Answer
$ \displaystyle \frac{(x+2)^{2}}{16}-\frac{(y+1)^{2}}{9}=1$
New foci:$\quad (3, -1)$ and $(-7,-1)$
New vertices:$\quad (2, -1)$ and $(-6,-1)$
New center:$\quad (-2,-1)$
New asymptotes:$\displaystyle \quad y-1=\pm\frac{3}{4}(x-2)$
Work Step by Step
Shift left by $2 \Rightarrow$ in the equation, replace $x$ with $x+2$
Shift down by $1 \Rightarrow$ in the equation, replace $y$ with $y+1$
New equation:$\displaystyle \quad \frac{(x+2)^{2}}{16}-\frac{(y+1)^{2}}{9}=1$
The hyperbola $\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ has the form for which:
foci are on the x-axis$, \quad a=4, b=2,$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{16+9}=5$
Foci: $\quad (\pm c, 0)= \quad (\pm 5, 0)$
Vertices: $\quad (\pm 4, 0)$
Asymptotes:$\quad y=\displaystyle \pm\frac{b}{a}x\Rightarrow \quad y=\displaystyle \pm\frac{3}{4}x$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x-2 & \text{... shift left}\\
y'=y-1 & \text{... shift down}
\end{array}\right.$
New foci:$\quad (\pm 5-2, 0-1)\Rightarrow\quad (3, -1)$ and $(-7,-1)$
New vertices:$\quad (\pm 4-2, 0-1)\Rightarrow\quad (2, -1)$ and $(-6,-1)$
New center:$\quad (0-2,0-1)\quad =(-2,-1)$
New asymptotes:$\displaystyle \quad y'=\pm\frac{3}{4}x'$
$y-1=\displaystyle \pm\frac{3}{4}(x-2)$
