# Maths in Focus - Margaret Grove - ch5

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TERMINOLOGY

## 5

Arc of a curve: Part or a section of a curve between two points

Asymptote: A line towards which a curve approaches but

never touches

Cartesian coordinates: Named after Descartes. A system of

locating points (x, y) on a number plane. Point (x, y) has Cartesian coordinates x and y

Curve: Another word for arc. When a function consists

of all values of x on an interval, the graph of y=f x] g is called a curve y=f x] g

Dependent variable: A variable is a symbol that can

represent any value in a set of values. A dependent variable is a variable whose value depends on the value chosen for the independent variable

Direct relationship: Occurs when one variable varies

directly with another i.e. as one variable increases, so does the other or as one variable decreases so does the other

Discrete: Separate values of a variable rather than a

continuum. The values are distinct and unrelated

Domain: The set of possible values of x in a given domain

for which a function is defi ned

Even function: An even function has line symmetry

(refl ection) about the y-axis, and f]-xg=-f x] g

Function: For each value of the independent variable x,

there is exactly one value of y, the dependent variable. A vertical line test can be used to determine if a relationship is a function

Independent variable: A variable is independent if it may

be chosen freely within the domain of the function

Odd function: An odd function has rotational symmetry

about the origin (0, 0) and where f]-xg=-f x] g

Ordered pair: A pair of variables, one independent and

one dependent, that together make up a single point in the number plane, usually written in the form (x, y)

Ordinates: The vertical or y coordinates of a point are

called ordinates

Range: The set of real numbers that the dependent

variable y can take over the domain (sometimes called the image of the function)

Vertical line test: A vertical line will only cut the graph of

a function in at most one point. If the vertical line cuts the graph in more than one point, it is not a function

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### INTRODUCTION

FUNCTIONS AND THEIR GRAPHS are used in many areas, such as mathematics, science and economics. In this chapter you will study functions, function notation and how to sketch graphs. Some of these graphs will be studied in more detail in later chapters.

### DID YOU KNOW?

The number plane is called the Cartesian plane after Rene

Descartes (1596–1650). He was known as one of the fi rst

modern mathematicians along with Pierre de Fermat (1601–1665). Descartes used the number plane to develop analytical geometry. He discovered that any equation with two unknown variables can be represented by a line. The points in the number plane can be called Cartesian coordinates.

Descartes used letters at the beginning of the alphabet to stand for numbers that are known, and letters near the end of the alphabet for unknown numbers. This is why we still use x and y so often!

Do a search on Descartes to fi nd out more details of his life and work.

Descartes

### Deﬁ nition of a function

Many examples of functions exist both in mathematics and in real life. These occur when we compare two different quantities. These quantities are called variables since they vary or take on different values according to some pattern. We put these two variables into a grouping called an ordered pair.

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### EXAMPLES

1. Eye colour

Name Anne Jacquie Donna Hien Marco Russell Trang

Colour Blue Brown Grey Brown Green Brown Brown Ordered pairs are (Anne, Blue), (Jacquie, Brown), (Donna, Grey), (Hien, Brown), (Marco, Green), (Russell, Brown) and (Trang, Brown).

2. y= +x 1

x 1 2 3 4

y 2 3 4 5

The ordered pairs are (1, 2), (2, 3), (3, 4) and (4, 5).

3. A B C D E 1 2 3 4

The ordered pairs are (A, 1), (B, 1), (C, 4), (D, 3) and (E, 2).

Notice that in all the examples, there was only one ordered pair for each variable. For example, it would not make sense for Anne to have both blue and brown eyes! (Although in rare cases some people have one eye that’s a different colour from the other.)

A relation is a set of ordered points ( x , y ) where the variables x and y are related according to some rule.

A function is a special type of relation. It is like a machine where for every INPUT there is only one OUTPUT.

INPUT PROCESS OUTPUT

The ﬁ rst variable (INPUT) is called the independent variable and the second (OUTPUT) the dependent variable. The process is a rule or pattern.

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For example, in y= +x 1, we can use any number for x (the independent variable), say x=3. Whenx y 3 3 1 4 = = + =

As this value of y depends on the number we choose for x , y is called the dependent variable.

A function is a relationship between two variables where for every independent variable, there is only one dependent variable. This means that for every x value, there is only one y value.

While we often call the independent variable x and the dependent variable y, there are other pronumerals we could use. You will meet some of these in this course.

### Investigation

When we graph functions in mathematics, the independent variable (usually the x -value) is on the horizontal axis while the dependent variable (usually the y -value) is on the vertical axis.

In other areas, the dependent variable goes on the horizontal axis. Find out in which subjects this happens at school by surveying teachers or students in different subjects. Research different types of graphs on the Internet to ﬁ nd some examples.

Here is an example of a relationship that is NOT a function. Can you see the difference between this example and the previous ones?

A B C D E 1 2 3 4

In this example the ordered pairs are (A, 1), (A, 2), (B, 1), (C, 4), (D, 3) and (E, 2).

Notice that A has two dependent variables, 1 and 2. This means that it is NOT a function.

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Here are two examples of graphs on a number plane.

1. x y 2. x y

There is a very simple test to see if these graphs are functions. Notice that in the ﬁ rst example, there are two values of y when x=0 . The y -axis passes through both these points.

x y

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If a vertical line cuts a graph only once anywhere along the graph, the graph is a function.

y

x

If a vertical line cuts a graph in more than one place anywhere along the graph, the graph is not a function.

x y

There are also other x values that give two y values around the curve. If we drew a vertical line anywhere along the curve, it would cross the curve in two places everywhere except one point. Can you see where this is?

In the second graph, a vertical line would only ever cross the curve in one place.

So when a vertical line cuts a graph in more than one place, it shows that it is not a function.

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### EXAMPLES

1. Is this graph a function?

### Solution

A vertical line only cuts the graph once. So the graph is a function.

2. Is this circle a function?

### Solution

A vertical line can cut the curve in more than one place. So the circle is not a function.

You will learn how to sketch these graphs later in this chapter.

