mathematics workshop functions

With the Educators, for the Educators

MATHEMATICS WORKSHOP

FUNCTIONS

GRADE 10 TEXTBOOK

(Chapter 6)

Presented by: Jurg Basson

MIND ACTION SERIES

Attending this Workshop = 10 SACE Points

1

1−2− 1 21−

2

1

2−

( 1;1)−

(0 ; 0)

(1;1)

y x=

CHAPTER 6 FUNCTIONS In this chapter, you will revise the graphs of linear functions (straight lines). Then you will explore the graphs of quadratic functions (parabolas), hyperbolic functions (hyperbolas) and exponential functions (exponential graphs). LINEAR FUNCTIONS (SKETCHING STRAIGHT LINE GRAPHS) Consider the graph of y x= We can select a few input values (x-values) and hence determine the corresponding output values (y-values). These values will be represented in a table.

x 1− 0 1 y 1− 0 1

The graph of this line is obtained by plotting the points on the Cartesian plane and drawing a solid line through the points. This graph is referred to as the “mother” graph of straight lines and based on this graph, we can generate different types of straight lines depending on the value of a and q in the general equation of a line, which is y ax q= + . Investigation of the effect of the value of a In Grade 9, you learnt about the gradient and steepness of a straight line. Let’s briefly summarise these concepts. The gradient of a line represents the ratio of the change of the y-values with respect to the x-values. In other words, gradient tells us the direction (or slope) of the line. We will revise the concept of gradient later on in this chapter. However, the focus now will be on the steepness of a line, which is the way that the line slants upwards or downwards from left to right. We will compare the steepness of different lines to the mother graph and show how the mother graph is transformed by changing the value of a (the coefficient of x).

2

Let’s investigate this by comparing the graphs of the following straight lines:

A: y x= B: 12

y x= C: 2y x= D: 3y x=

Lines A, B, C and D pass through the origin since for all of the given graphs, the y-value is 0 if 0x = . To sketch the graphs of these lines, select one x-value and then determine the corresponding y-value. For all four graphs, choose 1x = . A: (1) 1y = = Line A passes through the points (0 ;0) and (1 ;1)

B: 1 1(1)2 2

y = = Line B passes through the points (0 ;0) and 11;2

C: 2(1) 2y = = Line C passes through the points (0 ;0) and (1 ; 2)

D: 3(1) 3y = = Line D passes through the points (0 ;0) and (1 ; 3) We will now plot the points and then draw the lines on the same set of axes. Notice that line A (the mother graph) is closer to the y-axis than line B. We say that line A is steeper than line B. The coefficient of x in the equation of line A is greater than the coefficient of x in line B ( 1

21 > ). Line C is steeper line A ( 2 1> ) and line D is steeper than line C (3 2> ). Line D is the steepest of all the lines.

1−2− 1 2

1−

2

1

2−

12(1; )(0 ; 0)

(1;1)

1y x=3

3

3−

3−

2y x=

12

y x=(1; 2)

3y x=

(1; 3)

3

1−2− 1 21−

2

1

2−

(1; 1)−

(0 ; 0)( 1;1)−

y x= −

Let’s now consider what happens if the value of the coefficient of x is negative. (a) Consider the graph of y x= −

We can select a few input values (x-values) and hence determine the corresponding

output values (y-values). These values will be represented in a table.

x 1− 0 1 y 1 0 1−

The graph of this line is obtained by plotting the points on the Cartesian plane and

drawing a solid line through the points.

If you now compare the mother graph 1y x= to the graph of 1y x= − , it is interesting

to note that the graph of y x= − is the reflection of the mother graph in the x-axis. The

negative sign therefore causes a reflection in the x-axis.

(b) Consider the graph of 12

y x= −

We already know that the number 12

affects the steepness of the line. The mother

graph y x= transforms into a line that is not as steep as the mother graph line.

As already seen, the negative sign causes a reflection in the x-axis.

To draw this graph involves transforming the mother graph into the less steep line

12

y x= and then reflecting this newly-formed graph about the x-axis.

To sketch the graph of this line, select one x-value, determine the corresponding y-

value, plot the point formed and the y-intercept (0 ;0)

Choose 1x =

1 1(1)2 2

y = − = −

4

The line passes through the point 12(1; )−

Now plot this point and the y-intercept and draw the graph of the line.

Notice that the mother graph y x= is transformed into 12

y x= and then this new line

is reflected in the x-axis to form the graph of the line 12

y x= − .

Conclusion:

The value of a in the equation y ax q= + (ignoring negative signs), determines the steepness of the line (closeness to the y-axis). The larger the value of a, the steeper the line.

A negative sign will cause a reflection in the x-axis. Investigation of the effect of the value of q (a) Consider the graphs of the following: 3y x= + and 2y x= −

For 3y x= + : If 0x = then 0 3 3y = + = If 1x = then 1 3 4y = + = The line passes through the points (0 ; 3) and (1 ; 4) For 2y x= − : If 0x = then 0 2 2y = − = − If 1x = then 1 2 1y = − = − The line passes through the points (0 ; 2)− and (1 ; 1)−

1−2− 1 2

1−

2

1

2−

12(1; )

(1;1)

1y x=

33−

12

y x=

12

y x= −

12(1; )−

5

1−2− 1 21−

2

1

2−

( 1;1)−

(0 ; 0)(1;1)

y x=3

3−

3y x= +

2y x= −

(1; 4)

(1; 1)−

4

(0 ; 3)

(0 ; 2)−

1−2− 1 21−

2

1

2−

y x= − 3

3−

1y x= − +

2y x= − −

3−

y x=

It should be clear to you from the graphs that 3y x= + is the mother graph y x= shifted 3 units up and 2y x= − is the mother graph y x= shifted 2 units down. Also, the y-intercept of 3y x= + is 3 and the y-intercept of 2y x= − is 2− . (b) Consider the graphs of the following: 1y x= − + and 2y x= − −

We will first draw the graph of y x= − which is the reflection of the mother graph y x= in the x-axis and then based on this graph, we will form 1y x= − + by shifting y x= − one unit up and 2y x= − − by shifting y x= − two units down. It should be clear to you from the graphs that 1y x= − + is the line y x= − shifted 1 unit up and 2y x= − − is the line y x= − shifted 2 units down. Also, the y-intercept of 1y x= − + is 1 and the y-intercept of 2y x= − − is 2− .

6

1−2− 1 21−

2

1

2−

(0 ; 0)

(0 ; 4)

3

2 4y x= − +

4

(1; 2)

(1; 2)−

2y x=2y x= −

Conclusion:

The value of q in the equation y ax q= + determines the shift of the graph of y ax= up or down. It also represents the y-intercept of the graph of y ax q= + . EXAMPLE 1 Draw a neat sketch graph of 2 4y x= − + Solution First draw the graph of 2y x= , reflect this graph in the x-axis to form 2y x= − and then shift

2y x= − four units up to form 2 4y x= − + .

The line 2y x= cuts the y-axis at 0. Now choose 1x =

2(1) 2y∴ = = Plot the point (1 ; 2) and draw the line 2y x= Now reflect 2y x= in the x-axis to form

2y x= − . The point (1 ; 2) transforms into

the point (1 ; 2)− .

Then shift 2y x= − four units up. Draw the graph of 2 4y x= − + .

Alternatively, you can make use of the dual-intercept method that you studied in Grade 9.

This method involves determining the intercepts with the axes algebraically. Let’s use this

method for the line 2 4y x= − +

y-intercept: Let 0x = x-intercept: Let 0y = 2(0) 4 4y = − + = 0 2 4x= − +

2 42

xx

∴ =∴ =

The coordinates of the y-intercept are (0 ; 4) and the coordinates of the x-intercept are (2 ; 0).

You would now plot these points and draw the straight line.

7

1−2− 1 21−

2

1

2−

( 2 ; 2)− (1; 2)

2y =

1−2− 1 21−

2

1

2−

(1; 1)−

(1; 2)

1x =

Revision of horizontal and vertical lines In Grade 9 you learnt that horizontal lines have the general equation y n= where n is any real number. Vertical lines have the general equation x n= where n is any real number. For example, the graphs of the lines 2y = and 1x = are shown below. For the line 2y = , the y-values will be constant but the x-values will vary. For the line 1x = , the x-values will be constant but the y-values will vary. EXERCISE 1

(a) Given: y x= 32

y x= 14

y x= (1) Which of the three lines is the steepest? Explain (2) Sketch the three graphs on the same set of axes.

(b) Given: y x= − 32

y x= − 14

y x= − (1) Which of the three lines is the steepest? Explain (2) Sketch the three graphs on the same set of axes. (c) Given: 4y x= − + 3 6y x= − (1) Describe the transformation of y x= into the graph of 4y x= − + (2) Describe the transformation of y x= into the graph of 3 6y x= − (3) Sketch the graphs of these two functions on the same set of axes using

transformations or the dual-intercept method.

8

1−2− 1 21−

2

1

3

2y x=

(1;1)

4

( 1;1)−

(2 ; 4)( 2 ; 4)−

0

(d) Match the equations on the left to the graphs on the right. (1) 2y x= −

(2) 1 24

y x= +

(3) 2y x= − −

(4) 4y x=

(5) 14

y x=

(6) 4y x= − (7) 4x =

(8) 4y = QUADRATIC FUNCTIONS (SKETCHING PARABOLAS) Consider the graph of 2y x= We can select a few input values (x-values) and hence determine the corresponding output

values (y-values). These values will be represented in a table.

x 2− 1− 0 1 2 y 4 1 0 1 4

The graph of 2y x= is obtained by plotting the points on the Cartesian plane and drawing a

curve through the points.

