# Maths in Focus - Margaret Grove - ch5

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Descripción: Mathematics Preliminary Course - 2nd Edition...

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5 Functions and Graphs TERMINOLOGY Arc of a curve: Part or a section of a curve between two points

Even function: An even function has line symmetry (reflection) about the y-axis, and f ] - x g = - f ] x g

Asymptote: A line towards which a curve approaches but never touches

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

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 defined

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 ] - x g = - 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

Chapter 5 Functions and Graphs

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 first 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 find out more details of his life and work. Descartes

Functions Deﬁnition 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

Colour Blue

Jacquie Donna Hien

Marco

Russell

Trang

Brown

Green

Brown

Brown

Grey

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

1

B

2

C 3 D 4

E

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

Chapter 5 Functions and Graphs

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

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 ﬁnd 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

1 2

C 3 D E

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.

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.

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Here are two examples of graphs on a number plane. 1.

y

x

2.

y

x

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

x

Chapter 5 Functions and Graphs

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.

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. y

x

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EXAMPLES 1. Is this graph a function?

Solution

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

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.

Chapter 5 Functions and Graphs

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

3

x

Solution y

3

x

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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5.1 Exercises Which of these curves are functions? 1.

6.

2.

7.

3.

8.

4.

9.

^ 1, 3 h, ^ 2, -1 h, ^ 3, 3 h, ^ 4, 0 h

10. ^ 1, 3 h, ^ 2, -1 h, ^ 2, 7 h, ^ 4, 0 h 11.

5. 12.

13.

1

1

2

2

3

3

4

4

5

5

1

1

2

2

3

3

4

4

5

5

1

1

2

2

3

3

4

4

5

5

Chapter 5 Functions and Graphs

14. Name Ben Paul Pierre Hamish Jacob Lee Pierre Lien Sport Tennis Football Tennis Football Football Badminton Football Badminton 15. 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.

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

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

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

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

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 = x 2 + 3x + 1, ﬁnd f ] - 2 g .

Solution This is the same as finding y when x = - 2.

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

Solution f (x) = x 3 - x 2 f (- 1) = ] - 1 g 3 - ] - 1 g 2 = -1 - 1 = -2 3. Find the values of x for which f ] x g = 0, given that f ] x g = x 2 + 3x - 10.

Solution f (x) = 0 Putting f (x) = 0 is different from finding f (0) . Follow this example carefully.

i.e.

x + 3x - 10 = 0 ( x + 5 ) ( x - 2) = 0 x + 5 = 0, x-2=0 x = -5 x=2 2

4. Find f ] 3 g, f ] 2 g, f ] 0 g and f ] - 4 g if f ] x g is deﬁned as 3x + 4 when x \$ 2 f ]x g = ) - 2x when x 1 2. Use f (x) = 3x + 4 when x is 2 or more, and use f (x) = - 2x when x is less than 2.

Solution f (3 ) = 3 ( 3) + 4 = 13 f (2 ) = 3 ( 2) + 4 = 10 f (0) = - 2 (0) =0 f (- 4) = - 2 ( - 4) =8 5. Find the value of x2 g ] x g = * 2x - 1 5

since

3\$2

since

2\$2

since

012

since -4 1 2

g ] 1 g + g ] - 2 g - g ] 3 g if when x 2 2 when - 1 # x # 2 when x 1 - 1

Chapter 5 Functions and Graphs

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Solution g (1 ) = 2 ( 1) - 1 =1 g (- 2) = 5

since -1 # 1 # 2 since - 2 1 - 1

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

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.

5.2 Exercises 1.

Given f ] x g = x + 3, ﬁnd f ] 1 g and f ]-3 g.

10. If f ] x g = 2x - 9, ﬁnd f ^ p h and f ]x + h g.

2.

If h ] x g = x 2 - 2, ﬁnd h ] 0 g, h ] 2 g and h ] - 4 g .

11. Find g ] x - 1 g when g ] x g = x 2 + 2x + 3.

3.

If f ] x g = - x 2, ﬁnd f ] 5 g, f ] - 1 g, f ] 3 g and f ] - 2 g .

12. If f ] x g = x 3 - 1, ﬁnd f ] k g as a product of factors.

4.

Find the value of f ] 0 g + f ] - 2 g if f ] x g = x 4 - x 2 + 1.

5.

Find f ] - 3 g if f ] x g = 2x 3 - 5x + 4.

13. Given f ] t g = t 2 + 2t + 1, ﬁnd t when f ] t g = 0. Also ﬁnd any values of t for which f ] t g = 9.

6.

If f ] x g = 2x - 5, ﬁnd x when f ] x g = 13.

7.

Given f ] x g = x + 3, ﬁnd any values of x for which f ] x g = 28.

15. f ] x g = )

8.

If f ] x g = 3 x, ﬁnd x when 1 f ]x g = . 27

9.

Find values of z for which f ] z g = 5 given f ] z g = 2z + 3 .

Z 2x - 4 if x \$ 1 ] 16. f ] x g = [x + 3 if -1 1 x 1 1 ] 2 x if x # -1 \ Find the values of

2

14. Given f ] t g = t 4 + t 2 - 5, ﬁnd the value of f ] b g - f ] - b g . x3 for x 2 1 x for x # 1 Find f ] 5 g, f ] 1 g and ] - 1 g .

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

We can use pronumerals other than f for functions.

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17. Find g ] 3 g + g ] 0 g + g ] - 2 g if x+1 when x \$ 0 g ]x g = ) - 2x + 1 when x 1 0 18. Find the value of f ] 3 g - f ] 2 g + 2f ] - 3 g when x

for x 2 2

f ]x g = * x 4

2

for -2 # x # 2 for x 1 -2

19. Find the value of f ] - 1 g - f ] 3 g if f (x) = * 20. If f ] x g =

x3 - 1 2x 2 + 3x - 1

for x \$ 2 for x 1 2

x 2 - 2x - 3 x-3

(a) evaluate f (2) (b) explain why the function does not exist for x = 3 (c) by taking several x values close to 3, ﬁnd the value of y that the function is moving towards as x moves towards 3.

21. If f ] x g = x 2 – 5x + 4, ﬁnd f ] x + h g - f ] x g in its simplest form. f ]x + h g - f ]x g 22. Simplify where h f ] x g = 2x 2 + x 23. If f ] x g = 5x - 4, ﬁnd f ] x g - f ] c g in its simplest form. 24. Find the value of f ^ k 2 h if 3x + 5 for x \$ 0 f ]x g = * 2 x for x 1 0 Z 3 25. If when x \$ 3 ]x f ] x g = [5 when 0 1 x 1 3 ] 2 x - x + 2 when x # 0 \ evaluate (a) f (0) (b) f ] 2 g - f ] 1 g (c) f ^ - n 2 h

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 ﬁnd the x- and y-intercepts.

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

For x-intercept, y = 0 For y-intercept, x = 0

Chapter 5 Functions and Graphs

213

EXAMPLE Find the x- and y-intercepts of the function f ] x g = x 2 + 7x - 8.

