分类：教育/科学-学习帮助
问题：求抛物线顶点x=
x=-2y^2+2my+1(m属于R)
回答：x=-2y^2+2my+1(m属于R) 

x = -2y^2 + 2my + 1
x = -2(y^2 - my - 1/2)
x = -2[y^2 - my + (-m/2)^2 - (-m/2)^2 - 1/2]
x = -2[(y - m/2)^2 - (1/4)(m^2 + 2)]

顶点坐标为: (Xo，Yo) = ((1/2)(m^2 + 2)，m/2)