已知二次函数y=x²+2mx-m+1(m为常数)
(1)求证:不论m为何值,该二次函数图像的顶点P都在函数y=-x²+x+1的图像上;
(2)若顶点P的横、纵坐标相等,求P点坐标.
人气:175 ℃ 时间:2019-10-01 09:31:38
解答
﹙1﹚二次函数y=x²+2mx-m+1(m为常数)
即是y=﹙x+m﹚²-m²-m+1
∴它的顶点是:P﹙-m,-m²-m+1﹚
不论m为何值,满足函数:y=-x²+x+1
∴不论m为何值,该二次函数图像的顶点P都在函数y=-x²+x+1的图像上
﹙2﹚
∵顶点P的横、纵坐标相等
∴-m=-m²-m+1 ∴m=±1
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