Answer
Vertex: $(-1,-1)$
Focus: $\left(-0.75,-1\right)$
Directrix: $x=-1.25$
See graph
Work Step by Step
We are given the parabola:
$y^2+2y-x=0$
Put the equation in standard form:
$(y^2+2y+1)-1-x=0$
$(y+1)^2=x+1$
The standard equation is:
$(y-k)^2=4p(x-h)$
Determine $h,k,p$:
$h=-1$
$k=-1$
$4p=1\Rightarrow p=0.25$
Determine the vertex:
$(h,k)=(-1,-1)$
Determine the focus:
$(h+p,k)=\left(-1+0.25,-1\right)=\left(-0.75,-1\right)$
Determine the directrix:
$x=h-p$
$x=-1-0.25$
$x=-1.25$
Determine the two points defining the latus rectum:
$x=-0.75$
$(y+1)^2=-0.75+1$
$(y+1)^2=0.25$
$y+1=\pm 0.5$
$y+1=-0.5\Rightarrow y_1=-1.5$
$y+1=0.5\Rightarrow y_2=-0.5$
$\Rightarrow (-0.75,-1.5),(-0.75,-0.5)$
Plot the points, draw the directrix, and graph the parabola:
