

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

10 9 8 7 6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1

x 1 2 3 4

Reasoning 11

11, 13

12, 13

Consider the circle with centre (−8, −6) and radius 10. a

Write the equation of the circle.

Does the circle pass through (0, 0)? Justify your answer.

A circle has the equation:

y 6 4 2

Find the radius of the circle.

The graph of the circle has the same centre as the circle in part(a) but not the same radius.

−2 −2

x 2

10 12

−4

Would the circle with the equation lie inside or outside the

−6 −8

circle? Explain your reasoning. 13

A circle has the equation (x + 8)2 + ( y + 1)2 = 1. a

How many x-intercepts does the circle have?

Use translations to explain why there are no y-intercepts.

Problem-solving 14ab, 15, 16 14

14cd, 15, 17–19

Find the new equation after the graph of the circle with equation x2 + y2 = 1: a

Dilated by a scale factor of 3, then translated right 5 units and down 2 units.

Dilated by a scale factor of 6, then translated left 1 unit and up 4 units.

Dilated by a scale factor of 4, then translated down 3 units.

Dilated by a scale factor of 5, then translated right half a unit and up

units.

When expanded, the equation of a circle is x2 − 6x + y2 + 8y = 9. a

14bc, 15–17

Complete the square for the expression x2 − 6x then fill in the blanks: (x − ⬚)2 − ⬚ + y2 + 8y = 9

Complete the square for the expression y2 + 8y then fill in the blanks: x2 − 6x + ( y + ⬚)2 − ⬚ = 9

Use your answers from parts (a) and (b) to rewrite the equation in the form (x − h)2 + ( y − k)2 = r2.

Find the centre and radius of the circle.

The circle with equation (x − 3)2 + ( y + 1)2 = 9 is inscribed (ﬁtted exactly) inside a square. What are the coordinates of the vertices of the square?

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A wind turbine has 20 m long blades that are attached to a tower 30 m high. The distance from the origin to the base of the wind turbine is 25 m. Find the equation of the circle represented by the path of the blades of the wind turbine.

20 m

30 m x 25 m

A soccer match is being televised. One of the cameras is mounted on a drone which is programmed to zoom in on the ball only when it is inside the centre circle. The drone uses a coordinate system to track the position of the ball, where the origin is at the bottom left corner of the ﬁeld and each unit corresponds to 1 m. a

The centre of the “centre circle” is in the exact centre of the ﬁeld. What are the coordinates of the circle’s centre?

What is the radius of the centre circle?

State the equation for the centre circle.

The drone can be programmed to only check a rectangular area of the ﬁeld for whether the ball is within the centre circle or not. If the whole circle must be visible, what is the smallest area that the drone could check for this task?

Centre Circle 75 m

9.15 m

99 m

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Aarushi needs to use a laser cutter to make a cork component for her design project. The intersection of two different-sized circles inscribed on a piece of cork gives the shape that she wants. Using technology to draw a diagram, she creates two circles with the equations (x − 12)2 + ( y − 8)2 = 49 (x − 20)2 + ( y − 8)2 = 36 where each unit is 1 cm and (0, 0) is the bottom left corner of the rectangle. In the diagram, the rectangle represents the piece of cork, and the shaded area represents the desired component. a

What is the width of the component that Aarushi wants to make?

Find the least area of material that could be used to make Aarushi’s shape, considering these constraints: • The laser cutter will only accept rectangular pieces of material. • The height of the component is 10.17 cm. • There must be a gap of at least 1 cm between any cut and the edge of the piece.

If the bottom left corner of the rectangular piece in part (b) is now (0, 0), state the new equations of the two circles.

Investigation: Equation of any circle Investigate online

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10 Chapter review 1

For the equation x2 + y2 = 49: a

Identify the radius of the circle.

10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

Sketch the graph of the circle.

Write the equation of each circle shown: a

y 8 6 4 2

−8 −6 −4 −2 −2

2 4 6 8 10

x 2

−4 −6 −8

For circle with centre (0, 0) that passes through the point given: i

Find the radius of the circle.

(6, 0)

(5, −12)

Find the equation of the circle.

(4, −2)

Determine whether the following points are inside, outside, or on the circle with equation x2 + y2 = 100. a

(7, 2)

(−8, 6)

(9, −5)

(−5, −5)

A circle with its centre at the origin has an equation of x2 + y2 = 64. Find the area of the part of the circle that is above the x-axis.

Write the equation of the circle given its centre and radius: a

544

Centre: (4, 9) ; r = 1

Centre:

State the radius of the circle.

x2 + (y − 4)2 = 49

For each equation: i

State the centre of the circle.

iii

Sketch the graph of the circle.

(x − 3)2 + (y + 1)2 = 25

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For each graph: i

Describe how the graph has been translated from x2 + y2 = 16.

Write the equation of the circle.

9 8 7 6 5 4 3 2 1

−9 −8 −7 −6 −5 −4 −3 −2 −1 −1

x 1

2 1 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

y x 1 2 3 4 5 6 7 8

Find the equation after the graph of x2 + y2 = 1 is: a

Translated left 1 unit and down 7 units.

Translated right 5 units.

For the circle (x − 8)2 + (y + 6)2 = 100: a

What are the coordinates of the x-intercepts?

What are the coordinates of the y-intercepts?

Did you know?

Circles are one of the most important shapes in nature and mathematics! They are the only shape with constant curvature, meaning every point on the edge is the same distance from the centre. This property makes circles incredibly efficient and strong, which is why you’ll find them in everything from wheels to atoms. Even planets and stars take on a circular shape due to the force of gravity pulling evenly in all directions! Chapter 10 review mathspace.co

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Big ideas Understanding and analysing the properties and transformations of hyperbolas provide insight into inverse relationships, allowing for the exploration and prediction of real-world scenarios involving indirect proportions. Curriculum Outcomes VC2M10A11 explore the connection between algebraic and graphical representations of relations such as simple quadratic, reciprocal, circle and exponential, using digital tools as appropriate VC2M10AA05 describe, interpret, and sketch parabolas, hyperbolas, circles and exponential functions and their transformations

Hyperbolas Chapter outline Investigation: Hyperbolas 11.01 Direct and indirect proportion 11.02 Hyperbolas 11.03 Optional Transformations of hyperbolas Chapter 11 review

548 559 570 582

Hyperbolas are used in the design of reflector telescopes for sharp, focused images of space.

Investigation: Hyperbolas

Investigate online

11.01

mathspace.co

irect and indirect D proportion

After this lesson, you will be able to… • identify the relationship as a direct or indirect proportion. • calculate the constant of proportionality in a direct proportion. • express an inverse relationship mathematically and interpret its graph. • apply the knowledge of direct and indirect proportions to solve related problems.

Direct proportion Two quantities are said to be proportional (or have direct variation) if they increase or decrease together at the same rate. For example, if you earn $18 per hour, your income is directly proportional to the number of hours worked: income = $18 × hours worked, or y = 18x where y is the income and x is the number of hours worked. The constant of proportionality is the value that relates the two amounts. In that example, the constant of proportionality is 18. Notice that this constant is the gradient of the line. The y-intercept (c value) of a proportional relationship is zero. The general equation for variables that are directly proportional is:

y = kx k is the constant of proportionality We can write “y is directly proportional to x” as y ∝ x. The constant of proportionality is also called the unit rate. While it can be any non-zero number, positive or negative, we generally only consider relationships with a positive value of k.

Interactive exploration

Discover this concept in action online

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

mathspace.co

Example 1 Consider the values in each table. Which of them could represent a directly proportional relationship between x and y? Table A: x y

1 2

Table B: 2 8

3 18

4 32

x y

1 2

2 4

3 6

4 8

Create a strategy Determine which table shows that the y-values are a constant multiple of the x-values.

Apply the idea In table A, the ﬁrst y-value is equal to the corresponding x-value multiplied by 2: 1 × 2 = 2. The second y-value is equal to the corresponding x-value multiplied by 4: 2 × 4 = 8. The multiplier has changed from 2 to 4. In table B, the ﬁrst y-value is equal to the corresponding x-value multiplied by 2: 1 × 2 = 2. The second y-value is equal to the corresponding x-value multiplied by 2: 2 × 2 = 4. The multiplier has stayed the same. Table B could represent a directly proportional relationship between x and y.

Example 2 Consider the equation P = 90t. a State the constant of proportionality.

Create a strategy

Apply the idea

For equations in the form y = kx, k is the constant of proportionality.

The constant of proportionality in P = 90t is 90.

b Find the value of P when t = 2.

Create a strategy Substitute the given value into the equation.

Apply the idea P = 90 × 2 = 180

Substitute t = 2 Evaluate

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Example 3 Oliver is making cups of fruit smoothie. The amount of bananas and strawberries he uses is shown in the proportion table.

Strawberries Bananas

8 5.5

16 11

24 16.5

32 22

40 27.5

a Sketch the graph of this proportional relationship.

Create a strategy Use two pairs of values from the table as coordinates of points in the graph.

Apply the idea 11 10 9 8 7 6 5 4 3 2 1

The points (8, 5.5) and (16, 11) have been plotted, and a line drawn through them.

Bananas

Strawberries 10 12 14

Reﬂect and check Conﬁrm that the other points (24, 16.5), (32, 22), and (40, 27.5) lie on the same line you have created.

b What is the unit rate of this relationship?

Create a strategy The unit rate is the gradient or constant of proportionality.

Apply the idea 11 10 9 8 7 6 5 4 3 2 1

5.5 units

8 units

550

We can see in the graph that rise = 5.5 and run = 8.

Bananas

Strawberries 10 12 14

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

= 0.6875 bananas per strawberry

c Select ALL the statements that describe the proportional relationship. A

For every 5.5 bananas Oliver uses, he adds 8 strawberries.

The unit rate of bananas with respect to strawberries is

For every 8 bananas, William uses 5.5 strawberries.

The unit rate of bananas with respect to strawberries is

. .

Apply the idea From the table, when William uses 5.5 bananas, he adds 8 strawberries. So option A is correct. The unit rate is

bananas per strawberry. So option B is correct.

Reﬂect and check 11 10 9 8 7 6 5 4 3 2 1

We can also see in the graph that for every 5.5 bananas Oliver uses, he increases or adds 8 strawberries.

Bananas

5.5 units

This also shows why option A is correct.

8 units

Strawberries 10 12 14

Idea summary A directly proportional relationship is a linear relationship with a y-intercept of zero:

y = kx k is the constant of proportionality

Indirect proportion Some quantities have what we call an inverse relationship. As one of the quantities gets closer to zero, the other one gets further from zero, and vice versa. The relationship between speed and time is inversely proportional.

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For example, if a truck driver needs to drive 1000 km, the faster the driver travels, the less time it will take to cover this distance. Speed (km/h) Time (hours)

60 16.7

70 14.3

80 12.5

90 11.1

100 10

110 9.1

120 8.3

Plotting these values, the graph of the curve is given by: Each point on the curve corresponds to a different combination of speed and time, and all the points on the curve satisfy the equation

Time (hours) 35 30 T=

1000 S

, or TS = 1000, where T is the time

taken and S is the speed travelled.

(60, 16.7)

(100, 10)

10 5

Speed (km/h) 20 40 60 80 100 120 140 160

We can see from both equations that as average speed increases, the time taken will decrease. This is what we call inverse proportion, also known as inverse variation. Mathematically, we write this as y ∝

to say “y is inversely proportional to x”.

Two quantities x and y that are inversely proportional have an equation of the form: or

xy = k

where k ≠ 0. When we graph this relationship, it will be non-linear.

Example 4 The equation y =

represents an inverse relationship between x and y.

a Complete the table of values: x y

−5

−4

−3

−2

−1

Create a strategy Substitute the values of x into y =

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Apply the idea

Reﬂect and check

For x = −1, y = 6.

We could check our values by sketching a graph of the relationship using technology and see

For x = −2, y = 3. Similarly, substituting the other values of x, the complete table of values is given by: x

−5

−4

−3

−2

−1

that y =

looks like this: 6 5 4 3 2 1 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4

b Is y =

x 1

2 3 4

increasing or decreasing when x < 0?

Create a strategy

Apply the idea

Consider the table of values from part (a). Look what happens to y as the values of x increase.

Based on the table of values, y values increase as the values of x increase. y=

is increasing when x < 0.

c Describe the rate of change of the function when x < 0.

Apply the idea Let’s consider some values from the table: x y

−4

−3

−2

Based on the table, y increases by

from x = −4 to x = −3. Then y increases by 1 from x = −3 to x = −2.

This means that as x increases, y increases at a faster and faster rate.

Idea summary Two quantities, x and y, that are inversely proportional have an equation of the form: y=

or xy = k or x = ,

where k ≠ 0.

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11.01 Practice questions Understanding 1

Explorer 1–3, 4a–f

Adventurer 1–4

Is each statement true or false? a

In a directly proportional relationship, the constant of proportionality represents the gradient of the line.

The graph of any directly proportional relationship will go through the point (0, 0).

The graph of a directly proportional relationship can be curved.

A directly proportional relationship can be increasing or decreasing.

The equation y =

For an inversely proportional relationship, as x gets closer to zero, y gets further from zero.

represents a direct proportionality between two quantities, y and x.

Which graph indicates that y is directly proportional to x? A

−4 −3 −2 −1

−1

−4 −3 −2 −1

−2

−3

−4

−4 −3 −2 −1

−1

−2

x 1

−1

−2

554

3 1

Trailblazer 1–4

−4 −3 −2 −1

−1

x 1

−2

−3

−4

Determine whether each of the following is an example of a direct variation or an inverse variation: a

The variation relating the distance between two locations on a map and the actual distance between the two locations.

The variation relating the number of workers hired to build a house and the time required to build the house.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Does each relationship between x and y show direct proportion, inverse proportion, or neither? a

Ex 1

y = −x

x2 = y

y = 5x

y = 2x2 − 7x − 4

xy = −7

j 5–7, 8ab, 9, 11, 13a, 15, 16ab, 17, 18ab

5, 6, 8, 10–12, 13ab, 14, 15, 16cd, 17, 18

x = −2y

5, 6, 8cd, 10–15, 16cd, 17, 18

Given that y = 0.4x, ﬁnd the missing values in the table.

x y

10 8

Does each table represent a direct proportional relationship between x and y? a

x 1 3 5 7 y 20 16 12 8

x 1 2 3 4 y 5 20 45 80

x y

1 16

5 12

6 8

20 4

x y

1 5

2 10

3 15

4 20

q is directly proportional to r, and q = 10 when r = 20. a

Graph the relationship between q and r.

Find the gradient of the line.

Express q in terms of r.

Find the value of q when r = 30.

Find the value of r when q = 22.

Find the equation relating the directly proportional relationship using variables in each table. a

a 0 1 2 3 b 0 9 18 27

n 0 3 6 9

r Ex 2

y = 6x

Fluency 5

x y

0 0

1 −2

2 −4

3 −6

Given A = 70w: a

What is the constant of proportionality?

Find the value of A when w = 4.

Given M =

What is the constant of proportionality?

Find the value of M when n = 50.

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555

William is making rice. The amount of rice and water he uses is in the table. Rice (g) Water (mL)

Ex 3

100 210

110 231

120 252

130 273

140 294

Sketch the graph of the proportional relationship.

Find the unit rate of the relationship between rice and water.

Write two statements that describe the proportional relationship.

The table shows the proportional relationship between hours spent ﬁshing and the number of ﬁsh caught: a

Sketch the graph of the proportional relationship.

Find the unit rate of the relationship.

By considering parts (a) and (b), make a conclusion about the number of ﬁsh caught.

Hours Fish caught

4 9

556

12 27

For each table: i

Sketch the graph.

x 10 20 30 y 7.5 15 22.5

x 15 25 35 45 y 36 60 84 108

40 30

If s varies directly with t, and s =

50 37.5 55 132

Find the unit rate of the relationship. x y

5 9.5

10 19

x y

1 1.25

3 3.75

15 28.5

20 38

5 6.25

7 8.75

25 47.5 9 11.25

when t = 6:

Find the constant of proportionality, k.

Write the equation for s in terms of t.

If y is inversely proportional to x, and y = 20 when x = 10: a

Find the constant of variation, k.

Express y in terms of x.

Find the value of y when x = 5.

The equation xy = 7 can be described as a function where y varies inversely with x and has a constant of proportionality, k = 7. Find the constant of proportionality, k, for the following:

8 18

y = 4 when x = 3

when x = −2

y = −2 when x = 3

when x = −3

Does each table represent an inversely proportional relationship between x and y? a

x 1 2 3 4 y 3 1.5 1 0.75

x 1 5 6 10 y 3 75 108 300

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x y

1 36

2 18

3 12

4 9

x y

1 4

2 5

3 6

4 7

Ex 4

The equation

represents an inverse relationship between x and y.

Complete the table of values:

Describe the rate of change of the function when x < 0.

x y

increasing or decreasing when x < 0?

Reasoning

19, 20

−5

20, 21

−4

−3

−2

−1

21–23

Two quantities are directly proportional. If one of the quantities increases by a factor of 7, what happens to the other quantity?

Jimmy and Ben are both saving money for a vacation. Jimmy can represent his savings over time using the equation

, where x represents the number of days and y represents the

savings. Ben saves $7 every 4 days. a

Complete the table showing Ben’s savings over time. Days Amount saved

8 14

12 21

Graph the amount each of them have saved over time.

Explain how to determine who is saving more per day.

A real estate agent advertised a house for sale online and in the newspaper. The agent received 4 online enquiries each day and 2 enquiries each day through the newspaper advertisement. a

d 22

Let y represent the number of online enquiries received x days after initially advertising the house. Write the equation for y based on x.

The line representing the number of online enquiries has been graphed. On the same set of axes, graph the line representing the number of newspaper enquiries after x days.

Why is the line representing the number of online enquiries steeper than the line representing the number of newspaper enquiries? Why do both lines pass through the origin?

y 35 30 Online

20 15 10 5

x 2

, where f is the centripetal force of an object of mass m moving along a circle of

radius r at velocity v, describe the relationship between the centripetal force and the mass. 23

Boyle’s law describes the relationship between pressure and volume of a gas of ﬁxed mass under constant temperature. The pressure for a particular gas can be found using

where P has units kg/cm2 and V has units cm3. a

What is the pressure if the volume is 1 cm3?

What happens to the pressure as the volume increases? Explain.

Can the pressure ever get to 0 kg/cm2? Explain.

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Problem-solving 24–27 24

25, 28–30

28–31

Harry and Carl love reading. Harry reads 16 books every 12 weeks. Carl has kept a table of his reading habits: Number of weeks Number of books read a

12 20

24 40

36 60

48 80

12 16

Complete this table for Harry: Number of weeks Number of books read

60 80

Who reads faster?

Derek and Yuri are building robot kits. Derek can make 2 robots every 3 hours. Yuri can make 3 robots every 6 hours. a Sketch the graphs of both proportional relationships on the same Cartesian plane. b Who is the faster builder?

The height, h, of a regular tetrahedron varies directly with its side length, s. A particular tetrahedron with a side length of 3 cm has a height of 2.45 cm. a Find the constant of proportionality, k, rounded to two decimal places. b Express h in terms of s. c Find the height of a different tetrahedron with a side length of 2 cm.

The amount A (in grams) of a substance dissolved in water is directly proportional to the capacity C (in litres) of water used. When 5 litres of water is used, 15 grams of the substance is dissolved. a Find an equation relating the amount and capacity. b How much of the substance will be dissolved in 7 litres of water?

The time, t, taken by a typist to type up a document is inversely proportional to his typing speed, s. That is, the quicker the typing speed, the less time it will take. It takes a typist 20 minutes to type a particular document, typing at a speed of 61 words per minute. a Find the constant of variation, k. b How long, in minutes, will it take a typist with a typing speed of 30.5 words per minute to type up the document?

Write an equation involving constant k for the revenue of a company, r, that varies directly with the number of sales, n.

The values in the table represent the relationship between m and p, where m is inversely proportional to p. Find the values of a and b. p m

4 63

6 b

7 36

a 28

A particular shelter has cylindrical pillars that can hold a maximum load, L, that varies directly as the fourth power of the diameter, d, and inversely as the square of the height, h. If 6 m tall pillars with diameters of 1 m can support a maximum load of 18 metric tons, ﬁnd the maximum load that can be supported by a pillar 2 m tall and

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m in diameter.

11.02

Hyperbolas

After this lesson, you will be able to… • define what a hyperbola is and how it represents an inverse relationship. • identify key features of hyperbolas and use them to graph hyperbolas. • determine the domain and range of hyperbolas, and understand the restrictions on these. • apply the knowledge of hyperbolas to real-world situations.

Hyperbolas When the inverse relationship call a hyperbola.

is graphed on the Cartesian plane it produces a graph that we

Exploration Here are some hyperbolas with equations of the form 6 y y=

−3 x

−4

−1 y= x −6

. 5 x

2 x x

−2

Horizontal asymptote

2 −2 −4

Vertical asymptote

−6

What key features do the graphs have in common?

What key features are different?

What values of k result in a graph that appears in quadrants 1 and 3?

What values of k result in a graph that appears in quadrants 2 and 4?

Ignoring the sign, what happens when the size of the constant of variation increases?

This table shows the relationship x y

−4 −0.25

−2 −0.5

−1 −1

−0.5 −2

: −0.25 −4

0 undefined

0.25 4

0.5 2

1 1

2 0.5

4 0.25

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This shows the points from the table plotted to form the graph for 5 4 3 2 1 -5-4-3-2 -1 -1 -2 -3 -4 -5

Notice that for y=

1 x

1 2 3 4 5

. :

• If x is positive, y is positive. • If x is negative, y is negative. • As x gets further from zero (in either direction), y gets closer to zero. • As the value of one variable increases, the value of the other will decrease. • The graph is split in two pieces and not deﬁned at x = 0

Formally, we deﬁne that an inversely proportional relationship is modelled by:

where x and y are any variable, and k is the constant of proportionality. The graph y = 0.

has two asymptotes, a vertical asymptote at x = 0 and a horizontal asymptote at A vertical asymptote is represented by a dotted vertical line. As the y-values head towards positive or negative infinity, the graph approaches the asymptote but does not cross it.

Vertical asymptote

To find the vertical asymptote, examine the range to determine undefined values of x.

A horizontal asymptote is represented by a dotted horizontal line. As the x-values head towards positive or negative infinity, the graph approaches the asymptote but does not cross it.

Horizontal asymptote

To find the horizontal asymptote, examine the domain to determine non-existent values of y. We can also examine the end behaviour of the curve. Look at what value y approaches as x approaches positive or negative infinity. Let’s return to the graph of

is undefined

if x = 0. This means its domain is all real values of x, x ≠ 0, and that there is vertical asymptote at x = 0.

Vertical asymptote

Horizontal asymptote

560

. Notice that

Also notice that the domain excludes the value of 0. This means that there is a horizontal asymptote at y = 0. We can also find the horizontal asymptote by exploring the end behaviour. The end behavior is y → 0 as x → −∞ and x → ∞, confirming that there is a horizontal asymptote at y = 0.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For

: 6 −3 y= x

y = −1x

−6

−4

y y=

5 x

2 x x

−2

−2 Horizontal asymptote

Vertical asymptote

−4 −6

Key features of basic hyperbola

• Each hyperbola has two parts and they lie in diagonally opposite quadrants. • Vertical asymptote at x = 0 • Horizontal asymptote at y = 0 • The larger the size of the constant of variation (k), the further the graph is from the axes. • The domain is all real values of x, x ≠ 0. • The range is all real values of y, y ≠ 0. • The graphs are symmetrical about the lines y = x and y = −x. • The graphs have rotational symmetry about the origin.

Example 1 Consider the equation

a What is the value of the constant of variation?

Create a strategy

Apply the idea

We want to write this in the form the value of k.

and ﬁnd

We can write this as:

b Complete the following table of values:

x y

−3

, so

−2

−1

Apply the idea For x = −3: Write the equation Substitute x = −3

Evaluate

Similarly, by substituting the remaining x-values into x

−3

−2

−1

, we get:

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c Sketch the graph.

Apply the idea The ordered pairs of points to be plotted on the coordinate plane are

5 4

, and

3 2

, which are plotted on the graph.

1 -5 -4 -3 -2 -1 -1

-2 -3

The curve of the hyperbola

must pass

through each of the plotted points and approaches the horizontal asymptote y = 0 and the vertical asymptote x = 0.

-4 -5

Reﬂect and check Notice that the graph is quite close to the axes since the constant of variation is quite small. Also, the graph is in quadrants 2 and 4 since the constant of variation is negative. d In which quadrants does the graph lie?

Apply the idea From the graph in part (b), we see that the hyperbola lies in quadrant 2 and quadrant 4.

Example 2 5

Consider the graph of y = .

4 3 2 1 -5 -4 -3 -2 -1 -1

a For positive values of x, as x → ∞, y approaches what value? A C

−∞

-3

-4

∞

-5

Create a strategy Substitute a large value of x into the equation.

562

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea Let’s say we use x = 1000 so we have: y=

2 x

Write the equation

2 = 1000 = 0.002

Substitute x = 1000 Evaluate

2 If we keep increasing the value of x, the fraction y = becomes smaller and smaller, causing the value x of y to approach 0. So, the answer is option A. b As x takes small positive values approaching 0, (x → 0), what value does y approach? A

∞

−∞

Create a strategy Substitute a small value of x into the equation.

Apply the idea Let’s say we use x = 0.001 so we have: y=

2 x

Write the equation

2 = 0.001 = 2000

Substitute x = 0.001 Evaluate

As x approaches 0 from the positive side, the fraction value of y to approach ∞. So, the answer is option A.

becomes larger and larger, causing the

c What are the values that x and y cannot take?

Create a strategy Examine the equation and its graph to identify any restrictions on the values of x and y, that is, examine the domain and range.

Apply the idea The equation is

. Since a division by zero is undeﬁned, x cannot take the value 0.

Additionally, from the graph, we can observe that y never touches the x-axis, that is, y can never be equal to 0. Therefore, the values of x and y cannot be equal to 0.

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d The graph is symmetrical across two lines of symmetry. State the equations of these two lines.

Create a strategy Observe the graph and identify the lines along which the graph’s symmetry occurs.

Apply the idea The graph of

is symmetric with respect to the lines y = x and y = −x. These lines represent

the equations of the lines of symmetry.

Idea summary A reciprocal equation (inverse relationship) that models inverse variation is the .

hyperbola,

Key features of basic hyperbola • • • • • • • •

Each hyperbola has two parts and they lie in opposite quadrants. Vertical asymptote at x = 0 Horizontal asymptote at y = 0 The larger the size of the constant of variation (k), the further the graph is from the axes. The domain is all real values of x, x π 0. The range is all real values of y, y π 0. The graphs are symmetrical about the lines y = x and y = –x. The graphs have rotational symmetry about the origin.

11.02 Practice questions Understanding 1

Explorer 1–3

Which is a graph of the hyperbola in the form A

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

564

Adventurer 1–3 ? B

Trailblazer 1–3

x 1 2 3 4 5

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

For the given hyperbola, complete the statement:

5 4 3 2 1

“Every point (x, y) is on the hyperbola such that xy = ⬚.”

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

For equations of the form

Fluency 4

, what is the equation of the vertical asymptote?

4, 7, 8, 10ac, 11, 12ab

Consider the graph of

1 2 3 4 5

4, 5, 8, 9, 10bc, 11, 12bd

4–6, 8, 10cd, 11, 12b–d, 13

What are the equations of the horizontal and vertical asymptotes?

What are the domain and range?

In which quadrants does the graph lie?

Identify the intervals where the graph is increasing or decreasing.

Describe the end behaviour as x → ∞ and as x → −∞.

3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

5 For the equation y = − . x a For what value of x is the relationship undeﬁned? b

As x approaches 0 from the positive side, what does y approach?

What y-value does the graph get inﬁnitely close to, but never reach?

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The equation a

Which equations are equivalent to i

Ex 1

represents an inverse relationship between x and y.

x = 8y

xy = 8

iii

y = 8x

Can x or y be equal to 0?

When x = 2, what is the value of y?

If x is a positive value, must the corresponding y value be positive or negative?

If x is a negative value, must the corresponding y value be positive or negative?

In which quadrants does the graph of

lie?

For each equation: i

Complete the table of values. x

−4

−2

−1

Ex 2

Plot the points from the table of values on a Cartesian plane and sketch the graph of the hyperbola.

iii

In which quadrants does the graph lie?

Without graphing, determine the behaviour of each of the following hyperbolas: i

As x tends to 0 from the right (x → 0+)

As x tends to 0 from the left (x → 0−)

iii

As x tends to positive inﬁnity (x → ∞)

As x tends to negative inﬁnity (x → −∞)

Consider the equations

and

b .

For

, when x = 1, what is the y-value?

For

, when x = 1, what is the y-value?

Which graph lies further away from the axes?

For hyperbolas of the form

, what happens to the graph as the value of k increases?

Sketch the graph, label the asymptotes and two points on the curve. a

566

3 x

5 2x

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

y=−

4 x

y=−

2 3x

The graphs of y1 = , y2 = , and y3 =

are shown. 4

Match each graph with its equation.

3 2 1 −4 −3 −2 −1

−1

−2 −3 −4

Find the equation of each hyperbola. a

4 3

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−2

−3

−4

−4 −3 −2 −1

1 −1

−2

x 1

−1

−2

−4 −3 −2 −1

−1

x 1

−2

−3

−4

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Delano is sharing a jar of jelly beans evenly between his Jelly beans friends. The graph shows the relationship between the 50 45 number of people receiving jelly beans and the number 40 of jelly beans they receive. Determine whether there is an inverse variation between the number of people and the number of jelly beans they receive. If so, ﬁnd the constant of variation.

Reasoning 14

14–16

35 30 25 20 15 10 5

14–17

People 5 10 15 20 25 30 35 40 45 50

14, 16–18

Would an equation of the form xy = k be a reasonable model for the following relationships? Explain why or why not. a

The relationship between the number of people working on a job and how long it will take to complete the job.

The relationship between the number of sales and the amount of revenue.

The relationship between the minutes of cycling and the amount of calories burned.

Ursula wants to sketch the graph of y = , but knows that it will look similar to many other hyperbolas. What can she do to the graph to show that it is the hyperbola y = , rather than any other hyperbola of the form

For y = . a

Sketch a graph of the hyperbola on a Cartesian plane.

Substitute −x for x and simplify the expression.

How could you rotate or reﬂect the graph and have it look the same as the graph of the expression from part (b)?

For the graph of y = : a

Show that the line y = x is an axis of symmetry.

The graph has another axis of symmetry. What is the equation of the other one?

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

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x 1 2 3 4 5

Consider the hyperbola. a

Every point (x, y) is on the hyperbola such that xy = k. What is the value of k?

When x increases, does y increase or decrease?

If the graph was reﬂected over the x-axis, what would the new equation be? Explain.

Problem-solving 19, 20 19

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

20–23

x 2 4 6 8 10

20–23

The rent, electricity, telephone bill and other expenses for a ﬂat costs a total of $490 per week. These expenses are shared equally between the tenants of the ﬂat. a

How much will each occupant pay in dollars if the ﬂat is shared by two people?

Let the number of occupants be x, and the cost paid by each occupant be y. Write a formula that relates the two variables.

Use this equation to ﬁll in the following table of values. x y

10 8 6 4 2

Sketch the curve from your equation.

What type of relationship exists between x and y?

The relationship between the current, C, (in amperes) and resistance, R, (in ohms) in an electrical circuit is given by:

where the voltage provided to the circuit is 240 V.

Graph this relationship for 0 ≤ R ≤ 50.

What happens to the current as the resistance increases?

Ella is driving down the highway to her home. To drive safely, the speed limit is no more than 80 km/h and no less than 60 km/h. Ella is 30 kilometres from home and has already been driving for 0.4 hours. a

The relationship between speed, distance, and time is:

Write an equation relating y, the total time of Ella’s drive in hours, and x, Ella’s driving speed on the highway. b

State the domain of the relationship using interval notation.

State the range of the relationship using interval notation.

Determine the number of minutes Ella’s drive home will be if she drives at 68 km/h on the highway.

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Consider the hyperbola that has been graphed. Points A (4, 2), B(4, −2), C (−4, k) and D(−4, 2) form the vertices of a rectangle. a

Find the value of k.

Find the area of the rectangle with vertices ABCD.

y 6 4 D

x −8 −6 −4 −2 −2

−4 −6

The time it takes a commuter to travel 100 km depends on how fast they are going. We can write this using the equation hours.

where S is the speed in km/h and t is the time taken in

Graph the relationship

What is the time taken if the speed travelled is 10 km/h?

What is the time taken if the speed travelled is 50 km/h?

If we want the travel time to decrease, what must happen to the speed of travel?

11.03

Optional Transformations of hyperbolas

After this lesson, you will be able to… • explore the effects of parameters on the graph of the hyperbola . • identify the transformations that result from changes in the parameters. • determine the vertical and horizontal asymptotes of transformed hyperbolas. • apply the understanding of transformations to solve problems involving hyperbolas.

Transformations of hyperbolas The graph of

is the basic hyperbola which has a vertical asymptote at x = 0 and a horizontal

asymptote at y = 0.

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Interactive exploration

Discover this concept in action online

mathspace.co

Dilation, stretch and compression (a) The hyperbola y =

a has been dilated by a factor of a. x 4

y y=

4 x

3 y= - 1 x 2

2 1 −4 −3 −2 −1

x 1

−1

-4 -3 -2 -1

x 1

-1

−2

-2

−3 y = 1 x −4

-3

y= -

2 x

-4

stretched vertically away from x-axis by a factor of 4 to get y = 4 . x 2

1 stretched vertically away from the x 2 x-axis by a factor of 2 to get y = - . x

y=−

y y=

1 x

y y=−

1 2x x

x −2

−1

−2

−1

−1 −2

−1

1 y= 5x

compressed vertically towards the 1 x-axis by a factor of 1 to get y = . 5x 5

−2

y = − x1

1 compressed vertically towards the x x-axis by a factor of 1 to get y = − 1 . 2x 2

y=−

a : x • If a > 1, it is stretched vertically by a factor of a. This creates the effect of the graph being pulled away from the x-axis. • If a < -1, it is stretched vertically by a factor of a and reflected across the x-axis. • If -1 < a < 1, it is compressed vertically by a factor of a. This creates the effect of the graph being pulled toward the x-axis.

In general, for y =

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Horizontal translation (h) The hyperbola y =

1 has been horizontally translated h units from x−h 4

4 3

3 1 y= x+2

1 y= x

2 1 −4 −3 −2 −1

2 1

x 1

−1

−2 −3

1 x−3

−4 −3 −2 −1

1 x x

−1

−2 −3

−4

translated horizontally right by 3 units to get y = 1 . x−3 1 In general, for y = : x−h • If h > 0, it is translated h units to the right. • If h < 0, it is translated h units to the left.

translated horizontally left by 2 units to get y = 1 . x+2

Vertical translation (k) The hyperbola

+ k has been vertically translated up by k units from 4

y y=

3 2 1 -4 -3 -2 -1

-1

1 +2 x

1 x

4 2

-1

-2 -3

-4

+ k:

• If k > 0, it is translated up k units. • If k < 0, it is translated down k units.

Transformations A hyperbola in the form f ( x) =

572

-4 -3 -2 -1

-3

a + k: x−h

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 4

translated vertically up by 2 units

In general, for

-2

to get to

1 x

3 4 1 y= –3 x

translated vertically down by 3 units to get to

• If a > 1, is stretched vertically (away from the x-axis). • If a < -1, is stretched vertically and reflected across the x-axis. • If -1 < a < 1, is compressed vertically (toward the x-axis). • Has a vertical asymptote at x = h • Has a horizontal asymptote at y = k

Example 1 This is a graph of

3 2

−4 −3 −2 −1

1 x

−1

−2 −3 −4

a How do we shift the graph of

to get the graph of

Create a strategy

Apply the idea

A vertical translation occurs when a constant value is added or subtracted at the end of the equation.

In this case,

represents a graph that

has been translated vertically upwards by 3.

A horizontal translation occurs when a value is added or subtracted with the x variable in the denominator. b Draw

on the same graph as

Create a strategy

Apply the idea

Based on our answer to part (a), the graph

Each point on the graph of up by 3.

should show

translated vertically up by 3.

must be shifted

1 +3 x

1 x

3 2 1 −4 −3 −2 −1 −1

x 3

−2 −3

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Example 2 Describe the transformation that has occurred from

to the graph of g

Create a strategy Compare the function g

to the general form f ( x) =

a + k. x−h

Apply the idea There are no horizontal or vertical translations (h = 0 and k = 0). Since a = −3, the graph of across the x-axis.

has been stretched vertically by a factor of three and reﬂected

Reﬂect and check We can check our answer by sketching the graph of g x

−3

−2

−1

undefined

−3

6 5 4 3 2 1 −6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6

using a table of values.

3 −1 When the graph of g graph of

1 x

1 2 3 4 5 6

y=−

3 x

is compared to the

, we can see that for each x-value,

the point is 3 times as far away from the x-axis, but on the other side of the x-axis. This is most obvious if we compare (1, 1) on with (1, −3) on g

This conﬁrms that the graph has been stretched vertically by a factor of three and reﬂected across the x-axis.

Example 3 What is the equation of a hyperbola whose graph started out as

, was vertically translated

up by 6 units, horizontally translated to the left by 7 units, reflected across the x-axis and was 1 compressed vertically by a factor of ? 2

Create a strategy Identify all the components of transformations which are the values and the corresponding a parameter from the standard equation y = + k. x−h

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Apply the idea

1 The value of a represents the vertical dilation. The graph is “compressed vertically by a factor of ” 2 1 so a = . Additionally, since the graph is “reflected across the x-axis,” a becomes negative, 2 1 giving a = − . 2 The value of h corresponds to the horizontal translation. Since the graph is “horizontally translated to the left by 7 units” means that h = –7. The value of k represents the vertical translation. Since the graph is “vertically translated up by 6 units,” it means that k = 6. Putting this altogether, we get: a +k x−h

Write the equation

−1 2 = +6 x − ( −7)

Substitute values

1 +6 2( x + 7)

Simplify

Idea summary A hyperbola in the form f (x)y =

a + k, compared to x−h

• • • •

If a > 1, has been stretched vertically (away from the x-axis). If a < −1, has been stretched vertically and reﬂected across the x-axis. If −1 < a < 1, has been compressed vertically (toward the x-axis). Has been horizontally translated by h units, so has a vertical asymptote at x = h. • Has been vertically translated by k units, so has a horizontal asymptote at y = k.

11.03 Practice questions Understanding 1 2

Explorer 1–3

Adventurer 1–3

Trailblazer 1–3

What is the equation for a hyperbola with a vertical translation of c units up?

What is the equation for a hyperbola with a horizontal translation of b units to the right?

In each case, where would the point (1, 3) be translated to? a

Vertical translation up 4 units

Horizontal translation 2 units to the left

Vertical translation down 10 units

Vertical translation up 3 units, then 5 units to the right 11.03 Optional Transformations of hyperbolas mathspace.co

575

Is each statement true or false for a hyperbola in the form y = a

The vertical asymptote is y = h.

The horizontal asymptote is y = k.

If a < −1, the graph has been reﬂected across the x-axis.

The asymptotes cross at (k, h). 4ab, 5, 7ac, 9, 10, 12ab, 13, 16a, 18

Fluency Ex 1

This is a graph of

a + k? x−h

4bc, 6ab, 7bc, 8, 10, 12ac, 14, 16bc, 17a, 18

4cd, 6cd, 7bd, 8, 11, 12cd, 15, 16, 17b, 18, 19

. For each equation:

Describe the transformation from

Sketch the graph on the same Cartesian plane as

3 2

. a

For the function y = a

−4 −3 −2 −1 −1 −2

-2

−3

1 : x−4

Complete the table of values. x

7 2

9 2

Ex 2

Sketch the graph.

Describe the transformation from

576

For each equation: i

Describe the transformation from

Sketch the graph on the same Cartesian plane as

1 x+5

For each equation: i

Describe the transformation from

Sketch the graph on the same Cartesian plane as

1 x

−4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

y=−

. 1 x

x 3

For the function a

How can the graph of f (x) be obtained from the graph of

What is the equation of the vertical asymptote?

What is the equation of the horizontal asymptote?

Sketch the graph of y = f (x).

For the function

What is the equation of the vertical asymptote?

What is the equation of the horizontal asymptote?

Sketch the graph of the function.

For the hyperbola

How would the hyperbola relate to its asymptotes? A

What are the equations of the vertical and horizontal asymptotes of

Sketch the graph of

For the hyperbola a

. :

How would the hyperbola relate to its asymptotes? A

What are the equations of the vertical and horizontal asymptotes of

Sketch the graph of

For the following functions: i

Write the equation of the vertical asymptote.

Write the equation of the horizontal asymptote.

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For the graph of the hyperbola

What would be the new equation if the graph was shifted upwards by 4 units?

What would be the new equation if the graph was shifted to the right by 7 units?

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

For the graph of the hyperbola

What would be the new equation if the graph was shifted down by 5 units?

What would be the new equation if the graph was shifted to the left by 9 units?

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Ex 3

1 2 3 4 5

x 1 2 3 4 5

What is the equation of a hyperbola whose graph started out as , was vertically translated up by 3 units, horizontally translated to the right by 5 units, reflected across the 1 x-axis, and was compressed vertically by a factor of ? 3

For each graph: i

State the vertical and horizontal asymptotes.

Describe the transformation from

iii

Write the equation of the transformed function.

578

1 −4 −3 −2 −1

(shown in blue and dashed).

−1

x 1

−4 −3 −2 −1

−1

−2

−3

−4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1

1 −4 −3 −2 −1

−1

x 1

−4 −3 −2 −1

−2

−3

−4

−4 −3 −2 −1

−1

−4 −3 −2 −1

−2

−1

x 1

−2

−3

−4

Each table of values represents a transformation of

Determine the transformation of the graph of table.

which would give the values in the

Write an equation for f (x).

−3

-2

f (x) b

−7

−6

f (x) 18

x 1

−1

−2

-1

undefined

−5

−4

−3

−1

undefined

−2

−1

For the hyperbolas:

What value cannot be substituted for x in

In which quadrants does

What value cannot be substituted for x in

In which quadrants does

How can the graph of

lie? ?

lie? be transformed to create the graph of

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Without graphing, determine the behaviour of each of the following hyperbolas as: i

As x tends to 0 from the right, x → 0+

As x tends to 0 from the left, x → 0−

iii

As x tends to positive inﬁnity, x → ∞

Reasoning 20

20, 21

Arik is asked to draw the graph of graph shown.

20–22

. He drew the 4

Identify and correct his error.

2 1

−3 −2 −1 −1

−2 −3 −4

Determine whether each statement is true or false. Justify your answer. a

A hyperbola with a horizontal translation has the equation

If a hyperbola is translated vertically down by 4 units, the equation becomes

A horizontal translation of 2 units to the left affects the constant value in the numerator.

If a hyperbola has the equation

a + k: x−h What are the equations of the asymptotes of f (x)?

What are the domain and range of f (x)?

For a hyperbola in the form f ( x) =

24–26

Consider the table of values for f (x) and the graph of g(x) = f (x) + k. x f (x)

−3 −5

−1 −8

0 undefined

2 7

4 3

Determine the type of transformation applied to f (x) to obtain g (x).

Determine the value of k.

24–27

10 8 4

g (x)

2 -10 -8 -6 -4 -2 -2 -4 -6 -8

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

-10

580

, it has been vertically translated by 2 units.

Problem-solving 23, 25 23

x 2

8 10

Consider the table of values for f (x) and the graph of g (x) = f (x + k). x f (x)

−2 −6

−1 −12

0 undefined

1 12

2 6

g (x)

Determine the type of transformation applied to f (x) to obtain g (x).

Determine the value of k.

−12 −9 −6 −3 −3

9 12

−6 −9 −12

A graph of the hyperbola

is shown:

Given points C (−4, 0) and D (2, 0), ﬁnd the length of AB.

D B

Guido is organizing a yearbook order this year for his soccer club. There is a ﬁxed fee of $400 plus a cost of $25 per yearbook. The ﬁxed fee is evenly divided across everyone who orders a yearbook. a

Write the equation of a function that gives the cost per yearbook ( y) in terms of the number of yearbooks purchased (x).

There is a minimum purchase of 5 yearbooks. Draw the graph which represents this situation.

If the yearbook committee wants to get the price per yearbook down to $30, determine how many people would need to purchase a yearbook.

Explain why the cost will always be higher than $25.

A yoga studio offers classes on a pay what you can model. A third of the students paid $20 each for the class, a third of the students paid $7 each for the class, and a third of the students made various donations totalling $60. Determine the lower limit of the average amount paid per student for very large classes. Justify your answer.

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11 Chapter review 1

Determine whether the following is an example of a direct variation or an inverse variation: a

The time taken to dig a hole and the number of people digging

The number of macarons sold and the revenue earned

Does each table show a direct or inverse relationship between x and y? a

What is the constant of proportionality?

x y

1 6

3 2

4 1.5

Find the value of a when b = 6.

50 95

60 114

70 133

80 152

90 171

Sketch the graph of the proportional relationship.

Find the unit rate of the relationship between quinoa and water.

Write two statements that describe the proportional relationship.

If y is inversely proportional to x, and y = 30 when x = 5: a

Explain how we can ﬁnd the constant of proportionality, k.

Express y in terms of x.

Find the value of y when x = 10.

The equation

represents an inverse

relationship between x and y.

2 3

Rose is making quinoa. The amount of rice and water she uses is in the table. Quinoa (g) Water (mL)

Given a = 30b: a

x 1 2 3 4 y 6 12 18 24

Complete the table of values.

Describe the rate of change of the function when x < 0.

x y

−5

−4

−3

−2

−1

increasing or decreasing when x < 0?

At ProPainters, painters must paint at a minimum rate of 60 m2/hr. The equation represents the time, t, it takes p painters to paint the outside of a house with a surface area of 1200 m2. a b

582

Find the constant of proportionality, k.

Find the time it would take 4 painters to ﬁnish painting 1200 m2.

For the equation

For what value of x is the relationship undeﬁned?

As x approaches 0 from the positive side, what can we say about the value of y?

What y-value does the graph get inﬁnitely close to, but never reach?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

For each equation: i

Complete the table of values. x

−4

−2

−1

Plot the points from the table of values on a Cartesian plane and sketch the graph of the hyperbola.

iii

In which quadrants does the graph lie?

Without graphing, determine the behaviour of each hyperbola: i

In which quadrants does the graph lie?

As x tends to 0 from the right, x → 0+

iii

As x tends to 0 from the left, x → 0

As x tends to positive inﬁnity, x → ∞

−

Sketch each graph, labelling the asymptotes and two points. a

5 y = x

y=−

1 4x

Find the equation of each hyperbola. a

4 3

3 (1, 2)

1 −4 −3 −2 −1 (−1, −2)

−1

(1, 1.25)

x −4 −3 −2 −1

−2

(−1, −1.25)

−1

−3

−4

Determine whether or not there is an inverse variation between the number of winners and the prize money each receives. If so, ﬁnd the constant of variation.

55 50 45 40 35 30 25 20 15 10 5

−2

−3

In an art competition, the total prize money is distributed evenly among the winners. The graph shows the relationship between the number of winners and the prize money each winner receives.

x 1

Prize money

No. of winners 5 10 15 20 25 30 35 40 45 50 55

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The total costs of a seminar, including venue hire, materials, and refreshments, amount to $560. These costs are shared equally among the participants. a

What is the cost per participant if the seminar is attended by two people?

Let the number of participants be x, and the cost per participant be y. Formulate an equation to represent this relationship.

Complete the table with values based on the equation, rounding to two decimal places when needed. 1

x y

Sketch the graph based on the equation.

Describe the relationship between x and y.

(Optional) This is a graph of

For each equation:

Describe the transformation from

Sketch the graph on the same Cartesian plane as

−4 −3 −2 −1 −1

x 3

−3 −4

(Optional) Without graphing, determine the behaviour of each hyperbola: i

As x tends to 0 from the right, x → 0+

iii

As x tends to positive inﬁnity, x → ∞

As x tends to 0 from the left, x → 0−

(Optional) Tetay is asked to draw the graph of She drew the graph shown. Identify and correct her error.

. 4 2

−1 −1 −2 −3 −4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3 1

584

1 x

−2

c 16

. a

x 1 2 3 4 5 6 7 8 9 10 11 12 13

(Optional) Emilio is managing a custom footy guernsey order for his basketball team. There is an upfront design fee of $500 plus a cost of $30 per guernsey. The design fee is split equally among all orders. a

Write the equation of a function that represents the cost per guernsey, y, as a function of the number of guernseys ordered, x.

Assume the team decides on a minimum order of 10 guernseys. Draw the graph which represents this situation.

To lower the cost per guernsey to $35, how many guernseys need to be ordered?

Explain why the cost cannot go below $30.

Did you know?

Hyperbolas are more than just shapes in maths, they help guide space exploration! The paths of some spacecraft are actually hyperbolic, especially when they’re slingshotting around planets for speed. This hyperbolic trajectory allows them to escape a planet’s gravitational pull and head toward distant destinations. Hyperbolas also show up in everyday technology: they help GPS systems calculate exact locations by measuring distances from satellites. So, the next time you see a hyperbola in maths class, remember it’s a shape that helps us explore the universe and find our way on Earth!

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Big ideas • Right triangles have special properties that allow their side lengths to be determined using the Pythagoras’ theorem or ratios. • Trigonometry connects the angle measures and side lengths of a triangle and provides a set of functions that can be used to find unknown angle measures and side lengths.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10M03 solve practical problems by applying Pythagoras’ theorem and trigonometry to right-angled triangles, including problems involving direction and angles of elevation and depression VC2M10ASP02 establish the sine, cosine and area rules for any triangle and solve related problems VC2M10ASP05 apply Pythagoras’ theorem and trigonometry to solving three-dimensional problems in right-angled triangles

12 Trigonometry Chapter outline 12.01 Pythagoras’ theorem 12.02 Trigonometric ratios 12.03 Angles of elevation and depression 12.04 Bearings 12.05 Optional: Sine and cosine rules 12.06 Optional: Pythagoras in 3D 12.07 Optional: Trigonometry in 3D Chapter 12 review

588 597 611 620 636

656

Sundials, such as the famous Intihuatana of Machu Picchu, demonstrate the use of trigonometry to position the hour lines and determine the time of day!

12.01 Pythagoras’ theorem After this lesson, you will be able to… • apply Pythagoras’ theorem to identify right-angled triangles. • use Pythagoras’ theorem to find the length of the hypotenuse and other sides in a right-angled triangle. • apply knowledge of Pythagoras’ theorem to solve real-world problems.

Pythagoras’ theorem Hypotenuse

In a right-angled triangle, the largest angle in the triangle is 90°. The side opposite the right-angle will be the longest side. We call this side the hypotenuse.

Right angle

All three sides of a right-angled triangle are related by the equation shown. The two shorter sides will be called a and b, and the hypotenuse (the longest side) will be c.

a2 + b2 = c2 a, b, c

Sides of a right-angled triangle

a, b

The lengths of the shorter sides

The length of the hypotenuse

This relationship between sides in a right-angled triangle is called Pythagoras’ theorem. We can use this theorem to: • Check whether a triangle is a right-angled triangle • Find the length of the hypotenuse • Find the length of one of the shorter sides in a right-angled triangle

Interactive exploration

Discover this concept in action online

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mathspace.co

Example 1 Use Pythagoras’ theorem to determine whether this is a right-angled triangle.

16 9

a Let a and b represent the two shorter side lengths. First ﬁnd the value of a2 + b2.

Create a strategy Substitute the lengths of the two shorter sides and add them.

Apply the idea a2 + b2 = (9)2 + (16)2

Substitute the values of a and b

= 81 + 256

Evaluate the squares

= 337

Evaluate the sum

b Let c represent the length of the longest side. Find the value of c2.

Create a strategy

Apply the idea c2 = (18)2

Substitute the length of the longest side.

= 324

Substitute the value of c Evaluate

c Is the triangle a right-angled triangle?

Create a strategy

Apply the idea

Compare the results from parts (a) and (b) to determine if a2 + b2 = c2.

337 ≠ 324 The triangle is not a right-angled triangle because a2 + b2 ≠ c2.

Example 2 Which of the following three lengths represent the sides of a right-angled triangle? Select the two correct answers. A (6, 8, 13)

(5, 12, 13)

(2, 4, 6)

(3, 4, 5)

Create a strategy Add the squares of the two smaller numbers. If the result is the same as the square of the biggest number, then it’s a Pythagorean triple.

12.01 Pythagoras’ theorem mathspace.co

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Apply the idea The triple (a, b, c) is a Pythagorean triple if c2 = a2 + b2. 132 = 62 + 82

Substitute the sides from option A

169 ≠ 100

Evaluate both sides

132 = 52 + 122

Substitute the sides from option B

169 = 169

Evaluate both sides

62 = 22 + 42

Substitute the sides from option C

36 ≠ 20

Evaluate both sides

52 = 32 + 42

Substitute the sides from option D

25 = 25

Evaluate both sides

The correct answers are options B and D.

Example 3 Find the length of the unknown side as an exact value: a

Create a strategy We can use Pythagoras’ theorem: c2 = a2 + b2.

Apply the idea c 2 = a 2 + b2 2

Write the formula 2

c = 12 + 16

Substitute a and b

c = 144 + 256

Evaluate the squares

c = 400

Evaluate the sum

Take the square root of both sides

c = 20

590

Evaluate the square root

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Create a strategy Use Pythagoras’ theorem: a2 + b2 = c2.

Apply the idea We can substitute a = x, b = 10, and c = 20 into the formula: a2 + b2 = c2 2

Write the formula 2

x + 10 = 20

Substitute the values

x + 100 = 400

Evaluate the squares

x + 100 − 100 = 400 – 100

Subtract 100 from both sides

x = 300

Evaluate the difference

Take the square root of both sides

Reflect and check When finding a short side, the most important thing to remember is that the two given lengths need to go into different sides of the equation. If we substitute incorrectly, we will end up with the square root of a negative number (calculator error) or a side which is longer than the hypotenuse.

Example 4 A movie director wants to shoot a scene where the hero of the film fires a grappling hook from the roof of one building to the roof of another. If the first building is 37 m tall, the other building is 54 m tall and the street between them is 10 m wide, what is the minimum length, l, of rope needed for the grappling hook? Round your answer to two decimal places.

l 54 m 37 m

10 m

Create a strategy Use Pythagoras’ theorem: c2 = a2 + b2.

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Apply the idea Notice that we have a right-angled triangle, and the length of the grappling hook’s rope will be the hypotenuse. One of the short sides is the width of the street, 10, and the other short side is the difference in heights between the two buildings, 54 − 37. We can substitute a = 10, b = (54 − 37), and c = l into the formula: c2 = a2 + b2

Write the formula

l = 10 + (54 − 37)

Substitute the values

l = 10 + 17

Simplify the brackets

l = 100 + 289

Evaluate the indices

l = 389

Evaluate the addition

Take the square root of both sides

l ≈ 19.72

Evaluate the square root

Idea summary

Pythagoras’ theorem relates the three sides of a right-angled triangle. Sides a and b are the two shorter sides, and the longest side, called the hypotenuse, is c. We can also test to see if a triangle is right-angled by checking to see if its three sides satisfy a2 + b2 = c2.

To ﬁnd the hypotenuse: c2 = a2 + b2 To ﬁnd a shorter side, use: a2 = c2 − b2 or b2 = c2 − a2 We can take the square root of both sides to give us the following formulas:

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12.01 Practice questions Explorer

Understanding 1

1−3

Adventurer

Trailblazer

1−3

Write Pythagoras’ theorem in a form you could use to ﬁnd: a

The length of z.

The length of x.

Calculate the value of x, rounded to two decimal places. All values of x are positive. a

Ex 1

Ex 2

Ex 3

82 = 52 + x2

Use Pythagoras’ theorem to determine whether this is a right-angled triangle. a

Let a and b represent the two shorter side lengths. First ﬁnd the value of a2 + b2.

Let c represent the length of the longest side. Find the value of c2.

Is the triangle a right-angled triangle?

Fluency 4

x2 = 22 + 32

4a–c, 5a–d, 7ab, 9

4d–f, 5c–f, 6, 7bc, 8a, 9

4f–h, 5e–i, 6, 7cd, 8, 9

Determine whether each of the following sets of values are Pythagorean triples: a

(12, 16, 26)

(4, 8, 12)

(8, 15, 17)

(9, 12, 15)

(12, 16, 20)

(7, 7, 10)

(8, 11, 14)

(7, 24, 25)

Find the length of the unknown side as an exact value: a

d c

8 15

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f 5

A right-angled triangle has one side of length 6 and another side of length 4. Find the length of the hypotenuse.

A right-angled triangle has a hypotenuse of length 9 and another side of length 3. Find the length of the third side.

A right-angled triangle has a hypotenuse of length 17 and another side of length 14. Find the length of the third side.

Two poles, one of length 8 m, and the other of length 14 m, stand 8 m apart on level ground. Find the length of wire required to join the tops of the poles. 14 m 8m

Find the length of a diagonal of each shape, rounded to one decimal place where appropriate. a

10 cm

10 mm 5 mm

10 cm

21 m

10 cm 4 cm

20 m 16 cm

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Find the distance between the points P and Q, rounded to one decimal place. a

4 cm 4 cm 6 cm

20 mm

P Q

10 mm

4 cm

6 mm

9 Ex 4

18 mm Q

5 mm

Two buildings, one of height 300 m and the other of height 450 m, stand a distance apart on level ground. If a cable 300 m joining the tops of the buildings is 200 m long, ﬁnd, to the nearest metre, the distance between the buildings.

200 m 450 m ?

Reasoning

10, 12a, 12b(i)(ii)

10, 12

11, 12

Can a triangle with sides of length 3 cm, 5 cm and 7 cm be a right-angled triangle? Use Pythagoras’ theorem to justify your answer.

If AB = 10 cm, AD = 8 cm and ∠ADB is a right angle, show that D is the midpoint of BC.

A triangle with sides 3, 4, and 5 units is a right-angled triangle. a

If the triangle is enlarged by a factor of 3, does Pythagoras’ theorem still hold? Justify your answer.

Is a triangle with these sides a right-angled triangle? i

30, 40, 50

Problem-solving 13−16 13

35, 45, 55

iii

3k, 4k, 5k

13−17

Find the length of the diameter AB, of the circle.

Find the exact length x.

21a, 28a, 36a

iv 14−18

24 cm O

A x

7 cm B 9 cm

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595

Tricia’s house has outer dimensions as shown in the diagram:

A 5.6 m

A drainpipe needs to run from the vertex of the roof, indicated by point A in the diagram, vertically downwards until it is 30 cm above the ground. Find the length of pipe needed in metres, rounded to two decimal places.

5.6 m

3.6 m

10.2 m

Find the value of x, rounded to one decimal place. a

5 cm

Casper produces thin plastic triangles. a

If the triangles need to be edged by a material that costs 3 cents per cm, how much, in dollars, will it cost to edge 100 triangles?

If the triangles need to be painted with a coating that costs 2 cents per cm2, how much, in dollars, will it cost to coat 100 triangles.

20 cm

29 cm

The diagram shows four squares. The outer square has side lengths of 20 cm. The next square is formed by joining the midpoints of the outer square, the third square by joining the midpoints of the second square, and the fourth square by joining the midpoints of the third square. Tamsin is using wire to construct the shape in this diagram. Find, to the nearest centimetre, the length of wire she requires.

Angelia and Beth start their hikes from the same starting point. Angelia walks 900 m south and then 900 m east, before stopping for a lunch break. Beth stops for a break having walked 1200 m south and then 600 m east. If they both walked back to the start point in a straight line, who would walk furthest, and by how far to the nearest metre?

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12.02 Trigonometric ratios After this lesson, you will be able to… • identify and define sine, cosine, and tangent, and describe the relationship between angles and sides in right-angled triangles. • use trigonometric ratios to find an unknown side or angle in right-angled triangles. • use trigonometric ratios with other geometric principles and formulas to find missing lengths, distances, and angles.

Trigonometric ratios In right-angled triangles, we can use trigonometric ratios to describe the relationship between a given angle and two sides.

Hypotenuse

The trigonometric ratios are:

Opposite

θ A

Adjacent

Exploration We have three right-angled triangles with acute angles of 30° and 60°. 1. Is △ABC ∼ △DEF ∼ △GHI? Explain why or why not. G

2. Find the ratio for sin 30° for each of the triangles. 3. What do you notice about each ratio (use a calculator if you need to)? 4. Find the ratio for cos 30° for each of the triangles. 5. What do you notice about each ratio (use a calculator if you need to)?

30°

3 30°

3 30° 2 60°

1 2

2 3

60°

60° 1 F

6. Find the ratio for tan 30° for each of the triangles. 7. What do you notice about each ratio (use a calculator if you need to)? 8. Draw another similar triangle which would have the same ratios.

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The trigonometric ratios will be the same for similar right-angled triangles. This is because the size of the angles in a triangle determines the ratios between the side lengths of a triangle.

Example 1 Evaluate cos (θ ) within △ABC.

A α

Create a strategy Identify which sides we need for the cosine ratio, then ﬁll in the side lengths.

Apply the idea The angle θ is also called ∠B.

Hypotenuse 13

Opposite

so we need the lengths of the adjacent and hypotenuse sides.

θ B

We are using the cosine ratio

Adjacent

With respect to the angle θ, the adjacent side is 5, and the hypotenuse is the longest side of 13. Deﬁntion of cosine Substitute the sides

Idea summary

Hypotenuse

The trigonometric ratios are:

Opposite

θ A

598

Adjacent

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Find a side If we are given one side length and one angle, we can use the trigonometric ratios to ﬁnd one of the missing side lengths. We can follow these steps to ﬁnd a missing side in a right-angled triangle: 1. Highlight the reference angle. 2. Identify which sides are the hypotenuse, opposite, and adjacent — label if desired. 3. Choose the ratio which has one of the required sides given and one is the missing side: • Sine: opposite and hypotenuse • Cosine: adjacent and hypotenuse • Tangent: opposite and adjacent 4. Set up the chosen ratio:

5. Solve for the unknown. Remember the following: • For a given angle θ, the value of each trigonometric ratio stays the same no matter the size of the triangle. • Your calculator must be set to degree mode when working with degrees in trigonometric ratios.

Example 2 Find the value of the side length, rounded to two decimal places: a

f 47°

Apply the idea With respect to the angle 47°, the length of the adjacent side is f, and the length of the hypotenuse is 8, so we can use the cosine ratio.

Use the cosine ratio

Substitute the values and f

Multiply both sides by 8

Evaluate and round

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11 g

42°

Apply the idea With respect to the angle 42°, the length of the opposite side is 11, and the length of the hypotenuse is g, so we can use the sine ratio. Use the sine ratio Substitute the values and g Multiply both sides by g Divide both sides by sin 42° Evaluate and round

Reflect and check When the unknown is in the denominator, we need to take care to rearrange properly.

Idea summary We can use the trigonometric ratios to find an unknown side length of a rightangled triangle. Once we set up our equation with a variable representing the unknown side length, we can use inverse operations to make the variable the subject of the equation. Then we can evaluate the expression in our calculators to find the side length.

Find an angle If we are given two sides of a right-angled triangle, we can use the trigonometric ratios to find a missing angle measure. We can follow these steps: 1. Using the unknown angle as the reference angle, label the given sides as opposite, adjacent, or hypotenuse. 2. Fill in the trigonometric ratio with the given information. 3. Use the inverse ratio to solve for the measure of the angle.

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Example 3 Find the value of x to the nearest degree.

A 7

C 25

x B

Apply the idea With respect to the angle with a measure of x, the length of the adjacent side is 7, and the length of the hypotenuse is 25, so we can use the cosine ratio. Use the cosine ratio Substitute the values Apply the inverse ratio Evaluate and round

Reflect and check If we get an error on our calculator, it usually means we haven’t used the right ratio or have substituted incorrectly.

Idea summary To find the measure of an unknown angle of a right-angled triangle, we use a trigonometric ratio to set up an equation relating the angle and two given sides. Then, we can use the inverse ratio to ﬁnd the measure of the angle.

Multi-step problems In some cases, the application of a single trigonometric ratio on a right-angled triangle will not be enough to solve the question.

x cm 43°

In this example, to find the length of h, we must first use our knowledge of trigonometry to find the length of x. Once we find the length of x, we add 70 to x to get the length of h.

h cm 70 m

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13°

16° C 2000 m

Airport

In this example, to find the distance between the two planes, AB, we will need to apply our knowledge of trigonometric ratios to find the distance of the first plane to point C (AC) and then again to find the distance of the second plane to point C (BC). Once we find the distance of both planes to point C, we subtract the distance of the second plane (BC) from the distance of the first plane (AC) to find the distance between the two planes (AB).

Example 4 A girl is ﬂying a kite that is attached to the end of a 23.4 m length of string. The angle between the string and the vertical is 21°. If the girl is holding the string 2.1 m above the ground, ﬁnd the vertical distance from the ground to the place where the string attaches to the kite. Round your answer to two decimal places.

23.4 m 21°

2.1 m

Create a strategy Let’s call the vertical distance from the ground to the kite h. If we could ﬁnd the length of the vertical leg of the triangle, we could calculate h. Let’s call the length of the vertical leg of the triangle x. First, calculate an acute angle in the right-angled triangle to use as a reference angle, then use a trigonometric ratio to ﬁnd the value of x. Then, we can calculate the vertical distance from the kite to the ground.

23.4 m 21°

xm hm

2.1 m

Apply the idea Since the angle between the string and the horizontal is complementary to 21°, the bottom left angle in the triangle must be 69°.

Substitute the values Multiply both sides by 23.4 Evaluate and round

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The vertical distance, h, from the ground to the kite is the sum of the distance from the ground to where the string starts and the vertical distance, x, from the string to the kite. That means h = 2.1 + x, where x = 21.85. So, h = 2.1 + 21.85 = 23.95 The kite is 23.95 m from the ground.

Example 5 Find x and y, rounded to the nearest whole number.

y cm

x cm 64°

Create a strategy Use the tangent ratio to ﬁnd y, then use the cosine ratio to determine x.

Apply the idea With respect to the angle of 64°, the length of the opposite side is 8, and the length of the adjacent side is y, so we can use the tangent ratio.

Use the tangent ratio

Substitute the values

Multiply both sides by y

Divide both sides by tan 64°

Evaluate and round

With respect to the angle of 64°, the length of the adjacent side is 8, and the length of the hypotenuse is x, so we can use the cosine ratio. Use the cosine ratio

Substitute the values

Multiply both sides by x

Divide both sides by cos 64°

Evaluate and round

12.02 Trigonometric ratios mathspace.co

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Idea summary In multi-step problems, application of a single trigonometric ratio may not be enough to solve the problem. Extra addition, subtraction, or another application of trigonometry may also be needed.

12.02 Practice questions Understanding 1

Explorer 1, 2, 3a–c, 4

Adventurer 3b–d, 4

Trailblazer 3d–f, 4

Name the hypotenuse. a

A B

For the triangle ABC: a

Which side is opposite angle θ?

Which side is adjacent to angle θ?

Which side is opposite angle α?

Which side is adjacent to angle α?

Which angle is opposite the hypotenuse?

A α

θ B

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With reference to the angle θ, ﬁnd the ratio: i

iii b

34 5

θ 5

25 C

12 12

Write down the indicated ratios for each triangle using the side labels: a

tan θ

tan α

sin θ

A α

sin α

A α

θ B

cos θ

cos α

A α

θ B

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5a, 7−8, 10−12, 14, 16, 17a–c

Fluency Ex 1

5b, 7−8, 10−12, 14, 16, 17b–d

5b, 6−7, 9−11, 13, 15−16, 17c–e

Write down the indicated ratios for each triangle: a

tan θ

sin α

iii

cos θ

tan α

sin θ

cos α

iii

cos θ

sin α

B 17

θ°

C α°

Find the value of tan θ.

25p 24p

7 Ex 2

Find the side length, rounded to two decimal places: a

b 29° fm

7 cm

27°

f cm

4 cm

12 cm

44°

25° g cm 16.68 cm a cm

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, ﬁnd the value of b.

θ° 12 m

If tan (55°) = , ﬁnd the value of d.

A 2313 m 55° d

Calculate the size of angle θ.

Ex 3

For each triangle, ﬁnd the value of: i

θ 6

θ 3

If d is the distance between the base of the wall and the base of the ladder, ﬁnd the value of d, rounded to two decimal places.

1.88 m

38° dm

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607

A ladder is leaning at an angle of 44° against a 1.36 m high wall. Find the length of the ladder, rounded to two decimal places. 1.36 m

44°

A sand pile has base angles of 40° and is 10.6 m wide. Find the height of the sand pile, h, rounded to one decimal place.

h cm 40° 10.6 m

A string of lights of length 23.9 m is joined to the top of a tree, as shown in the diagram. If the angle that the string of lights makes with the ground is θ °, ﬁnd θ rounded to two decimal places.

23.9 m 4.1 m

θ°

16 Ex 4

There is a 39° angle from the top of the NJ building, which is 300 m high, to the top of the LS building. The diagonal distance between the tops of the two buildings is 200 m. a

Find x, rounded to two decimal places.

Find the height of the LS building, rounded to two decimal places.

200 m 39°

300 m

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17 Ex 5

Find the side lengths, rounded to the nearest whole number: a

54°

16 cm 42°

26 mm

f cm h mm

x cm

32 cm 52°

x cm

58° h cm

11 cm

82 cm

45 cm

A x

Reasoning 18

An isosceles triangle has equal side lengths of 10 cm and a base of 8 cm as shown. Explain how to calculate the size of angle A, rounded to one decimal place.

10 cm

A°

A° 8 cm

If tan θ = 0.4, explain how to ﬁnd the value of b. b cm

θ°

21.32 cm 7.9 cm

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Problem-solving 20−22

20−23

21−24

, ﬁnd the value of cos θ and tan θ, without calculating the size of angle θ.

, ﬁnd the value of cos θ and sin θ, without calculating the size of angle θ.

Consider the octagon:

Use trigonometry to calculate the size of each internal angle rounded to the nearest whole number.

Conﬁrm your answer to part (a) using the formula to ﬁnd an internal angle of a polygon with n sides:

10 cm

12 cm

A sporting ground is in the shape of an irregular hexagon. a

Find the perimeter of the ground.

Find the size of each of the four corner angles of the ground.

35 m 15 m

41 m

15 m 35 m

Calculate the size of angle DEC.

6 x+7 A

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12 − x C 4 B

12.03 Angles of elevation and depression After this lesson, you will be able to… • define angles of elevation and depression and understand their real-world applications. • model the relative position of two objects using a right-angled triangle. • identify and apply the three distances (horizontal, vertical, and direct) between two objects when given the angle of elevation or depression. • solve problems involving angles of elevation and depression using trigonometric ratios.

Angles of elevation and depression An angle of elevation is the angle from the lower object to the higher one. The angle of elevation from point A to B is the angle between the horizontal line at A and the line connecting the two points.

An angle of depression is the angle from the higher object to the lower one. B

The angle of depression from point B to A is the angle between the horizontal line at B and the line connecting the two points.

Both angles are measured with respect to the horizontal.

Exploration Rectangle ABCD has a diagonal drawn on it.

1. What is the value of x? What property did you use to ﬁnd this? Is there another way to conﬁrm this?

x 50°

2. What is the angle of elevation from point C? 3. What is the angle of depression from point A?

4. What do you notice about the angle of elevation and depression?

50° C

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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).

B θ θ A

When given the angle of elevation or depression between two objects, we will always be able to model their relative position using a right-angled triangle. B

Dir

ect

Horizontal

dis

Dir

tan

ect

θ Horizontal

dis

Vertical

There are three distances between two objects: horizontal distance, vertical distance, and direct distance. These will represent the adjacent, opposite, and hypotenuse sides respectively. If any two of these are given, then we can ﬁnd the angle of elevation or depression.

tan

Example 1 Find the angle of depression from point B to point D. Use x as the angle of depression and round your answer to two decimal places.

9 A

Create a strategy Identify where the angle of depression is and apply the appropriate trigonometric ratio.

Apply the idea The angle of depression is the angle ∠CBD, with adjacent side BC = 9 and hypotenuse BD = 19. We can let x be the unknown angle. Use the cosine ratio

Substitute the values

Apply inverse cosine on both sides

Evaluate using a calculator and round

The angle of depression from point B to point D is 61.73°.

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Example 2 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°

4000 m

Airport

a Find the distance AC, to the nearest metre.

Create a strategy Use a trigonometric ratio that relates the 13° and the opposite.

Apply the idea With respect to the given angle 13°, the opposite side is 4000, and the adjacent side is AC, so we can use the tangent ratio. Use the tangent ratio

Substitute the values

Multiply both sides by AC

Divide both sides by tan 13°

Evaluate using a calculator and round

The distance AC is 17 326 m.

b Find the distance BC, to the nearest metre.

Apply the idea With respect to the given angle 16°, the opposite side is 4000, and the adjacent side is BC, so we can use the tangent ratio.

Use the tangent ratio

Substitute the values

Multiply both sides by BC

Divide both sides by tan 16°

Evaluate using a calculator and round

The distance BC is 13 950 m.

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c Now ﬁnd the distance covered by the jet in that one minute.

Apply the idea The distance covered by the jet in one minute is given by the length of AB. AB = 17 326 − 13 950

Subtract BC from AC

= 3376 m

Simplify the difference

The distance covered by the jet is 3376 m.

Example 3 A room is rectangular and measures 5 metres in length and 4 metres in width. The angle of elevation from the bottom left corner to the top right corner of the room is 57°. a Draw a diagram of the room.

Create a strategy

Apply the idea

To draw a diagram of the room, we need to keep in mind that the room is rectangular in shape and the given dimensions are in metres. Additionally, the angle of elevation from one corner of the room to the other is given, which means we are considering a 3D model.

h 57°

b Find d, the distance from one corner of the ﬂoor to the opposite corner of the ﬂoor. Leave your answer in surd form.

Create a strategy Apply Pythagoras’ theorem, c2 = a2 + b2.

d 4 5

Apply the idea d2 = 52 + 42 2

d = 41 d=

Substitute the values

Take the positive square root of both sides

The distance from one corner of the floor to the opposite corner of the floor is m.

614

Evaluate the right-hand side

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c Find h, the height of the room. Give your answer to two decimal places.

Create a strategy Set up an equation involving the height, h, the diagonal on the ﬂoor, , and the angle of elevation, 57°.

h 57°

Apply the idea Use the tangent ratio h

Substitute the values

57° 41

Multiply both sides by Evaluate using a calculator and round The height of the room is 9.86 m.

Idea summary We can solve problems involving angles of elevation and depression using these steps: 1. Draw a diagram if one was not provided. 2. Label all known information on the diagram, including the right angle. 3. Identify which angles or sides need to be found and in what order. 4. Set up the correct trigonometric ratio and solve for the missing side or angle. 5. Give your answer as a sentence.

Angle of elevation

Horizontal

Angle of depression

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12.03 Practice questions Understanding 1

Explorer 1−3

Adventurer 1−3

Trailblazer 1−3

Is angle θ an angle of elevation or depression? Explain how you know. 4 2

A man standing at the top of the tower at point B is looking at the ground at point C. a

Identify the angle of depression in the ﬁgure.

Identify the angle of elevation in the ﬁgure.

What can be said about the size of the two angles?

θ A

Using the given diagram:

Identify the angle of elevation.

How could you draw an angle of depression from A to C?

Which angle in the diagram is the same size as the angle of depression from A to C?

A θ σ

α B

Fluency 4

4−7, 9, 10

4−10

5−11

x is the angle of elevation to point A from point C. a

Find x, rounded to two decimal places.

Find y, rounded to two decimal places.

y x C

5 Ex 1

x is the angle of depression from point B to point C. a

Find y, rounded to two decimal places.

Find x, rounded to two decimal places.

25 B

D y

x 5

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The person in the picture sights a pigeon above him. Find the angle of elevation, θ, rounded to two decimal places. 30 m θ 23 m

A helicopter is 344 m away from its landing pad. If the angle of depression to the landing pad is 32°, ﬁnd x, the height of the helicopter above the ground, rounded to the nearest metre.

32° xm

344 m

Landing pad

The ﬁnal approach of an aeroplane when landing requires the pilot to adjust the angle of depression to 3°, as shown in the diagram. If the plane is 12 m above the runway and has d m until touchdown, ﬁnd d rounded to the nearest metre.

3° 12 m

A man stands at point A looking 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 of 25 m and an angle of elevation of 57° from point A. a

Find the distance from A to B, rounded to two decimal places.

Find the distance from A to C, rounded to two decimal places.

10 Ex 2

Pole 2

25 m

Pole 1 8m A

Find BC, the distance between the two poles in metres. Round your answer to one decimal place.

A fighter jet, flying at an altitude of 2000 m is approaching an airport. The pilot measures the angle of depression to the airport to be 12°. One minute later, the pilot measures the angle of depression again and finds it to be 19°. a

Find the distance AC, rounded to the nearest metre.

Find the distance BC, rounded to the nearest metre.

Find the distance covered by the jet in that one minute.

12° B

19°

C 2000 m

Airport

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11 Ex 3

A room measures 6 m in length and 5 m in width. The angle of elevation from the bottom left corner to the top right corner of the room is 60°. Rounding all answers to two decimal places:

60°

Find the distance from one corner of the ﬂoor to the opposite corner of the ﬂoor.

Find the height of the room.

Find the angle of elevation from the bottom corner of the 6 m long wall to the opposite top corner of the wall.

Find the angle of depression from the top corner of the 5 m long wall to the opposite bottom corner of the wall.

Reasoning 12

12−13

13−14

An air traffic controller is communicating with a plane, which is flying at an angle of elevation of θ ° from the control tower. The plane is 10 369 m above the ground and is still 23 444 m from the runway. Viggo claims that the angle of depression that the pilot should fly is equal to 90° − θ °. Explain why Viggo is incorrect and correctly calculate the angle of depression that must be flown.

10 369 m θ° 23 444 m

Fiona says that the angle θ is not an angle of elevation as it is not measured from the ground. Is Fiona correct? Explain your answer. 18 m

27 m θ

The angle of elevation from the tip of a tree’s shadow to the sun is 43°. The tree’s shadow is 20 m long. a

Early in the morning, and late in the afternoon, when the angle of elevation of the sun is low, what happens to the length of the shadow?

What happens to the length of the shadow when the sun is directly overhead?

618

Find tan 90° on your calculator. How does this relate to your answer to part (b)?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

h 43° 20 m

Problem-solving 15, 17−21 15

15, 18−21, 23

16, 20, 22−23

Fred is on a ship and observes a lighthouse on a cliff in the distance. The base of the cliff is 651 m away from the ship, and the angle of elevation to the top of the lighthouse from Fred is 35°. a

If the top of the lighthouse is x m above sea level, ﬁnd x rounded to two decimal places.

If the lighthouse is 30 metres tall, how tall is the cliff that the lighthouse stands on?

Round your answer to two decimal places. 16

Buzz is standing 250 m from a building and measures the angle of elevation to the top of the building to be 27°. a

If the difference in height between the top of the building and Buzz’s eye is h m, ﬁnd h rounded to two decimal places.

If Buzz’s eye is 172 cm from the ground, ﬁnd the height of the building, rounded to one decimal place.

From the top of a rocky ledge that is 274 m high, the angle of depression to a boat is 15°. If the boat is d m from the foot of the cliff, ﬁnd d rounded to two decimal places.

Amelia measures the angle of elevation to the top of a tree from a point, 29 m away from the base, to be 31°. Find the height of the tree, h, rounded to the nearest metre.

The angle of elevation from an observer to the top of a tree is 18°. If the distance between the tree and the observer is d m and the tree is known to be 3.53 m high, ﬁnd the value of d rounded to two decimal places.

At a certain time of the day a light post, 6 m tall, has a shadow of 5.8 m. If the angle of elevation of the sun at that time is θ °, ﬁnd θ rounded to two decimal places.

A ramp of length 311 cm needs to ascend at an angle between 10° and 20° for it to be safe to use.

311 cm

If the height of the ramp is 152 cm, ﬁnd the angle the ramp makes with the ground, rounded to two decimal places.

If the height of the ramp is 100 cm, ﬁnd the angle the ramp makes with the ground, rounded to two decimal places.

If the height of the ramp is 25 cm, ﬁnd the angle the ramp makes with the ground, rounded to two decimal places.

At which height(s) is the ramp safe?

height

Two hot air balloons are floating in different locations, and each is tied to the ground directly below. An observer at each location measures the angle of elevation to the opposite balloon. The observers are 1600 m apart as shown in the diagram: a

Calculate the difference in height between the two balloons. Round your answer to the nearest metre.

Find the direct distance between the two balloons. Round your answer to the nearest metre.

3.38°

1.22° 1600 m

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Roald is standing at point P and observes two poles, AB and A CD, of different heights. P, B, and D are on horizontal ground:

From P, the angles of elevation to the top of the poles at A and C are 29° and 18° respectively. Roald is 16 m from the base of pole AB. The height of pole CD is 7 m. a

Calculate the distance from Roald to the top of pole CD, B rounded to two decimal places.

Calculate the distance from Roald to the top of pole AB, rounded to two decimal places.

29° 16 m

45°

18°

12.04 Bearings After this lesson, you will be able to… • use the starting direction and acute angle to describe the location of a point. • write bearings using mathematical notation. • apply bearings in 3D problems by drawing relevant triangles and using trigonometric ratios. • draw a diagram of a 3D situation, find distances and calculate angles related to bearings.

Compass bearings In surveying and navigation, bearings are used to identify the location of an object. The four main directions of a compass are known as cardinal directions. They are north (N), east (E), south (S), and west (W).

N NW

NNW

NNE

WNW

ENE

WSW

ESE

SSW

SSE

In between each of these cardinal directions are intermediate directions of northeast (NE), southeast (SE), southwest (SW), and northwest (NW) that are 45° from the two cardinal directions closest. There are smaller divisions shown on the compass as well.

S A compass bearing describes the location of a point using: • The starting direction of either north or south; • The acute angle needed to rotate • The direction to rotate, east or west.

Starting direction

Direction of rotation

Angle rotated by

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To describe the position of point B from A we would say: “Start at south, then rotate 53° towards east.”

North

We can write this mathematically as: S 53°E A

West

East 53°

South

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Example 1 N A

Consider the point A. 35° O

a For the compass bearing of point A from O, which option shows the correct letters that should be written either side of the angle? A

N⬚E

Apply the idea

N⬚W

S⬚E

S⬚W

Since A is closer to north, we start at north and rotate clockwise towards the east direction. So the letters that should be written are in option A, N ⬚ E. b Complete the compass bearing of point A from O. N⬚E

Create a strategy

Apply the idea

Write the angle measure shown in the ﬁgure. The angle given is measured from the north towards the east, therefore we can write in the angle given.

N 35°E

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Idea summary Starting direction

Direction of rotation

A compass bearing describes the location of a point using: • The starting direction of either north or south • The acute angle needed to rotate • The direction to rotate, east or west.

Angle rotated by

Three-figure (true) bearings A three-ﬁgure bearing is: • Measured from north (N) • Measured in a clockwise direction • Written using three ﬁgures T is often, but not always, used to describe a true bearing. If the angle measure is less than 100° it would be written as 040° or 040° T. To use true bearings to describe the location of a plane at point B from the airport at point A: • Place the centre of a compass on the starting point, in this case the airport. • Starting at north, rotate clockwise until we get to the line AB. • Write the angle as the true bearing of point B. The true bearing of B from A is 127° or 127° T.

North

West

127° East B

South

The diagram describes the bearing of P from O. Rotating clockwise from north, we get an angle of 55°.

North P

Since this measure is less than three digits, we put 0 in front so the true bearing of P is 055°.

55° West

East

South

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For the true bearing of O from P, the true bearing would be 180 + 55 = 235°.

North

West O

East 180°

55°

South

Example 2 Determine the true bearing of point A from point O. a

N A

35° O

Create a strategy

Apply the idea

Measure the bearing clockwise from north, and write it using three digits.

The angle between the north direction and the line OA measured clockwise is 35°. So the true bearing is 035° T.

42° O

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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Example 3 Consider the following diagram: A

B 42°

52°

a Find the true bearing of point A from point C.

Create a strategy We can find the angle between AC and north. Then, we can subtract that from 360°.

Apply the idea

52°

To find the angle that AC makes with true north, we can extend the north line from point C to segment AB. This will create a right-angled triangle.

Recall the acute angles in a right-angled triangle sum to 90°, so the angle that AC makes with true north is 90 - 42 = 48°.

B 42° 48°

Starting from point C and rotating clockwise from north gives us an angle of 360 - 48 = 312°. This means the true bearing is 312° T.

b Find the compass bearing of point A from point C.

Apply the idea

Reflect and check

From C to A it is north and west which means it is N 48°W.

Notice going from A to C would be the opposite direction and would be S 48°E.

Example 4 The position of a ship S is given to be 20 kilometres from P, on a true bearing of 049° T. The position of the ship can also be given by its (x, y) coordinates. Find the ship’s coordinates to one decimal place.

N x y P

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049° 20 km

Create a strategy For the x-coordinate, use the sine ratio given by sin θ = use the cosine ratio given by cos θ =

, and for the y-coordinate,

Apply the idea In the diagram, we are given θ = 49°, opposite = x, and hypotenuse = 20. Substitute the values Multiply both sides by 20 Evaluate and round In the diagram, we are given θ = 49°, adjacent = y, and hypotenuse = 20. Substitute the values Multiply both sides by 20 Evaluate and round The ship’s coordinates are (15.1, 13.1).

Idea summary True bearings are: • Measured from north (N) • Measured in a clockwise direction

• Written using three ﬁgures • Usually written with a T at the end

Bearings in 3D A problem in three dimensions may use bearings to describe the situation. It is helpful to ﬁrst draw a diagram if one is not given to us. Then, we can solve as usual for problems in 3D, by drawing out triangles in 2D.

Example 5 A miner is mining 500 m below the surface. If you walk 300 m north and then 400 m west from the oﬃce, you will be directly above the miner. a Draw a diagram of the situation in 3D and the related triangles in 2D.

Create a strategy We can start by drawing the surface using the bearings and then placing the miner below the surface. It is helpful to label the vertices.

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Apply the idea The whole scenario can be represented in a 3D diagram like the one shown.

400 m W

N 300 m

500 m

We can break this into two right-angled triangles which have a common side of WO. W

400 m

300 m 500 m O M

b Find the distance from the oﬃce to the point on the surface directly above the miner.

Apply the idea For triangle ONW, we know two sides and have one missing side, let’s use Pythagoras’ theorem to ﬁnd OW. Pythagoras’ theorem

Substitute known values

Evaluate the indices and addition

Take the square root of both sides

Simplify

The distance from the oﬃce to the point on the surface above the miner is 500 m.

Reflect and check We may have noticed this is the Pythagorean triple, 3, 4, 5 multiplied by 100.

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c Find the direct distance from the miner to the Oﬃce, to the nearest metre.

Apply the idea For triangle OWM, we know two sides and have one missing side, so we can use Pythagoras’ theorem to ﬁnd OM. Pythagoras’ theorem

Substitute known values

Evaluate the indices and addition

Take the square root of both sides

Simplify the surd

Evaluate using a calculator and round

The direct distance from the oﬃce to the miner is 707 m.

Reflect and check If we use this rounded value instead of the exact value of introduced.

, there will be some error

d At what angle of depression from the oﬃce would you dig to reach the miner?

Create a strategy We are now trying to ﬁnd ∠WOM. Since this is a right-angled triangle, we can use trigonometric ratios to ﬁnd this angle.

Apply the idea We know all the side lengths in this triangle, so we can use any pair. However, one of them is rounded, so it is best to use the two exact values.

500 m

θ 500 m

707 m

The sides with exact values are opposite and adjacent to the angle q. Use the tangent ratio

Substitute known values

Use inverse tan ratio

Evaluate using a calculator

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Reflect and check Instead of using trigonometric ratios, we could have recognised that this is an isosceles right-angled triangle, so both acute angles would be 45°.

Idea summary Although bearings refer to directions on the surface, we may use them in 3D problems where a given point is above or below ground. We can draw relevant triangles in 2D to solve one problem at a time.

12.04 Practice questions Understanding 1

Explorer

Adventurer

1−4

Write the compass bearing for point P from point O. a

628

Trailblazer

1−4

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Write the true bearing of P from point O. a

N 26° W P

S 64° W

N P

N 64° W

P S 26° W

Which of these diagrams correctly displays WNW? A

N NW

WNW

SE S

N NW

WNW E

WNW

What is the equivalent true bearing for these compass bearings? a

Southwest

Northwest

West-Northwest

South-Southeast

Northeast

Southeast

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5, 6, 8, 9, 11−13, 15, 18−20

Fluency Ex 1, 2

5, 6, 8, 9, 11−13, 16, 18−20

Find the true bearing of point A from O.

Find the compass bearing of point A from O.

N A

5, 7, 8, 10, 14, 16, 17, 20

30° O

41° A

A 40° O

O 35° A

Determine the true bearing for: a

Point B from point A.

Point C from point A.

Point C from point B.

B 37°

48°

Determine the true bearing for: a

Point A from point B.

Point A from point C.

Point B from point C.

B 47°

38°

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Ex 3

Find the true bearing of point A from point C.

Find the compass bearing of point A from point C.

B 49°

39°

Find the true bearing of point B from point C.

Find the compass bearing of point B from point C.

B 37°

45°

Find the true bearing of point C from point A.

Find the compass bearing of point C from point A.

B 42°

53°

Find the true bearing of point C from point B.

Find the compass bearing of point C from point B.

B 38°

49°

If a person walks at a compass bearing of SW, and then turns around to go back in the direction they came, what direction are they heading now?

An aeroplane sets off on a bearing of 051° T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

A boat sets off on a bearing of 125° T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

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If the bearing of the clearing from the town is a°, ﬁnd a rounded to the nearest degree.

N 5.9 km 3.2 km

Clearing

a°

O Town

Luke sailed for 116 km on a bearing of 231°.

Find w, the horizontal distance in kilometres he has sailed from his starting point, rounded to one decimal place. w 231° 116 km

Shortly after take-off, a plane is 42 km south and 57 km west of the airport in Sydney that it left from. a

Find the size of the angle marked b, rounded to one decimal place.

Find the compass bearing of the plane from the airport.

