To find the vertical and horizontal asymptotes of the function:

$f(x) = \frac{3 \cdot 5^x + 5}{5^x - 25}$

Step 1: Vertical Asymptote

Vertical asymptotes occur where the denominator is zero, as long as the numerator is not also zero at that point. So, we first set the denominator equal to zero and solve for $x$.

$5^x - 25 = 0$

This simplifies to:

$5^x = 25$

Since $25 = 5^2$, we can write:

$5^x = 5^2$

Therefore:

$x = 2$

So, there is a vertical asymptote at $x = 2$.

Step 2: Horizontal Asymptote

To find the horizontal asymptote, we examine the behavior of the function as $x$ approaches $+\infty$ and $-\infty$.

• As $x \to +\infty$, $5^x$ becomes very large. We can approximate the function for large $x$:

$f(x) \approx \frac{3 \cdot 5^x}{5^x} = 3$

Thus, as $x \to +\infty$, $f(x) \to 3$. Therefore, the horizontal asymptote is $y = 3$.

• As $x \to -\infty$, $5^x$ approaches zero. The function becomes:

$f(x) \approx \frac{5}{-25} = -\frac{1}{5}$

Thus, as $x \to -\infty$, $f(x) \to -\frac{1}{5}$. So, there is a horizontal asymptote at $y = -\frac{1}{5}$ as well.

Summary:

• Vertical asymptote: $x = 2$
• Horizontal asymptotes: $y = 3$ and $y = -\frac{1}{5}$

Would you like further details on any part of this process or have any questions?

Here are 5 related questions to expand on the topic:

1. How do you find the slant asymptote of a function?
2. What happens to the function near a vertical asymptote?
3. Can there be more than one vertical asymptote for a function?
4. How do transformations affect the asymptotes of a function?
5. How do logarithmic functions behave asymptotically?

Tip: When finding horizontal asymptotes, consider the dominant terms in the numerator and denominator for large values of $x$.