MIND ACTION SERIES
With the Educators, for the Educators
MATHEMATICS WORKSHOP
FUNCTIONS
GRADE 10 TEXTBOOK
(Chapter 6)
Presented by: Jurg Basson
Attending this Workshop = 10 SACE Points
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
y
−1
−1
0
0
1
1
The graph of this line is obtained by plotting the points on the Cartesian plane and drawing a
solid line through the points.
y=x
2
1
(1;1)
(0 ; 0)
−2 −1
(−1;1) −1
1
2
−2
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 xvalues. 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).
1
Let’s investigate this by comparing the graphs of the following straight lines:
A:
y=x
B:
y=
1
x
2
y = 2x
C:
y = 3x
D:
Lines A, B, C and D pass through the origin since for all of the given graphs, the y-value is 0
if x = 0 .
To sketch the graphs of these lines, select one x-value and then determine the corresponding
y-value. For all four graphs, choose x = 1 .
A:
y = (1) = 1
Line A passes through the points (0 ; 0) and (1 ; 1)
B:
1
1
y = (1) =
2
2
 1
Line B passes through the points (0 ; 0) and  1; 
 2
C:
y = 2(1) = 2
Line C passes through the points (0 ; 0) and (1 ; 2)
D:
y = 3(1) = 3
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.
y = 3x
3
y = 2x
(1; 3)
y = 1x
2
(1; 2)
1
(1;1)
−2
1
x
2
(1 ; 12 )
(0 ; 0)
−3
y=
−1
1
2
3
−1
−2
−3
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 > 12 ). 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.
2
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.
−1
1
x
y
1
−1
0
0
The graph of this line is obtained by plotting the points on the Cartesian plane and
drawing a solid line through the points.
y = −x
2
1
(−1;1)
−2
(0 ; 0)
1
−1
−1
2
(1; − 1)
−2
If you now compare the mother graph y = 1x to the graph of y = −1x , 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)
1
Consider the graph of y = − x
2
1
affects the steepness of the line. The mother
2
graph y = x transforms into a line that is not as steep as the mother graph line.
We already know that the number
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
y=
1
x and then reflecting this newly-formed graph about the x-axis.
2
To sketch the graph of this line, select one x-value, determine the corresponding yvalue, plot the point formed and the y-intercept (0 ; 0)
Choose x = 1
1
1
y = − (1) = −
2
2
3
The line passes through the point (1; − 12 )
Now plot this point and the y-intercept and draw the graph of the line.
y = 1x
2
1
y=
1
x
2
(1;1)
(1; 12 )
−3
−2
−1
1
−1
2
3
(1; − 12 )
1
y=− x
2
−2
Notice that the mother graph y = x is transformed into y =
1
x and then this new line
2
1
is reflected in the x-axis to form the graph of the line y = − x .
2
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)
y = x + 3 and y = x − 2
Consider the graphs of the following:
For y = x + 3 :
If x = 0 then y = 0 + 3 = 3
If x = 1 then y = 1 + 3 = 4
The line passes through the points (0 ; 3) and (1 ; 4)
For y = x − 2 :
If x = 0 then y = 0 − 2 = −2
If x = 1 then y = 1 − 2 = −1
The line passes through the points (0 ; − 2) and (1 ; − 1)
4
y = x+3
(1; 4)
4
(0 ; 3) 3
y=x
2
y = x−2
1
(1;1)
(0 ; 0)
−3 −2 −1
(−1;1) −1
1
2
(1; − 1)
−2
(0 ; − 2)
It should be clear to you from the graphs that y = x + 3 is the mother graph y = x
shifted 3 units up and y = x − 2 is the mother graph y = x shifted 2 units down.
Also, the y-intercept of y = x + 3 is 3 and the y-intercept of y = x − 2 is −2 .
(b)
y = − x + 1 and y = − x − 2
Consider the graphs of the following:
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 y = − x + 1 by
shifting y = −x one unit up and y = − x − 2 by shifting y = −x two units down.
y = −x +1
y = −x
y=x
3
2
y = −x − 2
−3 −2
1
−1
1
2
−1
−2
−3
It should be clear to you from the graphs that y = − x + 1 is the line y = −x
shifted 1 unit up and y = − x − 2 is the line y = −x shifted 2 units down.
Also, the y-intercept of y = − x + 1 is 1 and the y-intercept of y = − x − 2 is −2 .
5
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 y = −2 x + 4
Solution
First draw the graph of y = 2 x , reflect this graph in the x-axis to form y = −2 x and then shift
y = −2 x four units up to form y = −2 x + 4 .
The line y = 2 x cuts the y-axis at 0.
y = −2 x + 4
y = 2x
4 (0 ; 4)
y = −2 x
Now choose x = 1
∴ y = 2(1) = 2
3
Plot the point (1 ; 2) and draw the line y = 2 x
2
(1; 2)
1
Now reflect y = 2 x in the x-axis to form
(0 ; 0)
y = −2 x . The point (1 ; 2) transforms into
−2
the point (1 ; − 2) .
−1
1
2
−1
−2
Then shift y = −2 x four units up.
(1; − 2)
Draw the graph of y = −2 x + 4 .
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 y = −2 x + 4
y-intercept:
Let x = 0
x-intercept:
Let y = 0
0 = −2 x + 4
∴ 2x = 4
∴x = 2
y = −2(0) + 4 = 4
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.
6
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 y = 2 and x = 1 are shown below.
(−2 ; 2)
(1; 2)
2
y=2
1
−1
−2
2
1
(1; 2)
1
2
−2
−1
1
−1
−1
−2
−2
2
(1; − 1)
x =1
For the line y = 2 , the y-values will be constant but the x-values will vary.
For the line x = 1 , the x-values will be constant but the y-values will vary.
EXERCISE 1
(a)
(b)
(c)
Given:
y=x
3
y= x
2
1
y= x
4
(1)
Which of the three lines is the steepest? Explain
(2)
Sketch the three graphs on the same set of axes.
Given:
y = −x
3
y=− x
2
1
y=− x
4
(1)
Which of the three lines is the steepest? Explain
(2)
Sketch the three graphs on the same set of axes.
Given:
y = −x + 4
y = 3x − 6
(1)
Describe the transformation of y = x into the graph of y = − x + 4
(2)
Describe the transformation of y = x into the graph of y = 3 x − 6
(3)
Sketch the graphs of these two functions on the same set of axes using
transformations or the dual-intercept method.
7
(d)
Match the equations on the left to the graphs on the right.
(1)
y = −2 x
(2)
1
y = x+2
4
(3)
y = −x − 2
(4)
y = 4x
(5)
1
y= x
4
(6)
y = −4 x
(7)
x=4
(8)
y=4
QUADRATIC FUNCTIONS (SKETCHING PARABOLAS)
Consider the graph of y = x2
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
y
−2
4
−1
1
0
0
1
1
2
4
The graph of y = x 2 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
●
●
y = x2
(−2 ; 4)
(2 ; 4)
4
3
2
(−1;1)
1
−2
0
−1
−1
8
(1;1)
1
2
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 y = ax 2 + 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:
y = x2
y=
B:
1 2
x
2
C:
y = 2x2
D:
y = 3x 2
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
x =1.
A:
y = (1) 2 = 1
Parabola A passes through the points (0 ; 0) and (1 ; 1)
B:
1
1
y = (1)2 =
2
2
 1
Parabola B passes through the points (0 ; 0) and  1; 
 2
C:
y = 2(1)2 = 2
Parabola C passes through the points (0 ; 0) and (1 ; 2)
D:
y = 3(1) 2 = 3
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.
(1; 3)
3
2
(1; 2)
1
(1;1)
(1 ; 12 )
−2
−1 (0 ; 0)
1
2
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 x2 for
each parabola is positive and the graphs are concave up (happy!).
