1. Functions and Graphs (2)4
Functions and Graphs (2)
Case Study
4.1 Solving Inequalities by the Graphical Method
4.2 Finding Optimum Values of Quadratic Functions by the Algebraic Method
4.3 Transformation on the Graphs of Functions
Chapter Summary

2. Case Study A satellite dish is a
If you want to know the answer, you should consider the shape of the dish first.
How deep is this satellite dish?
A satellite dish is a
powerful antenna. Its cross-section is in the shape of a parabola.
To find the depth of the dish, we may make use of the quadratic graph.
Suppose the shape of the dish can be represented by a quadratic graph y  ax2  b with respect to the axes shown in the figure.
As y is minimum when x  0, we can find out the depth of the dish if we know the value of b.

3. 4.1 Solving Inequalities by the Graphical Method
If a function y  f (x) is given, then we can use the graph of y  f (x) to solve the inequalities f (x)  k and f (x)  k.
Consider a quadratic function y  f (x), whose graph opens upward.
If we add a straight line y  k onto the graph, we observe that the line divides the graph of  
y  f (x) into three parts.
 f (x)  k: the left part (where x  a) or
the right part (where x  b)
 f (x)  k: the middle part (where a  x  b)
For a quadratic function y  f (x) whose graph opens upwards,  
1. f (x)  k for x  a or x  b or 
2. f (x)  k for a  x  b,
if k is greater than the y-coordinate of the vertex of f (x).

4. 4.1 Solving Inequalities by the Graphical Method
If the graph of a quadratic function function y  f (x) opens downwards, we have similar conclusions:
For a quadratic function y  f (x) whose graph opens downwards,  
1. f (x)  k for a  x  b or
2. f (x)  k for x  a or x  b,
if k is less than the y-coordinate of the vertex of f (x).

5. Example 4.1T 4.1 Solving Inequalities by the Graphical Method
(a) Plot the graph of y  2x2  7x  3 for 2  x  6.
(b) Using the graph, solve the following inequalities:
(i) 2x2  7x  3  7 (ii) 2x2  7x  3  0 (iii) 2x2  7x  1  2
Solution:
y  2x2  7x  3
(a) x –2 –1 1 2 3 4 5 6 y 25 12 –3 7 18 33
(b) (i) From the graph, the solution is
0.5  x  4.
(ii) From the graph, the solution is
0.5  x  3.
y  7
(iii) 2x2  7x  1  2 can be written as
2x2  7x  3  0.
∴ The solution is x  0.5 or x  3.

6. 4.1 Solving Inequalities by the Graphical Method
The graphical method of solving quadratic inequalities is also applicable to solve inequalities involving other functions.
Consider the graph of y  f (x) in the following figure.
In order to solve f (x)  k or f (x)  k, we draw a line y  k on the graph first and then find the corresponding value of x (x  a).
For f (x)  k, the solution is x  a.
For f (x)  k, the solution is x  a.

7. Example 4.2T 4.1 Solving Inequalities by the Graphical Method
(a) Plot the graph of y  x3  2x for 2  x  2.
(b) Using the graph, solve x3  2x  0.
(c) Draw the line y  0.4 and hence solve x3  2x  0.4.
(Give the answer correct to 1 decimal place.)
y  x3  2x
Solution:
(a) x –2 –1 1 2 y 4 –4
(b) The solution is x  1.4 or
0  x  (cor. to 1 d. p.)
y  0.4
(c) The solution is x  1.3 or
0.2  x  (cor. to 1 d. p.)

8. Example 4.3T 4.1 Solving Inequalities by the Graphical Method
The figure shows the graph of y  x3  5x  k.
(a) Find the value of k.
(b) Using the graph, solve x3  5x  4. (Give the answers correct to 1 decimal place.)
y  4
Solution:
(a) Since the graph passes through (0, 8),
we have
8  03  5(0)  k
k  8
(b) x3  5x  4 can be rewritten as x3  5x  8  4  8,
that is, x3  5x  8  4.
From the graph, the solution is 1.7  x  1.0 or x  2.5.
(cor. to 1 d. p.)

