Check the picture below.
so, we know the parabola has a vertex at (0 , 50), we also know it passes through (60 , 0), so then
[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}[/tex]
[tex]\stackrel{vertex}{(0~~,~~50)}\implies y=a(x-0)^2+50\implies y = ax^2+50 \\\\\\ \begin{cases} x =60\\ y = 0 \end{cases}\implies 0=a6^2+50\implies -50=3600a\implies -\cfrac{50}{3600}=a \\\\\\ -\cfrac{1}{72}=a~\hfill therefore~\hfill \boxed{y=-\cfrac{1}{72}x^2+50}[/tex]
so then, what's "y" when x = 25?
[tex]y=-\cfrac{1}{72}25^2+50\implies y = -\cfrac{625}{72}+50\implies y = \cfrac{2975}{72}\implies y\approx 41.32[/tex]
