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

To find the x-intercepts of the graph of the function f(x) = x^2 + 4x - 12, use the quadratic formula to solve for x.

Step-by-step explanation:

To find the x-intercepts of the graph of the function f(x) = x2 + 4x - 12, we need to set the function equal to zero and solve for x. So:

x2 + 4x - 12 = 0

Now, we can use the quadratic formula to find the values of x. The quadratic formula is:

x = (-b ± √(b2 - 4ac)) / (2a)

For our quadratic equation, the values of a, b, and c are:

a = 1, b = 4, c = -12

Plugging in these values into the quadratic formula gives us:

x = (-4 ± √(42 - 4(1)(-12))) / (2(1))

x = (-4 ± √(16 + 48)) / 2

x = (-4 ± √64) / 2

x = (-4 ± 8) / 2

So the x-intercepts of the graph of the function f(x) = x2 + 4x - 12 are x = -6 and x = 2.