Answer
$ \displaystyle \frac{(x+1)^{2}}{1}-\frac{(y+1)^{2}}{1}=1$
New foci:
$(-1, -1\pm\sqrt{2})$
New vertices:
$(-1,0)$ and $(-1, -2)$
New center:
$(-1,-1)$
New asymptotes:
$y+1=\pm(x+1)$
Work Step by Step
Shift left by $1 \Rightarrow$ in the equation, replace $x$ with $x+1$
Shift down by $1 \Rightarrow$ in the equation, replace $y$ with $y+1$
New equation:$\displaystyle \quad \frac{(y+1)^{2}}{1}-\frac{(x+1)^{2}}{1}=1$
The hyperbola $\displaystyle \frac{y^{2}}{1}-\frac{x^{2}}{1}=1$ has the form for which:
foci are on the y-axis, $\quad a=b=1$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}\sqrt{1+1}=\sqrt{2}$
Foci: $\quad (0, \pm c)=\quad (0, \pm\sqrt{2})$
Vertices: $\quad (0, \pm a)=\quad (0, \pm 1)$
Asymptotes: $\quad y=\displaystyle \pm\frac{a}{b}x\Rightarrow\quad y=\pm x$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x-1 & \text{... shift left}\\
y'=y-1 & \text{... shift down}
\end{array}\right.$
New foci:
$\quad (0-1,\pm\sqrt{2}-1)\Rightarrow\quad (-1, -1\pm\sqrt{2})$
New vertices:
$\quad (0-1, \pm 1-1)\Rightarrow\quad$ $(-1,0)$ and $(-1, -2)$
New center:
$\quad (0-1,0-1)\quad =(-1,-1)$
New asymptotes:
$\quad$ (apply equation replacements as above)
$y+1=\pm(x+1)$
