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9

9

Functions

and graphs

ALGEBRAIC MODELLING

Algebraic modelling is the use of a mathematical formula to represent a relationship occuring in nature, industry and society.

Throughout the General Mathematics course, we have used the linear function

y=mx+b to model real-life situations, such as the cost of mobile phone usage under a particular monthly plan.

However, not all patterns and relationships follow straight line graphs and increase or decrease at a steady rate. For example, Australia’s population grows more rapidly over time because, like compound interest, as the population increases the rate of increase also increases. The future population of Australia can be modelled by the exponential function

P= 19(1.007)n

where P is the population in millions and n is the number of years after 2000. This model provides the following predictions:

In this chapter, we will examine non-linear algebraic models, whose graphs are smooth curves rather than straight lines.

In this chapter you will learn how to:

n generate tables of values and graph quadratic and cubic functions n apply quadratic and cubic functions to model and solve real-life problems

n use a quadratic graph to find maximum and minimum values in practical contexts n generate tables of values and graph exponential and hyperbolic functions

n apply exponential and hyperbolic functions to model and solve real-life problems n recognise exponential growth and decay by the value of a in y=b(ax)

n recognise the different types of functions and their applications and graphs

n use functions as models of physical phenomena and recognise their limitations when interpolating and extrapolating

n solve problems involving direct and inverse variation, including variation to the square and cube of a variable.

Year 2005 2010 2015 2020 2025 2030

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THE QUADRATIC FUNCTION

Not all functions are linear and of the form y=mx+b. Functions involving higher powers of x such as x2 and x3 are called non-linear functions because their graphs are not lines but

curves.

A function in which the highest power of x is 2 is called a quadratic function. Quad means ‘square’, and quadratic functions involve ‘x squared’. Some examples of quadratic functions are y= 5x2, y= 2x2+ 3x 5, d=− t2+ 1 and C=−v2+ 3v+ 7.

The general form of a quadratic function is y=ax2+bx+c

where a, b and c are constants. The graph of a quadratic function is a cup- or bowl-shaped curve called a parabola.

Use graphing software or a graphics calculator to graph parabolas. A spreadsheet can also be used to generate tables of values and graphs.

The graph of

y

=

x

2

The simplest quadratic function is y=x2. Its table of

values and graph are shown.

Can you see why the graph of y=x2 is symmetrical?

The graph of

y

=

ax

2

(a) y= 2x2

These y-values are twice as large as those for y= x2, so this parabola is taller and

steeper than y = x2.

x −3 −2 −1 0 1 2 3

y 9 4 1 0 1 4 9

x −3 −2 −1 0 1 2 3

y 18 8 2 0 2 8 18

1 2

---Technology idea:

Graphing quadratic functions

y

8

6

4

2

2 4

−2

−4 0

(3, 9)

(2, 4)

(1, 1) (−1, 1)

(−2, 4)

y = x2 (−3, 9)

x

18

16

14

12

10

8

6

4

2

y

x

−2 0

−4 2 4

(−3, 18) (3, 18)

(−2, 8) (2, 8)

(−1, 2) (1, 2)

y = x2

(3)

(b) y = x2

These y-values are half as large as those for y = x2, so this parabola is shorter and flatter

than y = x2.

(c) y =−2x2

These y-values are negative, so this parabola is ‘upside-down’—that is, the reverse of y = 2x2.

Features of a parabola

n Unlike a straight line, a parabola is symmetrical and both ends point in the same direction.

n The turning point of a parabola is called its vertex.

n If a is positive, the parabola is concave up and points upwards. If a is negative, the parabola is concave down and points downwards.

Memory aid: ‘When it’s positive, the parabola is smiling.

When it’s negative, the parabola is frowning.’

x −3 −2 −1 0 1 2 3

y 4.5 2 0.5 0 0.5 2 4.5

x −3 −2 −1 0 1 2 3

y −18 −8 −2 0 −2 −8 −18

y

8

6

4

2

2 4

−2

−4 0

(1, 0.5) (3, 4.5)

(2, 2) (−2, 2)

(−3, 4.5)

y = x2

(−1, 0.5)

x

y = x1 2

2

1 2

---−4

−6

−8

−10

−12

−14

−16

x

−2

−4 2 4

y

−18

(−1, −2) (1, −2)

y =−x2

y =−2x2 (−2, −8) (2, −8)

(−3, −18) (3, −18)

Vertex

(4)

n If a is large, the parabola is tall and steep. If a is small, the parabola is short and flat.

The graph of y

=

ax

2

+

c

(a) y = x2+ 4

Compared to y = x2, these y-values are 4 units

higher, so this parabola is shifted 4 units upwards (above the x-axis). The vertex is (0, 4).

(b) y = 2x2− 2

Compared to y = 2x2, these y-values are 2 units

lower, so this parabola is shifted 2 units downwards (below the x-axis). The vertex is (0,−2).

x −3 −2 −1 0 1 2 3

y 13 8 5 4 5 8 13

x −3 −2 −1 0 1 2 3

y 16 6 0 −2 0 6 16

When a is large When a is small

14

12

10

8

6

2

y

x

−2 0

−4 2 4

(−2, 8) (2, 8)

(−1, 5) (1, 5)

y = x2

4 (0, 4)

(3, 13) (−3, 13)

4 units above y = x2 + 4

x-axis

16

12

8

4

−4

x

−2

−4 2 4

(3, 16) (−3, 16)

y = 2x2 2

(2, 6) (−2, 6)

(−1, 0) (1, 0)

(0, −2)

y

2 units below

x-axis

The graph of y = ax2+ c is the parabola y = ax2 shifted upwards c units.

(5)

The graph of y

=

ax

2

+

bx

+

c

y = ax2+ bx + c is the general equation of a parabola. Its vertex is not so easily located.

y = x2− 2x 3

This parabola has vertex (1,−4), y-intercept −3 and two x-intercepts −1 and 3.

Notice that the axis of symmetry of the parabola is the line x = 1, halfway between the

x-intercepts −1 and 3.

The x- and y-intercepts of a parabola

A parabola has one y-intercept and up to two x-intercepts.

This parabola has x-intercepts −3 and 1, y-intercept 3, axis of symmetry x =−1 and vertex (−1,−4).

Graphing y

=

ax

2

+

bx

+

c

Step 1 Determine the shape of the parabola by examining the size and sign of a. If a is positive, the parabola is concave up.

If a is negative, the parabola is concave down. If a is large, the parabola is tall and steep. If a is small, the parabola is short and flat.

Step 2 Find the y-intercept. Note that when you substitute x = 0 into y = ax2+ bx + c, you

get y = c. Just as with a linear function, the constant term at the end is the y-intercept. For example, the y-intercept of y =−x2+ 4x 3 is 3.

Step 3 Complete a table of values.

