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# Functions and graphs - Grade 10

## Introduction to Functions and Graphs

Functions are mathematical building blocks for designing machines, predicting natural disasters, curing diseases, understanding world economies and for keeping aeroplanes in the air. Functions can take input from many variables, but always give the same answer, unique to that function. It is the fact that you always get the same answer from a set of inputs that makes functions special.

A major advantage of functions is that they allow us to visualise equations in terms of a graph. A graph is an accurate drawing of a function and is much easier to read than lists of numbers. In this chapter we will learn how to understand and create real valued functions, how to read graphs and how to draw them.

Despite their use in the problems facing humanity, functions also appear on a day-to-day level, so they are worth learning about. A function is always dependent on one or more variables, like time, distance or a more abstract quantity.

## Functions and Graphs in the Real-World

Some typical examples of functions you may already have met include:-

• how much money you have, as a function of time. You never have more than one amount of money at any time because you can always add everything to give one number. By understanding how your money changes over time, you can plan to spend your money sensibly. Businesses find it very useful to plot the graph of their money over time so that they can see when they are spending too much. Such observations are not always obvious from looking at the numbers alone.
• the temperature is a very complicated function because it has so many inputs, including; the time of day, the season, the amount of clouds in the sky, the strength of the wind, where you are and many more. But the important thing is that there is only one temperature when you measure it in a specific place. By understanding how the temperature is effected by these things, you can plan for the day.
• where you are is a function of time, because you cannot be in two places at once! If you were to plot the graphs of where two people are as a function of time, if the lines cross it means that the two people meet each other at that time. This idea is used in logistics, an area of mathematics that tries to plan where people and items are for businesses.
• your weight is a function of how much you eat and how much exercise you do, but everybody has a different function so that is why people are all different sizes.

## Recap

The following should be familiar.

### Variables and Constants

In Review of past work, we were introduced to variables and constants. To recap, a variable can take any value in some set of numbers, so long as the equation is consistent. Most often, a variable will be written as a letter.

A constant has a fixed value. The number 1 is a constant. Sometimes letters are used to represent constants, as they are easier to work with.

#### Investigation : Variables and Constants

In the following expressions, identify the variables and the constants:

1. 2 x 2 = 1 2 x 2 = 1
2. 3 x + 4 y = 7 3 x + 4 y = 7
3. y = - 5 x y = - 5 x
4. y = 7 x - 2 y = 7 x - 2

### Relations and Functions

In earlier grades, you saw that variables can be related to each other. For example, Alan is two years older than Nathan. Therefore the relationship between the ages of Alan and Nathan can be written as A=N+2A=N+2, where AA is Alan's age and NN is Nathan's age.

In general, a relation is an equation which relates two variables. For example, y=5xy=5x and y2+x2=5y2+x2=5 are relations. In both examples xx and yy are variables and 5 is a constant, but for a given value of xx the value of yy will be very different in each relation.

Besides writing relations as equations, they can also be represented as words, tables and graphs. Instead of writing y=5xy=5x, we could also say “yy is always five times as big as xx”. We could also give the following table:

 x x y = 5 x y = 5 x 2 10 6 30 8 40 13 65 15 75

#### Investigation : Relations and Functions

Complete the following table for the given functions:

 x x y = x y = x y = 2 x y = 2 x y = x + 2 y = x + 2 1 2 3 50 100

### The Cartesian Plane

When working with real valued functions, our major tool is drawing graphs. In the first place, if we have two real variables, xx and yy, then we can assign values to them simultaneously. That is, we can say “let xx be 5 and yy be 3”. Just as we write “let x=5x=5” for “let xx be 5”, we have the shorthand notation “let (x,y)=(5,3)(x,y)=(5,3)” for “let xx be 5 and yy be 3”. We usually think of the real numbers as an infinitely long line, and picking a number as putting a dot on that line. If we want to pick two numbers at the same time, we can do something similar, but now we must use two dimensions. What we do is use two lines, one for xx and one for yy, and rotate the one for yy, as in Figure 1. We call this the Cartesian plane.

### Drawing Graphs

In order to draw the graph of a function, we need to calculate a few points. Then we plot the points on the Cartesian Plane and join the points with a smooth line.

Assume that we were investigating the properties of the function f(x)=2xf(x)=2x. We could then consider all the points (x;y)(x;y) such that y=f(x)y=f(x), i.e. y=2xy=2x. For example, (1;2),(2,5;5),(1;2),(2,5;5), and (3;6)(3;6) would all be such points, whereas (3;5)(3;5) would not since 52×352×3. If we put a dot at each of those points, and then at every similar one for all possible values of xx, we would obtain the graph shown in Figure 2

The form of this graph is very pleasing – it is a simple straight line through the middle of the plane. The technique of “plotting”, which we have followed here, is the key element in understanding functions.

#### Investigation : Drawing Graphs and the Cartesian Plane

Plot the following points and draw a smooth line through them. (-6; -8),(-2; 0), (2; 8), (6; 16)

### Notation used for Functions

Thus far you would have seen that we can use y=2xy=2x to represent a function. This notation however gets confusing when you are working with more than one function. A more general form of writing a function is to write the function as f(x)f(x), where ff


is the function name and xx is the independent variable. For example, f(x)=2xf(x)=2x and g(t)=2t+1g(t)=2t+1 are two functions.

Both notations will be used in this book.

#### Exercise 1: Function notation

If f(n)=n2-6n+9f(n)=n2-6n+9, find f(k-1)f(k-1) in terms of kk.