9
Let’s now consider what happens if the value of the coefficient of x is negative.
(a)
Consider the graph of y = − x 2
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.
−2
x
y
−4
−1
0
0
−1
2
−4
1
−1
The graph of y = − x 2 is obtained by plotting the points on the Cartesian plane and
drawing a curve through the points.
y = x2
4
3
2
1
2
1
(1; − 1)
−2 −1 0
(−1; −1) −1
−2
−3
(−2 ; − 4)
(2 ; − 4)
−4
y = − x2
If you now compare the mother graph y = 1x 2 to the graph of y = −1x 2 , it is interesting
to
note that the graph of y = − x 2 is the reflection of the mother graph in the x-axis.
The negative sign therefore causes a reflection in the x-axis.
(b)
1
Consider the graph of y = − x2
2
1
affects the shape of the curve. The mother graph
2
y = x 2 transforms into a parabola with arms that arms that are less close to the y-axis.
We already know that the number
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 x = 1
1
1
∴ y = − (1)2 = −
2
2
10
1

The parabola passes through the point 1; − 
2

Now plot this point and the y-intercept and draw the graph of the parabola.
y = x2
y=
2
1
1 2
x
2
(1;1)
(1 ; 12 )
−2
−1
1
−1
2
(1; − 12 )
−2
1
y = − x2
2
−3
1 2
x and then this graph was
2
1
reflected in the x-axis to form the graph of the parabola y = − x 2
2
Notice that the mother graph y = x 2 transformed into y =
Conclusion:
The value of a in the equation y = ax 2 + 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 a > 0 , then the parabola is concave up (happy face!)
If a < 0 , then the parabola is concave down (sad face!)
The value of a is sometimes referred to as the vertical stretch factor
11
Investigation of the effect of the value of q
Consider the following graphs:
A:
y = x2 − 1
y = x2 + 2
B:
We will first draw the mother graph y = x 2 and then based on this graph, we will draw
parabola A and B.
The graph of A is the graph of y = x 2 shifted 1 unit down.
The graph of B is the graph of y = x 2 shifted 2 units up.
y = x2 + 2
5
4
3
y = x2 − 1
(1; 3)
2
1
(−1; 0)
−2 −1
(1;1)
(1; 0)
1
2
0
−1 (0 ; − 1)
Conclusion:
The value of q in the equation y = ax 2 + q determines the shift of the graph of y = ax 2 up
or down. It also represents the y-intercept of the graph of y = ax 2 + q .
EXAMPLE 2
Given:
y = −2 x 2 + 8 and y = −2 x 2 − 2
(a)
Sketch the graphs of y = −2 x 2 + 8 and y = −2 x 2 − 2 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 y = 2 x 2 in the x-axis to form y = −2 x 2 and then shift y = −2 x 2
eight units up to form y = −2 x 2 + 8 . Then shift the graph of y = −2 x 2 two units
down to form y = −2 x 2 − 2 .
12
For y = −2 x 2 , choose x = 1
∴ y = −2(1) 2 = −2
y = −2 x 2 passes through the origin (0 ; 0) and the point (1 ; − 2)
8
y = −2 x 2 + 8
6
4
2
−4 −2
−2
2
4
(1; − 2)
−4
−6
y = −2 x 2
−8
(b)
y = −2 x 2 − 2
Consider the graph y = −2 x 2 + 8 :
Let x = 0
y-intercept:
2
x-intercept:
Let y = 0
2
∴ y = −2(0) + 8 = 8
∴0 = −2x + 8
(0 ; 8)
∴ 2 x2 − 8 = 0
∴ x2 − 4 = 0
∴ ( x + 2)( x − 2) = 0
∴ x = −2 or x = 2
(−2 ; 0)
(2 ; 0)
Consider the graph y = −2 x 2 − 2 :
y-intercept: Let x = 0
∴ y = −2(0)2 − 2 = −2
(0 ; − 2)
x-intercept:
The graph doesn’t cut the x-axis
There are no x-intercepts
Also notice that the equation 0 = −2 x 2 − 2
has no real solutions:
0 = −2 x 2 − 2
∴ 2 x 2 = −2
∴ x 2 = −1
∴ x = ± −1 which is non-real
13
EXERCISE 2
(a)
(b)
(c)
(d)
(e)
(f)
Given:
y = x2
y = 3x 2
y = 4 x2
(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
Given:
y = − x2
1
y = − x2
2
1
y = − x2
4
(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
Given:
y=
1 2
x
4
y=
3 2
x
2
y = −4 x 2
(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
Given:
y = x 2 − 4 and y = x 2 + 3
(1)
Sketch the graphs of y = x 2 − 4 and y = x 2 + 1 on the same set of axes.
(2)
For these graphs, determine algebraically the coordinates of the intercepts with
the axes.
Given:
y = −4 x 2 + 4 and y = − x 2 − 2
(1)
Sketch the graphs of y = −4 x 2 + 4 and y = − x 2 − 2 on the same set of axes.
(2)
For these graphs, determine algebraically the coordinates of the intercepts with
the axes.
Given:
(1)
(2)
1
y = − x2 + 8
2
1
Sketch the graph of y = − x2 + 8
2
Determine algebraically the coordinates of the intercepts of this parabola with
the axes.
14
(g)
Match the equations on the left to the graphs on the right.
1 2
x
4
(1)
y=
(2)
1
y = − x2
3
(3)
1
y = − x2
5
(4)
y = −5x2 −
(5)
y = x2 − 1
(6)
y = 2 x2
1
2
HYPERBOLIC FUNCTIONS (SKETCHING HYPERBOLAS)
Consider the graph of y =
1
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
− 12
0
1
2
1
2
y
− 12
−1
−2
1
0
2
1
1
2
Notice from the table that if x = 0 then y =
1
which is undefined.
0
Since the y-value is undefined at x = 0 , this means that the graph has no y-intercept.
For the graph to have an x-intercept we let y = 0 and solve for x.
1
∴0 =
x
There is no value of x which satisfies this equation. Hence the graph has no x-intercept.
1
The graph of y = is obtained by plotting the points on the Cartesian plane and drawing a
x
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.
15
An asymptote is a horizontal or vertical line that a graph approaches but never touches.
The vertical line x = 0 (lying on the y-axis) is called the vertical asymptote of the graph.
The horizontal line y = 0 (lying on the x-axis) is called the horizontal asymptote of the
graph.
3
x=0
y=
1
x
( 12 ; 2)
2
(1;1)
1
−3
−2
−1
1
(−2 ; − 2 )
(−1; − 1)
0
1
(2 ; 12 )
2
3
y=0
−1
(− 12 ; − 2) −2
−3
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
a
equation of a hyperbola, which is y = + q .
x
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:
y=
1
x
B:
y=
2
x
C:
y=
4
x
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.
16
y=
For B:
x
y
−2
x
y
−1
−2
−1
y=
For C:
−4
−1
2
x
2
1
1
2
4
x
−1
−4
4
1
1
4
Let’s now plot the points and draw the graphs of A, B and C.
y=
4
y=
4
x
(1; 4)
2
x
3
−4
(−4 ; − 1)
−3
−2
−1
2
(1; 2)
1
1
y=
x
0
(4 ;1)
(2 ; 1)
1
2
3
4
−1
(−2 ; − 1)
(−1; − 2)
−2
−3
(−1; − 4)
−4
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 y =
stretched further away from the x-axis when compared to the graph of y =
17
2
.
x
4
are
x
Let’s now consider what happens if the value of the number in the numerator is negative.
