To find the horizontal asymptotes of the function $h(x) = \frac{4(2^x) + 16}{6 - 2^x}$, we need to evaluate the behavior of $h(x)$ as $x \to \infty$ and $x \to -\infty$.

Step 1: Find the limit as $x \to \infty$

As $x \to \infty$, the term $2^x$ grows exponentially, so we focus on the highest power of $2^x$ in both the numerator and the denominator.

$\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \frac{4(2^x) + 16}{6 - 2^x}$

In the numerator, $4(2^x)$ dominates, and in the denominator, $-2^x$ dominates as $2^x$ becomes very large. So, we approximate the function:

$\lim_{x \to \infty} h(x) \approx \lim_{x \to \infty} \frac{4(2^x)}{-2^x} = \lim_{x \to \infty} \frac{4}{-1} = -4$

Thus, as $x \to \infty$, the horizontal asymptote is:

$y = -4$

Step 2: Find the limit as $x \to -\infty$

As $x \to -\infty$, $2^x \to 0$ because exponential functions decay to zero for negative large values of $x$.

$\lim_{x \to -\infty} h(x) = \lim_{x \to -\infty} \frac{4(2^x) + 16}{6 - 2^x}$

Since $2^x \to 0$, the function simplifies to:

$\lim_{x \to -\infty} h(x) = \frac{4(0) + 16}{6 - 0} = \frac{16}{6} = \frac{8}{3}$

Thus, as $x \to -\infty$, the horizontal asymptote is:

$y = \frac{8}{3}$

Conclusion:

The horizontal asymptotes of the function $h(x)$ are:

• $y = -4$ as $x \to \infty$
• $y = \frac{8}{3}$ as $x \to -\infty$

Would you like further clarification or more details on this?

Here are 5 related questions to deepen your understanding:

1. How do vertical asymptotes differ from horizontal asymptotes?
2. What role does the exponential term play in determining horizontal asymptotes?
3. Can you find the oblique asymptote of a function?
4. How does the behavior of exponential functions differ as $x \to \infty$ and $x \to -\infty$?
5. How can limits be used to determine the asymptotes of rational functions?

Tip: Always compare the growth rates of the terms in the numerator and denominator to identify asymptotic behavior efficiently!