Step 4 Graph the parabola, showing the y-intercept, x-intercepts and vertex where possible. Sometimes you’ll need to add more values to your table to locate the vertex or x-intercepts. Sometimes a parabola has no x-intercepts and is completely above or below the x-axis.

Note: The axis of symmetry of a parabola is halfway between the two x-intercepts.

x −3 −2 −1 0 1 2 3

y 12 5 0 −3 −4 −3 0

x −3 −2 −1 0 1 2 3

y

6

4

2

−2

x

−2 2 4

−4 Vertex (1, 4)

Axis of symmetry

y = x2 2x 3

y

−2

−3

x

−1

−2 1 2

y

Vertex (−1, −4)

−1 1

−4

−3

y-intercept x-intercept x-intercept

(6)

Example 1

Graph y =−x2+ 5x 6.

Solution

a =−1 (negative), so the parabola is concave down. The y-intercept is −6.

The x-intercepts are 2 and 3.

The vertex must be at x = 2.5 (halfway between 2 and 3).

When x = 2.5, y =−(2.5)2+ 5(2.5) 6 = 0.25.

∴ The vertex is (2.5, 0.25).

Example 2

Graph y = x2− 4x + 1.

Solution

a = (positive), so the parabola is concave up. The y-intercept is 1.

One of the x-intercepts is somewhere between 0 and 1 (cannot be located).

By extending the table of values beyond x = 3, the vertex is found to be at (4,−7).

By symmetry, the other x-intercept is somewhere between 7 and 8.

x −3 −2 −1 0 1 2 3 4 5 6

y −30 −20 −12 −6 −2 0 0 −2 −6 −12

x −3 −2 −1 0 1 2 3 4 5

y 17.5 11 5.5 1 −2.5 −5 −6.5 −7 −6.5

−4

−6

−8

x

2 4

y

y =−x2+ 5x 6 6

−2

−12

−10

(2.5, 0.25)

1 2

---1 2

---6

2

−2

x

−2 2 4

y = x2 4x + 1

−4

(4, −7)

6 8 10 8

10

y

−6

1 2

(7)

1. Match each quadratic function with its graph below.

(a) y = 2x2− 2 (b) y =−x2+ 1 (c) y =−7x2

(d) y = x2− 1 (e) y = x2+ 1 (f) y = 2x2

2. Graph the following quadratic functions, showing the vertex and x- and y-intercepts where possible.

(a) y = 2x2+ 1 (b) y = (x + 1)2 (c) y =−4x2+ 4

(d) y = x2− 2x 3 (e) y =−2x2+ 12x 16 (f) y = x(x 6)

3. A ball is thrown up in the air from the balcony of Tom’s apartment. Its height above ground level after t seconds is given by the quadratic function

h =−5t2+ 20t + 25

where h is the height in metres.

(a) What is the independent variable of this function? (b) Graph h =−5t2+ 20t + 25 for values of t from 0 to 6.

(c) Write down the coordinates of the vertex of this graph. (d) What is the vertical intercept of this graph?

(e) What does the vertical intercept represent? (f) Calculate the height of the ball after 2.5 seconds.

(g) Find the maximum height of the ball and the time when this occurs. (h) When does the ball hit the ground?

(i) Hence, for what range of values of t is this quadratic model accurate? (j) Why is the quadratic model not valid for values of t outside this range?

4. Graph the following quadratic functions, showing the vertex and intercepts where possible.

(a) P = n2− 9n + 8 (b) w = 2r2− 4r + 5

5. (a) Construct a table of values for y = (x − 2)2+ 3 and graph the quadratic function.

(b) Construct a table of values for y = x2− 4x + 7 and compare your answers to those

from (a). What do you notice?

(c) By expanding y = (x − 2)2+ 3, prove that it is identical to y = x2− 4x + 7.

Exercise 9-01:

The quadratic function

1 2

---A B C

D E F

y

x

y

x

y

x

y

x

y

x y

x

(8)

---MAXIMUM AND MINIMUM PROBLEMS

The vertex of a parabola illustrates the maximum or minimum value (y-value) of the quadratic function y = ax2+ bx + c it represents. The function has a maximum value if a is

negative (when the parabola is concave down) and a minimum value if a is positive (when the parabola is concave up).

Example 3

Find the maximum value of y =−x2− 4x + 5.

Solution

a =−1 (negative), so parabola is concave down.

From the graph, the vertex is (−2, 9).

The maximum value of y =−x2− 4x + 5 is 9.

A formula for the vertex

There is a formula for the x-value of the vertex of a parabola, so we don’t need to use tables of values to find maximum or minimum values of quadratic functions.

x −3 −2 −1 0 1 2 3

y 8 9 8 5 0 −7 −16

y

x

Maximum value

y

x

Minimum value

6

4

2

−2

x

−2 2 4

−4 8

y

−6

−4 (−2, 9)

y =−x2 4x + 5

The x-value of the vertex of a parabola with equation y = ax2+ bx + c is x =

To find the y-value, substitute the x-value into the equation of the parabola.

b

(9)

---Example 4

(a) Find the maximum value of y =−x2− 4x + 5 (from Example 3).

(b) Find the minimum value of y = 2x2+ 3x + 1.

Solution

(a) y =−x2− 4x + 5

a=−1, b =−4, c = 5

x =

= = =−2

When x =−2, y =−(−2)2− 4(2) + 5

= 9

The maximum value of y =−x2− 4x + 5 is 9 (as found in Example 3).

(b) y = 2x2+ 3x + 1

a= 2, b = 3, c = 1

x =

= = −

When x =− , y= 2 2+3 + 1

=−0.125

The minimum value of y = 2x2+ 3x + 1 is 0.125.

Example 5

A farmer uses 18 m of fencing to build a rectangular pig pen. He wants to build the pen with the largest possible area. Here are some rectangles that can be built with 18 m of fencing:

Let x and y be the length and width of the rectangle respectively, in metres.

(a) Explain why 2x + 2y = 18. (b) Hence show that y = 9 − x.

(c) Hence show that the area of the pen, A (in m2), is the quadratic function A =−x2+ 9x.

(d) What is the area of the pen if its length is 5 m? (e) Find the largest possible area of the pen.

(f) What are the dimensions of the rectangle that give this maximum area?

a is negative, so the

vertex shows a maximum.

b2a

---– 4– 2×–1 --- 4

2 –

---a is positive, so the

vertex shows a minimum.

b2a

---3 – 2×2 --- 3

4

---3 4

--- 3

4 ---–

 

  3

4 ---–

   

7 m

2 m

4 m

5 m 5.5 m

3.5 m

A = 14 m2

A = 20 m2 A = 19.25 m2

x

(10)

Solution

(a) The perimeter of the rectangle must be 18 m. Perimeter= x + y + x + y = 18

2x + 2y= 18 (b) 2x + 2y= 18

2y= 18 − 2x We must make y the subject of the formula.

y= = 9 − x

(c) Area of the pen A= xy

= x(9 x) Substituting y = 9 − x from (b)

= 9x x2 Expanding

=−x2+ 9x

(d) When x = 5, A=−(52) + 9(5)

= 20

The area of the pen is 20 m2 when its length is 5 m (see the middle diagram on page 335).

