1

The exemplar questions provided in this document will be continuously upgraded every
year with new questions that align with the latest trends in final Maths exams.

Topic-based exam-type questions are grouped together under the headings:

QUESTION 1, QUESTION 2, QUESTION 3, …

This will help to provide an easy-to-use structure.

Teachers may use or adapt any of these questions in their own tests and exams.

This document is available in WORD format which will allow teachers to copy and paste
selected questions. To receive the questions in WORD format, please send an email to
Mark Phillips (mathmark@mweb.co.za).

The marking guidelines are provided at the end of each topic.

The following Paper 1 topics are included:

A Algebra, equations, and inequalities

B Patterns and sequences

C Financial mathematics

D Functions and graphs

E Differential calculus

F Probability

THIS DOCUMENT CONTAINS QUESTIONS ON THE FOLLOWING TOPIC:

All questions are adapted from past matric preliminary and final examinations or are
original.

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 2

QUESTION 1 Grade 11

1.1 Consider the function f ( x )  ( x  2) 2  1

1.1.1 Draw a neat sketch graph indicating the coordinates of the intercepts with
the axes, the coordinates of the turning point and the equation of the axis
of symmetry. (6)

1.1.2 Write down the range. (1)

1.2 Consider the following functions: f ( x )  x 2  2 x  3 and g ( x)  2 x2

1.2.1 Determine the coordinates of the turning point of f. (3)

1.2.2 Write down the x-intercepts of f. (3)

1.2.3 Draw a neat sketch graph of f. (3)

1.2.4 State the minimum value of f. (1)

1.2.5 Write down the equation of the axis of symmetry of g. (1)

1.2.6 If the graph of g is shifted 3 units left and 1 unit upwards, state the
equation of the newly formed graph. (2)

QUESTION 2 Grade 11

2.1 Consider the following functions:

f ( x)  2x1 1 g ( x)   x  1

2.1.1 Determine the coordinates of the y-intercept of f. (2)

2.1.2 Determine the coordinates of the x-intercept of f. (2)

2.1.3 Sketch the graph of f. (3)

2.1.4 Sketch the graph of g on the same set of axes. (2)

2.1.5 State the coordinates of the point of intersection of the graphs of

f and g. ` (1)

2
2.2 Given: f ( x) 
x
2.2.1 Sketch the graph of y  f ( x  2)  1 . (4)

2.2.2 State the domain of y  f ( x  2)  1 . (2)

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QUESTION 3 Grade 11
x 1
2 1
Given: f ( x)  x 2  4 x g ( x)  1 h( x )  2   2
x2 2

3.1 Sketch the graph of f showing intercepts with the axes, the turning point
and the axis of symmetry. (6)

3.2 Sketch the graph of g showing intercepts with the axes and the asymptotes. (6)

3.3 Sketch the graph of h showing the intercepts with the axes and the horizontal
asymptote. (4)

3.4 Write down the domain of g. (2)

3.5 Write down the range of h. (2)

3.6 Determine graphically the values of k for which f ( x)  x2  4x  k has

two real, unequal roots. (2)

QUESTION 4 Grade 11

Consider the following functions:

f ( x)  3x1 1 and g ( x)  2 x  2

4.1 Write down the equation of the horizontal asymptote of f. (1)

4.2 Determine the coordinates of the y-intercept of f. (2)

4.3 Determine the coordinates of the x-intercept of f. (3)

4.4 Sketch the graph of f . (3)

4.5 Sketch the graph of g on the same set of axes. (2)

4.6 Determine graphically the values of x for which f ( x)  g ( x) . (1)

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QUESTION 5 Grade 11
x 1
1
Given: f ( x)  4( x 1)  3 and g ( x)   
2
2
3

5.1 Write down the coordinates of the turning point of the graph of f. (2)

5.2 Write down the equation of the asymptote of the graph of g. (1)

5.3 Show that the graph of g passes through the turning point of f. (2)

5.4 Explain why the graph of f will never cut the x-axis. (2)

5.5 Determine the y-intercept of the graph of f. (2)

5.6 Determine the y-intercept of the graph of g. (2)

5.7 Sketch the graphs of f and g on the same set of axes. (7)

5.8 Write down the range of f . (2)

QUESTION 6 Grade 12

Given: f ( x)  2 x 2 for the domain x  0 .

6.1 Sketch the graph of f . (2)

6.2 Determine the equation of the inverse function of f . (2)

6.3 Sketch the graph of y  f 1 ( x) on the same set of axes as f . (2)

QUESTION 7 Grade 12

Given: f ( x)  3x

7.1 Write down the equation of f 1 in the form y  ... (1)

7.2 Sketch the graph of f and f 1 on the same set of axes. (3)

7.3 Solve the following using the graphs of f and f 1 :

7.3.1 f ( x). f 1( x)  0 (2)

7.3.2 x. f 1 ( x)  0 (1)

7.4 Sketch the graph of y  g ( x)   f ( x  1)  1 on a set of axes.

Show intercepts with the axes and the asymptote of the graph. (4)

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QUESTION 8 Grade 11 and 12

x 1
1
8.1 Given: g ( x)  2  2  
2

8.1.1 Write down the horizontal asymptote of g. (1)

8.1.2 Sketch the graph of g indicating the horizontal asymptote

as well as the intercepts with the axes. (6)

8.1.3 Write down the range of g. (1)

8.2 Given the function f ( x)  log 1 x

4

8.2.1 Sketch y  f ( x) indicating an intercept with the axes and ONE

other point on the graph. (3)

8.2.2 If f ( x ) is translated so that its new equation becomes h( x)  log 1 ( x  3) ,

4
state the domain of h( x ) . (1)

1
8.2.3 State the equation of f in the form g ( x)  ... (1)

QUESTION 9 Grade 11 and 12

9.1 Consider the function f ( x)  4 x  2 .

9.1.1 Determine the coordinates of the intercepts of f with the axes. (3)

9.1.2 Sketch the graph of f showing the intercepts with the axes and the horizontal
asymptote. (3)

9.1.3 Write down the equation of g if g is the graph of f shifted 3 units right and
2 units up. (2)

9.2 Given: f ( x)   4 x ; x  0 and g ( x)  3x  3

9.2.1 Draw a neat sketch graph of f. (2)

9.2.2 Consider the graph of y  f 1 ( x) .

Determine the equation of h, the reflection of f 1 in the x-axis.
Leave your answer in the form h( x )  ... (4)

9.2.3 Draw the graph of y  g ( x  1)  2 on the same set of axes as f.

Clearly indicate the intercepts with the axes as well as any asymptotes. (4)

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QUESTION 10 Grade 11

4
In the diagram below, the graphs of f ( x)   2 and g ( x )  a ( x  p ) 2  q are shown.
x 1
The graph of f and g both cut the y-axis at B and the graph of f has its turning point at A, the
point of intersection of the asymptotes of f . The graph of f cuts the x-axis at C.
The line h is an axis of symmetry of f which intersects the graph of f at the point ( 1 ; 4) and
passes through A.