−1
Consider the graph of y =
x
As with lines and parabolas, the negative sign indicates a reflection in the x-axis.
x=0
3
y=
(−2 ;
−3
−1
x (− 12 ; 2)
1)
2
(−1;1)
−2
−1
2
1
0
−1
−2
2
1
(1; − 1)
3
(2 ; − 12 )
y=0
( 12 ; − 2)
−3
As with the mother graph y =
1
−1
, the graph of y =
gets closer and closer to the axes but
x
x
never actually cuts them.
The vertical line x = 0 (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:
a
+ q (ignoring negative signs), determines the vertical
x
stretch of the branches of the hyperbola from the x-axis. The larger the value of a , the
The value of a in the equation y =
further the stretch away from the axes. A negative sign will cause a reflection in the x-axis.
18
Investigation of the effect of the value of q
2
+1
x
1
2
The branches of the mother graph y = stretch vertically to form the graph of y = and
x
x
2
then the newly-formed graph is shifted 1 unit up to form the graph of y = + 1 .
x
2
Let’s draw the graph of y = and then shift it up 1 unit and see what effect this shifting has.
x
2
2
y=
= −1
For x = −2
( −2 ; − 1) lies on y =
−2
x
Consider the following graph: y =
For x = −1
y=
2
= −2
−1
( −1 ; − 2) lies on y =
For x = 1
y=
2
=2
1
(1 ; 2) lies on y =
2
x
For x = 2
y=
2
=1
2
(2 ; 1) lies on y =
2
x
2
x
(1; 3)
3
(2 ; 2)
2
y=
(1 ; 2)
2
+1
x
y =1
1
(2 ;1)
y=
( −2 ; 0)
−4
−2
−3
−1
0
1
2
3
4
( −1 ; − 1)
−1
( −2 ; − 1)
( −1; − 2)
−2
−3
Notice that the graph of y =
2
+ 1 cuts the x-axis at ( −2 ; 0) .
x
The horizontal asymptote shifts 1 unit up and has the equation y = 1 .
2
+ 1 therefore represents the horizontal asymptote.
x
The vertical asymptote is still the line x = 0 (lying on the x-axis).
The constant in the equation y =
19
2
x
Conclusion:
a
a
+ q determines the shift of the graph of y = up or
x
x
a
down. It also represents the horizontal asymptote of the graph of y = + q .
x
The value of q in the equation y =
EXAMPLE 3
3
y = − −1
x
Given:
[Note: −
3
−3
is the same as
]
x
x
(a)
Determine algebraically the coordinates of the x-intercept for this graph.
(b)
Describe the different transformations of y =
(c)
3
3
to y = − − 1 .
x
x
3
Sketch the graph of y = − − 1 on a set of axes, clearly showing the asymptotes and
x
x-intercept.
Solutions
(a)
(b)
Let y = 0
−3
0=
−1
x
∴ 0 = −3 − x
∴ x = −3
( −3 ; 0)
3
3
y = − − 1 is the graph of y = reflected in the x-axis and then shifted 1 unit down.
x
x
y=
3
x
(Start with y =
y=
−3
x
(Reflect y =
y=
−3
−1
x
(Shift y =
20
3
)
x
3
in the x-axis)
x
−3
one unit down)
x
(c)
First draw the horizontal asymptote y = −1 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
3
corresponding y-values by substituting these x-values into y = − − 1
x
−3
y = −1 = 3 −1 = 2
For x = −1
( −1 ; 2) lies on one branch
−1
−3
y = − 1 = −3 − 1 = −4
For x = 1
(1 ; − 4) lies on the other branch
1
We already know the shape from (b) and we can now draw the graph as follows:
3
y = − −1
x
4
3
2
( −1; 2)
1
( −3 ; 0)
−4
−3
−2
−1
0
1
2
3
4
y = −1
−1
−2
−3
−4
(1; − 4)
21
EXERCISE 3
(a)
(b)
(c)
(d)
(e)
Given:
y=
1
x
and
y=
5
x
(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 y = −
Given:
y=
5
on the same set of axes.
x
2
+2
x
(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.
Given:
4
y = − −1
x
(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.
3
+ 2 on a set of axes. Indicate the coordinates of the
x
x-intercept as well as the asymptotes.
Sketch the graph of y =
Match the equations on the left to the graphs on the right.
(1)
(2)
(3)
(4)
2
x
4
y=
x
y=
3
x
3
y = − +1
x
y=−
22
EXPONENTIAL FUNCTIONS (SKETCHING EXPONENTIAL GRAPHS)
1
Consider y = 2 x and y =  
2
x
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.
y = 2x
y = 2x
x
−1
1
2
y
0
1
1
2
2
1
(−1;
1)
2
The x-intercept is calculated by letting y = 0 :
0 = 2x
1
2
(1; 2)
(0 ;1)
−1
But there are no real values of x for which
2x = 0 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.
1
−1
The graph will therefore not cut the x-axis. In fact, the graph has a horizontal asymptote
lying on the x-axis with equation y = 0 . The y-axis is not a vertical asymptote.
1
y= 
2
x
x
x
−1
0
y
2
1
1
y = 
2
(−1; 2)
1
1
2
2
1 (0 ;1)
1
2
The x-intercept is calculated by letting y = 0 :
1
0= 
2
−1
x
(1 ; 12 )
1
−1
But there are no real values of x for which
x
1
  = 0 and therefore there is no x-intercept.
2
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 y = 0 . The y-axis is not a vertical asymptote.
23
x
1
The graphs of y = 2 and y =   are called the “mother graphs” of the exponential
2
functions and based on these graphs, we can generate different types of exponential graphs,
x
depending on the value of a, b and q in the general equation of an exponential function, which
is y = a. b x + q where 0 < b < 1 or b > 1 . 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 b > 1 .
A:
y = 2x
y = 3x
B:
C:
y = 4x
For all of these graphs, the y-intercept is (0 ; 1) since b0 = 1
Select one other x-value and determine the corresponding y-value.
For A:
Choose x = 1
y = 21 = 2
(1 ; 2) lies on the graph of A
For B:
Choose x = 1
y = 31 = 3
(1 ; 3) lies on the graph of B
For C:
Choose x = 1
y = 41 = 4
(1 ; 4) lies on the graph of C
The graphs are sketched below:
x
4
y=
x
3
y=
x
2
y=
4 (1; 4)
3
(1 ; 3)
2
(1 ; 2)
1
−2
−1
1
0
2
Notice that if b > 1 , 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 .
24
(b)
Now let’s discuss exponential graphs where 0 < b < 1 .
Consider the following graphs:
D:
1
y = 
2
x
1
y= 
3
E:
x
F:
1
y = 
4
x
For all of these graphs, the y-intercept is (0 ; 1) since b0 = 1
Select one other x-value and determine the corresponding y-value. We will choose
negative x-values to avoid fractions.
For D: Choose x = −1
1
y= 
2
For E: Choose x = −1
1
y= 
3
Choose x = −1
1
y= 
4
For F:
−1
=2
( −1 ; 2) lies on the graph of D
=3
( −1 ; 3) lies on the graph of E
=4
( −1 ; 4) lies on the graph of F
−1
−1
The graphs are sketched below:
y=
y=
(1
2 )
x
(1
3 )
x
y=
(1
4 )
x
( −1; 4)
4
( −1; 3)
3
( −1; 2)
2
1
−2
−1
1
0
2
Notice that if 0 < b < 1 , 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
4
< 13 and E is steeper than D since
The value of b therefore affects the shape of the exponential graph.
25
1
3
< 12 .
Investigation of the effect of the value of a
Consider the following exponential graphs:
G:
y=2
x
H:
y = 2.2
x
I:
y = 3.2
x
J:
1
y = 2.  
2
We will calculate the y-intercept and one other point for each graph.