##### Solution
1. Step 1. Replace nn with k-1k-1 :
f ( n ) = n 2 - 6 n + 9 f ( k - 1 ) = ( k - 1 ) 2 - 6 ( k - 1 ) + 9 f ( n ) = n 2 - 6 n + 9 f ( k - 1 ) = ( k - 1 ) 2 - 6 ( k - 1 ) + 9
(1)
2. Step 2. Remove brackets on RHS and simplify :
= k 2 - 2 k + 1 - 6 k + 6 + 9 = k 2 - 8 k + 16 = k 2 - 2 k + 1 - 6 k + 6 + 9 = k 2 - 8 k + 16
(2)

We have now simplified the function in terms of kk.

#### Exercise 2: Function notation

If f(x)=x2-4f(x)=x2-4, calculate bb if f(b)=45f(b)=45.

##### Solution
1. Step 1. Replace xx with bb :
f ( b ) = b 2 - 4 but f ( b ) = 45 f ( b ) = b 2 - 4 but f ( b ) = 45
(3)
2. Step 2. f(b) = f(b) :
b 2 - 4 = 45 b 2 - 49 = 0 b = + 7 or - 7 b 2 - 4 = 45 b 2 - 49 = 0 b = + 7 or - 7
(4)

#### Recap

1. Guess the function in the form y=...y=... that has the values listed in the table.
 xx 1 2 3 40 50 600 700 800 900 1000 yy 1 2 3 40 50 600 700 800 900 1000
2. Guess the function in the form y=...y=... that has the values listed in the table.
 xx 1 2 3 40 50 600 700 800 900 1000 yy 2 4 6 80 100 1200 1400 1600 1800 2000
3. Guess the function in the form y=...y=... that has the values listed in the table.
 xx 1 2 3 40 50 600 700 800 900 1000 yy 10 20 30 400 500 6000 7000 8000 9000 10000
4. On a Cartesian plane, plot the following points: (1;2), (2;4), (3;6), (4;8), (5;10). Join the points. Do you get a straight line? Click here for the solution
5. If f(x)=x+x2f(x)=x+x2, write out:
1. f(t)f(t)
2. f(a)f(a)
3. f(1)f(1)
4. f(3)f(3)
6. If g(x)=xg(x)=x and f(x)=2xf(x)=2x, write out:
1. f(t)+g(t)f(t)+g(t)
2. f(a)-g(a)f(a)-g(a)
3. f(1)+g(2)f(1)+g(2)
4. f(3)+g(s)f(3)+g(s)
7. A car drives by you on a straight highway. The car is travelling 10 m every second. Complete the table below by filling in how far the car has travelled away from you after 5, 10 and 20 seconds.
 Time (s) 0 1 2 5 10 20 Distance (m) 0 10 20
Use the values in the table and draw a graph of distance on the yy-axis and time on the xx-axis. Click here for the solution

## Characteristics of Functions - All Grades

There are many characteristics of graphs that help describe the graph of any function. These properties will be described in this chapter and are:

1. dependent and independent variables
2. domain and range
3. intercepts with axes
4. turning points
5. asymptotes
6. lines of symmetry
7. intervals on which the function increases/decreases
8. continuous nature of the function

Some of these words may be unfamiliar to you, but each will be clearly described. Examples of these properties are shown in Figure 3.

### Dependent and Independent Variables

Thus far, all the graphs you have drawn have needed two values, an xx-value and a yy-value. The yy-value is usually determined from some relation based on a given or chosen xx-value. These values are given special names in mathematics. The given or chosen xx-value is known as the independent variable, because its value can be chosen freely. The calculated yy-value is known as the dependent variable, because its value depends on the chosen xx-value.

### Domain and Range

The domain of a relation is the set of all the xx values for which there exists at least one yy value according to that relation. The range is the set of all the yy values, which can be obtained using at least one xx value.

If the relation is of height to people, then the domain is all living people, while the range would be about 0,1 to 3 metres — no living person can have a height of 0m, and while strictly it's not impossible to be taller than 3 metres, no one alive is. An important aspect of this range is that it does not contain all the numbers between 0,1 and 3, but at most six billion of them (as many as there are people).

As another example, suppose xx and yy are real valued variables, and we have the relation y=2xy=2x. Then for any value of xx, there is a value of yy, so the domain of this relation is the whole set of real numbers. However, we know that no matter what value of xx we choose, 2x2x can never be less than or equal to 0. Hence the range of this function is all the real numbers strictly greater than zero.

These are two ways of writing the domain and range of a function, set notation and interval notation. Both notations are used in mathematics, so you should be familiar with each.

#### Set Notation

A set of certain xx values has the following form:

x : conditions , more conditions x : conditions , more conditions
(5)

We read this notation as “the set of all xx values where all the conditions are satisfied”. For example, the set of all positive real numbers can be written as {x:xR,x>0}{x:xR,x>0} which reads as “the set of all xx values where xx is a real number and is greater than zero”.

#### Interval Notation

Here we write an interval in the form 'lower bracket, lower number, comma, upper number, upper bracket'. We can use two types of brackets, square ones [;][;] or round ones (;)(;). A square bracket means including the number at the end of the interval whereas a round bracket means excluding the number at the end of the interval. It is important to note that this notation can only be used for all real numbers in an interval. It cannot be used to describe integers in an interval or rational numbers in an interval.

So if xx is a real number greater than 2 and less than or equal to 8, then xx is any number in the interval

( 2 ; 8 ] ( 2 ; 8 ]
(6)

It is obvious that 2 is the lower number and 8 the upper number. The round bracket means 'excluding 2', since xx is greater than 2, and the square bracket means 'including 8' as xx is less than or equal to 8.

