4ach02(functions And Graphs)
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2 Functions and Graphs
2 Functions and Graphs
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Activity
1.Xx x x) f(x) = x2 x)g(x) = x2 + 3
Activity 2.1 (p. 67)
−2 4
−1 1
0 0
1 1
2 4
3 9
12
7
4
3
4
7
12
2.
1.
Yes
2.
Yes, the graph has a minimum point.
3.
Any reasonable descriptions including the curve always lies on or above the x-axis, opening upwards, etc.
Activity 2.2 (p. 73) 1.
−3 9
3.
open upwards
4.
2.
A positive number or zero.
3.
(a) (b)
4.
(a) x y
0 2; (1, 2) −2 11
Yes The graph of y = x2 + 3 can be obtained by translating the graph of y = x2 upwards by 3 units.
Activity 2.4 (p. 101) −1 6
0 3
1 2
2 3
3 6
4 11
g ( x ) = f ( x + 2) = ( x + 2 ) 2 − 2( x + 2 ) + 1
1.
= x2 + 4x + 4 − 2x − 4 + 1 = x2 + 2x + 1 2.
x −4 g(x) 9
−3 4
−2 1
−1 0
0 1
1 4
2 9
3.
5.
(b) (c)
Yes x=1
(a) (b)
k (h, k); x = h 4.
Activity 2.3 (p. 95)
Yes
5.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) leftwards by 2 units.
6.
The graph of y = h(x) can be obtained by translating the graph of y = f(x) rightwards by 2 units.
30
2 Functions and Graphs 2
Follow-up Exercise
(b)
p. 56 1.x y
−4 −2 0 −17 −11 −5
2 1
4 7
6 13
2.
1 1 1 f = − − 1 3 3 3 1 1 = − −1 9 3 11 = − 9 ∴
The value of the function is − x=
11 when 9
1 . 3
−3 ( −3) 2 + 2 −3 = 9 +2 3 = − 11
f ( −3) = 2.
3.
4.
3.
(a) (b) (c) (d)
true true false true
(a)
x/cm 1 y/cm2 6
(a)
h ( −2) = ( −2) 2 − 5( −2) + 6 = 4 + 10 + 6 = 20 h ( 2) = 2 2 − 5( 2 ) + 6 = 4 − 10 + 6 =0
2 3 4 5 12 18 24 30
h ( 0) = 0 2 − 5( 0) + 6 =6
(b)
(b) (i)
h ( 2) + h ( −2) = 0 + 20 = 20
but h(0) = 6 i.e. h(2) + h(−2) ≠ h(0) (ii)
(c)
y = 6x
p. 60 1.
(a)
f ( −3) = ( −3) 2 − ( −3) − 1 = 9 + 3 −1 ∴
4.
(a)
∴
= 11
The value of the function is 11 when x = −3.
h ( 2) ⋅ h ( −2) = 0 ⋅ 20 = 0 h ( −4) = ( −4) 2 − 5( −4) + 6 but = 16 + 20 + 6 = 42 i.e. h ( 2) ⋅ h ( −2) ≠ h ( −4) g(4) = 0 4 2 − 2k = 0 16 − 2k = 0 k =8
(b) From (a), we have g ( x ) = x 2 − 16 ∴
g ( −2) = ( −2) 2 − 16 = 4 − 16 = − 12
31
Certificate Mathematics in Action Full Solutions 4A
5.
(a)
f ( −3a ) =2( −3a ) 2 −( −3a ) =18 a 2 +3a
2.
(a)
f (b −1) = 2(b −1) 2 −(b −1) = 2(b 2 −2b +1) −b +1
(b)
From the graph, y = ax + b cuts the x-axis and the y-axis at (−2, 0) and (0, −4) respectively. 2 and y-intercept = ∴ x-intercept = − −4 The line passes through (0, −4).
(b)
= 2b 2 −4b +2 −b +1 = 2b 2 −5b +3
∴
p. 66 1.
∴ (a)
x −1 f(x) 9
1 3
3 −3
−4 = a (0) + b b = −4
The line passes through (−2, 0). 0 = a ( −2 ) − 4 a = −2 a = −2 and
The graph of y = f(x) is:
b = −4
p. 70 Function
1.
y = −x2 − x + 1 y = −3 + x + 2x2 y = (1 − x)2 − 4 y = −(x + 1)2 + 2 y = (x + 2)(x − 5) + 9 2.
(a)x y
−4 8
−3 2
−2 0
Direction of opening open downwards open upwards open upwards open downwards open upwards −1 2
y-intercept 1 −3 −3 1 −1
0 8
From the graph, y = f(x) cuts the x-axis and the yaxis at (2, 0) and (0, 6) respectively. ∴ x-intercept = 2 and y-intercept = 6 (b)
x −3 h(x)−1
−1 3
1 7
The graph of y = h(x) is:
(b) (i)
Axis of symmetry is x = −2.
(ii) Coordinates of the vertex =
(− 2,0)
(iii) y-intercept = 8 (iv) The graph opens upwards. p. 76 1.
From the graph, y = h(x) cuts the x-axis and the yaxis at ( ∴
−
5 , 0) and (0, 5) respectively. 2
5 x-intercept = −2 and y-intercept =
(a)
Consider the equation y = 2x2 + 1, we have a = 2, h = 0 and k = 1. (i) a=2>0 ∴ The graph of y = 2x2 + 1 opens upwards. (ii) The vertex is at (0, 1) (iii) The axis of symmetry is x = 0.
(b) Consider the equation y = −2(x + 1)2 + 3, we have a = −2, h = −1 and k = 3.
5
(i)
32
a = −2 < 0
2 Functions and Graphs ∴
The graph of y = −2(x + 1)2 + 3 opens downwards. (ii) The vertex is at (−1, 3) (iii) The axis of symmetry is x = −1. (c)
y = 2 x 2 −4 x +3 (d)
Consider the equation y = 4(x −2) −1, we have a = 4, h = 2 and k = −1. (i) a=4>0 ∴ The graph of y = 4(x − 2)2 − 1 opens upwards. (ii) The vertex is at (2, −1). (iii) The axis of symmetry is x = 2.
= 2( x −1) 2 +1 a=2>0 ∴ The graph of the function opens upwards and it has a minimum point. ∴ The maximum value of y is 1. The axis of symmetry is x = 1.
(d) Consider the equation y = 3 −2(x −1)2, we have a = −2, h = 1 and k = 3. a = −2 < 0 The graph of y = 3 − 2(x − 1)2 opens downwards. (ii) The vertex is at (1, 3). (iii) The axis of symmetry is x = 1. 2.
∴
y = x 2 + 10 x + k = ( x 2 + 10 x + 25) − 25 + k = ( x + 5) 2 − 25 + k ∴ ∴
3. y = x + 4 x −1
= ( x + 2) −5 a=1>0 The graph of the function opens upwards and it has a minimum point. ∴ The minimum value of y is −5. The axis of symmetry is x = −2.
y = − x2 + 2x = −( x 2 − 2 x) = − ( x 2 − 2 x + 12 − 12 ) = − ( x 2 − 2 x + 1) + 1
y = − x + 2x − 2 2
= − ( x − 1) 2 + 1
= −( x − 2 x) − 2 2
∴ ∴
= −( x − 2 x + 1 − 1 ) − 2 2
2
= − ( x 2 − 2 x + 1) + 1 − 2 = − ( x − 1) 2 − 1
∴
a = −1 < 0 The graph of the function opens downwards and it has a maximum point. The maximum value of y is −1. The axis of symmetry is x = 1.
= −2( x 2 + 4 x) +1 2
2
The maximum value of y is 1. The maximum area of this rectangle is 1 m2.
p.85 1.
number of x-intercepts : 1 no axis of symmetry no maximum or minimum point
2.
number of x-intercepts : 2 axis of symmetry : x = 2 minimum point : (2, −5)
3.
number of x-intercepts : 3 no axis of symmetry no maximum or minimum point
y = −2 x 2 −8 x +1 (c)
y = x(2 − x )
By conpleting the square,
∴
∴
4 − x m i.e. (2 − x) m , and y m2 be 2
= −x2 + 2x
2
2
k = 33
Let x m be the length of the rectangle, then the width of
∴
= ( x 2 + 4 x + 4) − 4 −1
(b)
−25 + k = 8
the area of the rectangle.
= ( x 2 + 4 x + 2 2 − 2 2 ) −1
(a)
The minimum value of y is −25 + k.
the rectangle is
2
1.
= ( x 2 + 10 x + 52 − 52 ) + k
2.
(a) The optimum value (minimum value) of y is −1. (b) The optimum value (maximum value) of y is 5.
p. 79
= 2( x 2 − 2 x +12 −12 ) + 3 = 2( x 2 − 2 x +1) − 2 + 3
2
(i)
= 2( x 2 − 2 x ) + 3
2
= −2( x + 4 x + 2 − 2 ) +1 = −2( x 2 + 4 x + 4) +8 +1 = −2( x + 2) 2 + 9 ∴ a = −2 < 0 ∴ The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 9. The axis of symmetry is x = −2.
p. 90 1.
When x > −2.1 (cor. to 1d.p.), the corresponding portion
33
Certificate Mathematics in Action Full Solutions 4A of the graph of y =3 x lies above the straight line y = 0.1. Hence, the solution of 3 x >0.1 is x > −2.1. 2.
When 1 ≤ x ≤ 3 , the corresponding portion of the graph of y = x 2 −4 x +6 lies on or below the straight line y = 3. Hence, the solution of x 2 −4 x +6 ≤ 3 is
1 ≤ x ≤3 . 3.
When −2 < x <1 , the corresponding portion of the graph of y =−x 2 −x −2 lies above the straight line y = −4. Hence, the solution of − x 2 − x −2 > −4 is
From the graphs, the solutions of 2x2 − 5x − 8 > 4 is x < −1.5 or x > 4.
−2 < x <1 . 4.
p. 100
Draw y = 2 on the graph of y = x2 −5x −4.
1.
g( x) = x 3 + x − 2 = ( x 3 + x + 2) − 4 ∴
2.
From the y-intercepts of y =2 x and y = g(x), we find the number of units the graph of y =2 x has translated upwards. y-intercept of y = g(x) is 4, and y-intercept of y =2 x is 1. ∴ The difference in y-intercepts of the two graphs is
From the graphs, the solution of x2 −5x −4 < 2 is −1 < x < 6. 5.
= f ( x) − 4 The graph of y = f(x) is translated in the negative direction of the y-axis by 4 units.
Draw y = −1 on the graph of y = −x2 + 6x −9. ∴
3.
4 − 1 = 3. The graph of y = g(x) is obtained by translating the graph of y =2 x upwards by 3 units.
∴ ∴
g(x) = 2 x +3 The required symbolic representation is g(x) = 2 x +3 .
The graph of y = g(x) is obtained by translating the graph of y = f(x) in the negative direction of the y-axis by 4 units. g(x) = f(x) − 4
∴
x f(x) f(x) − 4 ∴ From the graphs, the solution of −x2 + 6x −9 ≥ −1 is 2 ≤ x ≤ 4.
The tabular representation of g(x) is:
x −2 g(x) 7
−1 0
0 −3
1 −2
2 1
3 12
p. 106 6.
Draw y = 4 on the graph of y = 2x −5x −8. 2
1.
34
(a)
The graph of y = f(x + 3) can be obtained by
2 Functions and Graphs translating the graph of y = f(x) in the negative directions of the x-axis by 3 units. (b) The graph of y = f(x − 2) can be obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 2 units. g ( x) = x 2 − 6 x + 9 2.
= x − 2(3)( x ) + 3
2
Level 1
= ( x − 3) 2 ∴
Exercise Exercise 2A (p. 61)
2
(a)
g(x) = (x − 5)2.
= f ( x − 3) The graph of y = f(x) is translated in the positive direction of the x-axis by 3 units.
1.
(a)
∴
g ( x ) = x 2 + 2 x +1 = x 2 + 2(1)( x) + 12
(b)
(b)
= ( x + 1) 2 ∴ 3.
−4 3
−3 0
−2 −1
−1 0
0 3
(c)
1 8
(a)
= 8 −1
=4 +1
(b) ∴ ∴ ∴
4.
Let
g(x) = f(x − 2) The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 2 units. The vertex of the graph of y = g(x) is also translated in the positive direction of the xaxis by 2 units. The coordinates of the vertex of y = g(x) are (0, −1).
=1+1
= ( x − 5) 2 The required symbolic representation is
The value of g(x) is
17 when x = −2. 16
f ( 30 °) = cos 30 ° 3.
(a)
= x 2 − 2(2)( x ) + 2 2
The graph of y = g(x) is obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 3 units. g ( x ) = f ( x − 3) ∴ = [( x − 3) − 2]2
= 2
The value of g(x) is 2 when x = 0.
g ( −2) = 4 −2 + 1 1 = +1 16 17 = 16 ∴
f ( x) = x 2 − 4 x + 4 = ( x − 2) 2
∴
(c)
=5
The value of g(x) is 5 when x = 1.
g ( 0) = 4 0 + 1
∴ From the graph, the coordinates of the vertex are (−2, −1).
=7
The value of the function is 7 when x = −2.
g (1) =41 +1
∴ (b)
= −1
The value of the function is −1 when x = 0.
f ( −2) = 2( −2) 2 − 1
∴
2.
The value of the function is 17 when x = 3.
f (0) = 2(0) 2 − 1 ∴
= f ( x +1) The graph of y = f(x) is translated in the negative direction of the x-axis by 1 unit.
x −5 f(x) 8
(a)
f (3) = 2(3) 2 −1 =18 −1 =17
=
3 2
f ( 45 °) = cos 45 ° (b)
(c)
=
2 2
f ( 60 °) = cos 60 ° 1 = 2
35
Certificate Mathematics in Action Full Solutions 4A
1 f = 3
4.
1 1 G = 2 2 1 = 2 = 0
1 2
1 9 + 1 3 1 = 1 9 + 1 9 1 = 1+1 1 = 2
(a)
1 G = 0 2 1 1 = but G ( 2) 6 i.e.
9(1) 2 + 1 1 = 9 +1 1 = 10
(b)
The relation
1 1 does not G = G ( 2) 2
hold. (a)
g ( 0) = ( 0 − 2)( 0 − 3)
(a)
1 1 G ≠ G ( 2) 2
∴
8. 5.
( 0)
(b)
1
f (1) =
2 1 − 1 2
2 f ( −4) = 2( −4 )( −4 + 1) = 2( −4 )( −3) = 24
= ( −2)( −3) =6
(b)
g (1) = (1 − 2 )( 1 − 3) = ( −1)( −2)
f ( 2) + [ f ( −3)] 2 = 2( 2 + 1) + [ −3( −3 + 1)] 2 = 2(3) + [ −3( −2)] 2
=2
= 6 + 62 = 42
g ( −1) = ( −1 − 2)( −1 − 3) = ( −3)( −4) = 12
(b)
9.
g (1) + g ( −1) = 2 + 12
(a)
(b)
f ( 2) =3( 2) −2
(c)
(a)
f (1) ⋅g ( −1) =(12 +1) ⋅[ 2( −1)] =2( −2) =−4
=6 −2 f ( 4) = 3( 4) − 2 = 12 − 2 = 10
7.
= 6 +1 = 7
=4
(b)
g (3) + f ( 0) = 2( 3) + (0 2 + 1)
= 14
but g(0) = 6 i.e. g(1) + g(−1) ≠ g(0) 6.
(a)
= 4 = 16 2 but f ( 2) = f ( 4 ) = 10 i.e. [ f ( 2)] 2 ≠ f ( 2 2 ) [ f ( 2)]
2
2
g ( 4) 2( 4 ) = 2 f (3) 3 +1 8 = 9 +1 8 = 10 4 = 5 h (3) = 1
G ( 2) = 2[ 2( 2) − 1] = 2( 3)
10.
3(3) + m = 1 27 + m = 1 2
m = − 26
=6
36
2 Functions and Graphs f (3) = 20 11.
32 + a ( 3) + 2 = 20 3a + 11 = 20
g (50 °) = sin(50 ° + 10 °) = sin 60 °
15. (a)
a = 3
=
g ( −1) = −1
g ( 35 °) = sin(35 ° + 10 °)
( −1 + k )( −1 − 2) = −1 ( −1 + k )( −3) = −1
12.
3 − 3k = −1 4 k = 3
=
g ( x ) = f (2 x ) = 2( 2 x ) + 3 ∴
(b)
h( x ) = = = =
∴
f ( x + 2) 2( x + 2 ) + 3 2x + 4 + 3 2x + 7
The symbolic representation of h(x) is h(x) = 2x + 7.
1 h( x ) − 1 2 1 = ( 2 x + 7) − 1 2 7 = x+ −1 2 1 = − 2 1 = − 6
H (0) =4(0) 2 −8(0) +1 =1
= 4( p 2 −4 p +4) −8 p +16 +1
(b)
= 4 p 2 −16 p +16 −8 p +16 +1 = 4 p 2 −24 p +33
f (8) = 6 g ( 4) 17.