What is the true bearing of the plane from the airport?

O b°

Plane

18 Ex 4

632

42 km

57 km

The position of a ship S is given to be 20 km from P, on a true bearing of 049° T. The position of the ship can also be given by its (x, y) coordinates.

Find the ship’s x-coordinate rounded to one decimal place.

Find the ship’s y-coordinate rounded to one decimal place.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Sydney

049°

20 km

20 Ex 5

On an orienteering course, Beth runs 1000 metres north from point A to point B, then turns east and runs to point C. If the true bearing of C from A is 062° T and the direct distance Beth must run to get back to point A is d metres, ﬁnd d to the nearest metre.

N B 1000 m

C 062° d m

From a point 15 m due north of a tower, the angle of elevation to the top of the tower is 32°. a

Find the height of the tower h. Round your answer to two decimal places.

Find θ, the angle of elevation to the top of the tower from a point 20 m due east of the tower. Round your answer to the nearest degree.

North

32° hm 15 m

20 m

East

Reasoning 21

21a–c

21b–d, 22

21cd, 22, 23

Given that a, b, c and d are angles between 0° and 90°: a

If a compass bearing is N a°W, write an expression for the true bearing.

If a compass bearing is S b°W, write an expression for the true bearing.

If a compass bearing is S c°E, write an expression for the true bearing.

If a compass bearing is S (90 − a) °E, write an expression for the true bearing.

xplain why the overall horizontal distance from A E to C is equal to 12 cos 45° + 16 cos 30°.

Write an expression using sin which will calculate the overall vertical distance from A to C.

Use Pythagoras’ theorem to calculate the direct distance from A to C.

N 45° 12 km

Find the bearing of A from C.

Find the direct distance from A to C.

Find the bearing of A from C.

N 60°

16 km

N N

B 72°

145°

600 m 800 m

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Problem-solving 24, 25, 27, 29, 30

24−27, 30

24, 26−28, 31, 32

During a group hike, a hiker walked 11 km NE of base camp. In the evening the hiker was separated from the group and after a night lost in the wilderness, the hiker was found 11 km NW of the base camp. What compass bearing must the hiker have followed during the night?

During a rescue search, a helicopter ﬂew west from point X to point Y, then changed course and ﬂew 10.7 km north to point Z. If point Z is on a bearing of 335° T from point X:

Find the size of ∠Y XZ.

Find b, the distance from point Y to point X, rounded to one decimal place.

Calculate d, the distance the helicopter must ﬂy between point Z and point X, rounded to one decimal place.

N Z

In remote locations, photographers must track their position from their base. One morning a photographer sets out from base, B, to the edge of an ice shelf, S, on a bearing of 055°. She then walked on a bearing of 145° to point P, which is 916 m due east of base.

b c

634

335°

145° S B

∠BAC

N B

∠DCE

Find the distance the boat is from its starting point as a straight line from A to E. Find θ °, the true bearing of the boat from its starting point, to the nearest degree.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

916 m

Find the size of the angles: i

A ﬁshing boat in search of large schools of ﬁsh sails from point A for 26 km on a bearing of 205° to point C. It then sails a further 24 km on a bearing of 226° to point E. Rounding all answers to one decimal place: a

X b km

Rounding all answers to one decimal place: a State the size of ∠BSP. b Find the distance BS. c Find the distance SP. d If the photographer walks directly back to her base from point P, determine the total distance travelled. 27

d km

10.7 km

A drone travels due east for 2.2 km and then travels on a bearing of S 32°E for 5.9 km. a

Given that the compass bearing is a°, write the bearing needed to return to the starting point, in terms Start of a.

Find x, the distance between the end point and the starting point of the drone’s ﬂight. Round your answer to two decimal places.

N 2.2 km

x km

32° 5.9 km

Find a rounded to the nearest degree.

A yacht sailed in a direction so that its ﬁnal position was 248 km west and 225 km south of its starting point. a

Draw a diagram to represent the situation.

Find the true bearing on which the yacht sailed, rounded to one decimal place.

Find the total distance the boat has sailed, rounded to one decimal place.

Three television presenters are practicing their navigation skills before heading off on an expedition to a remote location. Belinda, at point B, is positioned 17.6 m south of Amelia at point A. Carl, at point C, is due east of Belinda and on a bearing of S 38°E from Amelia. a

Draw a diagram to represent the situation.

If Amelia and Carl are d m apart, ﬁnd d to one decimal place.

A yacht travelled a distance of 30 km from port A at a bearing of 110°, then another 40 km to port C at a bearing of 130°. What is the distance between ports A and C?

A and B are two positions on level ground. From an advertising balloon at a vertical height of 740 m, A is observed at a bearing of 043° and B at a bearing of 133°. The angles of depression of A and B, as viewed from the balloon, are 30° and 20° respectively. Rounding all answers to the nearest metre: a

Find the distance from point A to the point directly below the advertising balloon.

Find the distance from point B to the point directly below the advertising balloon.

Find the distance between points A and B.

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12.05 Optional: Sine and cosine rules After this lesson, you will be able to… • utilise the sine rule to find unknown sides and angles. • distinguish between the two forms of the sine rule and when to use each. • identify the conditions that lead to multiple solutions when using the sine rule. • define and explain the cosine rule and its applications in solving problems related to non-right-angled triangles. • use the cosine rule to find unknown sides or angles. • apply the formula for the area of non-right-angled triangles to solve problems.

Sine Rule The sine rule is used in non-right-angled triangles that involve two sides and two angles. We are able to use this rule if we want to ﬁnd an unknown side or angle.

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The sine rule, sometimes called the law of sines or sine law, tells us that:

A, B, C the measures of the angles in the triangle a, b, c the lengths of the sides in the triangle We can also take the reciprocal of each fraction to give the alternate form:

B a

c A

This alternate form is used to find angles. Suppose we know the side lengths a and b and the angle B, and we want to ﬁnd the angle A. Using the form of the sine rule

, we can solve for A.

We can make sin A the subject by multiplying both sides by a:

Then, to ﬁnd A itself, we need to take the inverse sine of both sides:

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Example 1 Find the side length a using the sine rule. Round your answer to two decimal places.

a 69° 18

33°

Create a strategy Use the sine rule:

Apply the idea

Substitute b = 18, A = 33, B = 69

Multiply both sides by sin 33°

Evaluate

Example 2 Find the value of the acute angle A using the sine rule. Round your answer to two decimal places.

A° 81 50° 98

Create a strategy Use the sine rule:

Apply the idea Substitute a = 98, b = 81, B = 50

Multiply both sides by 98

Evaluate

Take the inverse sine of both sides

Evaluate

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Idea summary The sine rule can be written in two forms. The ﬁrst form is best when solving for an angle. The second form is best when solving for a side.

A is an angle of a triangle opposite side a B is an angle of a triangle opposite side b C is an angle of a triangle opposite side c We will generally only use two angle-side pairs when solving. We need to know the values of one side-angle pair and one other piece of information.

Ambiguous case of the sine rule When we are solving for angles using the sine rule, there may be two possible solutions.

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The ambiguous case occurs when it is possible to have multiple solutions with the information given. When using the sine rule, it is possible to have no solution, one solution, or two solutions: • Case 1: No triangle exists This is the case when an error occurs in the calculator while solving for the unknown angle. This means that no such triangle with the given side lengths and angle exists. • Case 2: Exactly one triangle exists This is the case when the sum of the given angle and the supplement to the solution calculated is equal to or exceeds 180°. • Case 3: Two possible triangles exist This is the case when the sum of the given angle and the supplement to the solution are less than 180°. Then, the possible triangles are as follows: • A triangle that includes the calculated solution • A triangle that includes the supplement to the calculated solution C1 b A

C2 b

a c1

Possibility 1

a B2 = 180° − B1

Possibility 2

To ﬁnd the second possible solution, subtract the ﬁrst value of B that you found from 180°.

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Example 3 Find the value of x in degrees, given that x is obtuse. Round your answer to two decimal places.

25° 18.8 x A 12.6 C

Create a strategy Use the sine rule:

Apply the idea Substitute A = x, a = 18.8, b = 12.6, B = 25

Multiply both sides by 18.8

Evaluate

Take the inverse sine of both sides

Evaluate and round

To ﬁnd the obtuse angle x we subtract the acute angle from 180°: x = 180 − 39.09 = 140.91°

Subtract 39.09 from 180 Evaluate

Example 4 If possible, solve the triangle where a = 85, b = 93, and ∠A = 61°. Round your answers to two decimal places.

Create a strategy Draw and label a diagram of the triangle; it does not need to be drawn to scale.

From this diagram we can see that we were given two sides and their nonincluded angle. This is an example of the ambiguous case and when we solve it we may ﬁnd one, two, or no solutions to this problem.

85 61° A

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We are given a pair of opposite sides and angles with side a and angle A in addition to side b. We will set up the equation:

to ﬁnd ∠B.

Apply the idea

Sine rule

Substitute given values

Multiply both sides by 93

Apply the inverse sine to both sides

Evaluate and round

Now that we’ve found an acute solution to the problem, we need to ﬁnd its supplement and check if this will produce a valid second solution. The supplement of ∠B1 = 73.12° is ∠B2 = 180° − 73.12° = 106.88°. If we add these two angles together we get 61° + 106.88° = 167.88° which does not exceed the maximum 180° possible in a triangle. That means this triangle has two solutions. Solution 1: ∠B1 = 73.12° So far we have a = 85, b = 93, ∠A = 61° and ∠B1 = 73.12°. Use the triangle angle sum to solve for the missing angle, ∠C1 = 180° − (61° + 73.12°) = 45.88° and use the sine rule to ﬁnd the missing side, c1:

Sine rule

Substitute known values

Multiply both sides by sin 45.88°

Evaluate and round

The three side lengths are a = 85, b = 93, and c1 = 69.77 and the three angles are ∠A = 61°, ∠B1 = 73.12° and ∠C1 = 45.88°. Solution 2: ∠B2 = 106.88°

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For the second solution we have a = 85, b = 93, ∠A = 61° and ∠B2 = 106.88°. Use the triangle angle sum to solve for the missing angle, ∠C2 = 180° − (61° + 106.88°) = 12.12° and use the sine rule to ﬁnd the missing side, c2: Sine rule Substitute known values Multiply both sides by sin 12.12° Evaluate and round The three side lengths are a = 85, b = 93, and c2 = 20.40 and the three angles are ∠A = 61°, ∠B2 = 106.88° and ∠C2 = 12.12°. The diagram shows a visual of both solutions: B1 c1

85 c2

61° A

B2 61°

85 93

Reflect and check There will be no solution, or a calculator error, if we attempt to take the inverse sine of a value greater than 1. There will be exactly one solution if the solved angle is acute, but its supplement is too large to create a second valid triangle.

Idea summary Sine rule (ambiguous case) If you are trying to find an angle using the sine rule (with two known sides and a known angle), use the following rule to find one value of B:

If the side opposite the known angle is the shorter side, you are in the ambiguous case. Subtract the ﬁrst value of B that you found from 180° to ﬁnd another possible solution.Check that the second solution is a valid solution using the sum of angles in a triangle.

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Cosine Rule The cosine rule rule is used in non-right-angled triangles that involve three sides and one angle. We are able to use this rule if we want to ﬁnd an unknown side or angle. Two side lengths and one angle, opposite the unknown side, are known. x 10

32° 4

Three side lengths are known, and one angle is unknown. x°

The cosine rule says: c2 = a2 + b2 − 2ab cos C

b2 = a2 + c2 − 2ac cos B

a2 = b2 + c2 − 2bc cos A

To ﬁnd a missing angle using the cosine rule, we need to know all three side lengths. When looking to solve for an unknown angle, the equation can also be rearranged and written as

Example 5 Find the length of a using the cosine rule. Round your answer to two decimal places.

C a

13 A

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33° 30

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Create a strategy Use the cosine rule a2 = b2 + c2 − 2bc cos A.

Apply the idea Write the formula

Substitute b = 13, c = 30, A = 33

Evaluate

Take the square root of both sides and round

Example 6 Find the value of angle w using the cosine rule. Round your answer to two decimal places.

32 25 w 19

Create a strategy , where w corresponds

Use the rearranged formula of the cosine rule given by to C.

Apply the idea Write the formula

Susbtitute the values

Evaluate

Take the inverse cosine of both sides

Evaluate and round

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Idea summary If we have a problem involving two sides and the included angle of a non-rightangled triangle, we can use the cosine rule to ﬁnd the unknown side length:

c2 = a2 + b2 − 2ab cos C c is the side opposite angle C a is another side length b is another side length If we have a problem involving 3 side lengths of a non-right-angled triangle, we can use the cosine rule to ﬁnd the unknown angle:

c is the side opposite angle C a is another side length b is another side length

The area formula The formula for the area of a triangle is

× base × height, which uses the base, b, and height,

h. Sometimes, the height is called the altitude of the triangle, and it must always be perpendicular to the base.

If the triangle is not right-angled, we don’t know the perpendicular height. However, we can still find the area, as long as we know two sides and the angle between them.

C b

The area of the triangle is half the base b times the height, h. a

Notice that the height divides the triangle into 2 right-angled triangles.

C b

One of the right-angled triangles has a hypotenuse a and short side h. So we can use trigonometric ratios, specifically the sine ratio, to find the value of sin C.

We can rearrange this to find h: h = a sin C 644

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Putting this all together with the area formula

× base × height, we obtain the formula:

Area of a triangle formula

Substitute the height and base

Simplify

Example 7 Calculate the area of the triangle. Round your answer to two decimal places.

73°

2.2 m

Create a strategy Use the area rule given by

Apply the idea We are given a = 6, b = 2.2, and C = 73°. Write the formula

Substitute the measurements

Evaluate and round

Example 8 △ABC has an area of 850 cm2. The side BC = 65 cm and ∠ACB = 31°. What is the length of b? Round your answer to the nearest centimetre.

A b cm B

31° 65 cm

Create a strategy Use the area rule given by

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Apply the idea We are given Area = 850, a = 65, and C = 31°. Write the formula Substitute the measurements Evaluate the multiplication Divide both sides by 16.7387 Evaluate and round

Idea summary To find the area of a non-right-angled triangle, we can use the formula:

a is the length of one side b is the length of another side C is the included angle between a and b

12.05 Practice questions Understanding 1

Explorer 1−4

Adventurer

Trailblazer

1−4

2−5

Which strategy would you use to ﬁnd the unknown: Pythagoras’ theorem, trigonometric ratios, sine rule or cosine rule? a

a 69° 18

33° 11

d 32 25 w 19

646

fm 27°

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Which angle would need to be given to find the length of AC using the cosine rule?

B 31.05 cm 38.44 cm

Write an equation relating the sides and angles using the sine rule.

Write an equation relating the sides and angles using the cosine rule.

R r P

For each triangle, suggest if you could use the sine, cosine, or both rules to find all remaining sides and angles in the triangle. a

Three sides are known.

wo of the angles and the side included T between them are known.

B a

A c

Two of the angles and a side not

Two of the sides and an angle included

included between them are known.

between them are known.

C b

A b 12.05 Optional: Sine and cosine rules mathspace.co

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6a–d, 7−9, 10ab, 11ab, 13a, 14ab, 16, 18ab, 19, 21a–d, 22, 23

Fluency

6 Ex 1

6−11, 13b, 14bc, 17, 18bc, 19, 21a–d, 22, 23, 25

6−12, 14, 15, 17, 18cd, 20, 21c–f, 22, 24, 25

Find the unknown side, rounded to two decimal places. a

b 13

64° 21

63°

28°

35°

d 42°

39°

64°

48° 15

16.76 D 41°

47°

12.4 mm 55°

a 35°

For triangle ABC, where BC = 37.8 m, ∠BAC = 57° and ∠ACB = 42°, calculate the length of side AB, rounded to two decimal places.

For triangle ABC with ∠C = 72.53° and ∠B = 31.69°, and one side length a = 5.816 m, find: a

∠A, rounded to two decimal places.

The length of side b, rounded to three decimal places.

The length of side c, rounded to three decimal places.

A c b

B a

Solve for x, given that all equations relate to acute-angled triangles, rounded to two decimal places. a

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10 Ex 2

For each triangle, ﬁnd the value of the angle using the sine rule. Round your answers to one decimal place. a

B 9 56°

A C

62° 11

B x C

6.58

39°

x x 11.2 cm

11 Ex 3

3.69

116°

9 cm

For each triangle, ﬁnd the angle using the sine rule. Round your answers to one decimal place. a

53°

75 m

41° 287

275 B

x 86 m

x 110°

30°

2.34 cm

5.52 cm

18.1 x A 12.4 C

For triangle MNP, where NP = 100 m, MN = 204 m and ∠NPM = 58°, calculate the size of the acute angle ∠NMP, rounded to the nearest degree.

For each triangle, calculate ∠B rounded to the nearest degree.

Ex 4

△ABC where ∠A = 57°, side a = 156 cm and side b = 179 cm

△ABC where ∠A = 48°, side a = 2.7 cm and side b = 1.9 cm

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14 Ex 5

For each triangle, ﬁnd the side length using the cosine rule. Round your answers to two decimal places. a

C 11

19.4

23.3

131

24°

54°

261

25°

33°

In △ABC,

, ﬁnd the exact length of side AB in centimetres.

A 7 cm

B 9 cm

Triangle QUV has side lengths v = 7, u = 10 and an angle at Q such that cos value of q, where q is the length of the side opposite angle Q.

Triangle QUV has side lengths q = 63, u = 36 and ∠V = 116°. Find the value of v, where v is the length of the side opposite ∠V. Round your answer to two decimal places.

For each triangle, ﬁnd the angle. Round your answers to two decimal places.

Ex 6

b 25

31 22

w°

B°

12 7

11.2

w° 18.1

A°

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5.2

. Find the

Triangle QUV has side lengths q = 5, u = 10 and v = 12. Find the size of ∠V, where V is the angle opposite the side of length v. Round your answer to the nearest degree.

Triangle QUV has side lengths v = 23.5, u = 15.6 and q = 10.0. Find the size of ∠Q, where Q is the angle opposite the side of length q. Round your answer to the nearest degree.

Calculate the area of the following triangles to two decimal places:

Ex 7

b 4m

50° 13.1 m

12 m

7.4 m

123°

29°

39° 25°

22 cm

107° 11.6 cm

33 cm

23.2 cm

24.7 cm

17 cm

54°

46°

11.6 cm 79°

155 mm

22 Ex 8

70°

The following triangle has an area of 520 cm2. Find the length of side b. Round your answer to the nearest centimetre.

A b cm B

35° 48 cm

For the following triangles, calculate: i

The size of angle x, rounded to the nearest degree.

The area of the triangle, rounded to one decimal place.

12.9 cm 6 cm

124°

9.2 cm

x 99° 12.9 cm x

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Consider the triangle shown:

Find the length of BC, rounded to two decimal places.

Find the area of the triangle, rounded to two decimal places.

7 43° B

For the triangle shown, find the following, rounded to one decimal place: a

The length of AC.

The area of △ACD.

The area of △ABC.

Reasoning 26

26, 27

A 9.2 cm

85° 32° B

26−28

31° 6.9 cm

27−29

Blain and Odett are trying to ﬁnd side c in △ABC where a = 15 m, b = 12 m and ∠C = 34°. Review the work, explain the errors (if any) and determine the correct value for the side c. Blain’s work:

Odett’s work:

A rhombus of side length 10 cm has a longer diagonal of length 16 cm. a

Describe two ways you could solve for d, the length of the shorter diagonal of the rhombus.

Solve for d.

θ° 16 cm

10 cm d x°

A teacher is writing exam questions for her maths class. She draws a triangle, labels the vertices A, B and C and labels the opposite sides a = 5, b = 8 and c = 15 respectively. She wants to ask students to ﬁnd the size of ∠A. Explain what the problem is with her question.

Use the cosine rule to prove that any angle in a triangle with three equal sides is equal to 60°.

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Problem-solving 30−32, 34, 41, 42ab 30

31−35, 39−41, 42ab

32, 35−39, 42−45

Find the value of θ, rounded to two decimal places.

15 cm 80°

62° 42°

35 cm

A radio signal is sent from a transmitter at tower T, via a satellite S, to a town W, as shown in the diagram. The town is 526 kilometres from the transmitter tower. The signal is sent out from the transmitter tower at an angle of 18°, and the town receives the signal at an angle of 26°.

18°

26°

Find ∠WST.

Find the distance, SW, that the signal travels from the satellite to the town. Round your answer to the nearest kilometres.

If the satellite is h kilometres above the ground, ﬁnd h. Round your answer to two decimal places.

526 km

The Bermuda triangle is an area in the Atlantic Ocean where many planes and ships have mysteriously disappeared. Its vertices are at Bermuda, Miami and Puerto Rico.

Bermuda Triangle 1789 km

Find the area taken up by the Bermuda Triangle. Round your answer to the nearest square kilometre.

Miami

64°

1548 km Puerto Rico

Two hikers walk 10 km from point D on a bearing of S 40°E to a point E. They then walk on a bearing of N 30°E to point F. If the distance DF is 15 km, find the bearing of D from F. Assume they walk along flat terrain throughout the hike.

Two wires help support a tall pole. One wire forms an angle of 36° with the ground and the other wire forms an angle of 70° with the ground. The wires are 29 m apart. a

Find a, the angle made between the two wires at the top of the pole.

a h

70°

36° 29 m

Find d, the length of the longest wire in metres. Round your answer to two decimal places.

Find h, the height of the pole. Round your answer to two decimal places.

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Mae observes a tower at an angle of elevation of 12°. The tower is perpendicular to the ground. Walking 67 m towards the tower, she ﬁnds that the angle of elevation increases to 35°. D

35°

12° A

Find ∠ADB.

Find the length of the side a. Round your answer to two decimal places.

Find the height, h, of the tower. Round your answer to one decimal place.

Prove the sine rule using these steps: a

For △ACD, ﬁnd an expression for sin A.

For △BDC, ﬁnd an expression for sin B.

Solve for x in the equation from part (b).

Substitute your expression for x into the equation from part (a) to prove the law of sines.

C b

How many triangles are possible given this information? Explain your reasoning. Assume a is the side length opposite ∠A, b is the side length opposite ∠B. a

a = 29, b = 28, and ∠A = 37°

a = 38, b = 29, and ∠B = 45°

a = 54, b = 65, and ∠A = 60°

The sides of a triangle are in the ratio 4 : 7 : 8. Find the size of x, the smallest angle in the triangle. Round your answer to the nearest degree.

For triangle △QUV where the side lengths q, u and v appear opposite the angles Q, U and V. If q = 16, sin V = 0.5 and sin Q = 0.8. Solve for v.

Two trains leave a station at the same time. Train H travels at an average speed of 80 km/h on a bearing of 035° T, while Train I travels at an average speed of 90 km/h on a bearing of 120° T. Find the distance between the two trains after 4 hours. Round your answer to the nearest kilometre.

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

During football training, the coach marks out the perimeter of a triangular course that players need to run around. The diagram shows some measurements taken of the course, where side length a = 14 m: a

Find ∠A.

Find the length of side b, rounding your answer to two decimal places.

Find the length of side c, rounding your answer to two decimal places.

55°

Each player must sprint one lap and then jog one lap around the triangle. This process is to be repeated 3 times by each player.

87°

a = 14 m

If Tara can run 280 m/min, and can jog at half the speed she runs, calculate the time this exercise will take her, rounding your answer to one decimal place. 42

A triangular-shaped field has sides of length 25 m, 29 m and 36 m. a

Find the value of the angle x opposite the side that is 25 m long. Round your answer to two decimal places.

Find the area of the field to two decimal places.

Rhett has been hired to plough the field and to erect fencing around its perimeter. If he charges $3 per square metre for ploughing and $7 per metre for fencing, how much does he charge in total? Round your answer to two decimal places.

Two straight paths to the top of a cliff are inclined at angles of 23° and 20° to the horizontal: a b c

If path A is 108 m long, ﬁnd the height h of the cliff, rounded to the nearest metre.

Find the length x of path B, rounded to the nearest metre.

xm 108 m Path A 23°

Let the paths meet at 49° at the base of the cliff. Find their distance apart, y, at the top of the cliff, rounded to the nearest metre.

Path B

49°

20°

From a point A, due north of a tower, the angle of elevation to the top of the tower is 51°. From a point B, which is on a bearing of 120° from the tower and is 100 m away from point A, the angle of elevation is 25°. Find the height of the tower rounded to the nearest metre.

A hiking trail is inclined at 15° to the horizontal. At the bottom of the trail, there is a starting point labelled as point A. A pine tree grows vertically from its base on the trail at point B. From point A, the angle of elevation to the top of the pine tree, labelled as point D, is 28°. After hiking 80 m along the trail and passing point B, a hiker arrives at point C. From point C, the angle of elevation to point D is 58°. a

Find the value of ∠ADC.

Compute the distance, DA, rounded to one decimal place.

Determine the height of the pine tree, DB, rounded to the nearest metre.

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12 Chapter review 1

Find the length of the unknown side as an exact value: a 17

A right-angled triangle has a hypotenuse of length 26 and another side of length 24.

A right-angled triangle has a one side of length 6 and another side of length 5.

Could a triangle with side lengths of 6 mm, 8 mm, and 13 mm be a right-angled triangle? Explain.

Rachel hikes south of her starting point for 834 m, then turns and walks 691 m east, before stopping for a lunch break. She then travels south again for 427 m before arriving at her ﬁnal destination.

What is the shortest distance between where Rachel started and where she stopped for lunch? Round your answer to two decimal places.

What is the shortest distance between where Rachel started and where she ﬁnished her journey? Round your answer to two decimal places.

Find the value of f, rounded to two decimal places: a

b 27°

10 cm

f cm

29° 5m

For △DEF, ﬁnd the exact value of : a

sin θ

θ 13 F

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Jason is ﬂying a kite that is attached to the end of a 25.2 m length of string. The angle between the string and the vertical is 40°. If Jason is holding the string 1.2 m above the ground: a

Find the angle of elevation from Jason to the kite.

Find x, rounded to two decimal places.

Find the height h of the kite above the ground, rounded to two decimal places.

25.2 m

40° hm

1.2 m

Find the value of the side length, rounded to the nearest whole number: a

20 cm

58° 23 mm

h mm

67°

x cm

10 cm

The person in the picture sights a pigeon above him. Find the angle of elevation, θ, to two decimal places.

24 m

θ 17 m

An aeroplane is ﬂying at an altitude of 3000 m and approaching an airport. The pilot measures the angle of depression to the airport to be 14°. One minute later, the pilot measures the angle of depression again and ﬁnds it to be 16°. a

Find the distance AC, to the nearest metre.

Find the distance BC, to the nearest metre.

Find the distance covered by the jet in that one minute, to the nearest metre.

14° B

16°

3000 m

Airport

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At a certain time of the day a light post, 5 m tall, has a shadow of 9.4 m. If the angle of elevation of the sun at that time is θ°, ﬁnd θ rounded to two decimal places.

Dave is standing 270 m from a building and measures the angle of elevation of the top of the building to be 19°.

If the difference in height between the top of the building and Dave’s eye is h m, ﬁnd h rounded to two decimal places.

If Dave’s eye is 161 cm from the ground, ﬁnd the height of the building rounded to one decimal place.

Find the true bearing of point A from O.

Find the compass bearing of point A from O.

A 35° O

O 38° A

Find the true bearing of point A from point C.

Find the compass bearing of point A from point C.

B 38°

49°

If a person walks at a compass bearing of NE, and then turns around to go back in the direction they came, what direction are they heading now?

A boat sets off on a bearing of 121° T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

Uri sailed for 155 km on a bearing of 215°.

Find w, the distance in kilometres he has sailed west from his starting point, rounded to one decimal place. w 215° 155 km

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

During a group hike, a hiker walked 12 km NE of base camp. In the evening the hiker was separated from the group and after a night lost in the wilderness, the hiker was found 12 km NW of the base camp. What compass bearing must the hiker have followed during the night?

(Optional) Find the unknown, rounded to two decimal places. a

20 D

41°

47°

b 13

a 63°

28°

(Optional) For each triangle, calculate ∠B to the nearest degree. a

△ABC where ∠A = 48° side a = 2.7 cm and side b = 1.9 cm

△ABC where ∠A = 57° side a = 156 cm and side b = 179 cm

(Optional) To use the cosine rule to ﬁnd the length of AC, which angle would need to be given?

B 31.05 cm 38.44 cm A

(Optional) Find a using the cosine rule. Round your answers to two decimal places.

C 11

33° A

(Optional) Triangle ABC has side lengths a = 83, b = 45 and c = 110. Find ∠C. Round your answer to the nearest degree.

(Optional) Calculate the area of the following triangles to two decimal places: a

5.5 m

44° 3m

b 101°

5.7 m

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(Optional) For this triangle, calculate: a

The size of angle x, to the nearest degree

The area of the triangle, to one decimal place

(Optional) An industrial site in the shape of a triangle is to take up the space between where three roads intersect. Calculate the area of the site, rounded to two decimal places.

15.3 m

115°

17.925 m

44°

50 m 80 m

Did you know?

Trigonometry plays a key role in medical imaging, like CT scans and MRIs! These machines use trigonometric calculations to create detailed cross-sectional images of the body. By measuring angles and distances within the scans, trigonometry helps doctors see precise internal views, making it easier to diagnose and treat patients accurately. Without trigonometry, it would be challenging to create such clear, layered images of tissues and organs. Thanks to this powerful tool, medical professionals can detect issues early and provide more targeted treatments.

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

“Truth is ever to be found in simplicity, and not in the multiplicity and confusion of things.” –Isaac Newton

Big ideas • Physical objects can be modelled with 3D geometric figures whose properties can be applied to solve real-world problems. • Estimation and accuracy are fundamental concepts for practical decision-making and interpreting measurements across diverse real-world contexts. • Scale factors and ratios provide a framework for comparing and relating measurements, dimensions and volumes across different scales and contexts.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10M01 solve problems involving the surface area and volume of composite objects using appropriate units VC2M10M04 use mathematical modelling to solve practical problems involving direct and inverse proportion and scaling of objects; formulate problems and interpret solutions in terms of the situation, including the impact of measurement errors on the accuracy of results; evaluate and modify models as necessary, and report assumptions, methods and findings VC2M10N01 recognise the effect of using approximations of real numbers in repeated calculations and compare the results when using exact representations VC2M10AM01 solve problems involving surface area and volume of right pyramids, right cones, spheres and related composite solids

Measurement Chapter outline 13.01 13.02 13.03 13.04

Surface area of prisms and cylinders Volume of prisms and cylinders Surface area and volume of composite solids Estimation and accuracy Investigation: Rainfall collection Investigation: Instruments and methods Investigation: Scale models 13.05 Optional: Spheres 13.06 Optional: Cones 13.07 Optional: Pyramids Chapter 13 review

664 674 684 695

704

If you unrolled a tennis ball, how big do you think its surface area would be? Roughly the same as a playing card, a piece of toast, or this textbook?

13.01 Surface area of prisms and cylinders After this lesson, you will be able to… • find the surface area of right prisms. • find the surface area of cylinders. • solve problems involving the surface area of right prisms and cylinders.

Surface area of right prisms Surface area is the total area of the surface of a solid. To ﬁnd the surface area of a prism, identify the shapes of the different faces, and then add the area of all faces.

Example 1 Consider the rectangular prism with a width, length and height of 5 m, 7 m and 15 m respectively.

5m 7m

Find the surface area. 15 m

Create a strategy We need to add the areas of the 3 pairs of equal rectangular faces: front and back, left and right, top, and bottom.

Apply the idea Front and back = 2 × (15 × 5) 2

= 150 m

Sides = 2 × (15 × 7) 2

= 210 m

Top and bottom = 2 × (7 × 5) 2

= 70 m

Evaluate Multiply the area by 2 Evaluate Multiply the area by 2 Evaluate

Surface area = 150 + 210 + 70 2

= 430 m

664

Multiply the area by 2

Add the areas Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 2 The dimensions of a trapezoidal prism are given in the diagram (not to scale):

10 cm 6 cm

6 cm

5 cm 8 cm

14 cm

Find the surface area of the trapezoidal prism.

Create a strategy A trapezoidal prism consists of two trapezium faces and four rectangular faces.

The total surface area of the prism is the sum of the areas of all the faces.

Apply the idea Surface area = 112 + 2 × 48 + 80 + 2 ×

Area of four rectangles then two trapeziums

= 112 + 96 + 80 + 24 × 5 cm2

Evaluate the brackets and product

= 112 + 96 + 200 cm2

Evaluate the product

= 408 cm

Evaluate

Idea summary To ﬁnd the surface area of a prism, identify the shapes of the different faces, then add the areas of the faces.

Surface area of cylinders A cylinder has a curved face so it’s useful to remember what the net of a cylinder looks like to work out the surface area. A cylinder has two ends in the shape of a circle, and a rectangular piece that creates the curved surface. The total surface area (SA) of a cylinder can be expressed as: SA = 2π r2 + 2π rh

2π r

where r is the radius and h is the height of the cylinder. 13.01 Surface area of prisms and cylinders mathspace.co

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Example 3 This cylinder has a height of 74 cm and base radius of 21 cm. Find the total surface area in square centimetres. Round your answer to two decimal places. 74 cm

21 cm

Create a strategy The total surface area of a cylinder is given by the formula 2π r2 + 2π rh, where r is the radius and h is the height of the cylinder. We can substitute the given radius and height into the formula to ﬁnd the total surface area.

Apply the idea Surface area = 2π r2 + 2π rh

Surface area of cylinder

= 2π × 21 + 2π × 21 × 74 = 12 534.95 cm

Substitute the values

Evaluate the expression

Idea summary

Surface area of a cylinder = 2π r2 + 2π rh r is the radius h is the height of the cylinder

13.01 Practice questions Understanding 1

Explorer 1−3

Adventurer 1−3

Trailblazer 1−3

A cube has a side length of x cm. a

How many faces does the cube have?

What is the area of one face?

Write an expression for the surface area of the cube.

x cm

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To ﬁnd the surface area of this pentagonal prism, the area of how many faces must be added together?

For each 3D shape, identify the number and type of 2D shapes that make up the net. a

Cube

Cylinder

Square-based pyramid

Hexagonal prism

4, 6acdf, 7a, 8ab, 10, 11ab, 15a, 17−21

Fluency

4, 6bd-f, 7b, 8cd, 10, 11cd, 12a, 13, 15b, 17, 21, 22

Find the surface area of a cube with side length 5 cm.

Find the side length of a cube that has a surface area of 384 cm2.

Find the surface area:

Ex 1

4, 5, 6c-f, 7b, 8cd, 9, 11d, 12b, 13, 14, 15b, 16, 21, 22

b 9.9 m

12 m 8.7 m

4m 7.5 m

5 cm

4 cm

23 cm 6 cm

4 cm

11 cm

9 cm

10 cm 5 cm

3 cm

5 cm

16 cm

4 cm

13.01 Surface area of prisms and cylinders mathspace.co

667

For each triangular prism: i

Find the value of x.

10 cm

x cm

4 cm

Find the surface area.

16 cm

18 cm 12 cm

3 cm

8 Ex 2

Find the surface area of the prisms. Round answers to two decimal places. a

10 cm 11 cm

5 cm

10 cm

3 cm

5 cm

13 cm

9 cm

4 cm

8 cm

12 m

16 cm

3 cm 17 cm 4m

For this triangular prism:

Find the exact length of the unknown side, k.

Find the exact surface area of the triangular prism.

If the value of k was rounded to the nearest metre before calculating the surface area, by how much would the surface area be overestimated? Round your answer to two decimal places.

10 m

For this rectangular prism: 5

5 5 5 5

5 8

b 668

Determine whether the net matches the rectangular prism.

iii

8 8 8

Find the surface area of the rectangular prism.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

7 5 7 5 7 8 7

7 5 7 7

11 Ex 3

Find the surface area of each of these cylinders rounded to two decimal places. a

b 6 cm

93 cm 12 cm

30 cm

9m 8m 10 m

Find the area of the curved rectangular face of these open-ended cylinders, rounded to two decimal places. a

2m 3m

For this cylinder:

Find the circumference of the circular base rounded to four decimal places.

Find the area of the curved face of the cylinder rounded to two decimal places.

4m 2m 4m

13.01 Surface area of prisms and cylinders mathspace.co

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Find the height, h, of this closed cylinder if its surface area is 25 547 m2 and the radius of each circular face is 38 m. Round your answer to the nearest whole number.

38 m

For each cylinder: i

Find the value of x, rounded to two decimal places.

Find the total surface area, rounded to two decimal places.

4 cm

xm 23 m

13.6 cm

25 m

The area of the circular face of a cylinder is 8281π m2. The total surface area of the cylinder is 25 662π m2. a

x cm

Find the radius of the cylinder.

Find the height of the cylinder.

A cylinder has a height of 20 cm and a diameter of 12 cm. a

Find the exact surface area of the cylinder.

If the value of π was rounded to 3.1 before calculating the surface area, by how much would the surface area be overestimated? Give your answer rounded to two decimal places.

A birthday gift is placed inside the box shown, which has the shape of a rectangular prism.

12 cm 14 cm

Find the minimum amount of wrapping paper needed to cover this box. 5 cm

The roof of a shed is constructed from two rectangular and two triangular sheets of steel, forming a triangular prism with dimensions as shown in the diagram: Find the surface area of the roof of the shed (do not include the underneath rectangular face of the prism). Give your answer rounded to the nearest square metre.

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10 m 6m

Laura is building a storage chest in the shape of a rectangular prism. The chest will be 55 cm long, 41 cm deep, and 39 cm high. Find the surface area of the chest.

Find the surface area of the brickwork for this cylindrical silo. Assume that there is a brick roof but no ﬂoor. Give your answer rounded to two decimal places.

2.24 m

6.68 m

A swimming pool has the shape of a trapezoidal prism 14 m long and 6 m wide. The depth of the water ranges from 1.2 m to 2.5 m, as shown in the diagram. The walls and base of the pool are going to be tiled. Calculate the total area inside the pool that is to be tiled. Give your answer rounded to one decimal place.

Reasoning 23

23−25

14 m

1.2 m

2.5 m

25−27

Here are three formulas: • SA = 2π r2 + 2π rh

• SA = π r2 + 2π rh

• SA = 2π rh

How do you know which formula to use to calculate the surface area of a cylinder? 24

Scott uses the formula SA = 2π r(r + h) to ﬁnd the surface area of a closed cylinder. Lachlan uses the formula SA = 2π r2 + 2π rh. Who is correct?

Quicker Oatmeal Company is choosing between two new packages for their chocolate chip oats. • Option 1 is a canister with a diameter of 6 cm and a height of 10 cm. • Option 2 is a box with dimensions 3 cm × 6.5 cm × 9.5 cm.

Which package should Quicker Oatmeal use in order to minimise the amount of materials used? Explain.

No matter which package Quicker Oats chooses, they will sell it for $4.99 per package. Which package should Quicker Oats use to maximise their proﬁts? Explain.

For the composite shape:

2 cm

List the faces and their dimensions that make up the total surface area.

3 cm

Calculate the total surface area of the composite shape.

3 cm

10 cm

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Write a simpliﬁed formula to ﬁnd the surface area of a cylinder that has its radius equal to its height.

Problem-solving 28, 30, 33, 35 28

28, 29, 31, 33−35

Find the surface area of the triangular prism, rounded to one decimal place.

29, 31, 32, 34−36

h cm

21 cm 46 cm

16 cm

The Super Cereal Company sells their Frosty O’s cereal in a rectangular box with dimensions 12 cm × 10 cm × 2.5 cm. In order to reduce their environmental impact, Super Cereal wants to reduce the surface area of their boxes while maintaining the proportions of the current packaging. a

List three different scale factors that Super Cereal could use to reduce the total surface area of Frosty O’s boxes by somewhere between 15 cm2 and 25 cm2.

Give the dimensions for a Frosty O’s box that satisﬁes the required reduction in surface area.

A paint roller is cylindrical in shape and has a diameter of 6.8 cm and a width of 31.2 cm.

31.2 cm

Find the area painted by the roller when it makes 300 revolutions. Give your answer in square metres rounded to two decimal places.

672

6.8 cm

This water trough in the shape of a half cylinder is to be painted on the outside with two coats of paint. a

Find the surface area of the outside rounded to two decimal places.

If paint costs $24.50 for a 1 L tin that covers 6.5 m2, ﬁnd the cost to paint the trough.

0.93 m

Both popcorn bags are designed to carry 1680 cm3 of popcorn. Assume there’s minimal wastage of space when the popcorn is packed into each bag. a

Find the height of bag A in centimetres.

Find the height of bag B in centimetres.

Find the surface area of bag A.

Find the surface area of bag B.

Which bag should be used to reduce the amount of paper?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

2.88 m

Bag A

7 cm

Bag B

8 cm 10 cm

10 cm

Find the surface area of the trapezoidal prism, rounded to one decimal place.

y cm

3 cm

10 cm

8 cm 13 cm

The square prism has a volume of 138.72 cm3 and the side length of the square base is 3.4 cm. Calculate the surface area of the prism.

Jenny wants to make several cans like the one shown. She plans to cut them out of a sheet of material that has an area of 1683 cm2. Determine the number of complete cans she can make.

5 cm

7 cm

If two identical spherical balls with radii of 2.6 cm ﬁt exactly inside a cylinder, ﬁnd the surface area of the closed cylinder. Give your answer rounded to one decimal place.

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13.02 Volume of prisms and cylinders After this lesson, you will be able to… • apply the formula to find the volume of various prisms. • apply the formula to calculate the volume of cylinders. • solve problems involving the volume of prisms and cylinders.

Volume of prisms We can ﬁnd the volume of any prism by multiplying the base area by the height of the prism. The height of the prism is the distance between the two base faces. It’s common to use the word height to represent this dimension as we can always reorient the shape. The formula used to ﬁnd the volume of a prism is given by V = Ah, where A is the base area and h is the height of the prism.

Interactive exploration

Discover this concept in action online

mathspace.co

Example 1 Find the volume of this triangular prism.

5 cm 10 cm 5 cm

Create a strategy

Apply the idea

Use the volume formula V = Ah, where the area of the base is given by

We are given base length, b = 5, height of the triangular base, H = 5; and height of prism, h = 10. Write the formula

674

Substitute the values

Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 2 A prism has a volume of 1080 cm3. If it has a base area of 120 cm2, ﬁnd the height of the prism.

Create a strategy

Apply the idea

Use the volume of a prism formula V = Ah.

We are given V = 1080 and A = 120. Write the formula Substitute the values Divide both sides by 120 Evaluate

Idea summary The volume of a prism is given by:

V = Ah A h

is the base area is the height of the prism

Volume of cylinders We can ﬁnd the volume of a cylinder using the same method, by multiplying the area of the circular base by the height of the cylinder. The formula for the area of a circle is A = π r2, therefore the formula for the volume of a cylinder is V = π r2h, where r is the radius and h is the height of the cylinder.

Interactive exploration

Discover this concept in action online

mathspace.co

Example 3 Calculate the volume of the quarter cylinder. Round your answer to one decimal place. 14 cm

5 cm

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Create a strategy Multiply the volume of a cylinder formula by a quarter.

Apply the idea We are given r = 5 and h = 14. Multiply the formula by Substitute the values Evaluate

Idea summary The volume of a cylinder is given by:

V = π r2h r h

is the radius of the cylinder is the height of the cylinder

13.02 Practice questions Understanding

Explorer 1−3

Adventurer 1−3

Trailblazer 1−3

Are these statements true or false? a The volume of a prism or cylinder is calculated by multiplying the area of its base by its height. b The volume of a triangular prism can be found by multiplying the base area by one third of the height. c The volume of a cylinder can be determined by multiplying the area of its circular base by its height. d The volume of a rectangular prism can be calculated by multiplying the length, width and height together. e The volume of a cylinder is always greater than the volume of a rectangular prism with the same base area and height.

Explain the similarities and differences in ﬁnding the volume of a rectangular prism and the volume of a cylinder.

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Which of these prisms would have a volume that would be found by calculating 4 × 3 × 5? A

B 4 cm 3 cm

5 cm

4 cm

D 5 cm

5 cm 4 cm

3 cm

Fluency 4

4, 5ace, 7, 8a-e, 9, 11, 12

What is the shape of the prism’s base?

What is the area of the prism’s base?

iii

What is the volume of the prism?

5ace, 7, 8b-f, 9, 11, 12

4 cm

5bdf, 6, 7, 8e-h, 10, 11

b 15 cm

10 cm

6 cm 5 cm 18 cm 6 cm

3 cm

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5 Ex 1

Find the volume of these prisms: a

A = 38 cm2

A = 52 cm2 10 cm

3 cm

7 cm

11 cm

3 cm

4 cm

3 cm

14 cm 5 cm

10 cm 12 cm

6 cm

A prism has a volume of 880 cm3. If it has a height of 11 cm, ﬁnd the base area of the prism.

A prism has a volume of 1200 cm3. If it has a base area of 240 cm2, ﬁnd the height of the prism.

Ex 2

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8 Ex 3

Find the volume of these solids, rounding your answers to one decimal place. a

5 cm

10 cm

5 cm

17 cm

6 cm 8m 12 cm

3 cm

10 cm 13 cm 3 cm

16 cm 7 cm 19 cm

8 cm

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Find the volume of each of the cylinders. Round your answers to one decimal place. a

A cylinder with a radius of 6 cm and height of 15 cm.

A cylinder with a radius of 7 cm and height of 15 cm.

A cylinder with a diameter of 2 cm and height of 19 cm.

Find the volume of the ﬁgure shown, rounded to two decimal places.

5 cm

15 cm

A cylindrical spa has a radius of 1.8 m and the water is 160 cm deep. Find the volume of water in the spa in litres. Round your answer to two decimal places.

A cylindrical candle has a radius of 46 mm and a height of 0.13 m. Find the volume of the candle in cubic centimetres. Round your answer to two decimal places.

Reasoning 13

13, 14

13−15

Oliver is helping his father build a wooden toy box in the shape of a rectangular prism for his younger brother. The planned dimensions for the box are 1.8 m in length, 1.1 m in width, and 0.9 m in height. They need to calculate the volume of the box to know how much space it will take up. a

Calculate the volume of the box using the exact dimensions.

Oliver’s father prefers to work with rounded dimensions to make the measurements easier. He decides to round each measurement to the nearest metre. Calculate the volume of the box using these rounded dimensions.

Compare the two volumes. Which is greater, and by how much?

A right prism has a rectangular base and three dimensions: length, width and height. How does the volume of a right prism change if we double the length and width while keeping the height the same?

Edrian’s science class is conducting an experiment that involves ﬁlling a cylindrical container with water. The container has a radius of 2.5 m and a height of 1.9 m. The students need to know the exact volume of the container to make sure they have enough water for the experiment.

680

Calculate the volume of the container using the exact value of π and the given dimensions rounded to two decimal places.

One of the students argues that using 3.14 for π will make the calculations easier. Calculate the volume of the container using 3.14 as the approximation for π rounded to two decimal places.

Compare the volumes calculated with the exact value of pi and the approximation. Which is greater, and by how much?

Discuss the potential implications of using rounded or approximated values in these real world situations.

Determine the capacity of the container in millilitres.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving 16, 17, 20, 23−26 16

16−18, 21, 22, 24, 26, 27

19, 25−30

A rectangular swimming pool has a length of 27 m, width of 14 m and depth of 3 m. Find the capacity of the swimming pool in litres, if 1 m3 = 1000 litres.

An ice cream store has two different types of cones. One type uses a pointy cone while the other type has a ﬂat base, made up of two cylinders. Both types have a hemispherical scoop on top and are ﬁlled with ice cream. Both cones have the same thickness. Types of ice cream cone Point base cone

Flat base cone 6 cm

3 cm

3 cm 1 cm

8 cm

6 cm

4 cm

Vidar wants to have as much ice cream as possible. Determine which type he should get.

Yolanda wants to have as much cone as possible. Determine which type she should get.

There are two types of cylindrical soup cans available for Ray to purchase at his local store. One has a diameter of 16 cm and a height of 18 cm while the other has a diameter of 18 cm and a height of 16 cm. Which type of can holds more soup?

Sanford Seminole Aquatic Center has a swimming pool that is 25 metres by 55 metres, and 2.5 metres deep. Morcom Aquatic Center has a swimming pool with a volume of 6187.50 cubic metres.

Which pool has the greater volume?

How many times bigger is the volume of the bigger pool than the smaller pool? Round your answer to one decimal place.

Find the exact scale factor between the pool at Morcom and the pool at Sanford Seminole.

Jack’s mother told him to drink 3 large bottles of water each day. She gave him a cylindrical bottle with height 17 cm and radius 5 cm. a

Find the volume of the bottle. Round your answer to two decimal places.

Assuming he drinks 3 full bottles as his mother suggested, calculate the volume of water Jack drinks each day. Round your answer to two decimal places.

If Jack follows this drinking routine for a week, how many litres of water would he drink altogether? Round your answer to the nearest litre.

5 cm

17 cm

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Find the capacity of the ﬁsh tank in litres: 50 cm 30 cm

25 cm

A cylindrical swimming pool has a diameter of 5 m and a depth of 1.8 m. a

How many litres of water can the pool contain? Round your answer to the nearest litre.

Express this amount of water in kilolitres.

Find the area of the trapezium-shaped block of land.

During a heavy storm, 63 mm of rain fell over the block of land. Find the capacity of water that landed on the property in litres.

10 m 17 m

25 m

The roof of a cottage has the shape of a triangular prism. Find the volume of the roof.

10 m

A ﬁsh tank designed to sit in the corner of a room has the shape of a right-angled triangular prism. Determine the capacity of the ﬁsh tank in litres.

50 cm 30 cm 90 cm

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Find the volume of the tent in cubic metres:

150 cm 290 cm 200 cm

The roof of a house is flat and rectangular, measuring 16 m by 11 m. a

Find the area of the roof.

Find the volume that could be collected from 20 mm of rain.

How many litres of water is equivalent to the volume collected in part (b)?

Once the water tank is full, a further 5 mm of rain falls on the roof. How many litres of rainfall are not collected as a result of the tank overflowing?

The outline of a rice ﬁeld is shown. a

Find the area of the rice ﬁeld.

Throughout the monsoon season, the rice ﬁeld receives 16 cm of rain. What volume of water, in cubic metres, has fallen on the rice ﬁeld?

35 m 36 m 52 m

45 m 16 m 80 m

The outline of a construction site for the foundations of a large oﬃce building is shown. a

Find the area of the construction site.

Prolonged rainstorms bring 0.21 m of rain to the region and the foundations of the building are ﬂooded. Determine the volume of water of the construction site, in cubic metres.

85 m 85 m

85 m

13 m 84 m

The ﬂoor plan for a house is shown, with measurements given in metres. The annual rainfall over the house is 700 mm. Calculate the potential amount of water that could fall on the house over one year, if 1 m3 holds 1000 litres. Give your answer in kilolitres.

3.00 m

4.00 m 2.60 m

3.20 m

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13.03 Surface area and volume of composite solids After this lesson, you will be able to… • recognise and define composite solids. • calculate the volume of composite solids by adding or subtracting individual volumes. • identify and compute the surface area of composite solids by summing the areas of their faces. • solve problems related to the volume and surface area of composite solids.

Volume of composite solids We can ﬁnd the volume of composite solids by ﬁnding the volume of each of the solids and adding them or subtracting them. To find the volume of this composite solid we can subtract the volume of a simple solid. This composite solid is a rectangular prism with half a cylinder subtracted from it.

If a composite solid is also a prism with a composite shape as its base, we can find its volume the same way we’d find the volume of a prism - by multiplying the area of the base by the height of the prism.

Example 1 This solid consists of a rectangular prism with a smaller rectangular prism cut out of it. Find the volume of the solid. 13 cm 2 cm

6 cm

11 cm

15 cm

Create a strategy Subtract the volume of the smaller rectangular prism from the volume of the larger rectangular prism.

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Apply the idea Large prism = 6 × 15 × 13 3

= 1170 cm

Small prism = 11 × 2 × 15 3

= 330 cm

V = 1170 − 330 3

= 840 cm

Multiply the dimensions Evaluate Multiply the dimensions Evaluate Subtract the volume of one prism from the other Evaluate

Example 2 A wedding cake with three tiers rests on a table. The layers have radii of 50 cm, 54 cm and 58 cm as shown in the ﬁgure. If each layer is 21 cm high, calculate the total volume of the cake in cubic 21 cm metres. Round your answer to two decimal places. 21 cm 21 cm

r1 = 50 cm r2 = 54 cm r3 = 58 cm

r2 r3

Create a strategy Add the three volumes and convert the units to cubic metres.

Apply the idea Add the volumes Substitute the radii and heights

Evaluate

To convert to cubic metres we need to divide by 100 . Convert to m3

Evaluate and round

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Example 3 Find the volume of the ﬁgure shown, rounded to two decimal places:

10 m 4m

Create a strategy Add the volume of the square prism to the volume of the half cylinder.

Apply the idea The half cylinder has a diameter of 4 m which means its radius is 2 m. Write the formula Substitute r = 2, h = 10 Evaluate and round Write the formula Substitute s = 4, h = 10 Evaluate Add the volumes Evaluate

Reflect and check Since the composite solid was made by adding simple solids together, its volume is the sum of the volumes of the simple solids.

Idea summary Some composite solids are made by joining simple solids together. We can find the volume of these composite solids by finding the volume of each of the pieces and adding them. In the case where the composite solid is made by subtracting one simple solid from another, we can subtract volumes to find the volume of the composite solid.

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Surface area of composite solids The surface area of a composite solid is the sum of the areas of its faces. If the composite solid is a prism, we can find its surface area in the same way that we’d find the surface area of a prism. If the composite solid is not a prism, we can find the surface area by considering the areas of each face of the solid and adding them up.

Example 4 Find the surface area of the house. 3m

6m 9m 5m

Create a strategy Add all the faces together. The diagram shows which faces are equal.

Use Pythagoras’ theorem, c2 = a2 + b2, to ﬁnd the width of the roof panels.

Apply the idea Add the areas and multiply by 2 Substitute the dimensions Evaluate Side walls = 9 × 6 × 2 Multiply the length and width and multiply by 2 = 108 m2 Base = 5 × 9 2

= 45 m

Evaluate Multiply the length and width Evaluate

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Using Pythagoras’ theorem, we can find the width of the roof panels. On the diagram, we can see two right-angled triangles where the shorter sides are a = 3 and

Use Pythagoras’ theorem Evaluate Square root both sides Find the area of the 2 roof sides Evaluate Add all the areas Evaluate

Idea summary The surface area of a composite solid is the sum of the areas of its faces.

13.03 Practice questions Understanding 1 2

Explorer 1, 2, 3a-e, 4ad

Adventurer 1, 2, 3a-e, 4ab

Trailblazer 1, 2, 3c-f, 4bc

What is a composite solid? Give an example. Explain what you would do to: a

Calculate the surface area of a composite shape.

Calculate the volume of a composite shape.

Austin is testing out his new 3D printer by creating some geometric objects. What basic shapes are used to make each of these composite solids? a

5 cm

7 cm

3 cm

8 cm 2 cm

7 cm 7 cm

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8 cm

2 cm

5 cm

2 cm

4 mm 4 mm

4 cm

7 cm

3 mm

4 mm

5 cm

10 cm

7 mm

4 km

5 mm 4 mm 3 mm 5 mm

8 km

Identify the solids used to make up each of these composite solids. a

b 4 cm 7 cm

15 m 7m

8 cm

11 cm 8 cm

9 mm 16 mm

11 cm

10 mm

6 cm 8 cm

4 cm

36 mm

53 mm

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Fluency 5 Ex 1

5, 6ad, 7, 8a-e, 9a-d

5, 6bd, 7, 8b-f, 8, 9c-f

5bc, 6bc, 8d-h, 9d-h, 10

Find the volume of these composite solids, rounded to two decimal places. a

8 cm

12 cm 3 cm

3 cm

4 cm

10 cm

7 cm 15 cm 3 cm

10 cm

12 cm 20 cm 7 cm

16 cm

8 cm

5 cm

Find the volume of these composite solids, rounded to two decimal places. a

4 cm 7 cm 8 cm

b 11 cm

8 cm 15 cm

5 cm

9 cm

4 cm 2 cm

12 cm 2 cm

690

2 cm

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

9 cm

2 cm

7 Ex 2

A wedding cake has three tiers. The layers have radii of 51 cm, 55 cm and 59 cm. If each layer is 20 cm high, calculate the total volume of the cake in cubic metres.

51 cm

20 cm 20 cm

Round your answer to two decimal places.

55 cm

20 cm

59 cm

8 Ex 3

Find the volume of these composite solids, rounded to two decimal places. a

b 15 m

10 m 8m

12 m

19.5 m

9m 11 cm

8m 6 cm

17 m

8 cm

4 cm

f 6 cm

3 cm

3 cm 12 cm

16 cm 12 m 6m

h 10 cm 14

5 cm 14 cm 5 cm

24 20

15 cm

10 cm

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9 Ex 4

Find the surface area of these composite solids, rounded to two decimal places. a

7 cm

12 m

16 cm 6m

1 cm

5 cm

13 cm

7 cm

4 cm 10 cm

9 mm

2 cm

16 mm 3 cm

10 mm

3 cm 53 mm

36 mm

10 cm

h 12 cm

1 cm

7 cm

4 cm

1 cm

15 cm

5 cm

By considering the water bottle as a composite solid, estimate its surface area.

2 cm 1 cm

4 cm

20 cm

3 cm 692

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Reasoning 11

11, 12

11, 13

Brenton wants to know the volume of the basketball hoop stand at school, including the base, pole, backboard, and hoop. He’s conﬁdent that he can measure the dimensions of the different parts of the stand, but isn’t sure how to break up the volume into simple solids.

Using the image of the basketball hoop stand, identify the simple solids that make up the stand and describe the dimensions that Brenton would need to measure to ﬁnd their volume. 12

A triangular tunnel is made through a rectangular prism. a

Find the volume of the solid formed using two different methods: i ii

ind the volume of the rectangular prism and subtract F the volume of the triangular prism. ind the area of the shaded face and use Volume of F Prism = A × h.

8 cm

3 cm 7 cm

Explain the process of finding the surface area of the solid formed, including the inside of the tunnel.

The larger prism has had two identical holes carved out of it, each of which is a rectangular prism. All measurements are in metres. a

6.5 5

Explain how to find, and then calculate, the volume of the remaining solid. Round your answer to two decimal places.

14 6.5

Explain how to find, and then calculate, the surface area of the remaining solid.

Problem-solving 14ab, 16, 17 14

3 cm 6 cm

14ab, 15−17

14, 15, 18−20

A swimming pool is the shape of a trapezoidal prism.

51 m

Find the volume of the pool in m .

If the pool is three quarters full, what is the volume of the unfilled space of the pool?

If the distance of the water level to the top of the pool is h m when it is is three quarters full, ﬁnd h.

1.4 m

10 m

3.3 m 50.7 m

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A hollow pipe needs painting on both inside and outside surfaces. a

Find the external surface area of the curved surface.

Find the total surface area of the two end pieces.

Find the internal surface area.

Calculate the total surface area of the pipe.

24 cm

11 cm 10 cm

The lid of this treasure chest is found to be exactly one half of a cylindrical barrel. Find the surface area of the chest. 30 cm

42 cm

The diagram shows the design for a marquee (tent). The material to make the marquee costs $44 per square metre.

Find the area of the front of the marquee.

Find the surface area of one of the side walls (not including the roof).

Find the surface area of the entire roof.

Find the total surface area. Do not include the ﬂoor.

What is the total cost of material to make the marquee?

10 m 8m

A company manufactures nuts shaped like regular hexagonal prisms, with cylindrical bolt holes cut out of the centre. a

694

59 cm

The total surface area of a nut before the bolt hole is drilled is 14.7 cm2. Find the surface area of a single nut after the bolt hole is drilled out, including the inside area of the hole. Each nut that is manufactured requires a zinc coating to prevent corrosion. If 1 kg of zinc is enough to coat a surface area of 1 m2, how many complete nuts can be coated with 1 kg of zinc?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

0.5 cm

1.0 cm 1.2 cm

A swimming pool is composed of a trapezoidal prism joined to a half cylinder.

7m 3m

Find the volume of the pool in cubic metres.

How many complete litres of water can fit in the pool?

If the pool is ﬁlled to a height 10 cm below the top, how many complete litres of water are in the pool?

After construction works at a neighbouring property, a crack opens in the bottom of the pool and water begins to leak from the pool. It is observed that the height of the surface of the water in the pool is decreasing by 7 cm each week. How much water is leaking out each week?

Assuming water continues to leak at this rate, at the end of which week will the pool become empty?

0.8 m

A steel shed is to be constructed with a rectangular cut-out at the front for the entrance. a

Determine the surface area of the shed, assuming the floor is separate.

Construction of the shed requires an additional 0.1 m2 of sheet metal for each 1 m2 of surface area, due to overlaps and wastage. How many complete square metres of sheet metal are required to construct the shed?

2.5 m 10 m 3m 4m

If the steel sheets cost $18 per square metre, calculate the total cost of the steel required to build the shed.

13.04 Estimation and accuracy After this lesson, you will be able to… • explore rounding and how it influences the accuracy of calculations, especially in large datasets or multi-step computations. • differentiate between rounding and truncating numbers. • define the term ‘accuracy’ in the context of measurement and discuss factors that can influence it.

Effects of rounding on accuracy Rounding numbers affects the accuracy of any calculation we do with those numbers. The effect of rounding becomes especially important when dealing with large sets of data or calculations involving multiple steps. Each instance of rounding can introduce a “rounding effect” that can cumulatively impact the ﬁnal result’s accuracy.

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When banks calculate how much interest to pay customers, it is routine to truncate rather than round. For example, they may calculate that you are due to receive $465.239 in interest. They would not round to $465.24 but would truncate to $465.23. Maintaining high precision in numbers can lead to time-consuming calculations. The minimal decrease in accuracy due to rounding is sometimes an acceptable trade-off for the simplicity it provides.

Example 1 The heights of five plants are measured. The measurements are as follows (in centimetres): Plant 1 2 3 4 5

Height 25.467 cm 30.234 cm 28.789 cm 32.512 cm 29.105 cm

a What is the total height of the plants without rounding or truncation?

Create a strategy Add the given heights of the plants.

Apply the idea 25.467 + 30.234 + 28.789 + 32.512 + 29.105 = 146.107 cm

b What is the total height truncating to 2 decimal places?

Create a strategy

Apply the idea

Truncate each height to 2 decimal places, then add the truncated values.

Truncate each measurement: 25.467 → 25.46 30.234 → 30.23 28.789 → 28.78 32.512 → 32.51 29.105 → 29.10 Add the truncated heights: 25.46 + 30.23 + 28.78 + 32.51 + 29.10 = 146.08 cm

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c What is the total height rounding to 2 decimal places?

Create a strategy

Apply the idea

Round each height to 2 decimal places, then add the rounded values.

Round each measurement: 25.467 → 25.47 30.234 → 30.23 28.789 → 28.79 32.512 → 32.51 29.105 → 29.11 Add the rounded heights: 25.47 + 30.23 + 28.79 + 32.51 + 29.11 = 146.11 cm

d How do the truncated and rounded values compare to the actual values?

Create a strategy Compare the values found from previous parts.

Apply the idea Exact total height: 146.107 cm Total height with truncation: 146.08 cm Total height with rounding: 146.11 cm The truncated value is further away from the actual total height than the rounded value is. Therefore, the rounded total height is a more accurate approximation of the exact total height compared to the truncated total height.

Idea summary Use the most accurate figures you have for calculations and perform any rounding necessary at the end. Rounding before a calculation can make a significant difference to the answers, especially if: • The rounding is all in the same direction; for example all rounded up. • There are several quantities involved in the calculation. The more rounded quantities that are used in calculations, the larger the potential variances that can arise. Truncation makes small differences to individual accounts, but over long periods of time, the amounts can become significant.

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Accuracy of measurement devices We use different devices for measurement, including a ruler for length, a weighing scale for weight, a thermometer for temperature and a clock for time. The accuracy of these devices is important in providing reliable results. Accuracy refers to how close a measured value is to the true or actual value. The quality and design of the measurement device can inﬂuence the accuracy but no measuring device is perfectly accurate, there will always be some degree of uncertainty in every measurement. This uncertainty can be due to factors such as the limitations of the device itself, environmental conditions or even human error in reading the measurements.

Example 2 Lila and Alex are working on a geometry project where they measure various geometric quantities using different tools. Their measurements sometimes lead to slight discrepancies. Calculate the percentage error of Lila’s measurements, to two decimal places, if Alex’s measurements are used as the actual values for the given scenarios. a The perimeter of a rectangular garden where Lila measures the sides as 68.2 m and 42.9 m, while Alex measures them as 68.4 m and 43.0 m.

Create a strategy Calculate the perimeter of the rectangular garden based on Lila and Alex’s measurements using the formula P = 2 × (length + width), then calculate the percentage error between their measurements.

Apply the idea Perimeter according to Lila: PLila = 2 × (68.2 + 42.9) = 222.2 m

Substitute the values Evaluate

Perimeter according to Alex: PAlex = 2 × (68.4 + 43.0) Substitute the values = 222.8 m

Evaluate

The percentage error is given by:

b The area of the circular pool where Lila measures the radius as 25.8 m and uses 3.14 as an estimate of π. Alex measures the radius as 26.0 m and uses the button π on the calculator.

Create a strategy Calculate the area of the circular pool based on Lila and Alex’s measurements using the formula A = π r2 then calculate the percentage error between their measurements. 698

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Apply the idea Area according to Lila: ALila = 3.14 × (25.8)2 2

= 2090.11 m

Substitute the values Evaluate

Area according to Alex: AAlex = π × (26.0)2

Substitute the values

= 2123.72 m

Evaluate

The percentage error is given by:

c The surface area of the rectangular box on which Lila measures the dimensions as follows: length 48.6 cm, width 37.5 cm, and height 29.8 cm. Alex’s measurements are slightly different: length 48.8 cm, width 37.6 cm, and height 29.9 cm.

Create a strategy Calculate the surface area of the circular pool based on Lila and Alex’s measurements using the formula SA = 2(wl + hl + hw) then calculate the percentage error between their measurements.

Apply the idea Surface area according to Lila: SALila = 2 × (37.5 × 48.6 + 29.8 × 48.6 + 29.8 × 37.5)

Substitute the values

= 8776.56 cm

Evaluate

Surface area according to Alex: SAAlex = 2 × (37.6 × 48.8 + 29.9 × 48.8 + 29.9 × 37.6)

Substitute the values

= 8836.48 cm

Evaluate

The percentage error is given by:

d The volume of a cubic container where Lila measures the length, width, and height as 71.4 cm each. Alex measures them as 71.6 cm each. Round your answer to two decimal places.

Create a strategy Calculate the volume of the cubic container based on Lila and Alex’s measurements using the formula V = side3, then calculate the percentage error between their measurements.

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Apply the idea Volume according to Lila: VLila = (71.4)3

Substitute the values 3

= 363 994.344 cm

Evaluate

Volume according to Alex: VAlex = (71.6)3

Substitute the values 3

= 367 061.696 cm

Evaluate

The percentage error is given by:

Idea summary Measuring devices can include rulers, weighing scales, thermometers, clocks, etc. These devices are not always accurate due to: • Limitations of the device • Environmental conditions • Human error

13.04 Practice questions Understanding 1

Explorer 1−3

Adventurer 1−4

Trailblazer 1−5

Are these statements true or false? a

Truncation and rounding errors can signiﬁcantly affect the ﬁnal results in a series of calculations.

In financial calculations, rounding or truncating repeatedly over time has no impact.

The greater the number of decimal places in a calculation, the more accurate the result.

Truncation and rounding are the same things when it comes to calculations involving real numbers.

List all numbers with two decimal places that round to 6.4.

What is the largest number with two decimal places that would round to 5.4?

Three two-digit numbers with two decimal places are first rounded to the nearest tenth and then added together. Describe the size of the error that could result.

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What is the smallest number with three decimal places that would give an answer of 12.72 when rounded to two decimal places?

Fluency 6

7 Ex 1

8 Ex 2

6ab, 7, 8

6−8

Tony is calculating the surface area of the solid shown, where the height and diameter of the cylinder are both 5 cm. a

Find the smallest value that each of the measurements could be.

Find the largest value that each of the measurements could be.

Find the smallest possible surface area for the solid, rounded to the nearest whole number.

Find the largest possible surface area for the solid, rounded to the nearest whole number.

5 cm

8 cm

6 cm

The weights of five boxes of fruits are measured (in kilograms).

Box

Weight

What is the total weight without rounding or truncation?

12.783

What is the total weight truncating to 2 decimal places?

15.482

What is the total weight rounding to 2 decimal places?

11.926

How do the truncated and rounded values compare to the actual values?

13.643

14.217

Jasper has a ruler that is marked in cm. When he 1 2 3 measures something with it, he looks carefully to see whether the object is less than halfway between two marks, in which case he rounds down, or more than halfway between two marks, in which case he rounds up.

When measuring the rectangle shown, he would record the length as 4 cm. Betty has a ruler that is marked in both cm and mm, so that she can get lengths in cm to one decimal place. Calculate the percentage error, to one decimal place, in Jasper’s measurement compared to Betty’s measurement for: a

The perimeter of a rectangle which Betty measures as being 5.3 cm long and 4.4 cm wide.

The area of a circle where Betty measures the radius as 4.3 cm, given that Betty uses the button π on the calculator and Jasper uses 3.1 as an estimate of π.

The surface area of a rectangular prism which Betty measures as 7.8 cm long, 5.7 cm wide and 4.1 cm high.

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Reasoning

9, 10

9−11

10−12

A company uses a faulty weighing scale that adds 0.1 kg to every measurement. How might this error inﬂuence their cost calculations if they are dealing with large quantities of materials?

Caleb measures the length of the classroom to be 7 metres. Ida says the same length is 6.8 metres and Lavinia gives her answer as 6.85 metres. a

Explain how all three people could have different answers for the same measurement.

Write down the level of accuracy being used by each person.

Are all their answers correct? Explain your reasoning.

A bank truncates interest rates at the third decimal place for daily calculations on a savings account. Over a year, how could this affect the total interest received by a large number of customers?

Alex and Liam are physics students conducting an experiment to measure the kinetic energy of a ball rolling down a ramp. They know the formula for kinetic energy is

mv2, where m is the mass

and v is the velocity. Alex has an old scale which only measures mass to the nearest 10 g (0.01 kg). Liam has an accurate scale which gives the mass in kilograms to three decimal places. To measure the velocity, Alex has a simple timer that gives the time it takes for the ball to roll a certain distance to the nearest second, whereas Liam has a radar gun that gives the velocity in m/s rounded to two decimal places. In one trial, where the ball rolls 5 metres, Liam measures the mass of the ball as 0.365 kg and the velocity as 2.24 m/s. a

Find the measurements that Alex will use in his calculations.

Calculate the kinetic energy of the ball for: i

Alex

Liam

Using the answers from (b), calculate the percentage error in Alex’s answer.

The students are then told that they can use the same data to calculate the momentum of the ball using the formula p = mv. Calculate the momentum of the ball for: i

702

Alex

Liam

Using the answers from (d), calculate the percentage error in Alex’s answer.

What do you notice about the answers to parts (c) and (e)? Explain why this is the case.

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Problem-solving 13 13

13, 14

13−15

Keepsafe Family Bank pays a small amount of interest on their customers’ accounts when the accounts are in credit (the account holder has money in the bank). The interest is added to the account daily, based on the balance in the account at the end of the day. Before they add the interest to the account, they truncate the amount, removing anything less than one cent, so if they calculated the interest to be $2.5387, the customer would receive $2.53, and the bank keeps $0.0087 or 0.87 cents. If the bank has 5 million customers with accounts in credit and the average amount truncated is $0.005, how much money does the bank keep from these interest payments in a standard 365-day year?

The dating of artefacts by Argon-Argon suggests that those found in volcanic deposits at a site called Tower Hill dated from 36.8 ± 3.8 thousand years ago. a

Find the earliest date (the longest time ago) that the artefacts can be from, measured in thousands of years.

Find the latest date (the shortest time ago) that the artefacts can be from, measured in thousands of years.

Express 3.8 as a percentage of 36.8, rounded to one decimal place.

Prior to 1999, the estimated age of the universe was 13.5 Byr ± 6.5 Byr (Byr means billion years). a

The absolute error compared to 13.5 Byr is 6.5 Byr. Find the percentage error, rounded to two decimal places.

Due to improvements in astronomical data, the current estimate of the age of the universe is 13.5 Byr, with an absolute error of just 200 million years. Find the percentage error in the current estimate, rounded to two decimal places.

Investigation: Rainfall collection

Investigate online

mathspace.co

Investigation: Instruments and methods

Investigate online

mathspace.co

Investigation: Scale models

Investigate online

mathspace.co

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13 Chapter review 1

Find the surface area: a

16 m 6 cm

d 13 cm

7 cm

5 cm

12 cm

26 cm 8 cm

10 cm

8 cm

Find the height, h, of a closed cylindrical tower given its surface area is 28 000 m2 and the radius of its circular base is 40 m. Round your answer to the nearest whole number.

Fruit and Nut Co. is deciding between two new packaging designs for their raisins. • Design A is a cylinder with a diameter of 10.2 cm and a height of 11.5 cm. • Design B is a rectangular prism with dimensions 4.4 cm × 11 cm × 12.9 cm. Which design should Fruit and Nut Co. select if they want to use the least amount of material for the packaging? Justify your answer.

For the cylinder: a

Find the value of x, rounded to two decimal places.

Find the total surface area, rounded to two decimal places.

5 cm

15.7 cm

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x cm

Find the volume of these prisms: a

b 5 cm

A = 35 cm2 10 cm

14 cm 4 cm

17 cm

4 cm

10 cm 8 cm

15 cm 7 cm

A storage box has a volume of 990 cm3. If its height is 9 cm, determine the base area of the storage box.

Find the volume of these solids, rounding your answers to two decimal places: a

11 cm 20 cm 12 cm

8 cm

A city park has a rectangular playground that measures 19 m by 12 m. a

If the playground is to be covered with a layer of sand for safety reasons, how many cubic metres of sand would be needed for a 15 mm thick layer across the entire playground?

If the workers mistakenly spread an additional 3 mm thick layer of sand across the playground, how many extra cubic metres of sand were used?

Find the volume. Round your answers to two decimal places where necessary. a

9 cm

12 cm

11 cm

9 cm

15 cm 5 cm

7 cm

5 cm

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For each ﬁgure, ﬁnd to two decimal places: i

The volume

6 cm

The surface area

9 cm 6 cm

6 cm

A wedding cake has three tiers. The layers have radii of 15 cm, 20 cm and 25 cm. Each layer is 15 cm high. Calculate the total volume of the cake, rounded to two decimal places.

15 cm

Find the amount of yellow icing needed to cover the top tier of the cake, rounded to two decimal places.

15 cm

Find the amount of white icing needed to cover the middle layer of the cake, rounded to two decimal places.

18 cm

6 cm

15 cm

20 cm 15 cm 25 cm

Find the amount of pink icing needed to cover the bottom layer of the cake, rounded to two decimal places.

The diagram shows the design for a shed. a

Find the total surface area of the shed. Do not include the ﬂoor.

4m 2

If paint costs $24 for a 1 L tin that covers 6.5 m , ﬁnd the cost to paint the shed, rounded to the nearest whole dollar.

8m 11 m 9m

Kareem, Alea, and Trina are calculating the volume of the solid shown. Kareem uses 3.14 for pi in his calculations, Alea uses

for pi in her

calculations, and Trina uses the exact value of pi in her calculations.

706

Whose estimation of the volume will be most accurate?

Whose estimation with be least accurate?

Find the difference between Alea’s estimation and Trina’s estimation, rounding each answer to three decimal places.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

10 cm 4 cm

An online streaming service tracks the amount of time a person spends watching a video, rounded down to the nearest minute. Over a month, how might this affect the total viewing time calculations for a large number of users?

A bakery uses a timer that is consistently 30 seconds slow. How might this error impact the quality of the baked items?

Before 2005, scientists believed the oldest fossilized plant was 500 Myr ± 25 Myr (Myr means million years). a

Find the oldest age of the plant fossil.

Find the youngest age of the plant fossil.

Given the absolute error in relation to 500 Myr is 25 Myr, determine the percentage error, rounding to two decimal places.

With the discovery of new fossils, the current estimation for the age of the oldest plant fossil is 500 Myr with an absolute error of just 10 million years. Determine the percentage error of this newer estimation, rounded to two decimal places.

Did you know?

Measurement accuracy has transformed sports! In events like swimming and sprinting, times are measured down to the millisecond, making it possible to determine winners by the tiniest margins. Technologies like laser timers and high-speed cameras ensure that every race is as fair and precise as possible. This level of precision also helps athletes improve by tracking their progress in minute detail, pushing the boundaries of human performance. Measurement advancements not only add excitement for fans but also give athletes the data they need to break records and set new standards.

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Big ideas Probability helps us analyse the chance that an event will occur and provides us with tools to make decisions about the future based on known information.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10P01 use the language of ‘if … then …’, ‘given’, ‘of’ and ‘knowing that’ to investigate conditional statements and identify common mistakes in interpreting such language, and describe and interpret situations involving conditional probability; design and conduct simulations using digital tools to model conditional probability and interpret results VC2M10P02 describe the results of two- and three-step chance experiments, both with and without replacements, assign probabilities to outcomes and determine probabilities of events; investigate the concept of independence VC2M10AA02 devise and use algorithms and simulations to solve mathematical problems VC2M10AP01 explore counting principles, and factorial notation as a representation that provides efficient counting in multiplicative contexts, including calculations of probabilities

Probability Chapter outline 14.A Review Venn diagrams and two-way tables 14.01 Independent and dependent events 14.02 Conditional probability 14.03 Mixed probability concepts 14.04 Optional: Geometric probability Investigation: Simulation Chapter 14 review

710 724 736

751

Scientists use probability methods to determine the likelihood of space debris hitting land.

14.01 Independent and dependent events After this lesson, you will be able to… • differentiate between independent and dependent events. • recognise the role of replacement in probability scenarios. • calculate probabilities of events occurring with and without replacement. • apply mathematical formulas to determine the probability of combined events. • analyse real-world examples to identify whether events are independent or dependent. • use tree diagrams to visualise sequences of events and calculate probabilities.

Independent and dependent events Two events are independent if the outcome of each event does not affect the outcome of the other event. For example, landing on heads and landing on tails when tossing a fair coin are independent events because the first toss has no effect on the second toss. If two events are independent, the probability of both events occurring is: P ( A ∩ B) = P ( A) × P (B) The converse statement is also true. This means that if P ( A ∩ B) = P ( A) × P (B), then A and B are independent. Two events that are not independent are called dependent. For example, receiving a first place prize and receiving a second place prize in a race are dependent events. Once someone has won the first place prize, the probability of winning the second place prize will increase for those who are still in the race.

Example 1 A fair standard six-sided die is rolled 691 times. a If it lands on a six 108 times, what is the probability that the next roll will land on a six?

Create a strategy We can use the formula:

Apply the idea The previous results do not affect the outcome of the next roll. There are 6 numbers on a die and 1 of them is a six.

710

Substitute the number of outcomes

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b Is the outcome of the next roll independent of or dependent on the outcomes of previous rolls?

Apply the idea The outcome of the next roll is not affected by the outcomes of previous rolls. So the outcome of the next roll is independent.

Example 2 Two events A and B are such that P ( A ∩ B) = 0.3 and P ( A) = 0.5. Calculate P (B) if events A and B are independent.

Create a strategy

Apply the idea

Use the fact that if the events A and B are independent, then P ( A) × P (B) = P ( A ∩ B).

Write the equation Substitute the values Divide both sides by 0.5 Evaluate

Idea summary Two events are independent if the outcome of each event does not affect the outcome of the other event. Two events that are not independent are called dependent. To check if two events are independent you can use: P ( A ∩ B) = P ( A) × P (B)

Replacement

Probability

A probability tree diagram is useful in tracking two events that happen one after the other. At each point, it branches out to all the possible events that could occur from that point. The probabilities of each event is written on the branches. For example, selecting two cards from a deck of cards, one after the other, are events that occur in sequence. There are two ways this can happen: • Draw a card, then put it back in the deck before selecting another card • Draw a card and keep it, then select another card from the deck

H Branch

0.1 Outcome 0.9 T

The ﬁrst method is described as “with replacement” because the card is placed back into the deck. The second method is called “without replacement” because the card is not placed back into the deck.

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Exploration The jar of marbles shown contains 7 red marbles, 3 blue marbles, and 10 green marbles. 1. A red marble is drawn from the jar, then placed back into the jar. How many marbles are in the jar? 2. A green marble is removed from the jar. How many marbles are left in the jar? 3. Suppose one marble is drawn and placed back into the jar, then a second marble is drawn. Let event A represent drawing a red marble and event B represent drawing a green marble in the second draw. Are these events independent or dependent? Explain. 4. Suppose one marble is drawn, then a second marble is drawn. The ﬁrst marble was not replaced. Are events A and B independent or dependent? Explain.

When selecting objects from a group “with replacement”, selections are independent. Each time you select a card, you have the same probabilities. R R

For example, selecting a red card from a standard deck will be

every time if you replace the card.

B R

B B

When selecting objects from a group “without replacement”, selections are dependent. Each time you select a card, you change the probabilities for the next selection. R R

For example, the probability of drawing a red card from a standard deck is

B R

If we draw a red card and do not replace it, there are only 25 red cards left and only 51 cards left in the deck. The probability of selecting a second red card is

. This can be seen in the top branches of the tree

diagram.

B B

This shows the probability of the second event is dependent on what colour card was drawn ﬁrst.

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Example 3 A number game uses a basket with 10 balls, all labelled with numbers from 1 to 10. One ball is drawn at random, then another ball is drawn at random. a What is the probability that the ball labelled 2 is picked once if the balls are drawn with replacement?

Create a strategy Find the probability for each way that this could happen, then add them together.

Apply the idea This could happen in two ways: 1. P (2) × P (2′): The ball labelled 2 is picked ﬁrst and a ball not labelled 2 is picked second. 2. P (2′) × P (2): A ball not labelled 2 is picked ﬁrst and the ball labelled 2 is picked second. Find the probability of picking the ball with 2

Find the probability of not picking the ball with 2 Add the probabilities of the two ways

Substitute the probabilities

Evaluate and simplify

b What is the probability that the ball labelled 2 is picked once if the balls are drawn without replacement?

Create a strategy Find the probability for each way that this could happen, then add them together. For the second event, the total number of balls will need to decrease.

Apply the idea Similarly to part (a) this could happen in two ways. However, for the probability of the second ball, the total number of outcomes will be 10 − 1 = 9 since the ball was not replaced. Add the probabilities of the two ways

Substitute the probabilities

Evaluate and simplify

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Example 4 Three marbles are randomly drawn without replacement from a bag containing 6 red, 6 yellow, 4 white, 6 black and 4 green marbles. Find: a The probability of drawing 3 white marbles.

Create a strategy Remember that the number of white marbles and the total number of marbles in the bag decrease by 1 every time a white marble is drawn.

Apply the idea There are 4 white marbles and a total of 26 marbles inside the bag. Multiply the probabilities

Evaluate

b The probability of drawing 3 black marbles.

Create a strategy Remember that the number of black marbles and the total number of marbles in the bag decrease by 1 every time a black marble is drawn.

Apply the idea There are 6 black marbles and a total of 26 marbles inside the bag. Multiply the probabilities

Evaluate

c The probability of drawing zero green marbles.

Create a strategy Remember that the number of non-green marbles and the total number of marbles in the bag decrease by 1 every time a non-green marble is drawn.

Apply the idea There are 22 non-green marbles inside the bag. Multiply the probabilities

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d The probability of drawing at least one red marble.

Create a strategy Use the fact that P (at least one red marble) is the complement of P (no red marbles).

Apply the idea There is a total of 20 non-red marbles inside the bag. Multiply the probabilities

Evaluate

We now calculate P (at least one red marble): Subtract P (at least one red marble) from 1

Evaluate

Example 5 Bart is purchasing a plane ticket online. He notices there are only 4 seats remaining, 1 of them is a window seat (W) and the other 3 are aisle seats ( A). Bart purchases the ticket first. His friend, Joh, gets on the computer and purchases a ticket immediately after. The seats are randomly allocated at the time of purchase. a Fill in the probabilities matching the edges of the probability tree for the seat Bart receives and the seat his friend receives:

A W

W d

b A

Create a strategy Use the fact that the ﬁrst event is Bart receiving his seat allocation and the second event is Bart’s friend (Joh) receiving a seat allocation.

Apply the idea

There are 3 aisle seats and 1 window seat. This means that when Bart is assigned a seat, the probability of getting a window seat is , and the probability of getting an aisle seat is .

d A

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A 1

d A

A A

If Bart gets a window seat, there are only aisle seats left when Joh purchases his ticket. So the probability that Joh gets an aisle seat is 1.

If Bart gets an aisle seat, there are 2 aisle seats and 1 window seat left when Joh purchases his ticket. The probability that Joh gets a window seat is , and the probability that he gets an aisle seat is .

A A

So, the probabilities matching the edges of the probability tree for the seats are , and

b What is the probability that Bart’s friend, Joh, has an aisle seat?

Create a strategy Find the probability for each way that this could happen, then add them together.

Apply the idea A W

A A

716

The second set of branches that finish on a node labelled A correspond to Joh having an aisle seat. That is, WA and AA. The probability of each event occurring is found by multiplying previous probabilities along each branch.

AW AA

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

This means that the probability that Joh has an aisle seat is given by:

Evaluate the multiplication Evaluate the addition

c What is the probability of Joh receiving an aisle seat if Bart has a window seat?

Create a strategy

Apply the idea

Use the probabilities found in part (a).

If Bart receives a window seat, only aisle seats are left. So, the probability of Joh receiving an aisle seat is 1.

Idea summary Events “with replacement” occur when the item drawn is placed back into the group before each selection. Each selection is independent of the others. Events “without replacement” occur when the item remains outside the group after selection. Each selection is dependent of the others. The probabilities of each selection will change depending on previous selections.

14.01 Practice questions Understanding 1

Explorer 1−3

Adventurer 1, 2, 4, 5

Trailblazer 1−5

Identify whether these events are independent or dependent: a

A card is randomly selected from a normal deck of cards, and then returned to the deck. The deck is shuﬄed and another card is selected.

Each student is allowed to randomly pick an item from the teacher’s prize bag.

A coin is tossed, the result noted, and then it is tossed again.

Ursula takes a bus to the station and then immediately gets on a train to work. Is the probability of her missing the train independent or dependent on her missing her bus?

Dean has a box that contains ﬁve blue marbles and four red marbles. He takes a marble from the box, then takes a second marble from the box. Determine whether these events are dependent or independent if: a

He replaces the ﬁrst marble before taking the second.

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A bag contains 3 pink balls and 2 orange balls. You draw one ball, note its colour, then draw another ball without putting the first ball back. a

Find the missing probability for drawing a second pink ball on the tree diagram.

Which outcome can be found by multiplying

Which outcome(s) has a probability of

Are the events of drawing the balls independent or dependent?

O O

P O

by ?

On the island of Guam, the probability that it rains during rainy season is P (R). This is shown in the tree diagram. a

What do the two sets of branches represent?

Does the weather on the first day affect the chance of rain on the second day?

Which two events have the same probability?

R 0.6

Ex 1

Ex 2

6, 7, 9, 11−13, 15−17, 19, 22

6, 7, 9−13, 15−17, 19, 22

0.6

0.4

PO PP

S R

6, 8−10, 12, 14−16, 18, 20, 21, 23

On a roulette table, a ball can land on one of 18 red or 18 black numbers. a

If it lands on a red number on the ﬁrst go, ﬁnd the probability that it will land on a red number on the second go.

Are the successive events of twice landing on a red number dependent or independent?

From a standard pack of cards, 1 card is randomly drawn and then put back into the pack. A second card is then drawn. a

Find the probability that neither of the cards are spades.

Find the probability that at least 1 of the cards is a spade.

Are the two draws dependent or independent?

Two standard dice are rolled. One is red and one is white. a

Are the results from the red die and the white die dependent or independent?

Calculate the probability that: i

The same number is rolled.

The sum of the two numbers exceeds 9.

iii

The red die is 4 and the white die is 6.

The red die is even and the white die is odd.

The sum of the two numbers is less than 2.

Two events A and B are such that P ( A ∩ B) = 0.1 and P ( A) = 0.2 Calculate P (B) if events A and B are independent.

718

0.6 0.4

Fluency

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Two events A and B are such that P ( A ∩ B) = 0.04 and P ( A) = 0.4 If P (B) = 0.1, state whether the events A and B are independent.

A number game uses a basket with 8 balls, all labelled with numbers from 1 to 8. Three balls are drawn at random, with replacement. Find the probability that the ball labelled 4 is picked:

Ex 3

Exactly once

Exactly twice

Exactly three times

Christa randomly selects two cards from a normal deck of cards. Calculate the probability, in each of the following, in the case that the cards are selected:

With replacement

Without replacement

The ﬁrst card is a Queen of diamonds and the second card is a 10 of spades.

Both cards are 10s.

The ﬁrst card is a Queen and the second card is a King

The ﬁrst card is red and the second card is not a spade.

Mae deals two cards from a normal deck of cards, without replacement. a b

Are the events of drawing Card 1 and drawing Card 2 dependent or independent? Calculate the probability that she deals: i

Two 10s

Two red cards

iii

Two diamonds

A 10 of spades and an Ace of diamonds, in that order

A 10 of clubs and an Ace of spades, in any order

Valentina randomly selects three cards from a normal deck of cards. Calculate the probability if the cards are selected: i

With replacement

The cards are ﬁve of clubs, King of clubs, and Jack of spades, in that order.

The cards are all red.

The ﬁrst card is the 2 of clubs, the second card is a spade and the third card is red.

The cards are all spades.

None of the cards is an 8.

Without replacement

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Ex 4

Three marbles are randomly drawn without replacement from a bag containing 5 red, 4 yellow, 5 white, 4 black and 5 green marbles. Find the probability of drawing: a

Three white marbles

Three black marbles

Zero green marbles

Zero yellow marbles

At least one red marble

In a game of cards, suppose a hand contains a 10, a Jack, a Queen, a King and an Ace. Two cards are drawn from the hand at random, in succession and without replacement. Find the probability that: a

The Ace is drawn.

