so hmmmm let's assume is a vertical parabola, so, the squared variable will then be the "x".
[tex]\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
\begin{cases}
h=5\\
k=-3
\end{cases}\implies y=a(x-\stackrel{h}{5})^2\stackrel{k}{-3}
\\\\\\
\textit{now, we also know that }
\begin{cases}
x=6\\
y=1
\end{cases}\implies 6=a(1-5)^2-3
\\\\\\
9=a(-4)^2\implies 9=16a\implies \cfrac{9}{16}=a\qquad thus\quad \boxed{y=\cfrac{9}{16}(x-5)-3}[/tex]
