Answer
center $(-2,1)$, foci $(-2-\sqrt 3,1),(-2+\sqrt 3,1)$, vertices $(-4,1),(0,1)$
See graph.
Work Step by Step
1. $x^2+4x+4y^2-8y+4=0\Longrightarrow (x+2)^2+4(y-1)^2=4$ or $\frac{(x+2)^2}{4}+\frac{(y-1)^2}{1}=1$,
2. we can find center $(-2,1)$ with a horizontal major axis, $a=2, b=1$, $c=\sqrt {a^2-b^2}=\sqrt {4-1}=\sqrt 3$,
3. thus foci $(-2\pm\sqrt 3,1)$ or $(-2-\sqrt 3,1),(-2+\sqrt 3,1)$, vertices $(-2\pm2,1)$ or $(-4,1),(0,1)$
4. See graph.
