F4 MATHS (20點)

(1)(a) If -x² - 8x + 12 = a(x - h )² + k , find the values of a,h and k
In -x² - 8x + 12 = a(x - h )² + k
R.H.S.
=a(x^2-2xh+h^2)+k
L.H.S.
=-1(x^2+8x-12)
=-1(x^2+8x+16)+28 (因為8/2=4, 所以我們要選擇4^2=16放在8x後。)
=-1(x+4)^2+28
By compare the R.H.S. and L.H.S., we have a=-1, h=4, k=28.

(b) find the maximum value or the minimum
value of the quadratic function y= -x² - 8x + 12 and the
corresponding value of x
The maximum value of the quadratic function y= -x² - 8x + 12 is 28.
The corresponding value of x is, (x+2)^2=0, x+2=0, x= -2

2(a)If 2x² + 8x = a(x - t)² + s ,find the values of a,t and s
In 2x² + 8x = a(x - t)² + s
R.H.S.
=a(x-2t+t^2)+s
L.H.S.
=2(x^2+4x)
=2(x^2+4x+4)-4*2
=2(x+2)^2-8
By comparing R.H.S. and L.H.S. , we have a=2, t=2, s=-8

(b) find the value of k if the minimum value of
the function y = 2x² + 8x + k is zero
Put 2x^2+8x+k=0 into B^2-4AC=0
8^2-4*2*k=0
64-8k=0
8k=64
k=8

1(a) R.H.S. = a(x² -2xh + h²) + k = ax² - (2ah)x + (ah² + k)

a = -1
- 2ah = -8
ah = 4
h = -4
ah² + k = 12
-1(-4)² + k = 12
-16 + k = 12
k = 28

(b) maximum value = k = 28
corresponding value of x = -4

第 2 題用同樣方法

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