Complete the square to find the vertex.
y^2+6y+8x+1=0

I have completed the square and got to this point: (y+3)^2=8x-1+9
I know the "y" cordinate of the vertex is -3 but how do I find the "x"?

Question

Complete the square to find the vertex.
y^2+6y+8x+1=0

I have completed the square and got to this point: (y+3)^2=8x-1+9
I know the "y" cordinate of the vertex is -3 but how do I find the "x"?

anonymous · Answer

some mistake here i think

anonymous · Answer

$$(y+3)^2=-8x-1+9$$or 
$$(y+3)^2=-8x+8$$

anonymous · Answer

rewrite as 
$$(y+3)^2=-8(x-1)$$ and eyeball the vertex as $(1,-3)$

anonymous · Answer

Now I see it! I forgot to combine like terms. Thank you!

anonymous · Answer

yw
you also need $-8x$ on the right hand side, not $8x$