The equation in your image is:

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

The instruction below seems to ask for the vertical and horizontal asymptotes of this function. Let’s determine those step by step.

1. Horizontal Asymptote:

• To find the horizontal asymptote, we need to analyze the behavior of the function as $x$ approaches infinity ($x \to \infty$) and negative infinity ($x \to -\infty$).

For large $x$: $y = \frac{2e^x}{e^x - 5} \approx \frac{2e^x}{e^x} = 2$ As $x \to \infty$, the horizontal asymptote is: $y = 2$

For large negative $x$ ($x \to -\infty$): $e^x \to 0, \quad y = \frac{2e^x}{e^x - 5} \to \frac{0}{-5} = 0$ Thus, as $x \to -\infty$, the horizontal asymptote is: $y = 0$

2. Vertical Asymptote:

Vertical asymptotes occur when the denominator equals zero, as long as the numerator is not zero at the same point.

Set the denominator to zero: $e^x - 5 = 0 \quad \Rightarrow \quad e^x = 5$ Taking the natural logarithm of both sides: $x = \ln(5)$ Thus, there is a vertical asymptote at $x = \ln(5)$.

Summary:

• Horizontal asymptotes: $y = 2$ as $x \to \infty$, and $y = 0$ as $x \to -\infty$.
• Vertical asymptote: $x = \ln(5)$.

Do you want further details or have any questions?

Related Questions:

1. How do we graph the function and its asymptotes?
2. What happens to the function near the vertical asymptote $x = \ln(5)$?
3. How do we confirm the horizontal asymptotes using limits?
4. Can you explain the behavior of the function around $x = 0$?
5. How do transformations affect the graph of $y = \frac{e^x}{e^x - 5}$?

Tip:

For rational functions, vertical asymptotes are found where the denominator is zero, and horizontal asymptotes describe end behavior.