Straight line functions

Contents

3.2 Functions and relations

3.3 Sequences and progressions

3.4 Gradient and intercepts of a linear function

3.5 Equations of linear functions

3.6 Distances and midpoints

3.7 Finding simultaneous solutions of linear functions Chapter summary Chapter review

Syllabus subject matter

■ Concepts of function, domain and range ■ Ordered pairs, tables, graphs and equations as representations of functions and relations ■ Graphs as a representation of the points whose coordinates satisfy an equation ■ Distinction between functions and relations ■ Distinctions between continuous functions, discontinuous functions and discrete functions ■ Solutions to simultaneous equations in two variables: graphically, using technology; algebraically (linear and quadratic equations only)

Quantitative concepts and skills

■ Calculation and estimation with and without instruments ■ Basic algebraic manipulations ■ Gradient of a straight line

Syllabus Guide Chapter 3

Straight line functions

3.1

Review of linear equations

An equation will usually only be true for some values in the stated range. It is even possible that there are no values in the stated range that make an equation true.

Because x1=_{x, a linear equation involving x would not contain x terms raised to any other} exponent.

So a linear equation in x cannot contain terms involving , x2_{, x}4_{, x}−1_{, x}−3_{, (or x}−1_{) and so on,} unless they cancel out on simplification.

Linear equations in one variable have, at most, one answer. To solve an equation in one variable, make the variable the subject of the equation—isolate the variable.

There are many kinds of equations and functions. Linear equations are among those that can be solved to give exact answers, and the skills used to solve these equations are very important to understanding how to solve other equations. The study of linear functions is sometimes called ‘coordinate geometry’, and a clear understanding of these functions is necessary before more complex functions or the study of calculus can be undertaken. Many of the mathematical puzzles that appear in newspapers and magazines are easily solved using linear equations.

The values that make an equation true are called roots or solutions of the equation. A linear equation is one that involves a pronumeral or variable raised to the power of 1 only.

!

x

1 2

--- ₁

x

---State whether or not each of the following is a linear equation.

a 3p − 5 = 6p + 7 b n(n + 8) = 16 c + 3v = 5v − 7

Solution

a 3p − 5 = 6p + 7 is linear as the variable has only the power 1.

b n(n + 8) = 16 is not linear because when you multiply out the brackets you get n2_.

c + 3v = 5v − 7 is not linear because it has the reciprocal or v−1_. 1

v

---1 v

--- 1

v

---Example

1

To solve a linear equation with one variable:

1 Remove divisions and fractions by multiplying each term by the lowest common

denominator (LCD).

2 Remove brackets by expanding.

3 Collect all variable terms on the left-hand side and all constant terms on the

right-hand side.

4 Undo operations (+, −, × and ) by performing the opposite operations. Remember that the bar (vinculum) of a fractional term implies brackets.

!

You can use a graphics calculator to solve equations. We will solve 3x+ 11 = 5x+ 3 − 4x+ 1 using a calculator. When solving simple equations on the graphics calculator, it is simplest to use the built-in X variable.

Remember: On all the calculators the key is used for negatives like −5, and the key is used for subtraction, like 4 − 5.

Casio fx-9860G AU

If the MAIN MENU screen is not displayed, press to

go into it.

Use the cursor keys to go to EQUA in the

main menu.

Solve the following.

a 4x − 9 = 15 b 4(y + 8) + 2y + 5 = y + 9 c 4 − = 2p − 8

Solution

a Write the equation. 4x − 9 = 15

Collect constant terms on the RHS. 4x = 15 + 9

Simplify. 4x = 24

Undo the multiplication using division. x =

Evaluate the RHS. x = 6

b Write the equation. 4(y + 8) + 2y + 5 = y + 9

Remove the brackets. 4y + 32 + 2y + 5 = y + 9

Simplify. 6y + 37 = y + 9

Collect variable and constant terms. 6y − y = 9 − 37

Simplify. 5y =−28

Divide both sides by 5. y =

Evaluate. y =−5.6

c Write the equation. 4 − = 2p − 8

Multiply all terms by the denominator. 3 × 4 − 3 × = 3 × 2p − 3 × 8

Simplify. 12 − ( p − 6) = 6p − 24

Remove the brackets. Watch the sign. 12 − p + 6 = 6p − 24

Simplify. 18 − p = 6p − 24

Collect variable and constant terms. −p − 6p =−24 − 18

Simplify. −7p =−42

Divide both sides by −7. p =

Evaluate. p = 6

p–6 3

---24 4

---28 –

5

---p–6 3

---p–6 3

---42 –

7 –

---Example

2

Technology

(–) −

MENU

Press to go into the equation menu and then press to go into the Solver mode.

If there is already an equation entered, delete it by pressing (DEL) and (YES).

Enter the equation 3X + 11 = 5X + 3 − 4X + 1 using the keys:

3 11 5 3

4 1

Press to store the equation. Press (SOLV) to solve

the equation.

Texas Instruments TI-84

Press the key to go into the maths menu and press 0 to go into the Solver menu. If a

screen with ‘bound = { …’ is displayed, press as many times as required to go into the

equation editor. If an equation is already displayed, press to get rid of it.

For the TI-84, equations are entered as 0 = LHS − RHS. To enter the equation 3X + 11 = 5X + 3 − 4X + 1 we actually need to enter 0 = (3X + 11) − (5X + 3 − 4X + 1). Do this as:

3 11 5

3 4 1 .

Press . Ignore what you see. Press to SOLVE the equation.

Note: You can change the bounds of the solution to make the calculator work more quickly.

Sharp EL-9900

See the instructions given on the CD-ROM.

Some types of real-world problems may be solved using linear equations. To express a problem algebraically, the variable must first be identified and named, and then the information must be translated into an equation.

EXE F3

F2 F1

X,θ,T + SHIFT . X,θ,T + – X,θ,T +

EXE F6

MATH

CLEAR

( X,T,θ,n + ) − ( X,T,θ,n + – X,T,θ,n + )

ENTER ALPHA ENTER

Calculator Instructions

Mandy, Andy and Sandy are candidates for a vacant position on the student council. In the election, 734 students each voted for one of the three candidates. Andy received 78 more votes than Mandy, while Sandy received two-thirds of Mandy’s total. How many votes did each candidate receive?

Solution

Identify the variable. Let the number of votes for Mandy be m. This means that Andy received (m + 78) votes and Sandy m votes.

Write an equation using m. m + (m + 78) + m= 734

Simplify. 2 m = 734 − 78

m = 656

Multiply both sides by . m = 656 × = 246

State the answer. Mandy received 246 votes, Andy received 324 votes

and Sandy received 164 votes.

2 3 ---2

3 ---2 3 ---8 3

---3 8

--- 3

8

René Descartes (1596–1650) is said to have been the first person to think of plotting points using rectangular coordinates. For this reason, the Cartesian coordinate system was named after Descartes.

Many relationships between real quantities show as straight lines when plotted on Cartesian coordinates. These are called linear relationships and graphs.

In Example 4, there are two variables: cost and the number of games played. Because the cost depends on the number of games, it is known as the dependent variable. The number of games is known as the independent variable. It is usual to show the independent variable on the horizontal (x) axis and the dependent variable on the vertical (y) axis. Points on the graph such as (1, 16) and (2, 22) are known as ordered pairs, with the x value followed by the y value.

Extra Material

Order and inequations

Extra Material

Tables and graphs

In Cartesian coordinate systems it is usual to:

■ make the first axis horizontal and refer to it as the x-axis, and refer to the vertical axis as the y-axis

■ show coordinates in parentheses—for example, (4, 7). The first coordinate is

sometimes called the abscissa and the second is called the ordinate

■ refer to the point (0, 0), where the axes cross, as the origin

■ consider the plane as being divided into four quadrants by the axes.

!

It costs $10 to enter an amusement parlour, and once inside it costs $6 for each game played. Draw a graph to show the cost of an outing to the amusement parlour where up to 6 games are played.

Solution

First construct a table to show the cost for the outing where 0, 1, 2, …, 6 games are played.

Now choose suitable scales to draw the axes and plot the data.

The points lie on a straight line, so the

relationship between the two quantities is linear.

Number of games played 0 1 2 3 4 5 6

Cost ($) 10 16 22 28 34 40 46

Outing to amusement parlour

Cost ($)

50 40 30 20 10 0

1 2 3 4 5 6 Number of games played

The graph in Example 4 shows a linear relationship between the variables. In this case, the number of games played is a discrete variable.

■ Discrete variables have only particular values and are often counted, such as the number of

stairs in a building and the number of people at a party. Typically, the values of discrete variables are integers or natural numbers.

■ Continuous variables are things that can be measured, such as height, weight and length.

Typically, the values of continuous variables are real numbers.

When there is a relationship between two variables, it is possible to find a rule to describe it. In Example 4, the initial cost was $10, with each game costing $6. The rule to describe the relationship between cost C and number of games g is C = 10 + 6g.

Draw a graph for each of the following using the Cartesian system.

a y = 4x − 5 where x = 1, 2, 3, 4, 5 b y =−3x + 8 where x is any real number

Solution

a First construct a table and show the x and y values.

The ordered pairs are (1,−1), (2, 3), (3, 7), (4, 11), (5, 15).

Construct suitable axes and plot the ordered pairs.

b First construct a table and show the x and y values.

The ordered pairs are (−2, 14), (−1, 11), (0, 8), (1, 5), (2, 2)

Construct suitable axes and plot the ordered pairs.

This time join the ordered pairs with a straight line to show that in-between values are possible.

x 1 2 3 4 5

y −1 3 7 11 15

y = 4x − 5

16 14 12 10 8 6 4

2 x

2

y

4 3 5 0

−2

y = 4x − 5

x −2 −1 0 1 2

y 14 11 8 5 2

y =−3x + 8

16 14 12 10 8 6 4

0 x

2

y

2 1 3

−1

−3 −2

y =−3x + 8

Graphics calculators are especially useful for drawing the graphs of data. In this activity, we will see how to graph the equation y = 3x − 6.

Casio fx-9860G AU

Use the cursor keys to go to GRAPH

in the main menu.

Press to go into the graph function menu and then enter

the equation Y1 = 3X − 6 by typing: 3 6.

Press to store the equation.

Next press (DRAW) to draw the graph of Y = 3X − 6.

Note: Once you have drawn the graph, you may need to change

the maximum and minimum values of X and Y by pressing (VIEW WINDOW) and setting the values as appropriate.

Use the TRACE, ZOOM and the cursor keys to investigate any points of

interest on the graph.

Can you find where the graph cuts the axes?

Press to return to the graph function menu and then

enter the equation Y2 =−3X − 6 by typing:

3 6.

Press to store the equation.

Press to draw the graph of Y = −3X − 6.

Note: The graphs of both equations are drawn.

Now use your graphics calculator to investigate the graphs of other linear equations.

Texas Instruments TI-84

Enter the equation Y1= 3X − 6 by typing: 3 6.

Press to store the equation.

Press to draw the graph.

Note: You may need to change the maximum and minimum

values of X and Y that are used to scale the axes in the display

of the calculator. Do this by pressing the key and

setting the values as appropriate.

Technology

EXE

X,θ,T – EXE

F6

SHIFT F3

AC/ON

(–) X,θ,T – EXE

F6

Y= X,T,θ,n – ENTER

GRAPH

Use the , and the cursor keys to investigate any points of interest on the graph.

Can you find where the graph cuts the axes? Now enter the expression Y2=−3X − 6 by typing:

3 6.

Press to store the equation.

Press to draw the graph.

Note that the graphs of both equations are drawn.

Now use your graphics calculator to investigate the graphs of other linear equations.

Sharp EL-9900

See the instructions given on the CD-ROM.

Exercise 3.1

Review of linear equations

1 State whether each of the following is a linear equation.

a 4x + 5 = 8 b + x = 3x + 4 c y + 9 = y2− ₃

d 5b2+ ₃ = _{0 e 5}x+ ₃ = _{11 f} + ₃ = ₅

g (v + 4)(v − 4) = v + 2 h 3(m + 2) − 5(m − 2) = 4(m + 4)

2 Solve the following.

a 2g − 8 = 28 b −9b + 6 =−40 c 2q + 8 = 3

d −6u − 5 = 13 e 6 − 3r = 17 f 11w + 40 = 3

3 Solve the following.

a k − 11 = 15 − k b 5 − 4a = 4a + 5

c 2u − 9 = 7u + 15 − u − 20 d 5 − 3x + 9 = x + 30 + 3x + 5

e 6t − 4t + 10 = 7t + 8 − 3t + 5 f −3 − 8e + 15 + 2e = 4e + 42

4 Solve the following.

a 5(r − 3) + 16 − 2r = 2(2r + 9) b 7(3q − 1) + 5(4 − 2q) = 6q + 3

c 4n − 3(4 − 5n) = 9 − (n − 19) d 3(2 − 3x) − 5(4 − 5x) = 11x + 6

e 8(w + 4) + 3(2w − 6) = 0 f 5(2d − 7) − 4(d + 5) =−1

5 Solve the following.

a (5k + 2) = b + + = 6

c (5 − 2x) − = 1 − x d = 2r −

e − = 2 f − = 2

6 Use your graphics calculator to solve the following.

a 3s − 8 − s = 42 − 5s + 19 b 3(2y − 4) − 4(5 − 3y) = 2(y − 4)

c 4j − 6(12 − 2j) = 4(2j + 6) d − =

TRACE ZOOM

Y= (–) X,T,θ,n –

ENTER

GRAPH

Calculator Instructions

Additional Exercise

3.1 x 1 x ---3 4

--- 4k–1

3

--- y–2

3

--- 3 y–1 2

--- 1+y 4 --- 1 2 ---1 3

--- 2 3 x( –4)

5

--- 4

5

--- 5 3r( –2)

7

--- 1–2 5( –2r) 3

---q–3 2

--- 4q–7 5

--- 1

2

--- 3 2w( –5)

4

--- 3w+1 6

---5 8( –c)

4

--- 2c–5 6

---7 Draw a graph of each of the following using the Cartesian system.

a y = 4x − 3 where x =−1, 0, 1, 2, 3 b y + 2x = 5 where x =−3, −2, −1, …, 3

c 3y = 2x − 3 where x =−6, −3, 0, …, 12 d y = 4x − 5 where x is any real number

e 2x = y + 8 where x is any real number f 3x + 4y = 8 where x is any real number

8 Use a graphics calculator to draw a graph of:

a y = 5x − 7 b y = 8 − 3x

c y = x + 6 d 4x + 2y = 9

e y − x +11 = 0 f y + 4x = 1

Modelling and problem solving

9 Matt’s mum told him that, if he paid the deposit on a skateboard, she would pay the rest. Matt paid a deposit of one-fifth of the price and left his mother $48 to pay. Write an equation to find the price of the skateboard and also calculate how much Matt paid.