For G:
For H:
For I:
y-intercept
y = 20 = 1
Let x = 0
1
(0 ; 1)
Select x = 1
y=2 =2
(1 ; 2)
y-intercept
Let x = 0
y = 2.20 = 2
Select x = 1
y = 2.21 = 4
(1 ; 4)
y-intercept
Let x = 0
y = 3.20 = 3
Select x = 1
y = 3.21 = 6
(1 ; 6)
y-intercept
Let x = 0
1
y = 2.   = 2
2
Select x = 1
1
y = 2.   = 1
2
(0 ; 2)
(0 ; 3)
0
For J:
(0 ; 2)
1
(1 ; 1)
The graphs have been drawn below on the same set of axes.
6 (1 ; 6)
5
(1 ; 4)
4
3
−2
−1
2
(1 ; 2)
1
(1;1)
1
0
2
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.
26
x
As with lines, parabolas and hyperbolas, the negative sign indicates a reflection in the x-axis.
x
1
x
The graphs of y = −2 and y = −   are reflections of the two mother graphs in the x-axis.
2
1
y = 
2
x
y = 2x
Note:
In the expression −2x , the value of b
is 2 and not −2 .
Remember that −2 x = −(2) x and
(0 ;1)
−1
(−1; − 12 )
(−1; − 2)
1
y = − 
2
−2 x ≠ (−2) x for all values of x
(not true for even values of x).
For example, if x = 2 , then:
1
− 12
−22 = −4 and (−2)2 = 4
(0 ; − 1)
∴−22 ≠ (−2)2
(1; − 2)
−2
x
y = −2 x
Investigation of the effect of the value of q
As with other graphs discussed thus far, the graph of y = a. b x + q is the graph of y = a. b x
shifted up or down by q units.
For example, y = 2 x + 1 is the graph of y = 2 x shifted 1 unit up. The horizontal asymptote is
indicated by the constant in the equation y = 2 x + 1 . The equation of the asymptote is y = 1 .
y = 2x + 1
4
y = 2x
3
(1 ; 3)
2
(1 ; 2)
y =1
1
−2
−1
1
0
27
2
Conclusion:
The value of b in the equation y = a. b x + q determines the shape and steepness.
If b > 1 , 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 < b < 1 , 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 y = a. b x 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
x
1
y = −3   + 1
3
Given:
(a)
Determine algebraically the coordinates of the intercepts with the axes for this graph.
(b)
Write down the equation of the horizontal asymptote.
(c)
1
1
Describe the different transformations of y =   to y = −3   + 1 .
3
3
x
x
x
(d)
1
Sketch the graph of y = −3   + 1 on a set of axes, clearly showing the horizontal
3
asymptote and intercepts with the axes.
Solutions
(a)
y-intercept:
Let x = 0
x-intercept:
1
y = − 3   + 1 = −3 + 1 = − 2
3
Let y = 0
0
(0 ; − 2)
x
1
0 = −3   + 1
3
x
1
∴3  = 1
3
x
1
1
∴  =
3
3
∴x =1
(b)
(1 ; 0)
Horizontal asymptote is y = 1
28
x
(c)
1
y =   undergoes a vertical stretch to form the
3
3
x
1
graph of y = 3   with a y-intercept of 3.
3
x
1
y = 3   is then reflected in the x-axis to form
3
x
1
y = −3   with a y-intercept of −3 .
3
−3
x
1
y = −3   is then shifted 1 unit up to form the graph
3
y =1
x
1
of y = −3   + 1 with a y-intercept of −2 , x-intercept
3
of 1 and a horizontal asymptote y = 1 .
(d)
1
−2
First draw the horizontal asymptote.
y =1
Then determine the intercepts with the
1
axes (if they exist).
(1; 0)
If there is no x-intercept, select one
−1
negative x-value and one positive
1
x-value and determine the corresponding
y-values. Plot these points and draw
−2 (0 ; − 2)
the graph.
x
Check the shape by considering the
1
y = −3   + 1
3
transformations of the graph from the
mother graph.
EXERCISE 4
(a)
(b)
Given:
y = 2x
y = 4x
y = 6x
(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.
(a)
Given:
(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.
1
y = 
2
x
1
y = 
4
29
x
1
y = 
6
x
(c)
(d)
Given:
y = 4.2
1
y = 3 
2
x
x
(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.
Given:
1
y = − 
2
x
1
y = −3  
2
x
(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.
x
x
1
1
Explain the transformations of y =   into the graph of y = −3  
2
2
(4)
(e)
y = 2.2
x
Given:
y = 2x − 2
y = 2x + 1
(1)
Determine the intercepts of y = 2 x − 2 with the axes.
(2)
Write down the equation of the horizontal asymptote of y = 2 x − 2 .
(3)
Sketch the graph of y = 2 x − 2 on a set of axes.
(4)
Explain the transformation of y = 2 x into the graph of y = 2 x − 2
(5)
Sketch the graph of y = 2 x + 1 on the same set of axes.
(6)
Explain algebraically why the graph of y = 2 x + 1 does not cut the x-axis.
x
(f)
Given:
1
y =  −4
4
(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)
1
1
Explain the transformation of y =   into the graph of y =   − 4
4
4
x
30
x
(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)
y = 2.4 x − 2
(3)
1
y = 2  − 2
4
(2)
y = 2.4 x + 2
(4)
1
y = 2  + 2
4
x
(h)
x
Match the equations on the left to the graphs on the right.
(1)
y = 4x
(2)
y = 2x
(3)
1
y = 
2
(4)
1
y = − 
2
(5)
1
y= 
3
(6)
1
y =  −4
2
(7)
1
y = 4 
2
x
x
x
x
x
For interest
Here is a short explanation as to why we restrict the base b to 0 < b < 1 or b > 1 for
exponential graphs. The values b = 0 , b = 1 and all negative values of b are excluded.
If b = 0 , then y = 0 x = 0 which is a horizontal line and not an exponential graph.
If b = 1 , then y = 1x = 1 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.
1
For example, if b = −2 , then (−2) 2 will not be possible to calculate since your calculator
will state Math ERROR.
1
In Grade 11 you will learn that (−2) 2 = −2 which is a non-real number.
31
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. a < 0 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:
a>0
q=0
a>0
q>0
a<0
q>0
a>0
q<0
a<0
q<0
a<0
How to sketch a straight line:
If q = 0 :
Plot (0 ; 0) and one other point and draw the graph of y = ax
Draw the graph of y = ax and shift it up or down. Find the x-intercept.
If q ≠ 0 :
Alternatively, use the dual-intercept method to sketch the graph.
Quadratic functions
( y = ax 2 + 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. a < 0 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:
a>0
q=0
a>0
q>0
a<0
q=0
a<0
q<0
a>0
q<0
a<0
q>0
How to sketch a parabola:
If q = 0 :
Plot (0 ; 0) and one other point and draw the graph of y = ax 2
Draw the graph of y = ax 2 and shift it up or down. Indicate the y-intercept.
Determine the x-intercepts, if they exist.
a
Hyperbolic functions
( y = +q)
(x is in the denominator)
x
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.
If q ≠ 0 :
a>0
a < 0 means a reflection in the x-axis
a<0
32
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
x = 0 is the equation of the vertical asymptote
All of the different types of hyperbolas are shown below:
a>0
q=0
a<0
q=0
a<0
q<0
a>0
q<0
a<0
q>0
a>0
q>0
How to sketch a hyperbola:
If q = 0 :
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: x = 0 ; horizontal: y = 0 )
If q ≠ 0 :
Draw the horizontal asymptote y = q . The vertical asymptote is x = 0 .
