To find the horizontal asymptotes of the function $y = \frac{3e^{4x}}{1 - e^{4x}}$, we need to analyze the behavior of the function as $x$ approaches infinity ($x \to \infty$) and as $x$ approaches negative infinity ($x \to -\infty$).

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

As $x$ approaches infinity:

• $e^{4x}$ becomes very large.
• Thus, $1 - e^{4x}$ approaches a large negative value.

The function simplifies to:

$y \approx \frac{3e^{4x}}{-e^{4x}} = \frac{3}{-1} = -3$

So, as $x \to \infty$, $y$ approaches $-3$.

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

As $x$ approaches negative infinity:

• $e^{4x}$ approaches zero.

The function simplifies to:

$y \approx \frac{3 \cdot 0}{1 - 0} = 0$

So, as $x \to -\infty$, $y$ approaches $0$.

Conclusion

The horizontal asymptotes are:

• $y = -3$ as $x \to \infty$
• $y = 0$ as $x \to -\infty$

Would you like more details or have any questions?

Here are 5 related questions you might consider:

1. How do you find vertical asymptotes for rational functions?
2. What is the significance of horizontal asymptotes in the context of a graph?
3. How does the behavior of exponential functions affect asymptotes?
4. Can a function have more than two horizontal asymptotes?
5. How do you differentiate between horizontal and slant asymptotes?

Tip: When analyzing horizontal asymptotes, always consider the behavior of the function as $x$ approaches both positive and negative infinity.