The King is not drawn.

The Queen is the second card drawn.

A number game uses a basket with 5 balls, all labelled with numbers from 1 to 5. Three balls are drawn at random, without replacement. Find the probability that: a

The ball labelled 4 is picked.

The ball labelled 4 is picked and the ball labelled 1 is also picked.

From a set of 10 cards numbered 1 to 10, two cards are drawn at random, in succession and without replacement. Find the probability that: a

Both numbers are even.

One is even and the other is odd.

The sum of the numbers is 12.

Three fair coins are tossed. Find the probability of obtaining: H H T H H T T H H T T

H T T

720

At least 1 head.

TTH in that order.

THH in that order.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

HHH HHT HTH HTT THH THT TTH TTT

A basketball player is practicing his free throws. He makes the shot with a probability of and misses with the probability of . The tree diagram shows all the possible outcomes and probabilities. Find the probability that he will: Hit Hit Miss Hit Hit Miss Miss Hit Hit Miss

Miss

Hit Miss Miss

Make all three free throws.

Miss at least one free throw.

Make at least one free throw.

James owns four green jackets and three blue jackets. He selects one of the jackets at random for himself and then another jacket at random for his friend. Find the probability that:

G G

James selects a blue jacket for himself.

Both jackets James selects are green.

B B Ex 5

Han owns four green ties and three blue ties. He selects one of the ties at random for himself and then another tie at random for his friend. a

Write the probabilities for the outcomes on the edges of the probability tree diagram.

Calculate the probability that: i

Han selects a blue tie for himself.

Han selects two green ties.

GG GB

BG BB

G G

B G

B B

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An ice cream shop offers one flavour of ice cream at a discounted price each day. There are 6 flavours for the owner to choose from that could be discounted. 3 of the flavours are sorbet and the other 3 are gelato. The three sorbet flavours are raspberry (R), lemon (L) and chocolate (C). The three gelato flavours are vanilla (V ), mint (M ) and chocolate (C). Each decision has equal probability. a

Write the sample space for which ice cream type and ﬂavour may be chosen.

Find the probability that the lemon sorbet is selected.

Determine the probability that the ﬂavour will be chocolate.

Reasoning 24

24, 25

R S

C V M G C

24, 26, 27

In a lottery there are 37 balls. a

Find the probability of the 22 ball being drawn ﬁrst.

A ball is discarded after it has been drawn. If ball number 22 is drawn on the ﬁrst go, ﬁnd the probability of ball number 15 being drawn next.

Is the probability of each successive ball drawn the same as the probability of the ﬁrst ball drawn?

Are the draws dependent or independent events? Explain your reasoning.

Two cards are to be chosen from a pack of cards numbered 1 to 11. Explain how to ﬁnd the probability of drawing a 1, 2 or 3 ﬁrst, and after replacing the card, drawing a 9, 10 or 11.

Recall that for mutually exclusive events, P ( A ∩ B) = 0. For two events A and B, where P ( A) ≠ 0 and P (B) ≠ 0, explain why the events cannot be both independent and mutually exclusive.

Using the Venn diagram shown, explain why events A and B are independent. A

6 4

Problem-solving 28−30, 33 28

722

29, 31, 33

28, 29, 31, 32, 34

Consider the word WOLLONGONG. If three letters are randomly selected from it without replacement, ﬁnd the probability that: a

The letters are W, O, L, in that order.

The letters are O, N, G, in that order.

All three letters are O.

None of the three letters is an O.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Sarah has a box of Favourites chocolates. In this box there are 30 chocolates, 5 of which are Turkish Delights. Sarah takes and eats a chocolate without looking until she gets a Turkish Delight. a

Find the probability she only eats one chocolate.

Find the probability she eats only two chocolates, rounded to the nearest percent.

Find the probability she eats ﬁve chocolates, rounded to the nearest percent.

As she eats more and more chocolates, state whether the probability of the next chocolate being a Turkish Delight is getting higher or lower.

How many chocolates must Sarah have to eat to be certain that the next chocolate will be a Turkish Delight?

Tom has shuﬄed a standard pack of 52 playing cards. He draws one card from the pack at a time, and then puts it aside. He will stop when he draws the 9 of diamonds. a

Find the probability he has to draw exactly 1 card.

Find the probability he has to draw exactly 2 cards.

Find the probability he has to draw exactly 3 cards.

Find the probability Tom will stop after the 44th card.

As more cards are drawn that aren’t the desired card, does the probability that the next card selected is the 9 of diamonds increase or decrease?

A bag contains 3 green marbles and 4 purple marbles. Two marbles are drawn at random from the bag. a

If the ﬁrst marble is replaced before the second marble is drawn, complete the table with the probabilities matching the edges of the probability tree. If the ﬁrst marble is not replaced before the second marble is drawn, complete the table with the probabilities matching the edges of the probability tree.

G A

P G

B P

GG GP

Determine the impact that replacing the marble before drawing a second has on the probability of these events: i

Drawing two green marbles

Drawing two purple marbles

iii

Drawing marbles of different colours

Edge

Probability

Sureya has three races to swim at her school swimming carnival. The probability that she wins a particular race is dependent on whether she won the previous races, as summarised: • The chance she wins the first race is 0.7 • If she wins the first race, the chance of winning the second is 0.8 • If she loses the first race, the chance of winning the second is 0.4 • If she wins the first two, the chance of winning the third race is 0.9 • If she lost the first two, the chance of losing the third race is 0.9 • If she won only one of the first two races, the chance of winning the third is 0.6 a

Construct a probability tree to represent all outcomes in this situation.

Calculate the probability Sureya won all three races, rounded to three decimal places.

Calculate the probability Sureya won the third race, rounded to three decimal places. 14.01 Independent and dependent events mathspace.co

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For breakfast, Moriah has something to eat and drinks a hot drink. She will either eat toast or cereal and then will drink tea or Milo.

• The chance of Moriah making toast is 0.7 • The chance of Moriah drinking Milo is 0.4 a

Construct a tree diagram illustrating all possible combinations of food and drink Moriah can have for breakfast and their associated probabilities.

Find the probability Moriah drinks tea and eats toast.

Find the probability Moriah drinks tea or eats toast.

The proportion of scholarship recipients at a particular university is

. The number of

students at the university is so large that even if a student is removed, we can say that the proportion of scholarship recipients remains the same. If three students are selected at random: a

Construct a probability tree showing all the possible combinations of recipients and non-recipients.

Find the probability that at least one of the students is a scholarship recipient.

Find the probability that at least one of the students is a non-recipient.

Find the probability there is at least one recipient and one non-recipient in the selection.

14.02 Conditional probability After this lesson, you will be able to… • calculate probabilities using the conditional probability formula. • interpret two-way tables, tree diagrams, and Venn diagrams in the context of conditional probability. • solve real-world problems involving conditional probability. • identify the mathematical properties of conditional probability in independent events.

Conditional probability We are often interested in probabilities under certain conditions. For example, if an even number is rolled on a die, what is the probability that it is a six? Conditions will reduce the sample space we are concerned with.

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Exploration A group of people were asked whether they went on holiday last summer. The results are provided in the given table: Teacher Not a teacher Total

Holiday 22 32 54

No holiday 26 20 46

Total 48 52 100

1. When randomly selecting a person from all those surveyed, how many people are we selecting from? 2. When randomly selecting a person from those who are not teachers, how many people are we selecting from? 3. Explain why your answers to the previous questions are different. 4. What is the probability of randomly selecting a person that is not a teacher and went on holiday? 5. Of those who are not teachers, what is the probability of randomly selecting a person that went on holiday? 6. Show that the probability from question 5 is equal to the probability from question 4 divided by the probability of randomly selecting someone who is not a teacher from the people surveyed.

For two events A and B, the probability of event A occurring given that event B has occurred has the notation P ( A∣B), which is read as the “probability of A given B”. This is known as conditional probability and is calculated by:

Note that if we have a Venn diagram or a two-way table, then we may use the number of favourable and total outcomes instead of their probabilities:

If we know the conditional probability, we can calculate the probability of A and B using: P ( A ∩ B) = P ( A∣B) × P (B) or P ( A ∩ B) = P (B∣A) × P ( A) For independent events, the probability of the second event does not depend on the first event, so P ( A∣B) = P ( A).

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Example 1 Find P( A ∩ B) given these probabilities for an experiment in which A and B are two possible events: P (B∣A) = 0.8 and P (A) = 0.5

Create a strategy Use the formula P (B∣A) =

Apply the idea Write the formula

Substitute known values

Multiply both sides by 0.5

Evaluate

Example 2 Use the two-way table to ﬁnd the probabilities. A A′ Total

B′

11 21 32

10 8 18

Total 21 29 50

a Find P ( A∣B).

Create a strategy Use the formula P ( A∣B) =

Apply the idea To use the formula for conditional probability, we need to ﬁnd P ( A ∩ B) and P (B). From the table, we see P ( A ∩ B) =

and P (B) =

. Write the formula

Substitute known values

Evaluate

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Reﬂect and check We can also ﬁnd P ( A∣B) without the formula. Since we are given B as the sample space, the total number of outcomes is 32. Since the total number of favourable outcomes in the sample space is A = 11, then P ( A∣B) =

b Find P ( A∣B′).

Create a strategy The formula for this conditional probability is P ( A∣B′) = P ( A ∩ B′) and P (B′).

, so use the table to ﬁnd

Apply the idea From the table, we see P ( A ∩ B′) =

and P (B′) =

Write the formula

Substitute known values

Evaluate

Simplify

Reﬂect and check Checking our answer, the total number of outcomes is B′ = 18 and there are A = 10 favourable outcomes in B′, so P ( A∣B′) =

c Find P ( A′∣B′).

Create a strategy From the previous part, we know the total number of outcomes is B′ = 18. Next, we use the table to ﬁnd how many outcomes of A′ are in B′.

Apply the idea From the table, we see there are A′ = 8 favourable outcomes. So, Substitute known values

Simplify

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Reﬂect and check Using the formula to check our answer: Write the formula

Substitute P ( A′ ∩ B′) =

Evaluate

Simplify

and P (B′) =

Example 3 A basketball team has a probability of 0.8 of winning its first season and 0.15 of winning its first season and its second season. What is the probability of winning the second season, given they won the first?

Create a strategy Use the formula P (B∣A) =

with:

• Event A: Winning the 1st season • Event B: Winning 2nd season

Apply the idea From the given information we know that: P (Win 1st ∩ Win 2nd) = 0.15 and P (Win 1st) = 0.8. We want to ﬁnd P (Win 2nd∣Win 1st): Start with the conditional probability formula

Substitute the probabilities

Simplify

Example 4 A flight departs from Melbourne to Sydney. The probability that the flight departs on time, given the weather is fine in Melbourne is 0.9, and the probability that the flight departs on time, given the weather is not fine in Melbourne is 0.7. The probability that the weather is fine on any particular day in July is 0.4.

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a By constructing a tree diagram or otherwise, ﬁnd the probability that the ﬂight from Melbourne to Sydney departs on time in a day in July.

Apply the idea We can construct the following tree diagram for this situation:

Fine 0.4

0.9 0.1

0.7

0.6 Not Fine

0.3

On time Late

The ﬂight can be on time in two different ways depending on whether the weather is ﬁne: P (On time) = P (Fine) × P (On time∣Fine) + P (Not fine) × P (On time∣Not fine) P (On time) = 0.4 × 0.9 + 0.6 × 0.7 = 0.78

Substitute the probabilities Evaluate

Reflect and check A two-way table could have been used instead, but the values in the table would need to be calculated first. From the information given, we know: • P (Fine) = 0.4 • P (Not fine) = 0.6 • P (On time∣Fine) = 0.9 • P (On time∣Not fine) = 0.7 The intersection of Fine and On time is: P (Fine ∩ On time) = P (On time∣Fine) × P (Fine) = 0.36 The intersection of Not fine and On time is: P (Not fine ∩ On time) = P (On time∣Not fine) × P (Not fine) = 0.42 Filling in the table and finding the total of the On time column gives us: Fine Not fine Total

On time 0.36 0.42 0.78

Late 0.04 0.18 0.22

Total 0.4 0.6 1

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b Find the probability that the weather is ﬁne in Melbourne given that the ﬂight departs on time on a day in July.

Create a strategy We can use the formula: P ( A∣B) =

Apply the idea Write the formula Substitute the probabilities Evaluate the numerator Multiply by Divide by common factor of 6

Idea summary For two events A and B, the probability of A occurring given that B has occurred is given by:

If we know the conditional probability, we can calculate the probability of A and B using either: P( A ∩ B) = P( A∣B) × P(B) or P( A ∩ B) = P(B∣A) × P( A).

14.02 Practice questions Understanding 1

Adventurer

1−3

Trailblazer 1−4

Are these statements true or false? a

If P ( A) = 0.5, P (B) = 0.3 and P ( A ∩ B) = 0.15, then the events are dependent.

Using the two-way table shown, P (Maths∣Fail) = Pass Fail Total

730

Explorer

Maths 25 15 40

Physics 7 12 19

Total 32 27 59

In a Venn diagram, if events A and B do not overlap, this means P ( A∣B) = 0.

The phrase “The probability of B given A” means that event A has already occurred and we want to know the probability that event B will occur next.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

If one event is “a student drives to school” and a second event is “the weather is rainy”, write the formula for ﬁnding the probability of selecting a student who drives to school given the weather is rainy. Fill in the blanks with “drive” or “rain”:

If P (B∣A) = 0.2, what additional piece of information is needed to ﬁnd P ( A ∩ B)? A

P ( A)

P (B)

P ( A ∪ B)

P ( A∣B)

There are two positions available at a company and the applicants have been shortlisted down to 6: Patricia, Nadia, Amy, Aaron, Lachlan, Jimmy. The two people to ﬁll the position will be picked randomly. • Event A: Patricia is chosen • Event B: Aaron is chosen a

Describe the probability P ( A∣B).

Is P ( A∣B) equal to, less than, or greater than P ( A ∩ B)?

Fluency

5−9, 11−13, 15, 18, 20

5−9, 11−14, 16, 18, 20

6−8, 10, 13, 14, 16, 17, 19−21

Find P (B∣A) given these probabilities for an experiment in which A and B are two possible events: P ( A ∩ B) = 0.48 and P ( A) = 0.6

Ex 1

Find P ( A ∩ B) given these probabilities for an experiment in which A and B are two possible events: P ( A∣B) = 0.6 and P (B) = 0.3

Ex 2

Use the two-way table to ﬁnd the probabilities.

P ( A)

P (B)

P ( A ∩ B)

P ( A ∪ B)

P ( A′ ∩ B′)

P ( A ∩ B′)

P ( A∣B)

P (B∣A)

P ( A′∣B)

P (B∣A′)

P ( A′∣B′)

P (B′∣A′)

Using the Venn diagram, ﬁnd: a

P ( A)

P (B′)

P ( A∣B)

P (B∣A)

P ( A∣B′)

A A′ Total

B′

5 7 12

6 2 8

Total 11 9 20

B 0.2

0.2

0.4 0.2

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A group of people were asked what type of exercise they like to do. The results are shown in the Venn diagram.

walk

A person is chosen from the group at random.

Find the probability that the person walks.

Find the probability that the person runs, knowing that they also walk.

At a university there are 816 students studying ﬁrst year engineering, 497 of whom say football is their favourite sport (set F ). Of the 348 students studying Civil Engineering (set C), 237 of them say football is their favourite sport.

Ex 3

run 10

C w

Find the value of: i

iii

Of the students whose favourite sport is not football, find the probability that a randomly selected student does not study Civil Engineering.

A player is rolling 2 dice and looking at their difference, that is, the largest number minus the smaller number. They draw a table of all the possible dice rolls for 2 dice and what their difference is: a

List the sample space for the difference of two dice.

Find the probability the dice will have a difference of 2.

Find the probability the dice will have a difference of 2 given that one of the dice rolled is a 3.

1 2 3 4 5 6

1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 0 1 2

5 4 3 2 1 0 1

6 5 4 3 2 1 0

A card is randomly drawn from a standard 52 pack of cards. Find the probability that it is a Jack, Queen, King or Ace if: a

No additional information is known.

We know it’s a 10, Jack or Queen.

We know it’s a 9 or a Queen.

We know it’s not 2, 3, 4 or 5.

A software development team has a probability of 0.8 of successfully completing their first project and 0.55 of successfully completing both their first and second projects. Find the probability of successfully completing the second project, given they completed the first one successfully.

In a population, 35% of people have brown eyes, 45% of people have blonde hair, and 10% of people have both brown eyes and blonde hair. A person is chosen randomly from the population. Find the probability that they have:

732

Blonde hair, knowing that they have brown eyes

Brown eyes, knowing that they have blonde hair

Two dice are rolled, one after another. Find the probability, in simplest form, of rolling: a

A pair of fours, given the ﬁrst die is a four.

A pair with a sum of 7 or more, given the ﬁrst die is a two.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Ex 4

Four identical balls labelled 1, 2, 3 and 4 are in a bag. Two balls are randomly drawn from the bag in succession and without replacement. Find the probability that: a

The ﬁrst ball is labelled 4 and the second ball is labelled 2.

The sum of the numbers on the two balls is 5.

The second ball drawn is 1 given that the sum of the numbers on the two balls is 5.

Two cards are randomly drawn in succession and without replacement from a deck of cards numbered from 1 to 20. Find the probability that the second card is: a

An even number given that the ﬁrst card is a 10.

Less than 5 given that the ﬁrst one is a 14.

A number divisible by 5 given that the ﬁrst card is a 15.

A bus service operates from Town A to Town B. The probabilities shown were found with regard to departure times and traffic in June: • The probability that the bus departs on time, given the traffic is light in Town A is 0.8 • The probability that the bus departs on time, given the traffic is heavy in Town A is 0.7 • The probability that the traffic is light on any particular day in June is 0.4 Find the probability that:

The bus service from Town A to Town B departs on time on a day in June.

The traﬃc is light in Town A given that the bus departs on time on a day in June. Give the answer as a fraction in simplest form.

A museum has a new software system that determines whether a painting is fake or authentic. The system has an accuracy of 75%. Art experts have previously determined that 5% of paintings that the museum examines are fakes. Find the probability that:

The software correctly identiﬁes a painting as fake.

The software incorrectly identiﬁes an authentic painting as fake.

The software will identify a painting as fake.

A painting that is identiﬁed as fake is actually fake, rounded to two decimal places.

In a survey, 531 people are asked whether they watch My Kitchen Rules (MKR) or Masterchef (MC). Of the total respondents, 177 people watch both and 65 watch neither. The number who watch MKR is twice the number who watch both. a

Find the number of people who only watch MKR.

Of the people who watch MC, ﬁnd the probability that a randomly selected person also watches MKR.

Of those who don’t watch MC, ﬁnd the probability that a randomly selected person watches neither.

The probability of two independent events, A and B are P ( A) = 0.8 and P (B) = 0.5 Determine the probability of: a

Both A and B occurring

Neither A nor B occurring

A or B or both occurring

B occurring but not A

A occurring, given B occurs

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Reasoning 22

22, 23

23, 24

Given the conditional probability formula,

show that, P ( A∣B) × P (B) = P (B∣A) × P ( A) 23

Use the Venn diagram shown, where B is a subset of A, to ﬁnd: a

P ( A∣B)

P (B∣A)

P ( A′∣B′)

P (B′∣A′)

Explain why the probability of A occurring given B has already occurred is always certain.

A B 5

A Venn diagram is shown in which the size of the regions are proportional to the number of elements in that region.

A B

Which conditional probability is greater, P ( A∣B) or P (B∣A)? Justify your answer.

Problem-solving 26−28 25

25, 26, 28

25, 26, 28, 29

Edwin watches three episodes of TV each night. He begins with News or Current Affairs, then the next two shows are either Comedy, Horror or Animation. He never watches more than one Horror show, and if he watched the News, he will follow this immediately with a Comedy. a

Construct a tree diagram of all possible options.

If, at every stage, the possible outcomes of each choice are equally likely, ﬁnd the probability Edwin watches two Comedies, given that he watched the News.

When Fred gets ready for work in summer, he ﬁrst decides whether it will be a hot day (H) or not (N ) and then decides whether to wear a tie (T ) or to just dress casually (C). The chance of Fred deciding it will be a hot day is 0.7. If he decides it will be a hot day, there is a 0.2 chance he will wear a tie. If he decides it will not be a hot day, there is a 0.85 chance he will wear a tie. a Construct a probability tree for this situation. b Calculate the probability Fred decided it was a hot day given that he wears a tie. Round your answer to two decimal places.

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For breakfast each morning, Marge eats porridge, toast or cereal. With that, she will either drink orange juice, tea or hot chocolate. Find the probability that Marge will: a Eat toast and drink tea b Eat cereal or drink orange juice c Eat porridge given that she drinks hot chocolate

Two hundred people were questioned about whether they voted for Labor, Liberal or Greens last election and who they’ll vote for this election.

Labor (last election) Liberal (last election) Greens (last election) Total a b c d

Labor (next election) 26

Liberal (next election) 40

1 5 10

Total 70 90 200

Complete the table. Find the probability that a randomly selected person will vote Labor next election given that they voted Liberal last election. Find the probability that a randomly selected person did not vote for the Greens last election given that they will vote Green next election. If a person voted Labor or Liberal last election, ﬁnd the probability they’ll vote Liberal next election.

A class of 13 students had their height, shoe size and 400 m race time collected. The results are shown in the table: a Let T be the event that a randomly chosen student from the class has a race time greater than 90. Find the value of P (T ). b Let H be the event that a randomly chosen student from the class has a height greater than 130. Find the value of P (H). c Find the value of P (T∣H). d Let S be the event that a randomly chosen student from the class has a shoe size greater than or equal to a size 10. Find the value of P (H∣S). e

140

Greens (next election)

Which of the events appear to be dependent, H and S or H and T?

Athlete Aaron Tobias Lisa Georgia Ben Patricia Vanessa Sandy Adam Amelia Rosey Neil Roald

400 m Height time (sec) (cm) 94 137 62 165 89 100 64 155 98 100 127 102 84 147 98 144 54 154 84 147 81 126 104 134 67 127

Shoe size 8 10 11 7 10 11 7 8 7 8 12 9 8

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14.03 Mixed probability concepts After this lesson, you will be able to… • use Venn diagrams, two-way tables, and tree diagrams for organising data. • perform probability calculations using appropriate formulas. • interpret results in the context of the problem.

Mixed probability concepts Many situations in probability can be modelled by a Venn diagram, an array or two-way table, and/or a tree diagram to organise the information, determine the size of different groups, and do calculations. Probability notation can also help us to organise the information given in a question, and these formulas can help us calculate different probabilities. If two events are independent: P ( A ∩ B) = P ( A) × P (B) When two events are dependent, the probability of each event will change, depending on the first event. P ( A ∩ B) = P ( A∣B) × P (B) For conditional probability:

For mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) For non-mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) − P ( A ∩ B)

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Example 1 A bag contains the letters A, B, and C. A letter is randomly drawn out of the bag without replacement and placed on the table. This is done three times.

ABC ACB BAC BCA CAB CBA

a Find the probability that A is drawn before B is drawn.

Create a strategy Count the number of branches where A appeared before B.

Apply the idea

Reﬂect and check

The sequences where A is drawn before B are ABC, ACB, and CAB. There are 3 desired outcomes and 6 total outcomes possible.

Notice each sequence has a probability of

Therefore, the probability is .

Since there are 3 possible sequences, the probability of drawing A before B is:

b Find the probability that two A’s are drawn.

Create a strategy

Apply the idea

Note that the letters are not replaced after being As there is only one A inside the bag, it is drawn. impossible to randomly draw two A ’s without replacement. So the probability is 0.

c Find the probability the word CAB is spelled out in that order.

Create a strategy Consider the number of outcomes that spell out CAB.

Apply the idea Only one of the six outcomes spells CAB. The probability is .

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Reflect and check We can check the answer by thinking about the probability of drawing C first, then A, then B. The probability of drawing C first is . For the second draw, there are 2 letters left. The probability of drawing A next is . There is only one letter left for the third draw, so the probability of drawing B third is 1. So the probability of drawing the letters CAB in that order is:

d Find the probability that B is drawn last, given that A is drawn ﬁrst.

Create a strategy The probability of selecting B last given A was drawn ﬁrst can be calculated using the formula P (B last∣A first) =

Apply the idea There is only one way to select B last and A ﬁrst, so P (B last ∩ A first) = . For the ﬁrst letter, A is one of the three letters, so P ( A first) = . Start with the conditional probability formula

Substitute known values

Evaluate the division

Simplify

Reﬂect and check With conditional probability, the given condition determines the total number of outcomes. If we know A was drawn ﬁrst, there are only 2 possible outcomes: drawing B last or drawing C last. Since drawing B last is one of those outcomes, the probability is .

Example 2 At a restaurant, 50 people are asked about whether they would like to add rice (R) or broccoli (B) to their meal. Some of the results are shown in the Venn diagram.

14 15

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a Find the number of people in R ∩ B′.

Create a strategy This is the blank section of the Venn diagram. We can subtract the amounts in the diagram from the total number of people surveyed.

Apply the idea

Reﬂect and check

The number of people in R ∩ B′ is

We can now complete the Venn diagram.

50 − 9 − 14 − 15 = 12

n (R ∩ B′) = 12 12

14 15

b Find P (B∣R).

Create a strategy Use the formula P (B∣R) =

Apply the idea Using the Venn diagram from part (a), we know the number of people in B ∩ R is 9 and the number of people in R is 9 + 12 = 21. Write the formula Substitute known values

Reﬂect and check We can check our answer using the conditional probability formula and

, so Start with the conditional probability formula

Substitute known values

Evaluate the division

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c Find P (R∣B′).

Create a strategy Use the formula

Apply the idea Using the Venn diagram from part (a), we know the number of people in R ∩ B′ is 12 and the number of people in B′ is 12 + 15 = 27. Start with the conditional probability formula

Substitute known values

Evaluate the division

Simplify

Reﬂect and check Checking our answer, we can look at the Venn diagram to conﬁrm that of those who did not choose broccoli (27 people), 12 of them chose rice.

Example 3 600 people were questioned as to whether they read novels in paperback form or on an e-reader. • 120 said they do both. • 180 said they read paperbacks. • The probability that a person who didn’t read paperbacks also didn’t read on an e-reader was 30%. a Complete the table for these results: Paperback

Not paperback

E-reader Not e-reader

Create a strategy Add a total row and column to the table to make it easier to ﬁnd missing values. Paperback

Not paperback

E-reader Not e-reader Total

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Total

Apply the idea The first piece of information goes in the first cell of the table. The second piece of information helps us fill in the first column and last row.

E-reader Not e-reader Total

Paperback 120 60 180

Not paperback

Total

420

600

The third piece of information can help us ﬁll in the rest of the table. Of the people who didn’t read paperback books, 30% did not read on an e-reader. 30% × 420 = 126

E-reader Not e-reader Total

Paperback 120 60 180

Not paperback 294 126 420

Total 414 186 600

b Calculate the probability that a randomly selected person reads novels on an e-reader but not in paperback.

Create a strategy Use the table from part (a) to ﬁnd the number in the row for E-reader and the column for Not paperback, E ∩ P′.

Apply the idea In the table from part (a), we see 294 people read novels on an e-reader but not in paperback. Divide desirable outcomes by total outcomes

Simplify

c Of those that read on e-readers, what percentage read in both formats? Round your answer to the nearest percent.

Create a strategy Divide the number of people who read on both formats by the number of people who read on an e-reader.

Apply the idea The number of people who read on both formats is 120. The number of people who read on an e-reader is 414. Use a calculator to estimate the probability as a percentage, and round to the nearest whole percentage:

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Reﬂect and check We can check this answer by using the formula for conditional probability: P (P∣E) = Start with the conditional probability formula

Substitute known values

Evaluate the division

Convert to a percentage

Evaluate using a calculator and round

Example 4 Vanessa has 12 songs in a playlist. Four of the songs are her favourite. She selects shuffle and the songs start playing in random order. Shuffle ensures that each song is played once only until all songs in the playlist have been played. Find the probability that: a The first song is one of her favourites.

Create a strategy The probability can be calculated using

Apply the idea Vanessa has 4 favourite songs and there are 12 songs in the playlist. The probability that the ﬁrst song is one of her favourites is

b Two of her favourite songs are the ﬁrst to be played.

Create a strategy The probability for the second song will be dependent on the ﬁrst song that is played.

Apply the idea From part (a), we know the probability that the ﬁrst song is one of Vanessa’s favourites is . This leaves 3 favourite songs out of 11 songs that have yet to play. So, the probability that the second song is also her favourite is

Multiplying these probabilities together gives us:

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c At least 1 of her favourite songs is played in the ﬁrst three.

Create a strategy The probability that at least 1 of the songs is her favourite is the complement of the probability that no songs are her favourite.

Apply the idea The probability that the ﬁrst song is not a favourite is

Since there are 7 non-favourite songs left and 11 songs left in the playlist, the probability that the second song is not a favourite is

For the third song, there are 6 non-favourites left and 10 songs left in the playlist. The probability the third song is not a favourite is

The probability that no songs are Vanessa’s favourite is:

The probability that at least one of the songs is Vanessa’s favourite is:

Idea summary Venn diagrams, arrays, and tree diagrams can be used to organise information and help solve problems in probability. We can also use the formulas for complementary events, independent events and conditional probability to solve problems. For independent events: P ( A ∩ B) = P ( A) × P (B) For dependent events: P ( A ∩ B) = P ( A∣B) × P (B) For conditional probability: P ( A∣B) = For mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) For non-mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) − P ( A ∩ B)

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14.03 Practice questions Understanding 1

Explorer 1−3

Adventurer

Trailblazer

1−3

Match each of the Venn diagrams to the correct statement. i

B A

ii A B

0.32

0.08

0.12 0.48

iii

A B

iv A

0.32

0.08

0.48 0.12

Events A and B are mutually exclusive.

Events A and B are independent.

P ( A∣B) = 1

P ( A ∩ B) ≠ P ( A) × P (B)

Match each formula to the probability described. i

P ( A ∪ B) = P ( A) + P (B) − P ( A ∩ B)

P ( A ∩ B) = P ( A) × P (B)

iii

P ( A) + P (B) = 1

P ( A ∩ B) = P ( A∣B) × P (B)

Conditional probability

Complementary events

Non-mutually exclusive events

Independent events

Which statement is equivalent to this statement? Find the probability that a randomly selected person is female, given they are a team leader. A

Of the females, ﬁnd the probability the person is a team leader.

If a person is a team leader, then ﬁnd the probability that they are female.

Find the probability that a randomly selected person is male knowing that they are a team leader.

Fluency 4

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6−8, 10, 11, 13, 15, 17, 18

4−6, 8, 10−13, 15−18

4−6, 9, 11, 12, 14, 16, 19, 20

Find the values of w, x, y, and z, given that P ( A ∩ B) = 0.3, P (B) = 0.6, and that the events A and B are independent.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A A′

B w x

B′ y z

Use the Venn diagram to ﬁnd P ( A∣B). Give the answer as a fraction in simplest form.

0.3

0.15

0.25 0.3

Ex 1

Three cards labelled 2, 3 and 4 are placed face down on a table. Two of the cards are selected randomly to form a two-digit number. The outcomes are displayed in the tree diagram. a b

3 2 4

List the sample space of two digit numbers produced by this process.

2 3

Find the probability that:

2 appears as a digit in the number.

The sum of the two selected cards is even.

iii

The number formed is greater than 40.

2 is the second digit, given 3 is the ﬁrst digit.

2 4 3

Students were asked what they are allergic to. The table shows the results: Allergic to dairy Not allergic to dairy

Allergic to nuts 14 25

Not allergic to nuts 12 25

If a student is chosen at random, ﬁnd the probability that:

Ex 2

The student is allergic to dairy.

The student has an allergy.

The student does not have an allergy.

The student is also allergic to dairy if he or she is allergic to nuts.

In a survey, 100 oﬃce workers were asked whether they buy a coffee (set C) or lunch (set L) during the work day. Some of the results are shown in the Venn diagram. Find: a

The number of workers in (L ∩ C′)

P ( C∣L)

P ( L∣C′)

L 43

47 7

Of the 100 students starting kindergarten surveyed, 80% believe in Santa, 41% believe in the Tooth Fairy and 31% believe in both. a

Complete the table: Believe in Santa Believe in Tooth Fairy Do not believe in Tooth Fairy Total

Do not believe in Santa

Total

100

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Ex 3

Of those who believe in Santa, ﬁnd the probability that they also believe in the Tooth Fairy. Write your answer as a percentage rounded to two decimal places.

Of those who believe in either Santa or the Tooth Fairy, what is the probability they believe in both?

Of those who believe in the Tooth Fairy, what is the probability they do not believe in Santa?

Two sets of numbers, A and B, are such that set A contains the even numbers from 1 to 20, inclusive, and set B contains the factors of 48 from 1 to 20, inclusive. a

List the numbers in set A.

List the numbers in set B.

How many numbers are unique (only appear in one set)?

Find the probability that a randomly selected number is in set B, given that it is in set A.

In a survey, 166 children were asked whether they had read or watched at least one of the stories in a popular children’s series: • 62 hadn’t watched any of the movies. • 39 had read and watched at least one of the stories. • 97 had read the book. • 4 had never read nor watched any of these. a

Complete the table for these results: Watched the movie 39

Read the book Didn’t read the book Total b

Didn’t watch the movie

Total 97

4 62

166

Find the probability that a randomly selected student had: i

Read the book but hadn’t watched the movie.

Read the book, given that they had watched the movie.

iii

Watched the movie, given that they hadn’t read the book.

In a survey, 1000 people were asked about whether they bought (set B) or grew (set G) their own herbs. The results are listed: • 4% said they do neither. • 40% said they do both. • Of those who bought their herbs, the probability that they also grew herbs was . In this Venn diagram, where x, y and z represent numbers of people: a

Find the value of: i

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iii

Find the number of people who only grew their own herbs.

Of those who grew herbs, what is the probability that they also bought herbs?

G z

y x

In a group of 69 people, 45 said they had been to Europe (E), 26 said they had been to Asia ( A) and 8 people said they had been to neither Europe nor Asia. Find: a

P ( A ∩ E′)

The number of people in (E ∩ A)

P ( E∣A)

P ( A∣E′)

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

At an Italian restaurant, Alessandra orders an entree, main and dessert. She has 2 entrees, 3 mains and 2 desserts to choose from:

• Entrees: calamari, soup • Mains: spaghetti, lasagna, risotto • Dessert: gelato, tiramisu

The possible combination of meals that Alessandra could order is shown in the tree diagram. Spaghetti

Calamari

Lasagna

Risotto

Gelato Tiramisu Gelato Tiramisu Gelato Tiramisu

Spaghetti

Lasagna Soup Risotto

Gelato Tiramisu Gelato Tiramisu Gelato Tiramisu

If each combination of dishes is equally likely, ﬁnd the probability that Alessandra orders lasagna given that she ordered calamari.

Find the probability that Alessandra ordered spaghetti and tiramisu, given that she ordered soup.

Find the probability that Alessandra ordered risotto or spaghetti, given that she ordered soup.

Two marbles are randomly drawn without replacement from a bag containing 1 blue, 2 red and 3 yellow marbles. a b

Construct a tree diagram to show the sample space. Find the probability of drawing these: i

A blue marble and a yellow marble, in that order

A red marble and a blue marble, in that order

iii

2 red marbles

No yellow marbles

2 blue marbles

A yellow marble and a red marble, in that order

vii A yellow and a red marble, in any order 16

In tennis, if the ﬁrst serve is a fault (out or in the net), the player takes a second serve. A player serves with these probabilities: • First serve in: 0.55 • Second serve in: 0.81 a

Construct a probability tree showing the probability of the ﬁrst two serves either being in or a fault.

Find the probability that the ﬁrst serve is a fault.

Find the probability that the player makes a double fault (both serves are a fault). 14.03 Mixed probability concepts mathspace.co

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Nine pilots from Star Jet and seven pilots from Air Tiger offer to take part in a rescue operation. If 2 pilots are selected at random: a

Construct a probability tree showing all possible combinations of airlines from which the pilots are selected.

Find the probability that the two pilots selected are from: i

Ex 4

Different airlines

A carton contains a dozen eggs, of which 4 contain a double yolk. If 3 eggs are chosen at random for a cake, find the probability that: a

The same airline

They all have a double yolk.

At least 1 has a double yolk.

From a standard pack of cards, 1 card is randomly drawn by Andrea and then put back into the pack. A second card is then drawn by Bridie. • Event A: Andrea selects a heart. • Event B: Bridie selects a heart.

Calculate the probability that neither of their cards are hearts.

Calculate the probability that at least 1 of the cards is a heart.

The probability that Joanna, Nadya and Adam get permission to go on their school trip are 0.8, 0.1 and 0.2, respectively. Find the probability that at least 1 of them gets permission.

Reasoning 21

21−23

23, 24

For P ( A ∩ B): a

Write the formula of independent events.

Write the formula of dependent events.

Which formula, the one from part (a) or the one from part (b), can be used for both independent and dependent events? Explain.

A sample of 20 students, A-T, from the same school were asked whether they had a haircut in the last month (H) and whether they had been sick in the last month (S). The results are shown in the given tables:

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Student Hair Sick

A ✓ −

B ✓ ✓

C − −

D ✓ ✓

E − −

F ✓ −

G ✓ ✓

H −

Student Hair Sick

K −

L −

M −

✓

O − −

P ✓ −

Q −

✓

N ✓ ✓

✓

I − −

J ✓ −

R ✓ ✓

S − −

T −

Find the probability that a student: i Had a haircut in the last month ii Had been sick in the last month iii Was sick and had a haircut in the last month

Find the value of P (H) × P (S).

Does the data suggest that getting haircuts and being sick in the last month are independent or dependent? Explain your reasoning.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

✓

In a memory game, 16 pairs of identical cards are randomly placed face down. When someone has a go, they turn one card over and then turn a second card over to try to ﬁnd an identical pair. If an identical pair is found, the two identical cards are removed from the pack. a

At the beginning of the game, Kathleen turns a card over. Find the probability of her selecting a card that makes an identical pair.

The ﬁrst two cards Kathleen picks are identical. Is the probability of picking any more pairs of identical cards independent or dependent on the number of pairs picked before? Explain your reasoning.

Candidates for a professional examination are allowed a maximum of two attempts at the exam. Over a period of time, it has been shown that 60% of the candidates pass on the ﬁrst attempt and, if a second attempt is required, 70% of the candidates pass. Assume that these probabilities apply to a randomly chosen candidate. a

Why does the second set of branches only have one fork?

Given that a candidate passes, calculate the probability that they passed on the ﬁrst attempt. Give the answer as a fraction in simplest form.

0.6

25, 27−29

0.7

0.4

Explain why it is inappropriate to assume that the probabilities apply to a randomly chosen candidate.

Problem-solving 25, 27, 28 25

0.3

26, 28−30

A school consisting of 410 primary students and 410 secondary students offers everyone an optional end of year project to improve their grade. There are various types of projects they can do. A summary of the number of student submissions are shown in the given table: Primary Secondary a b c d

Essay 18 76

Creative story 64 68

Art work 46 73

Other Project 24 45

Did not submit 258 148

Find the probability of selecting student that submitted a project. Find the probability of selecting student that submitted a project given that they are a secondary student. Find the probability that a submitted project is an essay. Find the probability that a submitted project is an essay given that the project was submitted by a primary student.

Find: a

P ( A ∩ B ∩ C′)

P (( A ∪ B) ∩ C′)

P ( B∣C)

P ( A′∣C)

P (( A ∪ B)∣C′)

B 0.17 0.18 0.1 0.07 0.11 0.12 0.2

0.05

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Yvonne is struggling to decide what to watch online. She decides to pick one movie at random from the streaming website. A Venn diagram of her options sorts movies into three categories based on their genre: Comedy, Action and Horror. a

How many of the movies are horror ﬁlms?

How many of the movies ﬁt into only one genre?

Find the probability of selecting a film that fits into only one genre given that it fits into the horror genre.

Comedy

Action

5 3

6 13 Horror

In a particular city, 55% of the population are adults. 8% of the children and 2% of the adults are left-handed. a

What percentage of the city’s population is left-handed?

If a person is chosen at random and is left-handed, what is the probability that the person is an adult? Give the answer rounded to two decimal places.

The probability of spotting a rare bird species in a national park on any given day is 0.3. If one is spotted one day, the probability of spotting another the next day increases to 0.6, due to ﬂock movement. However, if none are spotted, the probability of spotting one the next day decreases to 0.15. What is the probability of spotting exactly one rare bird over two consecutive days?

In a speciﬁc deck of 10 cards numbered from 1 to 10, the probabilities of drawing each number are in the ratio 1: 2: 3: 4: 5: 6: 7: 8: 9: 10. If two cards are drawn with replacement, what is the probability that they add up to 15?

Investigation: Simulation

Investigate online

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

mathspace.co

14 Chapter review 1

A spinner is divided into 6 equal sections: 3 sections are red and 3 sections are yellow. a

If the spinner lands on yellow on the ﬁrst spin, ﬁnd the probability that the second spin will be yellow.

Are the successive events of spinning yellow twice dependent or independent?

A photo album contains 5 photos, all numbered from 1 to 5. Four photos are drawn at random, with replacement. Find the probability that the photo with the number 2 is drawn: a

At least once

If three letters are randomly selected from the word CIRCUMFERENCE without replacement, ﬁnd the probability that: a

Exactly once

The letters are R, C, U in that order.

All three letters are E.

Find P (A ∩ B) given these probabilities for an experiment in which A and B are two possible events: P ( A∣B) = 0.4 and P (B) = 0.5

Use the two-way table to ﬁnd the probabilities. a

P (A)

P (A ∩ B)

P (B∣A′)

P (A′∣B′)

A A′ Total

Use the Venn diagram to ﬁnd: a

P (B′)

P (A ∪ B)

P ( A∣B)

B′

14 6 20

8 12 20

0.35

Total 22 18 40

0.1

0.2

0.35

Amy owns ﬁve necklaces and four earrings. She selects one of the items at random for herself and then another item at random for her sister: a

Find the probability that Amy selects a necklace for herself.

Find the probability that both items Amy selects are earrings.

N N

E N

E E

Chapter 14 review mathspace.co

NN NE

EN EE

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Find the values of j, k, l, and m, given that P (A ∩ B) = 0.4, P (B) = 0.5, and that the events A and B are independent.

A A′

B′ k m

In a survey, 640 participants were asked whether they enjoy watching basketball or volleyball. Of the total respondents, 120 enjoy watching both sports and 70 enjoy neither. The total number who enjoy watching basketball is three times the number who enjoy watching both sports. a

Complete the Venn diagram.

Of those who enjoy watching volleyball, ﬁnd the probability that a randomly chosen person also enjoys watching basketball.

If a randomly selected person does not enjoy watching volleyball, then ﬁnd the probability that they dislike watching basketball too.

120

In a group of 100 people, 60 said they went swimming, 40 said they went hiking, and 10 people said they neither swam nor hiked. a

Complete the table: Went hiking

Did not hike

Went swimming Did not swim Total b

B j l

Total

100

Find the probability that a randomly selected person: i

Went swimming and hiking

Went swimming given they went hiking

iii

Went hiking knowing they went swimming

For breakfast, Elijah has something to eat and something to drink. He will either eat an egg sandwich or a muﬃn and will drink orange juice or iced coffee. • The chance of Elijah eating an egg sandwich is 0.6 • The chance of Elijah drinking iced coffee is 0.5

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Construct a tree diagram showing all possible combinations of food and drink Elijah can have for breakfast and their associated probabilities.

Find the probability Elijah drinks orange juice and eats an egg sandwich.

Find the probability Elijah drinks orange juice or eats an egg sandwich.

Find the probability Elijah drinks iced coffee given he ate a muﬃn.

In a city, 40% of people prefer tea over coffee, 30% of people enjoy reading newspapers, and 15% of people both prefer tea and reading newspapers. A resident is chosen randomly. Find the probability that they: a

Enjoy reading newspapers given that they prefer tea

Prefer tea, knowing that they enjoy reading newspapers

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A box contains 20 candies, of which 6 are sour. If Jenna randomly chooses 3 candies to eat, ﬁnd the probability that: a

All are sour.

At least 1 is sour.

A small magazine asked people from different states to send in a vote on whether they supported Daylight Saving Time. The diagram shows the proportion of people that voted YES or NO in NSW, ACT and QLD: a

Find the probability that a randomly chosen person voted “YES”.

NSW YES NO

Of those that are not from NSW, ﬁnd the probability that a randomly chosen person voted “NO”.

YES

Considering just the voters from NSW and ACT, ﬁnd the probability that a randomly chosen person was from NSW and voted “YES”.

NSW

ACT NO

YES

NSW YES

QLD YES

YES

QLD

NSW NO

QLD YES NO

ACT

NSW YES

ACT YES NO

QLD NO

YES

Did you know?

Probability plays a huge role in baseball strategy! Coaches use it to decide when to take risks or play it safe during critical moments. For example, probability can help determine whether a player should attempt to steal a base or stay put, based on factors like the pitcher’s throw speed, the catcher’s reaction time, and the runner’s success rate in similar situations. By calculating these probabilities, teams make smarter, data-driven choices that can turn the game in their favour. It’s not just about instinct; probability helps teams predict which moves are most likely to score runs, making every decision a bit more strategic. In a close game, those small, calculated choices can make all the difference between a win and a loss! Chapter 14 review mathspace.co

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Big ideas • Different representations of data highlight different characteristics of the data. • Comparing and interpreting data sets involves analysing statistical measures and visual representations to gain meaningful insights, make informed conclusions and understand the variation and relationships within the data.

Curriculum Outcomes VC2M10ST01 compare data distributions for continuous numerical variables using quartiles and interquartile range and appropriate data displays including boxplots, histograms and dot plots; discuss the shapes of these distributions in terms of centre, spread, shape and outliers in the context of the data VC2M10ST04 analyse claims, inferences and conclusions of statistical reports in the media and other places, by linking claims to displays, statistics and representative data, including ethical considerations and identification of potential sources of bias VC2M10AST01 calculate and interpret the mean and standard deviation of data and use these to compare data sets; investigate the effect of individual data values, including outliers, on the standard deviation VC2M10AST02 identify measures of spread, and understand their interpretation and usefulness with respect to different data distributions

15 Univariate statistics Chapter outline 15.A Representations of data 15.01 Quartiles 15.02 Boxplots 15.03 Cumulative frequency 15.04 Compare data sets 15.05 Samples and populations 15.06 Optional: Standard deviation 15.07 Optional: Comparisons using standard deviation Chapter 15 review

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In medicine, scientists use standard deviation to assist in understanding drug effectiveness and side effects!

15.01 Quartiles After this lesson, you will be able to… • identify the lower quartile, median and upper quartile in a data set. • calculate the interquartile range in a data set. • apply the concept of quartiles and IQR to solve problems.

Quartiles The range is a measure of spread based on the minimum and maximum in a data set, but does not tell us about the spread of data within these two values. Range = Maximum − Minimum

Maximum : The highest value in the data set

Minimum : The lowest value in the data set

The median is a measure of centre, and tells us where the middle of the data set is. For a data set with n items, Median is the

value

To get a better picture of the internal spread in a data set, we can ﬁnd the quartiles of a data set, which can be used for a measure of spread called the interquartile range (IQR). First quartile

Second quartile Q1

2, 3, Minimum

Third quartile Q2 10, 11, Median

5+5 2 Q1 = 5 Q1 =

13,

Fourth quartile

16,

Q3 18,

22, 39, Maximum

16 + 18 2 Q3 = 17 Q3 =

Quartiles are scores at particular locations in the data set-similar to the median, but instead of dividing a data set into halves, they divide a data set into quarters.

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Minimum Lower quartile, Q1 Median, Q2 Upper quartile, Q3 Maximum

Lowest value At most 25% of the data is below this value 50% of the data lies on either side of this value At most 25% of the data is above this value Highest value

Let’s look at how we would divide up some data sets into quarters now. Make sure the data set is ordered before ﬁnding the quartiles or the median.

Median

First locate the median, between the 4th and 5th scores:

Now there are four scores in each half of the data set, so split each of the four scores in half to ﬁnd the quartiles. Q1

Median

We can see the ﬁrst quartile, Q1 is between the 2nd and 3rd scores. Similarly, the third quartile, Q3 is between the 6th and 7th scores.

For 9 scores, the 5th term is the median. There are four terms on either side of the median. So Q1 is still between the 2nd and 3rd scores and Q3 is between the 6th and 7th scores.

Median

For 10 scores, the median is between the 5th and 6th scores. This time, however, there are 5 scores on either side of the median. So Q1 is the 3rd term and Q3 is the 8th term.

Median

Each quartile represents 25% of the data set. For a data set with n items, Q1 is the Median is the Q3 is the

value value value

15.01 Quartiles mathspace.co

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Example 1 Here are Ray’s scores from his last 13 rounds of golf played: 66, 66, 68, 68, 70, 78, 80, 84, 106, 116, 126, 130, 132 a What is his median?

Create a strategy th where n is the total number of scores, to ﬁnd the position of the median.

Use the formula:

Apply the idea There are n = 13 scores in the list. Substitute n = 13

Position of median = = 7th Median = 80

Evaluate Choose the 7th score

b What is the lower quartile?

Create a strategy Find the median of the lower half of the scores excluding the median.

Apply the idea The lower half of the scores are: 66, 66, 68, 68, 70, 78. Lower quartile =

Average the middle scores

= 68

Evaluate

c What is the upper quartile?

Create a strategy Find the median of the upper half of the scores excluding the median.

Apply the idea The upper half of the scores are: 84, 106, 116, 126, 130, 132. Upper quartile =

758

= 121

Average the middle scores Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Idea summary • Q1 is the ﬁrst quartile (sometimes called the lower quartile). It is the middle score in the bottom half of the data set. • The median, solmetimes called Q2, is the second quartile. • Q3 is the third quartile (sometimes called the upper quartile). It is the middle score in the top half of the data set.

Interquartile range The interquartile range (IQR) is the difference between the third quartile and the ﬁrst quartile. 50% of scores lie within the IQR. Since it focuses on the middle 50% of the data set, the interquartile range often gives a better indication of the internal spread than the range does, and it is usually not affected by outliers.

IQR = Q3 − Q1 Q1

Lower quartile

Upper quartile

Interactive exploration

Discover this concept in action online

mathspace.co

Example 2 Use this frequency table to: Score 5 14 18 24 32 38 50

Frequency 1 1 3 2 1 5 5

a Find the number of scores.

Create a strategy Add all the frequencies.

Apply the idea Number of scores = 1 + 1 + 3 + 2 + 1 + 5 + 5 = 18

Find the sum Evaluate the addition 15.01 Quartiles mathspace.co

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b Find the median.

Create a strategy Use the formula:

th where n is the total number of scores, to ﬁnd the position of the median.

Apply the idea There are 18 scores in the list. Position of median = = 9.5

Substitute n = 18 Evaluate

Therefore the median is between the 9th and 10th score. Using the frequency table we can see that the 9th score is 38 and the 10th score is 38.

c Find the lower quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the lower half of the scores.

There are 9 scores in the lower half excluding the median. The lower quartile will be the 5th score from the frequency table. Lower quartile = 18

d Find the upper quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the upper half of the scores.

There are 9 scores in the upper half excluding the median. The upper quartile will be the 5th in the upper half, which is the 9 + 5 = 14th score overall. Upper quartile = 50

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e Find the interquartile range.

Create a strategy Use the interquartile range formula: IQR = Q3 − Q1

Apply the idea IQR = 50 − 18

Substitute the quartiles

= 32

Evaluate

Example 3 Answer the questions using the dot plot shown:

a Find the total number of scores.

Create a strategy

Apply the idea

Count the number of dots in the dot plot.

Number of scores = 15

Count the dots

b Find the median.

Create a strategy Use the formula:

th where n is the total number of scores, to ﬁnd the position of the median.

Apply the idea Position of median = = 8th

Substitute n = 15 Evaluate the division

The 8th score in the dot plot is on 15. Median = 15

c Find the lower quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the lower half of the scores.

There are 7 scores in the ﬁrst half excluding the median. The lower quartile will be the 4th score in the dot plot. Lower quartile = 9

15.01 Quartiles mathspace.co

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d Find the upper quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the upper half of the scores.

There are 7 scores in the second half excluding the median. The upper quartile will be the 8 + 4 = 12th score in the dot plot. Upper quartile = 16

e Find the interquartile range.

Create a strategy

Apply the idea IQR = 16 − 9

Use the interquartile range formula: IQR = Q3 − Q1

Substitute the quartiles

Evaluate

Idea summary To calculate the interquartile range:

IQR = Q3 − Q1 IQR Q1 Q3

is the interquartile range is the ﬁrst quartile is the third quartile

15.01 Practice questions Explorer

Understanding 1

1−4

Adventurer

Trailblazer

1−4

Find the median for each data set of scores: a

2, 10, 28, 35, 50

8, 9, 11, 12, 13, 15

41, 42, 43, 44, 49, 50, 51, 53, 58

3, 4, 6, 7, 7, 8, 12, 14

For each data set, the median is shown. Find the upper and lower quartiles. a

0, 1, 4, 6 8, 9, 11 Median

22, 24, 25, 29, 32, 34 Median: 27

5, 11, 19, 20, 22 24, 26, 27, 27

14, 15, 16, 16, 17, 18, 19, 20, 24, 25, 30, 36

Median Median: 18.5

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What is the interquartile range for this data set? Q1 = 4

Q3 = 14.5

1, 2, 3, 5, 5, 7, 11, 14, 15, 16 Minimum Median = 6

Match the term with the deﬁnition. i

Highest value

50% of the data lies on either side of this value

iii

At most 25% of the data is above this value

The lowest value

At most 25% of the data is below this value

Minimum

Lower quartile, Q1

Median

Upper quartile, Q3

Maximum

Fluency Ex 1

Maximum

5, 6, 7a–d, 8, 9, 11−14, 17

5, 7c–f, 8−11, 13, 15−17

5, 7e–h, 8, 10, 11, 15−18

The data set shows Rachelle’s scores from her last 13 rounds of golf played: 56, 58, 60, 62, 64, 72, 74, 78, 98, 108, 120, 124, 128

Find her median score.

Find the lower quartile score.

Find the upper quartile score.

Find the interquartile range.

The data set shows Luke’s scores from his last 17 exams: 42, 46, 48, 51, 52, 54, 56, 68, 72, 76, 78, 82, 85, 86, 88, 92, 96

Find his median score.

Find the lower quartile score.

Find the upper quartile score.

Find the interquartile range.

For each set of scores: i

Sort the scores in ascending order.

Find the number of scores.

iii

Find the median.

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

40, 39, 15, 17, 10, 6, 24

−4, −6, −1, 7, 9, 7, 9

8, 20, 19, 4, 15, 14, 10

42, 28, 22, 40, 20, 54, 32, 43

84, 85, 79, 71, 69, 88, 82, 78

102, 115, 110, 113, 100

228, 205, 198, 202, 207, 197

19.5, 29.6, 39.1, 22.4, 15.8, 46.2, 35.4

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8 Ex 2

The table shows the luggage weight, in kilograms, of 30 passengers: a

Find the mean check in weight, rounded to two decimal places.

In kilograms, ﬁnd the:

c 9

Ex 3

Median

Lower quartile

iii

Upper quartile

Frequency 4 5 5 3 4 6 3

Matches 45 46 47 48 49 50 51 52

Frequency 1 2 5 3 7 25 6 2

Phone calls 0 1 2 3 4 5 6 7

Frequency 8 5 10 6 6 7 3 2

In which quartile does the mean lie?

A sample of matchboxes were selected for quality control and the number of matches in each box recorded in a given frequency table. a

How many matchboxes were sampled?

Find the range of the data.

Find the interquartile range.

A group of students were asked how many phone calls they had made the previous day. The information was collected in a given frequency table. a

How many students were surveyed?

Find the range of the data.

Find the interquartile range.

For this dot plot:

764

Weight 16 17 18 19 20 21 22

9 10 11 12 13 14 15 16 17 18

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

Find the range.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The dot plot shows the ages of customers in a mobile phone store in one day:

Age of customers

Find the lower quartile.

Find the interquartile range.

Find the upper quartile.

The stem-and-leaf plot shows the number of hours students spent studying during an entire semester: a

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

Hours of study 6

2 7

0 1

2 2 4 7 9 2 5 7

Key: 5|2 = 52 For this bar chart: a

Organise the data into a frequency table.

Find the median score.

Find the lower quartile score.

Find the upper quartile score.

Find the interquartile range.

5 4 Frequency

3 2 1 0

The column graph shows the number of pets that each student in a class owns. a

Find the lower quartile.

Find the upper quartile of the set of scores.

Find the interquartile range.

Frequency

10 9 8 7 6 5 4 3 2 1 0

8 7 Score

2 3 4 5 6 Pets

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Dylan records the number of ﬁsh he catches for each ﬁshing trip over a period of time:

Frequency

20 18 16 14 12 10 8 6 4 2 0

Number of fish caught

How many ﬁshing trips did he go on?

Find the range in the number of ﬁsh caught.

Find the interquartile range.

This list shows the number of points scored by a basketball team in each game of their previous season: 59, 67, 73, 82, 91, 58, 79, 88, 69, 84, 55, 80, 98, 64, 82

766

Find the median value.

Find the lower quartile score.

Find the upper quartile score.

What percentage of games did the team score more than in the upper quartile?

What percentage of games did the team score less than in the lower quartile?

The stem-and-leaf plot shows the batting scores of two cricket teams, the Emerald Eagles and the Sapphire Sharks. Cricket team batting scores a Find the median score for the Emerald Emerald Eagles Sapphire Sharks Eagles. b

Find the median score for the Sapphire Sharks.

Find the range for the Emerald Eagles.

Find the range for the Sapphire Sharks.

Find the interquartile range for the Emerald Eagles.

Key: 6|1|2 = 16 and 12

Find the interquartile range for the Sapphire Sharks.

What are the differences between the distribution of scores for the Emerald Eagles and the Sapphire Sharks?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning

20−22

20−23

The bar graph shows the marks (out of 10) that students received on a spelling test: a

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

Explain the effect scores less than 7 have on the summary statistics.

28 24 Frequency

19, 22

20 16 12 8 4 0

7 8 Mark

These data sets each contain 10 scores:

Set 1: 10, 12, 14, 16, 18, 20, 22, 24, 26, 28

Set 2: 10, 12, 14, 16, 18, 40, 42, 44, 46, 48

Set 3: 10, 42, 44, 46, 48, 50, 52, 54, 56, 58

Which data set has the largest interquartile range and what does this tell you about the spread of data in this data set?

Identify if each set is symmetrical, skewed-left or skewed-right and justify your response.

There is a test to measure the Emotional Quotient (EQ) of an individual. Here are the EQ results for two groups of 21 people listed in ascending order: Group 1: 90, 90, 91, 92, 93, 94, 95, 95, 95, 97, 99, 100, 108, 114, 116, 116, 117, 118, 118, 122, 129 Group 2: 85, 85, 85, 86, 86, 88, 90, 95, 97, 98, 100, 100, 100, 100, 102, 105, 105, 105, 110, 110, 115

Calculate the range, interquartile range, median, and mean for both data sets.

Compare the mean of both data sets. What does this tell you about the EQ scores of the groups?

Compare the spread of both data sets. What does this tell you about the variability in EQ scores within the groups?

The dot plot shows the number of pets students have, for 25 students.

Number of pets

Find the lower quartile.

Find the percentage of students who have more pets than the upper quartile. Explain why it is not exactly 25%.

Find the upper quartile.

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Is the percentage of of data values between the upper and lower quartiles always exactly 50%? Explain why or why not, including an example to justify your conclusion.

Problem-solving 24, 25a–f 24

24, 25

24−26

In a competition, a contestant must complete 12 challenges earning as many points as possible. Her scores for the ﬁrst 11 challenges are: 38, 43, 45, 66, 67, 82, 92, 102, 105, 108, 119 Determine her score in the 12th round if the lower quartile of all of her 12 scores is 55.

For a week, the airline Fly Air decided to keep track of how many minutes behind schedule each ﬂight departed. 100 results are shown in the dot plot. a

Find the median. What is the meaning of the median in this scenario?

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

If a ﬂight is delayed for 10 minutes or more, the airline incurs a fee. Based on the dot plot, how many ﬂights did the airline incurred a fee for?

What percentage of ﬂights did the airline incur a fee for?

A rival airline, Fly High, had a median delay time during the same week of 45 minutes.

5 10 15 20 25 30 35 40 45 50 55 60 minutes

How many delayed ﬂights did Fly Air have that were longer than 45 minutes? h

What percentage of Fly Air’s ﬂights had delay times that were longer than Fly High’s median delay time?

To gain a place in the main race of a car rally, teams must compete in a qualifying round. The median time in the qualifying round determines the cut off time to make it through to the main race. Here are some results from the qualifying round: • 75% of teams finished in 159 minutes or less. • 25% of teams finished in 132 minutes or less. • 25% of teams finished between with a time between 132 and 142 minutes. a

Find the cut-off time required in the qualifying round to make it through to the main race.

Find the interquartile range in the qualifying round.

In the qualifying round, the ground was wet, while in the main race, the ground was dry. To make the times more comparable, the ﬁnishing time of each team from the qualifying round is reduced by 5 minutes. Find the new median time from the qualifying round.

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15.02 Boxplots After this lesson, you will be able to… • identify the five-number summary from a boxplot. • construct a boxplot using given data. • interpret outliers and their impact on data representation. • describe the shape of data distribution through a boxplot.

Boxplots Boxplots (also known as box-and-whisker plots) are a way of showing the five-number summary for a data set. The five-number summary is comprised of the following five statistics: Minimum Lower quartile Median Upper quartile Maximum

17 52 69 87 100

Minimum value

Median

Lower Quartile

Maximum value Upper Quartile

100

As you can see, the boxplot consists of a number line, a rectangle, and two horizontal lines (the whiskers). The box represents the middle 50% of the scores and its size tells us the interquartile range. 25% of the scores lie in each quartile. So for this example, 25% of the scores would lie between 17 and 52, 52 and 69, 69 and 87, and 87 and 100.

Minimum and outlier

Upper quartile Q3

Median Q2

Lower quartile Q1

Maximum ∼25%

∼25%

Interquartile range

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Outliers are data points which are very far away from the rest of the data set. On a boxplot, the outliers are shown as points or crosses. When there are outliers, the end of the whiskers may not be the minimum or maximum. Boxplot with outliers

Similar to other data displays, we can describe the shape by saying it is symmetrical, skewed right or left, or uniform. Symmetrical

A symmetrical data set is distributed around the centre with a similar frequency on the left and right. 0

10 15 20 25 30 35 Left skew

A left-skewed data set is where the majority of the data points have higher values, with some data points at lower values. It is sometimes called a negative skew. 0

10 15 20 25 30 35 Right skew

A right-skewed data set is where the majority of the data points have lower values, with some data points at higher values. It is sometimes called a positive skew. 0

10 15 20 25 30 35 Uniform

A uniform data set is evenly distributed across all values.

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10 15 20 25 30 35

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Keep in mind: • when describing skewed distributions, it’s better to use the median and interquartile range because they are less impacted by outliers. • when describing symmetrical or uniform distributions, it’s better to use mean and range because they take the values of all data points into account.

Interactive exploration

Discover this concept in action online

mathspace.co

Example 1 Create a boxplot to represent the data in the table. Minimum Lower quartile Median Upper quartile Maximum

10 20 40 55 75

Create a strategy

Apply the idea

Construct a boxplot using the data from the table.

0 10 20 30 40 50 60 70 80 90 100

Example 2 For the boxplot shown, fill in the table with the corresponding values: Minimum Maximum Outliers Range IQR

0 4 8 12 16 20 24 28 32 36 40 44

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Create a strategy Use the illustration shown to determine the values for the table. Minimum and outlier

Upper quartile Q3

Median Q2

Lower quartile Q1

Maximum ∼25%

∼25%

Interquartile range

The range is the difference between the minimum and maximum. For the outliers, determine the points beyond the whiskers. For the IQR, measure the distance from the lower to the upper quartile, or the length of the box.

Apply the idea Here’s the complete table:

Minimum Maximum Outliers Range IQR

1 43 1, 41, 43 42 7

Example 3 The stem-and-leaf plot displays the scores of students in a class on an exam.

Leaf 6

Key: 6|1 = 61

a Construct the five-number summary.

Create a strategy

Apply the idea

• To find the minimum value, locate the smallest score in the lowest stem. • To find the maximum value, locate the highest score in the highest stem. • Calculate the median by finding the middle score if the number of scores is odd, or by averaging the two middle scores if even. • Determine the lower quartile by identifying the middle score of the values below the median. • Identify the upper quartile as the middle score of the values above the median.

Here’s the five-number summary:

772

Minimum Lower quartile Median Upper quartile Maximum

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

61 69 73 80 85

b Construct a boxplot for the data.

Create a strategy

Apply the idea

Use the five-number summary in part (a) to construct the boxplot.

Exam scores

c Describe the shape of the boxplot.

Apply the idea The boxplot shows a slight right skew.

Example 4 The boxplot shows the mathematics test results for Mr. Smith’s and Ms. Johnson’s classes. Mr. Smith’s class

Ms. Johnson's class

100

a Calculate the range of Mr. Smith’s class.

Create a strategy

Apply the idea

Use the whiskers of the boxplot to calculate the range.

95 − 70 = 25 Mr. Smith’s class has range of 25.

b Calculate the interquartile range of Ms. Johnson’s class.

Create a strategy

Apply the idea

Use the edge of the boxes to calculate the interquartile range.

85 − 70 = 15 Ms. Johnson’s class has an interquartile range of 15.

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c How much higher was the median of Mr. Smith’s class than Ms. Johnson’s class?

Create a strategy

Apply the idea

Use the line inside the box to determine the median of both classes

85 − 80 = 5 Mr. Smith’s class had a median of 5 marks higher.

d Which class performed better overall? Explain your answer.

Create a strategy

Apply the idea

Use the measures of spread and central tendency to compare the class marks.

Mr. Smith’s class had a higher maximum, median and minimum score than Ms. Johnson’s. Mr. Smith’s class also had a smaller range, indicating more consistent results. Thus, Mr. Smith’s class performed better overall.

Idea summary Boxplots are a graphical representation showing the ﬁve-number summary for a data set.

Minimum value

Median

Lower Quartile

Maximum value Upper Quartile

100

Outliers can be shown as dots or crosses and are far away from the rest of the data set. We can describe the shape based on the distribution of the data set. • • • •

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Symmetrical boxplots are symmetrical about the median. Left-skewed boxplots have the majority of data points with higher values. Right-skewed boxplots have the majority of the data points with lower values. Uniform boxplots have all quartiles the same width.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

15.02 Practice questions Understanding 1

1−4

Adventurer

Trailblazer

1−4

Fill in the missing numbers for each boxplot. a

Explorer

Minimum 20 Lower quartile 28 Median

Upper quartile

Maximum

Minimum 16 Lower quartile 33 Median

Upper quartile

Maximum

Minimum 26 Lower quartile 33 Median

Upper quartile

Maximum

For each five-number summary, find the: i

Range

Interquartile range

Minimum 26 Lower quartile 33 Median 41 Upper quartile 44 Maximum 45

Minimum Lower quartile Median Upper quartile Maximum

16 27 35 42 50

Is each statement true or false for the data set shown in the boxplot? a

The maximum is 16

The minimum is 8

The median is 11

There is an error as the median is not exactly in the middle of Q1 and Q3

The interquartile range is 3

The value 1 is an outlier

2 4 6 8 10 12 14 16 18 20

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For the boxplot shown: a

What percentage of scores lie between each of these values? i

Q1 and Q3

Minimum and Q1

iii

Median and Maximum

Q1 and Maximum

Minimum and Q3

In which quartile(s) is the data the least spread out?

In which quartile(s) is the data the most spread out? 5, 6ab, 7, 8, 9abgi, 11a, 12

Fluency 5

5, 6bc, 7, 8, 9cdhi, 10, 11b, 13, 14

Minimum

Maximum

Range

Median

Interquartile range

6cd, 7, 8, 9efhj, 10−14

8 10 12 14 16 18 20

Construct a five-number summary for each boxplot: a

Scores

8 10 12 14 16 18 20

Scores

776

For the boxplot shown, ﬁnd the:

8 10 12 14 16 18 20

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

8 10 12 14 16 18 20

Scores

38 40 42 44 46 48 50 52 54 56 58 60

Ex 1

Construct a boxplot to represent the information in the tables: a

Ex 2

Minimum 5 Lower quartile 25 Median 40 Upper quartile 55 Maximum 65

Minimum Lower quartile Median Upper quartile Maximum

30 31 47 53 58

For the boxplot shown, ﬁnd: a

Minimum

Maximum

Outliers

Range

Interquartile range 0 4 8 12 16 20 24 28 32 36 40 44

Ex 3

For each of the data sets: i

Construct the ﬁve-number summary.

Draw a boxplot.

iii

Describe the shape of the boxplot.

14, 15, 20, 21, 23, 24, 27

21, 25, 29, 31, 32, 34, 39, 40

40, 12, 8, 20, 12, 28, 48, 36, 60

28, 4, 40, 16, 20, 48, 20, 44, 56

42, 43, 48, 54, 59, 51, 53, 50, 57, 45

60, 66, 67, 68, 69, 72, 73, 76, 77, 78, 86

Leaf

2 3 5 8 9

0 4 4 5 7 7

0 2 3 4 6 7

3 3 5 9

2 4 5

20 21 22 23 24 25 26 27 28 29 30 31

A cyclist measured her heart rate immediately after ﬁnishing each event in which she competed. The results are recorded in the boxplot. a

Find and interpret the range.

Find and interpret the interquartile range.

2 3 5 5 Key: 1|2 = 12

Key: 1|2 = 12 i

Leaf

10 11 12 13 14 15 16 17 18 19 20 21

Heart Rate

155 160 165 170 175 180 185 190 195 200

15.02 Boxplots mathspace.co

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Use the summary statistics to determine: i

The shape of each data set

The data set with the higher measure of centre

iii

The data set with the higher measure of spread

a Data set A Data set B

Mean 10 15

Median 12 15

IQR 2 5

Data set A Data set B

Minimum 1 10

Q1 4 18

Median 5 20

Range 5 10 Q3 10 21

Maximum 13 22

IQR 6 3

A mathematics test is given to two classes. The marks out of 20 received by students in each class are represented in the boxplots. a

Range 12 12

Class 10P

Complete the table: Class 10P

Class 10Q

Median Lower quartile Upper quartile Range Interquartile range Outliers b

Which class tended to score better marks?

Would the range or interquartile range be better for describing the spread of Class 10Q?

0 2 4 6 8 10 12 14 16 18 20 22

Class 10Q

0 2 4 6 8 10 12 14 16 18 20 22

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The boxplot shows the age at which a group of people obtained their driving licences: a

What is the oldest age?

What is the youngest age?

What percentage of people were aged from 18 to 22?

The middle 50% of responders were within how many years of one another?

In which quartile are the ages least spread out?

The bottom 50% of responders were within how many years of one another?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Age

Ex 4

The boxplots represent the daily sales made by Carl and Angelina over the course of one month. Angelina's sales

Carl's sales

Calculate the range for Angelina’s sales.

Calculate the range for Carl’s sales.

By how much did Carl’s median sales exceed Angelina’s?

By comparing the middle 50% of sales for each person, identify which salesperson had the more consistent number of sales.

Which salesperson had a more successful sales month? Explain your answer.

Reasoning 15

15, 17−19

15−17, 19, 20

16, 19−22

Ten participants had their pulse measured before and after exercise with results shown in different boxplots. Compare the pulse rates before and after exercise for these participants. Pulse before exercise

Pulse after exercise

45 50 55 60 65 70 75 80 85 90

80 85 90 95 100 105 110 115 120 125

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How does a boxplot help us to understand the centre and spread within a data set?

The test scores of 11 students in Science and Music are listed: • Science: 52, 59, 87, 66, 74, 76, 73, 63, 91, 86, 82 • Music: 87, 63, 60, 43, 74, 82, 66, 61, 88, 60, 84

Construct a boxplot for the Science scores.

Construct a boxplot for the Music scores.

In which subject did students perform better overall? Explain your answer with reference to measures of central tendency and spread.

These boxplots shows the number of points scored by two basketball teams in each of their matches. Gamma Geckos

30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70

Delta Dragons

30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70

Compare the median score of the Gamma Geckos and the Delta Dragons.

Compare the range of scores for the Gamma Geckos and the Delta Dragons.

Compare the interquartile range for the Gamma Geckos and the Delta Dragons.

If the two teams were to play against each other in the next game, determine which team you expect to win. Justify your answer.

The boxplots summarise results from a medical study. The treatment group received an experimental drug to relieve cold symptoms, and the control group received a placebo. The boxplots show the number of days each group continued to report symptoms:

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Control group

Treatment group

8 10 12 14 16 18 20

Describe the shape of the data from the control group.

Describe the shape of the data from the treatment group.

Does the drug have a positive effect on patient recovery? Explain your reasoning.

A boxplot was created for a data set. The value 32 was a typo and should have been 23. a

Can we draw a new boxplot with this information? Explain.

How does this change the range?

How does this change the interquartile range? 10

Create a context where a data set might meet the given conditions. Justify your answer. a

A range between 1 and 4.

A small interquartile range and a large range.

The marks in an end-of-year exam for a class of students are shown: 52, 56, 59, 64, 66, 77, 78, 80, 80, 80, 81, 84, 86, 90, 95, 96 a

Construct a boxplot for the data.

Calculate the interquartile range.

What percentage of marks lie in the range 85 to 96?

Which values do the lowest 75% of scores lie between?

If the score 91 was added to the data set, what percentage of the marks are below Q3? Why isn’t this 75%?

Problem-solving 23, 26bd 23

23−25, 26a–c

Construct a boxplot for the data set: • Range = 14 • Lowest score = 2 • IQR = 8

23, 25, 26a–c

• 25% of scores are below 4 • 50% of scores are above 8

For each of the conditions, use ﬁve of the digits from 1 to 9 with no repetition to complete a ﬁve-number summary: min: ⬚, lower quartile: ⬚, median: ⬚, upper quartile: ⬚, max: ⬚

The smallest possible interquartile range with the largest possible range.

The range is three times the length of the interquartile range.

15.02 Boxplots mathspace.co

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Match the histograms on the left to the corresponding boxplots on the right: Histogram A

Box plot 1

0 1

9 10

1 2 3 4 5 6 7 8 9 10

Histogram B

Box plot 2

0 1

2 3 4 5 6 7 8 9 10

Histogram C

Box plot 3

1 2 3 4 5 6 7 8 9 10

Histogram D

0 1

1 2 3 4 5 6 7 8 9 10

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Box plot 4

9 10

Create a boxplot for the data shown in each graph. Monthly salaries of employees 30

Number of employees

20 15 10 5 20 25 30 35 40 45 50 Salaries (in thousands)

Debbie’s blog visitors

Number of days

Scores of students

10 9 8 7 6 5 4 3 2 1 0

100

Number of pages read

10 0

50 100 150 200 Number of visitors

15.03 Cumulative frequency After this lesson, you will be able to… • calculate cumulative frequencies from a frequency table. • construct a cumulative frequency table from a given frequency distribution table. • interpret and solve problems involving cumulative frequencies. • construct and interpret cumulative frequency polygons. • apply knowledge of cumulative frequencies and their graphical representations to solve problems.

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Cumulative frequency We have seen that a frequency table allows us to group data into intervals and tells us how many data values are in that interval. Cumulative frequency is a ‘running total’ of the frequencies. To calculate it, we add an additional column to the frequency distribution table: Class interval 50 ≤ t < 55 55 ≤ t < 60 60 ≤ t < 65 65 ≤ t < 70 70 ≤ t < 75 75 ≤ t < 80 80 ≤ t < 85 Total

Frequency 5 10 25 26 40 49 28 183

Cumulative frequency 5 5 + 10 = 15 15 + 25 = 40 40 + 26 = 66 66 + 40 = 106 106 + 49 = 155 155 + 28 = 183

• The first value in the cumulative frequency column will always be the same as the first value in the frequency column. • To get the second cumulative frequency value we add the second frequency to the first cumulative frequency value, 5 + 10 = 15. • The second cumulative frequency value tells us there are 15 values in the interval 50 ≤ p < 60. • The third cumulative frequency value tells us there are 40 values in the interval 50 ≤ p < 65, and so on. • The final cumulative frequency value is always equal to the sum of the frequencies. In this case, there are 183 values in the entire data set, represented by 50 ≤ p < 85.

Example 1 A principal wants to investigate the performance of students at his school in Performing Arts. To do this, he has the marks of each student studying Performing Arts collected into groups and put into a frequency table. Each group of marks is assigned a grade. The frequency table for this is shown. a Complete the cumulative frequency column. Grade E D C B A

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Score (x) 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 100

Class centres

Frequency ( f ) 7 14 32 97 62

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Cumulative frequency (cf )

Create a strategy The cumulative frequency of the ﬁrst row is the frequency of that row. For the following rows, add the frequency of each row to the cumulative frequency of the previous row.

Apply the idea The ﬁrst value in the cumulative frequency column is the same as the ﬁrst value in the frequency column, so we have 7. For the second cumulative frequency value, we have 7 + 14 = 21. Repeating the process to get the next cumulative frequency value, we have the completed table: Grade E D C B A

Score (x) 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 100

Frequency ( f ) 7 14 32 97 62

Cumulative frequency (cf ) 7 21 53 150 212

b How many marks were recorded?

Create a strategy

Apply the idea

We can ﬁnd the total frequency in the last entry of the cumulative frequency column.

This value is the same as adding all the entries in the frequency column. Total frequency = 212

c Complete the sentence. ‘Approximately three quarters of the scores recorded are greater than ⬚’.

Create a strategy

This can be reworded to one quarter of the scores are below this value: ⬚’.

Apply the idea

There are 212 scores, so one quarter of that is

= 53.

The cumulative frequency reaches 53 in the interval 40 ≤ x < 60, so 60 is the corresponding upper boundary score. ‘Approximately three quarters of the scores recorded are greater than 60’.

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Example 2 For this table: Score (x) 10 11 12 13 14

Cumulative frequency (cf ) 7 15 18 20 26

a How many scores were there in total?

Create a strategy

Apply the idea

We can ﬁnd the total number of scores in the last entry of the cumulative frequency column.

This value is the same as adding all the entries in the frequency column. Number of scores = 26

b How many scores of 14 were there?

Create a strategy Find the difference between the cumulative frequency at 14 and the cumulative frequency at 13.

Apply the idea There were 26 scores after the 14s were counted and 20 before. Number of 14s = 26 − 20 =6

Subtract the values Evaluate

c How many scores were less than 13?

Create a strategy

Apply the idea

Find the corresponding cumulative frequency of The corresponding cumulative frequency of the the score 12. score 12 is 18, so there were 18 scores which are less than 13.

Idea summary Cumulative frequency is a ‘running total’ of the frequencies. To calculate it, we add an additional column to the frequency distribution table.

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Cumulative frequency graphs

Cumulative frequency

Cumulative frequency distribution Global life expectancy, 2016 200 180 160 140 120 100 80 60 40 20 0 50 55 60 65 70 75 80 85 Life expectancy (years)

A cumulative frequency histogram will always increase in size from left to right. The frequency represented by any particular column will be equal to the difference in height between that column and the one before it.

Cumulative frequency

Cumulative frequency distribution Global life expectancy, 2016 200 180 160 140 120 100 80 60 40 20 0 50 55 60 65 70 75 80 85 Life expectancy (years)

A cumulative frequency polygon, also known as an ogive, is a line representing the cumulative frequency and it joins the top right-hand corner of each column. Here is a frequency histogram and polygon displayed together.

Cumulative frequency distribution Global life expectancy, 2016 200 180 160 140 120 100 80 60 40 20 0 50 55 60 65 70 75 80 85 Life expectancy (years)

Here is a cumulative frequency polygon displayed on its own.

Cumulative frequency

We have already seen how frequency tables can be used to create a histogram. The cumulative frequencies can also be plotted to create another type of chart, called a cumulative frequency graph. This graph is made up of a cumulative frequency histogram and polygon.

We can use a cumulative frequency polygon to find values such as the median and quartiles which are at 25%, 50% and 75% of the cumulative frequencies.

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Example 3

Cumulative frequency

For this cumulative frequency histogram:

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

40 45 Class

a How many scores are there in total?

Create a strategy For a cumulative frequency histogram the total will be the cumulative frequency for the last column.

Apply the idea The 50 – < 55 column has the cumulative frequency of 14. So there 14 scores in total.

Reﬂect and check Notice this is different from a normal histogram where we would add up the frequencies of all of the columns. Since these columns represent cumulative frequency, this is already done.

b How many scores between 50 – < 55 occured?

Create a strategy Find the difference between the cumulative frequency at 50 – < 55 and at 45 – < 50.

Apply the idea There were 14 scores after the 50 – < 55s were counted and 12 before. Number of 50 – < 55s = 14 − 12 =2

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Subtract the values Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c Construct a cumulative frequency polygon for the data.

Create a strategy The cumulative frequency polygon connects the upper right-hand corner of each column.

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Reﬂect and check On its own, the ogive would appear as shown.

Cumulative frequency

Apply the idea

40 45 Class

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

40 45 Class

Idea summary A cumulative frequency polygon, also known as an ogive, is a line graph connecting the upper right-hand corner of each column of a cumulative frequency histogram.

15.03 Practice questions Understanding 1

Explorer 1−3

Adventurer 2−4

Trailblazer 2−4

Complete the frequency table based on the data set: 2.1, 5, 4, 5, 5.15, 5, 3.8, 4, 5.32, 5.5, 1.25, 3, 3.75, 3, 5, 2.82, 4.6, 1, 6, 5, 6.56, 3, 1, 1.6, 6 Score 1≤x<2 2≤x<3 3≤x<4 4≤x<5 5≤x<6 6≤x<7

Frequency

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Complete the cumulative frequency column. Score (x) 1≤x<5 5≤x<9 9 ≤ x < 13 13 ≤ x < 16 17 ≤ x < 20

790

Frequency ( f ) 8 5 8

Cumulative frequency (cf ) 8 21 26

Is each statement true or false? a

The height of the last column tells us the total.

The cumulative frequency must always add up to 100.

To find the cumulative frequency we add the frequency to the previous cumulative frequency.

To find the frequency we subtract the previous cumulative frequency from the cumulative frequency.

A cumulative frequency polygon may have segments that are decreasing.

Fluency Ex 1

Cumulative frequency (cf )

Complete the cumulative frequency table: Score (x) 12 ≤ x < 13 13 ≤ x < 14 14 ≤ x < 15 15 ≤ x < 16

Frequency ( f ) 4 5 9 5 4

5−10, 13

For the frequency table: a

Complete the cumulative frequency column.

Calculate the total frequency.

Approximately half of the scores recorded are greater than what score?

For the given frequency table: a

Complete the cumulative frequency column.

Calculate the total frequency.

Approximately one third of the scores recorded are greater than what score?

5−10, 12, 13

6−14

Score (x)

Frequency ( f )

1–5 6–10 11–15 16–20 21–25 26–30

15 26 18 14 7 2

Score (x)

Frequency ( f )

20–24 25–29 30–34 35–39 40–44 45–49 50–54

7 18 25 12 8 4 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Cumulative frequency (cf )

Ex 3

For this table:

Score (x)

Find the total number of scores recorded.

Find the number of times a score of 13 occurred.

Find the number of times a score less than 15 occurred.

11 12 13 14 15

Cumulative frequency (cf ) 9 13 19 27 30

For this cumulative frequency histogram: Score (x)

Frequency ( f )

20 ≤ x < 25 25 ≤ x < 30 30 ≤ x < 35 35 ≤ x < 40 40 ≤ x < 45 45 ≤ x < 50 50 ≤ x < 55 55 ≤ x < 60 60 ≤ x < 65 65 ≤ x < 70 70 ≤ x < 75

3 3 3 1 2 4 13 13 12 4 2

Cumulative frequency (cf ) 3 6 9 10 12 16 29 42 54 58 60

Class marks 70 60 50 40 30 20 10 0

20 25 30 35 40 45 50 55 60 65 70 75 Marks

Find the total number of scores recorded.

Find the number of times a score between 25 and 30 occurred.

Find the number of times a score between 50 and 55 occurred.

Find the percentage of scores that were less than 35.

For this cumulative frequency histogram: a

How many scores are there in total?

How many scores between 40– < 50 occurred?

Construct a cumulative frequency polygon for the data.

Cumulative frequency

Ex 2

100 90 80 70 60 50 40 30 20 10 0

20 30 40 50 60 70 80 Class

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The frequency table shows the time (t), in minutes, that it takes for people in a certain town to walk to the nearest park. Construct a cumulative frequency polygon for the data. Time (sec) 0 ≤ h < 10 10 ≤ h < 20 20 ≤ h < 30 30 ≤ h < 40 40 ≤ h < 50

Frequency 18 20 17 16 14

The graph shows the cumulative frequency polygon of the mass (in grams) of fish caught in a fishing competition: Construct a frequency histogram to depict the distribution of the mass of the fish.

170 Cumulative frequency 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10

Mass (g)

792

Construct a frequency table for the data represented in the cumulative frequency histogram.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Cumulative frequency

50 100 150 200 250 300 350 400

20 18 16 14 12 10 8 6 4 2 0

80 85 90 95 100 Score

A pair of dice are rolled 50 times and the numbers appearing on the uppermost face are added to give a score. The results are recorded in the given table. a

What is the lowest possible score when a pair of dice are rolled?

What is the highest possible score when a pair of dice are rolled?

Complete the table by ﬁnding the cumulative frequency values.

Find the number of times a score of 8 occurred.

Find the number of times a score more than 9 occurred.

Find the number of times a score of at most 6 occurred.

Score (x)

Frequency ( f )

2 3 4 5 6 7 8 9 10 11 12

1 2 5 5 5 9 7 5 8 1 2

Cumulative frequency (cf )

The number of sightings of the Northern Lights were recorded across various Canadian locations over a period of 1 month. The list represents the number of sightings at each location: 11, 10, 10, 9, 7, 8, 8, 12, 12, 12, 12, 12, 12, 9, 9, 12, 9, 9, 8, 8 a

Complete the table.

In how many locations were there at least 8 sightings?

In how many locations were there less than 11 sightings?

Number of sightings 7 8 9 10 11 12

Number of locations ( f )

Cumulative frequency (cf )

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Reasoning 15

15, 16

16, 17

A jump rope competition has routines which are judged on a scale of 0 to 10, and scores are rounded to one decimal place. The average of the scores from the middle three judges makes the score for the routine. 50

Cumulative frequency

45 40 35 30 25 20 15 10 5 0

Score

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Explain how the frequency of a certain score can be determined from a cumulative frequency histogram.

What is the most frequent range of scores?

Families were asked how many times they contracted the ﬂu during winter. The information has been partially ﬁlled out in the table. Explain your reasoning for each part of the problem. a

Complete the frequency and cumulative frequency values in the given table.

Find the total number of families that responded.

Find the median number of times a family contracted the ﬂu.

Find the mean number of times a family contracted the ﬂu. Round your answer to two decimal places.

Construct a cumulative frequency histogram for the data. Explain the signiﬁcance of the histogram.

Find the number of families who contracted the ﬂu more than twice.

Score (x) 0 1 2 3 4

Frequency ( f ) 0

Cumulative frequency (cf ) 2

3 2

To find each key feature, would you rather use a histogram or cumulative frequency histogram? Explain your reasoning. a

Total

Frequency for one interval or value

Shape of distribution

Number of data points below a particular value

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving 18

18−20

A 1500 m swimmer records her time (min:sec) over several training sessions. Her times are recorded in the histogram. Construct a cumulative frequency table for the data using the given intervals.

Find the total number of training sessions she completed.

Find the number of times she recorded a swim time faster than 16 : 40.

Find the percentage of swims that were less than 16 : 30.

10 9 8 7 6 5 4 3 2 1 0

16 :0 0 16 :10 16 :2 0 16 :3 0 16 :4 0 16 :5 0 17 :0 0

Cumulative frequency

18, 19

Time Recorded

The heights (cm) of 22 students in a class are listed: 164, 167, 158, 159, 150, 166, 150, 146, 149, 161, 164, 163, 152, 161, 157, 157, 153, 157, 156, 165, 162, 161

Construct a cumulative frequency histogram for the data. Use the discrete intervals of 146–150, 151–155, etc.

How many students are taller than 155 cm?

How many students are at most 155 cm tall?

How many students are taller than 150 cm but shorter than 156 cm?

Calculate the centre of the class 151–155.

For each graph, use the cumulative frequency curve to estimate: i

The median score

iii

The upper quartile

100 90 80 70 60 50 40 30 20 10

Score 1500 1900 2300 2700

The lower quartile

14 13 12 11 10 9 8 7 6 5 4 3 2 1

Score 550 600 650 700 750

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15.04 Compare data sets After this lesson, you will be able to… • interpret the five-number summary in a boxplot. • compare the range and interquartile range between two boxplots. • draw conclusions from real-world examples involving parallel boxplots. • compare data sets effectively. • understand the relationship between different graphical representations of the same data set.

Parallel boxplots We have seen that the range and interquartile range can be used to measure the spread of data and both can be seen on a boxplot. Parallel boxplots are used to compare two sets of data visually. When comparing boxplots, we compare the five-number summary and also the range and interquartile range. The two boxplots are drawn parallel to each other. It is important to clearly label each boxplot. Below are parallel boxplots comparing the time it took two different groups of people to complete an online task. 5

Under 30s

Over 30s

14 16 18 Seconds

20 22 24 26 28 30

The boxplots must be drawn on the same scale to compare them.

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 1 The two boxplots show the data collected by the manufacturers on the lifespan of light bulbs, measured in thousands of hours. Manufacturer A

Manufacturer B

3 4 5 Thousands of hours

a Complete this table using the two boxplots. Write each answer in terms of hours, remembering to multiply the values on the data display by 1000. Manufacturer A

Manufacturer B

Median Lower quartile Upper quartile Range Interquartile range

Create a strategy • To ﬁnd the lower quartile, median, and upper quartile, ﬁnd the corresponding values of the vertical lines of the box, respectively. • For the range, use the formula: Range = Highest score − Lowest score • For the interquartile range, use the formula: IQR = Q3 − Q1

Apply the idea Since the scale is in thousands of hours, we can multiply the numbers on the scale by a thousand to ﬁnd the number of hours. Median Lower quartile Upper quartile Range Interquartile range

Manufacturer A 4 × 1000 = 4000 2.5 × 1000 = 2500 4.5 × 1000 = 4500 (5 − 1) × 1000 = 4000 (4.5 − 2.5) × 1000 = 2000

Manufacturer B 5 × 1000 = 5000 3.5 × 1000 = 3500 6 × 1000 = 6000 (8 − 1.5) × 1000 = 6500 (6 − 3.5) × 1000 = 2500

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b Which manufacturer produces light bulbs with the best lifespan?

