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

##### 1 Answer

Jul 26, 2018

#### Explanation:

#"Since y is squared the parabola opens horizontally"#

#"and has an equation of the form"#

#•color(white)(x)(y-k)^2=4a(x-h)#

#"where "(h,k)" are the coordinates of the vertex"#

#"to obtain this form "color(blue)"complete the square"#

#x=y^2+2(3)y+9-9+1#

#x=(y+3)^2-8#

#(y+3)^2=x+8#

#"with vertex "=(-8,-3)#

graph{x=y^2+6y+1 [-10, 10, -5, 5]}