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

Answer:

(-6,0) and (2,0)

Step-by-step explanation:

The function with equation $f(x)=x^2 +4x-12$ has a graph that is a parabola, which

• opens up;
• goes trough (-6,0) and (2,0);
• has vertex (-2,-16);
• has y-intercept (0,-12);

x-intercepts are points where the graph intersects the x-axis. These points always have y-coordinate 0. Parabola can have at most 2 x-intecepts. As you can see, points (-6,0) and (2,0) have y-coordinates 0 and parabola passes through these points, so x-intercepts are (-6,0) and (2,0).

Answer 2

Answer:

The x-intercept to the graph of the given function f(x) is:

              (-6,0) and (2,0)

Step-by-step explanation:

We know that the x-intercept of a function are the points where the function is equal to zero.

i.e. the x-intercepts are of the type:

                       (x,0)

Here we have a function f(x) as:

            $f(x)=x^2+4x-12$

The x-intercept is calculated as follows:

            $x^2+4x-12=0\\\\x^2+6x-2x-12=0\\\\x(x+6)-2(x+6)=0\\\\(x-2)(x+6)=0\\\\x=2\ and\ x=-6$

Hence, the x-intercepts is given by:

             $(2,0)\ \text{and}\ (-6,0)$