10 Robyn and Jake go on a hiking trip. Robyn starts off carrying the pack with the tent and sleeping bags, and Jake starts with the one containing the food and cooking gear. After they have gone 15 km, they swap

packs for a change and discover that Jake’s original pack is twice as heavy. They transfer 6 kg of food from Robyn’s new pack to Jake’s new pack, so that when they have completed the final 25 km of the hike the total product of weight and distance will be the same for each person. Write an equation to find the original weight of each pack, then solve it to find the original weights.

11 A 5 m3 _{load of concrete is used to put a path around the outside of a square lawn at a} botanical garden. The path is 10 cm thick and 1 m wide. Draw a diagram, write an equation and find the dimensions of the lawn.

12 A physics student obtained the following data in an experiment to find the volume of dry air at different temperatures.

Plot a graph and use it to find the following.

a The volume at 80ºC

b The volume at 250ºC

c The temperature when the volume is 500 mL

d The temperature when the volume is 200 mL

e The temperature when the volume would be 0 mL

13 An investor has $5000 to invest in two companies:

Sun Minerals ($2.40/share) LGW Pork Processors ($4.20/share)

Draw a graph with Sun Minerals on the x-axis and LGW Pork on the y-axis to show the possible mixtures of shares that can be bought for $5000. Use the graph to find the total number of shares that would be purchased if equal numbers of each were bought.

Temperature (ºC) 0 40 100 120 160 180

Volume (mL) 400 460 545 575 635 665

1 3 ---1 2

---14 A furniture manufacturer must compete against local firms, so cannot fully recover the cost of delivery of furniture to retailers. The cost of delivery of a truckload of furniture is about $300 plus about $20/km. The profit on that quantity of furniture sold from the factory is about $3000. Work out the profit on loads of furniture delivered to retailers at distances of 10 km, 50 km, 100 km and 150 km, and draw a graph.

a Use the graph to find the profit on loads that must be delivered:

i 60 km ii 30 km iii 120 km iv 90 km.

b At what distance should the manufacturer not accept orders of furniture?

3.2

Functions and relations

The Cartesian plane is used in mathematics to show the relationships between many pairs of quantities. It is usually the case that only some pairs are allowed for a particular relationship. The possible pairs of numbers form a relation.

Consider the two sets of numbers shown below.

Now consider the relation defined as ‘is exactly divisible by’. One way of showing how this relation applies to the numbers in the sets is as follows.

The relation could also be written as the following ordered pairs of numbers: (4, 2), (4, 4), (10, 2), (10, 5), (12, 2), (12, 3), (12, 4).

The relation x2+ _y2 _{25 can be shown by the} shaded area of the Cartesian plane on the right. For any point (x, y) in this relation it is true that −5 x 5 and −5 y 5. Note that the dashed line on the outside means that x2+ _y2= _{25 is not} included.

2 3 4 5 4

10 12

2 3 4 5 4

10 12

−3 −2 −1 1 3

−3

−1

−2 1 2 3

y

2

−4 4 5

−5

4

−4

−5 5 x

In mathematics, a relation may be described as follows.

A relation may be shown by listing the ordered pairs or by giving a rule that enables them to be worked out. In mathematics, when the domain is not specifically stated, it is assumed that it consists of all possible real numbers. Sometimes only some real numbers make sense.

For example, the domain for y = 3x + 8 is all real numbers (R), but the domain for y = must

be non-negative real numbers, as only has meaning when x0.

In the same way, y= assumes that x≠3 (x is not equal to 3).

A relation is a set of ordered pairs of numbers that may be represented as part of the Cartesian plane. These pairs are sometimes called couples.

The domain of a relation is the set of numbers permitted for the first part, and the range (or co-domain) is the set of numbers permitted for the second part.

Elements of the domain are values of the independent variable, while elements of the range are values of the dependent variable.

!

x x

1

x–3

---What are the domain and range of each of these relations?

a (5, 8), (3, 4), (5, 10), (8, 4) (9, 7) b y = c (age of mother, age of daughter)

Solution

a The domain is: 3, 5, 8, 9. The range is: 4, 7, 8, 10.

b The domain is: any real number except 0. The range is: any real number except 0.

c If we accept that:

■ one cannot usually be a mother before the age of 12, and

■ the oldest women live to 110,

then the oldest age for a daughter would be 98.

So the domain is: [12, 110] (from 12 to 110 years of age) and the range is: [0, 98]. 1

x

---Example

6

Show these relations on the Cartesian plane.

a 1 x 5 and −2 y 2 b (3, 4), (3, 6), (4, 5), (4, 6), (2, 6), (1, 4)

Solution

a b

6 4 2 0

1 3 5 x

1 3 5

2 4

y

1

−1 3

−2 0 2

2 4

y

5 6 x

Note: Solid lines include values,

dashed lines are for or .

A relation with only a small number of ordered pairs can be shown as a table. This is often easier to understand than a list of ordered pairs.

A mapping is a diagram that shows the domain and range of a relation. In the following diagram the ellipse on the left represents the domain, and the one on the right shows the range. The arrow shows that an element of the domain is mapped to an element of the range. Mappings are used to show general features of relations.

Numbers are often rounded off to the nearest whole number. This is an example of a relation with real numbers in the domain and integers in the range. These are written in the ellipses of the mapping to emphasise the domain and range.

The most useful kind of relation is one where there is a rule to work out the range value from the domain value. This type of relation is called a function. Functions are the mathematical way to write down a rule. The domain is the set of independent values and the range is the set of dependent values.

Consider the relation y = 3x. The value of y is determined by the value of x, so x is the

independent variable and y is the dependent variable. This relation is a function. Another way to

write this is f (x) = 3x, which uses function notation. f (x) is usually said ‘f of x’, which is short for ‘a function of x’.

In this case you can see that y = f (x). This is said ‘y is a function of x’.

When calculating values of this function we can use function notation as follows. When x = 1, f (x) = 3 × 1 = 3.

A shorter way to write this is: f(1) = 3 × 1 = 3. In the same way, f (−2) =−6.

So the ordered pairs (1, 3) and (−2,−6) belong to the function described as f(x) = 3x.