(e) The largest possible area (i.e. A maximum) occurs when x = . In A =−x2+ 9x, a =−1, b = 9, c = 0.

x=

= = = 4.5

When x = 4.5, A=−(4.52) + 9(4.5)

= 20.25

The largest possible area is 20.25 m2.

(f) Length of rectangle x = 4.5 m

To find the width, y, substitute x = 4.5 into y = 9 − x. y = 9 − 4.5 = 4.5

The width of the rectangle is also 4.5 m. (Hence the rectangle that gives the maximum area is a square.)

Graphics calculators and graphing software

Use the TRACE and ZOOM functions of a graphics calculator or graphing software to locate the maximum and minimum values of a quadratic function (e.g. y = 2x2+ 3x + 1). Also note

the locations of the x- and y-intercepts. All of the questions in Exercise 9-02 which follows can be solved using a graphics calculator or graphing sofware.

Spreadsheets

Create a spreadsheet that calculates y-values of a quadratic function (e.g. y = 2x2+ 3x + 1),

and use a guess-check-and-refine method to find the maximum or minimum value. Use the charting feature of the spreadsheet to graph the function.

18–2x 2

---b2a

---b2a

---9 – 2×–1 --- –9

2 –

---4.5 m

4.5 m

To find the maximum/minimum value of a quadratic function y = ax2+ bx + c:

1. Evaluate x = .

2. Substitute the value of x into the function to find the maximum/minimum value. 3. If a is negative, the value is a maximum. If a is positive, the value is a minimum.

b

2a

(11)

1. Find the vertex of the parabola with equation:

(a) y = x2− 3x + 1 (b) y = x2+ 4x + 6 (c) y =−2x2+ 8x 2

(d) y =−x2− 5x (e) y = 2x2+ 7x + 10 (f) y = 0.4x2− 2x + 1

2. Find the maximum value of:

(a) y =− x2+ x + 5 (b) y =−5x2+ 20x + 11

3. Find the minimum value of:

(a) y = x2− 4x + 6 (b) y = 2x2+ 10x + 7

4. A golf ball is hit into the air and its height, h m, above the ground after time, t seconds, is given by the quadratic function

h =−5t2+ 24t

(a) What is the dependent variable?

(b) What is the vertical intercept and what does it represent? (c) Calculate the height of the ball after:

(i) 2 seconds (ii) 4 seconds

(d) Find the maximum height of the ball and the time when this is reached.

(e) Calculate the height of the ball after 1.8 and 3 seconds. Explain the pattern in your answers.

(f) Graph the quadratic function for values of t from 0 to 5.

(g) Why is this quadratic model not accurate for values of t greater than 5?

5. At an air show, the height of a plane in a power dive is given by the function h = t2− 10t + 80

where h is the height in metres and t is the time in seconds. (a) Graph the function.

(b) What was the height of the plane at the start of the dive? (c) What was the lowest height reached by the plane? (d) When was this lowest height reached?

6. The braking distance of a car travelling at a speed of v km/h is d = 0.01v2+ 0.7v

where d is measured in metres.

(a) Calculate the braking distance of the car if it is travelling at: (i) 86 km/h (ii) 58 km/h

(b) By a guess-check-and-refine method, find the speed that allows a stopping distance of 20 m.

7. A cricket ball is hit upwards and its height in metres is given by h =−8t2+ 40t

where t is the time in seconds.

(a) What is the independent variable of this quadratic function? (b) Graph h =−8t2+ 40t.

(c) Calculate the height of the ball after 3.5 seconds. (d) What is the maximum height reached by the ball? (e) When is this maximum height reached?

(f) For how long is the ball in the air?

Exercise 9-02:

Maximum and minimum problems

1 2

(12)

---8. Mary Contrary is a gardener who uses 18 m of fencing wire to border a rectangular garden in her backyard. She uses an existing brick wall as one side of the garden. The diagrams show some possible rectangular arrangements for the garden.

(a) Draw another possible garden that meets the requirements. (b) Let x and y be the length and width of the

garden in metres. Explain why x + 2y = 18.

(c) Hence show that y = 9 − x.

(d) Hence show that the area, A (in m2), of the garden is

A =− x2+ 9x

(e) Calculate the area of the garden with a length of 8 m. (f) Copy and complete this table:

(g) Find the maximum area of the garden and the dimensions that give this area.

9. The petrol consumption, P L/100 km, of a car travelling at different speeds, s km/h, is modelled by the function

P = 0.01s2− s + 33

(a) What is the dependent variable in this quadratic function?

(b) What is the petrol consumption of the car when it is travelling at 60 km/h? (c) What is the most economical speed—that is, the speed at which the petrol

consumption is lowest?

(d) What is the petrol consumption at the most economical speed? (e) Copy and complete this table of values:

(f) This function is a good model for speeds between 20 km/h and 80 km/h. Why would this formula not be useful for s = 0?

10. The temperature, T °C, of a town n hours after 6 am is given by the function T =− n2+ 2n + 14

(a) Graph this function for values of n from 0 to 15. (b) What was the temperature:

(i) at 6 am? (ii) at 12 noon?

(c) Find the time of day when the maximum temperature was reached, and hence the maximum temperature.

Length, x (m) 3 6 9 12 15 18

Area, A (m2)

s 20 30 40 50 60 70 80

P

6.5 m

5 m 14 m

2 m

8 m

5 m

A = 40 m2

A = 28 m2

A = 32.5 m2

y A = xy

x

1 2

---1 2

(13)

---(d) Estimate to the nearest half-hour the times when the temperature reached 19°C. Why are there two answers?

(e) For how many hours (to the nearest whole number) was the temperature 19°C or higher?

(f) This quadratic model is only accurate for values of n from 0 to 13. Why do you think the model does not hold for values of n above 13?

1. Five people meet and every person shakes hands with every other person. How many handshakes take place?

2. Twelve people meet and everyone shakes hands. How many handshakes take place now?

3. Copy and complete this table.

4. The formula for the number of handshakes, H, that occur when n people meet is H = n(n − 1)

(Can you work out why?)

(a) If this function were graphed, what would it look like? (b) Graph the function for values of n from 1 to 10.

5. Use the formula to calculate the number of handshakes that take place when: (a) 12 people meet (b) 22 people meet

6. By a guess-check-and-refine method, find how many people met if 666 handshakes took place.

THE CUBIC FUNCTION

A function in which the highest power of x is 3 is called a cubic function (involving ‘x cubed’) and its general equation is

y = ax3+ bx2+ cx + d

where a, b, c and d are constants. In this course, however, we will only examine simple cubic functions of the form y = ax3. Examples are y = 4x3, y = x3 andy =−3x3. The graph of a

cubic function is called a cubic curve.