(1; 4)

10.1 Show that the coordinates of A are (1 ; 2) . (2)

10.2 Determine the equation of g. (5)

10.3 Determine the equation of h and hence the equation of the other axis of
symmetry of h. (4)

10.4 Write down the range of g. (1)

10.5 Determine the values of x for which:

10.5.1 f ( x )  0 (3)

10.5.2 f ( x).g ( x )  0 (2)

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QUESTION 11 Grade 11 and 12

a
In the diagram below, the graphs of f ( x)  a( x  p)2  q and g ( x)   c are shown.
xb
The graph of f cuts the x-axis at C and D, the y-axis at B, has its turning point at A and the
line x  1 is the axis of symmetry. The line y   x  3 is an axis of symmetry of the graph
of g and this line passes through A and B. The graph of g passes through the point
(3 ;  2) .

(3;  2)

y  x  3
x 1

11.1 Show that the coordinates of A are (1;  4) . (1)

11.2 Determine the coordinates of B. (1)

11.3 Determine the equation of f . (3)

11.4 Determine the equation of g . (3)

11.5 Determine the equation of the other axis of symmetry of g . (2)

11.6 Calculate the length of CD. (3)

11.7 If h( x)   f ( x)  2 , determine the maximum value of h( x ) (2)

11.8 Determine graphically the values of x for which f ( x) . g ( x)  0 . (3)

11.9 If f represents h , the graph of the derivative of a function h, for which value
of x will h have a minimum turning point? (1)

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 8

QUESTION 12 Grade 11

Sketched below (not to scale) are the graphs of f ( x )  a x and g ( x )  a ( x  p ) 2  q .

The graphs of f and g intersect at A(1 ; 2) and B is the turning point of g.

A(1 ; 2)

12.1 Find the equation of f . (2)

12.2 Find the equation of g if BC  3 and BD  6 .

Please note that the given lengths are not drawn to scale. (4)

12.3 Show that the length of EF is equal to 2 6 units. (4)

12.4 Determine the range of h if h( x)  f ( x  1)  1 . (3)

12.5 Determine graphically the values of x for which f ( x).g ( x )  0 (1)

12.6 Determine the value of p for which g ( x )   x 2  6 x  k will

have non-real roots. (2)

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 9

QUESTION 13 Grade 11 and 12

3
Given: f ( x)  ax2  bx  c and g ( x)   q for all x  1 .
x p
The graph of f cuts the x-axis at A and B and the y-axis at 6. It has its turning point at
(2 ; 8) . The graph of g cuts the y-axis at 4 and intersects the line y  x at C.
The line y  1 is an asymptote of g.

(2 ; 8)
f yx
6

g
y 1

x  1

1
13.1 Show that a   , b  2 and c  6 by determining the equation of f. (3)
2
13.2 Determine the length of AB. (4)

13.3 Write down the equation of g. (1)

13.4 Determine the coordinates of C. (5)

13.5 Determine graphically the values of x for which

13.5.1 g ( x)  x  0 (2)

13.5.2 f ( x).g ( x)  0 (1)

1
13.6 For which values of k will the equation  x 2  2 x  k  6 have equal roots? (2)
2

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 10

QUESTION 14 Grade 11 and 12

The graphs of f ( x)  2 x 2  4 x  6 and g ( x)  a x are represented in the sketch below.

A and B are the x-intercepts of f and R is the turning point of f . The point C(  2 ; 4)
is a point on the graph of g.

C(2 ; 4)

1
14.1 Show that a  . (2)
2
14.2 Determine the length of AB. (3)

14.3 Determine the length of SR. (5)

14.4 Write down the equation of h, if h is the reflection of f in the y-axis.

Express your answer in the form h( x)  a( x  p)2  q . (2)

14.5 Write down the equation of g 1 in the form y  .... (2)

14.6 Sketch the graph of y  g 1 ( x) on a set of axes. (2)

14.7 Determine the values of x for which:

14.7.1 g 1 ( x)  2 (3)

14.7.2 x . f ( x)  0 (2)

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QUESTION 15 Grade 11
4
The diagram below shows the hyperbola defined by f ( x)  1
x p
The lines g and y   x  2 are axes of symmetry of f and intersect at B, the point of
intersection
of the asymptotes.

y  x  2

15.1 Determine the equation of f and hence the value of p . (4)

15.2 Determine the equation of g, the other axis of symmetry of f . (2)

15.3 Write down the domain of f . (2)

15.4 Suppose that the graph of f is shifted left so that A coincides with the origin.
Determine the equation of the vertical asymptote of the newly formed graph. (3)

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 12

QUESTION 16 Grade 11
a
In the diagram below, the graph of f ( x)   q cuts the y-axis at 3 and the x-axis at
x p
A. The graph of g ( x )  m( x  n) 2  c intersects f at A, cuts the y-axis at 3 and has a

turning point at B(2;1).

B(2;1)

y  2
3

x2

Determine:

16.1 the equation of f. (4)

16.2 the equation of g. (4)

16.3 the length of OA. (3)

16.4 the values of x for which f ( x).g ( x)  0. (3)

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QUESTION 17 Grade 11
a
In the diagram below, the graphs of f ( x)  dx2 and g ( x)   q intersect at P(  2 ; 2)
x p
and Q. The asymptotes of g intersect at A(  1;1) .

P(2 ; 2)

A(1;1)

17.1 Determine the value of a, p, q and d. (6)

17.2 Show that the coordinates of Q are (1; 12 ) . (4)

17.3 Determine graphically:

17.3.1 the values of x for which f ( x). g ( x)  0 . (2)

17.3.2 the values of k for which f ( x)  k  0 has real, unequal roots. (1)
1
17.3.3 the values of k for which  1  0 has a negative root. (1)
xk

17.4 Determine the length of BC if OD  4 units. (3)

17.5 If h( x)  g ( x  3)  2 , determine the equations of the asymptotes of h. (3)

17.6 If the graph of f is shifted 2 units left and 4 units down, determine the
x-intercepts of the newly-formed graph. (4)

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 14

QUESTION 18 Grade 11

In the diagram below are the graphs of f ( x)  ax2  bx  c and g ( x)  x  4 .

P is the turning point of f and R is the point of intersection of f and g.

18.1 If OA  4 units, OB  1 12 units and OC  1 unit, determine:

18.1.1 the coordinates of D. (1)

18.1.2 the value of a, b and c. (4)

18.1.3 the coordinates of R. (5)

18.1.4 the length of NP. (4)

18.2 Determine graphically the values of x for which f ( x) . g ( x)  0 (1)

18.3 Determine the values of k for which f ( x)  k  0 has two negative, unequal
roots. (2)

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QUESTION 19 Grade 11

The graphs of the following are given:

k
f ( x)  r g ( x)  ab x  q h( x )  d ( x  t ) 2
x p
The graphs of f and g intersect at (2 ; 5) . The graphs of f and h cut the y-axis at A.
The asymptotes of f intersect at (1 ; 2) . The graph of f cuts the x-axis at B.
The turning point of h lies on the x-axis.

