Question

二次函数F(X)=X^2+(2M+1)X+M^2-1,(M为实数)的顶点轨迹方程?

Answer 1

F(X)=X^2+(2M+1)X+M^2-1的顶点P(x,y)

通过配方

F(X)=X^2+(2M+1)X+M^2-1=[x+(2m+1)/2]^2-m-5/4

x=-(2m+1)/2=-m-1/2

y=-m-5/4 消去m,得

y=x-3/4

二次函数F(X)=X^2+(2M+1)X+M^2-1,(M为实数)的顶点轨迹方程y=x-3/4