let's put h(x) in vertex form, and then let's see if we can get f(x) from there.
[tex]~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{vertex}{\stackrel{h}{0}~~,~~\stackrel{k}{-3}}\qquad h(x)=a(x-0)^2-3 \\\\[-0.35em] ~\dotfill[/tex]
[tex]h(x+2)=a[(x+2)-0]^2-3\implies h(x+2)=\underline{a(x+2)^2-3} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llll} \underline{h(x+2)}-5\implies [a(x+2)^2-3]~~ - ~~5\implies &\boxed{a(x+2)^2-8=f(x)} \\\\ &a[x-\stackrel{h}{(-2)}]^2\stackrel{k}{-8}=f(x)\\\\ &\stackrel{vertex}{-2~~,~~-8} \end{array}[/tex]
