an x-intercept is namely a "solution" or "zero" or "root" often called, and when that happens, y = 0, just like with any other x-intercept.
[tex]\bf ~~~~~~\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~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
\begin{cases}
h=3\\
k=-2
\end{cases}\implies y=a(x-3)^2-2
\\\\\\
\textit{we also know that }
\begin{cases}
x=0\\
y=7
\end{cases}\implies 7=a(0-3)^2-2
\\\\\\
9=9a\implies \cfrac{9}{9}=a\implies 1=a\qquad therefore\qquad \boxed{y=(x-3)^2-2}[/tex]
so what is its x-intercept anyway?
[tex]\bf \stackrel{y}{0}=(x-3)^2-2\implies 2=(x-3)^2\implies \pm\sqrt{2}=x-3
\\\\\\
\pm\sqrt{2}+3=x\qquad therefore\qquad (\pm\sqrt{2}+3~~~~,~~~~0)[/tex]