2(8 + m ) = 6( 4 + m ) 16 + 2m = 24 + 6m 4m = −8 m = −2
g( x) = 4x + 3 (c)
4x + 3 3x x
18.
f ( 2) = 2 ( 2 + 2)( 2 − 2) + a ( 2) + b = 2 2a + b = 2 ……(1)
f ( −2) = 4
( −2 + 2)( −2 − 2) + a ( −2) + b = 4 − 2a + b = 4
……(2) (1) + (2), 2b = 6 b =3
Level 2 f ( 0) = 3[ 2 3( 0 ) ] 14. (a)
2 2
H ( p −2) = 4( p −2) 2 −8( p −2) +1
= 4x + 3
The symbolic representation of g(x) is g(x) = 4x + 3.
= sin 45 °
(b)
16. (a) 13. (a)
3 2
= 3( 2 0 ) = 3(1)
By substituting b = 3 into (1), we have 2a + 3 = 2
a = −
1 2
=3 2 3 − 3
(b)
2 f − = 3[2 ] 3 = 3(2 − 2 ) 1 = 3 4 3 = 4
19. (a)
2 f ( a ) − 3g ( a ) 2( 3a − 1) − 3( a − 2) 6a − 2 − 3a + 6 3a
= = = =
5 5 5 1 1 a = 3
37
Certificate Mathematics in Action Full Solutions 4A f ( 2b) − g (3b) 3( 2b) − 1 − (3b − 2) 6b − 1 − 3b + 2 3b
(b)
= = = =
5 5 5 4 4 b = 3
(b)
5 x 2 − x − 120 = 0 ( x − 5)(5 x + 24) = 0 x − 5 = 0 or 5 x + 24 = 0
( k + 3)( k + 2) − k 2 = 2k
S ( x ) = 60 x (5 x − 1) = 60 2 x (5 x − 1) = 120
∴
f ( k ) = 2k 20. (a)
The area of the trapezium is 60 cm2.
x = 5
k 2 + 5k + 6 − k 2 = 2k 3k = −6
3x − 2 = 13
∴ (b) From (a), we have f(x) = (x + 3)(x + 2) −4 Consider the equation 2 x − f ( x ) =0
23. (a)
2 x −[( x +3)( x +2) −4] =0
The lengths of AD and BC are 11 cm and 13 cm respectively.
F (50 ) = 500 + 4(50 ) = 500 + 200 = 700
∴
2 x −( x 2 +5 x +6 −4) =0
The cost for holding a party with 50 guests is $700.
2 x −x 2 −5 x −2 =0 (b)
x 2 +3 x +2 =0 ( x +1)( x +2) =0 x +1 = 0 or x + 2 = 0 x = −1 or x = −2
The cost for holding a party is $840.
F ( n ) = 840 500 + 4n = 840
∴
n = 85
∴
f ( x + 2) = ( x + 2) 2 − k ( x + 2)
(c)
= x 2 + 4 x + 4 − kx − 2k
The number of guests is 85. H ( n ) = 10n − F ( n ) = 10n − (500 + 4n ) = 10n − 500 − 4n = 6n − 500
(i)
= x 2 + ( 4 − k ) x + 4 − 2k f ( x −2) =( x −2) 2 −k ( x −2)
(ii)
= x 2 −4 x +4 −kx +2 k = x 2 −( 4 +k ) x +4 +2 k
H (150 ) = 6(150 ) − 500 = 900 − 500 = 400 ∴
(b)
The profit is $400 when there are 150 guests.
f ( x + 2) − f ( x − 2) = kx − 32 x
2
Exercise 2B (p. 71)
+ ( 4 − k ) x + 4 − 2k −
x 2 − ( 4 + k ) x + 4 + 2k ] = kx − 32 8 x − 4k = kx − 32
Level 1
By comparing coefficients,
1.
k =8
f ( x + 2) = f ( x − 2) + 40
(c)
f ( x + 2) − f ( x − 2) = 40 8 x − 32 = 40
(by (b))
x = 9
22. (a)
24 (rejected) 5
When x = 5, 2 x + 1 = 11
k = −2
21. (a)
x = −
or
[( 2 x +1) + (3 x − 2)] x 2 x(5 x −1) = 2
S ( x) =
38
(a)
x y
−1 −6
1 −2
3 2
2 Functions and Graphs 4.
(a)
From the graph, the line cuts the x-axis and the y-axis at (4, 0) and (0, 4) respectively. ∴ x-intercept = 4 and y-intercept = 4
(b) ∴
The line passes through (0, 4).
The line passes through (4, 0). 0 = a ( 4) + 4 a = −1 a = − 1 and b = 4
∴
5. (b) From the graph, y = 2x −4 cuts the x-axis and the y-axis at (2, 0) and (0, −4) respectively. 4 ∴ x-intercept = 2 and y-intercept = −
2.
(a)
−3 −6
x y
−1 2
(a)
4 = a ( 0) + b b = 4
For y = x2 −3x + 6, Coefficient of x2 = 1 > 0 ∴ The graph opens upwards. The y-intercept of y = x2 −3x + 6 is 6.
(b) For y = −2x2 + 6x −4, Coefficient of x2 = −2 < 0 ∴ The graph opens downwards. The y-intercept of y = −2x2 + 6x −4 is −4.
y = 5 − (2 − x ) 2
(c)
1 10
∴
= 5 − (4 − 4 x + x 2 ) y = − x2 + 4x + 1
Coefficient of x2 = −1 < 0 ∴ The graph opens downwards. The y-intercept of y = 5 −(2 −x)2 is 1.
y = ( x + 1)( 3 + x ) + 2
(d)
= x2 + 4x + 3 + 2 ∴
y = x2 + 4x + 5 Coefficient of x2 = 1 > 0 ∴ The graph opens upwards. The y-intercept of y = (x + 1)( 3 + x) + 2 is 5. 6.
∴ 3.
(a)
−
3 , 0) and (0, 6) respectively. 2
3 x-intercept = −2 and y-intercept =
7. 6
From the graph, the line cuts the x-axis and the y-axis at (−3, 0) and (0, 4) respectively. 3 and y-intercept = 4 ∴ x-intercept = −
(b) ∴
∴
0 = a ( −3) + 4 4 a = 3 4 a = and b = 4 3
(1, 3)
(c) (d) The graph opens downwards. 8.
(a) Axis of symmetry is x = −2. (b) Coordinates of the vertex =
(− 2, 1 0 )
(c) y-intercept = 6 (d) The graph opens downwards.
4 = a ( 0) + b
The line passes through (−3, 0).
(a) Axis of symmetry is x = 1. (b) Coordinates of the vertex = y-intercept = 1
The line passes through (0, 4).
b = 4
(− 1, − 2)
1 (c) y-intercept = − (d) The graph opens upwards.
(b) From the graph, y = 4x + 6 cuts the x-axis and the y-axis at (
(a) Axis of symmetry is x = −1. (b) Coordinates of the vertex =
9.
(a) Axis of symmetry is x = 2. (b) Coordinates of the vertex =
( 2, − 7)
(c) y-intercept = 5 (d) The graph opens upwards.
Level 2 10. ∴
The graph of y = ax2 + bx + c opens upwards. a is positive.
39
Certificate Mathematics in Action Full Solutions 4A
11.
∴
The y-intercept is negative. c is negative.
∴ ∴
The graph of y = ax2 + bx + c opens downwards. a is negative. The y-intercept is positive. c is positive.
12. ∴ ∴
y
−3
0
1
0
−3
The graph of y = ax2 + bx + c opens upwards. a is positive. The y-intercept is positive. c is positive.
13. The graph of y = ax2 + bx + c opens downwards. ∴ a is negative. The y-intercept is negative. ∴ c is negative. 14. (a) x 0 1 2 3 4 5 6 y 14 4 −2 −4 −2 4 14
(b) (i) Axis of symmetry is x = 0. (ii) Coordinates of the vertex =
( 0, 1 )
(iii) y-intercept = 1 (iv) The graph opens downwards. 17.
(b)
(i)
x y
−1 −1
0 2
1 3
2 2
3 −1
Axis of symmetry is x = 3.
(ii) Coordinates of the vertex =
( 3,− 4)
(iii) y-intercept = 14 (iv) The graph opens upwards. 15. (a)
−5 −4
x y
−4 1
−3 4
−2 5
−1 4
0 1
1 −4
(b) (i) Axis of symmetry is x = 1. (ii) Coordinates of the vertex =
(1, 3 )
y-intercept = 2
(iii) (iv) The graph opens downwards.
Exercise 2C (p. 80) Level 1
(b) (i) Axis of symmetry is x = −2. (ii) Coordinates of the vertex =
x
−2
−1
0
1
Consider the equation y = (x −2)2 −4, we have a = 1, h = 2 and k = −4. (a) a = 1 > 0 ∴ The graph of y = (x − 2)2 − 4 opens upwards. (b) The vertex is at (2, −4). (c) The axis of symmetry is x = 2.
2.
Consider the equation y = 2(x −3)2 −7, we have a = 2, h = 3 and k = −7. (a) a = 2 > 0 ∴ The graph of y = 2(x − 3)2 − 7 opens upwards. (b) The vertex is at (3, −7). (c) The axis of symmetry is x = 3. Consider the equation y = − (x − 1)2 − 2, we have a = −1, h = 1 and k = −2. (a) a = −1 < 0
(− 2, 5 )
1 (iii) y-intercept = (iv) The graph opens downwards.
16.
1.
3.
2
40
2 Functions and Graphs ∴
The graph of y = − (x − 1)2 − 2 opens downwards. (b) The vertex is at (1, −2). (c) The axis of symmetry is x = 1. 4.
∴ ∴
Consider the equation y = −3(x + 3)2 + 1, we have a = −3, h = −3 and k = 1. (a) a = −3 < 0 ∴ The graph of y = −3(x + 3)2 + 1 opens downwards. (b) The vertex is at (−3, 1). (c) The axis of symmetry is x = −3.
5.
The optimum value (minimum value ) of y is −1.
6.
The optimum value (minimum value) of y is 0.
7.
The optimum value (maximum value) of y is −5.
8.
The optimum value (maximum value) of y is
y = − x 2 + 6x + k = −( x 2 − 6 x) + k 13.
= − ( x 2 − 6 x + 32 − 32 ) + k = − ( x 2 − 6 x + 9) + 9 + k = − ( x − 3) 2 + (9 + k ) ∴ ∴
2
2
= 2( x 2 − 3 x ) + 6 − k
2
= −( x − 4 x + 2 − 2 ) = −( x 2 − 4 x + 4) + 4
14.
2
3 3 = 2 x 2 − 3x + − 2 2 9 9 = 2 x 2 − 3x + − + 6 4 2
= −( x − 2) 2 + 4 a = −1 < 0 ∴ The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 4. The axis of symmetry is x = 2.
y = x2 − 2x + 3
= ( x − 1) 2 + 2 a=1>0 The graph of the function opens upwards and it has a minimum point. The minimum value of y is 2. The axis of symmetry is x = 1.
∴
2
= ( x 2 + 6 x + 9) − 9 + 36
∴
∴
a=1>0 The graph of the function opens upwards and it has a minimum point. The minimum value of y is 27. The axis of symmetry is x = −3.
y = − x 2 − 8x + 7 = −( x 2 + 8 x) + 7 12.
3 −k. 2
3 − k = −8 2 19 k = 2
y = x (16 − x )
By completing the square,
y = − x 2 + 16 x = −( x 2 − 16 x )
= ( x + 3) 2 + 27
∴
The minimum value of y is
= − x 2 + 16 x 2
= x + 6 x + 3 − 3 + 36
11.
−k
15. Let x be one of the numbers, then the other number is 16 −x, and y be the product of these two numbers.
y = x 2 + 6 x + 36 2
+ 6−k
2
∴
= ( x 2 − 2 x + 1) − 1 + 3
2
3 3 = 2 x − + − k 2 2
= x 2 − 2 x + 12 − 12 + 3
∴
k =3
= 2x 2 − 6x + 6 − k
= −( x − 4 x )
∴
9 + k = 12
= 2x2 − 7x + 6 + x − k
2
10.
The maximum value of y is 9 + k.
y = ( x − 2)(2 x − 3) + x − k
3. 2
y = x 2 + 4x 9.
a = −1 < 0 The graph of the function opens downwards and it has a maximum point. The maximum value of y is 23. The axis of symmetry is x = −4.
= −( x 2 + 8 x + 4 2 − 4 2 ) + 7 = − ( x 2 + 8 x + 16) + 16 + 7
= −( x 2 − 16 x + 8 2 − 82 ) = −( x 2 − 16 x + 64) + 64 = −( x − 8) 2 + 64 ∴ ∴
The maximum value of y is 64. The maximum value of the product of these two numbers is 64.
16.
The maximum value of the quadratic function is −1. ∴ The quadratic function can be written as y = a(x − h)2 − 1, where a < 0 Let a = −1 and h = 1, we have
= − ( x + 4) 2 + 23
41
Certificate Mathematics in Action Full Solutions 4A y = 2( 2 − x )( x + 2) = 4 x
y = − ( x − 1) 2 − 1
= 2( − x 2 + 4 ) − 4 x
= − ( x 2 − 2 x + 1) − 1
= −2 x 2 − 4 x + 8
= − x2 + 2x − 2
19.
or let a = −1 and h = 2, we have
= − 2( x 2 + 2 x ) + 8 = −2( x 2 + 2 x + 12 − 12 ) + 8
y = −( x − 2) 2 −1
= −2( x 2 + 2 x + 1) + 2 + 8
= −( x 2 − 4 x + 4) −1
= −2( x + 1) 2 + 10 = −x 2 + 4 x − 5 (or any other reasonable answers) 17. Let y = a(x −h) + k be the quadratic function. The graph opens downwards and its axis of symmetry is x = −3. ∴ a < 0 and h = −3. Let a = −1 and k = 1, we have 2
∴
a = −2, h = −1 and k = 10
(a)
The optimum value (maximum value) of y is 10. a = −2 < 0 ∴ The graph of y = 2(2 −x)(x + 2) −4x opens downwards. (ii) The vertex is at (−1, 10). (iii) The axis of symmetry is x = −1.
(b) (i)
y = −( x + 3) 2 +1 = −( x 2 + 6 x + 9) +1 = −x 2 − 6 x −8 or let a = −2 and k = 3, we have
y = − ( x − 1)(2 x + 3) − 5 x = − ( 2 x 2 + x − 3) − 5x
y = −2( x +3) 2 +3
= − 2x 2 − 6x + 3
= −2( x 2 +6 x +9) +3 = −2 x
2
= −2 x
2
= − 2( x 2 + 3x ) + 3
−12 x −18 +3
20. −12 x −15 (or any other reasonable
Level 2
2
1 1 = x 2 − x + − − 6 2 2 1 1 = x 2 − x + − − 6 4 4
(a)
2
(b) (i)
(b) (i)
25 . 4 ∴
∴
a = −2 < 0 The graph of y = −(x −1)(2x + 3) −5x opens downwards.
(ii) The vertex is at
The optimum value (minimum value) of y is
−
The optimum value (maximum value) of y is
15 . 2
1 25 = x − − 2 4 1 and k = 25 ∴ a = 1, h = − 2 4 (a)
2
3 15 = − 2 x + + 2 2 3 and k = 15 ∴ a = −2, h = − 2 2
= x2 − x − 6 18.
2
2
y = ( x + 2)(x − 3) 2
2
3 3 + 3x + − + 3 2 2 9 9 = − 2 x 2 + 3x + + + 3 4 2
= −ers) 2 x answ
− 3 , 15 . 2 2
(iii) The axis of symmetry is a=1>0 The graph of y = (x + 2)(x − 3) opens upwards.
x = −
3. 2
y = ( x + 1) 2 + 2 x + 2( x − 1) = x2 + 2x + 1 + 2x − 2
1 25 (ii) The vertex is at , − . 4 2 1 . (iii) The axis of symmetry is x = 2
= x2 + 4x − 1
21.
= x 2 + 4 x + 22 − 22 − 1 = ( x 2 + 4 x + 4) − 4 − 1 = ( x + 2) 2 − 5
42
∴
a = 1, h = −2 and k = −5
(a)
The optimum value (minimum value) of y is −5.
2 Functions and Graphs ∴
(b) (i)
a=1>0 The graph of y = (x + 1)2 + 2(x − 1) opens upwards. (ii) The vertex is at (−2, −5). (iii) The axis of symmetry is x = −2.
y = 4 − ( x − 1) 2 = 4 − ( x 2 − 2 x + 1)
∴
= 2 x 2 − 40 x + 400
y = 2 x 2 − 40 x + 400 = 2( x 2 − 20 x ) + 400 = 2( x 2 − 20 x + 10 2 − 10 2 ) + 400
= −( x 2 − 2 x) + 3
= 2( x 2 − 20 x + 100 ) − 200 + 400
= − ( x 2 − 2 x + 12 − 12 ) + 3 = − ( x 2 − 2 x + 1) + 1 + 3 = − ( x − 1) 2 + 4
= 2( x − 10) 2 + 200 ∴ ∴
∴
The function attains its maximum value when x = 1. When x = 1, y = 4 ∴ The coordinates of C = (1, 4) (b) To find the coordinates of A and B, we have to solve y = 0.
y = 0
4 − ( x − 1)
2
= −5t 2 +10 t +15 26. (a)
= 0
∴
1 Area of △ABC = 2 [3 − ( −1)]( 4) =8
= −5(t 2 −2t ) +15 = −5(t 2 −2t +12 −12 ) +15 = −5(t 2 −2t +1) +5 +15
x − 2x − 3 = 0 ( x + 1)( x − 3) = 0 ∴
The minimum value of y is 200. The minimum value of the sum of the squares of these two numbers is 200. h =15 +10 t −5t 2
2
x + 1 = 0 or x − 3 = 0 x = −1 or x = 3 The coordinates of A and B are (−1, 0) and (3, 0) respectively.
