Answer
See below
Work Step by Step
Given $y^2+6x+4y+16=0$
Rewrite as: $(y^2+4y)+(6x+16)=00\\(y^2+4y+4)+(6x+12)=0\\(y+2)^2=-6(x+2)$
Where $(h,k)=(-2,-2)\\p=-\frac{3}{2}$
Compare the given equation to the standard form of an equation of a parabola.