Respuesta :
To obtain the vertex of the parabola, we are going to re-write the given parabola equation into its vertex form.
The vertex form of a parabola is given as:
[tex]\begin{gathered} y=a(x-h)^2+k \\ \text{where (h,k) is the vertex} \end{gathered}[/tex]Thus, by completing the square of the given parabola equation, we have:
[tex]\begin{gathered} y=x^2+4x-3 \\ y=x^2+4x+4-4-3 \\ y=(x+2)^2-7 \end{gathered}[/tex]Comapring this equation with the vertex form of a parabola;
Hence, the vertex of the parabola is:
[tex](h,k)=(-2,-7)[/tex]The focus of a parabola is at the point;
[tex]Focus=(h,k+\frac{1}{4a})[/tex]The parabola obtained opens up. An alternative equation for a parabola that opens up is:
[tex]y-k=4a(x-h)^2[/tex][tex]\begin{gathered} \text{ Rewriting }y=(x+2)^2-7\text{ to fit this form leads to} \\ y+7=4a(x+2)^2 \end{gathered}[/tex]We must find the value of a that makes the equation true at any point (x,y).
Suppose x=1;
[tex]\begin{gathered} y=x^2+4x-3 \\ y=1^2+4(1)-3 \\ y=1+4-3 \\ y=2 \end{gathered}[/tex][tex]\begin{gathered} y-k=4a(x-h)^2 \\ 2-(-7)=4a(1-(-2))^2 \\ 2+7=4a(1+2)^2 \\ 9=4a\times3^2 \\ 9=36a \\ a=\frac{9}{36} \\ a=\frac{1}{4} \end{gathered}[/tex]Hence, the focus of the parabola is:
[tex]\begin{gathered} F=(h,k+\frac{1}{4a}) \\ F=(-2,\text{ -7+}\frac{1}{4}) \\ F=(-2,-\frac{27}{4}) \end{gathered}[/tex]The directrix is:
[tex]y=-\frac{29}{4}[/tex]