Solution

This is the same as y = x 2 + 7x - 8.

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

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

y = ] 0 g2 + 7 ] 0 g - 8 = -8

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 deﬁned 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 = x 2 .

Solution You can see the domain and range from the graph, which is the parabola y = x 2 . y

x

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 ﬁnd the domain and range from the equation y = x 2. Notice that you can substitute any value for x and you will ﬁnd 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.

Chapter 5 Functions and Graphs

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

x2

x1

x3

x

The curve isn’t increasing or decreasing at x2. We say that it is stationary at that point. You will study stationary points and further curve sketching in the HSC Course.

Solution The left-hand side of the parabola is decreasing and the right side is increasing. So the curve is increasing for x 2 x2 and the curve is decreasing when x 1 x2. 2. y

x1

x2

x3

x

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 x3 and then increases again. So the curve is increasing for x 1 0, x 2 x 3 and the curve is decreasing for 0 1 x 1 x 3 .

Notice that the curve is stationary at x = 0 and x = x 3 .

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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 ﬁnd 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 (reﬂection) about the y-axis. This is an even function: y

x

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

x

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

Chapter 5 Functions and Graphs

217

EXAMPLES 1. Show that f ] x g = x 2 + 3 is an even function.

Solution f ] - x g = ] - x g2 + 3 = x2 + 3 = f ]x g ` f ] x g = x 2 + 3 is an even function 2. Show that f ] x g = x 3 - x is an odd function.

Solution f ] - x g = ] - x g3 - ] - x g = -x3 + x = - ^ x3 - x h = -f ]x g ` f ] x g = x 3 - x is an odd function

Investigation Explore the family of graphs of f ] x g = x n. 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 = kx n 2. f ] x g = x n + k 3. f ] x g = ] x + k gn

5.3 Exercises 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) (e) (f) (g) (h)

f ] x g = x 2 + 3x f ] x g = x2 - 4 p ] x g = x 2 + 5x + 6 y = x 2 - 8x + 15 p ] x g = x3 + 5

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.

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x+3 ]x ! 0 g x 2 (j) g ] x g = 9 - x (i) y =

2.

Show that f ] x g = f ] - x g where f ] x g = x 2 - 2. What type of function is it?

3.

If f ] x g = x 3 + 1, ﬁnd (a) f ^ x 2 h (b) 6 f (x) @ 2 (c) f ] - x g (d) Is it an even or odd function?

4.

Show that g ] x g = x 8 + 3x 4 - 2x 2 is an even function.

5.

Show that f (x) is odd, where f ] x g = x.

6.

Show that f ] x g = x 2 - 1 is an even function.

7.

Show that f ] x g = 4x - x 3 is an odd function.

8.

Prove that f ] x g = x 4 + x 2 is an even function and hence ﬁnd f ]x g - f ]-x g.

9.

Are these functions even, odd or neither? x3 (a) y = 4 x - x2 1 (b) y = 3 x -1 3 (c) f ] x g = 2 x -4 x-3 (d) y = x+3 x3 (e) f ] x g = 5 x - x2

10. If n is a positive integer, for what values of n is the function f ] x g = xn (a) even? (b) odd? 11. Can the function f ] x g = x n + x ever be (a) even? (b) odd?

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. y (a)

x

y

(b)

4

x

y

(c)

-2

2

x

Chapter 5 Functions and Graphs

y

(d)

(e)

y

4 2

-2

-1

1

2

x

x

-2 -4

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. 1. Sketch the families of graphs for these graphs with parameter k. (a) y = kx (b) y = kx 2 (c) y = kx 3 (d) y = kx 4 k (e) y = x What effect does the parameter k have on these graphs? Could you give a general comment about y = k f ] x g? 2. Sketch the families of graphs for these graphs with parameter k. (a) y = ] x + k g 2 (b) y = x 2 + k (c) y = x 3 + k (d) y = x 4 + k 1 (e) y = x + k What effect does the parameter k have on these graphs? Could you give a general comment about y = f ] x g + k? CONTINUED

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3. Sketch the families of graphs for these graphs with parameter k. (a) y = x + k (b) y = ] x + k g2 (c) y = ] x + k g3 (d) y = ] x + k g4 1 (e) y = x+k What effect does the parameter k have on these graphs? Could you give a general comment about y = f ] x + k g?

When k 2 0 , the graph moves to the left and when k 1 0 , the graph moves to the right.

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 + k g, as k varies, the graph moves left or right (horizontal translation). 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 ﬁnd the domain and range. You have already sketched some of these graphs in previous years.

Linear Function A linear function is a function whose graph is a straight line.

Gradient form: 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 ﬁnd an example of a straight line that is not a function? Are there any odd or even straight lines? What are their equations?

Chapter 5 Functions and Graphs

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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?

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 y For x-intercept: y = 0 6 0 = 3x - 5 5 5 = 3x 4

2 =x 3 For y-intercept: x = 0 1

y = 3 ]0 g - 5 = -5

3 2 1 -4 -3 -2 -1

-1

1 23 1

2

3

4

x

-2 -3 -4 -5

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}

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.

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.

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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. y 5 4 3 2 1 -4

-3

-2

-1 -1

1

2

3

4

x

-2 -3 -4 -5

Domain: " all real x , 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. y 5 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 -4 -5

Domain: " x: x = - 1 , Range: " all real y ,

1

3

4

x

Chapter 5 Functions and Graphs

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 ,

5.4 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 2x + 3y - 3 = 0 on the same set of axes, ﬁnd the point where they meet.

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f ] x g = ax 2 + bx + c is the general equation of a parabola. If a 2 0 the parabola is concave upwards

If a 1 0 the parabola is concave downwards

The pronumeral a is called the coefficient of x 2.

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.

Chapter 5 Functions and Graphs

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 ﬁnd 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 = kx 2 2. y = x 2 + k 3. y = ] x + k g2 4. y = x 2 + 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 = x 2 - 1, showing intercepts. (b) State the domain and range.

Solution (a) This is the graph of a parabola. Since a 2 0, it is concave upward For x-intercept: y = 0 0=x -1 1 = x2 !1 = x For y-intercept: x = 0 2

y = 02 - 1 = -1

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

1

3

4

5

-2 -3 -4 -5 -6

(b) From the graph, the curve is moving outwards and will extend to all real x values. The minimum y value is - 1. Domain: " all real x , Range: " y: y \$ -1 , 2. Sketch f ] x g = ] x + 1 g 2.

Solution This is a quadratic function. We ﬁnd the intercepts to see where the parabola will lie. Alternatively, you may know from your work on parameters that f ] x g = ] x + a g 2 will move the function f ] x g = x 2 horizontally a units to the left. So f ] x g = ] x + 1 g 2 moves the parabola f ] x g = x 2 1 unit to the left. For x-intercept: y = 0 0 = ]x + 1 g2 x+1=0 x = -1 For y-intercept: x = 0 y = ]0 + 1 g2 =1

x

Chapter 5 Functions and Graphs

227

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

1

2

3

4

x

-2 -3 -4 -5

3. For the quadratic function f ] x g = x 2 + x - 6 (a) Find the x- and y-intercepts (b) Find the minimum value of the function (c) State the domain and range (d) For what values of x is the curve decreasing?

