Answer
See graph; asymptote as $y=1$
Domain: $(-\infty,\infty)$ and range: $(1,\infty)$
Work Step by Step
Step 1. Start from the original function $f(x)=e^x$ shown as the green curve in the figure.
Step 2. To obtain the new curve of $h(x)=e^{2x}+1$, shrink the original horizontally by a factor of 2, then shift the result vertically 1 unit up to get the red curve as shown.
Step 3. We can identify the new asymptote as $y=1$
Step 4. We can determine the domain as $(-\infty,\infty)$ and range as $(1,\infty)$
