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Contents 1

Indices and surds 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08

Index laws Negative indices Optional: Simplify surds Optional: Add and subtract surds Optional: Multiply and divide surds Optional: Binomial expansions with surds Optional: Rationalise the denominator Optional: Fractional indices Chapter 1 review

4 13 20 27 33 41 45 54 65

Algebra 68 2.A 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08

Factorise algebraic expressions Expand binomial products Factorise binomial products Factorise monic quadratic trinomials Optional: Factorise non-monic quadratic trinomials Optional: Complete the square Optional: Add and subtract algebraic fractions Optional: Multiply and divide algebraic fractions Optional: Simplify algebraic fractions Chapter 2 review

70 76 81 87 93

Finance 102 3.A 3.01 3.02

Simple interest Compound interest Modelling investigation: Home battery storage system PAYG and tax Investigation: Superannuation (Optional) Chapter 3 review

104 119 121 134

Contents mathspace.co

iii

Linear relationships 4.A 4.B 4.01 4.02 4.03

5.04 5.05

6.04 6.05 6.06 6.07 6.08

Exponential relationships Characteristics of exponential graphs Exponential models Investigation: Half-life Transformations of exponentials Optional: Exponential equations Optional: Logarithms Optional: Order of magnitude and logarithmic scales Optional: Connect exponentials and logarithms Chapter 6 review

Simultaneous equations 7.01 7.02 7.03 7.04 7.05

Graphs and features of quadratics Vertical translations Horizontal translations Investigation: Transformations of quadratics (Optional) Optional: Quadratic graphs Optional: Quadratic equations Chapter 5 review

Exponentials and logarithms 6.01 6.02 6.03

138 146 162 172

Quadratics 174 5.01 5.02 5.03

Linear graphs Linear models Linear inequalities Optional: Graph linear inequalities Investigation: Experiment with linear inequalities (Optional) Optional: Parallel and perpendicular lines Chapter 4 review

136

Solve graphically Optional: Solve using substitution Optional: Solve using elimination Optional: Simultaneous inequalities Optional: Applications of simultaneous equations Chapter 7 review

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

176 186 193 199 222 228

234 236 246 260 275 299 304 312 321

326 328 336 344 352 362 374

Scale factor 8.01 8.02 8.03

380 387 392 399

Pythagoras’ theorem Trigonometric ratios Angles of elevation and depression Optional: Sine and cosine rules Optional: Bearings Chapter 9 review

404 418 428 448

Measurement 452 10.A 10.B 10.01

10.02 10.03 10.04 10.05 10.06

Scale and perimeter Scale and area Scale and volume Investigation: Volume versus surface area Chapter 8 review

Trigonometry 402 9.A 9.01 9.02 9.03 9.04

378

Surface area of prisms and cylinders Volume of prisms and cylinders Surface area and volume of composite solids Investigation: Rainfall collection Modelling investigation: Wind turbine design Estimation and accuracy Optional: Pyramids Optional: Spheres Optional: Cones Optional: Composite solids Investigation: Instruments and methods Chapter 10 review

Probability 11.A 11.01

11.02 11.03

454 466 468 476 488 495 504 521

526

eview Venn diagrams and two-way tables R Conditional probability 528 Modelling investigation: Animal tagging for environmental monitoring 539 Mixed probability concepts 541 Investigation: Simulation Optional: Geometric probability Chapter 11 review 558

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Univariate statistics 12.A 12.01 12.02 12.03 12.04 12.05 12.06 12.07

Bivariate statistics 13.01 13.02 13.03 13.04 13.05 13.06 13.07 13.08

Two-way tables Associations between categorical variables Bivariate numerical data Associations between numerical variables Line of good fit Interpolation and extrapolation Misrepresentations of data Investigation: Misrepresentations in the media Correlation coefficient Investigation: Conduct a statistical investigation Chapter 13 review

Optional: Further quadratics 14.01 14.02

14.03 14.04 14.05

epresentations of data R Quartiles Boxplots Compare data sets Modelling investigation: Ferry schedules Samples and populations Optional: Cumulative frequency Optional: Standard deviation Optional: Comparisons using standard deviation Chapter 12 review

Optional: Solve by factorisation Optional: Solve by completing the square Investigation: Use bisection to solve quadratics (Optional) Investigation: Derive the quadratic formula (Optional) Optional: The quadratic formula Optional: The discriminant Optional: Further quadratic equations Chapter 14 review

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

562 564 576 590 603 605

615

620 622 634 654 664 678 689 703

717

722 724 734

740

747

Optional: Cubic functions 15.01 15.02 15.03

16.03

776 778 786 797 807

810

Optional: Angles with exact values Optional: The unit circle Optional: Graphs of sine and cosine Optional: Trigonometric equations Chapter 17 review

812 819 837 847

Optional: Hyperbolas 18.01 18.02

752 760 768 773

Investigation: Circles (Optional) Optional: Circles at origin Optional: Translations of circles Investigation: Equation of any circle (Optional) Optional: Transform circle equations Chapter 16 review

Optional: Further trigonometry 17.01 17.02 17.03 17.04

Optional: Cubic function Optional: Factored form of cubic functions Optional: Cubic equations Chapter 15 review

Optional: Circles 16.01 16.02

750

Investigation: Hyperbolas (Optional) Optional: Hyperbolas Optional: Transformations of hyperbolas Chapter 18 review

Optional: Geometry and proofs 19.01 19.02 19.03 19.04 19.05 19.06 19.07 19.08

Optional: Deductive reasoning Optional: Proofs with parallel lines Optional: Congruent triangles Optional: Similar triangles Optional: Further geometric proofs Optional: Circles and chords Optional: Angles in circles Optional: Tangents Chapter 19 review Answers

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vii

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 • Extend and apply index laws with positive-integer indices and variable bases, to include negative-integer indices • (Optional) Establish the connection between fractional indices and surds. Perform the four operations with surds and rationalise the denominator if required • (Optional) Simplify algebraic products and quotients involving indices with integer and fractional indices

1 Indices and surds Chapter outline 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08

4 13 20 27 33 41 45 54 65

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 8

= 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 WA Curriculum – Year 10 Methods Pathway 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 WA Curriculum – Year 10 Methods Pathway mathspace.co

Consider the expression ( p3)2. a

Are these expressions equivalent to ( p3)2: i

( p × p × p) × ( p × p × p)

p3 × p3

iii

( p × p × p)2

p3 × p2

( p × p × p) × ( p × p)

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

x⬚ ÷ x3 = x7

(b8 × b2) ⬚ = b20

( y⬚ × y17)1 = y17

(3m2 × ⬚)4 = 81m24

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

Fluency Ex 1

Ex 2

a 3 × a5

6y2 × y5 2

u ×u ×u

a 2 × a8

3y6 × 4y

2w × w × w

b × b4

a3 × −2a8 5

5z × 3z × z

6bc4 × 8b6a × 0

5g4h2 × 8g3h4

8m2n4 × 7m6n

y3 × y7

a ma n

4y8 × 6y6 × 3y3

Simplify:

b a10 ÷ a5

c m5 ÷ m2

d j11 ÷ j4

h l

Simplify: a

Simplify:

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

Simplify:

i 6

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

4y4 × (−2y6) × 4y5 2

2 3

3 4

(−4y7) × (−6y2) × (−4y2)

2s t × (−5r s ) × 3r t

9y9 × 8 (−y)8 × 7y7

(−20b4) ÷ 2b4

(−9y9) ÷ (−3y5)

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9r8 ÷ (4r2)

Ex 3

Simplify: a

(j4)6

(w2)11

(5h7)2

(xy2)5

(−x9)4

(−5x4)3

Ex 4

Ex 5

(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

p9 ÷ p5 ÷ p2

(11x4)0 − (12x8)0

(3a2 b5 c)4

(15x3)0 + 150 − 15y0

(4u5 v2)2

m8 ÷ m2 × m5

(4h0)2 + 18 ÷ (3g0)

m ((x2)6)5

(10x2 × 10x5)0 − 10x0

p5 × ( p4)3

(4y4)3

(3y5)2 × (5y2)3

(c10)11 ÷ (c8)3

(−240u32) ÷ (−8u9) ÷ (−5u12)

(−2x )

s 13

Simplify:

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

−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 WA Curriculum – Year 10 Methods Pathway 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 using the division law, we subtract the powers and get a−2. By using expansion

By simplifying

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

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

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

Fluency

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

p−2 q3

m−5 n−4 p4

Simplify: a

Rewrite as terms with negative indices: a

Ex 2

7x−9

Simplify: a e

5y9 × 4y−3 2h

−4

× 4h

b f

7a4 × 4a−2 3y

−2

−5

× 2y

c g

5x4 × (−3x−8) 2

−5

−4y × (−4y )

3y−2 × 4y−3

(5mp)2 × mp−2

1.02 Negative indices mathspace.co

Evaluate: a

Ex 3

Ex 4

−1

2 × 3−2

10 ÷ 2 + 3

20 × 2−2 + 6

3−1 + 4−1

3 × 4−2 + 5 × 2−4

102 × 5−2

(7 + 32) × 2−2

Simplify: a

2y6 × 4y7 × 4y−5

6y7 × 2y−5 × 5y3

2x−2 × 5x5 × 4x−5

12x−9 × 2x4 × 3x−2

4y × 3y ÷ 2y

8y10 ÷ 2y−4 ÷ y3

40x−2 ÷ 5x8 ÷ 4x−9

14x−8 ÷ 2x × 3x6

(4m−6)4

−1

Simplify: (2m)−3

(3p−4)−2

(3y2)−2

Simplify: a

(10x3 y−4)2

(3x−4 × 2x2)2

(4x3)−1 × ( y−2)−3

(10x3)3 × (2x4 y3)−2

Reasoning 14

2−2 × 24

14, 16

14, 16, 18

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

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 WA Curriculum – Year 10 Methods Pathway 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?

as the y-value. What do you think

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

3. Use a calculator to evaluate

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 WA Curriculum – Year 10 Methods Pathway 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

1.03 Practice questions Understanding 1

Fill in the boxes:

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.

i m

4, 5abd, 6cd, 7e-l, 8acfg, 9bc

4, 5bc, 6d, 7i-p, 8dgh, 9cd

∞

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

Simplify: a

Ex 2

4, 6, 7a-h, 8bef, 9ac

Is each number rational, irrational, or neither?

a 6

Trailblazer 1–3

Fluency Ex 1

Adventurer 1, 2

Are these surds? a

Explorer 1, 2

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 WA Curriculum – Year 10 Methods Pathway mathspace.co

is positive

1.04 Optional: Add and subtract 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 simplify to , just as 2p + 3p = 5p.

can be rewritten as

. 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 WA Curriculum – Year 10 Methods Pathway 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

Add like surds

and

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 WA Curriculum – Year 10 Methods Pathway 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 WA Curriculum – Year 10 Methods Pathway 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.

A rectangle has length and width 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.

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

1.05 Optional: Multiply and 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

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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

Expand: 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.

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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 series has a width of . Each subsequent member times the lengths of of the series has lengths that are 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

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 WA Curriculum – Year 10 Methods Pathway 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 there would be four terms, provided c ≠ a and d ≠ b. for

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 WA Curriculum – Year 10 Methods Pathway 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?

. so that the expression is

1.07 Optional: Rationalise the 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

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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 WA Curriculum – Year 10 Methods Pathway 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

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

Simplify

Evaluate

Simplify

Does each product show a conjugate pair or not?

Fluency

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

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

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

13, 15

, if x and y are surds.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Explain why the rationalised form of

Xander says that to rationalise Justify your answer.

will contain a term of , 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.

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 WA Curriculum – Year 10 Methods Pathway 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

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 WA Curriculum – Year 10 Methods Pathway 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

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

18, 19

Given

Without calculating, which is greater:

Explain circumstances where it is easier to evaluate

, explain how this expression was simpliﬁed. or

? Explain your reasoning.

. Use a numeric example in your explanation. 62

20–22

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

by writing it as

, rather than

Blair attempted to fully simplify an expression: 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:

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

23, 24, 27b-j

and a width of

25, 26, 27e-p, 28, 29

. Find the area of the rectangle in:

Fractional index 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.

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

Surd form

1.08 Optional: Fractional indices mathspace.co

1 Chapter review 1

Simplify: a c

x3 × 6x2 7

a4 × −3a5

6m3 n7 × 9m6 n

(−2x3)4

−2y × 2y × −2y

Simplify: a

Find the value of a and b in the equation:

Simplify:

a 5

4y10 × 5y−4

3y−2 × 4y−3

(5mp)2 × mp−2

(6x−2 × 2x3)4

Given a = m3n2, b = m−1n4, and c = m5n−3, find: a×b

a÷c

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

(Optional) Simplify fully: a

5x4 × (−3x−8)

a 11

Fill in the blanks:

a 10

(4a)−3

a 9

Simplify, giving your answers with positive exponents: a

Fill in each box. Evaluate if possible. a

(Optional) Simplify fully, given the variable represents a positive value: Chapter 1 review mathspace.co

(Optional) Simplify fully: a

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

(Optional) Find the perimeter of the triangle in simpliﬁed surd form. 12

(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.

(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 length of of the garden in simplest form with a rational denominator.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

d . Find the length

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 • Extend and apply knowledge of the expansion of binomial products to explore the factorisation of monic quadratics • (Optional) Factorise monic and non-monic quadratic expressions using techniques, such as completing the square, perfect squares, difference of squares and grouping in pairs for four-term expressions • (Optional) Explore efficient strategies to simplify expressions that involve addition, subtraction, multiplication or division of algebraic fractions with an algebraic expression in the numerator and/or denominator, including the use of factorisation

2 Algebra Chapter outline 2.A 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08

70 76 81 87 93

2.01 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

5 × x = 5x

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Simplify the products Combine like terms

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

(x + 7)2

(7 − z)

(e + 5)2

( y − 5)2

(−6 − a)

(a + b)

(a − b)2

(3x + 7)2

(2y − 5)2

(2y − 3)2

(x2 + 1)2

Expand and simplify: 5(m − 5)2

6(n + 6)2

8(q + 3)2

(8 − m) (m + 8)

3x(7x − 5y)2

Expand and simplify: a

(c + 6)2

Expand and simplify:

a 11

Ex 3

Expand and simplify:

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

(x − 7) (x + 7)

(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) 2

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, 17

16, 17

Is (3x + 4)2 equal to 9x2 + 16? Explain your answer.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

15–18

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

b x

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.

2.01 Expand binomial products 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.

2.02 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 have 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)

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Ex 3

Factorise: b

64 + 16e + e2

49 − 14p + p

u2 − 2uq + q2

6c2 − 24c + 24

5q2 + 10qt + 5t2

16w2 − 40w + 25

64r2 + 48r + 9

b2 − 2br + r2

v2 − 1

121 − v2

16 − 9y2

x2y2 − 49

7x2 − 63

45t2 − 20

x2 + 6x + 9

i Ex 4

Factorise: a

x2 − y2

e i 10

n2 − 25

f 25m2 − 49

81x2 − 16y2

Factorise: a

3x2y − 6y4 + 9x2y3

16x2y2 + 24xyz + 9z2

121c2 − 49d2

3x2y + x2z + 3y2 + yz

8a b + 12bc

(b + 4)2 − (3 − b)2

14h3p2r + 21h5p2 + 49hp4

2 4

16w2 − 64wt2 + 6wt − 24t3

48df 2 − 3dh2

11, 12

12–14

16q + 24q r + 9r

Reasoning

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 + 3 = (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

n3 − 121n

2x2 − 6x + 5x − 15

9x2 − 12xy + 4y2

(x + 16)2 − y2

4x3 + 16x2 − x − 4

Explain why the trinomial 4a2−72a + 9 cannot be factorised using perfect squares. 15bd, 16–18

15d, 16–20

Using the difference of two squares, evaluate: a

x2 − 4x + 4 = (x + 2) (x + 2)

For each expression:

Problem-solving 15ac, 16 15

652 − 632

992 − 972

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

452 − 422

582 − 522

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:

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.

Area I

Area II

Area III

Area IV

2.03 Factorise monic quadratic 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

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 2.03 Factorise monic quadratic trinomials 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)

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Factorise

5ab, 6a-f, 7a-f, 8a-f, 9ab

Fluency 5

Ex 1

Ex 2

Ex 3

5cd, 6g-l, 7g-l, 8g-l, 9e-l

Factorise: a

x2 + 3x + 2x + 6

x2 + 4x + 5x + 20

y2 + 6y + 12y + 72

t + 7t + 8t + 56

Factorise: a

x2 + 11x + 24

x2 + 8x + 16

x2 + 17x + 72

x2 + 16x + 64

t2 + 7t + 10

t2 + 8t + 7

y2 + 9y + 20

x2 + 8x + 15

y + 9y + 14

40 + 13x + x

36 + 12x + x2

x + 7x + 12

Factorise: a

x2 − 4x + 3

x2 − 16x + 60

x2 − 19x + 84

x2 − 12x + 35

x − 8x + 7

y − 10y + 16

t − 13t + 42

x2 − 20x + 51

y2 − 18y + 77

t2 − 20t + 96

x2 − 12x + 11

x2 − 18x + 45

Factorise: a

x2 − x − 6

x2 − x − 30

x2 + x − 20

x2 − 3x − 54

x2 − 3x − 70

x2 + 4x − 117

y2 − 13y − 48

t2 + 8t − 84

y2 − 12y − 45

x2 + 10x − 56

x2 + 10x − 24

x2 + 3x − 28

Factorise: a

2x2 + 14x + 20

4x2 + 24x + 32

4x2 − 4x − 288

3x2 − 21x + 30

5x2 + 5x − 30

−2x2 − 18x + 44

−4x2 + 12x + 40

−3x2 + 12x − 12

2x3 + 16x2 + 30x

−8 − 6x − x

Reasoning 10

5bc, 6c-h, 7c-h, 8c-h, 9a-h

−12 + 7x − x

10–12

x + 7x + 12x

10–12, 14

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. 84

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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. x

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.

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.

Area I + Area II

x2− Area II

Area I

x2− Area I

Area I

Area II

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

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. x

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.

2.03 Factorise monic quadratic trinomials mathspace.co

Problem-solving 17–19 17

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. m

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

17–20, 21ac

Factorise u2 − 11u + 24.

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

10m

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

x4 − x2 − 12

y4 − 5y2 + 4

(a + 3)2 + 3(a + 3) − 10

(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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

2.04 Optional: Factorise nonmonic 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: ax2 + bx + c where a ≠ 1. 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). 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

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

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. 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) = 6x2 − 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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

Fluency Ex 1

Ex 2

Ex 3

4a-d, 5a-d, 6a, 7, 11

4c-f, 5b-e, 6a-e, 8, 9, 11

3x2

18x

4e-h, 5e-h, 6d-h, 8–10

Factorise fully: a

3y2 + 28y + 9

2x2 − 11x − 40

3x2 − 25x + 28

8x2 − 21x − 9

12t2 − 13t − 4

81x2 + 72x + 16

12x2 + 7x − 10

6x2 + 41x − 56

Factorise fully: a

3x2 − 21x − 54

4x2 + 40x + 100

8s2 + 6s − 54

9x2 − 42x − 15

8x2 − 20x + 8

72x2 − 6x − 6

105x2 − 50x + 5

168x2 − 88x − 16

Factorise fully: a

−6x2 + 5x + 6

−6x2 + 25x − 14

−12x2 − 12x + 45

−6x2 + 22x + 40

−4x2 + 11x − 6

−9x2 + 24x − 16

−5x2 − 14x − 8

−3x2 − 10x + 8

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

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

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

Gabriella:

16x + 48x + 36 = 16x + 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, 6x2 + 10x − 5, 12x2 − 12x + 1 a

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

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

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.

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

2.05 Optional: 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, for now we will just focus on the algebraic skills associated with completing the square. 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.

2.05 Optional: Complete the square mathspace.co

We will start with the ﬁrst two terms and build a perfect square using our visual. x x

−2

−2x

To show the −4x, we will have two rectangles each with an area of −2x. This is now a square, with a side length of (x − 2).

−2

−2x

−2 −2x

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

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 = (x − 2)2 + 23 − (−2)2

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

= (x − 2)2 + 19

Simplify the constant term

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

2.05 Practice questions Understanding 1

x2 + 6x + 36

x2 + 8x + 12

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

x2 − 14x − 72

3ac, 4, 5a-d, 6ac, 7a-h, 8a-d

3bd, 4, 5c-f, 6bc, 7c-j, 8b-e, 9a-e

3x2 + 6x − 8 = ⬚(x2 + 2x) − 8

4x2 + 12x − 6 = ⬚(x2 + ⬚) − 6

5x2 − 10x + 4 = ⬚(x2 − 2x) + 4

Complete each solution for completing the square: a

6x2 + 36x − 1 = ⬚(x2 + ⬚) − 1

Ex 4

= ⬚(x2 + 6x + ⬚) − 1 − ⬚ = ⬚(x + ⬚)2 − ⬚

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

x2 + 18x

x − 10x + 30

x2 − 8x

x2 + 10x + 31

x2 + 14x + 47

x − 18x + 77

x + 9x + 16

x2 − 7x + 15

(x + 4)2 − 9

(x + 5)2 − 49

(x − 1)2 – 3

Factorise fully: (x − 3)2 − 1

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

Ex 3

3bd, 4, 5e-h, 6cd, 7e-l, 8e-h, 9c-h

6x2 + 24x − 7 = ⬚(x2 + ⬚) − 7

= ⬚(x − ⬚) − 1

a Ex 2

x2 − 10x + 24

2x2 − 10x + 1 = ⬚(x2 − 5x) + 1

= ⬚(x2 − 2x + ⬚) + 4 − ⬚

Ex 1

Complete each equation: a

Trailblazer 1, 2

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.

Fluency 3

Adventurer 1, 2

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

Explorer 1, 2

2x2 + 8x − 7 2

5x2 + 20x − 9 2

4x2 − 16x + 3 2

10x2 + 20x − 11

7x − 14x + 1

9x + 54x − 2

3x + 33x + 88

5x2 + 5x + 1

4x2 − 11x + 7

3x2 + 9x + 8

2x2 − 7x + 6

4x2 + 20x + 16

Factorise fully using the method of completing the square: a

x2 + 6x + 4

x2 − 8x + 11

x2 − 6x + 5

x2 + 11x + 10

x2 − 11x + 30

x2 + 24x + 63

x2 − 20x + 19

x2 − 28x + 115

Factorise fully: a

3(x − 1)2 − 12

2(x + 4)2 − 50

4(x + 2)2 − 1

−2(x − 9)2 + 32

−4(x + 2)2 + 16

9(x + 2)2 − 4

4(x + 3)2 − 5

2(x − 1)2 − 3

2.05 Optional: Complete the square mathspace.co

Reasoning

11, 12, 14a

10–12, 14ab

13, 14

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

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

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?

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: x2 + ⬚x + ⬚ = (x + ⬚)2 − ⬚

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

2 Chapter review 1

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

(8 − 5p) (8 + 5p)

Expand and simplify: a

(m + 4) (m − 4)

(3x − 0.7) (3x + 0.7)

Expand and simplify: a

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

(x − 2) (x − 8)

−2 (3x − 4) (4x + 3)

Expand and simplify: a

(m + 5)2

(n − 4)2

(2p + 3q)2

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 by grouping:

a2 + 3a + 4a + 12

p2 − 2p + 6p − 12

xy + 5x + 3y + 15

mk − 4m − 3k + 12

5r2 + 10rs + 5s2

Factorise: a

k3 + 8k2 + k + 8

35m2 − 49

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

5x2 (x + 4y) − 5y(x + 4y)

Factorise: a

x2 − 10x + 25

x2 + 16x + 64

1 − 2x + x2

x2 + 17x + 72

x2 + 4x − 117

y2 − 13y − 48

x2 + 10x − 24

x2 + x − 20

Factorise each trinomial: a

y2 + 10y + 25

a2 − 12a + 36

b2 + 9b + 14

c2 − 11c + 18

d2 + 3d − 10

e2 − 5e − 24

Chapter 2 review mathspace.co

The area of a rectangular plot of land is given by the expression A = x2 + 12x + 35 square units. a

If the length and width are of the form (x + p) and (x + q), find the values of p and q.

If x = 5 units, calculate the perimeter of the plot.

The area of a rectangular banner is 20xy − 15x square metres. If its length is 5x metres, find an expression for its width.

The area of a square tile is x2 − 14x + 49 square centimetres.

Find an expression for the side length of the tile.

Find an expression for the perimeter of the tile.

A large rectangular garden measures (3x + 4) metres by (x + 5) metres. A square flower bed of side length (x − 1) metres is placed in the corner of the garden. a

Write an expression for the area of the large garden in expanded form.

Write an expression for the area of the square flower bed in expanded form.

Find an expression for the remaining area of the garden (the part not covered by the flower bed) in its simplest expanded form.

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

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

x2 − 8x

x2 − 4x + 6 2

2x + 8x − 7

c g

x2 + 10x + 21 2

4x + 20x + 28

x2 + 3x + 3

5x2 + 5x + 1

(Optional) Factorise fully using the method of completing the square: a

10x + 20x − 11

x2 + 6x + 4

x2 + 14x + 33

x2 + 42x + 185

x2 − 11x + 30

6x2 + 41x − 56

81x2 − 90x + 25

−6x2 + 22x + 40

(Optional) Factorise fully: a

12t2 − 13t − 4

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

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

(Optional) Find the highest common factor of each pair of terms. a

20x and 15x2n

16a2bx3y6 and 12b4m2n6x

(Optional) Factorise: a c

k2 − 8k 2

−4bc + 8bc + 4c

5y2 + 10y + 15

3p2q2 + 4p4q4 + 5p6q6

(Optional) A student was asked to factorise the expression 35x2y + 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.

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Big ideas Accurate calculation and interpretation of interest, earnings and profit are crucial for effective money management in earnings, investments and loans.

Curriculum Outcomes • Explore, explain and calculate income tax, including the use of tax tables • Apply repeated simple interest to develop the compound interest formula and solve problems that relate to saving and borrowing • Analyse, represent and interpret problems in real-world situations involving real numbers, absolute and percentage error, linear inequalities, simultaneous equations, real-world formulae, quadratic or exponential functions, taxation, and/or compound interest • (Optional) Use authenticated websites to investigate how changes to the principal, rate of return, voluntary contributions and time can affect superannuation balances or compare characteristics of insurance, such as young driver car insurance or holiday insurance and recognise that the cost is higher when the risk is higher

3 Finance Chapter outline 3.A 3.01 3.02

Simple interest Compound interest Modelling investigation: Home battery storage system PAYG and tax Investigation: Superannuation (Optional) Chapter 3 review

104 119 121 134

3.01 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.

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 = Prn I P r n

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 $1000 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

104

Opening balance 1000 1100 1200

Interest 100 100

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Closing balance 1100 1200

Exploration (Continued) 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?

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) 0

10 12

a Which investment has a higher principal amount?

Create a strategy Choose the graph with the higher y-intercept.

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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) 0

10 12

c After how many years are the investments equal in value?

Create a strategy Find the intersection point of the graphs.

Apply the idea

From 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) 0

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10 12

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

Balance + interest − $3200 + $192 $3392 + ⬚ ⬚

Total balance $3200 $3392 ⬚ ⬚

Interest earned $192 $203.52 ⬚ −

Create a strategy 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 ⬚

We calculate interest on $3595.52 and put it in the interest earned column.

Interest earned $192 $203.52 ⬚ −

$3595.52 × 0.06 = $215.73

First year Second year Third year Fourth year

Balance + interest − $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. First year Second year Third year Fourth year

Balance + interest − $3200 + $192 $3392 + $203.52 $3595.52 + $215.73

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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)t

Apply the idea We are given that: P = 9450

t = 14 × 12 months = 168 months

Substitute the values

Evaluate

The future value is $13 593.90

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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)t

Apply the idea We are given that: P = 35 000 t = 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.

Apply the idea Use the total amount that Victoria repaid after four months from part (a). Interest = 35 563.37 − 35 000 = 563.37

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Subtract the two values Evaluate

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

$1000 is invested for 10 years with an interest rate of 10% p.a., compounded annually. Which graph represents the growth of the investment? Graph 1 Graph 2 Investment ($)

Investment ($)

3000

2500

2000

1500

1000

500

500 T (years)

T (years) 0

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)

Balance ($)

Time (years)

Time (years) 0

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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) 0

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

116

$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 24 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 WA Curriculum – Year 10 Methods Pathway 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 the relationship between the number of years passed and 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 equivalent interest rate for an account which compounds yearly. 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.

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Represent with mathematics Use real numbers, linear inequalities, compound interest and taxation to model the battery system’s total cost, daily energy savings and loan repayment time.

Interactive exploration

Discover this concept in action online

mathspace.co

Calculate total cost (for example, 15 × 600 + 1500 − 1500 = $9000), daily savings (for example, 15 × 0.30 = $4.50) and repayment time using A = 9000 × (1 + 0.04)t with annual savings (for example, 4.50 × 365 = $1642.50). Check cost and capacity constraints. Use GeoGebra or manual calculations.

Investigate 1.

Calculate the total cost and daily savings using your values. Show your work.

Use the compound interest formula to find the time to repay the loan with annual savings. Verify capacity meets 8 kWh daily.

Use an inequality to check if the net cost is within $10 000.

Interpret your plan Analyse your model to ensure the battery system meets energy needs and is financially viable. Confirm capacity exceeds 8 kWh daily and repayment is under 5 years. Calculate a 5% cost increase error. Adjust one variable (for example, capacity) and recalculate.

Investigate 1.

Does your battery meet energy needs and repayment goals? Is the plan realistic? Why?

Calculate the absolute and percentage error for a 5% cost increase. How does adjusting capacity affect repayment time?

Present your battery plan to the class. How do assumptions (for example, fixed energy prices) affect it?

Discussion

120

What was the most challenging part of modelling the battery system’s finances? Why?

How could you improve your plan with more data or a larger budget?

Compare your plan with a classmate’s. How do your costs and repayment times differ?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

3.02 PAYG and tax After this lesson, you will be able to… • understand what is PAYG tax and how it is deducted from gross pay. • identify different types of allowable deductions and how they reduce taxable income. • calculate taxable income based on gross income and allowable deductions. • understand how to use tax brackets and rates to calculate income tax. • calculate total tax payable and net income.

PAYG tax Each pay period, whether weekly, fortnightly or monthly, an employer will deduct tax from their employee’s gross pay, and send it to the Australian Taxation Oﬃce (ATO). This is known as Pay As You Go (PAYG). Deducting or withholding tax in this way means that income earners are not faced with a single large tax bill at the end of the ﬁnancial year. The correct amount of tax to withhold will depend on the employee’s income. Net pay = Gross pay − Income tax − Other deductions • Net pay is the amount of pay the employee takes home • Gross pay is the amount of pay the employee earns • Income tax is the amount of money that goes to the government to pay for roads, schools, etc. • Other deductions include money that is deducted voluntarily for expenses like union fees

Example 1 The PAYG tax amount, y, that an employer must withhold each week, in January 2023, was given by the formula y = ax − b where x is the number of whole dollars of gross weekly income, plus 99 cents, and a and b are values obtained from this table: Weekly earnings (x) less than $355 $422 $528 $711 $1282 $1730 $3461 $3461 and over

0 0.19 0.29 0.21 0.3477 0.345 0.39 0.47

0 67.4635 109.7327 67.4635 165.4423 161.9808 239.8654 516.7885

Note: This table is based on tax withheld with the tax-free threshold. Jeremy’s gross weekly earnings are $959.50. 3.02 PAYG and tax mathspace.co

121

Allowable deductions Allowable deductions are a particular type of expense that can reduce the amount of income on which tax is calculated. This reduces the overall tax liability. To qualify as an allowable deduction, an expense must be directly related to earning assessable income. Here are some common allowable deductions for Australian income tax: • Work-related travel expenses: Costs incurred for travel between different work locations; this does not include the travel between work and home. • Uniform and protective clothing: Expenses for purchasing and maintaining required uniforms or protective clothing. • Self-education expenses: Costs associated with courses and training directly related to your current job. • Home office expenses: A portion of costs for utilities, phone, and internet if you work from home. • Union fees and subscriptions: Fees paid to unions and professional associations directly related to your employment. Taxable income = gross income − allowable deductions

Example 2 Paul has a job in a retail store where he earned $45 768 during the last financial year. He also did some weekend work as a DJ in a local club and received an additional $12 250 during the same period. Paul has the following allowable tax deductions: • $510 in union fees • $240 for the purchase of new DJ equipment • $55 in fees for his tax agent Calculate his taxable income for the financial year.

Create a strategy Use the formula: Taxable income = gross income − allowable deductions

Apply the idea First, calculate Paul’s gross income by adding his two incomes together. Gross income = 45 768 + 12 250 = $58 018 Next, we add up all of Paul’s allowable tax deductions. Allowable tax deductions = 510 + 240 + 55 = $805 Finally, we calculate his taxable income. Taxable income = gross income − allowable tax deductions = 58 018 − 805 = $57 213

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5. Calculate total tax payable The total tax payable is found by adding together the income tax and the Medicare levy. Total tax payable = income tax + Medicare levy = $5788 + $1000 = $6788 An individual’s net income can be calculated by subtracting their total tax payable from their taxable income. net income = taxable income − total tax payable Therefore, the net income of someone who has a taxable income of $50 000 is: $50 000 − $6788 = $43 212

Example 3 Last financial year, Roxanne earned $157 872. Using the income tax rates shown in the table, evaluate the following. Resident tax rates (2024-2025) Taxable income 0–$18 200 $18 201–$45 000 $45 001–$135 000 $135 001–$190 000 $190 001 and over

Tax on this income Nil 16c for each $1 over $18 200 $4288 plus 30c for each $1 over $45 000 $31 288 plus 37c for each $1 over $135 000 $51 638 plus 45c for each $1 over $190 000

a Calculate the income tax payable on Roxanne’s income.

Create a strategy Use the income bracket where Roxanne’s income lies, and use the necessary rule.

Apply the idea Income tax payable = 31 288 + (157 872 − 135 000) × 0.37

Substitute the values

= 31 288 + 22 872 × 0.37 Evaluate the subtraction = 31 288 + 8462.64 Evaluate the multiplication = $39 750.64

Evaluate

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Personal income tax The key ideas when it comes to personal income tax are: • Your hourly wage or annual salary is a gross income. • Your employer will generally withhold tax, called PAYG. Take home pay = Gross income – PAYG • The income you receive from your job is your net income. Net income = Gross income – Tax • You may incur costs associated with employment. These are called allowable deductions. Keep a record of these throughout the year. Taxable income = Gross income – Allowable deductions • Tax payable is calculated using your taxable income and the rates stated in the tax tables. Medicare levy of 2% Total liability = Tax payable + Medicare levy

= Tax payable + 2% × Taxable income

• The total tax payable is compared to the PAYG tax paid by your employer. PAYG − Tax payable = Tax refund If the tax payable is higher than the PAYG then you will have a tax debt.

Example 4 Jeremy has a gross income of $45 000, with allowable deductions totalling $1850. Total PAYG withholding was $4920. It was calculated that the total tax liability is $4740.50. a Will Jeremy get a tax refund? If so, what is the value of the refund?

Create a strategy Use the formula: Tax refund = PAYG − tax payable

Apply the idea Tax refund = PAYG − tax payable

Write the formula

= 4920 − 4740.50

Substitute the amounts

= $179.50

Evaluate

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3.02 Practice questions Understanding

Explorer 1–4

Adventurer 1–4

Trailblazer 1–4

What percentage of their taxable income is the Medicare levy for most Australian residents?

What is personal income tax?

For PAYG:

What does PAYG stand for?

What is PAYG?

Why is PAYG helpful?

What are deductions, and how do they affect an individual’s taxable income?

Fluency

5–11, 14–16, 18, 19a-d, 20, 21

6–10, 12, 14–21

6, 7, 9, 12–21

Each week, Jack has $195 in PAYG tax withheld from his pay. How much does Jack have withheld in a full ﬁnancial year? Assume that there are 52 weeks in a year.

In the last ﬁnancial year, Carl was paid a wage every month, and each week there was $61 deducted as PAYG tax by Carl’s employer.

7 Ex 1

How much did Carl’s employer deduct in total in PAYG tax last ﬁnancial year?

At the end of the ﬁnancial year, he calculates his total tax payable to be $3049. Will he receive a refund or have tax owing?

Tina wants to calculate how much PAYG tax to withhold from her employees’ gross weekly income. She uses the formula y = ax − b where y is the PAYG tax to withhold, x is the number of whole dollars of gross weekly income plus 99 cents, and a and b are values obtained from the table shown. Calculate how much PAYG tax Tina would withhold from her employee who has gross weekly earnings of $833.

8 Ex 2

Weekly earnings (x) less than $355 $422 $528 $711 $1282 $1730 $3461 $3461 and over

−

0.19 0.29 0.21 0.3477 0.345 0.39 0.47

67.4635 109.7327 67.4635 165.4423 161.9808 239.8654 516.7885

Emma’s total allowable deductions for the year are $493. If her gross annual income is $25 000, calculate her taxable income for the year. Hannah earns a salary of $86 220 and claims $1293 for work-related expenses. She also contributes $190 to charities throughout the year. Calculate her taxable income.

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This year, Victoria had a taxable income of $71 250. Her gross income is $74 000. a

State how much she claimed in allowable tax deductions.

What percentage of Victoria’s gross income were these tax deductions? Round your answer to one decimal place.

Yuri’s gross annual income is $51 120, but he wants to make superannuation contributions throughout the year so that his taxable income reduces to $50 820 (assuming no other deductions). Calculate the monthly superannuation contribution he will need to make.

Dylan has a gross income of $92 749 and can claim $877 in allowable tax deductions. Dylan wants to reduce his taxable income to $90 000 by making voluntary weekly contributions to his superannuation. Calculate how much he must contribute each week.

Luigi has a $49 per month mobile phone plan. He regularly uses his phone for work purposes and can claim his work-related phone expenses as an allowable tax deduction. Using his itemised bill each month, Luigi calculates that his work-related phone usage is consistently around 20% of his total phone usage. Luigi works for 48 weeks (11.1 months) of the year. Calculate the amount he can claim as a tax deduction for his mobile phone.

For a gross weekly income less than $528 but greater than or equal to $422, the amount of PAYG tax to withhold is given by the formula PAYG tax to withhold = 0.19x − 67.4635 where x is the number of whole dollars of gross weekly income plus 99 cents.

Calculate how much PAYG tax is withheld from a gross weekly income of $450.

Calculate how much PAYG tax is withheld from a gross weekly income of $511.40.

For gross weekly incomes that are $3461 or more, the amount of PAYG tax to withhold is given by the formula PAYG tax to withhold = 0.47x − 516.7885 where x is the number of whole dollars of gross weekly income plus 99 cents. Calculate how much PAYG tax is withheld from a gross weekly income of $3797.70.

The Medicare levy is charged at 2% of taxable income. Calculate the Medicare levy on a taxable income of $60 000.

For the last ﬁnancial year, Patricia claimed $1857 in total deductions and paid a Medicare levy of $1744. If her gross income for the year was $x, write an equation and solve for x. Note that the Medicare levy is charged at 2% of taxable income.

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Isabelle is a working holiday maker. She has a job in Australia and her taxable income for the year is $47 600. Use the table to ﬁnd how much income tax she is required to pay: Working holiday makers income tax rates for 2024 to 2025 Taxable income $0 - $45 000 $45 001 - $135 000 $135 001 - $190 000 $190 001 and over

Tax on this income 15c for each $1 $6750 plus 30c for each $1 over $45 000 $33 750 plus 37c for each $1 over $135 000 $54 100 plus 45c for each $1 over $190 000

Questions 19 – 24 will use the following table of tax rates for 2024-2025. Resident tax rates (2024-2025) Taxable income $0 – $18 200 $18 201 – $45 000 $45 001 – $135 000 $135 001 – $190 000 $190 001 and over 19

20 Ex 3

21 Ex 4

Nil 16c for each $1 over $18 200 $4288 plus 30c for each $1 over $45 000 $31 288 plus 37c for each $1 over $135 000 $51 638 plus 45c for each $1 over $190 000

Calculate the income tax payable for these taxable incomes: a

$8326

$36 810

$194 388

$225 810

$93 395

$146 664

Last financial year, Carmen’s taxable income was $197 955. Based on the Resident Tax Rate (2024-2025) table, calculate: a

The income tax payable on Carmen’s income.

The Medicare levy payable, given that it is 2% of taxable income.

The total amount Carmen needs to pay in income tax and Medicare levy.

Anna has a gross income of $52 000, with allowable deductions totalling $2400. Her PAYG withholding for the year was $5600. It was calculated that her total tax liability is $5550. a

Will Anna get a tax refund, and if so, what is the value of the refund?

Anna forgot to claim an additional $900 in deductions. Calculate her new taxable income and the tax payable on that amount.

Calculate the new refund based on the revised tax payable.

Reasoning 22

Tax on this income

22, 23

23, 24

Most people pay income tax and the Medicare levy, which is charged at 2% of taxable income. Janet has a taxable income of $17 000 and says that her total payment to the tax department is 2% of her income. Is Janet correct? Explain your answer.

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Stewie computed his tax using the tax rates for the financial year 2024-2025: Here is his work:

Tax = $4288 + 0.30 × ($65 000 − $45 000) Apply the corresponding tax formula

= $4288 + 0.30 × $20 000

Evaluate the brackets

= $4288 + $7400

ultiply 0.30 by $20 000 to find the M additional tax amount

= $11 688

dd the base tax amount and the A additional tax amount

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

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

A proposal suggested that to help those on low incomes the tax-free upper bound be increased from $18 200 to $20 000. Part of the proposal is that the tax paid by people earning more than $45 001 should remain the same. In order to achieve this, it is proposed that those earning between $20 000 and $45 000 should pay 18c for each $1 instead of the current 16c for each $1. a

Will this proposal achieve its stated aim? Explain.

An alternative proposal is to raise the rate for those earning between $20 000 and $45 000 to pay 18c for each $1 over $20 000. What effect would this have on people earning more than $45 000?

If the 18c for each $1 over $20 000 is adopted, what change would it make to the instructions for ‘Tax on this income’ for those earning between $135 001 and $190 000?

Problem-solving 25–27 25

132

Tobias works as a chef. Because he wears a uniform for his job, he is able to claim the cost and cleaning of his uniform as an allowable tax deduction. Tobias’s work-specific uniform expenses are shown in the table:

25–29

Uniform item Chef’s checked trousers Chef’s jacket Chef’s hat Apron Pair of protective shoes

25, 26, 28–30

Number of items 2 2 1 2 1

Price per item (dollars) 21.50 26.00 7.85 23.50 60.00

For his uniform, Tobias’s total cleaning expenses for the year come to $52. Calculate Tobias’s total clothing and cleaning expenses for the year.

Tobias also spent $295 on a set of knives and $17.25 on a knife pouch. He can claim these as a tools and equipment expense because he uses them only for work. Calculate his total tools and equipment expenses.

Calculate Tobias’s total allowable tax deductions.

Tobias earns $19.50 per hour and works 40 hours per week. Calculate his annual gross income if he works 48 weeks in a year.

Calculate Tobias’s taxable income.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Questions 26 – 30 will use the following table of tax rates for 2024-2025. Resident tax rates (2024-2025) Taxable income $0 - $18 200 $18 201 - $45 000 $45 001 - $135 000 $135 001 - $190 000 $190 001 and over 26

Tax on this income Nil 16c for each $1 over $18 200 $4288 plus 30c for each $1 over $45 000 $31 288 plus 37c for each $1 over $135 000 $51 638 plus 45c for each $1 over $190 000

Emma is a researcher with a taxable income of $97 589. Throughout the year, her employer has deducted $560 per week in pay-as-you-go (PAYG). a

Calculate the income tax payable using the Resident Tax Rate (2024-2025) table.

Calculate the Medicare levy, which is 2% of taxable income.

Since Emma’s employer has deducted more than is needed from her income, Emma is entitled to a refund for the year. Find the amount of this refund.

Jasmine, a marketing manager, has a gross income of $72 000 and allowable deductions of $3500. Her total PAYG withholding amounts to $9200. Will Jasmine receive a tax refund? If so, how much? (Don’t forget to account for the Medicare levy in the calculations.)

Michael, a teacher, receives a gross monthly salary of $5416.67 with PAYG withholding each month of $1842.25. He has allowable deductions totalling $1600 annually. What is Michael’s tax refund or liability? (Don’t forget to account for the Medicare levy in the calculations.)

Elias has just been hired in a new role where he will earn a fortnightly income of $5697 for an entire year.

Calculate how much Elias earns in a year. Assume 26 fortnights in a year.

Calculate the income tax payable on Elias’s annual income using the Resident Tax Rate (2024-2025) table.

Elias also needs to pay $113.94 alongside his income tax. How much PAYG tax would his employer need to withhold from his fortnightly pay so that Elias’s total PAYG tax instalments for the year exactly cover his total tax payable?

Cinta has a gross weekly income of $1023. Each week, her employer withholds $193 in PAYG tax. At the end of the financial year, Cinta has to submit a tax return. She has $842 in allowable tax deductions for the year. Assuming there are 52 weeks in a year, calculate: a

Her taxable income.

Her income tax payable using the Resident Tax Rate (2024-2025) table.

How much Cinta pays towards the Medicare levy, which is 2% of her taxable income.

How much Cinta has already paid in PAYG tax instalments throughout the year.

How much Cinta is entitled to as a tax refund.

Investigation: Superannuation (Optional)

Investigate online

mathspace.co

3.02 PAYG and tax mathspace.co

133

3 Chapter review 1

For each representation, identify whether it shows simple interest or compound interest. Explain how you know. (Assume A is the total amount in dollars after n time periods, such as months or years.) a

Opening Closing Month Interest Balance Balance 1 $300 $15 $315 2 $315 $15 $330 3 $330 $15 $345

b 1 2 3

A = 1000(1.03)n

A ($)

1800

A ($)

1600

1400

1200

1000

800

600

400

A = 1000 + 30n 1800

1600

200

Opening Closing Interest Balance Balance $300 $15 $315 $315 $15.75 $330.75 $330.75 $16.54 $347.29

Month

200

n (years) 2 4 6 8 10 12 14 16 18

Brad borrows $6000 at a rate of 5.5% p.a., compounded annually. a

After 3 years, Brad 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 three years?

Calculate the interest that each investment accumulates: a

A $6500 investment earns interest at 2.5% p.a., compounded annually over 18 years.

A $6000 investment earns interest at 3.0% p.a., compounded monthly over 11 years.

A $6800 investment earns interest at 2.4% p.a., compounded weekly over 7 years.

Jasmine is planning to make an investment which accumulates monthly compound interest at a rate of 5.2% p.a. If she aims to have her investment reach $120 000 after 10 years, find the principal amount that she should invest.

For a gross weekly income less than $528 but greater than or equal to $422, the amount of PAYG tax to withhold is given by the formula, where x is the number of whole dollars of gross weekly income plus 99 cents. PAYG tax to withhold = 0.19x − 67.4635 Calculate how much PAYG tax is withheld from each gross weekly income. Round your answer to the nearest dollar. a

134

$460

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

$510.50

Big ideas Equations can be solved more easily by manipulating algebraic expressions and applying a series of organised steps and techniques.

Curriculum Outcomes • Solve one-variable linear inequalities involving brackets and/or a variable on each side. Represent the solution on a number line and verify the solution by substitution • (Optional) Identify the region on the Cartesian plane defined by linear inequalities • (Optional) Use gradient and/or point/s to graphically and algebraically determine equations of parallel and perpendicular lines • Move flexibly between real number inequalities expressed as a worded statement, algebraically or on a number line

4 Linear relationships Chapter outline 4.A 4.B 4.01 4.02 4.03

138 146 162 172

Apply the idea

Reﬂect and check Test a value which satisﬁes our solution by substituting it into the original inequality.

Subtract 1 from both sides

Evaluate the subtraction

Divide both sides by 3

Evaluate

For example, choose x = 3 3×3+1>4 10 > 4 This is true, suggesting the solution is correct.

b Plot the solutions to the inequality 3x + 1 > 4 on a number line.

Create a strategy Plot the solution from part (a).

Apply the idea To graph the inequality x > 1, plot a hollow circle on point 1 with a ray towards positive inﬁnity. −5

−4

−3

−2

−1

Example 2 Solve: 5a > 15 (2 + a)

Create a strategy Use inverse operations to isolate a on one side of the inequality.

Apply the idea 5a > 30 + 15a –10a > 30 a < −3

Expand the brackets Subtract 15a from both sides Divide both sides by −10 (remembering to flip the inequality symbol)

Reﬂect and check Test a value which satisfies our solution by substituting it into the original inequality. For example, choose a = –4: 5 × (−4) > 15 (2 + (−4)) –20 > –30 This is true, suggesting the solution is correct.

4.01 Linear inequalities mathspace.co

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4.01 Practice questions Understanding 1

Explorer

Adventurer

1–3

Trailblazer

1–3

For each inequality: i

x ≥ 29

Plot the inequality on a number line.

Determine if each value satisfies them (makes them true). i

x = 27

x < −29

x = −28

iii

x = 29

x = −30

What happens to the inequality symbol when we multiply or divide both sides of an inequality by a negative number?

Is each statement true or false? a

The symbols ≤ and ≥ don’t include the endpoint, which we show with a hollow circle.

The symbols < and > don’t include the endpoint, which we show with a hollow circle.

The symbols < and > do include the endpoint, which we show with a ﬁlled-in circle.

The symbols ≤ and ≥ do include the endpoint, which we show with a ﬁlled-in circle.

Fluency 4 Ex 1

5 Ex 2

142

4, 5a–c, 7a–e, 8–10, 12, 13, 17, 18, 20

5, 6a, 7d–l, 8–10, 12, 15, 17, 19, 20

6, 7g–p, 8, 9, 11, 14–16, 18–20

Solve: a

6x − 54 ≥ 30

21 + 7v ≥ 21

8.4 < 1.9c + 2.7

3x + 27 > 3

4 (b + 3) ≤ 12

3.3g + 4.9 ≥ 34.6

24.15 > 1.5s − 5.55

19.8 > 33 (0.4y + 1.208)

0.25 (x − 8.2) + 1.13 ≤ 1.13

Solve: a

8 − y < 7y

4x ≤ 1 − x

−9 + e > 4e

3 − 2m < 5m

−7b − 7 ≥ 7b

−9h + 20 > −14h

2.4m − 34.33 > 7.19 + 3.6m

2.1 (0.37x − 4.63) ≤ 10.5x

Solve: a

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

−1.2(u + 3.7) ≥ −7.2

For each inequality: i

Solve.

Plot the solutions to the inequality on a number line.

3 < 4x − 1

7v + 4 > v + 16

0.75b − 1.3 ≤ 0.8

m Ex 3

−4y + 9 > 9

7.6 < 9.1 − 0.2s

3 (j − 1) < 9

−7a − 5 ≤ 2

2m + 6 > 30 − m

5.5a − 13.7 < −11.5

12.2y ≥ 15y − 14

Construct and solve an inequality for this situation: “The sum of 2 groups of x and 1 is at least 7.”

For “5 more than half of q is greater than 13”: a

Construct and solve the inequality.

Find the greatest value of q that satisﬁes this condition (if it exists).

Sophia has a budget for school stationery of $32, but has already spent $15.25 on books and folders. Write an inequality that shows how much Sophia can spend on other stationery, where p represents the amount she can spend, and solve for p.

Xavier struggles to save money and spent $33.40 last week. He needs $32.00 for a movie with friends on the weekend. Write an inequality representing the amount Xavier had at the beginning of the week, where A is the initial amount in his bank account, and solve for A.

To get a grade of C, John needs a total score of at least 300 over four exams. He scored 63, 73, and 98 on the ﬁrst three exams. If x represents what he must score on the last exam to get a C or better, write an inequality and solve for x.

Skye was given $72 for a birthday present. This present, along with earnings from a Saturday job, is being set aside for a mountain bike. The job pays $5.00 per hour, and the bike costs $379. a

Write an inequality to ﬁnd the minimum number of hours that Skye needs to work to afford to buy the mountain bike, where h represents the number of hours worked.

Solve for h. Round your answer to one decimal place.

If Skye can only work her job for a whole number of hours, ﬁnd the minimum number of hours she must work to afford her bike.

Lachlan is planning on going on holiday. He has saved $2118.40, and spends $488.30 on his aeroplane ticket. a

Let x represent the amount of money Lachlan spends on the rest of his holiday. Write an inequality to represent the situation.

Solve for x.

Find the highest amount that Lachlan could spend on the rest of his holiday.

4.01 Linear inequalities mathspace.co

143

James wants to order some books from an online bookstore. Each book costs $13.30 and shipping for the entire order is a ﬂat rate of $29.05. James can spend no more than $74.00. a

If B represents the possible number of books that James can buy, write an inequality and solve for B. Round your answer to one decimal place, if necessary.

Find the maximum number of books that James can afford to buy.

A farmer wants to construct a rectangular enclosure alongside a barn. One side of the enclosure will be the length of the barn, which measures 36 metres. The farmer has a maximum of 80 metres of fencing material available for the project. If the width of the roped area is W, form an inequality and solve for the range of possible widths.

Ex 4

For 21 + 7x > 21: Do these values of x satisfy the inequality 21 + 7x > 21? a

x=1

x = 14

x=7

x = −6

x=6

x=0

x = 12

x = 6.8

Do these values of x satisfy the inequality 3.3 − 5.2x < − 3.2x − 7.5? a

Do these values of x satisfy the inequality 6 (x + 2) ≤ 48? a

x=0

x = 5.4

x = −5.5

When breeding certain types of fish it is recommended that the number of female fish is more than double the number of male fish. This can be expressed as the inequality f > 2m, where f is the number of female fish and m the number of male fish. Determine whether these values satisfy the recommendation. a

f = 10, m = 8

Reasoning

f = 23, m = 9

21, 22

f = 13, m = 10

f = 7, m = 4

21–23

21–24

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

Isabelle tried to solve the inequality but made a mistake in her work.

25 − 2x < 31

Step 1:

2x < 6

Step 2:

x<3

Determine which step is incorrect and explain the error.

Write the correct working to solve the inequality.

Percy tried to plot the solution to the inequality 4x + 28 ≥ − 8 on a number line, however his answer is incorrect. −10 −9 −8 −7 −6 −5 −4 −3 −2 −1

Identify the error(s) and explain how to rectify them.

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Ryan wants to save up enough money to buy a new sports equipment set, which costs $40.00. Ryan has $22.10 that he saved from his birthday. To make more money, he plans to wash neighbours’ windows for $2 per window. The expression which represents the number of windows, w, required to be washed is w ≥ 8.95. Determine whether each statement is correct. Explain your reasoning. a

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

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

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

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

Problem-solving 25, 26 25

25–27

Jimmy spends $120 to produce some rockmelons. He is able to sell each one for $9. a

If n represents the number of rockmelons he must sell to make a proﬁt, write an inequality and solve for n.

Find the least number of rockmelons that Jimmy must sell to make a proﬁt.

The table shows the times of six swimmers in a 1500 m race. Only those who achieve a time of 22 minutes or below qualify for the final. Use the inequality t ≤ 22, where t is the time of each swimmer, to ﬁnd all of the swimmers who qualify for the ﬁnal.

27–29

Swimmer James Harry Lachlan Adam Glen Kenneth

Time (minutes) 21 24 22 27 31 23

Neville is saving up to buy an LED TV that is selling for $950. He has $650 in his bank account and expects a nice sum of money for his birthday next month. a

If the amount he is to receive for his birthday is represented by x, write an inequality that models the situation where he is able to afford the LED TV.

How much money would he have in total if his parents were to give him $310 for his birthday?

Would he have enough to buy the plasma TV if his parents were to give him $310 for his birthday?

A particular bus can carry a maximum weight of 4300 kg on board. If the average adult passenger weighs 75 kg and the driver weighs 81 kg, then the number of passengers on the bus at any one time must satisfy 75a + 81 ≤ 4300, where a represents the number of adult passengers. a

There are 58 adults already on board and 8 more at the bus stop waiting to get on. Find the combined weight of everyone on board if the driver lets all of the new passengers on.

Can the driver safely let all the passengers waiting at the bus stop to get on board?

A particular beef mince brand must contain between 10% and 15% fat before it is shipped. To make 1500 grams of mince which is acceptable to be shipped, the processors mix w grams of 18% fat mince with mince containing 3% fat. Write an inequality for the number of grams of fat in the mix, and solve this to ﬁnd the least and greatest acceptable values of w. 4.01 Linear inequalities mathspace.co

145

4.02 Optional: 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. 3

2 1 −5 −4 −3 −2 −1

−1

x 1

−2 −3 −4 −5 −6

The line indicates the set of all points on the Cartesian plane that make the equation true.

Interactive exploration

Discover this concept in action online

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mathspace.co

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. 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).

148

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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.