Create a strategy Choose the manufacturer with the greater median.

Apply the idea The data set for Manufacturer B has a median of 5000 hours, while the median of the data set for Manufacturer A is 4000 hours. So Manufacturer B produces light bulb with the best lifespan.

Reflect and check In fact, the best lightbulb produced by Manufacturer A has a lifespan of 5000 hours, which is the same as the median of Manufacturer B. This means that about half of the lightbulbs produced by Manufacturer B have a greater lifespan than all of the lightbulbs produced by Manufacturer A.

Example 2 This boxplots show the number of goals scored by two football players in each season. Sophie

Holly

15 Goals

a Who scored the most goals in a season?

Apply the idea By looking at the endpoints of the right whiskers, Sophie scored 18 goals, while Holly scored 19 goals. So Holly score the most goals in a season.

b How many more goals did Holly score in her best season compared to Sophie in her best season?

Apply the idea Number of goals = 19 − 18 =1

Subtract 18 from 19 Evaluate the subtraction

Holly scored 1 more goal in her best season.

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c What is the difference between the median number of goals scored in a season by each player?

Apply the idea Sophie’s median is 11 and Holly’s median is 10. Difference = 11 − 10 =1

Subtract 10 from 11 Evaluate the subtraction

d What is the difference between the interquartile range for both players?

Apply the idea Use the formula: IQR = Q3 − Q1 Sophie’s IQR = 14 − 7 =7 Holly’s IQR = 15 − 6 =9 Difference = 9 − 7 =2

Substitute the quartiles Evaluate Substitute the quartiles Evaluate Subtract the IQRs Evaluate

Idea summary When comparing boxplots, we compare the ﬁve-number summary values of each data set (the quartiles and minimum and maximum values) and the range and interquartile range.

Different graphs of the same data We can identify the centre and spread using key features. Measures of centre summarise the data set with a single value. We often use this to generalise which group performed better. • Mean • Median • Mode Measures of spread summarise the spread of a data set. We often use this to generalise which group was more consistent. The lower the spread, the more consistent the data. • Range • Interquartile range Different displays help us to identify different key features.

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From frequency displays like histograms, dot plots, and stem-and-leaf plots we can identify: • Shape • Frequencies, grouped or individual • Mode • Outliers From boxplots, we can identify: • Shape • Range and interquartile range • Median • Outliers We should expect then that the shape of data would be the same whether it is represented in a polygon, boxplot or histogram.

Frequency

All representations of symmetric data: 10 9 8 7 6 5 4 3 2 1 0

Histogram of approximately symmetric data

8 9 10 11 12 13 14 15 16 17 18 Score

Boxplot of symmetric data

Frequency

All representations of positively skewed data: 10 9 8 7 6 5 4 3 2 1 0

Histogram of positively skewed data

8 9 10 11 12 13 14 15 16 17 18 Score

Boxplot of positively skewed data

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All representations of negatively skewed data: Histogram of negatively skewed data

Frequency

10 9 8 7 6 5 4 3 2 1 0

8 9 10 11 12 13 14 15 16 17 18 Score

Boxplot of negatively skewed data

Looking at the diagrams above, can you see the similarities in the representations? We can see the skewed tails, where the bulk of the data sits, and general shape. These are some of the features you can use to match histograms and boxplots. We can also look at the data range.

Interactive exploration

Discover this concept in action online

mathspace.co

Example 3 Match the boxplot shown to the correct histogram.

9 10

1 2 3 4 5 6 7 8 9 10

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Create a strategy

Apply the idea

Look for characteristics of skewed and symmetrical distributions of data.

The given boxplot shows symmetrical data. Only option B shows symmetrical data. The correct answer is option B.

Idea summary We should expect that the shape of data would be the same whether it is represented in a boxplot or histogram. Remember that the shape of data can be symmetrical, left skewed or right skewed.

15.04 Practice questions Understanding 1

Adventurer 1−3

Trailblazer 1−3

The histogram and the boxplot display the same data. 10 9 8 7 6 5 4 3 2 1 0

802

Explorer 1, 2

0 10 20 30 40 50 60 70 80 90 100 10 20 30 40 50 60 70 80 90 100

Which data display shows the mode of the data set?

Which data display shows the median of the data set?

Which data display shows the range of the data set?

Which data display shows the number of values between 10 and 20?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Each pair displays the same data set. What is the skew for each pair of histograms and boxplots? a

10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 100

10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90

1 2 3 4 5 6 7 8 9 10

Which key features can be read off each display? a

Histogram

Boxplot

Dot plot

Cumulative frequency histogram and polygon

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Fluency Ex 1

4, 5, 8, 10

4−9

5−10

A class took an English test and a Mathematics test. Both tests had a maximum possible mark of 20. The parallel boxplots show the results of the tests:

English

Mathematics

Complete the table using the two boxplots. English

Mathematics

Median Lower quartile Upper quartile Range Interquartile range b Ex 2

In which test did the class tend to score better marks?

The parallel boxplots show the prices, in dollars, of the items on the menu of an upmarket restaurant and the menu of a fast food restaurant. Upmarket Restaurant

Fast Food

Price

804

Which restaurant has the higher median price for the items they sell?

What is the difference between the median prices of the items sold by each restaurant?

Which restaurant has a greater price range for the items on their menu?

What is the price difference between the most expensive items sold by each restaurant?

What amount of the cheapest item at the fast food restaurant could be bought for the same price as the most expensive item at the upmarket restaurant?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Two weightlifters both record their number of repetitions with a 70 kg bar over 30 days. The results are displayed in the boxplots:

Which weightlifter has the more consistent results?

Which statistics determine which is more consistent?

Which statistic is the same for each weightlifter?

Which weightlifter can do the most repetitions of 70 kg?

Repetitions

The residents of two blocks of townhouses were asked the number of pets they own. The frequency of various responses are presented in these dot plots. a

Is the pet ownership a little lower or higher in Block A than Block B?

In Block A, how many pets do most households have?

In Block B, how many pets do most households have?

What is the shape of the data for Block A?

Find the range of the number of pets in Block A.

Which block has more variability in the the number of pets?

Do either sets of scores have an outlier?

Block A

Block B

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Ex 3

Match the histograms to the corresponding boxplots. Histogram A Boxplot 1 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Boxplot 2 Histogram B 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Histogram C Boxplot 3 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Histogram D Boxplot 4 10 9 8 7 6 5 4 3 2 1 0

806

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Match each cumulative frequency graph to its corresponding boxplot: Cumulative Frequency Graph A Cumulative Frequency Graph B Cumulative Frequency

Cumulative Frequency

10 Mass (g)

Mass (g)

5 10 15 20 25 30 35 40 45 50 55 60

Cumulative Frequency Graph C

Cumulative Frequency Graph D

Cumulative Frequency

10 Mass (g) 5 10 15 20 25 30 35 40 45 50 55 60

Mass (g) 5 10 15 20 25 30 35 40 45 50 55 60

boxplot 1

boxplot 2

5 10 15 20 25 30 35 40 45 50

boxplot 3

5 10 15 20 25 30 35 40 45 50

boxplot 4

5 10 15 20 25 30 35 40 45 50

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The parallel boxplots show the distances, in centimetres, jumped by two high jumpers. a

Who had a higher median jump?

Who made the highest jump?

Who made the lowest jump?

Matt

Paul

140

150

160

170

180

Distance (cm)

Reasoning

808

11−14

12−14

Two History classes, each with 20 students, were given a 10 question True/False test. The results for each class are shown. 10 9 8 7 6 5 4 3 2 1 0

Class 1 Number of Students

Number of Students

11, 12

5 6 7 Questions Correct

10 9 8 7 6 5 4 3 2 1 0

Class 2

8 9 10 Questions Correct

Calculate the mean for Class 1 rounded to one decimal place.

Calculate the mean for Class 2 rounded to one decimal place.

Find the range for Class 1.

Find the range for Class 2.

Do you think Class 1 studied for their test? Justify your reasoning.

Do you think Class 2 studied for their test? Justify your reasoning.

Besides studying, what is another factor that may have had an effect on the test results for each class?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Two bookstores recorded the selling price of all their books. The results are presented in the parallel boxplots. Bookstore A prices ($)

Bookstore B prices ($) x 50

100

110

120

130

140

150

Which bookstore had the more consistent prices? Explain your answer.

Comparing the most expensive books in each store, how much more expensive is the one in store B? Is each statement true or false? i

5% of the books in Bookstore B are at least as expensive than the most expensive 2 book in Bookstore A.

5% of the books in Bookstore B are cheaper than the cheapest book in 2 Bookstore A.

Explain why each pair of histograms and boxplots do not match:

Salaries earned by employees at a software company are given in a histogram. a

Using the centres, construct a boxplot of the data.

Find an estimate for the interquartile range.

Frequency

10 9 8 7 6 5 4 3 2 1 0

65 000

75 000

85 000 Salaries ($)

95 000

105 000

Using the boxplot, approximately what percentage of salaries lie in the range $90 000 to $100 000?

Using the boxplot, what can be said about the highest 25% of salaries? Would this conclusion be different from using the histogram? Explain your answer. 15.04 Compare data sets mathspace.co

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Problem-solving 15 15

15, 16

Cooper and Marion are racing go-karts. The times (in seconds) for the 12 laps of their qualifying race are shown: • Cooper: 58.9, 46.5, 52.6, 66.6, 58.4, 53.1, 45.0, 52.1, 52.4, 52.7, 44.8, 51.7 • Marion: 47.8, 54.6, 68.5, 68.0, 62.8, 57.2, 54.8, 63.4, 58.1, 64.3, 66.2, 47.1 a

Construct the ﬁve-number summary for each set.

Are there any outliers?

Create a parallel boxplot of the two sets of times with the outlier displayed separately.

Which racer will be in pole position for the ﬁnal race, if it is given to the racer with the fastest qualifying lap time?

Does spinning out on a lap, causing a high outlier, impact the selection for pole position? Explain your answer.

Two friends compete in hammer throw competitions and train together over a season. They compete in 15 competitions and their ﬁnal throw for each competition is shown: • Tim: 29.8, 37.4, 33.9, 38.8, 34.3, 36.5, 34.5, 30.0, 35.2, 38.4, 33.0, 33.2, 39.6, 35.0, 36.9 • Odi: 32.2, 35.4, 34.8, 33.0, 38.4, 26.0, 40.0, 37.2, 39.5, 42.4, 38.6, 42.3, 38.4, 42.8, 37.2 a

Complete the table of statistics, rounded to one decimal place: Minimum Q1 Median Q3 Maximum Mean Range Interquartile range

810

Tim 29.8 33.2 35.0 37.4 39.6

Odi

37.2

Which competitor throws more consistently? Explain your answer.

Identify any outliers.

Who is the better hammer thrower? Explain your answer.

When considering Odi’s average throw is it reasonable to remove the outlier before calculating the mean? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

15.05 Samples and populations After this lesson, you will be able to… • distinguish between a population and a sample in statistics. • identify the key factors that make a sample representative of a population. • understand the differences between a parameter and a statistic. • recognise the importance of sample size in data collection. • distinguish between primary and secondary data, and assess the reliability of secondary data sources. • apply the concepts of samples and populations in practical scenarios.

Samples and populations In statistics, a population refers to every member within any particular group of interest. A survey conducted on every member of a population is called a census. Collecting data from every member of a population is the most accurate way of gathering information, but it is not always possible and can be very expensive. Collecting data from a sample, can be quicker and less expensive. A parameter is a number that summarises data from a population. • Population size: N • Population mean: m A statistic is a number that summarises data from a sample. • Sample size: n • Sample mean: A sample is representative if the characteristics of the sample are close to the population. Key factors that affect if a sample is representative include: • Demographic factors: If the population is diverse in terms of age, gender, ethnicity, or other factors, the sample should reflect this diversity. • Characteristics relevant to the study: If you’re studying a particular trait or behavior, individuals in the sample should exhibit a range of these characteristics. • Random selection: This minimises bias and increases the likelihood of the sample being representative of the population from which it was drawn. If a sample is unbiased and represents the population well, we can use sample survey data to make conclusions about the population. The size of the sample is important. If the sample is too large, it may be too expensive or timeconsuming to collect the data. If it is too small, the sample may not be representative of the population.

Interactive exploration

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Primary data is data that has been collected firsthand. If you are the researcher, primary data is data that you gather yourself, directly from observation, surveys, experiments, or interviews. Secondary data is data that has already been collected and analysed by someone else. This could be in the form of books, articles, reports, or statistics that you find online or in the library. When using secondary data: • Make sure the population from which the data was collected matches your target population. • Check whether a census or sample was used. If a sample was used, make sure it is representative of the population. • Analyse whether there is any source of bias.

Example 1 A sample of 25 people is drawn from a population. In this sample, the youngest is 18 years old, and the oldest is 64. Which two of these might be populations that this sample was drawn from? A Residents of a retirement home

Employees at a bank

C Students from a primary school

People stuck in traﬃc due to road works

Create a strategy Remember that a sample is a smaller group than the entire population but every member of the sample must be a member of the overall population.

Apply the idea

Reﬂect and check

Which of these populations would be likely to include people of this age group?

To check our answer we can conﬁrm that the other options are incorrect. Option A, residents of the retirement home must be of at least a retirement age (usually 65). While for Option C, the students attending a primary school will only range from ages 4 to 12

The answers are options B and D.

Example 2 A researcher captures 400 fish in a lake, tags them, then releases them. The next day, he captures 1200 fish, of which 100 have tags attached to them. Estimate the size of the fish population.

Create a strategy The proportion of tagged ﬁsh in the second sample should be equal to the proportion of 400 ﬁsh out of the total population.

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Apply the idea

Therefore, there is estimated to be 4800 ﬁsh in the lake.

Reflect and check To check our answer, we can multiply the total population by the proportion of tagged fish to see if it gives us the 400 fish originally tagged.

Example 3 Maryke is researching the monthly spending habits of students on school lunches at a local high school. She found two secondary sources of data: • Source A: This data set provides the average monthly spending of students on lunch for the entire school year. • Source B: This data set provides raw data for the monthly spending of each student for each month of the school year, broken down by demographics like age, sex, and postal code. a If Maryke wants to ﬁnd out in which month the spending was the lowest and in which month the spending was the highest, for year 12s, which source should she refer to? Explain.

Create a strategy Decide which source provides data on individual months of the school year and where she can separate out the year 12s.

Apply the idea Maryke should refer to Source B. Firstly, she can separate out the year 12 data. Also, she can determine in which month the spending was the lowest and the highest because it provides the monthly spending of each student for every month of the school year, allowing her to calculate the average for each speciﬁc month.

Reﬂect and check We some times call data which is all grouped together and summarised aggregate data and raw data disaggregated data. 15.05 Samples and populations mathspace.co

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b If Source A reports that the average monthly spending is $50 and there are 10 months in the school year, can Maryke determine if there was a month where students spent more than $100, on average? Why or why not?

Create a strategy Decide whether the monthly average can be used to ﬁnd the spending habits from an individual month.

Apply the idea

Reﬂect and check

No, Maryke cannot determine if there was a month where students, on average, spent more than $100 just from the average monthly spending data of Source A. The overall average does not provide details of spending in individual months.

Many times, it is best to ﬁnd secondary sources that report raw data values rather than data that has already been summarised. If you need to use the data for a different reason than the researcher who collected the data, the information in their report may not be useful for your statistical study.

Idea summary • In statistics, data can be gathered from every member of a group, known as a population, or from a subset of the population, known as a sample. • Gathering data from a sample instead of the entire population is often more practical and less expensive, however, the sample needs to be representative of the population for the data to be accurate. • A representative sample has the same demographic diversity as the population, relevant characteristics to the study, and random selection. • The size of the sample is crucial; a sample that’s too large may be too costly or time-consuming, while a sample that’s too small may not represent the population and is not useful to make conclusions.

15.05 Practice questions Understanding 1

814

Explorer 1−5

Adventurer 1−5

Trailblazer 1−4, 6

True or false: a

In a sample survey, information is obtained from the entire population.

A census of the whole population is always possible.

Using a sample can save time and money for the researcher.

A sample should have the same diversity as the population.

A convenience sample is a good sampling method to use.

The population and sample means will always be exactly the same.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

If the population has 42 000 members, would this sample size be large enough? a

4000

40 000

A study is being done on physical activity in a small town. Would this characteristic be important to consider when selecting the sample? a

Favourite colour

Gender

Age

Favourite sport

Preferred method of transportation

Name

Identify the population and sample in each scenario: a

The mayor of Anytown wants to determine the opinion of his residents on the new mall planned to be built. He surveys 500 residents at random.

A bookstore owner wants to know what genres of books are most popular amongst her customers. She decides to pull up the receipts for 20 purchases out of the 650 sales made over the weekend.

Antony wants to know what teams the students at his school support. He ﬁnds that 30 students support the Sydney Swans, and 55 students support the Melbourne Demons.

For a statistical survey, the population is deemed to be all the students that attend the local secondary school. Are these samples of that population? a

All of the seniors

The ﬁrst 30 students to arrive at school that day

All of the students that were in the library at lunch time

All of the teachers

Some of the teachers

For a statistical survey, the population is deemed to be people aged under 18 who live in Sydney, Australia. Are these samples of that population? a

40 professors chosen from University of Sydney

Every person who lives in Bondi

All elementary students who are visiting Taronga Zoo in a single day

Students enrolled in the on-campus preschools or high schools in Parramatta

100 children chosen from a Sydney junior rugby league team

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Fluency 7

7, 8, 9bc, 10, 11ab, 13, 15−17

7, 8, 9cd, 11cd, 13−17

Match each sample histogram to the population it was taken from. i

iii

10 9 8 7 6 5 4 3 2 1 0

9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

10 9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

iv 10

0 10 20 30 40 50 60 70 80

9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

9 8 7 6 5 4 3 2 1 0

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7, 8, 9ab, 10, 11ab, 12, 15−17

0 10 20 30 40 50 60 70 80

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10 9 8 7 6 5 4 3 2 1 0 10 9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

Ex 1

A sample of 50 people is drawn from a population. In this sample, the youngest person is 18 years old and the oldest person is 64 years old. Are these populations that the sample could have been chosen from? a

People in a grocery store

People who got tickets for speeding

People at a drop-in centre for seniors

Employees at a grocery store

For each population, give an example of a biased sample of the population. a

The population is all the students that attend the local high school.

The population is all people aged under 18, who live in Sydney.

The population is all people in a city who play in any organised sporting competition.

The population is all people who own a pet dog.

Sean wants to know what year 10s think of their English class, so he polls 70 random year 10s. Is the sample chosen biased or fair?

A school principal wants to estimate the number of students who ride a bicycle to school. Is each sample biased?

All students who are in the school band.

Eight students in the hallway.

Ten students from each year, chosen at random.

130 students during the lunch periods.

At a certain chocolate factory, 30% of products contain nuts. 400 chocolates are tested to check if they meet the required quality for sale. Of those tested, 61% contained nuts. a

What is the population?

What proportion of the population contains nuts?

What proportion of the sample contains nuts?

Are the 400 chocolates that were tested a representative sample?

At a popular orchard, 65% of apples are Gala apples. 500 apples are picked for a tastetesting event. Of those picked, 310 were Gala apples. a

What is the population?

What proportion of the sample is of the “Gala” variety?

Are the 500 apples that were picked a representative sample?

Lachlan asks 120 year 12 students at his school how much time they spend on homework per night. 78 year 12 students say they do more than 3 hours. At a meeting of the student council, Lachlan reports “65% of students at this school do too much homework.” For each statement, decide if it explains why Lachlan’s report is misleading. a

The survey does not represent the population of the school.

The question should have been multiple choice.

The question was biased.

The sample size was too small.

The conclusion does not align with the survey question.

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15 Ex 2

An ecologist collects 250 of a particular bee species in a meadow, marks them, and then releases them. In a follow-up collection, the ecologist gathers 200 bees, and 82 of them are found to be marked. Estimate the total bee population in the meadow. A television station wants to estimate the number of viewers it had for a new show. They know that the country’s population is 13 620 000. When they randomly selected 5000 people and asked them if they had watched the show, they found that 400 of them had. Estimate the number of people in the entire population who watched the show.

Ex 3

Yakov is studying the performance of students in two different classes over a term. He found two secondary sources of data:

• Source A: This data set provides the individual marks of every student from both classes. • Source B: This data set provides the average final mark of students from both classes. If Yakov wants to know how many students from one class scored above 90%, which source should he use?

Which source should Yakov use if he wants to calculate the median mark for only one class?

Source B indicates that the average mark for one class is 85%. Can Yakov determine how many students scored exactly 85% using Source B?

Reasoning 18

18, 19, 21, 22

19, 21, 22

20, 21, 23

Pablo has concluded that about 36% of people would drive an electric car based on his survey results: Would drive an electric car Would not drive an electric car

58 104

Jenny has concluded that about 44% of people would drive an electric car based on her survey results: Would drive an electric car Would not drive an electric car

224 285

Which claim do you think is more valid? Explain your reasoning. 19

For a statistical survey, the population is deemed to be all people who own a pet dog. Ariel decides to ﬁnd her sample by talking to every person at her local dog park. Describe the advantages and disadvantages of her sampling method.

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Explain why these samples are biased: a

Hannah is surveying customers at a shopping precinct. She wants to know which stores customers shop at the most. She walks around an entertainment store and chooses 30 customers from the store for the survey.

A TV station wants to know what the most popular type of music is, so they ask listeners to contact them and vote for their favourite type of music.

The community health nurse wants to survey the students in a school about their eating habits. At lunchtime, she stands by a vending machine and surveys every student who purchases something from the machine.

A political polling company calls 1000 people at home between 4:00 p.m. and 7:00 p.m. on week nights to ﬁnd out who they are most likely to vote for in an upcoming election. They publish their numbers based on the responses of only the 410 people who answered their call. a

How could they have used random sampling to choose the 1000 people to call?

Explain why the company’s published results will not be accurate.

A researcher captures 300 dingoes, tags them, then releases them. The next day, they capture 750 dingoes, of which 90 have tags. The day after that, they capture 100 dingoes, of which 6 have tags. a

Compare the sample proportions from the second and third day of research.

Explain why the sample proportions are different.

Explain which sample statistic you would use.

An environmentalist is interested in tracking the population of koalas in Australia. She begins by tagging 480 koalas. In a follow-up collection, she catches 820 koalas, of which 62 are tagged. In a final round of collection, she catches 820 koalas, of which 58 have tags. a

Explain why the sample proportions are different.

Explain which sample statistic they would use.

Problem-solving 25, 26 24

24−26

24−27

You are writing an essay on Australia’s role in World War II. While researching, you come across two sources of information: • Source A: A school textbook published in 1990 detailing Australia’s military engagements, key figures, and notable battles during the war. • Source B: A documentary released in 2015 that focuses on the personal stories of Australian soldiers, interviews with war historians, and unseen archival footage. a

If you find conflicting information between the two sources, how might you determine which one is more accurate?

Why is it important to use multiple secondary sources when trying to get a comprehensive understanding of a historical event?

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Marine biologists want to determine if a local species of ﬁsh is growing to a smaller size than it used to. They collect and measure 150 of the 450 ﬁsh known to exist in the area. a

Did they catch a large enough proportion of the species to be able to determine if the ﬁsh are growing to a smaller size?

What additional information would make us more conﬁdent that the sample is representative of the population?

For a statistical survey, the population is considered to be all of the athletes at an Australian Football League (AFL) game. a

Describe two different samples that could be taken at the game to represent this population.

Describe two samples that could be taken at this game that do not represent this population. Explain why these samples aren’t appropriate.

Farma, a newly established pharmaceutical company in Australia, has recently conducted clinical trials for a new medication. The trials involved 1000 participants across Sydney. According to the data they released, it appears that 10% of the trial participants experienced mild side effects such as headaches and fatigue. Farma plans to roll out this medication across the country, making it available to the 5 million residents of Sydney first. Consider the potential bias, accuracy, and statistical validity in your analysis.

820

Based on the trial data, estimate the number of Sydney residents who might experience side effects if the medication were given to the entire target population.

Discuss the ethical considerations of releasing this prediction to the public. How might the way the data is reported influence public opinion about the medication?

Do you think the sample size used in the initial trial is representative of the entire target population of Sydney? Why or why not?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

15 Chapter review 1

For the surveys, should the data be grouped or ungrouped when constructing a frequency table? a

A survey conducted of 3500 people, asking them how tall they are in centimetres.

A survey conducted of 1000 people, asking them how many books they read in the past year.

The scores for a recent mathematics test are displayed in the stem-and-leaf plot. The scores are out of 110. a

How many students took the test?

What test score occurred most frequently?

How many marks separate the highest score and the lowest score?

Stem

Leaf

5 7 7 9 9

0 3 5 7

0 3 3 3 3 5 8

0 2 2 4 5 7 7

0 2 5 7 9

0 Key: 10|0 = 100

The histogram displays amount of time that Aaliyah spent watching television on her streaming platform over different days.

Number of days

Minutes Frequency 0 ≤ Minutes ≤ 10 10 < Minutes ≤ 20 20 < Minutes ≤ 30 30 < Minutes ≤ 40 40 < Minutes ≤ 50 50 < Minutes ≤ 60 60 < Minutes ≤ 70 70 < Minutes ≤ 80

Time spent watching television 15

0 10 20 30 40 50 60 70 80 Time spent (in minutes)

Complete the frequency table based on the histogram.

Aaliyah’s subscription offers 1450 minutes of free streaming time, then charges $0.50 per minute for any additional time. How much was she charged for watching these episodes?

A musical has a large cast of students with various ages. The column graph shows the ages of all the students in the cast. a

How many students are in the cast?

What age is most common?

Find the range of the students’ ages.

Find the interquartile range.

Ages of students in a musical

20 Frequency

15 10 5 0

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These data sets each contain 10 movie ratings out of 10: Set A: 2, 3, 4, 5, 6, 7, 8, 9, 10, 10 Set B: 1, 2, 3, 4, 5, 9, 9, 9, 10, 10 Set C: 1, 1, 3, 4, 5, 7, 8, 8, 9, 10

Set B has the largest interquartile range. What does this tell you about the spread of ratings in this data set?

Based on the ﬁve-number summaries and the shapes of the data sets, identify which set is symmetric, which is skewed left, and which is skewed right.

In a game show, a contestant must complete 8 different mini games, and the time (in seconds) taken to complete each is recorded. One contestant’s times for the ﬁrst 7 games are: 47, 50, 52, 60, 66, 80, 82 Find their time in the 8th segment if the upper quartile of all 8 times is 75 seconds.

For the data set shown in the dot plot: a

Construct the ﬁve-number summary.

Draw a box plot.

Describe the shape of the box plot.

Stem

Leaf

5 8 8

0 2 5 5 6 8

0 1

2 3 3 5

Key: 1|2 = 12 8

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The box plot shows the number of books a group of students read in a year: a

Find and interpret the range.

What percentage of students read from 12 to 20 books?

The middle 50% of students read within how many books of each other?

In which quartile is the number of books read the most spread out?

The bottom 50% of students read within how many books of each other?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Number of Books

The box plots summarise the results from a botanical study. The treatment group received an experimental fertiliser to boost plant growth, and the control group received regular water. The boxplots show the number of weeks each group of plants took to reach maturity: Control group

12 15 18 21 24

Compare the medians of the control group and the treatment group.

Compare the ranges of the control group and the treatment group.

Compare the interquartile ranges of the control group and the treatment group.

Does the fertiliser have a positive effect on plant growth speed? Explain your answer.

For the cumulative frequency histogram:

300

Find the total number of scores recorded.

Find the number of times a score between 150 and 200 occurred.

Cumulative frequency

Treatment group

Find the percentage of scores that were 150 or more. Round your answer to one decimal place.

250 200 150 100 50 0

50 100 150 200 250 Scores

The number of attendees at various workshops during a tech conference was noted. The list represents the number of attendees in each workshop: 24, 23, 23, 22, 20, 21, 21, 25, 25, 25, 25, 25, 25, 22, 22, 25, 22, 22, 21, 21 a

Complete the table.

In how many workshops were there at least 24 attendees?

In how many workshops were there fewer than 23 attendees?

Approximately half of the data values recorded are greater than what number of attendees?

Number of attendees

Number of workshops (f )

Cumulative frequency (cf )

20 21 22 23 24 25

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Use the cumulative frequency histogram to estimate: a

The median

The lower quartile

The upper quartile

100 90 80 70 60 50 40 30 20 10

Score 44

Match the box plot shown to the correct histogram.

Score

7 8 9 10 11 12 13 14 15 16 17 18

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D Frequency

9 8 7 6 5 4 3 2 1 0

10 9 8 7 6 5 4 3 2 1 0

8 9 10 11 12 13 14 15 16 17 18 Score

10 Frequency

9 8 7 6 5 4 3 2 1

Frequency

8 9 10 11 12 13 14 15 16 17 18 Score

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

8 9 10 11 12 13 14 15 16 17 18 Score

10 9 8 7 6 5 4 3 2 1 0

8 9 10 11 12 13 14 15 16 17 18 Score

The residents of two apartment buildings were asked the number of indoor plants they own. The frequency of various responses are presented in the dot plots: a

Which building has more plants?

By how many plants does the mode of each building differ?

Which building has more variability in the number of plants?

Do either sets have an outlier?

2 3 4 5 Building X

2 3 4 5 Building Y

Two athletes compete in long jump events and train together over a season. They compete in 10 competitions and their ﬁnal jump (in metres) for each competition is shown: • Antoine: 6.8, 7.4, 6.9, 7.8, 7.3, 7.5, 7.1, 6.5, 7.2, 7.6 • Benji: 7.0, 7.2, 6.8, 6.7, 7.9, 6.0, 8.1, 7.5, 8.0, 7.7

Complete the table of statistics:

Which athlete jumps more consistently?

When considering Benji’s average jump, is it reasonable to remove the outlier before calculating the mean? Explain your answer.

Would removing the outlier from Benji’s data set change your answer to part (b)? Explain your answer.

Minimum Q1 Median Q3 Maximum Mean (1 d.p.) Range Interquartile range

Antoine 6.5 6.9 7.3 7.5 7.8

Benji

7.4

A fruit farm has 52 000 cherries of different varieties. During a day of harvest, 800 cherries are randomly picked. Of those picked, 468 were Bing cherries. a

What is the population?

What proportion of the sample is of the “Bing” variety?

Do the 800 cherries picked represent a random sample?

Estimate the number of cherries in the entire population that are Bing cherries.

A gym manager wants to estimate the number of members who attend yoga classes. Is each sample biased? a

All members who participate in the morning spin class.

Twelve members near the water reﬁll station.

Every 10th member that enters the gym.

Entomologists are tracking the habits of a local beetle species. At the beginning of their research project, they collect and tag 600 beetles. The next week, they collect 350 beetles, of which 98 had tags. The week after, they collect 200 beetles, of which 42 had tags. a

Why are the sample proportions different?

Explain which sample statistics you would use to estimate the population.

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Big ideas • Different representations of data highlight different characteristics of the data. • Many sets of bivariate data can be modelled using familiar functions. • Analysing data patterns and trends enables making informed estimations and predictions about values within or beyond the known data range. • Effective data representation requires the use of accurate and unbiased techniques to provide an accurate and unbiased understanding of the information being conveyed.

Curriculum Outcomes VC2M10AP02 investigate reports of studies in digital media and elsewhere for information on their planning and implementation VC2M10ST02 construct scatterplots and consider a line of good fit; comment on the association between the 2 numerical variables in terms of strength, direction and linearity VC2M10ST03 construct two-way tables and discuss possible relationship between categorical variables VC2M10ST04 analyse claims, inferences and conclusions of statistical reports in the media and other places, by linking claims to displays, statistics and representative data, including ethical considerations and identification of potential sources of bias VC2M10ST05 plan and conduct statistical investigations of situations that involve bivariate data, including where the independent variable is time; evaluate and report findings with consideration of limitations of any inferences VC2M10AST03 use digital tools to investigate bivariate numerical data sets; where appropriate use a straight line to describe the relationship allowing for variation, make predictions based on this straight line and discuss limitations

16 Bivariate statistics Chapter outline 16.01 Two-way tables 16.02 Associations between categorical variables 16.03 Bivariate numerical data 16.04 Associations between numerical variables 16.05 Line of good fit 16.06 Interpolation and extrapolation 16.07 Misrepresentations of data Investigation: Misrepresentations in the media Investigation: Conduct a statistical investigation 16.08 Optional: Correlation coefficient Chapter 16 review

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There’s a correlation between the number of honeybees and crop production!

16.01 Two-way tables After this lesson, you will be able to… • construct a two-way table using the given data. • identify and explain joint and marginal frequencies in a two-way table. • apply basic calculations to find percentages and totals in a two-way table. • interpret data presented in a two-way table.

Two-way tables Exploration Students were asked whether they prefer high-top sneakers or low-top sneakers, and they were also asked whether they prefer to shop online or shop in a store. Some of the results from the survey are: • The survey included 75 students. • 30 students said they prefer high-top sneakers. • 35 students said they prefer to shop online. • 7 students said they prefer to shop in a store and high-top sneakers. • 33 students said they prefer low-top sneakers and shopping in a store. 1. How many students prefer low-top sneakers? 2. How many students prefer low-top sneakers and shopping online? 3. Is there an easier way to display the data in order to answer the ﬁrst two questions?

A two-way table can be used to summarise two categorical variables or attributes. The cells in the table represent the intersections of the categories, displaying the frequency count for each combination of categories. This table contains information about the preference of art medium, and age. If we read across each row, we can tell how many people surveyed prefer drawing, painting or graphics and how many of those are children or adults. If we read down each column, we can tell how many children or adults surveyed prefer drawing, painting or graphics.

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Child

Adult

Total

Drawing

Painting

Graphics

Total

100

Joint frequencies

Add across

Add down

Marginal frequencies

Each joint frequency gives us speciﬁc combinations of conditions. Meanwhile, each marginal frequency gives us overall totals. Both types of information are important in understanding the data in a two-way table and performing calculations to draw conclusions.

Example 1 Some students in Ms. Rochelle’s class were polled on how many siblings they have. The results are provided in the table: 1st period 2nd period 3rd period

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 7 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 7 Period 1

Period 2

Period 3

Total

Less than two siblings Two or more siblings Total a Fill in the two-way table showing how many students have less than two siblings, and how many have two or more siblings.

Create a strategy Identify the joint frequencies of the number of students from each period who have less than two siblings or two or more siblings.

Apply the idea Here’s the complete two-way table: Less than two siblings Two or more siblings Total

Period 1 10 10 20

Period 2 11 9 20

Period 3 13 7 20

Total 34 26 60

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b How many of the students in Period 1 have two or more siblings?

Create a strategy

Apply the idea

Look at the ‘Period 1’ column and the row for ‘Two or more siblings’ in the table.

Number of students = 10

c What proportion of the students have less than two siblings?

Create a strategy

Apply the idea

Divide the marginal frequency of the number of students with less than two siblings by the total number of students polled.

Example 2 A university conducted a survey asking students to rate the effectiveness of online classes and traditional in-person classes. Students were divided into two categories: undergraduate and postgraduate students. The data from the survey is shown in the two-way tables. Undergraduate students

Online classes In-person classes

Not effective 27 32

Slightly effective 11 12

Somewhat effective 2 5

Effective 13 6

Very effective 31 29

Total 84 84

Postgraduate students

Online classes In-person classes

Not effective 9 3

Slightly effective 14 0

Somewhat effective 4 8

Effective 3 13

Very effective 5 11

Total 35 35

a Calculate the percentage of undergraduate students who found classes to be Effective or Very effective for online classes and for traditional in-person classes. Round your answers to one decimal place.

Create a strategy Add the joint frequencies for undergraduate students who rated classes as Effective or Very effective separately for both online and in-person classes, then divide each total by the respective class’s total number of undergraduate students.

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Apply the idea Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round

For undergraduate students, the percentages for Effective or Very Effective ratings are 52.4% for online classes and 41.7% for traditional in-person classes. b Calculate the percentage of postgraduate students who found classes to be Effective or Very effective for online classes and for traditional in-person classes. Round your answers to one decimal place.

Create a strategy Add the joint frequencies for postgraduate students who rated classes as Effective or Very effective separately for both online and in-person classes, then divide each total by the respective class’s total number of postgraduate students.

Apply the idea Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round

For postgraduate students, the percentages for Effective or Very effective ratings are 22.9% for online classes and 68.6% for traditional in-person classes. c What does this data suggest about the perception of online classes among different student levels? A

ndergraduate students slightly prefer online classes to in-person classes, while U postgraduate students strongly prefer in-person classes.

Both student groups greatly prefer online classes to in-person classes.

ndergraduate students have about equal preferences, while postgraduate students U strongly prefer online classes to in-person classes.

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Undergraduate students strongly prefer online classes to in-person classes, while postgraduate students have about equal preferences.

Create a strategy Look at the percentages that we calculated in parts (a) and (b).

Apply the idea Undergraduate students have a slightly higher percentage favouring online classes (52.4%) over in- person classes (41.7%). On the other hand, postgraduate students have a signiﬁcantly higher percentage preferring in-person classes (68.6%) to online classes (22.9%), so the correct answer is option A.

Idea summary Types of frequencies to consider in a two-way table are: • Joint frequency: How often a combination of two conditions occur • Marginal frequency: How often a particular category of data occurs (or the total row and column data) Both types of information are important in understanding the data in a two-way table and performing calculations to draw conclusions.

16.01 Practice questions Understanding 1

Explorer 1−3

Trailblazer 2−4

This table gives information on the treatment plans prescribed to cats with feline diabetes. Depending on the severity of the case, cats are either prescribed a special diet or insulin injections. Venetian Pet Clinic Mohawk Animal Clinic Outback Animal Care Centre Total

832

Adventurer 2−4

Special diet 14 11 15 40

Insulin injections 14 9 16 39

Total 28 20 31 79

How many cats from Venetian Pet Clinic were prescribed insulin injections?

How many cats from Outback Animal Care Centre were prescribed insulin injections?

Which animal clinic prescribed a special diet less frequently than insulin injections?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

This two-way frequency table displays the number of deaf-blind children in each state and their age group. Victoria New South Wales Queensland Total

0–5 year-olds 63 52 56 171

6–17 year-olds 388 143 193 724

Total 451 195 249 895

For each of the groups:

Classify as either a joint frequency or a marginal frequency.

Find the value.

The number of deaf-blind children in Victoria.

The number of deaf-blind 0–5 year-olds.

The number of deaf-blind 6–17 year-olds in New South Wales.

The number of deaf-blind children in Victoria who are more than 5 years old.

This table shows how much homework students do each night in different year levels. Use the information in the table to determine whether the statements are true or false. Year 8 students 7 18 25

1.5 hours or more Less than 1.5 hours Total

Year 9 students 28 6 34

Total 35 24 59

Most students in general do homework for less than 1.5 hours per night.

Most Year 8 students do homework for 1.5 hours or more per night.

There were more students surveyed who do homework for 1.5 hours or more per night than there were who do less than 1.5 hours per night.

There were more Year 8 students surveyed than Year 9 students.

This table shows the preferred flavour of milkshake for children and adults. Is each statement true or false based on the table? Child Chocolate Vanilla Strawberry Total

16 39 98

Adult 50 71 11

Total 93 50 230

43 children prefer chocolate.

A total of 50 adults were surveyed.

There were more adults surveyed than children.

87 people prefer vanilla.

230 people were surveyed.

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Fluency 5

5−9

5, 7, 9, 10

5, 7, 10, 11

Use the given information to complete each two-way frequency table. a

This table gives information about the number of sales of each type of shoe at Foot Swag, a shoe store in a shopping centre. Running shoes

Basketball slides

Work boots

43 85

13 36

29 48

Teenagers Adults Total b

Seventy-one students were asked if they prefer chocolate (C) or vanilla (V ) ice cream. Each of the students was also asked if they prefer sprinkles (S) or hot fudge (F ) as an ice cream topping. The results are displayed in the tree diagram: 12 C

30 71

41 71 V

Total 84 85 169

30 18 30

29 41

12 41

Sprinkles

Hot fudge

Total

Chocolate Vanilla Total

Laura asked some people in her community what their preference was between beef, chicken, and ﬁsh.

20 respondents said they preferred beef, 8 of those were children. 15 children said they preferred chicken, and 15 adults said they preferred ﬁsh. 35 children were surveryed, and there were 73 respondents overall. Beef

Chicken

Fish

Total

Children Adults Total 6

834

Special diet 14 11 15 40

What does 28 mean in this context?

What does 11 mean in this context?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Insulin injections 14 9 16 39

Total 28 20 31 79

Ex 1

Ex 2

Sophia asked 38 children and 54 adults in her community about their dietary preferences. The options given were: vegetarian, vegan or neither. 40 respondents said they were vegetarian, 15 of those were children. 18 children said they were vegan, and 22 adults said they were neither vegetarian nor vegan. a

Construct a two-way table based on the results of Sophia’s survey.

What is the proportion of respondents that are vegetarian?

What is the proportion of adults that are vegan?

Mr. Tobit asked the students on football teams to pick their favourite subject. He displayed the results in a two-way table. a

How many defence players did not pick Maths as their favourite subject?

How many players picked Music?

How many offence players are on the football teams?

At a local high school, a survey is conducted to gauge student satisfaction with canteen food. The responses are measured on a five-point Likert scale, with 1 being “very unsatisfied” and 5 being “very satisfied”. The survey also takes into account whether the student is a junior student or senior student.

Maths Music Science English

Offence 11 8 13 15

Defence 12 18 11 10

Rating 1 Rating 2 Rating 3 Rating 4 Rating 5

Junior students 23 35 50 70 22

Senior students 30 20 45 40 15

What percentage of juniors were very satisﬁed with the canteen food? What about the seniors?

Determine which rating is most common among junior students. What about senior students?

Comment on any patterns or discrepancies you observe between the two groups of students.

This table shows the number of trains arriving either on time or late at a particular station: a

How many trains were late on Friday?

How many trains passed through the station on Wednesday?

How many trains were on time for the entire week?

What fraction of trains were on time for the whole week?

Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday

Arriving on time 23 12 22 30 11 15 14

Arriving late 38 25 29 27 14 11 10

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At a local university, students were asked for their favourite subject at high school and what they had chosen to study at university. The results are shown in a table. Maths Science Music Art Total

Maths favourite 76 64 64 6 210

Science favourite 20 46 11 19 96

Music favourite 61 53 67 38 219

Art favourite 43 59 59 74 235

Total 200 222 201 137 760

Based on this data: a

What percentage of students surveyed said their favourite subject was Maths at high school? Round your answer to two decimal places.

What percentage of students are studying either Music or Arts at university? Round your answer to two decimal places.

What fraction of students had Music as their favourite subject in high school but decided to study a different ﬁeld at university?

What percentage of students have chosen to study in the same ﬁeld as their favourite subject from high school? Round your answer to two decimal places.

Reasoning 12

13, 14

A scientist recorded some data on the diet of two types of possums observed on Phillip Island: • 310 possums were observed. • 45 brushtail possums were observed to eat a diet of fruit. • 180 possums observed were honey possums. • 120 possums were observed to eat insects. • Of the 110 possums that were observed to eat nuts, 75 were honey possums. a

Complete the table shown based on the given information about possums: Brushtail possum

Honey possum

Total

Fruit Insects Nuts Total

836

Compare the total number of brushtail possums to the number of honey possums.

Compare the number of possums who were observed eating fruit to the total number of possums eating insects and nuts.

Compare the number of brushtail possums observed eating nuts to the number of honey possums observed eating insects.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Determine if each statement is true or false. Justify your response. a

The marginal frequencies for the columns must have the same sum as the marginal frequencies for the rows.

Each of the joint frequencies in a single row can be equal to the marginal frequency of that row.

A poll was given where adults and teenagers were asked if they like pineapple on their pizza. The possible responses were “yes”, “no” and “sometimes”. The results of the poll show that: • 60 people responded to the poll. • 18 adults responded “yes”. • 24 of the people who responded were adults. • The number of people who responded “yes” was equal to the number of people who responded “no” plus the number of people who responded “sometimes”. Find and explain the error made in each statement: a

Since 18 adults responded yes, there were 60 − 18 = 42 adults who responded either no or sometimes.

There were 18 adults who said yes and 24 of the people were adults so the total of adults was 42.

The number of people who responded “yes” was equal to the number of people who responded “no” plus the number of people who responded “sometimes”. So we can take the total number of people, 60, and divide by 3 to ﬁnd the number of people who said each response.

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Problem-solving 15, 16 15

16, 17

16−18

Yuri surveyed a group of people about the type of employment they had. He recorded the data in the graph. 40 35

Number of People

30 25 Men

20 15

Women

10 5 0

None

Part-time Casual Type of Employment

Full-time

Complete the two-way table with the information from the graph. No job

Casual

Part-time

Full-time

Men Women

How many people are unemployed?

What percentage of men are employed full-time? Round your answer to one decimal place.

What fraction of people employed casually are male?

If 111 people from the group are randomly selected, how many would you expect to be employed part-time?

In a biology study, the appearance of an animal carrying a certain gene is recorded. Assume that all animals must carry the MC1R gene or the Agouti gene but not both. a

Light fur Dark fur

If the proportion of all the animals recorded carrying the MC1R gene and having light fur is

838

MC1R gene 36 28

What does the value of x mean in this context?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

, ﬁnd the value of x.

Agouti gene 37 x

Lucinda sets out to see how many of her classmates use the term “soccer” or “football” to describe her favourite sport. Lucinda plans to survey her classmates from her ﬁrst three periods, and creates the following table for organising the data. Football

Soccer

Total

Period 1 Period 2 Period 3 Total Complete the frequency table so that it meets the conditions: • None of her class periods have the same number of students. • Overall, most students use the term “soccer”. • One of Lucinda’s class periods is equally divided between “soccer” and “football”. • The number of students who prefer “football” in one class period was equal to the number of students who prefer “soccer” in a different class period. 18

At a high school with 5000 students, a survey was conducted to investigate the participation of junior and senior students in school sports activities. Out of a sample of 100 students, a two-way table was produced. Always Often Sometimes Rarely Never Total

Junior students 12 9 10 7 11 49

Senior students 8 7 11 12 13 51

Total 20 16 21 19 24 100

Estimate the number of senior students in the entire school population who would say they “Never” participate in school sports.

On a particularly hot week, the school canteen had a sale on ice blocks. The sports coach believes that the sale might influence sports participation due to students preferring to sit out with an ice block instead of playing. How might you design a followup survey or data collection method to investigate this hypothesis using two-way tables?

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16.02 Associations between categorical variables After this lesson, you will be able to… • calculate and interpret row and column percentages from given data in a two-way table. • analyse data using row and column percentage tables to determine associations. • use 100% stacked column graphs to visually represent and analyse data. • describe the association between variables, if any, and provide reasoning based on the data. • differentiate between an association and causation between variables.

Row and column percentages Two-way tables display relationships between two categorical data sets. Categories label the table’s top and left side, while frequencies ﬁll the table. Row and column totals are usually included. This two-way table displays the statistics of the passengers and crew who sailed on the Titanic on its fateful maiden voyage in 1912: Survived Died Total

First class 202 123 325

Second class 118 167 285

Third class 178 528 706

Crew 212 696 908

Total 710 1514 2224

Although it is interesting to note that 202 First Class passengers survived, it is far more useful to know the percentage break up of survivors from each class. Row percentage tables and column percentage tables give us more useful information than the raw data alone. To know the proportion of survivors from each class, we need to calculate row percentages. To ﬁnd the row percentage, divide the value in the table by the row total and multiply by 100%. The row percentages are shown in this table with working shown in some of the cells: First class

Second class

Survived Died

11%

Third class

Crew

Total

25%

30%

100%

35%

46%

100%

This percentage frequency table now gives us more useful information than the raw data. The ﬁrst row gives us the percentage breakdown of survivors by class. We can now easily read that 46% of the people who died were crew members, whereas only 8% of the people who died were in ﬁrst class.

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

We can also calculate the percentage in each class type that survived or died. To do this, we calculate column percentages. To ﬁnd the column percentage, divide the value in the table by the column total and multiply by 100%. The column percentages are shown in this table with working shown in some of the cells: First class

Second class

Third class

Crew

25%

23%

59%

75%

77%

100%

Survived Died Total

100%

The first column gives us the percentage breakdown of survivals and deaths in first class. From the raw data, we can see that a similar number of first class (202) and third class (178) passengers survived, but this can be misleading. The percentage frequency table shows us that 62% of first class passengers survived, whereas only 25% of third class passengers survived.

Example 1 Maria surveyed a group of people according to their age about the type of job they had. She recorded the data in this graph: 40

Number of people

35 30 25 Under 50

Over 50

15 10 5 0

None

Casual Part-time Type of employment

Full-time

a Complete the two-way table displaying the row percentages, rounding your answers to one decimal place. Under 50 Over 50

None 17.6% 22.2%

Casual 27.8%

Part-time 29.4%

Full-time 41.2%

Total 100% 100%

Create a strategy Create a table for the data. Then for each missing value, divide the value in the table by the row total and multiply by 100%.

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Apply the idea From the graph, we can make the following two-way table: Under 50 Over 50 Total

None 15 20 35

Casual 10 25 35

Part-time 25 15 40

Full-time 35 30 65

Total 85 90

From the table, we can see that 10 of the casual employees are under 50, and the row total for under 50 is 85. Find the row percentage

Evaluate and round

From the table, we can see that 15 of the part-time employees are over 50, and the row total for over 50 is 90. Find the row percentage

Evaluate and round

From the table, we can see that there were 30 full-time employees that are over 50. Find the row percentage

Evaluate and round

The row percentage table is: Under 50 Over 50

None 17.6% 22.2%

Casual 11.8% 27.8%

Part-time 29.4% 16.7%

Full-time 41.2% 33.3%

Total 100% 100%

Reflect and check For casual employees that are over 50, we also could have subtracted the other values in the row from 100. Casual over 50 Percent = 100 − 17.6 − 29.4 – 41.2 = 11.8

b Complete the two-way table displaying the column percentages, rounding your answers to one decimal place. None Under 50 Over 50 Total

57.1% 100%

Casual 28.6% 100%

Part-time 62.5% 37.5% 100%

Full-time

100%

Create a strategy Use the table from (a) for each missing value, divide the value in the table by the column total and multiply by 100%.

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Apply the idea Under 50 Over 50 Total

None 15 20 35

Casual 10 25 35

Part-time 25 15 40

Full-time 35 30 65

Total 85 90

From the table, we can see that there were 15 under 50 in the None category, and the None column total is 35. Find the column percentage Evaluate and round From the table, we can see that there were 35 full-time under 50, and the full-time column total is 65. Find the column percentage Evaluate and round From the table, we can see that there were 25 casual employees that are over 50, and the casual column total is 35. Find the column percentage Evaluate and round From the table, we can see that there were 30 full-time employees that are over 50, and the fulltime column total is 65. Find the column percentage Evaluate and round The column percentage table is shown: Under 50 Over 50 Total

None 42.9% 57.1% 100%

Casual 28.6% 71.4% 100%

Part-time 62.5% 37.5% 100%

Full-time 53.8% 46.2% 100%

Reflect and check Notice that, when we have row percentage tables or column percentage tables, the values only sum to 100% in one of the margins. In this case, we would not get 100% when adding across the rows.

Idea summary To find the row percentage for a particular value, divide the value in the table by the row total and multiply by 100%. To find the column percentage for a particular value, divide the value in the table by the column total and multiply by 100%.

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Associations between categorical variables To find if there is an association between the variables in the Titanic table, we can ask the question, “Is the survival rate dependent on the class of the passenger?” In order to find this, we must first identify the independent variable. This is also known as the explanatory variable, which is the factor that is manipulated or controlled in an experiment or study to test its effects on the dependent variable. It explains, influences, or predicts changes in the response variable. In this situation, it is the class of passenger. Which percentage frequency table should we use? When to use a row percentage table Independent variable forms row headings

When to use a column percentage table Independent variable forms column headings

How do we tell if there is an association? For a row percentage table Look at values in the columns

For a column percentage table Look at values across the rows

The independent variable forms the heading of each column, therefore the column percentage frequency table will best indicate any patterns. Survived Died Total

First class 62% 38% 100%

Second class 41% 59% 100%

Third class 25% 75% 100%

Crew 23% 77% 100%

When we read across the first row in the column percentage frequency table and look at the numbers, we can see a clear difference in the percentages of passengers that lived or died in each class. This suggests there is an association between the class of passenger and the rate of survival. If the values are similar (±5%), we say there is no clear association. How do we describe the association? 1. State if there appears to be an apparent association or not. In this example, there appears to be an association between the variables. 2. Describe the association. In this example, we could suggest class of passenger affects the survival rate of the passengers. 3. Provide reasoning. We could say that based on the data, the higher the class of passenger, the more likely a passenger was to survive. Two-way tables show potential links between variables, with percentages making associations clear. Subtler links need deeper statistical study. Note: ‘association’ means one variable’s change seems to impact the other, not necessarily cause the change.

Example 2 Glen surveyed all the students in year 12 at his school and summarised the results in a table: Height ≥ 170 cm Height < 170 cm Total 844

Play netball 33 13 46

Do not play netball 72 30 102

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Total 105 43 148

a Which variable is the independent variable? A

Play netball

Height

Create a strategy

Apply the idea

Choose a variable which, when changed, would affect the other variable.

The decision of a student to play or not may depend on their actual height, whereas their height won’t change depending on their decision to play netball or not. So the height is the independent variable, option B.

b To examine if there is an association between height and playing netball, should Glen use a column or row percentage frequency table?

Create a strategy

Apply the idea

In a percentage frequency table, we want to sum We found in part (a) that the height is the the percentages of the independent variable. independent variable, which are the rows. To sum the percentages of heights, we should use a row frequency table. c Complete the row percentage frequency table for this data. Round your answers to the nearest percentage. Height ≥ 170 cm Height < 170 cm

Create a strategy

Play netball ⬚% ⬚%

Do not play netball ⬚% ⬚%

Total ⬚% ⬚%

Divide each number of students by the total number of students in that row and multiply the answer by 100%.

Apply the idea Here is the original table with the row totals: Height ≥ 170 cm Height < 170 cm

Play netball 33 13

Do not play netball 72 30

Total 105 43

Here is the table with the calculations for the row percentages: Play netball

Do not play netball

Total

Height ≥ 170 cm

100%

Height < 170 cm

100%

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d Looking at the columns of the completed table, does there appear to be an association between height and playing netball? A

No, there does not appear to be any association as the numbers are similar.

Yes, there appears to be an association as the numbers are quite different. It seems that taller people like to play netball.

Create a strategy In a row percentage table, if the values are similar, then we say there is no clear association.

Apply the idea The completed row percentage table is shown: Height ≥ 170 cm Height < 170 cm

Play netball 31% 30%

Do not play netball 69% 70%

Total 100% 100%

The numbers in each column are quite similar with just 1% difference in values. Therefore, we can say that there does not appear to be any association between height and playing netball, option A.

Idea summary When analysing associations between categorical variables, consider the type of percentage frequency table and how to identify and describe associations: • Which percentage frequency table should we use? When to use a row percentage table Independent variable forms row headings

When to use a column percentage table Independent variable forms column headings

• How do we tell if there is an association? For a row percentage table Look at values in the columns

For a column percentage table Look at values across the rows

If the values are similar, then we can say that there is no clear association. • How do we describe the association? 1. State if there is or is not an association apparent. For example: There appears to be an association between the variables. 2. Describe the association. For example: The class of passenger affects the survival rate of the passengers. 3. Provide reasoning. For example: Based on the data, the higher the class of passenger, the more likely the passenger were to survive.

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Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

100% stacked column graphs Association between variables can often be seen more clearly in a stacked column graph. Shown is a stacked column graph for the data from the Titanic table earlier. When we look at each column, we can see that the proportion of death to survival in each column is different. This indicates that there is an association between the variables. Death vs. survival: Titanic 1912 100% 90% 80%

Died, 38 Died, 59

70% 60%

Died, 75

Died, 77

Survived, 25

Survived, 23

Third class

Crew

50% 40% 30%

Survived, 62 Survived, 41

20% 10% 0% First class

Second class

To draw a stacked column graph: • Label the horizontal axis with the independent variables. • Label the vertical axis with percentages from 0% to 100%. • Draw a column for each independent variable that reaches the height of 100% on the vertical axis. • To divide each column into the percentages as shown in the frequency table, start from the bottom of the column, count to the ﬁrst percentage and draw a horizontal line to mark it off, then count up to the second percentage from the horizontal line and then mark off again, until all sections are complete. • Write the label and percentage in each section of the columns which indicates the dependent variables displayed or provide a key.

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Example 3 A group of year 12 students surveyed their class and recorded the hair colour and eye colour for each student. The results are displayed in the 100% stacked column chart shown: Hair colour for various eye colours 100% 80% Blonde

60%

Red Brown

40%

Black

20% 0% Blue

Green

Brown

Hazel

Eye colour

a What is the independent variable for this chart? A

Eye colour

Hair colour

Create a strategy In a stacked column graph, the independent variable is placed along the horizontal axis.

Apply the idea The stacked column graph has the eye colour placed along the horizontal axis. So it is the independent variable, option A.

b Does the chart suggest an association between eye colour and hair colour? A

Yes, as the corresponding segments are similar in size.

Yes, as the corresponding segments are of different sizes.

No, as the corresponding segments are similar in size.

No, as the corresponding segments are of different sizes.

Create a strategy Check whether the proportion of the segments in each column are similar sizes.

Apply the idea The stacked column graph segments differ in size, which suggests an association between eye colour and hair colour, option B.

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c Can we say that having blue eyes causes a high chance of having blonde hair? A

Yes. The data shows that students with blue eyes are more likely to have blonde hair.

No. There appears to be an association, but we cannot say that one causes the other.

Create a strategy Consider the difference between association and causation.

Apply the idea An association does not mean one variable causes the other variable to change, but that a change in one variable appears to affect the other. The chance of having blonde hair can be caused by many factors. Having blue eyes may affect the chance of having blonde hair, but we cannot say it is a cause. The correct answer is option B.

Idea summary When analysing a stacked column graph, we can determine if there is an association between the categorical variables by comparing the proportions of the segments in each column: • If the segments vary in size, we may say there is an association between the variables. • If the segments are similar in size, then there is no clear association between the variables. Association differs from causation. An association does not mean one variable causes the other variable to change, but that a change in one variable appears to affect the other.

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16.02 Practice questions Understanding 1

Explorer

Adventurer

1−4

Trailblazer

1−4

This table gives information on the treatment plans prescribed to dogs with canine diabetes. Depending on the severity of the case, dogs are either prescribed a special diet or insulin injections. Parkside Veterinary Clinic Riverbend Animal Hospital

Special diet 23 16

Insulin injections 15 12

Total 38 28

This table gives the row percentages on the treatment plans prescribed based on the previous table. Parkside Veterinary Clinic Riverbend Animal Hospital

Special diet 61% 57%

Insulin injections 39% 43%

Total 100% 100%

What percentage of dogs from the Riverbend Animal Hospital were prescribed with insulin injections?

Will the percentages remain the same when calculating the column percentages?

Is there an association between the pet clinic and the type of prescriptions for dogs with canine diabetes?

What type of percentage frequency table should be used when the explanatory variable forms the column headings?

When constructing a stacked column graph, is the explanatory variable labelled on the horizontal axis or the vertical axis?

Is there an association between household composition and distribution of money based on the data in the stacked column graph? Household expenditure by domestic situation 100% 90% 80% 70% 60% 50% 40% 30% 20% 10%

Housing 850

Single

Married

Transportation

Food

Married with kids Insurance

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Single with kids

Entertainment

Other

Fluency 5

5−8, 11, 12

Adults 30 45 75

Children 40 55 95

Total 70 100 170

A teacher recorded the hair type and colour of their students. The results are shown in a column graph. 10 9 8 Number of students

Ex 1

6−14

In a survey, people were asked about their favourite subjects. The results are shown in the table. Calculate the row percentages for each subject rounded to two decimal places. Math Science Total

5−9, 11, 12, 14

7 6

Straight

Curly

4 3 2 1 0

Brown

Blonde

Black

Complete the two-way table displaying the row percentages rounded to one decimal place. Straight Curly

Red

Red 5.9% 10%

Brown

Blonde 5.9%

Black 41.2%

40%

Total 100% 100%

Complete the two-way table displaying the column percentages rounded to one decimal place. Red Straight Curly Total

50% 100%

Brown 66.7% 100%

Blonde 25% 75% 100%

Black

100%

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A survey was conducted about people’s favourite movie genre. The results are shown in a column graph. 50 45

Number of people

40 35 30

Teens

Children

20 15 10 5 0

Comedy 30% 30%

Teens Children Total

Ex 2

Action

Drama 45%

Horror 5%

40%

Total 100% 100%

50% 100%

Action 33.3% 100%

Drama 81.8% 18.2% 100%

Horror

100%

Isaac surveyed all of the students in Year 12 at his school and summarised the results in a table. Height ≥ 170 cm Height < 170 cm

852

Horror

Complete the two-way table displaying the column percentages rounded to one decimal place. Comedy

Action Drama Movie genre

Complete the two-way table displaying the row percentages rounded to one decimal place. Teens Children

Comedy

Play tennis 62 60

Do not play tennis 138 140

Which variable is the explanatory variable?

Create a row percentage frequency table for this data. Round the values to the nearest percentage.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Members of a gym who are either currently pregnant or postpartum were asked what kind of training they do. Each of them only did one kind of training. The table shows the results.

Weight

Currently pregnant Currently postpartum

Which variable is the explanatory variable?

Create a row percentage frequency table for this data. Round the values to the nearest percentage.

Does there appear to be an association between the type of training and the pregnancy status of gym members?

Does a person’s pregnancy status cause them to choose a certain type of training?

Some students were asked if they are left-handed or right-handed. The results are provided in the table: Not female Female Total

Cardio

Left-handed 8 7 15

Right-handed 47 65 112

Total 55 72 127

Construct a row percentage two-way table. Round your answers to one decimal place.

Construct a column percentage two-way table. Round your answers to one decimal place.

The results of a survey of 279 people on their employment status and parenting status is shown in a row percentage table. People without kids People with kids

None 8% 10%

Part-time 17% 47%

Full-time 75% 43%

Total 100% 100%

Is employment status or parenting status the explanatory variable?

Does there appear to be an association between parenting status and employment status?

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A stacked column chart shows the ﬁtness levels for various categories of smokers. Fitness of smokers 100% 80% 60% 40% 20% 0%

Very heavy (1-5)

Heavy (16-25)

Below average fitness

Moderate (6-15) Normal fitness

Light (1-5)

Non-smoker (0)

Above average fitness

What is the response variable for this chart?

Does there appear to be an association between ﬁtness level and category of smoker?

Can we say that being a non-smoker causes a high chance of an above average ﬁtness level?

These tables and 100% stacked column graph shows the results of a survey on smoking: Introverts Extroverts

Smokers 37 51

Non-smokers 69 123

Introverts Extroverts

Smokers 35% 29%

Non-smokers 65% 71%

Total 100% 100%

Personality type of smokers 100 80 Percentage

Ex 3

Smokers Non-smokers

40 20 0

854

Extroverts Introverts Personality type

Is smoking status or personality type the explanatory variable?

Does the 100% stacked column graph indicate an association between the variables?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

170 people were surveyed about their music preference. The results have been recorded in a table. Music Preference Rock and Roll Classical Pop Rap Country and Western R and B Punk Total

Vegetarian 24 8 17 6 17 6 4 82

Non-vegetarian 19 15 17 2 24 9 2 88

Total 43 23 34 8 41 15 6 170

What is the explanatory variable in this data set?

Which of the 100% stacked column charts should be used to look for an association between the variables? A

Music preferences by dietary restriction 100% 80% 60%

Non-vegetarian Vegetarian

40% 20% 0%

Rock & Classical roll

Pop

Rap Country & R&B western

Punk

Music preference

Music preferences by dietary restriction 100% 80%

Punk R&B

60%

Country & western Rap

40%

Pop

20%

Rock & roll

Classical

0% Vegetarian

Non-vegetarian

Dietary restriction

Does this stacked column chart suggest that there is an association between music preference and dietary restriction? 16.02 Associations between categorical variables mathspace.co

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Reasoning

15, 16, 18

17, 19

What does it mean if the sections in each column of a stacked column graph are of similar size?

In Australia, 130 random people were surveyed, examining their carbon footprint and the city they lived in. The people were then categorised as living in either an urban or regional location, and whether they caused high carbon emissions or low carbon emissions. The results are displayed in a table. High emissions Low emissions Total

Urban 48 30 78

Is location or emission level the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Construct this table for the data, rounding your answers to the nearest whole number.

Does there appear to be an association between location and emission level? Explain your answer.

Mr. Tranor asked his class to pick their favourite subject. He displayed the results in a twoway table. Maths 13 20

Music 9 12

Science 11 18

English 11 10

How many left-handed children did not pick maths as their favourite subject?

How many students picked music as their favourite subject?

Is favourite subject or dominant hand the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this frequency table, rounding your answers to the nearest percentage.

Does there appear to be an association between a child’s dominant hand and their favourite subject? Explain your answer.

In a study of car accidents, data was found on the number of passengers in the car and whether or not the car rolled over. No. of passengers Roll over No roll over Total

856

Total 68 62 130

Right-handed children Left-handed children

Regional 20 32 52

<5 335 1622 1957

5—9 18 42 60

10—15 17 42 59

> 15 2 5 7

Total 372 1711 2083

Which variable is the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table for the data. Round your answers to one decimal place.

Does there appear to be an association between passengers and rolling over? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

At a local university, students were asked what their favourite subject at high school was and what they have decided to major in at university. The results are shown in the following table. Favourite subject

Maths major Science major Music major Art major Total

Maths 66 51 38 12 167

Music 21 43 68 48 180

Art 42 25 37 76 180

Total 185 159 176 153 673

Which variable is the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table for this data. Round your answers to two decimal places.

Does there appear to be an association between favourite subject at school and university major? Explain your answer.

A lecturer commented that “having mathematics as their favourite subject at high school causes many students to take mathematics at university”. Is he correct? Explain your answer.

Problem-solving 20, 21 20

Science 56 40 33 17 146

20, 21

The local library constructed divided bar graphs showing the last six months of book loans by category. Assume the total number of loans was exactly the same for each month.

20−22

Jun Jul Aug

In which month were business books the most borrowed type of book?

Sep

In which month were children’s books the most popular?

Oct

Which variable is the response variable?

Nov

Do the divided bar graphs suggest that there is an association between the variables?

Fiction Design Environmental

Children’s Business

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A group of people were asked if they are employed and if they have a smartphone. The results are shown in the table: Owns a smartphone Does not own a smartphone Total

Employed 34 71 105

Unemployed 18 87 105

Total 52 158 210

Is employment status or smartphone ownership the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table rounding your answers to the nearest percentage.

Does there appear to be an association between employment and owning a smartphone? Explain your answer.

Does a person’s employment status cause them to own a smartphone? Explain your answer.

Create a 100% stacked column graph for your percentage frequency table.

Does the 100% stacked column graph indicate an association between the variables? Explain your answer.

118 people were surveyed about their age and how many hours they watch television each week. The results are shown in a table.

Less than 3 hours 4 to 8 hours 9 to 15 hours Over 15 hours Total

858

10 years and under 9 2 7 10 28

11 to 20 years 8 10 8 9 35

21 to 30 years 2 2 8 9 21

31 years or older 5 9 10 10 34

Total 24 23 33 38 118

Is age or time spent watching TV the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table rounding your answers to two decimal places.

Create a 100% stacked column graph for this percentage frequency table.

Does the 100% stacked column graph indicate an association between the variables? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

16.03 Bivariate numerical data After this lesson, you will be able to… • identify what bivariate numerical data is. • identify the independent and dependent variables in a pair of bivariate data. • plot and interpret scatterplots to analyse bivariate data. • apply understanding of bivariate numerical data to solve real-world problems.

Bivariate numerical data Bivariate numerical data is the name for numerical data consisting of pairs of values. Each study participant will have two pieces of data collected. We use these pairs to find out whether there is a simple relation between variables.

Person 1 Person 2 Person 3 Person 4 Person 5

Age ( years) 30 40 50 60 70

Bone density (g/cm³) 1.35 1.28 1.22 1.10 0.97

For a study about bone health, a person’s bone density is measured against their age. Their age is the independent variable and can be any value. Their bone density is the dependent variable that is recorded against their age.

Each person’s age and bone density make a pair of values in the bivariate data set. A single data point in a bivariate data set is written in the form (x, y), with x being the independent variable and y being the dependent variable. We display bivariate data graphically by plotting the data points Bone density with the value of the independent variable on the horizontal Bone density ( g/cm3) axis and the value of the dependent variable on the vertical 1.4 axis. This is known as a scatterplot. We may then check whether age is related to bone density by analysing the relationship between the variables graphically. It’s important to note that the existence of a relationship between two variables in a scatterplot does not necessarily imply that one causes the other.

1.2 1

0.8 0.6 0.4 0.2

Age (years)

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Example 1 Consider the following variables: • Number of ice cream cones sold • Temperature (°C) a Which of the following statements makes sense? A

The temperature affects the number of ice cream cones sold.

The number of ice cream cones sold affects the temperature.

Create a strategy

Apply the idea

Determine which variable has an effect on the other.

When hot weather occurs, more ice cream cones will be sold. Less ice cream cones will be sold in cold weather. So, the correct answer is option A.

b Which is the dependent variable and which is the independent variable?

Create a strategy

Apply the idea

Recall that the independent variable is not affected by the other variable. In contrast, the dependent variable is affected or changed by the other variable.

Based on the previous part, we can see that the independent variable is the temperature and the dependent variable is the number of ice cream cones sold, since the temperature affects the number of ice creams sold.

Example 2 The prices of ten houses are graphed against the house’s Price (thousands) land area on the following graph: 1200 1000 800 600 400 200

Describe the relationship between land area and house price in the data.

Area (m2) 200 400 600 800 1000 1200

Create a strategy Describe what happens to the dependent variable ( price) as the independent variable (area) increases.

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Apply the idea The area increases from left to right. We can see from the graph that the price increases (rises) from left to right. So as the land area increases, the house price increases.

Example 3 Scientists were looking for a relationship between the number of hours of sleep we receive and the effect it has on our motor and process skills. Subjects were asked to sleep for different amounts of time, and they were all asked to undergo the same driving challenge in which their reaction time was measured. The table shows the results, which are to be presented as a scatterplot. Amount of sleep (hours) 9 6 4 10 3 7 2 5

Reaction time (seconds) 3 3.3 3.5 3 3.7 3.2 3.85 3.55

a Create a scatterplot for the observations in the table.

Create a strategy Plot the points on a Cartesian plane.

Apply the idea 3.9

Reaction time (seconds)

The scatterplot of the data is shown. Reaction time is dependent on how much sleep the person gets. So the values on the horizontal axis represent the number of hours of sleep and the values on the vertical axis represent the reaction time.

3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1

Sleep (hours) 1 2 3 4 5 6 7 8 9 10

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b According to the results, describe the relationship between amount of sleep and reaction time.

Create a strategy Describe what happens to the dependent variable (reaction time) as the independent variable (sleep) increases.

Apply the idea The sleep hours increase from left to right. We can see from the graph that the reaction time decreases ( falls) from left to right. So as the hours of sleep increase, the reaction time decreases.

Idea summary In statistics, we often study pairs of values to understand potential relationships between them. This is called bivariate numerical data. Each pair of values in bivariate data is known as a data point. These data points consist of an independent variable, which could be any value, and a dependent variable, which is the outcome we measure against the independent variable. To visualise these pairs of values, we plot the data on a scatterplot. By plotting these data points, we can observe any potential relationships between the variables.

16.03 Practice questions Understanding 1

862

Explorer 1, 2ac, 3, 4

Adventurer 1, 2bc, 3, 4

Trailblazer 1, 2cd, 3, 4

Is each an example of univariate data or bivariate data? a

A school collects data on the height of each student in two classes in order to compare the classes.

A scientist collects data on iron levels in soil and growth of a type of weed in order to investigate the relationship between them.

A school collects data on the shoe size of each student in the school.

A hockey club collects data on time travelled to get to training from members of three teams in order to compare them.

A political group collects data on the taxable income and addresses of the local electoral constituents in order to investigate if there is an association between them.

A psychologist collects data on the number of days in daycare and NAPLAN results in grade three in order to investigate the relationship between them.

John wishes to investigate whether there is an association between the amount of natural sunlight in a classroom and student’s exam results.

Emma wishes to investigate whether players in the A hockey team have higher average mathematics test results than players in the B hockey team.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Create a scatterplot for each set of data: a

x 1 3 5 7 y 2 7 11 15

9 17

x y

1 3

3 7

5 11

7 13

9 17

x 1 2 4 7 y 20 18 17 15

8 11

x y

1 7

3 3

5 11

9 15

12 19

Is there likely to be a relationship between each pair of variables? a

Study time and test score

Length of hair and number of pets

Typing speed and dancing ability

Driving time and petrol used

Identify whether the lines are increasing, decreasing, or neither. a

y 4

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

d 4

−4 −3 −2 −1 −1

−3

−4

6, 7, 10, 12, 14, 16

−2

Fluency

x 1

−4 −3 −2 −1 −1

6−8, 10−12, 14, 16

5, 7−11, 13−16

For each of the pairs of variables: i

What is the independent variable?

Number of dogs owned and number of destroyed shoes

Size of a pizza and cost of a pizza

Time spent studying and exam mark

What is the dependent variable?

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Ex 1

Consider the variables: • Ticket sales

• Revenue from a show

Which variable affects the other?

What are the dependent and independent variables?

A kite can rise to a height of 161 m when the wind is blowing at 6 kilometres per hour and to a height of 322 m when the wind is blowing at 12 kilometres per hour. a

What are the two variables?

From the two variables, state the dependent and independent variables.

Does one variable have a direct effect on the other? If yes, which variable affects the other? a

Temperature (°C), Distance from the equator (km)

Weekly income, Number of friends

Time of travel (minutes), Distance covered (m)

Age (years), Time spent watching television in a day

Time spent working, Wages earned

Time spent to ﬁnish a novel, Number of pages

Height (cm), Number of children

Number of pets, Time spent caring for pets

Time spent training, Athletic performance

The graph shows the height of a ball after it is dropped off the side of a building. a

Which variable is the dependent variable?

Which variable is the independent variable?

Height

Time

Ex 2

A cafe records the number of soups sold and the daily maximum temperature on a graph. What is the relationship between temperature and soups sold in the data?

Soups sold 60 48 36 24 12 4

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Temperature 16 20

A company records the height and annual salary ($’000) of ten employees on a graph. What is the relationship between height and salary in the data?

Annual salary 100 80 60 40 20 Height (cm) 40

Ex 3

120

160 200

The table presents data on the number of immigrants and the unemployment rate in Australia over a 10-year period: Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10

Number of immigrants (thousands) 250 265 275 280 290 295 300 305 310 320

Unemployment rate (%) 5.2 5.0 4.8 4.6 4.5 4.3 4.1 4.0 3.9 3.7

What are the independent and dependent variables?

Create a scatterplot for the set of data.

According to the results, what is the relationship between the number of immigrants and the unemployment rate?

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The table presents data on the global annual average carbon dioxide emissions and the average global temperature anomalies over a 10-year period: Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10

CO2 emissions (million metric tonnes) 35 000 35 500 36 000 36 500 37 000 37 500 38 000 38 500 39 000 39 500

Which are the independent and dependent variables?

Create a scatterplot for the set of data.

Based on the scatterplot, how do the CO2 emissions relate to the global temperature anomalies?

This table shows the number of apples eaten per week and the number of visits to the doctor per year by six students: Number of apples eaten per week Number of visits to the doctor per year

3 14

6 11

7 6

9 3

11 0

Create a scatterplot for the set of data.

Vanessa wants to visit the doctor as few times as possible each year. According to the plot, should she eat 0 or 11 apples per week?

The height of the tide at a beach is recorded at nine different hours in one day: 13 −0.21

24 −0.05

7 0.92

2 0.55

4 0.84

What are the dependent and independent variables?

Create a scatterplot for the set of data.

15 −0.76

6 1.01

11 0.31

21 −0.68

This table shows the age and height of six children in a family: Age ( years) Height (cm)

866

0 17

Time (hours) Height (metres)

Global temperature anomaly (°C) 0.6 0.65 0.68 0.7 0.75 0.8 0.83 0.85 0.88 0.9

3 90

4 105

9 130

11 145

13 160

16 165

Create a scatterplot for the set of data.

Bob is the second youngest child in the family. Use the plot from part (a) to ﬁnd out how tall he is.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning 17

17, 18

17−19

Six students were asked how many people there are in their family, and the number of trees planted in their garden. Their responses were plotted on a graph. Each dot represents one student.

One student, Jack, has 7 people in his family. How many trees are planted in his garden?

Do you think there is a connection between the number of people in a student’s family and the number of trees in their garden? Explain your answer.

This table shows the average time spent on a website (in minutes) and the average amount spent (in dollars) for a group of online shoppers in a particular month: Week 1 Week 2 Week 3 Week 4

32 Number of trees in the garden 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 Number of people in the family 3 4 5 6 7 8 9 10 11 12 13 14 15

Average time spent (minutes) 15 30 45 60

Average amount spent (dollars) 75 130 185 250

Create a scatterplot for the set of data.

A statement is made: “The more time a shopper spends on a website, the less they spend.” Critique this claim based on your scatterplot.

The table shows the number of books read per year by six students and the mark they received in their yearly maths exam. Number of books read per year Mark in yearly maths exam

5 50

15 65

20 70

30 80

40 90

50 100

Create a scatterplot for the set of data.

In this group of students, Valentina achieved the highest mark in the yearly maths exam. Use the plot from part (a) to ﬁnd out how many books she read throughout the year.

The teacher told the class: “Students who read more books also tend to score higher on maths exams.” According to the plot in part (a), is the teacher correct? Explain your answer.

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Problem-solving 20, 21

20−22

Georgia has a high income but does not like her job. Which dot would represent her? Explain your answer.

3 2

Job Satisfaction

20−22

4 1

0 0

The scatterplot represents the social media usage (in minutes per day) and the respective grades of ﬁve students: A, B, C, D, and E. a

Which student seems to spend the least amount of time on social media?

Which student has the lowest grades despite not spending the most time on social media?

Which student seems to manage their time best on social media while maintaining good grades?

Income

Student survey Grades

B B A A A E E E C C D D Social media usage (minutes per day)

868

I dentify a pair of variables in everyday life that you believe may have a relationship. Explain what the variables are and why you think there may be a relationship.

Identify which variable is the independent variable and which is the dependent variable in your chosen pair from part (a). Explain your choice.

Identify a pair of variables in everyday life that you believe do not have a relationship. Explain what the variables are and why you think they do not have a relationship.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

16.04 Associations between numerical variables After this lesson, you will be able to… • interpret scatterplots to describe the association between two numerical variables. • identify the form of a data set as linear or non-linear. • examine scatterplots to determine the form, direction and strength of data points. • recognise outliers in a data set. • distinguish between correlation and causation using real-world examples.

Associations between numerical variables We analyse how the independent variable (x) inﬂuences the dependent variable ( y) when determining associations between numerical bivariate variables. If an association is linear, then it can be referred to as a correlation. To identify an association between variables, we examine: • Form: The overall pattern or shape of the data points—linear or non-linear. • Direction (if linear): The trend of the data—positive (increasing y with increasing x, similar to a positive gradient) or negative (decreasing y with increasing x, similar to a negative gradient). • Strength (if linear): The extent to which points follow a straight line—strong (close to a line), moderate, weak, or no correlation. While data points might display non-linear forms like curves, we’ll focus on linear models. A perfect correlation means all points lie exactly on a straight line, and no correlation signifies a random scatter of points. Most plots fall between these extremes, showing weak, moderate, or strong correlation.

Interactive exploration

Discover this concept in action online

Perfect positive correlation

mathspace.co

Perfect negative correlation

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Strong positive correlation

Strong negative correlation

Moderate positive correlation

Moderate negative correlation

Weak positive correlation

Weak negative correlation

No correlation

We may also see outliers, which are data points drastically out of the pattern.

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Correlation means that the variables are related, and often does not mean that the changes in the dependent variable are caused by the independent variable. Causation means there is a cause and effect and must be shown with carefully designed studies. Two variables may be strongly correlated, but not necessarily causal for a few reasons. 1. Third (contributing) variable: A third variable is contributing to both x and y, so the relationship between x and y is not causal. • For example: There is a correlation between number of cats and number of street lights in a city, but cats don’t cause street lights. There is a third variable of population size. 2. Coincidence: It is purely coincidence that the two variables are related. • For example: There is a correlation between butter sales and number of female CEOs. 3. Unproven: In some cases, to be causation, it must be proven using proper and repeatable statistical studies. • For example: Cigarette smoking was not proven to cause cancer until the 1950-60s, though it was suspected and correlated before that.

Example 1 From the scatterplot given, we can conclude that:

Vision

x 130 140 150 160 170 180 Height (cm)

A There is no direct relationship between someone’s vision and their height. B Those who are taller have better vision. C The worse someone’s vision, the taller they are.

Apply the idea Examine the scatterplot and determine whether there is a discernable relationship between height and vision. Based on the given scatterplot, there is no direct relationship between someone’s vision and their height. So, the correct answer is option A.

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Example 2 The table shows the marks of 12 students in English and French. Student 1 2 3 4 5 6 7 8 9 10 11 12

Mark in English 85 71 57 60 79 76 71 91 50 49 66 92

Mark in French 89 71 56 62 86 76 77 86 90 47 67 92

a Which of the following scatterplots correctly represents the data? 100

100

90 80 70 60 50

40 40 50 60 70 80 90 100

Mark in English

100 Mark in French

B Mark in French

Mark in French

90 80 70 60 50 40 40 50 60 70 80 90 100 Mark in English

Apply the idea Draw your own scatterplot and plot each given point to identify the correct graph. The correct answer is A.

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b Is the correlation between students’ English and French marks positive or negative? A

Positive

Negative

Create a strategy

Apply the idea

Think of what happen to the students’ mark in French as their marks in English increase.

As the students’ marks in English increase, so too do their marks in French. So the answer is A: Positive

c Is the correlation between students’ English and French marks strong or weak? A

Strong

Weak

Apply the idea When the points on a scatterplot tend to follow a single line, the correlation is strong. When the points on a scatterplot are scattered greatly around a single line, the correlation is weak. So the answer is A: Strong d Which student’s scores represent an outlier? A

Student 10

Student 9

Student 12

None of them

Apply the idea Which point, if any, on the scatterplot is positioned away from the rest of the data? The answer is B: Student 9

Example 3 This table of values that show four excerpts from a database comparing the income per capita of a country and the child mortality rate of the country. If a scatterplot was created from the entire database, what relationship would you expect it to have? A Strongly positive

No relationship

Income per capita 1465 11 428 2621 32 468 C

Child mortality rate 67 16 35 9

Strongly negative

Create a strategy

Apply the idea

Look at how the mortality rate changes as the income increases.

As the income per capita increases, the child mortality rate decreases, so there is a negative correlation between the two sets of data. The correct answer is C.

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Example 4 A study found a strong positive association between the temperature and the number of beach drownings. a Does this mean that the temperature causes people to drown? A

Yes

Apply the idea The statement that there’s a strong positive association between temperature and beach drownings does not mean temperature causes people to drown. This simply indicates a correlation, not causation. So, the correct answer is B: No. b Which of the following is the most likely explanation for the strong positive association? A

Contributing variables - there are other causal relationships and variables that come into play, and these may lead to an indirect positive association between the temperature and the number of beach drownings.

Coincidence - there are no other contributing factors or reasonable arguments to be made for the strong positive association between the temperature and the number of beach drownings.

Create a strategy Is there a common factor that has a direct relationship with both temperature and the number of beach drownings?

Apply the idea A plausible argument would be that the increase in temperature could lead to more people visiting the beach and therefore more chances of drowning incidents. There might be other variables at play, like stronger currents during warmer periods or increased alcohol consumption. So, the correct answer is A.

Idea summary To identify an association or a correlation between variables, we examine: • Form: The overall pattern or shape of the data points—linear or non-linear. • Direction (if linear): The trend of the data—positive (increasing y with increasing x, similar to a positive gradient) or negative (decreasing y with increasing x, similar to a negative gradient). • Strength (if linear): The extent to which points follow a straight line—strong (close to a line), moderate, weak, or no correlation. • Presence of outliers However, correlation doesn’t always imply causation. Apparent relationships between x and y don’t necessarily mean changes in one cause changes in the other. There could be an unobserved variable indirectly linking x and y.

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16.04 Practice questions Understanding 1

Explorer

Adventurer

1−3

Trailblazer

1−3

Four different classes with four different professors had the same final exam. The exam results and number of classes attended by each student is displayed for each class. Does each graph suggest that there is an association between number of classes attended and final exam mark? a

b 100 Final exam mark

Final exam mark

100

x 0

10 20 Number of class attended

100

100 Final exam mark

Final exam mark

Number of classes attended

x 0

Number of classes attended

Each point on the scatterplot shows the time (in weeks) Sumon spent training for a half marathon and the corresponding kilometres they were able to run: Using the scatterplot, are these statements true or false?

Running distance (km) 20 15

The number of weeks that Sumon trained for the half marathon is the independent variable.

The y-coordinates of the points represent the time spent by Sumon training.

There is evidence to suggest that the longer Sumon trains, the further they can run.

No. of weeks 5

Training for a half marathon causes further running distance over time.

The correlation between the number of weeks training and the number of kilometres Sumon is able to run is positive. 16.04 Associations between numerical variables mathspace.co

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Complete each sentence with one of the words from this list. • association • causation • coincidence a b c d

The strength and direction of a scatterplot describe the ⬚. When two variables are related, we say there is an ⬚. To show ⬚, we must do a carefully designed study.

When correlation is not causation, some possible explanations are ⬚ and ⬚.

Fluency 4

• contributing (third) variables • correlation

4−6, 8, 9, 12−14, 16

4−8, 10, 12, 14, 16

4−8, 10, 11, 15, 16

Does each scatterplot show a linear or non-linear relationship? a

5 x

x 5

5 x

x 5

5 x 5

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Describe the strength and direction for each linear association. a

5 x 5

5 x

x 5

This scatterplot shows the relationship between air and sea temperature: a

Describe the relationship between air and sea temperature.

Describe the correlation between air and sea temperature.

35 Sea temperature

Ex 1

30 25

Air temperature

Scientists conducted a study where each person was asked to read a paragraph then recount as much information as they could remember. They found that the longer the paragraph, the less information each person could retain. If the length of the paragraph were plotted (on the horizontal axis) against the amount of information retained (on the vertical axis), would the correlation be negative or positive?

16.04 Associations between numerical variables mathspace.co

877

Ex 2

This table shows the marks of 12 students in Maths and Sports: a

Construct a scatterplot for the students’ marks in Maths versus their marks in Sports.

Describe the correlation between students’ marks in Maths and marks in Sports.

Which student’s scores appear to represent an outlier?

878

Sports 44 74 52 70 67 60 73 86 84 93 57 92

20 41

25 44

30 39

35 34

40 30

45 25

50 22

55 18

60 19

Construct a scatterplot for this data.

Describe the correlation between a person’s age and the number of accidents.

Does this data set contain any outliers?

65 17

This table shows the number of hours of sleep a person gets per week associated with a sample of persons of different age groups. Age Hours (per week)

Maths 63 92 60 79 88 81 61 91 72 42 66 92

This table shows the number of traﬃc accidents associated with a sample of drivers of different age groups. Age Accidents

Student 1 2 3 4 5 6 7 8 9 10 11 12

20 42

25 39

30 45

35 37

40 38

45 41

50 36

55 34

60 33

Construct a scatterplot for this data.

What is the correlation between a person’s age and the hours of sleep per week?

Does this data set contain any outliers?

The market price of bananas varies throughout the year. Each month, a consumer group compared the average quantity of bananas supplied by each producer to the average market price ( per unit). The results are shown in a table. a

Construct a scatterplot using the data from the table.

Describe the correlation between the supply quantity and the market price of bananas.

According to this data, when will a supplier of bananas receive a higher price per banana?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Supply (kg) 550 600 650 700 750 800 850 900 950 1000

65 30

Price (dollars) 16.25 15.75 15.75 14.75 14.50 14.00 13.75 12.75 12.50 12.00

This table lists the time taken to sprint 400 m by runners who all ran in different temperatures as part of a study. a

Construct a scatterplot to represent the data in the table.

How many runners were tested in the study?

Describe the correlation between temperature and sprint time for the data.

Which data point represents an outlier?

Time (sec) 60 67 48 69 65 49 57 53

A researcher is studying the relationship between the number of passers-by in an emergency, and the time taken (in seconds) before a passer-by helps a stranger during an emergency. The data is recorded in this table. Number of passers-by (n) Time until help is offered (t)

Ex 3

Temperature (°C) 5 2 10 8 1 7 6 4

1 8

2 19

3 26

4 37

5 51

6 65

Construct a scatterplot using the data from the table.

As more passers-by are present what happens to the time taken until help is offered?

Describe the correlation between the number of passers-by and the time until assistance is offered.

Does the scatterplot contain any outliers?