Determine the x-intercept by letting y = 0 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
( y = a .b x + q )
(x is the exponent)
The value of b affects the shape and steepness of the graph:
b >1
0 < b <1
For b > 1 :
As the value of b increases, the steeper the graph.
For 0 < b < 1 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.
a < 0 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:
a>0
b >1
q=0
a>0
0 < b <1
q=0
a<0
b >1
q=0
a<0
0 < b <1
q=0
a>0
b >1
q<0
a>0
b >1
q>0
a>0
0 < b <1
q<0
a>0
0 < b <1
q>0
33
y
a<0
b >1
q<0
x
y
y
x
a<0
0 < b <1
q>0
x
a<0
b >1
q>0
y
a<0
0 < b <1
q<0
x
How to sketch an exponential graph:
If q = 0 :
Determine the y-intercept (let x = 0 ).
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 y = 0 .
Draw the horizontal asymptote y = q
If q ≠ 0 :
Determine the y-intercept (let x = 0 ).
Determine the x-intercept (let y = 0 ).
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)
(b)
Identify the type of graph and then draw a neat sketch of the graph:
(1)
y = 2x
(2)
y=
2
x
(3)
y = 2 x2
(4)
y=2
(5)
y = 2x
(6)
y = x2 + 2
(9)
1
y = −   +1
2
(12)
y = 2.2 x − 2
(7)
y = −2 x + 2
(8)
2
y = − +1
x
(10)
y = −2 x − 2
(11)
x=2
2
x
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.
(3)
y=
x
3
(6)
y=
3− x
x
y = (3 − x )(3 + x )
(9)
y = 3− x + 3
y = 2 x+1 + 2 x − 3
(12)
y=
(1)
x = 3y
(2)
xy = 3
(4)
y=
x2
3
(5)
y=
(7)
y=
3x + 3
3
(8)
(10)
y = 3(3x − 1)
(11)
34
3− x
3
22 x − 4
2x + 2
FUNCTIONAL NOTATION
A function may be represented by means of functional notation.
Consider the function f ( x ) = 3 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 x = 4 , then the corresponding y-value is obtained by substituting x = 4 into
3x . For x = 4 , the y-value is f (4) = 3(4) = 12 .
The brackets in the symbol f (4) do not mean f multiplied by 4, but rather the y-value when
x = 4 . Also, f ( x ) = 3 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 f ( x) = 3x 2 − 1 , determine the value of:
(a)
f (2)
(b)
f ( −3)
(c)
(d)
f (3 x )
(e)
3 f ( x) + 1
(f)
Solutions
(a)
f ( x ) = 3x 2 − 1
(b)
f ( −3) = 3( −3) 2 − 1
= 3(9) − 1
f (2) = 3(2) 2 − 1
= 3(4) − 1
= 11
(c)
f ( x ) = 3x 2 − 1
f ( x ) = 3x 2 − 1
= 26
(d)
f (a ) = 3(a ) 2 − 1
f ( x ) = 3x 2 − 1
f (3 x ) = 3(3 x ) 2 − 1
= 3a 2 − 1
= 3(9 x 2 ) − 1
= 27 x 2 − 1
(e)
f ( x ) = 3x 2 − 1
(f)
f ( x) = 3x 2 − 1
∴ 3 f ( x) = 3(3x 2 − 1)
∴ 2 = 3x 2 − 1
∴ 3 f ( x) = 9 x 2 − 3
∴ 0 = 3x 2 − 3
∴ 3 f ( x) + 1 = 9 x 2 − 3 + 1
∴ 0 = x2 − 1
∴ 0 = ( x + 1)( x − 1)
∴ x = −1 or x = 1
∴ 3 f ( x) + 1 = 9 x 2 − 2
35
f (a)
x if f ( x ) = 2
EXERCISE 6
(a)
(b)
(c)
(d)
If f ( x) = 2 x 2 − x + 1 , determine the value of:
(1)
f (1)
(2)
f ( −1)
(3)
f (2)
(4)
f ( −2)
(5)
1
f 
2
(6)
 1
f − 
 2
(7)
f (a)
(8)
f (2 x )
(9)
2 f ( x)
(10)
f ( x) + 2
(11)
f (− x)
(12)
f ( x − 1)
(13)
2 f ( x − 1) − 3
(14)
f ( x + h)
(15)
f ( x + h) − f ( x)
If g ( x ) = x 2 − 5 , determine the value(s) of:
(1)
x if g ( x ) = 4
(2)
x if g ( x ) = 20
(3)
x if g ( x ) = 1 − 5 x
(4)
x if g ( x ) = 6 x − 14
Given:
g ( x ) = −2 x − 4
(1)
Write down an expression for g ( p − 5)
(2)
If g (2 p ) = 10 , determine the value of p.
Describe the transformation from f to g if:
(1)
f ( x) = x 2
and
g ( x) = 3 f ( x)
(2)
f ( x) = 2 x
and
g ( x) = − f ( x)
(3)
f ( x) =
and
g ( x) = f ( x) + 2
(4)
f ( x) = x 2 + 5 and
g ( x) = x 2 + 1
(5)
f ( x) = 2 x 2
g ( x) = 8 x 2
(6)
f ( x) =
2
x
and
6
+ 1 and
x
6
g ( x) = − − 1
x
36
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 y = 2 x .
When applied to a set of input x-values x ∈ {−2 ; − 1; 0 ;1; 2; 3} , the output y-values can be
obtained by substituting the given x-values into the equation (rule) y = 2 x . The output values
are therefore y ∈ {−4 ; − 2 ; 0 ; 2 ; 4; 6}
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; 2)
1 (0 ;1)
There are four y-values in the range:
(−1; 0)
−2
(−2 ; − 1)
1
−1
x ∈ {−2 ; − 1; 0 ;1}
2
y ∈ {−1; 0 ;1; 2}
−1
For each point on the graph, there is a
−2
y-value corresponding to an x-value.
(b)
There are an infinite number of x-values in
2
(1; 2)
excluding 1). The domain is written as:
1
−2
(−2 ; − 1)
−1
1
−1
−2
the domain from −2 to 1 (including −2 but
2
x ∈ [ −2 ;1) or −2 ≤ x < 1
There are an infinite number y-values in
the range from − 1 to 2 (including − 1 but
excluding 2). The range is written as:
y ∈ [ −1; 2 ) or −1 ≤ y < 2
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.
37
(c)
There are an infinite number x-values in
2
the domain (as indicated by the arrows).
1
We write the domain as follows:
x ∈ ( −∞ ; ∞ ) or x ∈ 
−2
−1
1
2
−1
There are an infinite number y-values in
the range (as indicated by the arrows).
−2
We write the range as follows:
y ∈ ( −∞ ; ∞ ) or y ∈ 
(d)
There are an infinite number x-values in
9
the domain (as indicated by the arrows).
We write the domain as follows:
x ∈ ( −∞ ; ∞ ) or x ∈ 
There are an infinite number y-values in
the range from 9 downwards. The range is
written as:
y ∈ ( −∞ ; 9] or y ≤ 9
(e)
There are an infinite number x-values in
the domain (as indicated by the arrows).
We write the domain as follows:
x ∈ ( −∞ ; ∞ ) or x ∈ 
2
There are an infinite number y-values in
the range from 2 upwards (excluding 2).
The range is written as:
y ∈ ( 2 ; ∞ ) or y > 2
EXERCISE 7
State the domain and range of the following functions:
(a)
(b)
(c)
(−3 ; 2)
0
−3
(1; − 4)
38
(d)
(e)
(f)
2
2
1
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 x < 0 . As we move from left to
right, the graph moves upwards. The graph is
decreasing for all values of x > 0 . 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 x = 0 , the gradient is 0 and therefore we
exclude 0 in the solutions.
x<0
x>0
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 x + 2 y = 6 and x − 4 y = 16
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 x + 2 y = 6
(c)
the gradient and y-intercept of x − 4 y = 16
(d)
the coordinates of C
(e)
the values of x for which the lines are
increasing or decreasing.
