Answer
See graph: foci $(1\pm\sqrt {2},-1)$, asymptotes $y=\pm(x-1)-1$
Work Step by Step
Step 1. Rewrite the given equation as $(x^2-2x+1)-(y^2+2y+1)=1+1-1$ or $(x-1)^2-(y+1)^2=1$. We have $a=1, b=1$. Thus $c=\sqrt {1^2+1^2}=\sqrt {2}$ and the center is at $(1,-1)$ with foci on $y=-1$.
Step 2. See graph; we can locate the foci at $(1\pm\sqrt {2},-1)$
Step 3. The asymptotes can be found as $y+1=\pm\frac{1}{1}(x-1)$ or $y=\pm(x-1)-1$