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

150

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

Apply the idea x of the 200 mL bottles contains 200x mL. y of the 500 mL bottles contains 500y mL.

he amount in bottles must be less than or equal to the amount T 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

152

Substitute y = 0

2x = 40

Simplify

x = 20

Divide both sides by 2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

Is the inequality represented using a solid or dashed line on the graph? a

y ≤ 6x − 3

y ≤ 5x − 3

y < −2x − 7

6ab, 7a, 8, 10–13, 14a-f, 15, 16ab, 17ab, 19

Fluency 6

y < 5x − 7

y ≥ 3x − 2

y>8

y ≥ 6x − 2

y < −9

Is the shaded region above or below the line? a

Ex 1

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

(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

(15, 20)

iii

(16, 7)

(10, −6)

(−2, −3)

iii

(0, 0)

(2, 7)

(−2, 4)

iii

(−4, 29)

(2, −13)

y > 3x i

(−2, −9)

(12, 20)

y ≥ −8x − 3 i

(6, −50)

4.02 Optional: Graph linear inequalities mathspace.co

155

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)

Graph the boundary line.

Does the point (0, 0) satisfy the inequality?

iii

Graph the inequality.

y≥x+5

Ex 4

158

(2, 2)

y > 2x − 4

2x + 3y < 12

6x + 3y > 12

y > 3x − 5

y < 4x − 7

y ≥ −3x + 5

Graph the inequality:

(5, −5)

Do the points on the line satisfy the inequality y ≤ −3x + 4?

y ≤ 4x − 3

iii

For each inequality:

Ex 3

(4, −8)

−4x + y < 5

g −3x + y < 7

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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?

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.

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.

4.02 Optional: Graph linear inequalities mathspace.co

159

4.03 Optional: 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.

162

Equation form

Characteristic of parallel lines

y = mx + c

Parallel lines have the same m value.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Examples y = 2x – 1 y = 4 + 2x

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. –3y –x = 5

Write the equation

–3y = x + 5

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

Write the equation

y = –3x + 7

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

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:

164

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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. 7–9, 10ab, 11–13, 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

A line with gradient

y=8−x

x + 3y = −10

y = −9x + 5

Are these lines perpendicular to y = 3x + 2? a

y = −3x

3y − x = 5

Are these lines perpendicular to y = 9x + 2? a

168

Find the equation of line L.

Find the equation of the line described:

8–11, 14–18

Which of these lines are parallel to y = 7x + 3?

a Ex 2

8–13, 14a–d, 15–18

y = 9x

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

Ex 3

Find the gradient of line L.

Find the equation of line L.

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

19, 20

and passes through (−8, 5).

and goes through the point (0, 6).

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.

4.03 Optional: Parallel and perpendicular lines mathspace.co

169

4 Chapter review 1

Does the coordinate pair satisfy the inequality? a

(4, 2); 3x − y ≥ 7

(−3, 5); y < −x + 2

A student earns $12 per hour tutoring and $8 per hour babysitting. They need to earn at least $100 per week to cover expenses. a

Write an inequality relating tutoring hours t and babysitting hours b.

Will working 5 hours tutoring and 4 hours babysitting meet the goal?

A bakery sells cakes for $25 each and cookies for $2 each. They need to sell at least $150 worth of goods daily. a

Write an inequality for cakes c and cookies k.

Does selling 4 cakes and 20 cookies satisfy the requirement?

Find the minimum number of cakes needed if no cookies are sold.

(Optional) The graph shows the distances covered by two runners over time, where distance is in kilometres and time is in hours: 35

Distance (km)

30 25

Runner A

Runner B

15 10 5

Time (h) 0

Graph the inequality

At what speed is Runner A travelling for the first 2 hours?

At what speed is Runner B travelling for the first 2 hours?

Test the point (4, −3) for the inequality.

(Optional) A linear inequality has boundary points (2, 5) and (−1, −1) and is satisfied by (0, 0). a

Find the equation of the boundary line.

Write the inequality.

Graph the inequality.

(Optional) Graph the inequality: a

172

4x − 3y ≤ 8

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

y ≥ −x + 4

(Optional) Are these lines parallel to y = 3x + 5? a

y = 3x + 2

3x + y = −5

y = 3x

4y = x

(Optional) Are these lines perpendicular to y = −4x + 1? a

y = −3x + 5

y = −4x + 5

(Optional) Find the equation of the line described. a

Parallel to y = −5x + 4 and passes through the point (−1, 3).

Perpendicular to y = 2x + 7 and passes through the point (6, −3).

(Optional) This pair of lines is parallel. Find the value of g. 0 = 6x − 5y − 2 0 = gx + 2y − 1

(Optional) The graph shows a hiker’s distance from a starting point over time: Distance (km) 25 20 15 10 5 0

Time (h) 4 5

Determine the equation of the line for the segment from 1 to 2 hours.

Find the equation of a parallel line passing through (3, 5).

Find the equation of a perpendicular line passing through (2, 10).

(Optional) A line passes through (−2, 4) and (3, −4). a

Find the equation of the line.

Write the equation of a perpendicular line through (2, 1).

(Optional) Two lines are parallel with equations

and 3x − 7y = 14.

Verify the slopes are equal.

Find the perpendicular slope and write an equation through (0, 0).

Chapter 4 review mathspace.co

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Big ideas Understanding quadratic relationships requires identifying their key features which highlight various characteristics of the function.

Curriculum Outcomes • (Optional) Solve monic and non-monic quadratic equations graphically and algebraically, including the use of the quadratic formula, factorising techniques and digital tools and verify the solution/s by substitution • Use a table of values to plot points and graph quadratic functions of the form y = ax2 + c. Identify and relate key graphical and algebraic features and make connections to the graphical and algebraic solution/s of ax2 + c = k. Use digital tools to explore the shapes, features and related solutions to more complex quadratic functions • (Optional) Graph monic and non-monic quadratics of the form y = ax2 + bx + c or y = a(x − p)2 + q and their transformations, manually and using digital tools. Identify and connect key graphical and algebraic features and make connections to the algebraic solution/s

5 Quadratics Chapter outline 5.01 5.02 5.03 5.04 5.05

176 186 193 199 222 228

Parabolas are symmetrical. The axis of symmetry is the dotted line pictured here. It has an equation of the form x = ⬚, and you can see that this x-value will be the same as the x-value for the maximum or minimum turning point.

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

Example 1 For each graph, identify: 1. the turning point 2. whether the turning point is a maximum or minimum 3. the equation of the axis of symmetry 4. any intercepts that can be identiﬁed a 16

14 12 10 8 6 4 2 −4

−2

x 2

Apply the idea 1. The graph has a turning point at (0, −2). 2. This graph has a minimum turning point because it is concave up. 3. The equation of the axis of symmetry is the vertical line through the turning point. The equation is x = 0 4. The graph has x-intercepts at (−1, 0) and (1, 0) and a y-intercept at (0, −2).

5.01 Graphs and features of quadratics mathspace.co

177

Plot from table of values If we have an equation for a quadratic function we can create a table of values to create a series of points to plot.

Example 2 Consider the quadratic function y = 2x2 + 1. a Complete the table of values from x = −2 to x = 2 and graph the parabola. −2

x y

−1

Create a strategy Substitute each x-value into y = 2x2 + 1 to complete the table. Plot the points and draw a smooth curve.

Apply the idea Calculate y for each x: y = 2(−2)2 + 1 Substitute x = −2 = 2 × 4 + 1 Evaluate the square = 9 Multiply Similarly, compute for x = −1, 0, 1, 2: x y

−2 9

−1 3

0 1

1 3

2 9

Plot these points and draw the parabola: 20 18 16 14 12 10 8 6 4 2 −4 −3 −2 −1

x 1

Key features: The vertex is at (0, 1), the parabola opens upward (since a = 2 > 0), and it is symmetric about the y-axis.

5.01 Graphs and features of quadratics mathspace.co

179

Determine if the statements are true or false. a

The axis of symmetry is a vertical line passing through the turning point.

A minimum turning point is the highest point on the graph.

The x-intercepts are the points where the graph cuts the y-axis.

The y-intercept occurs where x = 0.

In quadratic equations, there are always two x-intercepts since the degree is 2.

Describe the relationship between the turning point and the axis of symmetry.

Fluency 4

4, 5, 6abeg, 7

4, 5, 6acfg, 7, 8

4, 6c-eh, 7, 8

For each graph: i

Describe the graph as concave up or concave down.

Is the turning point a maximum or minimum point?

x −4 −3 −2 −1

4 20

−5 15

−10

−15

5 x −2

2 −1

−2

x 1

−4 −6 −8

182

8 4

−3 −2

−1

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

−2 −4 −6 −8 −10 −12 −14

y x 1

Reasoning

9, 10, 12, 13

9–11, 13

Explain why the graph of y = ax2 + c has a minimum value of c when a > 0.

Determine whether this statement is always true, sometimes true, or never true. Justify your answer with examples or counter-examples. “The graph of y = ax2 + c has either a maximum or minimum value, but not both.”

Justify the signiﬁcance of the y-value of the turning point in determining the range of values of the y-values for the quadratic function y = ax2 + c.

Match each equation to its graph. A

y = 16x2 −36

40 y

y = −4x2 + 25

35 30 25 20 15 10 5

−2

iii

−1 −5 −10 −15 −20 −25 −30 −35 −40

184

−2

−1 −5 −10 −15 −20 −25 −30 −35 −40

x 1

40 y 35 30 25 20 15 10 5

x 1

y = −9x2 + 25

D 40 y 35 30 25 20 15 10 5

x 1

35 30 25 20 15 10 5

−1 −5 −10 −15 −20 −25 −30 −35 −40

y = 49x2 − 36

40 y

−2

−1 −5 −10 −15 −20 −25 −30 −35 −40

Explain how the values of a and c affect the solutions to ax2 + c = 20 and their positions on the graph of y = ax2 + c.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Match the given quadratic equations with their corresponding graphs and features. Equations: A

y = −20x2 − 20

y = 8x2 + 4

y = 4x2 + 4

Parabola concave up, turning point at (0, 4), symmetric about the y-axis.

Parabola concave down, turning point at (0, −20), symmetric about the y-axis.

y = −12x2 − 20

Graphs and features: a

−2

−20

−30

−40

−50

x 1

−10

−2

−1

Parabola concave up, turning point at (0, 4), symmetric about the y-axis.

Parabola concave up, turning point at (0, −20), symmetric about the y-axis.

y −2

−1

x 1

−10 40 −20 30 −30 20

−40

−50

x −2

−1

Problem-solving 16 14

14–16

A small bakery’s weekly profit is modelled by P = 20x2 − 80, where P is the profit in dollars and x is the number of dozen cupcakes sold. a

Does the parabola open upward or downward? Does it have a minimum or maximum value?

Find the coordinates of the vertex of the parabola.

What is the minimum value, and what does it represent in this context?

Solve 20x2 − 80 = 420 and interpret the solutions.

5.01 Graphs and features of quadratics mathspace.co

185

A tutoring service charges a fee that increases with the square of the hours booked. It charges $25 for 1 hour and $100 for 2 hours. a

Use the information provided to complete the following table of fees. Hours Fee

1 $25

2 $100

Plot the graph of the fee against the number of hours.

Find the equation for the fee y as a function of hours x.

How much would a 8-hour session cost?

If a client paid $2500, how many hours were booked?

The distance a toy rocket travels upward is modelled by D = 9x2 + 16, where D is the distance in meters and x is the time in seconds after launch. a

Create a table of values for x from 0 to 5 and sketch the graph.

Solve 9x2 + 16 = 160 and explain what the solutions mean. Verify using the graph below.

What does the vertex represent in this context?

D 150

100

50 x −5 −4 −3 −2 −1

1 2 3 4 5

5.02 Vertical translations After this lesson, you will be able to… • interpret and describe the effects of vertical translations on quadratic graphs. • identify the equation of a quadratic function given certain points or intercepts. • understand and apply combinations of transformations including vertical translations, compressions, enlargements, and reflections. • find the equation of a specific quadratic function using given transformations.

Vertical translations Interactive exploration

Discover this concept in action online

186

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

mathspace.co

The features of y = x2 + q and y = −x2 + q are: y = x2 + q Turning point Minimum turning point at (0, q) Axis of symmetry x=0 For q > 0, no x-intercepts x-intercept(s), also called For q ≤ 0, the x-intercepts are the roots or solutions to the quadratic

Description

y = −x2 + q Maximum turning point at (0, q) x=0 For q ≥ 0, the x-intercepts are

For q < 0, no x-intercepts This is the graph of y = −x2, that has been vertically translated by q units: up (for positive q) and down (for negative q)

This is the graph of y = x2, that has been vertically translated by q units: up (for positive q) and down (for negative q)

Example 1 Identify the vertical translation amount and direction from y = x2, for each of these: a y = x2 + 2

Create a strategy The vertical translation is the amount up or down that the graph has shifted from the base y = x2. 4

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

Apply the idea The graph has been vertically translated up by 2 units.

5.02 Vertical translations mathspace.co

187

Reﬂect and check Show the graph of y = x2 also on it. y 4 y = x2 + 2

3 2

y = x2

x −4 −3 −2 −1 −1

−2

We can see the turning point has moved up 2 units from (0, 0) to (0, 2).

b y = x2 − 3

Create a strategy The vertical translation is the amount up or down that the graph has shifted from the base y = x2.

Apply the idea In equations of the form y = ax2 + q, the q value is the vertical translation. This equation has a vertical translation of 3 units downwards.

Reﬂect and check Show the graph with the graph of y = x2 on it. y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3

y = x2 1 2 3

x 4

y = x2 − 3

−4

We can see the turning point has moved down by 3 units from (0, 0) to (0, −3).

188

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

5.02 Practice questions Explorer

Understanding

Trailblazer

1–5

List two equations for a quadratic that has been vertically translated upwards from y = x2.

List two equations for a quadratic that has been vertically translated downwards from y = x2.

What changes and what stays the same, when comparing the graphs of y = x2 and y = x2 + 7?

What happens to the x-intercepts when the graph of y = −x2 + 1 is vertically translated downwards by 2 units?

What is the connection between the y-intercept and the value of q in y = x2 + q? 6, 7, 8ab, 9ab, 10ab, 12

Fluency Ex 1a

Adventurer

1–5

6, 7, 8bc, 9bc, 10bc, 11, 12

Identify the vertical translation amount and direction from y = x2, for these graphs: a

−6

−3

6 3

x 3

−3

−6

−3

y 4

1 −4 −3 −2 −1 −1

x 3

−6

x 1

−2

190

6, 7, 8cd, 9cd, 10cd, 11, 12

1 −4 −3 −2 −1 −1 −2

−3

−4

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1

Ex 1b

Identify the vertical translation amount and direction from y = −x2, for these equations: a

Ex 2

Ex 3

y = −x2 + 2

y = −x2 − 3

y = −x2 − 1

For each of these quadratic equations: i

Sketch the graph on a Cartesian plane alongside the graph of y = x2.

Find the value of the y-intercept.

iii

Find the turning point.

y = x2 − 5

y = x2 + 3

y = x2 − 12

y = 2x2 + 10

Find the equation of the quadratic in the form y = x2 + q that passes through the point: a

y = −x2 + 4

(1, 1)

(−4, 5)

(7, 0)

(2, −4)

Find the equation of the quadratic in the form y = ax + q that has the y-intercept provided and passes through the point given. Then describe the transformations that have occurred from the base graph of y = x2. a

y-intercept at y = 3, passes through (1, 5)

y-intercept at y = −1, passes through (3, 8)

y-intercept at y = −4, passes through (−6, −184)

y-intercept at

, passes through

A stone is thrown vertically into the air. Its height above the ground is given by the equation h = −5t2 + 20, where h is the height in metres, and t is the time in seconds. Determine the maximum height the stone reaches and the time it takes to reach the ground.

The height of a rocket above the ground is described by the equation h = 4t2 + 10, where h is the height in metres, and t is the time in seconds since the rocket was launched. a

Calculate the rocket’s height 3 seconds after it was launched.

Sketch the rocket’s trajectory for the ﬁrst 3 seconds of ﬂight.

What was the initial height of the rocket above the ground? What feature of the graph does this relate to?

Reasoning 13

13–15

With each pair of quadratic equations, identify the quadratic that has the largest maximum value: a

y = −x2 + 3, y = −x2 − 3

y = −2x2 + 8, y = −2x2 − 10

y = −x2 − 1, y = −5x2 − 2

y = −x2, y = −x2 − 1

With each pair of quadratic equations, identify the quadratic that has the smallest minimum value: a c

13, 14

y = x2, y = x2 + 1 2

y = x − 3, y = 5x + 1

y = 3x2, y = 3x2 − 4

y = x2 + 12, y = x2 − 12

Do vertical translations always result in a new turning point?

5.02 Vertical translations mathspace.co

191

Problem-solving 16, 17 16

192

16, 17, 19

16–19

A toy rocket is launched vertically from an elevated platform. The height of the rocket above the ground can be modelled by the equation h = −4t2 + 30, where h is the height in metres and t is the time in seconds. The rocket launches from a platform that is 10 metres above the ground. a

How does the vertical translation of 10 metres affect the equation modelling the toy rocket’s height? What is the new equation?

Using the new equation, ﬁnd the maximum height the toy rocket reaches above the ground.

Calculate the time it takes for the toy rocket to return to the ground. Round your answer to two decimal places.

Sketch a graph showing the movement of the toy rocket.

Water is sprayed into the air from a fountain, forming a parabolic arc. The height of the water above the fountain can be described by the equation h = −2t2 + 8, where h is the height in metres, and t is the time in seconds. a

What do the values of the coeﬃcient −2, and the constant term +8 reﬂect in the context of the problem?

Determine the maximum height the water reaches.

What is the time it takes for the water to hit the ground?

Another fountain nearby has a similar equation, h = −2t2 + 10. Describe the similarities and differences between the two fountain arcs.

The elevation of a hill can be modelled by the equation h = Ax2 + B, where h is the elevation in metres and x is the horizontal distance from the base of the hill in metres. a

5 metres from the base of the hill the elevation is 65 metres, and the elevation is 10 metres at the base, what is the equation?

As we move further and further away from the hill, what happens to the elevation? Is there a problem with the model?

A water tank is being drained, and the water level y in centimetres after x minutes is described by the quadratic equation y = −3x2 + 1500. a

What is the initial water level in the tank?

Find the time it takes for the tank to be completely drained.

What parameter in the equation (the dilation or the vertical translation) affects: i

The amount of water in the tank

The rate the water drains

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

5.03 Horizontal translations After this lesson, you will be able to… • define horizontal translation and explain its effect on a graph. • understand and interpret the equation y = (x − p)2 to denote a quadratic function that has been horizontally translated by p units. • identify the direction of horizontal translation based on the sign of p. • explain how the turning point and axis of symmetry are affected by horizontal translation. • apply the concept of horizontal translation to find the equation of a translated quadratic function.

Horizontal translation Interactive exploration

Discover this concept in action online

mathspace.co

The features of y = (x − p)2 are: y = (x − p)2 Minimum turning point at ( p, 0) x=p

Turning point Axis of symmetry x-intercept(s), also called the roots or solutions to ( p, 0) the quadratic This is the graph of y = x2 which has been horizontally translated by p units: Description

– right for p > 0, (x − p)2 – left for p < 0, (x − p)2 – for p = 0, no horizontal translation

y = −(x − p)2 Maximum turning point at ( p, 0) x=p ( p, 0) This is the graph of y = x2 which has been vertically reflected across the x-axis and horizontally translated by p units: – right for p > 0, (x − p)2 – left for p < 0, (x − p)2 – for p = 0, no horizontal translation

5.03 Horizontal translations mathspace.co

193

Example 1 For this graph:

6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1

x 1 2 3

−2

a Identify the horizontal translation amount and direction from y = x2.

Create a strategy The horizontal translation is the amount left or right that the graph has shifted from the base y = x2. Count from the vertex. y 4 3 2 1 −6 −5 −4 −3 −2 −1 −1

x 1

2 3 4

−2 −3 −4

Apply the idea The graph has been horizontally translated left by 4 units.

b Identify the horizontal translation amount and direction from y = x2 to y = (x − 7)2.

Create a strategy In equation y = (x − p)2, the p value is the horizontal translation.

Apply the idea Notice that in the translated equation, the p term inside the brackets has been replaced by 7. This equation has a horizontal translation of 7 units to the right.

194

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

5.03 Practice questions Understanding

Adventurer 1–5

Trailblazer 1–5

List two equations for a quadratic that has been horizontally translated left from y = x2.

List two equations for a quadratic that has been horizontally translated right from y = x2.

What changes and what stays the same when comparing the graph of y = x2 and the graph of y = (x − 2)2?

What happens to the y-intercept when the graph of y = −x2 is horizontally translated left?

What is the connection between the x-value of the turning point and the value of p in the equation y = (x − p)2? 6, 7, 8a–d, 9ab, 10ab, 11, 12a–c

Fluency 6

6, 7, 8c–f, 9bc, 10bc, 11, 12b–d, 13, 14

y = (x − 8)2

y = (x + 9)2

y = −(12 − x)2

y = (0.75 + x)2

Identify the horizontal translation amount and direction from y = x2 for these graphs: a

7 6

7 5

2 1

−5 −4 −3 −2 −1 −1

1 2 3 4 5 6 7

−2

−3 −2 −1 −1

x 1

2 3 4

−2

7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3

196

6, 7, 8e–h, 9cd, 10bd, 11, 12cef, 13, 14

Has the given equation been translated up, down, left or right from y = x2? a

Ex 1a

Explorer 1–5

x 1

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

7 6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3

x 1 2 3

Ex 1b

Ex 2

Identify the horizontal translation amount and direction from y = x2 for these equations: a

y = (x − 1)2

y = (3 + x)2

y = (x − 10)2

y = (x + 5)2

y = (−6.5 + x)2

y = −(5 − x)2

For each of these quadratic equations: i

Sketch the graph on a Cartesian plane alongside the graph of y = x2.

Find the value of the y-intercept.

iii

Find the turning point.

y = (x + 1)2

y = (x − 1)2

y = (x + 3)2

y = −(x + 2)2

For each of these quadratic equations: i

Sketch the graph on a Cartesian plane.

Identify the axis of symmetry.

iii

Find the turning point.

y = −(x + 1)2

y = −(x − 5)2

y = −(6 + x)2

Write the equation of the quadratic function that has the form y = (x − p)2 using the description of the horizontal translation from the base y = x2. a

Horizontally translated left 7 units

Horizontally translated right 13 units

Horizontally translated left

Horizontally translated right 4.6 units

Horizontally translated 3 units left and has a vertical reﬂection across the x-axis

Horizontally translated 2 units right and has a vertical reﬂection across the x-axis

units

For these equations, describe the transformations that have taken place from the base y = x2. a e

y = −(x + 3)2

y = (x + 6)2

y = (x − 9)2

y = (x − 0.15)2

y = −(x − 3)2

y = (13 + x)2

Find the equation of the quadratic that: • Passes through the point (5, −1) • Is concave down • Is in the form y = (x − h)2

A photographer is taking a picture of a tall building, and the camera lens is positioned at H = (D − 1)2 metres above the ground. If the height of the building (H) is 40 metres, ﬁnd the horizontal distance from the photographer to the building.

5.03 Horizontal translations mathspace.co

197

Reasoning

15, 16

16, 17

Explain why with a quadratic of the form y = (x − p)2, the y-values can never be negative.

What is the maximum y-value the quadratic y = −(x − p)2 + 5 can have? Explain why this is the maximum y-value for any value of p.

When subtracting 3 on a number line we move our starting point 3 units to the left. But when we go from y = x2 to y = (x − 3)2 we translate the graph 3 units to the right. Explain why these are different.

Problem-solving 19 18

18, 19

A marksman is practising at a shooting range with a target positioned at y = (x − 3)2 metres, where y is the height above the ground, and x is the horizontal position from the shooter. a

If the bullseye is at a height of 16 metres, ﬁnd the horizontal position from the shooter to the centre of the bullseye.

If the marksman wants to hit a target 25 metres away, at what height should the target be placed?

A hiker travels along a straight path in the forest for 6 hours at a speed modelled by s = (t − 2)2, where s represents the speed in km/hour and t is the time in hours. a

What is the hiker’s initial speed?

At what time is the hiker travelling at a speed of 9 kilometres/hour?

When does the hiker stop momentarily?

Draw a graph to determine the hiker’s fastest speed.

A company’s monthly sales data is modelled by the equation R = (m − 4)2, where R represents the sales revenue in thousands of dollars, and m is the month. The model is only accurate for months 1 through 12. a

Determine the month in which the company had its highest sales revenue

Calculate that maximum revenue.

Find the month when sales ﬁrst reached $6000.

What happens to the revenue at month 4?

Graph the model and describe what happens to revenue over the 12-month period.

Investigation: Transformations of quadratics (Optional)

Investigate online

198

18–20

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mathspace.co

5.04 Optional: 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.

Interactive exploration

Discover this concept in action online

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 q units gives the equation y = x2 + q. y 4 y = x2 + 2

3 2

y = x2

x −4 −3 −2 −1

1 −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

−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 p units we get y = (x + p)2.

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199

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

3 2 1 −4 −3 −2 −1

x2 2

This graph shows y = x2 vertically enlarged by a scale factor of 2 to get y = 2x2 and compressed by x

−1

a scale factor of 2 to get

−2

y 4 3 2 y = x2

1 −4 −3 −2 −1

−1

x 3

−2

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).

−3 −4

y = −x2

All of these transformations can be performed on a quadratic and are represented by the equation y = a(x − p)2 + q.

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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 p and q in y = a(x − p)2 + q. 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.

Example 2 For the quadratic equation y = −2 (x − 3)2 + 8: a Is the parabola concave up or concave down?

Create a strategy

Apply the idea

Concavity depends on the sign of a: positive for concave up, negative for concave down.

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

Apply the idea

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.

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

Apply the idea 2

A quadratic of the form y = a(x − p) + q has a turning point at ( p, q).

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 2

Write the equation

= −2(0 − 3) + 8

Substitute x = 0

= −2(9) + 8

Evaluate the power

= −10

Evaluate

The y-intercept is at (0, −10).

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203

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 = p, 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 − p)2 + q 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

204

Adding and subtracting 4

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

5.04 Optional: Quadratic graphs mathspace.co

205

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

−2

x 1

−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: 12

10 8 6 4 2 −7 −6 −5 −4 −3 −2 −1 −2

x 1

−4 −6

5.04 Optional: Quadratic graphs mathspace.co

207

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).

b Find the equation in the form y = a(x − p)2 + q.

Create a strategy Use the coordinates of the turning point found in part (a) to fill in p and q in the quadratic equation of the form y = a(x − p)2 + q. 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).

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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.

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).

5.04 Optional: Quadratic graphs mathspace.co

209

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 2

=4×0+0

Substitute x = 0

Evaluate

5.04 Optional: Quadratic graphs mathspace.co

211

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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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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).

5.04 Optional: Quadratic graphs mathspace.co

213

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 3

What feature is at the point ( p, q) on the parabola deﬁned by y = a (x − p)2 + q?

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

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

c Ex 2

Fluency

Ex 1

x 1

y = 3 (x + 1) + 4

y = −(x + 3)2 − 6

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 + 5)2 − 4

y = 3 (x + 1)2 + 2

y = −x2 + 4x − 3

y = x2 + x + 3

For each quadratic equation: i

Write in the form y = a (x − p)2 + q.

Find the coordinates of the turning point.

iii

Sketch the graph of the parabola.

y = x2 + 6x

y = −x − 8x − 18

5.04 Optional: Quadratic graphs mathspace.co

215

Ex 4

Each of the graphs has an equation of the form, y = a(x − p)2 + q. Find the values of a, p and q for each graph and write the equation. a

2 −3

−2

−1

x 1

−2

−1

y 12

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

−1

−3

−2

−4

−3

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1

−2

216

−2

−4

x 1

2 −3

−1

−4

x 1

The parabola y = x2 has been translated 2 units to the right and 6 units down. a

Ex 5

Find the equation of the parabola.

Sketch the graph of the parabola.

y = x (4 − x)

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 − 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)

Find the equation of the parabola in factorised form.

What is the turning point? y 14 12 10 8 6 4 2 −2 −1 −2

x 1

−4

5.04 Optional: Quadratic graphs mathspace.co

217

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 − p)2 + q? a

7 6 5 4 3 2 1

−4 −3 −2 −1−1

x 1

−4 −3 −2 −1

−1 −2 −3 −4 −5 −6 −7

Ex 6

9 8 7 6 5 4 3 2 1

x 1

−4 −3 −2 −1−1 −2 −3 −4 −5

For each quadratic: i

Factorise to ﬁnd the x-intercepts.

Express the equation in the form y = a (x − p)2 + q 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

Describe the concavity of the parabola.

Factorise, and ﬁnd the x-intercepts.

For each quadratic: i

218

7 6 5 4 3 2 1

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

−2 −3 −4 −5 −6 −7

7 6 5 4 3 2 1

iii

Find the y-intercept.

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

Find the axis of symmetry.

y = 4 − x

y = x2 + 4x + 3

y = x2 + 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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Ex 7

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

y = x2 + 6x + 7

y = −4x2 − 4x + 1

y = −(x2 + 8x + 11)

y = x + 6x − 2

y = −x2 + 10x − 23

y = −x2 + 8x − 9

y = x2 + 2x − 11

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 − p)2 + q, 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 determined by finding the x-intercepts of the corresponding quadratic graph.

Given a quadratic of the form y = a (x − p)2 + q, what are the conditions on a, p, and q 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) a

Which key features are the same for A and B?

Which key features are different for A and B?

5.04 Optional: Quadratic graphs mathspace.co

219

Marvin wants to understand how turning point form and standard form of a quadratic equation are related. Consider the turning point form: y = a(x − p)2 + q a

Expand a(x − p)2 + q and write the equivalent equation to y = a(x − p)2 + q 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.

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 p, in metres, reached by a ball thrown in the air after t seconds is given by the equation: p = 10t − t2 a

Complete the table of values for p = 10t − t2: t

Graph the relationship p = 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: p = 163 + 38s − s2 where s is its horizontal distance from where it was thrown.

220

Graph the relationship p = 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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

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.

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. y

(150, 10) x

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. 29

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 equation 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.

5.04 Optional: Quadratic graphs mathspace.co

221

5.05 Optional: Quadratic equations After this lesson, you will be able to… • recognise that quadratic equations can have zero, one, or two solutions. • apply the null factor law to solve quadratic equations.

Quadratic equations A quadratic equation is an equation where the highest power of the variable is 2. It can be written in the form ax2 + bx + c = 0, where a ≠ 0.

ax2 + bx + c = 0 a b c

the coefficient of x2, not equal to 0 the coefficient of x the constant term

For example, in the equation 2x2 − 3x + 5 = 0, the values are a = 2, b = −3, and c = 5. Quadratic equations can have 0, 1 or 2 solutions. depending on the equation. Consider these examples: 1. x2 = −4 has 0 solutions because a squared number cannot be negative. 2. x2 = 0 has 1 solution: x = 0. 3. x2 = 9 has 2 solutions: x = 3 and x = −3, or x = ± 3. The symbol ± means “plus or minus,” representing both the positive and negative square roots.

Exploration Explore how many solutions these equations have by attempting to solve them: • x2 = 25 • x2 + 1 = 0 What do you notice about the number of solutions based on the value of the constant?

222

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Example 1 Solve for x: a x2 = 16

Create a strategy Reverse the square operation by taking the square root of both sides.

Apply the idea Start with the given equation

Take the square root of both sides

Simplify

The solutions are x = 4 and x = −4.

Reﬂect and check Check: 42 = 16 and (−4)2 = 16, so both solutions satisfy the equation.

b 81x2 − 64 = 0

Create a strategy Isolate the x2 term and take the square root.

Apply the idea Add 64 to both sides

Divide both sides by 81

Take the square root of both sides

Simplify

The solutions are

and

5.05 Optional: Quadratic equations mathspace.co

223

The null factor law The null factor law states that if the product of two or more factors equals 0, then at least one factor must be 0. For example, if

ab = 0 a b

first factor second factor

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.

Example 3 Solve: a (x − 8) (x + 11) = 0

Create a strategy Apply the null factor law by setting each factor to 0 and solving.

Apply the idea x−8=0 x=8

Set the ﬁrst factor to 0 Add 8 to both sides

x + 11 = 0

Set the second factor to 0

x = −11

Subtract 11 from both sides

The solutions are x = 8 and x = −11.

b 6n(n + 12) = 0

Create a strategy Use the null factor law to set each factor to 0 and solve.

Apply the idea 6n = 0

Set the ﬁrst factor to 0

n=0

Divide both sides by 6

n + 12 = 0 n = −12

Set the second factor to 0 Subtract 12 from both sides

The solutions are n = 0 and n = −12.

5.05 Optional: Quadratic equations mathspace.co

225

Ex 3

Solve: a

n (n + 9) = 0

−3p ( p + 6) = 0

(6x − 5)2 = 0

y (3y − 10) = 0 (5k − 7)2 = 0

(2 − 5x) = 0

(r − 3) (r − 8) = 0

(t + 4) (t + 10) = 0

(7x − 4) (2x − 9) = 0

(4w + 9) (3w − 5) = 0

(5x − 11) (3x − k) = 0

Reasoning

9, 10, 12

11–13

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.

Mark used the null factor law to solve the equation (x − 9) (x + 13) = 0 and found the solutions x = −9 and x = 13. Is Mark’s answer correct? Explain why.

Explain and justify why the equation x2 + 4 = 0 has no solutions.

Problem-solving 14–16 14

14, 16

16–18

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 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.

The equation px2 + qx − 20 = 0 has solutions coeﬃcients p and q.

Given ax2 − c = 0:

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

5.05 Optional: Quadratic equations mathspace.co

227

5 Chapter review 1

For each graph: i

Identify the turning point.

State whether the turning point is a maximum or minimum.

iii

Identify the equation of the axis of symmetry.

Identify any intercepts that can be identiﬁed.

5 x

−1.5 −1 −0.5

−5

−10 x

−1.5 −1 −0.5

0.5

1.5

−15

1.5

Create a table of values from x = −2 to x = 2.

Plot the points and join them with a smooth parabolic curve.

y = 7x2 + 2

y = −4x2 + 5

y = −10x2 − 3

y = −6x2 + 9

y = 5x2 − 4

y = −13x2 − 2

y = 2x2 − 6

y = 11x2 + 4

This diagram shows only part of the graph of a quadratic relationship. Complete the table of values for the graph. x y

−3

−2

−1

Does the vertex represent a minimum or a maximum point?

If we knew the equation of the curve, what would be the sign of the coeﬃcient of x2?

16 14 12 10 8 6 4 2 −2 −4 −6

228

For each equation:

0.5

For the equation y = −17x2 + 6: i

State whether the graph is concave up or down.

Find the turning point.

iii

Sketch the graph using at least ﬁve points.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1 2 3 4 5 6 7 8 9 10

Identify the vertical translation amount and direction from y = x2, for these graphs: a

5 x

−6

−3

y 8

2 1

−4 −3 −2 −1 −2

−4

−6

−8

x 1

y = −x2 + 6

y = −x2 + 3

y = −x2 − 5

y = −x2 − 2

Find the equation of the quadratic in the form y = x2 + q that passes through the point: a

Identify the vertical translation amount and direction from y = −x2, for these equations: a

−10

−4 −3 −2 −1 −2

−5

5 −6

(2, 5)

(−5, 12)

(8, 15)

(3, −10)

Find the equation of the quadratic in the form y = ax2 + q that has the y-intercept provided and passes through the point given. Then describe the transformations that have occurred from the base graph of y = x2. a

y-intercept at y = 4, passes through (2, 12)

y-intercept at y = −2, passes through (4, 14)

y-intercept at y = −5, passes through (−7, −250)

y-intercept at y = 3, passes through (3, −24)

Chapter 5 review mathspace.co

229

Identify the horizontal translation amount and direction from y = x2 for these graphs: a

3 x

−2

x −6

−4

−2

x −4

−2

−10 −8 −6 −4 −2

−2

−2 −4

−5

Write the equation of the quadratic function that has the form y = (x − p)2 using the description of the horizontal translation from the base y = x2. a

Horizontally translated left 8 units

Horizontally translated right 10 units

Horizontally translated left 6.4 units

Horizontally translated right

Horizontally translated 4 units left and has a vertical reﬂection across the x-axis

Horizontally translated

units

units right and has a vertical reﬂection across the x-axis

For these equations, describe the transformations that have taken place from the base y = x2. a

y = (x + 8)2

y = −(x − 9)2

y = −(x − 5)2

y = (x − 11)2

y = (15 + x)2

The path of water from an irrigation sprinkler forms a parabolic arch. The height of the water, H metres, at a horizontal distance of D metres from the sprinkler, is modelled by the equation H = − 0.5( D − 4)2 + 8. Find the horizontal distances from the sprinkler at which the water is 6 metres high.

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(Optional) Each of the graphs has an equation of the form, y = a(x − p)2 + q. Find the values of a, p and q for each graph and write the equation. a

16 14

4 2 1

y 20

x 1

−5 −4 −3 −2 −1 −2

−1 −2

−6 −5 −4 −3 −2 −1

y 10

−4

−6 −8

−10

−12

10 8 6 4 2

−2 −4 −6 −8 −10

−7 −6 −5 −4 −3 −2 −1

2 −8

−6

−4

−2

x 2

−4

−6 −8 −10 −12

(Optional) 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 + 6)

y = x(5 − x)

y = (x − 4)(x − 8)

y = (x − 5)(x + 2)

y = (6 − x)(x + 3)

y = (2 − x)(7 − x)

Chapter 5 review mathspace.co

231

(Optional) A parabola has x-intercepts at x = −5 and x = 9. 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, −45)

The turning point is a maximum turning point and the y-intercept is at (0, 22.5)

The turning point is a minimum turning point and the y-intercept is at (0, −90)

The turning point is a maximum turning point and the y-intercept is at (0, 11.25)

(Optional) 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 − 6x + 5

y = x2 + 8x + 12

y = −2x2 − 8x − 3

y = −(x2 + 10x + 16)

y = x2 + 12x − 7

y = −x2 + 14x − 45

y = x2 + 10x − 15

z2 = 144

w2 = 169

64 = 16y2

−25 = −100x2

g 17

232

y = −x + 12x − 20

(Optional) Solve: a

b2 = 16

m2 = 36

x − 196 = 0

9y = 225

−5k2 = −45

49k2 + 15 = 64

20 = x − 16

(Optional) Solve: a

(z + 5)2 = 64

(m − 8)2 = 100

(x − 9)2 = 16

(−4 + p)2 = 81

144 = (3 − n)2

(10 − x)2 = 121

(6l + 11)2 = 169

12( p2 − 4) = 924

(Optional) Solve: a

m(m + 12) = 0

w(5w − 14) = 0

−4p( p + 8) = 0

(q + 10) = 0

(7x − 9)2 = 0

(6k − 11)2 = 0

(3 − 8x)2 = 0

(r − 5)(r − 10) = 0

(t + 6)(t + 15) = 0

(9x − 5)(3x − 8) = 0

(5w + 12)(4w − 7) = 0

(6x − 13)(2x − m) = 0

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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. • 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 • Use a table of values to plot points and graph exponential functions of the form y = ax where a > 0. Identify and relate key graphical and algebraic features and use these to determine graphical solutions of related equations. Use digital tools to explore the shapes, features and related solutions to more complex exponential functions • Identify and distinguish between linear, quadratic and exponential functions represented by equations, tables of values and graphs • (Optional) Use algebraic techniques to solve exponential equations that involve terms with related bases • (Optional) Interpret and use base-ten logarithmic scales on graphs of real-life contexts • Analyse, represent and interpret problems in real-world situations involving real numbers, absolute and percentage error, linear inequalities, simultaneous equations, real-world formulae, quadratic or exponential functions, taxation, and/or compound interest

6 Exponentials and logarithms Chapter outline 6.01 6.02 6.03 6.04 6.05 6.06 6.07 6.08

236 246 260 275 299 304 312 321

6.01 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 is 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 a 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

236

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 WA Curriculum – Year 10 Methods Pathway 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 = 0.1 100 10 1 100

If we plot the table of values on a graph, we get a curve that is decreasing at a 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.

6.01 Exponential relationships mathspace.co

237

Example 1 For y = 2x: a Complete this table of values. −3

x y

−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 =2

Write the equation

−3

Substitute the ﬁrst value of x from the table

= 0.125

Express as a decimal

−3

−2

−1

−3

−2

−1

23 8

2 y

2 0.125

2 0.25

2 0.5

2 1

2 2

2 4

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=2

−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.

238

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

6.01 Exponential relationships mathspace.co

241

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 1 2 3 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

6.01 Exponential relationships mathspace.co

243

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

Time (minutes) 1 2 3 4 5 6

9, 10

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.

Volume (mL) 3 9 27 81

10, 11

The cumulative rainfall over a series of 4 days is represented on the graph. a

Step 3

Volume (mL)

32 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

244

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6.02 Characteristics of exponential graphs After this lesson, you will be able to… • recognise the standard form of exponential relationships as y = ax. • identify the y-intercept and horizontal asymptote of exponential curves. • determine the behaviour of exponential curves based on the value of a. • 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 a.

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 = ax where a is the growth or decay factor. 8

All equations of the form y = ax have these features in common: • The growth or decay factor is a. • The y-intercept is at (0, 1). • There is a horizontal asymptote at y = 0.

7 6 5 0<a<1

a>1

The value of a determines whether the curve increases or decreases. • If 0 < a < 1, the curve decreases. • If a > 1, the curve increases.

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 a. As x increases in exponential decay relationships, the graph decreases faster for smaller values of a. 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

6.02 Characteristics of exponential graphs mathspace.co

247

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

For equations in the form y = ax, the curve is increasing if a > 1 and the curve is decreasing if 0 < a < 1.

Since

248

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

, the curve is decreasing.

b Identify the equation of the horizontal asymptote.

Create a strategy For any relationship in the form y = ax, 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 = ax, 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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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 = ax is at (0, 1). For more points, substitute x = 1, 2 into y = 2x: y = 21

Substitute x = 1

Evaluate the exponent

y = 22

Substitute x = 2

Evaluate the exponent

So we can plot the points (0, 1), (1, 2), and (2, 4).

y 4 3 2 1 −4 −3 −2 −1

−1

x 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.

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Apply the idea Recall that the y-intercept for curves of the form y = ax is at (0, 1). For more points, substitute x = 1, 2 into

Substitute x = −1

Apply the negative exponent law

Substitute x = 1

Evaluate the exponent

So we can plot the points (−1, 2), (0, 1), and

y 4 3 2 1 −4 −3 −2 −1

−1

−2 −3 −4

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.

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x 1

Describe what happens to y as the value of x increases. a

y 5

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

Determine whether these are increasing or decreasing exponential curves: a

y = 3x

Ex 1

256

b f

Fluency 6

y = 0.97x

6a-d, 7abf, 8, 10, 11

y = 0.5x

y = 1.5−x

6b-e, 7c-f, 8, 9, 11

y = 1.05x

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

7 Ex 2

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

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.

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

11 Ex 4

Consider the graphs of y = 5x and a

Sketch the graph of y = 5x.

Sketch the graph of

Compare the graphs of y = 5x and

−2

−1

. .

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Reasoning 12

12, 13

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

, which is decreasing more rapidly for x > 0? Justify your answer.

and

Problem-solving 16, 18, 19 16

16cd, 17–19

16cd, 17, 19–21

Find the missing coordinate in each ordered pair so that the pair is a solution to y = 5x: a

(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.

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The graph 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

−5 −4 −3 −2 −1

x Q (0, 1) 1 2 3 4 5

−x

For the exponential equation y = 3 : 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.

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6.03 Exponential models After this lesson, you will be able to… • differentiate between exponential growth and exponential decay. • identify and define key features of exponential graphs. • express exponential relationships algebraically, highlighting either the growth or decay factor or rate. • apply knowledge of exponential models to solve real-world problems. • interpret the implications of the growth or decay rate in practical scenarios.

Exponential growth

Exponential growth happens when a quantity increases by a certain percentage over time. We can describe this kind of growth with an exponential formula:

y = abx

y 10

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.

If b > 1, we have growth. If 0 < b < 1, it is decay.

7 6 5

y = 3(1.6x)

1 0

x 1

The compound interest formula: A = P(1 + r)t 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

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 0

1 1.15

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 a 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

Reﬂect and check

The growth factor, b, is 1.725

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.

1.725 = 1 + r r = 0.725 The y-intercept is (0, 3), and the growth rate is 72.5%.

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.

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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. 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

1 0

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.

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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 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

Apply the idea x

This equation is in the form y = ab . 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.

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

Apply the idea x

This equation is in the form y = ab . 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.

We see a = 1 and for the decay rate.

. Next, we need to solve

Equation for decay rate

Substitute

Evaluate the subtraction

The y-intercept is at (0, 1), and the decay rate as a percentage is 75%.

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Reﬂect and check Notice that the original equation tells us that we are keeping losing 75%, as shown with the decay rate.

. 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

Reﬂect and check

When n = 0, F = 480, so the population was 480.

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

Apply the idea F = 480(0.952)n

We can substitute n = 3, and then evaluate using a calculator.

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

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 0

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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6.03 Practice questions Understanding 1

Explorer

of the previous week’s sales.

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

y = 2500 (0.877)x

y = 5.9 (1.025)x

y = 10 000 (2)x

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. a

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: x y

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

1–4

A store owner checked the sales report for the previous month and found that each week the sales were

Trailblazer

1–4

Are these situations exponential growth, exponential decay, or neither? a

Ex 1, 3

Adventurer

1–4

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)

Sketch a graph of M. 6.03 Exponential models mathspace.co

269

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% per annum due to poor maintenance. Calculate its depreciated value. Round your answer 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

270

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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, rounded 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)

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.

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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. Copy and complete the table showing the number of newly infected people for that day. Day Number of people infected

272

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 were 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?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

Pedro

Find the rate, as a percentage, that the blood sugar level decreased each hour for: i

Juanita

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)

0 38 200

1 37 818

2 37 439.82

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?

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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 0

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 0

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.

Investigation: Half-life

Investigate online

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mathspace.co

6.04 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 a 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.

Interactive exploration

Discover this concept in action online

mathspace.co

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

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Connect the points to draw the curve y = 3x − 4.

4 3 2 1

y = 3x

−5 −4 −3 −2 −1 −1 −2

x 1

2 3 4 5 y = 3x − 4

−3 −4 −5

Example 2 The graph of

is shown.

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3

a Describe a transformation of the graph of that would obtain

−4

Create a strategy

Apply the idea

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.

Since the new equation is

k = 3 > 0, then it means that the transformation of the graph of g (x) is a translation up 3 units from

b Sketch the graph of

and

on the same set of axes as

Create a strategy Plot the graph of

. Use the transformation found from part (a) to plot the graph of

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Vertical dilation of an exponential graph We will now look at exponential functions of the form y = a (bx).

Interactive exploration

Discover this concept in action online

mathspace.co

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

y = 2x

4 2

y = 3 (2 )

The y-intercept for y = 2x is at (0, 1), and the y-intercept for y = 3 (2x) is at (0, 3).

−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 = 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 4 x

y = 0.5(2 )

−6 −5 −4 −3 −2 −1 −2

280

y = 2x x

Both curves have y = 0 as the horizontal asymptote.

1 2 3 4 5 6

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Example 4 Consider the function

a Find the coordinates of the y-intercept.

Create a strategy Substitute 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.

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

282

, the function is decreasing.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

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

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Consider graphing y = 3 (2x) − 4. y 12 10 8

We can start by graphing y = 2x with: • y-intercept at (0, 1) • Horizontal asymptote at y = 0 • Another point at (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)

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 = 2x

6 4 2 −4 −3 −2 −1 −2

−4

y 12

y = 3(2x)

10 8 6 4

y = 2x

2 −4 −3 −2 −1 −2

−4

y 12 10 8 6 4 x

y = 3(2 )

−4 −3 −2 −1 1 2 3 −2 y = 3(2x) − 4

−4

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.

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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

6.04 Transformations of exponentials mathspace.co

287

Reﬂect and check We can see each transformation applied one at a time as well. 4

3 2 1 −4 −3 −2 −1

x 1

−1

Original function of y = 2x.

−2 −3 −4

3 2 1 −4 −3 −2 −1

x 1

−1

Compression with the scale factor of 0.5.

−2 −3 −4

3 2 1 −4 −3 −2 −1

−1

x 1

Reﬂection across the x-axis.

−2 −3 −4

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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

iii b 8

x 1

−4

−6

−8

−4 −3 −2 −1 −2

290

x 1

−4 −3 −2 −1 −2

y = −2x

−4 −3 −2 −1 −2

−4

−6

−8

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 WA Curriculum – Year 10 Methods Pathway mathspace.co

Fluency Ex 1

5–11, 13, 14c-f, 16, 18, 19bc

y = 2x + 4

The graph of a

y = 2x − 6

is shown.

Describe a transformation of the graph of that would obtain

. on the same

Sketch the graph of .

set of axes as

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.

6 4 2 −4 −3 −2 −1 −2

f ( x) = 8x

x 1

−4 −6 −8

6.04 Transformations of exponentials mathspace.co

291

The graphs of A, B, and

are shown. B

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

292

−4 −3 −2 −1

y 11 10 9 8 7 6 5 4 3 2 1

−4

x 1

Mathspace WA Curriculum – Year 10 Methods Pathway 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

6.04 Transformations of exponentials mathspace.co

293

Ex 5

The equation of the graph is of the form y = a (b)x. What is the y-intercept? What does this tell us about the equation?

1 −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.

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.

How could we use the graph of y = 7x to draw the graph of y = 7 − x?

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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. 10 8 6

y y = 3(2x)

4 2 −10 −8 −6 −4 −2 −2 −4 −6 −8 −10

x 2 4 6 8 10

Describe a transformation of the graph of y = 3 (2x) that would obtain y = −3 (2x).

Sketch the graph of y = −3 (2x) 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.

6.04 Transformations of exponentials mathspace.co

295

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

x 1

−1

−3

y-intercept: (0, 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.

−1

x 1

−2 −3

Problem-solving 26, 27

26, 28

27, 29–31

The graph of f (x) = 2x is shown. 5

f (x) = 2x

3 2 1 −5 −4 −3 −2 −1 −1

x 1

2 3 4 5

−2 −3 −4 −5

296

−2

Yalile graphed y = 2 (3x) − 1 by ﬁnding some key features. Her work is shown.

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?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

6.04 Transformations of exponentials mathspace.co

297

The equation of the graph is of the form y = a (bx) + k. y (2, 8)

8 6 4 (1, 2)

x −2

−1

1 (0, −1)

−2

−1

x y = a (bx) + k y = a (bx) y = bx

⬚ ⬚

−1

⬚ ⬚

What is the asymptote? What does this tell us about the equation?

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?

Find the equation of the graphs below in the form y = a (bx) + k. a

8 6 4 2

10 5 −3

−2

(1, 0) 1 2

−1

−5 −10

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−2

−1

−2 −4 −6 (−1, −8) −8 −10 −12

x 1

6.05 Optional: 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.

Exponential equations An exponential expression is an expression of the form ax where the base, a, 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. The key property we need is that: If ax = ay, then x = y To solve an exponential equation, we can: 1. Write in the form ax = c. 2. Rewrite in the form ax = a⬚.

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

6.05 Optional: Exponential equations mathspace.co

299

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.

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 3(2) − 4

= 125

Substitute x = 2

25 = 125

Simplify the exponents

15 625 = 15 625

Evaluate on a calculator

300

Write the equation

2+1

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Ex 1

Ex 2

Solve: a

3x = 36

7x = 1

3x = 27

2x = 64

5n = 625

Rewrite both sides of the equation with a base of 2.

Solve for x.

8x = 4

5x = 0.2

35x − 10 = 1

4x + 4 = 8

81x − 1 = 93x + 5

Solve:

Solve: 9y = 27 2 x+7

b 3

25y = 125 10x + 33

a x − 1 = a4 −3x − 1

(2 )

= 125

9x + 3 = 27x

8x + 3 = 322x − 1

36x − 1 = 2162x + 5

= 3125

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

i 12

6x = 6–3

For each equation:

a Ex 3

9y = 81

Solve: a

2x × 2x + 2 = 16

Reasoning

13, 15

13–15

14, 16

Explain why 23x − 1 = 27x − 9 cannot be solved using index laws, while 23x − 1 = 8x − 9 can.

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Explain the reasoning for each step of this solution.

Given equation

1 2 3 4 5 6 7

Lorena was trying to ﬁnd the solution to the equation Her working is shown. a

Find and explain Lorena’s mistake.

Correct her solution.

Explain how to solve the equation roots.

Problem-solving 17, 19, 20acf 17

using your knowledge of squares and square

18, 19, 20b-f

18, 19, 20b-h, 21

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

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.

6.05 Optional: Exponential equations mathspace.co

303

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.

6.06

Optional: Logarithms

After this lesson, you will be able to… • represent numbers as powers of ten and find their base ten logarithms. • 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

304

Base 10 logarithm

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

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

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

6.06 Optional: Logarithms mathspace.co

305

Problem-solving 12, 13 12

13, 14

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

12–14

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.

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?

6.06 Optional: Logarithms mathspace.co

311

6.07 Optional: 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

102

101

0.1

10−1

−1

0.01

10−2

−2

0.001

−3

0.000 1

10−4

−4

0.000 01

−5

0.000 001

10−6

−6

The order of magnitude forms a logarithmic scale of base 10.

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Increasing order of magnitude

Decreasing order of magnitude

Similarly, we can use logarithmic scales to compare numbers with a large range, like the distance of the planets from the sun. Distance from the sun, d (km) 57 910 000 108 200 000 149 600 000 227 900 000 778 500 000

Planet Mercury Venus Earth Mars Jupiter

log10 (d) 7.76 8.03 8.17 8.36 8.91

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 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. 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.

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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.

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.

6.07 Optional: Order of magnitude and logarithmic scales mathspace.co

315

9 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

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 b

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 0

Richter scale 9.1 7.9 7.0 6.3

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?

6.07 Optional: Order of magnitude and logarithmic scales mathspace.co

317

12 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.

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?

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

lawn mower hair dryer normal conversation refrigerator library

80 70 60 50 40 30 20 10 0

−4

10 10−5 10−6 10−7 10−8 10−9 10−10 10−11 10−12

318

Type of noise Jet plane Pneumatic drill Industrial Stereo music Inside a car Office Household Wind turbine Bedroom Falling leaves

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 WA Curriculum – Year 10 Methods Pathway 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

15, 17

15–17

15–18

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.

A student notices that a logarithmic scale on a graph increases from 2 to 4 over a certain interval. If the scale is base 10, explain why this does not mean the measured quantity increased by 2 times. Then, determine how many times the quantity actually increased.

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.

A scientist measures the intensity of light from two sources on a base 10 logarithmic scale. Source A has a log intensity of 3.2, and Source B has a log intensity of 1.8. Explain how the difference in their log values relates to their actual intensity ratio. Then, calculate how many times brighter Source A is than Source B, rounding to the nearest whole number.

6.07 Optional: Order of magnitude and logarithmic scales mathspace.co

319

Problem-solving 19, 22 19

0 50

1 500

2 5000

3 50 000

4 500 000

What is the order of magnitude of the algae population after 4 days?

Express the algae population after each day on a base 10 logarithmic scale, rounding to one decimal place.

If the pond’s capacity is 108 algae, how many days will it take for the population to reach this limit, assuming the same growth pattern continues? Explain your answer.

A telecommunications company measures signal strength in decibels-milliwatts (dBm), a logarithmic scale where 0 dBm corresponds to 1 milliwatt (mW), and every 10 dBm increase represents a 10-fold power increase. A cellphone signal is initially 20 dBm. a

What is the signal strength in milliwatts at 20 dBm?

If the signal drops to −10 dBm due to interference, how many times weaker is the new signal compared to the initial signal?

If the company boosts the signal from −10 dBm to 15 dBm, by how many orders of magnitude does the signal strength increase?

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

19–23

A biologist tracks the population of algae in a pond, which grows rapidly due to nutrient influx. The population is recorded over days as follows: Days Population

19, 20, 22

linear measure (x) 5 50 500 5000 50 000

= = = = =

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

320

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 10 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 10 logarithmic scale.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

6 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

1 2 3 4 5

50 40 30

−6

−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

−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 y = 6x

−1

Chapter 6 review mathspace.co

321

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

−1

x 1

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.

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.

Describe how the graph of y = 10x is transformed to get each graph: a

322

y = 10x + 2

y = −10x

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

y = 3 (10)x

The graph of f (x) = 7x is shown.

y 7 6 5 4 3 2 1 f (x) = 7x

Describe a transformation of the graph of f (x) = 7 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 horizontal asymptote of g (x) = 7x − 4.

−5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

Two exponential relationships, X and Y, are graphed.

1 2 3 4 5

One of the graphs is of y = 3 and the other graph is of

Which is the graph of

For the exponential function

Find the coordinates of the y-intercept.

What is the equation of the horizontal asymptote?

Is the function increasing or decreasing?

Graph the function.

(Optional) The graph of y = 4x is shown. a

Find the value of y when x = 4.