3x + 2 y = 6
x − 4 y = 16
39
Solutions
(a)
(b)
We can use the dual-intercept method:
x-intercept: Let y = 0
y-intercept:
∴ 3 x + 2(0) = 6
B(2 ; 0)
A(0 ; 3)
In Grade 9, you learnt that the gradient of a line between any two points on the line is
change in y -values
given by the ratio:
change in x-values
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 ).
−3
The gradient is therefore
(0 ; 3)
2
We can also determine the gradient as follows:
−3
3x + 2 y = 6
(2 ; 0)
∴ 2 y = −3x + 6
∴y =
(c)
∴ 3x = 6
∴x = 2
Let x = 0
∴ 3(0) + 2 y = 6
∴2y = 6
∴y =3
−3
x + 3 [The coefficient of x is the gradient]
2
x − 4 y = 16
∴−4 y = − x + 16
1
∴y = x−4
4
The gradient is
(d)
1
and the y-intercept is −4
4
In order to determine the coordinates of C, we need to solve simultaneous equations:
3x + 2 y = 6
x − 4 y = 16
(1)
(2)
6 x + 4 y = 12
x − 4 y = 16
(1) ×2
(2)
∴7 x = 28
∴x = 4
Add like terms and constants
∴ 4 − 4 y = 16
∴−4 y = 12
Substitute x = 4 into (2) to get the corresponding y-value
∴ y = −3
The coordinates of C are C(4 ; − 3)
(e)
+2
3 x + 2 y = 6 decreases for all real values of x.
x − 4 y = 16 increases for all real values of x.
40
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 q = 3 .
∴ y = ax + 3
Substitute the point (8 ; − 1) to get a:
−1 = a(8) + 3
−1 = 8a + 3
−8a = 4
1
a=−
2
1
Therefore the equation is f ( x) = − x + 3
2
(b)
3
(8 ; − 1)
Determine the equation of the following line in the form g ( x ) = ax + q .
Solution
Method 1
The y-intercept is 4.
Therefore q = 4 .
∴ y = ax + 4
Substitute the point ( −2 ; 0) to get a:
0 = a (−2) + 4
4
−2
0 = −2a + 4
2a = 4
a=2
Therefore the equation is g ( x ) = 2 x + 4
Method 2
+4
=2
+2
The y-intercept is 4
+2
The gradient is
+4
Therefore, the equation is g ( x ) = 2 x + 4
41
−2
4
(c)
Determine the equation of the following line in the form f ( x ) = ax + q .
Solution
Method 1
(−1; 3)
Substitute the two points into y = ax + q and
solve simultaneous equations:
For ( −1 ; 3) : 3 = a ( −1) + q
∴3 + a = q
For (3 ; − 1)
−1 = a (3) + q
∴ −1 − 3a = q
∴3 + a = −1 − 3a
(both equal q)
∴ 4a = −4
∴ a = −1
∴ q = 3 + (−1) = 2
(3 ; − 1)
Therefore the equation is f ( x ) = − x + 2
Method 2
The gradient is a =
−4
= −1
+4
(−1; 3)
∴ y = − 1x + q
Substitute either of the two points:
into the equation y = − x + q to get q:
( −1 ; 3) :
3 = −( −1) + q
∴q = 2
or
(3 ; − 1) :
−1 = −(3) + q
∴q = 2
The y-intercept is 2
−4
+4
Therefore the equation is f ( x ) = − x + 2
Method 3
Use the Analytical Geometry formula to find the gradient (see Chapter 8)
−1 − 3 −4
Gradient =
=
= −1
3 − ( −1) 4
Therefore a = −1 .
∴ y = − 1x + q
Substitute one of the points into the equation y = − x + q to get q:
( −1 ; 3) :
3 = −( −1) + q
∴q = 2
The y-intercept is 2
Therefore, the equation is f ( x ) = − x + 2
42
(3 ; − 1)
EXERCISE 8
(a)
(b)
(c)
In the diagram, line 4 x − 2 y = 8 cuts the axes at A and B. Line x + y = −1 cuts the
axes at C and D. The two lines intersect at E.
Determine:
4x − 2 y = 8
(1)
the coordinates of A and B.
(2)
the coordinates of C and D.
(3)
the gradient of 4 x − 2 y = 8
(4)
the gradient of x + y = −1
(5)
the coordinates of E
(6)
the values of x for which the lines are
increasing or decreasing.
Given:
x + y = −1
3 x − y = 4 and 2 x − y = 5
(1)
On the same set of axes, draw neat sketch graphs of the two functions.
(2)
Determine the coordinates of the point of intersection.
Determine the equations of the following lines in the form f ( x ) = ax + q :
(1)
(2)
(3)
6
(−7 ; 2)
3
−3
−5
(−4 ; − 1)
(4)
(5)
(6)
(2 ; 4)
−4
−6
(d)
(−2 ; 7)
(2 ; 5)
(−3 ; − 1)
(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 (−1; 0) and
parallel to the y-axis. Do you remember what the gradient of this line is?
43
THE QUADRATIC FUNCTION
EXAMPLE 9
In the diagram, the graphs of f ( x ) = 4 − x 2 ,
2
g ( x) = x 2 + 6
2
g ( x ) = x + 6 and h( x) = 3x 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
h( x ) = 3 x 2
f ( x) = 4 − x 2
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 y = 0
0 = − x2 + 4
∴ x2 − 4 = 0
∴ ( x + 2)( x − 2) = 0
∴ x = −2 or x = 2
C(−2 ; 0)
(b)
D(2 ; 0)
E and F are the points of intersection between f and h
(y-values are equal at the points of intersection)
h( x) = f ( x)
∴ 3x2 = 4 − x2
∴ 4x2 − 4 = 0
∴ x2 − 1 = 0
∴ ( x + 1)( x − 1) = 0
∴ x = −1 or x = 1
Now determine the corresponding y-values by substituting into either of the two
equations:
f (−1) = 3(−1)2 = 3
f (1) = 3(1)2 = 3
E( −1 ; 3)
F(1 ; 3)
(c)
f increases for all x < 0
(d)
g decreases for all x < 0
44
(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 g is (0 ; 6)
The turning point of f is (0 ; 4)
(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
x =0.
Finding the equation of a quadratic function
EXAMPLE 10
(a)
Determine the equation of the following graph in the form f ( x) = ax 2 + q .
Solution
The y–intercept is 3
∴q = 3
(2 ; 7)
∴ y = ax 2 + 3
Substitute the point (2 ; 7)
3
2
into y = ax + 3 to get the value of a
∴ 7 = a (2) 2 + 3
∴ 7 − 3 = 4a
∴ 4 = 4a
∴a = 1
∴ f ( x) = 1x 2 + 3
(b)
Determine the equation of the following graph in the form f ( x) = ax 2 + q .
Solution
The x-intercept form of the equation of
a parabola can be used:
y = a ( x − x1 )( x − x2 )
where x1 and x2 represent the x-intercepts.