Notice that: ● all output values are positive ● the graph is not linear but rather a curve referred to as the graph of a parabola

9

1−2− 1 2

2

1

12(1; )

(0 ; 0)

(1;1)

3

(1; 2)

(1; 3)

This graph is referred to as the “mother” graph of parabolas and based on these graphs, we

can generate different types of parabolas depending on the value of a and q in the general

equation of a parabola, which is 2y ax q= + .

Investigation of the effect of the value of a Let’s investigate the effect of value of a on the shape of different parabolas by comparing the

graphs of the following parabolas:

A: 2y x= B: 212

y x= C: 22y x= D: 23y x=

Parabolas A, B, C and D pass through the origin. To sketch the graphs of these parabolas,

select one x-value and then determine the corresponding y-value. For all four graphs, choose

1x = .

A: 2(1) 1y = = Parabola A passes through the points (0 ;0) and (1 ;1)

B: 21 1(1)2 2

y = = Parabola B passes through the points (0 ;0) and 11;2

C: 22(1) 2y = = Parabola C passes through the points (0 ;0) and (1 ; 2)

D: 23(1) 3y = = Parabola D passes through the points (0 ;0) and (1 ; 3) The points are plotted and the parabolas have been drawn on the same set of axes. Notice that the arms of the mother graph parabola (A) are closer to the y-axis than those of B.

The arms of parabola C are closer to the y-axis than those of A. The arms of parabola D are

closer to the y-axis than those of C. The value of the coefficient of x affects the shape of the

parabola (or what is called its vertical stretch). The greater the value of this number, the closer the arms of the parabola will be to the y-axis. Also note that the coefficient of 2x for each parabola is positive and the graphs are concave up (happy!).

10

Let’s now consider what happens if the value of the coefficient of x is negative.

(a) Consider the graph of 2y x= −

We can select a few input values (x-values) and hence determine the corresponding

output values (y-values). These values will be represented in a table.

x 2− 1− 0 1 2 y 4− 1− 0 1− 4−

The graph of 2y x= − is obtained by plotting the points on the Cartesian plane and

drawing a curve through the points.

If you now compare the mother graph 21y x= to the graph of 21y x= − , it is interesting to note that the graph of 2y x= − is the reflection of the mother graph in the x-axis.

The negative sign therefore causes a reflection in the x-axis.

(b) Consider the graph of 212

y x= −

We already know that the number 12

affects the shape of the curve. The mother graph

2y x= transforms into a parabola with arms that arms that are less close to the y-axis.

To sketch the graph of this parabola, select one x-value, determine the corresponding

y-value, plot the point forms and the y-intercept (0 ; 0) .

Choose 1x =

21 1(1)2 2

y∴ = − = −

1−2− 1 2

1

2y x= −

(1; 1)−( 1; 1)− −

(2 ; 4)−( 2 ; 4)− −

1−

2−

3−

4−

0

2

3

42y x=

11

The parabola passes through the point 11;2

−

Now plot this point and the y-intercept and draw the graph of the parabola.

Notice that the mother graph 2y x= transformed into 212

y x= and then this graph was

reflected in the x-axis to form the graph of the parabola 212

y x= −

Conclusion:

The value of a in the equation 2y ax q= + (ignoring negative signs), determines the closeness of the arms of the parabola to the y-axis. The larger the value of a , the closer the

arms are to the y-axis. A negative sign will cause a reflection in the x-axis.

If 0a > , then the parabola is concave up (happy face!)

If 0a < , then the parabola is concave down (sad face!) The value of a is sometimes referred to as the vertical stretch factor

1−2− 1 2

2−

1−

3−

(1;1)1

2

12(1; )

2y x=21

2y x=

212

y x= −

12(1; )−

12

1−2− 1 2

1−

2

1

3

2 1y x= −

(1;1)

4

(0 ; 1)−

0(1; 0)( 1; 0)−

5

2 2y x= +

(1; 3)

Investigation of the effect of the value of q Consider the following graphs: A: 2 1y x= − B: 2 2y x= + We will first draw the mother graph 2y x= and then based on this graph, we will draw parabola A and B. The graph of A is the graph of 2y x= shifted 1 unit down. The graph of B is the graph of 2y x= shifted 2 units up.

Conclusion:

The value of q in the equation 2y ax q= + determines the shift of the graph of 2y ax= up or down. It also represents the y-intercept of the graph of 2y ax q= + . EXAMPLE 2 Given: 22 8y x= − + and 22 2y x= − −

(a) Sketch the graphs of 22 8y x= − + and 22 2y x= − − on the same set of axes. (b) For these graphs, determine algebraically the coordinates of the intercepts with the axes. Solutions (a) Reflect the graph of 22y x= in the x-axis to form 22y x= − and then shift 22y x= − eight units up to form 22 8y x= − + . Then shift the graph of 22y x= − two units down to form 22 2y x= − − .

13

2−4− 2 42−

4

2

6

22 8y x= − +8

4−

6−

8−22 2y x= − −

(1; 2)−

22y x= −

For 22y x= − , choose 1x = 22(1) 2y∴ = − = − 22y x= − passes through the origin (0 ; 0) and the point (1 ; 2)− (b) Consider the graph 22 8y x= − + :

y-intercept: Let 0x = x-intercept: Let 0y = 22(0) 8 8y∴ = − + = 20 2 8x∴ = − + (0 ; 8) 22 8 0x∴ − =

2 4 0( 2)( 2) 0

2 or 2( 2 ; 0) (2 ; 0)

xx x

x x

∴ − =∴ + − =∴ = − =−

Consider the graph 22 2y x= − − :

y-intercept: Let 0x = x-intercept: 22(0) 2 2y∴ = − − = − The graph doesn’t cut the x-axis (0 ; 2)− There are no x-intercepts Also notice that the equation 20 2 2x= − − has no real solutions:

2

2

2

0 2 2

2 2

1

x

x

x

= − −

∴ = −

∴ = −

1x∴ = ± − which is non-real

14

EXERCISE 2 (a) Given: 2y x= 23y x= 24y x= (1) Which parabola has arms that are closest to the y-axis? (2) Sketch the graphs of these parabolas on the same set of axes. (3) Are the parabolas concave up or down? Explain

(b) Given: 2y x= − 212

y x= − 214

y x= −

(1) Which parabola has arms that are closest to the y-axis? (2) Sketch the graphs of these parabolas on the same set of axes. (3) Are the parabolas concave up or down? Explain

(c) Given: 214

y x= 232

y x= 24y x= −

(1) Which parabola has arms that are closest to the y-axis? (2) Sketch the graphs of these parabolas on the same set of axes. (3) Are the parabolas concave up or down? Explain (d) Given: 2 4y x= − and 2 3y x= + (1) Sketch the graphs of 2 4y x= − and 2 1y x= + on the same set of axes. (2) For these graphs, determine algebraically the coordinates of the intercepts with the axes. (e) Given: 24 4y x= − + and 2 2y x= − − (1) Sketch the graphs of 24 4y x= − + and 2 2y x= − − on the same set of axes. (2) For these graphs, determine algebraically the coordinates of the intercepts with the axes.

(f) Given: 21 82

y x= − +

(1) Sketch the graph of 21 82

y x= − + (2) Determine algebraically the coordinates of the intercepts of this parabola with the axes.

15

(g) Match the equations on the left to the graphs on the right.

(1) 214

y x=

(2) 213

y x= −

(3) 215

y x= −

(4) 2 152

y x= − −

(5) 2 1y x= −

(6) 22y x= HYPERBOLIC FUNCTIONS (SKETCHING HYPERBOLAS)

Consider the graph of 1yx

= We can select a few input values (x-values) and hence determine the corresponding output values (y-values). These values will be represented in a table.

x 2− 1− 1

2− 0 12 1 2

y 12− 1− 2− 1

0 2 1 12

Notice from the table that if 0x = then 10

y = which is undefined.

Since the y-value is undefined at 0x = , this means that the graph has no y-intercept. For the graph to have an x-intercept we let 0y = and solve for x.

10x

∴ =

There is no value of x which satisfies this equation. Hence the graph has no x-intercept.

The graph of 1yx

= is obtained by plotting the points on the Cartesian plane and drawing a

curve through the points. From the above discussion, it should be clear that the graph has no

intercepts with the axes.

The graph gets closer and closer to the axes but never actually cuts them.

16

An asymptote is a horizontal or vertical line that a graph approaches but never touches. The vertical line 0x = (lying on the y-axis) is called the vertical asymptote of the graph. The horizontal line 0y = (lying on the x-axis) is called the horizontal asymptote of the

graph.

This graph is referred to as the “mother” graph of hyperbolas and based on this graph, we can

generate different types of hyperbolas depending on the value of a and q in the general

equation of a hyperbola, which is ay qx

= + .

Investigation of the effect of the value of a Let’s investigate the effect of value of a on the shape of different hyperbolas by comparing

the graphs of the following hyperbolas:

A: 1yx

= B: 2yx

= C: 4yx

= The graph of A is the mother graph. In order to sketch the graph of B and C, we can select a

few input values (x-values) and hence determine the corresponding output values (y-values)

for each graph.

