Question

How do you find the axis of symmetry and vertex point of the function: `y=x^2+6x-2`?

Answer 1

The method is called completing the square

Explanation:

We have `x^2+6x-2`
in the form `ax^2+bx+c`
where `a=1`, `b= 6`, and `c=-2`

So we divide the `b` value by `2` to get `3`, and square it but keep it as `3^2`, and add it to the `b` value and minus it from the `c` value.

`x^2+6x+3^2-2-3^2`

so `x^2+6x+3^2` is basically `(x+3)^2`. So shorten it that way and we have..

`(x+3)^2` and `-2-(9)`
So that is, `(x+3)^2 - 11`
From this, `x+3=0` and `x=-3`. and `y=-11`.
Line of symmetry is equal to `x=-3`.

Answer 2

Find vertex of `y = x^2 + 6x - 2`

Ans: (-3, -11)

Explanation:

`y = x^2 + 6x - 2.`
x-coordinate of vertex:
`x = (-b/(2a)) = (-6)/(2) = -3`
y-coordinate of vertex:
y = y(-3) = 9 - 18 - 2 = - 11
Vertex (-3, -11)
graph{x^2 + 6x - 2 [-40, 40, -20, 20]}