Answer
Vertex: $(2,-3)$
Focus: $\left(4,-3\right)$
Directrix: $x=0$
See graph
Work Step by Step
We are given the parabola:
$(y+3)^2=8(x-2)$
The standard equation is:
$(y-k)^2=4p(x-h)$
Determine $h,k,p$:
$h=2$
$k=-3$
$4p=8\Rightarrow p=2$
Determine the vertex:
$(h,k)=(2,-3)$
Determine the focus:
$(h+p,k)=\left(2+2,-3\right)=\left(4,-3\right)$
Determine the directrix:
$x=h-p$
$x=2-2$
$x=0$
Determine the two points defining the latus rectum:
$x=4$
$(y+3)^2=4(2)(4-2)$
$(y+3)^2=16$
$y+3=\pm 4$
$y+3=-4\Rightarrow y_1=-7$
$y+3=4\Rightarrow y_2=1$
$\Rightarrow (4,-7),(4,1)$
Plot the points, draw the directrix, and graph the parabola:
