已知二次函数y=x²-2(m-1)+2m²-2

(1)y=x²-2(m-1)x+2m²-2=[x-(m-1)]²+m²+2m-3所以顶点的坐标（m-1,m²+2m-3)所以顶点的轨迹是x=m-1,y=m²+2m-3,消去m,得y=x²+4x所以不论m为何值,二次函数图象的顶点均在某一函数图象...所以顶点的轨迹是x=m-1,y=m²+2m-3,消去m，得y=x²+4x怎么消得x=m-1m=x+1y=(x+1)^2+2(x+1)-3=x²+4x