[tex]\text{The vertex form:}\ y=a(x-h)^2+k\\\\f(x)=ax^2+bx+c\to h=\dfrac{-b}{2a},\ k=f(h)\\\\\text{We have}\ y=-x^2+6x+7\to f(x)=-x^2+6x+7\\\\a=-1,\ b=6,\ c=7\\\\h=\dfrac{-6}{2(-1)}=\dfrac{-6}{-2}=3\\\\k=f(3)=-3^2+6(3)+7=-9+18+7=16\\\\Answer:\ \boxed{y=-(x-3)^2+16}[/tex]
