The vertex form of the equation of a parabola with the vertex (h,k):
\(y=a(x-h)^2+k\)
The parabola has the vertex at (6,27).
\(y=a(x-6)^2+27\)
The y-intercept is (0,-81).
\(x=0, \ y=-81 \\
\Downarrow \\
-81=a(0-6)^2+27 \\
-81=a(-6)^2+27 \\
-81=36a+27 \\
-81-27=36a \\
-108=36a \\
\frac{-108}{36}=a \\
a=-3\)
The equation of the parabola is:
\(y=-3(x-6)^2+27\)
y=0 at the x-intercepts.
\(0=-3(x-6)^2+27 \\
-27=-3(x-6)^2 \\
\frac{-27}{-3}=(x-6)^2 \\
9=(x-6)^2 \\
\sqrt{9}=\sqrt{(x-6)^2} \\
3=|x-6| \\
x-6=3 \ \lor \ x-6=-3 \\
x=3+6 \ \lor \ x=-3+6 \\
x=9 \ \lor \ x=3\)
The x-intercepts are x=3 and x=9.
