a.maths

2007-09-16 6:11 pm
y=2x^2+ax+b的頂點是(3,5),求a和b的值。

回答 (2)

2007-09-16 6:31 pm
✔ 最佳答案
2x2 + ax + b
= 2 (x2 + ax/2) + b
= 2 (x + a/4)2 - a2 / 16 + b
頂點 = (-a/4 , b - a2 / 16)
所以 :
-a / 4 = 3 ... (1)
b - a2 / 16 = 5 ... (2)
由 (1): a = -12
將 a = -12 代入 (2):
b - 144/16 = 5
b = 13


2007-09-16 10:36:03 補充:
全錯了: 所有 a^2 / 16 都應該係 a^2 / 8即: 頂點 = (-a/4 , b - a^2 / 8)最後: 將 a = -12 代入 (2):b - 144 / 8 = 5b = 23
2007-09-16 6:16 pm
y=2x^2+ax+b的頂點是(3,5),求a和b的值。
5 = 8 + 3a + b
3a + b = -3 ------------ (1)
dy/dx = 4x + a
(4)(3) + a = 0
a = -12
b = 33


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