### Intercepts with the Axes

The intercept is the point at which a graph intersects an axis. The xx-intercepts are the points at which the graph cuts the xx-axis and the yy-intercepts are the points at which the graph cuts the yy-axis.

In Figure 3(a), the A is the yy-intercept and B, C and F are xx-intercepts.

You will usually need to calculate the intercepts. The two most important things to remember is that at the xx-intercept, y=0y=0 and at the yy-intercept, x=0x=0.

For example, calculate the intercepts of y=3x+5y=3x+5. For the yy-intercept, x=0x=0. Therefore the yy-intercept is yint=3(0)+5=5yint=3(0)+5=5. For the xx-intercept, y=0y=0. Therefore the xx-intercept is found from 0=3xint+50=3xint+5, giving xint=-53xint=-53.

### Turning Points

Turning points only occur for graphs of functions whose highest power is greater than 1. For example, graphs of the following functions will have turning points.

f ( x ) = 2 x 2 - 2 g ( x ) = x 3 - 2 x 2 + x - 2 h ( x ) = 2 3 x 4 - 2 f ( x ) = 2 x 2 - 2 g ( x ) = x 3 - 2 x 2 + x - 2 h ( x ) = 2 3 x 4 - 2
(7)

There are two types of turning points: a minimal turning point and a maximal turning point. A minimal turning point is a point on the graph where the graph stops decreasing in value and starts increasing in value and a maximal turning point is a point on the graph where the graph stops increasing in value and starts decreasing. These are shown in Figure 4.

In Figure 3(a), E is a maximal turning point and D is a minimal turning point.

### Asymptotes

An asymptote is a straight or curved line, which the graph of a function will approach, but never touch.

In Figure 3(b), the yy-axis and line hh are both asymptotes as the graph approaches both these lines, but never touches them.

### Lines of Symmetry

Graphs look the same on either side of lines of symmetry. These lines may include the xx- and yy- axes. For example, in Figure 5 is symmetric about the yy-axis. This is described as the axis of symmetry. Not every graph will have a line of symmetry.

### Intervals on which the Function Increases/Decreases

In the discussion of turning points, we saw that the graph of a function can start or stop increasing or decreasing at a turning point. If the graph in Figure 3(a) is examined, we find that the values of the graph increase and decrease over different intervals. We see that the graph increases (i.e. that the yy-values increase) from - to point E, then it decreases (i.e. the yy-values decrease) from point E to point D and then it increases from point D to +.

### Discrete or Continuous Nature of the Graph

A graph is said to be continuous if there are no breaks in the graph. For example, the graph in Figure 3(a) can be described as a continuous graph, while the graph in Figure 3(b) has a break around the asymptotes which means that it is not continuous. In Figure 3(b), it is clear that the graph does have a break in it around the asymptote.

#### Domain and Range

1. The domain of the function f(x)=2x+5f(x)=2x+5 is -3; -3; -3; 0. Determine the range of ff. Click here for the solution
2. If g(x)=-x2+5g(x)=-x2+5 and xx is between - 3 and 3, determine:
1. the domain of g(x)g(x)
2. the range of g(x)g(x)
3. On the following graph label:
1. the xx-intercept(s)
2. the yy-intercept(s)
3. regions where the graph is increasing
4. regions where the graph is decreasing
4. On the following graph label:
1. the xx-intercept(s)
2. the yy-intercept(s)
3. regions where the graph is increasing
4. regions where the graph is decreasing

## Graphs of Functions

x + q

### Functions of the form y=ax+qy=ax+q

Functions with a general form of y=ax+qy=ax+q are called straight line functions. In the equation, y=ax+qy=ax+q, aa and qq are constants and have different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 8 for the function f(x)=2x+3f(x)=2x+3.

#### Investigation : Functions of the Form y=ax+qy=ax+q

1. On the same set of axes, plot the following graphs:
1. a(x)=x-2a(x)=x-2
2. b(x)=x-1b(x)=x-1
3. c(x)=xc(x)=x
4. d(x)=x+1d(x)=x+1
5. e(x)=x+2e(x)=x+2
Use your results to deduce the effect of different values of qq on the resulting graph.
2. On the same set of axes, plot the following graphs:
1. f(x)=-2·xf(x)=-2·x
2. g(x)=-1·xg(x)=-1·x
3. h(x)=0·xh(x)=0·x
4. j(x)=1·xj(x)=1·x
5. k(x)=2·xk(x)=2·x
Use your results to deduce the effect of different values of aa on the resulting graph.

You may have that the value of aa affects the slope of the graph. As aa increases, the slope of the graph increases. If a>0a>0 then the graph increases from left to right (slopes upwards). If a<0a<0 then the graph increases from right to left (slopes downwards). For this reason, aa is referred to as the slope or gradient of a straight-line function.

You should have also found that the value of qq affects where the graph passes through the yy-axis. For this reason, qq is known as the y-intercept.

These different properties are summarised in Table 7.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

#### Domain and Range

For f(x)=ax+qf(x)=ax+q, the domain is {x:xR}{x:xR} because there is no value of xRxR for which f(x)f(x) is undefined.

The range of f(x)=ax+qf(x)=ax+q is also {f(x):f(x)R}{f(x):f(x)R} because there is no value of f(x)Rf(x)R for which f(x)f(x) is undefined.

