Answer
The graph of $y=\left(\frac{1}{2}\right)^{x+1}$ involves a $1$-unit shift to the left of the parent function $y=\left(\frac{1}{2}\right)^{x}$
The given function has:
domain: $(-\infty, +\infty)$;
range: $y\gt 0$; and
horizontal asymptote: $y=0$
Work Step by Step
Recall:
The graph of the function $y=f(x+h)$ involves a horizontal shift ($h$ units to the left when $h \gt0$, $|h|$ units to the right when $h\lt0$) of the parent function $y=f(x)$.
The given function has $f(x)=\left(\frac{1}{2}\right)^x$ as its parent function and can be written as $y=f(x+1)$.
With $h=1$, the graph of the given function involves a $1$-unit shift to the left of the parent function $f(x)=\left(\frac{1}{2}\right)^x$.
Recall:
The function is $y=a^x$ has:
(1) domain: $(-\infty, +\infty)$;
(2) range: $y\gt 0$; and
(3) horizontal asymptote: $y=0$
With a horizontal shift of the parent function $f(x)=\left(\frac{1}{2}\right)^x$ of $1$ units upward, the function $y=\left(\frac{1}{2}\right)^{x+1}$ has:
(1) domain: $(-\infty, +\infty)$;
(2) range: $y\gt 0$; and
(3) horizontal asymptote: $y=0$
