Question

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

Answer 1

`(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]}