For example, the domain of g(x)=x-1g(x)=x-1 is {x:xR}{x:xR} because there is no value of xRxR for which g(x)g(x) is undefined. The range of g(x)g(x) is {g(x):g(x)R}{g(x):g(x)R}.

#### Intercepts

For functions of the form, y=ax+qy=ax+q, the details of calculating the intercepts with the xx and yy axis are given.

The yy-intercept is calculated as follows:

y = a x + q y i n t = a ( 0 ) + q = q y = a x + q y i n t = a ( 0 ) + q = q
(8)

For example, the yy-intercept of g(x)=x-1g(x)=x-1 is given by setting x=0x=0 to get:

g ( x ) = x - 1 y i n t = 0 - 1 = - 1 g ( x ) = x - 1 y i n t = 0 - 1 = - 1
(9)

The xx-intercepts are calculated as follows:

y = a x + q 0 = a · x i n t + q a · x i n t = - q x i n t = - q a y = a x + q 0 = a · x i n t + q a · x i n t = - q x i n t = - q a
(10)

For example, the xx-intercepts of g(x)=x-1g(x)=x-1 is given by setting y=0y=0 to get:

g ( x ) = x - 1 0 = x i n t - 1 x i n t = 1 g ( x ) = x - 1 0 = x i n t - 1 x i n t = 1
(11)

#### Turning Points

The graphs of straight line functions do not have any turning points.

#### Axes of Symmetry

The graphs of straight-line functions do not, generally, have any axes of symmetry.

#### Sketching Graphs of the Form f(x)=ax+qf(x)=ax+q

In order to sketch graphs of the form, f(x)=ax+qf(x)=ax+q, we need to determine three characteristics:

1. sign of aa
2. yy-intercept
3. xx-intercept

Only two points are needed to plot a straight line graph. The easiest points to use are the xx-intercept (where the line cuts the xx-axis) and the yy-intercept.

For example, sketch the graph of g(x)=x-1g(x)=x-1. Mark the intercepts.

Firstly, we determine that a>0a>0. This means that the graph will have an upward slope.

The yy-intercept is obtained by setting x=0x=0 and was calculated earlier to be yint=-1yint=-1. The xx-intercept is obtained by setting y=0y=0 and was calculated earlier to be xint=1xint=1.

##### Exercise 3: Drawing a straight line graph

Draw the graph of y=2x+2y=2x+2

###### Solution
1. Step 1. Find the y-intercept :

To find the intercept on the y-axis, let x=0x=0

y = 2 ( 0 ) + 2 = 2 y = 2 ( 0 ) + 2 = 2
(12)
2. Step 2. Find the x-intercept :

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

0 = 2 x + 2 2 x = - 2 x = - 1 0 = 2 x + 2 2 x = - 2 x = - 1
(13)
3. Step 3. Draw the graph by marking the two coordinates and joining them :

##### Intercepts
1. List the yy-intercepts for the following straight-line graphs:
1. y=xy=x
2. y=x-1y=x-1
3. y=2x-1y=2x-1
4. y+1=2xy+1=2x
2. Give the equation of the illustrated graph below: Click here for the solution
3. Sketch the following relations on the same set of axes, clearly indicating the intercepts with the axes as well as the co-ordinates of the point of interception of the graph: x+2y-5=0x+2y-5=0 and 3x-y-1=03x-y-1=0

### Functions of the Form y=ax2+qy=ax2+q

The general shape and position of the graph of the function of the form f(x)=ax2+qf(x)=ax2+q, called a parabola, is shown in Figure 16. These are parabolic functions.

#### Investigation : Functions of the Form y=ax2+qy=ax2+q

1. On the same set of axes, plot the following graphs:
1. a(x)=-2·x2+1a(x)=-2·x2+1
2. b(x)=-1·x2+1b(x)=-1·x2+1
3. c(x)=0·x2+1c(x)=0·x2+1
4. d(x)=1·x2+1d(x)=1·x2+1
5. e(x)=2·x2+1e(x)=2·x2+1
Use your results to deduce the effect of aa.
2. On the same set of axes, plot the following graphs:
1. f(x)=x2-2f(x)=x2-2
2. g(x)=x2-1g(x)=x2-1
3. h(x)=x2+0h(x)=x2+0
4. j(x)=x2+1j(x)=x2+1
5. k(x)=x2+2k(x)=x2+2
Use your results to deduce the effect of qq.

Complete the following table of values for the functions aa to kk to help with drawing the required graphs in this activity:

 x x - 2 - 2 - 1 - 1 0 0 1 1 2 2 a ( x ) a ( x ) b ( x ) b ( x ) c ( x ) c ( x ) d ( x ) d ( x ) e ( x ) e ( x ) f ( x ) f ( x ) g ( x ) g ( x ) h ( x ) h ( x ) j ( x ) j ( x ) k ( x ) k ( x )

This simulation allows you to visualise the effect of changing a and q. Note that in this simulation q = c. Also an extra term bx has been added in. You can leave bx as 0, or you can also see what effect this has on the graph.

Figure 17
Phet simulation for graphing

From your graphs, you should have found that aa affects whether the graph makes a smile or a frown. If a<0a<0, the graph makes a frown and if a>0a>0 then the graph makes a smile. This is shown in Figure 18.

You should have also found that the value of qq affects whether the turning point is to the left of the yy-axis (q>0q>0) or to the right of the yy-axis (q<0q<0).

These different properties are summarised in (Reference).

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

#### Domain and Range

For f(x)=ax2+qf(x)=ax2+q, the domain is {x:xR}{x:xR} because there is no value of xRxR for which f(x)f(x) is undefined.