∴ y = a( x − (−2))( x − 2)
∴ y = a( x + 2)( x − 2)
Substitute (1 ; 6) to get the value of a
6 = a (1 + 2)(1 − 2)
∴ 6 = a(3)(−1)
∴ 6 = −3a
∴ a = −2
∴ y = −2( x + 2)( x − 2)
(1; 6)
−2
2
Notice that the x-intercept
form of y = −2 x 2 + 8 is:
y = −2 x 2 + 8
= −2( x 2 − 4)
= −2( x + 2)( x − 2)
= −2( x − (−2))( x − 2)
∴ y = −2( x 2 − 4)
∴ f ( x ) = −2 x 2 + 8
45
EXERCISE 9
(a)
(b)
In the diagram, the graph of f ( x) = 2 x 2 − 2
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
f ( x) = 2 x 2 − 2
In the diagram, the graphs of f ( x) = − x 2 + 9
g ( x) = − x 2 , 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:
(c)
f ( x) = − x 2 + 9
(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 h ( x ) = − g ( x ) + 12 . Describe the transformations.
(9)
the equation of p and the value of its gradient.
(10)
the domain and range of p
Given:
g ( x) = − x 2
f ( x) = x 2 − 9 and g ( x ) = − x 2 − 1
(1)
Determine the coordinates of A, B, C and D
(2)
Determine the coordinates of E and F.
f ( x) = x 2 − 9
g ( x) = − x 2 − 1
46
(d)
Determine the equation of each of the following parabolas in the form f ( x) = ax 2 + q.
(1)
(−1; 6)
(2)
(3)
−1
(1; − 2)
(4)
(5)
(6)
25
(1;15)
−2
−4
3
−3
(−2 ; − 5)
2
4
2
5
−5
−8
THE HYPERBOLIC FUNCTION
An interesting characteristic of the hyperbola is that it has two axes of symmetry.
y = −x
y=x
y=
y = x+q
y = −x + q
a
x
y=
q
a
+q
x
y=q
Each axis of symmetry has a gradient of 1 or −1 and a y-intercept of q.
EXAMPLE 11
Given: g ( x) =
3
+2
x
Determine:
y=
(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
47
3
+2
x
Solutions
Vertical asymptote is x = 0
(a)
Horizontal asymptote is y = 2
(b)
y = − x + 2 and y = x + 2
(c)
The graph of g is decreasing on the interval x ∈ ( −∞ ; 0) as well as the interval
x ∈ (0 ; ∞ ) .
(d)
Domain of g: x ∈  x ≠ 0
Range of g:
y∈ y ≠ 2
Finding the equation of a hyperbola
EXAMPLE 12
Determine the equation of the following graph in the form f ( x) =
a
+q.
x
Solution
The horizontal asymptote is y = −3
∴ q = −3
a
∴ y = −3
x
Substitute the point (3 ; − 4) :
a
−4 = − 3
3
a
∴−1 =
3
∴−3 = a
−3
∴ f ( x) =
−3
x
y = −3
(3 ; − 4)
EXERCISE 10
(a)
The line y = x + 4 is an axis of symmetry of the
−2
+ q . The graph of f cuts the
graph of f ( x) =
x
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.
48
y = x+4
(b)
Determine the equation of each of the following hyperbolas in the form f ( x) =
(1)
(2)
a
+ q.
x
(3)
(−3 ; 5)
(4 ; 3)
y=2
(−2 ; − 6)
(4)
(5)
(6)
3
−4
−2
(−8 ; − 8)
(1; − 2)
(7)
(2 ; − 6)
(8)
(9)
y = x −3
−3
−4
(−1; − 8)
y = −x +1
(10)
y = x−2
(11)
(2 ; 4)
( 14 ; 54 )
−1
−1
1
1
2
(2 ; − 6)
(c)
For each function below, state the domain and range and determine the equation:
(1)
y
(2)
(3)
(−4 ; 2)
x
y = −3
(−2 ; − 7)
49
y = x+5
(2 ; − 4)
THE EXPONENTIAL FUNCTION
Finding the equation of an exponential graph
EXAMPLE 13
(a)
Determine the equation of the given
graph in the form f ( x) = a .2 x + q
−4
Solution
−8
The horizontal asymptote is y = −8
x
∴ y = a .2 − 8
Substitute (0 ; − 4) :
−4 = a .20 − 8
∴ −4 = a − 8
∴a = 4
∴ f ( x) = 4.2 x − 8
(b)
Determine the value of b and q if the
equation of the given graph is
g ( x) = −b x + q
3
−1
Solution
The horizontal asymptote is y = 3
∴ y = −b x + 3
Substitute ( −1 ; 0) :
0 = −b −1 + 3
1
∴0 = − + 3
b
∴ 0 = −1 + 3b
∴−3b = −1
∴b =
1
3
x
1
∴ g ( x) = −   + 3
3
1
b = and q = 3
3
(c)
Determine the equation of the given
graph in the form y = a . b x + q
(2 ;1)
Solution
1
3
The graph passes though the origin.
y = a .bx −
− 13
50
Substitute the point (0 ; 0) :
1
0 = a . b0 −
3
1
∴0 = a −
3
1
∴a =
3
1
1
∴ y = . bx −
3
3
Substitute the point (2 ; 1) :
1
1
1 = .b2 −
3
3
2
∴3 = b −1
(2 ;1)
− 13
∴ 0 = b2 − 4
∴ 0 = (b + 2)(b − 2)
∴ b = −2 or b = 2
Choose b = 2 since b ≠ −2
1
1
∴ y = .2 x −
3
3
EXERCISE 11
(a)
Determine the equation of the given
graph in the form f ( x) = a .2 x + q
−3
−4
(b)
Determine the equation of the given
graph in the form g ( x ) = b x + q
(−1; 2)
−1
(c)
Determine the equation of the given
graph in the form h( x) = −b x + q
3
1
(d)
Determine the equation of the given
graph in the form f ( x) = a . b x + q
3
51
(2 ;12)
(e)
Determine the equation of the given
graph in the form f ( x) = a . b x + q
18
16
2
(f)
Determine the equation of the given
graph in the form g ( x) = a . b x + q
(−2 ; 60)
−4
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.
xA = 1 and OA = 1 unit
xB = −1 and OB = 1 unit
yC = 1 and OC = 1 unit
yD = −1 and OD = 1 unit
●
●
1
−1
1
−1
To determine a length OA along the y-axis, let x = 0
The y-value will be the length of OA.
To determine a length OB along the x-axis, let y = 0
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.
52

Let x = 0
Let y = 0 



●

●
A vertical length between two graphs can be
calculated using the formula:
ytop graph − ybottom graph

 AB = yA − yB

(substitute the x-value into this formula to
get the required length)






x
A horizontal length between two graphs can be
calculated using the formula:
xright end point − yleft end point
CD = xD − xC
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.
g ( x) = − x − 1
f ( x) = − x 2 + 1
−1
1
2
We can use the graphs to solve the following inequalities graphically:
(a)
For which values of x is f ( x ) > 0 ?
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 < x < 1 or we can write x ∈ ( −1 ; 1)
(b)
For which values of x is f ( x ) ≤ 0 ?
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.
∴ x ≤ −1 or x ≥ 1 or we can write x ∈ ( −∞ ; − 1] ∪ [1 ; ∞ )
(c)
For which values of x is g ( x ) > 0 ?
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.