9. 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
In Chapter 2, we learnt how to use the method of completing square to solve a quadratic equation ax2  bx  c  0.
Now, we consider a quadratic function in the form y  ax2  bx  c:
Leave the constant term outside the brackets.
Take out the factor a such that the coefficient
of x2 inside the brackets is 1.
Add to complete the square;
Then minus the same number.
x2  2px  p2  (x  p)2
, where and

10. 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Once we obtain y  a(x  h)2  k from y  ax2  bx  c, we can
 locate the vertex
 find the axis of symmetry
Optimum value is either a minimum or a maximum value.
 find the optimum value
of the quadratic function.
Notes:
 If a > 0:
the graph opens upwards
optimum value  minimum value
 If a < 0:
the graph opens downwards
optimum value  maximum value

11. 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Consider the quadratic function y  a(x  h)2  k.
Since (x  h)2  0 for all values of x,
 If a  0,
a(x  h)2  0
∴ Minimum value  k
at x  h
a(x  h)2  k  k for all values of x
 If a  0,
a(x  h)2  0
∴ Maximum value  k
at x  h
a(x  h)2  k  k for all values of x

12. 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
From the quadratic graph of y  a(x  h)2  k,
1. Vertex  (h, k)  
2. Axis of symmetry: x  h
3. If a  0, then the minimum value of y is k.
If a  0, then the maximum value of y is k.
Both values are attained at x  h.

13. Example 4.4T 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Example 4.4T
Find the minimum value of y  2x2  8x  21 using the method of completing the square.
Solution:
y  2x2  8x  21
 2(x2  4x)  21
Leave the constant term outside the brackets.
 2(x2  4x  22  22)  21
Add 22 to complete the square, then minus 22.
 2(x2  4x  22)  2(22)  21
Take the term  22 out of the brackets by
multiplying 2.
 2(x  2)2  13
∴ The minimum value of the function is 13.

14. Example 4.5T 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Example 4.5T
The graph of y  2x2  bx  12 passes through the point (2, 8).
(a) Find the value of b.
(b) Find the coordinates of the vertex and the axis of symmetry.
(c) Find the x-intercepts and y-intercept.
(d) Sketch the graph and label the axis of symmetry and vertex on the graph.
Solution:
(a) Substituting x  2 and y  8 into the given equation,
we obtain
8  2(2)2  b(2)  12
8  4  2b
b  2

15. Example 4.5T 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Example 4.5T
The graph of y  2x2  bx  12 passes through the point (2, 8).
(a) Find the value of b.
(b) Find the coordinates of the vertex and the axis of symmetry.
(c) Find the x-intercepts and y-intercept.
(d) Sketch the graph and label the axis of symmetry and vertex on the graph.
Solution:
(b) y  2x2  2x  12
 2(x2  x)  12
∴ The vertex is
and the axis of symmetry
is x 

16. Example 4.5T 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Example 4.5T
The graph of y  2x2  bx  12 passes through the point (2, 8).
(a) Find the value of b.
(b) Find the coordinates of the vertex and the axis of symmetry.
(c) Find the x-intercepts and y-intercept.
(d) Sketch the graph and label the axis of symmetry and vertex on the graph.
Solution:
(c) When y  0,
2x2  2x  12  0
(d)
x2  x  6  0
(x  3)(x  2)  0
x  3 or 2
∴ The x-intercepts are 3 and 2.
y-intercept  2(0)2  2(0)  12
 12

17. 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
B. Practical Problems Related to Optimum Values
In real life situation, we can find the solutions of many problems by solving relevant quadratic functions, as illustrated in Example 4.6T and Example 4.7T.

18. Example 4.6T 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
B. Practical Problems Related to Optimum Values
Example 4.6T
A man throws a stone upwards from the roof of a building. The height of the stone after t seconds from the ground s m can be expressed as
s  40  12t  2t2.
(a) When s  8, find the value of t.
(b) What is the maximum height that the stone can reach?
Solution:
(a) When s  8,
(b) s  40  12t  2t2
 2(t2  6t)  40
8  40  12t  2t2
2t2  12t  32  0
 2(t2  6t  32  32)  40
t2  6t  16  0
 2(t  3)2  2(32)  40
(t  8)(t  2)  0
 2(t  3)2  58
t  8 or 2 (rejected)
∴ The maximum height
that the stone can reach
is 58 m.