The range of f(x)=ax2+qf(x)=ax2+q depends on whether the value for aa is positive or negative. We will consider these two cases separately.

If a>0a>0 then we have:

x 2 0 ( The square of an expression is always positive ) a x 2 0 ( Multiplication by a positive number maintains the nature of the inequality ) a x 2 + q q f ( x ) q x 2 0 ( The square of an expression is always positive ) a x 2 0 ( Multiplication by a positive number maintains the nature of the inequality ) a x 2 + q q f ( x ) q
(14)

This tells us that for all values of xx, f(x)f(x) is always greater than qq. Therefore if a>0a>0, the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)[q,)}{f(x):f(x)[q,)}.

Similarly, it can be shown that if a<0a<0 that the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)(-,q]}{f(x):f(x)(-,q]}. This is left as an exercise.

For example, the domain of g(x)=x2+2g(x)=x2+2 is {x:xR}{x:xR} because there is no value of xRxR for which g(x)g(x) is undefined. The range of g(x)g(x) can be calculated as follows:

x 2 0 x 2 + 2 2 g ( x ) 2 x 2 0 x 2 + 2 2 g ( x ) 2
(15)

Therefore the range is {g(x):g(x)[2,)}{g(x):g(x)[2,)}.

#### Intercepts

For functions of the form, y=ax2+qy=ax2+q, the details of calculating the intercepts with the xx and yy axis is given.

The yy-intercept is calculated as follows:

y = a x 2 + q y i n t = a ( 0 ) 2 + q = q y = a x 2 + q y i n t = a ( 0 ) 2 + q = q
(16)

For example, the yy-intercept of g(x)=x2+2g(x)=x2+2 is given by setting x=0x=0 to get:

g ( x ) = x 2 + 2 y i n t = 0 2 + 2 = 2 g ( x ) = x 2 + 2 y i n t = 0 2 + 2 = 2
(17)

The xx-intercepts are calculated as follows:

y = a x 2 + q 0 = a x i n t 2 + q a x i n t 2 = - q x i n t = ± - q a y = a x 2 + q 0 = a x i n t 2 + q a x i n t 2 = - q x i n t = ± - q a
(18)

However, Equation 18 is only valid if -qa0-qa0 which means that either q0q0 or a<0a<0. This is consistent with what we expect, since if q>0q>0 and a>0a>0 then -qa-qa is negative and in this case the graph lies above the xx-axis and therefore does not intersect the xx-axis. If however, q>0q>0 and a<0a<0, then -qa-qa is positive and the graph is hat shaped and should have two xx-intercepts. Similarly, if q<0q<0 and a>0a>0 then -qa-qa is also positive, and the graph should intersect with the xx-axis.

If q=0q=0 then we have one intercept at x=0x=0.

For example, the xx-intercepts of g(x)=x2+2g(x)=x2+2 is given by setting y=0y=0 to get:

g ( x ) = x 2 + 2 0 = x i n t 2 + 2 - 2 = x i n t 2 g ( x ) = x 2 + 2 0 = x i n t 2 + 2 - 2 = x i n t 2
(19)

which is not real. Therefore, the graph of g(x)=x2+2g(x)=x2+2 does not have any xx-intercepts.

#### Turning Points

The turning point of the function of the form f(x)=ax2+qf(x)=ax2+q is given by examining the range of the function. We know that if a>0a>0 then the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)[q,)}{f(x):f(x)[q,)} and if a<0a<0 then the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)(-,q]}{f(x):f(x)(-,q]}.

So, if a>0a>0, then the lowest value that f(x)f(x) can take on is qq. Solving for the value of xx at which f(x)=qf(x)=q gives:

q = a x t p 2 + q 0 = a x t p 2 0 = x t p 2 x t p = 0 q = a x t p 2 + q 0 = a x t p 2 0 = x t p 2 x t p = 0
(20)

x=0x=0 at f(x)=qf(x)=q. The co-ordinates of the (minimal) turning point is therefore (0,q)(0,q).

Similarly, if a<0a<0, then the highest value that f(x)f(x) can take on is qq and the co-ordinates of the (maximal) turning point is (0,q)(0,q).

#### Axes of Symmetry

There is one axis of symmetry for the function of the form f(x)=ax2+qf(x)=ax2+q that passes through the turning point. Since the turning point lies on the yy-axis, the axis of symmetry is the yy-axis.

#### Sketching Graphs of the Form f(x)=ax2+qf(x)=ax2+q

In order to sketch graphs of the form, f(x)=ax2+qf(x)=ax2+q, we need to determine five characteristics:

1. sign of aa
2. domain and range
3. turning point
4. yy-intercept
5. xx-intercept

For example, sketch the graph of g(x)=-12x2-3g(x)=-12x2-3. Mark the intercepts, turning point and axis of symmetry.

Firstly, we determine that a<0a<0. This means that the graph will have a maximal turning point.

The domain of the graph is {x:xR}{x:xR} because f(x)f(x) is defined for all xRxR. The range of the graph is determined as follows:

x 2 0 - 1 2 x 2 0 - 1 2 x 2 - 3 - 3 f ( x ) - 3 x 2 0 - 1 2 x 2 0 - 1 2 x 2 - 3 - 3 f ( x ) - 3
(21)

Therefore the range of the graph is {f(x):f(x)(-,-3]}{f(x):f(x)(-,-3]}.

