MEMBY 2025

GRADE 11
FUNCTIONS
We will focus on dealing with the same four functions that were dealt with in grade
10. Namely: the Straight Line Graph, the Parabola, the Hyperbola and the
Exponential Graph. In this booklet, we will deal with how to sketch the functions.

SKETCHING FUNCTIONS

Straight Line Graphs

The general equation for the straight line graph is: y = mx + c .

• The variable m refers to the gradient. If m > 0 then the graph will slope
upwards: and if m < 0 then the graph will slope downwards:
.
• The variable c refers to the y–intercept, the place where the graph will cut or
intercept the y–axis.

In order to sketch the graph:

• find the co–ordinates of the y–intercept (by substituting x = 0 and solving the
resulting equation)
• find the co–ordinates of the x–intercept (by substituting y = 0 and solving the
resulting equation).
• Once the two intercepts have been found, plot them onto the Cartesian Plane
and join the intercepts. Ensure that you extend the graph past these two
intercepts and label the graph and the intercepts.)

Example.
3
Sketch the following function: g ( x ) = − x + 6 .
2

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Answer:
3
g (x) = − x + 6
2
Y − int ( x = 0 )
y = 6
Y − int : ( 0;6 )
X − int ( y = 0 )
3
0=− x+6
2
0 = −3 x + 12
3 x = 12
x = 4
X − int : ( 4;0 )
Parabolas

The general equation for the parabola can be written in three forms. The standard
form is y = ax 2 + bx + c and the turning point form is y = a ( x − p ) + q and the
2

intercept form is y = a ( x − x1 )( x − x2 ) , where x1 and x2 are the x-intercepts of the

parabola. You will need to be able to work with all three forms of the equation.
• In all forms of the equation, if a > 0 then the parabola will have a minimum
turning point (a smiley face) and if a < 0 then the parabola will have a
maximum turning point (a sad face).

• A point to note is that the x co–ordinate of the turning point is also the
equation of the Axis of Symmetry. e.g. if the x co–ordinate of the turning
point is 4, the equation
x = 4 is the equation of the axis of symmetry. The axis of symmetry in any
graph is a line that the graph can be folded along that results in the one half of
the graph falling on top of the other half. i.e. every point on the one half of the
graph can be mirrored (or reflected) around the axis of symmetry and will fall
on the other half of the graph.

Standard form:

In order to sketch the graph:

• find the co–ordinates of the turning point, (the x co–ordinate is found by using
the formula and the y co–ordinate is found by substituting this x co–ordinate
that you have just found into the original equation)
• find the co–ordinates of the y–intercept (by substituting x = 0 and solving the
resulting equation)

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 • find the co–ordinates of the x–intercept (by substituting y = 0 and solving the
resulting equation).
• plot these points onto the Cartesian Plane and join them being mindful of the
shape of the parabola according to the a value. This will be a good way to
check that the working is correct. Ensure that you label all these points on the
graph and label the graph itself.)

Example:
Sketch the graph of f ( x ) = −2 x 2 + 2 x + 12 , labelling all relevant points.
Answer:

f ( x ) = −2 x 2 + 2 x + 12 (a  0;  sad face )
TP : ( x; y )
b
x = −
2a
2
x = −
2 ( −2 )
1
x =
2
 1
y = f 
2
2
 1  1
y = −2   + 2   + 12
 
2 2
1
y = 12
2
 1 1
TP =  ; 12 
2 2
Y − int : x = 0
 y = 12
X − int : y = 0
0 = −2 x 2 + 2 x + 12
0 = x2 − x − 6
0 = ( x − 3) ( x + 2)
 x = 3 or x = −2

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 y = a ( x − p) + q
2
Turning Point Form

In order to sketch the graph:

• find the co–ordinates of the turning point, (the x co–ordinate is equal to the p
value in the formula and the y co–ordinate is equal to the q value in the
formula i.e. the
• TP = (p ; q)
• find the co– ordinates of the y–intercept (by substituting x = 0 and solving the
resulting equation) and then the co–ordinates of the x–intercept (by
substituting y = 0 and solving the resulting equation)
• Plot these points onto the Cartesian plane and join them being mindful of the
shape of the parabola according to the a value. This will be a good way to
check that the working is correct. Ensure that you label all these points on the
graph and label the graph itself.)

Example:
1
Sketch the graph of h ( x ) = ( x + 3 ) + 1 , label all the relevant points.
2

Answer:
1
h (x) = ( x + 3) + 1
2

2
TP = ( −3; 1)
Y − int : x = 0
1
y = (0 + 3) + 1
2

2
11
y =
2
X − int : there are no x − ints.

(The reason there are no x–intercepts is that the TP has a y co–ordinate of 1; i.e. a
minimum value greater than 0 which is above the x–axis and the graph has a
minimum value i.e. a smiley face.

Sneaky hint: if a and q have the same sign then there are no x–intercepts.)

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Intercept Form y = a ( x − x1 )( x − x2 )

In order to sketch the graph:

• write down the co-ordinates of the x-intercepts namely ( x1;0 ) and ( x2 ;0 ) .

• find the co–ordinates of the y-intercept (by substituting x = 0 and solving the
resulting equation)
• find the co–ordinates of the turning point, (the x co–ordinate is equal to
x2 + x1
. The y co–ordinate is found by substituting this x co–ordinate that you
2
have just found into the original equation)
• plot these points onto the Cartesian Plane and join them being mindful of the
shape of the parabola according to the a value. This will be a good way to
check that the working is correct. Ensure that you label all these points on the
graph and label the graph itself.)

