Page 21 - Ranger SPM 2022 - Additional Mathematics
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Additional Mathematics SPM Chapter 2 Quadratic Functions
5. Given a quadratic function Solution
f(x) = –3x + px – 8 has one maximum
2
point (4, q). Find the value of p and (a) When t = 2,
2
of q. h(2) = –4(2) + 24(2) + 28
= 60 m
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Solution
(b) h(t) = –4t + 24t + 28
2
p
2
f(x) = –3 x – x – 8 = –4(t – 6t) + 28 2 2
2
3
Form 4 = –3 x – x + – p 2 – – p 2 – 8 = –4[(t – 3) – 9] + 28
= –4[t – 6t +(–3) – (–3) ] + 28
2
p
2
2
3
6
6
= –4(t – 3) + 36 + 28
2
p
2
= –3 x – p 2 – 36 2 – 8 = –4(t – 3) + 64
6
p
= –3 x – p 2 + 12 2 – 8 The height when the ball hits the
ground, h(t) = 0.
6
2
p = 4 –4(t – 3) + 64 = 0
2
6 4(t – 3) = 64
2
p = 24 (t – 3) = 16
t – 3 = +4
2
2
q = p – 8 = 24 – 8 = 40 t = –1 or t = 7
12 12
6. Find the values of m if the straight line Thus, the ball will hit the ground at
y = mx + 8 is the tangent to the curve the 7 second.
th
y = 2x + 3x + 10. (c) Maximum height = 64 m
2
Solution
HOTS Example 1
mx + 8 = 2x + 3x + 10
2
2x + (3 – m)x + 2 = 0 Muzafar has a rectangular piece of paper
2
b – 4ac = 0 Tangent to the with a length of (3x + 2) cm and a width of
2
3x cm. He cuts and removes a square with
curve means
(3 – m) – 4(2)(2) = 0 the straight sides of x cm from the paper. Find the range
2
9 – 6m + m – 16 = 0 line (tangent) of values of x if the area of the remaining
2
touches the
m – 6m – 7 = 0 curve at one paper is at least 90 cm .
2
2
(m + 1)(m – 7) = 0 point only.
m = –1 or m = 7 Solution
2
7. A ball is tossed upwards 3x(3x + 2) – x 90
2
2
from a position as in 9x + 6x – x – 90 0
2
the diagram beside. The 8x + 6x – 90 0
2
height h, in metres, of the 4x + 3x – 45 0
ball at t seconds is given Ground (4x + 15)(x – 3) 0
by a function + +
h(t) = –4t + 24t + 28. x
2
(a) What is the height of the ball when – 15 – 3
t = 2? 4
(b) In which second will the ball hit the 15
ground? x – 4 (ignored) or x 3
(c) Calculate the maximum height of the Thus, x 3
ball from the horizontal ground.
28
02 Ranger Add Mathematics Tg4.indd 28 25/02/2022 9:10 AM

