Shatin Pui Ying College
F.4 Mathematics
Revision Exercise Ch.2 (ANS)
1.
Determine the directions of opening and find the y-intercepts of the graphs of the following functions.
(a) y = x2 – 4x + 6
(b) y = (1 – 2x)2 – 3
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(a) For y = x2 – 4x + 6,
∵ Coefficient of x2 = 1 > 0
∴ Its graph opens upwards.
∴ The y-intercept of its graph is 6.
(b)
y  (1  2 x) 2  3
 4x2  4x  2
∵ Coefficient of x2 = 4 > 0
∴
∴
2.
Its graph opens upwards.
The y-intercept of its graph is –2.
The figure shows the graph of y = 2x2 + 8x + 6. State the following features of the graph:
(a) Axis of symmetry
(b) Coordinates of the vertex
(c) y-intercept
(d) Direction of opening
From the graph,
(a) Axis of symmetry is x = –2.
(b) Coordinates of the vertex  (2,  2)
(c) y-intercept  6
(d) The graph opens upwards.
3.
The figure shows the graph of y = x2 – 4x + c. D(d, –16) is the minimum point of the graph.
(a) Find the values of c and d.
(b) State the following features of the graph:
(i) Axis of symmetry
(ii) Coordinates of the vertex
(iii) y-intercept
(iv) Direction of opening
(a) From the graph, the parabola y = x2 – 4x + c cuts the y-axis at
(0, –12).
∴ c   12 
∵
The parabola passes through D(d, –16).
By substituting c = –12 and (d, –16) into y = x2 + 4x + c, we have
1
 16  d 2  4d  12
d 2  4d  4  0
(d  2) 2  0
d 2
(b) From the graph,
(i) Axis of symmetry is x = 2.
(ii) Coordinates of the vertex  (2,  16)
(iii) y-intercept   12
(iv) The graph opens upwards.
4.
∵
∴

The figure shows the graph of y = x2 + bx + c which cuts the x-axis at
A(–6, 0) and B(1, 0). Find the values of b and c.
The graph of y = x2 + bx + c cuts the x-axis at A(–6, 0) and B(1, 0).
The roots of the quadratic equation x2 + bx + c = 0 are –6 and 1.
x  6 or
x  6  0 or
( x  6)( x  1)  0
x 1
x 1  0
x2  5x  6  0
∴ b  5 and c   6
5.
A marble is projected vertically upwards to the ceiling of a house from the floor. After t seconds, its height
(h m) above the ground is given by:
h = –5t2 + 5t + 1
(a) When will the marble attain its maximum height?
(b) If the ceiling is 3 m above the ground, will the marble hit the ceiling?
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(a) h  5t 2  5t  1
 5(t 2  t )  1
2
2
2
1 1 
 5t  t         1
 2   2  

1 5

 5 t 2  t     1
4 4

2
 1 9
 5 t   
 2 4
1
, h attains its maximum value.
2
∴
When t 
∴
The marble will attain its maximum height after
1
seconds.
2
2
 1 9
(b) From (a), h  5 t   
 2 4
2
6.
9
m.
4
∴
The maximum height reached is
∴
The marble will not hit the ceiling.
Given that the maximum value of the function y = –3x2 + 6x + p is 11.
(a) Find the value of p.
(b) State the coordinates of the vertex of its graph.
(a)
y  3x 2  6 x  p
 3( x 2  2 x)  p
 3( x 2  2 x  12  12 )  p
 3( x 2  2 x  1)  3  p
 3( x  1) 2  (3  p)
∴
The maximum value of y is 3 + p.
∴
3  p  11
p8
(b) From (a), we have
y = –3(x – 1)2 + 11
∴
Coordinates of the vertex  (1, 11)
7. If the sum of two numbers is 12, find the maximum value of the product of these two numbers.
Let x be one of the numbers, then the other number is 12 – x, and y be the product of these two numbers.
∴
y  x(12  x)
  x 2  12 x
 ( x 2  12 x)
 ( x 2  12 x  62  62 )
 ( x 2  12 x  36)  36
 ( x  6) 2  36
∴
∴
The maximum value of y is 36.
The maximum value of the product of these two numbers is 36.
8.
(a) Solve (x – 1)2 + 4x < 0 graphically.
(b) Find the largest integer x that satisfies x(x + 2)  4.
(a)
( x  1) 2  4 x  0
x2  2x  1  4x  0
x2  2x  1  0
 x2  2x  1
 x2  2x  1  2
Draw the straight line y = 2 on the graph of y = x2 – 2x + 1.
From the graphs, there is no solutions for  x 2  2 x  1  2 .
∴ There is no solutions for ( x  1) 2  4 x  0 .
3
x( x  2)  4
(b)
x 2  2x  4
 x 2  2 x  4
 x 2  2 x  1  3
Draw the straight line y = 3 on the graph of y = x2 – 2x + 1.
From the graphs, the largest integer x that satisfies  x 2  2 x  1  3 is 1.
∴ The largest integer x required is 1.
9.
In the figure, the graph of y = g(x) is obtained by translating the graph of y = 2x2  8x + 10.
(a) Describe the effect of the transformation on the graph of y =
2x2  8x + 10.
(b) Find the symbolic representation of g(x).
(a) From the graphs, the graph of y = 2x2 – 8x + 10 is translated
in the negative direction of the x-axis by 2 units.
2
(b) g ( x)  2( x  2)  8( x  2)  10
 2 x 2  8 x  8  8 x  16  10
 2x2  2
∴
The required symbolic representation is g(x) = 2x2 + 2.
10. It is given that f(x) = 2x2  4x  1 is transformed to g(x). If the coordinates of the vertex of the graph of y =
g(x) are (1, 3), find the symbolic representation of g(x).
f ( x)  2 x 2  4 x  1
 2( x 2  2 x)  1
 2( x 2  2 x  12  12 )  1
 2( x 2  2 x  1)  2  1
 2( x  1) 2  3
∴
∵
The coordinates of the vertex of the graph of y = f(x) are (1, –3).
The coordinates of the vertex of the graph of y = g(x) are (1, 3).
∴
The graph of y = g(x) is obtained by translating the graph of y = f(x) in the negative direction of the x-axis
by 2 units.
∴
g ( x)  f ( x  2)
 2[( x  2)  1]2  3
 2( x  1) 2  3
 2x2  4x  1
∴
The required symbolic representation is g(x) = 2x2 + 4x – 1.
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