This table of values shows the time taken to finish a lap on a race track for ten average speeds. If a scatterplot was created from the entire database, would you expect the relationship to be strongly positive, strongly negative or that there was no relationship? Speed (x) Time ( y)

20 85

25 87

30 75

35 82

40 69

45 73

50 60

55 57

60 45

65 49

This table of values shows four excerpts from a database comparing the literacy rate of a country and the gross domestic product (GDP) per capita of the country. If a scatterplot was created from the entire database would you expect the relationship to be strongly positive, strongly negative or that there was no relationship? Literacy rate 85% 92% 78% 95%

GDP per capita $5 000 $10 000 $3 000 $15 000

16.04 Associations between numerical variables mathspace.co

879

Ex 4

For each data example, is there likely a causal relationship between the variables? a

Number of times a coin lands on heads and the likelihood that it lands on heads on the next ﬂip

Amount of weight training a person does and their strength

Hair length and gender

Marital status and ﬁlm preference

Height and number of children

Number of pets and time spent caring for pets (hours)

Time spent to ﬁnish novel and number of pages in the novel

Time spent training and athletic performance

Reasoning

17, 18

18, 20

18−20

A student collects some data that shows there is a positive increase between height and the ability to play soccer. Does that mean if you are tall, you are better at soccer? Justify your answer.

As preparation for a Science test a group of 10 students was given a revision sheet containing 60 questions. The table shows the number of questions from the revision sheet successfully completed by each student and the mark, out of 100 of that student on the test.

880

Which variable is independent and which variable is dependent?

Construct a scatterplot of the data.

Describe the correlation between the two variables suggested by the scatterplot. Explain your reasoning.

Explain why the relationship is not perfectly linear.

Number of questions 11 14 36 60 57 42 20 27 50 59

For the scatterplot: a

Explain why the correlation between the variables is weak.

Determine whether the pairs of variables could be represented by the data set. i

arks in an English examination and distance M travelled from home to school

Cost of cars and cost of petrol

iii

istance travelled in a car and the cost of a D driver’s licence

Height and weight of students at school

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Test result 20 23 62 97 100 66 35 52 87 99

A researcher is studying the relationship between the average temperature of a city and the number of people migrating to that city. She collected data from 10 different cities and plotted the data on a scatterplot. Describe a possible pattern that might be observed in the scatterplot if there is a relationship between the average temperature and migration.

Problem-solving 21, 23

21−24

A chemical company is testing the effect of different chemicals on slowing the melting of snow. They are currently testing four different chemicals with the results showing how much ice is remaining after 3 hours at a certain temperature. A

B 100 Snow remaining (%)

Snow remaining (%)

100

x 0

Temperature outside (°C)

100

100 Snow remaining (%)

Snow remaining (%)

21−24

x 0

Temperature outside (°C)

A snow sculpture company wants a chemical that will keep snow from melting. a

Which chemical should they choose to ensure that at least half of the snow is still there after 3 hours?

If the company operates at a temperature of 3°C to 4°C, which chemical should they choose?

16.04 Associations between numerical variables mathspace.co

881

After a maths exam, Uther commented that he left out Question 10 because it was not relevant to the topics that were assessed, but that the other questions were ﬁne. The teacher felt that Question 10 and Question 7 were assessing the same skills, but decided to look at the results of students across all 8 classes. She plotted the percentage of students who had answered Question 10 correctly against the percentage of students who had answered Question 7 correctly. a

% that got Q10 correct 4

% that got Q7 correct

Determine whether the statements describe the data. i

There is no relationship between students’ performance in Q7 and Q10.

Students performed similarly in Q7 and Q10.

iii

Students who performed poorly in Q7 performed well in Q10.

Let’s assume that Uther performed exceptionally well in every other question. Which region should the teacher expect Uther’s data point to be in?

Determine whether each statement is true or false. Provide an example to support your claim. a

If event A has a causal relationship with event B, then there is a correlation between event A and event B.

If there is a correlation between event A and event B, then there is a causal relationship between event A and event B.

If event A is positively associated with event B, then there is a causal relationship between event A and event B.

Construct a scenario with data that suggests a weak positive relationship between two variables. Justify why the relationship may be weak.

882

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

16.05 Line of good fit After this lesson, you will be able to… • draw a line of good fit for a given data set. • interpret the gradient and y-intercept of a line of good fit. • use the equation of the line of good fit to make predictions about one variable based on the other. • evaluate the correlation between two variables using a line of good fit.

Line of good fit Exploration This scatterplot shows the car value in thousands of dollars and the number of years since the car was purchased for 20 different cars. 1. Is there a correlation between the years since purchased and the value in thousands of dollars? 2. Which of the lines on the graph ﬁts the data best - dashed or solid?

Car value Value (in thousands of dollars) y 30 25 20 15 10 5

Time since purchase (years) x 1

A line of good fit is a straight line drawn by eye that best represents the data on a scatterplot. It is sometimes called a trend line. We can use lines of good fit to help us make predictions or conclusions about the data. To draw a line of good fit, we want: • a line that passes through the centre of the points. Unless the correlation is perfect, it cannot pass through all of the points. • approximately the same number of points above and below the line • to minimise the vertical distances from the points to the line

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Given a set of data relating two variables x and y, it may be possible to form a linear model. This model can then be used to understand the relationship between the variables. A line of good ﬁt has an equation of the form y = mx + c. The m value shows the gradient, identifying whether the correlation is positive or negative. • If the gradient is positive, when the independent variable increases by 1 unit, the dependent variable increases by m units. The correlation in this case is positive. • If the gradient is negative, when the independent variable increases by 1 unit, the dependent variable decreases by m units. The correlation in this case is negative.

Height (cm) 14 12 10 8 6 4 2

Weekly growth 1

2 3 4 5 6

8 9

The c value shows the vertical intercept (also known as the y-intercept): • When the independent variable is 0, the value of the dependent variable is c. When we interpret the vertical intercept at (0, c), we need to consider if it makes sense for the independent variable to be zero and the dependent variable to have a value of c.

Example 1 The scatterplot shows the data for two variables, x and y:

10 9 8 7 6 5 4 3 2 1

x 1 2 3 4 5 6 7 8 9 10

Draw a line of good ﬁt for the data.

Create a strategy Draw a line that follows the trend of the points, and has the same number of points above and below the line.

Apply the idea 10 9 8 7 6 5 4 3 2 1

Here is an example of a line of good ﬁt that follows the trend of the data and has the same number of points above and below the line.

x 1 2 3 4 5 6 7 8 9 10

884

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 2 The number of hours spent watching TV each evening, h, is measured against the percentage results, m, achieved in the Economics exam. The following linear regression model is ﬁtted to the data: m = −10h + 97 a The vertical intercept indicates that when a student watched no TV, the mark in the exam, on average, was ⬚%.

Create a strategy

The vertical intercept occurs when x = 0. In this case, the variable h is used instead of x. The value h = 0 represents a student that watched 0 hours of TV.

Apply the idea When a student watched no TV, h = 0. m = −10h + 97

Given equation

m = −10(0) + 97

Set h = 0

m = 97

Evaluate

So, the vertical intercept indicates that when a student watched no TV, the mark in the exam, on average, was 97%.

b Does the interpretation in the previous part make sense in this context? A

es, it is reasonable that a student who spends no time watching TV could get an exam Y mark of 97%.

o, it does not make sense because the exam mark is not within the valid range for the N dependent variable.

Create a strategy

Apply the idea

In this case, the independent variable is “hours watching TV”.

It is possible to watch no TV and earn a mark of 97, so the answer is option A.

Is watching no TV possible in the context? Is earning a mark of 97 possible in this context?

c State the gradient of the line.

Create a strategy

Apply the idea

When an equation is of the form y = mx + c, the gradient of the line is m, the number in front of x. In this case, h is used instead of x, so the gradient will be the number in front of h.

The number in front of the independent variable (h) is −10, so the gradient is −10.

16.05 Line of good fit mathspace.co

885

d If a student increases the amount of TV they watch by 3.5 hours, how much can we expect their Economics exam mark to drop by?

Create a strategy The gradient indicates how much their exam mark will decrease by for each additional hour they spend watching TV. The gradient of −10 indicates that their exam mark will decrease by 10% for each additional hour they spend watching TV.

Apply the idea Exam mark drop = 10% × 3.5 = 35%

Multiply the decrease percentage by hours Evaluate

Idea summary To draw a line of good ﬁt, we want to minimise the vertical distances from the points to the line. This will roughly create a line that passes through the centre of the points. The number of points above and below the line should be equal. A line of good ﬁt has an equation of the form y = mx + c. The m value shows the gradient, and whether the correlation is positive or negative. • If the gradient is positive, when the independent variable increases by 1 unit, the dependent variable increases by m units. The correlation in this case is positive. • If the gradient is negative, when the independent variable increases by 1 unit, the dependent variable decreases by m units. The correlation in this case is negative. The c value shows the vertical intercept (also known as the y-intercept): • When the independent variable is 0, the value of the dependent variable is c.

16.05 Practice questions Understanding 1

Explorer 1−4

Adventurer 1−4

1−4

The equation for the line of good ﬁt is given by P = −4t + 116, where t is time. Would the correlation be increasing, decreasing, or neither?

886

Trailblazer

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Choose the line of good fit for this scatterplot. 9

8 7

Line 2

Line 3

6 5 4

Line 1

3 2 1 1

2 3 4 5 6

8 9

Does there appear to be a linear relationship between the variables? a

5 x 5

16.05 Line of good fit mathspace.co

887

Looking at the scatterplot: Is the relationship between the x and y-variables positive or negative?

Which could be the equation for the line of good ﬁt?

y = −4x − 4

y = 44 + 4x

y = −4x + 44

y = 4x − 4

Fluency Ex 1

35 25 20 15 10 5

5, 6a–c, 7, 9ac, 10−12

x 1

2 3 4 5 6

5, 6c–e, 8, 9bd, 10−12

5, 6d–f, 8−12

This scatterplot shows the data for two variables, x and y.

8 9

19 y 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Draw a line of good ﬁt for the data.

1 2 3 4 5 6 7 8 9 10 11 1213 14 1516 17 1819

For each scatterplot: i

Draw a line of good fit by hand.

Write the equation of the line of good fit.

5 x 5

888

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 5

5 x 5

2 x 12

This scatterplot graphs data for the number of people in a room and the room temperature. Draw a line of good ﬁt for this scatterplot.

38 36 34 32 30 28 26 24 22 20

Number of people

Room temperature 10 20 30 40 50 60 70 80 90

This scatterplot graphs data for the number of copies of a particular book sold at various prices. Draw a line of good ﬁt for this scatterplot.

190 180 170 160 150 140 130 120 110 100 90

Copies sold

Price 18 20 22 24 26 28 30 32 34 36

16.05 Line of good fit mathspace.co

889

Ex 2

For each line of good fit: i

What is the gradient of the line?

Describe the correlation of the data set as positive or negative.

iii

Describe the change in y as x increases by 1 unit.

What is the value of the y-intercept?

y = 4.62x + 7.58

y = −7.68x + 5.49

y = −5.69x + 7.84

y = 3.16x − 2.72

The amount of money households spend on dining out each week, D, is measured against their weekly income, I. This linear model D = 0.3I + 27 is fit to the data.

Explain the meaning of the vertical intercept.

Does the interpretation in the previous part make sense in this context?

State the gradient of the line.

If the weekly income of a family increases by $200, by how much can we expect their spending on dining out to increase?

This table shows the price of various used cars. Age ( years) Price (thousands of dollars)

1 16

2 13

0 21.99

Draw an approximate line of good ﬁt.

Find the y-intercept of the line of good ﬁt.

What does the y-intercept represent?

At what age would a car be worth $0?

5 10

7 8.6

4 12.5

3 11

4 11

8 4.5

2 14.5

Price (thousands of dollars) 24 20 16 12 8 4 Age (years) 2

Scientists record the number of aphids in areas with different numbers of ladybirds and graph the data in the scatterplot shown.

A 3200

A = −3.82L + 3865. 21 represents the line of good fit. a

How much does the average aphid population change by with each extra ladybird? Round your answer to the nearest aphid. Find the average aphid population of a region with no ladybirds. Round your answer to the nearest aphid.

2400 1600 800 L 200 400 600 800 1000

890

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning 13

13, 14

13−15

The heights (in centimetres) and the weights (in kilograms) of 9 primary school children is shown on the scatterplot:

Weight (kg) 60 50

State the y-value of the y-intercept.

Describe the meaning of the y-intercept in context.

Does the interpretation in the previous part make sense in this context? Explain your answer.

(0, 27.7)

20 10 Height (cm) 20 40 60 80 100 120 140

Concern over student use of the social media app LookAtMe leads to a study of student grades in Maths versus minutes spent using the app. The results are shown in the table: Minutes, M Grade, P %

292 26

153 63

354 13

253 37

11 97

42 89

195 51

7 98

162 59

254 36

Predict the grade of a student who spends no time on the LookAtMe app. Use a scatterplot and its line of good ﬁt to justify your response.

Use your model to explain and interpret the relationship between minutes spent on the LookAtMe app and Maths grades.

Lorena and Frasier each draw a line of good fit. a

Frasier said he noticed that the there was a point that was far away, so he moved his line closer to it. Does Frasier’s line meet the requirements for a line of good fit? Explain. Lorena said she noticed that the there was a point that was far away, but decided to ignore it. Does Lorena’s line meet the requirements for a line of good fit? Explain.

y 7

Frasier

6 5 4 3

Lorena

2 1

x 2 4 6 8 10 12 14 16 18

16.05 Line of good fit mathspace.co

891

Problem-solving 16−18 16

16−19

This scatterplot graphs data for the number of ice blocks sold at a shop on days with different temperatures. a

Sketch the line of good ﬁt for this data.

Use your line of good ﬁt to estimate the number of ice blocks that will be sold on a:

31°C day

42°C day

Does the number of ice blocks sold increase or decrease as the temperature increases?

17−19

Ice blocks 140 120 100 80 60 40 20 Temperature (°C) 14 16 18 20 22 24 26 28 30 32 34 36 38 40

The distance of several locations from the equator and their temperature on a particular day is measured and their values are presented on the scatterplot. a

Which equation could represent the line relating distance (x) and temperature ( y)? A

y = −0.005x + 49

y = −0.005x – 49

y = 0.005x + 49

y = 0.005x − 49

Temperature (°C)

20 Distance (1000 km)

Estimate the distance from the equator, x, if the temperature is 30.59°C.

For the scatterplot shown: a

Find the equation of the line of good ﬁt.

Use the line of good ﬁt to approximate the value of y for x = 6.9.

11 10 9 8 7 6 5 4 3 2 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1

892

2 3

4 5 6

8 9

Beekeepers and scientists study a phenomenon known as colony collapse disorder (CCD), where the majority of worker bees in a hive disappear, leaving behind the queen and immature bees. The percentage of beehive losses that can be attributed to CCD each year, since 2005, is shown in a table.

Year 2005 2006 2007 2008 2009 2010

Construct a scatterplot and find a line of good fit for this data.

More data has been collected, and the results are listed:

Y 0 1 2 3 4 5

Hives lost to CCD, (H)% 18 13 23 26 28 30

• In 2011, 30% of hives were lost to CCD. • In 2012, 33% of hives were lost to CCD. • In 2013, 36% of hives were lost to CCD. • In 2014, 40% of hives were lost to CCD. How would this new data affect the line of good fit?

16.06 Interpolation and extrapolation After this lesson, you will be able to… • understand and differentiate between interpolation and extrapolation. • predict values using both the equation of a line and a graph of a line of good fit. • evaluate the reliability of predictions. • apply their understanding of interpolation and extrapolation to solve real-world problems.

Interpolation and extrapolation We can draw lines of good fit by hand, but by using technology, we can find a regression line which is the best line that summarises the data set. The least-squares regression line is a line of fit that is calculated using technology.

Interactive exploration

Discover this concept in action online

mathspace.co

16.06 Interpolation and extrapolation mathspace.co

893

Once we have the equation of a line of fit, we can use it to make predictions. We can: • Substitute the x-value into the equation of the line to calculate the predicted y-value. • Evaluate whether the prediction is reliable. Predictions can also be made using the graph of the line of good fit. We go up from the x-axis to the line of fit and then across to the y-axis to find the predicted value. S

Ice Cream Sales

400

300

200

100

T 10

20 Temperature (°C)

These terms describe the range in which we make predictions: • Interpolation: Prediction within the range of x-values in the data • Extrapolation: Prediction outside the range of x-values in the data y

Interpolations

Extrapolations

The reliability of predictions depends on the correlation strength, whether the data is interpolated or extrapolated, and the number of points in the data set. • Strong correlation + interpolation = reliable prediction • Moderate/weak correlation + interpolation = less reliable prediction • Extrapolation = generally unreliable prediction • More data points increase reliability

894

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 1 The equation y = −8.71x + 6.79 is a line of good fit for a bivariate data set. When x = 3.49 predict the value of y.

Create a strategy Substitute the value of x into the equation for the line of good ﬁt.

Apply the idea y = −8.71x + 6.79

Write the equation

y = −8.71 × 3.49 + 6.79

Substitute x = 3.49

y = −23.6079

Evaluate

Example 2 Chirping crickets can be an excellent indication on how hot or cool it is outside. Different species of crickets have different chirping rates, but for a particular species the following data was recorded. Temperature (°C) Number of chirps per minute

14 77

17 115

21 150

24 176

220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

a According to the graph, what is the temperature when the crickets make 140 chirps each minute?

Create a strategy If we move from the frequency of chirps to the temperature, we find the x-value for a given y-value. Start with y = 140 and find the x-value.

Apply the idea We can draw a line from our y-value, 140, horizontally to the line of the relationship, then another line vertically down to the x-axis.

16.06 Interpolation and extrapolation mathspace.co

895

What is the x-value we end up at? 220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

The answer is 20°C.

b According to the graph, how many chirps per minute will the crickets make if the temperature is 27°C?

Create a strategy When x = 27, find the value of y.

Apply the idea We can draw a line from our x-value, 27, vertically to the line of the relationship, then another line horizontally left to the y-axis. What is the y-value we end up at? 220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

The answer is 210 chirps per minute.

896

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c According to the graph, how many chirps are the crickets making each minute if the temperature is 19°C?

Create a strategy When x = 19, find the value of y.

Apply the idea We can draw a line from our x-value, 19, vertically to the line of the relationship, then another line horizontally left to the y-axis. What is the y-value we end up at? 220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

The answer is 130 chirps per minute.

d Which prediction is more reliable, part (b) or (c)?

Create a strategy Remember that for reliability of predictions: • Strong correlation + interpolation = reliable prediction • Moderate/weak correlation + interpolation = less reliable prediction • Extrapolation = generally unreliable prediction • More data points increase reliability

Apply the idea On the initial scatterplot, the minimum number of chirps per minute is 77 and the maximum number of chirps per minute is 176. This means that part (b) uses extrapolation and part (c) uses interpolation. Part (c) is more reliable than part (b).

Reflect and check There are only four points, so none of the predictions are incredibly reliable. However, the correlation is strong, so that improves the reliability somewhat.

16.06 Interpolation and extrapolation mathspace.co

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Idea summary Predictions can be made either using a graph of the line of good ﬁt or substituting into the equation of the least-squares regression line. The reliability of predictions depends on correlation strength, interpolation or extrapolation, and the number of data points. Interpolation: Prediction within the range of x-values in the data Extrapolation: Prediction outside the range of x-values in the data Reliability of predictions: • • • •

Strong correlation + interpolation = reliable prediction Moderate/weak correlation + interpolation = less reliable prediction Extrapolation = generally unreliable prediction More data points increase reliability

16.06 Practice questions Understanding 1

Explorer 1, 2, 3a

Adventurer 1, 2, 3b

Is each statement true or false? a

A line of good ﬁt can only be used for interpolation, not extrapolation.

Extrapolation using a least-squares regression line is generally more reliable than interpolation.

Interpolation can lead to inaccurate predictions if the relationship changes beyond the range of known data points.

Scatterplots may not accurately represent the relationship between variables if there are outliers, non-linear relationships, or if the data is too sparse.

It is not important to consider data validity and limitations when using interpolation or extrapolation.

Considering data validity and limitations helps ensure that the predictions made are accurate and reliable, and that they do not lead to incorrect conclusions or decisions based on the data.

Does each graph show interpolation or extrapolation? a

90 80 70 60 50 40 30 20 10

80 70 60 50 40 30 20 x 10 20 30 40 50 60 70 80 90

898

Trailblazer 1, 2, 3c

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 10 20 30 40 50 60 70 80 90

10 20 30 40 50 60 70 80 90

Find the y-value for the given least-squares regression line and x-value. a

y = 0.45x + 7.1, at x = 3

y = 1.75x - 3.2, at x = -2.4

Fluency 4

4a, 5, 6, 8, 9, 11, 13

y = -2x - 7, at x = 5

4b, 5, 6, 8, 9, 11, 13

4c, 5−7, 10, 12

A prediction for the y-value when x = 5 is made from the data set: x y

4 0

7 2

8 4

11 7

12 6

13 4

17 8

18 8

19 11

20 8

A prediction for the y-value when x = 33 is made from the data set:

x y

37 72

54 53

58 26

59 21

43 73

55 47

60 12

38 102

64 10

35 112

A prediction for the x-value when y = 95.69 is made from the data set:

Ex 1

Are these predictions an extrapolation or an interpolation? a

10 20 30 40 50 60 70 80 90

94.4

96.4

94.4

97.8

94.9

95.9

94.4

A line of good fit for this data set is y = 1.26x − 57.01: When x = 3.49, predict the value of y. Is this an example of interpolation or extrapolation? Round your answer to two decimal places. x y

93 51.2

57 25.4

86 38.9

97 58.6

78 38.2

96 60.8

68 26.3

69 28.5

54 5.4

92 92

16.06 Interpolation and extrapolation mathspace.co

899

Scientists conducted a study to see people’s reaction times after having different amounts of sleep. The results are recorded in the table: Number of hours of sleep (x)

1.1

1.5

2.1

2.5

3.5

Reaction time in seconds ( y)

4.66

4.1

4.66

3.7

3.6

3.4

The data has been graphed along with a line of good fit found by eye. a b

Reaction time 5

Predict the reaction time for someone who has slept for 5 hours.

Predict the number of hours someone sleeps if they have a reaction time of 4 seconds.

3 2 1 1

Number of hours of sleep 4 5 2 3

The number of ﬁsh in a river is measured over a ﬁve year period: Time in years (t) Number of fish (F )

0 1903

1 1994

2 1995

The data has been graphed along with a line of good fit found by eye. a

Predict the number of years until there are no ﬁsh left in the river.

Predict the number of ﬁsh remaining in the river after 7 years.

According to the line of good ﬁt, how many years are there until there are 900 ﬁsh left in the river?

3 1602 2000 1800 1600 1400 1200 1000 800 600 400 200

4 1695

5 1311

t 2 4 6 8 10 12 14 16 18 20

Ex 2

The number of people at the beach can be affected by the temperature. After observing a particular beach, the data was recorded in a table. Temperature (°C) Number of people at the beach

15 80

18 120

Sketch a line of good fit by eye.

Approximately, how many people are at the beach when the temperature is 20°C?

c d e

22 180

Approximately, what is the temperature when there are 200 people at the beach?

25 230 225

People

200 175 150 125

Estimate how many people are at the beach when the temperature is 19°C?

100

Is the prediction for 30°C or 19°C more reliable?

75 50 25

900

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Temperature °C 12 14 16 18 20 22 24 26 28

One litre of gas is raised to various temperatures and its pressure is measured is shown in a table. Temperature (K) Pressure (Pa)

300 2400

302 2416

304 2434

308 2462

310 2478

312 2496

314 2512

316 2526

318 2546

The data has been graphed along with a line of good ﬁt:

Use the line of good ﬁt to predict the pressure when the temperature is 306 K.

Is the prediction in part (a) an example of interpolation or extrapolation?

Is the prediction in part (a) reliable?

Is it reasonable to use the line of good fit to predict pressure within each of these range of temperatures? i

300 ≤ Temp ≤ 320

300 ≤ Temp ≤ 600

iii

0 ≤ Temp ≤ 320

280 ≤ Temp ≤ 340

2580 2560 2540 2520 2500 2480 2460 2440 2420 2400

Pressure (Pa)

Temperature (K) 296 300 304 308 312 316

An ice cream shop records the number of ice creams sold and the maximum temperature of each day for two weeks: Temp (°C) Sales

35 70

29 45

34 71

33 67

25 30

25 28

34 65

34 75

The data has been graphed along with a line of good ﬁt: a

Predict the number of ice creams sold if the maximum temperature for a day is 38°C.

Predict the maximum temperature if 10 ice creams were sold.

Is it reasonable to use the line of good ﬁt to predict ice cream sales within each of these range of temperatures? i

0 ≤ Temp ≤ 40

20 ≤ Temp ≤ 40

iii

20 ≤ Temp ≤ 40

0 ≤ Temp ≤ 60

32 60

100 90 80 70 60 50 40 30 20 10

35 74

33 63

28 35

33 62

32 65

Sales

Temp (°C) 22 24 26 28 30 32 34 36 38 40

16.06 Interpolation and extrapolation mathspace.co

901

A plane’s altitude (A) is measured at several times (t) during its descent. Time (t seconds) Altitude (A metres)

0 9000

200 7815

400 7092

The data has been graphed along with a line of good fit drawn by eye. a

Predict the altitude of the plane 500 seconds into the descent.

For how many seconds has the plane been descending when it is at an altitude of 7500 metres?

A 9000

Predict the altitude of the plane 100 seconds into the descent.

How many seconds did the plane take to descend to the ground?

1700 593

8000 7000 6000 5000 4000 3000 2000 1000 t 200 400 600 800 1000 1200 1400 1600

A car company looked at the relationship between how much it had spent on advertising (A) and the amount of sales (S) each month over several months. The data has been plotted on a graph and a line of good ﬁt drawn. a

Two points on the line are (3200, 300) and (5600, 450). Find the gradient of the line of good ﬁt. The line of good ﬁt can be written in the form S = mA + c, where m is the gradient, c is the vertical intercept, S is sales in thousands of dollars, and A is the amount spent advertising costs.

S 800 700 600 500 400 300 200 100 A 1000 2000 3000 4000 5000 6000 7000 8000

Find the value of c, the vertical intercept of the line. c

902

Use the line of good ﬁt to estimate the number of sales next month if $4800 is to be spent on advertising.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The depth a diver, x, has descended below the surface of the water is plotted against her lung capacity, y.

y 105

Does the line of good ﬁt have a positive or negative gradient?

100

Find the gradient of the line.

Find the equation of the line of good ﬁt.

Use the line of good ﬁt to estimate the lung capacity, y, at a depth of 4 metres.

85 80 x 1 2 3 4 5 6 7 8 9 10

Reasoning 14

15−17

16, 17

14, 16, 17

Several cars underwent a brake test and their age, x, in years was measured against their stopping distance, y, in metres. The scatterplot shows the results and a line of good ﬁt of y = 3x + 30 that approximates the positive correlation.

y 70 60 50

Use the line of good ﬁt to estimate the stopping distance of a car that is 7.5 years old.

Is the predicted value in part (a) reliable? Explain your answer.

The land size of a sample of properties and their selling price were graphed on the scatterplot: Another property that is not represented in the scatterplot is sold for $670 000 and measured 3400 square metres in size. Is this consistent with the relationship shown in the graph? Explain your answer.

Price (100 000s of dollars)

1 2 3 4 5 6 7 8 9 10 11 12

12 10 8 6 4 2

x 00 00 00 00 00 00 00 10 20 30 40 50 60 70 Size (square metres)

16.06 Interpolation and extrapolation mathspace.co

903

Are these predictions examples of interpolations or extrapolations? Justify your answer. a

Using the provided table that shows a car’s fuel eﬃciency at various speeds, a prediction for the fuel eﬃciency at 70 km/h is made from the data set.

Using the table of a company’s annual revenue over the past 5 years, a prediction for the next year is made from the data set.

Using this scatterplot that shows a student’s test scores at different hours of study, a prediction for the test score at 15 hours of study is made from the data set.

40 50 60 80 90 100

Fuel efficiency (L/100 km) 6.5 5.8 5.2 4.9 5.4 6.1

Year 2015 2016 2017 2018 2019

Revenue ($ million) 10 12 15 18 22

Speed (km/h)

Test score

80 70 60 50 40 30 20 10

Hours of study 2 4 6 8 10 12 14 16 18

Discuss the limitations of using interpolation and extrapolation for making predictions in the context of climate change.

904

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving 19−22 18

x 1000

2000

3000

4000

5000

6000

7000

8000

9000

Find the equation of the line of good ﬁt shown.

If your distance from the equator is 4000 km, estimate the temperature, y, using the equation in the previous part.

Is the predicted value in part (b) reliable? Explain your answer.

A line of good ﬁt has been drawn to approximate the relationship between sea temperature, T in °C, and the area of healthy coral, A in hectares, in a particular location. Two particular points, (2, 700) and (23, 175), lie on the line. a

Find the equation of the line of good ﬁt.

Using the line of good ﬁt, ﬁnd the area A of coral expected when the sea temperature is 26°C.

18, 20, 22−24

The distance in kilometres, x, of several locations from the equator and their temperature, y in °C, is measured and graphed on a scatterplot. 50 45 40 35 30 25 20 15 10 5

19, 21−23

Is the predicted value in part (b) reliable? Explain your answer.

A 800 700 600 500 400 300 200 100

T 5

5 3.1

6 2.9

The table shows data on the number of kilograms of litter collected each week in a national park x weeks after the park managers started an antilittering campaign: Weeks (x) Kilograms of litter collected ( y)

1 3.9

2 3.7

3 3.5

4 3.1

Construct a scatterplot for this data.

Sketch a line of good ﬁt on your scatterplot.

Find the equation for your line of good ﬁt.

Use your equation of the line of good ﬁt to ﬁnd the number of kilograms of litter collected 8 weeks after the start of the anti-littering campaign.

Is the predicted value in part (d) reliable? Explain your answer.

7 2.7

16.06 Interpolation and extrapolation mathspace.co

905

The ﬂeet manager for the Australian Automotive C Association wants to estimate how car 550 maintenance costs, C (in hundreds of dollars), 500 are related to the distance, K (in thousands of kilometres), driven each year. 450 The data collected and the least-squares regression line are shown on a scatterplot.

400

The equation of the least-squares regression line for the graph is C = 9.9K + 146.2

300

Interpret what the coeﬃcient of K represents in this context.

Explain what the constant term in the equation represents in this context.

Is this interpretation of the constant term reasonable given the context? Explain your answer.

350

250 200 150 100 50 K 5

Use the least-squares equation C = 9.9K + 146.2 to predict the annual maintenance cost for a car that is driven 36 000 km per year.

Is this predicted value reliable? Explain your answer.

Is there any reason to believe that there is a causal relationship between distance driven and maintenance costs? Explain your answer.

The height of a particular projectile y in metres was measured at different times t in seconds and the curve of good ﬁt was drawn. a

Using the curve of good ﬁt, what is the predicted height of the projectile after 1 second?

Using the curve of good ﬁt, what is the predicted height of the projectile after 4 seconds?

The height at a given time has the following relationship y = −5t2 + 100. Which of these statements about interpolation and extrapolation are true?

y 100 80 60 40 20 t 1

xtrapolation is less reliable since we should use a E line of good ﬁt instead.

xtrapolation is less reliable since we make assumptions about the relationship E outside of the observed data.

Draw a scatterplot with a line of good fit which would not be suitable for interpolation. Explain.

906

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Based on the given scatterplot and line of good fit, are either interpolation or extrapolation reliable? Explain.

8 7 6 5 4 3 2 1

x 1

2 3 4 5 6

8 9

16.07 Misrepresentations of data After this lesson, you will be able to… • understand what data misrepresentation is and its implications. • explore common strategies of data manipulation and identify them in realworld examples. • develop skills to critically evaluate data displays and graphs for potential misrepresentation.

Data manipulation strategies Statistics are often used in the media to support facts, strengthen arguments or provide insights. However, they can also be manipulated to inﬂuence opinions. It’s vital to interpret statistics with caution as they can be skewed intentionally or unintentionally. Common ways of data manipulation include: • Cherry-picking data: selectively using favourable results • Strategic choice of measures of centre and spread: measures of centre and measures of spread may give different impressions if outliers are present • Biased surveys: question phrasing and interviewer influence can affect responses • Sampling technique: poor methods can introduce bias; small or non-random samples are typical culprits • Over-generalisation: asserting conclusions for a larger population based on a nonrepresentative group • Overstating significance of findings: unclear sample sizes or error margins can mislead • Correlation and causation: mistaking observed relationships for causality • Comparing apples and oranges: meaninglessly comparing dissimilar data sets • Misinterpreting statistics or probability: common in media and legal contexts Misrepresentation may be accidental or deliberate, and critical evaluation of ﬁgures is essential in our information-rich society. 16.07 Misrepresentations of data mathspace.co

907

Example 1 A company that sells a weight-loss supplement and their website says that “this product led to an average weight loss of 5 kilograms over a month”. This is based on the results of two studies of 10 people whose weight loss/gain is shown in this table. The results of the second study were excluded as they said there were too many outliers. Study 1 Study 2

+2 +2

−4 +1

−6 −1

−8 −3

−4 −2

−5 −5

−6 −2

−7 +2

−6 +3

−6 +1

Which option(s) explains why this is misleading? A Cherry-picking data

Comparing apples and oranges

C Biased survey question

Poor sampling technique

Create a strategy Decide which one most accurately describes the scenario: • Cherry-picking data: selectively using favourable results • Comparing apples and oranges: meaninglessly comparing dissimilar data sets • Biased surveys: question phrasing attempts to inﬂuence responses • Poor sampling technique: poor methods can introduce bias; small or non-random samples are typical culprits

Apply the idea

Reﬂect and check

The company only chose to use a study that showed favourable results. They failed to reference other studies that could provide a more complete and accurate picture of the supplement’s effectiveness.

Using the process of elimination: • The answer is not B because the study is relevant to the claim. • The answer is not C because they are not using a survey to collect data.

Both studies only had 10 participants which is too small to draw any conclusions from. The answer is A: Cherry-picking data and D: Poor sampling technique.

Idea summary Common ways that data may be manipulated during reporting: • Cherry-picking data • Strategic choice of measures of centre and spread • Biased surveys • Sampling technique

908

• • • • •

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Over-generalisation Overstating significance of findings Correlation and causation Comparing apples and oranges Misinterpreting statistics or probability

Misleading graphs Charts are often used in the media to show comparisons and describe trends.

Exploration

78 77 76 75 74 73 72 71 70 69 68

In favour of bill X

80 70 Percent in favour

Percent in favour

Graph 1 and Graph 2 show which political party is in favour of a particular bill.

60 50 40 30 20 10

Liberal

Labor

Other

Liberal

Labor

Political party

Graph 1

Graph 2

Other

1. Are they showing the same or different data? Are they conveying the same message? Explain. 2. What do you notice that is different about the graphs? Explain. 3. When would each graph be useful for a media source to show to viewers? Why or why not?

Sometimes, graphs are also used to misrepresent information and promote a biased view. Some of the ways in which a chart can be misleading include: Misleading axes or scale: • Biased labelling • Missing a scale on one of the axes • Using a scale that isn’t uniform • Having an inappropriate scale for the given data (making variations seem less important or more important than they are) • Using inappropriate scaling on chart elements like bars or sectors • Using a logarithmic scale without specifying Incorrect display creation • Using unconventional methods or using a display incorrectly • Only displaying a select portion of the data • Using the wrong type of chart With any set of data, it is important to analyse the data within the correct context.

16.07 Misrepresentations of data mathspace.co

909

Example 2

Women in cabinet around the world

12 10

Tony Abbott’s new cabinet

8 6 4

Cuba

Uganda

Indonesia

New Zealand

Liberia

Zimbabwe

Afghanistan

Egypt

India

United States

China

Malaysia

Australia

Rwanda

2 Canada

Authorised by G. Wright, Australian Labor Party 5/9 Sydney Ave, Barton ACT 2600

The Australian Labour Party released this graph after Tony Abbott was elected as Prime Minister:

Source: https://alp.org.au/our-history/labor-governments-and-women/

Which of the following comments apply? A This graph is misleading because the scale on the vertical axis is not uniform. B This graph is misleading because it assumes that there are always an equal number of male and female cabinet members to choose from. C This graph is misleading because it assumes all cabinet sizes are the same. D The graph is misleading because there are no European countries included.

Create a strategy Check if each statement matches what the graph shows.

Apply the idea The vertical axis has a uniform value of 2, so option A is incorrect. There is no assumption on male cabinet members since their data is not shown, so option B is incorrect. The cabinet sizes of each country are not the same, so the amount of women in Tony Abbott’s cabinet may be proportionally more comparable to the amount of women in other countries’ cabinets. Option C is correct. The title is “Women in cabinet around the world” and there are no mentioned European countries, so option D is correct.

Reflect and check Instead of having the total number on the vertical axis, a proportion would be more appropriate. If a country has 10 members in the parliament and one is female, then that is more representative than a country that has 60 members in the parliament and only three are female.

910

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example 3 Which feature of this graph is misleading? Risk of dying from COVID-19 versus normal annual risk

50% 10% 1% 0.5% 0.1% 0.05% 0.01% 0.005% 0.001%

Normal male death rate Normal female death rate Death rate with COVID-19

10 20 30 40 50 60 70 80 90 100 Age

Create a strategy

Apply the idea

Many of the issues relate to the scale and type The y-axis does not use a linear scale. Instead, of graph. Analyse the axes to see what might be a logarithmic scale was used but it was not misleading. specified.

Reflect and check Even if you specify that a logarithmic scale was used, like the image shown, it can still mislead someone who does not know how to interpret a logarithmic scale. It would be best to provide an additional explanation with the graph that explains how the data changes. Risk of dying from COVID-19 versus normal annual risk 10% 1%

Normal male death rate Normal female death rate Death rate with COVID-19

0.1% 0.01% 0

10 20 30 40 50 60 70 80 90 100 Age

*Log scale used to see differences in rates at younger ages

16.07 Misrepresentations of data mathspace.co

911

Idea summary Some of the ways in which the axes and scale can be misleading include: • • • •

Biased labelling Missing a scale on one of the axes Using a scale that isn’t uniform Having an inappropriate scale for the given data (making variations seem less important or more important than they are) • Using inappropriate scaling on chart elements like bars or sectors • Using a logarithmic scale without specifying Some of the ways in which a display can be misleading include: • Using unconventional methods or using a display incorrectly • Only displaying a select portion of the data • Using the wrong type of chart

16.07 Practice questions Understanding 1

Explorer 1−3

Adventurer

300 people were asked if they would support any of the three candidates for an election. a

What should the percentages in a conventional pie chart add up to?

What do the percentages in this pie chart add up to? Supports Huckabee Is this an appropriate data display?

Trailblazer

1−4

1−4 Presidential run 2012

Supports Palin 70% 63% 60% Supports Romney

912

A survey is being conducted to make a conclusion about mobile phone use in Australia. Which samples would likely be representative of the population? a

All the people who respond to a survey sent out on a popular social media platform.

All researchers at all universities in Australia.

Adding a question to the census.

Every tenth person going into a shopping centre in a randomly selected suburb or area for each town or city.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Determine whether each statement, that describes how this graph might misrepresent the statistics, is true or false. a

The wrong graph type was used.

The graph doesn’t start at zero.

The scale on the vertical axis is not uniform.

Line graphs are often helpful to show trends over time.

Line graph 16 8 4 2 1 0

Fluency 5

The data in the table is based on a base 10 logarithmic scale. It shows the number of influenza cases at the beginning of a flu season. Convert the data to a normal scale. Round each answer to the nearest whole number. Date October 15 October 18 October 21 October 24 October 27 October 30

Ex 1

Cases on log10 scale 1.5 2.3 2.24 2.53 2.4 2.6

5, 6, 9−11, 13, 14, 16

5, 6, 8, 10, 11, 13, 14, 16

5, 7, 8, 11, 12, 14−16

State the type of data manipulation described in each example. a

There is a strong correlation between temperature and ice cream consumption, therefore rising temperatures cause higher ice cream consumption.

75% of a high school class surveyed said they like sport, therefore all high school students like sport.

Asking an air traﬃc controller about his stress levels and comparing that to the response of a four-year-old child.

A politician cites a single year’s crime statistics to argue that crime has dramatically decreased under their 4-year tenure.

A newspaper states: “Oil prices have risen by 33% over the last two years”. The article states that in 2018, the oil prices increased by 10% and in 2019, the oil price increased by 23%. Are these statements true or false? a

The headline is true and accurate.

We need to know the prices to determine whether this statement is accurate.

An increase of 10% followed by an increase of 23% does not give an increase of 33% overall.

A graph is required for more complete analysis.

16.07 Misrepresentations of data mathspace.co

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In NSW, data collected on assaults occurring in hospitals showed that from 2006 to 2011, the number of assaults increased, from 2011 to 2014, the number of assaults remained constant and in 2015, there was a drop in the number of assaults. Based on this, a newspaper reported “Assaults in hospitals have risen dramatically from 2006 to 2015”. Are these statements accurate?

The reporter has not included the data since 2011 accurately in their representation.

The reporter has represented the trend correctly.

The reporter has misinterpreted the data, as the number of assaults has decreased from 2006 to 2015.

The reporter has misrepresented the ﬁndings by not reporting a drop in assaults in 2015.

A new affordable housing initiative requires 10% of units in new buildings to be $250 000 or less, so the units available for purchase are priced at: $250 000, $250 000, $250 000, $600 000, $605 000, $620 000, $630 000, $640 500, $645 000, $650 000, $660 000, $662 500, $668 000, $675 000, $680 000, $695 000, $700 000, $710 000, $715 000, $725 000, $740 000, $744 000, $750 000, $755 000, $759 000, $760 500, $765 000, $770 000, $772 000, $775 000 The marketer for the building says “Most units cost $250 000.”

Do the prices of the units meet the affordable housing initiative requirement?

Is the marketer’s statement accurate?

Which data manipulation strategy is the marketer using?

The marketer decides to change his statement to “Units starting at $250 000.” Is this new statement accurate?

A survey sent to dentists and hygienists asked for their preference on brands of toothpaste. The participants were allowed to choose more than one brand if they wished. The results of the survey are shown in the table. Percentage

Brand A 65%

Brand B 75%

Brand C 72%

Brand D 79%

Brand B advertises the claim “Most dentists recommend Brand B.”

Which data manipulation strategy is the toothpaste company using?

Which statement describes why the claim is misleading? A

The statement is inaccurate.

The survey was conducted by the company that owns toothpaste Brand B.

The statement implies that dentists recommend Brand B over the other brands.

The advertisement does not include the hygienists that were surveyed.

A statement on the box of an anti-wrinkle face cream reads: Noticeably firmer, younger feeling skin in just 4 weeks. Below the statement in small print, it says: *Reported by 25 women. Why is this claim misleading?

914

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

A restaurant is encouraging its staff to promote a new menu item, mushroom risotto. The servers are encouraged to ask each customer this question to improve sales: “Do you want a nutritious risotto for lunch or the standard sandwich?” a

Which manipulation strategy are the servers using?

Suggest a way for the servers to promote the new item menu that is less manipulative.

What feature is misleading about the line graph? 60 000

Turnover million

50 000 40 000 30 000 20 000 10 000 0 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 Year

The ﬁgure shows the number of tourists from each country to the United States:

IE CO IL

DE FR

Brazil: 613 347

France: 930 265

Germany: 1 263 344

Japan: 2 169 716

UK X

Canada: 14 043 658 Mexico: 4 295 124 United Kingdom: 2 905 909

Top 20 tourist-generating countries to United States of America

South Korea: 560 405 Italy: 558 594

India: 447 079

Colombia: 288 439

Australia: 526 441

Netherlands: 416 870

Argentina: 271 737

China: 488 484

Venezuela: 344 802

Switzerland: 262 595

Spain: 451 530

Ireland: 296 771

Israel: 237 818

What is the issue with the length of the line segments representing each country?

16.07 Misrepresentations of data mathspace.co

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Ex 2

Consider the graph:

Average house price ($)

Massive increase in house prices this year 82 000 81 000 80 000 79 000

2008

Year

2009

What two features of this graph might cause it to misrepresent the statistics? 15

When evaluating new drugs, health officials assess the drug’s potential for addiction, among other effects. One aspect of this evaluation involves studying how the drug levels change in the blood over time. Drugs that quickly peak and then decrease rapidly in the blood could be at a higher risk for abuse. Note: while drugs that maintain relatively stable blood levels might have a reduced risk of causing peaks and valleys associated with abuse patterns, this alone does not ensure that the drug won’t be addictive. The line graph shows the amount of a drug that remains in the blood over time. Medication in blood over time Medication concentration in blood (ng/mL)

Ex 3

100 40 mg 80 mg single dose 10 mg q12h steady-state

0 1 2 3 4 5 6 7 8 9 10 11 12 Hours from dosing

The manufacturer claims this drug has a very slow decrease, so it won’t be addictive. This is not an accurate claim. Which aspects of the graph allows for this misleading claim?

916

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The pie chart shows the number of nuclear weapons possessed by each country. Estimated nuclear stockpiles, 2005 Russia U.S. 7,200 10,315 deployed 8,800 inactive Russia has about 8,800 warheads in reserve or awaiting dismantling

In addition to known nuclear powers, North Korea says it has nuclear weapons. Iran could be next. Est. warheads China

410

India

75-110

France

350

Pakistan

150-110

Britain

200

N. Korea

7-15?

Israel

100-170

Source: Time Magazine a

Considering only the size of the slices (i.e. ignoring the numbers), which country has the largest nuclear stockpile?

Looking only at the numbers listed for each country, which country has the largest nuclear stockpile?

Has the pie chart been drawn accurately?

Reasoning 17

17, 18, 20

17, 19, 20

19−21

In the polls for an upcoming election, various polls are analysed to see who is likely to win the election. Most polls put Kodos ahead of Kang by between 10% and 12%. A few polls have Kang ahead of Kodos by a few percentage points. Kang tells his followers that they are sure to win the election based on the latest polls. Explain why this statement misrepresents the statistics.

16.07 Misrepresentations of data mathspace.co

917

Consider the line graphs of the same set of sales ﬁgures: 9000

Sales ($)

8000 7000 6000 5000 4000 3000 2000 1000

Month 1

Describe how the difference in scale between the two Sales ($) 9000 graphs might be misleading. 8000 7000 6000 5000 4000 3000 2000 1000 1

Consider the given column graph: a

Why is this a particularly poor example of a graph with broken axes?

Explain why it is not generally good practice to use a vertical scale with a broken axis for column graphs.

Describe a graphical display that would be better suited to this situation.

Month 5

Quarterly operating deficit 55 50 $ (million)

45 40 35 0

918

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Sept - 94 Dec - 94 Mar - 95

The same information is displayed in two different ways. 15

Average global temperatures Degrees (Celsius)

Average global temperatures Degrees (Celsius) 20 18 16 14 12

10 8 6 4 13 1920

2 Years

2020

1920

Years

2020

Which graph could be used to support the claim that the global temperature is increasing significantly?

Which graph could be used to support the claim that the global temperature is not increasing significantly?

At the United Nation’s climate summit, it was stated that limiting the planet’s temperature to a 1.5 °C overall increase by 2100 could avoid catastrophic events. How does this claim help you understand what is considered a “significant increase” in temperature?

Read the hypothetical article: Your mobile phone is ageing you - Research indicates that smartphones are causing premature skin ageing in over half of all users. ‘Smooth as Silk’, an anti-ageing skin product company, does a survey which reveals that more than 50% of frequent smartphone users display early signs of ageing such as ‘tech-neck’ (wrinkles from device use). The study also identifies ‘screen face’ ( pigmentation from screen’s blue light) among significant users. Dr. Louise Fairweather’s survey involving more than 2000 participants suggests blue light exposure accelerates ageing and recommends reducing screen time or using blue light skin products for skin protection. a

Was the survey’s sample selection well done or not?

Based on the company doing the study, do these results surprise you?

Do you find the article’s headline fitting? Why or why not?

16.07 Misrepresentations of data mathspace.co

919

Problem-solving 22, 23 22

22, 23

23, 24

Compare the two graphs and explain which graph is misleading and why. Graph 1

Graph 2 100 Days of freezing temps

Days of freezing temps

85 84 83 82 81 2017

80 60 40 20

2018

2017

This graph was shown on an Argentinian news service with the comment that ‘Argentina’s testing rate for Covid-19 is almost as high as that of the United States’. COVID-19 testing by country

14 100 330

258

Brazil

Argentina

15 700

22 300

7000

USA

Italy

Germany

Number of COVID-19 tests per million people

Source: Reddit May 2020

920

2018

Explain why the graph and comment are misleading.

Redraw the graph correctly.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Norway

Consider the line graph: The Main NAEP shows dramatic gains in math. From 1990 to 2000, fourth grade math scores increased fifteen points - equivalent to about 1.2 years of learning Scale score

240 238 236 234 232 230 228 226 224 222 220 218 216 214 212 210 1990

1992

2000

1996

Source: Brookings report on American education a

The labels on the horizontal axis are not evenly spaced. Explain why this may be misleading.

Is the scale on vertical axis reasonable or misleading? Justify your reasoning.

Investigation: Misrepresentations in the media

Investigate online

mathspace.co

Investigation: Conduct a statistical investigation

Investigate online

mathspace.co

16.07 Misrepresentations of data mathspace.co

921

16 Chapter review 1

Emrik conducted a survey in his neighbourhood about the type of car people preferred. The options given were: Truck, Sedan, or SUV. 36 respondents said they preferred trucks, of which 20 were under 30 years old. Among 22 respondents who preferred sedans, 14 people were older than 30 years old while among 28 respondents who preferred SUVs, 18 people were under 30 years old.

Construct a two-way table based on the results of Emrik’s survey.

What is the proportion of respondents that prefer trucks?

What is the proportion of people older than 30 that prefer sedans?

At a university with 32 000 students, a survey was conducted, asking students what they were majoring in and what team or club they are currently a member of. The results are shown in the table. Physical therapy Education Software engineering Graphic design Total

Sports team 90 58 60 10 218

Dance club 55 60 47 35 197

Art club 27 45 38 47 157

Total 190 225 158 107 680

How many students are members of either the dance or art club?

What fraction of students studying software engineering are members of a team or club other than the dance club?

What percentage of students surveyed are majoring in physical therapy? Round your answer to two decimal places.

Estimate the number of students in the entire university population who are majoring in physical therapy.

The results of a survey of 379 people on their transportation preference based on their living area is shown in the row percentage table: People living in urban areas People living in suburban areas

922

Book club 18 62 13 15 108

Public transport 25% 5%

Bike 20% 15%

Private car 55% 80%

Total 100% 100%

Which variable is the explanatory variable, living area or transportation preference?

Does there appear to be an association between living area and transportation preference?

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

The 100% stacked column graph show the results of Beverage preferences a survey on beverage preferences: 100% a Which is the explanatory variable, beverage 75% preference or preference on time of day? b

Does the 100% stacked column graph indicate an association between the variables?

25% 0%

Morning Evening person person

210 people were surveyed about their education level and how many hours they read each week. The results are shown in the table:

Less than 2 hours 3 to 6 hours 7 to 12 hours Over 12 hours Total

Tea Coffee

50%

High school or less 25 20 7 0 52

Some university 20 20 12 3 55

Bachelor’s degree 3 10 25 8 46

Graduate degree 2 12 31 12 57

Total 50 62 75 23 210

Which variable is the explanatory variable, education level or time spent reading?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table. Round your answers to two decimal places.

Does there appear to be an association between the variables?

The local music store constructed divided bar graphs showing the last six months of instrument sales by category (there are ﬁve categories: S P W K B Jan String, Percussion, Wind, Keyboard, and Brass). Assume that S P W K B Feb the total number of sales was consistent for each month. a

In which month were bass instruments the most purchased type of instrument?

In which month were string instruments the highest selling type of instrument?

Which variable is the response variable?

Does the divided bar graph suggest that there is an association between the variables?

Mar

Apr

May

Jun

B B

Percussion Wind String Keyboard Brass

Does one variable have a direct effect on the other? If yes, identify the independent and dependent variables. a

Time spent studying a subject, Number of chapters covered

Shoe size, Number of siblings

Duration of sunlight exposure, Plant growth

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923

The table shows the average number of days of exercise and the average number of sick days of students over a six month period: Number of days of exercise 2 7 12 15 20 24

Month 1 Month 2 Month 3 Month 4 Month 5 Month 6

Number of sick days 7 5 4 3 1 0

Create a scatterplot for the set of data.

A statement is made: “As the number of days a student exercises increases, the number of days they are sick decreases.” Is this claim correct? Explain your answer.

Describe the strength and direction for each linear association. a

5 x 5

924

x 5

The table represents the monthly sales ﬁgures of a new software product in its ﬁrst year in the market. a

Construct a scatterplot for this data.

Describe the correlation between the month and the sales ﬁgures.

Does this data set contain any outliers?

Does the data show that the month of the year causes an increase in sales? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Sales (in thousands) 15 18 20 21 22 22 23 22 23 24 24 26

For each scatterplot: i

Draw a line of good ﬁt by hand.

Write the equation of the line of good ﬁt.

iii

Describe the correlation of the data set as positive or negative.

ii b

16 14

10 8

6 4

x 5

2 4 6 8 10 12 14 16 18

The height above the ground (in centimetres) of a radish sprout over time is shown on the scatterplot: Height (cm) 25 20 15 10 5 Age (days) 5

20 25

Describe the meaning of the y-intercept in context.

Does the interpretation in the previous part make sense in this context? Explain your answer.

After how many days will the sprout be 16 cm tall?

A line of good ﬁt for this data set is y = 2.5x + 20.5 x y a

2 25.5

2.7 28.1

3 28.4

3.5 29.2

Predict the value of y when x = 5.3

4 30.0

4.2 30.6 b

5 32.45

6 35.5

6.7 38.0

Is this interpolation or extrapolation?

Chapter 16 review mathspace.co

925

A line of good ﬁt has been drawn to approximate the relationship between the number of days Charlee cooks at home, C and the amount of money she spends at restaurants in a month, M in dollars. A line of good ﬁt, M = −28.6C + 970, approximates the correlation. a

Using the line of good ﬁt, ﬁnd the amount of money Charlee would spend if she only cooks at home 1 day of the month.

Is the predicted value in part (a) reliable? Explain your answer.

Predict the number of days Charlee would cook at home if she spends $250 at restaurants in a month.

Is the predicted value in part (c) reliable? Explain your answer.

900 800 700 600 500 400 300 200 100

M (in dollars)

C (in days) 4

12 16 20 24 28

State the type of data manipulation described in each example: a

Chaysten’s car has broken down exactly 3 times, and all 3 times were during rainy days. He claims that the rain caused his car to break down.

68% of university students surveyed said they would not sign up for a drama class. The university removed the drama class from the next term’s schedule, stating the reason as “No student interest”.

A company received the ratings shown during their opening day of business. 5.0

4.5

4.8

4.0

0.0

4.5

5.0

4.4

4.7

5.0

4.4

The next day, they posted a sign in their window that said “Customer rating: 5 stars”. 16

400 people were surveyed on new models of smartphones, and which ones they preferred. Participants were allowed to choose more than one smartphone. a

The company marketing GammaDevice claims “Most people prefer GammaDevice”. Which data manipulation strategy are they using?

Why are the sizes of the sectors misleading?

Describe a graphical display that would be more appropriate for this data. Prefers BetaPhone 70%

Prefers GammaDevice

926

70%

67%

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Prefers AlphaModel

“If one proves the equality of two numbers, a and b, by showing first that a is less than or equal to b, and then that b is less than or equal to a, one has actually found a new equality—one that is often very useful.” –Emmy Noether

Answers

g 4p9

i 16u10v4

1.01 Index laws

k m11

Understanding

ii Yes

iii Yes

iv No

r 1125y16

b i False

c 2p14

d −5x2y14z11

f 9v5y11

g −6u11

h 4x2

c 6 b 9

e 11

f m

c 2

d 0

Fluency 5 a a8

b a10

c b5

e 6y

f 12y

g −2a

i u

j 2w

k 15z

b 0

8 5

c 56m n 7 a u11

h am+n

7 6

6 a 40g h

d y10

l 72y17

c 3m7

d k13

e a5

f m3

g j7

k 9j2k3

8 a −32y15

b −96y11

5 5 5

d 504y24

c −30s t r

f 3y4

e −10 9 a j24 e 25h14 i

b w22

c y12

d 81a8

f x5y10

k −125x

b k31

c 24r58

d x67y15

j x

10 a w36v24

39 4

45 60

e 160r s f a b

g p

b 1

c 1

d 1

e 11

f 1

h 6

12 a x28y8

Reasoning 14 Yes. The expression in the numerator allows the fraction to be separated into two parts, each with a denominator of a2. The division rule can be applied to both terms. 15 Indira and Ernestino are both correct. Chad is incorrect. Answers may vary. The expression in the numerator is a product, so the division rule does not get applied to both parts of the numerator. We can check by substituting, for example, a = 2. By simplifying each expression, we see that Indira and Ernestino’s first lines of work are equivalent to the original expression, while Chad’s is not. 16 a The coefficients of each term must multiply to 9, and the indices should add together when multiplying the same base. The solution to this problem does not multiply the coefficients and instead multiplies the indices. The solution is 9s8.

h x60

11 a 24

b p2

c 0

d 81a8b20c4

f y7

t c86

b 10x2y11z15

ii True

4 a 10

p 64y12

13 a n20r20

v No

928

q 16x12 s x

3 a i Yes

l 22 n −9

j 81x9

m x60 o p17

1 Subtraction

h −13

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b When raising a product to a power, each part of the product must be raised to the power. Therefore, 2 must also be raised to the power of 5, leading to the solution 32x15. c When dividing terms with the same base, subtract the indices, so not t1. The solution is 6.

d When raising a power to a power, we multiply the indices rather than add them which is done here. Then, we will have a different answer when we divide like bases. The solution is x2y22.

e The entire expression 4a must be raised to the power of zero, leading the expression to simplify to 1. Problem-solving 17 There will be 26 = 64 bacteria after 6 hours. 18 We know that 3x is not the same as x3, in general, but the three values x = 0, x = , and x= will make 3x = x3 be a true statement. 19 a = 3, b = 6 20 a True

b False

c False

21 16p = 24p b 343a−21

22 a 3a+6

1.02 Negative indices

b False

c False

d True

2 Less than 1. can be expressed as as a−1.

Fluency

l − b an

7 a u−4

b 2r−6

c xy−2

d 4p−6q−7

8 a

9 a 20y6

b 28a2

h 25m3

10 a 6

c 23

d 11

g 4

h 4

11 a 32y8

b 60y5

e 6y12

f 4y11

12 a

o 3x8

13 a

6 a

e 8h7

Understanding 1 a True

4 a 2−4

10−3 10−2 10−1 100 101 102

103

104

2−3

2−2

2−1

b 10−4

10 100 1000 10 000

c −4

3−3

3−2

3−1

5 a

Reasoning

14 Both Oreste and Donovan are correct. Using

the index law

we can show that their

answers are equivalent.

Answers mathspace.co

929

15 Using the negative index law, we would get and dividing by zero has no

Problem-solving 19 There are many different representations, such as:

meaning as there is no number that, when multiplied by zero, gives 1.

•

16 The −1 index only applies to the variable directly in front of it, since there are no brackets. The

correct working is

17 The value of x−a would get very small and stay positive (x−a gets closer to 0). We know that

, since x > 1 and a is a

21 a n = −3

Therefore,

• 2

−p

× 4p : −32, 0, 1,

b • For 2p and 4p: Since these expressions are directly proportional to p, as p increases, the values of 2p and 4p will also increase linearly. Thus, they will become very large as p becomes significantly large.

• For 2−p + 4p: As p increases, 2−p = approaches zero because the denominator becomes very large. Therefore, 2−p + 4p will be dominated by the 4p term and will increase linearly with p. • For 2−p × 4p: Although 4p increases linearly, the term 2−p decreases exponentially towards zero. The product of a linearly increasing term and an exponentially decreasing term will approach zero. Therefore, 2−p × 4p will get closer and closer to zero as p becomes very large.

, which is less than the , which we

know is greater than the base . There are many examples of this same pattern, such as 3−1 < 3 and

23 4.1 minutes 24 a 3 × 10−6 Farads c

b 5 × 10−6 Farads

Volts or 0.03 Volts

1.03 Optional: Simplify surds Understanding 1 a No

b Yes

c Yes

d No

2 a 4

b 25

c 9

d 1

e 8

f 64

3 a True

b False

c False

d True

e False Fluency 4 a Rational b Rational c Neither d Irrational e Irrational f Irrational g Rational h Neither i Rational j Rational k Irrational l Neither m Rational n Rational o Rational p Irrational 5 a

6 a

7 a 1, rational

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

, irrational

e 9, rational

, irrational

h 11, rational

930

c n = −3

base 2. However,

18 a • 2p : −4, 0, 8, 20 • 2−p + 4p : −4, 1,

b n = −4

• 4

22 The negative index law requires us to write the reciprocal of the base value and apply the positive index to the expressions.

getting very large, x gets very large. When we take the reciprocal, we are dividing 1 into a very large number of pieces, so each of the pieces would be very small and get closer and closer to 0 in size. For example, 10−10 = 0.0000000001. • 4p : −8, 0, 16, 40

• 22

•

, irrational , irrational

, irrational

Problem-solving

n 2, rational

, irrational

8 a

g 6

b 4

c 1

d 6

18 a 8x

e 60 9 a

e 6h

Reasoning

b 9z5

c 4v6

f 2x

g 3a

d 12b3

h 5a7b3

1.04 Optional: Add and subtract surds

10 a Yes b No, it is an approximate value and not exact because is an irrational number that cannot be expressed as a terminating decimal. 11 Jordan’s error was that he raised 27 to the power of 3 using the yx button instead of taking its cube root using the button; the correct calculation should result in an answer of 3. 12 a Yes, because 17 is not a perfect square and there are no further simplifications or common factors in the square root expression. b No, because 50 can be simplified further by factoring the perfect square 25. c No, because 125 can be simplified by factoring out the perfect square 25. d Yes, because 21 is not a perfect square and there are no further simplifications or common factors in the square root expression. 13 a No.

15 a

m 3, rational

can be simpliﬁed further.

Using 20 = 4 × 5, we can simplify , then = becomes

b We could use 16 × 5 and that would mean that we can fully simplify in one step. 14 a False, the square root of a negative number is not a real number.

Understanding 1 a like surds b a power that matches the degree of the radical. e.g. no square powers if a square root, no cube powers if it is a cube root and so on 2 a Not like surds

b Like surds

c Like surds

d Not like surds

e Not like surds

f Like surds

g Like surds

h Not like surds

3 a False

b True

c False

d False

e True

f True

g False

h True

4 a

Fluency

5 a

6 a

b True, the square root of a square is the absolute value of the original number, so this would be ∣p∣, which would also be rational.

c True, the square root of a square is the absolute value of the original number, so this would be ∣p∣, which would be positive.

7 a

d False, the square root of a negative number is not a real number, so we can’t square an undeﬁned number. This is also not a real number.

Answers mathspace.co

931

Reasoning

2 a b

8 a 5x + 3y b We can simplify both by collecting like terms. They have the same coeﬃcients, but one has variables and the other has surds. c 9 a For example, k = 16, m = 9, gives on the left hand side, and on the right hand side. b Yes, k = m = 0 c The equation is not generally true, but is true if k = m.

c d Fluency 3 a 75

b 288

c 108

d 80

e 50

f 35

g 6

h 5

4 a 5 a

6 a

b These are not like surds, so no simplification is possible. We cannot add the radicands; they must be like surds in order to add their coefficients.

11 We simplify surds to make them easier to understand, compare, and work with. Moreover, to add or subtract surds, they must have the same radicand and the same index. Working with surds in their simplest form makes it easier to see how different terms combine.

7 a

8 a

10 a and are not like surds, so they cannot be added or subtracted.

Problem-solving 12

units

13 a

b 218 m

9 a 2

b 4

i 2

14 a 34 cm b It would be an overestimate. is a bit less than , and is a bit less than . c

10 a 6

d 2.51% 15 a

b 10

g 16

h 10

11 a 5

f 2

c 1830

1.05 Optional: Multiply and divide surds

12 a Sometimes a surd. For example,

Understanding 1 a True

b True

c False

d False

e True

f True

g True

h True

i True

j False

932

Reasoning

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

a surd, but

is not

is a surd.

b Always a surd. For example, is a surd.

, which

c Never a surd. This is the deﬁnition of a rational number, so it can’t be a surd. For example, , which is not a surd. are and 13 She is incorrect. For example, is not a surd. surds, but 14 Yes. For example,

and are surds, but is not a surd.

15 a For example, a = 2, b = 3, gives b For example,

, gives c = 5

c For example,

, gives

16 a Prime numbers cannot be square, so is a surd as it is the square root of a number that is not square.

, which is an integer.

b Since a and b are different prime numbers, ab cannot be square. So would be a surd. 17 All papers in this series will have an area of the form , where n is an integer. If n is even, then it will be (an integer multiple of three), which would be a surd. If n is odd, then it will be of the form (an integer multiple of three), which would also be a surd. Problem-solving

units

c Answers will vary. Answers should consider what makes a reasonable height and weight for an adult in conjunction with the requirements to meet the 2m2 surface area. For example, a height of 165 cm and weight of 90 kg supports the formula and is around 2m2 surface area, however a height of 1 cm and weight of 14 400 kg does solve the question mathematically but is quite unreasonable contextually. 21 22 a

c 40 m

b 0.04

c 0.63

d 1.44

25 21 m, rounded to the nearest metre.

1.06 Optional: Binomial expansions with surds Understanding 1 a

2 a ac + ad + bc + bd

b x2 + 3x − 2x − 6 = x2 + x − 6

c x2 − 5x + 5x − 25 = x2 − 25

d x2 + 2x + 2x + 4 = x2 + 4x + 4 3 a

4 a A mix of rational and surd terms b All rational terms c All surd terms d All rational terms

5 a b

b 1.69 m2

20 a

24 a 0.14

Fluency b 100 cm2

18 a

b 46 m2

d The difference is 6 m2, approximately 13% of the area which is a signiﬁcant underestimate.

c d e f g h 6 a 12 e 532

b −110

c 284

d 2

f 30

g 7

h 212

7 a

f 2

Answers mathspace.co

933

14 a

8 a

b m = −4

1.07 Optional: Rationalise the denominator

c d

Understanding 1 a 5

Reasoning

b 1

The middle terms were removed, leaving us with a2 − b, so two terms.

d Because we have multiplied by 1.

The other expansion gives . If c ≠ a and d ≠ b, there are no like terms. There are no like terms, so no simpliﬁcation is possible. 10 a For example

2 a True

b False

c True

d True

3 a

b 6

c 10

d 45

4 a

5 a 7

. The terms are all unlike, so no simpliﬁcation is possible and so we have four terms.

b 1

6 a Conjugate pair

b Conjugate pair

c Conjugate pair

d Conjugate pair

e Not a conjugate pair f Conjugate pair

b For example

g Not a conjugate pair, but simpliﬁes to an integer

This simpliﬁes to

c This is not possible. The expansion either has unlike terms, resulting in four terms, or there are like terms which reduce to two terms. It is not possible to just have two of the terms being like, which is what would be necessary to get three terms.

h Not a conjugate pair Fluency 7 a

Problem-solving

11 a

units units

b c

units2

12 a

square units

units

13 a b c

934

units

8 a

9 a

i −

units units2

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

10 a

11 a

12 a

Reasoning 13 a Both terms of the expression are squares and we are subtracting, so ﬁnding the difference between them. b (x − y) (x + y) c By multiplying by , we are multiplying by 1, so we do not change the value of the expression, but the denominator becomes x2 − y2, which is rational if x and y are surds.

14 The rationalisation process involves multiplying the numerator and the denominator by the conjugate of the denominator. In this case, the conjugate is , so when we multiply we get a2 − b. The new numerator will be contains the term

, which .

15 Xander is correct. , and we can . Multiplying see that this is conjugate to them will produce 9 − 2 = 7, so the denominator will be rational.

is equal to 1, so the value of

is not

changed, only the form in which it written has changed. Problem-solving 16 Area of the rectangle 17 18 a The eighth term is rational. b Yes. It is the only term where the and the simplify to a rational value. 19 20 21 22

1.08 Optional: Fractional indices Understanding 1 2 a

Answers mathspace.co

935

3 a Yes

b Yes

c No

d Yes

e No

f Yes

g No

h No

14 a

d 3

4 a 2.15

b 1.82

c 2.83

d 2

15 a 2

b 4

c 81

d 2

h 243

16 a

17 a

g x2a

Fluency 5 a i

ii 4

b i

ii 8

c i

ii 9

d i

ii 5

e i

ii 7

f i

ii 10

g i

ii 11

h i

ii 16

i i

ii 10

Reasoning

6 a

18 This simpliﬁcation used the product rule of indices which states that when multiplying two expressions with the same base, you add the indices. This rule applies to all indices, not just integers.

7 a

8 a

9 a

10 a 4

b 32

c 81

d 25

e 16

f 128

g 256

h 64

11 a

12 a

instead of raising it to the power of , thus

13 a

c 128

d 9

incorrectly writing it as . This misapplication of the fractional index law made all the subsequent steps incorrect.

g 9

936

j i

ii 4

k i

ii 4

l i

ii 2

is greater because it represents the square root of 64, while represents the cube root of 64. The square root of a number greater than 1 is always greater than the cube root of the same number.

20 By writing as the numbers remain smaller so they are easier to work with. For example

is much easier

to do than ﬁnding

21 a Blair made an error in the first step when attempting to rewrite the cube root in index form. The correct method to rewrite

is:

However, Blair multiplied the exponent by 3

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b 1

Rewrite using the fractional index law

Rewrite the indices with a common denominator

Multiplication law

Power of a power law

Simplify

22 a The working up to and including Step 2 is correct. In going from Step 2 to Step 3, the numbers have been multiplied correctly, but the multiplication index law states that we should add indices. For the algebraic parts, she has multiplied the indices when she should have added them. b Applying the index law correctly, Step 3 , leading to

should be

Chapter 1 review 1 a 6x5

c 8y15

2 a u8

b 4v13 y

5 a

6 a 20y6

7 a

d 25m3

c 20736x4

8 a

b , irrational

, irrational

d -5, rational

b 4

d 6

25 Many solutions are possible, for example or

12 a 13 a

26 3

c x

d m

e b4

g x30y18

i 2a4

j 25u8v6 k n jx

10 a 54

24 25 cm3

c 16x12

4 a 53a−2

9 a

28 a 54y

b 25y8

c 9, rational

23 a

d 54m9 n8

3 a = 5, b = 4

Problem-solving

27 a m

b −3a9

6 4

o 9x y b 36y

units

15 a

16 a

17 a b

l 2x + 9

d 18 a Perimeter b Area

Answers mathspace.co

937

19 a

7 a

8 a

20 21 a The ﬁfth term is rational. b Yes. It is the only term where the and the

simplify to a rational value.

22 a

23 a i

ii 10

b i

ii 4

c i

9 Since the denominators are the same, we just need to add the numerators. The numerators are like terms, so we can add their coeﬃcients. We get: 10 Example answers:

ii 25

d i

Reasoning

24 a 4

b 2

d 81

25 a

b 10

c 2

2.01 Add and subtract fractions with numeric denominators Understanding 1 a

c 1

2 a 6

b 6

c 6

d 12

3 a False

b True

c True

d False

Fluency

11 Simplifying fractions before adding or subtracting them in algebraic equations can make the process easier and the calculations less complex. It also reduces the chance of errors. Additionally, it can help with understanding the problem and the relationship between the numbers. 12 No, we cannot directly add or subtract fractions with different denominators. The denominators represent the total number of equal parts the whole is divided into. Different denominators mean the fractions are divided into different sized parts. Therefore, to add or subtract fractions, we ﬁrst have to ﬁnd a common denominator, which changes the fractions so that they are divided into the same sized parts.

4 a 20

5 a

13 a

d 2k

e 4p

14 a

k 0

6 a

j 0

938

Problem-solving

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

16 a b

7 y 6 5 4 3 2 1 −15 −10 −5 −1 −2 −3 −4 −5 −6 −7

10 a

x 5

Reasoning 11 Yes. The denominator of the fraction (x) is unknown, but each fraction has the same denominator. So when the fractions are added

together, we add the numerators to get .

2.02 Add and subtract fractions with algebraic denominators Understanding 1 a True

b False

c True

2 a

d 0

3 a 120a

b 60x

c 25c

d 42yx

4 a False

b True

c True

d False

Fluency 5 a

6 a

7 a

8 a e

11w 18 y

9 a

12 The denominators represent the total number of equal parts the whole is divided into. Different denominators mean the fractions are divided into different sized parts. Therefore, to add or subtract fractions, we first have to find a common denominator, which changes the fractions so that they are divided into the same sized parts. Therefore, without a common denominator, the fractions cannot be added or subtracted correctly. 13 a Rather than finding a common denominator, both the numerators and denominators have been combined (incorrectly) by adding coefficients and simply writing the pronumerals together. The correct common denominator should be 8xy. b

6 x2 + 7 y2 8 xy

Problem-solving 14 15 a

17 a i

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939

2.03 Optional: Add and subtract further algebraic fractions

13 a

14 a

Understanding 1 a True

b False

c True

d False

e True 2 B 3 C 4 D 5 a ii

b i

c iii

Reasoning

Fluency 6 a

15 First, ﬁnd a common multiple of 4 and 5, which is 20 in this case. Then, convert each fraction to have this denominator and add the numerators.

The result is

Here is the detailed worked solution:

, which simpliﬁes to

. Start with the equation

7 a Yes

b Yes

c No

d No

8 a

Distribute 5 in (x + 1) and 4 in (2x − 3) in the numerator

9 a

Simplify the numerator

10 a

16 a Yes, the equation is correct. When the denominators of the fractions are the same, we can simply add the numerators. So,

11 a

12 a

940

Adjust the fractions to have the same denominator

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

simpliﬁes to

b No, the equation is not correct. When the denominators of the fractions are the same, we can simply subtract the numerators. So, simpliﬁes to

, which is , not

. c Yes, the equation is correct. When the denominators of the fractions are the same, we can simply subtract the numerators. So, simpliﬁes to

d No, the equation is not correct. When the denominators of the fractions are not the

same, we have to ﬁnd a common denominator, add the numerators and then

21 a

23 a

simplify the fraction. So, simpliﬁes to

17 a Joanne got the correct answer. She correctly found the common multiple of the denominators and adjusted the numerators accordingly before subtracting. b Edmon made a mistake on the second step where he forgot to distribute the negative sign to −8, thus obtaining the wrong sign for his terms in the numerator. 18 Both Jonathan and Jefferson are correct. Jonathan’s working: Start with Jonathan’s equation Adjust the fractions to have the same denominator Distribute 2 in (2x + 3)

c 24 0

2.04 Multiply and divide algebraic fractions with numeric denominators Understanding 1 a

2 a True

b False

c False

d True

4 a

2 5 a y 3

d ab

3 a

Simplify the numerator Recognise that the result is equal to d

Jefferson’s working: Start with Jefferson’s equation

Adjust the fractions to have the same denominator Distribute −2 to (2x + 3)

Simplify the numerator Recognise that the result is equal to b

We can conclude that the algebraic fractions b, c and d are related in such a way that the sum of b and c is equal to d, and the difference between d and c is equal to b. Problem-solving 19 a 2 20 a

b 2 b

Fluency xy 18

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941

6 a

l −3c2

Problem-solving 10 a

11 a

Reasoning 7 a

was not multiplied by the reciprocal of

12 a 14k

Start with the equation

Multiply by the reciprocal of

Simplify

2.05 Optional: Multiply and divide algebraic fractions with algebraic denominators

8 Both Amarion and Dean could go on to get the correct answer. Amarion needs to simplify the two fractions and then multiply them. Dean needs to simplify the fraction by removing common factors from the numerator and denominator to get the ﬁnal answer.

Understanding 1 B 2 a True

b False

3 a

b 7

4 a

b 7

5 a a5

b x14

c b2

d c3

6 a

7 a 4

b ab

c cd

Amarion: 1 2 3

Fluency

4 Dean:

3 4 9 a Sometimes true. For example, fraction, while

a numeric ,

an algebraic fraction. b Never true. An algebraic fraction has at least one variable in the numerator or denominator. Multiplying two fractions with no variables, cannot result in any variables.

8 a 10y3

b 48y5

c −8x3

9 a

c Never true. We cannot divide by 0, so this expression is undeﬁned, not 1.

942

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

10 a

11 a

5 a

6 a

Reasoning

7 a

8 a

12 13 a

Problem-solving 15 a

9 a

10 a

11 a 12 a

b 3

32 x y

32 jk3

18 19

Chapter 2 review

1 2 a

3 4 a Waheed got the correct answer. She correctly found the common multiple of the denominators and adjusted the numerators accordingly before subtracting.

15 a

b Azize made a mistake on the ﬁnal step. He didn’t simplify the fraction properly, thus obtaining the wrong result for the whole expression.

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943

3.01 Factorise algebraic expressions Understanding 1 a nm

b 5x

c 4ab

d 2bn

e 4bx

f 3ab

g (x + 1)

h (4 − x)

2 a 2x (x2 + 2x + 1) 2

c a (x − 3x + 6)

b 5y ( y2 + 3y + 2) d 4k (x2 + 3x + 4)

3 3pr Fluency 4 a 3a (x2 + 2x − 2)

b 2x (x2 − 8x + 1)

11 When factorising an algebraic expression, you can check your answer by expanding the factors and verifying that they multiply back to the original expression. Consider the algebraic expression 16x3 + 12x2 + 8x. Its factorised form is 16x3 + 12x2 + 8x = 4x(4x2 + 3x + 2). We can check if our factorisation is correct by multiplying the factors back together 4x(4x2 + 3x + 2) = 4x(4x2) + 4x(3x) + 4x(2) = 16x3 + 12x2 + 8x. The result matches the original expression, confirming that our factorisation is correct.

c 2ax (x + x + 2)

d 5x ( y2 + 2y + 3)

Problem-solving

e 4b (b − 4b + 5)

f 6q (r − 6r − 2)

g 3xz ( yz + 7y + 3)

h 4ac (−bc + 2b + 1)

12 a 3 (x2 + 2x + 1)

i −6ehg ( fh + 5f + 2)

j 10rtu (su + 7s − 4)

5 a (2y − 3) (x + 2)

c 2 (n − 3) (2m2 + 3) 3

b The entire expression 3(x2 + 2x + 1), is the length of the rectangle. It is 3 times the width. This means that the term x2 + 2x + 1 represents the width of the rectangle.

b (3m2 + 5) (n + 2)

d (5y3 + 7) (z + 5)

e q (r + 4) (2p + 5)

f 2b (c − 2) (3a2 − 2)

g ( y + 4 + x) ( y + 7)

h (g + h) (2f + g + h)

i −(7 − x )(2y + 1)

j (8 − c2)(7 − 5d)

k (4m + 7)(m − 5)

l 2(z + 2)(10z − 3)

13 a Let the width of the smallest field be x so that its area is x2. The largest field has an area of 9x2 and the middle field has an area of x(x + 5). b The area of the entire field is the sum of each smaller field: x2 + 9x2 + x (x + 5). The factored form of this expression is x (11x + 5) so the width of the field is x and the length is 11x + 5.

6 2x − 1 7 a 5j (4 − 3k − 11jk)

b p2 q2 (3 + 4p2 q2 + 5p4 q4) c 2y (z − 5x + 6xyz)

d pqr (1 + pq + p2 q2) e h (−f − j + g) f 5m (5 − 3n − 2mn) Reasoning 8 The expression has not been fully factorised, as the terms of 2x3 + 3x2 − 4x contain a common factor of x. 9 Xander’s error lies in factoring the highest common factor. He hasn’t considered the last term −5xy in the original expression. Consequently, the correct factored form of the expression should be 5xy(7x + 2y − 1). 10 The highest common factor will have the lowest possible index of the terms because if it were any higher, not all the terms could be divided by that factor.

14 It is a square when x = . For all other values of x, the rectangle is just a rectangle. 15 Example answer: 8x + 30 = 2(4x + 15)

3.02 Expand binomial products Understanding 1 a 6x2 2

c −21z

d −10p2

2 a (x + 2) (x + 1) = x2 + 3x + 2 b (x − 3) (x + 2) = x2 − x – 6

c (x + 2) (x − 5) = x2 − 3x – 10 d (x − 3) (x − 4) = x2 − 7x + 12 3 a (x + 3)2 = x2 + 6x + 9 b (x − 3)2 = x2 − 6x + 9

c (x + 6)2 = x2 + 12x + 36

d (3x − 8)2 = 9x2 − 48x + 64 4 a (x + 5) (x − 5) = x2 – 25 b (x + 2) (x − 2) = x2 − 4

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b 20y2

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c (x − 8) (x + 8) = x2 − 64

Reasoning

d (x + 14) (x − 14) = x2 − 196 Fluency 5 a x2 − 9x + 8

b x2 + 5x + 4

c x2 − 9x + 18

d x2 − x − 12

e x − 12x + 35 g x2 − x − 30

6 a 6x2 + x − 12

f x2 − 6x + 5

h x2 − 7x − 18

b 12x2 + 7x − 10

d 12x2 − x − 20

c 4x − 7x − 2 2

e 2x + 2x − 12

g 10x2 − 15x − 10

f 6x2 + 9x − 60

h −18x2 − 3x + 6

7 a

8 a x2 + 14x + 49 c e2 + 10e + 25 2

e 49 − 14z + z 2

g a + 2ab + b

9 a 9x2 + 42x + 49 2

b c2 + 12c + 36

d y2 − 10y + 25

f 36 + 12a + a2

h a2 − 2ab + b2 b 4y2 − 20y + 25

c 4y − 12y + 9

d x4 + 2x2 + 1

10 a 5m2 − 50m + 125 b 6n2 + 72n + 216 c 8q2 + 48q + 72

d 147x3 − 210x2 y + 75xy2 2

11 a x − 49

c 49 − 36p2

e 81x2 − 4y2

b 64 − m

d 81y2 − 1

f 0.04x2 − 9

12 a

13 a 75x3 − 108xy2

b 243x2 − 192y2 c 71

d 25t2 + 90t − 36r2 + 81 e 384x2 − 486y2

f 20x3 − 405xy2

14 No. Explanations may vary. One possible answer: To expand perfect squares, we square the first term, square the last term and double the product of the terms. 15 36 × 24 can be written as (30 + 6)(30 − 6). We can apply the formula of difference of two squares which states that for any two numbers a and b, (a + b)(a − b) = a2 − b2. In our case, a = 30 and b = 6. So, (30 + 6)(30 − 6) simplifies to 302 − 62. This gives us 900 − 36 = 864. 16 (x + 3)(x + 3) = x2 + 6x + 9 and 13 × 13 = 169. As we can see, the structure of the results in both cases follows a similar pattern, reflecting the general formula for squaring a binomial. The pattern emerges when a binomial expression (in the first case, a mathematical expression, and in the second case, a number) is squared. The first term in the result is the square of the first term of the binomial, the last term is the square of the second term of the binomial, and the middle term is twice the product of the terms of the binomial. 17 The student’s expression is incorrect because they didn’t apply the correct formula for squaring a binomial. The correct formula for the square of a binomial is (a + b)2 = a2 + 2ab + b2, not a2 + b2. Therefore, when you square a binomial, you must remember to include the 2ab term which represents the twice the product of the two terms. 18 a We can expand the expression using the square of a binomial formula, which is (A − B)2 = A2 − 2AB + B2. Our expression is in this form, with A being and B being 3b. Let’s start by squaring A, which is

Then, we calculate 2AB, which is : Lastly, B2 is (3b)2 B2 = (3b)2 = 9b2

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945

Now, substituting A2, 2AB, and B2 back into the formula, we get:

v No. This creates one of the side lengths to be negative, which creates a negative volume. c Maximum occurs at length = 7.145 and Volume = 525.365, so any answer around length = 7 would be acceptable.

Simplifying, we get:

3.03 Factorise binomial products Understanding 1 a 2

b The resuting expression is not a trinomial because it contains terms with variables in the denominator and the expression has more than 3 terms.

2 a 81 = 92

d 2d2

d 25 = 52

c 16 = 4 3 x+4 4 n+6 5 a Yes

19 a P = 4x + 16 metres

Fluency

c The length of the ﬁeld would be (6 + 10) which equals 16 metres.

c (c + 5) b 100 = 102

Problem-solving b A = x2 + 8x − 20 square metres.

b 5x

b No

6 a (m − 6) (x + y) c (4w + 3x) (5y − z) e (g + h) (3f + g + h)

The width of the ﬁeld would be (6 − 2) which equals 4 metres.

g (2c − d) (c + 5d + 3)

c Yes

d No

b ( y − z) (x − w) d ( y − 4) (8y − 3)

f (2x2 + 3z) (2y − 1)

h (3x2 − 5y) (x + 4y)

The perimeter of the ﬁeld would be 24 + 16 which equals 40 metres.

7 a (x + 2) (x + 5)

The area of the ﬁeld would be 16 × 4 which equals 64 square metres.

e (3 + x) (8 + y)

f (x + z) (7y + w)

g (p + 2) (2m + 3)

h (4x + 3z) (1 + 8y)

20 a 6x2 + 36x + 54 units2

i (2x + 3z) (1 + 6y)

j (df + rz) (3f − 11y)

b x + 9x + 27x + 27 units

c (a + 8) (a2 + 1)

8 a (x + 3)2

22 a 24x2 − 49 3

c (7 − p)

e 6 (c − 2)

b 62x2 − 25y2

23 n + 3n + 2n 3

b (8 + e)2

21 (4x2 − 8x + 4) cm2

24 a −y + 4y + 96y

b (x − 3) (x + 8) d (x − 2y) (8 + z)

d (u − q)2

f 5 (q + t)2

g (4w − 5)2

h (8r + 3)2

j (b − r)2

9 a (x + y) (x − y)

b (n − 5) (n + 5)

c (v − 1) (v + 1)

d (11 − v) (11 + v)

ii No, because you cannot have a zero length, and zero length yields zero volume.

g (4 + 3y) (4 − 3y)

h (xy − 7) (xy + 7)

iii Yes, all the side lengths are viable and the volume is 200 cubic units.

i (5m + 7) (5m − 7)

j (9x + 4y) (9x − 4y)

k 7 (x − 3) (x + 3)

l 5 (3t + 2) (3t − 2)

b i N o, because we cannot have a negative length and y = −10 would create 2 negative lengths, even though the volume calculation would be viable.

iv No, because the side length 12 − y becomes zero and zero length yields zero volume.

946

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

10 a 3y (x2 − 2y3 + 3x2 y2)

20 a

b (4xy + 3z)2

c (11c − 7d) (11c + 7d)

d (x + y) (3y + z)

e 4b (2a2 b3 + 3c2)

f 7 (2b + 1)

g 7hp2 (2h2 r + 3h4 + 7p2)

b (x + 6) (x + 4)

h 2 (8w + 3t) (w − 4t2)

3.04 Factorise monic quadratic trinomials

i (4q2 + 3r)2

j 3d (4f − h) (4f + h)

Understanding

Reasoning 11 No, if we multiply (a + b) (a + b), the result is a2 + 2ab + b2. The sum of two squares is not equal to (a + b) (a + b). 12 a The error is in the constant term of the factors. The correct factorisation should be x2 + 4x + 3 = (x + 1)(x + 3). b The error is in the signs of the factors. The correct factorisation should be x2 − 4x + 4 = (x − 2)(x − 2). 13 a i Perfect square trinomial ii (x + 1)2

1 a 9 and 10 c −6 and −4 2 a • (1 × 18)

(2 × 32) (−2 × −32)

• (3 × 6) •

(4 × 16)

• (−3 × −6) •

(−4 × −16)

•

(8 × 8)

•

(−8 × −8)

(−1 × 30) (1 × −30)

•

(−2 × 15)

c i Factoring by grouping ii (2x + 5)(x − 3)

•

(2 × −15)

•

(−3 × 10)

d i Perfect square trinomial ii (3x − 2y)2

•

(3 × −10)

e i Difference of squares

•

(−5 × 6)

•

(5 × −6)

14 The trinomial cannot be factorised using perfect squares because the middle term, −72a, is not equal to twice the product of the square roots of the ﬁrst and last terms.

15 a 256

b 392

16 a t + 7 units

c 261

d 660

b 4 (t + 7) units

17 x2 + 12x + 36 2

18 a (x + 4)

3 a m+6

b m + 10

c y+2

d x−4

4 a = 12, b = 7 Fluency 5 a (x + 2)(x + 3)

b (x + 4)(x + 5)

c (t + 7)(t + 8)

d ( y + 6)( y + 12)

6 a (x + 8) (x + 3)

Problem-solving

b (a − 3) (a + 3)

(1 × 64) (−1 × −64)

• (2 × 9) •

c •

ii (x + 4) (2x + 1) (2x − 1)

b •

• (−2 × −9) •

•

f i F actor by grouping then difference of squares

d −8 and 11

• (−1 × −18) •

b i Common factor then difference of squares ii n (n + 11) (n − 11)

ii (x + 16 + y) (x + 16 − y)

b −9 and 7

c (x + 9) (x + 8)

b (x + 4)2

d (x + 8)2

e (t + 5) (t + 2)

f (t + 7)(t + 1)

g ( y + 4)( y + 5)

h (x + 3)(x + 5)

i (x + 3)(x + 4)

j ( y + 7)( y + 2)

k (x + 5) (x + 8)

l (x + 6)2

19 (x − 1) (x + 1) (x2 + 1)

Answers mathspace.co

947

13 a Exactly one of p or q must be negative.

7 a (x − 3) (x − 1)

b (x − 6) (x − 10)

c (x − 7) (x − 12)

d (x − 5) (x − 7)

b Both p and q must be positive.

e (x − 7) (x − 1)

f ( y − 2) ( y − 8)

c Both p and q must be negative.

g (t − 6) (t − 7)

h (x − 3) (x − 17)

i ( y − 7) ( y − 11)

j (t − 12) (t − 8)

k (x − 11) (x − 1)

l (x − 15) (x − 3)

8 a (x + 2) (x − 3)

b (x + 5) (x − 6)

c (x + 5) (x − 4)

d (x + 6) (x − 9)

e (x + 7) (x − 10)

f (x + 13) (x − 9)

g ( y − 16) ( y + 3)

h (t + 14) (t − 6)

i ( y − 15) ( y + 3)

j (x + 14) (x − 4)

k (x + 12) (x − 2)

l (x + 7) (x − 4)

9 a 2 (x + 2) (x + 5)

b 4 (x + 2) (x + 4)

c 4 (x − 9) (x + 8)

d 3 (x − 5) (x − 2)

e 5 (x − 2) (x + 3)

f −2 (x − 2) (x + 11)

g −4 (x − 5) (x + 2)

h −3 (x − 2)2

i −(x + 4) (x + 2)

j −(x − 3) (x − 4)

k x (x + 3) (x + 4)

l 2x (x + 5) (x + 3)

14 a No, this would be the area of the whole square and would be equivalent to x2.

b Yes. x2− Area II is the same as Area I. Area I has length of x and width of x − 4, so it would have an area of x (x − 4). c Yes. Area I has length of x and width of x − 4, so it would have an area of x (x − 4). d No, this would be Area II which would be equivalent to 4x.

