Answer
vertex $(-3,-2)$, focus $(-3,-1)$, directrix $y=-3$
See graph.
Work Step by Step
1. We can identify that $x^2+6x-4y+1=0\Longrightarrow (x+3)^2=4(y+2)$ is a parabola opens upward,
2. Compare with the standard form to get:
vertex $(-3,-2)$, $p=1$, focus $(-3,-1)$, directrix $y=-3$
3. Use symmetry and a few test points such as at $x=0,-1,-2,-3$ to get the graph.
