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