Question

What are the coordinates of the vertex for f(x) = x2 + 6x + 13?

(4,4)
(-4,4)
(3,4)
(-3, 4)

Answer 1

(-3,4)

Explanation:

The axis of symmetry, or X coordinate for a parabolic function is
`(-b)/(2a)`

By plugging in 6 for b, and 1 for a, you have -3.

By then plugging back into the function you can get the Y coordinate of the vertex.

`f(x)=(-3)^(2)+6(-3)+13`

Or 4.

The answer is (-3,4)

Answer 2

(-3, 4)

EXPLANATION

The given function is

`f(x) = {x}^(2) + 6x + 13`

We complete the square to obtain the vertex form:

We add and subtract the square of half the coefficient of x.

`f(x) = {x}^(2) + 6x + {3}^(2) + 13 - {3}^(2)`

The first three terms form a perfect square trinomial.

`f(x) = {(x + 3)}^(2) + 13 - 9`

`f(x) = {(x + 3)}^(2) + 4`

Comparing this to the vertex form;

`f(x) = {(x - h)}^(2) + k`

h=-3 and k=4