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3. Does this set of ordered pairs represent a function? -2 3, , -1 4, , 0 5, , 1 3, , 2 4,

^ h ^ h ^ h ^ h ^ h

### Solution

For each x value there is only one y value, so this set of ordered pairs is a function. 4. Is this a function? y x 3

### Solution

y x 3

Although it looks like this is not a function, the open circle at x=3 on the top line means that x=3 is not included, while the closed circle on the bottom line means that x=3 is included on this line.

So a vertical line only touches the graph once at x=3 . The graph is a function.

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1. 2. 3. 4. 5. 6. 7. 8. 9. ^ 1 3, h,^2,-1h,^3 3, h,^4 0, h 10. ^1 3, h,^2,-1h,^2 7, h,^4 0, h 11. 1 2 3 4 5 1 2 3 4 5 12. 1 2 3 4 5 1 2 3 4 5 13. 1 2 3 4 5 1 2 3 4 5

### Exercises

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14. Name Ben Paul Pierre Hamish Jacob Lee Pierre Lien

A 3 B 4 C 7 D 3 E 5 F 7 G 4

### Function notation

If y depends on what value we give x in a function, then we can say that y is a function of x . We can write this as y=f x] g .

Notice that these two examples are asking for the same value and f (3) is the value of the function when x=3 .

### EXAMPLES

1. Find the value of y when x=3 in the equation y= +x 1 .

### Solution

Whenx : y x 3 1 3 1 4 = = + = + = 2. If f x] g= +x 1 , evaluate f (3).

### Solution

f x x f 1 3 3 1 4 = + = + = ] ] g g

If y=f x] g then f ( a ) is the value of y at the point on the function where x=a

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### EXAMPLES

1. If f x] g=x2+3x+1, ﬁ nd f]-2g.

### Solution

( ) ( ) f 2 2 3 2 1 4 6 1 1 2 - = - + - + = - + = -] g 2. If f x] g=x3-x2, ﬁ nd the value of f]-1g.

### Solution

( ) ( ) f x x x f 1 1 1 1 1 2 3 2 3 = -- = -- = = -2 ] g ] g

3. Find the values of x for which f x] g=0, given that f x] g=x2+3x-10.

### Solution

( ) i.e. ( ) ( ) , f x x x x x x x x x 0 3 10 0 5 2 0 5 0 2 0 5 2 2 = + - = + - = + = - = = - =

4. Find f]3g,f]2g,f]0g and f]-4giff x] g is deﬁ ned as

when when . f x x x x x 3 4 2 2 12 \$ = + -] g )

### Solution

since -412 ( ) ( ) since ( ) ( ) since ( ) ( ) since ( ) ( ) f f f f 3 3 3 4 3 2 13 2 3 2 4 2 2 10 0 2 0 0 2 0 4 2 4 8 1 \$ \$ = + = = + = = -= - = -=

5. Find the value of g]1g+ - -g] 2g g]3g if

when when when x x x 2 1 2 1 2 1 # # -g x x x 2 1 5 2 = -] g

### *

This is the same as fi nding y when x=-2.

Putting (x)f =0 is different from fi nding (0) .f Follow this example carefully.

Use f (x)=3x+ when 4 x is 2 or more, and use f (x)=- when x is less 2x than 2.

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### Solution

( ) ( ) ( ) ( ) g g g 1 2 1 1 1 1 2 1 2 5 2 1 3 3 3 2 9 since since since 2 1 2 # # = - -= - = - -= = g( )1 g( 2) g( )3 1 5 9 3 So + - - = + -= -

### DID YOU KNOW?

Leonhard Euler (1707–83), from Switzerland, studied functions and invented the term f (x) for function notation. He studied theology, astronomy, medicine, physics and oriental languages as well as mathematics, and wrote more than 500 books and articles on

mathematics. He found time between books to marry and have 13 children, and even when he went blind he kept on having books published.

1. Given f x] g= +x 3, ﬁ nd f 1] g and . f]-3g 2. If h x] g=x2-2, ﬁ nd h]0g,h]2g and h]-4g. 3. If f x] g= -x2, ﬁ nd f]5g,f]-1g,f]3g and f]-2g.

4. Find the value of f]0g+ -f] 2g if

. f x] g=x4-x2+1 5. Find f]-3g if f x] g=2x3-5x+4. 6. If f x] g=2x-5, ﬁ nd x when . f x] g=13 7. Given f x] g=x2+3, ﬁ nd any

values of x for which f x] g=28. 8. If f x] g=3x, ﬁ nd x when . f x 27 1 = ] g

9. Find values of z for which f z] g=5 given f z] g= 2z+3 . 10. If f x] g=2x-9, ﬁ nd f p^ h and . f x] +hg 11. Find g x] -1g when . g x] g=x2+2x+3 12. If f x] g=x3-1, ﬁ nd f k] g as a product of factors. 13. Given f t] g= +t2 2t+1, ﬁ nd

t when f t] g=0. Also ﬁ nd any values of t for which f t] g=9. 14. Given f t] g= + -t4 t2 5, ﬁ nd the value of f b] g- -f] bg. 15. f x x x x x 1 1 for for 3 2 # = ] g ) Find f]5g,f]1g and ]-1g. 16. f x x x x x x x 2 4 1 3 1 1 1 if if if 2 1 1 \$ # = -+ -] g Z [ \ ]] ]]

Find the values of

.

f]2g- - + -f] 2g f] 1g

### Exercises

We can use pronumerals other than f for functions.

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17. Find g]3g+g]0g+ -g] 2g if g x x x x x 1 0 2 1 0 when when 1 \$ = + - + ] g )

18. Find the value of

f]3g-f]2g+2f]-3g when f x x x x x x 2 2 2 4 2 for for for 2 2 1# # = -] g

### *

19. Find the value of f]- -1g f]3g

if ( ) 1 2 2 3 1 2 f x x x x x x for for 3 2 1 \$ = -+ -* 20. If f x x x x 3 2 3 2 = -- -] g evaluate (a) f (2)

explain why the function (b)

does not exist for x=3 by taking several

(c) x values

close to 3, ﬁ nd the value of y that the function is moving towards as x moves towards 3. 21. If f x] g=x2–5x+4 , ﬁ nd f x] +hg-f x] g in its simplest form. 22. Simplify h f x] +hg-f x] g where f x] g=2x2+x 23. If f x] g=5x-4 , ﬁ nd f x] g-f c] g in its simplest form.

