Recall that the equation of a vertical parabola in standard form is as follows:
[tex]\mleft(x-h\mright)^2=a\mleft(y-k\mright),[/tex]
where (h,k) is the vertex of the parabola.
Adding 4y-4 to the given equation we get:
[tex]\begin{gathered} x^2+8x-4y+4+4y-4=0+4y-4,^{} \\ x^2+8x=4y-4. \end{gathered}[/tex]
Adding 16 to the above equation we get:
[tex]\begin{gathered} x^2+8x+16=4y-4+16, \\ x^2+8x+16=4y+12. \end{gathered}[/tex]
Now, notice that:
[tex]\begin{gathered} x^2+8x+16=x^2+2\cdot4\cdot x+4^2=(x+4)^2, \\ 4y+12=4(y+3). \end{gathered}[/tex]
Therefore we can rewrite the given equation as follows:
[tex](x+4)^2=4(y+3)\text{.}[/tex]
Answer:
[tex](x+4)^2=4(y+3)\text{.}[/tex]