Using the fact that the maximum value that f(x)f(x) achieves is -3, then the yy-coordinate of the turning point is -3. The xx-coordinate is determined as follows:

- 1 2 x 2 - 3 = - 3 - 1 2 x 2 - 3 + 3 = 0 - 1 2 x 2 = 0 Divide both sides by - 1 2 : x 2 = 0 Take square root of both sides : x = 0 x = 0 - 1 2 x 2 - 3 = - 3 - 1 2 x 2 - 3 + 3 = 0 - 1 2 x 2 = 0 Divide both sides by - 1 2 : x 2 = 0 Take square root of both sides : x = 0 x = 0
(22)

The coordinates of the turning point are: (0;-3)(0;-3).

The yy-intercept is obtained by setting x=0x=0. This gives:

y i n t = - 1 2 ( 0 ) 2 - 3 = - 1 2 ( 0 ) - 3 = - 3 y i n t = - 1 2 ( 0 ) 2 - 3 = - 1 2 ( 0 ) - 3 = - 3
(23)

The xx-intercept is obtained by setting y=0y=0. This gives:

0 = - 1 2 x i n t 2 - 3 3 = - 1 2 x i n t 2 - 3 · 2 = x i n t 2 - 6 = x i n t 2 0 = - 1 2 x i n t 2 - 3 3 = - 1 2 x i n t 2 - 3 · 2 = x i n t 2 - 6 = x i n t 2
(24)

which is not real. Therefore, there are no xx-intercepts which means that the function does not cross or even touch the xx-axis at any point.

We also know that the axis of symmetry is the yy-axis.

Finally, we draw the graph. Note that in the diagram only the y-intercept is marked. The graph has a maximal turning point (i.e. makes a frown) as determined from the sign of a, there are no x-intercepts and the turning point is that same as the y-intercept. The domain is all real numbers and the range is {f(x):f(x)(-,-3]}{f(x):f(x)(-,-3]}.

The following video shows one method of graphing parabolas. Note that in this video the term vertex is used in place of turning point. The vertex and the turning point are the same thing.

Figure 24
Khan academy video on graphing parabolas - 1

##### Parabolas
1. Show that if a<0a<0 that the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)(-;q]}{f(x):f(x)(-;q]}. Click here for the solution
2. Draw the graph of the function y=-x2+4y=-x2+4 showing all intercepts with the axes. Click here for the solution
3. Two parabolas are drawn: g:y=ax2+pg:y=ax2+p and h:y=bx2+qh:y=bx2+q.
1. Find the values of aa and pp.
2. Find the values of bb and qq.
3. Find the values of xx for which ax2+pbx2+qax2+pbx2+q.
4. For what values of xx is gg increasing ?

### Functions of the Form y=ax+qy=ax+q

Functions of the form y=ax+qy=ax+q are known as hyperbolic functions. The general form of the graph of this function is shown in Figure 26.

#### Investigation : Functions of the Form y=ax+qy=ax+q

1. On the same set of axes, plot the following graphs:
1. a(x)=-2x+1a(x)=-2x+1
2. b(x)=-1x+1b(x)=-1x+1
3. c(x)=0x+1c(x)=0x+1
4. d(x)=+1x+1d(x)=+1x+1
5. e(x)=+2x+1e(x)=+2x+1
Use your results to deduce the effect of aa.
2. On the same set of axes, plot the following graphs:
1. f(x)=1x-2f(x)=1x-2
2. g(x)=1x-1g(x)=1x-1
3. h(x)=1x+0h(x)=1x+0
4. j(x)=1x+1j(x)=1x+1
5. k(x)=1x+2k(x)=1x+2
Use your results to deduce the effect of qq.

You should have found that the value of aa affects whether the graph is located in the first and third quadrants of Cartesian plane.

You should have also found that the value of qq affects whether the graph lies above the xx-axis (q>0q>0) or below the xx-axis (q<0q<0).

These different properties are summarised in Table 10. The axes of symmetry for each graph are shown as a dashed line.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

#### Domain and Range

For y=ax+qy=ax+q, the function is undefined for x=0x=0. The domain is therefore {x:xR,x0}{x:xR,x0}.

We see that y=ax+qy=ax+q can be re-written as:

y = a x + q y - q = a x If x 0 then : ( y - q ) ( x ) = a x = a y - q y = a x + q y - q = a x If x 0 then : ( y - q ) ( x ) = a x = a y - q
(25)

This shows that the function is undefined at y=qy=q. Therefore the range of f(x)=ax+qf(x)=ax+q is {f(x):f(x)(-;q)(q;)}{f(x):f(x)(-;q)(q;)}.

For example, the domain of g(x)=2x+2g(x)=2x+2 is {x:xR,x0}{x:xR,x0} because g(x)g(x) is undefined at x=0x=0.

y = 2 x + 2 ( y - 2 ) = 2 x If x 0 then : x ( y - 2 ) = 2 x = 2 y - 2 y = 2 x + 2 ( y - 2 ) = 2 x If x 0 then : x ( y - 2 ) = 2 x = 2 y - 2
(26)

We see that g(x)g(x) is undefined at y=2y=2. Therefore the range is {g(x):g(x)(-;2)(2;)}{g(x):g(x)(-;2)(2;)}.

#### Intercepts

For functions of the form, y=ax+qy=ax+q, the intercepts with the xx and yy axis is calculated by setting x=0x=0 for the yy-intercept and by setting y=0y=0 for the xx-intercept.

The yy-intercept is calculated as follows:

y = a x + q y i n t = a 0 + q y = a x + q y i n t = a 0 + q
(27)

which is undefined because we are dividing by 0. Therefore there is no yy-intercept.

