Question

On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 16), has a y-intercept at (0, negative 12), and goes through (2, 0).
What are the x-intercepts of the graph of the function f(x) = x2 + 4x – 12?

(–6, 0), (2,0)
(–2, –16), (0, –12)
(–6, 0), (–2, –16), (2, 0)
(0, –12), (–6, 0), (2, 0)

Answer 1

x^2 + 4x − 12 = 0 factors as (x + 6)(x − 2) = 0, so x = −6 or x = 2.
Thus the x-intercepts are (−6, 0) and (2, 0).