F.4 QUADRATIC FUNCTIONS

2010-10-12 1:49 am

1. If the minimum value of y = x^2+3x+k is -2 and k is a constant, find the value of k

2. if the maximum value of y =ax^2 +6x-2 is -1 abd a is a constant, find the value of a

更新1:

請問一下後面可寫成THE VERTEX= (-2,-4+K) Y=-4+K 0=-4+K K=4 嗎 THZ

更新2:

THE VERTEX= (-3/2,-9/4+K) Y=-9/4+k -8=-9+4K K=1/4 才對

回答 (1)

知識就是力量

2010-10-12 4:32 am

✔ 最佳答案

1. y = x^2 + 3x + k
= [x^2 + 3x + (3/2)^2] + k - (3/2)^2
= (x + 3/2)^2 + (k - 9/4)
When x = -3/2, it attains the min. value of y = k-9/4 = -2
k - 9/4 = -2
k = 1/4

2. y = ax^2 + 6x - 2
= a(x^2 + 6x/a) - 2
= a[x^2 + 6x/a + (3/a)^2 - (3x/a)^2] - 2
= a(x + 3/a)^2 - 9/a - 2
When x = -3/a, it attains the max. value of y = -9a - 2 = -1
-9a - 2 = -1
-9a = 1
a = -1/9

2010-10-11 20:33:19 補充：
Amendment:
Q2 Line 3 should be
"a[x^2 + 6x/a + (3/a)^2 - (3/a)^2] - 2"
Sorry!

2010-10-12 18:24:48 補充：
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