= x 2 + ( x 2 − 40 x + 400 )
By completing the square,
= − x2 + 2x + 3 22. (a)
y = x 2 + ( 20 − x ) 2
= −5(t −1) 2 + 20 ∴ When t = 1, h attains its maximum value. ∴ The object will attain its maximum height after 1 second. (b) From (a), h =−5(t −1) 2 +20 ∴ The maximum height reached is 20 m. h =30 t −5t 2 = −5t 2 +30 t
y = x 2 + 2kx + k = x 2 + 2kx + k 2 − k 2 + k
23.
27. (a)
= −5(t 2 −6t +3 2 −3 2 )
= ( x + k ) 2 − k ( k −1)
= −5(t 2 −6t +9) +45
=[ x −( −k )] 2 − k ( k −1) The axis of symmetry of y = x2 + 2kx + k is x + 4 = 0, i.e. x = −4. −k = −4 ∴
∴ 24. (a)
k =4
The optimum value (minimum value ) of the function is −4(4 −1) = −12. Let y cm2 be the area of the parallelogram. y =2 x ( 6 −x )
=− 2 x 2 +12 x =− 2( x 2 −6 x ) =− 2( x 2 −6 x +3 2 −32 ) =− 2( x 2 −6 x +9 ) +18 =− 2( x −3) 2 +18
∴ ∴
The maximum value of y is 18. The maximum area of this parallelogram is 18 cm2.
= −5(t 2 −6t )
= −5(t −3) 2 +45 ∴ When t = 3, h attains its maximum value. ∴ The ball will attain its maximum height after 3 seconds.
(b) From (a), h = −5(t −3)2 + 45 ∴ The maximum height reached is 45 m. ∴ Half of the maximum height reached is 22.5 m. By substituting h = 22.5 into the function h = 30t −5t2, we have
22.5 = 30t − 5t 2 2t 2 − 12t + 9 = 0
Using the quadratic formula,
(b) When y = 18, x = 3
2x = 6
t =
6 − x = 3
∴ The height and the base of the parallelogram are 6 cm and 3 cm respectively.
25. Let x be one of the numbers, then the other number is 20 −x, and y be the sum of the squares of these two numbers.
12 ±
( −12 ) 2 − 4( 2)( 9) 2( 2 )
12 ± 72 4 = 0.879 or 5.12 (cor. to 3 sig. fig.) =
∴
When t = 0.879 or 5.12, the ball reaches half of
43
Certificate Mathematics in Action Full Solutions 4A
28. (a)
the maximum height.
= DE ⋅ EF
Let x m be the width of the playground, then the length of the playground is (60 − 2x) m, and y m2 be the area of the playground.
= (c)
y = x (60 − 2 x )
Let y cm2 be the area of rectangle BDEF.
y=
= −2 x 2 + 60 x ∴
3 x (8 − x ) cm 2 2
= −2( x 2 − 30 x )
=
= −2( x 2 − 30 x + 15 2 − 15 2 ) = −2( x 2 − 30 x + 225) + 450 ∴ ∴
The maximum value of y is 450 m2. The maximum area of the playground is 450 m2.
= =
(b) From (a), y = −2(x − 15)2 + 450 ∴ When x = 15, the area of the playground is maximum. When x = 15, 60 −2x = 30 ∴ The dimensions of the rectangular playground is 15 m × 30 m. 29. (a)
= ∴
Let x cm be the length of a side of one of the squares, then the length of a side of the other square is
When x = 4, y attains its maximum value of 24. When x = 4,
∴
1 (80 − 4 x ) cm = (20 −x) cm 8
∴
=
∴
= −2( x − 15) 2 + 450
3 x (8 − x ) 2 3 − x 2 + 12x 2 3 − ( x 2 − 8 x) 2 3 − ( x 2 − 8x + 42 − 42 ) 2 3 − ( x 2 − 8 x + 1 6) + 24 2 3 − ( x − 4) 2 + 24 2
Total area of the two squares = [x2 + (20 −x)2] cm2
3 (8 − x ) = 6 2
The dimensions and the area of the largest rectangle that can be inscribed in △ABC are 4 cm × 6 cm and 24 cm2 respectively.
Exercise 2D (p.92)
(b) Let y cm2 be the total area of the two squares.
Level 1
y = x 2 + (20 − x ) 2 = x + ( x − 40 x + 400 ) 2
∴
2
1.
= 2 x 2 − 40 x + 400
x y
–5 0 5 –3 –3 –3
= 2( x 2 − 20 x ) + 400 = 2( x 2 − 20 x + 10 2 − 10 2 ) + 400 = 2( x 2 − 20 x + 100 ) − 200 + 400 = 2( x − 10) 2 + 200
∴
30. (a)
∴
When x = 10, y attains its minimum value. When x = 10 , 20 − x = 10 The lengths of a side of the two squares are both 10 cm, so that the total area of the two squares is a minimum.
△ABC ~ △AFE
∴
(AAA)
EF AF = CB AB AF EF = ⋅ CB AB 8− x = (12) cm 8 3 = ( 8 − x ) cm 2
From the graph, the number of x-intercepts is 0. 2.
(b) The area of rectangle BDEF
44
x y
–1 1 –5 –1
3 3
2 Functions and Graphs
From the graph, the number of x-intercepts is 1. 3.
x y
–3 –1 1 12 0 –4
3 0
5 12 From the graphs, the solution of 2x2 – 1 ≥ 1 is x ≤ –1 or x ≥ 1.
6.
Draw y = 1 on the graph of y = x2 + 4x + 4.
From the graph, the number of x-intercepts is 2.
From the graphs, the solution of x2 + 4x + 4 > 1 is x < –3 or x > –1. 7.
Draw y = 3 on the graph of y = 2x – 1.
From the graph, the number of x-intercepts is 1. 5.
Draw y = 1 on the graph of y = 2x2 – 1.
45
Certificate Mathematics in Action Full Solutions 4A
(any reasonable answers)
Level 2 11. (a) Draw y = 2 on the graph of y = x3 – 2x – 2. From the graphs, the solution of 2x – 1 ≤ 3 is x ≤ 2. 8.
Draw y = 21 on the graph of y = 2x + 5.
From the graphs, the solution of x3 – 2x – 2 ≥ 2 is x ≥ 2.0. From the graphs, the solution of 2x + 5 ≥ 21 is x ≥ 4. 9.
(b)
(a) The graph of y = f(x) has three distinct x-intercepts. ∴ The cubic function y = (x – a)(x – b)(x – c) has three distinct real roots. ∴ a, b and c are distinct. Put a = –1, b = 1, c = 2 or a = –2, b = –1, c = 1. (or any other reasonable answers)
x 3 > 2( x + 2) x3 > 2x +4 x3 −2 x −2 > 2
∴ ∴
x > 2.0 (By(a)) The smallest integer x required is 3.
12. (a) Draw y = 4 on the graph of y = 2x – x2.
(b) The graph of y = f(x) has two distinct x-intercepts. ∴ The cubic function y = (x – a)(x – b)(x – c) has two distinct real roots. ∴ Two of a, b and c are the same. Put a = –2, b = 1, c = 1 or a = –1, b = 2, c = 2. (or any other reasonable answers) 10.
46
2 Functions and Graphs
From the graphs, the solution of 2x – x2 ≥ 4 is x ≥ 4.7.
(b)
2 x + 4 x ≥ ( x + 2) 2 2x +4x ≥ x 2 +4x +4
From the graphs, the solution of 4 – x – 2x2 ≥ 3 is –1.0 ≤ x ≤ 0.5. 15. Plot the graphs of y = x2 + 2x – 2 and y = 5.
2x − x2 ≥ 4
∴ ∴
x ≥ 4.7 (By(a)) The smallest integer that x required is 3.
13. Plot the graphs of y = x2 + 2x – 6 and y = 2.
From the graphs, the solution of x2 + 2x – 2 < 5 is –3.8 < x < 1.8.
From the graphs, the solution of x2 + 2x – 6 < 2 is –4.0 < x < 2.0.
14. Plot the graphs of y = 4 – x – 2x2 and y = 3. 16. Plot the graphs of y = –x2 – 3x + 3 and y = 4.
47
Certificate Mathematics in Action Full Solutions 4A ∴
The solution of x2 – 2x – 1 < 0 is –0.4 < x < 2.4.
Exercise 2E (p. 107) Level 1 1.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the y-axis by 3 units.
2.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 3 units.
3.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the x-axis by 1 unit.
4.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the y-axis by 1 unit.
5.
The graph of y = q(x) is obtained by translating the graph of y = p(x) in the negative direction of the yaxis by 3 units. ∴ q(x) = p(x) – 3 x 0 1 2 3 4 p(x) –4 –3 0 2 2 p(x) – 3 –7 –6 –3 –1 –1 ∴ The tabular representation of q(x) is as follows: x 0 1 2 3 4 q(x) –7 –6 –3 –1 –1
6.
(a)
7.
(a)
From the graphs, the solution of –x2 – 3x + 3 ≤ 4 is x ≤ –2.6 or x ≥ –0.4. 17. We may rewrite x2 + x – 12 ≥ 0 as x2 + x – 7 ≥ 5. Hence, we can solve x2 + x – 12 ≥ 0 by solving x2 + x – 7 ≥ 5. In other words, x2 + x – 12 ≥ 0 can be solved by drawing y = 5 on the graph of y = x2 + x – 7.
From the graphs, the solution of x2 + x – 7 ≥ 5 is x ≤ –4 or x ≥ 3. ∴ The solution of x2 + x – 12 ≥ 0 is x ≤ –4 or x ≥ 3.
g ( x) = x + 2 = f ( x) + 2 ∴ The graph of y = f(x) is translated in the positive direction of the y-axis by 2 units. g ( x ) =( x −1) 2 ∴
18. We may rewrite x2 – 2x – 1 < 0 as x2 – 2x + 3 < 4. Hence, we can solve x2 – 2x – 1 < 0 by solving x2 – 2x + 3 < 4. In other words, x2 – 2x – 1 < 0 can be solved by drawing y = 4 on the graph of y = x2 – 2x + 3.
8.
(a)
= f ( x −1) The graph of y = f(x) is translated in the positive direction of the x-axis by 1 unit.
g ( x) = x 2 −10 x +25 = ( x −5) 2 = f ( x −5) ∴
The graph of y = f(x) is translated in the positive direction of the x-axis by 5 units.
f ( x) = x 2 − 4 x + 4 = ( x − 2) 2 9.
g ( x) = x 2 + 4 x + 4 = ( x + 2) 2 = [( x + 4) − 2] 2 = f ( x + 4) ∴
From the graphs, the solution of x2 – 2x + 3 < 4 is –0.4 < x < 2.4.
48
The graph of y = f(x) is translated in the negative direction of the x-axis by 4 units.
2 Functions and Graphs
Level 2 f ( x) = x 2 − 2 x = x( x − 2) 10. g ( x) = x 2 − 8 x + 15
= ( x − 3)( x − 5)
= ( x − 3)[ ( x − 3) − 2] = f ( x − 3) ∴
The graph of y = f(x) is translated in the positive direction of the x-axis by 3 units.
11. h( x) = g ( x ) + 2 = f ( x + 3) + 2 ∴ The graph of y = f(x) is translated in the negative direction of the x-axis by 3 units first, and then in the positive direction of the y-axis by 2 units.
From the graph, the roots of f(x) = 0 are 1 and 3. (b) g(x) = f(x + 2) ∴ The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the x-axis by 2 units. ∴ The x-intercepts of the graph of y = g(x) are also translated in the negative direction of the x-axis by 2 units. ∴ The roots of g(x) = 0 are –1 and 1. 14. Let f(x) = 3x2 – 4x + 2. The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of y-axis by 3 units. ∴
∴
From the graph, the coordinates of the vertex are (3, –5). (b) ∴ ∴ ∴
g(x) = f(x) + 5 The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the y-axis by 5 units. The vertex of the graph of y = g(x) is also translated in the positive direction of the y-axis by 5 units. The coordinates of the vertex of y = g(x) are (3, 0).
g ( x) = f ( x) − 3 = 3x 2 − 4 x + 2 − 3 = 3x 2 − 4 x − 1
The required symbolic representation is g(x) = 3x2 – 4x – 1.
15. Let f(x) = x2 – 6x + 8. The graph of y = g(x) is obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 2 units.
g ( x ) = f ( x − 2)
∴
= ( x − 2) 2 − 6( x − 2) + 8 = x 2 − 4 x + 4 − 6 x + 12 + 8
∴
= x 2 − 10 x + 24 The required symbolic representation is g(x) = x2 – 10x + 24.
16. (a) f(x) = 2x2 + 1 ∴ The coordinates of the vertex of y = f(x) are (0, 1).
g ( x) = 2 x 2 +12 x +19 = 2( x 2 + 6 x ) +19 (b)
= 2( x 2 + 6 x + 32 − 32 ) +19 = 2( x 2 + 6 x + 9) −18 +19
= 2( x + 3) 2 +1 = f ( x + 3) (c) The graph of y = f(x) is translated in the negative direction of the x-axis by 3 units. (d) The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the x-axis by 3 units.
49
Certificate Mathematics in Action Full Solutions 4A ∴
The vertex of the graph of y = g(x) is also translated in the negative direction of the x-axis by 3 units. The coordinates of the vertex of y = g(x) are (–3, 1).
∴
3.
(a)
∴
4.
5.
(a)
= x 2 − 4 x + 4 + x − 2 +1 = x 2 − 3x + 3
6.
t ( x) = s ( x ) + 2 = x 2 − 3x + 5
(c)
Level 1
2.
7.
f ( −2) =−( −2) 2 +3( −2) −4 =−4 −6 −4 =−4
(a)
f (1) +g ( 2) =12 −2(1) +3( 2) +5 =1 −2 +6 +5 =10
(b)
Coordinates of the vertex
(c)
3 y-intercept =−
(d)
The graph opens upwards.
y-intercept =3
x
−5
0
5
y
−15
−5
5
= −4
(b) [ g (1)] 2 ⋅ f ( −1) =[3(1) +5]2 ⋅[( −1) 2 −2( −1)] =82 ⋅3 =192
(c)
= (1, − 4)
Axis of symmetry is x = −2 .
(d) The graph opens downwards.
Revision Exercise 2 (p. 113)
(b)
(a)
Axis of symmetry is x =1 .
2, 11 ) (b) Coordinates of the vertex =(−
= x 2 − 3x + 3 + 2
f (0) = −(0) 2 +3(0) −4
2(3) +k = 2(1 +2k ) 6 +k = 2 +4k 4 k = 3
= ( x − 2) 2 + ( x − 2) + 1
(a)
f ( −3) =2( −3) +2 =−6 +2
f (3) = 2 g (1) ∴
s ( x) = r ( x − 2)
1.
2(0) +b =2 b =2
=−4
(b) The graph of y = t(x) is obtained by translating the graph of y = s(x) in the positive direction of the y-axis by 2 units. ∴ t(x) = s(x) + 2 x –1 0 1 2 3 4 5 s(x) 7 3 1 1 3 7 13 s(x) + 2 9 5 3 3 5 9 15 ∴ The tabular representation of t(x) is as follows: x –1 0 1 2 3 4 5 t(x) 9 5 3 3 5 9 15
(ii)
f (0) = 2
(b) From (a), we have f ( x ) =2 x +2
17. (a) The graph of y = s(x) is obtained by translating the graph of y = r(x) in the direction of the x-axis. From the tabular representations of the functions r(x) and s(x), we have s(x) = r(x – 2).
(c) (i)
∴
Number of x-intercepts: 1
8.
f ( 2) 2 2 −2( 2) = g ( −2) 3( −2) +5 =0
50
x
−4
−2
0
2
4
y
33
15
5
3
9
2 Functions and Graphs
Number of x-intercepts: 0 9.
x
y
−1
2
−
0
1 2
1
3 2
2
5 2
3
−2
−
0
11 8
2
9 8
−2
1 2 −
9 8
11 8
From the graph, the solution of 33 +1 > 10 is x >2 . 11.
Draw y = 7 on the graph of
y = x 3 − x 2 − 4 x +1 .
Number of x-intercepts: 3
From the graph, the solution of x 3 − x 2 −4 x +1 ≤ 7 is x ≤ 3 . 12. The optimum value (minimum value) of y is 1. The axis of symmetry is x = 2 . 13. The optimum value (maximum value) of y is −4. The axis of symmetry is x = 3 . 14. The optimum value (maximum value) of y is 3. The axis of symmetry is x = −4 .
10. Draw y =10 on the graph of y = 3x +1 .
51
Certificate Mathematics in Action Full Solutions 4A y = −x 2 + 6 x − 4 = −( x 2 − 6 x ) − 4 15.
19. Let x be one of the two numbers, then the other number is 8 − x , and y be the product of these two numbers.
= −( x 2 − 6 x + 32 − 32 ) − 4 = −( x 2 − 6 x + 9) + 9 − 4
y = x(8 − x)
= −( x − 3) 2 + 5 ∴
= −x 2 + 8 x
a = −1 < 0 The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 5. The axis of symmetry is x = 3 .