(2 ; 5)

(1; 2)

Determine:
19.1 the value of k, p and r. (5)

19.2 the equations of the axes of symmetry of f. (2)

19.3 the value of a, b and q. (5)

19.4 the value of d and t. (6)

19.5 the coordinates of B. (4)

19.6 the values of x for which f ( x)  g ( x) . (2)

19.7 the values of x for which f ( x).g ( x)  0 . (2)

19.8 the turning point of the graph of y  h( x  1)  4 (2)

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QUESTION 20 Grade 11

In the diagram below, the graphs of the following functions are shown:

x
 3 and g ( x)     2
6 1
f ( x)  
x 1 2

The graph of f cuts the x-axis at A and the y-axis at B. The graphs of f and g intersect at E.
C and D are points on the graphs of f and g respectively such that line segment CD
produced to H, a point on the x-axis, is parallel to the y-axis.

Determine:

20.1 the equations of the asymptotes of f. (2)

20.2 the equations of the axes of symmetry of f. (3)

20.3 the domain of g. (1)

20.4 the length of OA and OB. (5)

20.5 the distance CD if OH  2 units . (5)

20.6 f (3) and g ( 3) . Hence the coordinates of E. (3)

20.7 the value(s) of x for which f ( x).g ( x )  0 . (2)

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 17

QUESTION 21 Grade 11

The graphs of the following functions are drawn in the diagram below:

f ( x)  ax2  bx  c , g ( x )  4 and h( x)   x  5

The turning point of f is ( p ; q ) . The x-intercepts of f are (1 ; 0) and (5 ; 0) .

The point J is one of the points of intersection of f and h.

( p ; q)

21.1 Write down the value of p and q. (2)

21.2 Show that the equation of f is given by f ( x)   x 2  6 x  5 . (5)

21.3 Determine the coordinates of J. (6)

21.4 Hence, write down the length of JK, if JK is parallel to the y-axis. (2)

21.5 Use the graph to determine for which value(s) of k will  x2  6 x  k  0 have
real and unequal roots? (2)

f ( x)
21.6 For which values of x is  0? (3)
h( x )

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 18

QUESTION 22 Grade 11

The graphs of f ( x)  2 x 2  4 x  6 and g ( x)  ax  b are represented in the sketch below.

A and B are the x-intercepts of f and G is the turning point of f. The point B and C(  4 ;10)
are the points of intersection of the graphs of f and g. F is a point on the x-axis such that FG is
parallel to the y-axis. D lies on g and E lies on f such that DE is perpendicular to the x-axis.
DE cuts the x-axis at H.

C(4 ;10)

22.1 Determine the length of AB. (3)

22.2 Determine the length of FG. (3)

22.3 Determine the value of a and b. (3)

22.4 Determine the length of OH if DE  12 units. (4)

22.5 Determine the values of x for which f ( x).g ( x )  0 . (2)

22.6 Determine graphically the values of k for which y  2 x 2  4 x  k will

not cut the x-axis. (1)

22.7 Determine the coordinates of the turning point of the graph y  f ( x  2)  1 . (2)

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 19

QUESTION 23 Grade 11 and 12

Sketched below are the graphs of y  f ( x)  a( x  p)2  q and y  g ( x)  x  2 .

P(3 ; 2) is the turning point of f.
A and B are the x-intercepts of f.
D is the point of intersection of f and g.
C is the y-intercept of g.
The graph of f cuts the y-axis at (0 ;  16) .

P(3 ; 2)

16

23.1 Determine the equation of f in the form y  ax 2  bx  c . (4)

23.2 Calculate the coordinates of A and D. (5)

23.3 Write down the values of x for which f ( x).g ( x )  0 . (1)

23.4 Describe the transformation of f to p if p( x)  2 x2 . (2)

23.5 If f represents the graph of the gradient of a cubic function h, determine the
values of x for which the graph of h is concave up. (3)

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 20

QUESTION 24 Grade 11 and 12

8
Sketched below are the graphs of f ( x)   , x  0 and g ( x )  a x .
x
P is the reflection of the point Q(4 ;  2) in the line y  x .

yx

Q(4 ;  2)

24.1 Write down the coordinates of P. (2)

24.2 Write down the coordinates of A. (1)

24.3 Calculate the value of a. (2)

24.4 Determine the equation of g 1 ( x ) in the form y  ... (2)

24.5 For which values of x will f ( x)  g ( x) ? (1)

24.6 Write down the domain of k if k ( x)  f ( x  2) . (2)

QUESTION 25 Grade 11
a
The axes of symmetry of the graph of the hyperbola y   q are y  x and
x p
y  x  2 .

25.1 Determine the coordinates of the point of intersection of the two axes
of symmetry. (3)

25.2 The graph of the hyperbola cuts the x-axis at (1 ; 0) .

Determine the value of a, p and q. (4)

25.3 If the hyperbola is shifted 3 units left and 4 units up, write down the equation
of the newly formed graph. (2)

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 21

QUESTION 26 Grade 11 and 12

The graph of y  f ( x)  a( x  p)2  q where a, p and q are constants, is shown below.

Points E, F(1; 0) and C are its intercepts with the axes. A( 4 ; 5) is the reflection of C
across the axis of symmetry of f . B is the turning point of f . D is a point on the graph
such that the straight line through A and D has the equation g ( x)  2 x  3 .

A(4 ; 5)

F(1; 0)

26.1 Write down the coordinates of C. (1)

26.2 Show that the equation of the axis of symmetry is x  2 . (1)

26.3 Calculate the value of a, p and q. (6)

26.4 If f ( x )   x 2  4 x  5 , calculate the coordinates of D. (4)

26.5 If h( x)  f ( x)  1 , determine the coordinates of the turning point of h. (3)

26.6 Determine the maximum length of BG if it varies in length between A and D. (4)

26.7 26.7.1 Determine the equation of g 1 in the form g 1 ( x)  ... (1)

26.7.2 By drawing graphs, determine graphically the values of x for which

g 1 ( x)  g ( x) . (3)

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 22

QUESTION 27 Grade 11

Given: f ( x)  6  3 and g ( x)  6  q
x2 x p
27.1 Write down the equations of the asymptotes of f . (2)

27.2 State the domain of f . (1)

27.3 Write down the equations of the axes of symmetry of f . (2)

27.4 Sketch the graph of f . (4)

27.5 If the axes of symmetry of the graph of g are y  x  3 and y   x  1 ,

determine the value of p and q. (4)

27.6 Describe how the graph of f is translated to form the graph of g. (2)

QUESTION 28 Grade 12

The sketch below represents the inverses of f ( x)  ax2 ; x  0 and g ( x )  4 x .

The graph of f 1 and g 1 intersect at (4 ; 1) and (16 ; 2) . The graph of g 1
cuts the x-axis at 1.

y  f 1 ( x)

(16 ; 2) y  g 1 ( x)
(4 ;1)

28.1 Write down the coordinates of ONE point through which both the graphs of
f and g pass. (1)

28.2 Determine the equation of f. (3)

28.3 How has the domain of f been restricted? (1)

28.4 Determine the equation of g 1 in the form g 1 ( x)  (1)

28.5 If h( x )  g 1 ( x  2) , for which values of x will h( x)  0 ? (2)

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 23

QUESTION 29 Grade 11

In the diagram, the graphs of the following functions are shown:

f ( x)  x 2  3 x  4 g ( x)  2 x  8 h( x)  ab x  q

The graph of f cuts the x-axis at A and C and the y-axis at D and has a turning point at E.
The graph of f and g intersect at B and C. The graph of h cuts the x-axis at A and the y-axis
at F and its asymptote passes through D. F is the point (0 ; 1) .