24. Find the value of f k^ 2h if

f x x x x x 3 5 0 0 for for 2 1 \$ = + ] g * 25. If f x x x x x x 3 2 0 when when 3 2 \$ # = - + x 5 when01 13 ] g Z [ \ ]] ]] evaluate (a) f (0) (b) f]2g-f]1g (c) f^-n2h

### Graphing Techniques

You may have previously learned how to draw graphs by completing a table of values and then plotting points. In this course, you will learn some other techniques that will allow you to sketch graphs by showing their important features.

### Intercepts

One of the most useful techniques is to ﬁ nd the x- and y -intercepts.

For x -intercept, y=0 For y -intercept, x=0 Everywhere on the x-axis,

0 =

y and everywhere on the y-axis x=0 .

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### EXAMPLE

Find the x - and y -intercepts of the function f x] g=x2+7x-8.

### Solution

For x -intercept: y=0 , , x x x x x x x x 0 7 8 8 1 8 0 1 0 8 1 2 = + -= + -+ = - = = - = ] g] g For y -intercept: x=0 y 0 7 0 8 8 2 = + = -] g ] g

This is the same as

. y= +x2 7x-8

You will use the intercepts to draw graphs in the next section in this chapter .

### Domain and range

You have already seen that the x -coordinate is called the independent variable and the y -coordinate is the dependent variable.

The set of all real numbers x for which a function is deﬁ ned is called the domain .

The set of real values for y or f ( x ) as x varies is called the range (or image) of f .

### EXAMPLE

Find the domain and range of f x] g=x2.

### Solution

You can see the domain and range from the graph, which is the parabola y=x2.

x y

CONTINUED

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Notice that the parabola curves outwards gradually, and will take on any real value for x . However, it is always on or above the x -axis.

Domain: {all real x } Range: { y : y\$0 }

You can also ﬁ nd the domain and range from the equation y=x2 . Notice

that you can substitute any value for x and you will ﬁ nd a value of y . However, all the y -values are positive or zero since squaring any number will give a positive answer (except zero).

### Odd and even functions

When you draw a graph, it can help to know some of its properties, for example, whether it is increasing or decreasing on an interval or arc of the curve (part of the curve lying between two points) .

If a curve is increasing, as x increases, so does y , and the curve is moving upwards, looking from left to right.

If a curve is decreasing, then as x increases, y decreases and the curve moves downwards from left to right.

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### EXAMPLES

1. State the domain over which each curve is increasing and decreasing .

x x3 x2 x1 y

### Solution

The left-hand side of the parabola is decreasing and the right side is increasing.

So the curve is increasing for x 2 x 2 and the curve is decreasing when

x 1 x 2. 2. x x3 x2 x1 y

### Solution

The left-hand side of the curve is increasing until it reaches the y -axis (where x=0 ). It then turns around and decreases until x 3 and then increases again.

So the curve is increasing for x10,x2x3 and the curve is decreasing for 01 1x x3.

The curve isn’t increasing or decreasing at x 2 . We say that it is stationary at that point. You will study stationary points and further curve sketching in the HSC Course.

Notice that the curve is stationary at x= and 0 x=x3.

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Functions are odd if they have point symmetry about the origin. A graph rotated 180° about the origin gives the original graph.

This is an odd function:

x y

For even functions, f x] g= -f] xg for all values of x .

For odd functions, f]- = -xg f x] g for all values of x in the domain.

As well as looking at where the curve is increasing and decreasing, we can see if the curve is symmetrical in some way. You have already seen that the parabola is symmetrical in earlier stages of mathematics and you have learned how to ﬁ nd the axis of symmetry. Other types of graphs can also be symmetrical.

Functions are even if they are symmetrical about the y -axis. They have line symmetry (reﬂ ection) about the y -axis.

This is an even function:

x y

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### EXAMPLES

1. Show that f x] g=x2+3 is an even function.

### Solution

f x x x f x f x x 3 3 3 is an even function 2 2 2 ` - = - + = + = = + ] ] ] ] g g g g

2. Show that f x] g=x3-x is an odd function.

### Solution

f x x x x x x x f x f x x x is an odd function 3 3 3 3 ` - = - -= - + = - = -= -] ] ] ^ ] ] g g g h g g

### Investigation

Explore the family of graphs of f x] g=xn . For what values of n is the function even? For what values of n is the function odd?

Which families of functions are still even or odd given k ? Let k take on different values, both positive and negative.

1. f x] g=kxn 2. f x] g=xn+k 3. f x] g=]x+kgn

k is called a parameter. Some graphics calculators and computer programs use parameters to show how changing values of k change the shape of graphs .

1. Find the x - and y -intercept of each function. (a) y=3x-2 (b) 2x-5y+20=0 (c) x+3y-12 =0 (d) f x] g=x2+3x (e) f x] g=x2-4 (f) p x] g=x2+5x+6 (g) y=x2-8x+15 (h) p x] g=x3+5

### Exercises

ch5.indd 217 7/19/09 12:44:32 PM

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(i) y= +x x 3 ]x!0g

(j) g x] g= -9 x2

2. Show that f x] g= -f] xg where f x] g=x2-2 . What type of function is it? 3. If f x] g=x3+1 , ﬁ nd (a) f x^ 2h (b) 6f x( )@2 (c) f]-xg

Is it an even or odd function? (d)

4. Show that g x] g=x8+3x4-2x2 is

an even function .

