Answer
This is an ellipse, with the major axis on the line $x=-1$:
the center is $(-1,1)$
the foci are $(-1,1\pm\sqrt{3})$
the vertices are $(-1,3)$ and $(-1, -1)$.
Work Step by Step
We group like terms
$(4x^{2}+8x)+(y^{2}-2y)=-1 $
$4(x^{2}+2x)+(y^{2}-2y)=-1 \quad $
We complete the squares
$4(x^{2}+2x+1)+(y^{2}-2y+1)=-1+4(1)+1 $
$4(x+1)^{2}+(y-1)^{2}=4 \quad $
We divide with
$(x+1)^{2}+\displaystyle \frac{(y-1)^{2}}{4}=1$
This has the form $\quad \displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1 \quad (a\gt b)$
Foci on the y-axis: $\quad a=2,\quad b=1$
Center-to-focus distance: $c=\sqrt{a^{2}-b^{2}}=\sqrt{2-1}=1$
Foci: $(0, \pm 1)$
Vertices: $(0, \pm 2)$
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.
$(x,y)\rightarrow(x-1,y+1)$
(A horizontal shift $\Rightarrow$ the major axis moves left to the line $x=-1$.)
The center shifts to $(0-1,0+1)=(-1,1)$
The foci shift to $(-1,1\pm\sqrt{3})$
The vertices shift to $(0-1, 1\pm 2)\Rightarrow(-1,3)$ and $(-1, -1)$