F(0 ;1)

29.1 Determine the length of AC. (4)

29.2 Determine the coordinates of D. (1)

29.3 Determine the value of x for which the graph of f has a minimum value. (2)

29.4 Determine the value of a, b and q. (5)

29.5 For which value of k will x2  3x  4  2 x  k have two real roots that are
opposite in sign? (2)

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 24

MARKING GUIDELINES: FUNCTIONS AND GRAPHS

QUESTION 1

1.1.1 Intercepts with the axes:  y-intercept

y-intercept: Let x  0  x-intercepts
y  (0  2) 2  1  3  coordinates of TP
(2 ;1)
 axis of symmetry
(0 ;  3)  shape (6)
x-intercepts: Let y  0 (1; 0) (3 ; 0)

0  ( x  2) 2  1
 ( x  2)2  1 (0 ;  3)
 x  4x  4  1
2
x2
 x2  4 x  3  0
 ( x  1)( x  3)  0
x  1 or x  3
(1; 0) and (3 ; 0)
Axis of symmetry: x  2
Turning point: (2 ; 1)
1.1.2 Range: y  ( ; 1]  answer (1)
1.2.1 f ( x)  x 2  2 x  3  xTP  1
(2)  yTP  4 (3)
xTP   1
2(1)
yTP  (1) 2  2(1)  3  4
(1;  4)
1.2.2 0  x2  2 x  3  0  x2  2 x  3
 0  ( x  3)( x  1)  0  ( x  3)( x  1)
 x  3 or x  1  x  3 or x  1 (3)
1.2.3  shape
x 1  turning point
 intercepts with axes (3)

(1; 0) (3 ; 0)

(0 ;  3)

(1;  4)

1.2.4 minimum value  4 .  answer (1)

1.2.5 x0  answer (1)
1.2.6 y  2( x  3) 2  1  2( x  3) 2
 1 (2)

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 25

QUESTION 2

2.1.1 y  201  1  1  substitution

 (0 ; 1) (2)
(0 ;1)
2.1.2 0  2 x 1  1  substitution
 ( 1 ; 0) (2)
1  2 x 1
 20  2 x 1
 x  1 (1; 0)
2.1.3  intercepts with the axes
 asymptote
 shape (3)

(0 ;1)
(1; 0)

y  1

2.1.4  intercepts with the axes

 shape (2)

(0 ;1)
(1; 0) (1; 0)

y  1

2.1.5 (0 ; 1)  answer (1)

2.2.1 2  vertical asymptote
y  f ( x  2)  1  1  horizontal asymptote
x2
Vertical asymptote:  shape
x2  (0 ; 0) and any point on
Horizontal asymptote: right branch (4)
y 1 (3 ; 3)
y-intercept: y 1
y0
x-intercept:
x0 (0 ; 0)
Select x  3 .
2 x2
y  1  3
3 2

2.2.2 x  {2}  real numbers

 excluding 2 (2)

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 26

QUESTION 3

3.1  x-intercepts
 y-intercept
 axis of symmetry
 turning point (6)

(0 ;0) (4 ; 0)

(2 ;  4)

3.2  x  1
 y  2
 y-intercept
 x-intercept
 shape (6)

(2 ;1)

2 12
y  2

(2 ;  3)

(0 ;  5)
x  1

3.3  y  2
x 1
1  y-intercept
y  2  2
2  x-intercept
 shape (4)
(0 ; 2)

(1; 0)

y  2

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 27

3.4 Domain of g:  ( ; )

( ; )  {1}  excluding 1 (2)
3.5 Range of h:  (2 ; ) (2)
(2 ; )
3.6 0k 4  0  k  4 (2)

QUESTION 4

4.1 y  1  y  1 (1)
4.2 y2 01
1  Let x  0
 (0 ; 1) (2)
 y 1
(0 ;1)
4.3 0  2 x 1  1  Let y  0
1  2 x 1  0  2x1 1
 ( 1 ; 0) (3)
 20  2 x 1
0  x 1
 x  1
( 1; 0)
4.4  intercepts with axes
y  3x 1  1  asymptote
 shape (3)

(0 ; 2)

(1; 0) (1; 0)

y  1
y  2 x  2

4.5 See above.  intercepts with axes

 shape (2)
4.6 f ( x)  g ( x) for all x  0.  x0 (1)

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 28

QUESTION 5

5.1 (1 ; 3)  1
 3 (2)
5.2 y2  y2 (1)
5.3 Substitute x  1 in the equation of g:  substituting x  1
11  y3
1
y   23  stating that g passes
3
through f. (3)
The graph of g passes through the point (1 ; 3) which is
the turning point of f.
5.4 The graph of f has a minimum value of 3.  stating minimum value (2)
5.5 y  4(0  1)2  3  7  x0
 7 (2)
5.6 01  x0
1
y   25  5 (2)
3
5.7 For the graph of f:
x 1  shape
f  turning point
 y-intercept
For the graph of g:
(2 ;7)  shape
 asymptote
5  y-intercept
(1; 3) g  intersecting point (7)

y2

5.8 [3 ; )  minimum value 3

  (2)

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 29

QUESTION 6

6.1  shape
 one point indicated (2)

(1; 2)

6.2 y  2 x2 x  0 f  x  2 y2
x  2 y2 y  0 f 1  y x (2)
2
 x  y2
2
y  x (x  0)
2
6.3  shape
 one point indicated (2)
yx
(1; 2)

(2 ;1) f 1

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 30

QUESTION 7

7.1 y  log3 x  y  log3 x (1)

7.2  the graph of f
 the graph of f 1
yx  intercepts and one other
point on each graph (3)
(1; 3)

f 1
(0 ;1) (3 ;1)

(1; 0)

7.3.1 f ( x). f 1( x)  0 for all 0  x  1  boundaries

 inequalities (2)
7.3.2 x. f 1
( x)  0 for all x  1 .  x 1
7.4 y  g ( x )  3x 1  1  equation
 asymptote
 y-intercept
y 1  x-intercept (4)

(0 ; 23 )

(1; 0)

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 31

QUESTION 8

8.1.1 y2  y2 (1)

8.1.2 y -intercept:  x-intercept
0 1  y-intercept
1  y2
y  2  2 
2  shape (6)
 y  2  2(2)  2
x -intercept:
x 1
1
0  2  2 
2
x 1
1
2  2
2 y2
x 1
1
   1
2 (1; 0)
 x 1  0
x 1
(0 ;  2)

8.1.3 y  ( ; 2)  y  ( ; 2) (1)

8.2.1 y -intercept:  x-intercept
y  log 1 (0)  one other point
4  shape (3)
undefined
x -intercept:
0  log 1 x
4
0 (1; 0)
1
   x
4
(2 ;  12 )
x 1
For x  2
1
y  log 1 (2)  
4 2
8.2.2 x3 0  x  3 (1)
 x  3
8.2.3 x  log 1 y 1
x
4  g ( x)    (1)
x 4
1
   y
4
x
1
 g ( x)   
4

