求抛物线y=x^2+(2m+1)x+m^2-1(m是实数）的顶点的轨迹方程
问题描述：
求抛物线y=x^2+(2m+1)x+m^2-1(m是实数）的顶点的轨迹方程
答
y=2x^2+x-3/4
答
y=x^2+(2m+1)x+m^2-1={x^2+2*[(2m+1)/2]x+[(2m+1)/2]^2}-[(2m+1)/2]^2+m^2-1=[x+(2m+1)/2]^2+(2m^2+m-3/4)所以顶点[-(2m+1)/2,2m^2+m-3/4]即x=-(2m+1)/2,y=2m^2+m-3/4m=-x-1/2代入y=2m^2+m-3/4y=2x^2+x-3/4