[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-{{ h}})^2+{{ k}}\qquad
\begin{cases}
h=0\\
k=0
\end{cases}\implies y=a(x-0)^2+0
\\\\\\
y=ax^2\qquad
\begin{cases}
x=2\\
y=-2
\end{cases}\implies -2=a2^2[/tex]
solve for "a", to get the coefficient, and then plug it back in the equation.
