The given function is:

$h(x) = e^{-x} - 5$

Domain:

The domain of the function $h(x) = e^{-x} - 5$ refers to the set of all possible input values (values of $x$) for which the function is defined. Since $e^{-x}$ is defined for all real numbers, the domain of this function is:

$\text{Domain} = (-\infty, \infty)$

Range:

The range is the set of all possible output values (values of $h(x)$). Since the exponential function $e^{-x}$ always produces positive values (but never reaches 0), we know:

• $e^{-x} > 0$ for all real $x$,
• Therefore, $e^{-x} - 5 > -5$.

Thus, the range of the function is:

$\text{Range} = (-5, \infty)$

Horizontal Asymptote:

As $x \to \infty$, the value of $e^{-x}$ approaches 0. Therefore, $h(x) = e^{-x} - 5$ approaches:

$h(x) \to 0 - 5 = -5 \quad \text{as} \quad x \to \infty.$

Thus, the horizontal asymptote is:

$y = -5$

Would you like further details on graphing this function?

Here are 5 related questions:

1. What would happen to the graph if the function was $e^{x} - 5$ instead of $e^{-x} - 5$?
2. How does shifting the graph vertically by 5 units affect the domain and range?
3. What is the behavior of the function as $x \to -\infty$?
4. How would you compute the derivative of $h(x)$ and what does it tell us about the function's slope?
5. Can you find the x-intercept of the function?

Tip: When identifying horizontal asymptotes, remember to focus on the behavior of the function as $x \to \infty$ or $x \to -\infty$.