For example, the yy-intercept of g(x)=2x+2g(x)=2x+2 is given by setting x=0x=0 to get:

y = 2 x + 2 y i n t = 2 0 + 2 y = 2 x + 2 y i n t = 2 0 + 2
(28)

which is undefined.

The xx-intercepts are calculated by setting y=0y=0 as follows:

y = a x + q 0 = a x i n t + q a x i n t = - q a = - q ( x i n t ) x i n t = a - q y = a x + q 0 = a x i n t + q a x i n t = - q a = - q ( x i n t ) x i n t = a - q
(29)

For example, the xx-intercept of g(x)=2x+2g(x)=2x+2 is given by setting x=0x=0 to get:

y = 2 x + 2 0 = 2 x i n t + 2 - 2 = 2 x i n t - 2 ( x i n t ) = 2 x i n t = 2 - 2 x i n t = - 1 y = 2 x + 2 0 = 2 x i n t + 2 - 2 = 2 x i n t - 2 ( x i n t ) = 2 x i n t = 2 - 2 x i n t = - 1
(30)

#### Asymptotes

There are two asymptotes for functions of the form y=ax+qy=ax+q. Just a reminder, an asymptote is a straight or curved line, which the graph of a function will approach, but never touch. They are determined by examining the domain and range.

We saw that the function was undefined at x=0x=0 and for y=qy=q. Therefore the asymptotes are x=0x=0 and y=qy=q.

For example, the domain of g(x)=2x+2g(x)=2x+2 is {x:xR,x0}{x:xR,x0} because g(x)g(x) is undefined at x=0x=0. We also see that g(x)g(x) is undefined at y=2y=2. Therefore the range is {g(x):g(x)(-;2)(2;)}{g(x):g(x)(-;2)(2;)}.

From this we deduce that the asymptotes are at x=0x=0 and y=2y=2.

#### Sketching Graphs of the Form f(x)=ax+qf(x)=ax+q

In order to sketch graphs of functions of the form, f(x)=ax+qf(x)=ax+q, we need to determine four characteristics:

1. domain and range
2. asymptotes
3. yy-intercept
4. xx-intercept

For example, sketch the graph of g(x)=2x+2g(x)=2x+2. Mark the intercepts and asymptotes.

We have determined the domain to be {x:xR,x0}{x:xR,x0} and the range to be {g(x):g(x)(-;2)(2;)}{g(x):g(x)(-;2)(2;)}. Therefore the asymptotes are at x=0x=0 and y=2y=2.

There is no yy-intercept and the xx-intercept is xint=-1xint=-1.

##### Graphs
1. Using graph (grid) paper, draw the graph of xy=-6xy=-6.
1. Does the point (-2; 3) lie on the graph ? Give a reason for your answer.
2. Why is the point (-2; -3) not on the graph ?
3. If the xx-value of a point on the drawn graph is 0,25, what is the corresponding yy-value ?
4. What happens to the yy-values as the xx-values become very large ?
5. With the line y=-xy=-x as line of symmetry, what is the point symmetrical to (-2; 3) ?
2. Draw the graph of xy=8xy=8.
1. How would the graph y=83+3y=83+3 compare with that of xy=8xy=8? Explain your answer fully.
2. Draw the graph of y=83+3y=83+3 on the same set of axes.

### Functions of the Form y=ab(x)+qy=ab(x)+q

Functions of the form y=ab(x)+qy=ab(x)+q are known as exponential functions. The general shape of a graph of a function of this form is shown in Figure 32.

#### Investigation : Functions of the Form y=ab(x)+qy=ab(x)+q

1. On the same set of axes, plot the following graphs:
1. a(x)=-2·b(x)+1a(x)=-2·b(x)+1
2. b(x)=-1·b(x)+1b(x)=-1·b(x)+1
3. c(x)=-0·b(x)+1c(x)=-0·b(x)+1
4. d(x)=-1·b(x)+1d(x)=-1·b(x)+1
5. e(x)=-2·b(x)+1e(x)=-2·b(x)+1
Use your results to deduce the effect of aa.
2. On the same set of axes, plot the following graphs:
1. f(x)=1·b(x)-2f(x)=1·b(x)-2
2. g(x)=1·b(x)-1g(x)=1·b(x)-1
3. h(x)=1·b(x)+0h(x)=1·b(x)+0
4. j(x)=1·b(x)+1j(x)=1·b(x)+1
5. k(x)=1·b(x)+2k(x)=1·b(x)+2
Use your results to deduce the effect of qq.

You should have found that the value of aa affects whether the graph curves upwards (a>0a>0) or curves downwards (a<0a<0).

You should have also found that the value of qq affects the position of the yy-intercept.

These different properties are summarised in Table 11.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

#### Domain and Range

For y=ab(x)+qy=ab(x)+q, the function is defined for all real values of xx. Therefore, the domain is {x:xR}{x:xR}.

The range of y=ab(x)+qy=ab(x)+q is dependent on the sign of aa.

If a>0a>0 then:

b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q
(31)

Therefore, if a>0a>0, then the range is {f(x):f(x)[q;)}{f(x):f(x)[q;)}.

If a<0a<0 then:

b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q b ( x ) 0 a · b ( x ) 0 a · b ( x ) + q q f ( x ) q
(32)

Therefore, if a<0a<0, then the range is {f(x):f(x)(-;q]}{f(x):f(x)(-;q]}.

