# How do you find the VERTEX of a parabola y= x^2 + 6x + 5?

Jul 17, 2015

Complete the square to get the equation into vertex form:

$y = {\left(x - \left(- 3\right)\right)}^{2} + \left(- 4\right)$

Then the vertex can be read as $\left(- 3 , - 4\right)$

#### Explanation:

Given any quadratic: $y = a {x}^{2} + b x + c$, we can complete the square to get:

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

In our example, $a = 1$, $b = 6$ and $c = 5$, so we find:

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

$= {\left(x + 3\right)}^{2} - 4$

So $y = {\left(x + 3\right)}^{2} - 4$

Strictly speaking, vertex form is $y = a {\left(x - h\right)}^{2} + k$, from which you can read off the vertex $\left(h , k\right)$. So let's replace the $\left(x + 3\right)$ with $\left(x - \left(- 3\right)\right)$ and the $- 4$ with $+ \left(- 4\right)$ to get:

$y = {\left(x - \left(- 3\right)\right)}^{2} + \left(- 4\right)$