∴ x < −1 or we can write x ∈ ( −∞ ; 1)
53
(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 ≤ x ≤ 2 or we can write x ∈ [ −1 ; 2]
(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
∴ x < −1 or x > 2 or we can write x ∈ ( −∞ ; − 1) ∪ (2 ; ∞ )
EXAMPLE 14
Two lines cut the y-axis at A.
f ( x) = − x + 3
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 f ( x ) ≥ 0
(e)
the values of x for which g ( x ) < 0
(f)
the values of x for which f ( x ) ≤ g ( x )
g ( x) = 3x + 3
4
−3
Solutions
(a)
A(0 ; 3)
∴ OA = 3 units
For xC
0 = 3x + 3
∴−3x = 3
∴ x = −1
∴ OC = 1 unit
(b)
For xB
0 = −x + 3
∴x = 3
∴ OB = 3 units
BC = yB − yC
∴ BC = 3 − ( −1)
∴ BC = 4 units
OD = 4 units
∴ FE = 4 units (opp sides rectangle)
Substitute x = 4 into y = − x + 3
y = −4 + 3 = −1
∴ yE = y F = −1
∴ OF = 1 units and DE = 1 unit
54
(c)
OJ = 3 units
∴ GH = 3 units (opp sides rectangle)
Substitute y = −3 into y = 3 x + 3
−3 = 3x + 3
∴−3x = 6
∴ x = −2
∴OG = 2 units and JH = 2 units
(d)
f ( x) ≥ 0
∴x ≤ 3
(e)
g ( x) < 0
∴ x < −1
(f)
or write x ∈ ( −∞ ; 3]
or write x ∈ ( −∞ ; − 1)
f ( x) ≤ g ( x)
∴ x ≥ 0 or write x ∈ [0 ; ∞ )
EXAMPLE 15
The diagram below shows the graphs of f ( x) = 3x 2 − 12 , g ( x ) = 3 x − 6 and h ( x ) = −3 x + 6 .
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.
f ( x) = 3x 2 − 12
h( x) = −3x + 6
g ( x) = 3 x − 6
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 f ( x ) ≥ 0
(h)
the values of x for which f ( x ) < g ( x )
55
Solutions
(a)
AB = yA − yB = −6 − (−12) = 6 units
Let y = 0 in y = 3 x 2 − 12
0 = 3 x 2 − 12
∴ 0 = x2 − 4
∴ 0 = ( x + 2)( x − 2)
∴ x = −2 or x = 2
CD = xD − xC = 2 − (−2) = 4 units.
(b)
Determine the coordinates of E, the point of intersection of f and g
3 x 2 − 12 = 3 x − 6
∴ 3x 2 − 3x − 6 = 0
∴ x2 − x − 2 = 0
∴ ( x + 1)( x − 2) = 0
∴ x = −1 or x = 2
At E, x = −1
y = 3( −1) − 6 = −9
The coordinates of E are ( −1 ; − 9)
∴ OE = 1 unit and EF = 9 units
(c)
OG = 3 which means that x = −3 at G.
Substitute x = −3 into g ( x ) = 3 x − 6 to get y.
g ( −3) = 3( −3) − 6 = −15
∴ GH = 15 units
(d)
IJ = 6 which means that y = −6 at J.
Substitute y = −6 into h( x ) = −3 x + 6 to get x
−6 = −3x + 6
∴ 3x = 12
∴x = 4
∴ OI = 4 units
(e)
OR = 1 which means that x = 1
KL = yK − yL = (3x − 6) − (3x 2 − 12)
Substitute x = 1
∴ KL = (3(1) − 6) − (3(1) 2 − 12) = 6 units
(f)
MP = yM − yP = (3 x − 6) − (−3 x + 6) = 3 x − 6 + 3 x − 6 = 6 x − 12
∴18 = 6 x − 12 (since MP = 18 )
∴−6 x = −30
∴x = 5
∴ ON = 5 units
(g)
f ( x ) ≥ 0 for x ≤ −2 or x ≥ 2
or
x ∈ ( −∞ ; − 2] ∪ [2 ; ∞ )
(h)
f ( x ) < g ( x ) for −1 < x < 2
or
x ∈ ( −1 ; 2)
56
EXERCISE 12
(a)
(b)
Two lines f ( x ) = − x + 4 and g ( x ) = x − 2
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 f ( x ) ≥ 0
(7)
the values of x for which g ( x ) < 0
(8)
the values of x for which f ( x ) < g ( x )
g ( x) = x − 2
−2
−1
The diagram shows the line f ( x ) = x + 5
Determine:
(c)
f ( x) = − x + 4
f ( x) = x + 5
(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 f ( x ) < 0
Two lines f ( x ) = − x + 4 and g ( x ) = x + 2
intersect at E.
g ( x) = x + 2
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 HK = 1 12
(5)
the values of x for which f ( x ) > 0
(6)
the values of x for which g ( x ) ≤ 0
(7)
the values of x for which g ( x ) < f ( x )
57
f ( x) = − x + 4
(d)
The diagram shows the graphs of f ( x ) = x 2 − 9
and g ( x ) = 3 − x intersecting at E and D.
The graph of f cuts the axes at B, C and D.
f ( x) = x 2 − 9
Determine:
(e)
(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 f ( x ) > 0
(8)
the values of x for which f ( x ) ≥ g ( x )
The graph of f ( x) =
g ( x) = 3 − x
2
− 2 and g ( x ) = 2 is shown.
x
Determine:
g ( x) = 2
(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 f ( x ) ≥ 0
(6)
the values of x for which f ( x ) < 0
(7)
the values of x for which f ( x ) > −2
(8)
the values of x for which f ( x ) > g ( x )
f ( x) =
2
−2
x
x
(f)
1
The graph of f ( x ) =   − 2 is shown.
2
Determine:
(1)
the length of OA and CD.
(2)
the values of x for which f ( x ) > 0
(3)
the values of x for which f ( x ) ≤ 0
(4)
the values of x for which f ( x ) > −2
(5)
the values of x for which f ( x ) > −1
(6)
the values of x for which f ( x ) ≤ −1
58
x
1
f ( x) =   − 2
2
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(b)
(c)
Given:
f ( x) = x 2 − 1 and g ( x ) = x − 1
(1)
Sketch the graph of f and g on the same set of axes.
(2)
Determine the domain and range of f.
(3)
If h ( x ) = f ( x ) + 3 , 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 y = g ( x + 2) .
Given:
f ( x) = 3x and g ( x) =
3
x
(1)
Sketch the graph of h if h( x ) = 3 f ( x ) − 1 and describe the transformations.
(2)
Sketch the graph of p if p ( x ) = − g ( x ) + 1 and describe the transformations.
(3)
Write down the range of h.
(4)
Write down the domain of p.
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?
(1; 3)
x
(d)
1
The graph of f ( x) =   + q passes through
2
the origin. Determine:
(1)
the value of q
(2)
the length of AB of OA = 2 units
(3)
the length of OC if CD = 78 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.
59
f
(e)
(f)
The graph of f ( x ) = mx + c and g ( x) = ax 2 + b
intersect at D and on the x-axis at A.
H and T lie on line f . Determine:
(1)
the equation of f
(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 QR = 45
4
(8)
the values of x for which g ( x ) ≥ 0
(9)
the values of x for which g ( x ) < f ( x )
g
H(−1; 4)
T(4 ; − 1)
f
G(1; − 8)
a
+6
x
cut the x-axis at A. The graph of f passes through
The graph of f ( x) = a.b x + q and g ( x) =
the point ( −2 ; − 12) and cuts the y-axis at 4.
Determine:
(1)
the range of f
(2)
the equation of f
(3)
the equation of g
(4)
the equation of h if the graph of f is
reflected in the x-axis
(5)
(−2 ; − 12)
the values of x for which f is increasing
*(g) Given:
f ( x) = −3x 2 + 12 and g ( x ) = −3 x + 6
Determine:
(1)
the values of x for which f ( x ) ≥ 0
(2)
the values of x for which f ( x ) > g ( x )
(3)
the values of x for which f ( x ) ≥ 9
(4)
the values of x for which f ( x ). g ( x ) < 0
(5)
the values of x for which f ( x ). g ( x ) ≥ 0
(6)
the possible values of t if the graph of
f ( x) = −3x 2 + 12 − t
(i) does not cut or touch the x-axis.
(ii) cuts the x-axis in two distinct points.
60
f ( x) = −3x 2 + 12
g ( x) = −3x + 6