Solution (a) For x-intercept: y = 0 This means f ] x g = 0 0 = x2 + x - 6 = ]x + 3 g]x - 2 g x + 3 = 0, x - 2 = 0 x = - 3, x = 2 For y-intercept: x = 0 f ] 0 g = ] 0 g2 + ] 0 g - 6 = -6 (b) Since a 2 0, 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: -3 + 2 1 =2 2 1 Minimum value is f c - m 2 1 1 2 1 f c- m = c- m + c- m - 6 2 2 2 1 1 = - -6 4 2 1 = -6 4 1 So the minimum value is - 6 . 4 CONTINUED

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(c) Sketching the quadratic function gives a concave upward parabola. y 5 4 3 2 1 -4 -3 -2 -1 -1

1

3

4

5

x

-2 -3 -4 -5 -6 -1 , -6 14 2

From the graph, notice that the parabola is gradually going outwards and will include all real x values. Since the minimum value is - 6 1 , all y values are greater than this. 4 Domain: " all real x , 1 Range: ' y: y \$ -6 1 4 (d) The curve decreases down to the minimum point and then 1 increases. So the curve is decreasing for all x 1 - . 2 4. (a) Find the x- and y-intercepts and the maximum value of the quadratic function f ] x g = - x 2 + 4x + 5. (b) Sketch the function and state the domain and range. (c) For what values of x is the curve increasing?

Solution (a) For x-intercept: y = 0 So f ]x g = 0 0 = - x 2 + 4x + 5 x 2 - 4x - 5 = 0 ]x - 5 g]x + 1 g = 0 x - 5 = 0, x + 1 = 0 x = 5, x = -1 For y-intercept: x = 0 f ] 0 g = - ] 0 g2 + 4 ] 0 g + 5 =5

Chapter 5 Functions and Graphs

Since a 1 0, the quadratic function is concave downwards and has a maximum value halfway between the x-intercepts x = - 1 and x = 5. -1 + 5 =2 2 f ]2 g = -]2 g 2+ 4 ]2 g + 5 =9 So the maximum value is 9. (b) Sketching the quadratic function gives a concave downward parabola. y

9 8 7 6 5 4 3 2 1 -4

-3

-2

-1 -1 -2 -3 -4 -5

1

2

3

4

5

6

x

From the graph, the function can take on all real numbers for x, but the maximum value for y is 9. Domain: " all real x , Range: " y: y # 9 , (c) From the graph, the function is increasing on the left of the maximum point and decreasing on the right. So the function is increasing when x 1 2.

5.5 Exercises 1.

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

2.

Sketch (a) y = x 2 + 2 (b) y = - x 2 + 1 (c) f ] x g = x 2 - 4 2 (d) y = x + 2x (e) y = - x 2 - x (f) f ] x g = ] x - 3 g 2

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(g) (h) (i) (j) 3.

f ] x g = ] x + 1 g2 y = x 2 + 3x - 4 y = 2x 2 - 5 x + 3 f ] x g = - x 2 + 3x - 2

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

4.

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

5.

Find the range of each function over the given domain. (a) y = x 2 for 0 # x # 3 (b) y = - x 2 + 4 for -1 # x # 2 (c) f ] x g = x 2 - 1 for -2 # x # 5 (d) y = x 2 + 2x - 3 for -2 # x # 4 (e) y = - x 2 - x + 2 for 0 # x # 4

6.

Find the domain over which each function is (i) increasing (ii) decreasing (a) y = x 2 (b) y = - x 2 (c) f ] x g = x 2 - 9 (d) y = - x 2 + 4x (e) f ] x g = ] x + 5 g2

7.

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

8.

State whether these functions are even or odd or neither. (a) y = x 2 + 1 (b) f ] x g = x 2 - 3 (c) y = -2x 2 (d) f ] x g = x 2 - 3x (e) f ] x g = x 2 + x (f) y = x 2 - 4 (g) y = x 2 - 2x - 3 (h) y = x 2 - 5x + 4 (i) p ] x g = ] x + 1 g 2 (j) y = ] x - 2 g 2

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 deﬁnition of absolute value.

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 ﬁrst 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 ﬁnd other values.

Chapter 5 Functions and Graphs

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

-3

-2

-1

y

2

1

0

0

1

2

3

-1

0

1

2

This gives a v-shaped graph. y 5 4 3 2 1 -4

-3

-2

-1 -1

1

2

3

4

x

-2 -3 -4 -5

Method 2: Use the deﬁnition of absolute value when x \$ 0 x-1 y = | x | - 1 = &x - 1 when x 1 0 This gives 2 straight line graphs: y = x - 1 ]x \$ 0 g y

5 y=x-1

4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

CONTINUED

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y = -x - 1 ] x 1 0 g y y=-x-1

5 4 3 2 1

-4 -3 -2 -1 -1

1

2

3

x

4

-2 -3 -4 -5

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

5

y=x-1

4 3 2 1 -4 -3

-2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

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

Chapter 5 Functions and Graphs

The graph is V-shaped, passing through these intercepts. y 5 4 3 2 1 -4 -3

-2 -1 -1

1

2

3

4

5

x 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.

-2 -3 -4 -5

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 deﬁnition of absolute value. +2 when x + 2 \$ 0 y = | x + 2 | = 'x - (x + 2) when x + 2 1 0 This gives 2 straight lines: y = x + 2 when x + 2 \$ 0 x \$ -2 y 5

y=x+2

4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

CONTINUED

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y = - ] x + 2 g when x + 2 1 0 i.e. y = - x - 2 when x 1 -2 y 5 4

y = -x - 2

3 2 1

-4

-3

-2

-1 -1

1

2

3

4

x

-2 -3 -4 -5

Draw these on the same number plane and then disregard the dotted lines. y 5 y = -x - 2

y=x+2

4 3 2 1

-4 -3

-2 -1 -1 -2 -3 -4 -5

Method 2: Find intercepts For x-intercept: y = 0 So f ] x g = 0 There is only one solution for the equation | x + 2 | = 0. Can you see why?

0 =| x + 2 | 0=x+2 -2 = x For y-intercept: x = 0 f (0) = | 0 + 2 | =2

1

2

3

4

x

Chapter 5 Functions and Graphs

The graph is V-shaped, passing through these intercepts. y 5 4 3 2 1 -4

-3 -2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

Investigation Are graphs that involve absolute value always functions? Can you ﬁnd an example of one that is not a function? Can you ﬁnd 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.

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 | .

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EXAMPLES 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 graph shows that there is only one solution. Algebraically, you need to find the 2 possible solutions and then check them.

The solution is x = 3. 3. | x + 1 | 1 2

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

Chapter 5 Functions and Graphs

The solution of | x + 1 | 1 2 is where the graph y = | x + 1 | is below the graph y = 2, that is, - 3 1 x 1 1.

5.6 Exercises 1.

2.

3.

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 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 Find the domain and range of each function. (a) y = | x - 1 | (b) f ] x g = | x | - 8

(c) (d) (e)

f ] x g = | 2x + 5 | y = 2 | x |- 3 f ] x g = -| x - 3 |

4.

Find the domain over which each function is (i) increasing (ii) decreasing (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, ﬁnd the range of each function. (a) y = | x | for - 2 # x # 2 (b) f ] x g = - | x | - 4 for -4 # x # 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 |

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7.

(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

Solve graphically (a) | x | = 3 (b) | x | 2 1 (c) | x | # 2 (d) | x + 2 | = 1 (e) | x - 3 | = 0 (f) | 2x - 3 | = 1 (g) | x - 1 | 1 4 (h) | x + 1 | # 3 (i) | x - 2 | 2 2 (j) | x - 3 | \$ 1

The Hyperbola a A hyperbola is a function with its equation in the form xy = a or y = x .

EXAMPLE 1 Sketch y = x .

Solution 1 y = x is a discontinuous curve since the function is undeﬁned at x = 0. Drawing up a table of values gives: x

-3

y

-

1 3

1 2

-2

-1

-

1 2

-1

-2

-

1 4

0

1 4

1 2

1

2

3

-4

4

2

1

1 2

1 3

-

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 y is never 0. Why?

Chapter 5 Functions and Graphs

To sketch the graph of a more general hyperbola, we can use the domain and range to help ﬁnd 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 ﬁnd 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 k 1. y = x 1 2. y = x + k 3. y =

1 x+k

What effect does the parameter k have on these graphs?

EXAMPLES 3 . x-3 (b) Hence sketch the graph of the function.

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

Solution This is the equation of a hyperbola. To ﬁnd 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. CONTINUED

The denominator cannot be zero.

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To make the graph more accurate we can ﬁnd another point or two. The easiest one to ﬁnd is the y-intercept. For y-intercept, x = 0 3 y= 0-3 = -1 y 5 4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

-2 -3

Notice that this graph is 3 a translation of y = x three units to the right.

5

y=0

x

Asymptotes x=3

-4 -5

2. Sketch y = -

1 . 2x + 4

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 ﬁnd two or three points on the curve. To ﬁnd the domain, we notice that 2x + 4 ! 0. 2x ! - 4 x ! -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 ﬁnd the y-intercept. For y-intercept, x = 0 1 2 ( 0) + 4 1 =4

y=-

Chapter 5 Functions and Graphs

y

x

-2 - 14

a is a hyperbola with bx + c c domain & all real x: x ! - 0 and b

The function f ] x g =

range {all real y: y ! 0}

5.7 Exercises 1.

For each graph (i) State the domain and range. (ii) Find the y-intercept if it exists. (iii) Sketch the graph. 2 (a) y = x 1 (b) y = - x 1 (c) f ] x g = x+1 3 (d) f ] x g = x-2 1 (e) y = 3x + 6 2 (f) f ] x g = x-3 4 (g) f ] x g = x-1

2 x+1 2 (i) f ] x g = 6x - 3 6 (j) y = x+2

(h) y = -

2.

3.

2 Show that f ] x g = x is an odd function. Find the range of each function over the given domain. 1 (a) f ] x g = for -2 # x # 2 2x + 5 1 (b) y = for -2 # x # 0 x+3 5 (c) f ] x g = for - 3 # x # 1 2x - 4

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

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

(d) f ] x g = -

4.

Find the domain of each function over the given range. 3 (a) y = x for 1 # y # 3 2 1 (b) y = - x for - 2 # y # 2

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 x 2 + ax + y 2 + by + c = 0 has the shape of a circle. There is a special case of this formula:

The graph of x 2 + y 2 = r 2 is a circle, centre ^ 0, 0 h and radius r

Proof y

(x, y) r x

y x

Chapter 5 Functions and Graphs

243

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: c2 = a2 + b2 ` r2 = x2 + y2

EXAMPLE (a) Sketch the graph of x 2 + y 2 = 4. Is it a function? (b) State its domain and range.

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

2

-2

2

x

-2

The circle is not a function since a vertical line will cut it in more than one place. y

2

-2

2

x

-2

CONTINUED

4.

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Maths In Focus Mathematics Preliminary Course

(b) Notice that the 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 x 2 + y 2 = r 2 has domain: ! x: -r # x # r + and range: " y: -r # y # r ,

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

The equation of a circle, centre (a, b) and radius r is ] x – a g2 + ^ y – b h2 = r 2

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

(x, y)

y r b

y-b

x-a

(a, b) a

x

x

Notice that the small sides of the triangle are x – a and y – b and the hypotenuse is r, the radius. By Pythagoras’ theorem: c2 = a2 + b2 r 2 = ] x – a g2 + ^ y – b h2

Chapter 5 Functions and Graphs

EXAMPLES 1. (a) Sketch the graph of x 2 + y 2 = 81. (b) State its domain and range.

Solution (a) The equation is in the form x 2 + y 2 = r 2. This is a circle, centre (0, 0) and radius 9. y

9

-9

9

x

-9

(b) From the graph, we can see all the values that are possible for x and y for the circle. Domain: {x: -9 # x # 9} Range: {y: -9 # y # 9} 2. (a) Sketch the circle ] x – 1 g2 + ^ y + 2 h2 = 4. (b) State its domain and range.

Solution (a) The equation is in the form ] x – a g2 + ^ y – b h2 = r 2. ] x – 1 g 2 + ^ y + 2 h2 = 4 ] x – 1 g 2 + _ y – ] - 2 g i2 = 2 2

So a = 1, b = - 2 and r = 2 CONTINUED

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This is a circle with centre ^ 1, - 2 h and radius 2. To draw the circle, plot the centre point ^ 1, - 2 h and count 2 units up, down, left and right to ﬁnd points on the circle. y 5 4 3 2 1 x

-4 -3 -2

-1 -1 -2 -3

1

2

3

4

(1, -2)

-4 -5

(b) From the graph, we can see all the values that are possible for 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 – a g2 + ^ y – b h2 = r 2 where a = - 2, b = 1 and r = 3. Substituting: ] x – a g2 + ^ y – b h2 = r 2 You may need to revise this in Chapter 2.

] x - ] - 2 g g2 + ^ y – 1 h2 = 3 2 ] x + 2 g2 + ^ y – 1 h2 = 9 Remove the grouping symbols. ] a + b g2 = a 2 + 2ab + b 2 So ] x + 2 g2 = x 2 + 2 ] x g ] 2 g + 2 2 = x 2 + 4x + 4 2 ] a – b g = a 2 - 2ab + b 2 So ^ y – 1 h2 = y 2 - 2 ^ y h ] 1 g + 1 2 = y 2 - 2y + 1 The equation of the circle is: x 2 + 4x + 4 + y - 2y + 1 = 9 x 2 + 4 x + y - 2y + 5 = 9 x 2 + 4 x + y – 2y + 5 - 9 = 9 - 9 x 2 + 4x + y - 2y - 4 = 0

Chapter 5 Functions and Graphs

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?