1 2 31−2−3−

1

2

3

1−

2−

3−

0

0x =

0y =

(1;1)

12( ; 2)

12(2 ; )

( 1; 1)− −12( 2 ; )− −

12( ; 2)− −

1yx

=

17

For B: 2yx

=

x 2− 1− 1 2 y 1− 2− 2 1

For C: 4yx

=

x 4− 1− 1 4 y 1− 4− 4 1

Let’s now plot the points and draw the graphs of A, B and C. Notice that as the number in the numerator (a) gets larger, the branches of the hyperbolas

are stretched vertically away from the x-axis. The branches of the graph of 4yx

= are

stretched further away from the x-axis when compared to the graph of 2yx

= .

1 2 31−2−3−

1

2

3

1−

2−

3−

0

(2 ;1)

2yx

=(1; 4)4

(1; 2)

(4 ;1)

4−

4yx

=

( 2 ; 1)− −

( 1; 2)− −

( 4 ; 1)− −

1yx

=

( 1; 4)− −

4− 4

18

Let’s now consider what happens if the value of the number in the numerator is negative.

Consider the graph of 1y

x−=

As with lines and parabolas, the negative sign indicates a reflection in the x-axis.

As with the mother graph 1yx

= , the graph of 1y

x−= gets closer and closer to the axes but

never actually cuts them.

The vertical line 0x = (lying on the y-axis) is the vertical asymptote of the graph. The horizontal line (lying on the x-axis) is the horizontal asymptote of the graph.

Conclusion:

The value of a in the equation ay qx

= + (ignoring negative signs), determines the vertical

stretch of the branches of the hyperbola from the x-axis. The larger the value of a , the

further the stretch away from the axes. A negative sign will cause a reflection in the x-axis.

1 2 31−2−3−

1

2

3

1−

2−

3−

0

0x =

0y =

( 1;1)−

12( ; 2)−

12( 2 ; )−

(1; 1)−

12(2 ; )−

12( ; 2)−

1yx

−=

19

Investigation of the effect of the value of q

Consider the following graph: 2 1yx

= +

The branches of the mother graph 1yx

= stretch vertically to form the graph of 2yx

= and

then the newly-formed graph is shifted 1 unit up to form the graph of 2 1yx

= + .

Let’s draw the graph of 2yx

= and then shift it up 1 unit and see what effect this shifting has.

For 2x = − 2 12

y = = −−

( 2 ; 1)− − lies on 2yx

=

For 1x = − 2 21

y = = −−

( 1 ; 2)− − lies on 2yx

=

For 1x = 2 21

y = = (1 ; 2) lies on 2yx

=

For 2x = 2 12

y = = (2 ;1) lies on 2yx

=

Notice that the graph of 2 1yx

= + cuts the x-axis at ( 2 ; 0)− .

The horizontal asymptote shifts 1 unit up and has the equation 1y = .

The constant in the equation2 1yx

= + therefore represents the horizontal asymptote.

The vertical asymptote is still the line 0x = (lying on the x-axis).

1 2 31−2−3−

1

2

3

1−

2−

3−

0

(2 ;1) 2yx

=

(1; 2)

( 2 ; 1)− −

( 1; 2)− −

4− 4

(1; 3)

(2 ; 2)

1y =

2 1yx

= +

( 2 ; 0)−

( 1; 1)− −

20

Conclusion:

The value of q in the equation ay qx

= + determines the shift of the graph of ayx

= up or

down. It also represents the horizontal asymptote of the graph of ay qx

= + .

EXAMPLE 3

Given: 3 1yx

= − − [Note: 3x

− is the same as 3x

−]

(a) Determine algebraically the coordinates of the x-intercept for this graph.

(b) Describe the different transformations of 3yx

= to 3 1yx

= − − .

(c) Sketch the graph of 3 1yx

= − − on a set of axes, clearly showing the asymptotes and

x-intercept. Solutions (a) Let 0y =

30 1

0 33

( 3 ; 0)

xx

x

−= −

∴ = − −∴ = −

−

(b) 3 1yx

= − − is the graph of 3yx

= reflected in the x-axis and then shifted 1 unit down.

3yx

= (Start with 3yx

= )

3yx

−= (Reflect 3yx

= in the x-axis)

3 1yx

−= − (Shift 3yx

−= one unit down)

21

(c) First draw the horizontal asymptote 1y = − on a set of axes.

Then plot the x-intercept ( 3 ; 0)−

After this, select one negative x-value and one positive x-value. Find the

corresponding y-values by substituting these x-values into 3 1yx

= − −

For 1x = − 3 1 3 1 21

y −= − = − =−

( 1 ; 2)− lies on one branch

For 1x = 3 1 3 1 4

1y −= − = − − = − (1 ; 4)− lies on the other branch

We already know the shape from (b) and we can now draw the graph as follows:

1 2 31−2−3−

1

2

3

1−

2−

3−

0

4

( 1; 2)−

4−

( 3 ; 0)−4− 4

(1; 4)−

3 1yx

= − −

1y = −

22

EXERCISE 3

(a) Given: 1yx

= and 5yx

=

(1) Which graph has branches that have the furthest stretch away from the x-axis? Explain. (2) Sketch the graphs on the same set of axes.

(3) Now sketch the graph of 5yx

= − on the same set of axes.

(b) Given: 2 2yx

= +

(1) Write down the equations of the vertical and horizontal asymptotes. (2) Determine the coordinates of the x-intercept. (3) Sketch the graph on a set of axes.

(c) Given: 4 1yx

= − −

(1) Write down the equations of the vertical and horizontal asymptotes. (2) Determine the coordinates of the x-intercept. (3) Sketch the graph on a set of axes.

(d) Sketch the graph of 3 2yx

= + on a set of axes. Indicate the coordinates of the

x-intercept as well as the asymptotes. (e) Match the equations on the left to the graphs on the right.

(1) 2yx

=

(2) 4yx

=

(3) 3yx

= −

(4) 3 1yx

= − +

23

EXPONENTIAL FUNCTIONS (SKETCHING EXPONENTIAL GRAPHS)

Consider 2xy = and 12

xy =

A set of x-values { }1; 0 ;1− has been selected and the corresponding y-values have been calculated in each case. The graphs are shown on the right.

2xy =

x 1− 0 1

y 12

1 2

The x-intercept is calculated by letting 0y = : 0 2x=

But there are no real values of x for which 2 0x = and therefore there is no x-intercept. You might like to check this by substituting a few negative and positive values of x as well as 0 into the equation. The graph will therefore not cut the x-axis. In fact, the graph has a horizontal asymptote lying on the x-axis with equation 0y = . The y-axis is not a vertical asymptote.

12

xy =

x 1− 0 1

y 2 1 12

The x-intercept is calculated by letting 0y = :

102

x =

But there are no real values of x for which

1 02

x =

and therefore there is no x-intercept.

You might like to check this by substituting a few negative and positive values of x as well as 0 into the equation. The graph will therefore not cut the x-axis. In fact, the graph has a horizontal asymptote lying on the x-axis with equation 0y = . The y-axis is not a vertical asymptote.

1− 1

1−

(1; 2)

1

2

2xy =

(0 ;1)12( 1; )− 1

2

1− 1

1−

( 1; 2)−

1

2

12

x

y =

(0 ;1)12(1; )1

2

24

The graphs of 2xy = and 12

xy =

are called the “mother graphs” of the exponential

functions and based on these graphs, we can generate different types of exponential graphs,

depending on the value of a, b and q in the general equation of an exponential function, which

is . xy a b q= + where 0 1b< < or 1b > . These restrictions on b will be explained later in the

chapter.

Investigation of the effect of the value of b Let’s investigate the effect of b on the shape of different exponential graphs by comparing the

following graphs.

(a) We will first consider graphs where 1b > .

A: 2xy = B: 3xy = C: 4xy =

For all of these graphs, the y-intercept is (0 ;1) since 0 1b =

Select one other x-value and determine the corresponding y-value.

For A: Choose 1x = 12 2y = = (1 ; 2) lies on the graph of A

For B: Choose 1x = 13 3y = = (1 ; 3) lies on the graph of B

For C: Choose 1x = 14 4y = = (1 ; 4) lies on the graph of C

The graphs are sketched below:

Notice that if 1b > , all of the exponential graphs move upwards as the x-values

increase. Also as the value of b increases in value, the steeper the graph becomes.

C is steeper than B since 4 3> and B is steeper than A since 3 2> .

1 21−

1

2

3

0

4

(1; 2)

2−

(1 ; 3)

(1; 4)

2x

y =3x

y =4x

y =

25

(b) Now let’s discuss exponential graphs where 0 1b< < .

Consider the following graphs:

D: 12

x

y =

E: 13

x

y =

F: 14

x

y =

For all of these graphs, the y-intercept is (0 ;1) since 0 1b =

Select one other x-value and determine the corresponding y-value. We will choose

negative x-values to avoid fractions.

For D: Choose 1x = − 11 2

2y

− = =

( 1 ; 2)− lies on the graph of D

For E: Choose 1x = − 11 3

3y

− = =

( 1 ; 3)− lies on the graph of E

For F: Choose 1x = − 11 4

4y

− = =

( 1 ; 4)− lies on the graph of F

The graphs are sketched below:

Notice that if 0 1b< < , all of the exponential graphs move downwards as the x-values

increase. Notice that as the value of the base b decreases in value, the steeper the

graph becomes. F is steeper than E since 1 14 3< and E is steeper than D since 1 1

3 2< .

The value of b therefore affects the shape of the exponential graph.