5. Show that f ( x ) is odd, where

.

f x] g=x

6. Show that f x] g=x2-1 is an even

function. 7. Show that f x] g=4x-x3 is an odd function. 8. Prove that f x] g=x4+x2 is an

even function and hence ﬁ nd

.

f x] g- -f] xg

9. Are these functions even, odd or neither? (a) y x x x 4 2 3 = - (b) y x 1 1 3 = - (c) f x x 4 3 2 = -] g (d) y x x 3 3 = +- (e) f x x x x 5 2 3 = -] g

10. If n is a positive integer, for what values of n is the function f x xn = ] g even? (a) odd? (b)

11. Can the function f x] g=xn+x ever be even? (a) odd? (b)

12. For the functions below, state (i) the domain over which the graph is increasing

(ii) the domain over which the graph is decreasing (iii) whether the graph is odd, even or neither. x y (a) x 4 y (b) 2 -2 x y (c)

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### Investigation

Use a graphics calculator or a computer with graphing software to sketch graphs and explore what effect different constants have on each type of graph.

If your calculator or computer does not have the ability to use parameters (this may be called dynamic graphing), simply draw different graphs by choosing several values for k . Make sure you include positive and negative numbers and fractions for k .

Alternatively, you may sketch these by hand.

Sketch the families of graphs for these graphs with parameter

1. k. y kx y kx y kx y kx y xk (a) (b) (c) (d) (e) 2 3 4 = = = = =

What effect does the parameter k have on these graphs? Could you give a general comment about y=k f x] g ?

Sketch the families of graphs for these graphs with parameter

2. k. y x k y x k y x k y x k y x1 k (a) (b) (c) (d) (e) 2 2 3 4 = + = + = + = + = + ] g

What effect does the parameter k have on these graphs? Could you give a general comment about y=f x] g+k ?

-2 1 2 -4 -1 -2 2 4 y x (d) y x (e) CONTINUED ch5.indd 219 7/19/09 12:44:51 PM

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y=mx+b has gradient m and y -intercept b General form:

ax+by+ =c 0

### Investigation

Are straight line graphs always functions? Can you ﬁ nd an example of a straight line that is not a function?

Are there any odd or even straight lines? What are their equations? For the family of functions y=k f x] g , as k varies, the function changes its slope or steepness.

For the family of functions y=f x] g+k, as k varies, the graph moves up or down (vertical translation).

For the family of functions y=f x] +kg , as k varies, the graph moves left or right (horizontal translation).

Sketch the families of graphs for these graphs with parameter

3. k. y x k y x k y x k y x k y x 1k (a) (b) (c) (d) (e) 2 3 4 = + = + = + = + = + ] ] ] g g g

What effect does the parameter k have on these graphs? Could you give a general comment about y=f x] +kg ?

When 0 ,k2 the graph

moves to the left and when 0 ,

k1 the graph moves to

the right.

Notice that the shape of most graphs is generally the same regardless of the parameter k . For example, the parabola still has the same shape even though it may be narrower or wider or upside down.

This means that if you know the shape of a graph by looking at its

equation, you can sketch it easily by using some of the graphing techniques in this chapter rather than a time-consuming table of values. It also helps you to understand graphs more and makes it easier to ﬁ nd the domain and range.

You have already sketched some of these graphs in previous years.

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### EXAMPLE

Sketch the function f x] g=3x-5 and state its domain and range.

### Solution

This is a linear function. It could be written as y=3x-5. Find the intercepts

For x -intercept: y=0 0 3 5 5 3 1 x x x 3 2 = = = For y -intercept: x=0 3 5 5 y= 0 = -] g -1 -2 y 5 4 3 2 1 123 6 -3 -4 -5 1 4 -1 -2 2 3 -3 -4 x

Notice that the line extends over the whole of the number plane, so that it covers all real numbers for both the domain and range.

Domain: {all real x } Range: {all real y }

Notice too, that you can substitute any real number into the equation of the function for x, and any real number is possible for y.

The linear function ax+by+ =c 0 has domain {all real x } and range {all real y } where a and b are non-zero

### Special lines

Horizontal and vertical lines have special equations.

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a linear function, or choose

different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k

1. y=kx

2. y= +x k

3. y=mx+b where m and b are both parameters

What effect do the parameters m and b have on these graphs?

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### EXAMPLES

1. Sketch y=2 on a number plane. What is its domain and range?

### Solution

x can be any value and y is always 2.

Some of the points on the line will be (0, 2), (1, 2) and (2, 2). This gives a horizontal line with y -intercept 2.

-1 -3 y 4 3 2 1 5 -2 -4 -5 1 4 -1 -2 2 3 x -3 -4

Domain: "all realx,

Range: "y y =: 2,

2. Sketch x= -1 on a number plane and state its domain and range.

### Solution

y can be any value and x is always -1.

Some of the points on the line will be ^-1 0, h,^-1 1, h and ^-1 2, h. This gives a vertical line with x -intercept -1.

Domain: "x x = -: 1, Range: "all realy,

-1 -3 4 3 2 1 5 -2 -4 -5 1 4 --1 -2 3 -3 -4 y x

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x=a is a vertical line with x -intercept a Domain: !x x: =a+

Range: {all real y }

y=b is a horizontal line with y -intercept b Domain: {all real x }

Range: "y y: =b,

### Exercises

1. Find the x - and y -intercepts of each function. (a) y= -x 2 (b) f x] g=2x+3 (c) 2x+y-1=0 (d) x-y+ =3 0 (e) 3x-6y-2=0

2. Draw the graph of each straight line. (a) x=4 (b) x-3=0 (c) y=5 (d) y+ =1 0 (e) f x] g=2x-1 (f) y= +x 4 (g) f x] g=3x+2 (h) x+ =y 3 (i) x- -y 1=0 (j) 2x+y-3=0

3. Find the domain and range of (a) 3x-2y+ =7 0 (b) y=2 (c) x= -4 (d) x-2=0 (e) 3-y=0

4. Which of these linear functions are even or odd?

(a) y=2x (b) y=3 (c) x=4 (d) y= -x (e) y=x 5. By sketching x- -y 4=0 and x y

2 +3 -3=0 on the same set of axes, ﬁ nd the point where they meet.

