Answer
Foci: $(\pm\sqrt{2},1)$
Vertices: $(\pm 1,1)$
Asymptotes: $y=-x+1$ or $y=x+1$
Work Step by Step
Recall: The hyperbola $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ has the foci $(h\pm c,k)$ where $c^2=a^2+b^2$, the vertices $(h\pm a,k)$, and the asymptotes $y-k=\pm (b/a)(x-h)$.
We have $x^2-y^2+2y=2$.
Rewrite this equation:
$x^2-y^2+2y=2$
$x^2-y^2+2y-1=2-1$
$x^2-(y-1)^2=1$
$\frac{(x-0)^2}{1}-\frac{(y-1)^2}{1}=1$
Then,
$h=0$
$k=1$
$a=1$
$b=1$
$c=\sqrt{a^2+b^2}=\sqrt{1+1}=\sqrt{2}$
$b/a=1/1=1$
Find the foci:
$(h\pm c,k)=(0\pm \sqrt{2},1)=(\pm\sqrt{2},1)$
Find the vertices:
$(h\pm a,k)=(0\pm 1,1)=(\pm 1,1)$
Find the asymptotes:
$y-k=\pm (b/a)(x-h)$
$y-1=\pm 1(x-0)$
$y-1=\pm x$
$y=\pm x+1$
$y=-x+1$ or $y=x+1$
