# How do you find the vertex of y=x^2 + 4x + 2?

Jan 26, 2016

Convert the given equation into vertex form to find:
vertex at $\left(- 2 , - 2\right)$

#### Explanation:

The general vertex form for a parabola is
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with its vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 4 x + 2$
Completing the square
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 4 x \textcolor{g r e e n}{+ 4} + 2 \textcolor{g r e e n}{- 4}$

$\textcolor{w h i t e}{\text{XXX}} y = {\left(x + 2\right)}^{2} - 2$

$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - \left(\textcolor{red}{- 2}\right)\right)}^{2} + \left(\textcolor{b l u e}{- 2}\right)$

which is in vertex form with the vertex at $\left(\textcolor{red}{- 2} , \textcolor{b l u e}{- 2}\right)$

Just to help verify this result, here is the graph of $y = {x}^{2} + 4 x + 2$:
graph{x^2+4x+2 [-8.702, 2.397, -3.11, 2.437]}