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### Applications

The parabola shape is used in many different applications as it has special

properties that are very useful. For example if a light is placed inside the parabola at a special place (called the focus), then all light rays coming from this light and bouncing off the parabola shape will radiate out parallel to each other, giving a strong light. This is how car headlights work. Satellite dishes also use this property of the parabola, as sound coming in to the dish will bounce back to the focus.

The pronumeral a is called the coeffi cient of x2.

The quadratic function gives the graph of a parabola.

f x] g=ax2+bx+c is the general equation of a parabola.

If a20 the parabola is concave upwards

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The lens in a camera and glasses are also parabola shaped. Some bridges look like they are shaped like a parabola, but they are often based on the catenary. Research the parabola and catenary on the Internet for further information.

### Investigation

Is the parabola always a function? Can you ﬁ nd an example of a parabola that is not a function?

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a quadratic function, or choose different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k

### .

1. y=kx2 2. y=x2+k 3. y=]x+kg2 4. y=x2+kx

What effect does the parameter k have on these graphs?

Which of these families are even functions? Are there any odd quadratic functions?

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### EXAMPLES

1. (a) Sketch the graph of y=x2-1, showing intercepts.

(b) State the domain and range.

### Solution

This is the graph of a parabola. Since

(a) a20, it is concave upward

For x -intercept: y=0 x x x 0 1 1 1 2 2 ! = -= = For y -intercept: x=0 0 1 1 y= 2 -= -

From the graph, the curve is moving outwards and will extend (b)

to all real x values. The minimum y value is -1. Domain: "all realx,

Range: "y y: \$ -1,

2. Sketch f x] g=]x+1g2.

### Solution

This is a quadratic function. We ﬁ nd the intercepts to see where the parabola will lie.

Alternatively, you may know from your work on parameters that

f x] g=]x+ag2 will move the function f x] g=x2 horizontally a units to the

left.

So f x] g=]x+1g2 moves the parabola f x] g=x2 1 unit to the left.

For x -intercept: y=0 0 1 0 1 x x x 1 2 = + + = = -] g For y -intercept: x=0 1 y= 0+1 2 = ] g

-1 -3 4 3 2 1 5 -2 -4 -5 -6 1 4 -1 -2 3 5 -3 -4 y x

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3. For the quadratic function f x] g=x2+ -x 6

Find the

(a) x - and y -intercepts

Find the minimum value of the function (b)

State the domain and range (c)

For what values of

(d) x is the curve decreasing?

### Solution

For (a) x -intercept: y=0 This means f x] g=0 , , x x x x x x x x 0 6 3 2 3 0 2 0 3 2 2 = + -= + -+ = - = = - = ] g] g For y -intercept: x=0 f 0 0 0 6 6 2 = + = -] g ] g ] g Since

(b) a20, the quadratic function has a minimum value. Since the parabola is symmetrical, this will lie halfway between the

x -intercepts.

Halfway between x = -3 and x =2: 2 3 2 2 1 - + = - Minimum value is f 2 1 -c m f 2 1 2 1 2 1 6 4 1 2 1 6 6 4 1 2 - = - + - = = -c m c m c m

So the minimum value is 6 . 4 1 -

CONTINUED

-1 -3 4 3 2 1 5 -2 -4 -5 1 4 -1 -2 2 3 -3 -4 y x ch5.indd 227 7/31/09 4:30:36 PM

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Sketching the quadratic function gives a concave upward parabola. (c)

From the graph, notice that the parabola is gradually going outwards and will include all real x values.

Since the minimum value is 6

4 1

- , all y values are greater than this. Domain: "all realx,

Range: y:y 6 4 1 \$

-' 1

The curve decreases down to the minimum point and then (d)

increases. So the curve is decreasing for all x . 2 1 1-

4. (a) Find the x - and y -intercepts and the maximum value of the

quadratic function f x] g= - +x2 4x+5.

(b) Sketch the function and state the domain and range. (c) For what values of x is the curve increasing?

### Solution

For (a) x -intercept: y =0 So f x] g=0 0 4 5 4 5 0 0 x x x x x 5 x 1 2 2 = - + + = + = - -] g] g , , x x x x 5 0 1 0 5 1 - = + = = = For y -intercept: x =0 f 0 0 4 0 5 5 2 = - + + = ] g ] g ] g -1 -3 4 3 2 1 5 -2 -4 -5 -6 y 1 4 -1 -2 3 5 -3 -4 x -12,-614

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Since a10, the quadratic function is concave downwards and has a maximum value halfway between the x -intercepts x = -1 and x=5. 2 1 5 2 - + = f 2 2 4 2 5 9 = - + + = 2 ] g ] g ] g So the maximum value is 9.

Sketching the quadratic function gives a concave downward parabola. (b)

From the graph, the function can take on all real numbers for x , but the maximum value for y is 9.

Domain: "all realx,

Range: "y y #: 9,

From the graph, the function is increasing on the left of the (c)

maximum point and decreasing on the right. So the function is increasing when x 21 .

1. Find the x - and y -intercepts of each function. (a) y=x2+2x (b) y= - +x2 3x (c) f x] g=x2-1 (d) y=x2- -x 2 (e) y=x2-9x+8 2. Sketch (a) y=x2+2 (b) y= - +x2 1 (c) f x] g=x2-4 (d) y=x2+2x (e) y= - -x2 x (f) f x] g=]x-3g2

### Exercises

-1 9 8 7 5 4 3 2 6 1 -2 -3 -4 -5 y 2 5 1 3 4 6 -1 -2 -3 -4 x ch5.indd 229 7/19/09 12:46:30 PM

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### EXAMPLES

1. Sketch f x] g= x -1 and state its domain and range.

### Solution

Method 1: Table of values

When sketching any new graph for the ﬁ rst time, you can use a table of values. A good selection of values is -3# #x 3 but if these don’t give enough information, you can ﬁ nd other values.