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 32

QUESTION 9

9.1.1 y-intercept:  (0 ;  1)
y  40  2  40  2  1  0  4 x  2
(0 ;  1)  1 
   ; 0 (3)
x-intercept:  2 
0  4 x  2
 2  (22 )  x
 21  22 x
1  2 x
 2 x  1
1
x  
2
 1 
  ; 0
 2 
9.1.2  y  2
 intercepts shown
 shape (3)

( 12 ; 0)

(0 ;  1)
y  2

9.1.3 y  4( x 3)  2  2  4 ( x 3)

 2  2 (2)
 y  4 x 3
9.2.1 For the graph of f:
 shape
 at least one point (2)

(0 ; 3)

(1; 0)

y  1
(1;  2)

(4 ;  4)

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 33

9.2.2 y   4x ; x  0 f  x   4y
x   4y ; y  0 f 1  y  x2
1
4
 x2  4 y
1
1 2  h( x )   x 2
y  x ; x0 4
4  x0 (4)
1
 y  x2 h
4
1
 h( x )   x 2 ; x  0
4
9.2.3 See diagram in 6.1  shape
 x-intercept
 y-intercept
 asymptote (4)

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 34

QUESTION 10

10.1 The vertical asymptote x  1 intersects with the  stating the asymptotes
horizontal asymptote y  2 at the point (1 ; 2) .  point of intersection
 A is the point (1; 2) of asymptotes (2)
10.2 y  a( x 1)2  2  y  a( x 1)2  2
y-intercept of f :  y6
4  6  a (0  1) 2  2
y 2
0 1  a4
y 6  g ( x)  4( x 1)2  2
B(0 ; 6) (5)
Substitute B:
6  a(0  1) 2  2
a  4
g ( x)  4( x  1) 2  2
10.3 y  ( x  p )  q  y  ( x  1)  2
 y  ( x  1)  2  h( x )   x  3
 y  x 1 2  y  x 1 2
 h( x )   x  3  y  x 1 (4)
y  ( x  p)  q
y  x 1  2
 y  x 1
10.4 y  [2 ; )  y  [2 ; ) (1)
10.5.1 x-intercept of f : 4
4
 0 2
0 2 x 1
x 1  x3
 0  4  2( x  1)  1 x  3 (3)
 0  4  2 x  2
0  2x  6
 2 x  6
x  3
C(3 ; 0)
f ( x)  0 for all 1  x  3
10.5.2 f ( x).g ( x)  0 for all  x 1
x  1 or x  3  x3 (2)

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 35

QUESTION 11

11.1 For x  1  y  (1)  3 (1)

y  (1)  3  4
A(1 ;  4)
11.2 Let x  0  B(0 ;  3) (1)
 y  (0)  3  3
B(0 ;  3)
11.3 y  a ( x  1) 2  4  y  a( x 1)2  4
Substitute (0 ;  3)  a 1
 f ( x)  ( x 1)2  4 (3)
 3  a (0  1) 2  4
a 1
 f ( x )  ( x  1) 2  4
11.4 a a
y 4  y 4
x 1 x 1
Substitute (3 ;  2)  a4
4
2 
a
4  g ( x)  4 (3)
3 1 x 1
a
2 
2
a4
4
 g ( x)  4
x 1
11.5 y  ( x  1)  4  y  ( x  1)  4
 y  x5  y  x5 (2)

Alternatively:

Gradient of the line y   x  3 is  1

 Gradient of other axis of symmetry is 1
 y  1x  c
y  1x  c
 y  x5 (2)
Substitute (1;  4)
4  1  c
 c  5
 y  x5
11.6 0  ( x  1) 2  4  0  ( x 1)2  4
 0  x2  2 x  1  4  x  3 or x  1
 CD  4 units (3)
 0  x2  2 x  3
 0  ( x  3)( x  1)
x  3 or x  1
CD  4 units

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 36

Alternatively:
0  ( x  1) 2  4
 4  ( x  1) 2
2  x  1
 2  x  1 or  2  x  1
 x  3 or x  1
 CD  4 units
11.7 h( x)  [( x  1) 2  4]  2  ( x  1) 2  4  2  h( x)  ( x 1)2  6
 h( x)  ( x  1) 2  6  6 (2)
max value is 6
11.8 x-intercept of g: 4
 0 4
4 x 1
0 4  x2
x 1
 0  4  4( x  1)  x2 (3)
0  4  4x  4
 4x  8
x  2
 f ( x).g ( x)  0 for
x2

11 .9 x3  x2 (1)

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 37

QUESTION 12

12.1 y  ax  substitution
Substitute (1; 2) :  f ( x)  2 x (2)
2  a1
a  2
 f ( x)  2 x
12.2 y  a ( x  3) 2  6  y  a ( x  3) 2  6
Substitute: (1; 2):  substitution
2  a (1  3) 2  6  a  1
 equation (4)
4  4a
 a  1
g ( x )  ( x  3) 2  6
12.3 0  ( x  3) 2  6  0  ( x  3) 2  6
 ( x  3) 2  6  x  3 6
x 3   6  (3  6)  (3  6)
x  3 6  2 6 (4)
EF  xF  xE
 EF  (3  6)  (3  6)  2 6
12.4 h( x)  f ( x  1)  1  2 x1  1
 2 x 1  1  y  (1; ) (3)
Range: y  (1; )
12.5 f ( x).g ( x )  0 for all:  answer (1)
3 6  x  3 6
12.6 Graphically:  k  9 (2)
k  9
Algebraically:
(6) 2  4( 1)( k )  0
 k  9

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 38

QUESTION 13

13.1 y  a ( x  2) 2  8  substitution of point

 the value of a
Substitute (0 ; 6)  standard form and stating
 6  a(0  2) 2  8 the values a, b and c (3)
1
a  
2
1
 y   ( x  2) 2  8
2
1
 y   ( x 2  4 x  4)  8
2
1
 y   x2  2 x  2  8
2
1
 y   x2  2 x  6
2
1
a   b2 c6
2
13.2 1  letting y  0
0   x2  2 x  6
2  factors
 0  x 2  4 x  12  the values of x
 the length of AB (4)
 0  ( x  6)( x  2)
x  6 or x  2
 AB  8 units
13.3 3  equation (1)
g ( x)   1 for x  1
x 1
13.4 3  equating of functions
1  x  multiplying by LCD
x 1
 3  1( x  1)  x( x  1)  solving for x
 solving for y
 x  4  x2  x  coordinates of C (5)
 4  x2
 x  2 (x  0)
y 2
C(2 ; 2)
13.5.1 g ( x)  x  0  g ( x)  x
 g ( x)  x  x2 (2)
The values of x for which the hyperbola
is below the line y  x are:
x  2
13.5.2 f ( x).g ( x)  0 for all 1  x  2  1  x  2 (1)

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 39

13.6 1 1
 x2  2 x  k  6   x2  2 x  6  k
2 2
1  k 8 (2)
 x 2  2 x  6  k
2
This is where the parabola cuts the line
y  8 in one point only (two equal solutions)
k  8
Alternatively:
1    16  2k
 x2  2 x  6  k  0  k 8
2
 1
  (2) 2  4    (6  k )
 2
  4  2(6  k )
  4  12  2k
  16  2k
For equal roots,   0
 0  16  2k
k  8

