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

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Answer 1 · 2017-02-13T12:44:07

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

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]}

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