We are given the function:

$y = \frac{11 - e^x}{7 + e^x}$

1. Determine the Domain of the Function:

The domain of a rational function is all real numbers except where the denominator is zero. The denominator here is:

$7 + e^x$

Since $e^x > 0$ for all real $x$, the expression $7 + e^x$ will always be positive and never equal zero. Therefore, there are no restrictions on the domain from the denominator.

Domain: The domain is all real numbers $x \in (-\infty, \infty)$.

2. Find the Asymptotes Using Limits:

We need to find both horizontal asymptotes by taking limits as $x \to \infty$ and $x \to -\infty$.

a. Limit as $x \to \infty$:

As $x \to \infty$, $e^x$ grows exponentially large, so we can approximate the function by considering the behavior of the leading terms:

$\lim_{x \to \infty} \frac{11 - e^x}{7 + e^x} = \lim_{x \to \infty} \frac{-e^x}{e^x} = -1$

Thus, there is a horizontal asymptote at $y = -1$ as $x \to \infty$.

b. Limit as $x \to -\infty$:

As $x \to -\infty$, $e^x \to 0$, so the function simplifies to:

$\lim_{x \to -\infty} \frac{11 - e^x}{7 + e^x} = \frac{11 - 0}{7 + 0} = \frac{11}{7}$

Thus, there is a horizontal asymptote at $y = \frac{11}{7}$ as $x \to -\infty$.

Summary:

• Domain: $(-\infty, \infty)$
• Horizontal Asymptote as $x \to \infty$: $y = -1$
• Horizontal Asymptote as $x \to -\infty$: $y = \frac{11}{7}$

Would you like further details or have any other questions?

Here are some related questions:

1. How does the behavior of $e^x$ influence the function at different extremes?
2. What would the graph of this function look like based on the asymptotes?
3. Can this function have any vertical asymptotes? Why or why not?
4. How would you find the intercepts of this function?
5. How would the function change if you replaced $e^x$ with $e^{-x}$?

Tip: When analyzing rational functions, always check both the numerator and denominator to understand the asymptotic behavior.