Question 87090
{{{x^2+2x-8y+1=0}}} Start with the given equation


{{{x^2+2x-8y=-1}}} Subtract 1 from both sides


{{{2x-8y=-1-x^2}}} Subtract {{{x^2}}} from both sides


{{{8y=-1-x^2-2x}}} Subtract {{{2x}}} from both sides


{{{8y=-x^2-2x-1}}} Rearrange the terms


{{{y=(-x^2-2x-1)/-8}}} Divide both sides by -8


{{{y=(1/8)x^2+(2/8)x+(1)/8}}} Break up the fraction


{{{y=(1/8)x^2+(1/4)x+(1)/8}}} Reduce the middle term


Now lets complete the square to get the quadratic into vertex form


*[invoke completing_the_square "1/8", "1/4", "1/8"]


So the vertex is (-1,0)