= −( x 2 − 8 x + 4 2 − 4 2 ) = −( x 2 − 8 x + 16 ) +16 = −( x − 4) 2 + 16
y = 2x2 + 6x + 1
∴ ∴
= 2( x 2 + 3 x) + 1
16.
= −( x 2 − 8 x)
∴
2 2 3 3 = 2 x 2 + 3x + − + 1 2 2 9 9 = 2 x 2 + 3x + − + 1 4 2
g ( x ) = −2( x −1)( x +1) = −2( x 2 −1) 20.
= −2 x 2 +2 = −2 x 2 +3 −1 = f ( x ) −1
2
3 7 = 2 x + − 2 2 a =2 >0 ∴
∴
The graph of the function opens upwards and it has a minimum point.
7 ∴ The minimum value of y is − . 2 3 The axis of symmetry is x = − . 2
21.
= [( x − 3) − 2]3 = f ( x − 3)
= x 2 + 10 x + 52 − 52 + 28
= −( x 2 − 4 x + 22 − 2 2 ) + 5
= ( x 2 + 10 x + 25) − 25 + 28
22.
= −( x 2 − 4 x + 4) + 4 + 5
= ( x + 5) 2 + 3
= −( x − 2) 2 + 9
= [( x + 7) − 2]2 + 3
a =2 >0
= f ( x + 7)
∴
The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 9. The axis of symmetry is x = 2 .
∴
The graph of y = f ( x ) is translated in the negative direction of the x-axis by 7 units.
g ( x) = 3( x 3 + x +1)
y = x 2 −8 x − k
18.
The graph of y = f ( x ) is translated in the positive direction of the x-axis by 3 units.
g ( x) = x 2 + 10 x + 28
= −( x 2 − 4 x) + 5
The graph of y = f ( x ) is translated in the negative direction of the y-axis by 1 unit.
g ( x) = ( x − 5)3
∴
y = −x 2 + 4 x + 5 17.
The maximum value of y is 16. The maximum value of the product of these two numbers is 16.
= x 2 −8 x + 4 2 − 4 2 − k
= 3x3 + 3x + 3
23.
= 3 x 3 + 3 x + 2 +1
= ( x 2 −8 x +16 ) −16 −k
= f ( x ) +1
= ( x − 4) 2 −(16 + k )
∴
∴ The minimum value of y is −(16 +k ) . −(16 +k ) = −12 ∴ k = −4
24.
The graph of y = f ( x ) is translated in the positive direction of the y-axis by 1 unit.
The graph of y = g (x ) is obtained by translating the graph of y = f (x ) in the positive direction of the x-axis by 3 units.
52
2 Functions and Graphs g ( x) = f ( x − 3)
−3( 2) +b =−3 b =3
= 3( x − 3) + 1 2
∴
= 3( x 2 − 6 x + 9) + 1 ∴
28. (a)
= 3x 2 − 18 x + 28 The symbolic representation of g(x) is g ( x ) = 3 x 2 −18 x + 28 .
S ( 40 000 ) = 5000 +0.2 ( 40 000 ) =13 000 ∴ His salary in that month is $13 000.
S ( x) =11 600
(b) 25.
translating the graph of y = f ( x ) in the negative direction of the y-axis by 2 units. ∴
5000 +0.2 x =11 600
The graph of y = g ( x ) is obtained by
g ( x) = f ( x) − 2
∴
∴ 29. (a)
= 2x2 + x − 2 The symbolic representation of g(x) is g ( x) = 2 x 2 + x − 2 .
h( x) = g ( 2 x −1) = 2( 2 x −1) −1 = 4 x − 2 −1 = 4 x −3
∴
2
26. Let f(x) = ax + bx +c be the quadratic function. f (1) = 4 ∴ ∴
a (1) 2 +b(1) +c =4 a +b +c =4 f (3) =14 a (3) 2 +b(3) +c =14 9a +3b +c =14
(2) – (1): 8a + 2b = 10 4a + b = 5 Put a = 1 into (3),
(b) ∴
…… (1)
The symbolic representation of h(x) is h ( x ) = 4 x −3 .
g ( p − 2) = h( p ) 2( p −2) −1 = 4 p −3 2 p −5 = 4 p −3 p = −1
f ( x + 3) = ( x + 3) 2 + 3k
…… (2)
…… (3)
x = 33 000 His sales figure for that month is $33 000.
30.
= x 2 + 6 x + 9 + 3k
(a)
4(1) +b = 5
= x −6x +9 + k 2
b =1
Put a = 1 and b = 1 into (1),
(b)
1 +1 +c = 4 c =2
∴ f(x) = x2 + x + 2 or put a = 2 into (3),
4( 2) + b = 5
b = −3
put a = 2, b = –3 into (1),
∴
2 −3 + c = 4 c =5
f(x) = 2x2 – 3x +5 (or any other reasonable answers)
.
g ( x − 3) = ( x − 3) 2 + k
f ( x +3) − g ( x −3) =12 x +1
∴
x 2 +6 x +9 +3k −( x 2 −6 x +9 +k ) =12 x +1 12 x +2k =12 x +1 1 k= 2 31. (a)
x
−1
1
3
5
7
y
−16
−4
0
−4
−16
Level 2 27.
f (1) = −3 ∴ a (1 +2)(1 −2) +b = −3 −3a +b = −3 ...... (1) f (3) =13 ∴ a (3 +2)( 3 −2) +b =13 5a +b =13 ...... ( 2) (2) – (1), 8a =16 a =2 By substituting a = 2 into (1), we have (b) (i)
Axis of symmetry is x = 3 .
53
Certificate Mathematics in Action Full Solutions 4A (ii) Coordinates of the vertex
y = ( x − 2)( x − 4) + 3
= (3, 0)
9 (iii) y-intercept =−
= x2 − 6x + 8 + 3
(iv) The graph opens downwards.
= x 2 − 6 x + 11
34.
= x 2 − 6 x + 32 − 32 + 11 = x 2 − 6 x + 9 − 9 + 11 = ( x − 3) 2 + 2 (a)
∴
a =1 > 0 The graph opens upwards.
(b) y-intercept =11 32. (a)
x
−5
−3
−1
1
3
y
7
−5
−9
−5
7
(c)
Axis of symmetry is x = 3 .
(d) Coordinates of the vertex
=(3, 2)
y = 3((1 − x )( x + 5) − 6 x = 3( −x 2 − 4 x + 5) − 6 x = −3 x 2 −18 x +15 35.
= −3( x 2 + 6 x ) +15 = −3( x 2 + 6 x + 32 − 32 ) +15 = −3( x 2 + 6 x + 9) + 27 +15 = −3( x + 3) 2 + 42 (a)
(b) (i) Axis of symmetry is x = −1 . 1, (ii) Coordinates of the vertex =( −
− 9)
33. (a)
(c)
The graph opens upwards.
x
−2
0
2
4
6
y
10
−2
−6
−2
10
a = −3 < 0 The graph opens upwards.
(b) y-intercept =15
8 (iii) y-intercept =−
(v)
∴
Axis of symmetry is x = −3 .
(d) Coordinates of the vertex 36. (a)
= (− 3, 42 )
By substituting A(1, 0) into y = x 2 + px + q , we have 0 =1 + p +q
q =−1 − p
..... (1)
By substituting B(6, 0) into y = x 2 + px + q , we have 0 = 6 2 + p ( 6) + q
0 = 36 +6 p +q ..... ( 2) By substituting (1) into (2), we have 0 = 36 +6 p −1 − p p = −7
(b) (i) Axis of symmetry is x = 2 . (ii) Coordinates of the vertex =( 2,
By substituting p = −7 into (1) , we have q = −1 −( −7) =6 − 6)
(b) From (a), we have y = x 2 −7 x + 6 By completing the square,
2 (iii) y-intercept =−
(vi) The graph opens upwards.
54
2 Functions and Graphs
y = x2 − 7 x + 6 2
2
7 7 = x2 − 7 x + − + 6 2 2 49 49 = x2 − 7 x + +6 − 4 4 2
∴
7 25 = x − − 2 4 The optimum value (minimum value) of x 2 + px + q is −
25 . 4
37. Plot the graphs of y = 5 x − 2 x 2 and y = −12 .
39. Plot the graphs of y = ( x −1)( x 2 −3 x +1) and y =2 .
From the graphs, the solution of 5 x − 2 x 2 < −12 is x <1.5 or x > 4.0 . 38. Plot the graphs of y = ( x −5)( x − 2) and y =2 .
From the graphs, the solution of ( x −1)( x 2 −3 x +1) ≥ 2 is x ≥3.0 . 40. (a)
x
−3
−2
−1
0
1
2
3
y
19
8
1
−2
−1
4
13
From the graphs, the solution of ( x −5)( x − 2) ≤ 2 is 1.4 ≤ x ≤ 5.6 .
55
Certificate Mathematics in Action Full Solutions 4A by solving x 3 + x 2 − x +1 ≤ −1 . In other words, − x 3 − x 2 + x + 3 ≥ 5 can be solved by drawing y = −1 on the graph obtained in (a). From the graphs, the solution of x 3 + x 2 − x +1 ≤ −1 is x <−2.0 . ∴
42. (a)
The solution of − x 3 − x 2 + x +3 ≥ 5 is x ≤−2.0 .
Draw y = 2 on the graph of y = f (x ) .
(b) We may rewrite 2 x 2 − x − 3 > 0 as
2 x 2 − x − 2 > 1 . Hence, we can solve
2 x 2 − x − 3 > 0 by solving 2 x 2 − x − 2 > 1 . In other words,
2 x 2 − x − 3 > 0 can be solved by drawing y =1 on the graph obtained in (a). From the graphs, the solution of 2 x 2 − x − 2 > 1 is x < −1.0 or
x >1.5 . ∴
41. (a)
The solution of 2 x 2 − x − 3 > 0 is x <−1.0 or x >1.5 .
x
−4
−3
−2
−1
0
1
2
3
y
−43 −14
−1
2
1
2
11
34
From the graphs, the solution of f ( x ) ≥ 2 is
x ≤ −1 or x ≥ 2 . (b) ∴
From (a), the solution of f ( x −1) ≥ 2 is x −1 ≤ −1 or x −1 ≥ 2 . The solution of f ( x −1) ≥ 2 is x ≤ 0 or x ≥ 3 .
5 5 ≤ −2 as − ≥ 2 . Hence, we x x 5 5 ≤ −2 by solving − ≥ 2 . In other can solve x x 5 ≤ −2 can be solved by drawing y = 2 words, x
43. We may rewrite
on the given graph.
(b)
We may rewrite − x 3 − x 2 + x + 3 ≥ 5 as x 3 + x 2 − x +1 ≤ −1 .
Hence, we can solve − x 3 − x 2 + x + 3 ≥ 5
56
2 Functions and Graphs
From the graphs, the solution of −
− ∴
5 ≥ 2 is x
47. (a)
(i)
5 ≤ x <0 . 2 The solution of
x
1
2
3
4
5
6
7
y
8
3
0
−1
0
3
8
5 5 ≤ −2 is − ≤ x < 0 . x 2
g ( x) = ( x + 3) 2 = x2 + 6x + 9
44.
= x2 + 6x + 8 + 1 = f ( x) + 1 ∴
The direction of y = f (x ) is translated in the positive direction of the y-axis by 1 unit.
g ( x) = x 2 − 4 x + 5 = x 2 − 4 x + 2 2 − 22 + 5
45.
= ( x 2 − 4 x + 4) − 4 + 5
(ii) From the graph, the roots of f ( x ) = 0 are 3 and 5.
= ( x − 2) 2 + 1 ∴
The coordinates of the vertex of the graph of y = g ( x ) are (2, 1).
The vertex of the graph of y = f (x ) can be obtained by translating the vertex of the graph of y = g ( x ) in the negative direction of the x-axis by 4 units.
∴
The graph of y = f (x ) can be obtained
The graph of y = f ( x ) is translated in the positive direction of the x-axis by 1 unit. (ii) The graph of y = g ( x ) is translated in the negative direction of the x-axis by 3 units.
(b) (i)
(c)
by translating the graph of y = g ( x ) in negative direction of the x-axis by 4 units.
∴
x-intercepts of the graph of y = f ( x ) in the positive direction of the x-axis by 1 unit. ∴ The x-intercepts of the graph of y = g (x ) are 4 and 6.
f ( x ) = g ( x + 4)
= [( x + 4) − 2]2 + 1 = ( x + 2) 2 + 1 ∴
46. (a)
From (b), the x-intercepts of the graph of y = h(x) can be obtained by translating the
The symbolic representation of f(x) is f ( x ) = ( x + 2) 2 +1 .
x-intercepts of the graph of y = g ( x ) in the negative direction of the x-axis by 3 units. ∴ The x-intercepts of the graph of y = h(x) are 1 and 3.
AP = DS = CR = BQ = (10 − x ) cm
∴
∴
1 S ( x ) = (10 ×10 ) −4 ( x )(10 − x ) 2 =100 −20 x + 2 x 2 = 2 x −20 x +100
= −( x 2 + 4 x ) − 2 48.
= −( x 2 + 4 x + 2 2 − 2 2 ) − 2 = −( x 2 + 4 x + 4) + 4 − 2
S ( x) = 2 x 2 − 20 x + 100
= −( x + 2) 2 + 2
= 2( x 2 − 10 x) + 100
= f ( x + 2)
= 2( x 2 − 10 x + 52 − 52 ) + 100
∴
= 2( x 2 − 10 x + 25 ) − 50 + 100 = 2( x − 5) + 50 ∴ ∴
When x = 5 , S(x) attains its minimum value. The value of x is 5 so that the area of the remaining part attains its minimum.
The roots of h( x) = 0 are 1 and 3.
g ( x) = − x 2 − 4 x − 2
2
(b)
From (b), the x-intercepts of the graph of y = g (x ) can be obtained by translating the
49. (a)
The graph of y = f ( x ) is translated in the negative direction of the x-axis by 2 units.
∆AEF ~ ∆ACB
( AAA )
57
Certificate Mathematics in Action Full Solutions 4A
EF AF = CB AB AF EF = ⋅ CB AB 8− x = ⋅ 6 cm 8 3 = (8 − x) cm 4
∴
∴
(b)
f ( x −1) − f ( x ) = ( x −1) 2 +( x −1) +1 −( x 2 + x +1) = x 2 −2 x +1 + x −1 +1 −x 2 −x −1 = −2 x
5. Answer: C f ( 2 x +1) = 2( 2 x +1) +1 = 4 x +2 +1 = 4 x +3
3 = x (8 − x ) cm 2 4 Area of BDEF 3 = − x 2 + 6 x cm 2 4
6.
Answer: D The vertex of the parabola is at ( 2, −1) . ∴ The axis of symmetry is x = 2 .
7.
Answer: A The graph opens upwards. ∴ a >0 The graph has no x-intercepts. ∴ The quadratic equation ax 2 + bx + c = 0 has no real roots.
Let y cm2 be the area of BDEF.
3 2 x + 6x 4 3 = − ( x 2 − 8 x) 4 3 = − ( x 2 − 8 x + 42 − 42 ) 4 3 = − ( x 2 − 8 x + 16) + 12 4 3 = − ( x − 4) 2 + 12 4
y=−
∴
∴ ∴
∴
y = x2 − 4x + 8 = x 2 − 4 x + 22 − 22 + 8 = ( x 2 − 4 x + 4) − 4 + 8 = ( x − 2) 2 + 4
The maximum value of y is 12. The are of the largest rectangle that can be inscribed in triangle ABC is 12 cm2.
∴
9.
f ( x) = 2 + cos x
The symbolic representation of the function in the table is P =15 t .
HKMO (p.119) 1.
Answer: A
1 0 +2 1 = 2
3.
Answer: C The graph of y = f ( x ) is obtained by translating the graph of y = cos x in the positive direction of the y-axis by 2 units. f ( x) = 2 + cos x ∴ ∴ The required symbolic representation is
30 45 60 75 90 105 = = = = = = 15 2 3 4 5 6 7 P ∴ = 15 t ∴ P =15 t
f ( 0) =
The axis of symmetry of y = x 2 − 4 x +8 is
x =2 .
1. Answer: C From the table, we have
2.
b − 4ac < 0
8. Answer: C
Multiple Choice Questions (p. 118)
∴
∆<0
2
2
∴
Answer: B f ( x +1) = ( x +1) 2 −1 = x 2 + 2 x +1 −1 = x2 + 2x
4. Answer: A
58
Let x° be the angle of the sector of the circle. The perimeter of the sector is 18.
x ( 2πr ) + 2r = 18 360 x 9−r = 360 πr
2 Functions and Graphs
x (πr 2 ) 360 9−r = (πr 2 ) πr = (9 − r )(( r ) =
= −r 2 + 9r 2 Area of the sector = −(r − 9r ) 2 2 9 9 = − r 2 − 9 r + − 2 2 81 81 = − r 2 − 9r + + 4 4 2
9 81 = − r − + 2 4 ∴
2.
When the radius is r, the area of the sector attains the maximum value 9 r = 2
1 1 f x + = x2 + 2 x x 1 = x2 + 2 + 2 − 2 x 1 1 = x 2 + 2( x) + 2 − 2 x x = x−
2
1 −2 x
∴
f ( x ) = x 2 −2
∴
f (5) =52 −2 =23
59
2 Functions and Graphs
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Activity
1.Xx x x) f(x) = x2 x)g(x) = x2 + 3
Activity 2.1 (p. 67)
−2 4
−1 1
0 0
1 1
2 4
3 9
12
7
4
3
4
7
12
2.