Functions may be specified in the same kinds of ways as other relations: by listing the pairs or by writing a rule that shows how the range values are calculated from the domain values.

Domain Range

Rounding off

Domain Range

Real numbers Integers 5.3

14.8

5

15

Extra Material

Classifying relations

A function is a mathematical relation. For every number in the domain of a function there is at most one ordered pair. In other words, different ordered pairs must have different first numbers.

The graph of a relation can be any shape or area. However, in a function every x value has only one y value. This means that, in the graph of a function, there are no points that make a vertical line. This is called the vertical line test for functions. Among the following graphs, b and d are both functions and relations, but a, c and e are only relations.

Which of the following are functions?

a (3, 5), (4, 5), (5, 8), (7, 9), (7, 10) b (2, 3), (3, 4), (4, 4), (5, 4), (6, 5)

c f(x) = 3x4− _{5 d (x, ), where x is positive}

Solution

a Write down the ordered pairs. Check the domain and range values.

(3, 5), (4, 5), (5, 8), (7, 9), (7, 10)

Two different ordered pairs, (7, 9) and (7, 10),have the same first number.

This is not a function.

b Write down the ordered pairs. Check the domain and range values.

(2, 3), (3, 4), (4, 4), (5, 4), (6, 5)

No two pairs have the same first element. This is a function.

c Write down the rule. f(x)= 3x4− ₅

Try different values for x and look at the ordered pairs that result.

f(0)= 3 × 04− ₅ =−₅ ⇒ _(0,−₅₎

f(2)= 3 × 24− ₅ = ₄₃ ⇒ _{(2, 43)} The second number is calculated from the

first using the rule given.

f(−1)= 3 × (−1)4− ₅ =−2 ⇒ (−1,−2)

Every pair has only one first element. There is

only one pair possible for each domain value. This is a function.

d y = (where x is positive) is taken to mean the positive root of x.

Write down the rule. f(x) =

Try a domain value in the rule. f(4) = = 2 ⇒ (4, 2)

Try another domain value. f(25) = = 5 ⇒ (25, 5)

Every pair has only one first element. There is

only one pair possible for each domain value. This is a function.

x

x

x

4 25

Example

8

a b c

d e

Vertical

y

x

Vertical

y

x y

x

y

x

Vertical

y

You have seen how useful a graphics calculator can be to graph linear functions. One

shortcoming with these calculators is that the GRAPH mode uses the form Y = …

This means it is more difficult to graph non-functions with a graphics calculator. Consider the relation x = y2 _{or y}2= _{x. This can be written as y} =± _.

It is clear that x = y2 _{is not a function because more than one range value is possible for each} domain value. For example, 9 = 32 _{or (}−₃₎2_.

To graph these types of relations it is necessary to treat them as two separate graphs. In the case of y2= _{x, we need to look at the graphs of y} = _{and y} =− _.

Casio fx-9860G AU

Go to GRAPH in the main menu.

Press to go into the graph function menu and then enter

the function Y = .

Press to store the equation.

Next press (DRAW) to draw the graph of function Y = .

Press to return to the graph function menu and then

enter the function Y =− .

Press to store the equation.

Press to draw the function Y =− .

Note that the graphs of both functions are drawn, giving the graph of the relation X = Y2_.

Now use your graphics calculator to investigate the graphs of other relations, such as:

■ y2= _{ax where a is a constant} ■ x2+ _y2= _a2 _{where a is a constant.}

Texas Instruments TI-84

Enter the equation Y = .

Press to store the equation.

Press to draw the graph.

Technology

x

x x

EXE

X

EXE

F6 X

AC/ON

X

EXE

F6 X

X

ENTER

Now enter the equation Y =− .

Press to store the equation.

Press to draw the graph.

Note that the graphs of both functions are drawn, giving the graph of the relation X = Y2_.

Now use your graphics calculator to investigate the graphs of other relations, such as:

■ y2= _{ax where a is a constant} ■ x2+ _y2= _a2 _{where a is a constant.}

Sharp EL-9900

See the instructions given on the CD-ROM.

The domain of a function determines many properties of the function. One of the most important differences between domains is the nature of the variable. If the domain variable can take only certain specific values, it is called a discrete variable. If the domain variable can take

in-between values, it is called a continuous variable. For example, the height of a person is a continuous variable because it has in-between values, even if it is only measured to the nearest centimetre.

However, the number of children in a family is discrete because the values must be whole numbers, and in-between values make no sense. Shoe sizes are also discrete; even though there are sizes like 5 , only some sizes exist and there are no other in-between sizes.

X

ENTER

GRAPH

Calculator Instructions

1 2

---A function of a discrete variable has a discrete independent variable, so the domain is discrete.

A function of a continuous variable has a continuous independent variable, so the domain is continuous.

If no domain is specified for a function, it is assumed to be the real numbers, allowing any obvious exclusions.

!

What type of function is each of the following?

a The total cost of buying n towels is given by C(n) = 15n.

b The total cost of buying m kg of potatoes is given by C(m) = 1.2m.

c y = 3x − 2

d F(1) = 8, F(2) = 4, F(3) = 2, F(4) = 1, F(5) = 0.5, etc.

Solution

a Only whole numbers of towels are bought. Function of a discrete variable

b Potatoes may be bought in any quantity. Function of a continuous variable

c x can take any real value. Function of a continuous variable

d The domain consists only of 1, 2, 3, etc. Function of a discrete variable

Exercise 3.2

Functions and relations

1 Illustrate the relation ‘is less than’ between the sets of numbers {2, 3, 4} and {3, 4, 5}.

2 Illustrate the relation ‘when rounded to the nearest integer is’ between the sets of numbers {5.9, 3.2, 2.9, 5.1, 3.8} and {2, 3, 4, 5, 6}.

3 Describe the relation illustrated below in words.

4 Write down the domain and range of each of the following relations and show them on the

Cartesian plane. In some cases you will need to estimate what values are reasonable to use.

a (2, 7), (8, 19), (−1, 1), (5, 13), (−4, −5), (0, 3) b (human mass, human height)

c −3 x 7 and 2 y 8 d solutions of 2y = 4x − 7 for −4 x 4

5 Use the vertical line test to determine which of the following graphs represent functions.

6 State whether each of the following is a function.

a (4, 1), (5, 7), (−1, −6), (0, 2), (3, 8) b (−3, 7), (2, 5), (6, −3), (8, 5), (4, −7)

c (11, −1), (11, 4), (11, 0), (11, −3) d (7, 16), (−2, −2), (10, 22), (−5, −8), (1, 0)

e y = 4x − 11 f xy = 1 g 5x2= _2y

h 2y2= _{5x i f (x)} = _7x − _{10 j f (x)} = ₁₀ Additional

Exercise 3.2

15

−2 2 7 5

−4 13 0

y

x

y

x

y

x y

x y

x

y

x y

x

y

x

y

x

a b c

g h i

7 Each of the following is a function where x is a real number (R). Describe the greatest possible domain of each.

a y = 3x + 7 b f (x) = c y =

d y = x3+ _{8 e f (x)} = ₂₇ − _(x − ₄₎2 _{f y} =

g f (x) = h xy = 4

8 State whether each of the following is a function of a discrete or a continuous variable.

a The number of people entering a sports stadium per minute as a function of the number of times the turnstiles have rotated

b The volume of water passing through

a pipe as a function of the time in hours after midnight

c y = 5x + 7 where x is a real number (R)

d f (x) = 2x2+ _3x − _{9 where x is an} integer (J)

e The height of a ball as a function of the time in seconds after it is dropped

f The distance a runner has completed

as a function of the number of steps taken since the start of a race

3.3

Sequences and progressions

You will be familiar with the word sequence. In cooking, a recipe outlines a sequence of steps to be followed. In current affairs, we may be interested in the sequence of events that led to a government crisis. In sport, an athlete usually goes through a warm-up sequence before an event. In all cases where the word ‘sequence’ is used, a definite order is implied.