By rearranging the equation of a circle, we can also ﬁnd the equations of semi-circles. The equation of the semi-circle above the x-axis with centre (0, 0) and radius r is y = r 2 - x 2 The equation of the semi-circle below the x-axis with centre (0, 0) and radius r is y = - r 2 - x 2

Proof x2 + y2 = r2 y2 = r2 – x2 y = ! r2 - x2 This gives two functions:

y = r 2 - x 2 is the semi-circle above the x-axis since its range is y \$ 0 for all values. y

r

-r

r

x

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

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y = - r 2 - x 2 is the semi-circle above the x-axis since its range is y # 0 for all values. y

-r

r

x

-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 = r 2 - x 2 where r = 3. It is a semi-circle above the x-axis with centre (0, 0) and radius 3. y

3

-3

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

3

x

Chapter 5 Functions and Graphs

2. y = - 4 - x 2

Solution This is in the form y = - r 2 - x 2 where r = 2. It is a semi-circle below the x-axis with centre (0, 0) and radius 2. y

-2

2

x

-2

Domain: {x: -2 # x # 2} Range: {y: -2 # y # 0}

5.8 Exercises 1.

2.

For each of the following (i) sketch each graph (ii) state the domain and range. (a) x 2 + y 2 = 9 (b) x 2 + y 2 - 16 = 0 (c) ] x – 2 g2 + ^ y – 1 h2 = 4 (d) ] x + 1 g2 + y 2 = 9 (e) ] x + 2 g2 + ^ y – 1 h2 = 1 For each semi-circle (i) state whether it is above or below the x-axis (ii) sketch the function (iii) state the domain and range.

(a) (b) (c) (d) (e) 3.

y = - 25 - x 2 y = 1 - x2 y = 36 - x 2 y = - 64 - x 2 y = - 7 - x2

Find the length of the radius and the coordinates of the centre of each circle. (a) x 2 + y 2 = 100 (b) x 2 + y 2 = 5 (c) ] x – 4 g2 + ^ y – 5 h2 = 16 (d) ] x – 5 g2 + ^ y + 6 h2 = 49 (e) x 2 + ^ y – 3 h2 = 81

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4.

(e) (f) (g) (h) (i) (j)

Find the equation of each circle in expanded form (without grouping symbols). (a) Centre (0, 0) and radius 4 (b) Centre (3, 2) and radius 5 (c) Centre ^ -1, 5 h and radius 3 (d) Centre (2, 3) and radius 6

Centre ^ -4, 2 h and radius 5 Centre ^ 0, -2 h and radius 1 Centre (4, 2) and radius 7 Centre ^ -3, -4 h and radius 9 Centre ^ -2, 0 h and radius 5 Centre ^ -4, -7 h and radius 3

Other Graphs There are many other different types of graphs. We will look at some of these graphs and explore their domain and range. You will meet these graphs again in the HSC Course.

Exponential and logarithmic functions EXAMPLES 1. Sketch the graph of f ] x g = 3 x and state its domain and range.

Solution If you do not know what this graph looks like, draw up a table of values. You may need to revise the indices that you studied in Chapter 1. e.g. When x = 0: y = 3c =1 When x = -1: y = 3-1 1 = 1 3 1 = 3 x y

-3 1 27

-2 1 9

-1 1 3

0

1

2

3

1

3

9

27

If you already know what the shape of the graph is, you can draw it just using 2 or 3 points to make it more accurate.

Chapter 5 Functions and Graphs

You learned about exponential graphs in earlier stages of maths.

This is an exponential function with y-intercept 1. We can ﬁnd one other point. When x = 1 y = 31 =3

y

3 2 1 x

1

From the graph, x can be any real value (the equation shows this as well since any x value substituted into the equation will give a value for y). From the graph, y is always positive, which can be conﬁrmed by substituting different values of x into the equation. Domain: " all real x , Range: " y: y 2 0 , 2. Sketch f ] x g = log x and state the domain and range.

Solution Use the LOG key on your calculator to complete the table of values. Notice that you can’t ﬁnd the log of 0 or a negative number. x

−2

−1

0

0.5

1

2

3

4

y

#

#

#

−0.3

0

0.3

0.5

0.6

y

2 1

-1

1

2

3

4

x

From the graph and by trying different values on the calculator, y can be any real number while x is always positive. Domain: ! x: x 2 0 + Range: " all real y ,

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The exponential function y = a x has domain {all real x} and range {y: y 2 0} The logarithmic function y = log a x has domain ! x: x 2 0 + and range {all real y}

Cubic function A cubic function has an equation where the highest power of x is x 3 .

EXAMPLE 1. Sketch the function f ] x g = x 3 + 2 and state its domain and range.

Solution Draw up a table of values. x

−3

−2

−1

0

1

2

3

y

−25

−6

1

2

3

10

29

y 5 4 3 2 1

-4

-3

-2

-1

-1

1

-2 If you already know the shape of y = x 3, f (x) = x 3 + 2 has the same shape as f (x) = x 3 but it is translated 2 units up (this gives a y-intercept of 2).

-3 -4 -5

The function can have any real x or y value: Domain: " all real x , Range: " all real y ,

2

3

4

x

Chapter 5 Functions and Graphs

Domain and range Sometimes there is a restricted domain that affects the range of a function.

EXAMPLE 1. Find the range of f ] x g = x 3 + 2 over the given domain of -1 # x # 4.

Solution The graph of f ] x g = x 3 + 2 is the cubic function in the previous example. From the graph, the range is {all real y}. However, with a restricted domain of -1 # x # 4 we need to see where the endpoints of this function are. f ] -1 g = ] -1 g3 + 2 = -1 + 2 =1 f ] 4 g = ] 4 g3 + 2 = 64 + 2 = 66 Sketching the graph, we can see that the values of y all lie between these points. y

(4, 66)

(-1, 1) x

Range: " y: 1 # y # 66 ,

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You may not know what a function looks like on a graph, but you can still ﬁnd its domain and range by looking at its equation. When ﬁnding the domain, we look for values of x that are impossible. For example, with the hyperbola you have already seen that the denominator of a fraction cannot be zero. For the range, we look for the results when different values of x are substituted into the equation. For example, x2 will always give zero or a positive number.

EXAMPLE Find the domain and range of f ] x g =

x - 4.

Solution We can only ﬁnd the square root of a positive number or zero. So x – 4 \$ 0 x\$4 When you take the square root of a number, the answer is always positive (or zero). So y \$ 0 Domain: ! x: x \$ 4 + Range: " y: y \$ 0 ,

5.9 Exercises 1.

(c) f ] x g = | 2x - 3 |

Find the domain and range of (a) y = 4x + 3 (b) f ] x g = -4

(d) y = | x | - 2 (e) f ] x g = - 2x + 5

(c) x = 3 (d) f ] x g = 4x 2 – 1 (e) p ] x g = x 3 – 2

You may like to simplify the function by dividing by x.

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

(f) f ] x g = 12 - x - x 2 (g) x 2 + y 2 = 64 3 t-4 2 (i) g (z) = + 5 z (j) f ] x g = | x |

(h) f ] t g =

2.

Find the domain and range of (a) y = x (b) y =

x-2

3.

Find the x-intercepts of (a) y = x ] x - 5 g2 (b) f ] x g = ] x – 1 g ] x – 2 g ] x + 3 g (c) y = x 3 - 6x 2 + 8x (d) g ] x g = x 4 - 16x 2 (e) x 2 + y 2 = 49

Chapter 5 Functions and Graphs

4.

(a) Solve 1 - x 2 \$ 0. (b) Find the domain of f ] x g = 1 - x2 .

5.

Find the domain of (a) y = x 2 - x - 2 (b) g ] t g = t 2 + 6t

6.

Each of the graphs has a restricted domain. Find the range in each case. (a) y = 2x - 3 in the domain -3 # x # 3 (b) y = x 2 in the domain -2 # x # 3 (c) f ] x g = x 3 in the domain

9.

x Given the function f ] x g = x (a) ﬁnd the domain of the function (b) ﬁnd its range. Draw each graph on a number plane (a) f ] x g = x 4 (b) y = - x 3 (c) y = x 4 - 3 (d) p ] x g = 2x 3 (e) g ] x g = x 3 + 1 (f) x 2 + y 2 = 100 (g) y = 2 x + 1

-2 # x # 1 1 (d) y = x in the domain 1# x #5

10. (a) Find the domain and range of y = x - 1. (b) Sketch the graph of y = x - 1 .

(e) y = | x | in the domain 0#x#4

11. Sketch the graph of y = 5 x .

(f) y = x 2 - 2x in the domain -3 # x # 3

12. For each function, state (i) its domain and range (ii) the domain over which the function is increasing (iii) the domain over which the function is decreasing. (a) y = 2x - 9 (b) f ] x g = x 2 - 2 1 (c) y = x (d) f ] x g = x 3 (e) f ] x g = 3 x

(g) y = - x 2 in the domain -1 # x # 1 (h) y = x 2 - 1 in the domain -2 # x # 3 (i) y = x 2 - 2x - 3 in the domain -4 # x # 4 (j) y = - x 2 + 7x - 6 in the domain 0 # x # 7 7.

8.

(a) Find the domain for the 3 function y = . x+1 (b) Explain why there is no x- intercept for the function. (c) State the range of the function.

13. (a) Solve 4 - x 2 \$ 0. (b) Find the domain and range of (i) y = 4 - x 2 (ii) y = - 4 - x 2 .

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DID YOU KNOW? A lampshade can produce a hyperbola where the light meets the flat wall. • Can you find any other shapes made by a light?

Lamp casting its light

Limits and Continuity Limits

A line that a graph approaches but never touches is called an asymptote.

The exponential function and the hyperbola are examples of functions that approach a limit. The curve y = a x approaches the x-axis when x approaches very large negative numbers, but never touches it. That is, when x " - 3, a x " 0. Putting a - 3 into index form gives 1 a-3 = 3 a 1 =3 Z0 We say that the limit of a x as x approaches -3 is 0. In symbols, we write lim a x = 0. x " -3

EXAMPLES 1. Find lim x "0

x 2 + 5x . x

Solution 0 , which is undeﬁned. 0 Factorising and cancelling help us ﬁnd the limit. x 1 ]x + 5 g x 2 + 5x lim lim = x x "0 x "0 x1 = lim (x + 5) Substituting x = 0 into the function gives

x "0

=5

Chapter 5 Functions and Graphs

2. Find lim x "2

x-2 . x2 - 4

Solution Substituting x = 2 into the function gives

0 , which is undeﬁned. 0

x-2 x-2 = lim 2 1 x " 2 x -4 ^x + 2h _x - 2i 1 = lim x "2 x + 2 1 = 4 1

lim x "2

3. Find lim h "0

2h 2 x + hx 2 - 7h . h

Solution lim h "0

h ^ 2hx + x 2 - 7 h 2h 2 x + hx 2 - 7h = lim h "0 h h = lim 2hx + x 2 - 7 h "0

= x2 - 7

Continuity Many functions are continuous. That is, they have a smooth, unbroken curve (or line). However, there are some discontinuous functions that have gaps in their graphs. The hyperbola is an example. If a curve is discontinuous at a certain point, we can use limits to ﬁnd the value that the curve approaches at that point.

EXAMPLES 1. Find lim x "1

y=

x2 - 1 and hence describe the domain and range of the curve x-1

x -1 . Sketch the curve. x-1 2

Solution Substituting x = 1 into

x2 - 1 0 gives x-1 0 CONTINUED

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lim x "1

]x + 1 g]x - 1 g x2 - 1 = lim x-1 x-1 x-1 = lim (x + 1) x "1

=2 y=

x2 - 1 is discontinuous at x = 1 since y is undeﬁned at that point. x-1

This leaves a gap in the curve. The limit tells us that y " 2 as x " 1, so the gap is at ^ 1, 2 h . Domain: " x: all real x, x ! 1 , Range: " y: all real y, y ! 2 , y= =

x2 - 1 x-1 ^x + 1h ^x - 1h

x-1 =x+1 the graph is y = x + 1 where x ! 1

Remember that x ! 1.

`

2. Find lim

x " -2

x2 + x - 2 x2 + x - 2 and hence sketch the curve y = . x+2 x+2

Solution Substituting x = -2 into lim

x " -2

x2 + x - 2 0 gives x+2 0

^x - 1h ^x + 2h x2 + x - 2 = lim x " 2 x+2 ^x + 2h = lim ^ x - 1 h x " -2 = -3

y= y=

x2 + x - 2 is discontinuous at x = - 2 x+2 ^x + 2h ^x - 1h

x+2 =x-1 So the function is y = x - 1 where x ! -2. It is discontinuous at ^ -2, -3 h .

Chapter 5 Functions and Graphs

5.10 Exercises 1.

Find (a) lim x 2 + 5

2.

x "4

(b) lim t - 7 t " -3

(c) lim x 3 + 2x - 4 x "2

(d) lim

x 2 + 3x x

(e) lim

h2 - h - 2 h-2

(f) lim

y 3 - 125 y-5

(g) lim

x 2 + 2x + 1 x+1

(h) lim

x 2 + 2x - 8 x+4

x "0

h "2

y "5

x "-1

x " -4

Determine which of these functions are discontinuous and ﬁnd x values for which they are discontinuous. (a) y = x 2 - 3 1 (b) y = x+1 (c) f ] x g =

x-1 1 (d) y = 2 x +4 1 (e) y = 2 x -4 3.

Sketch these functions, showing any points of discontinuity. (a) y =

x 2 + 3x x

(i) lim

c-2 c2 - 4

(b) y =

(j) lim

x-1 x2 - x

x 2 + 3x x+3

(c) y =

x 2 + 5x + 4 x+1

(k) lim

h 3 + 2h 2 - 7h h

(l) lim

hx 2 - 3hx + h 2 h

(m) lim

2hx 3 - h 2 x 2 + 3hx - 5h h

c "2

x "1

h "0

h "0

h "0

x3 - c3 (n) lim x "c x - c

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Regions Class Investigation How many solutions are there for y \$ x + 2? How would you record them all?

Inequalities can be shown as regions in the Cartesian plane. You can shade regions on a number plane that involve either linear or non-linear graphs. This means that we can have regions bounded by a circle or a parabola, or any of the other graphs you have drawn in this chapter. Regions can be bounded or unbounded. A bounded region means that the line or curve is included in the region.

EXAMPLE Sketch the region x # 3.

Solution Remember that x = 3 is a vertical line with x-intercept 3.

x # 3 includes both x = 3 and x 1 3 in the region. Sketch x = 3 as an unbroken or ﬁlled in line, as it will be included in the region. Shade in all points where x 1 3 as shown. y 5 4 3 2 1 -4 -3 -2

-1 -1

1

2

3

4

-2 -3 -4 -5

x=3

x

Chapter 5 Functions and Graphs

261

An unbounded region means that the line or curve is not included in the region.

EXAMPLE Sketch the region y 2 -1.

Solution y 2 -1 doesn’t include y = -1. When this happens, it is an unbounded region and we draw the line y = -1 as a broken line to show it is not included. Sketch y = -1 as a broken line and shade in all points where y 2 -1 as shown. y 5 4 3 2 1 -4 -3 -2 -1 y = -1 -1

1

2

3

4

x

-2 -3 -4 -5

For lines that are not horizontal or vertical, or for curves, we need to check a point to see if it lies in the region.

Remember that y = -1 is a horizontal line with y-intercept -1.

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EXAMPLES Find the region deﬁned by 1. y \$ x + 2

Solution First sketch y = x + 2 as an unbroken line. On one side of the line, y 2 x + 2 and on the other side, y 1 x + 2. To ﬁnd which side gives y 2 x + 2, test a point on one side of the line (not on the line). For example, choose ^ 0, 0 h and substitute into y\$x+2 0\$0+2 0\$2 (false) This means that ^ 0, 0 h does not lie in the region y \$ x + 2. The region is on the other side of the line.

Any point in the region will make the inequality true. Test one to see this.

2. 2x - 3y 1 6

Solution First sketch 2x - 3y = 6 as a broken line, as it is not included in the region. To ﬁnd which side of the line gives 2x - 3y 1 6, test a point on one side of the line. For example, choose ^ 0, 1 h and substitute into 2x - 3y 1 6 2 ] 0 g - 3 (1 ) 1 6 -3 1 6 (true)

Chapter 5 Functions and Graphs

This means that ^ 0, 1 h lies in the region 2x - 3y 1 6.

2x - 3y = 6

3. x 2 + y 2 2 1

Solution The equation x 2 + y 2 = 1 is a circle, radius 1 and centre ^ 0, 0 h . Draw x 2 + y 2 = 1 as a broken line, since the region does not include the curve. Choose a point inside the circle, say ^ 0, 0 h x2 + y2 2 1 02 + 02 2 1 0 2 1 (false) So the region lies outside the circle.

4. y \$ x 2

Solution The equation y = x 2 is a parabola. Sketch this as an unbroken line, as it is included in the region.

CONTINUED

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Choose a point inside the parabola, say ^ 1, 3 h. y \$ x2 3 2 12 3 2 1 (true) So ^ 1, 3 h lies in the region. y = x2

Sometimes a region includes two or more inequalities. When this happens, sketch each region on the number plane, and the ﬁnal region is where they overlap (intersect).

EXAMPLE Sketch the region x # 4, y 2 -2 and y # x 2 .

Solution Draw the three regions, either separately or on the same set of axes, and see where they overlap.

Chapter 5 Functions and Graphs

Put the three regions together.

If you are given a region, you should also be able to describe it algebraically.

EXAMPLES Describe each region. 1.

y

6 5 4 3 2 1 -4 -3 -2

-1 -1

1

2

3

4

x

-2 -3 -4

Solution The shaded area is below and including y = 6 so can be described as y # 6. It is also to the left of, but not including the line x = 4, which can be described as x 1 4. The region is the intersection of these two regions: y # 6 and x 1 4

CONTINUED

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y

2. 2

-2

2

x

-2

Solution The shaded area is the interior of the circle, centre (0, 0) and radius 2 but it does not include the circle. The equation of the circle is x 2 + y 2 = 2 2 or x 2 + y 2 = 4. You may know (or guess) the inequality for the inside of the circle. If you are unsure, choose a point inside the circle and substitute into the equation e.g. (0, 0). LHS = x 2 + y 2 = 02 + 02 =0 1 4 ] RHS g So the region is x 2 + y 2 1 4.

5.11 Exercises 1.

Shade the region deﬁned by (a) x # 2 (b) x 2 1 (c) y \$ 0 (d) y 1 5 (e) y # x + 1

(f) (g) (h) (i) (j)

y \$ 2x - 3 x+y21 3x - y - 6 1 0 x + 2y - 2 \$ 0 2x - 1 1 0

Chapter 5 Functions and Graphs

2.

Write an inequation to describe each region. (a)

(d)

y 5

y

y = x2 - 4

4 3

6

2

5

1

4

-4 -3 -2 -1 -1

3 2

1

2

3

4

5

-2

1 -4 -3 -2 -1 -1

1

2

3

-3

x

4

-4 -5

-2 -3

(e)

-4

y y = 2x

(b)

y

3

6

2

5

1

4 3

1

2 1 -4 -3 -2 -1 -1

1

2

3

x

4

-2 -3

3.

Shade each region described. (a) y 2 x 2 – 1 (b) x 2 + y 2 # 9 (c) x 2 + y 2 \$ 1 (d) y # x 2 (e) y 1 x 3

4.

Describe as an inequality (a) the set of points that lie below the line y = 3x - 2 (b) the set of points that lie inside the parabola y = x 2 + 2 (c) the interior of a circle with radius 7 and centre (0, 0) (d) the exterior of a circle with radius 9 and centre (0, 0) (e) the set of points that lie to the left of the line x = 5 and above the line y = 2

-4

(c)

y

6 5

y=x+1

4 3 2 1 -4 -3 -2 -1 -1

-2 -3 -4

1

2

3

4

x

x

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The first quadrant is where x and y values are both positive.

5.

Shade the region (a) ] x - 2 g2 + y 2 # 4 (b) ] x - 1 g2 + ^ y - 2 h2 # 1 (c) ] x + 2 g2 + ^ y - 1 h2 2 9

6.

Shade the intersection of these regions. (a) x # 3, y \$ -1 (b) x \$ -3, y 2 x -3 (c) y # 1, y \$ 3x - 5 (d) y 2 x + 1, y # 3 - x (e) y # 1, x 2 + y 2 # 9 (f) x 2 -1, x 2 + y 2 1 4 (g) y # 4, y \$ x 2 (h) x 1 -2, y # 3, y 2 x 3 (i) y # 0, x 2 + y 2 \$ 1 (j) -1 1 x - y # 2

7.