1.
Yes
2.
Yes, the graph has a minimum point.
3.
Any reasonable descriptions including the curve always lies on or above the x-axis, opening upwards, etc.
Activity 2.2 (p. 73) 1.
−3 9
3.
open upwards
4.
2.
A positive number or zero.
3.
(a) (b)
4.
(a) x y
0 2; (1, 2) −2 11
Yes The graph of y = x2 + 3 can be obtained by translating the graph of y = x2 upwards by 3 units.
Activity 2.4 (p. 101) −1 6
0 3
1 2
2 3
3 6
4 11
g ( x ) = f ( x + 2) = ( x + 2 ) 2 − 2( x + 2 ) + 1
1.
= x2 + 4x + 4 − 2x − 4 + 1 = x2 + 2x + 1 2.
x −4 g(x) 9
−3 4
−2 1
−1 0
0 1
1 4
2 9
3.
5.
(b) (c)
Yes x=1
(a) (b)
k (h, k); x = h 4.
Activity 2.3 (p. 95)
Yes
5.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) leftwards by 2 units.
6.
The graph of y = h(x) can be obtained by translating the graph of y = f(x) rightwards by 2 units.
30
2 Functions and Graphs 2
Follow-up Exercise
(b)
p. 56 1.x y
−4 −2 0 −17 −11 −5
2 1
4 7
6 13
2.
1 1 1 f = − − 1 3 3 3 1 1 = − −1 9 3 11 = − 9 ∴
The value of the function is − x=
11 when 9
1 . 3
−3 ( −3) 2 + 2 −3 = 9 +2 3 = − 11
f ( −3) = 2.
3.
4.
3.
(a) (b) (c) (d)
true true false true
(a)
x/cm 1 y/cm2 6
(a)
h ( −2) = ( −2) 2 − 5( −2) + 6 = 4 + 10 + 6 = 20 h ( 2) = 2 2 − 5( 2 ) + 6 = 4 − 10 + 6 =0
2 3 4 5 12 18 24 30
h ( 0) = 0 2 − 5( 0) + 6 =6
(b)
(b) (i)
h ( 2) + h ( −2) = 0 + 20 = 20
but h(0) = 6 i.e. h(2) + h(−2) ≠ h(0) (ii)
(c)
y = 6x
p. 60 1.
(a)
f ( −3) = ( −3) 2 − ( −3) − 1 = 9 + 3 −1 ∴
4.
(a)
∴
= 11
The value of the function is 11 when x = −3.
h ( 2) ⋅ h ( −2) = 0 ⋅ 20 = 0 h ( −4) = ( −4) 2 − 5( −4) + 6 but = 16 + 20 + 6 = 42 i.e. h ( 2) ⋅ h ( −2) ≠ h ( −4) g(4) = 0 4 2 − 2k = 0 16 − 2k = 0 k =8
(b) From (a), we have g ( x ) = x 2 − 16 ∴
g ( −2) = ( −2) 2 − 16 = 4 − 16 = − 12
31
Certificate Mathematics in Action Full Solutions 4A
5.
(a)
f ( −3a ) =2( −3a ) 2 −( −3a ) =18 a 2 +3a
2.
(a)
f (b −1) = 2(b −1) 2 −(b −1) = 2(b 2 −2b +1) −b +1
(b)
From the graph, y = ax + b cuts the x-axis and the y-axis at (−2, 0) and (0, −4) respectively. 2 and y-intercept = ∴ x-intercept = − −4 The line passes through (0, −4).
(b)
= 2b 2 −4b +2 −b +1 = 2b 2 −5b +3
∴
p. 66 1.
∴ (a)
x −1 f(x) 9
1 3
3 −3
−4 = a (0) + b b = −4
The line passes through (−2, 0). 0 = a ( −2 ) − 4 a = −2 a = −2 and
The graph of y = f(x) is:
b = −4
p. 70 Function
1.
y = −x2 − x + 1 y = −3 + x + 2x2 y = (1 − x)2 − 4 y = −(x + 1)2 + 2 y = (x + 2)(x − 5) + 9 2.
(a)x y
−4 8
−3 2
−2 0
Direction of opening open downwards open upwards open upwards open downwards open upwards −1 2
y-intercept 1 −3 −3 1 −1
0 8
From the graph, y = f(x) cuts the x-axis and the yaxis at (2, 0) and (0, 6) respectively. ∴ x-intercept = 2 and y-intercept = 6 (b)
x −3 h(x)−1
−1 3
1 7
The graph of y = h(x) is:
(b) (i)
Axis of symmetry is x = −2.
(ii) Coordinates of the vertex =
(− 2,0)
(iii) y-intercept = 8 (iv) The graph opens upwards. p. 76 1.
From the graph, y = h(x) cuts the x-axis and the yaxis at ( ∴
−
5 , 0) and (0, 5) respectively. 2
5 x-intercept = −2 and y-intercept =
(a)
Consider the equation y = 2x2 + 1, we have a = 2, h = 0 and k = 1. (i) a=2>0 ∴ The graph of y = 2x2 + 1 opens upwards. (ii) The vertex is at (0, 1) (iii) The axis of symmetry is x = 0.
(b) Consider the equation y = −2(x + 1)2 + 3, we have a = −2, h = −1 and k = 3.
5
(i)
32
a = −2 < 0
2 Functions and Graphs ∴
The graph of y = −2(x + 1)2 + 3 opens downwards. (ii) The vertex is at (−1, 3) (iii) The axis of symmetry is x = −1. (c)
y = 2 x 2 −4 x +3 (d)
Consider the equation y = 4(x −2) −1, we have a = 4, h = 2 and k = −1. (i) a=4>0 ∴ The graph of y = 4(x − 2)2 − 1 opens upwards. (ii) The vertex is at (2, −1). (iii) The axis of symmetry is x = 2.
= 2( x −1) 2 +1 a=2>0 ∴ The graph of the function opens upwards and it has a minimum point. ∴ The maximum value of y is 1. The axis of symmetry is x = 1.
(d) Consider the equation y = 3 −2(x −1)2, we have a = −2, h = 1 and k = 3. a = −2 < 0 The graph of y = 3 − 2(x − 1)2 opens downwards. (ii) The vertex is at (1, 3). (iii) The axis of symmetry is x = 1. 2.
∴
y = x 2 + 10 x + k = ( x 2 + 10 x + 25) − 25 + k = ( x + 5) 2 − 25 + k ∴ ∴
3. y = x + 4 x −1
= ( x + 2) −5 a=1>0 The graph of the function opens upwards and it has a minimum point. ∴ The minimum value of y is −5. The axis of symmetry is x = −2.
y = − x2 + 2x = −( x 2 − 2 x) = − ( x 2 − 2 x + 12 − 12 ) = − ( x 2 − 2 x + 1) + 1
y = − x + 2x − 2 2
= − ( x − 1) 2 + 1
= −( x − 2 x) − 2 2
∴ ∴
= −( x − 2 x + 1 − 1 ) − 2 2
2
= − ( x 2 − 2 x + 1) + 1 − 2 = − ( x − 1) 2 − 1
∴
a = −1 < 0 The graph of the function opens downwards and it has a maximum point. The maximum value of y is −1. The axis of symmetry is x = 1.
= −2( x 2 + 4 x) +1 2
2
The maximum value of y is 1. The maximum area of this rectangle is 1 m2.
p.85 1.
number of x-intercepts : 1 no axis of symmetry no maximum or minimum point
2.
number of x-intercepts : 2 axis of symmetry : x = 2 minimum point : (2, −5)
3.
number of x-intercepts : 3 no axis of symmetry no maximum or minimum point
y = −2 x 2 −8 x +1 (c)
y = x(2 − x )
By conpleting the square,
∴
∴
4 − x m i.e. (2 − x) m , and y m2 be 2
= −x2 + 2x
2
2
k = 33
Let x m be the length of the rectangle, then the width of
∴
= ( x 2 + 4 x + 4) − 4 −1
(b)
−25 + k = 8
the area of the rectangle.
= ( x 2 + 4 x + 2 2 − 2 2 ) −1
(a)
The minimum value of y is −25 + k.
the rectangle is
2
1.
= ( x 2 + 10 x + 52 − 52 ) + k
2.
(a) The optimum value (minimum value) of y is −1. (b) The optimum value (maximum value) of y is 5.
p. 79
= 2( x 2 − 2 x +12 −12 ) + 3 = 2( x 2 − 2 x +1) − 2 + 3
2
(i)
= 2( x 2 − 2 x ) + 3
2
= −2( x + 4 x + 2 − 2 ) +1 = −2( x 2 + 4 x + 4) +8 +1 = −2( x + 2) 2 + 9 ∴ a = −2 < 0 ∴ The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 9. The axis of symmetry is x = −2.
p. 90 1.
When x > −2.1 (cor. to 1d.p.), the corresponding portion
33
Certificate Mathematics in Action Full Solutions 4A of the graph of y =3 x lies above the straight line y = 0.1. Hence, the solution of 3 x >0.1 is x > −2.1. 2.
When 1 ≤ x ≤ 3 , the corresponding portion of the graph of y = x 2 −4 x +6 lies on or below the straight line y = 3. Hence, the solution of x 2 −4 x +6 ≤ 3 is
1 ≤ x ≤3 . 3.
When −2 < x <1 , the corresponding portion of the graph of y =−x 2 −x −2 lies above the straight line y = −4. Hence, the solution of − x 2 − x −2 > −4 is
From the graphs, the solutions of 2x2 − 5x − 8 > 4 is x < −1.5 or x > 4.
−2 < x <1 . 4.
p. 100
Draw y = 2 on the graph of y = x2 −5x −4.
1.
g( x) = x 3 + x − 2 = ( x 3 + x + 2) − 4 ∴
2.
From the y-intercepts of y =2 x and y = g(x), we find the number of units the graph of y =2 x has translated upwards. y-intercept of y = g(x) is 4, and y-intercept of y =2 x is 1. ∴ The difference in y-intercepts of the two graphs is
From the graphs, the solution of x2 −5x −4 < 2 is −1 < x < 6. 5.
= f ( x) − 4 The graph of y = f(x) is translated in the negative direction of the y-axis by 4 units.
Draw y = −1 on the graph of y = −x2 + 6x −9. ∴
3.
4 − 1 = 3. The graph of y = g(x) is obtained by translating the graph of y =2 x upwards by 3 units.
∴ ∴
g(x) = 2 x +3 The required symbolic representation is g(x) = 2 x +3 .
The graph of y = g(x) is obtained by translating the graph of y = f(x) in the negative direction of the y-axis by 4 units. g(x) = f(x) − 4
∴
x f(x) f(x) − 4 ∴ From the graphs, the solution of −x2 + 6x −9 ≥ −1 is 2 ≤ x ≤ 4.
The tabular representation of g(x) is:
x −2 g(x) 7
−1 0
0 −3
1 −2
2 1
3 12
p. 106 6.
Draw y = 4 on the graph of y = 2x −5x −8. 2
1.
34
(a)
The graph of y = f(x + 3) can be obtained by
2 Functions and Graphs translating the graph of y = f(x) in the negative directions of the x-axis by 3 units. (b) The graph of y = f(x − 2) can be obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 2 units. g ( x) = x 2 − 6 x + 9 2.
= x − 2(3)( x ) + 3
2
Level 1
= ( x − 3) 2 ∴
Exercise Exercise 2A (p. 61)
2
(a)
g(x) = (x − 5)2.
= f ( x − 3) The graph of y = f(x) is translated in the positive direction of the x-axis by 3 units.
1.
(a)
∴
g ( x ) = x 2 + 2 x +1 = x 2 + 2(1)( x) + 12
(b)
(b)
= ( x + 1) 2 ∴ 3.
−4 3
−3 0
−2 −1
−1 0
0 3
(c)
1 8
(a)
= 8 −1
=4 +1
(b) ∴ ∴ ∴
4.
Let
g(x) = f(x − 2) The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 2 units. The vertex of the graph of y = g(x) is also translated in the positive direction of the xaxis by 2 units. The coordinates of the vertex of y = g(x) are (0, −1).
=1+1
= ( x − 5) 2 The required symbolic representation is
The value of g(x) is
17 when x = −2. 16
f ( 30 °) = cos 30 ° 3.
(a)
= x 2 − 2(2)( x ) + 2 2
The graph of y = g(x) is obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 3 units. g ( x ) = f ( x − 3) ∴ = [( x − 3) − 2]2
= 2
The value of g(x) is 2 when x = 0.
g ( −2) = 4 −2 + 1 1 = +1 16 17 = 16 ∴
f ( x) = x 2 − 4 x + 4 = ( x − 2) 2
∴
(c)
=5
The value of g(x) is 5 when x = 1.
g ( 0) = 4 0 + 1
∴ From the graph, the coordinates of the vertex are (−2, −1).
=7
The value of the function is 7 when x = −2.
g (1) =41 +1
∴ (b)
= −1
The value of the function is −1 when x = 0.
f ( −2) = 2( −2) 2 − 1
∴
2.
The value of the function is 17 when x = 3.
f (0) = 2(0) 2 − 1 ∴
= f ( x +1) The graph of y = f(x) is translated in the negative direction of the x-axis by 1 unit.
x −5 f(x) 8
(a)
f (3) = 2(3) 2 −1 =18 −1 =17
=
3 2
f ( 45 °) = cos 45 ° (b)
(c)
=
2 2
f ( 60 °) = cos 60 ° 1 = 2
35
Certificate Mathematics in Action Full Solutions 4A
1 f = 3
4.
1 1 G = 2 2 1 = 2 = 0
1 2
1 9 + 1 3 1 = 1 9 + 1 9 1 = 1+1 1 = 2
(a)
1 G = 0 2 1 1 = but G ( 2) 6 i.e.
9(1) 2 + 1 1 = 9 +1 1 = 10
(b)
The relation
1 1 does not G = G ( 2) 2
hold. (a)
g ( 0) = ( 0 − 2)( 0 − 3)
(a)
1 1 G ≠ G ( 2) 2
∴
8. 5.
( 0)
(b)
1
f (1) =
2 1 − 1 2
2 f ( −4) = 2( −4 )( −4 + 1) = 2( −4 )( −3) = 24
= ( −2)( −3) =6
(b)
g (1) = (1 − 2 )( 1 − 3) = ( −1)( −2)
f ( 2) + [ f ( −3)] 2 = 2( 2 + 1) + [ −3( −3 + 1)] 2 = 2(3) + [ −3( −2)] 2
=2
= 6 + 62 = 42
g ( −1) = ( −1 − 2)( −1 − 3) = ( −3)( −4) = 12
(b)
9.
g (1) + g ( −1) = 2 + 12
(a)
(b)
f ( 2) =3( 2) −2
(c)
(a)
f (1) ⋅g ( −1) =(12 +1) ⋅[ 2( −1)] =2( −2) =−4
=6 −2 f ( 4) = 3( 4) − 2 = 12 − 2 = 10
7.
= 6 +1 = 7
=4
(b)
g (3) + f ( 0) = 2( 3) + (0 2 + 1)
= 14
but g(0) = 6 i.e. g(1) + g(−1) ≠ g(0) 6.
(a)
= 4 = 16 2 but f ( 2) = f ( 4 ) = 10 i.e. [ f ( 2)] 2 ≠ f ( 2 2 ) [ f ( 2)]
2
2
g ( 4) 2( 4 ) = 2 f (3) 3 +1 8 = 9 +1 8 = 10 4 = 5 h (3) = 1
G ( 2) = 2[ 2( 2) − 1] = 2( 3)
10.
3(3) + m = 1 27 + m = 1 2
m = − 26
=6
36
2 Functions and Graphs f (3) = 20 11.
32 + a ( 3) + 2 = 20 3a + 11 = 20
g (50 °) = sin(50 ° + 10 °) = sin 60 °
15. (a)
a = 3
=
g ( −1) = −1
g ( 35 °) = sin(35 ° + 10 °)
( −1 + k )( −1 − 2) = −1 ( −1 + k )( −3) = −1
12.
3 − 3k = −1 4 k = 3
=
g ( x ) = f (2 x ) = 2( 2 x ) + 3 ∴
(b)
h( x ) = = = =
∴
f ( x + 2) 2( x + 2 ) + 3 2x + 4 + 3 2x + 7
The symbolic representation of h(x) is h(x) = 2x + 7.
1 h( x ) − 1 2 1 = ( 2 x + 7) − 1 2 7 = x+ −1 2 1 = − 2 1 = − 6
H (0) =4(0) 2 −8(0) +1 =1
= 4( p 2 −4 p +4) −8 p +16 +1
(b)
= 4 p 2 −16 p +16 −8 p +16 +1 = 4 p 2 −24 p +33
f (8) = 6 g ( 4) 17.
2(8 + m ) = 6( 4 + m ) 16 + 2m = 24 + 6m 4m = −8 m = −2
g( x) = 4x + 3 (c)
4x + 3 3x x
18.
f ( 2) = 2 ( 2 + 2)( 2 − 2) + a ( 2) + b = 2 2a + b = 2 ……(1)
f ( −2) = 4
( −2 + 2)( −2 − 2) + a ( −2) + b = 4 − 2a + b = 4
……(2) (1) + (2), 2b = 6 b =3
Level 2 f ( 0) = 3[ 2 3( 0 ) ] 14. (a)
2 2
H ( p −2) = 4( p −2) 2 −8( p −2) +1
= 4x + 3
The symbolic representation of g(x) is g(x) = 4x + 3.