15 a (x + 9)(x + 8) = x2 + 17x + 72 b (x + 2)(x + 1) = x2 + 3x + 2

c (x + 1)(x − 9) = x2 − 8x − 9 16 a a and b must be a factor pair of 36. To maximise the length of the entire ﬁeld you would increase the length by 36 and the width by 1. b a and b must be a factor pair of 36. We want to minimise the difference of the factor pair.

Reasoning 10 Octavia is correct, since the terms share the highest common factor of −3. The trinomial factorises to −3 (x2 + 5x + 6) = −3 (x + 3) (x + 2) first. 11 Omar used the fact that 144 is a perfect square and checked to see if it was a perfect square trinomial. Since it was, he efficiently factored the expression without needing to find a pair of numbers. Dima found two numbers that had a sum of 24 and a product of 144. She then used those to fill in the factors and then simplified.

The differences are: • 1 and 36 = 35 • 2 and 18 = 16 • 3 and 12 = 9 • 4 and 9 = 5 • 6 and 6 = 0 If both portions are extended by 6 then the areas in the bottom left and top right will be the same, which would minimise the difference in area. Problem-solving 17 (m + 5) (m + 9)

Both strategies work and lead to the same answer. Omar needed to recognise the pattern, while Dima needed to recognise that she could simplify since the factors were the same.

18 a (x + 6) (x + 3)

12 a The error is in the signs of the factors.

19 a (u − 8)(u − 3) b (x + 4)(x − 2)(x + 3)(x − 1)

The correct factorisation should be x2 + 7x − 18 = (x − 2)(x + 9).

b (m + 5) (m + 10)

20 a (x + 3) (x + 5)

b The error is in the signs of the factors. The correct factorisation should be x2 − 11x + 30 = (x − 5)(x − 6).

948

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b 2

c 9x2 + 72x + 135

21 a (x − 2) (x + 2) (x2 + 3) b ( y − 2) ( y + 2) ( y − 1) ( y + 1) c (a + 1) (a + 8) d (d − 7)2 22 To determine the values of c and d, use factoring to get the area into the form of ( y + c) ( y + d). The expression y2 + 8y + 15 can be factorised into ( y + 3)( y + 5). This means that the values are c = 5 m and d = 3 m. 23 x = 7

3.05 Factorise non-monic quadratic trinomials Understanding 1 Factorise −8 from the expression. 2 a 8x2 + 11x + 3 = (8x + 3) (x + 1)

b 2x2 + 3x − 20 = (2x − 5) (x + 4)

c 2x2 − 19x + 45 = (2x − 9) (x − 5)

d 10x2 + 29x + 21 = (2x + 3) (5x + 7)

Reasoning 12 a i 15 iii −16

ii The same iv Opposite

b i −45 iii −4

ii Opposite iv Opposite

13 Gabriella did not factorise the common factor correctly. Since there is a factor of 2 in both factors of their expression, they need to factor out a highest common factor of 4.The correct working would be: 16x2 + 48x + 36 = (4x + 6)2 = [2 (2x + 3)]2 = 4 (2x + 3)2 14 a Each polynomial has a leading coeﬃcient a ≠ 1 and none of the polynomials can be factored. b Find the product a × c, then look for factors of ac that can add or subtract to be equal to the value of b. If no such factors exist, the polynomial is not factorable.

3 (3x + 1) (x + 6)

Problem-solving

Fluency

15 a −5 (t + 1) (t − 2)

4 a (3y + 1)( y + 9) c (3x − 4) (x − 7) e (3t − 4)(4t + 1) g (4x + 5) (3x − 2)

b (2x + 5)(x − 8) d (8x + 3) (x − 3) f (9x + 4)2 h (6x − 7) (x + 8)

5 a 3 (x − 9) (x + 2)

b 4 (x + 5)2

c 2(s + 3)(4s − 9)

d 3 (3x + 1) (x − 5)

e 4 (2x − 1) (x − 2)

f 6 (4x + 1) (3x − 1)

g 5 (7x − 1) (3x − 1)

h 8 (3x − 2) (7x + 1)

6 a −(2x − 3) (3x + 2)

b −(2x − 7)(3x − 2)

c −3 (2x + 5) (2x − 3)

d −2(3x + 4)(x − 5)

b 0 c 0 d 2 seconds after jumping off the board, Tara hits the water. 16 a (4x2 − πx2) units2

b x2 (4 − π )

17 a 3x + 4, 2x + 3

e −(4x − 3) (x − 2)

f −(3x − 4)2

b To get the expression of the perimeter of the rectangular ﬁeld, we use the formula of the perimeter of the rectangle which is given by 2 × length + 2 × width. This means that the perimeter of the rectangular ﬁeld is P = 2(3x + 4) + 2(2x + 3) = 10x + 14.

g −(5x + 4) (x + 2)

h −(3x − 2)(x + 4)

c x = 10

7 a 10y − 112y + 22

b 2 (5y − 1) ( y − 11)

8 a 10x2 + 105x + 245

b 5 (2x + 7) (x + 7)

9 a 6 (x + 3)

10 2x + 5, 3x + 4 11 −10 (t + 2) (t − 7)

b (x + 3) units

18 a She might use a variable, such as x, to represent the width of her border if she did not know yet what border size she wanted. b The expression for area can be factorised as (2x + 5) (2x + 7) square units, so, the dimensions of the photograph are 2x + 5 units and 2x + 7 units.

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c If the border is a width of x units, you would subtract 2x from each dimension since there is a border on both sides. That leaves the dimensions as 5 units and 7 units, so, the area of the photograph is 35 square units. 19 4x2 + 12x + 5 = (2x + 1)(2x + 5) is one possible answer.

8 a b c (x − 1) (x − 5)

d (x + 10) (x + 1)

e (x − 5) (x − 6)

f (x + 21) (x + 3)

g (x − 1) (x − 19)

h (x − 5) (x − 23)

9 a 3 (x − 3) (x + 1)

b 2 (x + 9) (x − 1)

20 k = 9 21 m = −6, n = −8 or m = −8, n = −6

3.06 Complete the square

c 4 (x + 1.5) (x + 2.5) = (2x + 3) (2x + 5)

Understanding 1 a 9

b 16

c 49

d 25

2 a Yes

b No

c No

d Yes

d −2 (x − 5) (x − 13) e −4x (x + 4) f

Fluency 3 a 3x2 + 6x − 8 = 3 (x2 + 2x) − 8

c 4x2 + 12x − 6 = 4 (x2 + 3x) – 6

b 2x2 − 10x + 1 = 2 (x2 − 5x) + 1 2

d 6x + 24x − 7 = 6 (x + 4x) − 7 4 a 5x2 − 10x + 4 = 5 (x2 − 2x) + 4

Reasoning

10 When completing the square, we could get:

= 5 (x2 − 2x + 1) + 4 − 5

= 5 (x − 1)2 – 1

= 6 (x2 + 6x + 9) − 1 − 54

b 6x2 + 36x − 1 = 6 (x2 + 6x) − 1

5 a (x + 9) − 81 c (x + 5)2 + 6 e (x − 5)2 + 5 g

2 = (x + 3)2 − 32 + 13

b (x − 4)2 − 16

3 = (x + 3)2 + 4

Factor the perfect square trinomial

d (x + 7)2 − 2

f (x − 9)2 − 4 h

b (x + 7) (x + 1)

12 On line 3, she has expanded 3[(x + 3)2 − 9] − 7 incorrectly. She forgot to multiply the 9 by 3.

c (x + 12) (x − 2) d 2

7 a 2 (x + 2) − 15

b 5 (x + 2) – 29

e 7 (x − 1)2 − 6

f 9 (x + 3)2 – 83

c 4 (x − 2)2 − 13

950

Simplify the constant

This is a sum of squares, not a difference of squares. We cannot factor a sum of squares, so this quadratic is not factorisable. 11 In step 1, Alessandro added the 32 = 9 instead of subtracting it.

6 a (x − 4) (x − 2)

the coeﬃcient of x

= 6 (x + 3) − 55 2

1 x2 + 6x + 13 = x 2 + 6x + 32 − 32 + 13 dd and subtract the square of half of A

d 10 (x + 1)2 − 21

13 y = 3(x − 4)2 + 2 means that y is equal to 2 plus an expression, 3(x − 4)2, which cannot be negative. Therefore the minimum value of the function occurs when 3(x − 4)2 is equal to zero, which happens when x = 4.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

14 a The quadratic expression x2 − 18x + 81 can be factorised in two ways:

1. x2 − 18x + 81 = (x)2 + 2(x)(−9) + (−9)2

= (x − 9)2

2. x2 − 18x + 81 = x2 − 9x − 9x + 81

= (x2 − 9x) + (−9x + 81)

= x(x − 9) − 9(x − 9)

= (x − 9)(x − 9)

= (x − 9)2

b The quadratic expression 4c2 − 12c + 9 can be factorised in two ways:

1. 4c2 − 12c + 9 = (2c)2 + 2(2c)(−3) + (−3)2

= (2c − 3)2

2. 4c2 − 12c + 9 = 4c2 − 6c − 6c + 9

= (4c2 − 6c) + (−6c + 9)

Problem-solving

= 2c(2c − 3) − 3(2c − 3)

15 a x − 5 cm

= (2c − 3)(2c − 3)

= (2c − 3)2

c The quadratic expression 3a2 + 14a + 8 can be factorised in two ways:

1. 3a2 + 14a + 8 = 3a2 + 12a + 2a + 8

= (3a2 + 12a) + (2a + 8)

= 3a(a + 4) + 2(a + 4)

= (a + 4)(3a + 2)

b x > 5 since the side lengths must be positive. 16 x − 4 17 a y = −(x + 4)2 + 36 b y = 36 when x = −4. 18 Answers may vary.

x2 + 6x + 4 = (x + 3)2 − 5 and x2 + 8x + 7 = (x + 4)2 − 9 are two possibilities.

Chapter 3 review

1 a 5x

b 4xb

2 a k (k − 8)

b 5 (y2 + 2y + 3) c 4c (−bc + 2b + 1)

d p2 q2 (3 + 4p2 q2 + 5p4 q4) 3 The student’s error is that they haven’t considered the last term, −5xy in the original expression. Consequently, the correct factored form of the expression should be 5xy (7x + 2y − 1). d The quadratic expression 2x2 + 5x − 3 can be factorised in two ways:

1. 2x2 + 5x − 3 = (2x2 − x) + (6x − 3)

= x(2x − 1) + 3(2x − 1)

= (2x − 1)(x + 3)

4 a x2−10x + 16 2

c x − x − 30

b x2 + 4x + 3

d x2 − 5x − 36

5 a Yes

b No, (y − 3)2 = y2 − 6y + 9

6 a m2 – 6 c 9x2 − 0.49

b 64 − 25p2 d

Answers mathspace.co

951

7 a −24x2 + 14x + 24

Fluency

5 a

8 9x2 − 72x + 144 cm2 9 a (k + 8) (k2 + 1)

b y = −1

c x = −1

h c=1

6 a

g q = −1

h s=

7 a

d w = −9

f p=9

g q = −18

b 5 (r + s)2

c 7 (5m2 − 7)

e (a + b) (5 + x)

f 5 (x + 4y) (x2 − y)

10 a (x − 5)2

b (x + 8)2

c (1 − x)

d (x + 9) (x + 8)

e (x + 13) (x − 9)

f (y − 16) (y + 3)

g (x + 12) (x − 2)

h (x + 5) (x − 4)

11 a (3t − 4) (4t + 1)

b (6x − 7) (x + 8)

c (9x − 5)2

e n = −1 8 a e n = −4

d −2 (3x + 4) (x − 5)

Reasoning

12 5x + 6, 4x + 3 13 a 10x2 + 75x + 125

b 5 (2x + 5) (x + 5)

14 k = 9 15 a (x − 4)2 − 16

9 a Dylan made an error in Step 1. He missed the constant term and he should have made sure that had the same denominator. This

b (x − 2)2 + 2

c (x + 5)2 − 4

should be the Step 1 before proceeding to cross-multiplication. By writing the binomial

with the same denominator we have . Next should have multiplied 2 by the binomial 32 − x, giving us the answer of 64 − 2x.

e 10 (x + 1) − 21

f 2 (x + 2) − 15

b 1

16 a b (x + 11) (x + 3) c (x + 37) (x + 5) d (x − 5) (x − 6) 2

17 a y = 36 − (x + 4)

Cross multiply

Expand the brackets and simplify

Add 2x to both sides and subtract 48 from both sides

Divide both sides by 26 and simplify

b y = 36 when x = −4.

4.01 Equations involving fractions

Write the right-hand side as a fraction

Understanding 1 a 2

b 42

c 6

d 12

2 a Yes

b No

c No

d No

3 a True

b False

c True

d False

4 a

11 Many possible answers, for example: . When we multiply both sides of the equation by the common denominator, 12, we get 2 (3x + 6) = 3 (2x + 6). Expanding both sides gives 6x + 36 = 6x + 18. If we subtract 6x from both sides, we get 36 = 18. However, this equation is impossible to satisfy as no value for x can make both sides equal. 12 a Yes b Yes, the equation has inﬁnitely many solutions. c x=2

952

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving

13 a

b x = $7500

11 a y = x − 1

14 x = 20 15 u = −17 16 k = 60 17 r = 0 18 The equation for the total funds raised is . Solving for x gives

b y = −2x + 3

c y = −2x + 3

d y = 4x − 5

e y = 3x − 5

f y = −2x + 3

g y = 2x + 4 33 99 i yyy==−− xx+++ 22 22

h y = 2x + 3 4 j yy = x − 4 3

5 12 a m = − , y = 5 2 y

x = $1153.85.

4.02 Rearrange equations

5 4 3

Understanding

1 The purpose of rearranging equations is be able to evaluate for different variables when doing repeated calculations.

4 a False

c True

d False

b m = 4, y = −16 y 2

Fluency 5 a x=y−z

-3

b True

-2

2 We should make c the subject 3 C

-4 -3 -2 -1 -1

-2 -1 -2

b x = −8y

x 1

-4

-6

f x = −7y

-10

-8 -12 -14 -16

6 a m = ghy

-18

b y = 4 − 3x2

7 a c

c m = −2, y = 4 y 5 3 1

d y = 2x + 3

8 a

-5 -4 -3 -2 -1 -1

x 3

-2 -3 -4

-5

Answers mathspace.co

953

Bob, on the other hand, made a mistake in his rearrangement. He should have subtracted a2 from both sides to make b2 the subject, like this: c2 = a2 + b2 1

d m = 3, y = 6 y 7 6 5

2 2 2 2 2 2 c − a = a + b − a

4 3

2 1 -5 -4 -3 -2 -1 -1

x 1

However, he added a2 instead, yielding an incorrect equation: b2 = c2 + a2

So, Alice is correct, and Bob made a mistake in his rearrangement.

-2 -3

b2 = c2 − a2

22 Pamela’s mistake was in Step 2 of her algebraic working. When she divided both sides of the equation by L, she incorrectly wrote the

13 14 a

15 a

The equation for KE as given

g z = x2

i c = sn2 + 6

Multiplying both sides by 2

right-hand side of the equation as of

instead

. The correct equation should be

23 Ruﬁno has an error in his work.

Now he should have divided by v2

j 16

Problem-solving

18 19

24 a

b h = 3 cm

25 a

b r = 2.79 cm

26 a

b P = $2027

27 a

b r = 3 mm

Reasoning 20 Both equations can be rearrange to solve for x. When 3x + 8 = 2 is rearrange to solve for x the answer is numerical. Whereas when 3x + y = z is rearranged to solve for x the answer has unknowns, y and z, in it. However if you substitute y = 8 and z = 2, you get the same numerical answer as solving 3x + 8 = 2. 21 Alice is correct in her rearrangement, while Bob made a mistake. Alice rearranged the equation to make a2 the subject by subtracting b2 from both sides: c2 = a2 + b2 1 2 2 2 2 2 2 c − b = a + b − b

954

a2 = c2 − b2

4.03 Parallel and perpendicular lines Understanding 1 a Their gradients, the values of m, are the same. b All graphs have the same gradient or rate of increase. 2 −1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

20 y = 2x + 4

and

21 y = 2x + 4

b One pair: y = 3x + 15 and Another pair: y = 3x + 15 and

Problem-solving

4 Perpendicular

22 a

5 D

23 a −1

b y = −x + 18

6 D

24 a 6

Fluency

25 y = 0.5x + 5

7 a Yes

b No

c Yes

d Yes

e No

f No

g No

h No

8 a Yes

b Yes

c No

d Yes

e No

f No

g No

h No

9 a −5

b y = −5x + 40

10 a y = −2x − 5

b y = 8x + 5

c y = −3x – 4

d y=2

e x = −7 11 a

c −8

d 1

12 a No

b No

c Yes

d Yes

13 a No

b No

c No

d Yes

14 a No

b No

c Yes

d Yes

e No

f Yes

g Yes

h No

15 a

26 27 a

b k = 10

28 a

b (5, 10)

c 29 c = −2 30 a −1

b 1

c −1

d Yes

31 a 0

b −1

c No

d Yes

e A trapezium 32 a Gradient of: AB = , BC = 3, CD = , AD = 3 b Yes 33 a i

c No ii

16 a

c No

17 a

c Yes

4.04 Linear models

Understanding

c y = 2x + 6

d x = −8

e y = −8

1 A 2 B

Reasoning 19 T he ﬁrst two lines have the same gradient of 3, so they are parallel. The last two lines have the same gradient of

iv 1

b A parallelogram. This is because the opposite sides have the same gradient, making them parallel. It is not a rectangle, as the adjacent lines are not perpendicular.

18 a

iii 1

, so they are parallel.

3 a Horizontal

b Vertical

c Horizontal

d Vertical

4 a 250

Additionally, the product of the gradients of the

Fluency

ﬁrst and third lines is

5 a 115

, so they are

perpendicular. Since opposite sides are parallel and adjacent sides are perpendicular, the enclosure is a rectangle.

b 12.5

c 8

d −5

b B

6 a The straight line graph starting at the origin is the cost function. The curved graph with y-intercept of 50 is the revenue function.

Answers mathspace.co

955

b The intersection is the break-even point where cost is equal to revenue. 7 a Linear

b Non-linear

c Linear

d Non-linear

8 a

120 110 100 90 80 70 60 50 40 30 20 10

11 a

b C = 1.20x 12 a 5° C

M 2

Number of minutes 0 passed (x) 75

93 99 105 108

111 y 108 105 102 99 96 93 90 87 84 81 78 75 72

13 a

20 15

10 5 x

Number of minutes passed (x)

Depth of diver in metres ( y)

0 0.8 1.6 2.4 3.2

4 b 5 °C 14 a

3.2 2.4 2 1.6 1.2 0.8 0.4

x 1

c 0.8 m/min

c 5 °C

d T = 5x + 5

2.8

956

x 2

c 3 beats per minute

3.6

28 y 26 24 22 20 18 16 14 12 10 8 6 4 2 1

1 2 3 4 5 6 7 8 9 10 11

b 3° C

c y = 5x + 3

Heart rate ( y)

10 a

c $1.20

d $90.00

b No

5 10 15 20 25 30 35 40 45

9 a

48 C 45 42 39 36 33 30 27 24 21 18 15 12 9 6 3

d 19.2 m

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

28 T °C 26 24 22 20 18 16 14 12 10 8 6 4 2

20 40 60 80 100 120 140 160 180

b 21 °C

18 a

c 190 chirps per minute

d 120 chirps per minute

15 a

16 14

8 10 12 14 16

c 4 minutes

d 1

d C = 0.30t + 0.40

2 3 4 5 6 7 8 9

b 2 cm

c 26 cm

d 11

e h = 2d + 2

19 a The variables are income and number of hours worked. b

8500 8000 7500 7000 6500 6000 5500 5000 4500 4000 3500 3000 2500 2000 1500 1000 500

800 700 500 400 300 200 100

x 5

d A = −5t + 8500

d $625

b For example, y = 150x c $3300 d 36

1.6

Reasoning

1.4

21 a Litres of petrol used per kilometre

1.2 1

b Dollars earned per hour

0.8

c Heartbeats per minute

0.6

d Ascent or descent in metres per second

0.4 0.2

d 1 2 3 4 5 6 7 8 9 10 11

b F = 0.10d + 0.30 d 3 km

c For example, y = 25x

20 a The variables are operation cost and number of hours operated.

c 1700 seconds

c $1.70

10 15 20 25 30 35

e 5 hours

b 6000 m

1.8

600

200 400 600 800 1000 1200 1400 1600 1800

17 a

b $2.80

4 2

16 a

22 a 4 of B b 4 of A c Example answer: Exchange rates between different countries’ currencies change on a daily basis. So this graph may give accurate conversions one day, but not the next.

Answers mathspace.co

957

23 a

120

−8

e Example answer:

b The baseball is moving downwards instead of upwards at T = 4. c Example answer: The model is useful to track the velocity of the ball while it is in the air, but once it hits the ground or if it hits an object, the model will not be accurate. Also the velocity may change due to weather conditions. 24 a

Number of minutes passed, x

20 50

Amount of fuel 250 225 200 175 150 left in tank, y

The model is useful to track the temperature for a couple of hours, but eventually we can assume 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 by 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. Problem-solving 27 a 16 km 28 a 104 °F

b y = 250 − 5x

d Example answer:

25 a is the acceleration - that is, the rate of increase in the velocity. b 18 is the velocity at which the ball is initially rolled.

30 a 50 °F c Below

c −29 °C

b 1.8 °F d F = 1.8C + 32

e 95 °F 31 a −300 is the amount the ﬁsh population decreases by each year. b F = 6000 c F = −300t + 6000 d 3300 ﬁshes e 14 years 32 a 2 is the increase in earnings for each additional shirt ironed. b y = 4 is the earnings independent of how many shirts are ironed.

c $32 d 12 shirts

d 46.9 m/s e Example answer: The model is useful to track the velocity of the ball as it rolls down the hill, but once the hill ﬂattens the velocity will decrease as time increases. Therefore, the model will no longer be accurate. 26 a is the amount by which the temperature is rising every minute. b y = 3 is the temperature of the room before the heater has been turned on.

33 a 20 mi/hr 34 a

b 80 mi/hr

is the amount of water that is ﬂowing out

of the hole every minute. b 30 is the capacity of the bucket. c d 3L 35 a 10 is the variable cost of manufacturing per chair.

c d 13 °C

958

b 10 °C

29 $77.08

c The amount of fuel in the car is decreasing at a constant rate of 5 litres per minute. The model is useful to track the fuel consumption of the car while it is moving at a constant speed. But as the car changes speed the fuel consumption will change, and the model will not be accurate.

b 1.6 km

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b y = 85 is the ﬁxed cost of manufacturing. c For example, y = 10x + 85 d $335

4.05 Graph linear inequalities

11 a x = 4 c

Understanding

5 4 3 2 1

1 A 2 a False

b False

c True

d True

e False

f True

g True

h True

b Dashed c Solid

d Dashed

5 a Below

b Below

d Below

c Above

b y=4

12 a

6 a No

b Yes

c No

d Yes

7 a i No

ii No

iii No

iv Yes

b i No

ii Yes

iii No

iv Yes

c i Yes

ii No

iii Yes

8 a i y=2

ii y ≤ 2

b i x = −3

ii x ≥ −3

c i y = −3x − 5

ii y < −3x − 5

d i y = 4x + 7

ii y < 4x + 7

e i y = −2x

ii y > −2x

f i

ii b y < −2x − 6

c y ≤ −3

d y > −5x + 6 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 −4 − 4 − −11 −3 3 − −2 2 − −11

x 1

−2 2 −3 3 −4 4

ii Yes

iii No

iv No

e Yes 13 a i

9 8 7 6 5 4 3 2 1

−5 − 5− −4 4− −11 −3 3− −2 2− −11 − − 2 −2 −3 3 −4 − 4 −5 −5

5 4 3 2 1

d i No

9 a y ≥ 4x – 5

1 2 3 4 5

d 3x + 4y < 12 or 3x + 4y ≤ 12

Fluency

10 a

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

3 C 4 a Solid

b y=3 y

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

1 2 3 4 5

1 2 3 4

ii No iii

x 1 2 3 4 5

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

x 1 2 3 4

Answers mathspace.co

959

b i

4 3 2 1

d i

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6

2 3 4

4 3 2 1 −4 − 4 − −11 −3 3 − −2 2 − −11 −2 2 −3 − 3 − 4 −4 −5 − 5 −6

c i

6 5 4 3 2 1 −4−3−2−1 −1 −2 −3 −4

6 5 4 3 2 1 −4−3−2−1 − 4−3−2− −1 −11 − −2 − 2 −3 − 3 −4 − 4

960

−2 −3 −4

ii No

iii

9 8 7 6 5 4 3 2 1

x 1

−4 −3 −2 −1−1

−2 −3 −4

14 a

x 1 2 3 4 5 6 7 8 9

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3 −4 4 −5 5 −6 6 −7 7

x 1 2 3 4 5 6 7 8 9

ii Yes iii

−4 −3 −2 −1−1

ii Yes iii

9 8 7 6 5 4 3 2 1

5 4 3 2 1 −4 − 4 − −11 −3 3 − −2 2 − −11 − −2 2 −3 3 −4 4 −5 − 5 −6 − 6 −7 −

2 1

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 8 −9 9 − 10 −10 −11 − 11

7 6 5 4 3 2 1

y 1

x 1

5 4 3 2 1 −4 − 4 − −11 −3 3 − −2 2 − −11 − −2 2 −3 3 −4 −5 −6 −7

j y

x 1

5 4 3 2 1

−4 −3 −2 − −11 −11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6 −7 − 7

x 1

7 6 5 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5

−4 −3 −2 −1 −1 −2 −3 −4 −5 5 − −6 6 − 7 −7

−4 −3 −2 −1−1

i 5 4 3 2 1

−2 −3 −4 −5 −6 −7 −8 8 −9 9 −10 − 10 −11 − 11

−4 −3 −2 −1 −1 −2 −3 −4 −5

2 1

7 6 5 4 3 2 1

−4 −3 − −2 2 − −11 −11 − −2 − 2 −3 − 3

9 y 8 7 6 5 4 3 2 1 −4 − 4 − −3 3 − −2 2 − −11− −11 −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6 −7 − 7

Answers mathspace.co

961

Reasoning

11 y 10 9 8 7 6 5 4 3 2 1 −4 − 4 − −3 3 − −2 2 − −11− −11 −2 − 2 −3 − 3

20 a The inequality 4x + 5y ≥ 20 includes the line 4x + 5y = 20 as part of the solution set whereas 4x + 5y > 20 does not. In the graph, this will change the line from being dashed (exclusive) to being solid (inclusive). x 1

15 a i 2

b The region 4x + 5y ≤ 20 is on the other side of the line from the region 4x + 5y ≥ 20.

21 a No, the inequality x + y > 10 cannot have integer solutions where both x and y are less than 3. If x and y were both 3, the sum would be 6, which is less than 10.

b i

ii y = 2x − 3

ii y = 4x − 4

iii y ≤ 2x − 3

iii y ≥ 4x − 4

16 a y = 4

b y=3

17 a y = 2

c y = 13

18 a b

b Yes, for example x = 2 and y = −9. 22 a By substituting the values of 12 tool sets and none of the plant bundles into the inequality, we get 12(12) + 21(0) = 144 + 0 = 144, which is less than $150 leaving $6. Thus, the club can’t buy more than 12 tool sets. Similarly, by substituting the values of 0 tool sets and 7 of the plant bundles into the inequality, we get 12(0) + 21(7) = 0 + 147 = 147, which is less than $150 leaving $3. Thus, the club can’t buy more than 7 plants.

y 12 10 8

Therefore, the constraints of the budget make it impossible to exceed 12 tool sets or 7 plant bundles.

6 4 2 x 2

c i No

ii Yes

iii No

iv No

19 a b

b Substituting 5 tool sets into the inequality, we have 12 (5) + 21p ≤ 150, Simplifying, we get 60 + 21p ≤ 150, and thus 21p ≤ 90. Solving for , which is approximately p, we ﬁnd that 4.28. Since we can’t purchase a fraction of a bundle, the club can buy at most 4 plant bundles. c Let’s say the club buys the same number of tool sets and plant bundles, i.e., t = p.

42 36

Substituting this into the inequality, we get 12t + 21t ≤ 150, or 33t ≤ 150. Solving for t, we

30 24

. As we can’t purchase get fractions of tools or plants, the club can buy at most 4 tool sets and 4 plant bundles.

18 12 6

x 5

10 15 20 25 30 35

c i No

ii No

iii No

iv Yes

d 30 books

d If the budget is raised to $180, and they decide to buy an equal number of tool sets and plant bundles, the inequality becomes 12t + 21t ≤ 180, or 33t ≤ 180. Solving for t, . Since we can’t buy we get fractions of tools or plants, the club can afford at most 5 tool sets and 5 plant bundles.

962

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving

28 a 6x + 4y ≤ 48 b 14 y

23 x + y ≥ 43 24

13 12 11 10 9 8 7 6 5 4 3 2 1

y 70 60 50 40 30 20 10 −40−30−20 −10 −10

x 10 20 30 40

c There are 61 different ways, one for each integer coordinate inside the solution set of the inequality.

x + y ≥ 50 25 19x + 9y ≤ 855 26

29 a 2x + 5y ≤ 30 b

19 y 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 −8 − 8 − −6 6 − −4 4 − −2 2− −11 −2 − 2 −3 − 3

y 18 15 12 9 6 3 x

x 2

c The x-intercept of 15 represents the greatest number of games of Table Tennis that Oprah could play if she decides not to play any games of Frog Days. d The y-intercept of 6 represents the greatest number of games of Frog Days that Oprah could play if she decides not to play any games of Table Tennis.

27 a b

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14

y 45 40

30 a y ≥ 30 − 0.75x or 30x + 40y ≥ 1200

35 25

b She earns $30 per hour as a tutor and $40 per hour as a sales associate.

c She needs $1200 to cover expenses.

Chapter 4 review

10 5

x 5 10 15 20 25 30 35 40 45 50 55

c B

1 a

2 a c

Answers mathspace.co

963

3 a

b r = 2.52 cm

4 a No

b Yes

c No

d Yes

5 a No

b No

c Yes

d Yes

6 a 3m 7 a

120 110 100 90 80 70 60 50 40 30 20 10

11 a y > 0.2x

12 a (5, 0) c

−3 −2 −1 −1 −2 −3 −4 −5

13 a

b 5°C

c y = 3x + 5

2 1

d 28 y 26 24 22 20 18 16 14 12 10 8 6 4 2

−4 − 4 − −3 3 − −2 2 − −11 − −11

−2 − 2 −3 − 3 −4 − 4

b i No 2

14 a

9 a Number of litres (x)

20 30 40

Cost of rice (C)

64 96 128

130 C 120 110 100 90 80 70 60 50 40 30 20 10

ii Yes

iii No

7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3 −4 −5

x 5 10 15 20 25 30 35 40

d $76.80

iv No

c Yes

x 1

964

x 1 2 3 4 5 6 7 8

8 a 3°C

10 a No

d 3x + 5y < 15 or 3x + 5y ≤ 15

b No

c $3.20

b (0, 3) 6 5 4 3 2 1

or 3x + 4y ≤ −12

b 18 m

e 14 kg

b Yes

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5 4 3 2 1 −4 − 4 − −3 3 − −2 2 − −11 −11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 −7

15 a −4

b y = −4x − 3

c y ≥ −4x − 3 16 14x + 18y ≥ 210

5.01 Quadratic equations

12 Answer may look like:

Understanding 1 5 2 A quadratic equation can have 0, 1, or 2 solutions. 3 a a = 2, b = −7, c = 5

b a = −3, b = 8, c = −7

c a = 1, b = 0, c = −16

d a = −1, b = −4, c = 3

4 a a = 3, b = 5

11 Answer may look like:

The equation can be rewritten as x2 = 9. As both 32 and (−3)2 are equal to 9, there are two solutions.

b a = 6, b = −3

c a = 0, b = 4 5 x = −12

To write the equation in the form (x − a) (x − b) = 0, we can use the given solutions to ﬁnd the factors. Since x = 4 is a solution, one factor is x − 4. Similarly, since x = −3 is a solution, the other factor is x + 3. Therefore, the equation can be written as (x − 4)(x + 3) = 0. 13 Jade’s answer is incorrect. When using the null factor law, the correct equations should be x − 4 = 0 and x + 6 = 0. Solving these equations, we get x = 4 and x = −6 as the correct solutions, not x = −4 and x = 6. 14 Rearranging gives x2 = −4. There are no real numbers that will give −4 when squared.

Fluency 6 a Concave down

b Concave up

Problem-solving

c Concave down

d Concave up

15 a Concave down

7 a

c x = ±11

e x = ±11

f y = ±5

g x = ±5

16 k = −16

i k = ±2

k x = ±15

17 a = 13

8 a z = 4, −10

b m = 16, −2

d p = 11, −5

e n = −7, 11

f x = −2, 16

h p = ±10

9 a j = 0, −7

18 a x = 0, 13 19 p = 2, q = −3 20 a

b i a = 0, c = All real numbers and a > 0, c < 0 or a < 0, c > 0

d m = 0, −5

h e = 4, 2

5.02 Quadratic graphs

i s = −6, −7

Understanding

10 We can ﬁnd the solutions for x by taking the positive and negative square roots of k, that is, .

b Minimum

c x = 6.5 cm

c m = 0, −5

Reasoning

b Maximum

c t = 14, t = 10

b k = ±5

ii a ≠ 0, c = 0 iii a > 0, c > 0 or a < 0, c < 0

1 a A parabola is a ∪-shaped curve that opens either upward or downward. b i I f the coeﬃcient a is positive, the curve is ∪-shaped opening upward. If a < 0, then the curve is ∩-shaped, opening downwards.

Answers mathspace.co

965

ii Changing the constant term c in a quadratic equation shifts the entire graph vertically. If c is positive, the graph moves upward, and if c is negative, the graph moves downward. The magnitude of the shift depends on the value of c.

c i H orizontal shift 1 unit to the left, vertical shift 4 units up and dilated (compressed) by a factor of 3. ii (−1, 4) iii

11 10 9 8 7 6 5 4 3 2

2 The graph is concave upward.

axis of symmetry

10 8

y-intercept

6 4

x-intercepts

2 −2

ii (6, 1)

3 Turning point

iii

4 The x-intercepts Fluency 5 a i H orizontal shift 3 units to the right and vertical shift 4 units down y

−1 −2 −3 −4 −5

4 3 2 1

x 1

−1 −2 −3 −4 −5

ii (3, −4) 7 6 5 4 3 2 1

2 3 4 5 6 7 8 9

6 a i Concave up x 1

ii More steep

iii (1, −3)

iv (0, −1)

y 3 2 1 −3 −2 −1

b i R eﬂection about the x-axis, horizontal shift 3 units to the left and vertical shift 6 units down iii

−3

x −5

−2

b i Concave down

y 1

−5 −4 −3 −2 −1

x 1

−1

ii (−3, −6)

iii (2, 2)

−6 −7 −8 −9

966

1 2 3 4

d i H orizontal shift 6 unit to the right, vertical shift 1 unit up and dilated (compressed) by a factor of 0.5.

turning point

iii

−5 −4 −3 −2 −1

x 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

ii Less steep iv

b i y = −(x − 2)2 + 1

y 3

ii (2, 1)

iii

y 1

−1

−6 −5 −4 −3 −2 −1

−2

−1

−3

−2

c i Concave down iii (−5, −4)

ii More steep

−3

iv (0, −79)

−4

1 2 3 4 5 6 7

x y −9 −8 −7 −6 −5 −4 −3 −2 −1 −2

c i y = − (x + 4)2 – 2

−4

ii (−4, −2)

−6

iii

−8 −10 −12

d i Concave up

ii More steep

iii (−1, 2)

iv (0, 5)

18 16 14 12 10 8 6 4 2 −4 −3 −2 2

7 a i y = (x + 3) − 9

y x −9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

d i ii iii

y 5

−1

ii (−3, −9)

iii

1 −6 −5 −4 −3 −2 −1

−1 −2 −3 −4 −5 −6 −7 −8 −9

x 1

1 x −3 −2

−1

8 a y = x2 + 4 2

c y = −x + 6

e y = 0.5(x + 2)2 − 3

b y = (x − 1)2

d y = 3(x + 2)2

f y = −2(x − 3)2 + 4

Answers mathspace.co

967

9 a y = (x − 2)2 − 6 b

6 5 4 3 2 1 −1

d i y = −3

ii x = 3, −1

iii x = 1

iv (1, −4)

4 3 2

x 1

−1 −2 −3 −4 −5 −6

5 −2

−1

x 1

−1

−2 −3 −4

10 a i y = 0

ii x = 0, −4

iii x = −2

iv (−2, −4)

e i y=5

ii x = 5, −1

iii x = 2

iv (2, 9)

9 8 7 6 5 4 3 2 1

3 2 1

−4 −3 −2 −1 −1

−2 −3 −4

b i y=0

iv (2, 4)

x 1

−1

2 1 −1

−4

−1

x 1

−2

c i y = 12

ii x = 2, 6

iii x = 4

iv (4, −4)

11 a y = (x − 1) (x + 7)

y 2

10 8 6 4 2

x 1

−3

−4

968

iv (2, −1)

−2

ii x = 1, 3

f i y=3

iii x = 2

−2 −1

ii x = 0, 4

iii x = 2

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

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

x 1 2

b y = −2(x − 1) (x + 7)

28 24 20 12

−8 −7 −6 −5 −4 −3 −2 −1

1 2

x 1

−8 −12 −16 −20 −24 −28 −32 −36 −40 −44 −48 −52 −56

iii (−2, 18) iv

18 16 14 12 10 8 6 4 2 −6 −5 −4 −3 −2 −1

x 1

2 3

15 a i Concave up

d y = −0.5(x − 1) (x + 7) 32

ii (x − 1) (x − 3), (1, 0), (3, 0)

iii y = 3

iv x = 2

4 3

12 8

4 −8 −7 −6 −5 −4 −3 −2 −1

v (2, −1)

24 20

b (2.5, 12.25)

13 a

and y = x2 − 3

and y = −x2 + 3

and y = 2x2 − 6

and y = −3x2 + 9

14 a i y = (x − 4) (x − 8), (4, 0) and (8, 0)

−1

1 2

12 a y = −(x − 6) (x + 1)

iii (6, −4)

b i y = −2(x − 1) (x + 5), (1, 0) and (−5, 0)

y −8 −7 −6 −5 −4 −3 −2 −1−4

ii y = (x − 6)2 − 4

ii y = −2(x + 2)2 + 18

c y = 3 (x − 1) (x + 7)

−1 −2 −3 −4 −5

16 8 4

5 4 3 2 1

−1 −2

b i Concave up ii (x + 6) (x − 6), (−6, 0), (6, 0) iii y = −36

iv x = 0

v (0, −36)

y −6 −4 −2 −4 −8 −12 −16 −20 −24 −28 −32 −36

x 2

Answers mathspace.co

969

c i Concave down ii (2 + x) (2 − x), (−2, 0), (2, 0) iii y = 4

iv x = 0

v (0, 4)

y 4 3 2 1 −2

−2

−1

16 14 12 10 8 6 4 2

x 1

iii y = 24

iv x = −2

v (−2, −1)

y 4 3 2 x

1 2 3 4 5 6

ii −(x + 4) (x − 2), (−4, 0), (2, 0) iii y = 8

iv x = −1

e i Concave up iv x = −3

v (−3, −4) 7 6 5 4 3 2 1

−6 −5 −4 −3 −2 −1

−1 −2 −3 −4

i i Concave up ii (x + 2) (x − 4), (−2, 0), (4, 0) iii y = −8

iv x = 1

f i Concave up

970

−4 −3 −2 −1

ii (x − 3) (x − 5), (3, 0), (5, 0) iv x = 4

v (−1, 9) y

9 8 7 6 5 4 3 2 1

ii (x + 5) (x + 1), (−5, 0), (−1, 0) vi

v (1, 25)

h i Concave down

−1 −1

iii y = 15

−4 −3 −2 −1 −3

iii y = 5

24 21 18 15 12 9 6 3

ii (x + 3) (x + 1), (−3, 0), (−1, 0)

−2

iv x = 1

d i Concave up

−3

ii −(x + 4) (x − 6), (−4, 0), (6, 0)

−2

−4

g i Concave down

−1

iii y = 3

y −2

v (4, −1)

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

v (1, −9)

−1

−1 −2 −3 −4 −5 −6 −7 −8 −9

x 1

j i Concave down

ii −(x + 2) (x − 4), (−2, 0), (4, 0) iii y = 8

iv x = 1

9 8 7 6 5 4 3 2 1 −2

v (1, 9)

−4 −3 −2 −1 −1

−3

x 1

b i x = −3

ii (−3, −2)

iv x = −1 4

v (−1, −4)

−9−8 −7 −6−5−4 −3 −2 −1 −1 −2 −3 −4

2 1 −4 −3 −2 −1 −1

x 1

−2

c i

−3 −4

−4 −3 −2 −1−1

10 9 8 7 6 5 4 3 2 1 −4

16 a i x = 2

−3

−2

−1

ii (2, −3)

x 1

−2 −3 −4 −5 −6 −7

x 1 2 3 4

iii y = 1 4 3 2 1

l i Concave up ii (2x + 3) (x + 3), (−3, 0),

iii y = 7 9 8 7 6 5 4 3 2 1

k i Concave up

iii y = 9

−4

−1

−2

ii (x + 3) (x − 1), (−3, 0), (1, 0) iii y = −3

d i x = −4 iv x

−1

iii y = 1

ii (−4, 5)

iii y = −11 5 y 4 3 2 1

−9 −8 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12

x 1 2

Answers mathspace.co

971

e i x = −3

ii (−3, −11)

2 1

−7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 −14

f i x=5 iv

4 2

ii (5, 2)

−1 −2 −4 −6 −8 −10

h i x = −1

iii y = −23

17 For all forms, to find the y-intercept, we can substitute x = 0 and evaluate fully using the order of operations. It’s easiest to find the y-intercept using the standard form y = ax2 + bx + c since when x = 0, ax2 and bx become 0. This leaves just y = c and so the y-intercept is (0, c). 18 a False. The turning point of a quadratic graph can be located at any point in the Cartesian plane, not necessarily at the origin. It depends on the values of the coefficients in the quadratic equation. b False. If the coefficient of the x2 term is negative, the graph will open downwards. The sign of the coefficient determines the direction of the graph.

x 1 2 3 4 5 6 7 8 9 10 11

c True. All quadratic graphs have a line of symmetry that passes through the turning point. The line of symmetry divides the graph into two symmetric halves. d True. The constant term in the quadratic equation represents the y-intercept of the graph. It is the value of y when x equals zero.

g i x=4 8 6 4 2

1 2 3 4

Reasoning

−1 −2 −4 −6 −8 −10 −12 −14 −16 −18 −20 −22

iii y = −2 y

ii (4, 7)

iii y = −9

e True. The roots of a quadratic equation are the values of x where the graph intersects the x-axis. They can be found by locating the x-intercepts of the quadratic graph.

x 1 2 3 4 5 6 7 8 9

ii (−1, −12)

19 a a > 0 means the curve will be concave up and so has a minimum turning point, k < 0 means the turning point will be below the x-axis, so the curve will cut the x-axis at two points. h can take any value. b a < 0 means the curve will be concave down and so has a maximum turning point, k = 0 means the turning point will be on the x-axis, so the curve will touch the x-axis at one point. h can take any value.

iii y = −11

y 2 −4 −3 −2 −1 −2

x 1

c a > 0 means the curve will be concave up and so has a minimum turning point, k > 0 means the turning point will be above the x-axis, so the curve will not cut the x-axis at all. h can take any value.

−4 −6 −8 −10 −12

20 a (x − 2)2 + 5

b (x − 4)2 − 2

c The equation would become (x − (2 + a))2 + (5 + b). The turning point would move from (2, 5) to (2 + a, 5 + b)

972

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

21 a They have the same x-intercepts, both are concave up, and they share an axis of symmetry. b The y-intercept and turning points are different. 22 a y = ax2 − 2ahx + ah2 + k

b y = ax2 + (−2ah)x + (ah2 + k) so b = −2ah

c y = ax2 + (−2ah)x + (ah2 + k) so c = ah2 + k d Vertex form and standard form are equivalent quadratic equations in different forms. The equations give us different information more efficiently. We can find the vertex, (h, k), efficiently using vertex form. We can find the y-intercept, c, efficiently by plugging in 0 for x and evaluating the corresponding output using standard form. We can also find the axis of symmetry,

If the shared intercept is different to the turning point, then the three quadratics are the same. If the shared intercept is also the turning point, however, then there can still be differences in their shapes. For example, the curves represented by y = (x − 1)2, y = 2(x − 1)2 and y = −1(x − 1)2 share a single x-intercept of (1, 0) which is also their turning point, but they have different steepness and directions of their parabolas. They also have different y-intercepts. Problem-solving 25 a

9 16 21 24 25 24 21 16 9

b 25

and use it to ﬁnd the y-value of the vertex.

, 2

where a and b are the coeﬃcients of the x and x terms, respectively. 24 a If two quadratic curves have the same intercepts, it means they both pass through the same points on the x-axis and y-axis. If these are three different points, then the two quadratics are the same. If they have less than three intercepts, however, there can still be differences in their shapes, such as the steepness or direction of their parabolas. For example, the curves represented by y = x2 and y = 2x2 both have a single intercept at (0, 0), but the latter is a steeper parabola.

9 10 0

15 10

23 The turning point of a quadratic graph is the point where the graph reaches its minimum or maximum value. The x-coordinate of the turning point can be found using the formula

, in standard form

In both forms, the leading coeﬃcient, a, is the vertical stretch/compression factor and tells us if the parabola will open up or down.

5 t 1

2 3 4 5 6 7 8 9

c 24.75 m 26 a

500 450 400 350 300 250 200 150 100 50

d 25 m

s 5 10 15 20 25 30 35 40

b s = 19

c 524

27 a 6 m

b 3m

c x = −4

b If three quadratic curves have the same turning point, it means they all have the same vertex (the highest or lowest point of the parabola). Additionally, if they share one common intercept, it means they all pass through the same point on either the x-axis or y-axis.

5 4 3 2 1 x −5

−4

−3

−2

−1

Answers mathspace.co

973

Fluency

28 29 a Elise should expect to pay $900

6 a x = 0, x = 9

b m = 0, m = 8

c c = 0, c =

d y = 6, y = −11

e k = 3, k = 5

g x = 5, x = −5

b The price would decrease by $393.75 c

1000 900 800 700 600 500 400 300 200 100

cost ( y)

7 a x = ±8

metre (x) 2

d The decrease in price per square metre makes the parabola wider. The x and y-intercept, domain and range, turning point, axis of symmetry, end behaviour, and intervals of the domain for where the curve is positive/ negative and increasing/decreasing stay the same.

e x = −4, −1

f x = −5, −2

g x = 5, 3

h x = 6, −4

i x = 5, −4

j x = −5, 2

k x = 4, 10

l x=7 d x=5

e x = −2, 9

f x = 7, 8

g x = −2, 4

h x = −12, 11

9 a m = 0, 14

b m = 5, −4

c x = 9, −9

d m = 0, −2

e m = 14, 13

f m = 0, 10

g n = −4, −2

l x = 12

10 a x = 6, −8

5.03 Solve by factorisation

b x = −3, 4

c x = −5, −1

15 = a(8 − 4)(8 − 12)

b x = 2, −10

c x = 10, −10

e x=8

b a = −2, b = −9 or a = −9, b = −2

i y = 7, 1

j y = −1, 3

c a = 4, b = 4

Understanding 1 A and D 2 a a = 3, b = 2 or a = 2, b = 3

d a = −3, b = −3 e a = 1, b = −2 or a = −2, b = 1

11 a i x = −1, 6

f a = −3, b = −4 or a = −4, b = −3

iii y = −(x − 6) (x + 1) b i x = 2, 6

g a = 3, b = −3 or a = −3, b = 3 3 a 24 2

b 11

ii (0, 6) ii (0, 12)

iii y = (x − 2) (x − 6)

h a = 5, b = −3 or a = −3, b = 5 c 8 and 3

d x + 11x + 24 = (x + 8) (x + 3) 4 (m + 5) (m + 9)

c i x=4

ii (0, −16)

d i x = −5, 0

ii (0, 0)

iii y = −(x − 4)2 iii y = x(x + 5)

5 a x = −2, 3

b x = −4, 1

c x = −3, 3

d x = −1, 2

974

d x = 0, 8

8 a x = 0, 16

30 Since the turning point lies on the axis of symmetry, we know that 8 is the average of 12 and m. So then, we must have m = 4. If we substitute the coordinates of the turning point into the equation, we get: If we solve this equation we get

b x = ±3

c x = 0, −12

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning 2

Problem-solving 17 a x = 4, −7

b x = 8, −3

b k = 25, 14, 11, 10, −10, −11, −14, −25

c x = 7, −4

d x = 9, −5

c x = (−24, −1), (−12, −2), (−8, −3), (−6, −4), (4, 6), (3, 8), (2, 12), (1, 24)

h x = −7, 3

12 a x + 10x + 24 = (x + 4) (x + 6)

13 a When we factorise as (x − a) (x − b), the a and the b must be a factorisation of 28. So, as a pair we could have 28, 1, 14, 2, 7, 4. k is the sum of these factors, so the highest value is 28 + 1 = 29. b Using the value of k found in part (a) gives x2 + 29x + 28 = 0, with solutions x = −28, −1. The difference between the solutions is 27. 14 a To find the integer values of c that allow the expression to be factorised, we need to find all pairs of factors of c that have a sum of 4 (the coefficient of the x term). The possible pairs are (1, 3) and (2, 2), which correspond to the values of c = 1 × 3 = 3 and c = 2 × 2 = 4.

18 x = ±4 19 m = 16, n = 6 20 a t = 3, 9

22 a (x − 3) (x + 3) = y

16 When we have (ax + b) (cx + d) = 0, we know that zero can be produced by either (ax + b) = 0 or (cx + d) = 0. If (ax + b) (cx + d) is equal to a non-zero number, there are inﬁnite possibilities for two things multiplied together to give that number. We have to do further work to get a factorised equation equal to zero to get the correct values.

6 4 2 −8 −6 −4 −2 −2

x 2

−4 −6 −8

b (3x + 5) (x − 2) = y

d If c is negative, the y-intercept would be below the x-axis. Since the curve is concave up, or U-shaped. It must cross the x-axis. 15 If it can be factorised, there will be one or two x-intercepts. If it factorises to the form y = a(x − p)2, then there is one x-intercept, since there is only one value (x = p), that can satisfy the equation when y = 0. If it factorises to the form y = a(x − p) (x − q), then there are two x-intercepts, since x = p or q satisfy the equation when y = 0.

c 6 hours

21 t = 5 s

b If c = 4 then the equation is x2 + 4x + 4 = 0, leading to x = −2, −2. If c = 3 then the equation is x2 + 4x + 3 = 0, leading to x = −1, −3. c If c is negative, then we could use two integers, one positive and one negative, such that the sum of the two numbers is 4 to factor x2 + 4x + c. For example, 5 and −1, 6 and −2, 7 and −3 and so on. There are inﬁnite possibilities.

b 3 hours

4 y 3 2 1 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

x 1

c (5x − 4)2 = y

18 16 14 12 10 8 6 4 2

x 2

Answers mathspace.co

975

7 a x = −2, x = −16

d −(x + 2) (x − 7) = y 20 15

b x = 4, x = 2

c x = 8, x = 1

8 a x = 8, x = −2

9 a

5 x −8 −6 −4 −2

−5

e 3(x + 10) (x + 1) = y

Reasoning

20 10 −12 −10 −8 −6 −4 −2 −10

x 2

10 Halve the coeﬃcient of x to get

−20

square it to get

−30 −40

Understanding 1 a Error c Correct 2 a 25

b Correct d Error b 16

3 9 4 Fluency 5 a (x − 1)2 + 45 c (x − 7)2 + 7

b (x + 1)2 + 1

d (x − 3)2 − 4

h (x − 1)2 + 3

6 a x = −4, x = −8

c x = 1, x = 5

976

which would need to be

added to x2 + bx to make it a perfect square trinomial.

−50

5.04 Solve by completing the square

and then

11 We can use square roots to solve any quadratic equation of the form a (x − h)2 = k. We can solve these equations by taking the square root of both sides. Completing the square is a method of solving that allows us to write a quadratic equation as a perfect square, and we can then use the square roots to solve. 12 Some quadratics are not factorable, and if the solutions are not integers, and especially if they are irrational, then drawing the graph of the equation will not give exact solution, but only an approximation. For example: has solutions of

13 Tyler has made two mistakes. On the left hand

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

of Line 2, he should have added

but he has used subtraction sign before the expression. Going from Line 3 to Line 4 he has factorised x2 − 4x − 4 as (x − 4)2, but (x − 4)2 = x2 − 8x + 16.

Corrected solution: Given Complete the square Simplify fractions and exponents Factorise the perfect square part Evaluate and simplify

7 a x = 1.372, −4.372

b x = 0.405, −7.405 f x = 0.390, −3.279

Add 4 to both sides

g x = ±2.646

h x = −3.333, 3

i x = 5.589, −0.089

j x = 0.254, −5.254

16 The turning point form of the equation is . This means the turning point is at (9, 27), and the parabola has end points at (0, 54) and (18, 54), so the width is 18 centimetres, and the height is 27 centimetres. 17 a Two solutions

b Two solutions

c Zero solutions

d One solution

5.05 The quadratic formula Understanding c True

2 a a = 1, b = 7, c = 10

b a = 1, b = −3, c = −4

c a = 2, b = 9, c = 0

d a = 8, b = −3, c = 0

e a = −8, b = 3, c = 0

f a = −2, b = 9, c = 5

g a = 4, b = 3, c = −5

h a = 3, b = −17, c = 9

3 a = −9, c = 0

8 a

Reasoning 9 a x = −4, x = −5

b x = −4, x = −5

10 a Method 1: Add 10 to both sides to get x2 = 25. Then, take the square root of both sides to get x = 5, x = −5. Method 2: Rewrite the equation as x2 − 10 − 15 = 0. Then, factor the equation as (x − 5)(x + 5) = 0. Solve for x to get x = 5, x = −5. Efficiency: Both methods are efficient, but taking the square root is slightly faster. b Method 1: Factor the equation as (x − 6) (x − 1) = 0. Solve for x to get x = 6, x = 1. Method 2: Use the quadratic formula to find , which simplifies to x = 6, x = 1.

4 a = 1, c = −2

Eﬃciency: Factoring is more eﬃcient in this case, as it requires fewer steps.

c Method 1: Divide both sides by 25 to get

Fluency

c x = −4, 1

e x = 0.219, −0.965

6 a x = −2, −3

Take the square root of both sides

15 a 100 m

e False

d x = −0.260, −1.540

14 x = 12

d False

c x = −0.360, −1.390

Problem-solving

b True

Collect like terms

Evaluate 4 + 1 and 4 − 1

1 a True

b x = 2, 3

. Then, take the square root of both sides to get

Answers mathspace.co

977

Method 2: Rewrite the equation as 25y2 − 36 = 0. Then, factor the equation as (5y − 6)(5y + 6) = 0. Solve for y to get

Method 2: Complete the square by adding and subtracting

parentheses, then rewrite the equation as

. Efficiency: Taking the square root is more efficient in this case, as it requires fewer steps. d Method 1: Use the quadratic formula to find , which simplifies to

. Take the square root of both sides to get

and subtracting

h Method 1: Factor the equation as (x + 4) (x + 5) = 0. Solve for x to get x = −5, x = −4. Method 2: Use the quadratic formula to ﬁnd

inside the

, which simpliﬁes to

parentheses, then rewrite the equation as . Take the square root of both sides to get

Eﬃciency: The quadratic formula is more eﬃcient in this case, as completing the square requires more steps.

x = −5, x = −4. Eﬃciency: Factoring is more eﬃcient in this case, as it requires fewer steps. 11 a b The solution of m = 0 must be excluded. The denominator in the quadratic formula is equal to 2a and if m = 0 then 2a = 0 but we cannot divide by zero.

e Method 1: Complete the square by adding and subtracting

inside the

parentheses, then rewrite the equation as (x + 12)2 = 81. Take the square root of both sides to get x = −3, x = −21. Method 2: Use the quadratic formula to find , which simplifies to x = −3, x = −21. Efficiency: Both methods are efficient, but the quadratic formula is slightly faster. f M ethod 1: Factor the equation as x(x − 7) = 0. Solve for x to get x = 0, x = 7. Method 2: Use the quadratic formula to find , which simplifies to x = 0, x = 7.

12 a Negative numbers b i There will be two solutions in this case. Suppose that

, where d > 0.

The formula then gives

g Method 1: Use the quadratic formula to ﬁnd , which simpliﬁes to .

and

which are two different values. ii There will be one solution in this case. Suppose that then gives

. The formula , which is just

iii There will be no solutions in this case, because we cannot take the square root of a negative. 13 The equation has no real solution. Problem-solving

Eﬃciency: Factoring is more eﬃcient in this case, as it requires fewer steps.

14 Example answer: x2 − 6x + 1 = 0 15 a i a = −7, b = 5, c = 10

ii −7x2 + 5x + 10 = 0

b i a = 9, b = 22, c = 3

ii 9x2 + 22x + 3 = 0 16 x = 11.1 17 38 ms−1

978

Eﬃciency: The quadratic formula is more eﬃcient in this case, as completing the square requires more steps.

. Method 2: Complete the square by adding

inside the

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3 b2 − 4ac > 0

18 a 258.025 metres b 9.98 seconds 19 a 27.84 m/s b Ray was driving 10 m/s over the speed limit. 20 a We can use the formula l × w = A where l = 10 + 2x and w = 3 + 2x since there are 2 opposite pathways.

4 a Δ<0

b Δ=0

c Δ>0

d Δ>0

e Δ=0

f Δ<0

g Δ<0

h Δ=0

i Δ>0

Fluency 5 a i a = 1, b = 11, c = 28 ii 9

l×w=A

b i a = 1, b = −5, c = 6

(10 + 2x)(3 + 2x) = 120

30 + 20x + 6x + 4x2 = 120

ii 1

30 + 26x + 4x2 = 120

c i a = 4, b = −7, c = −15

30 + 26x + 4x − 120 = 120 − 120

ii 289

4x2 + 26x − 90 = 0

ii −7

−90 + 26x + 4x2 = 0

d i a = −1, b = −5, c = −8

b 2.5 m 21 a

e i a = 16, b = −24, c = 9 ii 0

Given

f i a = −2, b = 3, c = −2

Rewrite the fraction on the right hand side in expanded form Rewrite in terms of x and simplify Multiply both sides by x Subtract x + 1 from both sides

Use the quadratic formula

Understanding 1 None 2 One x-intercept

ii 16 h i ii 7 6 a i Δ=0

ii 1

b i Δ = −59

ii 0

c i Δ = 16

ii 2

d i Δ=0

ii 1

e i Δ = −23

ii 0

Simplify

f i Δ = 40

ii 2

g i Δ = 100

ii 2

h i Δ = −9

ii 0

i i Δ=1

ii 2

c The positive solution, 1.618. a and b are both lengths, and therefore positive, so their ratio,

5.06 The discriminant

g i

Substitute a = 1, b = −1, c = −1

Evaluate correct to three decimal places

must be positive.

ii −7

j i Δ = 76

ii 2

k i Δ = −59

ii 0

l i Δ = 17

ii 2

7 a i b i c i d i e i f i g i h i

Δ=0 Δ = 256 Δ = 112 Δ=0 Δ = 64 Δ=0 Δ = 65 Δ = 3.75

ii Rational ii Rational ii Irrational ii Rational ii Rational ii Rational ii Irrational ii Irrational

Answers mathspace.co

979

8 a i b i c i d i e i f i g i h i

a>0 a<0 a<0 a>0 a<0 a>0 a<0 a>0

9 a 9 + 8m

ii Δ > 0 ii Δ = 0 ii Δ < 0 ii Δ < 0 ii Δ < 0 ii Δ = 0 ii Δ > 0 ii Δ < 0

13 a No, this is not the only x-intercept of the quadratic. Since the quadratic equation has rational coeﬃcients and the axis of symmetry is at x = 3, the other x-intercept must be a conjugate pair of the given x-intercept. The given x-intercept is , so the other x-intercept must be , which is also units away from the axis of symmetry. b Since the quadratic has two distinct real roots (the x-intercepts), this tells us that the value of the discriminant is greater than 0.

b 16n2 + 8

c 28 − 4p

d 16 + 4p2

e 12m

Reasoning 10 The discriminant is the value under the square root symbol. This means that the properties of square roots determine the possibilities. A negative discriminant means that there will be no real solutions because we cannot take the square root of a negative number. We can take the square root of any positive number, so when the discriminant is positive there are two distinct real solutions because of the ± before the square root sign. More speciﬁcally, if the discriminant is a perfect square, then it will have rational solutions because when we take the square root of a perfect square we get an integer. However, we are dividing by 2a, so the resulting root may only be rational and not an integer. If the discriminant is equal to zero there will be one real solution because when we add and subtract zero, we don’t get different values. 11 The quadratic will have no real solutions when the discriminant is less than zero. This will occur when b2 − 4ac < 0 or b2 < 4ac. This gives us k2 < 144. Taking the square root of both sides gives us −12 < k < 12. 12 The discriminant of the first equation (Δ = −95) is negative, a is positive, so the concave up parabola has no x-intercepts. This means it is entirely above the x-axis. The discriminant of the second equation (Δ = −12) is negative, a is negative, so the concave down parabola has no x-intercepts. This means it is entirely below the x-axis.

14 No, it is not possible for a quadratic equation to have one rational root and one irrational root. If a quadratic equation has rational coefficients, then its irrational roots must come in conjugate pairs. This is because when we multiply the factors corresponding to the conjugate pair of roots, the irrational parts will negate each other out, resulting in rational coefficients for the quadratic equation. If a quadratic equation has one rational root and one irrational root, the coefficients of the equation would not be rational, which contradicts the given condition. Problem-solving 15 a n = ±9 c m = −2, m = 4 16 a k > 26

b k = 27

17 k = 16 18 19 m > −1 20 a i 64 − 8k

ii k = 8

iv k > 8

v k<8

b i 400 −8k iv k > 50 c i 52 − 12k iv 21 a 22 a 0 < p < q

Therefore, the two graphs do not intersect.

980

b k=1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

iii k ≤ 8

ii k = 50 v k < 50 ii

iii

v b m = −7

b x2 + 2(p − q) x + p2 − q2 = 0 c Δ = 8q (q − p)

iii k ≤ 50

d Two real solutions as q > p, therefore the discriminant is positive. e x = 3p

k l

23 a A = x − 5x − 50 b Δ = 225 c Two real solutions as the discriminant is positive. d The length has to be at least 10 m. e In part (c) we determined that there were 2 real solutions. This is true, the mathematical solutions are x = 10 and x = −5. But we cannot have a negative length, so in context of the problem being solved, there is only 1 solution. 2

24 a y = −ax + k

b Δ = 4ak

c Two real solutions, as the discriminant is positive, which means the quadratic equation has two distinct points where the water surface intersects the tank’s parabolic shape.

6 a b c d e f g h i j

5.07 Optional: Further quadratic equations

7 k = 24 8 a = −3 9 a i y=0

Understanding 1 a False

ii x = 0, x = −2

iii x = −1

iv (−1, −1)

b True

d False

c False, if b2 − 4ac is negative, we cannot take the square root.

3 1

2 9

−4 −3 −2 −1

3 D

b i y = −8

Fluency

−1

−2

4 C

5 a x = 0, x = −9

x 1

ii x = −2, x = 2

iii x = 0 b d x = 5, x = 3

iv (0, −8) 2 1

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

y x 1

Answers mathspace.co

981

c i y = −10

ii x = 1, x = 5

iii x = 3 v

8 6 4 2

1100 y 1000 900 800 700 600 500 400 300 200 100

−4 −2 −2 −4 −6 −8 −10

ii x = 4, x = 2

iii x = 3

iv (3, −9) 2 1

−2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

−20 −10 −100 −200 −300 −400

d i y = 72 v

iv (3, 8)

c i (0, 2) ii (2.25, −8.125) iii (0.234, 0), (4.266, 0) iv

x 1

10 20 30 40 50 60

2 3 4 5 6 7

4 2 −4 −2

−2

x 2

−4 −6 −8

d i (0, 29) 10 a i (0, −2)

ii (4.167, −23.083)

iii (1.393, 0), (6.941, 0)

iii (−3.562, 0), (0.562, 0) iv

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

45 40 35 30 25 20 15 10 5 −8 −6 −4 −2−5 −10 −15 −20 −25 −30 −35 −40 −45

x 2

b i (0, 945) ii (31, −16) iii (27, 0), (35, 0)

982

11 a

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Reasoning

u2 − 13u + 36 = (u − 4)(u − 9) Factoring the quadratic equation

12 The factorising steps and the solutions are correct. The solutions are x = 2 and x = 4. x2 − 6x + 8 = (x − 2)(x − 4)

x = 2, 4 x2 + 2x − 8 = 0

x2 + 2x − 8 = (x + 4)(x − 2)

x = −4, 2

Factoring the quadratic equation

u = 4, 9

Finding the solutions for u

x2 = 4, 9

Finding the solutions

Substituting back x2 for u

x = ±2, ±3

Taking the square root of both sides

Given Factoring the quadratic equation Finding the solutions

Verifying the solutions by substitution:

(−4)2 + 2(−4) − 8 = 16 − 8 – 8 Substituting x = −4

16 − 8 − 8 = 0

0=0

There are 4 solutions b There are two solutions. During the solution we arrive at x2 = −4 and x2 = 5. The ﬁrst of these has no real solutions and the second has two. Problem-solving 18 9 cm

Simplifying the expression

19 25 cm

Veriﬁed

20 a t = 2, t = 3

(2) + 2(2) − 8 = 4 + 4 – 8

Substituting x=2

4+4−8=0

Simplifying the expression

0=0

Veriﬁed

14 Yes, they will arrive at the same solutions, provided no mistakes are made, because all three methods are valid ways to solve quadratic equations. The solutions are x = 2 and x = 5.

b 5 seconds c No, we can factor or complete the square to ﬁnd that the turning point of this parabola would be at (2.5, 31.25), since the leading coeﬃcient is negative, it has a maximum turning point, so h ≤ 31.25. 21 a c

15 The values of x given in Step 3 should be the values that will make each bracket evaluate to 0. So the answers should be −9, −6.

22 24

16 The error in Tom’s solution is in Step 1, when substituting −2 into b2 − 4ac, Tom needed to appreciate that there is a difference between −22 = −4 and (−2)2 = 4. The latter version is correct.

24 a

17 a

x4 − 13x2 + 36 = 0

Given

u = x2

Substituting u for x2

u2 − 13u + 36 = 0

Replacing x2 with u

seconds

second

23 t = 7, t = 11 b x = 18, x = 21

c 36.83 cm

Chapter 5 review 1 a Concave up

b Concave down

c Concave up

d Concave down

2 a x = ±9 c n = −4, 8 3 a Maximum

b x = ±10 d x = 0, 8 b t = 6, t = 18

4 a i D ilated by a factor of 2 and horizontal shift 4 units to the right

Answers mathspace.co

983

ii (4, 0) iv

11 10 9 8 7 6 5 4 3 2 1

iii (0, 32)

7 a x = 12, 0

b x = 1, −10

c x = 12

8 a i x = −2, 4

ii y = −2(x + 2) (x − 4)

b i x = 0, 7

ii y = x (x − 7)

9 a x = 4, x = 10

b x=3

c x 1 2 3 4 5 6 7 8 9 10 11

b i R eﬂection about the x-axis, horizontal shift 1 unit to the left and vertical shift 7 units up. ii (−1, 7)

11 a

iii (0, 6)

11 10 9 8 7 6 5 4 3 2 1

5 a (0, −3)

12 a −x2 − 8x − 16 = 0 b The equation has 1 real solution.

−9−8−7−6−5−4−3−2 −1 −1 −2 −3 −4

1 2 3 4 5 6 7 8 9

b (1, 0), (−3, 0)

c x=1 e

10 a 144 m

d (−1, −4)

13 a i Δ = 16

ii 2

b i Δ=0

ii 1

c i Δ = −9

ii 0

d i Δ = 17

ii 2

14 a i Δ = 0 b i Δ = −144

ii No real roots

c i Δ = 76

ii Irrational roots

d i Δ = 3.75

ii Irrational roots

15 a i a > 0

ii Δ = 0

b i a<0

ii Δ > 0

c i a<0

ii Δ < 0

d i a>0

ii Δ > 0

ii Rational roots

3 2 1

−5 −4 −3 −2 −1 −1

16 17 a (0, 24)

−2

c x = −3

−3

d (−3, −3)

−4

y 25

6 a y = (x + 2)2 − 5 c

4 3 2 1 −6 −5 −4 −3 −2 −1

−1 −2 −3 −4 −5

984

b (−4, 0), (−2, 0)

b (−2, −5)

15 10 5 x

x 1

−7 −6 −5 −4 −3 −2 −1

18 10 19 a 3 seconds

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b No

6.01 Solve graphically

d i

Understanding 1 The x- and y-coordinates at the point of intersection satisfy both equations.

x −10−8 −6 −4 −2 2 4 6 8 10 −2 −4x + 2y = −8 2x − 4y = −8 −4 −6 −8 −10

2 The lines would be parallel and never intersect. 3 a 1

b 0

e 0

f 1

c 1

d Inﬁnite

Fluency

10 8 6 4 2

ii (4, 4)

4 a (5, −1)

b (−6, 4)

c (2, −4)

e (6, −3)

f (−4, −2)

5 a x = 6, y = 4

d (3, −6)

e i

b x = 0, y = −2

c x = 4, y = 0 6 a i

10 8 6 4 2

y y = 4x − 4 y = 8 − 2x

x 2 4 6 8 10

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

y 10 8 y = 3x + 6 6 4 2 y = −x + 2 x −10−8 −6 −4 −2 2 4 6 8 10 −2 −4 −6 −8 −10

ii (−1, 3) f i

ii (2, 4) b i

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

ii (−2, −1) g i

y 6

y = −x − 9

y = −x + 3

y = 4x − 3

4 2

ii (−9, 0) c i

x 2 4 6 8 10

−10−8 −6 −4 −2 −2 y = −1 −4 −6 −8 −10

y y=x+9

x 2 4 6 8 10

10 y = −x − 3 8 6 4 2

10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 y=x+3 −8 −10

−4 −2

x 2

−2

y = 4 − 3x

−4 −6 x 2 4 6 8 10

ii (1, 1)

ii (0, 3)

Answers mathspace.co

985

h i

l i

y 8

y = 3x + 2

y = 7 − 2x

2 x

−4 −2

−10−8 −6 −4 −2 −2

−2

−8

y = 4 − 0.5x 10 8 6 4 2

ii (1, 2)

7 a x

−10−8 −6 −4 −2 −2 −4 −6 −8 −10 y = 1.5x

2 4 6 8 10

500 400 300 y = 50x + 250 200 100 −5 −4 −3 −2 −1 −100 −200 −300 −400 −500

ii (2, 3) j i

−10−8 −6 −4 −2 −2 −4 −6 −8 y = 2.5x − 4 −10

y = 100x + 50 x 1 2 3 4 5

b (4, 450)

c Renting either studio A or B for 4 hours costs the same amount, $450.

16 y = 12 − 1.5x 14 12 10 8 6 4 2

8 a

x 2 4 6 8 10

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

y y = 2x + 10

y = 4x

x 1 2 3 4 5 6 7 8 9 10 11

b (5, 20)

ii (4, 6) 8 6

y y=

c Non-members and members pay the same amount for 5 classes, $20.

3 x−1 2

9 a (775, 2325)

4 2

−6 −4 −2 2 4 6 −2 y = 1 x − 2 2 −4 −6 −8

ii (-1, -2.5)

986

2 4 6 8 10

y = 5 x − 1 −6 2 2

ii (1, 5)

k i

y = 1 x+3 2 2

−4

i i

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b After making 775 items, the company’s revenue will equal their costs. This means they will not make a proﬁt and this is the break-even point.

10 a

1100 1000 900 800 700 600 500 400 300 200 100

b The value of m ≠ 5. If m = 5 and c = 3 then the second line is the same as the ﬁrst and there is an inﬁnite number of solutions.

Dollars

18 No, the lines have the same gradient, so they are parallel and will never intersect.

19 Non-integer solutions represent fractional quantities or parts of a product, which may not be practical in real-life scenarios. In such cases, the company might need to round to the nearest whole numbers. This may not result in the ideal way of distributing the goods, but it ensures that the goods are distributed in whole units, which is more practical.

n 50

100 150 200 250

b (200, 800) c This is the break-even point. For 200 toys, both the cost and revenue are $800. 11 a

30000 27000 24000 21000 18000 15000 12000 9000 6000 3000

Problem-solving 20 a y = x + 3, 2x + 2y = 18

y = 10 000 + 772.60x

2x + 2y = 18 10 8 6 4 2

y = 1200x x

−8 −6 −4 −2 −2 −4 −6 −8

2 4 6 8 10 12 14 16 18 20 22 24 26 28

b (23.4, 28 076.7) c After about 23 weeks, Laura will be spending more money than she has in her bank account. 12 a Yes

b No

c No

d Yes

13 a No

b No

c Yes

d Yes

Reasoning 14 No, it’s not possible. Simultaneous linear equations represent two lines. Two lines can either intersect at a single point (one solution), coincide resulting in infinite solutions, or be parallel (no solution). They can’t intersect at exactly two points. 15 These equations actually represent the same line, just written in different forms. This means there are infinite solutions. Every point on one line satisfies both equations. 16 No, because there are an infinite number of lines that go through that point. The gradient could be any real number. 17 a To intersect at (0, 3) the line must have a y-intercept of 3, therefore c = 3.

y y=x+3

x 2 4 6 8 10

c 6 m and 3 m 21 a Let x = Mohamad’s age and y = Rochelle’s age Equation 1: y + x = 21 Equation 2: y = 4x + 11 b

22 20 18 16 14 12 10 8 6 4 2

y = 4x + 11

y + x = 21

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

c Rochelle is 19 years old and Mohamad is 2 years old. 22 a Let x = the number of parts Elvia completes and y = the number of parts Kang completes. Equation 1: x + y = 9 Equation 2: y = 2x − 6

Answers mathspace.co

987

c x < 1 and y < 1 and y ≥ −x − 5

10 y 9 8 7 6 5 4 3 2 1 −2 −1 −1 −2 −3 −4

d y + 3x > 0, y + 3x ≤ 3

x+y=9

6 a

y = 2x − 6

3 2

1 6x + 3y = 0

1 2 3 4 5 6 7 8 9 10

b Yes c

Maria has 15 kgs of rice and cooks 5 kgs of rice every 7 days as represented by the purple line in the graph. Arjun has 12 kgs of rice and cooks 1 kg of rice every 2 days as represented by the green line in the graph.

7 6 5 4 3 2 1

7 a i 5

ii 3

iii

6 5 4 3 2 1 −4−3−2 − 4 −3 −2 − −1 11 −1 −

Understanding d No

e No

3 C b True

c True

Fluency 5 a x < 1, y ≤ 3 b y ≥ 2x − 3, y ≤ 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1 2 3 4 5 6 7 8

b i 0.714 iii

2 (1, 3), (1, 0), (4, 0)

−2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6

6.02 Simultaneous inequalities

c No

x 1 2 3 4 5 6 7

−2 −3 −4

Since the two lines do not intersect in the first quadrant, they will never reach the same height on the same day.

b Yes

−4 −3 −2 −1 −1

The solution (14, 5) means that after 14 days, Maria and Arjun have the same amount of rice remaining. 24 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.

2x − y = 4

−3 −4

Let x be the number of days and y be the weight of the rice in kilograms.

988

1 2 3 4 5 6 7

−2

23 Answers will vary. For example:

4 a False

−3 −2 −1 −1

c Kang did 4 parts of the assignment and Elvia did 5 parts.

1 a No

ii −1.857 6 5 4 3 2 1

−6−5−4−3−2 −1 −1

−2 −3 −4 −5 −6

x 1 2 3 4 5 6

c i No point of intersection

9 a

ii No point of intersection iii

7 6 5 4 3 2 1

2 1 −4 −3 −2 −1 −1

iii

6 5 4 3 2 1

10 a x ≥ 17, x + y ≥ 37 b

x 1

c i No

8 12 16 20 24 28 32 36

ii Yes

iii No

iv Yes

d 13

11 a x ≥ 10, y ≥ 7, x + y ≤ 23 b

4 2 2

−4 −6

y=x−5

−8

b Yes 7 y 6 5 4 3 2 1

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

y=0 −8 −6 −4 −2 −2

d 25 units2

36 32 28 24 20 16 12 8 4

−7−6−5−4−3−2−1 5− −4−3−2 −− 11 −1 −2 − 2 −3 − 3 −4 − 4 −5 −6 −7

−4

−5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

−3

ii 0

y = −x − 5

b 6 units2

d i 2

x 1

−2 x

−4 −3 −2 −1 −1 −2 −3 −4

8 a

x 2 4 6 8 10 12 14 16 18 2022

c 9 x

1 2 3 4 5 6 7

12 a x ≥ 0, y ≥ 0, 100x + 150y ≤ 3900 and y ≤ 32 − x b

y 30 25 20 15 10 5

x 5

10 15 20 25 30 35

c B, D

Answers mathspace.co

989

13 a 0 ≤ 5x + 8y ≤ 40

b 0 ≤ 10x + 4y ≤ 40

14 a 3x + 2y ≤ 18

b 4.5x + 1.5y ≤ 18

e i Yes

c x ≥ 0, y ≥ 0 d

12 y 11 10 9 8 7 6 5 4 3 2 1

b 6x + 1.5y ≤ 18

4 3 2

12 y 11 10 9 8 7 6 5 4 3 2 1

x 1

Reasoning 16 a No solutions. The regions are on opposite sides of the line y = 4.

5 4 3 2 1

c No solutions. The regions are on opposite sides of the boundary line.

−5 −4 −3 3− −2 2− −11 −11 − 2 −2 − 3 −3 − 4 −4 − 5 −5

d Inﬁnitely many solutions. The regions have the inﬁnite set of points on the boundary line in common. 17 a x ≥ 0, x ≤ 0.35

b y ≥ 0.2

c x+y≤1

20 a x ≥ 0, y ≥ 0

x 1 2 3 4 5

b x + y ≤ 200

c The number of ﬂuorescent light bulbs produced must be at least half as many as the number of incandescent light bulbs.

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

19 Yes, it is possible for a coordinate point to solve one part of the linear inequality pair but not the other. In the diagram, the dashed lines represent the two inequalities y = x + 2 and y = 3x + 2. The darker shaded region is the intersection of the regions that satisfy the inequalities y < x + 2 and y < 3x + 2. There are points like (−2, −2) that solves y < x + 2 but not y < 3x + 2.

b Inﬁnitely many solutions. The regions have the inﬁnite set of points on the line x = −2 in common.

x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

990

b 3x + 1.5y ≤ 12

x 2

c x ≥ 0, y ≥ 0

iv No

c No, x and y cannot have negative values as they are measuring number of pallets made in a day, of which the smallest value would be 0. This gives the inequalities x ≥ 0, y ≥ 0.

15 a 4x + 2y ≤ 18

iii Yes

18 a 2x + 4y ≤ 12

ii No

f Yes, because this would satisfy all of the inequalities and the point (0, 1) lies within the region.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving

25 a x ≥ 5, y ≥ x + 8, x + y ≤ 26, and x + y ≥ 20 b

21 a y ≥ 35, y ≤ 50 b

55 50 45 40 35 30 25 20 15 10 5

21 18 15 12 9 6 3

x 3 6 9 12 15 18 21 24 27

x 2

c Yes

c (5, 15), (5, 21), (6, 14), (9, 17)

d False

d 9 sessions

22 a x ≥ 0, y ≥ 0, y ≤ 12 − x

26 a x ≥ 5, y ≥ 0, y ≤ 2x, x + y ≤ 18, and x + y ≥ 13 b

y 12

10 8

x 2

c i Yes

ii No

x 2 4 6 8 10 12 14 16 18

iii No

iv Yes

d 6 cm × 6 cm

c (5, 8), (5, 10), (6, 12), (18, 0), (13, 0) d 12 lessons

23 a x ≥ 0, y ≥ 0, x + y ≤ 29, 15x + 18y ≥ 270 b

y 30

6.03 Solve using substitution Understanding

1 a Answer varies:

20 15 10 5

x 5

c 10 hours 24 a x ≥ 0, y ≥ 0 c 2x + 7y ≤ 200

20 25 30

d Yes b 55x + 65y ≤ 800

Tomi will get y = 2(6 − y) − 3, which simplifies to y = 12 − 2y − 3, then finally 3y = 9. This approach is efficient as it results in a simple equation where we can solve for y directly. b Answer varies: Carla will get x + 2x − 3 = 6, which simplifies to 3x − 3 = 6, then finally 3x = 9. This approach is also efficient because it results in a simple equation where we can solve for x directly. c Answer varies: Regie will get 6 − x = 2x − 3, which simplifies to 3x = 9. This approach is efficient because it results in a simple equation where we can solve for x directly.

Answers mathspace.co

991

d Answer varies:

Reasoning

Junard will get

, which simpliﬁes

to y + 3 + 2y = 12, then finally 3y = 9. This approach is slightly less efficient because the equation involves a fractional coefficient, which can be a bit harder to work with. Once the fraction is handled, the resulting equation is also simple. 2 Yes, it is a solution. 3 a True

b True

c False

d True

e False 4 No, it is not a solution. Fluency 5 a x = −8, y = 8

b x = −5, y = 3

c x = −8, y = −3

d x = −6, y = −3

e x = −3, y = 5

f x = 6, y = 4

g x = 1, y = 6

h x = 8, y = 4

i x = −5, y = −8

j x = 8, y = 1

l x = 12, y = 8

12 If the process of substitution leads to a false statement, it indicates that the simultaneous equations have no solutions. In other words, the lines represented by the equations do not intersect at any point. 13 If you get a true statement, it indicates that the simultaneous equations have inﬁnitely many solutions. This could mean that the equations represent the same line. 14 Example explanation: In this case, it may be easier to make y the subject in the ﬁrst equation and then substitute it into the second equation. This would leave us with an equation in terms of x, so we could then solve for x. We would then substitute the value of x and solve for y using either equation. Finally, we could substitute the values of x and y into both original equations to check our answer. This would avoid fractions and unnecessary rearranging. 15 No. If we substitute one equation into the other, we do not end up with a true statement: 3(x + 5) = 3x + 5 3x + 15 = 3x + 5 15 ≠ 5

6 a b Inﬁnitely many solutions 7 a x = 7, y = 4

b x = −6, y = 5

c x = 6, y = −5

d x = 2, y = 3

e f = 0.4, g = 1.2

f a = 1.5, b = 2.5

Since this is always false, the two equations would not have any solutions. 16 a x = 4, y = 2

8 The son’s current age is 4 and the mother’s current age is 40. 9 a a = $1.68 and k = $2.24 b The cost of one apple is $1.68 and the cost of one kiwi fruit is $2.24. 10 a x = 250 000 and y = 810 000 b In 2008, there were 250 000 jobs created and in 2013, there were 810 000 jobs created. 11 a x = 8, y = 14 b In the group, there are 8 adults and 14 children.

b Yes c Yes, this is the only solution to the simultaneous equations. The solution (x = 4, y = 2) is the point of intersection for all three lines represented by these equations. As these are linear equations, and each pair of them intersects at this point, there can’t be another solution for all three equations simultaneously. This means the simultaneous equations has exactly one solution. d

4 3 2 1 −2 −1 −1 −2 −3 −4

992

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1 2 3 4 5 6 7 8 9

Problem-solving

8 a x = 4, y = 5

17 a The fixed charge is $5 and the charge per km is $10. b $255 18 Each racket is $50 and each ball is $5 19 35 sausages and 52 soft drinks. 20 a (0, 0), (−5, 25)

b (8, 24), (−3, 24)

c (2, 14)

d (−5, 3)

e (3, −2)

g (−1, 1), (2, −8)

h (1, 0),

b x = 3, y = −1

c x = 3, y = 1

d x = 5, y = 15

e x = −3, y = 1

f x = 2, y = 2

g x = 5, y = 4

h x = 6, y = 4

9 a There are 17 four-passenger cabs and −2 six-passenger cabs. b The solution is not valid because it is impossible to have a negative value of a passenger cab. 10 a x = $2, y = $4 b One pen costs $2 and one ruler costs $4. 11 a R = 5, C = 4

b Marjan purchased 5 rhododendrons and 4 chrysanthemums.

6.04 Solve using elimination

Reasoning

Understanding 1 C 2 Multiply the ﬁrst equation by 7 and the second by 5. 3 a True

b True

c False

Fluency 4 a 7

b x = −3

5 a Multiply the ﬁrst equation by 5 and the second equation by 7. b x = −1, y = −3 6 a i 3x + 11y = −48 and −6x + 21y = −76 ii

and y = −4

b i 2x + 3y = 5 5x + 4y = 2 ii x = −2 and y = 3 c i 4x + 3y = 20 and 8x − 3y = 4 ii x = 2, y = 4 d i 6x + 5y = 240 and x + 3y = 66 ii x = 30, y = 12 7 a

b x = 2, y = 8

c x = 2, y = −9

d x = 8, y = 3

e x = 40, y = −2

f x = −8, y = −2

12 You would add the two equations. Because the y terms have opposite coeﬃcients, they would negate each other out, leaving you with an equation in x to solve. Once you’ve found the x-value, you can substitute it into either equation to find the y-value. 13 If you add the two equations, both variables would be eliminated, resulting in a true statement 0 = 0. This means there are infinitely many solutions. Each original equation represents a line, and these lines coincide (i.e., are the same line), hence there are infinitely many solutions. 14 Answer may look like: If the coefficients of one variable are the same or opposite in both equations, it’s often easier to use the elimination method because you can add or subtract the equations to eliminate that variable right away. With substitution, you would first have to solve one equation for one variable, which could be more work. 15 a When Elias substituted y to solve for x, he only found one value for x because he forgot that −1 could also be a solution to x2 = 1. b (−1, −3)

Answers mathspace.co

993

b i Elimination

Problem-solving

ii One variable already has the same coeﬃcient.

16 a x + y = 30 6x − 3y = 90

c i Elimination

b x = 20, y = 10

ii The coeﬃcients of the variables are multiples of each other.

17 x = 10, y = 7

d i Substitution

18 x = 5, y = 5 19 a 2016

b 68%

20 a (−1, 3) and (2, 6) b (−3, −4) and (2, 1)

ii One equation is already solved for a variable. 3 a False b True

c (−1, 1) and (1, 1)

c True

d (−8, −6) and (8, −6) e (−4, 3) and (4, 3) f (−6, 3), (0, 3) and (−3, −6)

6.05 Applications of simultaneous equations Understanding 1 a Answers may vary. For example, one could use technology to ﬁnd the accurate intersection point. Alternatively, one could use the substitution method as the ﬁrst equation is already in the form y = mx + c.

Fluency 4 a x + y = 6 y = 2x b $2 000 000 c $4 000 000 5 a x + y = 184 x − y = 22 b x = 103 c y = 81 6 a

b Answers may vary. For example, one can simply read the solution from the graph. Alternatively, one could use the substitution method as both equations are already in the form y = mx + c.

Call time (in minutes)

c Answers may vary. For example, one could use the substitution method as Equation 1 is already in the form y = mx + c. Alternatively, one could use the elimination method by multiplying Equation 1 by 2 and then adding the two equations to eliminate the y variable. d Answers may vary. For example, one could use the elimination method by multiplying Equation 2 by 3 and then adding the two equations to eliminate the y variable. Alternatively, one could use the substitution method by solving Equation 2 for x or y and then substituting it into Equation 1.

Total cost for GO MEDIUM plan

44.00

47.00

48.00

50.00

52.00

53.00

56.00

b 40 minutes 7 a

x + y = 70 5x + 10y = 385

b x = 63, y = 7 8 a T = 50n

b T = 25n + 1200

c 48 months 9 There are 10 part-time employees and 26 full-time employees. 10 a x + y = 34 x=y+6

2 a i Substitution

b 20

ii An equation is already solved for a variable.

c 14

994

Total cost for GO SMALL plan

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

18 a 190 student tickets and 160 general admission tickets

11 a

x + y = 4000 0.07x + 0.08y = 303

b y = $2300 12 a

x = y + 12 x + y = 28

b x = 20 units, y = 8 units 13 a x = 6, y = 4

b 396 units

c 108 units 14 0.9 km x + y = 16 000 0.14x − 0.06y = 700

b Answers may vary.

b x = $8300, y = $7700 c Qin invested $8300 in stock A, and $7700 in stock B. 4500 4000 3500

y Machine

3000

By choosing the substitution method, we can rewrite the ﬁrst equation as y = 38.49 − 2x and substitute this into the second equation, 5x + 3y = 99.22. This is an eﬃcient method to solve for x and once x is known, it can be substituted into either original equation to solve for y.

People

2500

Problem-solving

2000 1500 1000 500

x 10

b x = 60

By choosing the elimination method, we can multiply the equation x + y = 350 by 3 and combine it with the equation 3x + 5.5y = 1450 by means of subtraction. This is an eﬃcient method to solve for y and once y is known, it can be substituted into either original equation to solve for x. 19 a Oxtail stew costs $16.25 and sweet potato pie costs $5.99.

15 a

16 a

b Answers may vary.

c x = $1700

20 30 40 50 60

c x > 60

d 0 < x < 60

Reasoning 17 a C = 10 000 + 15x, R = 12x b No, Barry cannot break-even with this plan because the cost to make each meal is higher than the revenue he is earning from selling the meals. c Answers may vary. Under the current plan, Barry’s restaurant would not be successful because it cannot reach a break-even point or make a profit. Every meal sold would actually increase his losses due to the negative contribution margin. To turn this situation around, Barry might consider increasing the price he charges customers for the meals or finding ways to decrease the cost of producing each meal. This would help him cover his fixed costs and eventually reach a break-even point where the restaurant becomes more sustainable.

