[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}})\\\\
-------------------------------\\\\
y=a(x-(-4))^2-1\implies y=a(x+4)^2-1
\\\\\\
\textit{now, we also know that }
\begin{cases}
y=0\\
x=2
\end{cases}\implies \underline{0}=a(\underline{2}+4)^2-1[/tex]
solve for "a"