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二次函数y=ax^2+bx+c图像顶点为(1,-9/2),且过(-2,0),求该二次函数表达式
人气:278 ℃ 时间:2019-12-05 12:49:27
解答
y= ax^2+bx+c
y'= 2ax+b =0
x = -b/2a = 1
b = -2a
y(1) = a+b+c = -9/2
-a+c = -9/2 ( b=-2a)(1)
y(-2) = 4a-2b + c = 0
8a + c= 0(2)
(2)-(1)
9a= 9/2
a = 1/2
b =-2a = -1
8a + c = 0
4+c = 0
c = -4
y= (1/2)x^2-x-4
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