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 40

QUESTION 14

14.1 y  ax  substitution
 answer (2)
 4  a 2
1
4 
a2
 4a 2  1
1
 a2 
4
1
a 
2
14.2 0  2 x2  4 x  6  equating to 0
 x-values
 0  x2  2 x  3  length of AB (3)
 0  ( x  3)( x  1)
 x  3 or x  1
 AB  4 units
14.3 4  x-value of R
xR    1  y-value of R
2(2)
 length of SR (5)
 yR  2(1)2  4(1)  6  8
 SR  8 units
Alternatively:
f ( x)  4 x  4
0  4x  4
 x  1
14.4 h( x)  2( x  1) 2  8  answer (1)
14.5 x  interchanging x and y
1
y   equation (2)
2
y
1
x   
2
 y  log 1 x
2
14.6  shape
 x-intercept (2)
The other point is optional.

(1 ; 0)

(2 ;  1)
y  log 1 x
2

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 41

14.7.1  equation
 x-value
y  log 1 x  0 x4 (3)
2

y  2
(4 ;  2)

log 1 x  2
2
2
1
x    4
2
 g 1 ( x)  2 for all 0  x  4
14.7.2 1  x  0  answer (2)

QUESTION 15

15.1 y  1 is the horizontal asymptote.  y 1

Now substitute y  1 into y   x  2 :  substitution to get x
1   x  2 4
 f ( x)  1
x 1 x 1
 B is the point (1;1)  p  1
The vertical asymptote’s equation is x  1 (4)
Therefore, the equation of the hyperbola is
4
f ( x)  1
x 1
The value of p is p  1 .

15.2 The equation of the other axis of symmetry is:  y  ( x  1)  1

y  ( x  p)  q  g ( x)  x (2)
 y  ( x  1)  1
y  x
 g ( x)  x
15.3 Domain of f : 
x   1   1 (2)
15.4 4 4
0 1  0 1
x 1 x 1
 0  4  x  1  A(5 ; 0)
x  5  x  4 (3)
A(5 ; 0)
If the graph of f is shifted 5 units left, then the
newly formed graph’s asymptote will be x  4 .

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 42

QUESTION 16

16.1 a a
y 2  y  2
x2 x2
Substitute (0 ;  3) :  a2
2
3 
a
2  f ( x)  2 (4)
02 x2
a
1 
2
a  2
2
 f ( x)  2
x2
16.2 y  m( x  2)2  1  y  m( x  2)2  1
Substitute (0 ;  3):  m  1
3  m(0  2) 2  1  g ( x)  ( x  2) 2  1 (4)

4  m(2) 2
4  4m
 m  1
 g ( x)  ( x  2) 2  1
16.3 2  equating to 0
0 2  x3
x2
 0  2  2( x  2)  OA  3 units (3)
0  2  2x  4
 2x  6
x  3
 OA  3 units
Alternatively:
0  ( x  2) 2  1
 ( x  2) 2  1
 x2  4 x  4  1
 x2  4 x  3  0
 ( x  1)( x  3)  0
x  1 or x  3
 OA  3 units
16.4 f ( x).g ( x )  0 for all  1  x  2
1  x  2 or x  3  x 3
(3)

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 43

QUESTION 17

17.1 a a
y 1  y 1
x 1 x 1
Substitute the point (  2 ; 2) :  a  1
 p 1
a
2 1  q 1
2  1
1
a  d  (6)
1  2
1
 a  1
1
y 1
x 1
 p  1, q  1, a  1
y  dx 2
Substitute the point (  2 ; 2) :
2  d ( 2) 2
1
d 
2
1 2
y  x
2
17.2 1  substituting x  1
Substitute x  1 into the equation y  1 :
x 1  y  12
1 1  Q(1; 12 )
y 1  . (4)
11 2
OR
1 2
Substitute x  1 into the equation y  x :
2
1 2 1
y (1) 
2 2
Therefore, Q is the point Q(1; 12 ) .
Alternatively:
 equating
1 1
 1  x2  solving equation
x 1 2
 y  12
2  2 x  2  x3  x 2
 Q(1; 12 )
0  x  x  2x
3 2

 0  x( x 2  x  2)
 0  x( x  2)( x  1)
 x  0 or x  2 or x  1
Substitute x  1 into either of the two equations:
1 2 1
y (1) 
2 2
 Q is the point (1; 12 ).
17.3.1 f ( x).g ( x)  0 for  1  x  0.  1  x  0 (2)

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 44

17.3.2 k 0  k 0 (1)
17.3.3 k 1  k 1 (1)
17.4 1 2  1  1 2 1
BC  x   1  x  1
2  x 1  2 x 1
1 1  substituting x  4
 x2  1 1
2 x 1  7 (3)
Substitute x  4 : 5
1 1 1
BC  (4) 2  1  7
2 4 1 5
17.5 1 1
h( x )  1 2  3
( x  3)  1 x2
1  asymptotes (3)
 3
x2
Vertical asymptote: x2
Horizontal asymptote: y  3
17.6 1 1
y  ( x  2) 2  4  y ( x  2) 2  4
2 2
1  x  2  8
 0  ( x  2) 2  4
2  x  0,83
 ( x  2) 2  8  x  4,83 (4)
x  2   8
 x  2  8
 x  0,83 or x  4,83

Alternatively:

1 1
y ( x  2) 2  4  y ( x  2) 2  4
2 2
1  using quadratic formula
 0  ( x  2) 2  4  x  0,83
2
 x  4,83
 ( x  2) 2  8
 x2  4 x  4  8
 x2  4 x  4  0
4  (4) 2  4(1)(4)
x 
2(1)
4  32
x 
2
 x  2  2 2
 x  0,83 or x  4,83

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 45

QUESTION 18

18.1.1 D(  4 ; 0)  D(  4 ; 0) (1)
18.1.2 D(  4 ; 0) , C(1 ; 0) and A(0 ;  4) .  y  a ( x  4)( x  1)
y  a ( x  4)( x  1)  a 1
4  a (0  4)(0  1)  y  x2  3x  4
a  1  a  1, b  3 and c  4 (4)
 y  1( x  4)( x  1)
 y  x 2  3x  4
 a  1, b  3 and c  4
18.1.3 x 2  3x  4  x  4  x 2  3x  4  x  4
 x2  2 x  8  0  x2  2 x  8  0
 ( x  4)( x  2)  0
 ( x  4)( x  2)  0
 x  4 or x  2
 x  4 or x  2  R(4 ; 8) (5)
y  44 8
R(4 ; 8)
18.1.4 NP  ( x  4)  ( x 2  3 x  4)  NP  ( x  4)  ( x2  3x  4)
 NP  x  4  x 2  3 x  4  NP   x2  2 x  8
2
 NP   x 2  2 x  8  3  3
 NP       2     8
3  2  2
At x  1 12   3
2  NP  8 (4)
2 4
 3  3
NP       2     8
 2  2
9
 NP    3  8
4
3
 NP  8
4
18.2 f ( x) . g ( x)  0 for all x  1.  x  1. (1)
18.3  3
2
 3 1  4k
yP      3     4  6 1
 2  2 4  k 6 (2)
4
1
f ( x)  k  0 for all 4  k  6
4

