Parabola Standard Form & More

Section 2.1
Question 16

g(x) = x^2 + 2x +1

Note: I WILL NOT GRAPH g(x).

Express g(x) in standard form.

Factor g(x).

g(x) = (x + 1)(x + 1)

g(x) = (x + 1)^2 + 0

Find vertex (h, k).

Let h = -1.

Let k = 0

From the standard form, the vertex is clearly the point (-1, 0).

I now need the axis of symmetry.
The x-coordinate of the vertex is the axis of symmetry. So, it is x = -1.

To find the x-intercepts, set g(x) to 0 and solve for x.

Let g(x) = 0.

0 = (x + 1)^2 + 0

0 = (x + 1)^2

sqrt{0} = sqrt{(x + 1)^2}

0 = x + 1

- 1 = x

The only x-intercept occurs at the point (-1, 0).

You say?

 

correct

 

Cool.