### Absolute Value Function

You may not have seen the graphs of absolute functions before. If you are not sure about what they look like, you can use a table of values or look at the deﬁ nition of absolute value.

(g) f x] g=]x+1g2 (h) y=x2+3x-4 (i) y=2x2-5x+3 (j) f x] g= - +x2 3x-2

3. For each parabola, ﬁ nd the

(i) x - and y -intercepts the domain and range (ii) (a) y=x2–7x+12 (b) f x] g=x2+4x (c) y=x2-2x-8 (d) y=x2-6x+9 (e) f t] g= -4 t2

4. Find the domain and range of (a) y=x2-5 (b) f x] g=x2-6x (c) f x] g=x2- -x 2 (d) y= -x2 (e) f x] g=]x-7g2

5. Find the range of each function over the given domain.

(a) y=x2 for 0# #x 3 (b) y= - +x2 4 for -1# #x 2 (c) f x] g=x2-1 for -2# #x 5 (d) y=x2+2x-3 for -2# #x 4 (e) y= -x2-x+2 for 0# #x 4

6. Find the domain over which each function is increasing (i) decreasing (ii) (a) y=x2 (b) y= -x2 (c) f x] g=x2-9 (d) y= - +x2 4x (e) f x] g=]x+5g2

7. Show that f x] g= -x2 is an even function.

8. State whether these functions are even or odd or neither.

(a) y=x2+1 (b) f x] g=x2-3 (c) y= -2x2 (d) f x] g=x2-3x (e) f x] g=x2+x (f) y=x2-4 (g) y=x2-2x-3 (h) y=x2-5x+4 (i) p x] g=]x+1g2 (j) y=]x-2g2

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CONTINUED e.g. When x= -3: | | y 3 1 3 1 2 = - = -= x -3 -2 -1 0 1 2 3 y 2 1 0 -1 0 1 2

This gives a v-shaped graph.

y -2 4 3 2 1 5 -1 -3 -4 -5 1 4 -1 -2 2 3 -3 -4 x

Method 2: Use the deﬁ nition of absolute value | |

y= x - =1 &x- --x 11 whenwhenxx1\$00 This gives 2 straight line graphs: y= -x 1 ]x\$0g

-3 4 3 2 1 5 -2 -1 -4 -5 y 3 -1 -2 1 2 4 -3 -4 x y=x-1 ch5.indd 231 7/19/09 12:46:49 PM

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y= - -x 1 x] 10g

-3 4 3 2 1 5 -2 -1 -4 -5 y 3 -1 -2 1 2 4 -3 -4 x y= - x-1

Draw these on the same number plane and then disregard the dotted lines to get the graph shown in method 1.

-3 4 3 2 1 5 -2 -1 -4 -5 yy 3 -1 -2 1 2 4 -3 -4 x y=-x- 1 y=x-1

Method 3: If you know the shape of the absolute value functions, ﬁ nd the intercepts. For x -intercept: y =0 So f x] g=0 | | | | x x x 0 1 1 1 ` ! = -= = For y -intercept: x =0 ( ) | | f 0 0 1 1 = -= -

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The graph is V -shaped, passing through these intercepts.

-3 4 3 2 1 5 -2 -1 -4 -5 y 4 -2 -1 1 2 3 5 -3 -4 x

From the graph, notice that x values can be any real number while the minimum value of y is -1.

Domain: {all real x } Range: { y : y\$ -1 }

2. Sketch y=|x+2|.

### Solution

Method 1: Use the deﬁ nition of absolute value.

| | ( )

y= x+2 ='x- ++x2 2 whenwhenxx++221\$00 This gives 2 straight lines:

2 y= +x when x+2\$0 x\$ -2

-3 4 3 2 1 5 -2 -1 -4 -5 y 3 -1 -2 1 2 4 -3 -4 x y=x+2

If you already know how to sketch the graph of

y=| x | , translate the graph of y=| x |-1 down 1 unit, giving it a y -intercept of -1.

CONTINUED

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2 y= - +]x g when x+210 i.e. y= - -x 2 when x1 -2 -3 4 3 2 1 5 -2 -1 -4 -5 y 3 -1 -2 1 2 4 -3 -4 x y=-x- 2

Draw these on the same number plane and then disregard the dotted lines.

-3 4 3 2 1 5 -2 -1 -4 -5 y 3 -1 -2 1 2 4 -3 -4 x y=-x-2 y=x+2

Method 2: Find intercepts For x -intercept: y =0 So f x =] g 0 0 | 2 | 0 2 2 x x x = + = + - = For y -intercept: x =0 (0) | 0 2 | 2 f = + = There is only one

solution for the equation | x+2 |=0. Can you see why?

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The graph is V -shaped, passing through these intercepts.

-3 4 3 2 1 5 -2 -1 -4 -5 y 3 -1 -2 1 2 4 -3 -4 x

If you know how to sketch the graph of

y=| x | , translate it 2 places to the left for the graph of y=| x+2 |.

### Investigation

Are graphs that involve absolute value always functions? Can you ﬁ nd an example of one that is not a function?

Can you ﬁ nd any odd or even functions involving absolute values? What are their equations?

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on an absolute value function, or choose different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k 1. f x] g=k x| |

2. f x] g=|x|+k 3. f x] g=|x+k|

What effect does the parameter k have on these graphs?

The equations and inequations involving absolute values that you studied in Chapter 3 can be solved graphically.

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Solve

1. | 2x -1 |=3

### Solution

Sketch y=| 2x-1 | and y =3 on the same number plane.

The solution of | 2x -1 |=3 occurs at the intersection of the graphs, that is, x= -1 2, .

2. | 2x+1 |=3x-2

### Solution

Sketch y=| 2x+1 | and y=3x-2 on the same number plane.

The solution is x =3.

3. |x+1 |12

### Solution

Sketch y=|x+1 | and y =2 on the same number plane. The graph shows that

there is only one solution. Algebraically, you need to fi nd the 2 possible solutions and then check them.

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The solution of |x+1 |12 is where the graph y=|x+1 | is below the graph y =2, that is, -31 1x 1.