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 46

QUESTION 19

19.1 The horizontal asymptote is y  2 . k

 y  2
The vertical asymptote is x  1 . x 1
k  substitution of point
y 2
x 1  k 3
Substitute (2 ; 5) :  k  3, p  1, r  2 (5)
k
5  2
2 1
k  3
3
y  2
x 1
 k  3, p  1 and r  2.
19.2 Axes of symmetry:  y  x 1
 y  x  4 (2)
y  ( x  1)  2 y  ( x  1)  2
 y  x 1  y  x  4
19.3 The horizontal asymptote is y  2 .  q2
q  2  substituting (2 ; 5)
y  ab x  2  substituting (0 ; 3)
Substitute (2 ; 5) :  a 1
 b 3 (5)
5  ab2  2
 3  ab 2
Substitute (0 ; 3)
3  ab 0  2
a  1
 3  (1)b 2
b  3
 y  ( 3) x  2
 a  1, b  3 and q  2.

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 47

19.4 The axis of symmetry is x  1 .  y  d ( x  1) 2

 y  d ( x 1)2  obtaining (0 ;  1)
The graph of h passes through A, the y-intercept of f.  substituting (0 ;  1)
3  d  1
Let x  0 in y  2.
x 1  t  1
3 (6)
y   2  1
0 1
A is the point (0 ;  1).
Substitute (0 ;  1) :
1  d (0  1) 2
 d  1
 y  ( x  1) 2
 d  1 and t  1.
19.5 B is the x-intercept of f. 3
 0 2
0 
3
2 x 1
x 1  multiplying by LCD
 0  3  2( x  1) 1
 x
0  3  2x  2 2
2 x  1  1 
   ; 0 (4)
1  2 
x  
2
 1 
B is the point   ; 0  .
 2 
19.6 f ( x)  g ( x) for all x  0 or x  2  x0
 x2 (2)
19.7 1 1
f ( x).g ( x)  0 for all x   or x  1  x
2 2
 x 1 (2)
19.8 y  h( x  1)  y  ( x  2) 2  4
 ( x  1  1) 2  4  (2 ; 4)
(2)
 ( x  2) 2  4
Turning point is (2 ; 4)

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 48

QUESTION 20

20.1 x  1  x  1
y3  y3 (2)
20.2 y  ( x  1)  3 y  ( x  1)  3  y  x4
y  x4 y  x  4  y  ( x  1)  3
 y  x  4 (3)
20.3 x  ( ; )  answer (1)
20.4 6 6
0 3  0 3
x 1 x 1
 0  6  3( x  1)  x 1
 0  6  3 x  3  OA  1 unit
6
3 x  3  y  3  3
0 1
x 1  OB  3 units
OA  1 unit (5)
6
y  3  3
0 1
OB  3 units
 CD  f ( x)  g ( x)
  1  
20.5 x
 6
CD     3      2   x  2
 x  1   2    92
At x  2  7 units (5)
2
 6   1  
CD     3      2
 2  1   2  
 9   2
 7 units
20.6 6  f (3)  6
f (3)   3 6
3  1  g (3)  6
3  E(3 ; 6) (3)
1
g (3)    26
2
E(3 ; 6)
20.7 x 1  x  1 (2)

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 49

QUESTION 21

21.1 p3  p3

q4  q4 (2)
21.2 y  a( x  1)( x  5)  y  a( x  1)( x  5)
Substitute (3 ; 4):  Substitute (3 ; 4)
4  a(3  1)(3  5)  a  1
 y  1( x  1)( x  5)
 4  4a
 f ( x)   x 2  6 x  5 (5)
 a  1
y  1( x  1)( x  5)
 y  ( x 2  6 x  5)
 f ( x)   x 2  6 x  5
21.3  x2  6x  5   x  5  equation
 standard form
 0  x 2  7 x  10  factors
 0  ( x  2)( x  5)  x-values
 x  2 or x  5  y3
 J(2 ; 3) (6)
y  2  5  3
J(2 ; 3)
21.4 JK  yK  yJ  4  3  1 unit  JK  1 unit (2)
21.5 k  9  k  9 (2)
21.6 1  x  5 or x  5  1  x  5
 x5 (3)

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 50

QUESTION 22

22.1 2 x2  4 x  6  0  Let y  0
 x-intercepts
 x2  2 x  3  0
 AB  4 units
 ( x  3)( x  1)  0 (3)
 x  3 or x  1
AB  4 units
22.2 4  x  1
x  1  y  8
2(2)
 FG  8 units
y  2(1) 2  4(1)  6  8 (3)
G(  1;  8)
 FG  8 units
22.3 B(1; 0) and C(  4 ;10)  mBC  2
10  0  g ( x)  2 x  2
mBC   2
4  1  a  2 and b  2 (3)
y  2 x  c
0  2(1)  c
c2
g ( x)  2 x  2
a  2 and b  2
22.4 DE  (2 x  2)  (2 x 2  4 x  6)  2 x2  6 x  8
 2 x 2  6 x  8  2 x2  6 x  4  0
 x  2 or x  1
12  2 x 2  6 x  8
 OH  2 units (4)
 2x  6x  4  0
2

 x 2  3x  2  0
 ( x  2)( x  1)  0
 x  2 or x  1
 OH  2 units
22.5 f ( x).g ( x )  0 for all x  3 or x  1  x  3
 x 1
(2)
22.6 k 2  k 2 (1)
22.7 (1  2 ;  8  1)  1
 (1;  7)  7 (2)

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 51

QUESTION 23

23.1 y  a ( x  3) 2  2  y  a ( x  3) 2  2
Substitute (0 ;  16) :  value of a
16  a (0  3) 2  2  y  2( x  3)2  2
18  9a  y  2 x 2  12 x  16 (4)
 a  2
y  2( x  3)2  2
 2( x 2  6 x  9)  2
 2 x 2  12 x  16
23.2 x  2  2 x 2  12 x  16  equating
 standard form
 2 x 2  11x  14  0  factors
 (2 x  7)( x  2)  0  x-values
7  y-values (5)
x or x  2
2
7 3
y  2
2 2
7 3
D ; 
2 2
23.3 x4  answer (1)
23.4 2 units down.  2 units down
3 units left.  3 units left (2)
23.5 f ( x)  2 x 2  12 x  16  f ( x)  4 x  12
 f ( x)  4 x  12  4x  12  0
 x3 (3)
4 x  12  0
4 x  12
x  3

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 52

QUESTION 24

24.1 P(2 ; 4)  2
 4 (2)
24.2 A(0 ; 1)  A(0 ; 1) (1)
24.3 4  a 2  substitution
 value of a (2)
1
4  2
a
1
 a2 
4
1
a 
2
24.4
1
x
1
y  g  y 
2 2
y  y  log 1 x
1
g 1
2
x  (2)
2
 y  log 1 x
2
24.5 x  2  x  2 (1)
24.6 8 8
k ( x)    k ( x)  
x2 x2
x  2  x2 (2)

