The vertex of a parabola is at (12,-4) and its axis of symmetry is vertical. One of the x-intercepts is at (18,0). What is the x-coordinate of the other x-intercept?
You can use vertex form a(x - h)^2 + k, so that the vertex of the parabola is at (h, k) = (12, -4).
a(x - 12)^2 - 4 = y
Plug in 18
a(18 - 12)^2 - 4 = 0
36a = 4
a = 1/9
(x - 12)^2 - 36 = 9y
If y = 0
x - 12 = +-6
The other x - intercept is (6, 0)