1. Find the x - and y -intercepts of each function. (a) y=|x| (b) f x] g=|x|+7 (c) f x] g=|x|-2 (d) y=5 |x| (e) f x] g= -|x|+3 (f) y=|x+6 | (g) f x] g=|3x-2| (h) y=| 5x+4 | (i) y=| 7x-1 | (j) f x] g=|2x|+9

2. Sketch each graph on a number plane. (a) y=|x| (b) f x] g=|x|+1 (c) f x] g=|x|-3 (d) y=2 |x| (e) f x] g= -|x| (f) y=|x+1 | (g) f x] g= - -|x 1| (h) y=| 2x-3 | (i) y=| 4x+2 | (j) f x] g=|3x|+1

3. Find the domain and range of each function. (a) y=|x-1 | (b) f x] g=|x|-8 (c) f x] g=|2x+5| (d) y=2 |x|-3 (e) f x] g= - -|x 3|

4. Find the domain over which each function is increasing (i) decreasing (ii) (a) y=|x-2 | (b) f x] g=|x|+2 (c) f x] g=|2x-3| (d) y=4 |x|-1 (e) f x] g= -|x|

5. For each domain, ﬁ nd the range of each function. (a) y=|x| for -2# #x 2 (b) f x] g= -|x|-4 for x 4# #3 - (c) f x] g=|x+4| for -7# #x 2 (d) y=| 2x-5 | for -3# #x 3 (e) f x] g= -|x| for -1# #x 1

6. For what values of x is each function increasing? (a) y=|x+3 | (b) f x] g= -|x|+4 (c) f x] g=|x-9| (d) y=|x-2|-1 (e) f x] g= -|x+2|

### Exercises

ch5.indd 237 7/19/09 12:47:53 PM

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7. Solve graphically (a) | |x =3 (b) | |x 21 (c) | |x #2 (d) |x +2 |=1 (e) |x -3 |=0 (f) | 2x -3 |=1 (g) |x-1|14 (h) |x+1|#3 (i) |x-2|22 (j) |x-3|\$1 (k) |2x+3|#5 (l) |2x-1|\$1 (m) | 3x-1 |= +x 3 (n) | 3x-2 |= -x 4 (o) |x-1 |= +x 1 (p) |x+3 |=2x+2 (q) | 2x+1 |= -1 x (r) | 2x-5 |= -x 3 (s) |x-1 |=2x (t) | 2x-3 |= +x 3

### The Hyperbola

A hyperbola is a function with its equation in the form xy=aory=xa.

Sketch y= 1x.

### Solution

1

y= x is a discontinuous curve since the function is undeﬁ ned at x=0. Drawing up a table of values gives:

x -3 -2 -1 2 1 - -41 0 4 1 2 1 1 2 3 y 3 1 - -21 -1 -2 -4 — 4 2 1 2 1 3 1

### Class Discussion

What happens to the graph as x becomes closer to 0? What happens as x becomes very large in both positive and negative directions? The value of

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To sketch the graph of a more general hyperbola, we can use the domain and range to help ﬁ nd the asymptotes (lines towards which the curve approaches but never touches).

The hyperbola is an example of a discontinuous graph, since it has a gap in it and is in two separate parts.

### Investigation

Is the hyperbola always a function? Can you ﬁ nd an example of a hyperbola that is not a function?

Are there any families of odd or even hyperbolas? What are their equations?

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a hyperbola, or choose different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k 1. y= xk

2. y= +1x k

3. y 1 x k

= +

What effect does the parameter k have on these graphs?

### EXAMPLES

1. (a) Find the domain and range of f x .

x 3 3 =

-] g

Hence sketch the graph of the function.

(b)

### Solution

This is the equation of a hyperbola.

To ﬁ nd the domain, we notice that x-3!0. So x!3

Also y cannot be zero (see example on page 238). Domain: {all real x : x!3 }

Range: {all real y : y!0 }

The lines x =3 and y =0 (the x -axis) are called asymptotes.

The denominator cannot be zero.

CONTINUED

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To make the graph more accurate we can ﬁ nd another point or two. The easiest one to ﬁ nd is the y -intercept.

For y -intercept, x =0 1 y 0 3 3 = -= - -3 4 3 2 1 5 -2 -1 -4 -5 y -1 -2 1 2 4 5 -3 -4 x x=3 y=0 Asymptotes 3 2. Sketch y . x 2 4 1 = - +

### Solution

This is the equation of a hyperbola. The negative sign turns the hyperbola around so that it will be in the opposite quadrants. If you are not sure where it will be, you can ﬁ nd two or three points on the curve. To ﬁ nd the domain, we notice that 2x+4!0.

x x 2 4 2 ! !

For the range, y can never be zero. Domain: {all real x : x! -2 } Range: {all real y : y!0 }

So there are asymptotes at x= -2 and y=0 (the x -axis). To make the graph more accurate we can ﬁ nd the y -intercept. For y -intercept, x=0

( ) y 2 0 4 1 4 1 = - + = -Notice that this graph is

a translation of y 3 x

= three units to the right.

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y -2 x -14 The function f x bx c a = + ] g is a hyperbola with domain x x: b c all real ! -& 0 and range {all real y y: !0 }

1. For each graph

State the domain and range. (i)

Find the

(ii) y -intercept if it exists.

Sketch the graph. (iii) (a) y=2x (b) y= -1x (c) f x x 1 1 = + ] g (d) f x x 2 3 = -] g (e) 3 6 1 y x = + (f) f x x 3 2 = - -] g (g) f x x 1 4 = -] g (h) 1 2 y x = - + (i) f x x 6 3 2 = -] g (j) 2 6 y x = - +

2. Show that f x] g=2x is an odd function.

3. Find the range of each function over the given domain.

(a) f x x 2 5 1 = + ] g for -2# #x 2 (b) 3 1 y x = + for -2# #x 0 (c) f x x 2 4 5 = -] g for -3# #x 1

### Exercises

ch5.indd 241 7/19/09 1:26:13 PM

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(d) f x x 4 3 = - -] g for -3# #x 3 (e) 3 1 2 y x = - + for 0# #x 5

4. Find the domain of each function over the given range.

(a) y=3x for 1# #y 3 (b) y= -2x for 2 y 2 1 # # - - (c) f x x 1 1 = -] g for 1 y 7 1 # # - - (d) f x x 2 1 3 = - + ] g for 1 y 3 1 # # - - (e) 3 2 6 y x = - for 1 y 2 1 6 # #

### Circles and Semi-circles

The circle is used in many applications, including building and design.

