first off let's notice that hmmm the vertex is above the point (3,-3), so if we "assume" is a vertical parabola, then it'd be opening downwards.... anyhow that said, let's plug in those values.
[tex]\bf ~~~~~~\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}[/tex]
[tex]\bf vertex~~(\stackrel{h}{2}~~,~~\stackrel{k}{1})\qquad \implies \qquad y = a(x-2)^2+1 \\\\\\ \stackrel{\textit{we also know that }}{(3~~,~~-3)} \begin{cases} x = 3\\ y = -3 \end{cases}\qquad \implies -3=a(3-2)^2+1 \\\\\\ -3=a(1)^2+1\implies -3=a+1\implies -4=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = -4(x-2)^2+1~\hfill[/tex]
