The vertex form of a parabola:
[tex]y=a(x-h)^2+k[/tex]
(h,k) - the coordiantes of the vertex
The vertex is (0,36).
[tex]y=a(x-0)^2+36 \\
y=ax^2+36[/tex]
Now plug the coordinates (x,y) of one of the zeros into the equation and solve for a.
[tex]0=a \times 6^2+36 \\
0=36a+36 \\
-36=36a \\
\frac{-36}{36}=a \\
a=-1 \\ \\
\boxed{y=-x^2+36}[/tex]