Use graphing software or a graphics calculator to graph cubic curves. A spreadsheet can also be used to generate tables of values and graphs.

No. of people, n 1 2 3 4 5 6

No. of handshakes, H

Group modelling activity:

Quadratic handshakes

1 2

---1 2

(14)

The graph of y

=

x

3

The simplest cubic function is y = x3. Its table

of values and graph are shown.

The graph of y = x3 increases very sharply

(compared to y = x2) because when the

x-values are cubed the y-values are much higher.

The cubic curve has point symmetry around (0, 0). This means that if you spin the curve upside-down about (0, 0), it will map onto itself.

The graph of y

=

ax

3

(a) y = 2x3

Just as with a parabola, if a is large, the graph is taller and steeper than y = x3.

(b) y =− x3

If a is small, the graph is short and flat. Also, if a is negative, the graph is decreasing.

x −3 −2 −1 0 1 2 3

y −27 −8 −1 0 1 8 27

x −3 −2 −1 0 1 2 3

y −54 −16 −2 0 2 16 54

x −3 −2 −1 0 1 2 3

y 13.5 4 0.5 0 −0.5 −4 −13.5

x

−2 1 2

−4

−6

−8

−10

−12

−14

−16

y

16

14

12

10

8

6

4

2

−3 3

(2, 16)

(2, 8)

(1, 1)

(−1, −1) (−1, −2)

(−2, −16) (1, 2)

y = x3

y = 2x3

(−2, −8)

−2

1 2

---x

−2 2

−2

−4

−6

−8

y

8

6

4

2

−3 −1 3

(−2, 4)

(−1, 0.5) (1, −0.5)

(2, −4)

y =−x3

y =−1x3

(15)

The size and sign of a in y = ax3 affect the graph of the cubic curve.

Example 6

The volume of a sphere, V m3, can be approximated by the cubic function V = 4.19r3, where

r m is the radius of the sphere.

(a) Graph the cubic function for values of r from 0 to 5.

(b) Use the graph to estimate the volume of a sphere with a radius of 4.6 m. (c) Use the function to calculate the volume

of a sphere with a radius of 4.6 m. Answer to the nearest cubic metre. (d) Use the function to find the radius of the

sphere that has a volume of 7240 m2.

Answer to the nearest metre.

Solution

(a) Values of V calculated to 2 decimal places:

(b) From the graph, when r = 4.6, V ≈ 404 m3.

(c) V = 4.19r3

When r = 4.6, V = 4.19(4.6)3

= 407.837 …

≈ 408 m3

(d) When V = 7240, 7240= 4.19r3

= r3

r3= 1727.923 …

r=

= 11.999 …

≈ 12 m

r (m) 0 1 2 3 4 5

V (m3) 0 4.19 33.52 113.13 268.16 523.75

If a is small, the curve is short and flat. If a is positive, the curve is increasing. If a is negative, the curve is decreasing.

If a is large, the curve is tall and steep.

y

x

y

x

y

x

y

x

250

200

150

100

50

0

r (m)

1 2 3 4 5

V (m3)

Volume of a sphere

V = 4.19r3

450

400

350

300 550

500

V = 404 m3

r

=

4.6 m

7240 4.19

---1727.923…

(16)

What is the formula for the volume of a sphere?

Is V = 4.19r3 a good approximation?

1. Match each cubic function with its graph below.

(a) y =−x3 (b) y = x3 (c) y = 4x3

(d) y = x3− 1 (e) y =− x3 (f) y = x3+ 1

2. Graph y = x3.

3. Graph y =−3x3.

4. Graph y = (x + 1)3.

5. Is it possible for a person to lift a cubic metre of cork? Cork is fairly light material with 1 cm3 of it weighing only 0.25 g. The formula for the mass, M g, of a cork cube of length

s cm is

M = s3

(a) Copy and complete this table of values for M:

(b) Calculate the mass of a cork cube of side length 45 cm. Express your answer to the nearest:

(i) gram (ii) kilogram (c) (i) Convert 93.312 kg to grams.

(ii) Calculate the side length of a cork cube that has a mass of 93.312 kg. (d) Calculate the mass of cube of a cork with a side length of 1 m, to the nearest

kilogram.

(e) So is it possible for a person to lift a cubic metre of cork?

s (cm) 5 10 30 50 60 80

M (g)

Think:

Volume of a sphere

Exercise 9-03:

The cubic function

1 4 ---1 2

---A B C

D E F

y

x

y

x

y

x

y

x

y

x

y

x

1 3

(17)

---6. The mass, M kg, of a child at different heights, h cm, can be modelled by the cubic function

M =

(a) Calculate, to the nearest kilogram, the mass of a child at height:

(i) 100 cm (ii) 124 cm (iii) 165 cm (iv) 75 cm (b) Calculate, to the nearest centimetre, the height of a child with mass:

(i) 60 kg (ii) 32 kg (iii) 17 kg

(c) What do you think are the limitations of this model? Why isn’t it good for: (i) babies? (ii) adults?

THE EXPONENTIAL FUNCTION

What happens when x is the power of a function, as in y = 2x?

A function in which x is the power is called an exponential function (exponent is another name for power), and its general equation is

y = b(ax)

where a and b are constants and a is a positive number. The graph of an exponential function is called an exponential curve.

Use graphing software or a graphics calculator to graph exponential curves. A spreadsheet can also be used to generate tables of values and graphs.

h3

80 000

---Study tips

K

EEPING IT ALL IN PERSPECTIVE

There is life after the HSC. Despite what many people will tell you, it’s not the most important time of your life. Sure, it is one of the major turning points, but 12 months from now the HSC will be a distant memory and you’ll be laughing about how much stress it caused you. Don’t have a panic attack (I know this is easier said than done). Stress is a state of mind. Learn to live with it productively. Don’t get stressed about getting stressed.

Every year when the newspapers interview the high achievers of the HSC, there are always the same stories of hard work and determination. But these successful students also kept it all in perspective and remembered to maintain balance in their lives. They studied a lot but didn’t neglect their other needs: sport, recreation, going out, socialising, work, talking to their families and friends, watching movies, listening to music. As mentioned in a previous study tip, sometimes you need to take breaks and do something different so that you can return to your original task with a new perspective.

So don’t forget to have a life. Plan do to nothing once in a while. Look after yourself. Breathe, live, exercise, meditate. Enjoy.

(18)

The graph of y

=

a

x

(a) y = 2x

Note: 0.5 = , 0.25 = , 0.125 =

As the x-values increase by 1 unit, the y-values double. Because of this, the curve increases quickly. When x is positive, the curve is steep, pointing upwards. When x is negative, the curve is flat, close to the x-axis.

(b) y = 5x

Because a = 5 is larger than a = 2, this exponential curve increases faster. When x is positive, the curve is steeper. When x is negative, the curve is closer to the x-axis.

(c) y = 0.8x

y-values rounded to 3 decimal places where necessary:

When a is between 0 and 1, the curve decreases steeply, then becomes flatter as it approaches the x-axis.

x −3 −2 −1 0 1 2 3

y 0.125 0.25 0.5 1 2 4 8

x −3 −2 −1 0 1 2 3

y 0.008 0.04 0.2 1 5 25 125

x −3 −2 −1 0 1 2 3

y 1.953 1.563 1.25 1 0.8 0.64 0.512

8

6

2

y

x

−2 0

−4 2 4

4

y = 2x

(3, 8)

(2, 4)

(1, 2) (−3, 0.125) (0, 1)

(−2, 0.25) (−1, 0.5) 1 2 --- 1 4 --- 1 8 ---40 30 10 y x2 0

−4 2 4

20

y = 5x

(2, 25)

(1, 5)

(0, 1)

x

−3 −2 −1 0

1 (0, 1)

(−3, 0.008)

(−1, 0.2) (−2, 0.04)

y

x

−2 1 2

y

3

2

−3 −1 0 3

1

y = 0.8x

(−3, 1.953) (−2, 1.563)

(−1, 1.25)

(0, 1)

(19)

Features of the graph of y

=

a

x

n If a . 1, the exponential curve is increasing. It begins flat and close to the x-axis, but once it crosses the y-axis it rises steeply. The x-axis is called an asymptote because the curve approaches it but never touches it.

n If a is a large number, the graph increases even more quickly.

n The reverse occurs if 0 , a , 1. The curve begins by decreasing sharply, but once it crosses the y-axis it becomes flatter and approaches the x-axis.

n The entire graph is above the x-axis because ax (‘a to the power of x’) is always a positive

number.

n The y-intercept (when x = 0) of y = ax is always

1 because any number raised to the power of 0 is 1.

What does the graph of y = 1x look like?

The graph of y

=

b(a

x

)

(a) y = 8(2x)

These y-values are 8 times as large as those for y = 2x, so this curve is even taller and

steeper than that graph. The y-intercept is now 8.

(b) y = 0.2(2x)

These y-values are 0.2 times that of y = 2x, so

this curve is shorter and flatter than that graph. The y-intercept is now 0.2.

x −3 −2 −1 0 1 2 3

y 1 2 4 8 16 32 64

x −3 −2 −1 0 1 2 3

y 0.025 0.05 0.1 0.2 0.4 0.8 1.6

y

x

1

y = ax

(a > 1)

0

y

x

1

y = ax

(0 < a < 1)

0

Think:

y

=

1

x

x

−2 1 2

y

40

30

20

−3 −1 0 3

10 50

(2, 32)

(1, 16)

(0, 8) (−1, 4)

(−2, 2) (−3, 1)

y = 8(2x)

y = 2x

(0, 1)

x

−2 1 2

y

2

1.5

1

−3 −1 0 3

0.5

(1, 0.4) (2, 0.8) (3, 1.6)

(−1, 0.1) (0, 0.2)

y = 0.2(2x)

(20)

Features of the graph of y

=

b(a

x

)

n The graph is similar to y = ax, but the size of b affects the steepness and y-intercept of the

graph.

n The y-intercept (when x = 0) of y = b(ax) is always b, because b(a0) = b × 1 = b.

1. (a) Copy and complete the table for y = 3x, then graph the function.

(b) What is the y-intercept of this function?

2. (a) Graph y = (0.6)x.

(b) What is the y-intercept of this function? (c) What is the asymptote of the graph?

3. (a) Graph y = 80(1.3)x.

(b) What is the y-intercept of this function? (c) Is this function increasing or decreasing?

4. Match each exponential function with its graph below.

(a) y = 2x (b) y = 10x (c) y = (0.1)x

(d) y = 0.5(2x) (e) y = 2x 1 (f) y = 8(2x)

5. Is the graph of y = increasing or decreasing?

x −3 −2 −1 0 1 2 3

y

For the graphs of y = ax and y = b(ax):

n If a is greater than 1 (a . 1), the exponential curve is increasing.

n If a is between 0 and 1 (0 , a , 1), the exponential curve is decreasing.

n Large values of a and b cause steeper curves.

n Small values of a and b cause flatter curves.

n The x-axis is an asymptote: the graph gets close to it but does not touch it.

n The y-intercept is always 1 for y = ax and b for y = b(ax).

Exercise 9-04:

The exponential function

A. B. C.

D. E. F.

y

x y

x

y

x y

x

−1

y

x

1

y

x

1 1

0.5

8

1 4

(21)

6. Is the graph of y = (2.5)x increasing or decreasing?

7. A large sheet of paper is 0.1 mm thick. Suppose it can be folded repeatedly. After 1 fold, the thickness of the sheet will double to 0.2 mm, after 2 folds to 0.4 mm, after 3 folds to 0.8 mm, and so on. After n folds, the sheet has a thickness of

T = 0.1(2n)

where T is the thickness in millimetres. (a) Calculate the thickness of the sheet after:

(i) 5 folds (ii) 10 folds (iii) 12 folds

(b) By a guess-check-and-refine method, find the minimum number of folds required for the sheet to be:

(i) 400 mm thick (ii) 1 m thick (c) What are the limitations of this model?

EXPONENTIAL GROWTH AND DECAY

Exponential growth

Any quantity that increases according to the exponential function y = b(ax), where a . 1, is

said to experience exponential growth. Something that grows exponentially is increasing slowly at first, then very quickly. Examples of exponential growth are population size (people, animals, bacteria), an investment under compound interest, a flu or computer virus, the thickness of a tree trunk and the size of a bushfire.

Example 7

The population of Hicksville grows exponentially according to the function P = 60 000(1.08)t

where t is the number of years after 2000 and P is the population.

(a) Calculate the population of Hicksville in the year 2007 (to the nearest whole number). (b) What was Hicksville’s population in 2000?

(c) By a guess-check-and-refine method, find the year when Hicksville’s population will be double its population in 2000.

(d) Graph the population function for values of t from 0 to 8.

Solution

(a) When t = 7 (as 2007 is 7 years after 2000), P= 60 000(1.08)7

= 102 829.456 … ≈ 102 829

The population of Hicksville in the year 2007 will be 102 829. (b) When t = 0, P= 60 000(1.08)0 = 60 000

The population of Hicksville in the year 2000 was 60 000. (c) When P= 2 × 60 000 = 120 000,

120 000= 60 000(1.08)t

= (1.08)t

(1.08)t= 2

By guess and check, t ≈ 9.

The population will be doubled in the year 2009.

Checking: When t = 9, P = 60 000(1.08)9= 119 940.277 … 120 000

1 5

(22)

---(d) P-values to the nearest whole number:

The constant multiplication feature of a calculator with DAL can be used to display exponential growth. In Example 7 the population is increasing by a constant multiplier of 1.08 each year. That is, each new P-value is 1.08 times larger than the previous one. This can be keyed on the calculator like this:

Calculator keys Display

60 000 t = 0

1.08 t = 1

(Then keep pressing repeatedly.)

t = 2 t = 3 t = 4, etc.