Find the value of x when

14 12 10 8 6 4 2 −5 −4 −3 −2 −1

x 1 2 3 4 5

(Optional) Solve: a

8x = 88

6x−4 = 36

3x = 81

c g

d 5−4x−1 = 1

105x+18 = 100x−3 Chapter 6 review mathspace.co

323

(Optional) Solve: a

3x × 9x+1 = 27

d all lie on the curve with equation y = 3x,

(Optional) Given the points (0, n), (k, 243) and ﬁnd the value of: a

2x × 2x+4 = 8

10 000

(Optional) For each number: i

Express the number as a power of 10.

Find the base ten logarithm of the number.

0.01

(Optional) 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

(Optional) 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. 21

324

(Optional) 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.

(Optional) For the number 0.000 04: a

Write the number using scientiﬁc notation.

Identify the order of magnitude.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

7.01

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

mathspace.co

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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7.01 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

y 6 5 4 3 2 1

−6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

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

x 1 2 3 4 5 6

−6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

330

x 1 2 3 4 5 6

y 6 5 4 3 2 1

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

f 4

1 2 3 4 5

3 2 x 1 2 3 4 5 6

Mathspace WA Curriculum – Year 10 Methods Pathway 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

7.01 Solve graphically mathspace.co

331

Use the graph to solve the simultaneous equations: y

5 4 3 2 1 −2

−1

y=x−2 1 2

−1 −2 −3 −4 −5

x − 2y = 4 3

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

y = 3x + 2

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 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

332

• Studio A: y = 50x + 250

2x − 4y = −8

y = 4 − 3x y = 7 − 2x

y = 4 − 0.5x y = 12 − 1.5x 7

y = −x + 3 −4x + 2y = −8

• Members: y = 2x + 10

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 WA Curriculum – Year 10 Methods Pathway 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

What coordinates satisfy both equations?

2325

What do the coordinates from part (a) mean in this context?

1860

1395 930 465

x 155 310 465 620 775 930 1085 1240

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 (b) 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

14, 15, 17, 18

y = 4x + 5

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?

7.01 Solve graphically mathspace.co

333

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

334

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 WA Curriculum – Year 10 Methods Pathway 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

7.01 Solve graphically mathspace.co

335

7.02 Optional: 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. y = x + 11

y = 3x + 19

a First solve for x.

Create a strategy Substitute the expression for y from 1 , into the variable y in 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.

336

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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 as 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.

7.02 Optional: Solve using substitution mathspace.co

339

7.02 Practice questions Understanding 1

Explorer

Adventurer

1, 2

1–3

Trailblazer 1, 3, 4

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

y = −3x − 23

y = −5x − 22

y = 5x + 37

y = −4x – 27

y = 7x + 38 5y = 35x + 265 2y = 10x + 54 f

x+y=2

y = −3x + 22

y = −3x + 9

x + y = 10 6x − y = 0 j

y = 2x – 15

x + 4y = −37 3x − 4y = 20

5x − 2y = −10

y = 4x – 28 3x + 5y = 44

2x + 3y = 48

y = −5x + 3 4x − 5y = 8

7.02 Optional: Solve using substitution mathspace.co

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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 prices 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 WA Curriculum – Year 10 Methods Pathway 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 = −5 g

(x + 1) + ( y + 4) = 25 y = −3x − 2

y = 5x + 4 y = x2 + x + 8

y = 24

(x + 3)2 + ( y − 3)2 = 4 2

y = x2 − 5x y = (x − 5) (x − 2)

x2 + ( y + 7)2 = 7

y = −x + 1 y = −7 h

y = (x − 1)2 x − 3y = 1 7.02 Optional: Solve using substitution mathspace.co

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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

Ex 2

Use the elimination method to solve each pair of linear equations: a

4x − y = 13

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

1 h 2

1 2 1 2

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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.

7.03 Optional: Solve using elimination mathspace.co

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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

y = 3x2 − 6

Original equations

y = −x – 2 y = 3x2 – 6 3y = −3x2 – 6 Multiply equation 2 by 3

Explain the mistake Elias made that caused him to only ﬁnd one solution.

4y = −12

Add the equations, solve for y

y = −3

Find the second solution.

−3 = 3x2 − 6 3 = 3x

Substitute y, solve for x

1 = x2 1=x The solution is (1, −3).

Problem-solving 17, 20ab 16

16, 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.

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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.

What is this proportion (as a percentage) during the estimated year in part (a)?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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? 7.04 Optional: Simultaneous inequalities mathspace.co

357

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.

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 type of bike is shown below. Each worker can work on their own tasks simultaneously, but each is limited to 40 hours per week. Mechanic Painter Let x represent the number of Racing Bike 5 hours 10 hours racing bikes built and y represent Mountain Bike 8 hours 4 hours the number of mountain bikes built. Each worker can work at most 40 hours in a week.

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.

358

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 WA Curriculum – Year 10 Methods Pathway 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 proportions of the area planted with 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 inequalities.

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 b

Machine A 2 hours 4 hours

Machine B 3 hours 1.5 hours

Write an inequality for Machine A.

Can the values of x and y be negative? Explain your answer. Write any additional inequalities if necessary.

Write an inequality for Machine B.

Graph the region deﬁned by the inequalities for Machine A and Machine B on a Cartesian plane.

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.

7.04 Optional: Simultaneous inequalities mathspace.co

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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 and 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?

360

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 WA Curriculum – Year 10 Methods Pathway 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 sessions Eileen can attend. What is the maximum number of basketball sessions she can attend?

7.04 Optional: Simultaneous inequalities mathspace.co

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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, write 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.

7.05 Optional: 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:

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a Use the sum of the test scores to form an equation. We will refer to this as equation 1.

Create a strategy

Apply the idea

Write an equation in terms of x and y.

The sum of her two test scores is 172, where x and y are the scores. Equation 1: x + y = 172

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 = 95 x

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.

7.05 Optional: Applications of simultaneous equations mathspace.co

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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

364

Total cost for GO SMALL plan (S) 38.00 47.00 56.00 65.00

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Total cost for GO MEDIUM plan (M ) 40.00 47.00 54.00 61.00

b Sketch a graph for the two plans.

Create a strategy Use the points from the table to graph both plans.

Apply the idea 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.

Cost 65 60 55 50 45 40 35 30 M 25 20 S 15 10 5

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. 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).

Call time 0 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 65 60 55 50 45 40 35 30 25 20 15 10 5

Cost

The point of intersection is at (30, 47). This means that for a call time of 30 minutes, both plans will cost $47.

M S

Call time 0 5 10 15 20 25 30 35 40 45 50

7.05 Optional: Applications of simultaneous equations mathspace.co

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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

Multiply the selling price and the number of armchairs sold

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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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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

Subtract the cost from the revenue Evaluate

7.05 Optional: Applications of simultaneous equations mathspace.co

369

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. a

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. 7.05 Optional: Applications of simultaneous equations mathspace.co

371

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.

372

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?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

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. 24

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. 7.05 Optional: Applications of simultaneous equations mathspace.co

373

7 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 = −x − 3

−8

−7

5 4 3 2 1

y = −3x −13

−6

−5

−4

−3

−2

−1

x + 2y = −6

y = −x − 3

y = −3x − 13 x + 2y = −6

x 1

−1 −2 −3 −4 −5

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. y 3500 3000 2500 2000 1500

E S

1000 500 x 0

374

256 512 768 1024 1280 1536

What coordinates satisfy both equations?

What do the coordinates from part (a) mean in this context?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

(Optional) 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.

Milk Chocolate Dark Chocolate

Machine A

Machine B

5 hours

4 hours

3 hours

6 hours

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.

(Optional) 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?

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?

(Optional) Solve the simultaneous linear equations using substitution: a

y = 4x + 40

y = −2x − 11

y = 3x + 30 4x + 2y = −22 c

2a + b = 23

−3f + 2g = 6

b = −2a + 35

2f + 5g = 53

(Optional) Use the substitution method to find the solution(s) for each pair of simultaneous equations. a

(x + 2)2 + ( y − 4)2 = 10

376

y = 4x + 2

x = −1 y = −x2 + x + 6

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Big ideas Ratios, and scale factors are tools for identifying similar figures and calculating their areas and volumes.

Curriculum Outcomes • Investigate, explore and determine the effect on the perimeter and area of shapes when they are enlarged or reduced by a scale factor • Investigate in order to determine the effect on surface area and volume when objects are enlarged or reduced by a scale factor

8 Scale factor Chapter outline 8.01 8.02 8.03

Scale and perimeter Scale and area Scale and volume Investigation: Volume versus surface area Chapter 8 review

380 387 392 399

8.01 Scale and perimeter After this lesson, you will be able to… • explain how the scale factor impacts the perimeter of similar figures. • apply understanding of scale factors to calculate the perimeter of similar figures. • interpret and solve problems involving perimeters of similar figures and scale factors.

Scale and perimeter Since perimeter is the distance around the outside, the calculation of perimeter involves multiplying the same lengths based on the amount or adding different lengths together. For example: Square

Rectangle

l a

Triangle Parallelogram

Trapezium

P=4×s

P=a+b+c

b a

P = (2 × a) + (2 × b)

b d

P = 2(l + w)

P=a+b+c+d

Circle

C = 2π r

When adding lengths with the same units, the unit remains unchanged. For example, cm + cm = cm. This means that if all side lengths of a shape are scaled by k, the perimeter is also scaled by k.

Plarge = k × Psmall Plarge the perimeter of the larger ﬁgure Psmall the perimeter of the smaller ﬁgure k the scale factor

380

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Exploration Consider two similar triangles. If the smaller triangle has side lengths 3 cm, 4 cm and 5 cm, and the scale factor is 2, calculate the perimeter of both triangles. What do you notice about the ratio of their perimeters?

Example 1 The two rectangles shown in the diagram are similar. Find the value of x, the perimeter of the larger rectangle.

x mm 28 mm 10 mm

30 mm

Create a strategy Calculate the scale factor using the corresponding side lengths, then multiply the perimeter of the smaller rectangle by the scale factor.

Apply the idea Scale factor = =3

Find the ratio of the corresponding lengths Evaluate

Perimeter of smaller rectangle = 28 mm

Use the given perimeter from the diagram

x = 28 × 3

Multiply the perimeter by the scale factor

= 84 mm

Evaluate

The perimeter of the larger rectangle is 84 mm.

Reﬂect and check Check: The smaller rectangle has a perimeter of 28 mm and length 10 mm, so its width is 4 mm (since 2 × (10 + 4) = 28). The larger rectangle has length 30 mm and width 4 × 3 = 12 mm. Its perimeter is 2 × (30 + 12) = 84 mm, which matches.

8.01 Scale and perimeter mathspace.co

381

Fluency Ex 1

4–6, 7ab, 9, 10

4, 6, 7bd, 9–11

4, 6, 7cd, 8–11

Find the value of x for each pair of similar rectangles. a

72 m xm 4m

24 m

336 cm

13 cm

x cm 11 cm

The corresponding sides of two similar pentagons are 4 cm and 10 cm. The perimeter of the smaller pentagon is 25 cm. Find the perimeter of the larger pentagon.

A hexagon has a perimeter of 42 m. Another hexagon is created by scaling it up by a factor of 3.5. Find the perimeter of the larger hexagon.

Find the value of x for each pair of similar shapes. a

11 cm

164 cm

x cm

17 cm 13 cm

64 mm 16 mm 26 mm

26 mm

x mm

21 mm

384

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x mm 31 mm

580 mm

135 mm

34 m

106.81 m

Ex 2

Two similar trapeziums have corresponding sides in the ratio 3 : 8. The smaller trapezium is isosceles, with its two non-parallel sides (legs) equal, and its shorter base is 5 cm. The perimeter of the smaller trapezium is 33 cm. What is the length of the longer base of the larger trapezium?

A triangle has a perimeter of 35 cm with side lengths of 7 cm, 10 cm and 18 cm. A second similar triangle has a perimeter that is 4 times larger than the ﬁrst. a

Find the scale factor between the two triangles.

Find the perimeter of the second triangle.

A shape with a perimeter of 52 m is enlarged using the scale factor 2.5. Find the perimeter of the enlarged shape.

If a rectangle with a perimeter of 36 km is dilated by a factor of 0.25, what is the perimeter of the dilated rectangle?

Reasoning

12, 13

12–14

A circle has a circumference of approximately 25.12 m. Explain why a circle with a radius three times as large as a circumference of approximately 75.36 m.

Consider two similar garden plots: the ﬁrst with dimensions 6 m by 8 m, and the second with dimensions 15 m by 20 m. a

Find the length scale factor.

Find the perimeter of the smaller garden plot.

Find the perimeter of the larger garden plot.

Justify why the perimeter of the larger garden plot is 2.5 times that of the smaller garden plot.

8.01 Scale and perimeter mathspace.co

385

A pentagon has been enlarged with a scale factor of k. Describe the relationship between the perimeter of the two pentagons.

Problem-solving 15, 17, 19, 20 15

15–17, 19–22

15–18, 20–22

The ﬁgure shows a park trail used for a charity walk. To extend the event, the organisers dilate the trail boundaries by a factor of 2.2. What is the perimeter of the new trail?

12.2 m

A miniature racetrack has dimensions that are

of the actual racetrack. If the perimeter of

the actual racetrack is 750 m, ﬁnd the perimeter of the miniature racetrack. 17

A garden border has a perimeter of 12 m. In a blueprint, the corresponding perimeter is 60 cm. a

Find the scale factor from the blueprint to the garden border.

The longest side of the border in the blueprint is 25 cm. Determine the length of the longest side of the garden border in real life.

A triangular sculpture has side lengths of 3 m, 7 m and 9 m. a

Calculate the perimeter of the frame.

If the frame is scaled up by a factor of 4, give the side lengths of the enlarged frame.

Calculate the perimeter of the enlarged frame.

By what factor has the perimeter increased?

If a shape is enlarged with a scale factor of k, by what factor will its perimeter increase?

Leo designs custom picture frames. His standard frame has a perimeter of 80 cm. For a gallery exhibit, he creates a larger version with a perimeter of 240 cm. a

Find the scale factor between the two frames.

Leo uses 80 cm of wood to outline the standard frame. How much wood will he need for the larger frame?

A model of a historic castle wall is made in the ratio 1 : 1500. a

The perimeter of the actual castle wall is 900 m. Find the perimeter of the model in centimetres.

The model requires 24 cm of string to outline its perimeter. How much string, in metres, would be required to outline the actual castle wall?

A hexagon has a perimeter of 6y m. Determine the perimeter of the new hexagon if the original is dilated by a factor of: a

386

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Interactive exploration

Discover this concept in action online

mathspace.co

The ratio of areas in similar figures is equal to the square of the scale factor. For a scale factor of k, if the smaller figure has an area of A, then the area of the larger figure would be k2 × A.

Example 1 The two rectangles shown in the diagram are similar.

x cm2 15 cm2 12 cm

Find the value of x.

6 cm

Create a strategy Multiply the area of the smaller rectangle by the scale factor squared.

Apply the idea Find the ratio of the side lengths Evaluate

Multiply the area by the scale factor squared

Evaluate

Example 2 A triangle has side lengths of 6 cm, 12 cm and 16 cm. A second similar triangle has an area that is 9 times larger than the ﬁrst. a Find the scale factor between the two triangles.

Create a strategy Take the square root of the area scale factor.

Apply the idea Scale factor = =3

388

Take the square root Evaluate

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Fluency Ex 1

4–6, 8, 9

4, 6, 8–10

4, 6–8, 10

Find the value of x for each pair of similar rectangles. a

320 cm2

x cm2

x cm2 5 cm

20 cm

7 cm 9 cm

27 cm

c 10 cm

x cm2 6 cm

12 cm

The corresponding sides of two similar triangles are 6 cm and 12 cm. The area of the smaller triangle is 18 cm2. Find the area of the larger triangle.

A square has a side length of 8 cm. Another square is created by scaling it up by a factor of 7. Find the area of the larger square.

Two rectangles are similar, with matching sides in the ratio 6 : 7. The smaller rectangle has an area of 576 cm2 and its shortest side has length 8 cm. Find the length of the longest side of the bigger rectangle.

A triangle has side lengths of 3 cm, 12 cm and 14 cm. A second similar triangle has an area that is 25 times larger than the ﬁrst.

Ex 2

Find the scale factor between the two triangles.

Describe the relationship between the perimeters of the two triangles.

A shape with an area of 36 cm2 is enlarged using the scale factor 3. Find the area of the enlarged shape.

If a square with an area of 25 m2 is dilated by a factor of 0.4, what is the side length of the dilated square?

Reasoning

12, 13

11–13

A rectangle has been enlarged with a scale factor of k. Describe the relationship between the area of the two rectangles.

A circle has an area of approximately 78.5 cm2. Explain why a circle with a radius twice as large as an area of approximately 314 cm2.

390

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Consider two similar ceilings: the ﬁrst with dimensions 5 m by 4 m, and the second with dimensions 20 m by 16 m. a

Find the length scale factor.

Find the area scale factor.

The smaller ceiling took 1.5 L of paint to cover it. How many litres of paint would be required to paint the second ceiling?

Justify why the amount of paint needed for the second ceiling is not simply four times the amount needed for the ﬁrst ceiling.

Problem-solving 15, 16, 18–20 14

14–16, 18–20, 22

All dimensions of the miniature stadium are

14, 16, 17, 19–22

of the actual stadium sizes. If the pitch area

of the actual stadium is 7200 m , ﬁnd the area of the pitch in the miniature version. 15

The figure shows the field on which a game of tag is being played during a physical education class. To make it more of a fitness challenge, the teacher dilates the boundaries of the field by a factor of 1.5 What is the area of the new field?

12.2 m

A rectangular sign has a side length of 1.5 m. In a photograph, the corresponding side has a length of 12.5 cm. a

Find the scale factor from the photograph to the sign.

The area of the sign in the photograph is 75 cm2. Determine its area of the sign in real life.

A rectangular prism has dimensions of 2 cm, 4 cm and 5 cm. a

Calculate the surface area of the prism.

If the prism is scaled up by a factor of 3, give the dimensions of the enlarged prism.

Calculate the surface area of the enlarged prism.

By what factor has the surface area increased by?

If a three-dimensional shape is enlarged with a scale factor of k, by what factor will the surface area increase by?

Maria runs a business where she makes and sells soft toys. Her most popular item is a teddy bear that is 14 cm long. For a special promotion, she decides to make a larger version of the bear that is 56 cm long. a

Find the scale factor between the two bears.

Maria uses 224 cm2 of felt to make the small bear. How much felt will she need to make the big bear?

8.02 Scale and area mathspace.co

391

A model of the Eiffel tower is made in the ratio 1 : 12 960.

The height of the Eiffel tower is 324 m. Find the height of the model in centimetres.

The model requires 15 mL of paint to cover its entire surface area. How much paint, in litres, would be required to cover the entire surface area of the real Eiffel tower?

The Dominican Republic is shown on the map. According to the scale given, estimate the actual area of the Dominican Republic. Dominican Republic Santo Domingo 0 40 80 km

A square has sides of length x cm. Determine the area of the new square if the original is dilated by a factor of:

a 22

The ﬁgures shown are similar. What is the value of k?

k cm 6 cm Area = 300 cm2 Area = 75 cm2

8.03 Scale and volume After this lesson, you will be able to… • explain how the scale factor impacts the volume of similar objects. • identify and determine the scale factor based on the volumes of similar objects. • apply knowledge of scale factors and volume to find unknown sides in similar shapes. • solve problems involving volume of similar objects and scale factors.

392

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Scale and volume Since volume is a three-dimensional measurement, the calculation of volume involves three lengths multiplied together. For example: Vrectangular prism = length × width × height

V=l×w×h

Height Width

Length

Vcube = length3 V = s3

s s

Vtriangular prism = Area of triangle × prism height Height

This means that if the dimensions of an object are multiplied by a scale factor, the volume of that object will be multiplied by that scale factor three times (once for each dimension in the formula).

Exploration ×2

×2

Each side length of the cube in the image has been scaled by a factor of 2. 1.

If the original volume was 1 unit3, what is the volume of the new cube?

Did the volume also double? If not, how did it change?

If the original volume was 7 units3, what would the volume of the new cube be?

×2

This means that the ratio of volumes in similar objects is equal to the cube of the scale factor. For a scale factor of k, if the smaller ﬁgure has a volume of V, then the volume of the larger ﬁgure would be k3 × V.

8.03 Scale and volume mathspace.co

393

Example 1 The two rectangular prisms shown in the diagram are similar.

V = 11 cm3 2 cm V = x cm3 8 cm

Find the value of x.

Create a strategy Multiply the volume of the smaller rectangular prism by the scale factor cubed.

Apply the idea

Find the ratio of the side lengths

Evaluate

Multiply the volume by the scale factor cubed

704

Evaluate

Example 2 Spheres are always similar since they have only one deﬁning measurement (the radius). Two spheres have volumes of 40 cm3 and 625 cm3. Find the scale factor between the two spheres.

Create a strategy Find the volume scale factor, then take the cube root.

Apply the idea Find the ratio of the volumes

Evaluate

Take the cube root of the volume scale factor

Evaluate

394

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Problem-solving 15–19 15

16–20

18–22

For these two rectangular prisms: a

Find the length scale factor.

Find the surface area scale factor.

Find the volume scale factor.

Find the volume scale factor if the measurements of the smaller prism are doubled.

2.3 cm 1.9 cm

5.1 cm

3.22 cm 2.66 cm 7.14 cm

The surface areas of two similar triangular prisms are in the ratio 64 : 49. Give, as improper fractions: a

The scale factor.

The volume scale factor.

Consider the two spheres shown, where the smaller has radius 3 cm and the larger has a radius of 12 cm. a

Write the ratio (in the form m : n) of the radius of Sphere A to Sphere B.

Write the ratio of the volume of Sphere A to Sphere B.

If the matching dimensions of two similar ﬁgures are in the ratio m : n, what ratio are their volumes in?

Sphere B Sphere A 12 cm 3 cm

Two similar cylinders have volumes of 5760 cm3 and 90 cm3: a

What is the volume scale factor for the two cylinders?

How many times taller is the larger cylinder?

Find the value of h.

h cm 20 cm V = 5760 cm3 V = 90 cm3

The radii of two spherical balloons are 12 cm and 6 cm, respectively. a

What is the scale factor?

How many times greater is the surface area of the larger balloon?

How many times greater is the volume of the larger balloon?

If half the air is released from the smaller balloon, how many times more air does the larger now hold?

Valerie was making a trial birthday cake for her son. The trial cake dimensions are 10 cm for the length, 7 cm for the width and 3 cm for the height. The actual cake has dimensions 20 cm for the length, 14 cm for the width and 6 cm for the height. a

Find the ratio of the dimensions of the trial cake to those of the actual cake.

Deduce the ratio of the volume of the trial cake to that of the actual cake.

To make the actual cake, by what number must Valerie multiply each quantity of ingredients?

8.03 Scale and volume mathspace.co

397

8 Chapter review 1

A scale model of a building has a rectangular base with a perimeter of 80 cm. The actual building is 15 m tall and its model is 30 cm tall. a

Determine the scale factor used when creating the model.

What is the actual perimeter of the building’s base?

A model car is built using a scale of 1.20. The full-sized car has a rectangular base measuring 4 m long and 1.80 m wide. a

What are the dimensions of the model car?

What is the perimeter of the model car’s base in centimetres?

A square has sides of 6 cm. Give the side length of the new square if: a

The square was enlarged by a factor of 9.

A scale factor of 0.25 was applied to the square.

A rectangle has a length of 10 cm and a width of 4 cm. A new rectangle is created by applying a scale factor of 1.5 to the original. a

What is the perimeter of the original rectangle?

What is the perimeter of the scaled rectangle?

How does the perimeter of the scaled rectangle compare to the original?

Find the value of y for each pair of similar rectangles. a

450 m2 y cm2

32.5 cm

30 m y m2 10 m 13 cm 10 cm

The corresponding sides of two similar triangles are 8 cm and 16 cm. The area of the smaller triangle is 32 cm2. Determine the area of the larger triangle.

If a square has an area of 49 m2 and it is dilated by a factor of , how long is the side of the resulting square?

A triangle has been enlarged with a scale factor of p. How does the area of the dilated triangle relate to the original triangle’s area?

All dimensions of a toy house are

of the actual house sizes. If the living room area of the

actual house is 48 m , what is the area of the living room in the toy version? Chapter 8 review mathspace.co

399

A square has sides of length y cm. Calculate the area of the new square if the original is enlarged by a factor of: a

Find the value of x for the pair of similar figures. a

6.5 cm V = 14.5 cm3 4.2 mm 2.8 mm 13 cm

V = 243 mm3

V = x mm3

V = x cm3

Two similar cones have a radius length of 5 cm and 20 cm, respectively. a

State the scale factor of the height of the smaller cone to the height of the larger cone.

State the scale factor of the volume of the smaller cone to the volume of the larger cone.

Calculate the volume of the larger cone, if the volume of the smaller cone is 250 cm3.

Two similar prisms have volumes of 27 cm3 and 729 cm3. If two, corresponding sides in the prisms have lengths of x cm and 9 cm, where x represents the length in the smaller prism, find the value of x.

An office fish tank has a capacity of 400 L. The manager wants to downgrade to a new fish tank with a capacity of 50 L. The manager estimates that the dimensions of the new fish tank should be half of the old tank’s dimensions. Is the manager correct? Justify your answer.

For these two rectangular prisms: 1.8 m 6.4 m

2.4 m

2.88 m 3.84 m 10.24 m

400

Are the two rectangular prisms similar? Justify your answer.

Find the scale factor of the surface area.

Find scale factor of the volume.

Find the volume scale factor if the measurements of the larger prism are halved.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Big ideas • Right-angled 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 • Use Pythagoras’ theorem and/or trigonometry to determine unknown sides and angles in right-angled triangles involving angles of elevation and depression • (Optional) Apply right-angled trigonometry to two dimensional situations involving navigational bearings • (Optional) Explore to establish and use the sine, cosine and area rule to determine unknown sides and angles for any triangle

9 Trigonometry Chapter outline 9.A 9.01 9.02 9.03 9.04

Pythagoras’ theorem Trigonometric ratios Angles of elevation and depression Optional: Sine and cosine rules Optional: Bearings Chapter 9 review

404 418 428 448

9.01 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° 2 60°

1 2

30°

3 2

60° 1 F

60° H

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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2 3

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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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

8 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

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With reference to the angle θ, ﬁnd the ratio: i

iii b

34 5

13 B

Write down the indicated ratios for each triangle using the side labels: a

tan θ

tan α

sin θ

A α

θ B

cos θ

cos α

A α

θ B

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25 C

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

sin α

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°

α°

Find the value of tan θ. 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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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: 10 cm

12 cm

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:

A sporting ground is in the shape of an irregular hexagon. 35 m 15 m

41 m

15 m 35 m

Find the perimeter of the ground.

Find the size of each of the four corner angles of the ground.

Calculate the size of angle DEC. D

6 x+7 A

12 − x C 4 B

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9.02 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.

θ A

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?

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x B

50° C

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).

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

θ 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.

Apply the idea d2 = 52 + 42 2

d = 41 d=

Substitute the values

Evaluate the right-hand side

Take the positive square root of both sides

The distance from one corner of the floor to the opposite corner of m. the floor is

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9.02 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?

θ σ

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.

D y

x 5

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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 height

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?

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:

3.38°

1.22° 1600 m

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.

Roald is standing at point P and observes two poles, AB and CD, of different heights. P, B, and D are on horizontal ground: A C 7m B

29° 16 m

45°

18°

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, rounded to two decimal places.

Calculate the distance from Roald to the top of pole AB, rounded to two decimal places. 9.02 Angles of elevation and depression mathspace.co

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9.03 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.

Interactive exploration

Discover this concept in action online

mathspace.co

The sine rule, sometimes called the law of sines or sine law, tells us that:

A, B, C a, b, c

the measures of the angles in the triangle 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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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

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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. 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:

85 61° A

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.

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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° 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.

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Example 5 Find the length of a using the cosine rule. Round your answer to two decimal places.

C a

13 A

33° 30

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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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

9.03 Optional: Sine and cosine rules mathspace.co

437

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

included between them are known.

Two of the sides and an angle included between them are known.

C b

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439

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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441

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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443

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.

9.03 Optional: Sine and cosine rules mathspace.co

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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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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.

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.

108 m 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.

9.03 Optional: Sine and cosine rules mathspace.co

447

The person in the picture sights a pigeon above him. Find the angle of elevation θ, rounded 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, 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, rounded to the nearest metre.

14° B

16°

3000 m

Airport

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.

(Optional) Find the unknown, rounded to two decimal places. a

20 D

47°

41° x

b a

13 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

Chapter 9 review mathspace.co

449

(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

(Optional) Find a using the cosine rule. Round your answers to two decimal places.

C 11

33°

(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, rounded to two decimal places: a

b 5.5 m

44° 3m

101°

5.7 m

(Optional) For this triangle, calculate: a

The size of angle x, rounded to the nearest degree

The area of the triangle, rounded to one decimal place

15.3 m

115° x 17.925 m

(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.

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44°

50 m 80 m

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 • Use absolute and percentage error to compare the result of using approximate rather than exact real numbers on final calculations • Use efficient strategies and apply formulas to determine the volume, capacity and surface area of composite solids, using appropriate units • Investigate, calculate and identify the impact of errors on the accuracy and outcome of results in measurement situations • (Optional) Explore, explain and apply efficient strategies and formulas to determine the surface area and volume of right pyramids, right cones, spheres and related composite solids • Analyse, represent and interpret problems in real-world situations involving Pythagoras’ theorem, trigonometry and angles of elevation and depression in right-angled triangles, volume, capacity and surface area of composite objects, the impact of errors in measurement and/or the effect of enlargement and reduction on perimeter, area and volume of similar figures and objects

10 Measurement Chapter outline 10.A Surface area of prisms and cylinders 10.B Volume of prisms and cylinders 10.01 Surface area and volume of composite solids Investigation: Rainfall collection Modelling investigation: Wind turbine design 10.02 Estimation and accuracy 10.03 Optional: Pyramids 10.04 Optional: Spheres 10.05 Optional: Cones 10.06 Optional: Composite solids Investigation: Instruments and methods Chapter 10 review

454 466 468 476 488 495 504 521

10.01 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 = 1170 cm

Small prism = 11 × 2 × 15 = 330 cm

V = 1170 − 330 = 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

10.01 Surface area and volume of composite solids mathspace.co

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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

10.01 Surface area and volume of composite solids mathspace.co

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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

3 cm

2 cm 5 cm

4 cm

7 cm

8 cm 2 cm

5 cm

10 cm

2 cm

8 cm

c 5 mm 4 mm 3 mm 5 mm

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

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Mathspace WA Curriculum – Year 10 Methods Pathway 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

10.01 Surface area and volume of composite solids mathspace.co

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Reasoning 11

11, 12

11, 13

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 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 three quarters full, ﬁnd h.

1.4 m

10 m

3.3 m 50.7 m

10.01 Surface area and volume of composite solids mathspace.co

463

A hollow pipe needs painting on both inside and outside surfaces. a

Find the external surface area of the curved surface, rounded to two decimal places.

Find the total surface area of the two end pieces, rounded to two decimal places.

Find the internal surface area, rounded to two decimal places.

24 cm

11 cm

Calculate the total surface area of the pipe, rounded to two decimal places.

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, rounded to two decimal places. 30 cm

42 cm

The diagram shows the design for a marquee (tent). The material to make the marquee costs $44 per square metre. a

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?

3m 7m 10 m 8m

A company manufactures nuts shaped like regular hexagonal prisms, with cylindrical bolt holes cut out of the centre. a

464

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. Give your answer in exact form. 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 WA Curriculum – Year 10 Methods Pathway 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 2m

0.8 m

Find the volume of the pool in cubic metres, rounded to two decimal places.

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?

A steel shed is to be constructed with a rectangular cut-out at the front for the entrance. 2m

2.5 m 10 m 3m 4m

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?

If the steel sheets cost $18 per square metre, calculate the total cost of the steel required to build the shed.

Investigation: Rainfall collection

Investigate online

mathspace.co

10.01 Surface area and volume of composite solids mathspace.co

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Represent with mathematics Use mathematics to model the tower’s total height, volume, surface area, cost and stability, including the effect of errors or enlargement.

Interactive exploration

Discover this concept in action online

mathspace.co

Calculate the total height, volume, surface area, cost and the angle of elevation of the conical cap. If there were errors in the measurements for the height or radius explore the effect of these on the cost.

Investigate 1.

Calculate the total height, volume and surface area using your dimensions.

Calculate the cost and the angle of elevation for the conical cap.

Determine the impact of a 5% radius error on volume.

Interpret your plan Look at your model and explain what it means for the wind turbine tower design. Verify if the plan meets the original restrictions; ≤ 10 m, cost is ≤ $1500 and angle is ≤ 60°, then adjust one of your chosen variables to assess it’s impact on the model.

Investigate 1.

Does your tower’s height, cost and angle of elevation meet the limits? Is the volume practical? Why or why not?

Change one thing. How does it affect volume, cost or stability?

Explain your turbine tower plan to the class. How do your assumptions affect it?

Discussion 1.

What was the trickiest part of modelling the tower with mathematics? Why?

How could you improve your design with more materials or a taller limit?

Talk to a friend about their plan. How was their tower size or cost different from yours?

Modelling Investigation: Wind turbine design mathspace.co

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10.02 Estimation and accuracy After this lesson, you will be able to… • explore rounding and how it influences the accuracy of calculations, especially in large data sets 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. 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

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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

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.

10.02 Estimation and accuracy mathspace.co

469

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.

Apply the idea Area according to Lila: ALila = 3.14 × (25.8)2 = 2090.11 m

Substitute the values Evaluate

Area according to Alex: AAlex = π × (26.0)2 2

= 2123.72 m

Substitute the values 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.

10.02 Estimation and accuracy mathspace.co

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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.

Apply the idea Volume according to Lila: VLila = (71.4)3

Substitute the values

= 363 994.344 cm3

Evaluate

Volume according to Alex: VAlex = (71.6)3

Substitute the values 3

= 367 061.696 cm

Evaluate

The percentage error is given by:

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

7 Ex 1

8 Ex 2

The weights of five boxes of fruits are measured (in kilograms). a

What is the total weight without rounding or truncation?

What is the total weight truncating to 2 decimal places?

What is the total weight rounding to 2 decimal places?

How do the truncated and rounded values compare to the actual values?

Box 1 2 3 4 5

Weight 12.783 15.482 11.926 13.643 14.217

Jasper has a ruler that is marked in cm. When he 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. 1

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.

The volume of the rectangular prism in part (c).

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?

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

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).

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.

10.03 Optional: Pyramids After this lesson, you will be able to… • calculate the surface area of pyramids. • solve problems involving the surface area of pyramids. • apply the formula to find the volume of pyramids. • solve problems involving the volume of pyramids.

Surface area of a pyramid A pyramid can have any polygon as a base. We can have square bases, triangular bases or even hexagonal bases. Then connect every vertex of the base to an apex point above the base, and you have a pyramid.

Perpendicular height

Slant height

If the apex is directly above (perpendicular to) the centre of the base it is a right pyramid. We call the distance between the centre of the base and the apex the perpendicular height. The length of the triangular-shaped sides that slope towards the apex of the pyramid is called the slant height (or slope length). We can identify a pyramid by the shape of the base. Here are a few right pyramids with a triangular base, a rectangular base, and a square base.

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Interactive exploration

Discover this concept in action online

mathspace.co

When we unfold a pyramid to get its net, we can see that a pyramid is made up of one base face and a number of triangular faces.

As we found with other 3D shapes, calculating the surface area (SA) of a solid is done by adding the area of all faces. For right pyramids, we have the base and a number of triangular faces. Surface area of right pyramid = Area of base + Area of triangles We can calculate the area of the base using the appropriate formula for its shape. Finding the area of the triangular faces is not quite as straightforward. Since the triangular faces are tilted to meet at the vertex, the height of each triangle corresponds to the slope length (or slant height) of that face, not the height of the pyramid. If we know the base side lengths and the height of a pyramid, we can ﬁnd the slant heights of each triangular face using Pythagoras’ theorem:

BC2 = AB2 + AC2

When adding up the area of the triangle faces, we can also look for symmetries in the solid to make our calculations easier. For example, in the case of a square-based pyramid, each triangular face will also have the same area.

Example 1 Find the surface area of the following rectangular-based pyramid.

9.1 cm 8 cm

9.3 cm

7 cm

10.03 Optional: Pyramids mathspace.co

477

Create a strategy 9.3 cm

Remember the surface area of a rectangular-based pyramid consists of two pairs of identical triangles opposite each other and a rectangle.

8 cm 7 cm 9.1 cm

Apply the idea SA Start with the formulas

Substitute the values

Evaluate

Example 2 Consider the rectangular-based pyramid with rectangular base of dimensions 4 cm by 10 cm and perpendicular height 7 cm.

y cm x cm

7 cm

a Find the length of the slant height labelled x cm. Round your answer to two decimal places.

Create a strategy Use Pythagoras’ theorem given by c2 = a2 + b2.

Apply the idea We are given a = 7, and

478

c =a +b

Start with the formula

x =7 +2

Substitute the values

x = 53

Evaluate

x = 7.28 cm

Take the square root of both sides

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10 cm

4 cm

b Now ﬁnd the length of the other slant height, labelled y cm. Round your answer to two decimal places.

Apply the idea We are given a = 7, and

c =a +b

Start with the formula

y =7 +5

Substitute the values

y = 74

Evaluate

y = 8.60 cm

Take the square root of both sides

Reﬂect and check By rounding at this step, our answer will not be exact in the end.

c Find the surface area of the pyramid. Make sure to include all faces in your calculation. Round your answer to one decimal place.

Create a strategy The surface area of the pyramid is equal to the sum of the areas of four triangles and a rectangle.

Apply the idea Based from the diagram, two of the triangles have a base of 10 cm and a height of 7.28 cm, and the other two have a base of 4 cm and a height of 8.60 cm. The rectangle has a width of 4 cm and length of 10 cm. Rectangle = lw Start with the formula for the area of a rectangle = 4 × 10 = 40 cm

Substitute the dimensions 2

Evaluate Multiply the formula by 2

Substitute the dimensions

Evaluate Multiply the formula by 2

Substitute the dimensions

Evaluate SA = 40 + 34.4 + 72.8 2

= 147.2 cm

Add all the areas Evaluate

10.03 Optional: Pyramids mathspace.co

479

10.03 Practice questions Understanding 1

Explorer 1–4

Adventurer

Trailblazer

1–5

1–6

Match the label to the image. a

Oblique (non-right) pyramid

Right pyramid

B 2

A square-based pyramid has a base area of 20 cm and each triangular face has an area of 15 cm2. How would you ﬁnd the surface area of the pyramid?

Using Pythagoras’ theorem a2 + b2 = c2: a

Find the slant height of a pyramid with a base side length of 7 cm and perpendicular height of 24 cm.

Find the perpendicular height of a pyramid with a base side length of 5 m and slant height of 13 m.

A rectangular prism has a volume of 120cm3. What is the volume of a pyramid with the same base and perpendicular height?

What will the volume of a pyramid be if it has a base area of 2 cm2 and a height of 3 cm?

What will the volume of a square-based pyramid be if it has a base area of 8 m2 and a height of 9 m? 8, 9ab, 10, 12, 13abe

Fluency 7 Ex 1

8, 9cd, 10, 11, 13c-e, 14

7, 8, 9cd, 10, 11, 13b–f, 14–16

Find the surface area of the following pyramids, rounded to two decimal places if necessary: a

Square-based pyramid

entagonal-based pyramid with base P area of 440.44 mm2

14 cm 20 mm

8 cm 8 cm 16 mm 482

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Rectangular-based pyramid

exagonal-based pyramid with base area H of 314.37 cm2

19 cm

9 cm

8.8 cm 11 cm

8 cm

8 Ex 2

9 cm

The rectangular-based pyramid has base dimensions of 6 cm by 8 cm and a perpendicular height of 6 cm. a

Find the value of x, rounded to two decimal places.

Find the value of y, rounded to two decimal places.

Find the surface area of the pyramid, rounded to one decimal place.

x cm

y cm

6 cm 6 cm 8 cm

For each of these square-based pyramids: i

Find the length of the slant height, rounded to two decimal places.

Find the surface area of the square-based pyramid, rounded to one decimal place.

4 cm

9 mm 12 cm 11 mm 11 mm

6 cm

8 mm 10 cm

10 mm 10 mm

10.03 Optional: Pyramids mathspace.co

483

A regular hexagonal-based pyramid has a base area of 36 cm2 and each triangular face has an area of 14 cm2. Find the surface area of the pyramid.

Find the surface area of these pyramids:

A square-based pyramid with a base edge of 12 cm and a perpendicular height of 8 cm.

A rectangular-based pyramid with base dimensions of 4 mm × 12 mm and a perpendicular height of 8 mm.

A rectangular-based pyramid with base dimensions of 6 m × 8 m and a perpendicular height of 10 m.

Some very famous right square-based pyramids are the Egyptian Pyramids. The Great Pyramid which has a base of 230 m and a slant height of 216 m. Find the surface area of the Great Pyramid. Do not include the area of the base in your calculation.

13 Ex 3

Find the volume of the following pyramids, rounding your answers to two decimal places where necessary: a

b 6 cm

5 cm 10 cm

8 cm 12 cm

13 m 9 cm 6m

7 cm

80 mm

10 cm

12 cm 18 cm

8 cm 6 cm

24 cm 10 cm

484

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

In the triangular-based pyramid shown:

Find the length of RT in centimetres, rounding your answer to two decimal places.

Find the area of the base △RTU.

Find the volume of the pyramid rounded to two decimal places.

68 cm

95.9 cm

U 88 cm

Find the perpendicular height of the following pyramids: a

A right square-based pyramid with a base of length 24 cm and a volume 2304 cm3.

A rectangular-based pyramid with a volume of 234 cm3, a base width of 13 cm and base length of 9 cm.

Find the base length of these pyramids: a

A right square-based pyramid with a perpendicular height of 36 mm and a volume of 6912 mm3.

A rectangular-based pyramid with a volume of 180 cm3, a perpendicular height of 12 cm and a base width of 5 cm.

Reasoning 17

17, 18

17–19

17–20

Barney was demonstrating how to ﬁnd the surface area of the square-based pyramid. Explain where Barney has gone wrong.

4 cm

8 cm

+ 4 cm

4 cm 4 cm

+ 4 cm 4 cm

4 cm

8 cm

The base edge length and the height of a square-based pyramid are equal, and the volume of the pyramid is 243 m3. a

Find the length of each side of the square-base.

Explain how the volume of a square-based pyramid relates to the volume of a rectangular prism with the same dimensions.

A square-based pyramid has a surface area of 360 cm2 and a base edge length of 12 cm. Determine the slant height of the pyramid and explain your reasoning. 10.03 Optional: Pyramids mathspace.co

485

A rectangular-based pyramid has a volume of 160 m3 and a base with dimensions 8 m by 6 m. If the perpendicular height is doubled, what would the new volume be? Justify your answer.

Problem-solving 21, 22, 25ab 21

21–23, 25b–e

22–24, 25c–f

A paperweight is in the shape of a square-based pyramid with dimensions as shown. The paperweight is ﬁlled with solid glass. Find the volume of glass needed to make 3000 paperweights. 5 cm 4 cm 4 cm

Consider the following bottles of portable hand wash. Assume that their volumes are equivalent to pyramids of the same dimensions. a

By approximately what factor is the amount of hand wash in bottle B greater than in bottle A?

If bottle A costs $50.88 and bottle B costs $60.00, which is the better buy?

8 cm

6 cm 3 cm 4 cm 5 cm

Bottle A

Find the slant height of the triangle side with base length 12 cm. Round your answer to two decimal places.

Find the slant height of the triangle side with base length 10 cm. Round your answer to two decimal places. Find the surface area of the composite solid, rounded to two decimal places.

5 cm

10 cm 12 cm

A small square-based pyramid of height 5 cm was removed from the top of a large square-based pyramid of height 10 cm leaving the solid shown: a

Find the slant height of the trapezoidal sides of the new solid. Round your answer to two decimal places.

Find the surface area of the composite solid formed, rounded to one decimal place.

5 cm 4 cm 5 cm

8 cm 486

Bottle B

A pyramid has been removed from inside a rectangular prism, as shown in the ﬁgure:

4 cm

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Find the volume of these pyramids, rounding your answers to two decimal places where necessary: a

8 cm

80 mm

11.5 cm 15.2 cm

5 cm 3 cm 40 mm

12 mm

130 m 10 mm 50 m

8.7 mm

40 m

23 cm

f 13.5 mm

15.6 mm 9 mm

18 cm 15 cm 25 cm 18.5 mm

10.03 Optional: Pyramids mathspace.co

487

10.04 Optional: Spheres After this lesson, you will be able to… • apply the formula for calculating the surface area of a sphere. • apply the formula for calculating the volume of a sphere. • solve problems that involve finding the surface area and volume of spheres.

Surface area of a sphere A sphere is a solid consisting of all points at a ﬁxed distance from a central point. The central point of a sphere is often just called its centre. The distance of each point on the sphere from the centre is called the radius, just like for a circle. Unlike the solids we have seen so far, we cannot unwrap a sphere to get a 2D net to calculate its area, so the surface area of a sphere is approached in a differently way. r

Archimedes discovered that the area of the curved face of the cylinder is equal to the surface area of the sphere. We saw that the curved surface of a cylinder ﬂattens out into a rectangle. Since the height of this cylinder is twice the radius of the inscribed sphere, which is also the radius of the circular h = 2r ends, the area of the resulting rectangle is 2π r × 2r = 4π r2. We therefore have our formula for the surface area of a sphere. A = 4π r2

Example 1 Find the surface area of this sphere. Round your answer to two decimal places. 9 cm

488

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10.04 Practice questions Understanding 1

Explorer

Adventurer

1, 2

Trailblazer

1–3

2, 3

What fraction of a sphere is shown in this diagram? 6

If the radius of a sphere doubles, what happens to its surface area and volume?

Are these statements true or false? a

If two spheres have the same surface area, they also have the same volume.

The surface area of a sphere is always less than the square of its circumference.

If the volume of a sphere is 36π cubic units, the radius of the sphere is 3 units.

A sphere with radius r = 2 has a smaller volume than a sphere with r = 3.

Fluency 4 Ex 1, 3

4ace, 5ab, 6ac, 7, 8, 9a, 10a

4ace, 5ace, 6ac, 7, 8, 9a, 10a

4bdf, 5bdf, 6, 9, 10

For each sphere, find to two decimal places: i

The surface area

The volume

3 cm

8 cm

6 cm

9 cm

10.3 cm

30.04 mm

10.04 Optional: Spheres mathspace.co

491

Calculate the exact surface areas of the spheres. a

Radius = 2 cm

Radius = 5 cm

Radius =

Radius = 14.9 cm

Diameter = 10 cm

Diameter = 23 mm

Radius = 30 cm

Calculate the exact volume of the spheres. a

Radius = 2 m

Radius = 9 cm

Radius =

The diameter of a bowling ball is 8.6 cm. Find the volume of the ball, rounded to two decimal places.

The dome is a solid hemisphere. a

9 Ex 2

Find the area of the curved surface of the dome, rounded to three decimal places.

Find the area of the circular base of the dome, rounded to three decimal places.

Find the total surface area of the dome, rounded to two decimal places.

For each hemispheres, ﬁnd the total surface area rounded to three decimal places. a

74.74 m

28.04 cm

For each hemisphere, find the volume rounded to three decimal places. a

492

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

90.03 cm

Reasoning

11, 12

12, 13

How does changing the radius of a sphere affect its surface area and volume?

Explain how to ﬁnd the volume of this solid.

4 cm

14 cm

A student was calculating the volume of the sphere shown and has these workings.

9 cm

Explain what the student’s mistake was, then find the actual volume of the sphere.

Problem-solving 14, 18, 20, 21 14

14-16, 18, 19, 21

15, 17, 19, 21-23

The volume of a regulation adult beach volleyball is 106π cm3. The volume of a regulation youth volleyball is

π cm3. Find the approximate ratio of the radius of the youth volleyball

to the adult beach volleyball, rounded to two decimal places. 15

The density of a glass marble is 2.57 g/ml. If the mass of a single marble is 8 grams, ﬁnd its volume rounded to two decimal places.

A spherical chocolate truﬄe has a volume of 0.5625π cm3

Find the radius of a chocolate truﬄe.

If the truﬄe is dipped in sprinkles, ﬁnd the surface area of the sprinkles that will be used.

The chocolate truffles are packaged in boxes that are 1 truffle wide and 10 truffles long. Find the dimensions of each box.

A water droplet on a leaf is approximately shaped like a hemisphere. The density of water is 1 g/ml. a

If the mass of the water droplet is 0.002 g, ﬁnd the volume of the droplet in millilitres.

Using the conversion 1 ml = 1000 mm3, ﬁnd the radius of the droplet.

Find the surface area of the exposed portion of the droplet. Determine an appropriate degree of precision.

10.04 Optional: Spheres mathspace.co

493

The shape is constructed by removing a hemisphere from a cylinder. Report each calculated measurement to an appropriate degree of precision.

22 cm

15 cm

Find the surface area of the hemispherical part.

Find the surface area of the cylindrical part, including the circular base.

Find the total surface area of the shape.

The solid shown is constructed by cutting out a quarter of a sphere from a cube.

14.2 cm

Find its surface area if the side length is 14.2 cm and the radius of the sphere is half the side length. 20

The planet Jupiter has a radius of 69 911 km, and planet Mars has a radius of 3390 km. Approximately how many times bigger is the surface area of Jupiter compared to Mars?

Tennis balls are packaged and sold in cylindrical containers, with three balls per container. The height of a standard tennis ball container is 21 cm. a

Find the volume of a tennis ball.

Find the volume of the container.

Find the percentage of the container that is occupied by the tennis balls.

Find the radius of a sphere such that the ratio of the surface area in square metres to the volume (in cubic metres) is 4 : 1.

Water in a cylindrical vase reaches a height of 22 cm. Bianca pours this water into a new spherical vase. If both vases have a radius of 17 cm, determine how much empty space will be left empty in the spherical vase once the water is poured into it. Round your answer to one decimal place.

494

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10.05 Optional: Cones After this lesson, you will be able to… • define and describe the structure of a cone. • apply the formula for calculating the surface area of a cone. • apply the formula for calculating the volume of a cone. • solve problems that involve finding the surface area and volume of cones.

Surface area of a cone A cone is a solid with a circular base and a curved surface reaching to a point called the apex, from the base. If the apex is directly above the centre of the base, it is called a right cone. The slant height can be denoted by s and the radius of the base by r.

Interactive exploration

Discover this concept in action online

mathspace.co

The total surface area of a right cone is: Surface Area of Right Cone = Area of Base + Area of Sector SA = π r2 + π rs where r is the cone’s base radius and s is the slant height.

Example 1 Find the surface area of this cone and round your answer to two decimal places. 8 cm

3 cm

Create a strategy Use the formula: Surface area of a cone = π rs + π r2

Apply the idea Given r = 3 and s = 8 Surface Area = π × 3 × 8 + π × 32 = 103.67 cm

Substitute r and s Evaluate and round

10.05 Optional: Cones mathspace.co

495

Example 2 Find the surface area of this cone.

t 12

a Find the slant height. Give your answer in exact surd form.

Create a strategy Use Pythagoras’ theorem, where the square of the hypotenuse is equal to the sum of the squares of the remaining sides.

Apply the idea To ﬁnd the slant height, we use the Pythagoras’ theorem: t2 = 32 + 122 Solving for t, we are given

Take the square root on both sides Evaluate the power Evaluate the addition Simplify

b Find the surface area. Round your answer to two decimal places.

Create a strategy Use the surface area of a right cone of radius r and slant height h is π r(r + h). Use the value of h from part (a).

Apply the idea

Write the formula

Substitute the values

Evaluate

496

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10.05 Practice questions Understanding 1

Ex 1

Trailblazer 1–3 18 cm

Which part of the cone corresponds to the length of 18 cm?

30 cm

Which part of the cone corresponds to the length of 30 cm?

Using Pythagoras’ theorem a2 + b2 = c2: a

Find the slant height of a cone with radius of 7 cm and height of 24 cm.

Find the perpendicular height of a cone with radius of 5 m and slant of 13 m.

What will the volume of a cone be if it has a base area of 2 cm2 and a height of 3 cm?

Fluency 4

Adventurer 1–3

For this net of a cone: a

Explorer 1–3

4ab, 5a-c, 7, 8a-c, 9, 10, 11a

4ab, 5ace, 7, 8c-e, 9, 10, 11a

4, 5bdf, 6, 7, 8d-f, 11, 12

For each cone, ﬁnd the surface area rounded to two decimal places. a

12 cm

9 cm

6 cm 14 cm

d 9 km

18 km

7 cm

20 mm

10.05 Optional: Cones mathspace.co

499

5 Ex 2

Using the cone measurements: i

Find the slant height of the cone.

Calculate the surface area of the cone, rounded to two decimal places.

Radius = 3 cm, height = 4 cm

Radius = 5 cm, height = 12 cm

Radius = 12 cm, height = 5 cm

Radius = 15 cm, height = 8 cm

Radius = 1 m, height = 100 cm

Radius = 800 cm, height = 5 m

This cone has been sliced in half and has a diameter of 5 cm and a slant height of 11 cm. Find the surface area of the solid, rounded to two decimal places.

11 cm

5 cm

7 Ex 3

For each cone, ﬁnd the volume rounded to one decimal place. a

9 cm

11 cm 14 cm

2 cm

9 cm 12.4 cm

3 cm

Find the exact volume of the cones using the given measurements. a

500

2.5 cm

Radius = 3 cm, height = 8 cm

Radius = 4 cm, height = 6 cm

Radius = 10 cm, height = 3 cm

Radius = 12 cm, height = 15 cm

Radius = 5 cm, height = 4 cm

Radius = 10.8 cm, height = 7 cm

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Find the volume of a cone that has a diameter of 3 m and a height of 1.3 m, rounded to two decimal places.

Calculate the perpendicular height of this cone.

Find the volume of the cone rounded, to three decimal places.

26 m

20 m

11 Ex 4

Find the radius of these cones. a

A cone with volume of 12 441.02 m3 and a height of 30 m, rounded to one decimal place.

A cone with volume of 196 cm3 where the height and radius are equal, rounded to two decimal places.

A cone with a volume of 938.29 mm3 and a height of 14 mm, rounded to the nearest millimetre.

Find the height h of a cone that has a volume of 26 640.97 cm3 and a radius of 31.9 cm. Round your answer to the nearest whole number.

Reasoning 13

13, 14

13–15

The radius and the height of a cone are equal, and the volume of the cone is 72π cm3. a

Find the radius of the cone.

Explain how the volume of a cone relates to the volume of a cylinder with the same dimensions.

Show that

is the expression for the surface area of a cone when radius is

r and height is h. 15

A cylindrical cake is used to form a cone. The cone has the same base area and height as the cylindrical cake. What fraction of the cake is used to make the cone? Explain your reasoning.

Problem-solving 16, 18a, 20, 22, 24 16

16, 18, 19, 21, 22, 24

17, 18b, 19, 21, 23, 24

An ice cream cone has a surface area of 20 cm2. If the size of the cone is increased by a scale factor of 3.1, ﬁnd its new surface area, rounded to one decimal place.

10.05 Optional: Cones mathspace.co

501

A cone has a volume of V cm3. If the height of this cone is scaled by a factor of

and the radius is scaled

by a factor of , ﬁnd its new volume.

Find the volume of these solids rounded to two decimal places: a

11 cm 4 cm

12 cm

3 cm

A podium was formed by sawing off the top of a cone. Find the volume of the podium, rounded to two decimal places.

2m 6m 4m

Find the capacity of the ice cream cone in millilitres, rounded to one decimal place.

9 cm

13 cm

502

An ice cream cone is made by folding together a sector of pastry, with a small overlap. The dimensions of the cone are shown: a

Find the external surface area of the cone in square centimetres, rounded to one decimal place.

If the overlap adds an extra 5% to the area, determine how much pastry is required to produce the cone. Round your answer to the nearest square centimetre.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

4 cm

10 cm

A grain silo has the shape of a cylinder attached to a cone, with dimensions as shown: a b

Find the surface area of the silo, to the nearest square metre, assuming the top is closed.

10 m

The silo is manufactured out of sheet metal that has a mass of 2.4 kg/m2. Find the total mass of the silo to the nearest kilogram. 6m

The ice cream cones at an ice creamery have the dimensions indicated in the diagram.

8 cm

Determine how many millilitres of ice cream can ﬁt in each cone, including the hemisphere scoop on top. Round your answer to the nearest millilitre.

The ice cream is purchased in 10 L tubs. Determine how many whole cones can be made with a single tub of ice cream.

15 cm

Double cones are served with a second hemispherical scoop of the same dimensions as the ﬁrst scoop. Determine how many whole double cones can be made from a 10 L tub.

Before 1980, Mount St. Helens was a volcano approximately the shape of the first cone. a

What was the volume of the mountain, in cubic kilometres? Round your answer to two decimal places.

The tip of the mountain was in the shape of the bottom cone shown.

2.9 km

Find the volume of the tip in cubic kilometres, rounded to two decimal places. c

In 1980, Mount St. Helens erupted and the tip was destroyed.

4.82 km

Find the volume of the remaining mountain, in cubic kilometres, rounded to two decimal places. 0.4 km

0.6 km

10.05 Optional: Cones mathspace.co

503

10.06 Optional: Composite solids After this lesson, you will be able to… • calculate the surface area of composite solids. • solve problems related to the surface area of composite solids. • calculate the volume of composite solids by adding or subtracting individual volumes. • solve problems related to the volume of composite solids.

Surface area of composite solids A composite solid is made up of a combination of other solids. To ﬁnd the surface area of a composite solid, we must add the areas of its faces. If the composite solid is made up of a cylinder, a cone or a sphere, we will need to use parts of the formulas for the surface area of these solids. It’s important that we understand which part of the formula represents which face(s). The surface area of a sphere with radius r is: SA = 4πr2 The surface area of a cylinder with radius r and height h is: SA = Area of two circular ends + Area of curved surface = 2πr2 + 2πrh

The surface area of a right cone with radius r, perpendicular height h and slant height l is: SA = Area of circular base + Area of sector = π r2 + π rl r l

504

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

To ﬁnd the surface area of a pyramid, we add the area of the base and the area of each triangular face. To ﬁnd the height of each triangular face, we may need to use Pythagoras’ theorem. B

Example 1 The ice cream cones sold at an ice creamery have the dimensions indicated in this diagram: 8 cm

15 cm

a Calculate the slant height of the cone, giving your answer in exact form.

Create a strategy Use Pythagoras’ theorem 4 cm

15 cm

Apply the idea Write Pythagoras’ theorem

Substitute c = l, a = 4 and b = 15

Evaluate the exponents

Evaluate the sum

Take the square root of both sides to solve for l

The slant height of the cone is

cm.

10.06 Optional: Composite solids mathspace.co

505

b Find the surface area of the cone.

Apply the idea The surface area of a cone is SA = πr2 + πrl. Since the cone meets the hemisphere at its circular face, the circular face of the cone is no longer part of the surface. We need to calculate the area of the sector only. Start with the formula for the sector area

Substitute r = 4 and l =

Simplify

c Find the surface area of the hemisphere.

Create a strategy Use the formula for the surface area of a sphere, and divide it by 2.

Apply the idea

Start with the formula

Substitute r = 4

Simplify

d Find the total surface area of the ice cream cone, rounded to the nearest square centimetre.

Create a strategy Add the surface area of the cone and the hemisphere.

Apply the idea Add the surface areas

Evaluate and round

Reﬂect and check Note that in parts b and c, we left the answers as exact values. Evaluating too early can introduce rounding errors to the calculation of the total surface area.

506

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Volume of composite solids We can make composite solids out of spheres, hemispheres, cylinders and other curved-surface solids.

4 cm

5 cm

10 cm 5 cm 8 cm

As we have done with composite volumes in other years involving prisms, to ﬁnd the volume of composite solids we: • identify which smaller solids to break the composite into. • identify the dimensions of those components. • add or subtract volumes of the components as necessary to ﬁnd the composite volume.

Example 3 Find the volume of the composite ﬁgure shown, which consists of a cone and a hemisphere, joined at their bases. Round your answer to two decimal places.

10 cm 4 cm

Create a strategy Add the volume of the cone and the volume of the hemisphere.

Apply the idea Use the formula of the volume of the cone,

Start with the formula

Substitute the values

Evaluate

508

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

, where r = 4 cm and h = 10 cm.

, where r = 4 cm.

Use the formula of the volume of the hemisphere, Start with the formula

Substitute the radius

Evaluate

Add these volumes:

Write the sum

Evaluate and round

Example 4 This solid is a truncated cone, formed by cutting off a cone-shaped section from the top of a larger, original cone.

2 3 3

a What was the volume of the cone section that was cut off? Give your answer as an exact value.

Create a strategy Use the formula of the volume of the cone given by

Apply the idea We are given r = 2 and h = 6. Start with the formula

Substitute the values

Evaluate

10.06 Optional: Composite solids mathspace.co

509

What basic solids are used to make each of these composite solids? a

Describe how you would estimate the volume of the following objects using simple 3D solids. a

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.

10.06 Optional: Composite solids mathspace.co

511

State two simple 3D solids that could be averaged to estimate the volume of a six-sided die which has rounded corners.

7–10, 14a-e, 15a-e, 16, 18

Fluency Ex 3

Ex 4

7, 8, 10, 12, 13, 14b-f, 15b-f, 17, 19

7, 8, 10–13, 14d-h, 15d-h, 17–19

The solid shown is made up of a cone and a hemisphere. a

Calculate the slant height of the cone, giving your answer in exact form.

Find the surface area of the cone, rounded to two decimal places.

Find the surface area of the hemisphere, rounded to two decimal places.

Find the total surface area of the solid, rounded to the nearest square centimetre.

10 cm 4 cm

Find the surface area of these composite solids, rounded to two decimal places: a

4 cm 12 cm

3 cm 9 cm

11 cm

3 cm

0.2 cm

4 cm

8 mm

f 6m

5 cm 4.5 m 2 cm

512

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

The solid shown is a hemisphere on a cylinder. a

Find the exact surface area of each of these faces: i

The circular base.

The curved part of the cylinder.

iii

The hemisphere.

33.1 mm

Find the total surface area of the solid, rounded to two decimal places. 10.5 mm

This solid is made up of a cone and a hemisphere. a

6 cm

Find the surface area of each of these faces, rounding your answer to two decimal places: i

The hemisphere.

The curved surface of the cone.

iii

The torus (doughnut shape).

16.7 cm

Calculate the total surface area of the solid, rounding your answer to the nearest square centimetre.

14.8 cm

The solid shown is constructed by cutting out a quarter of a sphere from a cube. The side length of the cube and the diameter of the sphere are both 6 cm. a

List all of the faces of the solid.

Find the surface area of the solid.

6 cm

A pyramid has been removed from inside a cube as shown: a

Find the perpendicular height of the triangular faces, rounding your answer to two decimal places.

Find the surface area of the composite solid, rounded to one decimal place.

5 cm

5 cm 5 cm

A small square-based pyramid of height 6 cm was removed from the top of a large square-based pyramid of height 12 cm to form the solid shown. a

Find the length of the slant height of the sides of the new solid, rounded to two decimal places.

Find the surface area of the solid formed, rounded to one decimal place.

6 cm 4 cm 6 cm

8 cm

10.06 Optional: Composite solids mathspace.co

513

Ex 1

Find the volume of the solids rounded to two decimal places when necessary: a

4 cm

12 cm

3 cm 14 cm

11 cm

4 cm

8 cm

9 cm 15 cm

f 5 cm

4 cm

3 cm

5m 6m

4.5 cm

16 cm

6.5 m

48 cm

14 cm

514

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

19 m

16 m

Ex 2

Find the volume of the solids, rounded to two decimal places when necessary: a

12 m

6 cm

2 cm

3 cm 8m

3 cm

5 cm 9m

3 cm

5 cm

6 cm 5m

f 4m

4 cm 4m

4 cm

8 cm

6 cm

h 3m 5m 7m

6 cm 8 cm

17 m 14 cm

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The podium was formed by sawing off the top of a cone. Find its volume, rounded to two decimal places.

For the solid shown: a

Find the perpendicular height of the cone, rounded to two decimal place.

Find the volume of the solid, rounded to the nearest cubic centimetre.

6 cm

16.7 cm

14.8 cm

The diagram is a solid constructed from two identical frustums. Each frustum was made by removing the top third of the original square-based pyramid.

Find the volume of the solid, rounded to two decimal places. 4

An hourglass-like solid is constructed from two identical truncated cones. Each truncated cone is half the height of the original cone. 7

Find the volume of the solid, rounded to two decimal places.

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Problem-solving 26, 27, 30–32 25

27, 28, 32, 33

25, 27–29, 32–34

An eight-sided die is an octahedron, which can be thought of as a composite shape made of two square-based pyramids. Each face is an equilateral triangle with side lengths of 1.8 cm. a

Find the perpendicular height of the equilateral triangle, rounding your answer to two decimal places.

Find the surface area of the die. Round your answer to one decimal place.

Does the eight-sided die have a greater surface area than a six-sided die with side lengths of 1.8 cm?

1.8 cm

A spire is an architectural feature which originated in the 12th century in Germany and can be seen throughout Europe in Gothic and Baroque architecture. It has the shape of a cone or a pyramid.

A castle has two spires with square bases. The width of the base is 4 m and the slant height of each side of the spire is 35 m. Determine how much paint is required to cover both spires.

Another castle has two spires with the same slant height as in part (a), but the base of the spires is a hexagon with side length 2 m. Explain how will this building require more or less paint than the previous building.

Assuming they have equal slant heights, determine the side length that the spires with a hexagonal base need to have in order to require the same amount of paint as the spires with a square base in part (a).

A pedestal for an action ﬁgurine is the shape of the solid made by removing the top section from a cone. Find the surface area of the pedestal.

8 cm

2 cm

4 cm

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8 cm

Zane makes a frustum by cutting off a square-based pyramid a third of the way from the top, as shown: a

Find an expression for the exact volume of the original pyramid.

Find an expression for the volume of the pyramid section that was cut from the original pyramid.

What fraction of the original pyramid did Zane cut off?

If the original pyramid had a volume of 1161 cm3, ﬁnd the exact volume of Zane’s frustum.

Lucy makes a truncated cone by cutting off a smaller cone halfway from the top, as shown: a

Find an expression for the exact volume of the original cone.

Find an expression for the exact volume of the cone section that was cut from the original cone.

What fraction of the original cone did Lucy cut off?

If the original cone had a volume of 184 cm3, ﬁnd the exact volume of Lucy’s truncated cone.

Investigation: Instruments and methods

Investigate online

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mathspace.co

10 Chapter review 1

For each ﬁgure, ﬁnd to two decimal places: i

The volume

6 cm

The surface area

9 cm 6 cm

6 cm

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

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.

If paint costs $24 for a 1 L tin that covers 6.5 m2, ﬁnd the cost to paint the shed, rounded to the nearest whole dollar.

6m 4m 8m 11 m 9m

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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.

10 cm 4 cm

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.

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 fossilised plant was 500 Myr ± 25 Myr (Myr means million years).

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.

(Optional) Find the surface area of the pyramid.

14 m

8.7 m 10 m

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(Optional) For the pyramid shown: y cm x cm

8 cm 11 cm 15 cm

Find the slant height labelled x, rounded to two decimal places.

Find the slant height labelled y, rounded to two decimal places.

Find the surface area, rounded to one decimal place.

Calculate the volume.

(Optional) For each sphere, find to two decimal places: i

The surface area.

The volume.

4.7 cm

12 cm

(Optional) For the hemisphere:

4 cm

Find the surface area rounded to two decimal places.

Find the volume rounded to three decimal places.

Optional) For each cone, find to two decimal places: i

The surface area.

The volume.

8 cm 9 cm

5 cm

3 cm

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(Optional) Find the radius of a cone with volume of 232 cm3 and a height of 12 cm, rounded to two decimal places.

(Optional) Find the surface area of the composite shape rounded to two decimal places.

33.1 mm

10.5 mm

(Optional) For the composite figures: i

Find the surface area rounded to two decimal places.

Find the volume rounded to two decimal places.

b 10 m 19 m

11 cm 5 cm 16 m

2.5 m

10 cm

13 cm

(Optional) Find the remaining volume in the box after the pyramid is removed. 12 cm

12 cm 7 cm 18 cm

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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 • Choose and construct appropriate sample spaces to show outcomes for twoand three-stage chance experiments both with and without replacement. Assign probabilities to events involving conditional statements, such as ‘if ... then’, ‘given’, ‘of’, ‘knowing that’ • (Optional) Use weighted tree diagrams and/or formulas to assign probabilities to two- and three-stage chance events including situations involving conditional probability • Represent secondary data in two-way tables or Venn diagrams and assign probabilities to outcomes involving conditional statements • Analyse, represent and interpret problems in real-world situations involving two- and three-stage chance experiments both with and without replacement, conditional probability or statements, boxplots, bivariate data and/or two-way tables • Conduct repeated chance experiments and simulations to model conditional probability and produce datasets using digital tools. Discuss, compare and analyse variation and estimated probabilities for conditional events

11 Probability Chapter outline 11.A Review Venn diagrams and two-way tables 11.01 Conditional probability Modelling investigation: Animal tagging for environmental monitoring 11.02 Mixed probability concepts Investigation: Simulation 11.03 Optional: Geometric probability Chapter 11 review

528 539 541

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11.01 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.

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.

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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)

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).

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

11.01 Conditional probability mathspace.co

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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

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′) =

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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

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

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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. 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 (On time∣Fine) = 0.9 • P (Not fine) = 0.6 • 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

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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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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 positions 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:

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.

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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: 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

Liberal (next election) 40

10 140

Greens (next election) 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

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Labor (next election) 26

Which of the events appear to be dependent, H and S or H and T?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

Interactive exploration

Discover this concept in action online

mathspace.co

Simulate 15 sets of 3 catches with replacement (for example coin: heads = reef, tails = tiger heads = reef, tails = tiger) and without replacement (for example adjust probabilities after each catch), either in the applet or manually with a coin or die, recording species and lengths (reef: 1-2m, tiger: 2-4m).

Investigate 1.

Simulate your datasets with and without replacement. Record species outcomes and lengths. How did you ensure reliability?

Show a way to calculate the probability of one sequence and a conditional probability for both methods.

Make a two-way table for species outcomes and a boxplot for length distributions per species, comparing replacement scenarios.

Interpret your plan Look at your datasets and graphs to estimate the marine park’s shark population and explain what it means for conservation. Scale your results with and without replacement to 200 sharks, then adjust one factor (for example increase sample size by 5) either in the applet or by hand and compare estimates.

Investigate 1.

Predict the number of reef sharks in 200 for each method. Do the boxplots and tables look consistent? Why or why not?

Change one thing. How does it change your estimate?

Explain your shark tagging model to the class. How do chance and assumptions affect your conservation insights?

Discussion

540

What was the trickiest part of modelling shark tagging with mathematics? Why?

How could you improve your simulation with more samples or tools?

Talk to a friend about their plan. How was their reef shark proportion different from yours?

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11.02 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 .

11.02 Mixed probability concepts mathspace.co

543

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

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

544

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

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.

11.02 Mixed probability concepts mathspace.co

545

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

Total

E-reader Not e-reader 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

546

Paperback 120 60 180

Not paperback 294 126 420

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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 who read on e-readers, what percentage read in both formats? Round your answer to the nearest percentage.

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:

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

11.02 Mixed probability concepts mathspace.co

547

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:

548

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

11.02 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.

550

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

6−8, 10, 11, 13, 15, 17, 18

Fluency 4

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.

Use the Venn diagram to ﬁnd P ( A∣B). Give the answer as a fraction in simplest form.

A A′

B w x

B′ y z

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

11.02 Mixed probability concepts mathspace.co

551

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

Do not believe in Santa

Total

Believe in Tooth Fairy Do not believe in Tooth Fairy Total

Ex 3

100

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: Read the book Didn’t read the book Total

552

Watched the movie 39

Didn’t watch the movie 4 62

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Total 97 166

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

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′)

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. 11.02 Mixed probability concepts mathspace.co

553

Two marbles are randomly drawn without replacement from a bag containing 1 blue, 2 red and 3 yellow marbles. a

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

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).

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 b

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. a

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.

554

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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:

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.

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.

Explain why it is inappropriate to assume that the probabilities apply to a randomly chosen candidate.

P 0.6

0.7

0.4 F

0.3

11.02 Mixed probability concepts mathspace.co

F 555

Problem-solving 25, 27, 28 25

25, 27−29

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 is 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 0.05

0.2 C

556

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 10

Action 4

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

11 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 A′ Total

P (A)

P (A ∩ B)

P (B∣A′)

P (A′∣B′)

B′

14 6 20

8 12 20

Total 22 18 40

Use the Venn diagram to ﬁnd: A

0.35

0.1

0.2

0.35

558

P (B′)

P (A ∪ B)

P ( A∣B)

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

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 j l

EN EE

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. B

120

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.

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

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 Chapter 11 review mathspace.co

559

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

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 enjoy 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.

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.

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:

NSW YES NO

ACT YES NO

NSW

ACT

YES

NSW YES YES

YES

YES YES

YES

QLD NO

NSW NO

QLD YES NO QLD

ACT

NSW

560

At least 1 is sour.

YES

QLD NO

YES

Find the probability that a randomly chosen person voted “YES”.

Of those that are not from NSW, ﬁnd the probability that a randomly chosen person voted “NO”.

Considering just the voters from NSW and ACT, ﬁnd the probability that a randomly chosen person was from NSW and voted “YES”.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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 • Analyse, determine and interpret problems in real-world situations involving two- and three-stage chance experiments both with and without replacement, conditional probability or statements, boxplots, bivariate data and/or two-way tables • Represent and analyse boxplots. Explain differences between multiple boxplot datasets in terms of shape, spread and centre. Compare or match the shapes of boxplots to distributions depicting the same data • (Optional) Produce, organise and represent accurate and valid data in a cumulative frequency graph and use this to analyse quartiles and percentiles • (Optional) Determine the mean and standard deviation of a dataset. Investigate, analyse and interpret the effect of individual data values, including outliers, on the standard deviation • Critically analyse the claims, inferences and conclusions of statistical reports in the media and other real-life situations and identify potential sources of bias

12 Univariate statistics Chapter outline 12.A Representations of data 12.01 Quartiles 12.02 Boxplots 12.03 Compare data sets Modelling investigation: Ferry schedules 12.04 Samples and populations 12.05 Optional: Cumulative frequency 12.06 Optional: Standard deviation 12.07 Optional: Comparisons using standard deviation Chapter 12 review

564 576 590 603 605

615

12.01 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. Minimum Lower quartile, Q1 Median, Q2 Upper quartile, Q3 Maximum

564

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

value

Median is the

value

Q3 is the

value

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 Use the formula:

th where n is the total number of scores, to ﬁnd the position of the median.

Apply the idea There are n = 13 scores in the list. Position of median = = 7th Median = 80

Substitute n = 13 Evaluate Choose the 7th score

12.01 Quartiles mathspace.co

565

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

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.

12.01 Quartiles mathspace.co

567

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

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.

568

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Number of scores = 15

Count the dots

12.01 Practice questions Understanding 1

Explorer

Adventurer

1−4

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 3

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

570

Find her median score.

Find the lower quartile score.

Find the upper quartile score.

Find the interquartile range.

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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.

Find the number of scores.

For each set of scores: i

8 Ex 2

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

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

Sort the scores in ascending order.

Median

Lower quartile

iii

Upper quartile

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.

Weight 16 17 18 19 20 21 22

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

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571

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. a

Find the lower quartile.

Find the upper quartile.

Find the percentage of students who 0 have more pets than the upper quartile. Explain why it is not exactly 25%.

Number of pets

Is the percentage of data values between the upper and lower quartiles always exactly 50%? Explain why or why not, including an example to justify your conclusion.

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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.

12.01 Quartiles mathspace.co

575

12.02 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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∼25%

Outliers are data points that 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. Left skew

Symmetrical

10 15 20 25 30 35

A symmetrical data set is distributed around the centre with a similar frequency on the left and right.

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.

Right skew

Uniform

10 15 20 25 30 35

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.

10 15 20 25 30 35

A uniform data set is evenly distributed across all values.

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

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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

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.

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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

580

Ms. Johnson’s class has an interquartile range of 15.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

12.02 Practice questions Understanding 1

582

Adventurer

Trailblazer

1−4

Fill in the missing numbers for each boxplot. a

Explorer 1−4

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

2 4 6 8 10 12 14 16 18 20

4 a

For the boxplot shown: 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

6cd, 7, 8, 9efhj, 10−14

For the boxplot shown, ﬁnd the: a

Minimum

Maximum

Range

Median

Interquartile range 0

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

8 10 12 14 16 18 20

Scores

8 10 12 14 16 18 20

Scores

38 40 42 44 46 48 50 52 54 56 58 60

12.02 Boxplots mathspace.co

583

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

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?

Age

12.02 Boxplots mathspace.co

585

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.

586

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

12.02 Boxplots mathspace.co

587

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: 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, 25bd 23

23, 24, 25a-c

Construct a boxplot for the data set: • Range = 14 • Lowest score = 2 • IQR = 8 • 25% of scores are below 4 • 50% of scores are above 8

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23−26, 25a-c

Match the histograms on the left to the corresponding boxplots on the right: Histogram A

Boxplot 1

0 1

9 10

1 2 3 4 5 6 7 8 9 10

Histogram B

Boxplot 2

0 1 1

2 3 4 5 6 7 8 9 10

Histogram C

Boxplot 3

1 2 3 4 5 6 7 8 9 10

Histogram D

Boxplot 4

0 1

9 10

1 2 3 4 5 6 7 8 9 10

12.02 Boxplots mathspace.co

589

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

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.

12.03 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.

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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.

Example 1 These 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

12.03 Compare data sets mathspace.co

591

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

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 bulbs with the best lifespan.

Reflect and check In fact, the best light bulb 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 light bulbs produced by Manufacturer B have a greater lifespan than all the light bulbs produced by Manufacturer A.

Example 2 These boxplots show the number of goals scored by two football players in each season. Sophie

Holly

592

15 Goals

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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. From frequency displays like histograms, dot plots, and stem and leaf plots we can identify: • Shape • Mode • Frequencies, grouped or individual • 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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12.03 Practice questions Understanding 1

Trailblazer 1−3

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?

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

596

Adventurer 1−3

The histogram and the boxplot display the same data. 10 9 8 7 6 5 4 3 2 1 0

Explorer 1, 2

10 20 30 40 50 60 70 80 90

10 20 30 40 50 60 70 80 90 100

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10 9 8 7 6 5 4 3 2 1 0

1 2 3 4 5 6 7 8 9 10

Histogram

Fluency 4

Which key features can be read off each display? a

Ex 1

Boxplot

4, 5, 8, 9

Dot plot

4−8

5−8

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

In which test did the class tend to score better marks?

12.03 Compare data sets mathspace.co

597

Ex 2

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

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?

Two weightlifters both record their number of repetitions with a 70 kg bar over 30 days. The results are displayed in the boxplots: a

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

598

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 number of pets?

Does either set of scores have an outlier?

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Block A

Block B

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

Histogram B Boxplot 2 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

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

12.03 Compare data sets mathspace.co

599

The parallel boxplots show the distances, in centimetres, jumped by two high jumpers.

Matt

Paul

140

150

160

170

180

Distance (cm)

Who had a higher median jump?

Who made the highest jump?

Who made the lowest jump?

Reasoning

600

10−13

11−13

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

10, 11

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 WA Curriculum – Year 10 Methods Pathway 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 as the most expensive 2 book in Bookstore A.

25% of the books in Bookstore B are cheaper than the cheapest book in Bookstore A.

Explain why each pair of histograms and boxplots do not match: a

Salaries earned by employees at a software company are given in a histogram.

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 centres, construct a boxplot of the data.

Find an estimate for the interquartile range.

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. 12.03 Compare data sets mathspace.co

601

Problem-solving 14 14

14, 15

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

602

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 WA Curriculum – Year 10 Methods Pathway mathspace.co

12.04 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 behaviour, 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

Discover this concept in action online

mathspace.co

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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

Reﬂect and check

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.

We sometimes call data which is all grouped together and summarised aggregate data and raw data disaggregated data.

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607

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

610

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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The boxplot shows the number of books a group of students read in a year: Number of Books

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?

The boxplots 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

Treatment 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: a

Find the total number of scores recorded.

Find the number of times a score between 150 and 200 occurred.

Find the percentage of scores that were 150 or more. Round your answer to one decimal place.

300 Cumulative frequency

250 200 150 100 50 0

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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 Number of workshops ( f )

Number of attendees 20 21 22 23 24 25

Cumulative frequency (cf )

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?

Use the cumulative frequency polygon to estimate: 100 90 80 70 60 50 40 30 20 10

Score 44

The median

The lower quartile

The upper quartile

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Match the boxplot shown to the correct histogram.

Score

7 8 9 10 11 12 13 14 15 16 17 18

618

8 9 10 11 12 13 14 15 16 17 18 Score

9 8 7 6 5 4 3 2 1 0

D Frequency

10 Frequency

9 8 7 6 5 4 3 2 1 0

Frequency

10 Frequency

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:

2 3 4 5 Building X

2 3 4 5 Building Y

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?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

Minimum Q1 Median Q3 Maximum Mean (1 d.p.) Range Interquartile range

Antoine 6.5 6.9 7.3 7.5 7.8

Benji

7.3

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.

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 us to make 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 • Analyse, determine and interpret problems in real-world situations involving twoand three-stage chance experiments both with and without replacement, conditional probability or statements, boxplots, bivariate data and/or two-way tables • Analyse bivariate data represented in a two-way table, using proportions and comment on possible association between categorical variables • Represent the relationship between bivariate data in a scatterplot and draw a trend line by eye if appropriate. Use the graph and context to describe any association in terms of strength, direction, linearity and outliers. Make predictions and recognise and explain any limitations of the model • Critically analyse the claims, inferences and conclusions of statistical reports in the media and other real-life situations and identify potential sources of bias

13 Bivariate statistics Chapter outline 13.01 Two-way tables 13.02 Associations between categorical variables 13.03 Bivariate numerical data 13.04 Associations between numerical variables 13.05 Line of good fit 13.06 Interpolation and extrapolation 13.07 Misrepresentations of data Investigation: Misrepresentations in the media 13.08 Correlation coefficient Investigation: Conduct a statistical investigation Chapter 13 review

622 634 654 664 678 689 703

717

13.01 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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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

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13.01 Practice questions Understanding 1

Explorer

Adventurer

1−3

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

Trailblazer

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?

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. 1.5 hours or more Less than 1.5 hours Total

Year 8 students 7 18 25

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. 13.01 Two-way tables mathspace.co

627

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

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.

Fluency 5

Adult 50 71 11

5−9

5, 7, 9, 10

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

30 71

41 71 V

Total 84 85 169

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 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. 628

5, 7, 10, 11

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Beef Children Adults Total

Chicken

Fish

Total

Ex 1

Ex 2

Special diet 14 11 15 40

What does 28 mean in this context?

What does 11 mean in this context?

Insulin injections 14 9 16 39

Total 28 20 31 79

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.

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This table shows the number of trains arriving either on time or late at a particular station:

Day

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?

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

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

Fruit Insects Nuts Total

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Total

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

632

MC1R gene 36 28

What does the value of x mean in this context?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

, ﬁnd the value of x.

Agouti gene 37 x

13.02 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 and attaching the % symbol. 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. 634

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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 and attaching the % symbol. 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 and attaching the % symbol.

13.02 Associations between categorical variables mathspace.co

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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 and attaching the % symbol.

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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 638

Play netball 33 13 46

Do not play netball 72 30 102

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Total 105 43 148

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.

13.02 Associations between categorical variables mathspace.co

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13.02 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

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Single

Married

Transportation

Food

Married with kids Insurance

Mathspace WA Curriculum – Year 10 Methods Pathway 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%

13.02 Associations between categorical variables mathspace.co

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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

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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.

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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. Currently pregnant Currently postpartum

Weight

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?

13.02 Associations between categorical variables mathspace.co

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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

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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?

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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 Country & western Rap Pop Classical Rock & roll

60% 40% 20% 0%

Vegetarian Non-vegetarian Dietary restriction

Does this stacked column chart suggest that there is an association between music preference and dietary restriction? 13.02 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

Total 68 62 130

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 two- way table. Right-handed children Left-handed children

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Regional 20 32 52

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

<5 335 1622 1957

5—9 18 42 60

> 15 2 5 7

10—15 17 42 59

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.

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

Science 56 40 33 17 146

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.

13.02 Associations between categorical variables mathspace.co

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Problem-solving 20, 21 20

20, 21

20−22

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.

Aug 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

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.

652

Jul

In which month were business books the most borrowed type of book?

Jun

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

13.03 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

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Age (years)

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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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

−3

−4

d 4

Fluency 5

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−4 −3 −2 −1 −1

−2

−4 −3 −2 −1 −1

x 1

−4 −3 −2 −1 −1

−2

−3

−4

6, 7, 10, 12, 14, 16

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

What is the dependent variable?

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 0

Temperature 16 20

13.03 Bivariate numerical data mathspace.co

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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) 0

Ex 3

120

160 200

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?

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

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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

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

Global temperature anomaly (°C) 0.6 0.65 0.68 0.7 0.75 0.8 0.83 0.85 0.88 0.9

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?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

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)

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.

Reasoning 17

3 14

Time (hours) Height (metres)

0 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. 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

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.

13.03 Bivariate numerical data mathspace.co

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13.04 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

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mathspace.co

Perfect positive correlation

Perfect negative correlation

Strong positive correlation

Strong negative correlation

Moderate positive correlation

Moderate negative correlation

Weak positive correlation

Weak negative correlation

No correlation

13.04 Associations between numerical variables mathspace.co

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13.04 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

Number of classes attended

100

100 Final exam mark

Final exam mark

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:

Number of classes attended Running distance (km) 20

Using the scatterplot, are these statements true or false? a The number of weeks that Sumon trained for the half 15 marathon is the independent variable. 10 b The y-coordinates of the points represent the time spent by Sumon training. c There is evidence to suggest that the longer Sumon 5 trains, the further they can run. No. of weeks d Training for a half marathon causes further running 0 5 10 15 20 distance over time. e The correlation between the number of weeks training and the number of kilometres Sumon is able to run is positive. 670

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Complete each sentence with one of the words from this list. • contributing (third) variables • correlation

• 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

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 0

5 x

x 0

5 x 0

x 0

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Describe the strength and direction for each linear association. a

5 x 0

5 x

x 0

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?

672

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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?

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?

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

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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)

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%

674

20 85

GDP per capita $5 000 $10 000 $3 000 $15 000

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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 a 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.

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:

Test result 20 23 62 97 100 66 35 52 87 99

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

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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

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 (%)

Temperature outside (°C)

x 0

Temperature outside (°C)

A snow sculpture company wants a chemical that will keep snow from melting.

676

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?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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. 2

1 % that got Q10 correct

% 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.

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13.05 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 0 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.

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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

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%.

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Does there appear to be a linear relationship between the variables? a

5 x 0

Looking at the scatterplot:

x 0

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

35 25 20 15 10 5

x 0 1

2 3 4 5 6

8 9

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Fluency Ex 1

5, 6a-c, 7, 9ac, 10−12

5, 6c-e, 8, 9bd, 10−12

This scatterplot shows the data for two variables, x and y.

5, 6d-f, 8−12

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 0

x 0

5 x 0

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 12

2 x

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.

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.

38 36 34 32 30 28 26 24 22 20

190 180 170 160 150 140 130 120 110 100 90

Room temperature

Number of people 10 20 30 40 50 60 70 80 90 Copies sold

Price 18 20 22 24 26 28 30 32 34 36

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

13.05 Line of good fit mathspace.co

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Ex 2

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 0

686

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

200 400 600 800 1000

Reasoning 13

13, 14

13−15

The heights (in centimetres) and the weights Weight (kg) (in kilograms) of 9 primary school children are shown on 60 the scatterplot: 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 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.

y 7

Frasier

6 5

4 Lorena said she noticed that there was a point that was far away, but decided to ignore it. Does Lorena’s 3 line meet the requirements for a line of good fit? 2 Explain.

Lorena

x 0 2 4 6 8 10 12 14 16 18

13.05 Line of good fit mathspace.co

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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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

2 3 4 5 6

8 9

11 10 9 8 7 6 5 4 3 2 1

x 0

688

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. a b

Construct a scatterplot and find a line of good fit for this data.

Year 2005 2006 2007 2008 2009 2010

Y 0 1 2 3 4 5

Hives lost to CCD, (H)% 18 13 23 26 28 30

More data has been collected, and the results are listed: • 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?

13.06 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

13.06 Interpolation and extrapolation mathspace.co

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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

690

Mathspace WA Curriculum – Year 10 Methods Pathway 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 0 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.

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

The answer is 210 chirps per minute.

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Mathspace WA Curriculum – Year 10 Methods Pathway 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 0 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.

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x 0 10 20 30 40 50 60 70 80 90

y = 0.45x + 7.1, at x = 3

y = 1.75x - 3.2, at x = -2.4 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

x 0 10 20 30 40 50 60 70 80 90

Find the y-value for the given least-squares regression line and x-value.

Fluency 4

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

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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.

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.

4 3 2 1 0

Number of hours of sleep 2 3 4 5

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

22 180

25 230

Sketch a line of good fit by eye.

Approximately, how many people are at the beach when the temperature is 20°C?

200

Approximately, what is the temperature when there are 200 people at the beach?

150

Estimate how many people are at the beach when the temperature is 19°C?

100

c d e

Is the prediction for 30°C or 19°C more reliable?

225

People

175 125 75 50 25

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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

2580 2560 2540 2520 2500 2480 2460 2440 2420 2400

Pressure (Pa)

Temperature (K) 296 300 304 308 312 316

iii

0 ≤ Temp ≤ 320

280 ≤ Temp ≤ 340

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

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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 9000

Predict the altitude of the plane 100 seconds into the descent.

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?

How many seconds did the plane take to descend to the ground?

1700 593

8000 7000 6000 5000 4000 3000 2000 1000 t 0 200 400 600 800 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 the 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 0

1000 2000 3000 4000 5000 6000 7000 8000

Find the value of c, the vertical intercept of the line. c

Use the line of good ﬁt to estimate the number of sales next month if $4800 is to be spent on advertising.

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

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Reasoning 14

15−17

15, 17

14, 15, 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. a b

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.

x 0 1 2 3 4 5 6 7 8 9 10 11 12

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.

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)

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.

Test score

80 70 60 50 40 30 20 10

Hours of study 0 2 4 6 8 10 12 14 16 18

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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)

14 12 10 8 6 4 2

00 00 00 00 00 00 00 10 20 30 40 50 60 70 Size (square metres)

Discuss the limitations of using interpolation and extrapolation for making predictions in the context of climate change.

Problem-solving 19−22 18

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

x 0

19, 21−23

1000

2000

3000

4000

5000

6000

7000

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.

Is the predicted value in part (b) reliable? Explain your answer.

A 800 700 600 500 400 300 200 100

T 0

700

8000

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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 anti-littering campaign: Weeks (x) Kilograms of litter collected ( y)

1 3.9

2 3.7

3 3.5

4 3.1

5 3.1

6 2.9

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

The ﬂeet manager for the Australian Automotive Association wants to estimate how car maintenance costs, C (in hundreds of dollars), are related to the distance, K (in thousands of kilometres), driven each year. The data collected and the least-squares regression line are shown on a scatterplot. C 550 500 450 400 350 300 250 200 150 100 50 K 0

The equation of the least-squares regression line for the graph is C = 9.9K + 146.2

Interpret what the coeﬃcient of K represents in this context. b

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.

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.

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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. y 100 80 60 40 20 t 1

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? A

Extrapolation is less reliable since we should use a 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.

Based on the given scatterplot and line of good fit, are either interpolation or extrapolation reliable? Explain. 9

8 7 6 5 4 3 2 1

x 0 1

702

2 3

4 5 6

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8 9

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.

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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

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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.

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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?

710

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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:

IN CO IL

JP DE FR IT

Brazil: 613 347

France: 930 265

Japan: 2 169 716 Germany: 1 263 344

Mexico: 4 295 124 United Kingdom: 2 905 909

Canada: 14 043 658

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?

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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?

712

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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.

Consider the line graphs of the same set of sales ﬁgures: 9000

Sales ($)

8000 7000 6000 5000 4000 3000 2000 1000

Month 0

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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:

Month 5

Quarterly operating deficit 55

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.

50 $ (million)

45 40 35 0

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

714

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?

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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?

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

2018

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

Norway

Number of COVID-19 tests per million people

Source: Reddit May 2020 a

Explain why the graph and comment are misleading.

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Consider the shown 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

Investigation: Conduct a statistical investigation Investigate online

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mathspace.co

13 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

Book club 18 62 13 15 108

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

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?

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The 100% stacked column graph shows the results of a survey on beverage preferences: a

Which is the explanatory variable, beverage preference or preference on time of day?

Does the 100% stacked column graph indicate an association between the variables?

Beverage preferences 100% 75%

25% 0%

718

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 brass 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

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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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

x 0

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.

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

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For each scatterplot: i

Draw a line of good ﬁt by hand.

iii

Describe the correlation of the data set as positive or negative.

Write the equation of the line of good ﬁt.

ii b

16 14

10 8

6 4

x 0

x 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

720

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

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5 32.45

6 35.5

6.7 38.0

Is this interpolation or extrapolation?

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.

M (in dollars)

900 800 700 600 500 400 300 200 100

C (in days) 0

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

70%

67%

Prefers AlphaModel

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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 • (Optional) Solve monic and non-monic quadratic equations graphically and algebraically, including the use of the quadratic formula, factorising techniques and digital tools and verify the solution/s by substitution • (Optional) Factorise monic and non-monic quadratic expressions using techniques, such as completing the square, perfect squares, difference of squares and grouping in pairs for four-term expressions • (Optional) Graph monic and non-monic quadratics of the form y = ax2 + bx + c or y = a(x − p)2 + q and their transformations, manually and using digital tools. Identify and connect key graphical and algebraic features and make connections to the algebraic solution/s

14 Optional: Further quadratics Chapter outline 14.01 Optional: Solve by factorisation 14.02 Optional: Solve by completing the square Investigation: Use bisection to solve quadratics (Optional) Investigation: Derive the quadratic formula (Optional) 14.03 Optional: The quadratic formula 14.04 Optional: The discriminant 14.05 Optional: Further quadratic equations Chapter 14 review

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14.01 Optional: 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

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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 Evaluate the subtraction

14.01 Optional: Solve by factorisation mathspace.co

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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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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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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.

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

−4

−2

x −6

−4

−2

−4

−6

(x − 3) (x + 3) = 0

−6

−4

−2 −4 −6 −8

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 2

−2

(x + 1) (x − 2) = 0

x −6

−4

−2 −2 −4 −6

6a-d, 7a-h, 8a-d, 9a-d, 10a-c

Fluency 6

Ex 2

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

x + 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 + 40 = 0

x2 − 14x + 49 = 0

Solve: a 2x2 − 32x = 0 d 3x2 − 30x + 75 = 0 g −5x2 + 10x + 40 = 0

b e h

10x2 − 10x − 120 = 0 3x2 − 21x − 54 = 0 −3x2 − 3x + 396 = 0

c f

4x2 + 24x + 20 = 0 −2x2 + 30x − 112 = 0

Solve: a m2 = 14m

m2 = m + 20

x2 = 81

m2 − 27m = −182

h k

i l

d g j 10

6e-h, 7g-l, 8e-h, 9g-l, 10d-l, 11cd

Solve: a

Ex 1

6ef, 7d-h, 8c-f, 9d-g, 10a-f, 11ab

−3m (m + 2) = 0 2

3 (n + 8) = −18n 15 − 11b − 12b2 = 0

Solve: a x(x + 2) = 48 c 4x2 = 400 e 5x2 = −320 + 80x g −10x + x2 = 2 − 7x − x2 i ( y − 3)2 = 2y + 2 k 6x2 − 27x + 28 = −3x2 + 3x + 3

2y − 6y = 0 6k2 = −7 + 13k

b d f h j l

2y − 10y2 = 0 x2 = 24x − 144

x2 + 8x + 16 = 36 3x2 − 7x − 20 = 0 −5x2 = −53x + 72 −8x + x2 = −6 − x − x2 ( y + 1)2 = 4y + 4 (5x2 − 13x + 6) (2x2 − 13x + 20) = 0

14.01 Optional: Solve by factorisation mathspace.co

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Ex 3

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 −4

2 3 4 5 6 7

−2

−8 −6 −4 −2 −2

−10

x 2

−4

−12

732

−8

−6

−4

−14

−6

−16

−8

Reasoning

x 4

−2 −4

12, 15

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.

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

j 18

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)

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.

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 particle hits the ground.

14.01 Optional: Solve by factorisation mathspace.co

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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)

14.02 Optional: 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.

Interactive exploration

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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

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Add 72 − 72 Factorise the perfect square part Evaluate the constant

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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

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.

14.02 Optional: Solve by completing the square mathspace.co

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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

Solve: Collect like terms

Take the square roots of both sides

Evaluate

Subtract

Evaluate

Simplify using a common denominator

from both sides

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

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Ex 2

Ex 3

Solve: a

x2 + 18x + 32 = 0

x2 − 6x + 8 = 0

x2 − 2x − 32 = 0

6x2 + 48x + 24 = 0

x2 + 11x + 5 = 0

Solve: a

x − 9x + 8 = 0

2x2 − 12x − 32 = 0

Solve, leaving your answers in exact form: a

x2 + 24x + 5 = 0

x2 − 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.

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:

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Given

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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Problem-solving 14, 15 14

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 = 20x − x2 All distances are in metres. By completing the square:

Find the greatest 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. x

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

c is equal to 0

Investigation: Use bisection to solve quadratics (Optional) Investigate online

Investigation: Derive the quadratic formula (Optional) Investigate online

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14.02 Optional: Solve by completing the square mathspace.co

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14.03 Optional: 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:

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

14.03 Optional: The quadratic formula mathspace.co

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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 and

So the solutions are

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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and

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Fluency Ex 1

6a-f, 7a-e, 8a-d

x2 + 5x + 6 = 0 2

x2 − 5x + 6 = 0

2x + 6x − 8 = 0

4x2 − 10x + 4 = 0

2x2 + 7x + 3 = 0

4x2 − 17x − 15 = 0

5x + 7x − 6 = 0

5x2 − 13x − 6 = 0

3x2 − 11x − 20 = 0

5x2 + 21x − 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

7.1x + 5.3x − 1.5 = 0

1.8x2 + 5.2x − 2.3 = 0

2x2 = 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 −5x2 − 15x + 3 = 0

7x − 19x + 3 = 0

5x = (x − 5) (3x + 3)

3n2 = 2n2 − 2n + 7

Reasoning 9

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

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Given that it has two unique real 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 WA Curriculum – Year 10 Methods Pathway 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. b

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?

14.03 Optional: The quadratic formula mathspace.co

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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. is in the golden ratio if

a+b

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. 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 WA Curriculum – Year 10 Methods Pathway mathspace.co

14 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

x (2x − 16) = 0

A company’s revenue, R (in dollars), after t months is described by the equation: R = − (t − 12)2 + 36

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

Chapter 14 review mathspace.co

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For each graph: i

What are the solutions of the related quadratic equation?

Write the equation in factored form.

−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:

748

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.

(Optional) Solve by using the quadratic formula, leaving your answer in exact form: a

3x2 − 11x − 20 = 0

12 − 8m + 2m2 = m2 + 12m + 15

(Optional) 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?

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Big ideas Cubic equations can be represented graphically and solved through various methods, and their properties can be analysed using factored forms and roots.

Curriculum Outcomes (Optional) Solve cubic equations in the form ax3 = k or in factored form, algebraically or using digital tools

15 Optional: Cubic functions Chapter outline 15.01 Optional: Cubic function 15.02 Optional: Factored form of cubic functions 15.03 Optional: Cubic equations Chapter 15 review

752 760 768 773

15.01 Optional: Cubic function After this lesson, you will be able to… • graph cubic functions in the form y = ax3 and describe the effect of a on the graph’s shape. • identify the point of inflection and concavity changes in cubic graphs. • recognise the x-intercept at the origin and its significance in the graph of y = ax3.

Cubic Function A cubic function in the form y = ax3 is a type of polynomial function, where a is a non-zero constant known as the leading coefficient. This form is a simplified version of the general cubic function, and its graph exhibits specific characteristics based on the value of a. The coefficient a determines the steepness and direction of the curve. By exploring this form, we can understand the fundamental behaviour of cubic functions.

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All cubic graphs in the form y = ax3 can be obtained by dilations and reflections of the basic graph y = x3. These transformations depend on the value of a. Key Properties: • a dilates (or stretches) the graph by a factor of ∣a∣ from the x-axis. The larger the magnitude of a, the steeper the graph. • If a > 0, the graph is increasing, rising from the bottom left to the top right. • If a < 0, the graph is reflected across the x-axis and is decreasing, falling from the top left to the bottom right. • All cubic graphs in the form y = ax3 have apoint of inflection at the origin (0, 0). This is where the graph changes concavity. The point of inflection at (0, 0) is where the graph transitions from being concave up to concave down or vice versa. • a>0 • The left side of the graph (before the point of inflection) is concave down. • The right side of the graph (after the point of inflection) is concave up. • a<0 • The left side of the graph (before the point of inflection) is concave up. • The right side of the graph (after the point of inflection) is concave down.

752

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Example 1 Consider the graph of the function y = 2x3:

16 12 8 4 −5 −4 −3 −2 −1 −4

x 1 2 3 4 5

−8 −12 B

−16

a At which point is the curve concave up?

Create a strategy Determine where the graph is increasing at an increasing rate, forming a bowl shape, which indicates positive concavity.

Apply the idea y 15 12 9 6 3 A −5 −4 −3 −2 −1 −3 −6 −9 −12 B −15

x 1 2 3 4 5

For y = 2x3, the graph is increasing. It is concave up where x > 0, as the rate of increase accelerates. This corresponds to point C.

b Which is the point of inflection?

Create a strategy Identify the point where the concavity changes, which for y = ax3 is always at the origin.

Apply the idea The concavity changes at (0, 0), which is point A.

15.01 Optional: Cubic function mathspace.co

753

c At which point is the curve concave down?

Create a strategy Determine where the graph is increasing at a decreasing rate, forming an upside-down bowl shape, indicating negative concavity.

Apply the idea For y = 2x3, the graph is concave down where x < 0, as the rate of increase slows. This corresponds to point B.

Example 2 Consider the graph of the function:

7 y 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

x 1

a Determine whether the value of a is positive or negative.

Create a strategy Check the direction of the graph: if it increases (bottom left to top right), a is positive; if it decreases (top left to bottom right), a is negative.

Apply the idea The graph decreases from top left to bottom right, indicating that a is negative.

b For which value(s) of x is the graph concave up as an inequality?

Create a strategy Identify where the graph has a bowl shape, indicating positive concavity.

Apply the idea The graph has a bowl shape (concave up) to the left of the origin, where x < 0.

754

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15.01 Practice questions Understanding 1

Explorer

Adventurer

1–3

Trailblazer

1–3

True or false: a

The graph of the cubic function y = x3 has no turning points.

The graph of y = 2x3 is steeper than the graph of y = x3 for x > 0.

The graph of y = −x3 increases from left to right.

The graph of

has the same shape as y = x3 but is less steep.

For the cubic function

Consider the graph of y = x3: a

, is the graph increasing or decreasing from left to right?

As x becomes larger in the positive direction (i.e., x approaches inﬁnity), what happens to the corresponding y-values? As x becomes larger in the negative direction (i.e., x approaches negative inﬁnity), what happens to the corresponding y-values?

3 2 1

−4 −3 −2 −1 −1

−2 −3 −4

Fluency Ex 1

4–9

4–10

4–11

Consider the graph of the function y = −x3. For the points A, B, and C: a

At which point is the curve concave up?

Which is the point of inﬂection?

At which point is the curve concave down?

10 8 6 4 2

−2

−4 −6 −8 −10

756

Consider the cubic function

Determine whether the value of a is positive or negative.

What is the value of y when:

x = −3

iii

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x=4

−1 −2

x=3

Determine whether the statements are true of the graphs of a

One is a reﬂection of the other about the y-axis.

y increases more rapidly on

than on y = x3.

y increases more slowly on

For these equations: i

Complete the table of values: x y

−2 ⬚

−1 ⬚

0 ⬚

1 ⬚

2 ⬚

y = 2x3

Reasoning 12

than on y = x3.

is a horizontal shift of y = x3.

c d

and y = x3:

12, 13

Sketch the graph.

y = −3x3

12–14

12–15

Consider the graph of the cubic function y = 2x3. y 24 16 8 −3 −2

−1

x 1

−8 −16 −24

For what values of x is y ≥ 0? 13

For the cubic function y = ax3, describe the relationship between the sign of a and the direction of the graph.

Consider the cubic function y = ax3, where a ≠ 0. Suppose that when x = 2, the value of y is −8. Without calculating additional x-intercepts beyond the origin, determine how many x-intercepts the function has. Then, analyse the function’s behaviour for x = 0, x > 0, and x < 0.

Use the shape and symmetry of the graph to explain why the point of inﬂection for a cubic function in the form y = ax3, where a ≠ 0 is always at (0, 0).

758

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15.02 Optional: Factored form of cubic functions After this lesson, you will be able to… • identify x-intercepts and analyse graph behaviour using the factored form of cubic functions. • construct cubic equations in factored form given roots and additional conditions.

Factored form of cubic function The factored form of a cubic function is given by: y = a(x − x1) (x − x2) (x − x3) It is useful for identifying the x-intercepts of the function, much like the factored form for quadratic functions. The null factor law can be used or the x-intercepts can be read directly as x1, x2 and x3.

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• A dilation by a factor of |a|. The larger the magnitude of a the steeper the graph. • If a > 0, the graph is mostly increasing (with a positive slope from bottom left to top right) • If a < 0, the graph is mostly decreasing (with a negative slope from top left to bottom right) • For three distinct roots, there are three x-intercepts at: (x1, 0), (x2, 0) and (x3, 0) • For two distinct roots, that is a graph of the form y = a(x − x1)2 (x − x2), there are two x-intercepts: a touch at (x1, 0) and a cut at (x2, 0) • For one distinct root, that is graphs of the form y = a(x − x1)3, there is a single x-intercept and horizontal point of inflection at (x1, 0). This is the same as translated form with k = 0. An interesting fact is that the point of inflection will occur at the average of the roots. This is always true when the roots are real.

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Example 1 Consider the curve y = (x + 4) (x + 2) (x − 1): a Calculate the x-value(s) of the x-intercept(s).

Create a strategy Substitute y = 0 into the equation and then use the null factor law to solve for x.

Apply the idea 0 = (x + 4) (x + 2) (x − 1)

Substitute y = 0

0 = x + 4, x + 2, x − 1

Equate each factor to zero

x = −4, −2, 1 Solve for x So, the x-values of the x-intercepts are −4, −2 and 1.

Reflect and check The point of inflection occurs at the average of the roots. In this case:

Determine the average of the roots Evaluate the operations inside the brackets Substitute

Evaluate the operations inside the brackets Evaluate .

So, the coordinates of the point of inflection is

b Identify the y-value of the y-intercept.

Create a strategy Substitute x = 0 into the equation.

Apply the idea y = (x + 4) (x + 2) (x - 1) = (0 + 4) (0 + 2) (0 − 1)

Write the equation Substitute x = 0

= (4) (2) (-1) Evaluate the operations inside the brackets = −8 Evaluate

15.02 Optional: Factored form of cubic functions mathspace.co

761

Sketch the graph of f (x) = (x − 2)2(x − 4) and g (x) = −(x − 2)2(x − 4) on the same Cartesian plane.

Determine the equation of each cubic function, given the graph of the function: a

Cuts through the x-axis at (−3, 0), (1, 0) and (2, 0) and has a y-intercept of (0, 18)

Cuts through the x-axis at (−1, 0), touches at (5, 0) and has a y-intercept of (0, 2)

Reasoning

15–17

15–18

Consider the equation y = (x − 2) (x + 1) (x − 3). Find the roots and explain how you can verify them graphically without plotting the full graph.

Two cubic equations are given: y = 3x(x + 3)2 and y = (x − 2) (x − 3) (x + 1). Solve both equations and determine which has more distinct real roots. Explain why the number of real roots differs between the two.

Consider the cubic function y = (x + 2) (x − 1) (x − 4). Explain why the graph must have exactly three x-intercepts and how this relates to the number of real roots.

For the cubic function y = a(x + 3) (x − 2)2, explain why the graph touches the x-axis at x = 2 but crosses it at x = −3, regardless of the value of a (assuming a ≠ 0).

Problem-solving 19, 20 19

b c d

Determine the value of k.

Find the x-intercepts of the function.

Consider the equation y = (x − 4)3: a

19–22

A cubic function is given as y = (x + 1) (x − 3) (x − k), where k is a constant. The y-intercept of the graph is at (0, 6). a

19–21

Calculate the y-values for the given x-values to complete the points: A(3, ⬚), B(2, ⬚), C(5, ⬚), D(4, ⬚), E(6, ⬚) Plot the points on the Cartesian plane.

On the same axes, sketch the curve that results from the entire set of solutions for the equation being graphed. Consider x = 6.17. According to the points on the graph, between which two integer values should the corresponding y-value lie?

A box without a top cover is to be constructed from a rectangular cardboard, measuring 6 cm by 10 cm by cutting out four square corners of length x cm. Let V represent the volume of the box: a

Express the volume V of the box in terms of x, writing the equation in factorised form.

For what range of values of x is the volume function defined?

Sketch the graph of the volume function.

Determine the volume when the height equals the shorter base side length.

Given the equation (x − 2) (Ax2 + Bx + C) = x3 + 2x2 − 5x − 6, determine all the roots of the cubic polynomial x3 + 2x2 − 5x − 6.

15.02 Optional: Factored form of cubic functions mathspace.co

767

15.03 Optional: Cubic equations After this lesson, you will be able to… • solve cubic equations in the form ax3 = k using algebraic manipulation, including inverse operations like cube roots. • factorise cubic equations and apply the null factor law to find roots. • use technology to solve cubic equations graphically or algebraicall

Cubic equations Recall, we have several methods to solve cubic equations:

a (x − x1) (x − x2) (x − x3) = 0

a the leading coefficient x1, x2, x3 the roots or x-intercepts

• Solve using algebraic manipulation - for cubics such as 2x3 = −16 or 2(x − 2)3 = 0. • Factorise - Fully factorise the cubic and use the null factor law to solve. If A × B × C = 0, then either A = 0, B = 0 or C = 0. When factorising a cubic, recall we can: • use the highest common factor • factor special forms like sum and difference of cubes • identify a single factor, then use division to find the remaining quadratic • Technology - Use technology to solve graphically or algebraically after forming the equation. Solutions to cubic equations are called roots, zeros or solutions, and graphically, they are the x-intercepts of the curve.

Example 1 Solve these equations: a x3 = −125

Create a strategy

Apply the idea

Take the cube root of both sides.

768

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Take the cube root of both sides x = −5 Evaluate

b (x − 3)3 = 64

Create a strategy

Apply the idea

Isolate x using inverse operations.

Take the cube root of both sides Evaluate Add 3 to both sides Evaluate

Example 2 Solve for the roots of the cubic (x + 2)(x − 3)(x − 6) = 0.

Create a strategy Use the null factor law to solve for x.

Apply the idea (x + 2)(x − 3)(x − 6) = 0 x + 2 = 0, x − 3 = 0, x − 6 = 0

x = −2, 3, 6

Write the equation Use the null factor law Solve for x

Example 3 Use technology to solve: x3 = 20

Create a strategy

Apply the idea

Use the solve command on a graphics calculator.

The result gives x ≈ 2.71.

Reflect and check The solutions to the equation are the same as identifying the x-intercepts or roots of the graph of x3 − 20 = 0. An alternative to using the calculator is to graph y = x3 − 20 and identify the x-intercepts.

y 20 10 x 1

10 20

15.03 Optional: Cubic equations mathspace.co

769

6 Ex 1b

7 Ex 2

Solve the cubic equations: a

(x + 1)3 = 27

(x − 3)(x + 1)(x − 4) = 0

(x + 6)(x − 5)(x + 2) = 0

(x − 7)(x + 4)(x) = 0

3(x + 5)(x − 1)(x − 2) = 0

4(x − 1)2(x + 3) = 0

Solve the cubic equations: (x)2(x + 2) = 0 2

(x + 6) (x − 2) = 0

(x − 4)(x + 7)2 = 0

(x − 8)2(x + 5) = 0

(x + 10)(x − 3)2 = 0

(2x + 5) (5x − 1)2 = 0

Solve the cubic equations: a

(x + 7)(x + 3)(x − 1) = 0

(4x − 3)(x + 6)(x − 5) = 0

(3x + 1)(x − 8)(x + 4) = 0

(2x − 5)(3x + 2)(x − 7) = 0

Ex 3

Use technology to solve each cubic equation: a

(x − 6)(x + 2)(x − 3) = 0

(x + 5)(x − 4)(x + 1) = 0

(x − 8)(x + 3)(x − 2) = 0

(x + 7)(x − 6)(x + 4) = 0

Use technology to solve each cubic equation, expressing answers in exact form: a

(x + 3)(x − 5)(x + 2) = 0

(x − 7)(x + 1)(x − 4) = 0

(x + 6)(x − 3)(x + 5) = 0

(x − 2)(x + 8)(x − 1) = 0

e 12

2 (x + 3)3 = 16

Solve the cubic equations:

3 (x − 2)3 = 81

Use technology to solve for the roots of the cubic, expressing answers in exact form: a

Reasoning

13, 14

13–15

13–16

Explain why the cubic equation (x − 4)(x + 1)(x − 6) = 0 has exactly 3 real roots, and verify this using its factored form.

Determine whether each statement about cubic equations is always, sometimes, or never true. Justify your answer. a

A cubic equation has exactly 3 real roots.

The graph of a cubic equation intersects the x-axis at least 1 time.

