Answer
A parabola that opens down with
$\left\{\begin{array}{ll}
\text {Vertex:} & (-1,1)\\
\text {Focus:} & (-1,0)\\
\text {Directrix:} & y=2
\end{array}\right.$
Work Step by Step
Group like terms
$(x^{2}+2x)+4y=3$
Complete the square
$(x^{2}+2x+1)+4y=3+1$
$(x+1)^{2}=-4y+4$
$(x+1)^{2}=-4(y-1)$
This is the form of the equation of a parabola,
$ x^{2}=-4py\qquad$ (opens downward, $p=1$)
where $\left\{\begin{array}{ll}
\text {Vertex:} & (0,0)\\
\text {Focus:} & (0,-p)=(0,-1)\\
\text {Directrix:} & y=1
\end{array}\right.$
When $x$ is replaced with $(x+1)$, a shift to the left is made.
When $y$ is replaced with $(y-1)$, a shift upward is made.
$(x,y)\rightarrow(x-1,y+1)$
So the equation $(x+1)^{2}=-4(y-1)$ represents
a parabola that opens down with
$\left\{\begin{array}{ll}
\text {Vertex:} & (0-1,0+1)=(-1,1)\\
\text {Focus:} & (0-1,-1+1)=(-1,0)\\
\text {Directrix:} & y=1+1=2
\end{array}\right.$