1 21−

1

2

3

0

4

( 1; 2)−

2−

( 1; 3)−

( 1; 4)−

12( ) x

y = 13( ) x

y =1

4( ) x

y =

26

Investigation of the effect of the value of a Consider the following exponential graphs:

G: 2xy = H: 2.2xy = I: 3.2xy = J: 12.2

x

y =

We will calculate the y-intercept and one other point for each graph.

For G: y-intercept Let 0x = 02 1y = = (0 ;1) Select 1x = 12 2y = = (1 ; 2)

For H: y-intercept Let 0x = 02.2 2y = = (0 ; 2) Select 1x = 12.2 4y = = (1 ; 4)

For I: y-intercept Let 0x = 03.2 3y = = (0 ; 3) Select 1x = 13.2 6y = = (1 ; 6)

For J: y-intercept Let 0x = 012. 2

2y = =

(0 ; 2)

Select 1x = 112. 1

2y = =

(1 ;1)

The graphs have been drawn below on the same set of axes.

Notice that the value of a causes a vertical stretch of the mother graphs.

Also note that except for graphs H and J, the graphs have different y-intercepts.

1 21−

1

2

3

0

4

(1; 2)

2−

(1; 6)

(1; 4)

5

6

(1;1)

27

As with lines, parabolas and hyperbolas, the negative sign indicates a reflection in the x-axis.

The graphs of 2xy = − and 12

x

y = −

are reflections of the two mother graphs in the x-axis.

Investigation of the effect of the value of q

As with other graphs discussed thus far, the graph of . xy a b q= + is the graph of . xy a b= shifted up or down by q units.

For example, 2 1xy = + is the graph of 2xy = shifted 1 unit up. The horizontal asymptote is indicated by the constant in the equation 2 1xy = + . The equation of the asymptote is 1y = .

1−

(1; 2)−

1

2−

2xy = −

(0 ; 1)−12( 1; )− −

12−

2xy =

12

x

y = −

( 1; 2)− −

12

x

y =

(0 ;1)

Note: In the expression 2x− , the value of b is 2 and not 2− . Remember that 2 (2)x x− = − and

2 ( 2)x x− ≠ − for all values of x (not true for even values of x). For example, if 2x = , then:

22 4− = − and 2( 2) 4− = 2 22 ( 2)∴− ≠ −

1 21−

1

2

3

0

4

(1; 2)

2−

(1 ; 3)

1y =

2xy =

2 1xy = +

28

Conclusion:

The value of b in the equation . xy a b q= + determines the shape and steepness.

If 1b > , then the graph moves upwards from left to right as the x-values increase. As the value of b increases, the graphs get steeper.

If 0 1b< < , then the graph moves downwards from left to right as the x-values increase. As the value of b decreases, the graphs get steeper. The value of a determines the vertical stretch of the mother graph.

The value of q represents the shift of the graph of . xy a b= up or down. It also indicates where the horizontal asymptote cuts the y-axis. A negative sign will cause a reflection in the x-axis. EXAMPLE 4

Given: 13 13

x

y = − +

(a) Determine algebraically the coordinates of the intercepts with the axes for this graph. (b) Write down the equation of the horizontal asymptote.

(c) Describe the different transformations of 13

x

y =

to 13 13

x

y = − +

.

(d) Sketch the graph of 13 13

x

y = − +

on a set of axes, clearly showing the horizontal

asymptote and intercepts with the axes. Solutions (a) y-intercept: Let 0x =

013 1 3 1 2

3y = − + = − + = −

(0 ; 2)−

x-intercept: Let 0y =

10 3 13

x = − +

13 13

x ∴ =

1 13 3

x ∴ =

1x∴ = (1 ; 0) (b) Horizontal asymptote is 1y =

29

(c) 13

x

y =

undergoes a vertical stretch to form the

graph of 133

x

y =

with a y-intercept of 3.

133

x

y =

is then reflected in the x-axis to form

133

x

y = −

with a y-intercept of 3− .

133

x

y = −

is then shifted 1 unit up to form the graph

of 13 13

x

y = − +

with a y-intercept of 2− , x-intercept

of 1 and a horizontal asymptote 1y = .

(d) First draw the horizontal asymptote.

Then determine the intercepts with the

axes (if they exist).

If there is no x-intercept, select one

negative x-value and one positive

x-value and determine the corresponding

y-values. Plot these points and draw

the graph.

Check the shape by considering the

transformations of the graph from the

mother graph.

EXERCISE 4

(a) Given: 2xy = 4xy = 6xy =

(1) Which graph is the steepest? Explain. (2) Write down the equation of the horizontal asymptote for the three graphs. (3) Sketch the graphs on the same set of axes.

(b) (a) Given: 12

x

y =

14

x

y =

16

x

y =

(1) Which graph is the steepest? Explain. (2) Write down the equation of the horizontal asymptote for the three graphs. (3) Sketch the graphs on the same set of axes.

3

3−

2−

1y =

1

1− 1

2−

1y =

13 13

x

y = − +

(1; 0)

(0 ; 2)−

1

30

(c) Given: 2.2xy = 4.2xy = 132

x

y =

(1) Determine the coordinates of the y-intercept for each graph. (2) Write down the equation of the horizontal asymptote for all three graphs. (3) Sketch the graphs on the same set of axes indicating the y-intercept and one other point.

(d) Given: 12

x

y = −

132

x

y = −

(1) Determine the coordinates of the y-intercept for each graph. (2) Write down the equation of the horizontal asymptote for both graphs. (3) Sketch the graphs on the same set of axes indicating the y-intercept and one other point.

(4) Explain the transformations of 12

x

y =

into the graph of 132

x

y = −

(e) Given: 2 2xy = − 2 1xy = +

(1) Determine the intercepts of 2 2xy = − with the axes. (2) Write down the equation of the horizontal asymptote of 2 2xy = − . (3) Sketch the graph of 2 2xy = − on a set of axes. (4) Explain the transformation of 2xy = into the graph of 2 2xy = − (5) Sketch the graph of 2 1xy = + on the same set of axes. (6) Explain algebraically why the graph of 2 1xy = + does not cut the x-axis.

(f) Given: 1 44

x

y = −

(1) Determine the intercepts of this graph with the axes. (2) Write down the equation of the horizontal asymptote (3) Sketch the graph on a set of axes.

(4) Explain the transformation of 14

x

y =

into the graph of 1 44

x

y = −

31

(g) On different axes, draw neat sketch graphs of the following exponential graphs. Indicate the coordinates of intercepts with the axes as well as the horizontal asymptotes.

(1) 2.4 2xy = − (2) 2.4 2xy = +

(3) 12 24

x

y = −

(4) 12 24

x

y = +

(h) Match the equations on the left to the graphs on the right. (1) 4xy = (2) 2xy =

(3) 12

x

y =

(4) 12

x

y = −

(5) 13

x

y =

(6) 1 42

x

y = −

(7) 142

x

y =

For interest

Here is a short explanation as to why we restrict the base b to 0 1b< < or 1b > for

exponential graphs. The values 0b = , 1b = and all negative values of b are excluded.

If 0b = , then 0 0xy = = which is a horizontal line and not an exponential graph.

If 1b = , then 1 1xy = = which is a horizontal line and not an exponential graph.

If b is negative, and you raise b to a rational power, you may not get a real number.

For example, if 2b = − , then 12( 2)− will not be possible to calculate since your calculator

will state Math ERROR.

In Grade 11 you will learn that 12( 2) 2− = − which is a non-real number.

32

Here is a short summary of the main features of the four functions discussed, the different types of functions, how to recognise each function and the methods required to sketch the graphs of these functions. Linear functions ( y ax q= + ) The value of a determines the steepness of the line (ignoring negative signs). The greater the value of a, the steeper the line. 0a < means a reflection in the x-axis. The value of q is the vertical shift up or down.

All of the different types of lines are shown below:

How to sketch a straight line: If 0q = : Plot (0 ; 0) and one other point and draw the graph of y ax= If 0q ≠ : Draw the graph of y ax= and shift it up or down. Find the x-intercept. Alternatively, use the dual-intercept method to sketch the graph. Quadratic functions ( 2y ax q= + ) (x has been squared) The value of a determines the vertical stretch (ignoring negative signs). The greater the value of a, the closer the arms are to the y-axis. 0a < means a reflection in the x-axis. The value of q is the vertical shift up or down.

All of the different types of parabolas are shown below:

How to sketch a parabola: If 0q = : Plot (0 ; 0) and one other point and draw the graph of 2y ax= If 0q ≠ : Draw the graph of 2y ax= and shift it up or down. Indicate the y-intercept. Determine the x-intercepts, if they exist.

Hyperbolic functions (ay qx

= + ) (x is in the denominator)

Ignoring negative signs, a determines the vertical stretch of the branches away from the axes. The greater the value of a, the greater the stretch away from the axes.

0a < means a reflection in the x-axis

00

aq

><

00

aq

<>

00

aq

<<

0a >

0a <0q =

00

aq

>>

00

aq

>=

00

aq

>>

00

aq

<=

00

aq

><

00

aq

<>

00

aq

<<

0a >

0a <

33

The value of q is the vertical shift up or down. The hyperbola has two asymptotes: y q= is the equation of the horizontal asymptote

0x = is the equation of the vertical asymptote

All of the different types of hyperbolas are shown below: How to sketch a hyperbola: If 0q = : Select one negative x-value and one positive x-value and determine the corresponding y-values. Plot these points and draw the two branches. The asymptotes lie on the axes (vertical: 0x = ; horizontal: 0y = ) If 0q ≠ : Draw the horizontal asymptote y q= . The vertical asymptote is 0x = . Determine the x-intercept by letting 0y = and solving for x. Then select one negative x-value and one positive x-value and determine the corresponding y-values. Plot these points and draw the two branches. Exponential functions ( . xy a b q= + ) (x is the exponent) The value of b affects the shape and steepness of the graph:

For 1b > : As the value of b increases, the steeper the graph. For 0 1b< < As the value of b decreases, the steeper the graph. The value of a causes a vertical stretch of the mother graph and the y-intercept is affected.