15.03 Optional: Cubic equations mathspace.co

771

A cubic equation has roots 2, 2 and 5. a

Write the equation in its factored form.

Explain why the equation only has two distinct real roots, even though it is cubic.

Neil claims that the equation 5x3 = 40 might have three different real solutions. Is Neil correct? Explain your reasoning.

Problem-solving 17, 18ab 17

17, 18c-e, 19, 20

17, 18c-f, 19–21

One of the solutions of the equation (ax − 3) (x + 5)(x − 8) = 0 is x = 6. Solve for the value of a.

Graph these cubic equations using technology, and then determine the roots. Round your answer to 2 decimal places where necessary. a

There are just 2 solutions to the equation (ax − 2)(x + 4)(cx + 7) = 0, and they are x = 5 and x = 1. What are the possible values for a and c?

A biologist is studying the growth of a particular type of plant and ﬁnds that the height of the plant h, in centimetres, can be modelled by the equation: h = 5 (t−3) (t−2) (t−1) + 10 where t is the time in weeks since the plant started growing. At what times is the height of the plant exactly 10 centimetres?

A cube has side length 7 cm and a mass of 1715 g. The mass of the cube m is directly proportional to the cube of its side length x: a

Let k be the constant of proportionality for the relationship between the side length x, and the mass m of a cube. Solve for k.

Determine the equation relating the mass and side length of a cube.

Complete the table of values for m = 5x3: x m

772

1 ⬚

2 ⬚

3 ⬚

4 ⬚

Sketch the curve of m = 5x3.

From the equation, calculate the mass of a cube with side 8.5 cm, rounded to the nearest gram.

A cube has a mass of 1920 g. From the graph in part (d), determine what whole number value its side length is closest to.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

15 Chapter review 1

Consider the graph of the function y = −3x3. a

Determine whether the value of a is positive or negative.

What is the value of y when: x = −2

For the function

i 2

x=5

For what values of x is the cubic concave up?

For what values of x is the cubic concave down?

Complete the table of values. x y

iii

−3 ⬚

−1 ⬚

0 ⬚

2 ⬚

x=1

4 ⬚

Consider the graph of a cubic function:

Is the value of a positive or negative?

For which values of x is the graph concave up? Identify the cubic equation of the graph in the form y = ax3.

1 −3 −2

−1

−1 −2 −3

Consider the graph of the function y = 4x3.

For what values of x is the graph concave up?

30 20 10 −2

−1

x 1

−10 −20 −30

Identify whether each function has exactly two x-intercepts: a

y = (x − 1)3

y = (x − 4)2(x + 3)

y = (x +2)(x − 5)(x + 7)

y = x(x + 1)(x − 6)

Chapter 15 review mathspace.co

773

Consider the function y = (x − 3)3: a

Complete the table of values: x y

1 ⬚

2 ⬚

−5 −4 −3 −2 −1 1 2 3 4 5 −5 −10 −15 (0, −15) −20 −25 −30 −35 −40 −45 −50

x 3

−1 ⬚

0 ⬚

1 ⬚

y = (x + 2)(x − 1)2

Sketch the graph.

3 ⬚

y = (x − 3)(x + 1)(x − 5)

3x3 = 24

8x3 = 27

2(x − 1)3 = 54

4(x + 4)3 = 32

2 ⬚

x3 = 125

Solve the cubic equations: a

774

Solve the cubic equations: a

Complete the table of values: x y

35 30 (0, 30) 25 20 15 10 5

For each cubic function: i

5 ⬚

−3 −2 −1 −5 −10 −15 −20

4 ⬚

Sketch the graph.

For each graph of cubic functions, determine the equation of the graph in factored form: a

3 ⬚

(x + 2)3 = 64

Use technology to solve each cubic equation: a

(x − 5)(x + 3)(x − 1) = 0

(x + 4)(x − 7)(x + 2) = 0

(x − 3)(x + 6)(x − 4) = 0

(x + 8)(x − 2)(x + 5) = 0

Solve the cubic equations: a

(x + 5)(x − 2)(x + 1) = 0

(2x − 1)(x + 4)(x − 3) = 0

(3x + 2)(x − 5)(x + 7) = 0

(4x − 3)(x + 6)(x − 1) = 0

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

64x3 = −1

Big ideas The equations representing circles provide a systematic way to understand their geometry and position in a Cartesian plane.

Curriculum Outcomes (Optional) Explore, describe and interpret circles and rectangular hyperbolas where the asymptotes are parallel to the axes, and their transformations, using digital tools and make connections between the algebraic and graphical representations

16 Optional: Circles Chapter outline Investigation: Circles (Optional) 16.01 Optional: Circles at origin 16.02 Optional: Translations of circles Investigation: Equation of any circle (Optional) 16.03 Optional: Transform circle equations Chapter 16 review

778 786 797 807

Investigation: Circles (Optional)

Investigate online

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16.01 Optional: 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

778

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

16.01 Practice questions Understanding 1

Explorer

Adventurer

1–3

Trailblazer 2, 3

1–3

State the radius of each circle: a

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

10 8 6 4 2

d 16

8 4

x 3

−16 −12 −8 −4 −4

9 12

−8

−6

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

−9

−12

−16

Use the graph of the circle to complete the statement: 8

6 4 2 −8 −6 −4 −2 −2

x 2

−4 −6 −8

Every point on the circle is exactly ⬚ units away from the point (⬚, ⬚).

782

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x 4

8 12 16

Ex 2

Ex 3

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.

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.

784

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

16.02 Optional: 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

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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 x2 + y2 = 1

x2 + y2 = 1

1 x

−2

−1

−2

−3

The circle with equation x2 + y2 = 1 is translated up by 2.

786

x −2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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 −3 −2

(x − 2)2 + y2 = 1

−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

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).

16.02 Optional: Translations of circles mathspace.co

787

b Find the radius of the circle.

Create a strategy

Apply the idea 2

Equations of the form (x − h) + ( y − k) = r 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

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

−2 −3 −4 −5 −6 −7

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.

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.

788

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

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

16.02 Optional: Translations of circles mathspace.co

789

Find the centre and radius of each circle: a

1 −6 −5 −4 −3 −2 −1 −1

3 2

−2

−3

−4

−4 −3 −2 −1 −1

−5

−2

−6

−3

2 1 x −3 −2

−1

−1 −2 −3

8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2

−4

x 1

What are the coordinates of the centre?

What is the length of the diameter?

x 1 2 3 4 5

6 5 4 3 2 1 −6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6

Fluency

Using the graph of the circle: a

x 1 2 3 4 5 6

4ab, 5a-d, 6ab, 7, 8, 9a-c, 4bc, 5b-e, 6bc, 7, 8, 9b-d, 4cd, 5ef, 6cd, 7–9, 10ab 10bc 10cd

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

16.02 Optional: Translations of circles mathspace.co

791

Ex 1

For each equation: i

State the centre of the circle.

iii

Sketch the graph of the circle.

(x − 2)2 + ( y + 2)2 = 36

(x + 1) + ( y − 3) = 16

e 6

Ex 2

State the radius of the circle.

(x − 2)2 + ( y + 2)2 = 16

(x + 3)2 + ( y + 4)2 = 25

(x − 6)2 + y2 = 49

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

9 8 7 6 5 4 3 2 1

x 1 2 3 4 5 6 7

−5 −4 −3 −2 −1 −−1

y 3 2 1

−3 −2 −1 −1 −2 −3 −4 −5 −6 −7

792

x 1 2 3 4 5 6 7

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

11 10 9 8 7 6 5 4 3 2 1 −1 −1

x 1 2 3 4 5

x 1 2 3 4 5 6 7 8 9 10 11

Ex 3

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 the transformations required to get from the graph of the unit circle x2 + y2 = 1 to each of the circles shown. a

12 10

2 x

6 −4

4 2 −12 −10 −8 −6 −4 −2 −2

−2

1 −1

10 9 8 7 6 5 4 3 2 1

x 1

−2 −3 −4 −5 −6

−7 −6 −5 −4 −3 −2 −1 −1

−7

Reasoning 11

−6

y −3 −2 −1

−4

2 −2

x 2

11, 13

x 1 2 3 4

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.

16.02 Optional: Translations of circles mathspace.co

793

A circle has the equation:

y 6 4 2 −2 −2

x 2

10 12

−4 −6 −8

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. lie inside or outside the circle?

Would the circle with the equation 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

794

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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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?

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.

Centre Circle 75 m

9.15 m

99 m

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?

16.02 Optional: Translations of circles mathspace.co

795

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 (Optional)

Investigate online

796

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

mathspace.co

16.03 Optional: Transform circle equations After this lesson, you will be able to… • apply the distance formula to find the radius of a circle when given two points. • rewrite a given equation of a circle in its standard form using the process of completing the square.

The circle equation and distance formula Recall that 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.

Exploration Consider the circle with a radius of 13 units: 1. Find the distance from the point (4, 8) to the centre of the circle.

y 10 5 x −10 −5 −5

15 20

(9, −4)

−10 −15

2. Find the distance from any point, (x, y), on the circle to the centre of the circle. 3. Rewrite your equation to represent any circle with a centre at (h, k). 4. How does your equation relate to the equation of a circle?

−20

Recall the distance formula is

If the ﬁrst point (h, k) is the centre of a circle, the second point (x, y) is a point on the circle, then the distance between the two points is the radius, r.

Squaring both sides relates the distance between the points back to the equation of a circle. (x − h)2 + ( y − k)2 = r2

16.03 Optional: Transform circle equations mathspace.co

797

Example 1 The centre of a circle is (−1, −27) and the point (13, 21) is on the circle. a What is 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 the distance formula: distance

Apply the idea Substitute the coordinates

Evaluate the subtraction

Evaluate the indices

Evaluate the square root

b Write the equation of the circle in the form (x − h)2 + ( y − k)2 = r2.

Create a strategy Substitute the centre of a circle (−1, −27) and the value of radius found from part (a).

Apply the idea (x − (−1))2 + ( y − (−27))2 = 502 2

(x + 1) + ( y + 27) = 2500

Substitute the coordinates and the value of radius Evaluate the subtraction and index

Example 2 The endpoints of the diameter of a circle are (−5, −3) and (−15, −15). a What are the coordinates of the centre of the circle?

Create a strategy The centre of the circle is the midpoint of the diameter. Use the midpoint formula:

Apply the idea We have x1 = −5, y1 = −3, x2 = −15 and y2 = −15. Substitute x1, y1, x2, y2

Evaluate the numerators

Simplify

The centre of the circle is (−10, −9).

798

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b What is the exact value of the radius of the circle?

Create a strategy Use the distance formula with the centre and one of the endpoints of the diameter: distance

Apply the idea We can use a point on the circle (−5, −3) and the centre (−10, −9) in part (a). Substitute x2, x1, y2, y1

Evaluate the subtraction

Evaluate the indices

Evaluate the addition

Reﬂect and check We can also ﬁnd the radius by calculating half of the distance between the two endpoints of the diameter so we have x1 = −5, y1 = −3, x2 = −15 and y2 = −15. By using the distance formula, we have:

Substitute x2, x1, y2, y1

Evaluate the numerator

Simplify the numerator

Evaluate

c Find the equation of the circle which has (−5, −3) and (−15, −15) as endpoints of its diameter.

Create a strategy Remember the equation (x − h)2 + ( y − k)2 = r2 represents a circle, where (h, k) are the coordinates of the circle’s centre and r is its radius.

Apply the idea In part (a), we found the centre to be at (−10, −9). In part (b), we determined the radius to be

Write the equation

Substitute h = −10, k = −9, r =

Evaluate

16.03 Optional: Transform circle equations mathspace.co

799

Example 3 Consider the equation of a circle given by x2 + 4x + y2 + 6y − 3 = 0. a Rewrite the equation of the circle in the form (x − h)2 + ( y − k)2 = r2.

Create a strategy Complete the square on both the x-terms and the y-terms.

Apply the idea x2 + 4x + y2 + 6y − 3 = 0 2

Start with the equation

x + 4x + y + 6y = 3 2

Add 3 to both sides 2

x + 4x + 2 + y + 6y + 3 = 3 + 2 + 3

(x + 2)2 + ( y + 3)2 = 3 + 22 + 32 2

(x + 2) + ( y + 3) = 16

Add 22 and 32 to both sides Write in factorised form Evaluate

b What are the coordinates of the centre of this circle?

Create a strategy Equations of the form (x − h)2 + ( y − k)2 = r2 are circles with centre at (h, k).

Apply the idea In part (a), we have the equation (x + 2)2 + ( y + 3)2 = 16 which can be written as (x − (−2))2 + ( y − (−3))2 = 42 and it is now of the form (x − h)2 + ( y − k)2 = r2. So we have h = −2 and k = −3. The centre of the circle is (−2, −3).

c What is the radius of this circle?

Create a strategy Equations of the form (x − h)2 + ( y − k)2 = r2 are circles with radius of r.

Apply the idea In part (b) we rewrote the equation as (x + 2)2 + ( y +3)2 = 42. So we have r = 4, and therefore the radius of the circle is 4 units.

16.03 Optional: Transform circle equations mathspace.co

801

Example 4 A circle has its centre at (3, −2) and a radius of 4 units. a Plot the graph of the circle.

Create a strategy Plot the centre of the circle. Count 4 units in each direction from the centre: left, right, up, and down and draw a round curve passing through each these points.

Apply the idea The centre of the circle is at the point (3, −2). So we first plot this point (3, −2).

4 3 2 1

Now we count 4 units in all four directions to get the points (3, 2), (7, −2), (3, −6) and (−1, −2).

Connect the points using a round curve passing through the four points to draw the circle.

1 2 3 4 5 6 7 8

−2 −1 −1 −2 −3 −4 −5 −6

b Write the equation of the circle in the form x2 + y2 + ax + bx + c = 0.

Create a strategy Use the form (x − h)2 + ( y − k)2 = r2 and expand each perfect square.

Apply the idea We are given that the centre of the circle is (3, −2) and the radius is 4. So h = 3, k = −2 and r = 4. (x − h)2 + ( y − k)2 = r2 2

(x − 3) + ( y − (−2)) = 4 2

(x − 3) + ( y + 2) = 16 2

Expand the perfect squares Combine like terms

x − 6x + y + 4y + 13 − 16 = 16 – 16 2

Subtract 16 from both sides

x − 6x + y + 4y − 3 = 0

Evaluate

Write in general form

x + y − 6x + 4y – 3 = 0

802

Simplify

x − 6x + y + 4y + 13 = 16 2

Substitute h, k, and r

x − 6x + 9 + y + 4y + 4 = 16 2

Write the formula

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Fluency 6

Ex 1

Ex 2

Ex 3

Ex 4

804

6ab, 7a-c, 8a, 9ab, 10, 11abc, 12a, 14ac

6bc, 7b-d, 8a, 9bc, 10, 11a-d, 12ab, 14bc

6cd, 7d-f, 8ab, 9cd, 10, 11e-h, 12cd, 13, 14d-f

Using the centre and another point for each circle: i

Find the radius of the circle.

Find the equation of the circle in the form (x − h)2 + ( y − k)2 = r2.

Centre at (2, 4), passing through the point (3, 3)

Centre at (−2, 4), passing through the point (3, 5)

Centre at (−2, −5), passing through the point (−1, 0)

Centre at (−2, 5), passing through the point (3, −7)

Using the centre and another point for each circle: i

Find the radius of the circle.

Find the equation of the circle in the form x2 + y2 + ax + by + c = 0.

Centre at (1, 2), passing through the point (4, 2)

Centre at (−4, −3), passing through the point (−4, 1)

Centre at (5, 7), passing through the point (1, 10)

Centre at (1, 2), passing through the point (0, 1)

Centre at (3, −1), passing through the point (2, −5)

Centre at (−2, 3), passing through the point (−4, 4)

For each circle: i

Find the coordinates of the centre.

Find the exact value of the radius.

iii

Write the equation of the circle in the form (x − h)2 + ( y − k)2 = r2.

A circle with endpoints of the diameter at (10, 6) and (−12, −14).

A circle with the smallest radius that contains the points (1, 2) and (−5, 4) within or on its boundary.

For each equation: i

Rewrite the equation of the circle in the form (x − h)2 + ( y − k)2 = r2.

Find the coordinates of the centre of the circle.

iii

Find the radius of the circle.

x2 + 4x + y2 + 10y − 7 = 0

x2 + y2 − 14x − 10y + 73 = 0

x2 + y2 + 18x − 16y + 141 = 0

A circle has its centre at (4, −6) and a radius of 3 units. a

Plot the graph of the circle.

Write the equation of the circle in the form x2 + y2 + ax + by + c = 0.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

For each equation: i

Rewrite the equation in the form (x − h)2 + ( y − k)2 = r2 by completing the square.

Sketch the graph of this circle.

x2 + y2 − 12x + 2y + 1 = 0

x2 + y2 − 6x + 2y + 1 = 0

x + y + 4x − 14y = 11

x2 + y2 − 12x − 18y = −81

y2 + 2y + 3 = 10x − x2 + 58

y2 + 6y + 6 = −4x − x2 + 42

x2 + 3x + y2 + 6x − 10 = 0

x2 + 7x + y2 +

+1=0

For each equation: i

Rewrite the equation in the form (x − h)2 + ( y − k)2 = r2 by completing the square.

Sketch the graph of this circle.

2x2 + 2y2 + 8x − 12y + 20 = 0

2x2 + 2y2 − 8x + 4y + 5 = 0

4x2 + 4y2 + 12x − 36y + 41 = 0

6x2 + 6y2 − 36x + 24y + 54 = 0

For each equation: i

Rewrite the equation in the form (x − h)2 + ( y − k)2 = r2 by completing the square.

Sketch the graph of this circle.

Describe how each circle has been transformed from the graph of x2 + y2 = 1: a

x2 + y2 − 8y = 0

x2 + y2 − 10x − 75 = 0

x2 + y2 + 4x + 8y + 11 = 0

x2 + y2 − 8x + 14y + 29 = 0

x2 + y2 + 1.5x − 0.4y − 0.3975 = 0

Reasoning

15–17

16–18

Use the distance formula to explain why the equation (x − a)2 + ( y − b)2 = r2 represents a circle with its centre at (a, b) and radius r.

A circle has a diameter with endpoints A (−1, 1) and B (9, −7). a

Determine the equation of the circle.

Show that the point P (8, 3) is outside of the circle.

A circle with centre (−7, 4) passes through the point (1, 4). Explain how to ﬁnd the radius without using the distance formula.

Is it possible to ﬁnd the centre and radius of a circle in the form x2 + y2 + ax + by + c = 0 without completing the square? Explain your answer.

16.03 Optional: Transform circle equations mathspace.co

805

Problem-solving 19ab, 21ab 19

19cd, 20ab, 21ab

19d, 20cd, 21cd, 22–23

For each circle: i

Find the radius.

Find the circumference.

iii

Find the area.

x2 + y2 + 10x + 8y + 4 = 0

x2 + y2 − 10x + 8y + 40 = 0

x2 + y2 − 8x + 6y − 4 = 0

x2 + y2 + 4x − 10y − 20 = 0

For each circle: i

Find the radius.

Find the circumference.

iii

Find the area.

2x2 + 2y2 − 8x + 4y − 4 = 0

3x2 + 3y2 + 12x − 6y − 9 = 0

4x2 + 4y2 − 16x + 24y + 4 = 0

5x2 + 5y2 + 20x − 30y − 20 = 0

Does the point lie inside, outside, or on the given circle? a

Point: (0, 0) 2

Point: (−1, −4)

Equation: x + y − 2x + 4y + 3 = 0 Equation: x2 + y2 − 2x + 6y + 5 = 0 c

Point: Equation:

Point:

Equation:

Darnell shines a torch at a wall which lights up a circular region with a diameter of 4 metres. The centre of the light is positioned 3 metres above the ground, and 5 metres horizontally from the left side of the wall. a

Let the bottom left corner of the wall be the origin. Determine the equation of the circle which describes the edge of the lighted area in the form x2 + y2 + ax + by + c = 0.

Yvonne has a height of 1.66 metres and is standing against the wall, 5 metres from the left side. Will any part of Yvonne be in the lighted area?

Kayoko is standing against the wall, 6 metres from the left side of the wall. Determine the greatest height that Kayoko can be without being in the lighted area. Round your answer to the nearest centimetre.

A circle passes through the points (0,0), (4,0) and (0,7). Find the equation of the circle and state the radius, and centre.

806

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

16 Chapter review 1

(Optional) 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.

(Optional) 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

(Optional) 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)

(Optional) 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)

(Optional) 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.

(Optional) Write the equation of the circle given its centre and radius: a

Centre: (4, 9) ; r = 1

Centre:

State the radius of the circle.

x2 + (y − 4)2 = 49

(Optional) For each equation: i

State the centre of the circle.

iii

Sketch the graph of the circle.

(x − 3)2 + (y + 1)2 = 25

Chapter 16 review mathspace.co

807

(Optional) 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

2 1 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8

x 1

y x 1 2 3 4 5 6 7 8

(Optional) Find the equation after the graph of x2 + y2 = 1 is: a

Translated left 1 unit and down 7 units.

Translated right 5 units.

(Optional) 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?

(Optional) Using the given centre and another point for each circle: i

Find the exact radius of the circle.

Find the equation of the circle in the form x2 + y2 + ax + by + c = 0.

Centre at (−4, 4), passing through the point (1, 9)

Centre at (8, 10), passing through the point (6, 7)

(Optional) The endpoints of the diameter of a circle are (−9, 9) and (−1, 3). a

Find the coordinates of the centre of the circle.

Find the radius of the circle.

Write the equation of the circle in the form (x − h)2 + ( y − k)2 = r2.

(Optional) For each equation: i

Rewrite the equation in the form (x − h)2 + ( y − k)2 = r2.

Find the coordinates of the centre of the circle.

iii

Find the radius of the circle.

x2 + y2 + 14x + 24 = 0

3x2 + 3y2 − 12x + 6y − 33 = 0

(Optional) Describe how each of the circles has been transformed from the graph of x2 + y2 = 1. a

808

x2 + y2 + 16x − 6y + 24 = 0

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x2 + y2 − 0.5x − 1.5y − 0.375 = 0

Big ideas • The symmetry of the unit circle helps to define the properties of trigonometric functions. • Trigonometric functions are periodic functions that repeat over a constant interval. • Equations involving trigonometric functions require solving for the angles that satisfy a given trigonometric expression.

Curriculum Outcomes • (Optional) Use the unit circle and dynamic geometry software to explore and represent trigonometric functions graphically • (Optional) Solve simple trigonometric equations graphically, algebraically or using the unit circle and verify solution/s by substitution

17 Optional: Further trigonometry Chapter outline 17.01 17.02 17.03 17.04

Optional: Angles with exact values Optional: The unit circle Optional: Graphs of sine and cosine Optional: Trigonometric equations Chapter 17 review

812 819 837 847

17.01 Optional: Angles with exact values After this lesson, you will be able to… • calculate the exact values of trigonometric ratios. • use the angles with exact values to solve problems.

Exact value triangles The figure shown is a right-angled isosceles triangle, with two equal sides of length 1 unit. Using Pythagoras’ theorem, we can work out that the hypotenuse is

45° 1

units long.

The angles in a triangle add up to 180° and the base angles in an isosceles triangle are equal, so we can also work out that these two base angles are 45°.

45° 1

We can then determine exact values for the following trigonometric ratios:

We can also find some exact values by starting with an equilateral triangle:

60° 2

60°

To ﬁnd the exact ratios of 30 and 60 degree angles, we need to start with an equilateral triangle with side lengths of 2 units. Remember all the angles in an equilateral triangle are 60°.

60° 2

30° 30°

812

60°

We then draw a line that divides the triangle in half, into two congruent right-angled triangles. The base line is cut into two 1 unit lengths, and the opposite angle is cut into two 30° angles.

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Now let’s focus on just one half of this triangle. 30° 2

We can calculate the length of the centre line to be using Pythagoras’ theorem. We can then use our trigonometric ratios to determine the exact values.

60° 1

Now, not every isosceles right-angled triangle has sides measuring 1, 1 and , but no matter how large or small it is, the two base angles will always be 45° angles and therefore the ratios of the sides will always be the same. This also applies to the triangle with 60° and 30° angles. Any pair of similar triangles have the same side ratios, and so any triangle with the angles we have just seen will have the same exact value trigonometric ratios.

Interactive exploration

Discover this concept in action online

mathspace.co

Notice that: • sin θ = cos(90° − θ ) • cos θ = sin(90° − θ ) These angles are called complementary angles.

Example 1 Use the exact value triangles in the diagram to answer the following: 30° 2

45°

60° 1

1 45° 1

a What is the exact value of cos 45°?

Create a strategy Use the triangle with an angle of 45° in it.

Apply the idea Using the triangle on the right, we can ﬁnd cos 45°:

Use the cosine ratio

Substitute the values

17.01 Optional: Angles with exact values mathspace.co

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b What is the exact value of cos 60°?

Create a strategy

Apply the idea

Use the triangle with an angle of 60° in it.

Using the triangle on the left, we can ﬁnd cos 60°: Use the cosine ratio

Substitute the values

Example 2 θ is an angle in a right-angled triangle where

. Find the value of θ.

Create a strategy Use the exact value triangle with a side length of

Apply the idea Using this exact value triangle we can see that So θ = 30°

30° 2

60° 1

Reflect and check We may also be given ratios in rationalised form. In this case, if we were told still get that θ = 30° because

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, we would

Find the exact value of x.

45°

Find the exact value of y. 4 2

60° y

3 3

Find m.

Find t.

6 10

30° t Ex 2

θ is an angle in a right-angled triangle. Find the value of θ : a

Find the value of θ.

and b

Find the value of θ.

Find the value of tan θ.

θ is an angle in a right-angled triangle where a

and b

Find the value of tan θ.

Find θ : a

8 3 θ 12

4 2 4

Evaluate the expressions, leaving your answer in exact and rationalised form: a

sin 45° cos 30°

sin 45° + cos 30°

sin 45° cos 60° + tan 45°

sin 90° cos 0° + sin 270° cos 180°

sin2 90° + cos2 90°

17.01 Optional: Angles with exact values mathspace.co

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Reasoning

17, 18

17–19

Use exact values to show that sin2 45° + cos2 45° = 1.

Use the triangle to show that sin 30° = cos 60°.

sin θ is equal to the cosine of which angle?

sin(90° − θ ) is equal to the cosine of which angle?

What is the relationship between tan(90° − θ ) and tan θ?

17, 19, 20

30° 2

60° 1

Use

and exact values to explain why the tangent function has an undeﬁned

value at 90° and 270°. 20

for any angle θ, except 90°, in a right-angled triangle.

Prove that

Problem-solving 21–23

22–24

23–26

A right-angled triangle has one angle of 30° and a hypotenuse of length 10. Find the exact lengths of the other two sides and the area of the triangle.

Find the exact value of the angle θ if

A right-angled triangle has one angle of 45° and an adjacent side of length exact lengths of the other two sides and the area of the triangle.

Find the exact value of the expression

A right-angled triangle has one angle of 60° and an opposite side of length exact lengths of the other two sides and the area of the triangle.

If sin

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and

, ﬁnd the exact value of cos θ and tan θ.

. . Find the

. Find the

17.02 Optional: The unit circle After this lesson, you will be able to… • divide the unit circle into four quadrants and describe the unique trigonometric properties of each. • identify and use reference angles to find trigonometric ratios in any quadrant. • recognise and apply the signs of trigonometric ratios in different quadrants. • apply the knowledge of the unit circle to solve problems.

The unit circle In right-angled triangles, we can use trigonometric ratios to describe the relationship between a given angle and two sides. The trigonometric ratios are:

Opposite

Hypotenuse

θ A

Adjacent

Exploration Two special triangles are each drawn in a circle with a radius of 1. 1

θ = 45°

θ = 30° −1

1 0

−1

1 45°

60° 1

−1

1. Calculate the lengths of the legs of each triangle, keeping your answer in fraction form. 2. Write the coordinates of the point where each hypotenuse meets the circle. 3. Write the coordinates of the point where each hypotenuse meets the circle in terms of sine and cosine.

17.02 Optional: The unit circle mathspace.co

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If we place a right-angled triangle on the Cartesian plane at the origin and make the hypotenuse have a length of 1, we can write trigonometric ratios in terms of x and y. The trigonometric ratios become:

P (x, y) 1

θ −1

−1

If we look at any right-angled triangle with a hypotenuse of 1, the endpoint of the hypotenuse will lie on a circle with radius 1, known as the unit circle. The radius of the unit circle, which creates the angle with the positive x-axis, is called the terminal arm. unit circle

terminal arm −1

P (cos θ ,sin θ )

θ is the angle that the terminal arm makes with the positive x-axis when rotating anticlockwise.

Since cos(θ ) = x and sin(θ ) = y, the coordinates of the endpoint can be represented by

θ x

P(cos θ, sin θ )

The ratio , which is the gradient of the hypotenuse, can be represented by −1

This will be true in all quadrants.

Example 1 y

The graph shows an angle a with its terminal side intersecting the circle at

1 a −1 −1

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x 1

Interactive exploration

Discover this concept in action online

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We know that sin θ is equal to the y-coordinate (height), cos θ is equal to the x-coordinate (length), and tan θ is the gradient of the line from the origin to the point. y (cosθ , sinθ )

cosθ tanθ

sinθ

−1

Second quadrant: • y is positive: sin θ is positive. • x is negative: cos θ is negative. • Gradient is negative: tan θ is negative.

−1 y

θ −1

1 x

sinθ tanθ cosθ

(cosθ , sinθ )

−1 y

θ −1

sinθ tanθ cosθ −1

824

Third quadrant: • y is negative: sin θ is negative. • x is negative: cos θ is negative. • Gradient is positive: tan θ is positive.

Fourth quadrant: • y is negative: sin θ is negative. • x is positive: cos θ is positive. • Gradient is negative: tan θ is negative.

(cosθ , sinθ )

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S sin positive

A all positive (cosθ , sinθ )

θ −1

T tan positive

−1

C cos positive

A helpful mnemonic for positive trigonometric function values is “All Stations To Central” (ASTC). This can help you remember the signs of the trigonometric ratios in each quadrant. • In quadrant 1, all ratios are positive. • In quadrant 2, only sin θ is positive. • In quadrant 3, only tan θ is positive. • In quadrant 4, only cos θ is positive.

Example 3 Will tan 195° have a positive or a negative answer?

Create a strategy Think about which quadrant the angle is in and what the sign of tangent will be in that quadrant.

Apply the idea The angle 195° is between 180° and 270°, so it lies in quadrant 3. In quadrant 3, both x and y are negative, so

tan 195° will have a positive answer.

Reflect and check y 1 195°

−1

If we graph the angle and terminal arm, we will see that the terminal arm is increasing from left to right, so it has a positive gradient.

−1

17.02 Optional: The unit circle mathspace.co

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For any angle beyond quadrant 1, we can draw an angle in the ﬁrst quadrant that creates a congruent triangle to the one formed by the original angle. In the congruent triangles, the angles at the origin will be congruent. y

1 3 , 2 2

60°

60° angle

1 3 P − , 2 2 120°

60°

120° angle

210°

300°

60°

−

210° angle

300° angle

90° y 1 B

180° −1

The relative acute angle, or reference angle, is the smallest angle from the x-axis to the terminal arm. It is shown in each quadrant in the diagram.

180° − θ

180° + θ

360° − θ

0°

In each quadrant, the reference angle is equivalent to: • Quadrant 1: θ • Quadrant 2: 180 − θ • Quadrant 3: 180 + θ • Quadrant 4: 360 − θ

−1 270°

17.02 Optional: The unit circle mathspace.co

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We can use the relative acute angle to ﬁnd trigonometric ratios in any quadrant. We can use these steps to evaluate any trigometric ratio: 1. Sketch the angle on the unit circle. 2. Identify the sign of the ratio based on the quadrant the terminal arm is in. 3. Find the reference (relative acute) angle. 4. Rewrite the ratio using the sign and reference angle. 5. Evaluate using special angles or a calculator.

Example 4 y

Point P on the unit circle shows a rotation of 330°. What acute angle in the ﬁrst quadrant can 330° be related to?

Create a strategy We want to draw an acute angle with the closest x-axis.

Apply the idea y

330°

30° P

330° is between 270° and 360°, so it is in the fourth quadrant.

This diagram shows that an acute angle 360° − θ is related to θ in the fourth quadrant.

So the positive x-axis is closest.

360° − 330° = 30° which is the reference angle.

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Example 5 For sin 300°: a Write an equivalent ratio using an acute reference angle.

Create a strategy Find the quadrant the angle lies in and determine the sign of the ratio. Then we can ﬁnd the closest x-axis to ﬁnd the reference angle.

Apply the idea 90° y sin positive

all positive

θ θ

180°

θ θ

tan positive

Starting at the positive x-axis and rotating 300° anticlockwise, we move into the fourth quadrant. This diagram shows that an angle of 360° − θ in the fourth quadrant is equivalent to an acute angle of θ in the ﬁrst quadrant for the trigonometric ratios.

0° 360°

Sine is negative in the fourth quadrant.

cos positive 270°

sin 300° = −sin(360° − 300°) = −sin 60°

Find the acute angle and use the sign Evaluate the angle

b Evaluate sin 300°, without a calculator.

Create a strategy We can now use a special triangle to evaluate.

Apply the idea

30° 2

Using our special triangle, we can see that:

60° 1

So we have that: From above

Using special triangles

17.02 Optional: The unit circle mathspace.co

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17.02 Practice questions Explorer

Understanding 1

Adventurer

1–4

Trailblazer

1–4

1–5

In which quadrant is the angle located? a

299°

5°

160°

229°

40°

310°

138°

344°

Use the special triangles to complete the table. sin

cos

30°

tan

30° 45° 60°

45°

60° 1

45°

Use your calculator to evaluate each expression to three decimal places. a

tan 300°

sin 25°

cos 108°

tan 171°

cos 344°

sin 226°

sin 270°

cos 360°

Is each statement true or false? a

All points on the unit circle have positive coordinates.

We cannot evaluate trigonometric ratios for angles greater than 90°.

All trigonometric ratios are positive in the first quadrant.

The reference angle is always acute.

The reference angle is made with the closest x-axis.

A point P (x, y), on the unit circle makes an angle of θ with the positive x-axis. y 1

θ −1

x 1

−1

What is the radius of the unit circle?

What is cosθ?

What is sinθ?

What is tanθ?

Write the coordinates of P (x, y) in terms of θ. 17.02 Optional: The unit circle mathspace.co

831

6ab, 7, 8a-d, 9a-d, 10, 11a-f, 12–13, 16a-g

Fluency Ex 1

6bc, 7, 8c-g, 9c-g, 10, 11c-h, 14–15, 16e-l

For each diagram, find the value of: i

sin a

cos a

iii

tan a

P −

3 4 P  ,  5 5 θ

21 20 , 29 29

a −1

−1

1 a

−1

Find the coordinates of P (x, y), rounded to two decimal places. a

y 1

−1

40°

832

160°

−1

1 −1

Ex 3

x 1

−1

x 1

−1

Ex 2

6cd, 7, 8c-h, 9c-h, 10, 11e-l, 14–16

x 1

−1

Will the ratio have a positive or a negative answer? a

sin 143°

tan 284°

cos 143°

cos 284°

sin 284°

cos 269°

sin 170°

cos 320°

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Ex 4

Write the relative acute angle on the unit circle that is equivalent to: a

310°

100°

160°

245°

192°

138°

256°

289°

For each statement, fill in the blank with = or ≠. a d

Ex 5a

cos 300° ⬚ cos 60°

sin (150°) ⬚ sin 30°

−cos (110°) ⬚ cos 70°

tan 275° ⬚ tan 85°

tan 145° ⬚ −tan 35°

Write an equivalent ratio using an acute reference angle. a

sin 215° ⬚ sin 35°

sin 106°

cos 149°

tan 124°

sin 240°

cos 315°

tan 140°

sin 174°

cos 210°

tan 295°

sin 320°

cos 111°

tan 264°

Find the value of y.

Consider the point (x, y): i

Find the value of x.

(x, y) 1

135°

225°

x 1 (x, y)

The diagram shows a unit circle with point

marked on the circle. Point P represents a rotation of 45° anticlockwise around the origin from the positive x-axis: 45°

−1

1 x

−1

Find the exact value of sin 45°.

The 45° reference angle has been marked in all quadrants. For each quadrant, find the relative angle. i

Quadrant 2

Quadrant 3

iii

Quadrant 4

Find the exact coordinates of point S.

Write sin 315° in terms of an equivalent ratio of 45°.

Find the exact value of sin 135°.

y 1 Q −1

P 45° 45°

45° 45°

1 x S

−1

17.02 Optional: The unit circle mathspace.co

833

The diagram shows a unit circle with point

y 1

marked on the circle. Point P represents a rotation of 60° anticlockwise around the origin from the positive x-axis:

P 1

−1

60°

1 x

−1

Find the exact value of the ratio tan 60°.

The 60° reference angle has been marked in all quadrants. For each quadrant, find the relative angle. i

Quadrant 2

Quadrant 3

iii

−1

Quadrant 4

Find the exact coordinates of point R.

Write tan 240° in terms of an equivalent ratio of 60°.

Find the exact value of the ratio tan 240°.

60°

The diagram shows a unit circle with point

−1

1 x

y 1

marked on the circle. Point P represents a rotation of

30° anticlockwise around the origin from the positive x-axis: −1

30°

−1

834

Find the exact value of cos 30°

The 30° reference angle has been marked in all quadrants. For each quadrant, find the relative angle: i

Quadrant 2

Quadrant 3

iii

Quadrant 4

Find the exact coordinates of point Q.

Write cos 150° in terms of an equivalent ratio of 30°.

Find the exact value of cos 150°.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

y 1 Q −1

P 30° 30°

30° 30°

x S

−1

Ex 5b

Evaluate without using a calculator. a

sin 120°

tan 135°

cos 240°

cos 300°

sin 135°

cos 225°

tan 315°

sin 240°

cos 120°

tan 300°

sin 210°

tan 330°

Reasoning 17

17, 18, 20

18–20

In which quadrant is angle θ located? Explain. a

sin θ > 0 and cos θ < 0.

tan θ < 0 and sin θ > 0.

tan θ < 0 and cos θ < 0.

tan θ > 0 and sin θ > 0.

The points (0, 1), (0, 1), (−1, 0) and (0, −1) are signiﬁcant points on the unit circle. a

Explain how the points can be used to ﬁnd the value of sin 0°, sin 90°, sin 180°, and sin 270°.

Use the points to ﬁnd the value of cos 0°, cos 90°, cos 180°, and cos 270°.

Using the Pythagoras’ theorem and the unit circle, show that sin2 θ + cos2 θ = 1.

y 1

(cosθ , sinθ )

θ −1

x 1

−1

The sine, cosine and tangent ratios are all positive for angles in the ﬁrst quadrant.

y 1

This unit circle is labelled with the letters A, S, T, C in the quadrants. Explain how these letters help identify the sign of the ratios for angles in each quadrant.

1 −1

−1

17.02 Optional: The unit circle mathspace.co

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Problem-solving 21, 23, 25a 21

21–24, 25a

22–26

Find, rounded to two decimal places:

The coordinates of P (x, y)

The new coordinates of P if it was reﬂected across the y-axis.

sin 114°

66°

cos 114°

tan 114°

Suppose that a

Find sin θ.

, where 270° < θ < 360°. b

Find tan θ. θ 5

Given: w = 65°, x = 10°, y = 44°, z = 380°, a = 460°, b = 550° Is the statement true or false?

tan a, cos(90° + x) and sin(180° + y) are negative.

sin x, cos z, and tan b are all positive.

Angles w, x, and y are in Quadrant 1.

sin x, cos y, and tan w are all positive.

Angles z, a, and b are in Quadrant 1.

Given: w = 253°, x = 265°, y = 193°, z = 462°, a = 538°, b = 407° Is the statement true or false?

836

Angles z, a, and b are in Quadrant 3.

sin z and sin b are both negative.

tan(z − 90°) and tan(z + 90°) are both positive.

The angles w, x, and y all have negative cosine values.

Angles w, x, and y are in Quadrant 3.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

The diagram shows points P, Q, R and S, which represent rotations of 62°, 118°, 242° and 298° respectively around the unit circle. a

If point P has coordinates (a, b), write the coordinates of the following points in terms of a and b: i

iii

Q 180°

tan 118°

iii

tan 298°

P (a, b) 0°

−1

62° R −1

Is the ratio equivalent to −tan 62° or tan 62°?

tan 242° 270°

Express in terms of tan x, where x < 90°: i

90°

tan(180° − x)

tan(180° + x)

iii

tan(360° − x)

For the trigonometric expressions: i

Find the value of the relative acute angle.

Find the exact value of the trigonometric expression.

sin 480°

tan 540°

cos 780°

sin 720°

17.03 Optional: Graphs of sine and cosine After this lesson, you will be able to… • graph the values of sine and cosine functions on a Cartesian plane. • identify the key features of sine and cosine functions.

Graphs of trigonometric functions You may remember that the unit circle provides us with a visual understanding that the trigonometric functions of sin θ, cos θ and tan θ exist for angles larger than what can be contained in a right-angled triangle. The sine of the angle is defined to be the y-coordinate of the point on the unit circle. The cosine of the angle is defined to be the x-coordinate of the point on the unit circle. The tangent of the angle can be defined algebraically as the ratio

. This also represents

the gradient of the line that forms the angle θ to the positive x-axis. It can also be geometrically deﬁned to be the y-coordinate of a point Q, where Q is the intersection of the extension of the line OP and the tangent of the circle at (1, 0).

17.03 Optional: Graphs of sine and cosine mathspace.co

837

Properties of trigonometric graphs The graphs of y = cos θ and y = sin θ have certain common properties. Each graph demonstrates repetition. We call the graphs of y = cos θ and y = sin θ periodic or cyclic. We deﬁne the period as the length of one cycle. For both graphs, the period is 360°. Here is an example of a cycle. y 2 1 −540°

−360°

−180°

0 −1

180°

360°

540°

−2

Because of the oscillating behaviour, both graphs have regions where the curve is increasing and decreasing. Remember that we say the graph of a particular curve is increasing if the y-values increase as the x-values increase. Similarly, we say the graph is decreasing if the y-values decrease as the x-values increase. Here is an example of where y = sin θ is decreasing. y 2 1 −540°

−360°

−180°

0 −1

180°

360°

540°

−2

In addition, the height of each graph stays between y = −1 and y = 1 for all values of θ, since each coordinate of a point on the unit circle can be at most 1 unit from the origin. This means, the range of both the sin θ and cos θ functions is between −1 and 1.

Example 1 For the graph of y = sin x: y 1 x −360°−270° −180° −90°

90°

180° 270° 360°

−1

a What is the x-value of the x-intercept in the region −360° < x < 0°?

Create a strategy Use the fact that x-intercept is the point at which the curve intersects the x-axis.

17.03 Optional: Graphs of sine and cosine mathspace.co

839

For the curve y = sin x, are these statements true or false? a

The graph of y = sin x is periodic.

As x approaches inﬁnity, the height of the graph approaches inﬁnity.

The graph of y = sin x is increasing between x = −90° and x = 0°.

The graph of y = sin x is symmetric about the line x = 0.

The graph of y = sin x is symmetric with respect to the origin.

The y-values of the graph repeat after a period of 360°.

2 1 x −360° −180°

180° 360° −1 −2

For the curve y = cos x, are these statements true or false? a

The graph of y = cos x is cyclic.

As x approaches inﬁnity, the height of the graph approaches inﬁnity.

The graph of y = cos x is increasing between x = −180° and x = −90°.

The graph of y = cos x has reﬂective symmetry across the line x = 0.

The graph of y = cos x has rotational symmetry with respect to the origin.

The y-values of the graph repeat after a period of 180°.

Fluency 5

5–9, 12, 13

y 2 1 x −360° −180°

180° 360° −1 −2

7–14

For the graph of y = sin x: a

Determine the coordinates of the y-intercept.

Find the maximum y-value.

Find the minimum y-value.

y 2 1 x −360° −180°

180° 360° −1 −2

842

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

For the graph y = cos x:

Determine the coordinates of the y-intercept.

Find the maximum y-value.

Find the minimum y-value.

2 1 x −360° −180°

180° 360° −1 −2

For the equation y = cos x: a

Complete the table of values: x

0°

90°

180°

270°

360°

270°

360°

cos x b 8

Sketch the graph of y = cos x.

For the equation y = sin x: a

Complete the table of values: x

0°

90°

180°

sin x b 9

Sketch the graph of y = sin x.

Estimate these values using the graph of y = sin x: y 1 0.5 x 30°

60°

90°

120°

150°

180°

210° 240° 270° 300° 330° 360°

−0.5 −1

sin 120°

sin 300°

sin 40°

sin 340°

Two values of θ where sin θ = 0.9

Two values of θ where sin θ = −0.6

17.03 Optional: Graphs of sine and cosine mathspace.co

843

Estimate these values using the graph of y = cos x: y 1 0.5 x −180° −150° −120° −90° −60° −30°

30°

60°

90°

120° 150° 180°

−0.5 −1

cos 30°

cos(−100°)

cos 160°

cos(−20°)

Two values of θ where cos θ = −0.75

Two values of θ where cos θ = 0.3

For the graph of y = cos x: a

Is cos 162° positive or negative?

In which quadrant does an angle with measure 162° lie?

y 2 1 x −360° −180°

180° 360° −1 −2

For the graph of y = sin x: a

Is sin 200° positive or negative?

In which quadrant does an angle with measure 200° lie?

y 2 1 x −360° −180°

180° 360° −1 −2

844

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Ex 1

For the graph of y = sin x:

If one cycle of the graph of y = sin x starts at x = 0°, when does the next cycle start?

Find the x-value of the x-intercept in these region 0° < x < 360°.

True or false: As x approaches inﬁnity, the graph of y = sin x stays between y = −1 and y = 1.

Does the graph of y = sin x increase or decrease in these regions? i

90° < x < 270°

−270° < x < −90°

iii

−450° < x < −270°

−90° < x < 90°

270° < x < 450°

x −360° −180°

180° 360° −1 −2

For the graph of y = cos x:

If one cycle of the graph of y = cos x starts at x = −90°, when does the next cycle start?

Find the x-values of the x-intercepts in these regions:

0° < x < 360°

−360° < x < 0°

−540°−360°−180°

180° 360° 540°

True or false: As x approaches inﬁnity, the graph of y = cos x stays between y = −1 and y = 1.

−1

Does the graph of y = cos x increase or decrease in these regions?

−2

0° < x < 180°

−180° < x < 0°

iii

180° < x < 360°

−360° < x < −180°

Reasoning

15, 16

16, 17

Describe the relationship between the graphs of the sine and cosine functions and how they are interconnected.

Are these statements true or false? Justify your answer.

The sine and cosine functions have the same amplitude because they both oscillate between −1 and 1.

The sine function has a period of 180° because it starts at 0 and reaches its maximum value at 90°.

The graphs of the sine and cosine functions intersect at every multiple of 90°.

The sine and cosine functions have the same period because they are both based on the unit circle.

The functions y = sin q and y = cos q intersect at two points in the region 0° ≤ θ ≤ 360°. Use the unit circle to justify why these are the only two points of intersection in this region.

17.03 Optional: Graphs of sine and cosine mathspace.co

845

17 Chapter review 1

(Optional) Copy and complete: a

45° 1

b c

tan 45° = ⬚

45° 1

(Optional) Find the exact value of the pronumeral: a

b 20

w 60°

30° 2x

(Optional) Evaluate the expressions, leaving your answer in exact and rationalised form:

a c 4

sin 30° cos 30° + tan 45°

sin 45° tan 60°

sin2 60° + cos2 60°

(Optional) A right-angled triangle has one angle of 60° and an adjacent side of length a

Find the exact lengths of the other two sides.

Find the area of the triangle.

(Optional) Find the exact value of the expression:

(Optional) Use the diagram to ﬁnd the value of: a

sin b

cos b

tan b

1 b −1

x 1

−1

7 24 ,−  25 25 

Chapter 17 review mathspace.co

847

(Optional) Will the ratio have a positive or a negative answer? a

cos 309°

tan 225°

cos 92°

sin 318°

cos 315°

(Optional) Write an equivalent ratio using an acute reference angle: a

sin 200°

cos 230°

sin 115°

tan 171°

(Optional) For the point (x, y): y (x, y) 1 120° x

Find the value of x.

Find the value of y.

(Optional) Evaluate without using a calculator: a

sin 330°

cos 150°

tan 240°

(Optional) For the equation y = sin x: a

Complete the table of values. Round to two decimal places where necessary. x

0°

30° 90°

135°

180°

240° 270° 300° 360°

sin x

Write the coordinates of the y-intercept.

Write the coordinates of the x-intercepts from 0° ≤ x ≤ 360°.

Sketch the graph of y = sin x.

(Optional) Refer to the graph of y = cos x: a

Is the sign of cos 237° positive or negative?

In which quadrant of the unit circle is the angle 237°?

Find the x-value of the x-intercepts in these regions: i

0° < x < 360°

(−360) ° < x < 0°

y 2 1 x −360° −180°

180° 360° −1 −2

848

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Answers 1.01 Index laws Understanding

g 4p9

h −13

i 16u10v4

j 81x9

k m11

l 22

m x60

n −9

o p17

1 Subtraction

p 64y12 12

q 16x 3

3 a i Yes

ii Yes

iii Yes

iv No

13 a n20r20

b 10x2y11z15

c 2p14

d −5x2y14z11

f 9v5y11

g −6u11

h 4x2

v No b i False

ii True

c 6 4 a 10

b 9

e 11

f m4

c 2

d 0 Reasoning

Fluency 5 a a8

b a10 7

c b5 7

e 6y

f 12y

i u

j 2w 7 6

6 a 40g h

7 a u11

d y10 11

h am+n

l 72y17

g −2a k 15z b 0

c 56m8n5

d b

c 3m7

d k13

e a5

f m3

g j7

k 9j2k3

8 a −32y15

b −96y11

5 5 5

c −30s t r

d 504y24

e −10

f 3y4

9 a j24

b w22

c y12

d 81a8

e 25h14

f x5y10

j x36

k −125x12 l

b k31

c 24r58

10 a w36v24 39 4

45 60

e 160r s f a b 11 a 24 e 11 12 a x28y8

850

t c86

s x

b 1 f 1

r 1125y16

g p

c 1 g

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.

d x67y15

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.

h x60

c When dividing terms with the same base,

d 1 h 6

b p2

c 0

d 81a8b20c4

f y7

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

subtract the indices, so

not t . 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.

l −

k 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

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 will make 3x = x3 be a true statement. x= 19 a = 3, b = 6 20 a True p

b False

c False

21 16 = 2 22 a 3a+6

b 343a−21

1.02 Negative indices

e 8h7

Understanding 1 a True

b False

c False

d True

2 Less than 1. can be expressed as a−1.

Fluency 4 a 2−4

2−3

2−2

2−1

10−3 10−2 10−1 100 101 102

103

104

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

851

15 Applying the negative index law, 0−3 can be written as

, which equals

. This expression

Problem-solving 19 There are many different representations, such as:

is undefined, as there is no number that when multiplied by 0 will equal to 1.

•

16 The −1 index only applies to the variable directly in front of it, since there are no brackets. The

correct working is

21 a n = −3

17 As a becomes a very large positive number, the value of x−a approaches 0 while remaining positive. This is because the negative index law rewrites the expression as

. Since x > 1,

• 2

know is greater than the base . There are many examples of this same pattern, such as 3−1 < 3 and

• 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.

23 4.1 minutes 24 a 3 × 10−6 Farads

× 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.

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 c

, irrational

e 9, rational g

852

, which we

base 2. However,

18 a • 2p : −4, 0, 8, 20

−p

c n = −3

, which is less than the

Therefore,

the denominator x becomes an extremely large number as a increases. Dividing 1 by an increasingly large denominator results in a value that gets progressively closer to 0. For example, 10−10 = 0.000 000 000 1.

• 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.

• 4p : −8, 0, 16, 40

• 22

•

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

, irrational

h 11, rational

, irrational

m 3, rational

Problem-solving 15 a

n 2, rational

, irrational

8 a

g 6

b 4

c 1

d 6

18 a 8x

b 9z5

c 4v6

d 12b3

e 6h2

f 2x2

g 3a2

h 5a7b3

e 60 9 a

Reasoning

1.04 Optional: Add and subtract surds

10 a Yes b No, it is an approximate value and not exact is an irrational number that because 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 button; the correct its cube root using the 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.

can be simpliﬁed further.

Using 20 = 4 × 5, we can simplify , then = becomes

Understanding 1 a like surds b a power that matches the degree of the radical. For example, 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

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.

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. 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.

7 a

Answers mathspace.co

853

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

c d Fluency 3 a 75

b 288

c 108

d 80

e 50

f 35

g 6

h 5

4 a

5 a

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. are not like surds, so they cannot 10 a and be added or subtracted.

6 a

7 a

8 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. Problem-solving units

b 218 m

13 a

9 a 2

b 4

i 2

b 10

g 16

h 10

11 a 5

f 2

14 a 34 cm is a bit less b It would be an overestimate. , and is a bit less than than . c

10 a 6

d 2.51% 15 a b c 1830

1.05 Optional: Multiply and divide surds

Reasoning

Understanding

12 a Sometimes a surd. For example,

1 a True

b True

c False

d False

e True

f True

g True

h True

i True

j False

854

Mathspace WA Curriculum – Year 10 Methods Pathway 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. and are 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, would be a ab cannot be square. So surd. 17 The area of any paper is given by the formula Area = Width2 × . For every paper in this series, the Width2 simplifies to a whole number (an integer). Since the area is always an integer multiplied by the irrational number , the result is always a surd. Problem-solving 2

units

19 20 a

24 a 0.14

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 Fluency

b 100 cm

18 a

5 a b

b 1.69 m2

c d

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 2 m2 surface area. For example, a height of 165 cm and weight of 90 kg supports the formula and is around 2 m2 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

b 46 m2

c 40 m2

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

d The difference is 6 m , approximately 13% of the area which is a signiﬁcant underestimate.

Answers mathspace.co

855

14 a

8 a

b m = −4

1.07 Optional: Rationalise the denominator

c d

Understanding 1 a 5

Reasoning

b 1 9

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

units

11 a

units

b c

units

12 a

square units

units

13 a b

856

8 a

9 a

i −

units units units2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10 a

11 a

14 The rationalisation process involves multiplying the numerator and the denominator by the conjugate of the denominator. In this case, the , so when we multiply conjugate is we get a2 − b. The new numerator will be contains the term

, which .

, and we can 15 Xander is correct. . 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 12 a

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

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) , we are multiplying c By multiplying by 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.

21 22

1.08 Optional: Fractional indices Understanding 1 2 a

Answers mathspace.co

857

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

j i

ii 4

k i

ii 4

l i

ii 2

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

13 a

c 128

d 9

g 9

is greater because it represents the represents the square root of 64, while cube root of 64. The square root of a number greater than 1 is always greater than the cube root of the same number.

as the numbers remain 20 By writing smaller so they are easier to work with. For example

858

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 instead of raising it to the power of , thus

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

incorrectly writing it as . This misapplication of the fractional index law made all the subsequent steps incorrect.

Rewrite using the fractional index law Rewrite the indices with a common denominator Multiplication law Power of a power law

Chapter 1 review 1 a 6x5

b Applying the index law correctly, Step 3 , leading to

should be

2 a u8

b 25y8

c 16x12

4 a 53a−2

b 4v13 y

5 a

6 a 20y6

c 20 736x4

8 a

b b

, irrational

c 9, rational

c b

b 4

d 6

11 a 54

25 Many solutions are possible, for example

13 a

14 a

d units

c x2

d m

e b4

g x30y18

i 2a4

j 25u8v6 k

m 28 a 54y 29

n jx

l 2x + 9 6 4

o 9x y b 36y

, irrational

d -5, rational

24 25 cm3

27 a m3

d 25m3

7 a

10 a

3 a = 5, b = 4

9 a m2n6

Problem-solving 23 a

d 54m9 n8

c 8y

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 −3a9

16 a

17 a

18 a b c d 19 a b

Answers mathspace.co

859

20 a

6 a 6x2 + x − 12

b 12x2 + 7x − 10

c 4x − 7x − 2

d 12x2 − x − 20

e 2x2 + 2x − 12

f 6x2 + 9x − 60

g 10x − 15x − 10

21 22 a The fifth term is rational. b Yes. It is the only term where the and the simplify to a rational value.

7 a

8 a x2 + 14x + 49 b

23 a

24 a i

ii 10

b i

ii 4

c i

d i

ii 25

25 a 4

b 2

d 81

26 a

b 10

c 2

h −18x2 − 3x + 6

b c2 + 12c + 36

c e + 10e + 25

d y2 − 10y + 25

e 49 − 14z + z

f 36 + 12a + a2

g a2 + 2ab + b2

h a2 − 2ab + b2

9 a 9x2 + 42x + 49

b 4y2 − 20y + 25

c 4y2 − 12y + 9

d x4 + 2x2 + 1

10 a 5m2 − 50m + 125 b 6n2 + 72n + 216

2.01 Expand binomial products

c 8q2 + 48q + 72 d 147x3 − 210x2 y + 75xy2

Understanding 2

b 20y

1 a 6x

c −21z

d −10p

11 a x2 − 49

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

c 49 − 36p

d 81y2 − 1

e 81x2 − 4y2

f 0.04x2 − 9

12 a

d (x − 3) (x − 4) = x2 − 7x + 12 3 a (x + 3)2 = x2 + 6x + 9

b 64 − m2 2

13 a 75x3 − 108xy2

b (x − 3)2 = x2 − 6x + 9

b 243x2 − 192y2

c (x + 6)2 = x2 + 12x + 36

c 71

d (3x − 8)2 = 9x2 − 48x + 64

d 25t2 + 90t − 36r2 + 81

4 a (x + 5) (x − 5) = x – 25

e 384x2 − 486y2

b (x + 2) (x − 2) = x2 − 4

f 20x3 − 405xy2

c (x − 8) (x + 8) = x − 64 d (x + 14) (x − 14) = x2 − 196 Fluency 5 a x2−9x + 8 c x2 − 9x + 18

d x2 − x − 12

f x2 − 6x + 5

h x2 − 7x − 18

e x − 12x + 35 g x − x − 30

860

b x2 + 5x + 4

Reasoning 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.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

b The resulting expression is not a trinomial because it contains terms with variables in the denominator and the expression has more than 3 terms. Problem-solving 19 a P = 4x + 16 metres b A = x2 + 8x − 20 square metres. c The length of the field would be (6 + 10) which equals 16 metres. The width of the field would be (6 − 2) which equals 4 metres. The perimeter of the field would be 24 + 16 which equals 40 metres. The area of the field would be 16 × 4 which equals 64 square metres. 20 a 6x2 + 36x + 54 units2 b x3 + 9x2 + 27x + 27 units3 21 (4x2 − 8x + 4) cm2 22 a 24x2 − 49

b 62x2 − 25y2

23 n3 + 3n2 + 2n 24 a −y3 + 4y2 + 96y

Our expression is in this form, with A being and B being 3b. Let’s start by squaring A, which is

Simplifying, we get:

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. ii No, because you cannot have a zero length, and zero length yields zero volume.

Then, we calculate 2AB, which is : Lastly, B2 is (3b)2 B2 = (3b)2 = 9b2 Now, substituting A2, 2AB, and B2 back into the formula, we get:

iii Yes, all the side lengths are viable and the volume is 200 cubic units. iv No, because the side length 12 − y becomes zero and zero length yields zero volume. 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.

Answers mathspace.co

861

2.02 Factorise binomial products

Reasoning

Understanding

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).

1 a 2

b 5x

2 a 81 = 92

d 2d2

c (c + 5)

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 100 = 102

c 16 = 4

d 25 = 5

3 x+4

b The error is in the signs of the factors. The correct factorisation should be x2 − 4x + 4 = (x − 2)(x − 2).

4 n+6 5 a Yes

b No

c Yes

d No

13 a i Perfect square trinomial ii (x + 1)2

Fluency 6 a (m − 6) (x + y)

b ( y − z) (x − w)

c (4w + 3x) (5y − z)

d ( y − 4) (8y − 3)

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

f (2x2 + 3z) (2y − 1)

g (2c − d) (c + 5d + 3)

h (3x − 5y) (x + 4y)

7 a (x + 2) (x + 5)

b (x − 3) (x + 8)

c (a + 8) (a2 + 1)

d (x − 2y) (8 + z)

e (3 + x) (8 + y)

f (x + z) (7y + w)

g (p + 2) (2m + 3)

h (4x + 3z) (1 + 8y)

i (2x + 3z) (1 + 6y)

j (df + rz) (3f − 11y)

8 a (x + 3)2

b (8 + e)2

c (7 − p)

d (u − q)2

e 6 (c − 2)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)

g (4 + 3y) (4 − 3y)

h (xy − 7) (xy + 7)

i (5m + 7) (5m − 7)

j (9x + 4y) (9x − 4y)

k 7 (x − 3) (x + 3)

l 5 (3t + 2) (3t − 2)

10 a 3y (x2 − 2y3 + 3x2 y2)

b i Common factor then difference of squares ii n (n + 11) (n − 11) c i Factoring by grouping ii (2x + 5)(x − 3) d i Perfect square trinomial ii (3x − 2y)2 e i Difference of squares ii (x + 16 + y) (x + 16 − y) f i F actor by grouping then difference of squares ii (x + 4) (2x + 1) (2x − 1) 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. Problem-solving 15 a 256

b 392

c 261

16 a t + 7 units

b 4 (t + 7) units

17 x + 12x + 36 18 a (x2 + 4)2

b (a − 3)2 (a + 3)2

19 (x − 1) (x + 1) (x2 + 1) 20 a

b (4xy + 3z)

c (11c − 7d) (11c + 7d) 2

d (x + y) (3y + z) 2

+ 2

e 4b (2a b + 3c )

f 7 (2b + 1) g 7hp2 (2h2 r + 3h4 + 7p2) h 2 (8w + 3t) (w − 4t2) i (4q2 + 3r)2 j 3d (4f − h) (4f + h)

862

d 660

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b (x + 6) (x + 4)

2.03 Factorise monic quadratic trinomials

e (x + 7) (x − 10)

f (x + 13) (x − 9)

g ( y − 16) ( y + 3)

h (t + 14) (t − 6)

Understanding

i ( y − 15) ( y + 3)

j (x + 14) (x − 4)

1 a 9 and 10

b −9 and 7

k (x + 12) (x − 2)

l (x + 7) (x − 4)

d −8 and 11

9 a 2 (x + 2) (x + 5)

b 4 (x + 2) (x + 4)

(1 × 64)

c 4 (x − 9) (x + 8)

d 3 (x − 5) (x − 2)

c −6 and −4 2 a • (1 × 18)

b •

• (−1 × −18) •

(−1 × −64)

e 5 (x − 2) (x + 3)

f −2 (x − 2) (x + 11)

• (2 × 9) •

(2 × 32)

g −4 (x − 5) (x + 2)

h −3 (x − 2)2

• (−2 × −9) •

(−2 × −32)

i −(x + 4) (x + 2)

j −(x − 3) (x − 4)

k x (x + 3) (x + 4)

l 2x (x + 5) (x + 3)

• (3 × 6) •

(4 × 16)

• (−3 × −6) •

(−4 × −16)

•

(8 × 8)

Reasoning

•

(−8 × −8)

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.

c •

(−1 × 30)

• (1 × −30)

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.

• (−2 × 15) • (2 × −15) • (−3 × 10) • (3 × −10) • (−5 × 6) • (5 × −6) 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)

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. 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. 12 a The error is in the signs of the factors. The correct factorisation should be x2 + 7x − 18 = (x − 2)(x + 9).

6 a (x + 8) (x + 3)

b (x + 4)2

c (x + 9) (x + 8)

d (x + 8)2

e (t + 5) (t + 2)

f (t + 7)(t + 1)

The correct factorisation should be x2 − 11x + 30 = (x − 5)(x − 6).

g ( y + 4)( y + 5)

h (x + 3)(x + 5)

i (x + 3)(x + 4)

j ( y + 7)( y + 2)

13 a Exactly one of p or q must be negative.

k (x + 5) (x + 8)

l (x + 6)2

7 a (x − 3) (x − 1)

b (x − 6) (x − 10)

c (x − 7) (x − 12)

d (x − 5) (x − 7)

e (x − 7) (x − 1)

f ( y − 2) ( y − 8)

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)

b The error is in the signs of the factors.

b Both p and q must be positive. c Both p and q must be negative. 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.

Answers mathspace.co

863

15 a (x + 9)(x + 8) = x2 + 17x + 72 b (x + 2)(x + 1) = x2 + 3x + 2

2.04 Optional: Factorise non-monic quadratic trinomials

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.

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)

b a and b must be a factor pair of 36. We want to minimise the difference of the factor pair. 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)

3 (3x + 1) (x + 6) Fluency 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) c 2(s + 3)(4s − 9) e 4 (2x − 1) (x − 2) g 5 (7x − 1) (3x − 1)

b 4 (x + 5)2 d 3 (3x + 1) (x − 5) f 6 (4x + 1) (3x − 1) h 8 (3x − 2) (7x + 1)

6 a −(2x − 3) (3x + 2) c −3 (2x + 5) (2x − 3) e −(4x − 3) (x − 2) g −(5x + 4) (x + 2)

b −(2x − 7)(3x − 2) d −2(3x + 4)(x − 5) f −(3x − 4)2 h −(3x − 2)(x + 4)

7 a 10y2 − 112y + 22

b 2 (5y − 1) ( y − 11)

8 a 10x2 + 105x + 245

b 5 (2x + 7) (x + 7)

18 a (x + 6) (x + 3) b (m + 5) (m + 10) 19 a (u − 8)(u − 3) b (x + 4)(x − 2)(x + 3)(x − 1) 20 a (x + 3) (x + 5) 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 = 3 m and d = 5 m. 23 x = 7

9 a 6 (x + 3)

b (x + 3) units

10 2x + 5, 3x + 4 11 −10 (t + 2) (t − 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

864

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

14 a Each polynomial has a leading coefficient a ≠ 1, and none of the polynomials can be factored easily over integers. 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. Problem-solving 15 a −5 (t + 1) (t − 2)

3 a 3x2 + 6x − 8 = 3 (x2 + 2x) − 8 b 2x2 − 10x + 1 = 2 (x2 − 5x) + 1 c 4x2 + 12x − 6 = 4 (x2 + 3x) – 6 d 6x2 + 24x − 7 = 6 (x2 + 4x) − 7 4 a 5x2 − 10x + 4 = 5 (x2 − 2x) + 4

= 5 (x2 − 2x + 1) + 4 − 5

= 5 (x − 1)2 – 1

b 6x2 + 36x − 1 = 6 (x2 + 6x) − 1

b 0 c 0 d 2 seconds after jumping off the board, Tara hits the water. 16 a (4x2 − πx2) units2

Fluency

b x2 (4 − π )

17 a 3x + 4, 2x + 3 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. c x = 10 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. 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. 2

= 6 (x2 + 6x + 9) − 1 − 54

= 6 (x + 3)2 − 55

5 a (x + 9)2 − 81 2

b (x − 4)2 − 16

c (x + 5) + 6

d (x + 7)2 − 2

e (x − 5)2 + 5

f (x − 9)2 − 4

6 a (x − 4) (x − 2) b (x + 7) (x + 1) c (x + 12) (x − 2) d 7 a 2 (x + 2)2 − 15

b 5 (x + 2)2 – 29

c 4 (x − 2)2 − 13

d 10 (x + 1)2 − 21

e 7 (x − 1) − 6

f 9 (x + 3)2 – 83

8 a b

19 4x + 12x + 5 = (2x + 1)(2x + 5) is one possible answer.

c (x − 1) (x − 5)

d (x + 10) (x + 1)

e (x − 5) (x − 6)

f (x + 21) (x + 3)

20 k = 9

g (x − 1) (x − 19)

h (x − 5) (x − 23)

9 a 3 (x − 3) (x + 1)

21 m = −6, n = −8 or m = −8, n = −6

b 2 (x + 9) (x − 1)

2.05 Optional: Complete the square

c 4 (x + 1.5) (x + 2.5) = (2x + 3) (2x + 5) d −2 (x − 5) (x − 13)

Understanding

e −4x (x + 4)

1 a 9

b 16

c 49

d 25

2 a Yes

b No

c No

d Yes

Answers mathspace.co

865

c The quadratic expression 3a2 + 14a + 8 can be factorised in two ways:

Reasoning 10 When completing the square, we could get: 1 x2 + 6x + 13 = x 2 + 6x + 32 − 32 + 13 A dd and subtract the square of half of

1. 3a2 + 14a + 8 = 3a2 + 12a + 2a + 8

= (3a2 + 12a) + (2a + 8)

= 3a(a + 4) + 2(a + 4)

= (a + 4)(3a + 2)

the coeﬃcient of x

2 = (x + 3)2 − 32 + 13 Factor the perfect square trinomial

3 = (x + 3)2 + 4 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. 12 On line 3, she has expanded 3[(x + 3)2 − 9] − 7 incorrectly. She forgot to multiply the 9 by 3. 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. 14 a The quadratic expression x2 − 18x + 81 can be factorised in two ways:

1. 2x2 + 5x − 3 = (2x2 − x) + (6x − 3)

= x(2x − 1) + 3(2x − 1)

= (2x − 1)(x + 3)

1. x2 − 18x + 81 = (x)2 + 2(x)(−9) + (−9)2

d The quadratic expression 2x2 + 5x − 3 can be factorised in two ways:

= (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)

= 2c(2c − 3) − 3(2c − 3)

= (2c − 3)(2c − 3)

= (2c − 3)2

866

Problem-solving 15 a x − 5 cm b x > 5 since the side lengths must be positive. 16 x − 4

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

17 a y = −(x + 4)2 + 36

16 a (x − 4)2 − 16

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 2 review b x2 + 4x + 3

c x2 − x − 30

d x2 − 5x − 36

b No, ( y − 3)2 = y2 − 6y + 9 2

c 9x − 0.49

b (x + 11) (x + 3) c (x + 37) (x + 5)

b n2 − 8n + 16 2

b (6x − 7) (x + 8) d −2(3x + 4) (x − 5) 2

5 a m2 + 10m + 25

19 a y = 36 − (x + 4)

b y = 36 when x = −4.

20 a 5x

b 4xb

21 a k(k − 8) b 5( y2 + 2y + 3) c 4c(−bc + 2b + 1)

d p2q2(3 + 4p2q2 + 5p4q4)

6 9x2 − 72x + 144 cm2 7 a (a + 3) (a + 4)

b ( p − 2) ( p + 6)

c (x + 3) ( y + 5)

d (m − 3) (k − 4)

9 a (x − 5)2

c (9x − 5)

e (a + b) (5 + x)

f 2 (x + 2)2 − 15

4 a −24x2 + 14x + 24

c 7 (5m2 − 7)

e 10 (x + 1)2 − 21

18 a (3t − 4) (4t + 1)

b 64 − 25p

3 a m – 16

8 a (k + 8) (k2 + 1)

d (x − 5) (x − 6)

2 a Yes

c 4p + 12pq + 9q

c (x + 5)2 − 4

17 a

1 a x2 − 10x + 16

b (x − 2)2 + 2

b 5 (r + s)2 d

22 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).

3.01 Compound interest 2

f 5 (x + 4y) (x − y) b (x + 8)2

c (1 − x)2

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)

Understanding 1 a Always true b Never true c Always true

(a − 6)2

d Sometimes true, this is only for the ﬁrst period.

c (b + 2) (b + 7)

(c − 2) (c − 9)

e Always true

e (d + 5) (d − 2)

(e − 8) (e + 3)

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.

10 a (y + 5)2

11 a p = 5 and q = 7

b 44 units

12 4y − 3 metres 13 a x − 7 cm

b 4x − 28 cm

14 a 3x + 19x + 20 square metres b x2 − 2x + 1 square metres c 2x2 + 21x + 19 square metres 15 k = 9

2 Account B 3 Graph 1 4 a No 5 a $9000 c Quarterly

b No

c No

d Yes

b 7% d 3 years

Answers mathspace.co

867

Fluency 6 a Investment A

b Investment A

14 a $25 631.28

b $631.28

15 a $2529.86

b $1761.93

c 11 years

c $837.74

7 a

Balance at beginning of year

Interest earned

Reasoning

First year

$3200

$160

16 a $1794.85

Second year

$3360

$168

Third year

$3528

Fourth year

$3704.40

$176.40 −

Balance at beginning of year

Interest earned

First year

$3500

$133

Second year

$3633

$138.05

Third year

$3771.05

Fourth year

$3914.35

$143.30 −

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

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

Third year, 3

$2269.25 + $95.31

$2364.56

$99.31

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

−

c $1.35 10 a 500 000 × (1 + 0.012)24 b $165 736.40 11 a $7706.73

b $6617.25

12 a $5705.83

b $705.83

13 a $4477.00

b $1477.00

868

b $414.35 9 a

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”.

b $504.40 8 a

b $1797.60

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,

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. • 4.75% compounded annually: He would owe $26 187.50 after one year • 4.7% compounded quarterly: He would owe $26 195.87 after one year • 4.65% compounded monthly: He would owe $26 187.60 after one year 19 a $6367.25

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b The value in the table has been rounded to the nearest whole dollar, so the amount of interest will be slightly different.

6 a $3172

Problem-solving

b He will receive a refund.

20 $53 136.44

7 $124.54

21 $53 516.75 22 a 0.02 23 a

b $3750

8 $24 507 9 $84 737

Value ($) 3000

10 a $2750

2500

11 $25

2000

12 $36

1500

13 $108.78

1000

14 a $18.22

500 0

b 3.7%

No. of years 20 25

b 24 years 24 24 years 25 $60 803.96 26 5.116% 27 18.5 years

3.02 PAYG and tax

b $29.81

15 $1268.27 16 $1200 17 x = $89 057 18 $7530 19 a $0

b $2977.60

c $18 806.50

d $35 603.68

e $53 612.60

f $67 752.50

20 a $55 217.75

b $3959.10

c $59 176.85 Understanding

21 a $50

1 The Medicare levy is calculated as 2% of an individual’s taxable income for the ﬁnancial year.

b New taxable income: $48 700, Tax payable: $5398

2 Personal income tax is a tax that must be paid based on the income of individuals, households, or other legal entities.

c $202

3 a Pay As You Go b Pay As You Go is an amount of tax that is deducted from each pay weekly, fortnightly, monthly, or occasionally quarterly. c Pay As You Go ensures that there isn’t a huge sum due at the end of the ﬁnancial year. 4 Deductions are speciﬁc and approved expenses that can be subtracted from an individual’s gross income to reduce the amount of income subject to tax. They lower an individual’s taxable income, which, in turn, reduces their income tax owed. Fluency 5 $10 140

Reasoning 22 Yes, she is correct as she pays no income tax on $17 000, therefore, her only payment is the Medicare levy of 2%. 23 a Stewie made an error in Step 3. $7400 is incorrect. He must have overlooked and multiplied it by 0.37, which is in the formula of the next row. b Tax = $4288 + 0.30 × ($65 000 − $45 000) Apply the corresponding tax formula

= $4288 + 0.30 × $20 000

Evaluate the brackets

= $4288 + $6000

Multiply 0.30 by $20 000 to ﬁnd the additional tax amount

= $10 288

Add the base tax amount and the additional tax amount

Answers mathspace.co

869

6 a

n i q>1

c m ≥ −24

d y < −14

e a>4

f k ≤ 45

g y > 40

h x≤3

−10

−5

b i y<0 ii

−10

−5

c i j<4 ii

−10

−5

−10

−5

−15

−10

−5

−1

−0.5

0.5

ii 0

j i s < 7.5 ii 6

k i y≤5 ii

−10 −8 −6 −4 −2 0 2 4 6 8 10

l i x>5 ii

−5

8 x≥3 9 a q > 16 10 p + 15.25 ≤ 32 11 A − 33.40 ≥ 32 A ≥ 65.40 x ≥ 66 b h ≥ 61.4

c 62 14 a x + 488.30 ≤ 2118.40 15 a B ≤ 3.4

c $1630.10 b 3

16 W ≤ 22

i i b ≤ 2.8

−10

b x ≤ 1630.10

−5

h i a < 0.4 ii

p i x≤6 ii

13 a 5h + 72 ≥ 379

g i m>8 ii

12 x + (63 + 73 + 98) ≥ 300

−10

f i b ≥ −12 ii

p ≤ 16.75

e i v>2 ii

b There is no greatest value.

d i a ≥ −1 ii

o i u≥8 0

−10

7 a i x>1 ii

−5

0.5

1.5

mi ii

17 a No

b Yes

c Yes

d Yes

18 a Yes

b Yes

c Yes

d No

19 a No

b No

c No

d Yes

20 a No

b Yes

c No

d No

Reasoning 21 x < 4 means x is any number less than 4. It doesn’t include the endpoint (4). 0 ≤ x ≤ 4 means x is any number between 0 and 4. It does include the endpoints, so x can also be 0 or 4. 22 a Step 1 is incorrect. Isabelle has subtracted 25 from both sides but the left-hand side should now be −2x not 2x. b 25 − 2x < 31 Step 1: −2x < 6 Step 2: x>−3

Answers mathspace.co

871

9 a y = −5x − 2

6 a i

−2

−1

11 a y = −10x + 25

b y = −10x + 35

28 24

12 a

16 12

13 a First line:

Second line:

, equal −2

b Perpendicular slope: b i

Equation:

5.01 Graphs and features of quadratics

2 a True

c False

b False

−2

−1

−16

−1

−16

4 2 −2

y x

−1 −2

−4 −6 −8 −10 −12 −14 −16 −18

d True

e False 3 The turning point lies right on the axis of symmetry. Knowing the axis of symmetry makes it possible to ﬁnd the turning point, and vice versa. Both elements help in understanding the shape and position of the graph.

ii b Concave down

Understanding 1 a Concave up

−1

c i

−2

−1

−50

−14

−2

−14

−50

Fluency 4 a i Concave down

ii Maximum point

b i Concave up

ii Minimum point

c i Concave down

ii Maximum point

d i Concave up

ii Minimum point

5 a i The turning point is (0, 5). ii Minimum iii The equation of the axis of symmetry is x = 0.

y −2

−1

−5 −10 −15 −20 −25 −30 −35 −40 −45 −50

iv There are no x-intercepts and the y-intercept is at (0, 5). b i The turning point is (0, 3). ii Maximum iii The equation of the axis of symmetry is x = 0. iv The x-intercepts are at (−0.5, 0) and (0.5, 0), while the y-intercept is at (0, 3).

Answers mathspace.co

879

d i

−2

−1

−18

e i

−1 −2 −4 −6 −8 −10 −12 −14 −16 −18

f i

880

−3

−6

−7

−6

−3

y −1

x 1

−1

−3 −4 −5 −6 −7 −8

−5

−1 −2 −4 −6

h i

−2

−1

20 y 18 16 14 12 10 8 6 4 2

55 50 45 40 35 30 25 20 15 10 5

x 1

−2

−1

−61

−16

−1

−16

−61

y −1 −5 −10 −15 −20 −25 −30 −35 −40 −45 −50 −55 −60

−2

−1

−2

10 y 8 6 4 2 −2

g i

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

7 a

−2

−1

−3

−2

−1

−2

−5

−2

b Minimum c Positive 8 i Concave down ii (0, 4) iii

5 −2

−1 −5 −10 −15 −20 −25 −30 −35 −40 −45 −50 −55 −60 −65 −70 −75 −80

20. For example, y = −4x2 + 15 doesn’t hit 20 since its highest point is 15 and it goes down. The c moves the parabola up or down, deciding if 20 is even possible to reach.

Reasoning 9 Answers may vary. An explanation may be: When a > 0, the parabola opens upward, and the square of any number (positive or negative) multiplied by a is always greater than or equal to 0. Thus, the minimum value of y occurs at the vertex, where x = 0, and y = c. 10 Always true. 2

For y = ax + c, the parabola has a turning point determined by the sign of a. If a > 0, it opens upward and has a minimum value of c at x = 0. If a < 0, it opens downward and has a maximum value of c at x = 0:

4 3

maximum

1 minimum 1

b (0, −80) c The minimum value is −80 at x = 0, representing a $80 loss when no cupcakes are sold (for example, ﬁxed costs).

Hours

Fee

2 3 4

$25 $100 $225 $400 $625 $900

Fee 800

−4

11 The y-value of the turning point is signiﬁcant in determining the range of the quadratic function y = ax2 + c because it represents the minimum or maximum value of the function, depending on the sign of a. When a > 0, the parabola opens upward, and the y-value of the turning point, which is c at x = 0, is the minimum value. Thus, the range is all y-values greater than or equal to c. When a < 0, the parabola opens downward, and the y-value of the turning point, also c at x = 0, is the maximum value. Thus, the range is all y-values less than or equal to c. 2

d A

14 a The parabola opens upward because a = 20 is positive. It has a minimum value.

−3

ii A

c C

Problem-solving

15 a

−2

12 a i D

b D

d 20x2 = 500, so x2 = 25 and x = ±5. Only x = 5 is practical, meaning selling 5 dozen (60 cupcakes) yields a $420 proﬁt.

2 −4 −3 −2 −1 −1

13 a B

iii C

iv B 2

b For y = ax + c, we solve ax + c = 20 by rewriting it as ax2 = 20 − c. If a is positive, the parabola opens up, and there are solutions only if c is less than 20. A bigger a makes the parabola steeper, so the solutions (where the graph hits y = 20) are closer to the y-axis. For example, with y = 16x2 − 36, the solutions are farther apart than with y = 49x2 − 36. If a is negative, the parabola opens down, and solutions only happen if c is more than 20, because the graph needs to reach down to

600 400 200 0

c y = 25x2

Hours 6

d $1600

e 10 hours 16 a

160

241

D 250 200 150 100 50 x 1

Answers mathspace.co

881

b 9x2 = 144, so x2 = 16 and x = ±4. Only x = 4 makes sense, meaning the rocket reaches 160 metres after 4 seconds.

8 a i

4 3 2 1

c The vertex at (0, 16) represents the starting height of 16 metres when the rocket is launched at x = 0.

y = x2 1 2 3

−4 −3 −2 −1 −1 −2 −3 −4 −5

5.02 Vertical translations

x 4

y = x2 − 5

Understanding ii −5

1 Examples answers: y = x2 + 5, y = x2 + 10

iii (0, −5)

b i

5 4 3 2 1

2 Example answers: y = x2 − 4, y = x2 − 8 3 What changes: 2. y-coordinate of the vertex: Changes from 0 to 7 (vertex from (0, 0) to (0, 7)). 3. y-intercept: Changes from 0 to 7.

y = x2 1

ii 3

−4 −3 −2 −1

2. Axis of symmetry: Remains x = 0

−3

3. x-coordinate of the vertex: Remains 0

−6

4. Direction of opening: Upwards

−12

iii (0, −12)

d i

y 20 15

y = 2x2 + 10

Fluency

c 3 units downwards

b 3 units downwards

−4 −3 −2 −1

unit upwards

y = x2 1

x 4

−5

b 2 units upwards d 1 unit downward ii 10

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

iii (0, 10)

b y = x2 − 11

c y = x2 − 49

d y = x2 − 8

9 a y=x

882

xx 4

y = x2 − 12

ii −12

5 The connection between the y-intercept and the value of q in y = x2 + q is that the y-intercept is equal to q, because when x = 0, the equation becomes y = q.

7 a 4 units upwards

y = x2 2 3

−9

5. Domain: (−∞, ∞)

x 4

1. Shape and width: Congruent parabolas

c 1 unit upwards

What stays the same:

6 a 6 units upwards

iii (0, 3)

c i

5. Range: Changes from [0, ∞) to [7, ∞).

4 The entire parabola shifts below the x-axis so there are no longer x-intercepts.

y = x2 + 3

−4 −3 −2 −1 −1 −2 −3 −4 −5

1. Vertical position: Graph shifts 7 units up.

4. x-intercepts: y = x2 has one at x = 0; y = x2 + 7 has none.

10 a y = 2x2 + 3, The graph has been narrowed (vertically enlarged) by a factor of 2 and vertically translated up 3 units. b y = x2 − 1, The graph has been vertically translated down by 1 unit. c y = −5x2 − 4, The graph has been vertically reﬂected across the x-axis, narrowed (vertically enlarged) by factor of 5 and vertically translated down 4 units.

Problem-solving 16 a The vertical translation of 10 metres means that the initial height of the toy rocket is 10 metres above the ground. To include this in the equation, we need to add 10 to the original height equation: h = −4t2 + 30 + 10. The new equation becomes h = −4t2 + 40. b 40 metres d 40

, The graph has been vertically

vertically translated up

and

unit.

20 15

11 hmax = 20 metres and t = 2 seconds

12 a 46 metres

50 45 40 35 30 25 20 15 10 5

h (metres)

reﬂected across the x-axis, widened (vertically compressed) by a factor of

c 3.16 seconds

t (seconds) 1

17 a The coeﬃcient −2 in the equation h = −2t2 + 8 represents the effect of gravity on the water’s height, as it causes the parabolic arc to open downwards. The constant term +8 represents the initial height of the water above the fountain when t = 0. t

c The initial height of the rocket is given by the value of h when t = 0: h = 4(0)2 + 10 = 0 + 10 = 10 metres. This value corresponds to the y-intercept of the graph. Reasoning 13 a y = −x2 + 3 c y = −x2 − 1

b y = −2x2 + 8 d y = −x2

14 a y = x2

b y = 3x2 − 4

d y = x2 − 12

c y=x −3

15 Yes, vertical translations always result in a new turning point. When a function is vertically translated, every point on the graph, including the turning points, is shifted up or down by a constant value. This means that the turning points will have the same relative position to the rest of the graph, but their coordinates will be different due to the vertical shift.

b 8 metres c 2 seconds d The two fountains have similar parabolic arcs as both equations have the same coefficient of −2, indicating the same effect of gravity on the water’s height. However, the constant term differs, with +8 for the first fountain and +10 for the second fountain. This means that the second fountain has an initial height of 10 metres, which is 2 metres higher than the first fountain. Consequently, the maximum height of the water in the second fountain is also 2 metres higher than the first fountain. 18 a b As we move further and further away from the hill (increasing x), the elevation (h) will continue to increase indefinitely according to the equation , which is a parabola opening upwards. This indicates a problem with the model, as it does not realistically represent the elevation of a hill, which should have a maximum height and then decrease as we move further away.

Answers mathspace.co

883

19 a 1500 cm

9 a i

7 6 5 4 3 2 1

b It takes approximately 22.36 minutes for the tank to be completely drained. c i T he amount of water in the tank is affected by the vertical translation in the equation. In this case, the constant term 1500 represents the initial water level in the tank. ii The rate the water drains is affected by the dilation in the equation. In this case, the coeﬃcient of the x2 term, −3, represents the rate at which the water drains.

iii (−1, 0) 7 6 5 4 3 2 1

Understanding 1 Example answers: • y = (x + 2)2 • y = (x + 5)2

−5 −4 −3 −2 −1 −1 −2

2 Example answers: • y = (x − 3)2 ii 1

• y = (x − 8)2

4 When the graph of y = −x2 is shifted left, the y-intercept changes. For example, if we shift y = −x2 two units to the left, the new y-intercept will be (0, −4). 5 The relationship between the x-value of the turning point and the value of p in the equation y = (x − p)2, is that the x-value of the turning point is equal to p.

7 6 5 4 3 2 1

ii 9

b 1 unit left

c 1 unit right

d 5 units left

8 a 1 unit right

b 3 units left

c 10 units right

d 5 units left

e 3 units left

g 6.5 units right

h 5 units right

unit left

d i 7 6 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

1 2 3 4 5

iii (−3, 0)

d Left

7 a 3 units right

−5 −4 −3 −2 −1 −1 −2

Fluency c Right

1 2 3 4 5

iii (1, 0)

c i

3 When comparing the graph of y = x2 and y = (x − 2)2, the shape and width of the parabolas remain the same, but the vertex shifts from the origin (0, 0) to the point (2, 0) due to a horizontal translation of 2 units to the right.

884

1 2 3 4 5

ii 1

5.03 Horizontal translations

b Left

−5 −4 −3 −2 −1 −1 −2

b i

6 a Right

iii

x 1 2 3 4 5

10 a i

12 a The graph has been horizontally translated 6 units to the left.

7 y 6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

ii x = −1 b i

b The graph has been horizontally translated 9 units to the right. x

c The graph has been vertically reflected across the x-axis and horizontally translated 3 units to the right.

1 2 3

d The graph has been horizontally translated 13 units to the left. e The graph has been horizontally translated

iii (−1, 0)

unit to the left.

7 y 6 5 4 3 2 1 −2 −1−1 −2 −3 −4 −5 −6 −7

f T he graph has been horizontally translated 0.15 units to the right. 13 y = − (x − 6)2 or y = −(x − 4)2

x 1

14 There are two values for D that solve the equation approximately 7.32 metres and −5.32 metres, however negative distance doesn’t make sense in this context, so the answer is 7.32 m.

2 3 4 5 6 7

ii x = 5

iii (5, 0)

c i

Reasoning

7 y 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

ii x = −2

15 A quadratic of the form y = (x − p)2 always yields non-negative y-values because squaring any real number, whether positive, negative or zero, results in a positive value of zero.

x 1

16 The maximum y-value is 5. This is always the case because, for any value of p, (x − p)2 will always be zero or positive since it’s a square number. 17 Sample Answer:

iii (−2, 0)

d i

For a given value of x to get the same y-value as y = x2 for the function y = (x − 3)2 we need to add 3 to the x-value.

7 y 6 5 4 3 2 1 x −9 −8 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

ii x = −6 11 a y = (x + 7)

For example: For the quadratic equation y = x2 when x = 2, y = 4, to get a y-value of 4 for the quadratic y = (x − 3)2 we need x − 3 to be 2. So we add 3 in order to get the point (5, 4).

Problem-solving

iii (−6, 0) 2

b y = (x − 13)

d y = (x − 4.6)2

e y = −(x + 3)2

f y = −(x − 2)2

18 a There are two possible horizontal positions: 7 metres to the right of the shooter (positive direction) and 1 metre to the left of the shooter (negative direction).

Answers mathspace.co

885

b The target should be placed at a height of 484 metres above the ground if the marksman wants to hit a target 25 metres away. 19 a When t = 0, s = 4 km/hr. b When s = 9, t = −1, 5. We take positive values only so t = 5 hours.

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 s = 0 when t = 2 hours

2 The graph is concave upward.

d Fastest speed occurs at the end of the journey at 6 hours s = 16 km/hr

y 10

16 s(km/hr)

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

−2

x 1

turning point t(hours) 2

3 Turning point

b R = (12 − 4)2 = 64, which is $64 000 c Sales reach 6000, set R = 6. So, 6 = (m − 4)2. Solve for m we get two values as: ± + 4 = m the ﬁrst time the sales reach 6000 is in month 1.55, which we would say is month 2.

4 The x-intercepts Fluency 5 a i H orizontal shift 3 units to the right and vertical shift 4 units down ii (3, −4) iii

d At month 4, the revenue is 0. y (in dollars)

7 6 5 4 3 2 1

−1 −2 −3 −4 −5

5.04 Optional: Quadratic graphs

x 1

b i R eﬂection across the x-axis, horizontal shift 3 units to the left and vertical shift 6 units down

x (month) 1 2 3 4 5 6 7 8 9 10 11 12

ii (−3, −6) iii

Understanding 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.

886

x-intercepts

20 a The maximum revenue occurs at the end of the range, so at month 12.

72 64 56 48 40 32 24 16 8

y-intercept

axis of symmetry

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x y 1

−5 −4 −3 −2 −1 −5 −6 −7 −8 −9

c i H orizontal shift 1 unit to the left, vertical shift 4 units up and dilated (stretched) by a factor of 3.

y 3 2

ii (−1, 4)

iii

11 10 9 8 7 6 5 4 3 2

−1

x 1

−1 −2 −3

c i Concave down

ii More steep

iii (−5, −4) v

−5 −4 −3 −2 −1

1 2 3 4

d i R eflection across the x-axis, horizontal shift 6 unit to the right, vertical shift 1 unit up and dilated (compressed) by a factor of 0.5.

iv (0, −79)

x y −9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6

ii (6, 1)

−8

iii

−10

4 3 2 1

−12

x 1

−1 −2 −3 −4 −5

2 3 4 5 6 7 8 9

d i Concave up

ii More steep

iii (−1, 2)

iv (0, 5)

6 a i Concave up

18 16 14 12 10 8 6 4 2

ii More steep

iii (1, −3)

iv (0, −1)

y 3 −4 −3 −2

2 1 −3 −2 −1

x 1

−1

−3

iii (2, 2)

−1

ii Less steep iv

ii (−3, −9)

iii

1 −6 −5 −4 −3 −2 −1

−2

b i Concave down

7 a i y = (x + 3)2 − 9

−1 −2 −3 −4 −5 −6 −7 −8 −9

y x 1

Answers mathspace.co

887

b i y = −(x − 2)2 + 1

9 a y = (x − 2)2 − 6

ii (2, 1)

iii 1

x −6 −5 −4 −3 −2 −1

1 2 3 4 5 6 7

−1

−1 −2 −3 −4

6 5 4 3 2 1

x 1

−1 −2 −3 −4 −5 −6

10 a i y = 0

ii x = 0, −4 iv (−2, −4) 4

ii (−4, −2)

iii

y x

−9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

iii x = −2 v

c i y = − (x + 4)2 – 2

−4 −3 −2 −1 −1

−2 −3 −4

b i y=0

ii x = 0, 4

iii x = 2

iv (2, 4)

d i

3 2

iii

−2 −1

x 1

−1

−2

−3

−4

3 2 1 −3 −2

−1

ii x = 2, 6 iv (4, −4)

c i y = 12 iii x = 4

b y = (x − 1)

c y = −x2 + 6

d y = 3(x + 2)2 2

e y = 0.5(x + 2) − 3

f y = −2(x − 3)2 + 4

8 6 4 2 −2 −4

888

10 2

8 a y=x +4

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1

d i y = −3

ii x = 3, −1

iii x = 1

iv (1, −4)

b y = −2(x − 1) (x + 7)

28 24

1 −2

−1

x 1

−1

8 4

−2 −3

ii x = 5, −1

iii x = 2

iv (2, 9)

f i y=3

d y = −0.5(x − 1) (x + 7)

iv (2, −1) 6

24 20

1 −1

−1

ii x = 1, 3

iii x = 2

−8 −12 −16 −20 −24 −28 −32 −36 −40 −44 −48 −52 −56

x 1

y −8 −7 −6 −5 −4 −3 −2 −1−4

−2 −1

1 2

c y = 3 (x − 1) (x + 7)

e i y=5

9 8 7 6 5 4 3 2 1

−8 −7 −6 −5 −4 −3 −2 −1

−4

x 1

−8 −7 −6 −5 −4 −3 −2 −1

x 1 2

−2

11 a y = (x − 1) (x + 7)

12 a y = −(x − 6) (x + 1)

b (2.5, 12.25)

13 a

and y = x2 − 3

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

x 1 2

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) ii y = (x − 6)2 − 4 iii (6, −4)

Answers mathspace.co

889

5 4 3 2 1

c i Concave down

ii (2 + x) (2 − x), (−2, 0), (2, 0) iii y = 4 x 1

−1 −2 −3 −4 −5

iv x = 0

v (0, 4)

y 4

3 2 1 −2

b i y = −2(x − 1) (x + 5), (1, 0) and (−5, 0)

−1

ii y = −2(x + 2)2 + 18

−2

iii (−2, 18) iv

18 16 14 12 10 8 6 4 2 −6 −5 −4 −3 −2 −1

d i Concave up

ii (x + 3) (x + 1), (−3, 0), (−1, 0) iii y = 3

iv x = −2

y 4 3 2

x 1

2 3

x −4

15 a i Concave up

−3

−2

−1 −1

ii (x − 1) (x − 3), (1, 0), (3, 0) iii y = 3

iv x = 2

v (−2, −1)

v (2, −1)

e i Concave up

ii (x + 5) (x + 1), (−5, 0), (−1, 0)

iii y = 5

iv x = −3

v (−3, −4) 7 6 5 4 3 2 1

−1

−1 −2 −6 −5 −4 −3 −2 −1

b i Concave up ii (x + 6) (x − 6), (−6, 0), (6, 0) iii y = −36

iv x = 0

y −6 −4 −2 −4 −8 −12 −16 −20 −24 −28 −32 −36

890

v (0, −36) x 2

−1 −2 −3 −4

f i Concave up ii (x − 3) (x − 5), (3, 0), (5, 0) iii y = 15

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

iv x = 4

v (4, −1)

16 14 12 10 8 6 4 2

j i Concave down

−2

ii −(x + 2) (x − 4), (−2, 0), (4, 0) iii y = 8

iv x = 1

9 8 7 6 5 4 3 2 1

x 1

g i Concave down ii −(x + 4) (x − 6), (−4, 0), (6, 0) iii y = 24

iv x = 1

iii y = −3

iv x = −1

v (−1, −4)

3 2 1

x 1 2 3 4 5 6

−4 −3 −2 −1 −1

x 1

iv x = −1

−4

v (−1, 9) l i Concave up

9 8 7 6 5 4 3 2 1

ii (2x + 3) (x + 3), (−3, 0), iii y = 9

10 9 8 7 6 5 4 3 2 1

−4 −3 −2 −1

i i Concave up ii (x + 2) (x − 4), (−2, 0), (4, 0) iv x = 1

v (1, −9)

y −2

−1

−1 −2 −3 −4 −5 −6 −7 −8 −9

−3

ii −(x + 4) (x − 2), (−4, 0), (2, 0)

−2

h i Concave down

iii y = −8

ii (x + 3) (x − 1), (−3, 0), (1, 0)

−4 −3 −2 −1 −3

iii y = 8

−1

k i Concave up

24 21 18 15 12 9 6 3

−2

v (1, 25)

v (1, 9)

x 1

−4

−3

−2

−1

Answers mathspace.co

891

16 a i x = 2

ii (2, −3)

iii y = 1

e i x = −3

3 1

x 1

−2 −3 −4

b i x = −3

ii (−3, −2)

iii y = 7 9 8 7 6 5 4 3 2 1

−9−8 −7 −6−5−4 −3 −2 −1 −1 −2 −3 −4

f i x=5

iii y = 1 4 3 2 1

−4 −3 −2 −1−1

x 1

−1 −2 −4 −6 −8 −10

−2 −3 −4 −5 −6 −7

d i x = −4 iv

x 1 2 3 4 5 6 7 8 9 10 11

iii y = −9

ii (4, 7) y

x 1 2 3 4 5 6 7 8 9

ii (−1, −12)

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

892

h i x = −1

iii y = −11

ii (−4, 5)

8 6 4 2

iii y = −23

ii (5, 2)

g i x=4 iv

x 1 2 3 4

−1 −2 −4 −6 −8 −10 −12 −14 −16 −18 −20 −22

1 2 3 4

c i

4 2

iii y = −2 y

2 1 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 −14

−4 −3 −2 −1 −1

ii (−3, −11)

iii y = −11

y 2

x 1 2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

−4 −3 −2 −1 −2 −4 −6 −8 −10 −12

x 1

Reasoning 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. 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. 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. 19 a a > 0 means the curve will be concave up and so has a minimum turning point, q < 0 means the turning point will be below the x-axis, so the curve will cut the x-axis at two points. p can take any value. b a < 0 means the curve will be concave down and so has a maximum turning point, q = 0 means the turning point will be on the x-axis, so the curve will touch the x-axis at one point. p can take any value. c a > 0 means the curve will be concave up and so has a minimum turning point, q > 0 means the turning point will be above the x-axis, so the curve will not cut the x-axis at all. p can take any value. 20 a y = (x − 2)2 + 5

b y = (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)

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 − 2apx + ap2 + q b y = ax2 + (−2ap)x + (ap2 + q) so b = −2ap c y = ax2 + (−2ap)x + (ap2 + q) so c = ap2 + q 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, (p, q), 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,

, in standard form

and use it to ﬁnd the y-value of the vertex. In both forms, the leading coeﬃcient, a, is the vertical stretch/compression factor and tells us if the parabola will open up or down. 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 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. 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.

Answers mathspace.co

893

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.

27 a 6 m

Problem-solving

25 a

9 16 21 24 25 24 21 16 9

9 10

4 3 2 1 x −5

20 15 10 5 t

500 450 400 350 300 250 200 150 100 50

2 3 4 5 6 7 8 9

b s = 19

c 524

−2

−1

cost ( y)

metre (x) 2

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: 15 = a(8 − 4)(8 − 12) If we solve this equation we get

5.05 Optional: Quadratic equations Understanding 1 5 2 A quadratic equation can have 0, 1, or 2 solutions.

894

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.

d 25 m

5 10 15 20 25 30 35 40

1000 900 800 700 600 500 400 300 200 100 0

−3

b The price would decrease by $393.75

26 a

−4

29 a Elise should expect to pay $900 c

c 24.75 m

y 5

c x = −4

b 3m

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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 = 2, b = 7

b a = −5, b = 1

c a = 0, b = −8 5 x = −12

12 Mark’s answer is incorrect. When using the null factor law, the correct equations should be x − 9 = 0 and x + 13 = 0. Solving these equations, we get x = 9 and x = −13 as the correct solutions, not x = −9 and x = 13. 13 Rearranging gives x2 = −4. There are no real numbers that will give −4 when squared.

Fluency b k = ±5

c x = ±11

Problem-solving

e x = ±11

f y = ±5

g x = ±5

14 t = 14, t = 10

i k = ±2

k x = ±15

6 a

15 k = −16 16 a = 13

7 a z = 4, −10

b m = 16, −2

d p = 11, −5

e n = −7, 11

f x = −2, 16

h p = ±10

8 a n = 0, −9

17 p = 2, q = −3 18 a b i ac < 0 or a = 0, c ≠ 0 ii a ≠ 0, c = 0 iii a > 0, c > 0 or a < 0, c < 0

c p = 0, −6

d q = 0, −8

h r = 3, 8

ii Maximum

i t = −4, −10

iii The equation of the axis of symmetry is x = 0.

iv The x-intercepts are at (−1.1, 0) and (1.1, 0), while the y-intercept is at (0, 22).

Reasoning 9 We can ﬁnd the solutions for x by taking the positive and negative square roots of k, that is, . 10 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. 11 Answer may look like: 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.

Chapter 5 review 1 a i The turning point is (0, 22).

b i The turning point is (0, −15). ii Minimum iii The equation of the axis of symmetry is x = 0. iv The x-intercepts are at (−1.5, 0) and (1.5, 0), while the y-intercept is at (0, −15). 2 a i

−2

−1

y 30 25 20 15 10 5 −2

−1

x 1

Answers mathspace.co

895

b i

−2

−1

−11

−2

−1

x −2

f i

−2

−1

−43

−13

−3

−13

−43

y −2

−1

−5 −10 −15 −20 −25 −30 −35 −40 −45

−1

−15

8 y 6 4 2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

−2

−1

−54

−15

−2

−15

−54

−2

−1

−4

−6

−4

4 y 3 2 1

−2

−1

−4

−1 −1 −2 −3 −4 −5 −6 −7 −8 −9

−2

−1

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1

45 40 35 30 25 20 15 10 5 −2

896

−5 −10 −15 −20 −25 −30 −35 −40 −45 −50 −55

−1

h i

−2 1

g i

−2

−1

e i

15 10

−2 −4 −6 −8 −10 −12

d i

y 4 2

c i

−1

x 1

3 a

−3

−2

−1

−4

−6

−4

b Minimum c Positive 4 i Concave down x

−2

−1

−62

−11

−62

5 −2

−1

−5 −10 −15 −20 −25 −30 −35 −40 −45 −50 −55 −60 −65

x 1

e y = −(x + 4)2

c The graph has been horizontally translated 11 units to the right. d The graph has been horizontally translated 15 units to the left.

units to the right. 12 The water is 6 metres high at horizontal distances of 2 m and 6 m.

c 5 units upwards

d 4 units downwards

6 a 6 units upwards

b 3 units upwards

c 5 units downwards

b The graph has been vertically reﬂected across the x-axis and horizontally translated 5 units to the right.

f The graph has been horizontally translated

b 7 units downwards

c y = x2 − 49

c y = (x + 6.4)2

e The graph has been vertically reﬂected across the x-axis and horizontally translated 9 units to the right.

5 a 8 units upwards

7 a y = x2 + 1

b y = (x − 10)2

11 a The graph has been horizontally translated 8 units to the left.

ii (0, 6) iii i

10 a y = (x + 8)2

13 a y = (x + 3)2 + 5 2

b y = (x − 4)2 − 2

c y = −2(x + 2) + 3

d y = 3(x + 5)2 − 4

e y = −(x − 6)2 + 8

f y = 2(x + 4)2 − 10

d 2 units downwards

14 a i y = 0

ii x = 0, −6

iii x = −3

iv (−3, −9)

b y = x2 − 13

4 y 3 2 1

d y = x2 − 19

8 a y = 2x2 + 4. The graph has been narrowed (vertically enlarged) by a factor of 2 and vertically translated up 4 units. b y = x2 − 2. The graph has been vertically translated down by 2 units. c y = −5x2 − 5. The graph has been vertically reﬂected across the x-axis, narrowed (vertically enlarged) by a factor of 5 and vertically translated down 5 units.

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

x 1

d y = −3x2 + 3. The graph has been vertically reﬂected across the x-axis, narrowed (vertically enlarged) by a factor of 3 and vertically translated up 3 units. 9 a 5 units right

b 4 units left

c 2 units right

d 6 units left

Answers mathspace.co

897

f i y = 14

b i y=0

ii x = 0, 5

iii x = 2.5

iv (2.5, 6.25)

7 6 5 4 3 2 1

iii x = 4.5 v

−1−1 −2 −3 −4

ii x = 2, 7

iv (4.5, −6.25)

18 y 16 14 12 10 8 6 4 2 −2 −1 −2 −4 −6 −8

1 2 3 4 5 6 7 8 9

c i y = 32

ii x = 4, 8

15 a y = (x + 5)(x − 9)

iii x = 6

iv (6, −4)

35 30 25 20 15 10 5 −2 −5

x 2

y −4 −2 −5 −10 −15 −20 −25 −30 −35 −40 −45

8 10 12

x 2

b y = −0.5(x + 5)(x − 9) d i y = −10

ii x = 5, −2

iii x = 1.5

iv (1.5, −12.25)

y 2 −3 −2 −1 −2

y 30

1 2 3 4 5 6

−4

−6 −8 −10 −12

e i y = 18

ii x = 6, −3

iii x = 1.5

iv (1.5, 20.25) y 25 20 15 10 5 x

−4−3−2 −1

898

c y = 2(x + 5)(x − 9)

−4 −2

1 2 3 4 5 6 7

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

y −4 −2 −10 −20 −30 −40 −50 −60 −70 −80 −90 −100

d y = −0.25(x + 5)(x − 9)

d i x = −5

iii y = −16

ii (−5, 9)

10 y 8 6 4 2 x

20 15 −10 −8

−6

−4

−2 −2 −4 −6 −8 −10 −12 −14 −16 −18

10 5 x −4 −2

16 a i x = 3 iv

ii (3, −4)

iii y = 5

e i x = −6

−2

x 1

−4

b i x = −4

ii (−4, −4)

iii y = 12

6 3 x 1 2

iii y = −3

ii (−2, 5)

iv 6

−4 −6 −8

10 12

iii y = −20

ii (6, 16) y

10 5

2 −2

x 2

g i x=6 15

−5 −4 −3 −2 −1

iii y = −45

ii (7, 4) y

−5 −10 −15 −20 −25 −30 −35 −40 −45

c i x = −2

f i x=7 iv

−8 −7 −6 −5 −4 −3 −2 −1 −3

iii y = −7

−14 −12 −10 −8 −6 −4 −2 −5 −10 −15 −20 −25 −30 −35 −40 −45

ii (−6, −43)

x 1

−5

x 2

−10 −15 −20

Answers mathspace.co

899

h i x = −5

ii (−5, −40)

iii y = −15

Fluency

y x

4 a

−12 −10 −8 −6 −4 −2 −5 −10 −15

5x 50 51 52 53 y 1 5 25 125

−20

b Exponential

−25

5 a Exponential

−30 −35 −40

625 3125

b Linear

c Quadratic

d Exponential

e Exponential

f Exponential

g Linear

h Exponential

17 a b = ±4

b m = ±6

6 a i 3.5

ii 1500.625

c z = ±12

d w = ±13

b i 15

ii 101 250

e x = ±14

f y = ±5

c i 2

ii 1.968

g x = ±6

h y = ±2

d i 0.25

ii 16

i k = ±3

j k = ±1

k x = ±16

7 a 5

b 125 triangles

8 a The volume increases by a factor of 3 each minute.

18 a z = 3, −13

b m = 18, −2

c x = 13, 5

d p = 13, −5

e n = −9, 15

f x = −1, 21

h p = ±9

243

729

g 19 a m = 0, −12 c p = 0, −8

d q = 0, −10

h r = 5, 10

i t = −6, −15

Reasoning 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. They do not lie on a straight line.

ii 80, 160

c 256 mL

b i Adding 5

ii 25, 30

c i Multiplying by 3

ii 202.5, 607.5

10 a graph B =

d i Dividing by 10

ii 4, 0.4

1 a i Multiplying by 2

2 a Multiplying by 3

b 108 circles

3 a Multiplying by 4

b 128 hexagons

900

Volume (mL)

c 6561 mL

6.01 Exponential relationships Understanding

Time (minutes)

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b graph A = 2x

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 increases at a constant rate of .

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.

6.02 Characteristics of exponential graphs

Problem-solving

2 Asymptote

12 a Points A, C, F

b 6 units

c Points B, D, E 13

d 2

1 The base, a.

3 a True b False, if 0 < a < 1, then it is exponential decay. c False, if 0 < a < 1, then it is exponential decay.

Time

Population

160

320

d False, if a > 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.

The relationship between time and population is exponential because the ratio between consecutive values of the population is constant 2. 14

Understanding

5 a Increasing

b Decreasing

c Decreasing

d Increasing

e Increasing

f Decreasing

g Decreasing

h Increasing

Time

Mass

Fluency

100

6 a i

51.2

40.96

32.768

26.2144

−1

x y = 4x

4 16 64

5 25 125

y 30 25 20

The relationship between time and mass is exponential because the ratio between consecutive values of the mass is constant 0.8 15

15 10 5 x

Number of pallets Number of bricks 0

20 000

19 500

19 000

18 500

18 000

17 500

−2 −1

iii Increasing b i

−1

x x

y=5

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.

Answers mathspace.co

901

y 4 1 3 2 3 1 2 2 2 1 1 2 1 1 2

150 100 50 x −2 −1

iii Increasing c i

−1

x y = 10x

−2

100

1000

−1

iii Decreasing f i

−1 0

1000

800

600

y 5 2

400 200

x −2 −1

3 2

iii Increasing d i

−1 0

2 ii

1 2

−2

ii y = 0

7 a i Increasing iii (0, 1)

iv Answers may vary, for example: (2, 4) v

7 6 5

−1

3 2

iii Decreasing e i

iii Decreasing

2 3 1 4 1 1 2 1 1 4 1 3 4 1 2 1 4 −2

−1

−1 0 4

−4 −3 −2 −1

b i Increasing

x 1

ii y = 0

iii (0, 1) iv Answers may vary, for example:

902

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

7 6

iv Answers may vary, for example: (−2, 9)

4 3

1 1

3 2

ii y = 0

c i Increasing

iii (0, 1) y

ii y = 0

g i Decreasing

iii (0, 1)

iv Answers may vary, for example: (−1, 5)

4 3

1 1

ii y = 0

d i Increasing

−4 −3 −2 −1

2 1

iii (0, 1)

−4 −3 −2 −1

iv Answers may vary, for example: (2, 64) v

−4 −3 −2 −1

iv Answers may vary, for example: (1, 7) 8

−4 −3 −2 −1

ii y = 0

f i Decreasing iii (0, 1)

ii y = 0

h i Decreasing

iii (0, 1)

iv Answers may vary, for example:

−4 −3 −2 −1

4 3

ii y = 0

e i Increasing

iii (0, 1)

iv Answers may vary, for example: v

−4 −3 −2 −1

x 1

8 a No. For a > 0, the expression ax is greater than zero for all real x.

7 6

b y tends towards ∞.

c 0

d 1

e 0

3 2 1 −4 −3 −2 −1

x 1

Answers mathspace.co

903

f 7

Reasoning

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.

5 4 3 2 1

−5 −4 −3 −2 −1 −1

1 2 3 4 5

b 1

9 a

13 No, both functions are always greater than zero, because a positive base to any power is always greater than zero. c 16

d None. For all values of x, it will have a positive value.

14 a i False

ii False

iii True

iv True

b Shared y-intercept of (0, 1)

e It gets larger, approaching inﬁnity.

c As x gets larger, the y-values approach 0.

f It gets smaller, approaching zero. 15

10 a Positive b y = 0 is a horizontal asymptote of the curve. 11 a

Example justiﬁcation: When graphed on the interval 0 < x < 2, we can see that the two curves start at the same point but,

7 6

ends below the graph of

5 4

is approaching y = 0 faster,

that

. This tells us

2 1 −5 −4 −3 −2 −1 −1

b 7

which means it is decreasing more rapidly. The pattern continues even for larger values of x.

x 1 2 3 4 5

y1 y2

6 5 4

2 1 −5 −4 −3 −2 −1 −1

x 1 2 3 4 5

c The two graphs are reﬂections across the y-axis.

Problem-solving 16 a 125

b 1

d −2

17 a 1

b −2

c 3

d 3

18 m = 8 19 a y = 128 20 80 units

904

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b x=7

21 a b

27 24 21 18 15 12 9 6 −x 3 y=3 −4 −3 −2 −1

Mass (kg)

3 2 1 1

0 x

c The two graphs are reﬂections across the y-axis.

