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"?

Answer 1

some mistake here i think

Answer 2

\[(y+3)^2=-8x-1+9\]or
\[(y+3)^2=-8x+8\]

Answer 3

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

Answer 4

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

Answer 5

yw
you also need \(-8x\) on the right hand side, not \(8x\)