= sin 45 °
(b)
16. (a) 13. (a)
3 2
= 3( 2 0 ) = 3(1)
By substituting b = 3 into (1), we have 2a + 3 = 2
a = −
1 2
=3 2 3 − 3
(b)
2 f − = 3[2 ] 3 = 3(2 − 2 ) 1 = 3 4 3 = 4
19. (a)
2 f ( a ) − 3g ( a ) 2( 3a − 1) − 3( a − 2) 6a − 2 − 3a + 6 3a
= = = =
5 5 5 1 1 a = 3
37
Certificate Mathematics in Action Full Solutions 4A f ( 2b) − g (3b) 3( 2b) − 1 − (3b − 2) 6b − 1 − 3b + 2 3b
(b)
= = = =
5 5 5 4 4 b = 3
(b)
5 x 2 − x − 120 = 0 ( x − 5)(5 x + 24) = 0 x − 5 = 0 or 5 x + 24 = 0
( k + 3)( k + 2) − k 2 = 2k
S ( x ) = 60 x (5 x − 1) = 60 2 x (5 x − 1) = 120
∴
f ( k ) = 2k 20. (a)
The area of the trapezium is 60 cm2.
x = 5
k 2 + 5k + 6 − k 2 = 2k 3k = −6
3x − 2 = 13
∴ (b) From (a), we have f(x) = (x + 3)(x + 2) −4 Consider the equation 2 x − f ( x ) =0
23. (a)
2 x −[( x +3)( x +2) −4] =0
The lengths of AD and BC are 11 cm and 13 cm respectively.
F (50 ) = 500 + 4(50 ) = 500 + 200 = 700
∴
2 x −( x 2 +5 x +6 −4) =0
The cost for holding a party with 50 guests is $700.
2 x −x 2 −5 x −2 =0 (b)
x 2 +3 x +2 =0 ( x +1)( x +2) =0 x +1 = 0 or x + 2 = 0 x = −1 or x = −2
The cost for holding a party is $840.
F ( n ) = 840 500 + 4n = 840
∴
n = 85
∴
f ( x + 2) = ( x + 2) 2 − k ( x + 2)
(c)
= x 2 + 4 x + 4 − kx − 2k
The number of guests is 85. H ( n ) = 10n − F ( n ) = 10n − (500 + 4n ) = 10n − 500 − 4n = 6n − 500
(i)
= x 2 + ( 4 − k ) x + 4 − 2k f ( x −2) =( x −2) 2 −k ( x −2)
(ii)
= x 2 −4 x +4 −kx +2 k = x 2 −( 4 +k ) x +4 +2 k
H (150 ) = 6(150 ) − 500 = 900 − 500 = 400 ∴
(b)
The profit is $400 when there are 150 guests.
f ( x + 2) − f ( x − 2) = kx − 32 x
2
Exercise 2B (p. 71)
+ ( 4 − k ) x + 4 − 2k −
x 2 − ( 4 + k ) x + 4 + 2k ] = kx − 32 8 x − 4k = kx − 32
Level 1
By comparing coefficients,
1.
k =8
f ( x + 2) = f ( x − 2) + 40
(c)
f ( x + 2) − f ( x − 2) = 40 8 x − 32 = 40
(by (b))
x = 9
22. (a)
24 (rejected) 5
When x = 5, 2 x + 1 = 11
k = −2
21. (a)
x = −
or
[( 2 x +1) + (3 x − 2)] x 2 x(5 x −1) = 2
S ( x) =
38
(a)
x y
−1 −6
1 −2
3 2
2 Functions and Graphs 4.
(a)
From the graph, the line cuts the x-axis and the y-axis at (4, 0) and (0, 4) respectively. ∴ x-intercept = 4 and y-intercept = 4
(b) ∴
The line passes through (0, 4).
The line passes through (4, 0). 0 = a ( 4) + 4 a = −1 a = − 1 and b = 4
∴
5. (b) From the graph, y = 2x −4 cuts the x-axis and the y-axis at (2, 0) and (0, −4) respectively. 4 ∴ x-intercept = 2 and y-intercept = −
2.
(a)
−3 −6
x y
−1 2
(a)
4 = a ( 0) + b b = 4
For y = x2 −3x + 6, Coefficient of x2 = 1 > 0 ∴ The graph opens upwards. The y-intercept of y = x2 −3x + 6 is 6.
(b) For y = −2x2 + 6x −4, Coefficient of x2 = −2 < 0 ∴ The graph opens downwards. The y-intercept of y = −2x2 + 6x −4 is −4.
y = 5 − (2 − x ) 2
(c)
1 10
∴
= 5 − (4 − 4 x + x 2 ) y = − x2 + 4x + 1
Coefficient of x2 = −1 < 0 ∴ The graph opens downwards. The y-intercept of y = 5 −(2 −x)2 is 1.
y = ( x + 1)( 3 + x ) + 2
(d)
= x2 + 4x + 3 + 2 ∴
y = x2 + 4x + 5 Coefficient of x2 = 1 > 0 ∴ The graph opens upwards. The y-intercept of y = (x + 1)( 3 + x) + 2 is 5. 6.
∴ 3.
(a)
−
3 , 0) and (0, 6) respectively. 2
3 x-intercept = −2 and y-intercept =
7. 6
From the graph, the line cuts the x-axis and the y-axis at (−3, 0) and (0, 4) respectively. 3 and y-intercept = 4 ∴ x-intercept = −
(b) ∴
∴
0 = a ( −3) + 4 4 a = 3 4 a = and b = 4 3
(1, 3)
(c) (d) The graph opens downwards. 8.
(a) Axis of symmetry is x = −2. (b) Coordinates of the vertex =
(− 2, 1 0 )
(c) y-intercept = 6 (d) The graph opens downwards.
4 = a ( 0) + b
The line passes through (−3, 0).
(a) Axis of symmetry is x = 1. (b) Coordinates of the vertex = y-intercept = 1
The line passes through (0, 4).
b = 4
(− 1, − 2)
1 (c) y-intercept = − (d) The graph opens upwards.
(b) From the graph, y = 4x + 6 cuts the x-axis and the y-axis at (
(a) Axis of symmetry is x = −1. (b) Coordinates of the vertex =
9.
(a) Axis of symmetry is x = 2. (b) Coordinates of the vertex =
( 2, − 7)
(c) y-intercept = 5 (d) The graph opens upwards.
Level 2 10. ∴
The graph of y = ax2 + bx + c opens upwards. a is positive.
39
Certificate Mathematics in Action Full Solutions 4A
11.
∴
The y-intercept is negative. c is negative.
∴ ∴
The graph of y = ax2 + bx + c opens downwards. a is negative. The y-intercept is positive. c is positive.
12. ∴ ∴
y
−3
0
1
0
−3
The graph of y = ax2 + bx + c opens upwards. a is positive. The y-intercept is positive. c is positive.
13. The graph of y = ax2 + bx + c opens downwards. ∴ a is negative. The y-intercept is negative. ∴ c is negative. 14. (a) x 0 1 2 3 4 5 6 y 14 4 −2 −4 −2 4 14
(b) (i) Axis of symmetry is x = 0. (ii) Coordinates of the vertex =
( 0, 1 )
(iii) y-intercept = 1 (iv) The graph opens downwards. 17.
(b)
(i)
x y
−1 −1
0 2
1 3
2 2
3 −1
Axis of symmetry is x = 3.
(ii) Coordinates of the vertex =
( 3,− 4)
(iii) y-intercept = 14 (iv) The graph opens upwards. 15. (a)
−5 −4
x y
−4 1
−3 4
−2 5
−1 4
0 1
1 −4
(b) (i) Axis of symmetry is x = 1. (ii) Coordinates of the vertex =
(1, 3 )
y-intercept = 2
(iii) (iv) The graph opens downwards.
Exercise 2C (p. 80) Level 1
(b) (i) Axis of symmetry is x = −2. (ii) Coordinates of the vertex =
x
−2
−1
0
1
Consider the equation y = (x −2)2 −4, we have a = 1, h = 2 and k = −4. (a) a = 1 > 0 ∴ The graph of y = (x − 2)2 − 4 opens upwards. (b) The vertex is at (2, −4). (c) The axis of symmetry is x = 2.
2.
Consider the equation y = 2(x −3)2 −7, we have a = 2, h = 3 and k = −7. (a) a = 2 > 0 ∴ The graph of y = 2(x − 3)2 − 7 opens upwards. (b) The vertex is at (3, −7). (c) The axis of symmetry is x = 3. Consider the equation y = − (x − 1)2 − 2, we have a = −1, h = 1 and k = −2. (a) a = −1 < 0
(− 2, 5 )
1 (iii) y-intercept = (iv) The graph opens downwards.
16.
1.
3.
2
40
2 Functions and Graphs ∴
The graph of y = − (x − 1)2 − 2 opens downwards. (b) The vertex is at (1, −2). (c) The axis of symmetry is x = 1. 4.
∴ ∴
Consider the equation y = −3(x + 3)2 + 1, we have a = −3, h = −3 and k = 1. (a) a = −3 < 0 ∴ The graph of y = −3(x + 3)2 + 1 opens downwards. (b) The vertex is at (−3, 1). (c) The axis of symmetry is x = −3.
5.
The optimum value (minimum value ) of y is −1.
6.
The optimum value (minimum value) of y is 0.
7.
The optimum value (maximum value) of y is −5.
8.
The optimum value (maximum value) of y is
y = − x 2 + 6x + k = −( x 2 − 6 x) + k 13.
= − ( x 2 − 6 x + 32 − 32 ) + k = − ( x 2 − 6 x + 9) + 9 + k = − ( x − 3) 2 + (9 + k ) ∴ ∴
2
2
= 2( x 2 − 3 x ) + 6 − k
2
= −( x − 4 x + 2 − 2 ) = −( x 2 − 4 x + 4) + 4
14.
2
3 3 = 2 x 2 − 3x + − 2 2 9 9 = 2 x 2 − 3x + − + 6 4 2
= −( x − 2) 2 + 4 a = −1 < 0 ∴ The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 4. The axis of symmetry is x = 2.
y = x2 − 2x + 3
= ( x − 1) 2 + 2 a=1>0 The graph of the function opens upwards and it has a minimum point. The minimum value of y is 2. The axis of symmetry is x = 1.
∴
2
= ( x 2 + 6 x + 9) − 9 + 36
∴
∴
a=1>0 The graph of the function opens upwards and it has a minimum point. The minimum value of y is 27. The axis of symmetry is x = −3.
y = − x 2 − 8x + 7 = −( x 2 + 8 x) + 7 12.
3 −k. 2
3 − k = −8 2 19 k = 2
y = x (16 − x )
By completing the square,
y = − x 2 + 16 x = −( x 2 − 16 x )
= ( x + 3) 2 + 27
∴
The minimum value of y is
= − x 2 + 16 x 2
= x + 6 x + 3 − 3 + 36
11.
−k
15. Let x be one of the numbers, then the other number is 16 −x, and y be the product of these two numbers.
y = x 2 + 6 x + 36 2
+ 6−k
2
∴
= ( x 2 − 2 x + 1) − 1 + 3
2
3 3 = 2 x − + − k 2 2
= x 2 − 2 x + 12 − 12 + 3
∴
k =3
= 2x 2 − 6x + 6 − k
= −( x − 4 x )
∴
9 + k = 12
= 2x2 − 7x + 6 + x − k
2
10.
The maximum value of y is 9 + k.
y = ( x − 2)(2 x − 3) + x − k
3. 2
y = x 2 + 4x 9.
a = −1 < 0 The graph of the function opens downwards and it has a maximum point. The maximum value of y is 23. The axis of symmetry is x = −4.
= −( x 2 + 8 x + 4 2 − 4 2 ) + 7 = − ( x 2 + 8 x + 16) + 16 + 7
= −( x 2 − 16 x + 8 2 − 82 ) = −( x 2 − 16 x + 64) + 64 = −( x − 8) 2 + 64 ∴ ∴
The maximum value of y is 64. The maximum value of the product of these two numbers is 64.
16.
The maximum value of the quadratic function is −1. ∴ The quadratic function can be written as y = a(x − h)2 − 1, where a < 0 Let a = −1 and h = 1, we have
= − ( x + 4) 2 + 23
41
Certificate Mathematics in Action Full Solutions 4A y = 2( 2 − x )( x + 2) = 4 x
y = − ( x − 1) 2 − 1
= 2( − x 2 + 4 ) − 4 x
= − ( x 2 − 2 x + 1) − 1
= −2 x 2 − 4 x + 8
= − x2 + 2x − 2
19.
or let a = −1 and h = 2, we have
= − 2( x 2 + 2 x ) + 8 = −2( x 2 + 2 x + 12 − 12 ) + 8
y = −( x − 2) 2 −1
= −2( x 2 + 2 x + 1) + 2 + 8
= −( x 2 − 4 x + 4) −1
= −2( x + 1) 2 + 10 = −x 2 + 4 x − 5 (or any other reasonable answers) 17. Let y = a(x −h) + k be the quadratic function. The graph opens downwards and its axis of symmetry is x = −3. ∴ a < 0 and h = −3. Let a = −1 and k = 1, we have 2
∴
a = −2, h = −1 and k = 10
(a)
The optimum value (maximum value) of y is 10. a = −2 < 0 ∴ The graph of y = 2(2 −x)(x + 2) −4x opens downwards. (ii) The vertex is at (−1, 10). (iii) The axis of symmetry is x = −1.
(b) (i)
y = −( x + 3) 2 +1 = −( x 2 + 6 x + 9) +1 = −x 2 − 6 x −8 or let a = −2 and k = 3, we have
y = − ( x − 1)(2 x + 3) − 5 x = − ( 2 x 2 + x − 3) − 5x
y = −2( x +3) 2 +3
= − 2x 2 − 6x + 3
= −2( x 2 +6 x +9) +3 = −2 x
2
= −2 x
2
= − 2( x 2 + 3x ) + 3
−12 x −18 +3
20. −12 x −15 (or any other reasonable
Level 2
2
1 1 = x 2 − x + − − 6 2 2 1 1 = x 2 − x + − − 6 4 4
(a)
2
(b) (i)
(b) (i)
25 . 4 ∴
∴
a = −2 < 0 The graph of y = −(x −1)(2x + 3) −5x opens downwards.
(ii) The vertex is at
The optimum value (minimum value) of y is
−
The optimum value (maximum value) of y is
15 . 2
1 25 = x − − 2 4 1 and k = 25 ∴ a = 1, h = − 2 4 (a)
2
3 15 = − 2 x + + 2 2 3 and k = 15 ∴ a = −2, h = − 2 2
= x2 − x − 6 18.
2
2
y = ( x + 2)(x − 3) 2
2
3 3 + 3x + − + 3 2 2 9 9 = − 2 x 2 + 3x + + + 3 4 2
= −ers) 2 x answ
− 3 , 15 . 2 2
(iii) The axis of symmetry is a=1>0 The graph of y = (x + 2)(x − 3) opens upwards.
x = −
3. 2
y = ( x + 1) 2 + 2 x + 2( x − 1) = x2 + 2x + 1 + 2x − 2
1 25 (ii) The vertex is at , − . 4 2 1 . (iii) The axis of symmetry is x = 2
= x2 + 4x − 1
21.
= x 2 + 4 x + 22 − 22 − 1 = ( x 2 + 4 x + 4) − 4 − 1 = ( x + 2) 2 − 5
42
∴
a = 1, h = −2 and k = −5
(a)
The optimum value (minimum value) of y is −5.
2 Functions and Graphs ∴
(b) (i)
a=1>0 The graph of y = (x + 1)2 + 2(x − 1) opens upwards. (ii) The vertex is at (−2, −5). (iii) The axis of symmetry is x = −2.
y = 4 − ( x − 1) 2 = 4 − ( x 2 − 2 x + 1)
∴
= 2 x 2 − 40 x + 400
y = 2 x 2 − 40 x + 400 = 2( x 2 − 20 x ) + 400 = 2( x 2 − 20 x + 10 2 − 10 2 ) + 400
= −( x 2 − 2 x) + 3
= 2( x 2 − 20 x + 100 ) − 200 + 400
= − ( x 2 − 2 x + 12 − 12 ) + 3 = − ( x 2 − 2 x + 1) + 1 + 3 = − ( x − 1) 2 + 4
= 2( x − 10) 2 + 200 ∴ ∴
∴
The function attains its maximum value when x = 1. When x = 1, y = 4 ∴ The coordinates of C = (1, 4) (b) To find the coordinates of A and B, we have to solve y = 0.
y = 0
4 − ( x − 1)
2
= −5t 2 +10 t +15 26. (a)
= 0
∴
1 Area of △ABC = 2 [3 − ( −1)]( 4) =8
= −5(t 2 −2t ) +15 = −5(t 2 −2t +12 −12 ) +15 = −5(t 2 −2t +1) +5 +15
x − 2x − 3 = 0 ( x + 1)( x − 3) = 0 ∴
The minimum value of y is 200. The minimum value of the sum of the squares of these two numbers is 200. h =15 +10 t −5t 2
2
x + 1 = 0 or x − 3 = 0 x = −1 or x = 3 The coordinates of A and B are (−1, 0) and (3, 0) respectively.
