The given function is:

$y = \frac{5}{e^x - 2}$

Let's find both the vertical and horizontal asymptotes.

1. Finding the Vertical Asymptotes:

A vertical asymptote occurs when the denominator of the function equals zero (as long as the numerator is non-zero).

For this function, we need to solve: $e^x - 2 = 0$ $e^x = 2$ $x = \ln(2)$

So, there is a vertical asymptote at: $x = \ln(2)$

2. Finding the Horizontal Asymptotes:

To find the horizontal asymptote, we examine the behavior of $y$ as $x \to \infty$ and $x \to -\infty$:

• As $x \to \infty$, $e^x$ grows exponentially, so $e^x - 2 \approx e^x$. Therefore: $y = \frac{5}{e^x - 2} \approx \frac{5}{e^x} \to 0$
• As $x \to -\infty$, $e^x \to 0$, hence $e^x - 2 \to -2$. Therefore: $y = \frac{5}{e^x - 2} \approx \frac{5}{-2} = -\frac{5}{2}$

So, the horizontal asymptotes are: $y = 0 \quad \text{and} \quad y = -\frac{5}{2}$

Answer:

• Vertical Asymptote: $x = \ln(2)$
• Horizontal Asymptotes: $y = 0$ and $y = -\frac{5}{2}$

Would you like more details or have any questions?

Related Questions:

1. How do you determine asymptotes for rational functions in general?
2. What happens to the asymptotes if the numerator was a function involving $e^x$?
3. How do vertical asymptotes differ from holes in a function's graph?
4. Can a function have multiple vertical asymptotes? If so, under what conditions?
5. How can you confirm the horizontal asymptotes using limits?

Tip: Remember that horizontal asymptotes describe end behavior, while vertical asymptotes describe undefined points within the domain.