Answer
asymptotes: $y=\pm(x-1)-2$
foci: $(1\pm\sqrt {2},-2)$
Work Step by Step
Step 1. Rewriting the equation as $(x^2-2x+1)-(y^2+4y+4)=4+1-4$ or $\frac{(x-1)^2}{1}-\frac{(y+2)^2}{1}=1$, we have $a=b=1, c=\sqrt {a^2+b^2}=\sqrt {2}$ centered at $(1,-2)$ with a horizontal transverse axis.
Step 2. We can find the vertices as $(0,-2),(2,-2)$ and asymptotes as $y+2=\pm\frac{b}{a}(x-1)$ or $y=\pm(x-1)-2$
Step 3. We can graph the equation as shown in the figure with foci at $(1\pm\sqrt {2},-2)$
