Answer
Vertex: $(-4,-2)$
Focus: $(-4,2)$
Directrix: $y=-6$
See graph
Work Step by Step
We are given the parabola:
$(x+4)^2=16(y+2)$
The standard equation is:
$(x-h)^2=4p(y-k)$
Determine $h,k,p$:
$h=-4$
$k=-2$
$4p=16\Rightarrow p=4$
Determine the vertex:
$(h,k)=(-4,-2)$
Determine the focus:
$(h,k+p)=(-4,-2+4)=(-4,2)$
Determine the directrix:
$y=k-p$
$y=-2-4$
$y=-6$
Determine the two points defining the latus rectum:
$y=2$
$(x+4)^2=16(2+2)$
$(x+4)^2=64$
$x+4=\pm 8$
$x+4=-8\Rightarrow x_1=-12$
$x+4=8\Rightarrow x_2=4$
$\Rightarrow (-12,2),(4,2)$
Plot the points, draw the directrix, and graph the parabola:
