Question

Find the x-intercept of the parabola with vertex (6,27) and y-intercept (0,-81)

Answer 1

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 \\ (-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 \\ (-27)/(-3)=(x-6)^2 \\ 9=(x-6)^2 \\ √(9)=√((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.