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 53

QUESTION 25

25.1 x  x  2  x  x  2
 2 x  2  x  1
 ( 1 ;  1) (3)
 x  1
y  1
The axes of symmetry intersect at ( 1 ;  1) .
25.2 The asymptotes also intersect at the point of a
intersection of the axes of symmetry.  y 1
x 1
 a2
Vertical asymptote: x  1
 x 1  0  p 1
Horizontal asymptote: y  1  q  1 (4)
a
y 1
x 1

Substitute the point (1 ; 0) :

a
0 1
11
a
0  1
2
a
1 
2
a  2
2
y 1
x 1
 a  2; p  1 and q  1
25.3 2  x4
y 1 4  3
x 1 3 (2)
2
y  3
x4

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 54

QUESTION 26

26.1 C(0 ; 5)  C(0 ; 5) (1)

x  4  0  2  4  0
26.2
(1)
2 2
26.3 y  a( x  p)2  q  p2
Axis of symmetry is x  2.  y  a( x  2)2  q
x  2  0  5  4a  q
 0  9a  q
p2
 a  1
 y  a ( x  2) 2  q  q9 (6)
Substitute A( 4 ; 5) :
5  a (4  2) 2  q
 5  4a  q
Substitute F(1; 0) :
0  a (1  2) 2  q
 0  9a  q
Solving simultaneously: a  1 and q  9
Alternative:
By symmetry, E is the point  E(5 ; 0)
y  a ( x  5)( x  1)  y  a( x  5)( x  1)
Substitute C(0 ; 5) :  completing the square
5  a(0  5)(0  1)  a  1
 a  1  p2
 q9 (6)
 y  1( x  5)( x  1)
E(5 ; 0) .   x2  4 x  5
 ( x 2  4 x )  5
 ( x 2  4 x  4  4)  5
 [( x  2) 2  4]  5
 ( x  2) 2  9
 a  1; p  2 and q  9
26.4  x  4 x  5  2 x  3
2  equating
 standard form
 0  x2  2 x  8  x-values
 0  ( x  4)( x  2)  D(2 ;  7) (4)
 x  4 or x  2
For x  2 :
y  2(2)  3  7
D(2 ;  7)

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 55

26.5 h( x )  f (  x )  1  (2  x) 2  8
 ( x  2) 2  9  1  (2 ; 8)
(3)
 (2  x) 2  8
Axis of symmetry of h is:
2 x  0
x  2
The turning point of h is (2 ; 8) .
Alternative:  reflection in y-axis and
The graph of h is the reflection of f in the y-axis and a shift of 1 down.
shift of 1 unit down.  (2 ; 8)
The turning point of f is (2 ; 9) . (3)
Therefore, the turning point of h is (2 ; 8) .
26.6 BG  ( x2  4x  5)  (2 x  3)   x2  2 x  8
There are three methods of finding the maximum
length.
Method 1 Completing the square   x2  2 x  8
BG   x 2  2 x  8  completing the square
 9 (4)
 ( x 2  2 x )  8
 ( x 2  2 x  1  1)  8
 [( x  1) 2  1]  8
 ( x  1) 2  9
Maximum length of BG is 9.
  x2  2 x  8
Method 2 The formula  substitution in formula
 x  1
(2)
x  1  substitution to get 9 (4)
2(1)
Maximum BG  (1)2  2(1)  8  9
  x2  2 x  8
Method 3 Calculus  2x  2  0
d (BG)  2 x  2  0  x  1
dx  substitution to get 9 (4)
 x  1
Maximum BG  (1) 2  2(1)  8  9

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 56

26.7.1 y  2 x  3 g  g 1 ( x)   1 x  3 (1)
1 2 2
x  2 y  3 g
2 y  x  3
y  1 x 3
2 2
 g 1 ( x)   1 x  3
2 2
26.7.2  sketching two graphs
2 x  3   1 x  3
2 2  x  1
4 x  6   x  3  x  1 (3)
3x  3
 x  1
y  g ( x)  2 x  3

1 12 1
3

1 12
1 3
y  g 1 ( x)   x 
2 2
3

g 1 ( x)  g ( x) for all x  1.

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 57

QUESTION 27

27.1 Vertical asymptote: x2  x2

Horizontal asymptote: y3  y3 (2)
27.2 Domain: x  2 or x  2 Alt: x   {2}  answer (1)
27.3 y  ( x  2)  3 y  ( x  2)  3  y  x 1
 y  x 1  y  x  5  y  x  5 (2)
Alternative:
The axes of symmetry intersect at the point of intersection
of the asymptotes, which is (2 ; 3) .
y  xc y  x  c
Substitute (2 ; 3): Substitute (2 ; 3):
3  2  c 3  2  c
c  1 c5
 y  x 1  y  x  5
27.4  asymptotes
 intercept at the origin
(3 ; 9)  two branches
 one point on right
branch (4)

y 3

x2

27.5 The point of intersection of the axes of symmetry is the  x  1

same as the point of intersection of the asymptotes.  y2
x  3  x 1  p 1
 2 x  2  q2
 x  1 (4)
y 2
(1; 2)
Vertical asymptote: x  1  x 1  0
Horizontal asymptote: y2
y  6 2
x 1
 p  1 and q  2

27.6 3 units left and 1 unit down.  3 left

 1 down (2)

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 58

QUESTION 28

28.1 0  x 2  3x  4  equating to 0
 factors
 0  ( x  4)( x  1)
 x-values
 x  4 or x  1  length of AC (4)
AC  5 units
28.2 D(0 ;  4)  answer
28.3 f ( x)  x  3x  4
2  0  2x  3
3
 f ( x)  2 x  3  x (2)
2
0  2x  3
3
x 
2
Alternatively:
3 3  substitution in formula
x  3
2(1) 2  x (2)
2
28.4 y  ab x  q  q  4
The horizontal asymptote is y  4 .  Substitute (0 ; 1)
 q  4  a 5
 Substitute ( 1 ; 0)
 y  ab x  4
5
Substitute (0 ; 1) :  b (5)
4
1  ab0  4
 5  a(1)
a  5
 y  5b x  4
Substitute ( 1 ; 0) :
0  5b 1  4
5
4 
b
 4b  5
5
b 
4
28.5 k  4  answer (2)

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 59

QUESTION 29

29.1 (1 ; 4) or (2 ; 16)  one point (1)

29.2 y  ax 2  substituting point
Substitute (1 ; 4) :  a4
 f ( x)  4 x 2 (3)
4  a(1)2
a  4
f ( x)  4 x 2
Alternatively:
Substitute (2 ; 16) :
16  a(2) 2
a  4
f ( x)  4 x 2
29.3 x0  answer (1)
29.4 g 1 ( x)  log 4 x  answer (1)
29.5 The graph of y  log 4 x has been shifted 2 units right.  shift 2 units right.
h( x)  log 4 ( x  2)  0 x3 (2)
The x-intercept of h is:
0  log 4 ( x  2)
 40  x  2
1  x  2
x  3
h( x)  0 for all 0  x  3 .
See diagram below.

y  h( x)  g 1 ( x  2)

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