Shade the region bounded by (a) the curve y = x 2, the x-axis and the lines x = 1 and x = 3 (b) the curve y = x 3, the y-axis and the lines y = 0 and y = 1

(c) the curve x 2 + y 2 = 4, the x-axis and the lines x = 0 and x = 1 in the ﬁrst quadrant 2 (d) the curve y = x , the x-axis and the lines x = 1 and x = 4 1 , the x+2 x-axis and the lines x = 0 and

(e) the curve y = x=2 8.

Shade the regions bounded by the intersection of (a) x 1 2, y 1 5 and y # x 2 (b) x 1 3, y \$ -1, y # x - 2 (c) y # 1 - x, y # 2x + 1, 2x - 3y # 6 (d) x \$ -3, y # 2, x 2 + y 2 \$ 9 (e) x 1 2, y # 3, y \$ | x |

Application Regions are used in business applications to find optimum profit. Two (or more) equations are graphed together, and the region where a profit is made is shaded. The optimum profit occurs at the endpoints (or vertices) of the region.

EXAMPLE A company makes both roller skates (X ) and ice skates (Y ). Roller skates make a \$25 profit, while ice skates make a profit of \$21. Each pair of roller skates spends 2 hours on machine A (available 12 hours per day) and 2 hours on machine B (available 8 hours per day). Each pair of ice skates spends 3 hours on machine A and 1 hour on machine B. How many skates of each type should be made each day to give the greatest profit while making the most efficient use of the machines?

Chapter 5 Functions and Graphs

SOLUTION Profit P = \$25 X + \$21Y Machine A: 2X + 3Y # 12 Machine B: 2X + Y # 8 Sketch the regions and find the point of intersection of the lines.

The shaded area shows all possible ways of making a profit. Optimum profit occurs at one of the endpoints of the regions. (0, 4): P = \$25 ] 0 g + \$21 ] 4 g = \$84 (4, 0): P = \$25 ] 4 g + \$21 ] 0 g = \$100 (3, 2): P = \$25 ] 3 g + \$21 ] 2 g = \$117

^ 3, 2 h gives the greatest profit, so 3 pairs of roller skates and 2 pairs of ice skates each day gives optimum profit.

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Test Yourself 5 1.

If f ] x g = x 2 - 3x - 4, ﬁnd (a) f ] -2 g (b) f ] a g (c) x when f ] x g = 0

2.

Sketch each graph (a) y = x 2 - 3x - 4 (b) f ] x g = x 3 (c) x 2 + y 2 = 1 (d) y = 1 - x 2 (e) y = - 1 - x 2 2 (f) y = x (g) 2x - 5y + 10 = 0 (h) y = | x + 2 |

(b)

3.

Find the domain and range of each graph in question 2.

4.

If f ] x g = *

2x

if x \$ 1

x -3

if x 1 1

2

11. Describe each region (a)

ﬁnd f ] 5 g - f ] 0 g + f ] 1 g 3 5.

if x 2 3

Given f ] x g = * x if 1 # x # 3 2 - x if x 1 1 ﬁnd (a) f ] 2 g (b) f ] -3 g (c) f ] 3 g (d) f ] 5 g (e) f ] 0 g 2

6.

Shade the region y \$ 2x + 1.

7.

Shade the region where x 1 3 and y \$ -1.

8.

Shade the region given by x 2 + y 2 \$ 1.

9.

Shade the region given by 2x + 3y - 6 # 0 and x \$ -2.

10. Shade the region y 2 x + 1 and x + y # 2.

(c)

12. (a) Write down the domain and range of 2 the curve y = . x-3 2 (b) Sketch the graph of y = . x-3

Chapter 5 Functions and Graphs

13. (a) Sketch the graph y = | x + 1 |. (b) From the graph, solve (i) | x + 1 | = 3 (ii) | x + 1 | 1 3 (iii) | x + 1 | 2 3

17. Find (a) lim

x 2 - 2x - 3 x-3

(b) lim

2x x + 5x

(c) lim

x3 + 1 x2 - 1

x "3

x "0

14. If f ] x g = 3x - 4, ﬁnd (a) f ] 2 g (b) x when f ] x g = 7 (c) x when f ] x g = 0

2

x " -1

(d) lim h "0

15. Find the x- and y-intercepts of (a) 2x - 5y + 20 = 0 (b) y = x 2 - 5x - 14 16. State which functions are (i) even (ii) odd (iii) neither even nor odd. (a) y = x 2 - 1 (b) y = x + 1 (c) y = x 3 (d) y = x 4 (e) y = 2 x

2xh 2 + 3h h

18. Sketch y = 10 x, y = log x and y = x on the same number plane. 19. (a) State the domain and range of y = 2x - 4 . (b) Sketch the graph of y = 2x - 4 . 20. Show that (a) f ] x g = x 4 + 3x 2 - 1 is even (b) f ] x g = x 3 - x is odd.

Challenge Exercise 5 1.

Find the values of b if f ] x g = 3x 2 - 7x + 1 and f ] b g = 7.

2.

6.

Sketch y = ] x + 2 g2 - 1 in the domain -3 # x # 0.

Find the domain and range of 1 y= 2 . x -1

7.

Sketch the region x 1 y, x + 2y 1 6, x + 2y - 4 \$ 0.

3.

Sketch the region y 2 quadrant.

8.

Find the domain and range of x 2 = y in the ﬁrst quadrant.

4.

Draw the graph of y = | x | + 3x - 4.

9.

If f ] x g = 2x 3 - 2x 2 - 12x, ﬁnd x when f ] x g = 0.

5.

4 - x 2 in the ﬁrst

Z 2x + 3 when x 2 2 ] f ] x g = [1 when -2 # x # 2 ] 2 x when x 1 -2 \ Find f ] 3 g, f ] -4 g, f ] 0 g and sketch the curve.

1 10. Sketch the region deﬁned by y 2 x+2 in the ﬁrst quadrant.

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11. If h ] t g = )

1 - t2 t2 - 1

(b) Find the domain and range of

if t 2 1 if t # 1

y=

ﬁnd the value of h ] 2 g + h ] -1 g - h ] 0 g and sketch the curve. 12. Sketch y =

1 - x in the ﬁrst quadrant. 2

13. Sketch the region y \$ x - 5, y 1 x + x. 2

14. If f ] x g = 2x - 1, show that f ^ a 2 h = f _ (-a)2 i for all real a. 15. Find the values of x for which f ] x g = 0 when f ] x g = 2x 2 - x - 5 (give exact answers). 16. (a) Show that

2x + 7 1 =2+ . x+3 x+3

2x + 7 . x+3

(c) Hence sketch the graph of y=

2x + 7 . x+3

17. Sketch y = 2 x - 1 . 18. Sketch y =

|x |

. x2 19. Find the domain and range of f ] x g = 2x - 6 . 20. What is the domain of y = 21. Sketch f ] x g = 1 -

1 . x2

1 4 - x2

?