Answer :
The vertex form of the equation of a parabola with the vertex (h,k):
[tex]y=a(x-h)^2+k[/tex]
The parabola has the vertex at (6,27).
[tex]y=a(x-6)^2+27[/tex]
The y-intercept is (0,-81).
[tex]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[/tex]
The equation of the parabola is:
[tex]y=-3(x-6)^2+27[/tex]
y=0 at the x-intercepts.
[tex]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[/tex]
The x-intercepts are x=3 and x=9.
[tex]y=a(x-h)^2+k[/tex]
The parabola has the vertex at (6,27).
[tex]y=a(x-6)^2+27[/tex]
The y-intercept is (0,-81).
[tex]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[/tex]
The equation of the parabola is:
[tex]y=-3(x-6)^2+27[/tex]
y=0 at the x-intercepts.
[tex]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[/tex]
The x-intercepts are x=3 and x=9.
