# Program Latih Tubi SPM Matematik Tambahan - PowerPoint

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1.
(a) At y-axis, x = 0, f(x) = −0 + k(0) – 5 = − 5
A (0, −5) √ A1
(b) f(x)= −[x2 − kx + (−k/2)2 – (−k/2)2] – 5
= −[(x – k/2)2 – (k2/4)] – 5
√ M1 = −(x – k/2)2 + (k2/4) – 5 √ K1
k/2 = 2          ==> k = 4   √ A1
p = (42/4) – 5 ==>p = −1 √ A1     Refer to graph

(c) − x2 + 4x – 5 ≥ −5
x
x2 − 4x ≤ 0 ==> x(x – 4) ≤ 0 √ M1 0           4

0 ≤ x ≤ 4 √ A1
dy
2. (a)   dx
= hx2 – kx
dy
At turning point (3, −4);    = 0; √ K1
dx
h(3)2 – k(3) = 0     9h – 3k = 0 ……(1)
dy
At x = −1;   dx   = 8 √ K1
h(−1)2 – k(−1) = 8        h + k = 8 ……(2)
(2) x 3:   3h + 3k = 24 …..(3)
(1) + (3):     12h = 24     √ M1
h = 2 √ A1
h = 2  (2):     k = 6 √ A1
dy
(b)   dx   = 2x2 – 6x

 ( 2x  6x) d x
2
y=
y = (⅔)x3 – 3x2 + c √ K1
At (3, −4);     −4 = (⅔)(3)3 – 3(3)2 + c √ M1
−4 = 18 – 27 + c
c=5
     y = (⅔)x3 – 3x2 + 5 √ A1

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