0a < means a reflection in the x-axis. The value of q is the vertical shift up or down. The exponential graph has one horizontal asymptote: y q=

All of the different types of exponential graphs are shown below:

00

aq

<=

00

aq

>=

00

aq

><

00

aq

<<

00

aq

>>

00

aq

<>

010

abq

<>=

010

abq

>>=

010

abq

>>>

00 1

0

ab

q

>< <>

00 1

0

ab

q

>< <<

010

abq

>><

00 1

0

ab

q

>< <=

00 1

0

ab

q

<< <=

1b > 0 1b< <

34

How to sketch an exponential graph: If 0q = : Determine the y-intercept (let 0x = ). Then select one negative x-value and one positive x-value and determine the corresponding y-values. Plot these points and draw the graph. The horizontal asymptote will be the line 0y = . If 0q ≠ : Draw the horizontal asymptote y q= Determine the y-intercept (let 0x = ). Determine the x-intercept (let 0y = ). If the graph has no x-intercept, select one negative x-value and one positive x-value and determine the corresponding y-values. Plot these points and draw the graph. The following exercise is a mixed exercise in which you will be required to identify and sketch the graphs of all functions studied thus far. Use the above summary to guide you. EXERCISE 5 (a) Identify the type of graph and then draw a neat sketch of the graph:

(1) 2y x= (2) 2yx

= (3) 22y x=

(4) 2y = (5) 2xy = (6) 2 2y x= +

(7) 22 2y x= − + (8) 2 1yx

= − + (9) 1 12

x

y = − +

(10) 2 2y x= − − (11) 2x = (12) 2.2 2xy = − (b) Sketch the graphs of the following functions on different axes. You will need to re-work the equations algebraically into the standard equations for the functions you have studied.

(1) 3x y= (2) 3xy = (3) 3xy =

(4) 2

3xy = (5)

33

xy −= (6) 3 xy

x−=

(7) 3 33

xy += (8) (3 )(3 )y x x= − + (9) 3 3xy −= +

(10) 3(3 1)xy = − (11) 12 2 3x xy += + − (12) 22 4

2 2

x

xy −=+

00 1

0

ab

q

<< <>

y

x

00 1

0

ab

q

<< <<

y

x

010

abq

<><

y

x010

abq

<>>

y

x

35

FUNCTIONAL NOTATION A function may be represented by means of functional notation. Consider the function ( ) 3f x x=

The symbol ( )f x is used to represent the value of the output given an input value.

In other words, the y-values corresponding to the x-values are given by ( )f x , i.e. ( )y f x= .

For example, if 4x = , then the corresponding y-value is obtained by substituting 4x = into

3x . For 4x = , the y-value is (4) 3(4) 12f = = .

The brackets in the symbol (4)f do not mean f multiplied by 4, but rather the y-value when

4x = . Also, ( ) 3f x x= is read as “ f of x is equal to 3x ”.

We can also use other letters to name functions. For example, ( ), ( )g x h x and ( )p x may be

used.

EXAMPLE 5 If 2( ) 3 1f x x= − , determine the value of: (a) (2)f (b) ( 3)f − (c) ( )f a (d) (3 )f x (e) 3 ( ) 1f x + (f) x if ( ) 2f x = Solutions (a) 2( ) 3 1f x x= − (b) 2( ) 3 1f x x= −

2(2) 3(2) 13(4) 111

f = −= −=

2( 3) 3( 3) 13(9) 126

f − = − −= −=

(c) 2( ) 3 1f x x= − (d) 2( ) 3 1f x x= −

(e) 2( ) 3 1f x x= − (f)

2

2

2

2

3 ( ) 3(3 1)

3 ( ) 9 3

3 ( ) 1 9 3 1

3 ( ) 1 9 2

f x x

f x x

f x x

f x x

∴ = −

∴ = −

∴ + = − +

∴ + = −

2

2

2

(3 ) 3(3 ) 1

3(9 ) 1

27 1

f x x

x

x

= −

= −

= −

2

2

( ) 3( ) 1

3 1

f a a

a

= −

= −

2

2

2

2

( ) 3 1

2 3 1

0 3 3

0 10 ( 1)( 1)

1 or 1

f x x

x

x

xx x

x x

= −

∴ = −

∴ = −

∴ = −∴ = + −∴ = − =

36

EXERCISE 6 (a) If 2( ) 2 1f x x x= − + , determine the value of:

(1) (1)f (2) ( 1)f − (3) (2)f

(4) ( 2)f − (5) 12

f

(6) 12

f −

(7) ( )f a (8) (2 )f x (9) 2 ( )f x (10) ( ) 2f x + (11) ( )f x− (12) ( 1)f x − (13) 2 ( 1) 3f x − − (14) ( )f x h+ (15) ( ) ( )f x h f x+ −

(b) If 2( ) 5g x x= − , determine the value(s) of:

(1) x if ( ) 4g x =

(2) x if ( ) 20g x =

(3) x if ( ) 1 5g x x= −

(4) x if ( ) 6 14g x x= − (c) Given: ( ) 2 4g x x= − − (1) Write down an expression for ( 5)g p −

(2) If (2 ) 10g p = , determine the value of p. (d) Describe the transformation from f to g if:

(1) 2( )f x x= and ( ) 3 ( )g x f x=

(2) ( ) 2xf x = and ( ) ( )g x f x= −

(3) 2( )f xx

= and ( ) ( ) 2g x f x= +

(4) 2( ) 5f x x= + and 2( ) 1g x x= +

(5) 2( ) 2f x x= and 2( ) 8g x x=

(6) 6( ) 1f xx

= + and 6( ) 1g xx

= − −

37

FURTHER CHARACTERISTICS OF THE FOUR FUNCTIONS Domain and range In Grade 8 and 9, you learnt that a function is a rule which when applied to a given set of input values, produces a set of output values. For example, suppose that the rule is 2y x= . When applied to a set of input x-values { }2 ; 1; 0 ;1; 2; 3x∈ − − , the output y-values can be obtained by substituting the given x-values into the equation (rule) 2y x= . The output values are therefore { }4 ; 2 ; 0 ; 2 ; 4; 6y ∈ − −

The domain is simply all of the inputs (x-values) that are used.

The range is the set of outputs (y-values) obtained from the input values. EXAMPLE 6 Determine the domain and range for the following functions: (a) There are four x-values in the domain:

{ }2 ; 1; 0 ;1x∈ − −

There are four y-values in the range:

{ }1; 0 ;1; 2y ∈ −

For each point on the graph, there is a

y-value corresponding to an x-value.

(b) There are an infinite number of x-values in

the domain from 2− to 1 (including 2− but

excluding 1). The domain is written as:

[ )2 ;1x ∈ − or 2 1x− ≤ <

There are an infinite number y-values in

the range from 1− to 2 (including 1− but

excluding 2). The range is written as:

[ )1; 2y ∈ − or 1 2y− ≤ <

You can use the vertical ruler method to obtain the domain and range: For domain: Keep the edge of the ruler vertical and slide it across the graph from

left to right. Where the edge starts cutting the graph, the domain starts (as read off from the x-axis). Where it stops cutting the graph the domain ends.

For range: Keep the edge of the ruler horizontal and slide it across the graph from bottom to top. Where the edge starts cutting the graph, the range starts (as read off from the y-axis). Where it stops cutting the graph the range ends.

1−

2−

1 22− 1−

(0 ;1)1

2 (1; 2)

( 1; 0)−

( 2 ; 1)− −

1−

2−

1 22− 1−

1

2 (1; 2)

( 2 ; 1)− −

38

1−

2−

1 22− 1−

1

2

9

(c) There are an infinite number x-values in

the domain (as indicated by the arrows).

We write the domain as follows:

( ; )x ∈ −∞ ∞ or x∈


the range (as indicated by the arrows).

We write the range as follows:

( ; )y ∈ −∞ ∞ or y ∈

(d) There are an infinite number x-values in



( ; )x ∈ −∞ ∞ or x∈


the range from 9 downwards. The range is

written as:

( ]; 9y ∈ −∞ or 9y ≤

(e) There are an infinite number x-values in



( ; )x ∈ −∞ ∞ or x∈


the range from 2 upwards (excluding 2).

