Question

How do you find the standard form given `y^2+8y-4x+16=0`?

Answer 1

`x=1/4y^2+2y+4`

Explanation:

This is the equation of a parabola, whose standard form is

`x=ay^2+by+c`

Here we have `y^2+8y-4x+16=0`

Therefore `4x=y^2+8y+16`

or `x=1/4y^2+2y+4`
graph{y^2+8y-4x+16=0 [-7.05, 32.95, -13.84, 6.16]}