Answer
vertex $(-1,-1)$, focus $(-\frac{3}{4},-1)$, directrix $x=-\frac{5}{4}$
See graph.
Work Step by Step
1. We can identify that $y^2+2y-x=0\Longrightarrow (y+1)^2=(x+1)$ is a parabola opens to the right,
2. Compare with the standard form to get:
$p=\frac{1}{4}$, vertex $(-1,-1)$, focus $(-\frac{3}{4},-1)$, directrix $x=-\frac{5}{4}$
3. Use symmetry and a few test points such as at $x=-1,0,1,2$ to get the graph.