The range is written as:

( )2 ;y ∈ ∞ or 2y >

EXERCISE 7 State the domain and range of the following functions:

(a) (b) (c)

2

( 3 ; 2)−

(1; 4)−

03−

39

0x < 0x >

(d) (e) (f) Increasing and decreasing functions For the graph of a given function, if the y-values increase as the x-values increase, then the graph is said to be increasing. If the y-values decrease as the x-values increase, then the graph is said to be decreasing. In short, a graph is increasing if it moves upwards when moving from left to right and decreasing if it moves downwards when moving from left to right. For example, the graph on the right is increasing for all values of 0x < . As we move from left to right, the graph moves upwards. The graph is decreasing for all values of 0x > . As we move from left to right, the graph moves downwards. Note to the educator: Increasing and decreasing depends on how we define these concepts. For school purposes, we will define increasing and decreasing in terms of the gradient of the curve (positive or negative). At 0x = , the gradient is 0 and therefore we exclude 0 in the solutions. THE LINEAR FUNCTION We will now briefly revise lines of the form ax by c+ = , the gradient of a line and points of intersection of lines. EXAMPLE 7 In the diagram, two lines are drawn: 3 2 6x y+ = and 4 16x y− = The first line cuts the y-axis at A and the x-axis at B. The two lines intersect at C. Determine: (a) the coordinates of A and B. (b) the gradient of 3 2 6x y+ = (c) the gradient and y-intercept of 4 16x y− = (d) the coordinates of C (e) the values of x for which the lines are increasing or decreasing.

2 2

1

3 2 6x y+ =

4 16x y− =

40

Solutions (a) We can use the dual-intercept method: x-intercept: Let 0y = y-intercept: Let 0x =

3 2(0) 63 6

2B(2 ; 0)

xx

x

∴ + =∴ =∴ =

3(0) 2 62 6

3A(0 ; 3)

yy

y

∴ + =∴ =∴ =

(b) In Grade 9, you learnt that the gradient of a line between any two points on the line is

given by the ratio: change in -valueschange in -values

yx

We can determine the gradient by referring to the diagram or by re-writing the equation of the line in the form y ax q= + , bearing in mind that the value of a represents the gradient of the line. Between the points (0 ; 3) and (2 ; 0) ,

the y-values decrease ( 3− ) as the x-values increase ( 2+ ).

The gradient is therefore 3

2−

We can also determine the gradient as follows:

3 2 6

2 3 6x y

y x+ =

∴ = − +

3 3

2y x−∴ = + [The coefficient of x is the gradient]

(c) 4 16x y− =

4 16

1 44

y x

y x

∴− = − +

∴ = −

The gradient is 14

and the y-intercept is 4−

(d) In order to determine the coordinates of C, we need to solve simultaneous equations: 3 2 6x y+ = (1) 4 16x y− = (2) 6 4 12x y+ = (1) 2× 4 16x y− = (2)

7 28x∴ = Add like terms and constants 4x∴ = 4 4 16y∴ − = Substitute 4x = into (2) to get the corresponding y-value

4 12

3y

y∴− =∴ = −

The coordinates of C are C(4 ; 3)− (e) 3 2 6x y+ = decreases for all real values of x. 4 16x y− = increases for all real values of x.

(0 ; 3)

3−

(2 ; 0)

2+

41

Finding the equation of a linear function EXAMPLE 8 (a) Determine the equation of the following line in the form ( )f x ax q= + . Solution The y-intercept is 3. Therefore 3q = . 3y ax∴ = + Substitute the point (8 ; 1)− to get a:

1 (8) 31 8 38 4

12

aa

a

a

− = +− = +− =

= −

Therefore the equation is 1( ) 32

f x x= − +

(b) Determine the equation of the following line in the form ( )g x ax q= + . Solution Method 1

The y-intercept is 4. Therefore 4q = . 4y ax∴ = + Substitute the point ( 2 ; 0)− to get a:

0 ( 2) 40 2 42 4

2

aa

aa

= − += − +

==

Therefore the equation is ( ) 2 4g x x= +

Method 2

The gradient is 4 22

+ =+

The y-intercept is 4 Therefore, the equation is ( ) 2 4g x x= +

(8 ; 1)−

3

4

2−

4

2−

4+

2+

42

(c) Determine the equation of the following line in the form ( )f x ax q= + . Solution Method 1 Substitute the two points into y ax q= + and solve simultaneous equations: For ( 1 ; 3)− : 3 ( 1)a q= − + 3 a q∴ + = For (3 ; 1)− 1 (3)a q− = + 1 3a q∴ − − = 3 1 3a a∴ + = − − (both equal q)

4 4

13 ( 1) 2

aaq

∴ = −∴ = −∴ = + − =

Therefore the equation is ( ) 2f x x= − + Method 2

The gradient is 4 14

a −= = −+

1y x q∴ = − + Substitute either of the two points: into the equation y x q= − + to get q: ( 1 ; 3)− : 3 ( 1) q= − − + 2q∴ = or (3 ; 1)− : 1 (3) q− = − + 2q∴ = The y-intercept is 2 Therefore the equation is ( ) 2f x x= − + Method 3 Use the Analytical Geometry formula to find the gradient (see Chapter 8)

Gradient 1 3 4 13 ( 1) 4− − −= = = −− −

Therefore 1a = − . 1y x q∴ = − + Substitute one of the points into the equation y x q= − + to get q: ( 1 ; 3)− : 3 ( 1) q= − − + 2q∴ = The y-intercept is 2 Therefore, the equation is ( ) 2f x x= − +

( 1; 3)−

(3 ; 1)−

( 1; 3)−

(3 ; 1)−

4−

4+

43

EXERCISE 8 (a) In the diagram, line 4 2 8x y− = cuts the axes at A and B. Line 1x y+ = − cuts the axes at C and D. The two lines intersect at E.

Determine: (1) the coordinates of A and B. (2) the coordinates of C and D. (3) the gradient of 4 2 8x y− = (4) the gradient of 1x y+ = − (5) the coordinates of E (6) the values of x for which the lines are increasing or decreasing. (b) Given: 3 4x y− = and 2 5x y− = (1) On the same set of axes, draw neat sketch graphs of the two functions. (2) Determine the coordinates of the point of intersection. (c) Determine the equations of the following lines in the form ( )f x ax q= + : (1) (2) (3) (4) (5) (6) (d) (1) Determine the equation of the line passing through the point (0 ; 1)− and

parallel to the x-axis. Do you remember what the gradient of this line is?

(2) Determine the equation of the line passing through the point and parallel to the y-axis. Do you remember what the gradient of this line is?

( 1; 0)−

1x y+ = −

4 2 8x y− =

3

( 4 ; 1)− − 5−

( 7 ; 2)−

3−

6

6−

4−

( 3 ; 1)− −

(2 ; 4) ( 2 ; 7)−(2 ; 5)

44

2( ) 6g x x= +

2( ) 4f x x= −

2( ) 3h x x=

THE QUADRATIC FUNCTION EXAMPLE 9

In the diagram, the graphs of 2( ) 4f x x= − , 2( ) 6g x x= + and 2( ) 3h x x= are shown.

The graph of f cuts the axes at A, C and D.

The graph of g cuts the y-axis at B. Determine:

(a) the coordinates of A, B, C and D (b) the coordinates of E and F. (c) the values of x for which f is increasing (d) the values of x for which g is decreasing (e) the maximum or minimum values of f and g (f) the turning points of f and g (g) the equation of the axis of symmetry for f and g Solutions (a) The y-intercept of f is (0 ; 4) and the y-intercept of g is (0 ; 6) A(0 ; 4) and B(0 ; 6) Let 0y = 20 4x= − +

2 4 0( 2)( 2) 0

2 or 2C( 2 ; 0) D(2 ; 0)

xx x

x x

∴ − =∴ + − =∴ = − =

−

(b) E and F are the points of intersection between f and h ( ) ( )h x f x= (y-values are equal at the points of intersection)

2 2

2

2

3 4

4 4 0

1 0( 1)( 1) 0

1 or 1

x x

x

xx x

x x

∴ = −

∴ − =

∴ − =∴ + − =∴ = − =

Now determine the corresponding y-values by substituting into either of the two equations: 2( 1) 3( 1) 3f − = − = 2(1) 3(1) 3f = = E( 1 ; 3)− F(1 ; 3)

(c) f increases for all 0x <

(d) g decreases for all 0x <

45

3

(2 ; 7)

2−

(1; 6)

2

(e) The maximum value refers to the largest y-value on a graph and the minimum value refers to the smallest y-value on a graph. The maximum value of f is 4 and the minimum value of g is 6 (f) A turning point on a graph is the point at which the graph changes from increasing to decreasing or decreasing to increasing. The turning point of f is (0 ; 4) The turning point of g is (0 ; 6) (g) The axis of symmetry of a parabola is a vertical line that divides the parabola into two congruent halves. The y-axis is the axis of symmetry for both graphs. The equation is 0x = . Finding the equation of a quadratic function EXAMPLE 10

(a) Determine the equation of the following graph in the form 2( )f x ax q= + .

Solution The y–intercept is 3 3q∴ = 2 3y ax∴ = + Substitute the point (2 ; 7) into 2 3y ax= + to get the value of a

27 (2) 37 3 44 4

1

aa

aa

∴ = +∴ − =∴ =∴ =

2( ) 1 3f x x∴ = +

(b) Determine the equation of the following graph in the form 2( )f x ax q= + .

Solution The x-intercept form of the equation of a parabola can be used: 1 2( )( )y a x x x x= − − where 1x and 2x represent the x-intercepts.

