so hmm let's see if we can get the parabola's equation
if we assume a vertical parabola, "y" in x-terms.... then
[tex]\bf \begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\[/tex]
[tex]\bf \textit{we know that }
\begin{cases}
h=6\\
k=27
\end{cases}\implies y=a(x-6)^2+27
\\\\\\
\textit{we also know the y-intercept }
\begin{cases}
x=-81\\
y=0
\end{cases}\implies 0=a(-81-6)^2+27
\\\\\\
-27=a(-87)^2\implies \cfrac{-27}{(-87)^2}=a\implies -\cfrac{3}{841}=a\\\\
-------------------------------\\\\
\boxed{y=-\cfrac{3}{841}(x-6)^2+27}[/tex]
now, to get the x-intercepts, we simply set y = 0
[tex]\bf 0=-\cfrac{3}{841}(x-6)^2+27\implies -27=-\cfrac{3}{841}(x-6)^2
\\\\\\
\cfrac{-27\cdot 841}{-3}=(x-6)^2\implies 7569=(x-6)^2\implies \pm\sqrt{7569}=x-6
\\\\\\
\pm 87=x-6\implies \pm 87+6=x\implies x=
\begin{cases}
93\\
-81
\end{cases}[/tex]
and... since the y = 0, then (93, 0) and (-81, 0)
