Question

The graph of the function f(x) = x2 + 8x + 12 is shown. Which statements describe the graph? Check all that apply.

Answer 1

The axis of symmetry is x = –4.
The domain is all real numbers.
The x-intercepts are at (–6, 0) and (–2, 0).

Answer 2

Answer:

The correct options are 2, 3 and 6.

Step-by-step explanation:

The given function is

$f(x)=x^2+8x+12$

It is an upward parabola because the leading coefficient is positive.

The vertex of an upward parabola is the minimum value, therefore option 1 is incorrect.

Write the above function in vertex form.

$f(x)=(x^2+8x)+12$

If a polynomial is $g(x)=x^2+bx$ then we add $(\frac{b}{2})^2$, to make it perfect square.

Here b=8 so we have to add $4^2$.

$f(x)=(x^2+8x+4^2)+12-4^2$

$f(x)=(x+4)^2-4$              ... (1)

the vertex form of a parabola is

$g(x)=a(x+h)^2+k$              ... (2)

Where (h,k) is vertex, axis of symmetry is x=h, domain all real numbers, range y>k. Function is decreasing over (–∞, h) and the function is increasing over (h,∞).

From (1) and (2), we get

$a=1,h=-4,k=-4$

The vertex of the parabola is (-4,-4), axis of symmetry is x=-4, domain all real numbers, range y>-4. Function is decreasing over (–∞, -4) and the function is increasing over (-4,∞).

Substitute f(x)=0 in the given function, to find the x-intercepts.

$0=x^2+8x+12$

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

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

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

$x=-2,-6$

Therefore the  x-intercepts are at (–6, 0) and (–2, 0).

Thus the correct options are 2, 3 and 6.