6.03 Exponential models

6 a y = 1300(1 + 0.08)v

Months 6

b $2806.60

7 41 199 people 8 6591 bees 9 $503 579 10 $5.20

Understanding 1 a Exponential decay c Neither

b Exponential growth d Exponential decay

11 $12 072 12 a

e Neither 2 a i Exponential decay ii 0.723 b i Exponential decay

Number of players (p)

c p = 2r d 256 players

iii 12.3%

c i Exponential growth ii 5.9

b Any power of 2 such as 2, 4, 8, 16, 32, 64 … iii 5%

ii 2500

Number of rounds (r)

iii 2.5%

d i Exponential growth ii 10 000

iii 100%

3 a 840

b 4%

4 a 1.4

b Increase by 40%

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 d

N (’000s)

80 70

Fluency

60 50

5 a 2.4 kg b

Months

Mass (kg)

1.8000

1.9800

2.1780

2.3958

2.6354

2.8989

3.1888

30 20 10 0

t 30

120 150

e Approximately 100 minutes

Answers mathspace.co

905

14 a

Week number

Weight of treats (g)

2000

1000

500

250

125

62.5

31.25

It will take 6 weeks to get the dog down to less than 50 g of treats per week. b

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

, where w is the number of

weeks on the plan and T (w) is the weight of treats in grams.

Seconds passed (t)

Undissolved sugar in grams ( y)

Reasoning

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.

c It is decreasing at a decreasing rate.

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.

16 a a represents the initial temperature of the substance, and r represents the rate at which it cools down.

e If the sugar dissolves twice as quickly, it can ,

be modelled by the equation

b T = 145(1 − 0.4)t which is equivalent to

d It will never reach 0 because there is no time that could make (0.6)t equal to zero.

The amount of sugar that dissolves between the second and third second is

Exponential relationships have a horizontal asymptote at y = 0.

, which is less than .

e Substance P 17 a

(N)

(C)

Problem-solving 19 a 256

480

2880

17 280

103 680

622 080

3 732 480

22 394 880

20 a

N−1

b C = 480 × 6

c 29 023 764 480 sales

906

c 0.88 °C

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b The 8th stage Day

Number of people infected

5 25 125 625

b 5n c 8 days d 210 = 1024 people

21 a

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.

Value ($) 3500 3000 2500 2000 1500 1000 500 0

No. of years 2 4 6 8 10 12 14 16 18 20

b 13 years 22 a

b 3 bounces

23 2020 24 24 hours 25 a i 179

ii 132

b i 70%

ii 40%

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 d The Motorbike 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. 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 to 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. Example reasoning: From the graph, we see that Raymond remembers about 85% of the

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. • 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.

6.04 Transformations of exponentials Understanding 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. c If k is negative, the graph of y = 2x + k moves downward from y = 2x. 2 a i Different

ii The same

iii Different

iv The same

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.

Answers mathspace.co

907

b y = −2x

3 a c y = 2x

9 a A vertical dilation by a factor of 0.5, this is a compression. b

4 a No c No

b Yes

d No

5 4

Fluency

5 a The graph is translated up 4 units.

b The graph is translated down 6 units.

f (x) = 8x

c The graph is dilated (compressed) by a factor

−3 −2

g(x) = 0.5(8x)

−1

of . d The graph is dilated (stretched) by a factor of 5. 6 a A translation down 5 units b

y 5 4 3 x 2 f ( x) = 1 3  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. 11 a Translated down 2 units b Reﬂection across the y-axis

−5 −4 −3 −2 −1 1 2 3 4 5 −1 x −2 g( x ) = 1 − 5 3   −3 −4 −5

c Translation up 3 units d Dilation by a factor of 2 12 Graph R 13 a A dilation by a factor of 3 b

c y = −5 7 a A reﬂection about the x-axis b

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

f (x) = 4

12 48

g(x) = 3(4)x

f (x) = 5x

c 16

1 2 3 4 5

12 10

g(x) = −5x

c As x increases, the function values get increasingly large in the negative direction. 8 y = 4x

908

−2 −1

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

8 6 g(x) = 3(4)x −3 −2 −1

4 2

f (x) = 4x x 1

14 a i (0, −4) iv

ii y = −5 6 5 4 3 2 1

b i (0, 3)

18 16 14 12 10 8 6 4 2

1 2 3 4 5

ii y = 2 8 7 6 5 4 3 2 1

ii y = 0

e i

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6

iii Increasing

−4 −3 −2 −1

iii Decreasing

f i (0, 2)

iii Increasing

ii y = 0

iv 4

iii Decreasing

3 2 1 −4 −3 −2 −1

−4 −3 −2 −1 −1 −2

2 3

x 1

−1

−2 −3 −4

c i (0, −3) iv

ii y = −4 3

iii Decreasing g i (0, 2)

2 1

18 16 14 12 10 8 6 4 2

−4 −3 −2 −1 −1

2 3 4

−2 −3 −4 −5

d i (0, 4) iv

ii y = 3 8 7 6 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2

h i (0, −3) iv

iii Increasing

−4 −3 −2 −1

iii Increasing

ii y = 0

3 4

ii y = 0 1

−4 −3 −2 −1 −1

iii Decreasing

y x 1

−2 x 1 2 3 4 5

−3 −4 −5 −6 −7

Answers mathspace.co

909

15 a y = 5 which tells us a = 5.

b The growth factor is 4 which tells us b = 4.

c y = 5(4x)

2 1

16 a y = −4 which tells us a = −4. which tells us

b The decay factor is

−4 −3 −2 −1 −1

−2

−3 −4

b y=3

17 a

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. ii y = 1

19 a i (0, 4)

Translated 1 unit upwards. iv

y −4 −3 −2 −1 −1

iii Vertical dilation by a factor of 3, a stretch.

x 1

−2

−3 −4

−5

−6

−7

−8

3 2 1

−4 −3 −2 −1

3 4

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 x = 0 into the transformed equation, we will get y = a.

ii y = 4

b i (0, 3)

iii Reﬂected across the x-axis. Translated 4 units upwards. iv

21 By reﬂecting the graph across the y-axis

3 2

22 For x < 0, the graph of

1 −4 −3 −2 −1 −1

Reasoning

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.

−2 −3

23 a Yes. For any value of x, 10x is always positive. Therefore, −10x is always negative.

−4

c i (0, −2.5)

ii y = −3

iii Vertical dilation by a factor of 0.5, a compression. Translated 3 units downwards.

910

is below the

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Problem-solving

y 50 −3 −2

−1 −50

26 a A reﬂection about the y-axis b g(x) = 2−x

−100 −150 −200

c There is no x-value for which y = 0.

27 a

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. 0

y = 3x

y = 2(3x)

y = 2(3x) − 1

1 2 3 4 5

−3 −2 −1

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

16 64

x 1 2 3 4 5

d The two graphs are the same.

e The graph of

4 3

can be reﬂected

across the y-axis to get the graph of

2 1 −4 −3 −2 −1 −1

g(x)

25 a Yalile’s graph is correct, Emir’s is incorrect.

x 1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −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).

−1

5 4 3 2 1

−5 −4 −3 −2 −1 1 2 3 4 5 −1 −2 x −3 y = −3(2 ) −4 −5

f (x) = 2x

c Decreasing

5 4 x 3 y = 3(2 ) 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−250

24 a A reﬂection across the x-axis

5 4 3 2 1

x 1

−2 −3

Answers mathspace.co

911

28 a

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.

Point on (−1, 2) (0, 1) original graph Point on new graph

(−1, 6) (0, 3)

3 a 53 e 6

b c 7

−2

b 25 f 10 b 7

c 3

d 0

e −3

f −2

g 2

h 2

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

Fluency

4 y=

1  2

3 2

y=3

1  2

1 −4 −3 −2 −1

−1

x 1

d Above

e x=0

e Above

29 30 a y = −4. Tells us k = −4. b

−1

−1 2

⬚

⬚ ⬚

⬚

−1

y = a(bx) + k y = a(bx) y = bx c a=3 d

x y = a(bx) + k

y = a(b )

y = bx

b i 24x = 2−1

c i 2−3x = 25

d i 2−x = 2−2

ii 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

e x=3

g x = −1

h x=4

j x=1

12 a x = 1

b x = −3

6.05 Optional: Exponential equations

7 a i 23x = 22

8 a x = −2

e y = 3(2x) − 4 31 a y = 5(2x) − 10

d 34

4 a 9

c 43

−3

Understanding

c x=3

1 a False

b True

e x=0

c True

d False

912

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Reasoning

b x=3

20 a

13 In 23x−1 = 27x−9 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.

e x = 4, −1

g y=4

21 Answers may vary. For example, (29−5)1 = 48−6.

6.06 Optional: Logarithms

Given equation

Understanding 2

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

15 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). b 16 When is squared, it equals 7. So the equation can be solved as shown: y = 4

Given equation Square both sides

1 a 100

b 0.1

c 10 000

d 1000

e 1

f 10

g 0.01

h 100 000

2 a The base ten logarithm of 100 is 2 because 102 = 100 b The base ten logarithm of 1000 is 3 because 103 = 1000 c The base ten logarithm of 0.01 is -2 because 10-2 = 0.01 d The base ten logarithm of 1 is 0 because 100 = 1 e The base ten logarithm of 65 is 1.8129 because 101.8129 = 65 f T he base ten logarithm of 224 is 2.3502 because 102.3502 = 224 Fluency 3 a i 106

ii 6

b i 10−4

ii −4

c i 100

ii 0

−3

ii −3

d i 10 2

Evaluate

b 100 = 1

4 a 10 = 100

Rewrite 49 with base 7 Power of a power law

Problem-solving 17 a 5 hours

b 10 hours

c 12 hours

18 a n = 8

b k=4

c m = −2

19 a 440 Hz

b 262 Hz

c n = 73

c 104 = 10 000

d 10−2 = 0.01

5 a 1.6532

b 3.8860

c 0.7126

d −2.2373

6 a 37.154

b 0.063

c 6.310

d 204.174

7 a B

b A

c A

d B

8 a 2 countries

b 10 countries

9 a 13

c 6

b 25

Answers mathspace.co

913

Reasoning

City

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

Frequency

4–5

5–6

6–7

11 a B

b C

Frequency

10 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

3 2 1 1 2 3 4 5 Log of lifespan

14 a R = 3.8

b Minor

6.07 Optional: Order of magnitude and logarithmic scales Understanding 1 a 102

b 103

c 106

d 101

e 100

f 10−2

g 10−1

h 10−3

2 a 6 × 102

b 7 × 103 6

c 4.8 × 10

d 3.5 × 101

e 2.3 × 102

f 7.6 × 10−3

−2

h 9 × 10−5

g 1.2 × 10

c A

3 a 3

b −5

c 0

d −1

Problem-solving Fluency

12 a $79 432.82

4 −7

b $100 000 to $1 000 000 13 a

Lifespan in a log10 scale

1.30

African elephant

25 500

4.41

Bowhead whale

71 400

4.85

Bumblebee

1.32

Chicken

2920

3.47

Dog

5110

3.71

9 a 16

Emperor penguin

8500

3.93

10 a 5.2

1.26

Animal/Insect Ant

Fly

914

5 a i 8 × 103

Lifespan (days)

ii 3

b i 9.644 × 105

ii 5

c i 8 × 10−4

ii −4

d i 2.574 × 10−1

ii −1

6 3

Giraffe

10 120

4.01

Hamster

780

2.89

Ladybug

365

2.56

Little penguin

2190

3.34

Monkey

7300

3.86

Shark

10 980

4.04

7 4 8 2 b 631

b 100.1 or approximately 1.26 times larger 11 a The concentration of hydrogen ions in the pH 8 solution is 100 times greater than in the pH 10 solution.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

c The population increases by a factor of 10 each day (e.g., 50 ×10n). Starting at 50 = 5 × 101, solve 50 × 10n = 108 : 10n = 2 × 106, so n = log10(2 × 106) ≈ 6.3. Thus, it takes 6.3 days, or 7 days when rounded up.

12 a 1 000 000 b 100 000 times louder 13 a 104 times

b 103 times

c 106 times

d 107 times

14 a 102.8 times

b 103.6 times

20 a 100 mW b 103 or 1000 times weaker c 2.5 21

c 106.4 times

22 R = 8.7

Reasoning 15 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.

23 a

16 On a base 10 logarithmic scale, the values 2 and 4 represent 102 = 100 and 104 = 10 000. The difference is not 2 times but a ratio of 4−2

= 10

1000 900 800 700 600 500 400 300 200 100

x 1

= 10 = 100 times, so the quantity b

increased by 100 times.

Base 10 log scale 0.602

1.204

1.806

256

2.408

1024

3.010

y 3

18 On a base 10 logarithmic scale, the difference 3.2 − 1.8 = 1.4 means the intensity ratio is 101.4. Calculating 101.4 ≈ 25.12 or 25.

Therefore, source A is 25 times brighter than Source B.

1 x

Problem-solving

19 a 5 b

b 101.5 times

Population of shrimp

17 a 102 times 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.

Chapter 6 review Population

Base 10 log scale

1 a Linear

b Quadratic

1.7

c Linear

d Exponential

500

2.7

5000

3.7

e Exponential

f Quadratic

50 000

4.7

500 000

5.7

2 a i 0.25

ii 7.8125

b i 2.5

ii 312.5

3 a

b 2 cubes

Answers mathspace.co

915

4 a i

−1 0

x y = 6x

6 36

216

7 a y = 7700 (1 + 0.053)n b $16 707.69 8 $12 068

9 a 10 300

b Decreasing

c 9.8%

d 6818 bees

200 160

10 a The graph is translated up 2 units.

120

b The graph is reﬂected across the x-axis.

c The graph is dilated (enlarged) by a factor of 3.

40 x −1

11 a A translation down 4 units

5 4 3 2 1

iii Increasing b i

−1

f (x) = 7x −5 −4 −3 −2 −1 −1 −2 −3 g (x) = 7x − 4 −4 −5

y 4 1 3 2 3 1 2 2 2 1 1 2 1 1 2

x 1 2 3 4 5

c y = −4 12 Graph Y b y=0

13 a d

−1

c Increasing

12 10

iii Decreasing 5 a y=0

b (0, 1)

6 4

c One point is

−8 −6 −4 −2

15 a x = 8

e x=6

5 x

6 a None

−1

b Positive

16 a 17 a n = 1

b x=4

c x = −2

d y = −3

h x = −8

c x=0

b k=5

c m = −2 ii −2

b i 104

ii 4

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b x = −2

18 a i 10−2 19 5 cities

916

14 a y = 256

−2

x 2

Animal

Weight (in kg)

Weight in log10 scale

Ferret

0.8

−0.10

Cat

3.2

0.51

Whale

32 840

4.52 3.44

Rhinoceros

2754.23

Arabian wolf

25.12

1.40

Hamster

0.15

−0.82

Shetland pony

127

2.10

21 a 16 to 32 years old

b 9 years old

−5

b −5

22 a 4 × 10

b i

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

c i y = −x + 3

b 10

24 a 1000

times

25 20

Understanding 1 The x- and y-coordinates at the point of intersection satisfy both equations. 2 The lines would be parallel and never intersect. b 0

e 0

f 1

c 1

b (−6, 4)

e (6, −3)

f (−4, −2)

c (2, −4)

5 a x = 6, y = 4

d i

x 2 4 6 8 10

x −10−8 −6 −4 −2 2 4 6 8 10 −2 −4x + 2y = −8 2x − 4y = −8 −4 −6 −8 −10

d (3, −6)

ii (4, 4) 10 8 6 4 2

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

ii (2, 4)

10 8 6 4 2

d Inﬁnite

b x = 0, y = −2

c x = 4, y = 0 6 a i

10 8 6 4 2

ii (0, 3)

Fluency 4 a (5, −1)

y = −x − 9

−10−8 −6 −4 −2 −2 −4 −6 y=x+3 −8 −10

7.01 Solve graphically

3 a 1

x 2 4 6 8 10

ii (−9, 0)

23 3 5.1

y y=x+9

e i

y y = 4x − 4 y = 8 − 2x x 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)

Answers mathspace.co

917

b (4, 450)

b (23.4, 28 076.7)

c Renting either studio A or B for 4 hours costs the same amount, $450.

c At approximately 23.4 weeks, Laura’s total living expenses will equal her total available funds. After this point, her cumulative expenses will exceed her funds.

8 a

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

y y = 2x + 10

x 1 2 3 4 5 6 7 8 9 10 11

c Non-members and members pay the same amount for 5 classes, $20. 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.

13 a No

b No

c Yes

d Yes

16 No, because there is an inﬁnite number of lines that go through that point. The gradient could be any real number.

Dollars

17 a To intersect at (0, 3) the line must have a y-intercept of 3, therefore c = 3. C

b The value m = 5 is not possible for a unique point of intersection. If m = 5, the second line y = 5x + 3 becomes identical to the first, resulting in an infinite number of solutions.

18 No, the lines have the same gradient, so they are parallel and will never intersect.

100 150 200 250

b (200, 800)

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.

c This is the break-even point. For 200 toys, both the cost and revenue are $800. 11 a

d Yes

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.

9 a (775, 2325)

1100 1000 900 800 700 600 500 400 300 200 100

c No

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.

b (5, 20)

10 a

b No

Reasoning

y = 4x

12 a Yes

30000 27000 24000 21000 18000 15000 12000 9000 6000 3000 0

y y = 10 000 + 772.60x

y = 1200x x 2 4 6 8 10 12 14 16 18 20 22 24 26 28

Answers mathspace.co

919

Problem-solving

23 Answers will vary. For example:

20 a y = x + 3, 2x + 2y = 18

Let x be the number of days and y be the weight of the rice in kilograms.

2x + 2y = 18 10

8 6 4 2 −8 −6 −4 −2 −2 −4 −6 −8

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.

y=x+3

x 2 4 6 8 10

The solution (14, 5) means that after 14 days, Maria and Arjun have the same amount of rice remaining.

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 0

y = 4x + 11

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. Since the two lines do not intersect in the first quadrant, they will never reach the same height on the same day.

7.02 Optional: Solve using substitution

y + x = 21

Understanding 1 a Answer varies:

x 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

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:

Equation 2: y = 2x − 6 b

10 y 9 8 7 6 5 4 3 2 1 −2 −1 −1 −2 −3 −4

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.

x+y=9 y = 2x − 6

d Answer varies: x

Junard will get

1 2 3 4 5 6 7 8 9 10

c Kang did 4 parts of the assignment and Elvia did 5 parts.

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.

920

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

, which simpliﬁes

3 a True

b True

14 Example explanation:

c False

d True

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.

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

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

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

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 Since this is always false, the two equations would not have any solutions. 16 a x = 4, y = 2 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

b In 2008, there were 250 000 jobs created and in 2013, there were 810 000 jobs created.

3 2

11 a x = 8, y = 14 b In the group, there are 8 adults and 14 children. Reasoning 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.

1 −2 −1 −1

x 1 2 3 4 5 6 7 8 9

−2 −3 −4

Problem-solving 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.

Answers mathspace.co

921

20 a (0, 0), (−5, 25)

b The solution is not valid because it is impossible to have a negative value of a passenger cab.

b (8, 24), (−3, 24)

c (2, 14)

d (−5, 3)

e (3, −2)

g (−1, 1), (2, −8)

h (1, 0),

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.

7.03 Optional: 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 b x = −3

4 a 7

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 b x = 2, y = 8

7 a

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)

c x = 2, y = −9

d x = 8, y = 3

e x = 40, y = −2

f x = −8, y = −2

Problem-solving

8 a x = 4, y = 5

b x = 3, y = −1

c x = 3, y = 1

d x = 5, y = 15

16 a x + y = 30 6x − 3y = 90

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.

922

b x = 20, y = 10 17 x = 10, y = 7 18 x = 5, y = 5 19 a 2016

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b 68%

14 The enlarged pentagon has a perimeter that is k times the perimeter of the original pentagon.

13 a 4

15 83.6 m 16 30 m b 5m

18 a 19 m

b 12 m, 28 m, and 36 m

c 76 m

d 4

e k 19 a 3

b 240 cm

20 a 60 cm

b 360 m

21 a 18y m

b 6y2 m

22 a k = 13

15 334.89 m2

b False

b 10 800 cm2

16 a 12 17 a 76 cm2

b 6 cm, 12 cm and 15 cm c 684 cm2 d 9 e k2

c True b 1 cm

3 0.04

18 a 4

b 3584 cm2

19 a 2.5 cm

b 2 519 424 L

20 48 400 km2 21 a 4x2 cm2 b x4 cm2

8.03 Scale and volume b x = 567

c x = 30

Understanding 1 a False

5 72 cm2

c 512 cm

10 2 m

d Multiplies it by 8

3 1.5 = 2.25 more times Fluency 4 a x = 1280

b x=4

c x = 36 000

5 a 4

b 64

c 54 528 cm3

6 a 4

b 64 3

Reasoning

7 1080 cm

11 The enlarged rectangle has an area that is k2 times the area of the original rectangle.

8 5

12 The circle has been dilated with a scale factor of 2. This means that its area will increase by a factor of 22 = 4, and 4 × 78.5 = 314 cm2.

c False 3

8 a 5

9 324 cm2

b 64 cm

2 a 8 cm

7 84 cm b The larger triangle’s perimeter is 5 times the smaller triangle’s perimeter.

b False

6 3136 cm2

930

c 1 cm2

22 k = 12

Fluency 4 a x = 20

14 18 m2

b l = 11 m

Understanding

2 a 60 cm

Problem-solving

c 6m

8.02 Scale and area

1 a True

c 24 L

d The amount of paint needed is not simply four times the amount for the ﬁrst ceiling because the paint covers area, not length. The area scale factor is the square of the length scale factor, so the amount of paint needed is sixteen times the amount for the ﬁrst ceiling.

Problem-solving

17 a 20

b 16

9 x = 8.2 10 11

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Reasoning 12 a The surface area will increase by factor of 32 = 9. b The volume will increase by factor of 33 = 27. 13 Capacity is the amount that an object can hold, given in litres in this case. The required volume (which is the space the esky takes up) of the new esky is eight times the old esky. This means that the required dimensions of the times the old dimensions. new esky are If Basili buys an esky with dimensions 8 times his old one he will end up with an esky with a capacity of 20 × 83 = 10 240 L, which is much larger than he needs. 14 a Since all sides of a cube are the same length, the side lengths of two different cubes are always in the same ratio. All the faces are also squares, and so all adjacent sides are perpendicular to each other. This means any cube is similar to another cube. b

c The perimeter increased by the same factor as the scale factor 1.5, because perimeter is a linear measurement. 5 a y = 50 6 128 cm 7 3.5 m 8 The area of the dilated triangle is p2 times the area of the original triangle. 9 0.75 m2 10 a 9y2 cm2

c 1 cm2

b x = 72

12 a 4

c 16 000 cm3

b 64

13 x = 3 cm 14 Capacity refers to the volume an object can hold (expressed in litres or similar units). The desired volume of the new ﬁsh tank is 8 times smaller than that of the old tank. This implies that the required dimensions of the new tank = 0.5, which is half the dimensions of

the old one. This means the manager’s estimation is correct.

Problem-solving b 1.96

16 a

c 2.744

d 0.343

15 a Yes because all corresponding sides are in the same ratio:

17 a 1 : 4

b 1 : 64

c m3 : n3

18 a 64

b 4

c h = 80

19 a 2

b 4

c 8

20 a 1 : 2

b 1:8

c 8

21 a 20 c 0.6429 cm3 22 a 1.5

d 16

b 7.6 cm d 5143 cm3 b 2.25 times greater

b 2.56

b 40 m

2 a 20 cm long and 9 cm wide b 58 cm 3 a 54 cm

b 1.5 cm

4 a 28 cm

b 42 cm

c 4.096

d 0.512

9.01 Trigonometric ratios Understanding 1 a AB 2 a AC

Chapter 8 review 1 a 0.02

b y4 cm2

11 a x = 116

are

15 a 1.4

b y = 812.5

b BC b BC

c BC

3 a i

iii

b i

iii

c i

iii

d AC

e ∠ACB

Answers mathspace.co

931

Reasoning

d i

iii

e i

iii

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.

f i

iii

Using the formula:

4 a i

b i

Simplify the numerator and make A the subject of the identity. Take the inverse cos of both sides.

c i

19 Using the formula:

Fluency 5 a i

iii

Multiply both sides of the equation by b, and make the subject of the equation by dividing both sides by 0.4.

b i

iii

Problem-solving

6 7 a f = 4.08

b f = 6.12

c g = 9.46

d a = 11.59

20 21

8 b=8 9 d = 3084 10 29.0° 11 a i x = 10

ii 53.1°

b i x=4

ii 53.1°

12 d = 1.48

22 a 135°

b I = 135°

23 a 131.2 m

b 101.3°

24 31.8°

9.02 Angles of elevation and depression Understanding

13 1.96 m 14 h = 4.4 m

1 It is an angle of elevation as it is the angle formed from the horizontal looking upwards.

15 θ = 9.88°

2 a σ or ∠CBD

16 a x = 155.43 m

b 425.86 m

17 a h = 21

b f = 22

c h = 154, x = 72

d x = 34

e x = 12

c They are the same size. 3 a α or ∠BCA 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

932

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b θ or ∠ACB

9.03 Optional: Sine and cosine rules

Fluency 4 a x = 37.23°

b y = 52.77°

5 a y = 36.87°

b x = 53.13°

Understanding 1 a Sine rule

6 θ = 52.52°

c Cosine rule

7 x = 182

b Pythagoras’ theorem d Trigonometric ratio

2 ∠ABC

8 d = 229

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

4 Possible answers: • p2 = q2 + r2 − 2qr cos P • q2 = p2 + r2 − 2pr cos Q • r2 = p2 + q2 − 2pq cos R

11 a d = 7.81 m c 66.08°

b h = 13.53 m

5 a Cosine rule

b Sine rule

c Sine rule

d Cosine rule

d 69.72°

Reasoning

Fluency

12 Viggo is incorrect as the angle of depression must be measured from the horizontal. The correct angle of depression is 23.9°.

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

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.

7 30.16 m 8 a ∠A = 75.78°

b b = 3.152 m

c c = 5.723 m

b The length of the shadow is zero.

9 a x = 71.71

b x = 45.45

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.

10 a x = 26.0°

b x = 32.7°

c x = 30.4°

d x = 30.3°

11 a x = 44.1°

b x = 43.2°

c x = 133.1°

d x = 23.5°

Problem-solving

12 25°

15 a x = 455.84 m

b 425.84 m

16 a h = 127.38

b 129.1 m

13 a ∠B = 74° or ∠B = 106° 14 a a = 19.71

17 d = 1022.58

c a = 212.35

18 h = 17 m

19 d = 10.86

16 q = 11

20 θ = 45.97

17 v = 85.17

21 a 29.26°

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

b B = 32°

b a = 10.00 d c = 10.61

18 a B = 43.38

b w = 31.69

c w = 41.49

d A = 27.33

19 V = 101° 20 Q = 18°

Answers mathspace.co

933

21 a 60.21 m2

b 12.41 m2 2

c 326.26 cm

d 99.52 cm

e 110.23 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 cm2

Simplify the equation Isolate cos A

So,

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. c = 8.40 m

Evaluate

. To ﬁnd the ∠A, we take the

inverse cosine of :

Since the triangle is equilateral and all sides are equal, the other ∠B and ∠C will also be equal to 60°. Problem-solving 30 θ = 33.84° 31 a ∠WST = 136°

b SW = 234 kilometres

c h = 102.58

27 a We could use the fact that diagonals of a rhombus bisect each other at right angles. using the

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. Another way to solve for d would be to use the law of cosines. A rhombus is a 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. b d = 12 cm 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.

934

Substitute c = a, b = a, C = A

Reasoning

So we could solve for

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:

32 1 244 548 km2 33 248.8°T 34 a a = 74°

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

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°.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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°. 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 2

42 a x = 43.60

b 360 m

c 1710 dollars

43 a h = 42

b x = 123

c y = 97

b 68 m

c 31.9 m

44 38 m 45 a 94°

Chapter 9 review 1 a f = 2.77

b f = 8.91

2 a

b θ = 30°

3 a 50°

b x = 19.30 m

c h = 20.50 m 4 a h = 20

b x = 26

5 θ = 54.69° 6 a AC = 12 032 m

b BC = 10 462 m

10.01 Surface area and volume of composite solids Understanding 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. 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 Two rectangular prisms b A trapezoidal prism and a triangular prism c A rectangular prism with a cylindrical hole 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

b 171 cm3 d 3136 cm3

6 a 1608.50 cm3

b 883.57 cm3

d 609.47 cm3

c 263.89 cm

7 θ = 28.01°

7 0.57 m3

8 a h = 92.97 m

b 94.6 m

9 a x = 13.13

b a = 6.85 b ∠B = 74° or ∠B = 106°

8 a 1140 m3

b 424 m3

c 1203.75 m3

d 272 cm3

e 982.35 cm

f 601.65 m3

h 1800 cm3

g 7164 units 9 a 1512 cm2

11 ∠ABC

c 228 cm

12 a = 19.71

b 429.37 m2

d 365.45 cm2 2

13 ∠C = 115° 14 a 5.94 m2

16 1389.32 m2

c 1920 cm3

c 1570 m

10 a B = 32°

b A = 33.2 m2

15 a x = 51°

b 24.30 m2

e 6755.25 mm

f 867.08 cm2

g 120 cm2

h 621.24 cm2

10 157π cm2

Answers mathspace.co

935

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.

Reasoning 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

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 Volume = 300 cm

14 a 1198.5 m3

b 70 m2

d 376 m2

c 100 m

e $16 544 18 a 14.7 + .05 p cm2

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:

b 614 d 1717 L

e 21

2 × 7 × 6 + 2 × 6 × 8 + 2 × 7 × 8 = 292 cm2 Subtract the triangle at each end of the triangular hole:

20 a 127.1 m2

b 140 m2

c $2520

10.02 Estimation and accuracy Understanding

1 a True

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

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

cm2

= 361 cm2 (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.

5 12.715 Fluency 6 a • Length of the rectangular prism: 7.5 cm • Height of the rectangular prism: 7.5 cm • Width of the rectangular prism: 5.5 cm • Diameter of the cylinder: 4.5 cm

Area of Rectangle = l × w Area of Square = side × side Volume = (14 × 14 − 2 × 6.5 × 5) × 14 m3 Volume = 1834.00 m3

936

b • Length of the rectangular prism: 8.5 cm • Height of the rectangular prism: 8.5 cm • Width of the rectangular prism: 6.5 cm • Diameter of the cylinder: 5.5 cm

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

directly proportional to velocity, so errors in the velocity measurement have a less drastic impact on the momentum calculation.

c Total surface area = 341 cm2 2

d Total surface area = 461 cm 7 a Total weight = 68.051 kg

b Total truncated weight = 68.05 kg

Problem-solving

c Total rounded weight = 68.05 kg

13 $9 125 000

d In this case, truncating and rounding the total weight to two decimal places both result in 68.05 kg. This value is 0.001 kg less than the actual total of 68.051 kg. 8 a 7.2%

b 14.6%

c 4.2%

d 5.3%

14 a 40.6 thousand years ago b 33.0 thousand years ago c 10.3% 15 a 48.15%

b 1.48%

10.03 Optional: Pyramids

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 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. 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 taken is

, which Alex

will measure as 2 seconds, giving a velocity of . b i 1.16 Joules

1 a B

b A

2 Add the area of the square base to four times the area of the one triangular face to get 20 + 4 × 15 = 80 cm2. 3 a 25 cm

b 12 m

4 40 cm3 5 2 cm3 6 24 m3 Fluency 7 a 288 cm2

b 1240.44 mm2 2

c 223.20 cm 8 a x = 6.71

d 941.37 cm2 b y = 7.21

c 144.9 cm

ii 317.1 cm2

9 a i 7.21 cm b i 10.55 mm

ii 353.1 mm2

c i 7.81 cm

ii 256.2 cm2

d i 9.43 mm

ii 288.7 mm2

10 120 cm2 ii 0.92 Joules

c 26.09% d i 0.93 kg m/s

Understanding

11 a 384 cm2

b 201 mm2 2

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

c 196.80 mm 12 99 360 m2 13 a 160 cm3 c 208 m3 3

e 80 cm

14 a RT = 38.12 cm

b 200 cm3 d 154.78 cm3 f 864 cm3 b 1677.28 cm2

c 38 018.35 cm3

Answers mathspace.co

937

15 a 12 cm

b 6 cm

16 a 24 mm

b 9 cm

25 a 16 cm3 or 16 000 mm3 b 466.13 cm3 c 85 634.88 m3 d 119.84 mm3

Reasoning 17 Barney’s mistake is using right-angled triangles with height 4 cm for the sides of the pyramid instead of triangles with the correct slant height.

e 2472.5 cm3 f 1497.6 mm3

10.04 Optional: Spheres

18 a s = 9 m b The volume of a rectangular prism is calculated by using the product of the area of its base by its height. The volume of a square-based pyramid is exactly one third the volume of a rectangular prism formed from the same square base with the same perpendicular height. Since the squarebased pyramid and the rectangular prism have the same base edge length and height, the volume of three of the square-based pyramids is equal to the volume of one rectangular prism. 19 The slant height is 9 cm. To find this, subtract the base area (122 = 144 cm2) from the total surface area (360 cm2), leaving 216 cm2 for the four triangular faces. Each triangular face has an area of 216 ÷ 4 = 54 cm2. Since the area of a triangular face is

2 The surface area is multiplied by 4 and the volume is multiplied by 8 3 a True

4 a i 113.10 cm2

× 48 × h, h = 10 m.

d True

ii 2144.66 cm3

c i 254.47 cm

ii 381.70 cm3

d i 113.10 cm2

ii 113.10 cm3 2

ii 4577.20 cm3

e i 1333.17 cm

f i 11 339.91 mm 5 a 16π cm2

ii 113 550.33 mm3 b 100π cm2

d 888.04π cm2

e 100π cm2

f 529π mm2

6 a c 36 000π cm3

b 972π cm3 d

7 333.04 cm3

× base area × h. Base area is

8 × 6 = 48 m2, so 160 =

c True

ii 113.10 cm3

b i 804.25 cm2

20 The new volume would be 320 m3. The original height is found using the volume formula V=

b True

Fluency

× base × slant height, we

× 12 × s = 54, so 6s = 54 and s = 9 cm.

solve

8 a 760.265 units2

b 380.133 units2

c 1140.40 units2

Doubling the height to 20 m gives × 48 × 20 = 320 m3. Since volume is

9 a 52 647.447 m2

b 7410.153 cm2

directly proportional to height, doubling the height doubles the volume.

10 a 452.389 units3

b 1 528 341.353 cm3

Reasoning Problem-solving 21 80 000 cm3 22 a 1.25

b Bottle B

23 a 7.81 cm

b 7.07 cm

c 502.94 cm2 24 a 5.39 cm

938

11 Changing the radius of a sphere will affect both its surface area and volume. Increasing the radius will increase both the surface area and volume, while decreasing the radius will decrease both.

b 209.4 cm2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

12 The composite solid consists of two hemispheres stuck together. We can ﬁnd the volume of the composite solid by adding the volumes of the two hemispheres together. The volume of a hemisphere is half that of a sphere. We can ﬁnd the volume of a sphere using the formula: Volume of a sphere = Notice that we are given the diameters of each hemisphere rather than their radii.

5 a i 5 cm

ii 75.40 cm2

b i 13 cm

ii 282.74 cm2

c i 13 cm

ii 942.48 cm2

d i 17 cm

ii 1507.96 cm2

e i

ii 7.58 m2

f i

ii 438.16 m2

6 79.79 cm2 7 a 1187.5 cm3

b 45.3 cm3

c 80.0 cm3

d 79.4 cm3

8 a 24π cm3

b 32π cm3

Volume = 735.13 cm

c 100π cm

d 720π cm3

f 272.16π cm3

13 The student’s mistake was substituting the diameter for the radius. The correct volume volume is 381.70 cm3

10 a h = 24

b 2513.274 m3

Problem-solving

11 a 19.9 m

b 5.72 cm

14 About 0.93

12 25 cm

15 About 3.11 cm3

9 3.06 m3

c 8 mm

Reasoning b 2.25π cm2

16 a 0.75 cm

c 1.5 cm × 1.5 cm × 15 cm 17 a 0.002 mL

13 a r = 6 cm b The volume of the cylinder is calculated by using the product of the area of its base by its height. The volume of a cone is exactly one third the volume of a cylinder formed from the same base with the same perpendicular height. Since the cone and the cylinder have the same diameter and height, the volume of three of the cones is equal to the volume of one cylinder.

b 2

c About 6.09 mm

18 a 1413.72 cm2 b 2780.31 cm2 c 4194.03 cm2 19 1209.84 cm2 20 The surface area of Jupiter is approximately 425 times greater than the surface area of Mars. b 257.25π cm3

21 a

22 0.75 m

c 66.6%

, cake discarded = volume of the cylinder - volume of the cone

23 605.3 cm3 Problem-solving

10.05 Optional: Cones

16 192.2 cm2

1 a The radius

b The slant height

2 a 25 cm

b 12 m

17 18 a 169.65 cm3

3 2 cm3

b 221.74 cm3

19 43.37 m3

Fluency

20 1102.7 mL

4 a 365.99 cm2 2

c 823.48 cm

b 725.05 cm2 d 58.31 cm

21 a 135.3 cm2

b 142 cm2

Answers mathspace.co

939

22 a 273 m2

b 656 kg

23 a 385 mL 24 a 70.55 km

b 25 3

b 0.15 km

c 19 3

c 70.40 km3

• 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.

10.06 Optional: Composite solids

6 A cube and a sphere

Understanding

Fluency b 135.34 cm2

7 a

1 A composite solid is a 3D ﬁgure that is made up of simpler solids. Examples could include a house shape (a rectangular prism with a triangular prism on top) or a cone with a hemisphere on top (like an ice cream cone).

8 a 245.04 cm2

b 173.13 cm2

d 62.83 mm2

f 136 m2

2 Calculate the area of all of the faces and add these.

9 a i 110.25π mm2

3 a A cone and a hemisphere b A cylinder and a cone c A rectangular prism and a pyramid d A cube and a cone 4 a We can estimate the volume of the cupcake using a truncated cone for the bottom part of the cupcake, and a cone for the icing on the cupcake. b We can estimate the volume of the tree using cylinders for the trunk and branches, and either hemispheres or cones (or an average of the two) for the clumps of leaves. c We can estimate the volume of the twentysided die using a sphere. Another possible estimation could use two pyramids and a cylinder.

d 236 cm2

c 100.53 cm

c 307.57 cm e 100.53 cm

ii 695.1π mm2

iii 220.5π mm b 3222.80 mm2 10 a i 56.55 cm2 iii 143.76 cm

ii 388.24 cm2

b 589 cm2 11 a • Four square faces with side length of 6 cm. • Two faces which are squares of side length 6 cm, with semicircles of radius 3 cm cut out of them. • A quarter of a sphere with radius 3 cm. b 216 cm2 12 a 5.59 cm

b 180.9 cm2

13 a 6.32 cm

b 231.7 cm2

14 a 169.65 cm3

b 735.13 cm3

c 221.74 cm3

d We can estimate the volume of the tent by taking the average of a pyramid and a hemisphere.

d 720 cm3

f 81.69 cm3

e 170 m

g 8928.41 cm

h 1417.19 m3

5 The simple solids and their required dimensions to ﬁnd the volume are:

15 a 59.69 cm3

b 619.94 m3 3

• A rectangular prism for the base. We need to know the length, width, and height.

c 263.89 cm

d 290.27 m3

e 149.33 cm3

f 221.67 m3

h 938.33 m3

• A cylinder for the pole. We need to know the radius or diameter of the pole, and its height. • 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.

940

g 842.67 cm 16 43.37 m3 17 a 14.97 cm 18 188.74 units3 19 160.36 units3

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b 915 cm3

Reasoning 20 a The first step is identifying all the faces. In this case there are three: the curved face of a cylinder and the curved faces of two identical cones. Calculate the slant height of the cone using Pythagoras’ theorem, then calculate the surface area of the curved part of the cone. Double this value (since there are two identical cones). Find the surface area of the curved face of the cylinder. Finally, add the surface areas of the faces. b 193.07 cm2 21 The type with the most waffle will be the type with the greatest surface area, ignoring the hemispherical ice cream scoop on top. The slant and height of the pointy waffle ‘cone’ is its radius is 3 cm, so its surface area is 80.5 cm2. The cylindrical ‘cone’ has a surface area of 106.8 cm2. A customer who wants the most waffle should buy the cylindrical ice cream ‘cone’. 22 The volume of the composite solid is the sum of the volumes of the individual solids. The volume of a cone is given by Vcone = π r2 h, while the volume of a hemisphere is given by Vhemisphere = π r3. To find the volume of the composite solid, we add the volumes of the cone and the hemisphere. This requires the radius and height of the cone, and the radius of the hemisphere. Note that in this case, since the cone is placed on top of the hemisphere with the same radius, the radius is the same for both solids, simplifying the calculation. 23 a The volume of a cone is dependent on its height. Since James cuts the cone a third of the way from the top, the cut section has a third of the height of the original cone. Therefore, its volume is a third cubed, or a twenty-seventh, of the original cone’s volume, because the volume of a cone is proportional to the cube of its dimensions. b Since the cut section is a twenty-seventh of the original cone’s volume, the volume of the truncated cone is the original volume minus a twenty-seventh of the original volume. This means the truncated cone has original volume.

of the

24 The volume of a pyramid is proportional to the cube of its height. Since Laura cuts the pyramid halfway from the top, the height of the frustum is half the height of the original pyramid. Therefore, the volume of the frustum is of the original pyramid’s volume because the volume of a pyramid is proportional to the cube of its height. Therefore, the relationship between V and F is

Problem-solving b 11.2 cm2

25 a 1.56 cm

c No, a six-sided die with side lengths of 1.8 cm has a surface area of 19.44 cm2. 26 a 560 m2 of paint b The building with hexagonal spires will require less paint than the building with the rectangular spires. The spires with the square base require 560 m2 of paint while the spires with the hexagonal base require 420 m2 of paint. c The base of the hexagonal spire would need .

to have a side length of 27 91.11 cm2 28 a 30.14 m2

b $6449.96

29 157π cm ≈ 493.23 cm 30 a 1 528 341.353 cm3

b 1528 L

b 30 cm

d 27 075 cm3

31 a 38 475 cm c 11 400 cm 32 a 385 mL

b 25

c 19 33 a 9xy2 units3

units3

d 1118 cm3

c 34 a

units3

d 161 cm3

Chapter 10 review 1 a i 864 cm3 b i 1526.81 cm3 2 a 180 cm3

ii 648 cm2 ii 894.58 cm2 b 824.67 cm3

Answers mathspace.co

941

3 a 58 904.86 cm3 2

c 2434.73 cm 4 a 488 m

b 2120.58 cm2 d 3063.05 cm

b Greater than

b $1824

5 a Trina’s

b Kareem’s

c 0.025 cm

Fluency

5 0.8

6 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. 7 The products may be overcooked if the timer is consistently slow, which can affect the texture and taste. 8 a 525 Myr

b 0.4

8 a 0.4 e

b 10.97 cm 2

ii 904.78 cm3

b i 277.59 cm2

ii 434.89 cm3

12 a 150.80 cm2

b 134.041 cm3

13 a i 240.26 cm2

ii 235.62 cm3

b i 103.67 cm2

b ii 237

11 a {0, 1, 2, 3, 4, 5}

15 3222.80 mm2 16 a i 346.88 cm2

ii 549.78 cm3

b i 769.45 m2

ii 1675.52 m3

c i 1155.10 cm2

ii 2760.33 cm3 ii 150.53 m3

17 1008 cm3

12 a

Understanding c True

15 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). Out of them, there are only two possibilities with sum of 7 or more: (2, 5) and (2, 6). The answer is

11.01 Conditional probability

b True

14 a

d i 153.15 m

iv 208

13 0.6875

ii 69.9 cm3

14 4.30 cm

iii 111

d 440 cm3

11 a i 452.39 cm2

d True

16 a

17 a

18 a (0.4 × 0.8) + (0.6 × 0.7) = 0.74

3 A

942

10 a i 260

10 a 9.71 cm

7 a

9 a

d 2%

9 253.5 m2

1 a False

6 0.18

b 475 Myr

c 5%

c 431.3 cm

4 a Patricia is chosen second, after Aaron has been chosen ﬁrst.

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Problem-solving

19 a 0.75 × 0.05 = 0.0375 b 0.25 × 0.95 = 0.2375

25 a

Comedy News

c 0.0375 + 0.2375 = 0.275

Comedy

Horror Animation

Comedy Comedy

20 a 177

Horror

Animation

Current Affairs

21 a 0.8 × 0.5 = 0.4

Comedy

Horror

Animation

b (1 − 0.8) × (1 − 0.5) = 0.1

Comedy

c 1 − 0.1 = 0.9

Horror

d (1 − 0.8) × 0.5 = 0.1 e P (A∣B) = P (A) = 0.8 (A and B are independent events)

b 26 a

Reasoning 22

0.2

H 0.7

0.8

Since P (B ∩ A) = P (A ∩ B), therefore

23 a

0.15

e Since B is inside of A, if the event B has occurred, the event A must have occurred. So, P (A∣B) = 1. 24 P (A∣B) is greater because: and

0.85

0.3

P (A∣B) × P (B) = P (B∣A) × P (A)

Animation

27 a

28 a

Labor (next election)

Liberal (next election)

Greens (next election)

Total

Labor (last election)

Liberal (last election)

Greens (last election)

Total

140

200

But according to the size of the circles on the Venn diagram:

29 a

e H and S

Answers mathspace.co

943

11.02 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

14 a

5 6 a {23, 24, 32, 34, 42, 43} b i

iii

7 a

b 10

13 a

4 w = 0.3, x = 0.3, y = 0.2, z = 0.2

15 a

Y B

8 a 3

9 a

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

944

Mathspace WA Curriculum – Year 10 Methods Pathway 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

25 a

Reasoning 21 a P (A ∩ B) = P (A) × P (B) b P (A ∩ B) = P (A∣B) × P (B)

26 a 0.18

c The formula P (A ∩ B) = P (A∣B) × P (B) can be used for both independent and dependent events.

27 a 27

iii

c Two events, H and S, are considered independent if P (H ∩ S ) = P (H) × P (S ). From the previous parts, we have

Since

b 43

d 0.64

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 11 review

and

c 0.38

e 0.9

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

b 0.45

, the condition for

independence is not met. Therefore, the data suggests that getting a haircut and being sick in the last month are dependent events.

1 a

b Independent

2 a

3 a

4 0.2 5 a

6 a 0.7

b 0.65

7 a

8 j = 0.4, k = 0.4, l = 0.1, m = 0.1

Answers mathspace.co

945

9 a

d iii: Upper quartile, Q3: At most 25% of the data is above this value

e i: Maximum: Highest value 240

120

210

Fluency 70

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

10 a

Went hiking

Did not hike

Total

Went swimming

Did not swim

Total

100

b i

iii

ii 7

iii 17

iv 10

v 39

vi 29 b i −6, −4, −1, 7, 7, 9, 9 ii 7

iii 7

iv −4

v 9

vi 13 11 a

Egg sandwich 6 10

5 10

5 10 5 10

4 10

Muffin

b 0.3

5 10

c 0.8 b

13 a

14 a

Iced coffee Orange juice

Iced coffee

2 a Q1 = 1, Q3 = 9 c Q1 = 15, Q3 = 26.5

v 19

vi 11 d i 20, 22, 28, 32, 40, 42, 43, 54 ii 8

iii 36

iv 25

v 42.5

ii 8

iii 80.5

iv 74.5

v 84.5

vi 10 f i 100, 102, 110, 113, 115 c

ii 5

iii 110

iv 101

v 114

vi 13 g i 197, 198, 202, 205, 207, 228

Understanding b 11.5

iii 14

iv 8

e i 69, 71, 78, 79, 82, 84, 85, 88

12.01 Quartiles

1 a 28

ii 7

vi 17.5

12 a

c i 4, 8, 10, 14, 15, 19, 20

Orange juice

c 49

d 7

b Q1 = 24, Q3 = 32 d Q1 = 16, Q3 = 24.5

ii 6

iii 203.5

iv 198

v 207

vi 9 h i 15.8, 19.5, 22.4, 29.6, 35.4, 39.1, 46.2 ii 7

iii 29.6

3 10.5

iv 19.5

v 39.1

4 a iv: Minimum: Lowest value

vi 19.6

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

946

8 a 18.93 kg b i 19 kg

ii 17 kg

iii 21 kg

b 7

c 1

c 2 9 a 51

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10 a 47

b 7

11 a 5

b 16

c 4 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)

Freq ( f )

Totals

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.

b 7

c 6

d 8

15 a 1

b 3

c 2

16 a 75

b 8

c 3

17 a 79

b 64

c 84

d 20%

e 2

18 a 74

b 79.5

c 24

d 35

e 16.5

f 18.5

e 20%

g The Sapphire Sharks’ scores are generally higher and more spread out than the Emerald Eagles’ scores, indicated by their higher median (79.5 vs 74) and larger range (35 vs 24). In terms of shape, the Emerald Eagles’ scores are approximately symmetric, while the Sapphire Sharks’ scores are slightly skewed to the left. Reasoning 19 a 8 marks

Set 2 is symmetrical. This can be justified by observing that the mean (29) and the median (29) are equal. Alternatively, the distances from the median to the quartiles are equal (15.5) and the distances from the quartiles to the extremes are also equal (3.5).

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.

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. 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. 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 No, the percentage of data values strictly between the upper and lower quartiles is not always exactly 50%. The interquartile range is defined as the range that contains the middle 50% of the data, but individual data points may fall on the quartiles themselves. For example, in the data set 10, 20, 30, 40, 50, 60, 70, the lower quartile (Q1) is 20 and the upper quartile (Q3) is 60. The values strictly between them are 30, 40, and 50. This represents 3 out of 7 data points, which is approximately 43%, not 50%.

Answers mathspace.co

947

Problem-solving

Fluency

24 65

5 a 3 minutes. This means that about half of

25 a

the ﬂights had a delay less than

minutes

6 a

minutes.

and half had a delay of more than b 10 minutes

c 45 minutes

d 35

e 83

f 83%

g 22 b 27

c 137 minutes

12.02 Boxplots

Minimum

1 8

Median

Upper quartile

Maximum

Minimum

Lower quartile

Median

Upper quartile

Maximum

Minimum

Understanding 1 a

Lower quartile

Median

Upper quartile

Maximum

Minimum

d 20

Lower quartile

Median

51.5

Upper quartile

Maximum

7 a 16

c 15

Lower quartile

h 22% 26 a 142 minutes

b 18

d 10

e 7

Data

b 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

d False

iii 50%

iv 75%

20 30 40 50 60 70 80 90 100 Data

v 75% b Second c First and third

948

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

20 30 40 50 60 70 80 90 100

8 a 2

b 40

d 38

e 6

9 a i

c 2, 3, 40

Minimum

Lower quartile

Median

Upper quartile

Maximum

Data

iii Somewhat symmetrical, with slight right skew d i

20 22 24 26 28 30

Minimum

Lower quartile

Median

31.5

Upper quartile

36.5

Maximum

4 18

Median

Upper quartile

Maximum

iii Left skewed b i

Minimum Lower quartile

Data

Data 0

iii Somewhat symmetrical, with slight right skew e i

20 22 24 26 28 30 32 34 36 38 40

iii Symmetrical c i

Minimum

Lower quartile

Median

50.5

Upper quartile

Maximum

59 Data

Lower quartile

Median

Upper quartile

Maximum

40 42 44 46 48 50 52 54 56 58 60

iii Somewhat symmetrical, with slight left skew

Answers mathspace.co

949

f i

Minimum

Lower quartile

Median

Upper quartile

Maximum

Data

iii Somewhat symmetrical, with slight right skew i i

iii Symmetrical g i

Minimum

Lower quartile

Median

Upper quartile

Maximum

Minimum

Lower quartile

Median

Upper quartile

Maximum

Data

iii Right skew j i

iii Somewhat symmetrical, with slight right skew h i

Minimum

Lower quartile

Median

Upper quartile

Maximum

Minimum

Lower quartile

Median

15.5

Upper quartile

Maximum

Data

iii Symmetrical

950

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

10 a The cyclist’s heart rates varied by 36 beats per minutes (bpm).

17 a

Science Scores

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 Data set B

b i D ata set A - skewed right, data set B - skewed left ii Data set B

iii Data set A

12 a

Class 10P

Class 10Q

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.

100

Music Scores

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 on 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.

Answers mathspace.co

951

19 a Negatively skewed

Problem-solving

b Positively skewed c Yes. The group that received treatement 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.

b This changes the range from 21 to 14, a decrease by 7.

24 a Any ﬁve-number summary with an interquartile range of 2 and a range of 8. For example:

c Because the positions of Q1 and Q3 were unchanged, there is no impact on the IQR.

• Minimum: 1

• Lower quartile: 2

• Median: 3

• Upper quartile: 4

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.

• 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

22 a

• Boxplot 4/Histogram C 26 a

Monthly salaries (in thousand dollars)

15 50

95 100

Scores of students

b 20 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.

952

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

100

Debbie’s blog visitors

5 a Upmarket restaurant

b $22

c Upmarket restaurant

d $36

e 26 items 6 a Weightlifter A 0

100

150

200

b The range and interquartile range c The median

Number of pages read

d Weightlifter B 7 a Lower

b 0

c 2

d Positively skewed

e 4

f Block B

g No

12.03 Compare data sets

8 • Histogram A/Boxplot 2 • Histogram B/Boxplot 4

Understanding

• Histogram C/Boxplot 1

1 a The histogram

b The boxplot

c The boxplot

d The histogram

2 a Negatively skewed

• Histogram D/Boxplot 3

10 a 5.2

3 a • Frequencies for each group

c Matt

b 8.9

c 3

d 3

e No. The results are mostly 4 or 5 out of 10 and the mean is only 5.2

• Mode • Shape

f Y es. The results are mostly 9 and 10 out of 10, and the mean is 8.9

• Outliers b • Quartiles

g Answers may vary. One possible answer is that one class may have covered more material than the other class.

• Median • Shape • Outliers

11 a Bookstore A, it has a lower range and interquartile range.

• Range and interquartile range c • Frequencies for each group

b $10

• Mode

c i True

• Shape • Outliers Fluency English Mathematics Median

b Matt

Reasoning

c Symmetrical

4 a

9 a Matt

b Bimodal

Lower quartile

Upper quartile

Range

Interquartile range

ii True

12 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. b The boxplot indicates that the data is widely spread, while the histogram indicates that most of the data occurs near the median.

b Mathematics

Answers mathspace.co

953

b Tim throws more consistently, his range, interquartile range and standard deviation are all signiﬁcantly smaller.

13 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. 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.

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.

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.

Problem-solving 14 a

Cooper

Marion

Minimum

44.8

47.1

49.1

54.7

Median

52.5

60.45

55.75

65.25

Understanding

Maximum

66.6

68.5

1 a False

b False

e False

f False

12.04 Samples and populations

b Yes, Cooper has an outlier of 66.6. c

Cooper Marion

50 55 60 Time (seconds)

954

2 a No

b No

c Yes

d Yes

3 a No

b Yes

c Yes

d Yes

e Yes

f No

b The population is all sales at the bookstore, and the sample is the 20 receipts she looks at.

e No, a high outlier does not change what the minimum score is, hence would not alter the selection for pole position.

c The population is all students at Antony’s school. The sample is the 85 students he surveyed.

Tim

Odi

Minimum

29.8

26.0

33.2

34.8

Median

35.0

38.4

37.4

40.0

6 a No e Yes

Maximum

39.6

42.8

Mean (1 d.p.)

35.1

37.2

Range

9.8

16.8

Interquartile range

4.2

5.2

d True

4 a The population is all the residents of the town, and the sample is the 500 residents surveyed.

d Cooper

15 a

c True

5 a Yes

b Yes

c Yes

d No

b No

c No

d Yes

7 a ii

b iii

c i

d iv

8 a Yes

b Yes

c No

d Yes

e No

Fluency

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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:

dog such as a professional dog walker or a friend of a dog owner. There is also the possibility that someone with accessibility needs would not be able to access the park so would be excluded. 20 a The sample is not representative of the target population.

• All students in Year 7 in a particular suburb

b The sample is from self-selecting participants, i.e. only those that made an effort to respond.

• 100 children chosen from the Sydney under 15s baseball tournament

c The sample is not representative of the target population.

• All students of a primary school in Sydney

c Example answers: • The members of 3 teams chosen from the local hockey tournament • All students from a local school who compete in a school sports competition • All the active members of a local football club d Example answers: • 30 people walking their dog at the local park • All the students from a local school who own a dog 10 Fair 11 a Biased

b Biased

c Not biased

d 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.

21 a Example answer: Assign a number to everyone on the electoral roll 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.

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

Answers mathspace.co

955

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.

Problem-solving 24 a Example answers: • Check additional reputable secondary sources to see which information is corroborated by other authors or creators. • 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. • 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.

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. 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.

• 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.

Chapter 12 review

• 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.

1 a 101

b 16

c 8

d 3

2 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. 3 70 seconds 4 a

Minimum

Lower Upper Median Maximum quartile quartile

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.

956

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

60 65 70 75 80 85 90

c Somewhat symmetrical, with a slight left skew. 5 a The number of books read by students varied by 43 books. b 25%

c 13

d 4th quartile

e 10

6 a The median of the control group is 5 weeks higher than the median of the treatment group. 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 interquartile 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. 7 a 290 8 a

b 60

c 31.0%

Number of attendees

Number of workshops ( f)

Cumulative frequency (cf )

b 8 workshops

c Building Y 12 a

d 30 420

14 a Biased

b Biased

c Not biased 15 a The sample proportions are different because the researchers used samples of very different sizes. 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.

13.01 Two-way tables

b 16

c Outback Animal Care Centre

c 55

10 A 11 a Building X

b 58.5%

c Yes

1 a 14

c 10 workshops b 45

13 a All cherries in the farm.

Understanding

d 22 attendees 9 a 49

d No, Antoine’s range and interquartile range are still smaller than Benji’s, even after removing the outlier.

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

3 a False

b False

c True

d False

4 a True

b False

c True

d True

e True

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

Fluency 5 a

Work boots

Total

Teenagers

Adults

Total

169

Sprinkles Hot fudge

Total

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 more representative average of his usual performance in competitions.

Running Basketball shoes slides

Chocolate

Vanilla

Total

c Children

Beef

Chicken

Fish

Total

Adults

Total

Answers mathspace.co

957

6 a This value represents the number of cats with feline diabetes treated at Venetian Pet Clinic. b This value represents the number of cats with feline diabetes who received a special diet at Mohawk Animal Clinic. 7 a

Vegetarian

Vegan

Neither

Total

Children

Adults

Total

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.

b 8 a 39

b 26

c 47

9 a 11% of junior students and 10% of senior students were very satisﬁed with the canteen food.

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. 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”.

b The most common rating among junior students is 4, while for senior students it is 3. 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

c The error is assuming the three responses are equal. The condition that the number of “yes” responses equals the sum of the “no” and “sometimes” responses means that “yes” accounts for exactly half of the total responses 30 out of 60, not one-third. The total cannot simply be divided by 3.

b 51 trains

c 127 trains

11 a 27.63% c

b 44.47%

Problem-solving

d 34.61%

15 a

No Casual Part-time Full-time job

Reasoning 12 a

Men Brushtail possum

Honey possum

Total

Fruit

Insects

120

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. c The number of possums eating insects and nuts (230) was significantly higher than the number of possums eating fruit (80).

958

Women 25

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. 17 Answers will vary. The table provided is one such solution:

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

Football

Soccer

Total

Period 1

Period 2

Period 3

Total

18 a 650 students

6 a

b Answers may vary. One possible answer: To investigate the hypothesis, you could design a follow-up survey that looks at two primary variables:

Red

Black

Total

47%

5.9%

41.2%

100%

Curly

40%

30%

20%

100%

10%

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)

Brown Blonde

Straight 5.9%

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

Your two-way table would look something like:

Teens

30%

20%

45%

100%

Children

30%

40%

10%

20%

100%

Bought ice block

Did not buy ice block

Total

Always Often Sometimes Rarely

Comedy

Action

Drama

Horror

Teens

50%

33.3%

81.8%

20%

Children

50%

66.7%

18.2%

80%

Total

100%

Play tennis

Do not play tennis

Total

Height ≥ 170 cm

31%

69%

100%

Height < 170 cm

30%

70%

100%

8 a Height

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.

13.02 Associations between categorical variables Understanding 1 a 43%

b No

c No

2 A column percentage frequency table 3 The horizontal axis

9 a Pregnancy status b

Weight

Total

30%

70%

100%

Currently postpartum

73%

27%

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. 10 a

4 Yes Fluency 5

Cardio Currently pregnant

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%

Left-handed

Right-handed

Not female

53.3%

42.0%

Female

46.7%

58.0%

Total

100%

Answers mathspace.co

959

11 a Parenting status

b Yes

12 a Fitness level

b Yes

13 a Personality type

b No

14 a Dietary restriction

b B

19 a Favourite subject at school b Column percentage table

c No

Maths

Music

Art

11.67% 23.33%

Science major 30.54% 27.40% 23.89% 13.89%

c No

Music major 22.75% 22.60% 37.78% 20.56%

Reasoning 15 It means that there is no clear association between the variables.

Urban

Regional

High emissions

62%

38%

Low emissions

38%

62%

Total

100%

b June

Maths Music Science English Total

b Column percentage frequency table c

30%

Lefthanded children

33%

c Book loans

21 a Employment status

d Row percentage table Righthanded children

100%

d Yes, the proportion of Children’s books decreases from June to November, while the proportion of Business books increases. This shows the pattern of loans changes with the month.

b 21

20%

25%

20%

30%

25%

17%

Employed

Unemployed

Owns a smartphone

32%

17%

Does not own a smartphone

68%

83%

Total

100%

f N o, there does not appear to be an association as the percentages for each subject are quite similar for both right-handed and left-handed children. 18 a Number of passengers b Column percentage table

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. f

No. of passengers

5−9

Roll over

17.1%

30.0% 28.8% 28.6%

10−15 71.2%

Smartphone owners employment

> 15 71.4%

100.0% 100.0% 100.0% 100.0%

d Yes, the percentage of cars that roll over is significantly lower for those with fewer than 5 passengers (17.1%) compared to those with 5 or more passengers (around 29 – 30%).

Owns smartphone

Doesn’t own smartphone

100 Percentage

No roll over 82.9% 70.0% Total

26.67% 42.22%

100%

20 a November

c Dominant hand e

11.64%

100%

Problem-solving

d Yes, there does appear to be an association as the numbers are signiﬁcantly different. 17 a 40

7.19%

Total

e No, association is not causation. There appears to be an association but we cannot say that one causes the other.

b Column percentage table c

Art major

d Yes, because values across the rows are different.

16 a Location

Science

Maths major 39.52% 38.36%

80 60 40 20 0

Employed Unemployed Employment

g Yes it does as the segments are different size in each column.

960

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

22 a Age

b Column percentage frequency table c

10 years 11 to 20 21 to 30 31 years and under years years or older Less than 3 hours

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%

Percentage

40 20 0 11 to 20 21 to 30 31 & older Age group

13.03 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

2 a

2 4 6 8 10 12 14 16 18 y

18 16 14 12 10 8 6 4 2 0

x 2 4 6 8 10 12 14 16 18

3 a Yes

b No

4 a Neither c Decreasing

18 16 14 12 10 8 6 4 2 0

d 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.

2 4 6 8 10 12 14 16 18

18 16 14 12 10 8 6 4 2

10 & under

100%

TV watching per age group 3 or less 4 to 8 9 to 15 15 or over 100

y 18 16 14 12 10 8 6 4 2

c No

d Yes

b Increasing d Decreasing

Fluency 5 a i Number of dogs owned ii Number of shoes destroyed b i Size of a pizza ii Cost of pizza x 2 4 6 8 10 12 14 16 18

c i Time spent studying ii Exam mark 6 a Ticket sales affect the revenue made from a show. b Dependent variable - revenue from the show, Independent variable - number of ticket sales

Answers mathspace.co

961

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

13 a The independent variable is CO2 emissions and the dependent variable is the global temperature anomaly. b Climate change Global temperature anomaly (°C)

8 a Distance from the equator affects temperature.

0.9

b No c Both time spent travelling and distance covered affect each other, with speed playing a crucial role in their relationship.

0.8

0.7

d No e Time spent working affects wages earned.

0.6

f N umber of pages affects time spent to ﬁnish a novel. g No

CO2 emissions (million MT) 35 000 36 000 37 000 38 000 39 000

h Number of pets affects time spent caring for pets (hours). i T ime spent training (hours) affects athletic performance.

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

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. 12 a The independent variable is the number of immigrants and the dependent variable is the unemployment rate.

b 11

b Unemployment rate in Australia

15 a Dependent variable - height, Independent variable - time

Unemployment rate (%) 5.5

b 5

4 3.5 Number of immigrants (thousands) 250

275

300

325

c As the number of immigrants increases, the unemployment rate appears to decrease.

962

Height 0.8 0.6 0.4 0.2

4.5

x 1 2 3 4 5 6 7 8 9 10 11 12 13

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

−0.2 −0.4 −0.6 −0.8

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

b 50

16 a 180 y 175 170 165 160 155 150 145 140 135 130 125 120 115 110 105 100 95 90

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. Problem-solving

x 2

b 105 cm 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)

20 Looking at the scatterplot, the number that fulﬁls the required characteristics is 4. 21 a Student B

b Student C

c Student B

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. 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.

Average time spent (minutes)

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 0

13.04 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.

Exam mark

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.

Answers mathspace.co

963

Fluency

10 a

Hours 45

4 a Linear relationship b Non-linear relationship

c Non-linear relationship

d Non-linear relationship

e Linear relationship

f Linear relationship

Age

5 a Positive and strong

b Positive and weak

b Moderate, negative, linear

c Positive and moderate-weak

c Yes, (30, 45)

d Negative and strong 6 a As air temperature increases, sea temperature increases.

11 a

Price 16

b Strong, positive 15

7 Negative 8 a

Sport 90

Supply (kg)

600 700 800 900 1000

b Strong, negative, linear

c When very few bananas are available to be sold.

40 Maths 40 50 60 70 80 90

12 a

b Weak, positive, linear

c Student 10’s 9 a

Time 65

Accidents 40

45 Temp.

1 2 3 4 5 6 7 8 9 10

b 10

c Weak, negative, linear d 8 °C, 69 seconds

Age 25

b Strong, negative, linear

13 a

c No

t 70 60 50 40 30 20 10 0

964

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

n 1

b Increases

Problem-solving

c Strong, positive, linear

21 a D

d No

22 a i No

14 Strongly negative

16 a No

b Yes

c No

d No

e No

f Yes

g Yes

h Yes

Reasoning 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. 18 a The explanatory variable is the number of questions and the response variable is the test result. b 100 Test result

Number of questions 10 15 20 25 30 35 40 45 50 55

c Strong, positive, linear correlation; the larger the number of completed revision questions, the higher the mark on the test. d The relationship is not perfectly linear because of the different abilities of the students. 19 a It appears that y can change signiﬁcantly without much corresponding change in x. b i Yes

ii Yes

iii No

b Region 3

15 Strongly positive

95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15

b B

iii Yes

iv No

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.

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. 24 Answers may vary. One possible answer: Time spent studying and test scores could lead to a weak positive relationship. 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.

Answers mathspace.co

965

Here is a table of values and scatterplot that would lead to a weak positive association:

Hours Studied

Test Score

Student A

Student B

Student C

Student D

Student E

Student F

Student G

Student H

Student I

Student J

6 a i

y 20 15 10 5 x 0

ii y = −x + 21 b i

y 20 15

Test score

10 65 5 x

60 0

ii y = 0.56x + 5

c i Hours studied 1

y 20

13.05 Line of good fit

Understanding 5

1 Decreasing

2 Line 2 3 a No 4 a Negative

b No

c Yes

d Yes

ii y = 0.9x + 1.5 d i

b C

5 19 y

10 5 x 0

ii y = −1.3x + 29.5

x 1 2 3 4 5 6 7 8 9 10 11 1213141516171819

966

y 20

Fluency 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

e i

b i −7.68

ii Negative

iii Decreases by 7.68 units iv 5.49

c i −5.69

ii Negative iii Decreases by 5.69 units

4 x 0

d i 3.16

ii y = 1.5x − 17.5 f i

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.

6 4

b No

2 x 0

190 180 170 160 150 140 130 120 110 100 90

Price (thousands of dollars) 24

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. x 10 20 30 40 50 60 70 80 90

d $60

7 Example answer: 38 36 34 32 30 28 26 24 22 20

11 a

c 0.3

ii y = 0.9x + 0.5

iv 7.84

d Answers should be around 11 years. 12 a −4

b 3865

Copies sold

Reasoning 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. Price 18 20 22 24 26 28 30 32 34 36

9 a i 4.62 ii Positive iii Increases by 4.62 units iv 7.58

Answers mathspace.co

967

14 a A scatterplot can help visualise the relationship between the quantities, but is not necessary for a complete response. 100

Problem-solving 16 a

Ice blocks 140

Grades

120

100

80 70

60 50

20 10

Temperature 14 16 18 20 22 24 26 28 30 32 34 36 38 40

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. 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%.

b i Approx. 91 c Increases 17 a A

b x = 3682 km

18 a y = 0.5x + 4

b 7.45

19 a A line of good fit: y = 3.5x + 15

H% 30 25 20 15 10

15 a No. A line of good fit should pass through the middle of the data. Frasier’s line is skewed by the outlier, resulting in only 3 points above it and 16 points below it. This makes it a poor representation of the overall trend. b An outlier should not be completely ignored, but the line should go through the majority of the data. Lorena’s line has 8 points above it and 11 points below it, and goes through the middle of the data, so it is a line of good fit.

ii Approx. 149

5 Y 1

b The new line of good fit would have a smaller slope and a higher y-intercept.

13.06 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

968

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b Extrapolation

5 y = −52.61, Extrapolation 6 a 3 seconds 7 a 20 years 8 a

225 200 175 150 125 100 75 50 25

b 2.5 hours b 1300 ﬁsh

c 11 years

People

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

c Yes, because it is using interpolation.

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

19 a A = −25T + 750

20 a

b Interpolation ii No

10 a 90 c i No

iii No

11 a 8500 m

iv No

iii No

iv No

b 6500 m

c 300 seconds

Litter collected (kg) 5

b 22°C ii Yes

1 Weeks

d 1800 seconds 0

12 a

b c = 100

b A = 100 hectares

c No, because it is using extrapolation.

c Yes d i Yes

b y = 29 °C

c $400 000

13 a Negative gradient

b −2.5

c y = −2.5x + 105

d y = 95

Reasoning 14 a 52.5 m b Yes, because the prediction is determined using interpolation. 15 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. 16 Yes, if we examine the scatterplot and compare with the other points. It follows the same relationship.

2 3 4 5 6 7 8 9

b Example answer:

Litter collected (kg) 5 4 3 2 1 Weeks 0

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. 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.

Answers mathspace.co

969

d $50 260

Fluency

e No, the prediction is less reliable as it is an extrapolation. The prediction is for K = 36, which is outside the observed data range of [2, 35].

5 a Causation and correlation b Over-generalisation c Comparing apples and oranges d Cherry-picking

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

b 50 m

c Option B

c True

d False

7 a Yes

b No

c No

d Yes

d Yes 9 a Cherry-picking

x 1

2 3 4 5 6 7 8 9

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. b Not representative d Representative b False

Date

Cases

October 15

October 18

200

October 21

174

October 24

339

October 27

251

October 30

398

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 numerical increments on the vertical axis are not uniform. This distorts the visual representation of the data, making changes in turnover appear larger or smaller than they actually are. 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.

c Representative 3 a False

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.

13.07 Misrepresentations of data

c True

d True

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 initial increase would be much more signiﬁcant. 16 a Russia

970

b No

c Strategic choice of measure of centre (mode)

24 Neither interpolation, nor extrapolation would be overly reliable. This data set is more nonlinear, so a non-linear model would be more appropriate for making predictions.

b False

8 a Yes

23 Any scatterplot with weak correlation or very few points. For example: 9 8 7 6 5 4 3 2 1

6 a False

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b U.S.

c No

Reasoning

Problem-solving

17 Kang has used only the statistics that are favourable to him.

22 Graph 1 is misleading because the vertical axis does not start at zero therefore the difference in the number of days is exaggerated.

19 a The graph is a very poor example because it incorrectly plots a data point. The bar for March 1995 represents a deficit of $20 million, but the scale on the vertical axis breaks at 0 and resumes at $35 million. It is mathematically impossible to plot a value of 20 on an axis that jumps from 0 to 35, making the graph nonsensical and highly misleading. 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.

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. b

Number of COVID-19 tests per million people

18 By compressing or expanding the horizontal axis it can appear as if the trend is gradual or quite abrupt.

COVID-19 Testing by Country

25000

22 300

20000

15 700 14 100

15000 10000

7000

5000 0

330

258

Brazil Argentina

USA

Italy

Germany Norway

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). 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 13 review 1 a

Truck

Sedan

SUV

Total

Under 30

Over 30

Total

b 2 a 354 students c 27.94%

c b d 8941 students

3 a Living area b Yes, there appears to be an association. 4 a Time of day preference

b Yes

c One possible answer: The article’s headline is sensationalised, potentially causing unnecessary worry as it lacks context on the study’s broader use.

Answers mathspace.co

971

5 a Education level

10 a

b Column percentage frequency table c

High School or less

Sales (in thousands) 25

Some Bachelor’s Graduate university degree degree

Less than 48.08% 2 hours

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

5 0

b Strong, positive, linear c No d No, correlation does not imply causation. 11 a i

6 a January

b February

d Yes, there is an association because the proportions of instrument sales change each month, indicating a different sales distribution over time. 7 a Yes, time spent studying a subject is the independent variable and the number of chapters is the dependent variable. b No

y 25

c Instrument sales

15 10 5 x 0

ii y = x + 4 b i

c Yes, the duration of sunlight exposure is the independent variable, and plant growth is the dependent variable. 8 a

Month 1 2 3 4 5 6 7 8 9 10 11 12

iii Positive

16 14 12

Sick days

10 8

6 5

x 2 4 6 8 10 12 14 16 18

3 2

1 5

Days of exercise 15 20 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. 9 a Positive and strong

972

b Negative and weak

iii Negative

12 a The y-intercept suggests that when the radish is 0 days old, it has no height above the ground. 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

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

b Interpolation

14 a M = $941.40

7 a x = ±8

b x = ±3

b No, because it is using extrapolation.

c x = 0, −12

d x = 0, 8

c 25 days

e x = −4, −1

f x = −5, −2

d Yes, because there is a strong correlation and interpolation was used for the prediction.

g x = 5, 3

h x = 6, −4

i x = 5, −4

j x = −5, 2

k x = 4, 10

l x=7

15 a Mistaking causation and correlation b Over-generalisation

8 a x = 0, 16

c Strategic choice of measure of centre 16 a Cherry-picking 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.

14.01 Optional: Solve by factorisation

1 A and D 2 a a = 3, b = 2 or a = 2, b = 3 b a = −2, b = −9 or a = −9, b = −2 c a = 4, b = 4

d a = −3, b = −3

e a = 1, b = −2 or a = −2, b = 1 f a = −3, b = −4 or a = −4, b = −3 g a = 3, b = −3 or a = −3, b = 3

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

b 11

c 8 and 3

d x2 + 11x + 24 = (x + 8) (x + 3)

b x = 2, −10

c x = 10, −10

e x=8

i y = 7, 1

j y = −1, 3

11 a i x = −1, 6

h a = 5, b = −3 or a = −3, b = 5 3 a 24

c x = −5, −1

10 a x = 6, −8

Understanding

b x = −3, 4

ii (0, 6)

iii y = −(x − 6) (x + 1) b i x = 2, 6

ii (0, 12)

iii y = (x − 2) (x − 6)

4 (m + 5) (m + 9)

c i x=4

5 a x = −2, 3

b x = −4, 1

iii y = −(x − 4)2

c x = −3, 3

d x = −1, 2

d i x = −5, 0

ii (0, −16) ii (0, 0)

iii y = x(x + 5) Fluency 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

Reasoning 12 a x2 + 10x + 24 = (x + 4) (x + 6) b k = 25, 14, 11, 10, −10, −11, −14, −25 c x = (−24, −1), (−12, −2), (−8, −3), (−6, −4), (4, 6), (3, 8), (2, 12), (1, 24)

Answers mathspace.co

973

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.

Problem-solving 17 a x = 4, −7

b x = 8, −3

c x = 7, −4

d x = 9, −5

h x = −7, 3

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. 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.

18 x = ±4 19 m = 16, n = 6 20 a t = 3, 9

22 a (x − 3) (x + 3) = y

6 4 2 −8 −6 −4 −2 −2

x 2

−4 −8

b (3x + 5) (x − 2) = y

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

974

−6

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.

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 function equal to zero to get the correct values.

c 6 hours

21 t = 2 s

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.

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.

b 3 hours

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

d −(x + 2) (x − 7) = y

8 a x = 8, x = −2

9 a

15 10

5 x −8 −6 −4 −2

−5

Reasoning

e 3(x + 10) (x + 1) = y

10 Halve the coeﬃcient of x to get

20 10 −12 −10 −8 −6 −4 −2 −10

square it to get x 2

−20 −30 −40 −50

14.02 Optional: Solve by completing the square Understanding 1 a Error

b Correct c Correct d Error

2 a 25

b 16

3 9

and then

which would need to be

added to x + bx to make it a perfect square trinomial. 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 has approximation. For example: solutions of

4 13 Tyler made several key errors: Fluency Line 2: The term

should be added and

5 a (x − 1)2 + 45

b (x + 1)2 + 1

c (x − 7)2 + 7

d (x − 3)2 − 4

Line 4: The factorisation is incorrect as x2 − 4x − 4 is not a perfect square.

h (x − 1)2 + 3

Line 6: The expression was incorrectly changed from (x2 − 4)2 to (x − 4)2.

6 a x = −4, x = −8

c x = 1, x = 5

7 a x = −2, x = −16 c x = 8, x = 1

then subtracted, not subtracted twice.

b x = 4, x = 2 d

Answers mathspace.co

975

The correct solution is:

Given equation Add and subtract

complete the square Simplify the squared term Factorise the perfect square trinomial Combine the constant terms

7 a x = 1.372, −4.372

Move the constant to the right-hand side Take the square root of both sides Add 2 to both sides

b x = 0.405, −7.405

c x = −0.360, −1.390

d x = −0.260, −1.540

e x = 0.219, −0.965

f x = 0.390, −3.279

g x = ±2.646

h x = −3.333, 3

i x = 5.589, −0.089

j x = 0.254, −5.254

Evaluate both solutions

8 a

Problem-solving 14 x = 12 15 a 100 m b 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

14.03 Optional: The quadratic formula

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.

c True

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.

2 a a = 1, b = 7, c = 10

b a = 1, b = −3, c = −4

Eﬃciency: Both methods are eﬃcient, but taking the square root is slightly faster.

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

Understanding 1 a True

b True

d False

e False

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

3 a = −9, c = 0 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 6 a x = −2, −3 c x = −4, 1

976

b x = 2, 3 d

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

. Then, take the square root of both sides to get

Method 2: Rewrite the equation as 25y2 − 36 = 0. Then, factor the equation as (5y − 6)(5y + 6) = 0. Solve for y to get

Move the constant term to the right side: . Add

to both

. sides. The equation becomes

Eﬃciency: Taking the square root is more eﬃcient in this case, as it requires fewer steps.

, which simplifies to

d Method 1: Use the quadratic formula to ﬁnd

. Take the square root of both

, which simpliﬁes to

sides to get

. Method 2: Factor the equation by grouping. Find two numbers that multiply to ac = 4 × 1 = 4 and add to b = 5. The numbers are 4 and 1. Rewrite the middle term: 4x2 + 4x + x + 1 = 0. Factor by grouping: 4x(x + 1) + 1(x + 1) = 0, which gives (4x + 1) (x + 1) = 0. Solve for x to get .

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 , which simplifies to x = −5, x = −4. Efficiency: Factoring is more efficient in this case, as it requires fewer steps.

Eﬃciency: The quadratic formula is more eﬃcient in this case, as completing the square requires more steps. e Method 1: Complete the square by adding inside the

and subtracting

Eﬃciency: The quadratic formula is more eﬃcient in this case, as completing the square requires more steps.

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

11 a b The value 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. 12 a Negative numbers. b i There will be two solutions in this case. , where d > 0.

Suppose that

and

The formula then gives

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. Efficiency: Factoring is more efficient in this case, as it requires fewer steps. g Method 1: Use the quadratic formula to find , which simplifies to . Method 2: Complete the square. First, divide the equation by 5 to get

which are two different values. ii There will be one solution in this case. . The formula

Suppose that then gives

, which is just

iii There will be no real solutions in this case, because we cannot take the square root of a negative. 13 The equation has no real solution. Problem-solving 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

Answers mathspace.co

977

17 38 ms−1

2 a x = ±9

b x = ±10

18 a 258.025 metres

c n = −4, 8

b 9.98 seconds

3 a Maximum

19 a 27.84 m/s

ii (4, 0) iv

l×w=A (10 + 2x)(3 + 2x) = 120 30 + 20x + 6x + 4x2 = 120 30 + 26x + 4x2 = 120 30 + 26x + 4x2 − 120 = 120 − 120 −90 + 26x + 4x2 = 0

11 10 9 8 7 6 5 4 3 2 1

4x2 + 26x − 90 = 0

iii (0, 32)

x 1 2 3 4 5 6 7 8 9 10 11

b 2.5 m 21 a

b t = 6, t = 18

4 a i D ilated by a factor of 2 and horizontal shift 4 units to the right

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.

d x = 0, 8

b i R eﬂection about the x-axis, horizontal shift 1 unit to the left and vertical shift 7 units up. Given

ii (−1, 7)

Rewrite the fraction on the right hand side in expanded form

iii (0, 6)

Rewrite in terms of x and simplify Multiply both sides by x Subtract x + 1 from both sides

−9−8−7−6−5−4−3−2 −1 −1 −2 −3 −4

Use the quadratic formula Substitute a = 1, b = −1, c = −1

11 10 9 8 7 6 5 4 3 2 1

5 a (0, −3)

Simplify Evaluate correct to three decimal places

b (1, 0), (−3, 0)

c x = −1 e

1 2 3 4 5 6 7 8 9

d (−1, −4) 4

3 2 1

c The positive solution, 1.618. a and b are both lengths, and therefore positive, so their ratio, must be positive.

−5 −4 −3 −2 −1 −1 −2 −3 −4

Chapter 14 review 1 a Concave up

b Concave down

c Concave up

d Concave down

978

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1

6 a y = (x + 2)2 − 5 c

b (−2, −5) 4 3 2 1

8 a x>0 c

192

9 a Negative

−6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

b x<0

10 a No 11 a i

b No

c No

c x = 12

d Yes

−2

−1

−16

−2

ii 7 a x = 12, 0

c y = −4x3

b x<0

b x = 1, −10

10 5

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

−5

−10

b x=3

−15

c 10 a 144 m

11 a

b i

12 a −x2 − 8x − 16 = 0 b The equation has 1 real solution.

c False

−2

−1

−3

−24

25 20 15 10 5 −5 −10 −15 −20 −25

Understanding b True

−3 −2 −1

15.01 Optional: Cubic function

1 a True

−3 −2 −1

d True

x 1

2 Decreasing Reasoning

3 a y-values approach infinity. b y-values approach negative infinity. Fluency 4 a C

b A

c B

5 a Positive b i −13.5 iii 32

ii 0.0625 iv 13.5

6 x>0 7 a Negative c (0, 0)

b x<0

12 For y = 2x3, y ≥ 0 when x ≥ 0 because the function is positive for positive x and zero at x = 0. 13 If a > 0, the graph increases from bottom left to top right. If a < 0, the graph decreases from top left to bottom right, as it is reflected across the x-axis. 14 The function has one x-intercept. Using x = 2 and y = −8, we find −8 = a(2)3, so a = −1, and the function is y = −x3.

d y = −2x3

Answers mathspace.co

979

2 3

A cubic function of the form y = ax3 has exactly one x-intercept because it is a continuous, monotonic function (always increasing if a > 0 or always decreasing if a < 0), crossing the x-axis only at x = 0 (since ax3 = 0 implies x = 0). For y = −x3: at x = 0, y = −03 = 0; for x > 0, y is negative and decreases (e.g., y = −8 at x = 2); for x < 0, y is positive and increases (e.g., y = 1 at x = −1). 15 For a cubic function y = ax3, the point of inflection is where the graph changes concavity, from curving downward to upward or vice versa.

3 a No

c No

d Yes

4 a x = −3, 2, 5 b y = −(x + 3) (x − 2) (x − 5) Fluency 5 a No

b Yes

c No

d No

6 a i x = −3, x = −2, x = 2 ii y = −12 iii

y 6 3

When a > 0, the graph is concave down for x < 0 (forming an upside-down bowl) and concave up for x > 0 (forming a regular bowl), with the transition happening at x = 0, where y = a(0)3 = 0.

−4 −3 −2 −1 −3

x 1

2 3 4

−6 −9

Similarly, when a < 0, the graph is concave up for x < 0 and concave down for x > 0, again transitioning at x = 0, where y = 0. This change in concavity always occurs at the origin (0, 0) because it’s the only point where the graph is neither increasing nor decreasing at an accelerating rate, and the symmetry of the function around this point ensures it’s the consistent location for this transition, regardless of the value of a (as long as a ≠ 0).

b No

−12

b i x = −4, x = −2, x = 1 ii y = 8 iii

y 8 6 4 2 −6 −5 −4 −3 −2 −1 −2

Problem-solving 16 r = 3 m

x 1

17 625 g 18 y = 5x3 19 a a = 4, so the function is P = 4t3.

c i x = 2, x = −5 ii y = 20 iii

b The population is increasing as time progresses because a = 4 is positive. For P = 4t3, as t increases, P grows larger, indicating a rising population trend.

c 2022

40 30 20

−7−6−5−4−3−2 −1 −10

15.02 Optional: Factored form of cubic functions

−20

Understanding d i

1 3

ii y = 108

980

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1 2 3 4

iii

162 144 126 108 90 72 54 36 18 −7 −6 −5 −4 −3 −2 −1 −18

7 a

x −4 −3 −2 −1

-5

-4

-3

-2

-1

-8

-1

8 y

11 a y = (x + 4)(x + 1)(x − 1) b y = (x + 3)(x − 1)2 c y = −x(x − 2)2 d

e y = −0.25(x − 1) (x − 6) (x − 2)

4 2

−2

12 a i

−4 −6 −8

-8

-1

y 4 2 −2

x 1

−1

6 4 2

b i

−6 −8

-3

-2

-1

-8

-1

6 4

−2

2 3 4

−2

−1

y 30 27 24 21 18 15 12 9 6 3 −4 −3 −2 −1−3 −6

2 −4 −3 −2 −1

x 1

−4

9 a

−2 −16

−4 −3 −2 −1−2 −4 −6 −8 −10 −12 −14 −16 −18

−2 −1

x y

2 3 4

8 a

1 2 3

−6 −5 −4 −3 −2 −1

x 1

2 3 4

x 1

−4 −6 −8

Answers mathspace.co

981

21 a V(x) = x(6 − 2x) (10 − 2x) or V = 4x(3 − x) (5 − x) cm3

9 a x = −7, −3, 1

b 0<x<3 c

36 32 28 24 20 16 12 8 4

−1 −4 −8 −12 −16 −20

10 a x = 6, −2, 3

b x = −5, 4, −1

c x = 8, −3, 2

d x = −7, 6, −4

11 a x = −3, 5, −2

b x = 7, −1, 4

c x = −6, 3, −5

d x = 2, −8, 1

d 24 cm3

12 a

22 x = 2, x = −1, x = −3 Reasoning

15.03 Optional: Cubic equations Understanding 1 a True

b True

c False

d True

2 a Cubic

b Not cubic

c Cubic

d Not cubic

3 a 1 real root

b 3 real roots

Fluency 4 (x − 5)(x + 2)(x − 1) = 0

x = 5, −2, 1

5 a x=2

b x = 10

c x=3

e 6 a x=2 c x = −1 7 a x = 3, −1, 4 c x = 7, −4, 0

b x=5 d b x = −6, 5, −2 d x = −5, 1, 2

8 a x = 0, −2

b x = 1, −3

c x = −6, 2

d x = 4, −7

e x = 8, −5

f x = −10, 3

13 The cubic equation (x − 4)(x + 1)(x − 6) = 0 has 3 real roots because it is already in factored form, giving roots x = 4, −1, 6. Since it’s a cubic, it has at most 3 roots, and all are real and distinct here. 14 a Sometimes true. A cubic equation always has 3 roots (by the Fundamental Theorem of Algebra), but they may not all be real or distinct. For example, x3 = 8 has 1 real root (x = 2) and 2 complex roots, while (x − 1) (x + 2)(x − 3) = 0 has 3 real roots (x = 1, −2, 3). b Always true. Since a cubic polynomial is continuous and has odd degree, it must cross the x-axis at least 1 time (going from negative to positive or vice versa as x goes from −∞ to ∞). 15 a Since the root 2 is repeated, the equation in factored form is (x − 2)2(x − 5) = 0. b The equation has only two distinct real roots because the root x = 2 has multiplicity 2 (it is repeated), while x = 5 has multiplicity 1. A cubic equation has exactly three roots in total, counting multiplicity, but here the repetition of 2 means there are only two unique real values, 2 and 5, that satisfy the equation.

Answers mathspace.co

983

16 Neil is incorrect. For the equation 5x3 = 40, you can simplify it to x3 = 8, which gives x = 2 as the only real solution. A single-term cubic function like 5x3 is strictly increasing, so it crosses the x-axis only once and does not have three distinct real solutions.

y 150 100 50

Problem-Solving x −4

−2

x = 0.67, −5, 1 18 a

y 200

30 20 10 −5 −4 −3 −2 −1 −10

x = 8, −0.43, −4 f

y 400

x 2

300

−50

200

−100

100 x

−150

−4

x = 6, −3, 0.75

−100

, and c = −2.

If x = 5 is the double root, then either

−150 −200

20 t = 1, 2, 3

−300

19 If x = 1 is the double root, then c = −2 or

−250

−2

x = −6, 0.63, 3 x

y −2 −50

and

21 a k = 5

−350

b m = 5x3

−400

x = −2, 7, −0.80

984

−1000

−800

−2

−600

2 3

x = −4, 2, −0.33

−4

−400

−20

−6 −4 −2 −200

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

135

320

and

d 2560 m 2240 1920 1600 1280 960 640 320

y 20 15 10

x 0

x −3 −2 −1

e m = 3071 g

f Side length = 7 cm b i

Chapter 15 review 1 a Negative b i 24

iii −375

2 a x>0 c

iv −3

-1

y 15

b x<0

-3

-18

-1

5 x

−2 −1

2 3 4 5 6

−5

b x>0

3 a Positive

−10

4 x>0 5 a No 6 a

b Yes

c No

d No

9 a x=5

-8

-1

10 a x = 2

b x=4

c x = -2

11 a x = 5, −3, 1

b x = −4, 7, −2

c x = 3, −6, 4

d x = −8, 2, −5

2 −1

b x=2 d

−2

x 1

12 a x = −5, 2, −1

−4

b d

−6 −8

16.01 Optional: Circles at origin

7 a y = (x + 3) (x − 2) (x − 5)

Understanding

b y = −(x + 5) (x − 1) (x − 3) 8 a i

1 a 4 units

-1

b 8 units

c 9 units

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.

Answers mathspace.co

985

Fluency

units

e i

4 x2 + y2 = 25

5 4 3 2 1

5 9 units 6 x2 + y2 = 64 7 a i 6 units ii

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

6 4 2

−8 −6 −4 −2 −2

x 1 2 3 4 5

f i 0.8 units ii

−4

−6

0.8

−8

0.6 0.4

b i 4 units ii

0.2

−0.8 −0.6 −0.4 −0.2 −0.2

6 4

0.2 0.4 0.6 0.8

−0.4

−8 −6 −4 −2 −2

−0.6

−0.8

−4 −6 −8

b x2 + y2 = 36

d x2 + y2 = 100

9 a i 5 units

ii x2 + y2 = 25

b i 10 units

ii x2 + y2 = 100

c i

−8 −6 −4 −2 −2

11 a Inside

b On

c Outside d On

13 The point is on the circle.

−6 −8

Reasoning

d i 13 units y 12 8 4

x 4

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.

−4 −8 −12

986

ii x2 + y2 = 125

12 The point lies inside the circle.

−4

−12 −8 −4

units

10 x2 + y2 = 100

4 2

c x + y = 121

c i 3 units ii

8 a x2 + y2 = 9

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

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.

5 a i (2, −2) iii

8 4 2

−8 −6 −4 −2 −2

−4 −6 −8

b i (2, −2) iii

ii 4 units 8

17 Using the distance formula, we can ﬁnd the radius of the circle:

r2 = (5 − 0)2 + (6 − 0)2 = 25 + 36 = 61. Therefore, the equation of the circle is x2 + y2 = 61.

4 x

−8 −6 −4 −2 −2

−4

Problem-solving 18 a x2 + y2 = 36

As both points satisfy the same equation, they are on the same circle. 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.

ii 6 units

−6

b 36π units2

19 x2 + y2 = 64

−8

c i (−1, 3)

20 x2 + y2 = 36

iii

ii 4 units 8

21 x2 + y2 = 32

6 4

23 (−2.846, −8.538) and (2.846, 8.538) units

−4 −6

16.02 Optional: Translations of circles Understanding 1 a No

−8 −6 −4 −2 −2

−8

d i (−3, −4) b Yes

c Yes

d Yes

2 a Centre: (−3, −3), r = 3

iii

ii 5 units 8 6

b Centre: (0, 1), r = 4

c Centre: (2, −3), r = 1

d Centre: (−2, 4), r = 3

−8 −6 −4 −2 −2

3 a (1, −2)

b 8 units

x 2

−4 −6 −8

Fluency 4 a i (5, −6)

units

b i (5, 5)

units

c i (0.4, −3.8)

units

d i

units

Answers mathspace.co

987

iii

5 4 3 2 1

8 a (x − 2)2 + y2 = 16

b x2 + ( y − 4)2 = 25 c (x − 3)2 + ( y + 3)2 = 9 d (x − 5)2 + ( y − 5)2 = 25 x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

9 a x2 + ( y − 10)2 = 25 b x2 + ( y + 10)2 = 25 c (x − 10)2 + y2 = 25 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

ii r = 7

f i (6, 0) iii

ii (x − 3)2 + ( y + 2)2 = 16

units

e i

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

4 2 −2

x 2

d Scaled by a factor of 4.5, then translated left 1.5 units and up 4.5 units

8 10 12 14 16

−4 −6

Reasoning

−8

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:

6 a (x − 7)2 + ( y + 1)2 = 9 b (x + 2)2 + ( y + 3)2 = 25

LHS = (0 + 8)2 + (0 + 6)2 = 64 + 36 = 100 = RHS

c (x + 4.3)2 + y2 = 36 d (x + 10)2 + ( y − 2)2 = 144 7 a i

10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

Since the point satisﬁes the equation, it must lie on the circle. x 2 4 6 8 10

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.

4 2

−4

988

will lie inside

the graphed circle.

−8 −6 −4 −2 −2

units

b Inside. The circle on the given graph has a is smaller radius of 7.5. A radius of than the graphed circle, so the circle with the equation

ii (x − 1)2 + ( y − 2)2 = 16 b i

12 a

x 2

Problem-solving 14 a (x − 5)2 + ( y + 2)2 = 9

−6

b (x + 1)2 + ( y − 4)2 = 36

−8

c x2 + ( y + 3)2 = 16

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

7 a i 3 units

ii x2 + y2 − 2x − 4y − 4 = 0

15 a (x − 3)2 − 9 + y2 + 8y = 9 2

b i 4 units ii x2 + y2 + 8x + 6y + 9 = 0

b x − 6x + ( y + 4) − 16 = 9 c (x − 3)2 + ( y + 4)2 = 34

c i 5 units

d The centre of the circle is (3, −4) and the . radius is r =

units

e i

17 (x − 25)2 + ( y − 30)2 = 400

ii x + y2 − 6x + 2y − 7 = 0

b 9.15 m

c (x − 49.5)2 + ( y − 37.5)2 = 83.7225 d (9.15 × 2)2 = 334.89 m2

units

f i

ii x2 + y2 + 4x − 6y + 8 = 0 8 a i (−1, −4)

b 85.19 cm2

19 a 5 cm

units

d i

ii x2 + y2 − 2x − 4y + 3 = 0

16 (0, 2), (6, 2), (0, −4), (6, −4)

18 a (49.5, 37.5)

ii x2 + y2 − 10x − 14y + 49 = 0

c Circle one: (x + 1)2 + ( y − 6.085)2 = 49, Circle two: (x − 7)2 + ( y − 6.085)2 = 36

16.03 Optional: Transform circle equations

units

iii (x + 1)2 + ( y + 4)2 = 221 b i (−2, 3) units

iii (x + 2)2 + ( y − 3)2 = 10 9 a i (x + 2)2 + ( y + 5)2 = 36

Understanding

ii (−2, −5)

1 a 4 units

units

2 a A translation 3 units to the right. b A translation 6 units downwards. c A translation 6 units to the left and 2 units downwards. d A dilation 3 a (x + 9)2 − 81

b (x − 4)2 − 16

c (x + 5)2 + 6

d (x + 7)2 − 2

4 a Centre: (4, −2), radius: 5 , radius:

b Centre: 5 a 10 units

b 15 units

c 8.54 units

d 5.83 units

iii 6 units b i (x + 9)2 + ( y − 8)2 = 4 ii (−9, 8) iii 2 units c i (x − 7)2 + ( y − 5)2 = 1 ii (7, 5) iii 1 unit d i ii iii

units

10 a

x 2

Fluency 6 a i

−2

units

ii (x − 2)2 + ( y − 4)2 = 2

−4 −6

b i 2

ii (x + 2) + ( y − 4) = 26 c i

−8

units

ii (x + 2)2 + ( y + 5)2 = 26 d i 13 units

b x2 − 8x + y2 + 12y + 43 = 0

ii (x + 2)2 + ( y − 5)2 = 169

Answers mathspace.co

989

11 a i (x − 6)2 + ( y + 1)2 = 36 ii

e i (x − 5)2 + ( y + 1)2 = 81 ii

y 6 4 2

x 2

−4 −2 −2 −4 −6 −8 −10

−2 −4 −6

b i (x − 3)2 + ( y + 1)2 = 9 ii

8 6 4 2

x 2 4 6 8 10 12 14

f i (x + 2)2 + ( y + 3)2 = 49 ii

−1 −1

−10 −8 −6 −4 −2 −2

−2

−4

−3

−6

−4

−8

−5

−10

x 2

c i (x + 2)2 + ( y − 7)2 = 64 ii

g i

10 8

4.5

4 2 −10 −8 −6 −4 −2 −2

1.5

x 2

−10−9 −8 −7 −6 −5 −4 −3 −2 −1 −1.5

d i (x − 6)2 + ( y − 9)2 = 36 ii

−3 −4.5

y 14 12 10

h i

8 6 4 2

x 2

990

10 12

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

x 1 2

d i (x − 3)2 + ( y + 2)2 = 4 ii

2 1

−2 −1

−7 −6 −5 −4 −3 −2 −1 −1

−2 −1

−2

x 1

−1

−2

−3

−4

−4 −5

12 a i (x + 2)2 + ( y − 3)2 = 3 ii 5

13 a i

y 2

2 1 −5 −4 −3 −2 −1

x 1

−1

2 −2

−1

−2

−1 −2

b i ii

y 2

b i

1 −3 −2

−1

−2 1

−3 −4

x −1

−1

c i ii

c i

3 2

1 −5 −4 −3 −2 −1

x 1

x −2

−1

−1 −2

Answers mathspace.co

991

The distance between the centre and P(4, −3) is:

d i ii

Distance formula

Substitution

Evaluate the subtraction

x −1

Evaluate the indices

−1

Evaluate the addition

−2

Since this distance is greater than the radius, 14 a A vertical translation 4 units up, and a dilation by a factor of 4. b A horizontal translation 5 units right, and a dilation by a factor of 10. c A horizontal translation 2 units left, a vertical translation 4 units down, and a dilation by a factor of 3. d A horizontal translation 4 units right, a vertical translation 7 units down, and a dilation by a factor of 6. e A horizontal translation

units left, a vertical

translation 3 units up, and a dilation by a factor of . f A horizontal translation 0.75 units left and a vertical translation up 0.2 units.

, P is outside the circle. 17 Since both points have the same y-value, they lie on the same horizontal line. The point is directly to the right of the centre. To find the radius, we can find the horizontal distance between the points by subtracting the smaller x-coordinate from the larger x-coordinate: r = 1– (−7) = 8 18 Yes, we can graph the circle with equation in the form x2 + y2 + ax + by + c = 0 using technology to find the centre and radius of the circle. Problem-solving 19 a i

. In this case, the

centre of the circle is (a, b) and a point on the circle is (x, y). The distance between these two points is the radius r. Squaring both sides of the distance formula, we get (x − a)2 + ( y − b)2 = r2, which is the equation of a circle with centre (a, b) and radius r. Since (x, y) represents any point, this equation represents all the points that are r units from the centre point, and these points form a circle. 2

16 a (x − 4) + ( y + 3) = 41 b From the equation in part (a), we know the centre of the circle is (4, −3) and the radius is .

units

iii 37π units

ii 2π units

b i 1 unit

15 The distance formula states that the distance between two points (x1, y1 ) and (x2, y2 ) is given

Reasoning

992

units

iii π units2 c i

units

iii 29π units2 d i

units

iii

units2

20 a i

units

iii 7π units2 b i

units 2

iii 8π units c i

units

iii 12π units2 d i

units

iii 17π units2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

21 a Outside the circle

b On the circle

c Outside the circle

ii r = 7

b i (0, 4)

d Inside the circle

iii

22 a x2 + y2 − 10x − 6y + 30 = 0

b Yes

c 127 cm

23 x 2 + y2 − 4x − 7y = 0, or (x − 2)2 + ( y − 3.5)2 = The radius is

which is approximately equal

−8 −6 −4 −2 −2

to 4.03. The centre is (2, 3.5).

x 2

−4

Chapter 16 review 8 a i Translated left 4 units and up 5 units

1 a 7 units b

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

4 2 −8 −6 −4 −2 −2

x 2

−4

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)

units

11 a i

−6

ii x + y2 + 8x − 8y − 18 = 0

−8

units

b i 2 a x2 + y2 = 64

b x2 + y2 = 25

ii x2 + y2 − 16x − 20y + 151 = 0

3 a i 6 units

ii x2 + y2 = 36

12 a (−5, 6)

b i 13 units

ii x2 + y2 = 169 ii x2 + y2 = 20

units

c i 4 a Inside

b On

c Outside d Inside

13 a i (x + 7)2 + y2 = 25

ii (−7, 0)

iii 5 units b i (x − 2)2 + ( y + 1)2 = 16

5 32π units

ii (2, −1)

iii 4 units

6 a (x − 4)2 + (y − 9)2 = 1

14 a A horizontal translation 8 units left, a vertical translation 3 units up, and a dilation by a factor of 7.

b 7 a i (3, −1) iii

b 5 units

c (x + 5)2 + ( y − 6)2 = 25

ii 5 units 5 4 3 2 1

−3−2 −1 −1 −2 −3 −4 −5 −6 −7

b A horizontal translation 0.25 units right and a vertical translation up 0.75 units.

15 a i (x − 2)2 + ( y + 7)2 = 36 x 1 2 3 4 5 6 7 8 9

y −4 −2

x 2

−2 −4 −6 −8 −10 −12

Answers mathspace.co

993

7 a h=1

b i ii

c w = 28

y 3 1

x 1

−1

9 a

10 a i

b i

11 a m = 5

−2 −3 −4

16 a Outside the circle

b Inside the circle

c Outside the circle

d Inside the circle

12 a θ = 60°

17 x2 + y2 − ax − by = 0 or . The radius is

and the centre is

17.01 Optional: Angles with exact values Understanding b 0°, 90°, 180°, 270°

2 a 60°

c 48°

d (90 − x)°

14 a θ = 30°

15 a θ = 45°

b θ = 60°

16 a 1

d θ = 60°

f 1

Reasoning and

therefore

18 a

4 a

∴ sin 30° = cos 60° b cos (90° − θ )

c cos θ

d tan (90° − θ ) =

Fluency 5 a

h 1

i b x = 30°

6 a c i

iii

d i

iii

19 At 90° and 270°, the cosine function has a value of 0, which makes the denominator of the tangent function 0. Since division by 0 is undeﬁned, the tangent function has an undeﬁned value at 90° and 270°. 20 In a right-angled triangle, sin θ = cos θ =

and tan θ =

expression, we get

994

c θ = 45° b tan θ = 1

1 a 30°, 45°, 60°

b θ = 60°

13 a θ = 45°

e 2

b 18°

8 2

−5 −4 −3 −2 −1

b x=2

Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

. and simplifying this , which is equal to tan θ.

Therefore, tan θ =

for any angle θ, except

Fluency 6 a i

iii

Problem-solving

b i

iii

21 • Opposite side (30° angle) = 5

c i

iii

d i

iii

90°, in a right-angled triangle.

• Adjacent side (30° angle) = • Area = 22 θ = 30° 23 • Opposite side = • Hypotenuse = 8 • Area = 16 24 3 25 • Adjacent side = 3 • Hypotenuse = 6 • Area =

17.02 Optional: The unit circle

8 a Positive

b Negative

c Negative

d Positive

e Negative

f Negative

g Positive

h Positive

9 a 50°

b 80°

c 20°

d 65°

e 12°

f 42°

g 76°

h 71°

b cos 300° = cos 60°

c tan 275° ≠ tan 85°

d sin 150° = sin 30°

e −cos 110° = cos 70°

f tan 145° = − tan 35°

11 a sin 74°

b −cos 31° c −tan 56° d −sin 60°

e cos 45° f −tan 40° g sin 6°

Understanding 1 a Quadrant 4

b Quadrant 1

c Quadrant 2

d Quadrant 3

e Quadrant 1

f Quadrant 4

g Quadrant 2

h Quadrant 4

sin

b (−0.94, 0.34)

10 a sin 215° ≠ sin 35°

7 a (0.77, 0.64)

cos

tan

12 a i

b i

13 a

30°

b i 135°

45°

h −cos 30°

i −tan 65° j −sin 40° k cos 69° l tan 84°

ii 225°

iii 315°

ii 240°

iii 300°

c d −sin 45°

60°

e 3 a −1.732

b 0.423

c −0.309

d −0.158

e 0.961

f −0.719

g −1

h 1

4 a False

a False

c True

d True

e True 5 a 1

14 a b i 120° c

b x

e (cos θ, sin θ )

c y

d tan 60° e

Answers mathspace.co

995

20 A represents all ratios being positive in the ﬁrst quadrant.

15 a b i 150°

ii 210°

iii 330°

S represents the sine ratio being positive in the second quadrant.

T represents the tangent ratio being positive in the third quadrant.

d −cos 30°

C represents the cosine ratio being positive in the fourth quadrant.

e 16 a e i

b −1

g −1

Problem-solving 21 a P (0.41, 0.91)

b (−0.41, 0.91) d −0.41

c 0.91 e −2.25

22 a Reasoning

23 a True

17 a Quadrant 2

b Quadrant 2

e False

c Quadrant 2

d Quadrant 1

24 a False

18 a The points are the coordinates where the angles 0°, 90°, 180° and 270° intersect the unit circle. The y-coordinate is the sine of the angle.

b True

c True

d True

b False

c True

d True

e True ii R(−a, −b)

iii S(a, −b)

b i −tan 62°

ii tan 62°

iii −tan 62°

c i −tan x

ii tan x

iii −tan x

25 a i Q(−a, b)

0° intersects the unit cirle at (1, 0) therefore sin 0° = 0.

26 a i 60°

90° intersects the unit cirle at (0, 1) therefore sin 90° = 1.

b i 0°

ii 0

180° intersects the unit cirle at (−1, 0) therefore sin 180° = 0.

c i 60°

d i 0°

ii 0

270° intersects the unit cirle at (0, −1) therefore sin 270° = −1. b 0° intersects the unit cirle at (1, 0) therefore cos 0° = 1. 90° intersects the unit cirle at (0, 1) therefore cos 90° = 0.

17.03 Optional: Graphs of sine and cosine Understanding

180° intersects the unit cirle at (−1, 0) therefore cos 180° = −1. 270° intersects the unit cirle at (0, −1) therefore cos 270° = 0. 19 We can draw a vertical line from the point to the x-axis to form a right-angled triangle with one side cos θ and the other side sin θ. Since the point is on the unit circle, the hypotenuse of the triangle is 1. Using the Pythagoras’ theorem, we have x2 + y2 = 12, or cos2 θ + sin2 θ = 1.

1 a The sine function graph has a wave-like shape that oscillates between −1 and 1, repeating every 360 degrees. b The cosine function graph has a wave-like shape similar to the sine function, but it starts at its maximum value of 1 and oscillates between −1 and 1, repeating every 360 degrees. 2 a i 20°

c 360°

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

ii 120°

iii 220°

b The function values and shape of the graph repeat at regular intervals.

3 a True

b False

c True

d False

e True

f True

4 a True

b False

c True

d True

e False

f False

5 a (0, 0)

b 1

c −1

6 a (0, 1)

b 1

c −1

Fluency

7 a

x cos x

0° 90° 180° 270° 360° 1

−1

90 180 270 360

−1

c A decimal between −0.95 and −0.9 is acceptable. d A decimal between 0.9 and 0.95 is acceptable.

• A number between −75° and −70° is acceptable.

−2

0° 90° 180° 270° 360° 0

b A decimal between −0.2 and −0.15 is acceptable.

f • A number between 70° and 75° is acceptable.

x −360 −270 −180 −90

sin x

10 a A decimal between 0.85 and 0.9 is acceptable.

• A number between −140° and −135° is acceptable.

• A number between 320° and 325° is acceptable.

e • A number between 135° and 140° is acceptable.

y 2

8 a

f • A number between 215° and 220° is acceptable.

−1

11 a Negative

b Second quadrant

12 a Negative

b Third quadrant

13 a 360° b 180°

c True

2 1 x −360 −270 −180 −90

90 180 270 360

−1 −2

9 a A decimal between 0.85 and 0.9 is acceptable. b A decimal between −0.9 and −0.85 is acceptable. c A decimal between 0.6 and 0.7 is acceptable. d A decimal between −0.4 and −0.3 is acceptable. e • A number between 60° and 70° is acceptable.

d i Decreases

ii Decreases

iii Increases

iv Increases

v Increases 14 a 270° b i 90°, 270°

ii −270°, −90°

c True d i Decreases

ii Increases

iii Increases

iv Decreases

Reasoning 15 The sine and cosine functions are related in that the cosine function is a horizontal shift of the sine function by 90 degrees to the left. This means that the cosine function can be obtained by shifting the sine function 90 degrees to the left.

• A number between 110° and 120° is acceptable.

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997

16 a True. The amplitude of both the sine and cosine functions is 1 because their maximum and minimum values are 1 and −1, respectively.

b i x = 0°, ±180°

22 a

b False. The sine function has a period of 360° because it completes a full cycle of its wave-like shape every 360°, not 180°. c False. The graphs of the sine and cosine functions intersect at points separated by 180°, not 90°. They intersect at 45°, 225°, 405° and so on. d True. The sine and cosine functions have the same period of 360° because they are both based on the unit circle, which has a circumference of 360°. 17 For points on the unit circle, the x-coordinate is equal to cos θ and the y-coordinate is equal to sin θ. So these two functions intersect at points on the unit circle where the x- and y-coordinates are equal.

ii x = ±90°

iii x = ±60°, ±120°

iv x = ±135°, ±45°

tan θ

0°

45°

90°

undefined

135°

−1

180°

225°

270°

undefined

315°

−1

360°

b We can sketch the graph by plotting the points and connecting them, taking note of the undeﬁned values at 90°, 270°, etc., which would indicate vertical asymptotes. Here’s the graph:

y 2

That is, they intersect at points which also lie on the diagonal line y = x. This line passes through the origin at an angle of 45°, and crosses the circle exactly twice. These points occur in the ﬁrst and third quadrants, at angles of 45° and 180° + 45° = 225°.

1 x −360 −270 −180 −90

90 180 270 360

−1 −2

Problem-solving 18 a 1

b 0

c 0

d −1

e −1

f 0

g 0

h 1

19 a 30°

b 120°

c 210°

20 a

c Upon research, the graph of tan θ is found to have a periodic wave-like shape with vertical asymptotes at every odd multiple of 90°. This behaviour is because tan θ goes to inﬁnity when cos θ approaches zero. Compare your sketch with the correct shape and identify where you correctly captured these characteristics and where you might have missed them.

y 1

y = cos x

−150 −120 −90 −60 −30

90 120 150

−1

b i x = ±90°

ii x = 0°, ±180°

iii x = ±60°, ±120°

iv x = ±135°, ±45°

21 a

Chapter 17 review 1 a

c tan 45° = 1

y = sin x

x −150 −120 −90 −60 −30

90 120 150

−1

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Mathspace WA Curriculum – Year 10 Methods Pathway mathspace.co

2 a w = 32

3 a

d 1

12 a Negative c i 90°, 270° 13 a 0° c i Decreasing

4 a • Opposite side = 6

b Third quadrant ii −90°, −270° b False ii Decreasing

iii Increasing

• Hypotenuse = units2

b Area = 5 2

6 a

7 a Negative

b Positive

c Positive

d Negative

8 a − cos 50°

b sin 65°

c − tan 9°

d − sin 42° b

9 a 10 a

11 a

sin x

0°

30° 90°

135°

0.71

180°

240°

−0.87

270°

−1

300°

−0.87

360°

b (0, 0) c (0, 0), (180, 0), (360, 0) d

y 2

1 x −360−270 −180 −90

90 180 270 360 −1

−2

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999