Question

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:

x-intercepts are (-6,0 and (2,0)

First option is correct.

Step-by-step explanation:

The given equation is $y=x^2+4x-12$

For x-intercept, y=f(x)=0. Thus, we have

$0=x^2+4x-12$

We can split the middle term as 4x = 6x-2x

Thus, the equation becomes

$x^2+6x-2x-12=0$

Now, we make group

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

Take GCF from each of the group

$x(x+6)-2(x+6)=0$

Factored out the common term

$(x+6)(x-2)=0$

Apply the zero product property,

$x+6=0,x-2=0$

Solve for x

$x-6,x=2$

Thus, x-intercepts are (-6,0 and (2,0)

First option is correct.

Answer 2

Answer:

A

Step-by-step explanation: