let's take a peek at the parabola in the graph, hmmmm the vertex is at 0,0 the origin and we also know that it passes through (3 , -2), ok
[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] ~\dotfill[/tex]
[tex]\stackrel{vertex}{(0~~,~~0)}\qquad y=a(x-0)^2+0~\hfill \textit{we also know that} \begin{cases} x = 3\\ y = -2 \end{cases} \\\\\\ -2=a(3-0)^2+0\implies -2=9a\implies -\cfrac{2}{9}=a \\\\\\ y = -\cfrac{2}{9}(x-0)^2+0\implies \boxed{y=-\cfrac{2}{9}x^2}[/tex]
