Question

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

Answer 1

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)