Example:
Sketch the graph of g ( x ) = −2 ( x + 1) ( x − 3 ) , label all the relevant points.

Answer:

g ( x ) = −2 ( x + 1) ( x − 3 )
x intercepts : ( −1; 0 ) and ( 3; 0 )
y − int : x = 0
y = −2 ( 0 + 1) ( 0 − 3 )
y = 6 ( 0; 6 )
T.P.
x co-ordinate:
x2 + x1 3 −1
= =1
2 2
y co-ordinate:
y = −2 ( 1 + 1) ( 1 − 3 )
y = 8
T.P. (1 ; 8 )

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Hyperbolas
a
The general equation for the hyperbola can be written as y = + q.
x−p
The equation x = p is the equation of the vertical asymptote and the equation y = q is
the equation of the horizontal asymptote.

To sketch the graph:

• first sketch the vertical and horizontal asymptotes

• then plot a set of points containing x co – ordinates that are on either side of
the vertical asymptote. Join up the points on either side of the vertical
asymptote and this will produce the graph which is in two parts. Use your
calculator in table mode to calculate the set of points that you will need to plot.
If a > 0 then the graph will appear with one section in the top right
and the other in the bottom left when using the asymptotes as a
reference.
If a < 0 then the graph will appear with one section in the top left
and the other in the bottom right when using the asymptotes as a
reference.

Example.
12
Sketch the graph of f ( x ) = − + 3,
x −2
label all relevant points.

Answer:
Vertical Asymptote:
x=2

Horizontal Asymptote:
y=3

Points to plot from calculator:

(Label one or two points)
(–4 ; 5) , (–2 ; 6) , (0 ; 9) and
(4 ; –3) , (6 ; 0) , (8 ; 1)

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Exponential Graphs

The general equation for the exponential graph can be written as y = a  b x − p + q .

The equation y = q is the equation of the horizontal asymptote.
There is no vertical asymptote.
b is a positive number.

To sketch the graph:

• first sketch the horizontal asymptote
• then plot the set of points and join up the points. Use your calculator in table
mode to calculate the set of points that you will need to plot.

The shape of the graph is dependent on the values of a and b.

If a > 0 and b > 1 then the graph will increase and be positioned above
the asymptote. e.g.

0 < b < 1 then the graph will decrease and be positioned above the
asymptote. e.g.

If a < 0 and:
b > 1 then the graph will decrease and be positioned below the
asymptote. e.g.

0 < b < 1 then the graph will increase and be positioned below the
asymptote. e.g.

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Example.
Sketch the graph of g ( x ) = −2− x + 3 , label all relevant
points.

Answer:
(A point to note here is that the equation of g can be
x
 1
rewritten in the following form: g ( x ) = −1    + 3 .
2
1
i.e. a = −1, b = , p = 0 and q = 3 .)
2

Horizontal Asymptote:
y = 3.

Points to plot from calculator:

(Label one or two points)
(–1 ; 2½) , (0 ; 2), (1 ; 1)
and (2 ; –1)

General Points to Note

You need to be able to tell the difference easily between the four equations so that,
when you are sketching the graphs, you know the general shape of the graph before
you start.

The positioning of the x variable and its exponent in each equation points to the type
of graph that is being referred to.

• For a straight line graph, the variable x has an exponent to the power of 1.
• For a parabola, the variable x has an exponent to the power of 2.
• For a hyperbola, the variable x is in the denominator and has an exponent of
1.
• For an exponential graph, the variable x is in the exponent and has an
exponent of 1.

The variable a (and b where applicable) affect the shape of the graph.

The variables p and q affect the positioning of the graph and do not change the
shape in any way.
• The more positive the value of p is, then the further the graph shifts (or is
translated) to the right and the more negative the value of p is, then the further
the graph shifts to the left.
• The more positive the value of q is, then the further the graph shifts upwards
and the more negative the value of q is, then the further the graph shifts
downwards.

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EXERCISE 1

Sketch the following graphs, labelling all relevant points. (Note, relevant points are
the x and y intercepts, turning points and asymptotes where applicable.) You may
sketch each pair of graphs on one set of axes.

1. f ( x ) = 2 x 2 − 4 x − 6 and g ( x ) = 2x − 6

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 8
2. h ( x ) = 3 x − 2 + 2 and p ( x ) = −3
x −2

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3. g ( x ) = − ( x − 4 ) ( x + 1) and q ( x ) = − 3 x + 2

1 6
4. f (x) = − x + 4 and d ( x ) = − +2
2 x +3

x 1
p (x) = − + 2 and q ( x ) = − ( x − 4 ) − 1
2
5.
3 2

g ( x ) = 2 ( x − 3 ) − 8 and h ( x ) = 2.31− x − 4
2
6.

4
7. f (x) = − + 3 and g ( x ) = − x + 1
x +2

8. p ( x ) = −2 x 2 − 4 x − 5 and k ( x ) = −2 x − 5

6
9. f ( x ) = 32 − x + 1 and q ( x ) = +1
x −3

−2
10. h (x) = − 1 and p ( x ) = −2x − 3 − 1
x +3

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