已知抛物线y=-2分之一x²+（5-m）x+m-3的对称轴是y求抛物线顶点的坐标 有过程给好评
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亲，求采纳~
y= - 0.5x^2+(5-m)x+(m-3) 对称轴：x=0
y= -0.5 [ x^2 - 2(5-m)x ] +(m-3)
y= -0.5 [ x^2 - 2(5-m)x + (5-m)^2 - (5-m)^2 ] + (m-3)
y= -0.5 [ x - (5-m) ]^2 + 0.5(5-m)^2 + (m-3)
y= -0.5 [ x - (5-m) ]^2 + 0.5m^2 - 4m + 9.5
对称轴 x = 5-m = 0 m = 5
顶点坐标（0，0.5m^2 - 4m + 9.5 ) = (0,2)
y= - 0.5x^2+(5-m)x+(m-3) 对称轴：x=0
y= -0.5 [ x^2 - 2(5-m)x ] +(m-3)
y= -0.5 [ x^2 - 2(5-m)x + (5-m)^2 - (5-m)^2 ] + (m-3)
y= -0.5 [ x - (5-m) ]^2 + 0.5(5-m)^2 + (m-3)
y= -0.5 [ x - (5-m) ]^2 + 0.5m^2 - 4m + 9.5
对称轴 x = 5-m = 0 m = 5
顶点坐标（0，0.5m^2 - 4m + 9.5 ) = (0,2)
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