求二次函数y=-2x^2+6x在下列定义域上的值域 定义域为[-2,1]
求二次函数y=-2x^2+6x在下列定义域上的值域 定义域为[-2,1]
数学人气：903 ℃时间：2019-10-03 10:11:47
优质解答
y = -2x^2 + 6x = -2(x^2 - 3x)= -2( x^2 - 3x + 9/4) + 9/2= -2(x - 3/2)^2 + 9/2在定义域为[-2,1]内,当x = -2时,y取最小值.此时 y = -2*（-7/2)^2 + 9/2 = -20当x = 1时,y取最大值.此时y = -2(-1/2)^2 + 9/2 = 4所...
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