19. Example 4.7T 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
B. Practical Problems Related to Optimum Values
Example 4.7T
The base and the height of a triangle are (x  20) cm and (50  x) cm respectively. Let A cm2 be the area of the triangle.
(a) Express A in terms of x.
(b) Find the maximum area of the triangle.
Solution:
(a) A
(b) A
∴ The maximum area of the
triangle is cm2.

20. 4.3 Transformation on the Graphs of Functions
A. Translation
In junior forms, we learnt that an object could be translated.
In the figure, translating triangle (I) 5 units to the right and 3 units upwards will give a congruent triangle (II).
We can also translate a graph horizontally or vertically.

21. 4.3 Transformation on the Graphs of Functions
A. Translation
The figure shows the graph of y  x2.
Let f (x)  x2,
g1(x)  x2  3 and
g2(x)  x2  2.
y  x2  3 4 1
f (x) 2 –1 –2 x 7 4 3
g1(x) 2 –1 –2
g2(x)
From the graph,
y  f (x) is translated 3 units upwards to become y  g1(x).
y  x2  2
y  f (x) is translated 2 units downwards to become y  g2(x).

22. 4.3 Transformation on the Graphs of Functions
A. Translation
The figure shows the graph of y  x2.
Let f (x)  x2,
h1(x)  (x  1)2 and
h2(x)  (x  1)2. 4 1
f (x) 2 –1 –2 x 1 4 9
h1(x)
y  (x  1)2
y  (x  1)2 9 4 1
h2(x)
From the graph,
y  f (x) is translated 1 unit to the right to become y  h1(x).
y  f (x) is translated 1 unit to the left to become y  h2(x).

23. 4.3 Transformation on the Graphs of Functions
A. Translation
We observe that if we translate the graph of y  f (x)
 vertically k units (upwards for k  0 and downwards for k  0), we will get the graph of y  f (x)  k.
k  0
k  0
 horizontally h units (towards to the right for h  0 and towards to the left for h  0), we will get the graph of y  f (x  h).
h  0
h  0

24. 4.3 Transformation on the Graphs of Functions
A. Translation
In the figure, the graph of y  g(x) is obtained by translating the graph of y  f (x) 3 units upwards.
∴ g(x)  f (x)  3
In general, we can translate the graph of y  f (x) vertically by k units and horizontally by h units to get the graph of y  f (x  h)  k.

25. Example 4.8T 4.3 Transformation on the Graphs of Functions Solution:
A. Translation
Example 4.8T
Describe the transformation between the graph of y  x2  6x  10 and y  x2  6x  14.
Solution:
y  x2  6x  14
 (x2  6x  10)  4
∴ A translation of 4 units upwards of the graph of y  x2  6x  10 will obtain the graph of y  x2  6x  14.

26. Example 4.9T 4.3 Transformation on the Graphs of Functions Solution:
A. Translation
Example 4.9T
Consider the graph of f (x)  2x2  16. Describe the transformation between the graph of y  f (x) and each of the following functions.
(a) y  2(x  1)2  16
(b) y  2(x  3)2  16
Solution:
(a) 2(x  1)2  16  f (x  1)
∴ A translation of 1 unit to the right of the graph of y  f (x) will obtain the graph of y  2(x  1)2  16.
(b) 2(x  3)2  16  f (x  3)
∴ A translation of 3 units to the left of the graph of y  f (x) will obtain the graph of y  2(x  3)2  16.