To solve the equation (1.08)t= 2 in Example 7(c), we can evaluate t = instead of

using a guess-check-and-refine method

Calculator keys Display

2 1.08

(or 2 1.08 )

The solution is t = 9 (as before).

t 0 1 2 3 4 5 6 7 8

P 60 000 64 800 69 984 75 583 81 629 88 160 95 212 102 829 111 056

120

110

100

90

80

70

60

0

t (years after 2000)

1 2 3 4 5 6 7 8

P

(

×

1000)

Population of city

P = 60 000(1.08)t

Technology:

Constant multiplication on a calculator

= 60000

× = 64800

=

= 69984

= 75582.72

= 81629.3376

Technology:

The

log

key on your calculator

log 2 log 1.08

---log 4 log = 9.0064…

(23)

Exponential decay

Any quantity that decreases according to the exponential function y = b(ax), where a is

between 0 and 1, is said to experience exponential decay. Something that decays

exponentially is decreasing quickly at first, then more slowly. Examples of exponential decay are radioactive decay, the cooling of substances, the intensity of light in water and the dampening of vibrations.

Example 8

A superball is bounced from a fourth-floor balcony and its height after each bounce is modelled by the exponential function

h = 12(0.6)b

where b is the number of bounces and h is the height from the ground in metres. (a) Calculate the height of the ball (to the nearest centimetre) after the 5th bounce. (b) Graph the function for values of b from 0 to 8.

(c) What is the height from which the ball was first dropped? (d) Estimate after which bounce the ball rises to 0.14 m.

(e) What are the limitations of this exponential model for large values of b?

Solution

(a) When b = 5, h= 12(0.6)5

= 0.933 12 m

= 93.312 cm

≈ 93 cm

(b) h-values correct to 1 decimal place:

(c) When b = 0, h = 12.

The ball was dropped from a height of 12 m.

b 0 1 2 3 4 5 6 7 8

h 12 7.2 4.3 2.6 1.6 0.9 0.6 0.3 0.2

12

10

8

6

4

2

0

Bounces, b

1 2 3 4 5 6 7 8

Height,

h

(m)

Height of a bouncing superball

(24)

(d) When h= 0.14 m, 0.14= 12(0.6)b

= (0.6)b

(0.6)b= 0.0116 …

By guess and check, b ≈ 9.

The ball will reach 0.14 m (14 cm) after the 9th bounce.

(e) The exponential model assumes the ball will continue bouncing forever and never reach a ‘height’ of 0. In reality, it will stop bouncing after some time.

1. A bushfire is spreading through a forest and the amount of area burnt is modelled by the exponential function

A = 5(1.22)h

where A is in hectares and h is the time in hours. (a) Graph the function for values of h from 0 to 6. (b) What was the initial amount of area burnt?

(c) What area was burnt after 3.5 hours (to the nearest hectare)? (d) Estimate when 20 ha will be burnt.

(e) What are the limitations of this exponential model?

2. The compound interest formula A = P(1 + r)n is an example of exponential growth.

Suppose a principal of $8000 is invested at 6.5% p.a. interest over n years. The amount of the investment after n years is

A = 8000(1.065)n

where A is in dollars.

(a) What is the independent variable for this exponential function? (b) What is the amount of the investment after:

(i) 5 years? (ii) 10 years?

(c) By a guess-check-and-refine method, find (to the nearest year) when the principal of $8000 will be doubled.

3. A bowl of boiling hot soup is allowed to cool. Its temperature, T°C, after m minutes is given by the formula

T = 100(0.93)m

(a) Is this function an example of exponential growth or exponential decay? (b) Graph this function for values of m from 0 to 15.

(c) Calculate the temperature of the soup after 2.5 minutes (correct to 1 decimal place). (d) What is the vertical intercept of this function and what does it represent?

(e) Estimate when the soup’s temperature will reach 27°C.

(f) Why is this model probably inaccurate for values of m greater than 15? 0.14

12

---For y = b(ax):

n Exponential growth occurs when a . 1.

n Exponential decay occurs when 0 , a , 1.

n b represents the initial value of y (when x = 0) and is the y-intercept of the exponential curve.

(25)

4. The thickness, T cm, of a tree trunk grows according to the function T = 16(1.11)n

where n is the time in years after it was first measured. (a) What is the independent variable of this function?

(b) Calculate (to the nearest centimetre) the thickness of the trunk: (i) after 6 years (ii) at the start of the measuring period (c) Estimate when the tree trunk will reach a thickness of 24 cm.

5. The number of bacteria in a particular culture is given by N = 1000(6.9)t

where t is the time in minutes.

(a) Calculate the number of bacteria that exist after: (i) 10 minutes (ii) 1 hour

Answer in standard notation correct to 2 decimal places. (b) What was the initial number of bacteria?

(c) Estimate when the number of bacteria will pass 2 000 000.

6. A water tank is being emptied and the volume of water remaining in the tank is V = 1400(0.97)m

where V is in litres and m is the time in minutes. (a) How much water was in the tank at the start?

(b) How much water (to the nearest litre) is in the tank after:

(i) 42 minutes? (ii) 1 hour? (iii) 1 hours? (c) Estimate when the tank will be half full.

(d) What are the limitations of this exponential decay model?

7. The number of computers, C, affected by a virus after n hours is given by the function C = 5(3.03)n

(a) Is this an example of exponential growth or exponential decay? (b) What is the independent variable in this function?

(c) What is the vertical intercept and what does it represent? (d) How many computers are affected by the virus after:

(i) 3 hours? (ii) 8 hours?

(e) Estimate when the number of computers affected by the virus will grow to 300 000.

8. The population of the world in 1965 was 3 billion and each year it has been growing by 2%. This can be modelled by the function

P = 3(1.02)n

where P is the population in billions and n is the number of years after 1965. (a) What is the dependent variable in this function?

(b) Copy and complete this table of values, expressing P correct to 2 decimal places.

(c) Graph P against n.

(d) Calculate the world population (to the nearest 0.01 billion) in: (i) 1993 (ii) 2040

(e) Estimate when the world population was/will be: (i) 4.5 billion (ii) 10 billion

Year 1965 1970 1975 1980 1985 1990 1995 2000

n (years) 0 5 10 15

P (billions)

(26)

---9. Carbon-14, a radioactive isotope of carbon, decays according to the function P = 100(0.999 88)n

where P is the percentage of the original carbon mass that remains and n is time in years. (a) What percentage (correct to 1 decimal place) of carbon-14 remains after:

(i) 200 years? (ii) 500 years?

(b) Estimate how long it will take the carbon-14 to decay to half its original mass—this is called the half-life of the isotope. Answer to the nearest 10 years.