= x 2 + ( x 2 − 40 x + 400 )
By completing the square,
= − x2 + 2x + 3 22. (a)
y = x 2 + ( 20 − x ) 2
= −5(t −1) 2 + 20 ∴ When t = 1, h attains its maximum value. ∴ The object will attain its maximum height after 1 second. (b) From (a), h =−5(t −1) 2 +20 ∴ The maximum height reached is 20 m. h =30 t −5t 2 = −5t 2 +30 t
y = x 2 + 2kx + k = x 2 + 2kx + k 2 − k 2 + k
23.
27. (a)
= −5(t 2 −6t +3 2 −3 2 )
= ( x + k ) 2 − k ( k −1)
= −5(t 2 −6t +9) +45
=[ x −( −k )] 2 − k ( k −1) The axis of symmetry of y = x2 + 2kx + k is x + 4 = 0, i.e. x = −4. −k = −4 ∴
∴ 24. (a)
k =4
The optimum value (minimum value ) of the function is −4(4 −1) = −12. Let y cm2 be the area of the parallelogram. y =2 x ( 6 −x )
=− 2 x 2 +12 x =− 2( x 2 −6 x ) =− 2( x 2 −6 x +3 2 −32 ) =− 2( x 2 −6 x +9 ) +18 =− 2( x −3) 2 +18
∴ ∴
The maximum value of y is 18. The maximum area of this parallelogram is 18 cm2.
= −5(t 2 −6t )
= −5(t −3) 2 +45 ∴ When t = 3, h attains its maximum value. ∴ The ball will attain its maximum height after 3 seconds.
(b) From (a), h = −5(t −3)2 + 45 ∴ The maximum height reached is 45 m. ∴ Half of the maximum height reached is 22.5 m. By substituting h = 22.5 into the function h = 30t −5t2, we have
22.5 = 30t − 5t 2 2t 2 − 12t + 9 = 0
Using the quadratic formula,
(b) When y = 18, x = 3
2x = 6
t =
6 − x = 3
∴ The height and the base of the parallelogram are 6 cm and 3 cm respectively.
25. Let x be one of the numbers, then the other number is 20 −x, and y be the sum of the squares of these two numbers.
12 ±
( −12 ) 2 − 4( 2)( 9) 2( 2 )
12 ± 72 4 = 0.879 or 5.12 (cor. to 3 sig. fig.) =
∴
When t = 0.879 or 5.12, the ball reaches half of
43
Certificate Mathematics in Action Full Solutions 4A
28. (a)
the maximum height.
= DE ⋅ EF
Let x m be the width of the playground, then the length of the playground is (60 − 2x) m, and y m2 be the area of the playground.
= (c)
y = x (60 − 2 x )
Let y cm2 be the area of rectangle BDEF.
y=
= −2 x 2 + 60 x ∴
3 x (8 − x ) cm 2 2
= −2( x 2 − 30 x )
=
= −2( x 2 − 30 x + 15 2 − 15 2 ) = −2( x 2 − 30 x + 225) + 450 ∴ ∴
The maximum value of y is 450 m2. The maximum area of the playground is 450 m2.
= =
(b) From (a), y = −2(x − 15)2 + 450 ∴ When x = 15, the area of the playground is maximum. When x = 15, 60 −2x = 30 ∴ The dimensions of the rectangular playground is 15 m × 30 m. 29. (a)
= ∴
Let x cm be the length of a side of one of the squares, then the length of a side of the other square is
When x = 4, y attains its maximum value of 24. When x = 4,
∴
1 (80 − 4 x ) cm = (20 −x) cm 8
∴
=
∴
= −2( x − 15) 2 + 450
3 x (8 − x ) 2 3 − x 2 + 12x 2 3 − ( x 2 − 8 x) 2 3 − ( x 2 − 8x + 42 − 42 ) 2 3 − ( x 2 − 8 x + 1 6) + 24 2 3 − ( x − 4) 2 + 24 2
Total area of the two squares = [x2 + (20 −x)2] cm2
3 (8 − x ) = 6 2
The dimensions and the area of the largest rectangle that can be inscribed in △ABC are 4 cm × 6 cm and 24 cm2 respectively.
Exercise 2D (p.92)
(b) Let y cm2 be the total area of the two squares.
Level 1
y = x 2 + (20 − x ) 2 = x + ( x − 40 x + 400 ) 2
∴
2
1.
= 2 x 2 − 40 x + 400
x y
–5 0 5 –3 –3 –3
= 2( x 2 − 20 x ) + 400 = 2( x 2 − 20 x + 10 2 − 10 2 ) + 400 = 2( x 2 − 20 x + 100 ) − 200 + 400 = 2( x − 10) 2 + 200
∴
30. (a)
∴
When x = 10, y attains its minimum value. When x = 10 , 20 − x = 10 The lengths of a side of the two squares are both 10 cm, so that the total area of the two squares is a minimum.
△ABC ~ △AFE
∴
(AAA)
EF AF = CB AB AF EF = ⋅ CB AB 8− x = (12) cm 8 3 = ( 8 − x ) cm 2
From the graph, the number of x-intercepts is 0. 2.
(b) The area of rectangle BDEF
44
x y
–1 1 –5 –1
3 3
2 Functions and Graphs
From the graph, the number of x-intercepts is 1. 3.
x y
–3 –1 1 12 0 –4
3 0
5 12 From the graphs, the solution of 2x2 – 1 ≥ 1 is x ≤ –1 or x ≥ 1.
6.
Draw y = 1 on the graph of y = x2 + 4x + 4.
From the graph, the number of x-intercepts is 2.
From the graphs, the solution of x2 + 4x + 4 > 1 is x < –3 or x > –1. 7.
Draw y = 3 on the graph of y = 2x – 1.
From the graph, the number of x-intercepts is 1. 5.
Draw y = 1 on the graph of y = 2x2 – 1.
45
Certificate Mathematics in Action Full Solutions 4A
(any reasonable answers)
Level 2 11. (a) Draw y = 2 on the graph of y = x3 – 2x – 2. From the graphs, the solution of 2x – 1 ≤ 3 is x ≤ 2. 8.
Draw y = 21 on the graph of y = 2x + 5.
From the graphs, the solution of x3 – 2x – 2 ≥ 2 is x ≥ 2.0. From the graphs, the solution of 2x + 5 ≥ 21 is x ≥ 4. 9.
(b)
(a) The graph of y = f(x) has three distinct x-intercepts. ∴ The cubic function y = (x – a)(x – b)(x – c) has three distinct real roots. ∴ a, b and c are distinct. Put a = –1, b = 1, c = 2 or a = –2, b = –1, c = 1. (or any other reasonable answers)
x 3 > 2( x + 2) x3 > 2x +4 x3 −2 x −2 > 2
∴ ∴
x > 2.0 (By(a)) The smallest integer x required is 3.
12. (a) Draw y = 4 on the graph of y = 2x – x2.
(b) The graph of y = f(x) has two distinct x-intercepts. ∴ The cubic function y = (x – a)(x – b)(x – c) has two distinct real roots. ∴ Two of a, b and c are the same. Put a = –2, b = 1, c = 1 or a = –1, b = 2, c = 2. (or any other reasonable answers) 10.
46
2 Functions and Graphs
From the graphs, the solution of 2x – x2 ≥ 4 is x ≥ 4.7.
(b)
2 x + 4 x ≥ ( x + 2) 2 2x +4x ≥ x 2 +4x +4
From the graphs, the solution of 4 – x – 2x2 ≥ 3 is –1.0 ≤ x ≤ 0.5. 15. Plot the graphs of y = x2 + 2x – 2 and y = 5.
2x − x2 ≥ 4
∴ ∴
x ≥ 4.7 (By(a)) The smallest integer that x required is 3.
13. Plot the graphs of y = x2 + 2x – 6 and y = 2.
From the graphs, the solution of x2 + 2x – 2 < 5 is –3.8 < x < 1.8.
From the graphs, the solution of x2 + 2x – 6 < 2 is –4.0 < x < 2.0.
14. Plot the graphs of y = 4 – x – 2x2 and y = 3. 16. Plot the graphs of y = –x2 – 3x + 3 and y = 4.
47
Certificate Mathematics in Action Full Solutions 4A ∴
The solution of x2 – 2x – 1 < 0 is –0.4 < x < 2.4.
Exercise 2E (p. 107) Level 1 1.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the y-axis by 3 units.
2.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 3 units.
3.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the x-axis by 1 unit.
4.
The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the y-axis by 1 unit.
5.
The graph of y = q(x) is obtained by translating the graph of y = p(x) in the negative direction of the yaxis by 3 units. ∴ q(x) = p(x) – 3 x 0 1 2 3 4 p(x) –4 –3 0 2 2 p(x) – 3 –7 –6 –3 –1 –1 ∴ The tabular representation of q(x) is as follows: x 0 1 2 3 4 q(x) –7 –6 –3 –1 –1
6.
(a)
7.
(a)
From the graphs, the solution of –x2 – 3x + 3 ≤ 4 is x ≤ –2.6 or x ≥ –0.4. 17. We may rewrite x2 + x – 12 ≥ 0 as x2 + x – 7 ≥ 5. Hence, we can solve x2 + x – 12 ≥ 0 by solving x2 + x – 7 ≥ 5. In other words, x2 + x – 12 ≥ 0 can be solved by drawing y = 5 on the graph of y = x2 + x – 7.
From the graphs, the solution of x2 + x – 7 ≥ 5 is x ≤ –4 or x ≥ 3. ∴ The solution of x2 + x – 12 ≥ 0 is x ≤ –4 or x ≥ 3.
g ( x) = x + 2 = f ( x) + 2 ∴ The graph of y = f(x) is translated in the positive direction of the y-axis by 2 units. g ( x ) =( x −1) 2 ∴
18. We may rewrite x2 – 2x – 1 < 0 as x2 – 2x + 3 < 4. Hence, we can solve x2 – 2x – 1 < 0 by solving x2 – 2x + 3 < 4. In other words, x2 – 2x – 1 < 0 can be solved by drawing y = 4 on the graph of y = x2 – 2x + 3.
8.
(a)
= f ( x −1) The graph of y = f(x) is translated in the positive direction of the x-axis by 1 unit.
g ( x) = x 2 −10 x +25 = ( x −5) 2 = f ( x −5) ∴
The graph of y = f(x) is translated in the positive direction of the x-axis by 5 units.
f ( x) = x 2 − 4 x + 4 = ( x − 2) 2 9.
g ( x) = x 2 + 4 x + 4 = ( x + 2) 2 = [( x + 4) − 2] 2 = f ( x + 4) ∴
From the graphs, the solution of x2 – 2x + 3 < 4 is –0.4 < x < 2.4.
48
The graph of y = f(x) is translated in the negative direction of the x-axis by 4 units.
2 Functions and Graphs
Level 2 f ( x) = x 2 − 2 x = x( x − 2) 10. g ( x) = x 2 − 8 x + 15
= ( x − 3)( x − 5)
= ( x − 3)[ ( x − 3) − 2] = f ( x − 3) ∴
The graph of y = f(x) is translated in the positive direction of the x-axis by 3 units.
11. h( x) = g ( x ) + 2 = f ( x + 3) + 2 ∴ The graph of y = f(x) is translated in the negative direction of the x-axis by 3 units first, and then in the positive direction of the y-axis by 2 units.
From the graph, the roots of f(x) = 0 are 1 and 3. (b) g(x) = f(x + 2) ∴ The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the x-axis by 2 units. ∴ The x-intercepts of the graph of y = g(x) are also translated in the negative direction of the x-axis by 2 units. ∴ The roots of g(x) = 0 are –1 and 1. 14. Let f(x) = 3x2 – 4x + 2. The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of y-axis by 3 units. ∴
∴
From the graph, the coordinates of the vertex are (3, –5). (b) ∴ ∴ ∴
g(x) = f(x) + 5 The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the positive direction of the y-axis by 5 units. The vertex of the graph of y = g(x) is also translated in the positive direction of the y-axis by 5 units. The coordinates of the vertex of y = g(x) are (3, 0).
g ( x) = f ( x) − 3 = 3x 2 − 4 x + 2 − 3 = 3x 2 − 4 x − 1
The required symbolic representation is g(x) = 3x2 – 4x – 1.
15. Let f(x) = x2 – 6x + 8. The graph of y = g(x) is obtained by translating the graph of y = f(x) in the positive direction of the x-axis by 2 units.
g ( x ) = f ( x − 2)
∴
= ( x − 2) 2 − 6( x − 2) + 8 = x 2 − 4 x + 4 − 6 x + 12 + 8
∴
= x 2 − 10 x + 24 The required symbolic representation is g(x) = x2 – 10x + 24.
16. (a) f(x) = 2x2 + 1 ∴ The coordinates of the vertex of y = f(x) are (0, 1).
g ( x) = 2 x 2 +12 x +19 = 2( x 2 + 6 x ) +19 (b)
= 2( x 2 + 6 x + 32 − 32 ) +19 = 2( x 2 + 6 x + 9) −18 +19
= 2( x + 3) 2 +1 = f ( x + 3) (c) The graph of y = f(x) is translated in the negative direction of the x-axis by 3 units. (d) The graph of y = g(x) can be obtained by translating the graph of y = f(x) in the negative direction of the x-axis by 3 units.
49
Certificate Mathematics in Action Full Solutions 4A ∴
The vertex of the graph of y = g(x) is also translated in the negative direction of the x-axis by 3 units. The coordinates of the vertex of y = g(x) are (–3, 1).
∴
3.
(a)
∴
4.
5.
(a)
= x 2 − 4 x + 4 + x − 2 +1 = x 2 − 3x + 3
6.
t ( x) = s ( x ) + 2 = x 2 − 3x + 5
(c)
Level 1
2.
7.
f ( −2) =−( −2) 2 +3( −2) −4 =−4 −6 −4 =−4
(a)
f (1) +g ( 2) =12 −2(1) +3( 2) +5 =1 −2 +6 +5 =10
(b)
Coordinates of the vertex
(c)
3 y-intercept =−
(d)
The graph opens upwards.
y-intercept =3
x
−5
0
5
y
−15
−5
5
= −4
(b) [ g (1)] 2 ⋅ f ( −1) =[3(1) +5]2 ⋅[( −1) 2 −2( −1)] =82 ⋅3 =192
(c)
= (1, − 4)
Axis of symmetry is x = −2 .
(d) The graph opens downwards.
Revision Exercise 2 (p. 113)
(b)
(a)
Axis of symmetry is x =1 .
2, 11 ) (b) Coordinates of the vertex =(−
= x 2 − 3x + 3 + 2
f (0) = −(0) 2 +3(0) −4
2(3) +k = 2(1 +2k ) 6 +k = 2 +4k 4 k = 3
= ( x − 2) 2 + ( x − 2) + 1
(a)
f ( −3) =2( −3) +2 =−6 +2
f (3) = 2 g (1) ∴
s ( x) = r ( x − 2)
1.
2(0) +b =2 b =2
=−4
(b) The graph of y = t(x) is obtained by translating the graph of y = s(x) in the positive direction of the y-axis by 2 units. ∴ t(x) = s(x) + 2 x –1 0 1 2 3 4 5 s(x) 7 3 1 1 3 7 13 s(x) + 2 9 5 3 3 5 9 15 ∴ The tabular representation of t(x) is as follows: x –1 0 1 2 3 4 5 t(x) 9 5 3 3 5 9 15
(ii)
f (0) = 2
(b) From (a), we have f ( x ) =2 x +2
17. (a) The graph of y = s(x) is obtained by translating the graph of y = r(x) in the direction of the x-axis. From the tabular representations of the functions r(x) and s(x), we have s(x) = r(x – 2).
(c) (i)
∴
Number of x-intercepts: 1
8.
f ( 2) 2 2 −2( 2) = g ( −2) 3( −2) +5 =0
50
x
−4
−2
0
2
4
y
33
15
5
3
9
2 Functions and Graphs
Number of x-intercepts: 0 9.
x
y
−1
2
−
0
1 2
1
3 2
2
5 2
3
−2
−
0
11 8
2
9 8
−2
1 2 −
9 8
11 8
From the graph, the solution of 33 +1 > 10 is x >2 . 11.
Draw y = 7 on the graph of
y = x 3 − x 2 − 4 x +1 .
Number of x-intercepts: 3
From the graph, the solution of x 3 − x 2 −4 x +1 ≤ 7 is x ≤ 3 . 12. The optimum value (minimum value) of y is 1. The axis of symmetry is x = 2 . 13. The optimum value (maximum value) of y is −4. The axis of symmetry is x = 3 . 14. The optimum value (maximum value) of y is 3. The axis of symmetry is x = −4 .
10. Draw y =10 on the graph of y = 3x +1 .
51
Certificate Mathematics in Action Full Solutions 4A y = −x 2 + 6 x − 4 = −( x 2 − 6 x ) − 4 15.
19. Let x be one of the two numbers, then the other number is 8 − x , and y be the product of these two numbers.
= −( x 2 − 6 x + 32 − 32 ) − 4 = −( x 2 − 6 x + 9) + 9 − 4
y = x(8 − x)
= −( x − 3) 2 + 5 ∴
= −x 2 + 8 x
a = −1 < 0 The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 5. The axis of symmetry is x = 3 .
= −( x 2 − 8 x + 4 2 − 4 2 ) = −( x 2 − 8 x + 16 ) +16 = −( x − 4) 2 + 16
y = 2x2 + 6x + 1
∴ ∴
= 2( x 2 + 3 x) + 1
16.
= −( x 2 − 8 x)
∴
2 2 3 3 = 2 x 2 + 3x + − + 1 2 2 9 9 = 2 x 2 + 3x + − + 1 4 2
g ( x ) = −2( x −1)( x +1) = −2( x 2 −1) 20.
= −2 x 2 +2 = −2 x 2 +3 −1 = f ( x ) −1
2
3 7 = 2 x + − 2 2 a =2 >0 ∴
∴
The graph of the function opens upwards and it has a minimum point.
7 ∴ The minimum value of y is − . 2 3 The axis of symmetry is x = − . 2
21.
= [( x − 3) − 2]3 = f ( x − 3)
= x 2 + 10 x + 52 − 52 + 28
= −( x 2 − 4 x + 22 − 2 2 ) + 5
= ( x 2 + 10 x + 25) − 25 + 28
22.
= −( x 2 − 4 x + 4) + 4 + 5
= ( x + 5) 2 + 3
= −( x − 2) 2 + 9
= [( x + 7) − 2]2 + 3
a =2 >0
= f ( x + 7)
∴
The graph of the function opens downwards and it has a maximum point. ∴ The maximum value of y is 9. The axis of symmetry is x = 2 .
∴
The graph of y = f ( x ) is translated in the negative direction of the x-axis by 7 units.
g ( x) = 3( x 3 + x +1)
y = x 2 −8 x − k
18.
The graph of y = f ( x ) is translated in the positive direction of the x-axis by 3 units.
g ( x) = x 2 + 10 x + 28
= −( x 2 − 4 x) + 5
The graph of y = f ( x ) is translated in the negative direction of the y-axis by 1 unit.
g ( x) = ( x − 5)3
∴
y = −x 2 + 4 x + 5 17.
The maximum value of y is 16. The maximum value of the product of these two numbers is 16.
= x 2 −8 x + 4 2 − 4 2 − k
= 3x3 + 3x + 3
23.
= 3 x 3 + 3 x + 2 +1
= ( x 2 −8 x +16 ) −16 −k
= f ( x ) +1
= ( x − 4) 2 −(16 + k )
∴
∴ The minimum value of y is −(16 +k ) . −(16 +k ) = −12 ∴ k = −4
24.
The graph of y = f ( x ) is translated in the positive direction of the y-axis by 1 unit.
The graph of y = g (x ) is obtained by translating the graph of y = f (x ) in the positive direction of the x-axis by 3 units.
52
2 Functions and Graphs g ( x) = f ( x − 3)
−3( 2) +b =−3 b =3
= 3( x − 3) + 1 2
∴
= 3( x 2 − 6 x + 9) + 1 ∴
28. (a)
= 3x 2 − 18 x + 28 The symbolic representation of g(x) is g ( x ) = 3 x 2 −18 x + 28 .
S ( 40 000 ) = 5000 +0.2 ( 40 000 ) =13 000 ∴ His salary in that month is $13 000.
S ( x) =11 600
(b) 25.
translating the graph of y = f ( x ) in the negative direction of the y-axis by 2 units. ∴
5000 +0.2 x =11 600
The graph of y = g ( x ) is obtained by
g ( x) = f ( x) − 2
∴
∴ 29. (a)
= 2x2 + x − 2 The symbolic representation of g(x) is g ( x) = 2 x 2 + x − 2 .
h( x) = g ( 2 x −1) = 2( 2 x −1) −1 = 4 x − 2 −1 = 4 x −3
∴
2
26. Let f(x) = ax + bx +c be the quadratic function. f (1) = 4 ∴ ∴
a (1) 2 +b(1) +c =4 a +b +c =4 f (3) =14 a (3) 2 +b(3) +c =14 9a +3b +c =14
(2) – (1): 8a + 2b = 10 4a + b = 5 Put a = 1 into (3),
(b) ∴
…… (1)
The symbolic representation of h(x) is h ( x ) = 4 x −3 .
g ( p − 2) = h( p ) 2( p −2) −1 = 4 p −3 2 p −5 = 4 p −3 p = −1
f ( x + 3) = ( x + 3) 2 + 3k
…… (2)
…… (3)
x = 33 000 His sales figure for that month is $33 000.
30.
= x 2 + 6 x + 9 + 3k
(a)
4(1) +b = 5
= x −6x +9 + k 2
b =1
Put a = 1 and b = 1 into (1),
(b)
1 +1 +c = 4 c =2
∴ f(x) = x2 + x + 2 or put a = 2 into (3),
4( 2) + b = 5
b = −3
put a = 2, b = –3 into (1),
∴
2 −3 + c = 4 c =5
f(x) = 2x2 – 3x +5 (or any other reasonable answers)
.
g ( x − 3) = ( x − 3) 2 + k
f ( x +3) − g ( x −3) =12 x +1
∴
x 2 +6 x +9 +3k −( x 2 −6 x +9 +k ) =12 x +1 12 x +2k =12 x +1 1 k= 2 31. (a)
x
−1
1
3
5
7
y
−16
−4
0
−4
−16
Level 2 27.
f (1) = −3 ∴ a (1 +2)(1 −2) +b = −3 −3a +b = −3 ...... (1) f (3) =13 ∴ a (3 +2)( 3 −2) +b =13 5a +b =13 ...... ( 2) (2) – (1), 8a =16 a =2 By substituting a = 2 into (1), we have (b) (i)
Axis of symmetry is x = 3 .
53
Certificate Mathematics in Action Full Solutions 4A (ii) Coordinates of the vertex
y = ( x − 2)( x − 4) + 3
= (3, 0)
9 (iii) y-intercept =−
= x2 − 6x + 8 + 3
(iv) The graph opens downwards.
= x 2 − 6 x + 11
34.
= x 2 − 6 x + 32 − 32 + 11 = x 2 − 6 x + 9 − 9 + 11 = ( x − 3) 2 + 2 (a)
∴
a =1 > 0 The graph opens upwards.
(b) y-intercept =11 32. (a)
x
−5
−3
−1
1
3
y
7
−5
−9
−5
7
(c)
Axis of symmetry is x = 3 .
(d) Coordinates of the vertex
=(3, 2)
y = 3((1 − x )( x + 5) − 6 x = 3( −x 2 − 4 x + 5) − 6 x = −3 x 2 −18 x +15 35.
= −3( x 2 + 6 x ) +15 = −3( x 2 + 6 x + 32 − 32 ) +15 = −3( x 2 + 6 x + 9) + 27 +15 = −3( x + 3) 2 + 42 (a)
(b) (i) Axis of symmetry is x = −1 . 1, (ii) Coordinates of the vertex =( −
− 9)
33. (a)
(c)
The graph opens upwards.
x
−2
0
2
4
6
y
10
−2
−6
−2
10
a = −3 < 0 The graph opens upwards.
(b) y-intercept =15
8 (iii) y-intercept =−
(v)
∴
Axis of symmetry is x = −3 .
(d) Coordinates of the vertex 36. (a)
= (− 3, 42 )
By substituting A(1, 0) into y = x 2 + px + q , we have 0 =1 + p +q
q =−1 − p
..... (1)
By substituting B(6, 0) into y = x 2 + px + q , we have 0 = 6 2 + p ( 6) + q
0 = 36 +6 p +q ..... ( 2) By substituting (1) into (2), we have 0 = 36 +6 p −1 − p p = −7
(b) (i) Axis of symmetry is x = 2 . (ii) Coordinates of the vertex =( 2,
By substituting p = −7 into (1) , we have q = −1 −( −7) =6 − 6)
(b) From (a), we have y = x 2 −7 x + 6 By completing the square,
2 (iii) y-intercept =−
(vi) The graph opens upwards.
54
2 Functions and Graphs
y = x2 − 7 x + 6 2
2
7 7 = x2 − 7 x + − + 6 2 2 49 49 = x2 − 7 x + +6 − 4 4 2
∴
7 25 = x − − 2 4 The optimum value (minimum value) of x 2 + px + q is −
25 . 4
37. Plot the graphs of y = 5 x − 2 x 2 and y = −12 .
39. Plot the graphs of y = ( x −1)( x 2 −3 x +1) and y =2 .
From the graphs, the solution of 5 x − 2 x 2 < −12 is x <1.5 or x > 4.0 . 38. Plot the graphs of y = ( x −5)( x − 2) and y =2 .
From the graphs, the solution of ( x −1)( x 2 −3 x +1) ≥ 2 is x ≥3.0 . 40. (a)
x
−3
−2
−1
0
1
2
3
y
19
8
1
−2
−1
4
13
From the graphs, the solution of ( x −5)( x − 2) ≤ 2 is 1.4 ≤ x ≤ 5.6 .
55
Certificate Mathematics in Action Full Solutions 4A by solving x 3 + x 2 − x +1 ≤ −1 . In other words, − x 3 − x 2 + x + 3 ≥ 5 can be solved by drawing y = −1 on the graph obtained in (a). From the graphs, the solution of x 3 + x 2 − x +1 ≤ −1 is x <−2.0 . ∴
42. (a)
The solution of − x 3 − x 2 + x +3 ≥ 5 is x ≤−2.0 .
Draw y = 2 on the graph of y = f (x ) .
(b) We may rewrite 2 x 2 − x − 3 > 0 as
2 x 2 − x − 2 > 1 . Hence, we can solve
2 x 2 − x − 3 > 0 by solving 2 x 2 − x − 2 > 1 . In other words,
2 x 2 − x − 3 > 0 can be solved by drawing y =1 on the graph obtained in (a). From the graphs, the solution of 2 x 2 − x − 2 > 1 is x < −1.0 or
x >1.5 . ∴
41. (a)
The solution of 2 x 2 − x − 3 > 0 is x <−1.0 or x >1.5 .
x
−4
−3
−2
−1
0
1
2
3
y
−43 −14
−1
2
1
2
11
34
From the graphs, the solution of f ( x ) ≥ 2 is
x ≤ −1 or x ≥ 2 . (b) ∴
From (a), the solution of f ( x −1) ≥ 2 is x −1 ≤ −1 or x −1 ≥ 2 . The solution of f ( x −1) ≥ 2 is x ≤ 0 or x ≥ 3 .
5 5 ≤ −2 as − ≥ 2 . Hence, we x x 5 5 ≤ −2 by solving − ≥ 2 . In other can solve x x 5 ≤ −2 can be solved by drawing y = 2 words, x
43. We may rewrite
on the given graph.
(b)
We may rewrite − x 3 − x 2 + x + 3 ≥ 5 as x 3 + x 2 − x +1 ≤ −1 .
Hence, we can solve − x 3 − x 2 + x + 3 ≥ 5
56
2 Functions and Graphs
From the graphs, the solution of −
− ∴
5 ≥ 2 is x
47. (a)
(i)
5 ≤ x <0 . 2 The solution of
x
1
2
3
4
5
6
7
y
8
3
0
−1
0
3
8
5 5 ≤ −2 is − ≤ x < 0 . x 2
g ( x) = ( x + 3) 2 = x2 + 6x + 9
44.
= x2 + 6x + 8 + 1 = f ( x) + 1 ∴
The direction of y = f (x ) is translated in the positive direction of the y-axis by 1 unit.
g ( x) = x 2 − 4 x + 5 = x 2 − 4 x + 2 2 − 22 + 5
45.
= ( x 2 − 4 x + 4) − 4 + 5
(ii) From the graph, the roots of f ( x ) = 0 are 3 and 5.
= ( x − 2) 2 + 1 ∴
The coordinates of the vertex of the graph of y = g ( x ) are (2, 1).
The vertex of the graph of y = f (x ) can be obtained by translating the vertex of the graph of y = g ( x ) in the negative direction of the x-axis by 4 units.
∴
The graph of y = f (x ) can be obtained
The graph of y = f ( x ) is translated in the positive direction of the x-axis by 1 unit. (ii) The graph of y = g ( x ) is translated in the negative direction of the x-axis by 3 units.
(b) (i)
(c)
by translating the graph of y = g ( x ) in negative direction of the x-axis by 4 units.
∴
x-intercepts of the graph of y = f ( x ) in the positive direction of the x-axis by 1 unit. ∴ The x-intercepts of the graph of y = g (x ) are 4 and 6.
f ( x ) = g ( x + 4)
= [( x + 4) − 2]2 + 1 = ( x + 2) 2 + 1 ∴
46. (a)
From (b), the x-intercepts of the graph of y = h(x) can be obtained by translating the
The symbolic representation of f(x) is f ( x ) = ( x + 2) 2 +1 .
x-intercepts of the graph of y = g ( x ) in the negative direction of the x-axis by 3 units. ∴ The x-intercepts of the graph of y = h(x) are 1 and 3.
AP = DS = CR = BQ = (10 − x ) cm
∴
∴
1 S ( x ) = (10 ×10 ) −4 ( x )(10 − x ) 2 =100 −20 x + 2 x 2 = 2 x −20 x +100
= −( x 2 + 4 x ) − 2 48.
= −( x 2 + 4 x + 2 2 − 2 2 ) − 2 = −( x 2 + 4 x + 4) + 4 − 2
S ( x) = 2 x 2 − 20 x + 100
= −( x + 2) 2 + 2
= 2( x 2 − 10 x) + 100
= f ( x + 2)
= 2( x 2 − 10 x + 52 − 52 ) + 100
∴
= 2( x 2 − 10 x + 25 ) − 50 + 100 = 2( x − 5) + 50 ∴ ∴
When x = 5 , S(x) attains its minimum value. The value of x is 5 so that the area of the remaining part attains its minimum.
The roots of h( x) = 0 are 1 and 3.
g ( x) = − x 2 − 4 x − 2
2
(b)
From (b), the x-intercepts of the graph of y = g (x ) can be obtained by translating the
49. (a)
The graph of y = f ( x ) is translated in the negative direction of the x-axis by 2 units.
∆AEF ~ ∆ACB
( AAA )
57
Certificate Mathematics in Action Full Solutions 4A
EF AF = CB AB AF EF = ⋅ CB AB 8− x = ⋅ 6 cm 8 3 = (8 − x) cm 4
∴
∴
(b)
f ( x −1) − f ( x ) = ( x −1) 2 +( x −1) +1 −( x 2 + x +1) = x 2 −2 x +1 + x −1 +1 −x 2 −x −1 = −2 x
5. Answer: C f ( 2 x +1) = 2( 2 x +1) +1 = 4 x +2 +1 = 4 x +3
3 = x (8 − x ) cm 2 4 Area of BDEF 3 = − x 2 + 6 x cm 2 4
6.
Answer: D The vertex of the parabola is at ( 2, −1) . ∴ The axis of symmetry is x = 2 .
7.
Answer: A The graph opens upwards. ∴ a >0 The graph has no x-intercepts. ∴ The quadratic equation ax 2 + bx + c = 0 has no real roots.
Let y cm2 be the area of BDEF.
3 2 x + 6x 4 3 = − ( x 2 − 8 x) 4 3 = − ( x 2 − 8 x + 42 − 42 ) 4 3 = − ( x 2 − 8 x + 16) + 12 4 3 = − ( x − 4) 2 + 12 4
y=−
∴
∴ ∴
∴
y = x2 − 4x + 8 = x 2 − 4 x + 22 − 22 + 8 = ( x 2 − 4 x + 4) − 4 + 8 = ( x − 2) 2 + 4
The maximum value of y is 12. The are of the largest rectangle that can be inscribed in triangle ABC is 12 cm2.
∴
9.
f ( x) = 2 + cos x
The symbolic representation of the function in the table is P =15 t .
HKMO (p.119) 1.
Answer: A
1 0 +2 1 = 2
3.
Answer: C The graph of y = f ( x ) is obtained by translating the graph of y = cos x in the positive direction of the y-axis by 2 units. f ( x) = 2 + cos x ∴ ∴ The required symbolic representation is
30 45 60 75 90 105 = = = = = = 15 2 3 4 5 6 7 P ∴ = 15 t ∴ P =15 t
f ( 0) =
The axis of symmetry of y = x 2 − 4 x +8 is
x =2 .
1. Answer: C From the table, we have
2.
b − 4ac < 0
8. Answer: C
Multiple Choice Questions (p. 118)
∴
∆<0
2
2
∴
Answer: B f ( x +1) = ( x +1) 2 −1 = x 2 + 2 x +1 −1 = x2 + 2x
4. Answer: A
58
Let x° be the angle of the sector of the circle. The perimeter of the sector is 18.
x ( 2πr ) + 2r = 18 360 x 9−r = 360 πr
2 Functions and Graphs
x (πr 2 ) 360 9−r = (πr 2 ) πr = (9 − r )(( r ) =
= −r 2 + 9r 2 Area of the sector = −(r − 9r ) 2 2 9 9 = − r 2 − 9 r + − 2 2 81 81 = − r 2 − 9r + + 4 4 2
9 81 = − r − + 2 4 ∴
2.
When the radius is r, the area of the sector attains the maximum value 9 r = 2
1 1 f x + = x2 + 2 x x 1 = x2 + 2 + 2 − 2 x 1 1 = x 2 + 2( x) + 2 − 2 x x = x−
2
1 −2 x
∴
f ( x ) = x 2 −2
∴
f (5) =52 −2 =23
59