27. 4.3 Transformation on the Graphs of Functions
B. Reflection
If the graph of y  f (x) is reflected about the x-axis, then we will get the graph of y  f (x).
Let f (x)  x2  10x  26.
Then f (x)  x2  10x  26. 5 2 1
f (x) 7 6 4 3 x 5 2 1
f (x)
Moreover, if the graph of y  f (x) is reflected about the y-axis, then we will get the graph of y  f (x).
If f (x)  x(x  3),
x-intercepts: 0 and 3
then f (x)  x(x  3)
 x(x  3)
x-intercepts: 0 and 3

28. Example 4.10T 4.3 Transformation on the Graphs of Functions Solution:
B. Reflection
Example 4.10T
Consider the graphs of f (x)  x3  6, g(x)  x3  6 and h(x)  x3  6. Describe the transformation between the graphs of
(a) y  f (x) and y  g(x).
(b) y  f (x) and y  h(x).
Solution:
(a) g(x)  x3  6
 (x)3  6
 f (x)
The graph of y  g(x) is the reflected image of the graph of y  f (x) about the y-axis.
(b) h(x)  x3  6
 (x3  6)
 f (x)
The graph of y  h(x) is the reflected image of the graph of y  f (x) about the x-axis.

29. 4.3 Transformation on the Graphs of Functions
C. Enlargement or Reduction
The figure shows the graph of y  f (x), where f (x)  2x2  10x
 2x(x  5)
Let g(x)  2 f (x) and h(x)  0.5 f (x)
y  g(x)
 4x(x  5)  x(x  5) 12 –8
–12
f (x) 3 4 5 6 2 1 –1 x
y  h(x) 24
–16
–24
g(x) –6 –4 6
h(x)
The graph of y  f (x) enlarged to y  g(x) or reduced to y  h(x) along the y-axis (the vertical direction) to give y  kf (x) for different values of k (for k  0).

30. 4.3 Transformation on the Graphs of Functions
C. Enlargement or Reduction
The figure shows the graph of y  f (x), where f (x)  2x2  8x
 2x(x  4)
Let g(x)  f (2x)
y  g(x)
 2(2x)2  8(2x)
 8x2  16x
 8x(x  2)
Let h(x)  f (0.5x)
y  h(x)
 2(0.5x)2  8(0.5x)
 0.5x2  4x
 0.5x(x  8)
The graph of y  f (x) enlarged to y  h(x) or reduced to y  g(x) along the x-axis (the horizontal direction) to give y  f (kx) for different values of k (for k  0).

31. 4.3 Transformation on the Graphs of Functions
C. Enlargement or Reduction
Consider the graphs of y  f (x) and y  kf (x), where k  0.
 Enlarge (for k  1) or reduce (for k  1) the graph of y  f (x) along the y-axis k times to obtain y  kf (x).
fixed point(s) at
x-intercept(s)
 Enlarge (for k  1) or reduce (for k  1) the graph of y  f (x) along the x-axis times to obtain y  f (kx).
fixed point at
y-intercept

32. Example 4.11T 4.3 Transformation on the Graphs of Functions Solution:
C. Enlargement or Reduction
Example 4.11T
The figure shows the graph of y  f (x).
(a) Sketch the graph of each of the following functions:
(i) y  f (x) (ii) y  f (2x)
(b) Describe the transformations between the graph
of y  f (x) and each of the graphs in (a).
y  f(2x)
y   f(x) 1 4
Solution:
(a) Refer to the figure on the right.
(b) The graph of y  f (x) is reduced to times of the original
graph along the y-axis to give the graph of y  f (x).
The graph of y  f (x) is reduced to half of the original graph along the x-axis to give the graph of y  f (2x).

33. Chapter Summary 4.1 Solving Inequalities by the Graphical Method
To solve the inequality of f (x)  k (or f (x)  k) by the graphical method, we can draw the line y  k on the same diagram of the graph of y  f (x). From the points of intersection of the graphs, determine the range of values of x such that f f (x)  k (or f (x)  k).

34. Chapter Summary 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
For y  a(x  h)2  k,
if a  0, then y has the minimum value, that is k;
if a  0, then y has the maximum value, that is k.

35. Chapter Summary 4.3 Transformation on the Graphs of Functions
1. If we translate the graph of y  f (x)
 vertically k units (upwards for k  0; downwards for k  0), we will get the graph of y  f (x)  k.
 horizontally h units (to the right for h  0; to the left for h  0), we will get the graph of y  f (x  h).
2. If we reflect the graph of y  f (x) about
 the y-axis, we will get the graph of y  f (x).
 the x-axis, we will get the graph of y  f (x).
3. For k  0, the graph of y  f (x) is
 enlarged (k  1) or reduced (k  1) k times of the original one along the y-axis to give the graph of y  kf (x).
 enlarged (k  1) or reduced (k  1) times of the original one along the x-axis to give the graph of y  f (kx).

36. Follow-up 4.1 4.1 Solving Inequalities by the Graphical Method
The figure shows the graph of y  x2  3x  4.
(a) Add the line y  6 on the same graph and solve x2  3x  4  6.
(b) If you want to solve x2  3x  0, which line should you add?
(c) Hence write down the solution of x2  3x  0.
Solution:
(a) From the graph, the
solution is 2  x  5.
y  4
(b) x2  3x  0 can be written
as x2  3x  4  4.
y  6
∴ The line y  4 should
be added.
In order to make use of the given graph of y  f (x), we have to rewrite the inequality into the form f (x)  k.
(c) From the graph, the solution is 0  x  3.

37. Follow-up 4.2 4.1 Solving Inequalities by the Graphical Method
The figure shows the graph of y  x3  x2  2x.
(a) Using the graph, solve x3  x2  2x  0.
(b) Draw a straight line y  1 on the same coordinate plane.
(c) Hence solve x3  x2  2x  1  0.
(Give the answers correct to 1 decimal place.)
y  1
Solution:
(a) From the graph, the solution is x  1
or 0  x  2.
(b) Refer to the figure on the right.
(c) x3  x2  2x  1  0 can be written as
x3  x2  2x  1.
∴ The solution is x  (cor. to 1 d. p.)

38. Follow-up 4.3 4.1 Solving Inequalities by the Graphical Method
The figure shows the graph of y  x3  4x  k.
(a) Find the value of k.
(b) Using the graph, solve
(i) x3  4x  12  0, (ii) x3  4x  2.
(Give the answers correct to 1 decimal
place.)
y  10
Solution:
(a) Since the graph passes through (0, 12), we have
12  03  4(0)  k
k  12
(b) (i) From the graph, the
solution is x  2.9.
(ii) Since x3  4x  2, x3  4x  2.
∴ x3  4x  12  10
∴ The solution is x  1.7 or
0.5  x  2.2.

39. Follow-up 4.4 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Follow-up 4.4
Find the maximum value of y  2x2  8x  5 using the method of completing the square.
Solution:
y  2x2  8x  5
 2(x2  4x)  5
Leave the constant term outside the brackets.
 2(x2  4x  22  22)  5
Add 22 to complete the square, then minus 22.
 2(x2  4x  22)  (2)(22)  5
Take the term  22 out of the brackets by
multiplying 2.
 2(x  2)2  3
∴ The maximum value of the function is 3.

40. Follow-up 4.5 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Follow-up 4.5
The graph of y  (x  3)(x  k) passes through the point (2, 15).
(a) Find the value of k.
(b) Find the coordinates of the vertex and the axis of symmetry.
(c) Find the x-intercepts and y-intercept.
(d) Sketch the graph and label the axis of symmetry and vertex on the graph.
Solution:
(a) Substituting x  2 and y  15 into the given equation,
we obtain
15  (2  3)(2  k)
3  2  k
k  5

41. Follow-up 4.5 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Follow-up 4.5
The graph of y  (x  3)(x  k) passes through the point (2, 15).
(a) Find the value of k.
(b) Find the coordinates of the vertex and the axis of symmetry.
(c) Find the x-intercepts and y-intercept.
(d) Sketch the graph and label the axis of symmetry and vertex on the graph.
Solution:
(b) y  (x  3)(x  5)
 x2  2x  15
 (x2  2x)  15
 (x2  2x  12  12)  15
 (x2  2x  12)  12  15
 (x  1)2  16
∴ The vertex is (1, 16) and the axis of symmetry is x  1.

