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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.

Find the vertical and horizontal asymptotes (if any) for the following function: y = 5 / (e^x - 2). If they don’t exist, write 'none'.

Learn how to find the vertical and horizontal asymptotes for the function y = 5 / (e^x - 2). This detailed step-by-step solution explains the key concepts of exponential functions and asymptotes. Suitable for Grades 11-12.

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