taking a peek at the picture, we can see that the U-turn or vertex occurs at the point (2 , -9), and we can also see that the graph passes through the point say (5 , 0).
[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}{(2~~,~-9)}\qquad y=a(x-2)^2-9\qquad \qquad \textit{we know that} \begin{cases} x=5\\ y=0 \end{cases} \\\\\\ 0=a(5-2)^2-9\implies 9=9a\implies \cfrac{9}{9}=a\implies 1=a~\hfill \\\\\\ y=1(x-2)^2-9\implies \boxed{y=(x-2)^2-9}[/tex]