20 This equation cannot be solved algebraically, so we can use technology to ﬁnd the approximate solution: x ≈ 0.458, y ≈ 1.373. 21 (1.29, 0.29) and (2.71, 1.71) 22 a

23 24 (−5.11, 7.30) and (22.93, 16.64)

Chapter 6 review 1 a i

3 2 1 −6 −5 −4 −3 −2 −1 −1

x 1

−2 −3 −4

ii (−4, 3)

Answers mathspace.co

995

b i

6 5 4 3 2 1

y ≤ −2x + 7 y≤5

9 a x

−6 −5−4−3−2 −1 −1 −2 −3 −4 −5 −6

1 2 3 4 5 6

y=1

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

ii (2, −1) 2 a x = −5, y = 2

b x = 0, y = −3

c x = −4, y = −1 3 a (1280, 2500)

5 No, the lines have the same gradient, so they are parallel and will never intersect. 6 a y = x + 2, x + 2y = 22 y x + 2y = 22

10 a 5x + 3y ≤ 24

y=x+2

y=x−4

y = −x − 4

x 1 2 3 4 5 6 7

2 x 2

c Base is 6 m and the other side lengths are 8 m each. 7 a −3

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

7 6 5 4 3 2

b −7 3 2 1

x 1

b 4x + 6y ≤ 24

c x ≥ 0, y ≥ 0

996

x 1 2 3 4 5 6 7

d 25 units2

5 4 3 2 1 −7−6−5−4−3−2 4− −3 −2 − −11 −11 − −2 − 2 −3 −3 −4 4 −5 −6 −7 −8 −9

4 D

b Yes

b After selling 1280 dresses, the boutique’s revenue from sales is equal to their expenses. This is their break-even point, implying any dresses sold beyond this quantity will lead to a proﬁt.

5 4 3 2 1

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3 f (3) represents the output value of the dependent variable when the input value of the independent variable is equal to 3.

11 a x ≥ 0, y ≥ 0, y ≤ 8 − x b

4 g (7) = −6

6 5

5 f (7) = 22

4 3

Fluency

2 1

x 1

c i Yes

ii No

iii No

iv Yes

6 a Yes

b No

c Yes

d No

7 a Yes

b Yes

c No

d Yes

8 a i

d 4 cm × 4 cm

12 a x = −10, y = 0

2 1

b Inﬁnitely many solutions c No solution

x 1 2 3 4

−4 −3 −2 −1 −1

d f = 4, g = 9

−2

13 a (−1, 7) and (−1, 1)

b (1, 6) and (−4, −14)

14 a x = −2, y = 4

c x = 6, y = −15

d x = −14, y = 8.5

15 a (4, 7) and (−3, 0)

b (−4, 8) and (4, 8)

−3 −4

ii No b i

6 5 4 3 2 1

16 a x + y = 10 000 y = 3x b $2500 c $7500 17 a

−4 −3 −2 −1 −1 −2 −3 −4

x + y = 20 000 0.10x + 0.04y = 1340

b x = $9000, y = $11 000 c Alfred invested $9000 in bond C, and $11 000 in bond D.

ii Yes 9 a

18 1.8 km

7.01 Optional: Define relations and functions

1 2 3 4 5 6

−4 −3 −2 −1

−1

−1 4

−1

c Yes

3 2

Understanding

1 a True

b False

c True

d True

e False

f False

g True

h True

2 No, for a machine that represents a function, you cannot get different outputs for the same input. This is because by deﬁnition, a function assigns each input to one and only one output.

−4 −3 −2 −1 −1

x 1

−2 −3 −4

Answers mathspace.co

997

10 a

Reasoning

7 y 6 5 4 3 2 1

−9 −8 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

b No

21 The values of a, c, and e must be distinct from each other. In a function, each element from the ﬁrst set (input) can be associated with only one element in the second set (output).

Having repeated values in the ﬁrst component of the ordered pairs would mean that a single input corresponds to multiple outputs, which goes against the deﬁnition of a function. 22 a W (19) represents the amount of money Sharon makes if she works 19 hours. b The equation W (t) = 165 represents that Sharon makes $165 after working t hours.

c (−5, −10) or (−5, −4)

c W (26) = $390

11 D

23 Yes, the relation between the number of books rented and the total rental cost is a function. For each speciﬁc number of books rented, there is a clear cost associated. For instance, if a non-member rents 4 books, the cost will be $20. On the other hand, for a member with the yearly membership, irrespective of the number of books they rent, the cost remains $120 for that year. Given a speciﬁc number of books rented, the cost is deterministic, making it a function.

12 8 or 6 or 1 13 a Yes

b Yes

c Yes

d No

e No

f Yes

g Yes

h Yes

14 a 8

b −13

15 16 a i f (2) = 9

ii f (−3) = −11

b i g(2) = 12

ii g(−3) = 2

c i h(2) = −4

ii h(−3) = 16

d i k(2) = 13

ii k(−3) = −12

17 a Yes

b Yes

c Yes

d No

e Yes

f No

g No

h Yes

i Yes

j No

18 a

Call length (minutes)

24 No two people can sit in the same seat at the same time. This is akin to the vertical line test; a seat (or ticket number) can only be associated with one person at a time. 5

International call cost 1.80 2.90 4.00 5.10 6.20 (dollars) b Yes 19 a

Number of T-shirts (x)

Total cost

b Yes 20 P (2) = 14 000 represents the number of ﬁsh in the lake after 2 years.

998

25 Yes, this represents a function. Each pizza type (from the domain) has one specific topping (from the co-domain) that John likes. 26 Yes, this machine operates as a function. Each button (input) gives a specific type of coffee (output). Each button has only one corresponding coffee type. 27 No, it’s not a function. Since multiple families can own the same type of car, a family could also own multiple types of cars. Therefore, one family could be associated with multiple car types. 28 The relation is not a function. Even though each pastry type uses all three ingredients, the same ingredient (e.g. flour) is used in making each of cake, cookie, and muffin.

Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

8 a i

Problem-solving 29 a f (1) + g(2) = 20 c f (0) × g(0) = 30 30 a f (a + 1) = 4a + 5 2

c h(a + 1) = a − a − 4 31 a f (2) = 8 + k

b f (−2) − g(7) = −13

3 2

1 2

b f (a + 1) = a + 5a + 6

−2

b k=7

−4

−3

ii All real x b i

33 Yes, it’s a function since each input is associated with exactly one output. The relation could represent the areas of squares with side lengths from the ﬁrst set.

iii All real y 4

3 2 1

−4 −3 −2 −1 −1

7.02 Optional: Domain and range

−2 −3

Understanding

−4

1 The domain is all real numbers. 2 Fractions (denominator cannot be zero), square roots (cannot have negative values under them), logarithms (argument must be positive), and any other restrictions given in the function’s deﬁnition. b True

ii All real x c i

iii All real y 4

3 2 1

c False

−4 −3 −2 −1 −1

Fluency 4 a x ≠ −4

−4 −3 −2 −1 −1

32 A function can be: (1, 4), (2, 5), (3, 6). Adding the ordered pair (1, 6) makes it no longer a function.

3 a False

−2

b x≠9

c x ≠ −7

5 a x≥1

b No

6 a x ≥ −3

b All real numbers

c x≠2

d x<3

7 a i All real numbers

ii y ≥ 0

b i All real numbers

ii y ≥ −2

c i All real numbers

ii y ≤ 2

−3

−4

ii All real x d i

iii All real y 4

3 2 1 −4 −3 −2 −1 −1

d i x > −3

ii All real numbers

e i All real numbers

ii y > 1

f i x≤4

ii y ≥ 0

−3

g i x≠4

ii y ≠ 2

−4

h i x ≠ −6, x ≠ 6

ii y ≠ 0

x 1

−2

ii All real x

iii y = −4

9 a 16

b y ≤ 16

c All real x

10 a 0

b y≥0

c All real x

11 a i Minimum

ii −1

iii y ≥ −1

b i Maximum

ii −3

iii y ≤ −3

Answers mathspace.co

999

12 a

c i 4 3

b y

c No

2 1

3 2 1 −4 −3 −2 −1 −1

x 1

-4 -3 -2 -1 -1

-2

-3

−2

-4

−3 −4

13 a Concave up

ii Domain: all real x, Range: all real y

b 5

c 5

d i

d y≥5

14 a y ≥ 1

b y ≥ −7

c y≥9

d y≤0

e y≥0

f y ≤ −5

g y≤4

h y≥0

15 a i 4

4 3 2

1 -3 -2 -1

3 2

x 1

-1

-2

-3

-4 -3 -2 -1 -1

-4

-2 ii Domain: x > 0, Range: all real y

-3 -4

e i 4 3

ii Domain: x ≥ 0, Range: y ≥ 0 b i

5 4

-1

-2

-3

1 -4 -3 -2 -1 -1

-2

8 10 12 14 16

-4

x 1

4 ii Domain: x ≥ 7, Range: y ≥ 0

-3 ii Domain: all real x, Range: y > 0

1000 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

f i 14

b y

18 a Domain: all real x. Range: all real y.

c Domain: x ≠ 0. Range: y > 0. d Domain: all real x. Range: 0 < y ≤ 1.

Reasoning

19 If the range of a function consists of only non-negative values, then its graph will lie on or above the x-axis, since y-values represent the output of the function.

x 2 4

−1

20 The denominator x2 + 1 is always positive, regardless of the value of x, since squaring any real number gives a non-negative result. Hence the denominator will never be zero, allowing all real numbers in the domain.

−2 −3 −4

ii Domain: x ≠ 3, Range: y ≠ 0

21 After 10 hours, x2 becomes greater than 100, resulting in a negative number of items, which is not feasible in a real-world context.

y 5 4 3 2 1 -1

-1

b Domain: x ≠ 2 and x ≠ −2. Range: y > 0 or y ≤ −1 or (−∞, −1]∪(0, ∞)

h i

−4

x 2

ii Domain: all real x, Range: y > 4

−6 −4 −2

−3

−2

g i

−4 −3 −2 −1 −1

-10 -8 -6 -4 -2 -2

c Yes

x 1

3 4

22 The function 2x is always increasing; the smallest value is when x = 0, and it increases exponentially thereafter. This means the popularity can only go up. Problem-solving b n2

23 a A

-2

-3

f (n)

ii Domain: x > 1, Range: all real y 16 a −6 ≤ x ≤ 6 17 a

b 0≤y≤6

−2 −1

−9 −2 −1

5 4 3 2 1 −1

n 1

Answers 1001 mathspace.co

24 x ≥ 0

7 a i Odd

ii Decreasing

iii Increasing

25 All non-negative integers 26 a y ≤ 125

b 0 ≤ y ≤ 125

27 a x ≥ 0

b y ≥ 0.14

ii Increasing

iii Increasing c i Odd

ii Increasing

iii Decreasing

28 a 0 ≤ x ≤ 7

b 30

29 a 0 ≤ t ≤ 5

b 25 ≤ T ≤ 80

c 0 ≤ t ≤ 3, 31 ≤ t ≤ 34 d 0 ≤ t ≤ 3

7.03 Optional: Graphs of x

b i Even

d i Even

ii Decreasing

iii Decreasing e i Even

ii Increasing

iii Increasing f i Odd

Understanding

ii Increasing

iii Decreasing

1 n represents the degree of the polynomial. It determines the highest power of x in the equation and inﬂuences the overall shape and behaviour of the graph.

8 a As x → ∞, y → ∞. b

5 4 3 2 1

2 Increasing 3 a False

b True

e True

f True

c False

d False

9 a

y 5

12 8

y = 2x + 4

x 1

y = x2

4 -4

1 2 3 4 5

-4 -3 -2 -1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Fluency

As x → −∞, y → −∞.

-8 -12

−3 −2

y = x4

−1

5 a i Odd

b The graph of the function becomes narrower as y increases more rapidly.

ii Increasing for: x < 0 and x > 0 iii As x → −∞, y → −∞ and as x → ∞, y → ∞. b i Even ii Increasing for: x < 0 and decreasing for x > 0

10 a −1 < k < 0

iii As x → −∞, y → −∞ and as x → ∞, y → −∞. 6 a As x → −∞, y → ∞ and as x → ∞, y → ∞. b As x → −∞, y → ∞ and as x → ∞, y → −∞. c As x → −∞, y → −∞ and as x → ∞, y → −∞. d As x → −∞, y → −∞ and as x → ∞, y → ∞. e As x → −∞, y → −∞ and as x → ∞, y → −∞. f As x → −∞, y → −∞ and as x → ∞, y → ∞. g As x → −∞, y → ∞ and as x → ∞, y → −∞. h As x → −∞, y → ∞ and as x → ∞, y → ∞.

1002 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b g ( x) =

1 3 x 4

45 y 40 35 30 25 20 15 10 5

−5 −4 −3 −2 −1−5 −10 −15 −20 −25 −30 −35 −40 −45

f (x) = x3 x 1

11 a

1 1

−3

−4

3 2

−3

−4

−3

−4

−4 y

12 a

3 2 1

−4 −3 −2 −1 −1

−2 −3 −4 y

13 y 12 11 10 9 8 7 6 5 4 3 2 1

3 2

−4

−4 −3 −2 −1−1

−3

−2

−3

−2

−4 −3 −2 −1 −1

−2

−4 −3 −2 −1 −1

−2

−3

−4 −3 −2 −1 −1

−2

−4 −3 −2 −1 −1

−2

−3

−4 −3 −2 −1 −1

−2

−4 −3 −2 −1 −1

x 1

1 −4 −3 −2 −1 −1

−2 −3 −4

Answers 1003 mathspace.co

−4 −3 −2 −1 −1

−3

−4

13 a

13 y 12 11 10 9 8 7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3

x 2

−4

4 3

−1 −1

−2

−2 −3

−3

−4

4 y 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 −14 −15

−4 −3 −2 −1 −1

−2

−3

−4 −3 −2 −1−1

−4 −3 −2 −1 −1

−2

3 x 1

1 −10 −8 −6 −4 −2

−1

x 2

−2 −3 −4 y

x 1

1004 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3 2 1 −1 −1 −2 −3 −4

x 1

2 3 4 5 6 7 8

15 a Increasing for x > 3, so after 3 months.

b Revenue is negative for 2 < x < 4, so between 2 and 4 months.

3 2 1

−4 −3 −2 −1 −1

−3

16 a Decreasing approximately between weeks 3 and 7.

−4

b Approximately 32

c After 12 weeks

3 2 1 −2 −1 −1

x 1 2 3 4 5 6 7 8 9

−2 −3 −4

3 1

−7 −6 −5 −4 −3 −2 −1 −1

−2 −3 −4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

5 4 3 2 1 −5 −4 −3 −2 −1 −1 y = x3 −2 −3 −4 −5

d As the number of weeks increases the number of crows begins to increase rapidly, so y → ∞. This model likely won’t be accurate too much beyond week 15 as the park will not have an inﬁnite number of crows. 17 The end behaviour of a polynomial is determined both by the leading coeﬃcient and also the degree of the leading term.

d Just after 10 months, around 10.2 months. Reasoning

−2

c $40 000

If the degree of the leading term is even then the end behaviour is the same at both x → ∞ and x → −∞, otherwise they will be opposite. If the leading coefficient is positive then as x → ∞ we get y → ∞, with the end behaviour as x → −∞ depending on the degree of the leading term. If the leading coefficient is negative then as x → ∞ we get y → −∞, with the end behaviour as x → −∞ again depending on the degree of the leading term. 18 Florin is incorrect. We can use information about the end behaviour to determine the sign of the leading coefficient of the polynomial function and whether the degree of the polynomial is even or odd. Since the end behaviour falls as x → −∞ and x → ∞, we can conclude that the sign of the leading coefficient must be negative and that the degree of the polynomial is an even number. Problem-solving 19 a

x 1 2 3 4 5 y = x3 − 4

4.19 33.51 268.08 523.60 1436.76

b 36π < V < 85π

c 6m

Answers 1005 mathspace.co

f Horizontal stretch

b m = 3x3

20 a k = 3

g Vertical compression

1536

h Reﬂection across x-axis

1344 1152

6 a (−9, 0) and (2, 0)

b (−1, 0) and (10, 0)

c (−5, 0) and (6, 0)

d (−6, 0) and (5, 0)

960 768

7 a i Vertical translation

576

ii g (x) = f (x) + 2

384 192

b i Vertical stretch

x 1

d 824 g

ii g (x) = 3f (x)

c i Vertical translation

e 6 cm

ii g (x) = f (x) − 2

21 a 0.28 g b

Side length (cm)

d i V ertical compression and reﬂection across the x-axis 1

Mass 0.28 2.24 7.56 17.92 35 60.48 96.04 (g) c

Mass (g)

ii 8 9 a (2, −3)

d (2, 18)

10 B

140

11 a Reﬂection across the x-axis or y-axis

112

b Horizontal translation of 2 units to the left

c Vertical stretch by a factor of 3

d Reﬂection across the y-axis 12 When a is negative, the transformation will involve a reﬂection across the x-axis.

28 Side length (cm) 1

13 a i Translated up 4 units

7.04 Optional: Transformation of functions Understanding

ii g (x) = f (x) + 4 b i Vertically stretched by a factor of 4 ii g (x) = 4f (x) c i H orizontally compressed by a factor of or Vertically stretched by a factor of 2

1 A

ii g (x) = f (2x) or g (x) = 2f (x)

2 A

d i Translated 5 units to the left

3 A

ii g (x) = f (x + 5)

4 a False

b False

e True

f False

c True

d True

14 a y = 10x2 c y = x + 10

e y = −x2 − 7

Fluency 5 a Horizontal translation left b Vertical translation down c Horizontal translation right d Vertical translation up

b y = −5x2

d y = (x − 2)2

f y = 8 (x + 4)2

g y = (x − 6) + 5 15 a i f (x) was translated right 1 unit and down 1 unit. ii g (x) = f (x − 1) − 1

e Vertical stretch

1006 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b i f (x) was translated left 2 units or vertically stretched by a factor of 9 ii g (x) = f (x + 2) or g(x) = 9f (x) c i f (x) was vertically stretched by a factor of 4 ii g (x) = 4f (x) 16 a

y 10 8 6 4

20 a The intervals where the function is increasing or decreasing and the intervals where the function values are positive, negative, or zero will change. These intervals will be shifted to the right or left as the function is translated. For example, if the interval of increase was x > 0 and it is translated 2 units to the right, then the new interval of increase will be x > 2. End behaviour will not change.

2 −4 −2 −2

Reasoning

x 2

b f (x) is reﬂected across the y-axis. 17 a

c The intervals where the function is increasing or decreasing and the intervals where the function values are positive or negative will change. The intervals of increase and decrease will swap. The intervals of positive and negative will swap. The intervals or x-values where the function is zero will not change.

4 2 -6 -4 -2 -2

b None of these key features will change.

x 2 4

-4

The end behaviours will change. If the behaviour was y → a, this will become y → −a, for example y → −∞ will become y → ∞.

19 a i 0 units up or down

21 Even if Stavros changed the amount he was depositing each week, his initial bank balance at Week 0 will not change. However, when a graph is transformed by a vertical stretch, the only point of intersection of the two graphs is the x-intercept. Since this is not the case for when Stavros changes the amount he deposits, this change cannot be represented by a vertical stretch of the current graph.

ii 0 units left or right

22 a k = −4

-6 b f (x) is horizontally stretched by a factor of 2. 18 a Reﬂection across the x-axis and vertical stretch by a factor of 2 b a = −2

iii No

b Horizontal translation of 4 units to the right

iv Yes

23 a Oreste reﬂected the graph of f (x) across the y-axis instead of the x-axis when drawing the graph of g (x).

b i 2 units down ii 0 units left or right iii Yes

b If f (x) is a linear function passing through the origin, then reﬂecting across the y-axis has the same effect as reﬂecting across the x-axis. If this is the case, then Oreste’s error would still give him the correct answer.

iv No c i 0 units up or down ii 4 units right iii Yes iv No d i 4 units down iii No

ii 3 units right iv Yes

Answers 1007 mathspace.co

24 a 2

Distance (km) 4

1.5

0.5

Time (minutes) 4

16 20 24 28 32 36 40 44 −1

b The gradient of the function will be twice as steep, they will reach the lake after 10 minutes, meaning the vertex will be at (10, 2) instead of (20, 2).

b To transform f (x) into g(x), the following transformations would be applied: 1. Reﬂect f (x) across the x-axis because of the negative sign. 2. Stretch the graph vertically by a factor of 4 due to the

coeﬃcient.

3. Shift the graph to the right by 3 units to account for the (x − 3) term. 4. Shift the graph upwards by 5 units to account for the +5. These steps will take the standard cubic function f (x) = x3and transform it into g(x). 27 a Yasushi is correct. If a function is transformed by only a horizontal and a vertical translation, it doesn’t matter in which order the two translations are applied. For example, if we start with the function y = x2 and translate horizontally by 3 units to the right, we get the function y = (x − 3)2. If we then translate vertically by 2 units upwards, we get y = (x − 3)2 + 2. Doing these translations in the other order will result in the same function.

y 4 3 2

25 a Answers may vary. One possibility is a horizontal translation of 2 units to the right, followed by a vertical stretch by a factor of 3.

26 a Both graphs are cubic functions, but they have different shapes and positions. The graph of g(x) is a transformation of the graph of f (x).

−1

c The graph would be reﬂected across the x-axis and translated up 2 kilometres.

b Answers may vary. A different possibility is a vertical translation of 2 units downwards, followed by a vertical stretch by a factor of 3.

1 x −1 −1

b Peter is incorrect. For example, if we start with the function y = x2 and vertically stretch by a factor of 2, we get the function y = 2x2. If we then translate vertically by 1 unit downwards, we get y = 2x2 − 1. If we perform the vertical translation ﬁrst, however, we get y = x2 − 1. Now performing the vertical stretch by a factor of 2 instead results in the function y = 2x2 − 2. Problem-solving 28 100 dB. The graph is shifted 5 units to the right and stretched vertically by 2. 29 150 cups 30 3 p.m. 31 t = 14 hours 32 5.33 hours

Chapter 7 review 1 a Yes

b No

c No

d Yes

e No

f Yes

g Yes

h Yes

2 a −20

b 4

3 a Yes

b No

4 a Yes

1008 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b Yes

c No

d Yes

5 a x≠5

b x ≠ −7

c x≠5

6 a All real numbers

b x = −3 and x = 2

13 a A reﬂection of the graph across the x-axis, followed by a vertical translation of 4 units upwards. This is equivalent to a vertical translation of 4 units downwards followed by a reﬂection across the x-axis.

7 a Domain: All real numbers Range: y ≤ 4 b Domain: x ≥ −8

Range: y ≥ 0

5 3

Range: y ≥ −2

d Domain: All real numbers

Range: y ≥ 0

−4 −3 −2 −1 −1

8 a y≥0

b y≥4

c y≤0

d y ≥ −2

−1

−0.5

y = 3x3 −3 −0.375 b

0.5

0.375

x 4

−2 −3

y = 5x

−4

10 a As x → ∞, y → ∞. y 3 2 1 −3 −2 −1

8 7 6 5 4 3 2 1

−5−4−3−2−1 −1 −2 −3

As x → −∞, y → ∞. b

x 1

y = 4 − x2

14 a A reﬂection of the graph across the y-axis, followed by a horizontal translation of 3 units to the right and then, a vertical translation of 1 unit downwards. b

−3

x 1

−2

−4 −3 −2 −1 y = 3x 132 −1

y = x2

c Domain: All real numbers

9 a

−1 −2 −3

11 Velin is incorrect. The graph shows that as x → −∞, y → −∞ and as x → ∞, y → ∞. This indicates that the leading coeﬃcient of the polynomial is positive and its degree is odd. 12 Vertical stretch by a factor of 2 and vertical translation of 4 units upwards

y = 53−x − 1

x 1 2 3 4 5 6 7 8

15 a Reﬂection across the x-axis and a vertical 1 compression by a factor of . 4 1 b a=− 4 16 a −3x + 4 + 1. The domain remains the same and the range shifts up and becomes y < 1. b y = (x − 3)2 + 6. The domain remains the same, and the range changes by shifting 6 units up.

8.01 Exponential relationships Understanding 1 a i Multiplying by 2

ii 80, 160

b i Adding 5

ii 25, 30

c i Multiplying by 3

ii 202.5, 607.5

d i Dividing by 10

ii 4, 0.4

2 a Multiplying by 3

b 108 circles

3 a Multiplying by 4

b 128 hexagons

Answers 1009 mathspace.co

Fluency 4 a

5x 50 51 52 53 y 1 5 25 125

625 3125

Problem-solving

b Exponential 5 a Exponential

b Linear

c Quadratic

d Exponential

e Exponential

f Exponential

g Linear

h Exponential

6 a i 3.5

ii 1500.625

b i 15

ii 101 250

c i 2

ii 1.968

d i 0.25

ii 16

7 a 5

b 125 triangles

8 a The volume increases by a factor of 3 each minute. b

Time (minutes)

11 No, y2 > y1 for some values of x ≥ 1. However, as y1 is increasing at an increasing rate while y2 increases at a constant rate, at some point y1 will become greater in value.

Volume (mL)

243

729

12 a Points A, C, F 13

9 a The volume increases by a factor of 2 each day. b The points represent an exponential relationship as they increase by a constant ratio of 2.

160

320

Time

Mass

100

51.2

40.96

32.768

26.2144

Number of pallets Number of bricks 0

c 256 mL 10 a graph B = b graph A = 2x

increases at a constant rate of .

Population

The relationship between time and mass is exponential because the ratio between consecutive values of the mass is constant 0.8

They do not lie on a straight line.

c Graph A is an exponential relationship as it is increasing at an increasing rate and has a constant base of 2. Graph B is linear as it

Time

d 2

The relationship between time and population is exponential because the ratio between consecutive values of the population is constant 2.

c 6561 mL Reasoning

b 6 units

c Points B, D, E

20 000

19 500

19 000

18 500

18 000

17 500

The relationship between number of pallets and number of bricks is linear because there is a constant difference of −500 between the consecutive values of the pallets which is equal to the gradient of the line.

1010 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

8.02 Characteristics of exponential graphs

y 150

Understanding

100

1 The base, b. 50

2 Asymptote

3 a True

−2 −1

100

1000

b False, if 0 < b < 1, then it is exponential decay. c False, if 0 < b < 1, then it is exponential decay. d False, if b > 1, then it is exponential growth. e True 4 a As x increases, y increases at an increasing rate. b As x increases, y decreases at a decreasing rate.

iii Increasing c i

−1

x y = 10x

y 1000

5 a Increasing

b Decreasing

800

c Decreasing

d Increasing

600

e Increasing

f Decreasing

400

g Decreasing

h Increasing

200 x

Fluency 6 a i

−2 −1

−1

x x

y=4 ii

4 16 64

iii Increasing d i

−1 0

y 2 3 1 4 1 1 2 1 1 4 1 3 4 1 2 1 4

20 15 10 5 x −2 −1

iii Increasing b i

−1

x x

y=5

3 −2

5 25 125

−1

iii Decreasing e i

−1 0 4

Answers 1011 mathspace.co

y 4 1 3 2 3 1 2 2 2 1 1 2 1 1 2 −2

7 5 4 3 2 1

x 1

c i Increasing

ii y = 0

iii (0, 1) iv Answers may vary, for example: (1, 7) v

−1 0

6 5 4

5 2

2 1

−4 −3 −2 −1

3 2

d i Increasing

ii y = 0

iii (0, 1)

iv Answers may vary, for example: (2, 64)

1 2

x −2

−4 −3 −2 −1

−1

iii Decreasing f i

−1

7 6 5

iii Decreasing

7 a i Increasing

ii y = 0

3 2

iii (0, 1)

iv Answers may vary, for example: (2, 4) v

ii y = 0

iii (0, 1)

iv Answers may vary, for example:

b i Increasing

e i Increasing

−4 −3 −2 −1

x 1

5 4

ii y = 0

iii (0, 1)

iv Answers may vary, for example:

1 −4 −3 −2 −1

1012 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1

f i Decreasing

ii y = 0

iii (0, 1)

iv Answers may vary, for example: (−2, 9) v

4 3

−5 −4 −3 −2 −1 −1

4 3

1 2 3 4 5

2 1

−4 −3 −2 −1

g i Decreasing

iv Answers may vary, for example: (−1, 5) 8

e It gets larger, approaching inﬁnity. f It gets smaller, approaching zero. b y = 0 is a horizontal asymptote of the curve.

11 a

y 7

−4 −3 −2 −1

h i Decreasing

ii y = 0

iv Answers may vary, for example: y

1 2 3 4 5

−5 −4 −3 −2 −1 −1

1 −4 −3 −2 −1

−5 −4 −3 −2 −1 −1

iii (0, 1)

c 16

10 a Positive

b 1

d None. For all values of x, it will have a positive value.

ii y = 0

iii (0, 1) v

9 a

x 1

8 a No. For b > 0 the expression bx is greater than zero for all real x.

x 1 2 3 4 5

c The two graphs are reﬂections across the y-axis.

b y tends towards ∞. c 0

d 1

e 0

Answers 1013 mathspace.co

Reasoning

y 27 24 21 18 15 12 9 6 −x 3 y=3

12 a Graph R b For x < 0, the graph of y = (5)x is below the graph of y = (3)x. This is because negative values of x result in fractional y-values, and for any negative value of x, (5)x will result in a smaller value than (3)x. 13 No, both functions are always greater than zero, because a positive base to any power is always greater than zero. 14 a i False

ii False

iii True

iv True

−4 −3 −2 −1

c As x gets larger, the y-values approach 0.

Example justiﬁcation: When graphed on the interval 0 < x < 2, we can see that the two

1 a i The same

ii Different

iii Different

iv The same

v The same b If k is positive, the graph of y = 2x + k moves upward from y = 2x.

. This tells us

c If k is negative, the graph of y = 2x + k moves downward from y = 2x.

is approaching y = 0 faster,

which means it is decreasing more rapidly. The pattern continues even for larger values of x. y y1 y2

2 a i Different

ii The same

iii Different

iv Different

v The same b If a > 1, the graph will be stretched and have a higher y-intercept. c If 0 < a < 1, the graph will be compressed and have a lower y-intercept. d If a is negative, the graph will be reﬂected across the x-axis.

17 a 1

b 1 b −2

c y = 2x c c 3

18 m = 8 19 a y = 128

b y = −2x

3 a

Problem-solving 16 a 125

d −2

c The two graphs are reﬂections across the y-axis.

curves start at the same point but,

Understanding

that

8.03 Optional: Transformations of exponentials

b Shared y-intercept of (0, 1)

ends below the graph of

4 a No

b Yes

c No

d No

d 3

Fluency b x=7

5 a The graph is translated up 4 units.

20 80 units 21 a

b The graph is translated down 6 units. c The graph is dilated (compressed) by a factor of . d The graph is dilated (stretched) by a factor of 5.

1014 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

11 a Translated down 2 units

6 a A translation down 5 units b

b Reﬂection across the y-axis

5 4 3 x 2 f ( x) = 1 3  1

c Translation up 3 units d Dilation by a factor of 2 x

−5 −4 −3 −2 −1 1 2 3 4 5 −1 x −2 g( x ) = 1 − 5 3 −3 −4 −5

12 Graph R 13 a A dilation by a factor of 3 b

7 a A reﬂection about the x-axis 5 4 3 2 1

f (x) = 4

g(x) = 3(4)x

12 48

c y = −5 b

12 10 8

1 2 3 4 5 g(x) = 3(4)x g(x) = −5x

−3 −2 −1

c As x increases, the function values get increasingly large in the negative direction.

14 a i (0, −4) iv

9 a A vertical dilation by a factor of 0.5, this is a compression.

4 2

f (x) = 4x x 1

7 6

iii Increasing

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

ii y = −5 6 5 4 3 2 1

8 y = 4x

14 f (x) = 5x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−2 −1

1 2 3 4 5

b i (0, 3)

4 3

2 1

f (x) = 8x −3 −2

−1

g(x) = 0.5(8x) 1

10 a Vertical dilation by a factor of , which is a compression. b Vertical dilation by a factor of 2, which is a stretch.

ii y = 2 8 7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2

iii Decreasing

x 1

2 3

Answers 1015 mathspace.co

c i (0, −3)

ii y = −4

g i (0, 2)

iii Decreasing

18 16 14 12 10 8 6 4 2

2 1

−4 −3 −2 −1 −1

2 3 4

−2 −3 −4 −5

d i (0, 4)

8 7 6 5 4 3 2 1

ii y = 0 1

iii Decreasing

y x 1

−4

−5

1 2 3 4 5

−6 −7

iii Increasing

15 a y = 5 which tells us a = 5. b The growth factor is 4 which tells us b = 4. c y = 5(4x) 16 a y = −4 which tells us a = −4. b The decay factor is

which tells us

c 17 a x

−4 −3 −2 −1

ii y = 0

iii Decreasing

b y=3

19 a i (0, 4)

iii Vertical dilation by a factor of 3, a stretch.

1 −4 −3 −2 −1

3 4

−3

−2

ii y = 0

f i (0, 2)

h i (0, −3) iv

e i

x 1

−4 −3 −2 −1 −1

−5 −4 −3 −2 −1 −1 −2

18 16 14 12 10 8 6 4 2

iii Increasing

−4 −3 −2 −1

ii y = 3

ii y = 0

−1

x 1

ii y = 1

Translated 1 unit upwards. iv

−2

−3

−4

5 4 3 2 1 −4 −3 −2 −1

1016 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1

3 4

b i (0, 3)

x = 0 into the transformed equation, we will get y = a.

ii y = 4

iii Reﬂected across the x-axis. Translated 4 units upwards. iv

21 By reﬂecting the graph across the y-axis

22 For x < 0, the graph of

2 1

−4 −3 −2 −1 −1

graph of y = (3) . This is because the y-values will be half of the y-values of 3x, which are smaller in value than (3)x. 23 a Yes. For any value of x, 10x is always positive. Therefore, −10x is always negative.

−2

−3 −4

y 50

c i (0, −2.5)

−1 −50

−3 −2

ii y = −3

iii Vertical dilation by a factor of 0.5, a compression.

x 1

−100

Translated 3 units downwards.

−150

−200

is below the

−250

3 2 1

−4 −3 −2 −1 −1

c There is no x-value for which y = 0. 24 a A reﬂection across the x-axis

−2

−3

y 5 4 x 3 y = 3(2 ) 2 1

−4

d i (0, −4)

ii y = −1

iii Vertical dilation by a factor of 3, a stretch. A reﬂection across the x-axis. Translated 1 unit downwards. iv

y −4 −3 −2 −1 −1

x 1

−2 −3 −4

−5 −4 −3 −2 −1 1 2 3 4 5 −1 −2 x −3 y = −3(2 ) −4 −5

c y = 3(2x). The number of bacteria can’t be negative and all values of y = −3(2x) are negative, so it must be y = 3(2x).

−5

25 a Yalile’s graph is correct, Emir’s is incorrect.

−6

b Emir should have gone in the order of operations, vertical dilation, reflection, and then translation, but he did translation ﬁrst and then vertical dilation.

−7 −8

Reasoning 20 The coordinates of the y-intercept are (0, a). When a dilation is applied to f (x) = bx, the equation becomes f (x) = a (b)x. If we substitute

y=3

y = 2(3x)

y = 2(3x) − 1

−1

x x

Answers 1017 mathspace.co

d The two graphs are the same.

across the y-axis to get the graph of

2 1

−4 −3 −2 −1 −1

28 a

−2 −3

g(x) = 2−x

5 4

x 1 2 3 4 5

1 y=  2

3 2

y=3

−4 −3 −2 −1

−1

x 1

e Above

30 a y = −4. Tells us k = −4. b

x 1 2 3 4 5

−1 2

⬚

⬚ ⬚

⬚

−1

y = a(b ) + k

−1

y = a(bx)

y = bx

−1

x x

y = a(b ) + k y = a(bx) y=b

g(x)

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−3 −2 −1

5 4 3 2 1

1  2

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

d Above 5 4 3 2 1

c Decreasing

f (x) = 2x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

27 a

(−1, 6) (0, 3)

26 a A reﬂection about the y-axis 5 4 3 2 1

Point on (−1, 2) (0, 1) original graph Point on new graph

Problem-solving b

can be reﬂected

e The graph of

16 64

c a=3

x x

e y = 3(2x) − 4

x 1 2 3 4 5

31 a y = 5(2x) − 10

1018 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

8.04 Exponential equations

e x=3

Understanding

g x = −1

h x=4

j x=1

12 a x = 1

b x = −3

1 a False

b True

c True

d False

2 If the bases on both sides of the equation are the same, then the exponents will be equal. By setting the exponents equal to each other, we can solve for the variable. 3 a 53

b 25

−2

c 43

d 34

c x=3

e x=0

−3

e 6

f 10

4 a 9

b 7

c 3

d 0

e −3

f −2

g 2

h 2

13 a

9 8 7 6 5 4 3 2 1

Fluency 5 a x=2

b x = −1

c x = −2

d x = −4

6 a x=6

c x = −3

d x=3

f y=2

g x=6

h n=4

e x=0

7 a i 23x = 22

b i 24x = 2−1

−3x

c i 2

d i 2−x = 2−2

y = 2x

−5 −4 −3 −2 −1 −1

1 2 3 4 5

b 3

c x=3 x

14 a i y = 2 and y = 5 ii

9 8 7 6 5 4 3 2 1

ii x = 2

8 a x = −2

b y = −3

c x = −5

d x = −1

9 a

c x=5

d x=2

f x = −3

g x = −2

i x=6

j x=2

l x = −7

10 a x = −3

b x = −2

c x=9

d x = −7

e x=1

11 a x = 2

b x=3

c x=5

d x=4

−4 −3 −2 −1

−1 −2 −3 −4

3 4

x = 2.32 b i One possible answer: y = 3x and y = 5 ii

9 8 7 6 5 4 3 2 1 −4 −3 −2 −1−1

x 1

−2 −3 −4

x = 1.47

Answers 1019 mathspace.co

c i One possible answer: y = 7x and y = 4 ii

Given equation

Rewrite 16 with a base of 4

Distributive property

Quotient exponent law

Since 24 = (22)2, we can rewrite 24 with a base of 4

ax = ay, so x = y

Solve the linear equation

6 5 4 3 2 1

−4 −3 −2 −1

x = 0.71 x

d i One possible answer: y = 5 − 2 and ii

19 a When rewriting the right-hand side of the equation with a negative exponent in step 2, Lorena did not apply the negative to both terms. Step 2 should have been 42p = 4−(2p+1).

1 x −4 −3 −2 −1

−1

20 When is squared, it equals 7. So the equation can be solved as shown: y = 4

−2

Square both sides Evaluate

x = 0.32

Rewrite 49 with base 7

15 There are two solutions, −3.5 and 3.9

Power of a power law

16 There are two solutions, −0.7 and 1.5 Reasoning 3x−1

x−9

17 In 2 = 27 cannot solve using exponent laws because the bases 2 and 27 are not powers of the same number, so we cannot write them in the form ax = ay.

Solving 23x−1 = 8x−9, can be solved because the bases 2 and 8 can both be rewritten as powers of 2. The equation can then be solved for x by comparing the exponents directly.

Given equation

21 To estimate the solution to 200(1.02)x = 220, Julian needs to ﬁnd the x-value that results in a y-value that is close to 220. According to the table, the approximate solution is x = 5. Problem-solving 22 a 5 hours

b 10 hours

c 12 hours

23 a n = 8

b k=4

c m = −2

24 a 440 Hz

b 262 Hz

c n = 73

25 a

b x=3

e x = 4, −1

g y=4

26 Answers may vary. For example, (29−5)1 = 48−6.

1020 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

8.05 Compound interest

9 a

Understanding 1 a Always true

Year

Balance + interest

Value

Interest earned

Start of investment, 0

−

$2090

$87.78

First year, 1

$2090 + $87.78

$2177.78

$91.47

Second year, 2

$2177.78 + $91.47

$2269.25

$95.31

d Sometimes true, this is only for the ﬁrst period.

Third year, 3

$2269.25 + $95.31

$2364.56

$99.31

e Always true

Fourth year, 4

$2364.56 + $99.31

$2463.87

$103.48

Fifth year, 5

$2463.87 + $103.48

$2567.35

−

Year

Balance + interest

Value

Interest earned

Start of investment, 0

−

$2090

$88

First year, 1

$2090 + $88

$2178

$91

Second year, 2

$2178 + $91

$2269

$95

Third year, 3

$2269 + $95

$2364

$99

Fourth year, 4

$2364 + $99

$2463

$103

Fifth year, 5

$2463 + $103

$2566

−

b Never true c Always true

f S ometimes true, it depends on the interest rates and principal. For the same principal and interest rate, this is true after the ﬁrst period. 2 Account B 3 Graph 1 4 a No

b No

5 a $9000 c Quarterly

c No

d Yes

b 7% d 3 years

Fluency 6 a Investment A

b Investment A

c 11 years 7 a

c $1.35 10 a 500 000 × (1 + 0.012)24 b $165 736.40

Balance at beginning of year

Interest earned

11 a $7706.73

b $6617.25

First year

$3200

$160

12 a $5705.83

b $705.83

Second year

$3360

$168

Third year

$3528

13 a $4477.00

b $1477.00

Fourth year

$3704.40

$176.40 −

14 a $25 631.28

b $631.28

15 a $2529.86

b $1761.93

b $504.40 8 a

c $837.74 Balance at beginning of year

Interest earned

Reasoning 16 a $1794.85

First year

$3500

$133

Second year

$3633

$138.05

Third year

$3771.05

Fourth year

$3914.35

$143.30 −

b $414.35

b $1797.60

c $2.75

d The more frequently interest is compounded, the greater the total interest earned will be if all other conditions are the same. This is because more frequent compounding results in more periods where interest is earning “interest on interest”.

Answers 1021 mathspace.co

17 a Jadria should choose Plan 2, because she would have to pay back less interest. For Plan 1, the amount of interest she would have to pay is $9000(0.029) (3) = $783. For Plan 2, the amount of interest she would have to pay

23 a

Value ($) 3000 2500 2000 1500

1000

b No, Plan 2 will not always be the better choice. As Plan 2 involves compound interest, it is in the form of an exponential relationship, and Plan 1 is in the form of a linear relationship. If we were to draw the graphs of both plans, then the graph of Plan 2 will eventually exceed the graph of Plan 1. In fact, after only 4 years, the interest needing to be repaid on Plan 2,

500 5

No. of years 20 25

b 24 years 24 24 years 25 $60 803.96 26 5.116% 27 18.5 years

will exceed that of Plan 1, $9000(0.029) (4) = $1044

18 4.75% compounded annually. Although the other rates are lower, they are compounded more often which works out to accruing slightly more interest than 4.75% compounded annually.

8.06 Exponential models Understanding 1 a Exponential decay c Neither e Neither 2 a i Exponential decay

• 4.75% compounded annually: He would owe $26 187.50 after one year

ii 0.723

• 4.7% compounded quarterly: He would owe $26 195.87 after one year

ii 2500

• 4.65% compounded monthly: He would owe $26 187.60 after one year

ii 5.9

iii 5%

b i Exponential decay iii 12.3%

c i Exponential growth iii 2.5%

d i Exponential growth

19 a $6367.25 b The value in the table has been rounded to the nearest whole dollar, so the amount of interest will be slightly different. Problem-solving

ii 10 000

iii 100%

3 a 840

b 4%

4 a 1.4

b Increase by 40%

Fluency

20 $53 136.44

5 a 2.4 kg

21 $53 516.75 22 a 0.02

b Exponential growth d Exponential decay

b b $3750

1022 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Months

Mass (kg)

1.8000

1.9800

2.1780

2.3958

2.6354

2.8989

3.1888

14 a

Mass (kg)

Week number

Weight of treats (g)

2000

1000

500

250

4 3 2 1 1

6 a y = 1300(1 + 0.08)v

Months 6

31.25

, where w is the number of

weeks on the plan and T (w) is the weight of treats in grams.

9 $503 579 10 $5.20

Reasoning

11 $12 072 Number of rounds (r)

Number of players (p)

b Any power of 2 such as 2, 4, 8, 16, 32, 64 … c p = 2r

15 The decay rate is the percentage decrease from one term to the next while the decay factor is the number being multiplied to the previous term. For example, if 20% is the decay rate, 0.8 is the decay factor. 16 a a represents the initial temperature of the substance, and r represents the rate at which it cools down.

d 256 players 13 a 5000 b The constant percent rate of change is 2.3%. This means that the population will increase by 2.3% every minute. c 299 629 90

62.5

8 6591 bees

125

It will take 6 weeks to get the dog down to less than 50 g of treats per week.

b $2806.60

7 41 199 people

12 a

b T = 145(1 − 0.4)t c 0.88 °C d It will never reach 0 because there is no time that could make (0.6)t equal to zero. Exponential relationships have a horizontal asymptote at y = 0.

N (’000s)

e Substance P

17 a

60 50 40

(N)

(C)

480

2880

17 280

103 680

120 150

622 080

e Approximately 100 minutes

3 732 480

22 394 880

t 30

b C = 480 × 6N−1 c 29 023 764 480 sales

Answers 1023 mathspace.co

d No, this is not a reasonable goal. It is impossible for them to have approximately 29 billion sales in one quarter. To achieve this, they would need to sell more than three phones to every single person in the world. This level of growth is unsustainable. They need to abandon the assumption that growth will change in exactly the same way in the coming quarters and try a more sophisticated model. 18 a

21 a

Value ($) 3500 3000 2500 2000 1500 1000 500

No. of years 2 4 6 8 10 12 14 16 18 20

b 13 years

Seconds passed (t)

Undissolved sugar in grams ( y)

22 a

b 3 bounces

23 2020 24 24 hours

25 a i 179

ii 132

c It is decreasing at a decreasing rate.

b i 70%

ii 40%

d No, the model shows the amount of undissolved sugar is decreasing at a decreasing rate and has an asymptote of y = 0, so theoretically some sugar will always remain undissolved.

c Juanita’s blood sugar level peaked at a higher amount and decreased more rapidly. 26 a 0.01 or 1% b A = 38 200(1 − 0.01)n c $35 964.54

e If the sugar dissolves twice as quickly, it can be modelled by the equation which is equivalent to

The amount of sugar that dissolves between the second and third second is which is less than

Problem-solving

20 a

b The 8th stage Day

Number of people infected

5 25 125 625

b 5n c 8 days

d 210 = 1024 people

27 a The graph represents the content Maximilian and Lucy remember. After 1 day, it appears that Maximilian remembers 52% which means he forgets 48%. After 1 day, Lucy remembers 62% which means she forgets 38%. An equation that models the percentage Maximilian recalls is P = 100(1 − 0.48)t. An equation that models the percentage Lucy recalls is Q = 100(1 − 0.38)t.

19 a 256

d The Motorbike

b Example explanations: No. As the models are equations for exponential decay, there is a horizontal asymptote of y = 0 that the curve will get closer and closer too but never reach.

In reality, the amount they remember will probably not continue to decrease at a percent rate. They may always remember a small percentage or they may forget all of it by a certain point. c Both equations do accurately model the amount of content remembered.

1024 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Example reasoning: From the graph, we see that Raymond remembers about 85% of the previous day’s content, which means he forgets 15%. That is consistent with the given equation. Sunyi remembers about 87%, according to the graph, which means she forgets 13%. This is also consistent with the given equation. d Maximilian and Lucy forget things a lot quicker than Raymond and Sunyi. Possible reasons may include some of the following ideas: • One factor may be the type of information used in the studies. Maximilian and Lucy were given information to memorise. Raymond and Sunyi are given content to learn in school, which means they are most likely reviewing that content regularly. Raymond and Sunyi know that the information given to them needs to be retained as they will likely be assessed on it in school.

y 200 160 120 80 40 x −1

Chapter 8 review

b i

−1

y 4 1 3 2 3 1 2 2 2 1 1 2 1 1 2

−1

iii Decreasing 5 a y=0

b (0, 1)

c One point is

1 a Linear

b Quadratic

c Linear

d Exponential

e Exponential

f Quadratic

2 a i 0.25

ii 7.8125

b i 2.5

ii 312.5

3 a 4 a i

iii Increasing

• Maximilian and Lucy may not have known about the design of the study, so they may not have tried to actively retain the information they were given. • Finally, if the information given to Maximilian and Lucy was not very interesting, it would be easier to forget.

y 25

b 2 cubes

5 x

x y = 6x

−1 0 1

6 36

216

−2

−1

6 a None

b Positive

7 a y = 256

b x = −2

8 a The graph is translated up 2 units. b The graph is reﬂected across the x-axis. c The graph is dilated (enlarged) by a factor of 3.

Answers 1025 mathspace.co

9 a A translation down 4 units b

5 4 3 2 1

18 a 0.02 or 2%

19 P = $33 195.04 20 a y = 7700 (1 + 0.053)n b $16 707.69

f (x) = 7x −5 −4 −3 −2 −1 −1 −2 −3 g (x) = 7x − 4 −4 −5

21 $12 068

1 2 3 4 5

22 a 10 300

b Decreasing

c 9.8%

d 6818 bees

9.01 Logarithms

c y = −4

Understanding

10 Graph Y 11 a

b y=0

c Increasing

10 6 2

x 2

e 1

f 10

g 0.01

h 100 000

b x=4

c x = −2

d y = −3

e x=6

h x = −8 b

f T he base ten logarithm of 224 is 2.3502 because 102.3502 = 224 3 a The base two logarithm of 8 is 3 because 23 = 8.

c x=0 b k=5

e The base ten logarithm of 65 is 1.8129 because 101.8129 = 65

12 a x = 8

b The base two logarithm of 16 is 4 because 24 = 16 c The base two logarithm of 0.5 is -1 because 2-1 = 0.5

d The base two logarithm of 1 is 0 because 20 = 1 e The base two logarithm of 0.125 is -3 because 2-3 = 0.125

c m = −2

Balance at beginning of year

Interest earned

First quarter

$4500

$180

Second quarter

$4680

$187.20

Third quarter

$4867.20

Fourth quarter

$5061.89

$194.69 −

b $561.89

Fluency 4 a i 106

ii 6

b i 10−4

ii −4

c i 100

ii 0

−3

d i 10

ii −3

5 a i 2

ii 5

b i 2

ii 0

16 a 30 000 × (1 + 0.062)25

c i 2−3 d i 28

b $104 969.04 17 a $1612.89

d 1000

d The base ten logarithm of 1 is 0 because 100 = 1

15 a

c 10 000

c The base ten logarithm of 0.01 is -2 because 10-2 = 0.01

14 a n = 1

b 0.1

b The base ten logarithm of 1000 is 3 because 103 = 1000

−8 −6 −4 −2

1 a 100

2 a The base ten logarithm of 100 is 2 because 102 = 100

13 a

b $6250

b $10 226.30

6 a 102 = 100 c 104 = 10 000

1026 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

ii −3 ii 8 b 100 = 1 d 10−2 = 0.01

7 a 1.6532

b 3.8860

c 0.7126

d −2.2373 b 0.063

c 6.310

d 204.174

b 2 =1

c 24 = 16

10 a B

d 2−1 = 0.5 b A

Lifespan (days)

Animal/Insect Ant

8 a 37.154 9 a 2 =4

18 a

c C

1.30

African elephant

25 500

4.41

Bowhead whale

71 400

4.85

Bumblebee

d B

Lifespan in a log10 scale

1.32

Chicken

2920

3.47

11 a 2 countries

b 10 countries

Dog

5110

3.71

12 a 13

b 25

c 6

8500

3.93

13 a 11

b 17

c 7

Emperor penguin

1.26

Fly

Reasoning 14 a The data is very large and very spread out. Trying to ﬁt the information on the same scale would be diﬃcult and would not show us the trends in the data. b

Population

Log scale

Melbourne

4 929 200

6.693

Adelaide

1 396 783

6.145

Wollongong

310 479

5.492

Geelong

278 708

5.445

Hobart

231 707

5.365

Toowoomba

146 955

5.167

Ballarat

105 924

5.025

Bunbury

80 106

4.904

Warragul

44 625

4.650

Log scale 4–5

2 5

6–7

15 a B

b C

365

2.56

Little penguin

2190

3.34

Monkey

7300

3.86

Shark

10 980

4.04

3 2

1 2 3 4 5 Log of lifespan

19 a R = 3.8

c A 1

Number of bacteria in a log2 scale

c The bacteria double each hour.

Ladybug

20 a

b 27 = 128 bacteria

2.89

16 a Number of hours passed 0

Actual number of bacteria

4.01

780

Frequency

5–6

10 120

Hamster

Frequency

City

Giraffe

10 9 8 7 6 5 4 3 2 1

b Minor

x 1

b The x-value of the point represents the paper being folded 5 times, and the y-value represents the layers as an exponent of two: 25 layers. c 10 layers

d 11

Problem-solving 17 a $79 432.82 b $100 000 to $1 000 000

Answers 1027 mathspace.co

9.02 Order of magnitude and logarithmic scales Understanding 1 a 102

e 100

b 103

c 106

d 101

f 10−2

g 10−1

h 10−3

2 a 6 × 102

17 The order of magnitude for DNA is −9, and the order of magnitude for a hydrogen nucleus is −15. These magnitudes have a difference of −9 − (−15) = 6, so they differ by 6 orders of magnitude. 18 a 102 times

b 7 × 103

c 4.8 × 106

Reasoning

b 101.5 times

d 3.5 × 101

e 2.3 × 102

c Yes, if houses collapse for earthquakes of magnitude 6, then it would take an earthquake of smaller magnitude to cause only a crack. A magnitude of 6.2 would cause more than just a crack in a building.

f 7.6 × 10−3

−2

−5

g 1.2 × 10

h 9 × 10

3 a 3

b −5

c 0

d −1

4 a 3

b 4

c −1

d −2

19 a 965 years

Fluency

b 883 years

c R1 : 2326 years, R2 : 2128 years

5 −7 3

6 a i 8 × 10

d No, at twice the half-life there will be one-quarter of the substance left. As the amount of substance approaches zero the time taken for a unit of substance to decay grows at an increasing rate.

ii 3

b i 9.644 × 10

ii 5

c i 8 × 10−4

ii −4

d i 2.574 × 10−1

ii −1

7 3

Problem-solving

8 4

9 2 10 a 8

b 0.25

11 a 16

b 631

d 25 = 32

c 1

b 10

ii 9.01 years

b 16.09 years 22 R = 8.7

12 a 5.2

0.1

21 a i 10.24 years iii 8.04 years

or approximately 1.26 times larger

23 a

13 a The concentration of hydrogen ions in the pH 8 solution is 100 times greater than in the pH 10 solution. b The concentration of hydrogen ions in the pH 5.4 solution is 104.6 times greater than in the pH 10 solution. c The concentration of hydrogen ions in the pH 5.4 solution is 102.6 times greater than in the pH 8 solution. 14 a 1 000 000 b 100 000 times louder 4

15 a 10 times

b 10 times

c 106 times

d 107 times

16 a 102.8 times

b 103.6 times

c 106.4 times

1028 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

1000 900 800 700 600 500 400 300 200 100

x 1

Population of shrimp

16 64 256 1024

Base 4 log scale

8 a x = 100

b x = 87

9 a 3 and 4

b 2 and 3

c 1 and 2

d −3 and −2

Reasoning

3 2

10 a i Undeﬁned

ii Undeﬁned

iii Undeﬁned

iv Undeﬁned

b 10x is a positive number for all values of x, therefore it’s not possible to ﬁnd the logarithm of numbers less than or equal to zero.

x 1

9.03 Optional: Other bases

11 logx y = 2 implies that y = x2 so

Understanding 1 a 4

b −1

c 4

d 3

e 1

f 3

g −2

h −3

c a

d 10

2 2 to what power is 8? 3 a 5

b 9

Fluency

This means that

12 log7 x = y implies that x = 7y so x2 = (7y)2. That is, x = 72y which means log7 x2 = 2y. 13 a log3 a = log3 x b . When the exponential equation is written in logarithmic form, we see that a = x. Since the original expression was set equal to a and a = x, this means the original expression equals x.

4 a log5 25 = 2

b log3 81 = x

c log6 6 = 1

d log2 1 = 0

e log4 0.0625 = −2

g log9 81 = x

h log4 0.25 = −1

j log4 16 = 2

14 The equivalent exponential statement is y = 4x, and we know that 4x is greater than zero for all values of x. Therefore, y can be any number greater than zero.

k log3 1 = 0

l log9.8 1 = 0

Problem-solving

b 51 = 5

15 a 12

5 a 42 = 16 c 2−3 = 0.125

e 62 = 36

f 21 = 2

g 8 =1

i (5.8)2 = 33.64

k (8.4)2 = 70.56

b 2

c 10

d 9

16 a i 4x = 64

ii x = 3

b i 3x = 243

ii x = 5

c i 125x = 5

d i

ii x = −3

e i x2 = 64

ii x = 8

c 0

d 3

f i x3 = 125

f 1

g −1

h −2

g i 2 =x

j −2

k −3

l −4

e 2

f 5

g −2

i −4

j −2

k −4

6 a 3

b 2

e 4 i −1 7 a

h i 34 = x

ii x = 5 ii x = 32 ii x = 81

i i 25x = 125

h 2

j i 4x = 32

l −3

k i x1.5 = 64

ii x = 16

l i x

ii x = 9

2.5

= 243

Answers 1029 mathspace.co

17 a x = 2

b x = 58

c x=9

e x = ±11

f x = 12

g x = 11

h x = 27

18 a x = 16

b x=7

8 a 6 log2 (x + 7)

b 6 log (2x + 5)

c −log (2x + 3)

d −8 log (4x + 3)

9 a log2 (5) + log2 (a) − log2 (11)

9.04 Optional: Logarithm laws

b log2 (3) − log2 (7) − log2 (a)

Understanding

d log 2 + log u − log 3 − log v

1 a +

b −

c log p + log q − log r e 3 log u + 5 log v

c ×

f 3 log a − 2 log b

2 a Power law: logB pn = n logB p

g 6 log(7) + 6 log(x)

b Product law: logB (pq) = logB p + logB q c Quotient law:

h log 7 + 4 log x

= logB p − logB q

i −log x − log y j −log 2 − log x

d Product law: logB (pq) = logB p + logB q

e Power law: logB pn = n logB p

k log 5 − log x

f Quotient law:

l −7 log 3 − 7 log x

= logB p − logB q

m log 7 − 6 log x

Fluency

3 a No

b Yes

c No

d Yes

4 a No

b Yes

c Yes

d No

e Yes

f No

g Yes

h Yes

5 a x

b a

c 3

d 1

e 2

f 1

g 0

h 0

i 0

j −10

k 56

o p 10 a log 84

6 a log7 ( f ) + log7 ( g) b log8 x + log8 y c log5 (x) − log5 ( y)

b log 165

c log10 6

e log10 2

g 5 log10 18

h 3 log10 2

i 3 log10 2

j −35 log10 5

11 a 1

b 3

c 9

d 2

e log2 (3) + log2 (s)

e 2

f −1

g 2

h −3

f log2 (11) + log2 (c) + log2 (d)

j 2

k 2

l 1

d log4 a − log4 9

g log2 (11) − log2 (r) h logb 20 − logb 7

12 a log (x6 y5)

i logb 5 + logb 4

j log2 (3) + log2 (13) k log3 7 − log3 11 l log 7 − log 2 7 a 3 log2 ( y)

b 6 log2 (r)

c 7 log4 y

d 4 log x

e −3 log5 (w)

f −7 log5 (r)

g −5 loga B

1030 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c 6 log x

d log (100xy)

e 7 log x

g log (10x)

13 a 3

b 4

c 2

d −2

e 5

f 2

g 9

h 5

Given Switch to exponential form

Reasoning 14 a Pei has made an error in Step 1. He performed the product property ﬁrst instead of the quotient property, which is incorrect. When condensing with the product and quotient properties, they should be done in order from left to right. b 1 log3 (72) − log3 (8) + log3 (3) = log3 (9) + log3 (3)

2 = log3 (27) 3 = log3 (33) 4 = 3 log3 (3) 5 =3

uotient property of logarithms Q Product property of logarithms Write 27 as a power of 3 Power property of logarithms Inverse property of logarithms

15 Consider that 10

2 = 1024 ≈ 1000 = 10 This means that

10 log10 (2) = log10 (210) = log10 (1024) ≈ log10 (1000) = log10 (103) = 3

20 The product property of logarithms states that logB (p) + logB (q) = logB (pq) Using this property, we can add n terms of logB (p) to get

n logB (p) = logB (p) + … + logB (p) = logB (pn)

which gives us the equality of the power property of logarithms. We can see this more clearly when adding on terms of logB (p) one at a time. 2 logB (p) = logB (p) + logB (p) = logB (p2) 3 logB (p) = logB (p) + 2 logB (p) = logB (p) + logB (p2) = logB (p3) 4 logB (p) = logB (p) + 3 logB (p) = logB (p) + logB (p3) = logB (p4) n logB (p) = logB (p) + (n − 1) logB (p) = logB (p) + logB (pn−1) = logB (pn)

Cinzia’s method is: 1

21 The power property of logarithms states that logB (pn) = n logB (p)

2 3 Uri’s method is: 1 log10 (x7) − log10 (x4) = 7 log10 (x) − 4 log10 (x) 2 = 3 log10 (x) b

c 3.17 18 Step 1: Substitution Step 2: Quotient property of exponents Step 3: Inverse property of logarithms Step 4: Re-substitution

Inverse property of logarithms

⋮

16 Both methods are valid.

17 a

Product property of exponents

Consider that, for every non-zero value of x, it is true that x0 = 1. So then, using the power property of logarithms, we can show that:

loga (1) = loga (x0) = 0 loga (x) = 0

Problem-solving 22 a 0.78

b 1.40

c 1.62

d 2.70

e 1.21

f 3.03

g 1.32

h 2.11

23 a 7.56

b 10.2

c 1.41

d 1.48

24 a x = 5

d x = 45

e x=5

f x = 16

g x=8

h x = 12

i x = 12

k x=9

l x=1

25 a x2 = 6xb − 8b2

b x = 2b, 4b

Answers 1031 mathspace.co

9.05 Optional: Connect exponentials and logarithms

3 1

1 The inverse of an exponential function is a logarithmic function of the same base. 2 a i No

ii No

iii Yes

b i No

ii Yes

iii No

−1

c No

d No

e Yes

f No

g Yes

h No

b log4 (64) = 3

e x=1

7 a

5 4 3 2 1

d 7 =1

c g (x) = 7x

f (x)

−3

−2 −1

4 1 3

−2 −1 0

g(x)

x −2 −1

1032 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3 2 1

x 1

f ( x ) = log 1 x 3

−3 −4

e Decreasing

f (x)

y is undeﬁned

9 a g (x) = 2x b

−2

−4 −3 −2 −1 −1

8 a g (x) = log5 (x)

−3

8 10 12 14

d −∞

x 1

−2

g ( x) =

c Increasing

g Increasing

−4

x 4

y−1

−4 −3 −2 −1 −1

d y = −2

−2 −1 −2 −3 −4 −5

c y=0

4 16 64 256

y −2 −1 0 b

b y=4

f x=0

Fluency 6 a y=3

10 a

d b

−3

b No

−4

3 a Yes

4 a log5 (25) = 2

x 1

−2

c A vertical line.

5 a 4 = 16

Understanding

f y is undeﬁned.

11 a i

c i g (x) = 5x

3 y

1 −4 −3 −2 −1 −1 −2

y 5 3

x 1

−1

g (x) = 5x

−5 −4 −3 −2 −1 −1

−3

−2

−4

−3

f (x) = log5 x

−4 −5

ii y = log4 x b i

4 3 2

y y

d i −1

1 −4 −3 −2 −1 −1

x 3

g ( x) =

−2 −3

1  2

2 1

−4 −3 −2 −1 −1

−4

x 1

−2 −3

−4

12 a i g (x) = 10x ii

13 a x = 1

5 4 3 2 g(x) = 10x 1

7 6 5 4 3 2 1 −2 −1 −2 −3

g(x) = 4x

3 2 1 −4 −3 −2 −1 −1 −2 −3

x 1

f (x) = log4 x

5 25

0 −1 −2

c x=0

b i g (x) = 4x 4

x y

−5 −4 −3 −2 −1 1 2 3 4 5 −1 f (x) = log10(x) −2 −3 −4 −5

x 2

8 10 12 14

e Decreasing 14 a k = 3

b f (x) = log3 x

15 a Yes

−4

Answers 1033 mathspace.co

Reasoning 16 Example answer: The exponential function f (x) = 2x raises the base 2 to the power of x, while the logarithmic function g (x) = log2 x determines the exponent needed to obtain a given value of x with base 2. Since these functions “undo” each other, they are considered inverse functions. 17 a 0 < x < 1

b x>1

c No, because the value of loga x can take any value, given that a > 1.

18 a The function f (x) = log3 x will have larger function values for values of x that are greater than 1. When looking at the graphs of the inverses, y = 3x and y = 5x, the graph of y = 5x increases faster and is more steep for small values of x than y = 3x. When reﬂected across the line y = x, this will mean g (x) = log5 x will have smaller function values than f (x) = log3 x as the values of x increase. b A table of values or graphs can be used. The graph is shown:

If we graph the function log2 x, we will see that the curve extends indefinitely in both the positive and negative directions along the y-axis. As x approaches 0, the curve decreases without bound, and as x approaches infinity, the curve increases without bound. Thus, there is no fixed value that loga x will always be greater than.

4 3 2 1

8 10 12 14

−2

From the graph, we can see f (x) = log3 x is higher than g (x) = log5 x. So, f (x) = log3 x has larger y-values for values of x that are greater than 1.

19 h = 2

−1 −2 −3 −4 −5

g(x) = log5 x

−2 −1

Problem-solving

d No, because the value of loga x can take any value, given that a > 1.

−2 −1

−3

−1 −2 −3 −4 −5

4 3 2 1

2 f (x) = log3 x

−2 −1

y 3

20 a n = 3 21 a

x 1

1034 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b k = 16 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

y y = 2x

x 1 2 3 4 5

5 4 3 2 1 −8 −6 −4 −2 −1 −2 −3 −4 −5

b i

3 1

x 2

−2 −1

y = log2 x

5 4 3 2 1

y = −1

−4

ii x = −0.667 c i

4 3 2 x

c 3

y = 2.4

1 y = −log2 (x + 3) −4 −3 −2 −1

−1

x 2

−2 −3

ii x = −2.811

−4 −3 −2 −1 −1 −2 −3 −4 −5

23 a i

−1

−3

1 2 3 4 5 −5 −4 −3 −2 −1 −1 y = log9 x −2 −3 −4 −5

y = log3 (x + 1) x 1

−2

d x=3 5 4 3 2 1

c 3 22 a

d i

y = 9x

4 3 2

x 1

1 y = 0.3

−1

x 3

y = −log4 (x − 3)

−2 −3 −4

d x=3 5 4 3

ii x = 4.516

24 a y = 4.2

2 1 −1

y = log2(x)

b x

2 4 6 8 10 12 14 16 18 20

−2

ii x = 18.379

0.01 0.1

−4

100

−2 0

3 2 1 −1

x 1

2 3 4 5 6 7 8 9 10

−2 −3 −4

Answers 1035 mathspace.co

g Rewrite 9 and 27 as powers of 3: 32x = 33(x−1)

y 3

h Write both sides as logarithms of base 10 or set the exponent equal to 0: 2x + 3 = 0

2 y = 2 log10 x 1

Fluency

x 1 2 3 4 5 6 7 8 9 10 11 y = log10 x

4 a i

ii 0.627

b i

ii −1.490

c i

ii 0.631

d i

ii 1.339

e i

ii 1.594

f i

ii 0.478

g i

ii 1.318

h i

ii −1.756

ii When g (4) = 7, g (128) = 12, g (21000) = 1005

i i

ii −1.792

iii When h (4) = 10, h (128) = 35, h (21000) = 5000

j i

ii 5

k i

ii 2.807

l i

ii 2.294

−1 −2 −3

d i The y-values of both functions are negative, but the y-values of y = 2 log10 x are twice the y-values of y = log10 x. So, the y-values of y = log10 x are greater. ii The y-values of both functions are equal; y = 0. iii The y-values of both functions are positive, but the y-values of y = 2 log10 x are twice the y-values of y = log10 x. So, the y-values of y = 2 log10 x are greater. 25 a i f (4) = 2, f (128) = 7, f (21000) = 1000

b As x increases, the function value of g (x) will always be 5 greater than f (x), and the function value of h (x) will always be 5 times greater than f (x).

9.06 Optional: Solve exponential equations with logarithms

5 a x = 0.57

Understanding 1 a Yes

b No

c No

d Yes

c x = 0.54

d x = 5.53

e x = 0.64

f x = 1.70

g x = 2.30

h x = 5.35

i x = 1.59

j x = 1.42 l x = −0.51 b x = 1.37

2 a x = 5.860

b x = 0.119

k x = 0.83

c x = 25.875

d x = 0.849

6 a x = 1.26

3 a Rewrite both sides as logarithms of base 10: log 7x = log 2 b Divide both sides by 5: 4x = 20 c Divide or multiply both sides by

b x = 1.60

c x = 4.19

d x = −4.92

e x = 5.91

f x = 0.87

g x = −1.90

h x = −8.39

7 a

d Subtract 4 from both sides: e Multiply both sides by 4: 3x = 36 f Add 12 to both sides: 10(6)x = 20

1036 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

8 a i

ii x = 1.020

b i

ii x = 5.322

c i

ii x = 0.823

15 By the deﬁnition of logarithms, Bx = A can be rewritten in terms of x as x = logB A. We can use this to rearrange the equation Bx = A using base 10 logarithms: Definition of logarithms

d i

Take the logarithm base 10 of both sides Use the log law logBn = n log B

ii 5.532 9 a x = 0.33

b x = 0.40

c x = −0.52

d a = 1.09

10 a

11 a y = −8.87

b y = 62.91

c y = 3.19

d y = 1.88

12 Akiko’s solution is correct. An alternative method is graphing with technology and estimating the solution. A third method is shown: Original equation

Divide both sides by log B

So,

for A, B ≥ 0.

16 a i S tep 2. When applying the power law of logarithms, a should be multiplied to the logarithm and not added to it. ii

iii 1.679

b i S tep 3. The numerator should be log 89 and the denominator should be log 2. ii

iii 6.476

17 We can let log3 16 equal a to show that:

Add 7 to both sides Power law of logarithms Distributive property Subtract 11 log 3 from both sides Divide both sides by −4 log 3

We can then do a substitution to show that:

Estimate to 4 decimal places

Reasoning 13 For 23x−1 = 27, logarithms are needed because the bases 2 and 27 are not powers of the same number. However, in 23x−1 = 4x−9, both bases can be expressed as powers of 2, allowing for direct power comparison to solve for x. 14 The exponential function y = 3x can only have positive outputs. If we consider the graph of y = 3x, it is always above the x-axis. This means it will never be equal to a negative number, so there is no solution to the equation 3x = −2.

All 3 expressions are written in base 10. Any of them can be entered into a calculator to estimate the solution. Problem-solving 18 a 500 mice

b 20%

c 6.03 months

Answers 1037 mathspace.co

19 a 800 wallabies

b 15%

15 a

c 8.53 years 20 5.03 weeks

21 6.18 hours

22 16 years 23 32.87 years

16 a 32

Chapter 9 review 1 a i 10−2

ii −2

b i 104

2 a i 2

ii 6

b i 2−1

ii −1

c g (x) = 6x

b Weight in log10 scale

Weight (in kg)

Ferret

0.8

Cat

3.2

0.51

Whale

32 840

4.52

Rhinoceros

2754.23

3.44

Arabian wolf

25.12

1.40

Hamster

0.15

−0.82

Shetland pony

127

2.10

6 a 4 × 10

−0.10

−2 −1 0

g (x)

5 25

−1

8 10 12 14

f (x)

−2

d x=0 e

b 9 years old b −5

7 6 5 4 3 2 1 −2 −1 −2 −3

7 3 b 105.1 times

8 a 1000

18 a g (x) = 5x

Animal

−5

c 1 b

3 5 cities

5 a 16 to 32 years old

b 7

17 a g (x) = log4 (x)

ii 4

9 20

x 2

f Increasing

10 a 3

b 0

11 a 3 and 4 x

12 a i 3 = 27 b i x4 = 16

c i 42 = x + 4

d i 71 = 3 − 2x

c −2

d −1

b −2 and −1

19 a k = 4 20 a i

ii x = 3 ii x = 2

ii x = −2

−4 −3 −2 −1 −1 −2 −3

b log5 3 − log5 8

−4

c log4 (7) + log4 (11)

c 8 log6 (x − 10)

ii x = 12

b −5 log3 (p)

13 a log7 (6) + log7 (s)

14 a 12 log (a)

b f (x) = log4 x

ii y = log3 x

1038 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

x 1 y−1

b i

7 a i 6 units

1 −4 −3 −2 −1 −1 −2

x 1

y−1

−8 −6 −4 −2 −2

−3

−4

−6 −8

ii b i 4 units 21 a i

ii x = 0.388

b i

ii x = 2.854

6 4

c i

ii x = −7.213

d i

ii x = 3.973

−8 −6 −4 −2 −2 −4 −6

22 a x = 0.42

b x = 2.67

c x = 3.46

d x = −0.54

e x = 0.69

f x = −9.06

g x = 5.60

h −2.37

−8

c i 3 units ii

23 48 years

10.01 Circles at origin

b 8 units

c 9 units

−4

d 12 units

2 Every point on the circle is exactly 7 units away from the point (0, 0). 3 The distance between the point (x, y) and the centre of the circle is equal to the radius r, since the point lies on the circle.

−6 −8

d i 13 units ii

y 12 8 4

Fluency 4 x2 + y2 = 25

−8 −6 −4 −2 −2

Understanding 1 a 4 units

−12 −8 −4

x 4

−4

5 9 units

−8

6 x2 + y2 = 64

−12

Answers 1039 mathspace.co

e i

15 Points (7, 24) and (18, 15) both satisfy the equation x2 + y2 = 252, since 72 + 242 = 49 + 576 = 625 and 202 + 152 = 400 + 225 = 625.

units

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

As both points satisfy the same equation, they are on the same circle. x

16 The equation represents the distance formula between two points (x, y) and (0, 0) in the Cartesian plane. Since the distance between the centre of the circle (0, 0) and any point on the circle (x, y) is equal to the radius r, the equation x2 + y2 = r2 represents all the points on the circle.

1 2 3 4 5

f i 0.8 units ii

17 Using the distance formula, we can ﬁnd the radius of the circle:

0.8 0.6

r2 = (5 − 0)2 + (6 − 0)2 = 25 + 36 = 61. Therefore, the equation of the circle is x2 + y2 = 61.

0.4 0.2

−0.8 −0.6 −0.4 −0.2 −0.2

0.2 0.4 0.6 0.8

18 a x2 + y2 = 36

−0.4

20 x2 + y2 = 36

−0.8 2

8 a x +y =9

c x2 + y2 = 121

9 a i 5 units b i 10 units c i

units

b x + y = 36

b On

21 x2 + y2 = 32

d x2 + y2 = 100

ii x2 + y2 = 25

23 (−2.846, −8.538) and (2.846, 8.538)

ii x2 + y2 = 100 ii x2 + y2 = 125

units

10.02 Translations of circles

10 x2 + y2 = 100 11 a Inside

b 36π units2

19 x2 + y2 = 64

−0.6

Problem-solving

c Outside d On

Understanding

12 The point lies inside the circle.

1 a No

13 The point is on the circle.

2 a Centre: (−3, −3), r = 3

c Yes

d Yes

b Centre: (0, 1), r = 4

Reasoning 14 To determine if a point (a, b) lies inside, on, or outside the circle with equation x2 + y2 = r2, calculate the value of a2 + b2. If a2 + b2 < r2, the point is inside the circle. If a2 + b2 = r2, the point is on the circle. If a2 + b2 > r2, the point is outside the circle.

b Yes

c Centre: (2, −3), r = 1 d Centre: (−2, 4), r = 3 3 a (1, −2)

b 8 units

Fluency 4 a i (5, −6)

units

b i (5, 5)

units

c i (0.4, −3.8)

units

d i

units

1040 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5 a i (2, −2) iii

iii

ii 6 units 8

5 4 3 2 1

6 4 2 2

−4 −6 −8

iii

ii 4 units 8

2 x

−8 −6 −4 −2 −2

−2

c i (−1, 3)

8 10 12 14 16

6 a (x − 7)2 + ( y + 1)2 = 9

ii 4 units

b (x + 2)2 + ( y + 3)2 = 25 c (x + 4.3)2 + y2 = 36

d (x + 10)2 + ( y − 2)2 = 144

6 4

7 a i

−8 −6 −4 −2 −2

−4 −6

d i (−3, −4)

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

−8

ii 5 units y

x 2 4 6 8 10

ii (x − 1)2 + ( y − 2)2 = 16

6 4

b i

2 −8 −6 −4 −2 −2

−8

−4

x 2

−6

iii

ii r = 7

iii

1 2 3 4 5

f i (6, 0)

b i (2, −2) iii

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−8 −6 −4 −2 −2

x 2

−4

6 4 2

−6

−8 −6 −4 −2 −2

−8

x 2

−4

e i

units

−6 −8

ii (x − 3)2 + ( y + 2)2 = 16

Answers 1041 mathspace.co

8 a (x − 2)2 + y2 = 16 2

b x2 + ( y − 4)2 = 25

15 a (x − 3)2 − 9 + y2 + 8y = 9

b x2 − 6x + ( y + 4)2 − 16 = 9

c (x − 3) + ( y + 3) = 9

c (x − 3)2 + ( y + 4)2 = 34

d (x − 5)2 + ( y − 5)2 = 25 9 a x2 + ( y − 10)2 = 25 2

d The centre of the circle is (3, −4) and the . radius is r =

b x2 + ( y + 10)2 = 25

c (x − 10) + y = 25

16 (0, 2), (6, 2), (0, −4), (6, −4)

d (x + 10)2 + ( y − 4)2 = 25 10 a Scaled by a factor of 7, then translated left 4 units and up 5 units b Scaled by a factor of 4, then translated right 2 units c Scaled by a factor of 3, then translated right 1.5 units and down 4 units d Scaled by a factor of 4.5, then translated left 1.5 units and up 4.5 units

17 (x − 25)2 + ( y − 30)2 = 400 18 a (49.5, 37.5) b 9.15 m

c (x − 49.5)2 + ( y − 37.5)2 = 83.7225 d (9.15 × 2)2 = 334.89 m2 19 a 5 cm

b 85.19 cm2

c Circle one: (x + 1)2 + ( y − 6.085)2 = 49, Circle two: (x − 7)2 + ( y − 6.085)2 = 36

Reasoning 11 a (x + 8)2 + ( y + 6)2 = 100 b Yes. When we substitute x = 0 and y = 0 into the equation from part (a), we ﬁnd:

Chapter 10 review 1 a 7 units

LHS = (0 + 8)2 + (0 + 6)2 = 64 + 36 = 100 = RHS

4 2 −8 −6 −4 −2 −2

b Inside. The circle on the given graph has a radius of 7.5. A radius of is smaller than the graphed circle, so the circle with the will lie inside

the graphed circle.

−6 −8

2 a x2 + y2 = 64

b x2 + y2 = 25

3 a i 6 units

ii x2 + y2 = 36

b i 13 units

13 a One b The centre of the circle has been translated left 8 units from the origin. Since the radius is 1 unit, the circle does not intercept the y-axis. Problem-solving

x 2

−4

units

equation

Since the point satisﬁes the equation, it must lie on the circle. 12 a

c i

units

4 a Inside

b On

5 32π units2 6 a (x − 4)2 + (y − 9)2 = 1

14 a (x − 5)2 + ( y + 2)2 = 9

b (x + 1)2 + ( y − 4)2 = 36 c x2 + ( y + 3)2 = 16 d

1042 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

ii x2 + y2 = 169

ii x2 + y2 = 20

c Outside d Inside

7 a i (3, −1)

6 a No

ii 5 units

iii

5 4 3 2 1 −3−2 −1 −1 −2 −3 −4 −5 −6 −7

7 a

b No

c No

d Yes

x 1 2 3 4 5 6 7 8 9

r 10

b i (0, 4)

ii r = 7

iii

d q = 15

8 a b = 9a

b y = −2x

e r = 44

4 2 −8 −6 −4 −2 −2

x 2

−4

9 a 70

b 280

10 a

b 5

11 a 300 Water 270 240 210 180 150 120 90 60 30

8 a i Translated left 4 units and up 5 units

ii (x + 4)2 + (y − 5)2 = 16

b i T ranslated right 3.5 units and down 2.5 units

ii (x − 3.5)2 + (y + 2.5)2 = 16

Rice 10

9 a (x + 1)2 + (y + 7)2 = 1

b (x − 5)2 + y2 = 1

10 a (0, 0), (16, 0)

b (0, 0), (0, −12)

90 100 110 120 130 140

b 2.1 c For every 100 g of rice William uses, he adds 210 mL of water.

11.01 Direct and indirect proportion

The unit rate of water with respect to rice is 2.1 mL/g.

Understanding

12 a

1 a True

b True

e False

f True

c False

d True

2 A 3 a Direct

b Inverse

4 a Inverse b Direct

c Inverse d Neither

e Neither f Direct

g Neither h Neither

i Inverse j Inverse k Direct

Fish caught

Hours 4

l Direct

Fluency 5

30 27 24 21 18 15 12 9 6 3

10 20

c 9 ﬁsh are caught every 4 hours

Answers 1043 mathspace.co

13 a i y

14 a

15 a k = 200

22.5

c y = 40

16 a k = 12

37.5

7.5 x 10

17 a Yes

18 a

ii b i

b k = −6

b Yes

d No

−5 −4 −3 −2 −1

c No

57 47.5

b Increasing

c As x increases, y increases at a faster and faster rate.

28.5 19

Reasoning

9.5

19 It increases by a factor of 7.

y 5

20 25 30

20 a

ii c i

144 132 120 108 96 84 72 60 48 36 24 12

21 28

30 27 24 21 18 15 12 9 6 3

Jimmy Ben

Days 8

c By comparing the constant of proportionality or unit rate. The higher unit rate would be the steeper graph, which represents the person who is saving more quickly.

21 a y = 4x b

y 40 35 30 25

Online

20 x 1

Amount saved

ii 11 10 9 8 7 6 5 4 3 2 1

x 5 10 15 20 25 30 35 40 45 50 55

d i

Days Amount saved

2 3 4 5 6 7 8 9

15 Newspaper

10 5

x 2

1044 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c There were more online enquiries each day than there were newspaper enquries. d Both lines pass through the origin because, on day 0, the number of online enquiries and newspaper enquiries is both 0. There are no enquiries before advertising begins. 22 The centripetal force of an object varies proportionally to its mass and the square of its velocity, and inversely to the radius of the circle it moves along.

Understanding 1 B and C 2 6 3 x=0 Fluency 4 a Horizontal: y = 0 Vertical: x = 0

23 a 6000 kg/cm

b The pressure decreases. Since we dividing 6000 by the volume, as the volume increases we are dividing 6000 into more and more pieces, so each piece will be smaller and smaller. c No, the pressure cannot ever reach zero. When dividing 6000 by a very large number we can get closer to zero, but it will never actually get to zero. Problem-solving

b Domain: All real values of x, x ≠ 0 Range: All real values of y, y ≠ 0 c Quadrants 1 and 3 d Increasing: None Decreasing: x < 0 and x > 0 e As x → ∞, y → 0 As x → −∞, y → 0 5 a 0

b −∞

c y=0

6 a i No

ii Yes

iii No

iv Yes

b No

c y=4

12 24 36 48 60

d Positive

e Negative

Number of books read 16 32 48 64 80

f 1 and 3

24 a

Number of weeks

7 a i

b Carl 25 a

11.02 Hyperbolas

−4

−2

−1

7 6 Derek

x 1

−3

b Derek

−4

b h = 0.82s

iii Quadrants 1 and 3

c h = 1.64 cm 27 a A = 3C

b 21 grams

28 a k = 1220

b 40 minutes

29 r = kn

−2

2 3 4 5 6 7 8 9

26 a k = 0.82

−4 −3 −2 −1 −1

Time (hours) 1

−2

Yuri

−1

Robots built

b i

−4 −2 −1

1 2

−2 −1

30 a = 9, b = 42 31 32 metric tons

Answers 1045 mathspace.co

9 a y=4

b y=3

2 1

10 a

−4 −3 −2 −1 −1

−4

−3

y=0

−4

−4 −3 −2 −1 −1 (−3, −1) −2

x 1

−4

1 2 3  1, − 2   3   x=0

−4

4 b

 1, 5   2  

y=0

3 1

x 1

−4

−2

−3

−4

(−4, 1)

y=0

iii Quadrants 1 and 3 d i

−4 −3 −2 −1 1 2 −1  − 5 , − 1 −2  2    −3 x = 0

−4 −3 −2 −1 −1

−4 −2 −1

−4 −3 −2 −1 −1

−2 x = 0

−3 (1, −4)

−4

−3 x = 0

−2 −1 −1 −2

(1, 3)

iii Quadrants 2 and 4 x

−2

c i

d The graph moves further away from the axes.

 − 2 , 1 2  3    1

−4 −3 −2 −1 −1

x 1

y=0

−2

−4 −3 −2 −1 −1

−3

−2 −3

−4

iii Quadrants 2 and 4 8 a i y→∞

ii y → −∞

iii y → 0

iv y → 0

b i y → −∞ iii y → 0

11 y1

is A

ii y → ∞

is B

iv y → 0

is C

1046 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

12 a c

or xy = 3

or xy = −2

or xy = −4

13 Yes, since the product of the number of people and jelly beans is always the same. This product is the constant of variation, 200. Reasoning 14 a Yes. Because if the number of workers increases, the number of hours decreases. b No. Because if the number of sales increases, the number of revenue increases as well. c No. Because as the minutes of cycling increases the amount of calories being burned also increases. 15 She can label a point on the graph. 16 a

6 5 4 3 2 1 −6−5−4−3−2 −1 −1

In general, when we reﬂect over the line y = x, the coordinates are swapped. The points are point

the graph is symmetrical across the line y = x. b y = −x 18 a k = 8 b Decrease c xy = −8, if we reﬂect over the x-axis the point (x, y) would become (x, −y). This means that the product xy will switch signs, so instead the coordinates of a point having a product of 8, they will have a product of −8. Problem-solving

b x

1 2 3 4 5 6

490 245 163.33 122.5

81.67

y 500 400

b y

300

c We could rotate 180° about the origin, or we could reﬂect over the x-axis and then reﬂect over the y-axis.

200 100 x 1 2 3 4 5 6 7 8 9 10

e Inverse variation

20 a

3 2

C 50

x −1

is also on the graph. This means

19 a $245.00

−2 −3 −4 −5 −6

17 a

, however, the reﬂected

−1

If we sketch the line y = x, we can see that it cuts each piece of the graph into two. Let’s look at the piece in the first quadrant. Each point below the line of symmetry is reflected to a point above the line of symmetry. For example, (2, 1) is reflect to (1, 2).

20 10 R 10

b The current decreases. 21 a c [0.775, 0.9]

b [60, 80] d 50 minutes

Answers 1047 mathspace.co

b 32 units2

22 a −2 23 a

c i Translated up 5 units ii

5 4 3 2 1

40 30

S 10

b 10 hours

5 4 3 2 1

11.03 Optional Transformations of hyperbolas

b y=

k b (−1, 3)

1 x−h

c (1, −7)

d (6, 6)

3 a False (the vertical asymptote is x = h) c True

5 a

d False

4 a i Translated up 1 unit y

1 y= x +1 1 y= x

-5 -4 -3 -2 -1 -1 -2 -3 -4 -5

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1 -1 -2 -3 -4 -5

1 2 3 4 5

1 y= x

1 2 3 4 5 1 y = 3- 2 x

4 2

−1 −2

x 1

c Translated right 4 units

b i Translated down by 4 units ii

x y

Fluency

5 4 3 2 1

-5 -4 -3 -2 -1 -1 -2 -3 -4 -5

Understanding

x 1 2 3 4 5

d i Translated down 2 units

c 2 hours

d The speed must increase.

b True

1 y= x

-2 -3 -4 -5

2 a (1, 7)

1 y= x +5

-5 -4 -3 -2 -1-1

1 a

6 a i Translated left by 5 units. ii

1 x

1 2 3 4 5

1 −4 x

1048 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

5 4 3 1 2 y= x+5 1 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5

1 x

1 2 3 4 5

b i Stretched vertically by a factor of 2.

b i Translated right by 4 units. ii

5 4 3 2 1

y y=

1 x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 x−4

5 4 3 2 1

ii y=

1 x

1 x+1 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

1 x x

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

7 a i Stretched vertically by a factor of 3. 5 4 3 2 1

1 2 3 4 5

1 x

1 2 3 4 5 y=

-1 x

1 x x

1 2 3 4 5

−4 x

8 a Translate to the left by 4 units. b x = −4

3 x

1 y= x

d i S tretched vertically by a factor of 4 and reﬂected across the x-axis.

2 x

1 2 3 4 5

-5 -4 -3 -2 -1 -1 -2 -3 -4 -5

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

1 x−5

d i Translated left by 1 unit. ii

c i Reﬂected across the x-axis.

