Answer:
[tex](1,-3)[/tex]
Step-by-step explanation:
Complete the squares
[tex]y^2+6y+8x+1=0\\\\y^2+6y=-8x-1\\\\y^2+6y+9=-8x-1+9\\\\(y+3)^2=-8x+8\\\\(y+3)^2=-8(x-1)[/tex]
We now compare to the equation [tex](y-k)^2=2p(x-h)[/tex] where [tex](h,k)[/tex] is the vertex of the parabola. Hence, the vertex of the parabola is [tex](1,-3)[/tex].