For example, the domain of g(x)=3·2x+2g(x)=3·2x+2 is {x:xR}{x:xR}. For the range,

2 x 0 3 · 2 x 0 3 · 2 x + 2 2 2 x 0 3 · 2 x 0 3 · 2 x + 2 2
(33)

Therefore the range is {g(x):g(x)[2;)}{g(x):g(x)[2;)}.

#### Intercepts

For functions of the form, y=ab(x)+qy=ab(x)+q, the intercepts with the xx and yy axis is calculated by setting x=0x=0 for the yy-intercept and by setting y=0y=0 for the xx-intercept.

The yy-intercept is calculated as follows:

y = a b ( x ) + q y i n t = a b ( 0 ) + q = a ( 1 ) + q = a + q y = a b ( x ) + q y i n t = a b ( 0 ) + q = a ( 1 ) + q = a + q
(34)

For example, the yy-intercept of g(x)=3·2x+2g(x)=3·2x+2 is given by setting x=0x=0 to get:

y = 3 · 2 x + 2 y i n t = 3 · 2 0 + 2 = 3 + 2 = 5 y = 3 · 2 x + 2 y i n t = 3 · 2 0 + 2 = 3 + 2 = 5
(35)

The xx-intercepts are calculated by setting y=0y=0 as follows:

y = a b ( x ) + q 0 = a b ( x i n t ) + q a b ( x i n t ) = - q b ( x i n t ) = - q a y = a b ( x ) + q 0 = a b ( x i n t ) + q a b ( x i n t ) = - q b ( x i n t ) = - q a
(36)

Which only has a real solution if either a<0a<0 or q<0q<0. Otherwise, the graph of the function of form y=ab(x)+qy=ab(x)+q does not have any xx-intercepts.

For example, the xx-intercept of g(x)=3·2x+2g(x)=3·2x+2 is given by setting y=0y=0 to get:

y = 3 · 2 x + 2 0 = 3 · 2 x i n t + 2 - 2 = 3 · 2 x i n t 2 x i n t = - 2 3 y = 3 · 2 x + 2 0 = 3 · 2 x i n t + 2 - 2 = 3 · 2 x i n t 2 x i n t = - 2 3
(37)

which has no real solution. Therefore, the graph of g(x)=3·2x+2g(x)=3·2x+2 does not have any xx-intercepts.

#### Asymptotes

There is one asymptote for functions of the form y=ab(x)+qy=ab(x)+q. The asymptote can be determined by examining the range.

We saw that the function was undefined at y=qy=q. Therefore the asymptote is y=qy=q.

For example, the domain of g(x)=3·2x+2g(x)=3·2x+2 is {x:xR}{x:xR} because g(x)g(x) is defined for all xx. We also see that g(x)g(x) is undefined at y=2y=2. Therefore the range is {g(x):g(x)(-;2)(2;)}{g(x):g(x)(-;2)(2;)}.

From this we deduce that the asymptote is at y=2y=2.

#### Sketching Graphs of the Form f(x)=ab(x)+qf(x)=ab(x)+q

In order to sketch graphs of functions of the form, f(x)=ab(x)+qf(x)=ab(x)+q, we need to calculate determine four characteristics:

1. domain and range
2. yy-intercept
3. xx-intercept

For example, sketch the graph of g(x)=3·2x+2g(x)=3·2x+2. Mark the intercepts.

We have determined the domain to be {x:xR}{x:xR} and the range to be {g(x):g(x)[2,)}{g(x):g(x)[2,)}.

The yy-intercept is yint=5yint=5 and there are no xx-intercepts.

##### Exponential Functions and Graphs
1. Draw the graphs of y=2xy=2x and y=(12)xy=(12)x on the same set of axes.
1. Is the xx-axis and asymptote or and axis of symmetry to both graphs ? Explain your answer.
2. Which graph is represented by the equation y=2-xy=2-x ? Explain your answer.
3. Solve the equation 2x=(12)x2x=(12)x graphically and check that your answer is correct by using substitution.
4. Predict how the graph y=2.2xy=2.2x will compare to y=2xy=2x and then draw the graph of y=2.2xy=2.2x on the same set of axes.
2. The curve of the exponential function ff in the accompanying diagram cuts the y-axis at the point A(0; 1) and B(2; 4) is on ff.
1. Determine the equation of the function ff.
2. Determine the equation of hh, the function of which the curve is the reflection of the curve of ff in the xx-axis.
3. Determine the range of hh.

## End of Chapter Exercises

1. Given the functions f(x)=-2x2-18f(x)=-2x2-18 and g(x)=-2x+6g(x)=-2x+6
1. Draw ff and gg on the same set of axes.
2. Calculate the points of intersection of ff and gg.
3. Hence use your graphs and the points of intersection to solve for xx when:
1. f(x)>0f(x)>0
2. f(x)g(x)0f(x)g(x)0
4. Give the equation of the reflection of ff in the xx-axis.
2. After a ball is dropped, the rebound height of each bounce decreases. The equation y=5·(0,8)xy=5·(0,8)x shows the relationship between xx, the number of bounces, and yy, the height of the bounce, for a certain ball. What is the approximate height of the fifth bounce of this ball to the nearest tenth of a unit ?
3. Mark had 15 coins in five Rand and two Rand pieces. He had 3 more R2-coins than R5-coins. He wrote a system of equations to represent this situation, letting xx represent the number of five rand coins and yy represent the number of two rand coins. Then he solved the system by graphing.
1. Write down the system of equations.
2. Draw their graphs on the same set of axes.
3. What is the solution?

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