Question

How do you find the vertex and the intercepts for `y=x^2+6x+5`?

Answer 1

Here is what I would do:

Explanation:

`y` intecept

When `x = 0`, we get `y = 5`

Vertex and `x` Intercepts

Note that the following method will not work if the equation has no `x` intercepts.

Also note that there are other ways to find the vertex.

`x` Intercepts

x^2+6x+5 = 0#

`(x+1)(x+5) = 0`

`(x+1) = 0` `" "` OR `" "` `(x+5) =0`

`x=-1` or `-5`

the `x` intercepts are `-1` and `-5`.

Vertex

The vertex is midway between the `x` intercepts. So it is at the average of the `x` intercepts.

The average is `1/2` of the sum, so the vertex is at

`x = 1/2((1-)+(-5)) = (-6)/2 = -3`

To find the `y` coordinate of the vertex, plug `-3` in for `x` in the equation

`y = x^2+6x+5` at `x = -3` is

`(-3)^2+6(-3)+5 = 9-18+5 = -9+5=-4`

The vertex is `(-3,-4)`