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

Aug 7, 2017

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#

$\left(x + 1\right) \left(x + 5\right) = 0$

$\left(x + 1\right) = 0$ $\text{ }$ OR $\text{ }$ $\left(x + 5\right) = 0$

$x = - 1$ 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 $\frac{1}{2}$ of the sum, so the vertex is at

$x = \frac{1}{2} \left(\left(1 -\right) + \left(- 5\right)\right) = \frac{- 6}{2} = - 3$

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

$y = {x}^{2} + 6 x + 5$ at $x = - 3$ is

${\left(- 3\right)}^{2} + 6 \left(- 3\right) + 5 = 9 - 18 + 5 = - 9 + 5 = - 4$

The vertex is $\left(- 3 , - 4\right)$