This table shows the maximum temperature each day in Brisbane for the first week of July.

The following diagram shows the information as a function, with the domain values (date in July) mapped onto the range values (maximum temperature).

Date in July 1 2 3 4 5 6 7

Maximum temperature (ºC) 16 18 12 13 17 20 11

16_–x2 1

x+5

---x2₊₉

x x–3

---16 12 18

13 17

20 11 1

5 6

7 3 2

This function is an example of a sequence whose domain is the positive integers. This discrete function can be represented in a variety of alternative ways as follows:

f(1) = 16, f (2) = 18, f (3) = 12, f(4) = 13, f (5) = 17, f (6) = 20, f (7) = 11 or (1, 16), (2, 18), (3, 12), (4, 13), (5, 17), (6, 20), (7, 11)

However, it is usual to write a sequence just showing the range elements in order. In this way, the sequence of maximum daily temperatures for the first week in July would be written:

16, 18, 12, 13, 17, 20, 11

This sequence is finite. Also, there is no relationship between successive terms; that is, there is no pattern.

On the other hand, consider this sequence:

1, 3, 5, 7, 9, … or (1, 1), (2, 3), (3, 5), (4, 7), (5, 9), …

It represents the odd numbers and is an infinite sequence. In this sequence it is easy to work out the pattern.

Each number in a sequence is called a term. Rather than use function notation, the terms of a sequence are shown using subscript notation. For example, in the previous sequence the 3rd term is 5, and this would be shown as:

t3=5 or a3= 5 instead of (3, 5) or f (3) = 5

The nth term of a sequence, tn, is referred to as the general term. In the above sequence the general term would be:

tn= 2n − 1 where n is the term number, i.e. f(n) = 2n − 1 The general term gives a way of working out the other terms of the sequence.

Certain sequences have general terms that are reasonably easy to calculate. Consider the following:

2, 4, 6, 8, … 1, 4, 7, 10, … 3, 5, 7, 9, …

However, some are more complicated—especially when their general term is a non-linear function. For example:

4, 7, 12, 19, … has the general term t_n= n2+ _3. Work in groups to determine the general terms of the following sequences.

1 2, 8, 18, 32, 50, 72, … 2 0, 2, 6, 12, 20, 30, …

3 0, 4, 10, 18, 28, 40, … 4 3, 10, 23, 42, 67, 98, …

Investigation

General terms

Find the next two terms in each of the following sequences.

a 2, 5, 8, 11, … b 3, 4, 6, 9, 13, … c , , , , …

Solution

a Write down the sequence. 2 5 8 11 …

In this pattern, each term after the 1st

is 3 more than the term before. The next two terms are 14 and 17. 1

4 --- 1

9 --- 1

16 --- 1

25

---+3 +3 +3 +3

b Write down the sequence. 3 4 6 9 13 …

The difference between successive

terms increases by 1. The next two terms are 18 and 24.

c Write down the sequence. , , , , …

The denominators form a sequence of consecutive square numbers.

4, 9, 16, 25, … ⇒ 22_{, 3}2_{, 4}2_{, 5}2_{, …}

The next two terms are and .

1 2 3 4 5

1 4 --- 1

9 --- 1

16 --- 1

25

---1 36

--- 1

49

---Write down the first four terms of the sequence whose nth term is tn= 1 + .

Solution

Write down the general term. t_n = 1 +

To find the 1st term, let n = 1. t₁ = 1 + = 1 + 1 = 2

Now, let n = 2. t2 = 1 +

= 1

Let n = 3. t3 = 1 +

= 1

Let n = 4. t4 = 1 +

= 1

Write the answer. The first four terms are 2, 1 , 1 and 1 .

1

n

---1

n

---1 1

---1 2

---1 2

---1 3

---1 3

---1 4

---1 4

---1 2 --- 1

3 --- 1

4

---Example

11

For the sequence where tn= 3n + 2, which term is equal to 50?

Solution

Write the general term. tn = 3n + 2

We want the term that equals 50. So let t_n= 50. 50 = 3n + 2

Now solve the resulting equation. 3n = 48

n = 16

Write the answer. The 16th term is 50.

One famous sequence came from the work of the Italian mathematician Leonardo of Pisa (1170–1250), who is better known as Fibonacci. He posed the following problem in 1202:

How many pairs of rabbits will be produced in a single year if we begin with a single pair, and if in every month each pair bears a new pair, which then becomes productive from the second month onwards? (None of the rabbits die.)

The problem can be represented in the following way.

Let’s look at the number of pairs of mature rabbits and immature rabbits, and the total number of pairs of rabbits, at the beginning of each month, as shown in the table.

The sequence 1, 1, 2, 3, 5, 8, … is known as the Fibonacci Sequence.

■ What is the solution to Fibonacci’s problem?

■ How many pairs of rabbits will be present after 2 years if no rabbits die?

Now consider the following problem. (You may find a spreadsheet helpful here.)

A vending machine accepts only $1 and $2 coins. In how many ways can the amounts $1, $2, $3, $4, … be fed into the machine? Complete the following table to answer the problem.

Month

1 2 3 4 5 6

Mature 1 1 2 3 5 8

Immature 0 1 1 2 3 5

Total 1 2 3 5 8 13

Amount Arrangements of coins Number of arrangements

$1 $1 1

$2 $1 then $1 OR $2 2

$3 $1 then $1 then $1 OR $2 then $1 OR $1 then $2 3

$4 … …

… … …

Beginning of month 2 2 pairs

3 pairs

5 pairs

13 pairs 8 pairs 1 pair Beginning of month 1

Beginning of month 3

Beginning of month 4

Beginning of month 5

Beginning of month 6

Jeans R Us is a jeans retailer that has an incentive system for staff. Staff get 10 points for the first sale of the day and 2 points for every subsequent sale.

Staff accumulate points in the following way: One sale for the day = 10 points

Two sales for the day = 10 + 1 × 2 = 12 points

Three sales for the day = 10 + 2 × 2 = 14 points and so on This information can be summarised in the table below.

The number of points shown in the table forms the sequence:

10, 12, 14, 16, 18, 20, … The graph of jeans sales and points is shown.

The sequence of incentive points (10, 12, 14, 16, …) is an example of a special pattern called an

arithmetic progression (AP) or arithmetic sequence.

