Given,
The equation of the parabola is y^2+6y+8x+1=0
Required:
The vertex of the parabola.
The equation of the parabola is taken as:
[tex]\begin{gathered} y^2+6y+8x+1=0 \\ y^2+6y+1=-8x \\ y^2+6y+9-9+1=-8x \\ (y+3)^2-9+1=-8x \\ (y+3)^2-8=-8x \\ (y+3)^2=8-8x \\ -8x=(y+3)^2-8 \\ x=\frac{-(y+3)^2}{8}+1 \end{gathered}[/tex]
The standard form of the equation is,
[tex]x=a(y-k)^2+h[/tex]
Here, h and k are the vertex of the parabola.
On comparing the standard form with given vertex form of the parabola.
[tex](h,k)=(1,-3)[/tex]
Hence, the vertex of the parabola is (1, -3).
