确定下列二次函数的对称轴和顶点坐标.（1）y=2x²-8x-6 （2）y=4x²+x-8 （3）y=
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(1) y = 2x² -8x + 8 -8 -6 = 2(x²-4x+4) -14 = 2(x-2)² -14
对称轴:x = 2; 顶点(2,-14)
(2) a = 4,b = 1,c = -8
对称轴:x = -b/(2a) = -1/(2*4) = -1/8
顶点的纵坐标= (4ac - b²)/(4a) = (-128 -1)/(4*4) = -129/16
顶点:(-1/8,-129/16)
(3) y = (x+3)*(x-1) = x² + 2x -3 = x² + 2x + 1 -1 -3 = (x+1)² -4
对称轴:x = -1; 顶点(-1,-4)
(4) y = -½x² + 2x - 3/2
a = -½,b = 2,c = -3/2
对称轴:x = -b/(2a) = -2/(-2*½) = 2
顶点的纵坐标= (4ac - b²)/(4a) = [4(-½)(-3/2) -1²)/[4*(-½)] = 1/2
顶点:(2,1/2)
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