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

c i Translated right by 5 units. ii

c y=0

y 6

4 2 −8 −7 −6 −5 −4 −3 −2 −1 −2

x 1

−4 −6

Answers 1049 mathspace.co

9 a x = −3

b i x = 0 and y = 3

b y=0

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

10 a A

ii Translated up 3 units

iii c i x = 1 and y = 0 x 1

iii d i x = 0 and y = 0 ii Reﬂected across the x-axis

b x = 0, y = −3

ii Translated 1 unit to the right

iii

y x

−8 −6 −4 −2 −1

e i x = 0 and y = −2 ii Translated down 2 units iii

−2 −3

f i x = −3 and y = 0

−4

ii Translated 3 units to the left

−5 −6

iii

−7

11 a B

b x = 0, y = 2

7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3

17 a i Translate up by 6 units

b i Translate left by 4 units ii

18 a x = 0

b 1 and 3

c x = −3

d 1, 2 and 3

e Translated 3 units left x 1 2 3 4 5

ii y = 0

b i x=0

ii y = −5

c i x=0

ii y = −9

d i x=4

ii y = 0

14 a

ii y → ∞

iii y → −4

b i y→∞

ii y → −∞

iii y → 3

Reasoning

12 a i x = −1

13 a

19 a i y → −∞

20 Arik has forgotten to reﬂect the graph over the x-axis. The correct graph is:

b b

1 15 y = − 3( x − 5) −+ 13 16 a i x = 0 and y = 0 ii Vertical stretch by a factor of 2 iii

1050 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

3 2 1 −3 −2 −1 −1 −2 −3 −4

x 1

21 a False. A hyperbola with a horizontal 1 translation has the equation y = . x−h The variable h must be in the denominator of the fraction. b True. The vertical translation down by 4 units corresponds to a change of −4 in the y-value, so the equation becomes

c False. A horizontal translation affects the expression in the denominator, not the numerator. For example, a horizontal translation of 2 units left is represented by y = d False. The equation

represents a

vertical stretch by a factor of 2, not a vertical translation. 22 a x = h and y = k b Domain is x ≠ h Range is y ≠ k Problem-solving 23 a Vertical translation down by 3 units

d The cost per yearbook without splitting up the ﬁxed fee is $25, so even if the $400 fee was split over thousands of people, it would still add some small amount of money to the $25, making the total per yearbook over $25. On the graph, we can see that the horizontal asymptote is y = 25, so for a positive number of yearbooks, the cost must always be greater than $25. 27 Suppose we have 3x students in class, this means that x students pay $20 each, x students pay $7 each, and the remaining x students pay a total of $60 between all x of them. This makes the total earned by the studio 20x + 7x + 60 = 27x + 60, so the average per student, let’s call it y, would be This can also be written as

If we were to graph this or examine the equation, we notice that the horizontal asymptote is y = 9. This means that as x → ∞, or the classes get very large, y → 9, so the average per student will approach $9.

b k = −3

y 30

24 a Horizontal translation by 3 units to the right.

b k = −3 25

units

15 10

26 a

Cost per yearbook ($) 120 100

(5,105)

80 60 40 20 Number of yearbooks 5 10 15 20 25 30 35 40 45 50 55

c 80 people

x 5 10 15 20 25 30 35 40 45

Chapter 11 review 1 a Inverse

b Direct

2 a Direct

b Inverse

3 a 30

b 180

4 a

171 152 133 114 95 76 57 38 19

Water (mL)

Quinoa (g) 10

Answers 1051 mathspace.co

b 1.9

3 2

• The unit rate of water with respect to quinoa is 1.9 mL/g.

6 a

−5

−4

−3

−2

−1

10 a i Quadrants 1 and 3 ii y → ∞

b 5 hours

8 a 0

b y tends to −∞.

ii y → −∞ 11 a

−2

(1, 5)

4 2

−4 −3 −2 −1 −2

(−1, −5)

−1

iv y → 0

−4

x 1

4, −

1 16 x

x=0

−6 −8

−4 −3 −2 −1 −1

y=0

−4 −3−1−2 −1 −1 −4, 16 −2

−2 −3 −4

−4 −2

−3 x = 0 −4

iii Quadrants 2 and 4 x

b i

iv y → 0

iii y → ∞ 8

y=0

iii y → −∞

b i Quadrants 2 and 4

c y=0 −1

iii Quadrants 1 and 3

7 a k = 20

−4 −2

−3

c As x increases, y increases at a faster and faster rate.

−4

b Increasing

9 a i

−2

c y = 15 x

−4 −3 −2 −1 −1

5 a For an inverse relationship, the product of the coordinates gives us the constant of proportionality, so k = 150. b

c • For every 100 g of quinoa Rose uses, she adds 190 mL of water.

−1 −3

12 a

13 Yes, since the product of the number of winners and the prize money each receives is constant. This product, the constant of proportionality, is 275. 14 a $280 c

1052 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

560

280 186.67 140

112

93.33

y 500

16 a i y → −∞

ii y → ∞

iii y → −8

b i y→∞

ii y → −∞

iii y → 5

17 Tetay has forgotten to reﬂect the graph over the x-axis. The correct graph is:

400 300

200

100

2 x

1 2 3 4 5 6 7 8 9 10

−1 −1

e x is inversely proportional to y.

−2

15 a i Stretched vertically by a factor of 5 ii

5 4 3 2 1

x 1 2 3 4 5 6 7 8 9 10 11 12 13

−3 −4

5 x

18 a x

b 180

−5 −4 −3 −2 −1 1 2 3 4 5 −1 1 −2 y= x −3 −4 −5

Cost per guernsey ($)

160 140 120 100

b i Translated down 4 units ii

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 x

1 2 3 4 5

6 5 4 3 2 1 −6−5−4−3−2 −1 −1

1 x−2 x

1 2 3 4 5 6

No. of guernseys 5 10 15 20 25 30 35 40 45 50 55

c 100 guernseys

1 −4 x

c i Translated horizontally right by 2 units ii

(10, 80)

d The base cost of each guernsey is $30, so even if the $500 design fee were distributed among a very large number of orders, it would still add a small amount to the base cost of $30, ensuring that the total cost per guernsey is always more than $30. In the graph, the horizontal line y = 30 acts as an asymptote; thus, as the number of guernseys increases, the cost approaches but never falls below $30.

1 −2 x −3 −4 −5 −6

Answers 1053 mathspace.co

12.01 Pythagoras’ theorem

14 h = 5.61 m

Understanding

15 a x = 4.9 m

b x = 1.8 cm

16 a $210

b $420

1 a z2 = x2 + y2

b x2 = z2 − y2

2 a x = 8.49

b x = 3.61

c x = 6.24

d x = 8.06

18 Beth, by 69 m

b 49

12.02 Trigonometric ratios

3 a 45

17 205 cm

c No, not a right-angled triangle, 45 ≠ 49.

Understanding

Fluency 4 a No

b No

c Yes

d Yes

e Yes

f No

g No

h Yes

5 a c=5

d c = 17

1 a AB 2 a AC

b BC

c BC

3 a i

iii

b i

iii

c i

iii

d i

iii

e i

iii

f i

iii

e ∠ACB

g Hypotenuse: h Side: i Side: 6 10 m 7 a 14.1 cm

b 11.2 mm

c 29 m

d 13.6 cm

8 a 10.8 cm

b 19.2 mm

9 132 m

4 a i

b i

c i

Reasoning 10 The triangle can be a right-angled triangle since 32 + 52 = 72 (9 + 25 = 34 ≠ 49). Hence, by the Pythagoras’ theorem, it’s not a right-angled triangle. 11

102 = 82 + BD2

∴ BD = 6 cm 2

10 = 8 + AC

Fluency 5 a i

iii

b i

iii

∴ AC = 6 cm ∴ BD = DC = 6 cm, so D is the midpoint of BC. 12 a 152 = 225

92 + 122 = 225

∴ 152 = 92 + 122 and Pythagoras’ theorem holds. b i Yes

ii No

iii Yes

iv No

6 7 a f = 4.08

b f = 6.12

c g = 9.46

d a = 11.59

Problem-solving 13 a 25 cm

b BC

8 b=8

1054 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

d AC

12.03 Angles of elevation and depression

9 d = 3084 10 29.0° 11 a i x = 10

ii 53.1°

b i x=4

ii 53.1°

Understanding

12 d = 1.48

1 It is an angle of elevation as it is the angle formed from the horizontal looking upwards.

13 1.96 m

2 a σ or ∠CBD

14 h = 4.4 m

3 a α or ∠BCA

15 θ = 9.88° 16 a x = 155.43 m 17 a h = 21 c h = 154, x = 72

b θ or ∠ACB

c They are the same size.

b 425.86 m b f = 22 d x = 34

e x = 12

b Draw a horizontal line from A to the left of A and the angle of depression is from the horizontal line to the line AC. c α Fluency

Reasoning

4 a x = 37.23°

b y = 52.77°

18 An isosceles triangle can be split into two right-angled triangles. Therefore, the adjacent side can be seen as half the length of the base.

5 a y = 36.87°

b x = 53.13°

Using the formula:

7 x = 182

6 θ = 52.52°

8 d = 229 Simplify the numerator and make A the subject of the identity. Take the inverse cos of both sides.

9 a AB = 11.86 m

b AC = 16.24 m

c BC = 4.4 m 10 a AC = 9409 m

b BC = 5808 m

c 3601 m

11 a d = 7.81 m

19 Using the formula:

c 66.08°

Multiply both sides of the equation by b, and make b the subject of the equation by dividing both sides by 0.3.

b h = 13.53 m d 69.72°

Reasoning 12 Viggo is incorrect as the angle of depression must be measured from the horizontal. The correct angle of depression is 23.9°. 13 Fiona is not correct as an angle of elevation is measured from the horizontal.

14 a The shadow gets longer as the angle of elevation gets smaller.

Problem-solving 20

b The length of the shadow is zero.

21 22 a 135°

b I = 135°

23 a 131.2 m

b 101.3°

c tan 90° is undeﬁned. The sun directly overhead means the angle of elevation is 90° and it’s not possible to draw a triangle, therefore the shadow has no length.

24 31.8°

Answers 1055 mathspace.co

Problem-solving

18 a x = 15.1

15 a x = 455.84 m

b 425.84 m

19 d = 2130

16 a h = 127.38

b 129.1 m

20 a h = 9.37

17 d = 1022.58

b θ = 25°

Reasoning

18 h = 17 m

21 a The true bearing is (360 − a)°T

19 d = 10.86

b The true bearing is (180 + b)°T c The true bearing is (180 − c)°T

20 θ = 45.97 21 a 29.26°

b y = 13.1

b 18.76°

c 4.61°

d 100 cm

22 a 60 m

b 1601 m

23 a PC = 22.65 m

b PA = 18.29 m

d The true bearing is (90 + a)°T 22 a Sample answer: At point A use the bearing of 45° to draw a right-angled triangle. The horizontal distance, a, is found using cos 45° =

12.04 Bearings

At point B use the bearing of 60° to draw a right-angled triangle. The horizontal distance,

Understanding 1 a NW

b NE

c SE

d SW

2 a 244° T

b 334° T

c 296° T

d 206° T

3 C

b, is found using cos 30° =

Therefore the overall distance is a + b = 12 cos 45° + 16 cos 30°. b 12 sin 45° + 16 sin 30°

4 a 225° T

b 315° T

e 045° T

f 135° T

c 27.76 km

d 233.6°T

23 a 1131.7 m

b 294.5°T

c 292.5° T d 157.5° T

Fluency

Problem-solving

5 a i 030° T

ii N 30° E

b i 221° T

ii S 41° W

c i 320° T

ii N 40° W

d i 125° T

ii S 55° E

6 a 090° T

b 127° T

c 222° T

7 a 270° T

b 317° T

c 052° T

24 Due west 25 a 65°

b b = 5.0

26 a ∠BSP = 90°

c d = 11.8

b BS = 750.3 m

c SP = 525.4 m

d 2191.7 m

27 a i ∠BAC = 65°

ii ∠DCE = 44°

b 49.2 km

8 a 319° T

b N 41°W

9 a 053° T

b N 53°E

28 a N a°W

10 a 132° T

b S 48°E

c a = 47°

11 a 221° T

b S 41°W

c θ = 215°T

29 a

b x = 7.31

12 NE 13 231° T 248 km

14 305° T

15 a = 62°

225 km

16 w = 90.1 km 17 a b = 53.6°

Start

Finish

b S 53.6°W

c 233.6°T

1056 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b 227.8°T

c 334.9 km

30 a

13 a ∠B = 74° or ∠B = 106°

14 a a = 19.71

b a = 10.00

c a = 212.34 A

17.6 m

38°

16 q = 11

dm C

31 68.96 km b 2033 m

c AB = 2403 m

12.05 Optional: Sine and cosine rules

1 a Sine rule

b Pythagoras’ theorem

c Cosine rule

d Trigonometric ratio

2 ∠ABC 3 4 Possible answers: 2

17 v = 85.17 18 a B = 43.38

b w = 31.69

c w = 41.49

d A = 27.33

19 V = 101° 20 Q = 18° 21 a 60.21 m2

c 326.26 cm e 110.23 cm2

Understanding

d c = 10.60

b d = 22.3

32 a 1282 m

b B = 32°

• p = q + r − 2qr cos P

• q2 = p2 + r2 − 2pr cos Q

b 12.41 m2

d 99.52 cm2

f 109.23 cm2 or 10 922.57 mm2

22 b = 38 cm 23 a i x = 23°

ii A = 21.1 cm2

b i x = 45°

ii A = 34.9 cm2

24 a BC = 10.24

b 24.44 units2

25 a AC = 4.8 cm

b 16.3 cm2

c 19.3 cm

• r2 = p2 + q2 − 2pq cos R

Reasoning

5 a Cosine rule

b Sine rule

c Sine rule

d Cosine rule

26 Blain’s ﬁrst mistake is making the equation in Step 1 equal to c instead of c2. Blain’s second mistake is in Step 3, when Blain does not follow order of operations and incorrectly subtracts 360 from 369 instead of subtracting 360 cos 34° from 369. Odett’s mistake is in Step 1, as the set up of the law of cosines is incorrect. The ﬁrst line should be c2 = 152 + 122 − 2(12)(15) cos 34°. Odett then solves for c from an incorrect initial equation.

Fluency 6 a a = 13.40

b a = 6.85

c x = 12.92

d a = 16.66

e x = 11

f a = 17.71 mm

7 30.16 m 8 a ∠A = 75.78°

b b = 3.152 m

c = 8.40 m

9 a x = 71.71

b x = 45.45

27 a We could use the fact that diagonals of a rhombus bisect each other at right angles.

10 a x = 26.0°

b x = 32.7°

So we could solve for

c x = 30.4°

d x = 30.3°

Pythagoras’ theorem (using the right-angled triangle with hypotenuse of length 10 and one side of the triangle with length 8). We can then solve for d.

c c = 5.723 m

11 a x = 44.1°

b x = 43.2°

c x = 133.1°

d x = 23.5°

12 25°

using the

Another way to solve for d would be to use the law of cosines. A rhombus is a

Answers 1057 mathspace.co

parallelogram with four congruent sides, so we can solve for θ and x using the law of cosines. Then use this information to ﬁnd the angle needed in order to use law of cosines. Once the angle is found, law of cosines can be used to solve for d.

37 a There is one possible triangle from the given information. Using the law of sines, ∠B = 35.53°. The other possible value for ∠B would be 180° − 35.53° = 144.47°. However, this would not be possible in this case because then the sum of ∠A and ∠B is greater than 180°.

b d = 12 cm

b There are two possible triangles from the given information. Using the law of sines, ∠A = 67.90°. The other possible value for ∠A would be 180° − 67.90° = 112.10°. This is possible because the sum of ∠B and ∠A is 157.1° and this is less than 180°.

28 There is no solution because cos A = 1.1 and it is not possible for the cosine of an angle to be greater than 1. 29 Let’s consider an equilateral triangle with sides of equal length a. We will denote the opposite angles to these sides as A, B and C. By the cosine rule, let us ﬁnd one of the angles, say ∠A. In this case: Substitute c = a, b = a, C = A Simplify the equation Isolate cos A Evaluate

So,

. To ﬁnd the ∠A, we take the

inverse cosine of :

c There are no possible triangles from the given information. Using the law of sines, sin B = 1.042 and no angles between 0° and 180° satisfy this equation because the sine of any angle is always between −1 and 1. 38 x = 30° 39 v = 10 40 460 km 41 a ∠A = 38°

b b = 18.63 m

c c = 22.71 m

d 1.8 min

42 a x = 43.60

b 360 m2

c 1710 dollars 43 a h = 42

b x = 123

c y = 97

b 68 m

c 31.9 m

Since the triangle is equilateral and all sides are equal, the other ∠B and ∠C will also be equal to 60°.

45 a 94°

Problem-solving

Chapter 12 review

44 38 m

1 a b=8

30 θ = 33.84° 31 a ∠WST = 136°

b SW = 234 kilometres

c h = 102.58 32 1 244 548 km2

c hypotenuse = 2 No, to be a right-angled triangle we can check if (6, 8, 13) is a Pythagorean triple. 62 + 82 = 36 + 64

33 248.8°T 34 a a = 74°

b third side = 10

= 100

133 = 169

b d = 28.35 m

c h = 16.66 m 35 a ∠ADB = 23°

b a = 35.65 m

c h = 20.4 m 36 a c x = a sin B

b d

Sum of squares of shorter sides Square of longer side, the hypotenuse

≠ 100

Compare to sum of squares of shorter sides

62 + 82 ≠ 132

Not a Pythagorean triple

3 a 1083.07 m

b 1437.92 m

4 a f = 2.77

b f = 8.91

5 a

b θ = 30°

1058 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

6 a 50°

b x = 19.30 m

c 96 cm

7 a h = 20

b x = 26

8 θ = 54.69° 9 a AC = 12 032 m

b BC = 10 462 m

c 1570 m

b 94.6 m

12 a i 035° T

ii N 35° E

b i 218° T

ii S 38° W b N 41° W

f 204 cm2

7 a i 5 cm

ii 228 cm2

ii 608 cm2

b i 8 cm 8 a 216.49 cm2

9 a

16 w = 88.9 km 17 Due west 18 a x = 13.13

b a = 6.85

19 a B = 32°

b ∠B = 74° or ∠B = 106°

20 ∠ABC

10 a i No

ii No

iii Yes

iv No

b 262 square units b 452.39 cm2

c 207.35 m

d 439.82 m2

12 a 75.40 m2

b 75.40 m2

13 a 12.5664 m

b 50.27 m2

14 h = 69 15 a i x = 11.00 cm b i z = 4.90 m 16 a 91 m

ii 376.95 cm2 ii 858.76 m2

b 50 m 2

17 a 312π cm

21 a = 19.71

d 534 cm2

c 11.32 m2

15 301° T

b 380 cm2

11 a 23 184.95 cm2

14 SW

b 12.98 cm2

18 596 cm2

22 ∠C = 115° 23 a 5.94 m2

b 24.30 m2

24 a x = 51°

b A = 33.2 m2

25 1389.32 m2

13.01 Surface area of prisms and cylinders Understanding b x2 cm2

c SA = 6x2

2 7 faces 3 a 6 squares b 2 circles and a rectangle c 4 triangles and a square d 6 rectangles and two hexagons

4 150 cm2

e 380 cm

c 174.49 m

11 a h = 92.97 m

Fluency

d 392 cm2

10 θ = 28.01°

1 a 6 faces

b 451.26 m2

c h = 20.50 m

13 a 319° T

6 a 288 m2

19 103 m2 20 11 998 cm2 21 109.78 m2 22 158.4 m2 Reasoning 23 Example answer: The formula depends on how many circular faces the cylinder has as each circular face has an area of π r2. A closed cylinder has two circular faces so the formula is SA = 2π r2 + 2π rh. A cylinder open at one end has one circular face and the formula is SA = π r2 + 2π rh. A cylinder open at both ends has no circular faces and the formula is SA = 2π rh.

5 8 cm

Answers 1059 mathspace.co

24 They are both correct as the formulas are the same. Scott’s formula is the factorised version of Lachlan’s formula. 25 a Option 2 has a smaller surface area (219.5 cm2) than Option 1 (about 245 cm2) so Quicker Oats should choose Option 2 to minimise the amount of materials used. b Maximising profits means selling the smallest volume of chocolate chip oats per package. Option 2 has a smaller volume (185.25 cm3) than Option 1 (about 282.74 cm3) so to maximise profits, Quicker Oats should choose Option 2 to minimise the amount of product used. 26 a The surface area is made up of: • Two rectangles for the curved walls, both with heights of 3 cm and lengths equal to the circumference of circles of radii 2 cm and 10 cm respectively. • A small circular base with a radius of 2 cm that fits into the large circular base, making a complete circle with a radius of 10 cm. • A large circular base with a radius of 10 cm.

13.02 Volume of prisms and cylinders Understanding 1 a True

b False

3 Options B and D Fluency 4 a i Triangle

iii 90 cm3

b i Trapezium 3

5 a 520 cm3 c 64 cm

d 231 cm3

f 90 cm3

e 840 cm

6 Base Area = 80 cm2

28 2978.7 cm2

8 a 785.4 cm3

c 678.6 cm

29 a Any scale factor s which satisﬁes the equation s2(350) = a, where a is the new surface area of the box and 325 ≤ a ≤ 335. For example,

, and

b Any dimensions which result in a total surface area between 325 cm2 and 335 cm2. For example, 11.65 cm × 9.71 cm × 2.43 cm. 30 20.00 m2 b 4 L costs $98 2

d 836 cm

ii 99 cm2 b 114 cm3

7 Height = 5 cm

32 a h = 24

ii 9 cm2

iii 495 cm

Problem-solving

31 a 11.13 m2

d True

2 The volume of both a rectangular prism and a cylinder involves ﬁnding the base area and multiplying it by the height. However the methods of ﬁnding the base area, the interpretation of the height, and the visual representation of the volume differ between the two shapes due to their distinct geometric properties.

b Surface Area = 854.51 cm2

27 Surface Area = 4π r2

c True

e False

b h = 21 e Bag B

c 886 cm2

b 1335.2 cm3

d 226.2 m3

e 45.9 cm

f 70.7 cm3

9 a 1696.5 cm3

b 2309.1 cm3

g 307.9 cm3 3

h 955.0 cm3

c 59.7 cm

10 883.57 cm3 11 16 286.02 L 12 864.19 cm3 Reasoning 13 a 1.782 m3

33 495.1 cm

34 186.32 cm2 35 Number of cans = 9 36 212.4 cm2

1060 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b 2 m3

c Oliver’s father’s rounded measurements give a volume that is 0.218 m3 larger.

14 Doubling the length and width of the right prism while keeping the height the same results in a four-fold increase in the volume.

13.03 Surface area and volume of composite solids

15 a 37.31 m3

Understanding

b 37.29 m

c The exact value is larger by 0.02 m . d In these real-world situations, using rounded or approximated values can lead to small discrepancies in calculations. In the case of the cylindrical container, it could mean not having enough water for the experiment if a large number of cylinders needed to be filled. It is essential to understand the potential impact of approximations and make decisions accordingly. e 37 310 000 mL Problem-solving 16 1 134 000 L 17 a Vidar should get the flat base type. b Yolanda should get the flat base type. 18 The second can with a diameter of 18 cm and a height of 16 cm. 19 a Morcom Aquatic Center b The pool at Morcom is 1.8 times the volume of the pool at Sanford Seminole. c 20 a 1335.18 cm3

b 4005.54 cm3

c 28 L

2 a To calculate the surface area of a composite solid, we need to calculate the area of all of the faces that are on the outside of the composite solid. b To calculate the volume of a composite solid, we need to calculate the volume of each of the solids that make up the composite solid and combine them appropriately. Sometimes this requires addition, sometimes subtraction. 3 a A cube and a square pyramid b Two rectangular prisms c A trapezoidal prism and a triangular prism d A triangular prism and a cone e A cylindrical hole in a rectangular prism f Two hemispheres 4 a A rectangular prism and a triangular prism. b Three cylinders, two of which are identical. c Two trapezoidal prisms. d A rectangular prism and a cylinder. Fluency 5 a 672 cm3

21 37.5 L 22 a 35 343 L

b 35.343 kL

23 a 140 m2

b 8820 L

24 90 m

b 171 cm3

c 1920 cm3

d 3136 cm3

6 a 1608.50 cm3

b 883.57 cm3

d 609.47 cm3

c 263.89 cm 7 0.57 m3

25 67.5 L

8 a 1140 m3

27 a 176 m2 c 3520 L

b 424 m3

c 1203.75 m3

26 4.35 m

b 3.52 m3 d 880 L

28 a 2540 m2

b 406.4 m3

29 a 7771 m2

b 1631.91 m3

30 14 kL

1 A composite solid is a 3D ﬁgure that is composed of simpler geometric ﬁgures. Examples could include a house shape (a rectangular prism with a triangular prism for the roof) or a cone with a hemisphere (like an ice cream cone) on top.

d 272 cm3

f 601.65 m3

e 982.35 cm

g 7164 units

h 1800 cm3

9 a 1512 cm2

b 429.37 m2

c 228 cm

e 6755.25 mm g 120 cm2

d 365.45 cm2

f 867.08 cm2 h 621.24 cm2

10 157π cm2

Answers 1061 mathspace.co

Reasoning

b Calculate the surface area of the original cube, subtract the four rectangles which are the ends of the holes and add in the rectangular surfaces which are parts of the prism going through the shape.

11 The simple solids and their required dimensions to ﬁnd the volume are: • A rectangular prism for the base. We need to know the length, width, and height. • A cylinder for the pole. We need to know the radius or diameter of the pole, and its height.

6 × 142 − 4 × 5 × 6.5 + 4 × 14 × 6.5 + 4 × 14 × 5 = 1690 m2

• A rectangular prism for the backboard. We need to know the length, width, and height. One of these dimensions can be thought of as the thickness.

Problem-solving

• A cylinder that has had a cylinder cut out of it for the hoop. We need to know the inner and outer radii or diameters, and the vertical thickness of the hoop.

15 a 1658.76 cm2

14 a 1198.5 m3

b 299.625 m3

c h = 0.5875 m b 131.95 cm2

d 3298.67 cm2

c 1507.96 cm 16 13 815.88 cm2 17 a 68 m2

12 a i

b 70 m2

d 376 m2

c 100 m

Volume = 300 cm

e $16 544 18 a 16.3 cm2

b 614

19 a 36.47 m

Volume = 300 cm3

b 36 469 L

c 34 016 L

b Calculate the total surface area of the rectangular prism:

d 1717 L

e 21 20 a 127.1 m2

2 × 7 × 6 + 2 × 6 × 8 + 2 × 7 × 8 = 292 cm2 Subtract the triangle at each end of the triangular hole:

b 140 m2

c $2520

13.04 Estimation and accuracy Understanding

Add in the three rectangular faces that are part of the triangular prism going through the shape. Pythagoras’ theorem is needed to work out that the width of two of these rectangles is

. 2

cm 2

= 361 cm (nearest whole number) 13 a To ﬁnd the volume of prism, use the formula: Volume of Prism = A × h where A is the base area and h is the height. To ﬁnd the base area, subtract the area of the two rectangles from the square. Area of Rectangle = l × w Area of Square = side × side

Volume = (14 × 14 − 2 × 6.5 × 5) × 14 m3 Volume = 1834.00 m3

1 a True

b False

c True

d False

2 6.35, 6.36, 6.37, 6.38, 6.39, 6.40, 6.41, 6.42, 6.43, 6.44 3 5.44 4 0.15 5 12.715 Fluency 6 a • Length of the rectangular prism: 7.95 cm • Height of the rectangular prism: 7.95 cm • Width of the rectangular prism: 5.95 cm • Diameter of the cylinder: 4.95 cm b • Length of the rectangular prism: 8.05 cm • Height of the rectangular prism: 8.05 cm • Width of the rectangular prism: 6.05 cm • Diameter of the cylinder: 5.05 cm c Total surface area = 393 cm2

d Total surface area = 405 cm2

1062 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

measurement have a less drastic impact on the momentum calculation.

7 a Total weight = 68.051 kg b Total truncated weight = 68.03 kg c Total rounded weight = 68.05 kg d The truncated value is slightly below the actual total, while the rounded value is very close to the actual weight. 8 a 7.2%

b 14.6%

c 4.2%

d 5.3%

Problem-solving 13 $9 125 000 14 a 40.6 thousand years ago b 33.0 thousand years ago c 10.3%

Reasoning 9 The company would overestimate the weight of the materials, which could lead to overpayment if the materials are bought or sold based on weight. They could reduce this error by recalibrating or replacing the weighing scale. 10 a Different rounding and level of accuracies being used. b Caleb rounded his to the nearest metre, Ida used to the nearest tenth of a metre and Lavinia to the nearest hundredth of a metre. c Yes, they have all used different levels of accuracy.

15 a 48.15%

b 1.48%

Chapter 13 review 1 a 544 m2

b 216 cm2

d 608 cm2

c 612 cm 2 h = 71

3 Design B because it has a smaller surface area (494.12 cm2 ) than Design A (531.93 cm2 ). 4 a x = 12.10 cm

b 537.21 cm2

5 a 490 cm3

b 200 cm3

c 2040 cm

11 Truncating the interest rate could lead to slightly lower total interest payments. Over a large number of customers, this could result in signiﬁcant savings for the bank.

6 Base Area = 110 cm2

12 a Alex will use: Mass: 0.37 kg (the closest value to Liam’s measurement that Alex’s scale can measure) Velocity: Using Liam’s measurements, the time 5 taken is = 2.2321 s, which Alex will 2.24 measure as 2 seconds, giving a Velocity of 5 = 2.5 m/s m/s. 2 b i 1.16 Joules ii 0.92 Joules c 26.09% d i 0.93 kg m/s

ii 0.82 kg m/s

e 13.41% f T he percentage error in Alex’s kinetic energy calculation is higher than the percentage error in his momentum calculation. This is because kinetic energy is proportional to the square of the velocity, so errors in the velocity measurement are amplified in the kinetic energy calculation. In contrast, momentum is directly proportional to velocity, so errors in the velocity

d 140 cm3

7 a 552.92 cm3

b 2261.95 cm3

8 a 3.42 m3

b 0.684 m3

9 a 180 cm3

b 824.67 cm3

10 a i 864 cm3

ii 648 cm2

11 a 58 904.86 cm3

b 2120.58 cm2

12 a 488 m2

b $1802

b i 1526.81 cm3 c 2434.73 cm2

13 a Trina’s

ii 894.58 cm2 d 3063.05 cm2

b Kareem’s

c 0.025 cm

14 By rounding down, the total calculated viewing time will be slightly less than the actual time spent by users. Over a vast number of users, this could result in a signiﬁcant discrepancy between the tracked and the actual viewing times. 15 The products may be overcooked if the timer is consistently slow, which can affect the texture and taste. 16 a 525 Myr c 5%

b 475 Myr d 2%

Answers 1063 mathspace.co

14.01 Independent and dependent events

12 a i

b i

c i

d i

Understanding 1 a Independent

b Dependent

c Independent 2 Dependent 3 a Independent

b Dependent

13 a Dependent b i

iii

4 a b Drawing an orange ball, then a pink ball c • Drawing a pink ball, then an orange ball • Drawing a pink ball, then a pink ball

d Dependent 5 a Two days

14 a i

b No

c Rain on the ﬁrst day, sun on the second day (RS) and sun on the ﬁrst day, rain on the second day (SR). Fluency

ii b i ii

6 a

b Independent

c i

7 a

d i

c Independent

8 a Independent

e i

b i 6 possibilities out of 36: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

The answer is 15 a

ii 6 possibilities out of 36: (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6).

The answer is iii

v 0

9 0.5

10 Yes, events A and B are independent. 11 a

16 a

1064 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Since the card is replaced, there are always 11 cards to choose from.

17 a b 18 a b c 8 possibilities out of 90: (2, 10), (3, 9), (4, 8), (5, 7), (7, 5), (8, 4), (9, 3), (10, 2).

26 Independent would mean that P (A ∩ B) = P (A) × P (B) and since P (A) ≠ 0 and P (B) ≠ 0, it follows that P (A ∩ B) ≠ 0. For mutually exclusive events P (A ∩ B) = 0. 27 If the events A and B are independent, then P (A ∩ B) = P (A) × P (B). Here

The answer is 19 a

20 a

21 a

22 a

Problem-solving G

B G

B B

b i

28 a

29 a

d The probability is getting higher. e 25

23 a {SR, SL, SC, GV, GM, GC} b

30 a

e Increase Reasoning 24 a

31 a b

c No

d The draws are independent if the probability of each successive ball drawn is the same as the probability of the ﬁrst ball drawn. The draws are dependent if the probability of each successive ball drawn is not the same as the probability of the ﬁrst ball drawn. In this case, it is dependent. 25 3 of the cards are labelled 1, 2 or 3. 3 of the cards are labelled 9, 10 or 11.

Edge

C D

Probability b

Edge Probability

c i Higher probability by ii Higher probability by iii Lower probability by

Answers 1065 mathspace.co

32 a 1st race 0.8 W 0.7

0.2

2nd race W

0.9 0.1 0.6

0.4 0.4

0.3 L

0.6

0.6 0.4 0.1

0.9

b 0.504

3rd race W WWW L WWL W WLW L WLL W LWW L LWL W LLW L LLL

c 0.678

33 a

3 A 4 a Patricia is chosen second, after Aaron has been chosen ﬁrst. b Greater than Fluency 5 0.8 6 0.18 7 a

b 0.4

Tea Toast

8 a 0.4

Milo Tea

e 9 a

Cereal Milo

b 0.42

10 a i 260

c 0.88

34 a

Recipient

Not

11 a {0, 1, 2, 3, 4, 5}

Recipient Not

Recipient Not Not

Recipient Not

Not

iii 111

12 a

13 0.6875 14 a

b Given that the ﬁrst die is a two, the possibilities are six: (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6).

c d

Out of them, there are only two possibilities with sum of 7 or more: (2, 5) and (2, 6).

14.02 Conditional probability

The answer is

Understanding

16 a

17 a

15 a

1 a False

iv 208

Not

Recipient

ii 237

Recipient

b True

c True

d True

1066 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

Problem-solving

18 a (0.4 × 0.8) + (0.6 × 0.7) = 0.74

25 a

Comedy News

Comedy

Horror

19 a 0.75 × 0.05 = 0.0375

Animation

b 0.25 × 0.95 = 0.2375

Comedy Comedy

c 0.0375 + 0.2375 = 0.275 Current Affairs

d 20 a 177

Horror Animation Comedy

Horror

Animation

Comedy Horror

21 a 0.8 × 0.5 = 0.4

Animation

b (1 − 0.8) × (1 − 0.5) = 0.1 c 1 − 0.1 = 0.9

d (1 − 0.8) × 0.5 = 0.1

26 a

e P (A∣B) = P (A) = 0.8 (A and B are independent events)

0.7

0.8

Reasoning 22

C T

0.85

0.3

0.2

0.15

Since P (B ∩ A) = P (A ∩ B), therefore P (A∣B) × P (B) = P (B∣A) × P (A) 23 a

b 27 a

e Since B is inside of A, if the event B has occurred, the event A must have occurred. So, P (A∣B) = 1.

28 a

24 P (A∣B) is greater because:

and

But according to the size of the circles on the Venn diagram:

Labor (next election)

Liberal (next election)

Greens (next election)

Total

Labor (last election)

Liberal (last election)

Greens (last election)

Total

140

200

29 a

e H and S

Answers 1067 mathspace.co

14.03 Mixed probability concepts

11 a

Understanding 1 a Diagram iv

b Diagram i

c Diagram ii

d Diagram iii

2 a P (A ∩ B) = P (A∣B) × P (B) b P (A) + P (B) = 1

Watched Didn’t watch Total the movie the movie Read the book

Didn’t read the book

Total

104

166

b i

c P (A ∪ B) = P (A) + P (B) − P (A ∩ B) d P (A ∩ B) = P (A) × P (B)

12 a i x = 40

iii

ii y = 400

iii z = 200

b 360

3 B

Fluency

13 a

4 w = 0.3, x = 0.3, y = 0.2, z = 0.2

14 a

5 6 a {23, 24, 32, 34, 42, 43} b i

iii

7 a

b 10

15 a

8 a 3

9 a

Y B

Do not Believe believe in Total in Santa Santa Believe in Tooth Fairy

Do not believe in Tooth Fairy

Total

100

b 38.75%

10 a 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Y B R Y

b i

iii

v 0

vii

16 a

b 1, 2, 3, 4, 6, 8, 12, 16

BY RB RR RY YB YR YY

Serve in 0.55

c 4 numbers from set A: 10, 14, 18, 20 and 2 numbers from set B: 1, 3.

Serve in 0.81

In total, 6 numbers are unique.

0.45

Fault

0.19 Fault

b 0.45

1068 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c 0.0855

17 a Star Jet

Star Jet

Air Tiger

Star Jet Air Tiger

Air Tiger

23 a b The probability of picking an identical pair depends on how many cards are left. Since the number of cards decreases as pairs are picked, the probability of picking a pair is affected by how many cards have been removed. In this case, it is dependent. 24 a No second attempt is needed if the exam is passed on the ﬁrst attempt.

b i ii

18 a

19 a

c Candidates will have a variety of abilities and preparedness for the examination. Problem-solving

20 0.856 Reasoning 21 a P (A ∩ B) = P (A) × P (B) b P (A ∩ B) = P (A∣B) × P (B) c The formula P (A ∩ B) = P (A∣B) × P (B) can be used for both independent and dependent events. For dependent events, we have to use the formula from part (b). For independent events, P (A∣B) = P (A) since it doesn’t matter whether B occurred or not. So, we can use the formula from part (b) and still get the same answer as the formula from part (a). 22 a i

iii

26 a 0.18

b 0.45

c 0.38

b 43

d 0.64

e 0.9 27 a 27

28 a (0.55 × 0.02) + (0.45 × 0.08) = 0.047 The answer is 4.7%. b 0.23 29 22.5% 30 10.58%

Chapter 14 review

b c Two events A and B are considered independent if P (A∣B) = P (A). Using the fact that

25 a

, we

can rewrite this condition for independence as P (A ∩ B) = P (A) × P (B). We calculated P (S ∩ H) and P (H) × P (S), and with these results, we can conclude that these events are independent.

1 a

b Independent

2 a

3 a

4 0.2 5 a

6 a 0.7

b 0.65

7 a

Answers 1069 mathspace.co

d iii: Upper quartile, Q3: At most 25% of the data is above this value

8 j = 0.4, k = 0.4, l = 0.1, m = 0.1 9 a

e i: Maximum: Highest value

240

120

Fluency

210 70

120 4 = 330 11

5 a 74

b 61

c 114

d 53

6 a 72

b 51.5

c 85.5

d 34

7 a i 6, 10, 15, 17, 24, 39, 40

70 7 = 310 31

iii 17

Went hiking

Did not hike

Total

iii 7

Went swimming

Did not swim

Total

100

10 a

b i 11 a

Egg sandwich

6 10 4 10 Muffin

b 0.3

5 10 5 10

Orange juice

13 a

14 a

Iced coffee

2 a Q1 = 1, Q3 = 9 c Q1 = 15, Q3 = 26.5

iv 8

ii 7

v 19

vi 11

d i 20, 22, 28, 32, 40, 42, 43, 54 iii 36

iv 25

v 42.5

iv 74.5

v 84.5

iv 101

ii 8 vi 10 ii 5

v 114

vi 13

g i 197, 198, 202, 205, 207, 228 iii 203.5 iv 198

ii 8 vi 17.5

f i 100, 102, 110, 113, 115

v 207

ii 6 vi 9

h i 15.8, 19.5, 22.4, 29.6, 35.4, 39.1, 46.2 iii 29.6

iv 19.5

v 39.1

vi 19.6 b i 19 kg

ii 17 kg

iii 21 kg

9 a 51

b 7

c 1

10 a 47

b 7

c 4

c 2

11 a 5

c 49

vi 13

8 a 18.93 kg

Understanding b 11.5

iii 14

ii 7

15.01 Quartiles

1 a 28

ii 7

v 9

c i 4, 8, 10, 14, 15, 19, 20

iii 110

d b

vi 29

e i 69, 71, 78, 79, 82, 84, 85, 88

5 10 5 10

12 a

iv −4

iii 80.5

Orange juice

c 0.8

ii 7

v 39

b i −6, −4, −1, 7, 7, 9, 9

iii

iv 10

b 16

c 11

d 17

12 a 19.5

b 26.5

c 7

13 a 72 hours

b 85 hours

c 13

14 a

Score (x)

Frequency ( f )

d 7

b Q1 = 24, Q3 = 32 d Q1 = 16, Q3 = 24.5

3 10.5 4 a iv: Minimum: Lowest value b v: Lower quartile, Q1: At most 25% of the data is below this value c ii: Median: 50% of the data lies on either side of this value

1070 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b 7

Totals

15 c 6

d 8

e 2

15 a 1

b 3

c 2

16 a 75

b 8

c 3

17 a 79

b 64

c 84

d 20%

18 a 74

b 79.5

c 24

d 35

e 16.5

f 19.75

e 20%

g The Emerald Eagles’ scores are approximately symmetrically distributed around the median, while the Sapphire Sharks’ scores show a slight right skew, indicating the majority of scores on the lower side and some higher scores pulling the mean upwards. Reasoning 19 a 8 marks

b 9 marks

c 1

d In this data set, the scores less than 7, while constituting the minimum values, don’t significantly impact the mean and median due to their low frequency. However, they do increase the range and could potentially affect the first quartile by shifting it lower. 20 a Set 2 has the largest interquartile range. This indicates that the middle 50% of data in Set 2 is spread over a larger range of values compared to the other two data sets. b Set 1 appears to be symmetric as the differences between corresponding quartiles are nearly equal, indicating a symmetrical distribution. Set 2 is skewed right because the difference between the third quartile and the maximum is much larger than the difference between the minimum and the first quartile, indicating a few higher values causing a right skew. Set 3 is skewed left as the difference between the minimum and the first quartile is much larger than the difference between the third quartile and the maximum, indicating a few lower values causing a left skew. 21 a For Group 1, the range is 39, the interquartile range is 23 (116.5 − 93.5), the median is 99 and the mean is approximately 104.24.

b The mean of Group 1 is higher than that of Group 2, which suggests that Group 1, on average, scored higher on the EQ test. c The spread of Group 1 is larger than that of Group 2, indicating that there is a greater variability in EQ scores within Group 1 than within Group 2. 22 a 2.5

b 8

c There are only 2 students who have more than 8 pets (the upper quartile), which is equal to 8% of the total number of students (2 out of 25). It is not exactly 25% because the quartiles are determined by dividing the data into 4 equal parts, but this does not guarantee that the percentage of values above or below a quartile will always be 25%. 23 Yes, the percentage of data values between the upper and lower quartiles is always exactly 50%. This is because the lower quartile (Q1) represents the 25th percentile of the data and the upper quartile (Q3) represents the 75th percentile of the data. The difference between the 75th and 25th percentiles is 50% of the data. For example, consider the dataset: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The lower quartile is 3 (25th percentile) and the upper quartile is 8 (75th percentile). The data values between Q1 and Q3 are 4, 5, 6 and 7, which represent 50% of the total dataset. Problem-solving 24 65 25 a

minutes. This means that about half of

the ﬂights had a delay less than

minutes

and half had a delay of more than b 10 minutes

minutes.

c 45 minutes

d 35

e 83

f 83%

g 22

h 22% 26 a 142 minutes

b 27

c 137 minutes

For Group 2, the range is 30, the interquartile range is 18 (105 − 87), the median is 100 and the mean is approximately 97.48.

Answers 1071 mathspace.co

15.02 Boxplots

Minimum

Understanding 1 a

Lower quartile

Median

Upper quartile

Maximum

c 20

Minimum

Lower quartile

Median

Upper quartile

Maximum

Minimum

Lower quartile

Median

51.5

Upper quartile

Maximum

7 a

2 a i 19

ii 11

b i 34

ii 15

3 a True

b False

e True

f True

4 a i 50%

ii 25%

c True

Data

d False

20 30 40 50 60 70 80 90 100

b iii 50%

iv 75%

c 15

d 10

Data

v 75% b Second c First and third Fluency 5 a 3

b 18

e 7 6 a

Minimum

Lower quartile

Median

Upper quartile

Maximum

20 30 40 50 60 70 80 90 100

8 a 2

b 40

d 38

e 6

9 a i

1072 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c 2, 3, 40

Minimum

Lower quartile

Median

Upper quartile

Maximum

d i

Data

Minimum Lower quartile

Median

Upper quartile

Maximum

ii 10

Data

20 22 24 26 28 30

iii Left skewed b i

Minimum

Lower quartile

Median

31.5

Upper quartile

36.5

Maximum

iii Somewhat symmetrical, with slight right skew

Data

e i

Minimum

Lower quartile

Median

50.5

Upper quartile

Maximum

Data

20 22 24 26 28 30 32 34 36 38 40

iii Symmetrical c i

Minimum

Lower quartile

Median

Upper quartile

Maximum

40 42 44 46 48 50 52 54 56 58 60

iii Somewhat symmetrical, with slight left skew

Data

f i

Minimum

Lower quartile

Median

Upper quartile

Maximum

iii Somewhat symmetrical, with slight right skew

Answers 1073 mathspace.co

i i

Data

Minimum

Lower quartile

Median

Upper quartile

Maximum

ii 50

Data

iii Symmetrical g i

Minimum

Lower quartile

Median

Upper quartile

Maximum

j i

Minimum

Lower quartile

Median

15.5

Upper quartile

Maximum

ii 45

iii Right skew

Data

iii Somewhat symmetrical, with slight right skew h i

Minimum

Lower quartile

Median

Upper quartile

Maximum

iii Symmetrical 10 a The cyclist’s heart rates varied by 36 beats per minutes (bpm).

Data

b The middle half of the cyclist’s heart rates varied by 19 beats per minute. 11 a i D ata set A - skewed left, data set B - symmetrical ii Data set B

iii Somewhat symmetrical, with slight right skew

iii Data set B

b i D ata set A - skewed right, data set B - skewed left ii Data set B

1074 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

iii Data set A

12 a

Class 10P

Class 10Q 13

Median

Lower quartile

Upper quartile

Range

Interquartile range

Outliers

None

b Class 10P

c Interquartile range

13 a 31 years c 25% 14 a 49

b 17 years d 7

e 1st

f 5

b 50

c 12

d Carl

e Carl Reasoning 15 The typical pulse rates increases from about 68 to about 108 and are spread out over a larger range after exercise. 16 Answers may vary. One possible explanation: A boxplot visually displays the median, providing a measure of the central value of the data set. The interquartile range, represented by the box, indicates the variability within the central half of the data, offering insight into the spread of data. Lastly, the ‘whiskers’ of the boxplot highlight the overall range of data, contributing to our understanding of the data’s distribution. 17 a

Scores

100

c Science. The highest score is the maximum value of the data. Science has a highest score of 91, while music has a highest score of 88. Science has a minimum score of 52, while music has a minimum score of 43. Science has a median score of 74 and Music has a median score of 66. 18 a The median score of the Gamma Geckos is 5 points higher than the median of the Delta Dragons. b The range of the Delta Dragons is 12 points wider than the Gamma Geckos. c The interquartile range of the Delta Dragons is 19 points wider than the interquartile range of the Gamma Geckos. d Answers can support either team, but should use summary statistics to justify the response. For example: • We can expect the Gamma Geckos to win since the median number of points scored is 5 points higher than the median number of points scored by the Delta Dragons. • We can expect the Delta Dragons to win since in about 25% of games they scored more points than the highest number of points scored by the Gamma Geckos. 19 a Negatively skewed b Positively skewed

100

c Yes. The group that received treatment had a lower median and most people had symptoms for less than 9 days, whereas the control group had half the people having symptoms for more than 9 days. 20 a Yes, because correcting the error will not affect the positions of the Q1, Q2, and Q3 points. As the data point 23 is less than the maximum but greater than the Q3 division then the boxplot will remain unchanged.

Answers 1075 mathspace.co

b This changes the range from 21 to 14, a decrease by 7. c Because the positions of Q1 and Q3 were unchanged, there is no impact on the IQR. 21 Answers should specify the expected minimums, maximums, or differences to justify the context. a For example, the price of items at the discount store. Most items will be $1, but some discount stores have items that also cost up to $5 giving them a range of $4. b For example, the ages of every person in a kindergarten classroom. Most of the people are the kindergarteners whose ages are close together giving an interquartile range of 1–2 years old, but the teacher is also in the room so the range could be anywhere from 15–70 depending on how old the teacher is. 22 a

24 a Any ﬁve-number summary with an interquartile range of 2 and a range of 8. For example: • Minimum: 1

• Lower quartile: 2

• Median: 3

• Upper quartile: 4

• Maximum: 9 This has the largest possible range of 8 and the smallest possible interquartile range of 2. b Any ﬁve-number summary where the range is three times the interquartile range. For example: • Minimum: 1

• Lower quartile: 2

• Median: 3

• Upper quartile: 4

• Maximum: 7 This has a range of 6 and an interquartile range of 2. 25 • Boxplot 1/Histogram A • Boxplot 2/Histogram B • Boxplot 3/Histogram D • Boxplot 4/Histogram C 26 a

Monthly salaries (in thousand dollars)

15 50

95 100

b 20

Scores of students

c 25% d 52 and 85 e There are now 17 scores, so there are 13 values below Q3 and

= 76.5%. At least

75% of the data lies below Q3.

100

150

200

Debbie’s blog visitors

Problem-solving 23 0

Number of pages read

5 0

1076 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

100

15.03 Cumulative frequency

6 a

Score (x)

Understanding 1

Frequency

1≤x<2

2≤x<3

3≤x<4

4≤x<5

5≤x<6

6≤x<7

Score (x)

Frequency ( f )

Cumulative frequency (cf )

1≤x<5

5≤x<9

9 a 86

9 ≤ x < 13

13 ≤ x < 17

17 ≤ x < 20

Score (x)

Frequency ( f )

Cumulative frequency (cf )

12 ≤ x < 13

13 ≤ x < 14

15 ≤ x < 16

4 a True

b False

c True

d True

e False Fluency 5 a

b 82

20 → 24

25 → 29

30 → 34

35 → 39

40 → 44

45 → 49

50 → 54

b 75

14 ≤ x < 15

Score (x)

Frequency ( f )

Cumulative frequency (cf )

1→5

6 → 10

11 → 15

16 → 20

21 → 25

26 → 30

c 10

Cumulative frequency (cf )

c 34

7 a 30

b 6

c 27

8 a 60

b 3

c 13

100 90 80 70 60 50 40 30 20 10 0

d 15%

b 14

Cumulative frequency

Score

Frequency ( f )

20 30 40 50 60 70 80 Class

90 Cumulative frequency 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 10

Time (sec) 40

Answers 1077 mathspace.co

Reasoning

50 45

15 a The difference between the heights of two adjacent columns.

40 Frequency

b 4 ≤ Score < 5

30 25

16 a

20 15

Score (x)

Frequency ( f )

Cumulative frequency

10 5

50 100 150 200 250 300 350 400 Mass (g)

Score

Frequency

75 ≤ x < 80

80 ≤ x < 85

85 ≤ x < 90

90 ≤ x < 95

95 ≤ x < 100

13 a 2 c

14 a

d 7

e 11 Number of sightings

Cumulative frequency (cf )

b 19 locations

d 2.44, the mean is calculated by dividing the sum of the product of scores and their frequencies by the total number of families:

f 18

Number of locations ( f )

c 2, the median is the middle value when the data is ordered. Since there are 9 families, the median is the value corresponding to the 5th family in the cumulative frequency, which is 2.

Cumulative Frequency ( f ) frequency (cf )

b 9, the total number of families is the sum of the frequencies in the table.

b 12 Score (x)

1 2

Frequency values are ﬁlled based on the given data. Cumulative frequency is calculated by adding the frequency of the current row to the cumulative frequency of the previous row.

Cumulative Frequency

Flu Season

10 8 6 4 2 0

2 3 Score

The histogram shows the distribution of the number of times families got the ﬂu. It helps to visualise the data and identify patterns such as the median and range of the data.

c 12 locations

1078 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

f 4 , the number of families who got the flu more than twice can be found by looking at the cumulative frequency for scores greater than 2. There are 9 families in total, and 5 of them got the flu 2 times or less, so 9 − 5 = 4 families got the flu more than twice.

17 a Cumulative frequency histogram, this is the height of the last bar. We don’t need to add up the frequencies for each bar like we do for a histogram. b Histogram, this is the height of the bar. We don’t need to subtract the cumulative frequencies like we do for a cumulative frequency histogram. c Histogram, the shape tells us about how the data values are spread out. We would have to look at the rate of change for the cumulative frequency histogram. d Cumulative frequency histogram, this is the height of the bar with that value. We don’t need to add up the frequencies for each bar below like we do for a histogram.

Time recorded

Frequency (f)

d The histogram

2 a Negatively skewed

b Bimodal

c Symmetrical 3 a • Frequencies for each group • Mode

• Shape

• Outliers b • Quartiles • Shape

• Median • Outliers

• Range and interquartile range c • Frequencies for each group

4 a

16:20 ≤ x < 16:30

16:30 ≤ x < 16:40

16:40 ≤ x < 16:50

16:50 ≤ x < 17:00

c 16

d • Total Fluency

16:10 ≤ x < 16:20

Cumulative Frequency

b The boxplot

c The boxplot

Cumulative frequency (cf )

16:00 ≤ x < 16:10

22 20 18 16 14 12 10 8 6 4 2 0

1 a The histogram

• Shape

• Outliers

18 a

19 a

Understanding

• Mode

Problem-solving

b 25

15.04 Compare data sets

d 48%

• Rate of change

English Mathematics Median

Lower quartile

10 7

Upper quartile

Range

Interquartile range

b Mathematics 5 a Upmarket restaurant

b $22

c Upmarket restaurant

d $36

e 26 items 6 a Weightlifter A b The range and interquartile range 145 150 155 160 165 170 Height (cm)

b 16 students

c 6 students

d 2 students

e 153

20 a i 2100 ii 1900

iii 2450

b i 675

ii 640

iii 760

c The median d Weightlifter B 7 a Lower

b 0

c 2

d Positively skewed

e 4

f Block B

g No 8 • Histogram A/Boxplot 2 • Histogram B/Boxplot 4 • Histogram C/Boxplot 1 • Histogram D/Boxplot 3

Answers 1079 mathspace.co

9 • boxplot 1/Cumulative Frequency D • boxplot 2/Cumulative Frequency C • boxplot 3/Cumulative Frequency B

Problem-solving 15 a

Cooper

• boxplot 4/Cumulative Frequency A 10 a Matt

b Matt

c Matt

Reasoning 11 a 5.2

b 8.9

c 3

44.8

47.1

49.1

54.7

Median

52.5

60.45

55.75

65.25

66.6

68.5

Maximum

d 3

Marion

Minimum

b Yes, Cooper has an outlier of 66.6.

e No. The results are mostly 4 or 5 out of 10 and the mean is only 5.2

f Y es. The results are mostly 9 and 10 out of 10, and the mean is 8.9

Marion

g Answers may vary. One possible answer is that one class may have covered more material than the other class.

13 a The boxplot has a long tail to the right which indicates positive skew, while the histogram does not appear to be skewed. The median for the histogram is roughly in the middle, while the median of the boxplot is located further to the left.

50 55 60 Time (seconds)

e No, a high outlier does not change what the minimum score is, hence would not alter the selection for pole position.

b $10 ii True

d Cooper

12 a Bookstore A, it has a lower range and interquartile range. c i True

Cooper

16 a

b The boxplot indicates that the data is widely spread, while the histogram indicates that most of the data occurs near the median.

Tim

Odi

Minimum

29.8

26.0

33.2

34.8

Median

35.0

38.4

37.4

40.0

Maximum

39.6

42.8

Mean (1 d.p.)

35.1

37.2

Range

9.8

16.8

Interquartile range

4.2

5.2

b Tim throws more consistently, his range, interquartile range and standard deviation are all signiﬁcantly smaller.

14 a

c All of Tim’s scores are consistent, so no outliers for his data. Odi has an outlier of 26, which is far from the rest of his data set.

65 000 70 000 75 000 80 000 85 000 90 000 95 000 100 000 105 000

b 20 000 c 50% d From the boxplot it appears the highest 25% all earn $100 000. However, this is using class centres, from the data provided we can only conclude that the top 25% earn in the range $95 000 to $105 000.

1080 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

d Odi appears to be the better hammer thrower. His mean is higher and other than the outlier the other values of the 5 number summary are all larger for Odi. In particular, his top 25% of throws are larger than any throw from Tim. Odi’s median is also signiﬁcantly higher (3.4 m) than Tim’s.

e As the next lowest throw is more than 6 m larger than his outlier it appears this throw may have been a misadventure and it would be reasonable to remove it when looking at his mean. Including the outlier would distort how high he usually throws and what he would likely achieve in a competition.

e False

f False

c True

d True

2 a No

b No

c Yes

d Yes

3 a No

b Yes

c Yes

d Yes

e Yes

f No

4 a The population is all the residents of the town, and the sample is the 500 residents surveyed. b The population is all sales at the bookstore, and the sample is the 20 receipts she looks at. c The population is all students at Antony’s school. The sample is the 85 students he surveyed. 5 a Yes

b Yes

c Yes

d No

b No

c No

d Yes

e No 6 a No

• All the active members of a local football club • 30 people walking their dog at the local park

Understanding b False

• All students from a local school who compete in a school sports competition

d Example answers:

15.05 Samples and populations

1 a False

c Example answers: • The members of 3 teams chosen from the local hockey tournament

e Yes

• All the students from a local school who own a dog 10 Fair 11 a Biased

b Biased

c Not biased

d Not biased

12 a All products in the factory

b 30%

c 61%

d No

13 a All apples in the orchard

b 62%

c Yes 14 a Yes

b No

c No

d No

e Yes 15 Approximately 610 bees 16 1 089 600 17 a Source A

b Source A

c No

Reasoning 18 Jenny’s claim is more valid since she has surveyed more people.

Fluency 7 a ii

b iii

c i

d iv

8 a Yes

b Yes

c No

d Yes

9 a Example answers: • All the seniors • The ﬁrst 30 students to arrive at school that day • All the students that were in the library at lunch time b Example answers: • All students of a primary school in Sydney • All students in Year 7 in a particular suburb • 100 children chosen from the Sydney under 15s baseball tournament

19 This is a very convenient sample so would be very inexpensive and not take much time. However, a disadvantage is that there’s a chance that someone at the dog park doesn’t have a pet dog such as a professional dog walker or a friend of a dog owner. There is also the possibility that someone with accessiblity needs would not be able to access the park so would be excluded. 20 a The sample is not representative of the target population. b The sample is from self-selecting participants, i.e. only those that made an effort to respond. c The sample is not representative of the target population.

Answers 1081 mathspace.co

21 a Example answer: Assign a number to everyone on the electoral role and use a random number generator to decide who to call. b The sample will not include anyone who was out for the evening for work or social reasons, and the results are based on less than half of the people they tried to contact, so the sample is not representative of the populations. 22 a The proportion from the second day, 12% is twice the proportion from the third day, 6%. b The differences in sample proportions are due to the varying sample sizes (750 dingoes in the first sample and 100 in the second) and natural random variation. A larger sample size generally yields more representative results, and the uneven distribution of tagged dingoes across the area can lead to different proportions in each sample due to chance and spatial distribution factors. c The sample statistic from the second day is more accurate because sample size is larger, and thus more likely to be representative of the population of dingoes. 23 a The sample proportions are different due to sampling variability. It is impossible for every sample to have the exact proportions as the population. b Both samples are valid, so they could use either one. Often we will take the average of many sample proportions to get a more accurate estimate of the population parameter.

• Evaluate the intent and potential bias of each source. For instance, a textbook might aim for neutrality and comprehensive coverage, whereas a documentary might focus on specific stories or narratives. b Example answers: • Different sources may offer various perspectives or focus on different aspects of an event, giving a more rounded understanding. • Sources can sometimes contain errors, biases, or omissions, and crossreferencing multiple sources can help identify and rectify these issues. • Diverse sources can provide a richer context, helping one understand the broader implications and nuances of a historical event. • Relying on a single source may lead to a limited or skewed understanding, while multiple sources can fill gaps and provide a more in-depth exploration of a topic. 25 a Yes b The sample of fish were collected in several different places and one at a time, without large nets. 26 a Samples should focus on ways to reach only athletes without including non-athletes. For example, the sample could be all starting players from both teams or all of the players on the bench for one of the teams.

• Check additional reputable secondary sources to see which information is corroborated by other authors or creators.

b Samples that include non-athletes. For example, the ﬁrst 50 spectators would not necessarily be athletes, and a sample that includes all of the people in the stadium, including coaches, oﬃcials, and fans, would have a mixture of athletes and non-athletes.

• Look at the sources or references cited by each source. A well-cited claim with primary sources to back it up is generally more reliable.

27 a If 10% of people experience side effects, we might expect about 500 000 residents in Sydney to experience side effects with the medication.

Problem-solving 24 a Example answers:

• Check the date of publication. Newer sources, like the 2015 documentary, might have had access to more recent research or previously unavailable information. However, it’s essential to note that newer doesn’t always mean more accurate.

1082 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b One possible answer: The way we share this prediction is important because causing unnecessary panic could lead to people avoiding the medication, but we also need to be honest about potential risks.

c One possible answer: While 1000 people is a decent sample size, it may not be truly representative of Sydney’s diverse population of 5 million, so we should be cautious about making wide-reaching assumptions.

Chapter 15 review 1 a Grouped b 83

3 a

c 45

Minutes

Frequency

0 ≤ Minutes ≤ 10

10 < Minutes ≤ 20

20 < Minutes ≤ 30

30 < Minutes ≤ 40

40 < Minutes ≤ 50

50 < Minutes ≤ 60

60 < Minutes ≤ 70

70 < Minutes ≤ 80

b The treatment group’s range is 3 weeks wider than the control group’s range. c The treatment group’s interquartile range is 1 week wider than the control group’s range. d Yes. The group that received the experimental fertiliser matured faster since it had a lower median, and most plants reached maturity in less than 12 weeks. In the control group, half of the plants took more than 13 weeks.

b Ungrouped

2 a 29

9 a The median of the control group is 5 weeks higher than the median of the treatment group.

10 a 290 11 a

b 16

c 8

d 3

5 a This indicates that the middle 50% of ratings in Set B are spread over a larger range of values compared to the other two data sets. b Based on the ﬁve-number summaries, Set A appears to be symmetric, Set B is skewed right, and Set C is skewed left. 6 70 seconds 7 a

Minimum 65

60 65 70 75 80 85 90

c Somewhat symmetrical, with a slight left skew. 8 a The number of books read by students varied by 42 books. b 25%

c 13

d 4th quartile

e 10

Cumulative frequency (cf )

b 8 workshops

c 10 workshops

d 22 attendees 12 a 49

b 45

c 55

13 A 14 a Building X

Lower Upper Median Maximum quartile quartile

c 31.0%

Number of workshops (f )

Number of attendees

b $610.00 4 a 101

b 60

c Building Y 15 a

b 2 plants d Yes, Building Y Antoine

Benji

Minimum

6.5

6.0

6.9

6.8

Median

7.3

7.4

7.5

7.9

Maximum

7.8

8.1

Mean (1 d.p.)

7.2

7.3

Range

1.3

2.1

Interquartile range

0.6

1.1

b Antoine c Given that Benji’s lowest jump is signiﬁcantly shorter than his next lowest jump, it might be considered an off-performance. It would be reasonable to exclude this outlier to get a

Answers 1083 mathspace.co

more representative average of his usual performance in competitions.

d No, Antoine’s range and interquartile range are still smaller than Benji’s, even after removing the outlier. 16 a All cherries in the farm.

b 58.5%

c Yes

d 30 420

17 a Biased

Chicken

Fish

Total

Adults

Total

6 a This value represents the number of cats with feline diabetes treated at Venetian Pet Clinic.

b Biased

b This value represents the number of cats with feline diabetes who received a special diet at Mohawk Animal Clinic.

c Not biased 18 a The sample proportions are different because the researchers used samples of very different sizes.

7 a

b The sample statistic from the second week is more accurate because sample size is larger, and thus more likely to be representative of the population of beetles.

16.01 Two-way tables

Vegetarian

Vegan

Neither

Children

Adults

Total

b b 26

b 16

2 a i Marginal frequency

ii 451

b i Marginal frequency

ii 171

c i Joint frequency

ii 143

d i Joint frequency

ii 388

b The most common rating among junior students is 4, while for senior students it is 3.

3 a False

b False

c True

d False

4 a True

b False

c True

d True

e True

c The data suggests that satisfaction with canteen food tends to decrease as students progress from junior to senior years, indicating potential issues with food quality or variety for older students. 10 a 14 trains

b 51 trains

c 127 trains

Fluency

11 a 27.63% Running Basketball shoes slides

Work boots

Adults

Total

169

Sprinkles Hot fudge

b 44.47%

Total

Teenagers

c 47

9 a 11% of junior students and 10% of senior students were very satisﬁed with the canteen food.

c Outback Animal Care Centre

5 a

Total

8 a 39

Understanding 1 a 14

Beef Children

d 34.61%

Reasoning 12 a

Brushtail possum

Honey possum

Total

Fruit

Chocolate

Insects

120

Vanilla

Nuts

110

Total

130

180

310

b There were 130 brushtail possums and 180 honey possums, so there were more honey possums observed than brushtail possums.

1084 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

c There were 80 possums observed eating fruit, 120 possums observed eating insects, and 110 possums observed eating nuts. So, there were more possums eating insects than fruit or nuts. d There were 35 brushtail possums observed eating nuts and 70 honey possums observed eating insects, so there were signiﬁcantly more honey possums observed eating insects than brushtail possums observed eating nuts. 13 a True. Both sets of marginal frequencies will add to the total. b False. A single joint frequency can be equal to the marginal frequency, but only if all other joint frequencies in that row are equal to 0. This would only be possible if all values in a row were 0, in which case that row likely wouldn’t be included. 14 a The 60 people include teenagers not just adults so it can’t be concluded that all 42 are adults who said either no or sometimes.

17 Answers will vary. The table provided is one such solution:

Football

Soccer

Total

Period 1

Period 2

Period 3

Total

18 a 650 students b Answers may vary. One possible answer: To investigate the hypothesis, you could design a follow-up survey that looks at two primary variables: 1. Participation in school sports during the hot week (using the same Likert scale: Always, Often, Sometimes, Rarely, Never) 2. Purchase of an ice block from the canteen during the hot week (Yes or No) Your two-way table would look something like:

b The 24 adults represent the total of adults. Using this information with the 18 adults saying “yes” tells us that 6 adults responded either “no” or “sometimes”. c The statement only indicates the relative distribution among the “yes”, “no”, and “sometimes” responses, not their absolute values. We would need more information to precisely determine the number of each response. Problem-solving 15 a

No Casual Part-time Full-time job 15

Women 25

Men

b 40 people

c 38.9%

e 21 people

16 a x = 70 b There are 70 animals in the study that carry the Agouti gene and have dark fur.

Bought ice block

Did not buy ice block

Total

Always Often Sometimes Rarely Never Total Students would be surveyed on both their sports participation and their ice block purchases. By comparing the results, you could see if there’s a noticeable difference in sports participation between those who bought ice blocks and those who didn’t. If the coach’s hypothesis is correct, you might expect higher “Rarely” or “Never” responses in the “Bought ice block” column.

16.02 Associations between categorical variables Understanding 1 a 43%

b No

c No

2 A column percentage frequency table 3 The horizontal axis 4 No

Answers 1085 mathspace.co

Fluency

10 a

Adults

Children

Total

Math

42.86%

57.14%

100%

Science

45%

55%

100%

Lefthanded

Righthanded

Total

Not female

14.5%

85.5%

100%

Female

9.7%

90.3%

100%

6 a

Red

Brown Blonde

Black

Total

Straight 5.9%

47%

5.9%

41.2%

100%

Curly

40%

30%

20%

100%

10%

Red

Brown

Blonde

Black

Straight

50%

66.7%

25%

77.8%

Curly

50%

33.3%

75%

22.2%

Total

100%

7 a

Comedy Action Drama Horror Total Teens

30%

20%

45%

100%

Children

30%

40%

10%

20%

100%

Comedy

Action

Drama

Horror 20%

Teens

50%

33.3%

81.8%

Children

50%

66.7%

18.2%

80%

Total

100%

Play tennis

Total

Height ≥ 170 cm

31%

69%

100%

Height < 170 cm

30%

70%

100%

58.0%

Total

100%

b Yes

12 a Fitness level

b Yes

13 a Personality type

b No

14 a Dietary restriction

b B

Cardio

Weight

Total

30%

70%

100%

73%

27%

Reasoning 15 It means that there is no clear association between the variables. 16 a Location b Column percentage table Urban

Regional

62%

38%

Low emissions

38%

62%

Total

100%

d Yes, there does appear to be an association as the numbers are signiﬁcantly different. b 21

c Dominant hand d Row frequency table e

100%

c Yes, there appears to be an association as the numbers are quite different. Those who are pregnant seem to prefer weights, while people who are postpartum seem to prefer cardio. d No, association is not causation. There appears to be an association but we cannot say whether one variable causes the other.

c No

17 a 40

9 a Pregnancy status

Currently postpartum

46.7%

High emissions

Do not play tennis

Currently pregnant

42.0%

Female

Right-handed

53.3%

11 a Parenting status

8 a Height

Left-handed Not female

Maths Music Science English Total Righthanded children

30%

20%

25%

100%

Lefthanded children

33%

20%

30%

17%

100%

f N o, association does not exist because values down the column are the same. 18 a Number of passengers

1086 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b Column frequency table

c No. of passengers

5−9

Roll over

17.1%

30.0% 28.8% 28.6%

No roll over

82.9% 70.0%

Total

100.0% 100.0% 100.0% 100.0%

10−15

> 15

Smartphone owners employment Owns smartphone

71.4%

d No, because values across the rows are mostly the same. b Column frequency table c

Maths

Science

Maths major 39.52% 38.36%

Music

80 60 40 20 0

19 a Favourite subject at school Art

11.67% 23.33%

Doesn’t own smartphone

100 Percentage

71.2%

Employed Unemployed Employment

g Yes it does as the segments are different size in each column. 22 a Age

Science major 30.54% 27.40% 23.89% 13.89%

b Column percentage frequency table

Music major 22.75% 22.60% 37.78% 20.56%

Art major

7.19%

11.64%

26.67% 42.22%

Total

100%

Less than 3 hours

d Yes, because values across the rows are different. e No, association is not causation. There appears to be an association but we cannot say that one causes the other.

32.14%

22.86%

9.52%

14.71%

4 to 8 hours

7.14%

28.57%

9.52%

26.47%

9 to 15 hours

25.00%

22.86% 38.10%

29.41%

Over 15 hours

35.71%

25.71% 42.86% 29.41%

Total

100%

TV watching per age group 9 to 15 15 or over 4 to 8

3 or less

Problem-solving

100

b June

c Book loans

d No, there does not appear to be an association as the corresponding segments are similar in size. 21 a Employment

80 60 40 20 0

b Column percentage frequency table c

Percentage

20 a November

10 years 11 to 20 21 to 30 31 years and under years years or older

Employed

Unemployed

Owns a smartphone

32%

17%

Does not own a smartphone

68%

83%

Total

100%

d Yes, association exists because values across the rows are different. e No, association is not causation. There appears to be an association but we cannot say one variable causes another.

10 & under

11 to 20 21 to 30 31 & older Age group

e Yes it does as the segments in each column are quite different. It suggests that older people tend to watch more hours of TV.

16.03 Bivariate numerical data Understanding 1 a Univariate data

b Bivariate data

c Univariate data

d Univariate data

e Bivariate data

f Bivariate data

g Bivariate data

h Univariate data

Answers 1087 mathspace.co

2 a

Fluency

y 18 16 14 12 10 8 6 4 2

5 a i Number of dogs owned ii Number of shoes destroyed b i Size of a pizza ii Cost of pizza c i Time spent studying ii Exam mark

6 a Ticket sales affect the revenue made from a show.

2 4 6 8 10 12 14 16 18

b Dependent variable - revenue from the show, Independent variable - number of ticket sales

18 16 14 12 10 8 6 4 2

7 a The height of the kite and the speed of the wind. b Dependent variable - height of the kite, Independent variable - speed of the wind 8 a Distance from the equator affects temperature.

x 2 4 6 8 10 12 14 16 18

b No

c Both time spent travelling and distance covered affect each other, with speed playing a crucial role in their relationship.

18 16 14 12 10 8 6 4 2

d No e Time spent working affects wages earned. f N umber of pages affects time spent to ﬁnish a novel. x

g No

2 4 6 8 10 12 14 16 18

h Number of pets affects time spent caring for pets (hours).

y 18 16 14 12 10 8 6 4 2

i T ime spent training (hours) affects athletic performance. 9 a Height of the ball b Amount of time passed 10 As temperature increases, soups sold decreases.

11 Height has no effect on salary.

2 4 6 8 10 12 14 16 18

3 a Yes

b No

4 a Neither c Decreasing

c No

d Yes

b Increasing d Decreasing

12 a The independent variable is the number of immigrants and the dependent variable is the unemployment rate.

1088 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b Unemployment rate in Australia Unemployment rate (%)

15 a Dependent variable - height, Independent variable - time b

5.5

Height 0.8 0.6 0.4 0.2

5 4.5 4

−0.2 −0.4 −0.6 −0.8

3.5 Number of immigrants (thousands)

250

275

300

325

c As the number of immigrants increases, the unemployment rate appears to decrease.

16 a 180 y 175 170 165 160 155 150 145 140 135 130 125 120 115 110 105 100 95 90

13 a The independent variable is CO2 emissions and the dependent variable is the global temperature anomaly. b

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

Climate change Global temperature anomaly (°C) 0.9

0.8

x 2

0.7

b 105 cm 0.6 CO2 emissions (million MT)

35 000 36 000 37 000 38 000 39 000

c It appears that as CO2 emissions increase, global temperature anomalies also increase. 14 a 18 y 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Reasoning 17 a 10 trees b No. There is no connection between the number of people in a student’s family and the number of trees in their garden because there is no trend in the data in the scatterplot. 18 a Online shopper survey averages Average amount spent (dollars)

x 1 2 3 4 5 6 7 8 9 10 11 12 13

b 11

Average time spent (minutes)

Answers 1089 mathspace.co

b Based on the scatterplot, the statement does not seem to be accurate. In fact, the data suggests the opposite: as the average time spent on a website increases, the average amount spent also increases. 19 a

100 90 80 70 60 50 40 30 20 10

Exam mark

16.04 Associations between numerical variables Understanding 1 a No

b Yes

c Yes

d Yes

2 a True

b False

c True

d False

e True 3 a The strength and direction of a scatterplot describe the correlation. b When two variables are related, we say there is an association. c To show causation, we must do a carefully designed study.

Books read per year 10

d When correlation is not causation, some possible explanations are coincidence and contributing (third) variables.

b 50 c Yes, the teacher is correct. Notice how the points follow a line from the bottom left corner to the top right corner. This means that if a student reads more books they receive a higher mark in their yearly maths exam.

Fluency 4 a Linear relationship b Non-linear relationship c Non-linear relationship

Problem-solving

d Non-linear relationship

20 Looking at the scatterplot, the number that fulﬁls the required characteristics is 4.

e Linear relationship

21 a Student B

b Student C

c Student A

22 a Answers may vary. One possible answer is the relationship between the amount of study time and test scores. These variables may have a relationship because the more time a student spends studying, the better they might perform on a test. b Answers may vary. One possible answer is that the amount of study time is the independent variable because it can be controlled by the student, while the test scores are the dependent variable because they can change based on the amount of study time.

f Linear relationship 5 a Positive and strong b Positive and weak c Positive and moderate-weak d Negative and strong 6 a As air temperature increases, sea temperature increases. b Strong, positive, linear 7 Negative 8 a

c Answers may vary. One possible answer is the relationship between the colour of a car and its speed. These variables likely do not have a relationship because the colour of a car does not affect how fast it can go.

Sport 90 80 70 60 50 40 Maths 40 50 60 70 80 90

b Weak, positive, linear c Student 10’s

1090 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

9 a

b 10

Accidents 40

c Weak, negative, linear

d 8 °C, 69 seconds 13 a

t 70

60 50

20 Age 25

40 30

b Strong, negative, linear

c No

10 a

n 1

Hours

b Increases

c Strong, positive, linear

d No

14 Strongly negative

15 Strongly positive

Age 25

b Strong, negative, linear

b Yes

c No

d No

e No

f Yes

g Yes

h Yes

17 This data suggests that as height increases, there is an increase in the ability to play soccer. However, it does not necessarily mean that if you are tall, you are automatically better at soccer. Correlation does not imply causation, and other factors may also inﬂuence soccer skills.

Price 16 15 14 13 Supply (kg) 600 700 800 900 1000

18 a The explanatory variable is the number of questions and the response variable is the test result. b 100 Test result

b Strong, negative, linear c When very few bananas are available to be sold. 12 a

16 a No

Reasoning

c No 11 a

Time 65 60 55 50

95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15

Number of questions 10 15 20 25 30 35 40 45 50 55

45 Temp. 1 2 3 4 5 6 7 8 9 10

c Strong, positive, linear correlation; the larger the number of completed revision questions, the higher the mark on the test.

Answers 1091 mathspace.co

d The relationship is not perfectly linear because of the different abilities of the students.

24 Answers may vary. One possible answer: Time spent studying and test scores could lead to a weak positive relationship.

20 Answers may vary. For example, if there is a relationship between the average temperature and migration, the scatterplot might show a pattern where the points tend to increase or decrease together, indicating a positive or negative relationship between the two variables.

It could be that while additional study hours slightly benefit most students, other factors play a significant role in determining their final score. For instance, some students might have natural aptitude, meaning even with fewer study hours, they can achieve similar or even better scores than their peers who study longer. Others might benefit from private tutoring, have access to better resources, or engage in practical applications of their knowledge that aren’t directly measured by the study hours.

Problem-solving

Here is a table of values and scatterplot that would lead to a weak positive association:

19 a It appears that y can change signiﬁcantly without much corresponding change in x. b i Yes

ii Yes

21 a D 22 a i No

iii Yes

iv No

b B ii Yes

iii No

b Region 2 23 a True. For example, if we know that influenza causes patients to develop a fever then there is a correlation between having influenza and developing a fever. However, it is possible for the correlation between influenza and developing a fever to be a strong or weak correlation. b False. For example, if there is a correlation between the number of people drowning in pools and number of people going to the movie theatre, that does not mean that an increase in pool drownings causes more people to go to the movie theatre or vice versa. c False. A positive association means that as the value of the independent variable increases, the value of the dependent variable also increases, but does not mean that changes in one cause changes in the other. For example, suppose data collected from a study found that people who exercised more were also found to be at higher risk for skin cancer. These quantities demonstrate a positive association, but that does not mean exercising more causes a higher risk of skin cancer, there are likely other underlying factors influencing the relationship.

Hours Studied

Test Score

Student A

Student B

Student C

Student D

Student E

Student F

Student G

Student H

Student I

Student J

Test score

55 Hours studied 1

16.05 Line of good fit Understanding 1 Decreasing 2 Line 2 3 a No 4 a Negative

1092 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b No

c Yes b C

d Yes

d i

Fluency

5 19 y

18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

15 10 5 x 12

ii y = −1.3x + 29.5 x

e i

1 2 3 4 5 6 7 8 9 10 11 1213141516171819

6 a i

8 4

x 12

5 x 5

ii y = 1.5x − 17.5 f i

ii y = −x + 21 b i

4 2

x 2

5 x 5

ii y = 0.9x + 0.5

38 36 34 32 30 28 26 24 22 20

y 20 15 10 5 x 5

7 Example answer:

ii y = 0.56x + 5 c i

x 10 20 30 40 50 60 70 80 90

ii y = 0.9x + 1.5

Answers 1093 mathspace.co

190 180 170 160 150 140 130 120 110 100 90

Reasoning

Copies sold

13 a 27.7 kg b The y-intercept indicates that when a child’s height is 0 cm, their average weight is 27.7 kg. c No, a child cannot have a height of 0 cm. 14 a A scatterplot can help visualise the relationship between the quantities, but is not necessary for a complete response.

Price 18 20 22 24 26 28 30 32 34 36

9 a i 4.62

100

ii Positive

iii Increases by 4.62 units

iv 7.58

b i −7.68

ii Negative

iii Decreases by 7.68 units

iv 5.49

30 20

c i −5.69

ii Negative iii Decreases by 5.69 units iv 7.84 ii Positive iii Increases by 3.16 units iv −2.72 10 a The vertical intercept indicates that when a family has no income, $27 is its average spending on dining out. b No

c 0.3

A student who spends 0 minutes on LookAtMe is predicted to have a 99.6% grade in maths based on the line of good fit. b According to the linear model, for every additional minute spent on LookAtMe, a student’s maths grade is expected to drop 0.25% points. Or, for every additional 100 minutes spent on LookAtMe, a student’s grade is predicted to drop 25%.

d $60

Price (thousands of dollars) 24

15 a There are only 2 points above the line and 18 points below it, so this line does not go through the middle of the data. It is too heavily impacted by the outlier, so is not a line of good fit.

20 16 12 8 4

Age (years) 2

b Answers will vary if calculated by hand, but should be around eighteen. c A car that is 0 years old costs $18 400. d Answers should be around 11 years. 12 a −4

Minutes 50 100 150 200 250 300 350 400 450

There is a strong, negative, linear association between minutes spent on LookAtMe and maths grades.

d i 3.16

11 a

Grades

b 3865

1094 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b An outlier should not be completely ignored, but the line should go through the majority of the data. Lorena’s line has 9 points above it and 10 points below it, and goes through the middle of the data, so it is a line of good fit.

Problem-solving

5 y = −52.61, Extrapolation

16 a

6 a 3 seconds

Ice blocks 140

7 a 20 years

120

8 a

100 80 60 40 20 Temperature

225 200 175 150 125 100 75 50 25

ii Approx. 149

c Increases 17 a A

b x = 3682

18 a y = 0.5x + 4

b 7.45

c 11 years

Temperature °C 12 14 16 18 20 22 24 26 28

b 150 people

c 23 °C

d 140 people

e 19 °C

9 a 2448 Pa

b Interpolation

c Yes d i Yes

ii No

10 a 90

19 a A line of good fit: y = 3x + 15

b 1300 ﬁsh

People

14 16 18 20 22 24 26 28 30 32 34 36 38 40

b i Approx. 91

b 2.5 hours

c i No

H% 30

ii Yes

12 a

iii No

iv No

b 6500 m

c 300 seconds

iv No

b 22°C

11 a 8500 m

iii No

d 1800 seconds b c = 100

c $400 000

13 a Negative gradient

b −2.5

c y = −2.5x + 105

d y = 95

Y 1

Reasoning

b The new line of good fit would have a smaller slope and a higher y-intercept.

16.06 Interpolation and extrapolation Understanding 1 a False

b False

e False

f True

c False

d True

2 a Interpolation

b Extrapolation

c Extrapolation

d Interpolation

3 a y = 8.45

b y = −17

c y = −7.4

Fluency 4 a Interpolation c Interpolation

b Extrapolation

14 a 52.5 m b Yes, because the prediction is determined using interpolation. 15 Yes, if we examine the scatterplot and compare with the other points. It follows the same relationship. 16 a Interpolation. The speed of 70 km/h falls within the range of known data points, which is from 40 km/h to 100 km/h. b Extrapolation. The prediction for the revenue for the next year is beyond the range of known data points (2015 to 2019). c Extrapolation. The number of hours of study at 15 hours does not fall within the range of known data points, which spans from 2 to 14 hours.

Answers 1095 mathspace.co

17 Interpolation and extrapolation can be limited by the accuracy and validity of the data, as well as the assumption that the relationship between variables remains constant. In the context of climate change, factors such as human intervention, natural disasters, and technological advancements can inﬂuence the relationship between variables, making predictions less reliable. Problem-solving 18 a y = −0.005x + 49

b y = 29 °C

d $50 260 e Yes, because it is an extrapolation close to the data range where the cost is likely to increase along with distance driven, and the correlation coeﬃcient is high. f Y es, a car that is driven more will have more wear and tear and would be likely to have greater maintenance costs. 22 a 95 m

9 8 7 6 5 4 3 2 1

b A = 100 hectares

c No, because it is using extrapolation. 20 a

c Option B

23 Any scatterplot with weak correlation or very few points. For example:

c Yes, because it is using interpolation. 19 a A = −25T + 750

b 50 m

Litter collected (kg) 5 4

24 Neither interpolation nor extrapolation would be overly reliable. This data set is more non- linear, so a non-linear model would be more appropriate for making predictions.

1 Weeks 1

2 3 4 5 6 7 8 9

b Example answer:

2 3 4 5 6 7 8 9

16.07 Misrepresentations of data

Litter collected (kg) 5

Understanding

1 a 100%

b 193%

c No

2 a Not representative, convenience bias and likely missing certain age demographics or those without reliable internet.

3 2

b Not representative

c Representative

Weeks 1

2 3 4 5 6 7 8 9

c Example answer: y = −0.2x + 4 d Approximately 2.4 kg e No, because it is using extrapolation.

d Representative 3 a False 4

21 a For each 1000 km increase in distance driven, the least-square line predicts that the maintenance costs increase by $990. b A car that is not driven at all is predicted to cost $14 620 to maintain. c No, because for a car that hasn’t been driven, it should have little to no cost to maintain.

1096 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b False

Date

Cases

October 15

October 18

200

October 21

174

October 24

339

October 27

251

October 30

398

c True

d True

Fluency

Reasoning

5 a Causation and correlation

17 Kang has used only the statistics that are favourable to him.

b Over-generalisation c Comparing apples and oranges d Cherry-picking 6 a False

b True

c True

d False

7 a Yes

b No

c No

d Yes

8 a Yes

b No

c Strategic choice of measure of centre (mode) d Yes 9 a Cherry-picking

b C

10 The claim is directed to any person that uses the product. However, the sample size is small and not representative of all people that might use the product. There is no information on whether the product was properly tested. There is no information on the sampling method used or the method for collecting the data. 11 a Biased question b When giving customers the menu, the servers can simply inform the customers that the mushroom risotto is a new menu item. 12 The tick marks on the vertical axis are not evenly spaced. 13 They do not correspond to the tourist numbers they are supposed to represent. In particular, on the corners the sizes are exaggerated. 14 The title is persuasive and the vertical axis scale doesn’t start at 0. It also does not give an idea of the trend overall. 15 By using a logarithmic scale, the graph appears to show a slight increase after taking the drug, and a steady decline of the drug in the blood over time. This may lead people to think that the drug concentrations in the blood are relatively stable over time. When converted to a normal scale, the inital increase would be much more signiﬁcant. 16 a Russia

b U.S.

18 By compressing or expanding the horizontal axis it can appear as if the trend is gradual or quite abrupt. 19 a The axis is broken too close to the top of the Mar 95 column, so it is not clear where the scale break is. b It is not recommended for graphs such as column graphs where the viewer takes in the comparative areas of the different categories, using a truncated axis will exaggerate any differences in the data. c A line graph with clear indication of a broken axis below the lowest point in March. 20 a The graph with the vertical axis that goes from 13 to 15. b The graph with the vertical axis that goes from 0 to 20. c Since 1920, the temperature has increased by around 0.8 °C. The claim states that a further increase by 0.7 °C in the next few decades could lead to an increase in catastrophic events. This shows that increases by less than 1 °C are actually signiﬁcant increases in temperature. 21 a One possible answer: The survey claims to use over 2000 samples, but doesn’t clarify the selection process. This raises doubts about its accurate representation of all frequent smartphone users. b One possible answer: The survey, conducted by an anti-ageing skin company, suggests half of the regular smartphone users are showing early ageing signs. It’s unsurprising that the study recommends reduced screen time and speciﬁc skincare products. c One possible answer: The article’s headline is sensationalised, potentially causing unnecessary worry as it lacks context on the study’s broader use.

c No

Answers 1097 mathspace.co

Problem-solving

5 a Education level b Column percentage frequency table

22 Graph 1 is misleading because the vertical axis does not start at zero therefore the difference in the number of days is exaggerated.

High School or less

Number of COVID-19 tests per million people

23 a The y-axis scale is not shown and therefore the Argentinian column appears taller than it should be. There is actually a large difference in the rates of the USA and Argentina. COVID-19 Testing by Country

25000

22 300

20000

15 700 14 100

15000 10000

36.36%

6.52%

3.51%

3 to 38.46% 6 hours

36.36%

21.74%

21.05%

7 to 13.46% 12 hours

21.82%

54.35%

54.39%

Over 12 hours

0.00%

5.45%

17.39%

21.05%

Total

100%

d Yes

330

258

Brazil Argentina

USA

Italy

Germany Norway

6 a January

b Example answer: Although the scale on the y-axis does not begin from 0, it is reasonable because the difference in the data is small. If the scale began from 0, it would be diﬃcult to choose a scale that would allow the reader to identify the exact values for each year.

Chapter 16 review Sedan

SUV

Total

Under 30

d No, there does not appear to be an association as the corresponding segments are similar in size. 7 a Yes, time spent studying a subject is the independent variable and the number of chapters is the dependent variable. b No c Yes, the duration of sunlight exposure is the independent variable, and plant growth is the dependent variable. 8 a

Truck Over 30

Total

b February

c Instrument sales

24 a It has made the increases in the 4-year intervals 1992-1996 and 1996-2000 appear faster than they really are (relative to the rate in the 2-year interval 1990-1992).

1 a

Less than 48.08% 2 hours

7000

5000

Some Bachelor’s Graduate university degree degree

Days of exercise

8 7 6 5 4 3

b 2 a 354 students c 27.94%

2 1

3 a Living area b Yes, there appears to be an association. b Yes

Sick days 25

b Yes, the statement is correct. The scatterplot shows that the data forms a decreasing, linear relationship, meaning that as a student exercises more, they get sick less.

d 8941 students

4 a Time of day preference

9 a Positive and strong

1098 Mathspace Victorian Curriculum – Year 10 Pre-Methods mathspace.co

b Negative and weak

10 a

12 a The y-intercept suggests that when the radish is 0 days old, it has no height above the ground.

Sales (in thousands) 25 20 15

b Yes, it is plausible that a seed will not sprout above the ground in the same day it is planted.

c 19 days

13 a y = 33.75

b Interpolation

14 a M = $941.40

Month 1 2 3 4 5 6 7 8 9 10 11 12

b No, because it is using extrapolation.

b Moderate, positive, linear

c 25 days

c No

d Yes, because there is a strong correlation and interpolation was used for the prediction.

d No, correlation does not imply causation. 11 a i

15 a Mistaking causation and correlation

b Over-generalisation

c Strategic choice of measure of centre

16 a Cherry-picking

15 10 5 x 5

ii y = x + 4 b i

iii Positive

b The sector for Prefers AlphaModel is much smaller than the other two sectors, but it is only 3% smaller than the other two percentages. c A column chart would be more appropriate because it would show that all three phones are preferred by approximately the same percentage of people.

16 14 12 10 8 6 4 2

x 2 4 6 8 10 12 14 16 18

iii Negative

Answers 1099 mathspace.co