F.4 數學問題
2006-10-17 7:13 am
1)Given that the minimum value of the function y=(x-2)(2x-3)+ x - k is-8 , find the value of k.
回答 (3)
2006-10-17 7:18 am
✔ 最佳答案
y = (x - 2)(2x - 3) + x - k= 2x^2 - 4x - 3x + 6 + x - k
= 2x^2 - 6x + 6 - k
= 2(x^2 - 3x) + 6 - k
= 2[x^2 - 3x + (3 / 2)^2 - (3 / 2)^2] + 6 - k
= 2[(x - (3 / 2))^2 - 9 / 4] + 6 - k
= 2[x - (3 / 2)]^2 - 9 / 2 + 6 - k
The minumum value is given by -9 / 2 + 6 - k
Thus -8 = -9 / 2 + 6 - k
k = -9 / 2 + 6 + 8
= 14 - 9 / 2
= 19 / 2
2006-10-17 10:18 am
y=2x^2-7x+6+x-k
=2x^2-6x+6-k
=2(x^2-3x)+6-k
=2(x-3/2x)^2- 9/2+6-k
=2(x-3/2x)^2+3/2-k
-K+3/2=-8
-K=-9.5
K=9.5
=2x^2-6x+6-k
=2(x^2-3x)+6-k
=2(x-3/2x)^2- 9/2+6-k
=2(x-3/2x)^2+3/2-k
-K+3/2=-8
-K=-9.5
K=9.5
2006-10-17 7:53 am
Solution
y = (x-2)(2x-3)+ x – k
y = 2x^2 – 4x – 3x + 6 + x – k
y = 2x^2 + 6x + 6 – k
y = 2 (x^2 + 3x) + 6 - k
y = 2 [x^2 + 2(3/2)x + (3/2)^2 - (3/2)^2] + 6 – k
y = 2 [x^2 + 2(3/2)x + (3/2)^2] - 2(3/2)^2 + 6 – k
y = 2 [x+ (3/2)]^2 - 2(9/4) + 6 – k
y = 2 [x+ (3/2)]^2 - 9/2 + 6 – k
y = 2 [x+ (3/2)]^2 + 6 – 4.5 – k
y = 2 [x+ (3/2)]^2 + 1.5 – k
For minimum value,
[x+ (3/2)]^2 = 0
So
-8 = 0 + 1.5 – k
k = 1.5 + 8
∴ k = 9.5
y = (x-2)(2x-3)+ x – k
y = 2x^2 – 4x – 3x + 6 + x – k
y = 2x^2 + 6x + 6 – k
y = 2 (x^2 + 3x) + 6 - k
y = 2 [x^2 + 2(3/2)x + (3/2)^2 - (3/2)^2] + 6 – k
y = 2 [x^2 + 2(3/2)x + (3/2)^2] - 2(3/2)^2 + 6 – k
y = 2 [x+ (3/2)]^2 - 2(9/4) + 6 – k
y = 2 [x+ (3/2)]^2 - 9/2 + 6 – k
y = 2 [x+ (3/2)]^2 + 6 – 4.5 – k
y = 2 [x+ (3/2)]^2 + 1.5 – k
For minimum value,
[x+ (3/2)]^2 = 0
So
-8 = 0 + 1.5 – k
k = 1.5 + 8
∴ k = 9.5
參考: I got grade A in both A Maths and Maths
收錄日期: 2021-04-12 22:51:15
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