(c) At an archaeological dig, a bone was found to contain 96% of its original carbon-14. Estimate its age.

10. The number of days, D, that milk will stay fresh if stored at temperature T°C is D = 18(0.72)T

(a) Is this an example of exponential growth or exponential decay? (b) What is the number of days milk will stay fresh if stored at:

(i) 2°C? (ii) 5°C? (iii) 0°C?

(c) Estimate the temperature at which milk will keep for 5 days.

In 1999, Australia’s population reached 19 million and the world’s reached 6 billion. A population clock is a counter that displays the current population value of a country or the world using an exponential model of population growth. Several population clocks exist on the Internet, and these are updated every second or minute. For example, according to population clocks at the Australian Bureau of Statistics website www.abs.gov.au and the US Bureau of Census website www.census.gov, the populations of Australia and the world at 9:30 am (AEST) on 9 September 2000 were 19 218 387 and 6 074 954 798 respectively. Use an Internet search engine to locate and visit some population clock websites and compare their values for the current population of Australia or the world.

THE HYPERBOLIC FUNCTION

A hyperbolic function has the equation y =

where a is a constant. Note that x appears in the denominator of the function, so if x is large, y is small, and if x is small, y is large. The graph of a hyperbolic function is called an hyperbola, a symmetrical curve with two separate branches.

(a) y =

y has no value when x = 0 because is not defined.

We will add more values to our table to gain a better picture of the graph:

x −3 −2 −1 0 1 2 3

y −0.67 −1 −2 2 1 0.67

x −5 −4 −3 −2 −1 −0.5 −0.2 0 0.2 0.5 1 2 3 4 5

y −0.4 −0.5 −0.67 −1 −2 −4 −10 10 4 2 1 0.67 0.5 0.4

Technology:

Population clocks

a x

---y-values rounded to 2 decimal

places where necessary

2 x

(27)

---As x gets closer to 0, y gets larger and further from 0. ---As x gets further from 0, y gets smaller and closer to 0. The x- and y-axes are asymptotes. The hyperbola has point symmetry about (0, 0). If you spin it upside-down about (0, 0), it will map onto itself.

(b) y =−

Because in y =− the value of a is negative (−3), the branches of this hyperbola are

in the quadrants opposite to those of the graph of y = . Also note that this

graph is further away from the x- and y-axes than the graph of y = .

x −5 −4 −3 −2 −1 −0.5 −0.2 0 0.2 0.5 1 2 3 4 5

y 0.6 0.75 1 1.5 3 6 15 −15 −6 −3 −1.5 −1 −0.75 −0.6

6

4

2

−2

−6

−2

−6 4

2

6 y

x

y = 2x

(−0.5, −4) (−1, −2) (−5, −0.4)

(−4, −0.5) (−3, −0.67)

(−2, −1)

(0.5, 4)

(1, 2) (2, 1)

(3, 0.67) (4, 0.5)

(5, 0.4)

3 x

---−2

−6

−2

−4

−6

2

x

x

3

y = −

2 6

−4 4 y

(−0.5, 6)

(−1, 3)

(−2, 1.5) (−3, 1) (−4, 0.75) (−5, 0.6)

(0.5, −6) (1, −3)

(2, −1.5) (4, −0.75)

(5, −0.6) (3, −1)

3 x

---2 x

(28)

---Use graphing software or a graphics calculator to graph hyperbolic curves. A spreadsheet can also be used to generate tables of values and graphs.

Features of an hyperbola

n Both the x- and y-axes are asymptotes. There are no x- and y-intercepts.

n The hyperbola has two separate branches, which are symmetrical and in opposite quadrants.

n As x gets further from 0, y gets closer to 0.

n As x gets closer to 0, y gets further from 0.

n The higher the value of a, the further the hyperbola is from the x- and y-axes.

n When a is negative, the branches of the hyperbola are in different quadrants from those when a is positive.

Example 9

The time, T months, taken to pay off an interest-free loan is related to the size of each monthly repayment, $r, by the hyperbolic function

T =

(a) Graph the function for values of r from 100 to 900.

(b) What is the time taken to pay off the loan if the monthly repayments are $250? (c) What is the monthly repayment required to pay off the loan in 5 months? (d) Why is this model not useful for values of r greater than 1800?

Solution

(a) T-values rounded to 2 decimal places where necessary:

(b) When r = 250, T = = 7.2 months ≈ 8 months

It takes 8 months to pay off the loan.

r ($) 100 200 300 400 500 600 700 800 900

T (months) 18 9 6 4.5 3.6 3 2.57 2.25 2

Technology idea:

Graphing hyperbolic functions

1800 r

---20

16

12

8

4

0

r ($)

100 200 300 400 500 600 700 800

T

(months)

Time taken to pay off a loan

T =1800

r

900

(29)

---(c) When T = 5, 5=

5r= 1800

r=

= 360

The monthly repayment is $360.

(d) If r . 1800, T , 1. For example, if r = 2000, T = 0.9 (i.e. 0.9 months). It is not possible to pay off a loan in less than 1 month if monthly repayments are used.

1. Graph y = and y = on the same axes.

2. Match each hyperbolic function with its graph below.

(a) y = (b) y =− (c) y =−

(d) y = + 1 (e) y = (f) y = − 1

3. The time, T minutes, taken for Erika to drive to Nowra from her home depends on her average speed, s km/h, according to the function

T =

(a) What is the dependent variable? (b) Graph T for values of s from 40 to 110.

(c) Calculate the time taken to drive to Nowra at a speed of: (i) 65 km/h (ii) 92 km/h

Answer to the nearest minute.

(d) At what speed (to the nearest 1 km/h) must Erika travel in order to reach Nowra in: (i) 45 minutes? (ii) 84 minutes?

(e) How far is Nowra from Erika’s house?

4. Graph y = − .

1800 r

---1800 5

---Exercise 9-06:

The hyperbolic function

1 x

--- 5

x

---3 x

--- 1

x

--- 4

x

---1 x

--- 5

x

--- 1

x

---A. B. C.

D. E. F.

y

x

y

x

y

x

y

x

y

x

y

x

1

−1

1

−1

4800 s

(30)

---5. Nilshan, standing next to a spotlight, notices that the height of his shadow decreases as he moves further from the light, according to the formula

h =

where h cm is the height of his shadow and d cm is his distance from the spotlight. (a) Complete this table of values.

(b) What is the height of Nilshan’s shadow if he is 220 cm from the spotlight? (c) How far is Nilshan standing from the spotlight if his shadow is 860 cm? Express all answers to the nearest centimetre.

MORE APPLICATIONS OF FUNCTIONS

We have examined the following functions and graphs in this course:

n the linear function y = mx + b

n the quadratic function y = ax2+ bx + c

n the cubic function y = ax3

n the exponential function y = b(ax) n the hyperbolic function y =

and studied the different ways they are used as algebraic models of real-life situations. Now we will review these functions by examining further applications of them.