Circle gate

A graph whose equation is in the form x2+ax+y2+by+ =c 0 has the shape of a circle.

There is a special case of this formula:

The graph of x2+y2=r2 is a circle, centre ^0, 0h and radius r

### Proof

r y x (x, y) y x

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Given the circle with centre (0, 0) and radius r :

Let ( x , y ) be a general point on the circle, with distances from the origin x on the x -axis and y on the y -axis as shown.

By Pythagoras’ theorem: c a b r x y 2 2 2 2 2 2 ` = + = +

### EXAMPLE

Sketch the graph of

(a) x2+y2=4. Is it a function?

State its domain and range. (b)

### Solution

This is a circle with radius 2 and centre (0, 0). (a) y x -2 -2 2 2

The circle is not a function since a vertical line will cut it in more than one place. y x 2 2 -2 -2 The radius is 4. CONTINUED ch5.indd 243 7/19/09 12:48:53 PM

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Notice that the

(b) x -values for this graph lie between -2 and 2 and

the y -values also lie between -2 and 2. Domain: {x:-2# #x 2}

Range: {y:-2# #y 2}

The circle x2+y2=r2 has domain: !x:-r # #x r+ and

range: "y:-r# #y r,

The equation of a circle, centre ( a , b ) and radius r is ]x ag2+^y bh2=r2

We can use Pythagoras’ theorem to ﬁ nd the equation of a more general circle.

### Proof

Take a general point on the circle, ( x , y ) and draw a right-angled triangle as shown.

y x (a, b) x y r (x, y) a b x - a y - b

Notice that the small sides of the triangle are x a and y b and the

hypotenuse is r , the radius. By Pythagoras’ theorem: – – c a b r x a y b 2 2 2 2 2 2 = + =] g +^ h

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### EXAMPLES

1. (a) Sketch the graph of x2+y2=81.

(b) State its domain and range.

### Solution

The equation is in the form

(a) x2+y2=r2.

This is a circle, centre (0, 0) and radius 9.

y x 9 9 -9 -9

From the graph, we can see all the values that are possible for

(b) x

and y for the circle. Domain: {x:-9# #x 9} Range: {y:-9# #y 9}

2. (a) Sketch the circle ]x–1g2+^y +2h2=4.

(b) State its domain and range.

### Solution

The equation is in the form

(a) ]x ag2+^y bh2=r2. – – – x y x y 1 2 4 1 2 2 2 2 2 + + = + - = 2 2 ] ^ ] _ ] g h g gi So a=1,b= -2 and r =2 CONTINUED ch5.indd 245 7/19/09 12:49:12 PM

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This is a circle with centre ^1,-2h and radius 2.

To draw the circle, plot the centre point ^1,-2h and count 2 units up, down, left and right to ﬁ nd points on the circle.

y x 1 1 -2 -1 2 3 4 -3 -4 -5 2 3 4 5 -1 -2 -3 -4 (1, -2)

From the graph, we can see all the values that are possible for

(b) x

and y for the circle. Domain: {x: 1- # #x 3} Range: {y:-4# #y 0}

3. Find the equation of a circle with radius 3 and centre ^-2 1, h in

expanded form.

### Solution

This is a general circle with equation ]x ag2+^y bh2=r2 where

, a= -2 b=1 and r=3. Substituting: – – – – x a y b r x y x y 2 1 3 2 1 9 2 2 2 2 2 2 2 2 + = - - + = + + = ] ^ ] ] ^ ] ^ g h gg h g h

Remove the grouping symbols.

– – a b a ab b x x x x x a b a ab b y y y y y 2 2 2 2 2 4 4 2 1 2 1 1 2 1 So So 2 2 2 2 2 2 2 2 2 2 2 2 2 2 + = + + + = + + = + + = - + = - + = - + ] ] ] ] ] ^ ^ ] g g g g g h h g

The equation of the circle is:

x x y y x x y y x x y y x x y y 4 4 2 1 9 4 2 5 9 4 2 5 9 4 2 4 0 9 9 2 2 2 2 + + + - + = + + - + = + + + = + + - - = - -

You may need to revise this in Chapter 2.

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### Investigation

The circle is not a function. Could you break the circle up into two functions?

Change the subject of this equation to y .

What do you notice when you change the subject to y ? Do you get two functions? What are their domains and ranges?

If you have a graphics calculator, how could you draw the graph of a circle?

The equation of the semi-circle above the x -axis with centre (0, 0) and radius r is y= r2-x2

The equation of the semi-circle below the x -axis with centre (0, 0) and radius r is y= - r2-x2

y= r2-x2 is the semi-circle above the x -axis since its range is y \$ 0

for all values.

y

x r

r

-r

The domain is { :x -r # #x r } and the range is { :y 0# #y r }

### Proof

x y r y r x y r x 2 2 2 2 2 2 2 2 ! + = = =

This gives two functions:

By rearranging the equation of a circle, we can also ﬁ nd the equations of semi-circles.

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y= - r2-x2 is the semi-circle above the x -axis since its range is

y#0 for all values.

y

x r

-r -r

The domain is { :x -r# #x r } and the range is { :y -r# #y 0}

### EXAMPLES

Sketch each function and state the domain and range.

1. f x] g= 9-x2

### Solution

This is in the form f x] g= r2-x2 where r=3.

It is a semi-circle above the x -axis with centre (0, 0) and radius 3.

y x 3 3 -3 Domain: {x:-3# #x 3} Range: {y: 0# #y 3}

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## References

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