Answer
please see graph (blue curve),
asymptote of $h$:$\quad\quad y=-1$
domain of $h=(-\infty,\infty)$
range of $h=(-1,\infty)$
Work Step by Step
Graph $ f(x)=2^{x}\qquad$ (red, dashed)
by plotting the points from the table and connecting with a smooth curve.
$h(x)=2^{x+1}-1=f(x+1)-1$
so the graph of $h(x)$ (blue) is obtained by
shifting the graph of f(x) (red) left by one unit, $[2^{x+1}]$, (dashed green)
and then down by one unit, $[2^{x+1}-1]$, (blue)
Reading the graph,
asymptote of $h$:$\quad\quad y=-1$
domain of $h=(-\infty,\infty)$
range of $h=(-1,\infty)$