42. Follow-up 4.5 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
A. Properties of Quadratic Function y  a(x  h)2  k
Follow-up 4.5
The graph of y  (x  3)(x  k) passes through the point (2, 15).
(a) Find the value of k.
(b) Find the coordinates of the vertex and the axis of symmetry.
(c) Find the x-intercepts and y-intercept.
(d) Sketch the graph and label the axis of symmetry and vertex on the graph.
Solution:
(c) When y  0,
(x  3)(x  5)  0
(d)
x  3 or 5
∴ The x-intercepts are 3 and 5.
y-intercept  (0  3)(0  5)
 15

43. Follow-up 4.6 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
B. Practical Problems Related to Optimum Values
Follow-up 4.6
A cannon-ball is shot upwards, its distance above the ground s m is given by
s  10t  5t2,
where t seconds is the time after shooting.
(a) When will the cannon-ball reach the maximum height?
(b) Find the maximum height that the cannon-ball can reach.
Solution:
(a) s  10t  5t2
 5(t2  2t)
(b) When t  1,
s  10(1)  5(1)2
 5(t2  2t  12  12)
 10  5
 5
 5(t  1)2  5
∴ The cannon-ball will reach
the maximum height 1 second
after shooting.
∴ The maximum height
is 5 m.

44. Follow-up 4.7 4.2 Finding the Optimum Values of Quadratic
Functions by the Algebraic Method
B. Practical Problems Related to Optimum Values
Follow-up 4.7
The length and the width of a rectangular board are (40  2x) cm and x cm respectively. Let A cm2 be the area of the board.
(a) Express A in terms of x.
(b) Find the value of x so that the area of the board is maximum.
Solution:
(a) A  (40  2x)(x)
 40x  2x2
(b) A  40x  2x2
 2(x2  20x)
 2(x2  20x  102  102)
 2(x  10)2  2(102)
 2(x  10)2  200
∴ When the area of the board is maximum, the value of x is 10.

45. Follow-up 4.8 4.3 Transformation on the Graphs of Functions Solution:
A. Translation
Follow-up 4.8
Consider the graphs of y  x2  6 and y  x2  2. Describe the transformation between the graphs.
Solution:
y  x2  2
 (x2  6)  4
∴ A translation of 4 units downwards of the graph of y  x2  6 will obtain the graph of y  x2  2.

46. Follow-up 4.9 4.3 Transformation on the Graphs of Functions Solution:
A. Translation
Follow-up 4.9
Consider the function f (x)  x2  8. Describe the transformation between the graph of y  f (x) and each of the following functions.
(a) y  (x  2)2  8
(b) y  (x  1)2  8
Solution:
(a) (x  2)2  8  f (x  2)
∴ A translation of 2 units to the right of the graph of y  f (x) will obtain the graph of y  (x  2)2  8.
(b) (x  1)2  8  f (x  1)
∴ A translation of 1 unit to the left of the graph of y  f (x) will obtain the graph of y  (x  1)2  8.

47. Follow-up 4.10 4.3 Transformation on the Graphs of Functions Solution:
B. Reflection
Follow-up 4.10
Consider the functions f (x)  3x2  2x  4, g(x)  3x2  2x  4 and h(x)  3x2  2x  4. Describe the transformation between the graphs of
(a) y  f (x) and y  g(x).
(b) y  f (x) and y  h(x).
Solution:
(a) g(x)  3x2  2x  4
 3(x)2  2(x)  4
 f (x)
The graph of y  g(x) is the reflected image of the graph of y  f (x) about the y-axis.
(b) h(x)  3x2  2x  4
 (3x2  2x  4)
 f (x)
The graph of y  h(x) is the reflected image of the graph of y  f (x) about the x-axis.

48. Follow-up 4.11 4.3 Transformation on the Graphs of Functions Solution:
C. Enlargement or Reduction
Follow-up 4.11
The figure shows the graph of y  f (x), where f (x)  (x  1)2  3.
(a) Sketch the following graphs on the same graph of f (x).
(i) y  f (x) (ii) y  f
(b) Describe the transformations
between the graph of y  f (x) and
each of the graphs in (a).
y  f
Solution:
(a) Refer to the figure on the right.
y   f(x) 1 3
(b) The graph of y  f (x) is reduced to one third of the original
graph along the y-axis to give the graph of y  f (x).
The graph of y  f (x) is enlarged to two times of the original
graph along the x-axis to give the graph of y  f