The example above has a common difference of 2. Each term may be obtained from the 1st term by adding the common difference, as is shown below:

a₁ = 10

a₂ = 12 = 10 + 2

a₃ = 14 = 10 + 2 + 2

a4 = 16 = 10 + 2 + 2 + 2

a5 = 18 = 10 + 2 + 2 + 2 + 2

a 4 × 2 d

a6 = 20 = 10 + 5 × 2 = a + 5d

a7 = 22 = 10 + 6 × 2 = a + 6d In this case, the nth term is 10 + (n − 1) × 2.

Number of sales 1 2 3 4 5 6

Number of points 10 10 + 1 × 2 = 12

10 + 2 × 2 = 14

10 + 3 × 2 = 16

10 + 4 × 2 = 18

10 + 5 × 2 = 20

Extra Material

Describing sequences

1 2 3 4 5 6 7 8 4

8 12 16 20 24

0

Number of sales for day, n

Number of points

Jeans sales and points

A sequence is said to be an arithmetic progression if successive terms differ by the same amount. This amount is called the common difference, d.

!

⎧ ⎪ ⎨ ⎪ ⎩

In general, for an AP:

an= a + (n − 1)d

We can write the general form of an AP as:

Any AP forms a linear plot when graphed.

If we know the general term (an) of an AP, we can work out other terms in the sequence.

a, a + d, a + 2d, a + 3d, …, a + (n − 1)d, …

where a is the 1st term, d is the common difference and n is the term number.

!

1 2 3 4 5 n a

a + d

a + 2d

a + 3d

a + 5d

0

Any arithmetic progression

tn

6

a + 4d

State the 1st term and calculate the common difference of each of the arithmetic progressions that follow.

a 3, 8, 13, … b 7, 4, 1, −2, …

Solution

a Write down the terms. We can see that a = 3.

3 8 13

+5 +5

Calculate the common difference, d. Write the answer.

d = t₂− t₁= 8 − 3 = 5

t₁= 3 and d = 5

b Write down the terms.

We can see that a = 7. Calculate d.

7 4 1 −2

−3 −3 −3

Write the answer. t1= 7 and d =−3

Example

13

In the Jeans R Us incentive points sequence already discussed (10, 12, 14, 16, 18, 20, …), find the 13th term.

Solution

Write down the terms of the AP. We can see that a = 10 and d = 2.

10 12 14 16 18 20 …

+2 +2

Write the general term. an = a + (n − 1)d

We want the 13th term, so n = 13.

Substitute for a, d and n. a13 = 10 + (13 − 1)2

Evaluate. = 34

Write the answer. The 13th term is 34 points.

We can use the formula for the general term to calculate other information about the AP.

Exercise 3.3

Sequences and progressions

1 Work out the next three terms of each of the following sequences.

a 9, 15, 27, 51, … b 16, 37, 58, 79, … c 1, 3, 6, 10, …

d 4, 9, 13, 22, 35, … e 14, 56, 224, 896, … f 1, 1, 2, 3, 5, 8, …

2 Find the first four terms of the sequence where:

a t_n= 3n − 5 b t_n= 3n2 _{c t}

n= n3

d t_n= e u_n= n + f t_n= 4n−2

3 For the sequence un= n2− n, which term is equal to 272?

4 Which term of the sequence t_n= is equal to 8?

Find the number of terms in the arithmetic progression 11, 18, 25, …, 207.

Solution

Write down the terms of the AP. Here, a = 11 and d = 7.

11 18 25 … 207

+7 +7

The last term is 207. tn = 207 = a + (n − 1)d

Substitute for a and d. 207 = 11 + (n − 1) × 7

Expand and simplify. 207 = 7n + 4

Regroup. 7n = 203

Evaluate. n = 29

Write the answer. There are 29 terms in the sequence.

Example

15

Show that the sequence tn= 3n − 2 is an arithmetic progression.

Solution

For this sequence to be an AP, the difference between consecutive terms must be constant.

Write the nth term. tn = 3n − 2

Now write the next consecutive term. tn + 1 = 3(n + 1) − 2 = 3n + 1

Find the difference between these terms. tn + 1− tn = 3n + 1 − (3n − 2) = 3

Find the 1st term. t1 = 3 × 1 − 2 = 1

State the solution. The sequence tn= 3n − 2 is an AP because it

has common difference d = 3 (and a = 1).

Example

16

Additional Exercise

3.3

1

2n–1

--- 1

n

---3n+1

---5 Which of the following appear to be APs? Find the common difference where appropriate.

a 1, 5, 9, 13, … b 4, −1, −4, −8, …

c 1, 2, 4, 8, … d , , 1 , 1 , …

e 7, 3, −1, −5, … f 3, 13, 23, 33, …

g , , , , … h 1, 4, 9, 16, …

6 Show that the following sequences are arithmetic. Find the common difference and the

general term for each sequence.

a 1, 7, 13, 19, … b −2, −7, −12, −17, …

c −4, − , 3, 6 , … d 5.3, 9.5, 13.7, 17.9, …

e 5 , 8, 10 , 13 , … f 1 , , 0, − , …

g m, m + k, m + 2k, m + 3k, … h 2r + 3, 2r − 1, 2r − 5, 2r − 9, …

7 Find the first four terms of the each of the arithmetic progressions below.

a a = 3, d = 4 b a =−7, d = 3

c a =−6, d =−8 d a = 24, d =−4

e a = 2x, d = 9 f a =−13m, d = 3n

g t₄=−12, t₅=−16 h t₃=−1, t₇=−25

8 Find the 33rd term of the AP 67, 72, 77, …

9 Find the 23rd term of the AP 227, 209, 191, …

10 Find the number of terms in each of the following APs.

a 1, 4, 7, …, 61 b −2, −8, −14, …, −212

c 15, 23 , 32 , …, 198 d 46, 38.6, 31.2, …, −94.6

e 8.53, 14.95, 21.37, …, 91.99 f w, w + v, w + 2v, …, w + 23v

g 2x + 7, 3x + 11, 4x + 15, …, 20x + 79 h m, 3m + 3n, 5m + 6n, …, 35m + 51n

Modelling and problem solving

11 Here is a problem that you may like to solve by using a strip of paper. A strip of paper 0.01 cm thick is folded in half to make a crease. The thickness is then 0.02 cm. When it is folded in half again (in the same direction), it has 3 creases and is 0.04 cm thick. Complete the table below and use it to state the number of creases and the thickness after 10 such folds.

12 Find the (k + 1)th term of each of the following sequences.

a 21, 16, 11, … b −3, −11, −19, …

c 3, 9, 15, … d 14, 5, −4, …

e 6.7, 10.2, 13.7, … f 5, 8 , 12 , …

g h, h + 7, h + 14, … h p, q, 2q − p, …

13 Find m and p so that 7 −2 , m, p, − 2 are four consecutive terms of an AP.

Number of folds 1 2 3 4 5 6 …

Number of creases 1 3 …

Thickness (cm) 0.02 0.04 …

3.4

Gradient and intercepts of a linear

function

We have seen that the graph of a linear function is a straight line. As shown in the diagram below, you can fix the position of a straight line in a number of ways. If we know two points on the line, the line can only be drawn in one place. There is also only one way we can draw a line if we have a starting point and a direction.

The gradient of a line is measured using the amount the line rises for each step of 1 unit in the positive direction of the x-axis. If the line is falling, we simply make the ‘rise’ negative.

The gradient is calculated by dividing the rise between two points on the line by the horizontal distance (run) between them. The symbol m is used for the gradient.

The Greek symbol ∆ (delta) is used in mathematics to indicate a change. In this way, the gradient may also be written as:

m =

y

x

Point 1

Point 2

y

x

Direction Starting

point Only one

straight line is possible.

The direction of a line in the Cartesian plane is given by its gradient. The gradient of a line describes the direction and size of its slope or steepness.

Gradient = m =

or m =

y

x

Rise = y2− y1

Run = x2− x1

(x2, y2)

(x1, y1) rise

run

---y₂–y₁ x₂– x₁

---!

∆y

∆x

---a Calculate the gradient of the line that intercepts the x-axis at −2 and the y-axis at 3.

b Calculate the gradient of the line that passes through (2, 5) and (5,−7).

Solution

a Draw a sketch.

Find the rise and the run: rise = 3 and run = 2.

Use the formula to calculate m. m =

= = 1

y

x

Rise Run

−2 3

rise run

---3 2

--- 1

2

If a graph is drawn with the same scale for the x-axis and y-axis, the angle that a line makes with the positive direction of the x-axis can also be used to find its gradient.

Look at ABC in the diagram at right.

tan θ =

So tan θ =

= = m

b Draw a sketch.

From the sketch, the rise is −12 and the run is 3.

Calculate the gradient. m =

= =−4

We could also calculate the gradient using the formula as follows.

Write down the points. (2, 5) and (5,−7)

Now use the formula to calculate the gradient. m =

Let (x₁, y₁) = (2, 5) and (x₂, y₂) = (5,−7).

Substitute for the x and y values. = =

Evaluate. =−4

Write the solution. The gradient is −4.

The order of the points doesn’t matter for this formula. Let (x1, y1) = (5,−7) and (x2, y2) = (2, 5) to check this.

m =

Substitute and evaluate.

The gradient is the same. = = =− 4

y

x

Rise = −12

Run = 3 5

−7

5 2

(2, 5)

(5, −7)

rise run

---12 –

3

---y₂–y₁ x₂–x₁

---−7–5 5–2

--- –12 3

---y₂–y₁ x₂–x₁

---5––7

2–5

--- 12 3 –

---y

x

Run =∆x

Rise =∆y θ

opposite adjacent

---rise run ---∆y

∆x

---Assuming that both axes are drawn with the same scale, find the gradient of:

a a line making an angle of 60º with the positive direction of the x-axis

b the line shown in the diagram on the right.

y

x θ

45°

Horizontal and vertical lines are special cases.

For a horizontal line: For a vertical line:

m = m =

= = which is undefined ( by 0)

= 0

There are special relationships between the gradients of lines that are parallel and lines that are perpendicular.

Solution

a Use the formula for gradient. m = tan θ = tan 60º

≈ 1.73

b First calculate the value of the angle that the line makes with the positive direction of the x-axis (θ).

θ = 180º − 45º = 135º

Now use the formula. m = tan θ = tan 135º

=−1

The gradient (m) of a straight line can be measured in the following ways:

m = =

m =

where (x1, y1) and (x2, y2) are points on the line

m = tanθ

where θ is the inclination, the angle the line makes with the positive direction of the

x-axis.

θ y

x

Rise Run

∆y

∆x

--- rise run

---y₂–y₁ x₂–x₁

---!

y

x ∆y = 0

y

x ∆x = 0

∆y

∆x

--- ∆y

∆x

---0 ∆x

--- ∆y

0

---y

x

θ θ

Parallel lines

θ m1

B

θ

m2

A C

D

Perpendicular lines

m1

As the previous diagram shows, parallel lines make the same angle with the positive direction of the x-axis, so their gradients are the same size and have the same sign.

However, the gradients of perpendicular lines have opposite signs.

Now, as shown: ∠BAD =∠CBD =θ

So = (= tan θ)

Rearranging gives ÷ = × = 1 [1]

Using m = m1 = and m2=−

So −m₂ =

Substituting m1 and −m2 into [1] gives m1×−m2 = 1

So m₁m₂ =−1

BD AD --- DC BD ---BD AD --- DC BD --- BD AD --- BD DC ---rise run --- BD AD --- BD DC ---BD DC ---Extra Material Angles between lines

Parallel lines have the same gradient: m₁ = m₂

Perpendicular lines have gradients whose product is −1:

m1m2 =−1 or m2= –1

m₁

---!

Show that the triangle with vertices (1,−3), (3, 2) and (6,−5) is right-angled.

Solution

In this type of problem, it is a good idea to sketch the information before starting the algebra.

Name the vertices of the triangle A, B and C.

From the sketch it looks like AB AC.

Write the formula to find the gradient of AB. m_AB =

Substitute for x₁, y₁, etc.

Evaluate. = =

Write the formula to find the gradient of AC. mAC =

Substitute for x1, y1, etc.

Evaluate. = =−

Check whether the sides are perpendicular. m_ABm_AC = ×− =−1

Write the solution. There is a right angle at A, so

ABC is right-angled.

y

x B (3, 2)

A

C

(1, −3)

(6, −5)

y₂–y₁ x₂–x₁

---2––3 3–1 --- 5

2

---y₂–y₁ x₂–x₁

Gradients can also be used to determine whether three points are in a straight line. This is equivalent to finding whether a third ordered pair belongs to a linear function for which two other ordered pairs are already known.

Collinear points are in a straight line.

Three points A, B and C must be collinear if m_AB= m_AC. If A, B and C are collinear, AB and AC have the same equation.

y

x A

B C

!

Determine whether the points (−3,−2), (4, 5) and (1, 2) are collinear.

Solution

Draw a sketch and mark in the information. Label the points A, B and C.

Find the gradient of AB. m_AB =

Evaluate. = = 1

Find the gradient of AC. mAC =

Evaluate. = = 1

Compare the gradients. m_AB = m_AC

Write the solution. A, B and C are collinear.

y

x A

C B

(4, 5) (1, 2)

(−3, −2)

5––2 4––3

---7 7

---2––2 1––3

---4 4

---Example

20

The linear function g(x) has ordered pairs (4, 6) and (5, 3). Is (8,−3) also an ordered pair of the function?

Solution

Draw a sketch and label the points.

y

x

(5, 3)

A

B

C

(4, 6)

(8, −3)

The x- and y-intercepts of a straight line are the sections of the axes cut off by the line, measured from the origin. The terms ‘x-intercept’ and ‘y-intercept’ are also used to refer to points of intersection. These intercepts (where a line meets the axes) are the most significant points on a straight line.

If C does belong to the function, A, B and C must be collinear.

