Page 21 - Focus SPM KSSM Tg 4.5 - Add Maths

Additional Mathematics SPM Chapter 2 Quadratic Functions

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2
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21 = x – 8x + (–4) – (–4) + 7
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Express the quadratic function f(x) = 2x – 4x + 5 in = (x – 4) – 16 + 7
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the vertex form by using the completing the squares = (x – 4) – 9 ....................................1
method. Hence, determine the coordinates of the Compare 1 to y = (x – p) + q,
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minimum point. p = 4, q = –9.
Solution (b) The coordinates of the turning point are (p, q)
f(x) = 2x – 4x + 5 = (4, –9). Since a = 1 . 0, the turning point is a
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4

= 2 x – x + 5  minimum point.
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–2
–2
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
–
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= 2 x – 2x +     2 + 5  Try Questions 16 – 18 in ‘Try This! 2.3’
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2
2

= 2 (x – 1) – 1 + 5  23
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Form 4
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
= 2 (x – 1) + 3  The minimum value of the quadratic function
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= 2(x – 1) + 3 y = x + 2kx + 5k – 3 is –17, where is k a constant. Find
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The coordinates of the minimum point are (1, 3). the possible values of k.
Try Question 15 in ‘Try This! 2.3’ Solution
y = x + 2kx + 5k – 3
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2k
2k
2
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   
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22 = x + 2kx + —– – —– + 5k – 3
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= x + 2kx + k – k + 5k – 3
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Express the quadratic function y = x – 8x + 7 in the = (x + k) – k + 5k – 3
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form of y = (x – p) + q , where p and q are constants.
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Find The minimum value = –k + 5k – 3
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(a) the values of p and q, –k + 5k – 3 = –17
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(b) the coordinates of the turning point and hence, k – 5k – 14 = 0
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determine whether that point is a maximum or (k + 2)(k – 7) = 0
minimum point. k + 2 = 0 or k – 7 = 0
Solution k = –2 or k = 7
(a) y = x – 8x + 7 Try Questions 19 – 22 in ‘Try This! 2.3’
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–8
–8
2
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   
= x – 8x + —– – —– + 7
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2
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D Analysing the effects of the changes of a, h and k on the shape and position of the
graph of f(x) = a(x – h) + k
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a . 0 a  0
Only f(x) f(x)
the a > 1 a = 1
value 0 < a < 1 0 x
of a
changes 0 < a < 1
0 x a > 1
a = –1
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