hmmm the y-intercept is at -2? what does that mean? well, is where the graph "intercepts" or touches the y-axis, and when that happens, x = 0, so the point is really ( 0 , -2 ).
and we know where the vertex is at. Let's assume a vertical parabola, in which case the squared variable is 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}})\\\\
-------------------------------\\\\
vertex
\begin{cases}
h=1\\
k=-3
\end{cases}\implies y=a(x-1)^2-3
\\\\\\
\textit{now, we also know that }
\begin{cases}
x=0\\
y=-2
\end{cases}\implies -2=a(0-1)^2-3
\\\\\\
1=a(-1)^2\implies \boxed{1=a}\qquad thus\implies
\begin{cases}
y=1(x-1)^2-3\\
\textit{or just}\\
y=(x-1)^2-3
\end{cases} [/tex]