( ( 2))( 2)( 2)( 2)

y a x xy a x x

∴ = − − −∴ = + −

Substitute (1 ; 6) to get the value of a

6 (1 2)(1 2)6 (3)( 1)6 3

2

aa

aa

= + −∴ = −∴ = −∴ = −

2

2

2( 2)( 2)

2( 4)

( ) 2 8

y x x

y x

f x x

∴ = − + −

∴ = − −

∴ = − +

Notice that the x-intercept form of 22 8y x= − + is:

2

2

2 8

2( 4)2( 2)( 2)2( ( 2))( 2)

y x

xx xx x

= − +

= − −= − + −= − − − −

46

2( ) 2 2f x x= −

2( ) 9f x x= − +

2( )g x x= −

2( ) 9f x x= −

2( ) 1g x x= − −

EXERCISE 9

(a) In the diagram, the graph of 2( ) 2 2f x x= − is shown. The graph of f cuts the axes at

A, B and C. Line h is parallel to the x-axis and

passes through A. Determine:

(1) the coordinates of A, B and C (2) the values of x for which f is increasing (3) the values of x for which f is decreasing (4) the minimum value of f (5) the turning point of f (6) the equation of the axis of symmetry of f (7) the domain and range of f (8) the equation of h and the value of its gradient (9) the domain and range of h (b) In the diagram, the graphs of 2( ) 9f x x= − + 2( )g x x= − , h and p are shown. The graph of

f cuts the axes at A, B and C. Line p cuts the

x-axis at B and is parallel to the y-axis.

Determine:

(1) the coordinates of A, B, and C. (2) the values of x for which f is increasing (3) the values of x for which g is decreasing (4) the maximum value of f and g (5) the turning points of f and g (6) the equation of the axis of symmetry for f (7) the domain and range of f and g (8) the equation of h if ( ) ( ) 12h x g x= − + . Describe the transformations.

(9) the equation of p and the value of its gradient. (10) the domain and range of p (c) Given: 2( ) 9f x x= − and 2( ) 1g x x= − −

(1) Determine the coordinates of A, B, C and D (2) Determine the coordinates of E and F.

47

2

( 1; 6)−

3−( 2 ; 5)− −

3

1−(1; 2)−

4−

(1;15)

4 5− 5

25

y x q= +y x q= − +y x=y x= −

y q=q

ay qx

= +ayx

=

3 2yx

= +

(d) Determine the equation of each of the following parabolas in the form 2( ) .f x ax q= + (1) (2) (3) (4) (5) (6) THE HYPERBOLIC FUNCTION An interesting characteristic of the hyperbola is that it has two axes of symmetry. Each axis of symmetry has a gradient of 1 or 1− and a y-intercept of q. EXAMPLE 11

Given: 3( ) 2g xx

= +

Determine: (a) the equations of the asymptotes (b) the equations of the axes of symmetry (c) the values of x for which g is decreasing (d) the domain and range of g

2− 2

8−

48

3y = −

(3 ; 4)−

4y x= +

Solutions (a) Horizontal asymptote is 2y = Vertical asymptote is 0x = (b) 2y x= − + and 2y x= + (c) The graph of g is decreasing on the interval ( ; 0)x ∈ −∞ as well as the interval (0 ; )x ∈ ∞ . (d) Domain of g: 0x x∈ ≠ Range of g: 2y y∈ ≠ Finding the equation of a hyperbola EXAMPLE 12

Determine the equation of the following graph in the form ( ) af x qx

= + .

Solution The horizontal asymptote is 3y = −

3q∴ = −

3ayx

∴ = −

Substitute the point (3 ; 4)− :

4 33

13

33( ) 3

a

a

a

f xx

− = −

∴− =

∴− =−∴ = −

EXERCISE 10 (a) The line 4y x= + is an axis of symmetry of the

graph of 2( )f x qx

−= + . The graph of f cuts the

x-axis at A. (1) Write down the value of q (2) Write down the equation of f (3) State the domain and range of f (4) For which values of x is f increasing? (5) Write down the equations of the asymptotes (6) Write down the equation of the other axis of symmetry.

49

( 2 ; 6)− −

(4 ; 3)2y =

( 3 ; 5)−

(1; 2)−

3

( 8 ; 8)− −4−

(2 ; 6)−

2−

4−

1y x= − + 2y x= −

3−

( 1; 8)− −

3y x= −

( 2 ; 7)− −

3y = −(2 ; 4)−5y x= +

y

x( 4 ; 2)−

514 4( ; )

121−

(2 ; 6)−

1− 1

(2 ; 4)

(b) Determine the equation of each of the following hyperbolas in the form ( ) .af x qx

= +

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (c) For each function below, state the domain and range and determine the equation:

(1) (2) (3)

50

4−

8−

3

1−

13−

(2 ;1)

THE EXPONENTIAL FUNCTION Finding the equation of an exponential graph EXAMPLE 13 (a) Determine the equation of the given graph in the form ( ) .2xf x a q= + Solution The horizontal asymptote is 8y = − .2 8xy a∴ = − Substitute (0 ; 4)− :

04 .2 84 8

4

aa

a

− = −∴− = −∴ =

( ) 4.2 8xf x∴ = − (b) Determine the value of b and q if the

equation of the given graph is ( ) xg x b q= − + Solution The horizontal asymptote is 3y = 3xy b∴ = − + Substitute ( 1 ; 0)− :

10 310 3

0 1 33 1

13

b

bb

b

b

−= − +

∴ = − +

∴ = − +∴− = −

∴ =

1( ) 33

x

g x ∴ = − +

13

b = and 3q =

(c) Determine the equation of the given graph in the form . xy a b q= +

Solution

1.3

xy a b= −

The graph passes though the origin.

51

3−

4−

1−

( 1; 2)−

3

1

3(2 ;12)

13−

(2 ;1)

Substitute the point (0 ; 0) :

0 10 .3

103

13

a b

a

a

= −

∴ = −

∴ =

1 1.3 3

xy b∴ = −

Substitute the point (2 ;1) :

2

2

2

1 11 .3 3

3 1

0 40 ( 2)( 2)

2 or 2

b

b

bb b

b b

= −

∴ = −

∴ = −∴ = + −∴ = − =

Choose 2b = since 2b ≠ −

1 1.23 3

xy∴ = −

EXERCISE 11 (a) Determine the equation of the given graph in the form ( ) .2xf x a q= + (b) Determine the equation of the given graph in the form ( ) xg x b q= + (c) Determine the equation of the given graph in the form ( ) xh x b q= − + (d) Determine the equation of the given graph in the form ( ) . xf x a b q= +

52

18

2

16

4−

( 2 ; 60)−

11−

1

1−

Let 0x =Let 0y =

(e) Determine the equation of the given graph in the form ( ) . xf x a b q= + (f) Determine the equation of the given graph in the form ( ) . xg x a b q= + GRAPH INTERPRETATION This topic involves determining the lengths of line segments using the different functions you have studied thus far. It also discusses the graphical interpretation of inequalities, which is so important for matric. Determining the length of line segments using graphs The following information is extremely important for determining the length of line segments: ● The length of a line segment is always positive.

A 1x = and OA 1= unit B 1x = − and OB 1= unit C 1y = and OC 1= unit D 1y = − and OD 1= unit ● To determine a length OA along the y-axis, let 0x = The y-value will be the length of OA. To determine a length OB along the x-axis, let 0y = The positive x-value will be the length of OB. ● The point of intersection (B) is obtained by equating the equations of the functions ( ( ) ( )f x g x= ) and solving for x and hence for y. The value of x will give the horizontal length OA and the value of y will give the vertical length OB.

53

x

A BAB y y= −

D CCD x x= −

1− 1 2

2( ) 1f x x= − +( ) 1g x x= − −

● A vertical length between two graphs can be calculated using the formula: top graph bottom graphy y− (substitute the x-value into this formula to get the required length) A horizontal length between two graphs can be calculated using the formula: right end point left end pointx y− Using graphs to solve inequalities The diagram below shows the graph of a parabola and a line. A and B are the points of intersection of the two graphs. Both graphs cut the x-axis at 1− and the graph of the parabola cuts the x-axis at 1. We can use the graphs to solve the following inequalities graphically: (a) For which values of x is ( ) 0f x > ? We are required to determine the x-values for which the y-values of f are positive. Where the parabola lies above the x-axis will be where the y-values are positive. The solution lies between x-values of 1− and 1. 1 1x∴− < < or we can write ( 1 ;1)x ∈ − (b) For which values of x is ( ) 0f x ≤ ? We are required to determine the x-values for which the y-values of f are negative. Where the parabola lies below the x-axis will be where the y-values are negative. 1 or 1x x∴ ≤ − ≥ or we can write ( ; 1] [1 ; )x ∈ −∞ − ∪ ∞ (c) For which values of x is ( ) 0g x > ? We are required to determine the x-values for which the y-values of g are positive. Where the line lies above the x-axis will be where the y-values are positive. 1x∴ < − or we can write ( ;1)x ∈ −∞

54

( ) 3f x x= − +

( ) 3 3g x x= +

3−

4

(d) For which values of x is ( ) ( )f x g x≥ ? We are required to determine the values of x for which the y-values of f are greater than or equal to the y-values of g. This is where the graph of f is above the graph of g (between A and B). 1 2x∴− ≤ ≤ or we can write [ 1 ; 2]x ∈ − (e) For which values of x is ( ) ( )f x g x< ? We are required to determine the values of x for which the y-values of f are smaller than the y-values of g. This is where the graph of f is below the graph of g 1 or 2x x∴ < − > or we can write ( ; 1) (2 ; )x ∈ −∞ − ∪ ∞ EXAMPLE 14 Two lines cut the y-axis at A. Determine: (a) the length of OA, OB, OC and BC (b) the length of OD, FE, OF and DE (c) the length of OJ, GH, OG and JH (d) the values of x for which ( ) 0f x ≥ (e) the values of x for which ( ) 0g x < (f) the values of x for which ( ) ( )f x g x≤ Solutions

(a) A(0 ; 3) For Bx OA 3∴ = units 0 3x= − + 3x∴ = OB 3∴ = units For Cx B CBC y y= − 0 3 3x= + BC 3 ( 1)∴ = − − 3 3x∴− = BC 4∴ = units 1x∴ = − OC 1∴ = unit (b) OD 4= units FE 4∴ = units (opp sides rectangle) Substitute 4x = into 3y x= − + 4 3 1y = − + = − E F 1y y∴ = = − OF 1∴ = units and DE 1= unit

55

2( ) 3 12f x x= −

( ) 3 6g x x= −( ) 3 6h x x= − +

(c) OJ 3= units (d) ( ) 0f x ≥ GH 3∴ = units (opp sides rectangle) 3x∴ ≤ or write ( ; 3]x ∈ −∞ Substitute 3y = − into 3 3y x= + 3 3 3x− = + (e) ( ) 0g x < 3 6x∴− = 1x∴ < − or write ( ; 1)x ∈ −∞ − 2x∴ = − OG 2∴ = units and JH 2= units (f) ( ) ( )f x g x≤ 0x∴ ≥ or write [0 ; )x ∈ ∞

EXAMPLE 15

The diagram below shows the graphs of 2( ) 3 12f x x= − , ( ) 3 6g x x= − and ( ) 3 6h x x= − + . The graphs of f , g and h share a common x-intercept (D). The graph of f and g intersect

at D and F. the diagram is not drawn to scale. Determine:

(a) the length of AB and CD.

(b) the length of OE and EF.

(c) the length of GH if OG 3= units

(d) the length of OI if IJ 6= units

(e) the length of KL if OR 1= unit

(f) the length of ON if MP 18= units

(g) the values of x for which ( ) 0f x ≥

(h) the values of x for which ( ) ( )f x g x<

56

Solutions

(a) A BAB 6 ( 12) 6y y= − = − − − = units Let 0y = in 23 12y x= −

2

2

0 3 12

0 40 ( 2)( 2)

2 or 2

x

xx x

x x

= −

∴ = −∴ = + −∴ = − =

D CCD 2 ( 2) 4x x= − = − − = units. (b) Determine the coordinates of E, the point of intersection of f and g

2

2

2

3 12 3 6

3 3 6 0

2 0( 1)( 2) 0

1 or 2

x x

x x

x xx x

x x

− = −

∴ − − =

∴ − − =∴ + − =∴ = − =

At E, 1x = − 3( 1) 6 9y = − − = − The coordinates of E are ( 1 ; 9)− − OE 1∴ = unit and EF 9= units (c) OG 3= which means that 3x = − at G. Substitute 3x = − into ( ) 3 6g x x= − to get y. ( 3) 3( 3) 6 15g − = − − = − GH 15∴ = units

(d) IJ 6= which means that 6y = − at J. Substitute 6y = − into ( ) 3 6h x x= − + to get x

6 3 63 12

4

xx

x

− = − +∴ =∴ =

OI 4∴ = units (e) OR 1= which means that 1x = 2

K LKL (3 6) (3 12)y y x x= − = − − − Substitute 1x = 2KL (3(1) 6) (3(1) 12) 6∴ = − − − = units

(f) M PMP (3 6) ( 3 6) 3 6 3 6 6 12y y x x x x x= − = − − − + = − + − = − 18 6 12x∴ = − (since MP 18= ) 6 30x∴− = − 5x∴ = ON 5∴ = units

(g) ( ) 0f x ≥ for 2 or 2x x≤ − ≥ or ( ; 2] [2 ; )x ∈ −∞ − ∪ ∞ (h) ( ) ( )f x g x< for 1 2x− < < or ( 1 ; 2)x ∈ −

57

( ) 5f x x= +

( ) 2g x x= +

( ) 4f x x= − +

( ) 4f x x= − +

( ) 2g x x= −

2−1−

EXERCISE 12 (a) Two lines ( ) 4f x x= − + and ( ) 2g x x= − intersect at R. By using the information on

the diagram, determine: (1) the length of OA, OP, OB and OC (2) the length of AP and BC (3) the length of OD, EF, OF and DE (4) the length of OK, GH, OG and KH (5) the length of RS and OS (6) the values of x for which ( ) 0f x ≥ (7) the values of x for which ( ) 0g x < (8) the values of x for which ( ) ( )f x g x< (b) The diagram shows the line ( ) 5f x x= +

Determine: (1) the length of OP and OQ

(2) the length of AB if OB 2= units (3) the length of DC if OD 7= units

(4) the length of OF if EF 3= units

(5) the length of OH if HG 3= units

(6) the values of x for which ( ) 0f x < (c) Two lines ( ) 4f x x= − + and ( ) 2g x x= + intersect at E.

Determine: (1) the length of AB, CD and DF

(2) the length of PQ if OR 3= units

(3) the length of OU if ST 8= units

(4) the length of GH if 12HK 1=

(5) the values of x for which ( ) 0f x >

(6) the values of x for which ( ) 0g x ≤

(7) the values of x for which ( ) ( )g x f x<

58

2( ) 9f x x= −

( ) 3g x x= −

2( ) 2f xx

= −

( ) 2g x =

1( ) 22

x

f x = −

(d) The diagram shows the graphs of 2( ) 9f x x= − and ( ) 3g x x= − intersecting at E and D. The graph of f cuts the axes at B, C and D.

Determine: (1) the length of AB, CT and CD (2) the length of OF and EF (3) the length of GH if OH 1= unit

(4) the length of OV if VW 8= units (5) the length of JL if OK 1= unit

(6) the length of OQ if PR 8= units (7) the values of x for which ( ) 0f x >

(8) the values of x for which ( ) ( )f x g x≥

(e) The graph of 2( ) 2f xx

= − and ( ) 2g x = is shown.

Determine:

(1) the length of OA.

(2) the length of BC if OB 4 units=

(3) the length of OD and OE

(4) the coordinates of F

(5) the values of x for which ( ) 0f x ≥

(6) the values of x for which ( ) 0f x <

(7) the values of x for which ( ) 2f x > −

(8) the values of x for which ( ) ( )f x g x>

(f) The graph of 1( ) 22

xf x = −

is shown.

Determine:

(1) the length of OA and CD.

(2) the values of x for which ( ) 0f x >

(3) the values of x for which ( ) 0f x ≤



(6) the values of x for which ( ) 1f x ≤ −

59

(1; 3)

CONSOLIDATION AND EXTENSION EXERCISE

(a) Given: 2( ) 1f x x= − and ( ) 1g x x= −

(1) Sketch the graph of f and g on the same set of axes.

(2) Determine the domain and range of f.

(3) If ( ) ( ) 3h x f x= + , write down the coordinates of the turning point of h.

(4) Determine the maximum value of p if ( ) ( )p x h x= − .

(5) Sketch the graphs of h and p on the same set of axes.

(6) Determine the y-intercept of the graph of ( )y g x= − .

(7) Determine the x-intercept of the graph of ( 2)y g x= + .

(b) Given: ( ) 3xf x = and 3( )g xx

=

(1) Sketch the graph of h if ( ) 3 ( ) 1h x f x= − and describe the transformations.

(2) Sketch the graph of p if ( ) ( ) 1p x g x= − + and describe the transformations.

(3) Write down the range of h.

(4) Write down the domain of p.

(c) In the diagram below, the graphs of f and g are shown. The graphs intersect at (1 ; 3) . Determine:

(1) the equation of f

(2) the coordinates of the turning point of f.

(3) the equation of the axis of symmetry of f.

(4) the equation of the horizontal asymptote of g

(5) the equation of g and the coordinates of A

(6) the domain and range of f and g

(7) the values of x for which f is decreasing?

(d) The graph of 1( )2

x

f x q = +

passes through

the origin. Determine:

(1) the value of q

(2) the length of AB of OA 2= units

(3) the length of OC if 78CD = units

(4) the length of OT if MN 2= units

(5) the equation of g , the image when the graph of f is translated downwards so that the image of M lies on the x-axis.

f

60

H( 1; 4)−

f

g

T(4 ; 1)−

G(1; 8)−

( 2 ; 12)− −

2( ) 3 12f x x= − +

( ) 3 6g x x= − +

(e) The graph of ( )f x mx c= + and 2( )g x ax b= + intersect at D and on the x-axis at A.

H and T lie on line f . Determine:


(2) the coordinates of A and hence C

(3) the equation of g

(4) the length of AC and OB

(5) the length of DE and DF

(6) the length of ST

(7) the length of OM if 454QR =

(8) the values of x for which ( ) 0g x ≥

(9) the values of x for which ( ) ( )g x f x<

(f) The graph of ( ) . xf x a b q= + and ( ) 6ag xx

= +

cut the x-axis at A. The graph of f passes through the point ( 2 ; 12)− − and cuts the y-axis at 4.

Determine:

(1) the range of f


(3) the equation of g

(4) the equation of h if the graph of f is reflected in the x-axis

(5) the values of x for which f is increasing *(g) Given: 2( ) 3 12f x x= − + and ( ) 3 6g x x= − +

Determine:


(2) the values of x for which ( ) ( )f x g x>


(4) the values of x for which ( ). ( ) 0f x g x <

(5) the values of x for which ( ). ( ) 0f x g x ≥

(6) the possible values of t if the graph of

2( ) 3 12f x x t= − + −

(i) does not cut or touch the x-axis.

(ii) cuts the x-axis in two distinct points.

mathematics workshop functions

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