求y=1/2x^2-x+1和y=(x-1)(x-5)顶点坐标
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求y=1/2x^2-x+1和y=(x-1)(x-5)顶点坐标求y=1/2x^2-x+1和y=(x-1)(x-5)顶点坐标求y=1/2x^2-x+1和y=(x-1)(x-5)顶点坐标y=1/2x^2-x
求y=1/2x^2-x+1和y=(x-1)(x-5)顶点坐标
求y=1/2x^2-x+1和y=(x-1)(x-5)顶点坐标
求y=1/2x^2-x+1和y=(x-1)(x-5)顶点坐标
y=1/2x^2-x+1
=1/2(x-1)^2+1/2
所以顶点(1,1/2)
y=(x-1)(x-5)
=(x-3)^2-4
所以顶点(3,-4)
y=ax^2+bx+c
则顶点坐标为((-b/2a),(4ac-b^2)/4a)
y=(x-x1)(x-x2)
则顶点横坐标为(x1+x2)/2
纵坐标代入即可
第二个展开配方,第一个没顶点。
(1,1/2),(3,-4)
因为f(x)在(0,1)上到少有一个零点所以得出 f(0) * f(1) < 0 (因为两者肯定在Y轴两端) 又因为 f(0) =0^3 + (a-1) * 0 -1 =
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