# How to find the vertex of the parabola y^2+2y-2x+5=0?

Sep 11, 2015

The vertex is at $\left(x , y\right) = \left(2 , - 1\right)$

#### Explanation:

${y}^{2} + 2 y - 2 x + 5 = 0$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} x = \frac{{y}^{2} + 2 y + 5}{2}$

Note that this is a parabola with a horizontal axis of symmetry

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

Convert the equation into this form:
$x = \frac{{y}^{2} + 2 y + 5}{2}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{{y}^{2} + 2 y + 1}{2} + \frac{4}{2}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{2} {\left(y + 1\right)}^{2} + 2$

$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{2} {\left(y - \left(- 1\right)\right)}^{2} + 2$

which is the vertex form with $\left(x , y\right) = \left(2 , - 1\right)$
graph{y^2+2y-2x+5=0 [-3.57, 8.914, -4.64, 1.6]}