Find the gradient of AB. m_AB =

Evaluate. = =−3

Find the gradient of AC. m_AC =

Evaluate. = =−2

Compare the gradients. m_AB ≠ m_AC

Write the solution. A, B and C are not collinear, so (8,−3)

does not belong to g(x). 3–6

5–4

---3 –

1 ---−3–6

8–4

---9 –

4

--- 1

4

---The x-intercept of a straight line is the distance of its point of intersection with the x-axis measured from the origin. The y-intercept of a straight line is the distance of its point of intersection with the y-axis measured from the origin. The value of the y-intercept of a straight line is normally symbolised by the letter c, so the point where the line intercepts the y-axis is (0, c).

In this book, the term ‘intercept’ is used to mean both the point of intersection of a line with the axis and the distance of this point from the origin.

y

x x-intercept y-intercept

(0, c)

(a, 0)

!

Find the x- and y-intercepts of the line shown in the diagram.

Solution

Look at the graph.

The line cuts the x-axis at (3, 0). The x-intercept is 3.

The line cuts the y-axis at (0, 5). The y-intercept is 5.

6 5 4 3 2 1

−1

y

x

2 1 3

−1 4

7

Exercise 3.4

Gradient and intercepts of a linear function

1 Calculate the gradient of each of the following lines using the intercepts shown.

2 Find the gradients of the lines shown in the diagram below. Find the intercepts of the line 2x − 3y = 18.

Solution

Write the equation. 2x − 3y= 18

The x-intercept has a y-coordinate of 0, so let y = 0. 2x = 18

Solve for x. x = 9

The y-intercept has an x-coordinate of 0, so let x = 0. −3y = 18

Solve for y. y =−6

State the solution. The x-intercept is 9 and the

y-intercept is −6.

Example

23

Additional Exercise

3.4

a b c

d e f

x y

4 3

x y

−5 2

x y

−6 1

x y

−7

−3

x y

6

x y

−4

−3−2−1 1 3

−3

−2 1 2 3

y

2

−4 4 5 6

−5

−6

−4 4

7

−7

−8 8 9 10

9 10x

5 6 7 8

−7−6−5

−8

−10−9

a

b

c d

e

f g h

i

3 Find the gradients of the lines joining the following pairs of points.

4 Without drawing a sketch, find the gradient of the line passing through:

a (1, 6) and (3, 8) b (4, 3) and (8, 11) c (−5, 0) and (0, 4)

d (2, −8) and (−2, 12) e (−1, −7) and (2, −5) f (−6, −1) and (−8, −10)

g (9, −4) and (3, −15) h (11, −2) and (−3, 16) i (4, −9) and (−8, 15).

5 The following graphs have been drawn with the same scales on the x- and y-axes. Arrange

the gradients in numerical order from smallest to largest.

6 Find the gradient of the line that has an angle of inclination of:

a 30° b 85° c 120° d 10° e 15°

f 0° g 170° h 95° i 57° j 90°

7 Find the gradient of each of the following lines.

a b c

d e f

y

x

x

x

x

y y

y (4, 2) y

(4, 7) (0, 0)

(−3, −8)

(−2, −5) (−6, 1)

(−3, 5) (3, 5) (−2, 5) (−4, 3)

(5, −2) (2, −2)

y

x

x

a b c

d e f

y

x

y

x

y

x y

x

y

x

y

x

a b c

d e f

y

x

y

x

y

y

x y

x

x

55° 28°

45°

22°

y

x

8 What angle does a line passing through (1, 5) and (3, 10) make with the positive direction of the x-axis?

9 What angle does a line passing through (−1, 5) and (2,−4) make with the positive direction of the x-axis?

10 Calculate the value of b for each of the following.

a (2, 1) (b, 3); gradient = 1 b (5, b) (7, 3); gradient =

c (b, 3) (4,−1); gradient = d (−3, b) (2, 3); gradient = 1

11 Determine whether the following sets of points are collinear.

a (3, 7), (5, 4), (−1, 13) b (2, 2), (−2, 7), (4, 0)

c (3, 8), (1, 3), (7, 18) d (−2,−7), (−1,−4), (1, 2)

e (3,−4), (5, 1), (−1,−9)

12 The ordered pairs (4, 8) and (6, 12) are part of the linear function f(x). Is (9, 20) also part of f(x)?

13 The linear function h(x) has the following properties: h(2) = 8 and h(5) = 2. Is it true that

h(7) =−2?

14 Determine whether the line through AB is parallel or perpendicular to the line through XY in each of the following.

a A(2, −2), B(7, 3) and X(−6, −2), Y(1, 5)

b A(−7, 1), B(3, −9) and X(−4, −5), Y(6, 5)

c A(−4, 3), B(6, 8) and X(−2, −5), Y(10, 1)

d A(−7, 2), B(5, 6) and X(−9, −9), Y(6, −4)

e A(2, −3), B(6, −1) and X(7, −8), Y(2, 2)

15 Find the intercepts of each line drawn below.

1 2 ---2

3

---−3−2−1 1 3

−3

−2 1 2 3

y

2

−4 4 5 6

−5

−6

−4 4

7

−7

−8 8 9 10

9 10x

5 6 7 8

−7−6−5

−8

−10−9

−9

a b c

d

e

f g

h i

16 Find the intercepts of:

a 5x + y = 10 b x − 2y + 6 = 0 c 4x − 3y =−24 d 2x + y − 8 = 0

e y = 4x + 3 f x − y = 4 g y = 3x − 12 h 3y = x + 15

i y = 3x − 8 j 5y − 3x = 8

Modelling and problem solving

17 A ramp leads up to the front doors of an office block that are 520 mm higher than the ground. Workplace health and safety regulations state that the maximum safe inclination of the ramp in this situation is .

a What is the shortest distance that the ramp should start from the base of the platform?

b If the ramp had a gradient of would it be considered safe?

c If there is only room for a ramp to be built 8 m from the base of the platform, what is the greatest height that the platform can be for the ramp to be safe?

18 What name best describes the figure with vertices A(−3, −7), B(−7, 1), C(5, 3) and D(9, −5)?

19 What name best describes the triangle with vertices P(−3, 3), Q(11, −5) and R(−4, −2)?

20 What type of quadrilateral has vertices W(−9, 2), X(3, 8), Y(7, 0) and Z(−5, −6)?

3.5

Equations of linear functions

A straight line consists of a set of points. The equation that describes the points is the rule for a linear function.

A straight line must be one of the following: ■ parallel to the x-axis (horizontal, m = 0) ■ parallel to the y-axis (vertical, m not defined) ■ sloping (with a definite non-zero gradient).

Every point on a horizontal line has the same y-coordinate, so we can write the equation of the line as y = c (where c is the y-intercept).

Every point on a vertical line has the same x-coordinate, so we can write the equation of the line as x = a (where a is the x-intercept).

1 12

---520 mm

---A vertical line passing through the point (a, b) has the equation x = a.

A horizontal line passing through the point (a, b) has the equation y = b.

Of course, in this case the value of the y-intercept is b. In the general case, the y-intercept is c, so c