[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\ (-4,6)\qquad
\begin{cases}
h=-4\\
k=6
\end{cases}\implies y=a(x-(-4))^2+6
\\\\\\
y=a(x+4)^2+6
\\\\\\
\textit{we also know that }
\begin{cases}
x=-3\\
y=14
\end{cases}\implies 14=a(-3+4)^2+6[/tex]
solve for "a"
