To determine the hole in the function $f(x) = \frac{x+13}{x^2 - 169}$, we follow these steps:

Step 1: Factorize the denominator

The denominator can be factorized using the difference of squares: $x^2 - 169 = (x - 13)(x + 13)$

Step 2: Simplify the function

The function becomes: $f(x) = \frac{x+13}{(x-13)(x+13)}$

Cancel $x+13$ in the numerator and denominator, but this cancellation introduces a hole because the function is undefined where $x+13 = 0$. This leaves: $f(x) = \frac{1}{x - 13}, \quad \text{with a hole at } x = -13.$

Step 3: Confirm the hole

The hole occurs at $x = -13$ because:

1. The original denominator becomes 0 at $x = -13$, making the function undefined.
2. After simplification, the numerator and denominator both contain $x+13$, indicating a removable discontinuity.

Final Answer:

The hole is at $x = -13$. Let me know if you need a detailed graph or further clarifications!

Related Questions:

1. How do you distinguish between a hole and a vertical asymptote in a rational function?
2. What happens to the function's graph near the hole?
3. Can a rational function have multiple holes?
4. What is the domain of this function after simplifying?
5. How do you find the horizontal asymptote of this function?

Tip:

Always check for removable discontinuities by factoring the numerator and denominator fully!