# How do you write the Vertex form equation of the parabola x=y^2+6y+1?

Jul 26, 2018

${\left(y + 3\right)}^{2} = x + 8$

#### Explanation:

$\text{Since y is squared the parabola opens horizontally}$
$\text{and has an equation of the form}$

•color(white)(x)(y-k)^2=4a(x-h)

$\text{where "(h,k)" are the coordinates of the vertex}$

$\text{to obtain this form "color(blue)"complete the square}$

$x = {y}^{2} + 2 \left(3\right) y + 9 - 9 + 1$

$x = {\left(y + 3\right)}^{2} - 8$

${\left(y + 3\right)}^{2} = x + 8$

$\text{with vertex } = \left(- 8 , - 3\right)$
graph{x=y^2+6y+1 [-10, 10, -5, 5]}