now, assuming is a vertical parabola, so the squared variable is the "x", let's do some checking.
[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}})\\\\
-------------------------------\\\\
vertex
\begin{cases}
h=3\\
k=5
\end{cases}\implies y=a(x-3)^2+5
\\\\\\
\textit{we also know that }
\begin{cases}
y=6\\
x=-1
\end{cases}\implies 6=a(-1-3)^2+5
\\\\\\
1=a(-4)^2\implies 1=16a\implies \cfrac{1}{16}=a[/tex]
