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

1 Answer
Jul 26, 2018

(y+3)^2=x+8

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]}