y=-x²-6x+2 x∈〖-2,4〗求最值

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y = -(x^2+6x-2)
= -(x^2+6x+9-11) =11-(x+3)^2
由于：[-2,4] y的最大值：Ymax = y(-2)=10,在定义域上,y的最大值Ymax=10
由于二次曲线y开口向下,y的最小值：Ymin = min｛y(-2),y(4)｝
由于：y(-2)=-4+12+2=10
y(4)=-16-24+2=-38
因此y的最小值为：Ymin = -38
y的最大值为：Ymax = 10

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