Example 10

The running costs, $C per hour, of a ship travelling at an average speed of s knots, is given by the function

C = 2s2 128s + 2400

(a) What type of function is this?

(b) Calculate the running costs when the ship is travelling at 12 knots. (c) Calculate the most economical speed and the running costs at this speed. (d) Graph this function for values of s from 5 to 50.

(e) Estimate the average speeds that have a running cost of $800 per hour.

Solution

(a) It is a quadratic function.

(b) When s = 12, C = 2(122) 128(12) + 2400 = 1152

The running costs are $1152 per hour.

(c) The most economical speed occurs when s = .

In C = 2s2− 128s + 2400, a = 2, b =−128, c = 2400.

s=

= = 32

When s = 32, C = 2(322) 128(32) + 2400 = 352

The most economical speed is 32 knots, giving a running cost of $352 per hour.

d (cm) 50 100 150 200 250 300 350 400

h (cm)

56 000 d

---a x

---b2a

---– – 128

2×2

(31)

---(e) From the graph, s = 17 and 47 knots.

1. Match each of these graphs with an appropriate function below.

A y = 2(2x) B y =− x +1 C y = x3

D y = x2−2 E y =−x2+ 3x F y =

2. The hearing of people at different ages, A years, was tested by measuring the highest frequency, F kHz (kilohertz), they could hear.

(a) Graph the data tabulated.

(b) What type of function is this: linear, quadratic or exponential?

(d) s 5 10 15 20 25 30 35 40 45 50

C 1810 1320 930 640 450 360 370 480 690 1000

Age, A (years) 7 10 12 16 21 24 27 30

Frequency, F (kHz) 18.2 17.0 16.2 14.6 12.6 11.4 10.2 9.0

2000

1600

1200

800

400

0

s (knots)

10 20 30 40 50

C

($/hour)

Running costs of a ship

C = 2s2128s + 2400

Exercise 9-07:

More applications of functions

(a) (b) (c)

(d) (e) (f)

y

x

y

x y

x

−2

y

x

y

x

y

x

(1, 1)

1

3

2 3

1 3

--- 1

(32)

---(c) If the vertical intercept is 21, find the equation of the function.

(d) Use the function to calculate the highest frequency (correct to 1 decimal place) heard by an 18-year-old person. Check using your graph.

3. The U-shape of a skateboarding pipe is given by the formula y = x2− 7x

for values of x from 0 to 14, where both x and y are in metres. (a) What type of function is this?

(b) What type of graph illustrates this function? (c) Graph this function for values of x from 0 to 14. (d) What is the greatest depth of this pipe?

(e) What is the depth of this pipe 2 m from the side?

4. The number of days, D, it takes to build a house is related to the number of builders, b, working on the job, according to the function

D =

(a) What type of function is this?

(b) Graph this function for values of b from 2 to 10.

(c) How long (to the nearest day) will it take to build the house using: (i) 9 builders? (ii) 25 builders?

(d) How many builders are needed to build the house in: (i) 30 days? (ii) 5 days?

(e) What are the limitations to this model?

5. The volume of a cone whose height is the same as its diameter can be approximated by the function

V = 0.26d3

(a) What type of function is this?

(b) Graph this function for values of d from 10 to 80.

(c) Use the graph to estimate the volume of the cone that has a diameter of 45 cm.

(d) Use the formula to calculate the volume of the cone that has

a diameter of 45 cm. Answer to the nearest cm3 and compare it to your answer in (c).

(e) Use the function to find the diameter of the cone that has a volume of 4062.5 cm3.

Check using the graph that your answer is reasonable.

6. What type of function is represented by this table of values? Find using a graph or a guess-and-check method.

7. The relative intensity of light d m under water is given by I = 100(0.996)d

where I is the percentage of the original intensity of the light (at its source).

(a) What is the relative intensity of the light under 3 m of water (to 1 decimal place)? (b) What type of function is this?

(c) What is the independent variable?

(d) As d increases, does I increase or decrease?

(e) Estimate when the relative intensity will be half the original intensity.

x −2 −1 0 1 2 3 4 5

y −1.6 −0.2 0 0.2 1.6 5.4 12.8 25

1 2

---240 b

---d

(33)

8. Australia’s population can be modelled by the function P = 19(1.007)n

where P is the population in millions and n is the number of years after the year 2000. (a) What type of function is P?

(b) Use the function to predict Australia’s population (to the nearest 0.1 million) in the year:

(i) 2040 (ii) 2100

(c) Estimate when Australia’s population will be: (i) 20 million (ii) 25 million

9. A rock falls off a cliff. Its height, h m, above ground level after t seconds is shown in the table.

(a) Graph this table of values.

(b) What type of function can be used to model this situation?

(c) Why is this model not useful for values of t greater than or equal to 5? (d) Estimate when the rock hits the ground.

10. A bookcase is depreciating according to the following schedule.

(a) Graph this table of values.

(b) What type of function can be used to model this situation: linear, exponential or hyperbolic?

(c) Use the graph to estimate the value of the bookcase after: (i) 1.5 years (ii) 8 years

(d) Estimate when the value of the bookcase is $160. (e) What was the original price of the bookcase? (f) What is the annual percentage rate of depreciation? (g) What are the limitations of this model of depreciation?

t (s) 0 1 2 3 4 5

h (m) 85 80 65 40 5 −40

Year, y 0 1 2 3 4 5 6

Value, $V 289 254 224 197 173 153 134

Just for the record

P

OPULATION GROWTH

Australia’s population grows by 0.7% each year, while the world’s population grows by 1.3% each year. Why do you think Australia’s growth rate is lower than the world’s?

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DIRECT VARIATION

Linear variation

We learned about direct linear variation in the Preliminary Course, where two variables, say x and y, are related by the equation

y = kx and k is called the constant of variation.

Example 11

The weight of an astronaut on Mars is proportional to his weight on Earth. A 72 kg person weighs 27.4 kg on Mars.

(a) Calculate how much a 60 kg person weighs on Mars.

(b) If an astronaut weighs 32 kg on Mars, what is his weight on Earth? Give answers correct to 1 decimal place.

Solution

Let M be the weight of an astronaut on Mars, and E be his weight on Earth. The variation equation is M = kE.

To find the value of k, substitute E = 72, M = 27.4. 27.4= k(72)

= k k=

≈ 0.3806

M= 0.3806E

Note: Here we have rounded k to 4 decimal places, but you can leave in your calculator’s display or memory and use that value of k for more accurate answers. (a) Substitute E = 60.

M = 0.3806 × 60

= 22.836

≈ 22.8

A 60 kg person weighs 22.8 kg on Mars. (b) Substitute M = 32.

32= 0.3806E

= E

E= 84.0777 …

≈ 84.1

The astronaut weighs 84.1 kg on Earth.

Non-linear variation

A variable can also vary as the square, cube or square root of another variable. This is called non-linear variation.

References

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