Answer
Vertex: $(-3,-2)$
Focus: $\left(-3,-1\right)$
Directrix: $y=-3$
See graph
Work Step by Step
We are given the parabola:
$x^2+6x-4y+1=0$
Put the equation in standard form:
$(x^2+6x+9)-9-4y+1=0$
$(x+3)^2=4y+8$
$(x+3)^2=4(y+2)$
The standard equation is:
$(x-h)^2=4p(y-k)$
Determine $h,k,p$:
$h=-3$
$k=-2$
$4p=4\Rightarrow p=1$
Determine the vertex:
$(h,k)=(-3,-2)$
Determine the focus:
$(h,k+p)=\left(-3,-2+1\right)=\left(-3,-1\right)$
Determine the directrix:
$y=k-p$
$y=-2-1$
$y=-3$
Determine the two points defining the latus rectum:
$y=-1$
$(x+3)^2=4(-1+2)$
$(x+3)^2=4$
$x+3=\pm 2$
$x+3=-2\Rightarrow x_1=-5$
$x+3=2\Rightarrow x_2=-1$
$\Rightarrow (-5,-1),(-1,-1)$
Plot the points, draw the directrix, and graph the parabola: