To find the holes of the function \( f(x) = \frac{-3x + 15}{5x - 25} \), we need to check for any values of \( x \) that cause both the numerator and the denominator to be zero simultaneously, as these would indicate a removable discontinuity, or a "hole," in the function.

### Step 1: Factor the numerator and the denominator
First, factor both the numerator and the denominator:

\[
f(x) = \frac{-3(x - 5)}{5(x - 5)}
\]

### Step 2: Simplify the expression
Notice that \( (x - 5) \) is a common factor in both the numerator and the denominator, so we can cancel it out:

\[
f(x) = \frac{-3}{5}
\]

However, the function originally had \( x = 5 \) in the denominator, which means that \( x = 5 \) is a point where the function is undefined. The cancellation of \( (x - 5) \) in the simplification does not eliminate the fact that at \( x = 5 \), the function is not defined, resulting in a hole at \( x = 5 \).

### Conclusion
The function \( f(x) \) has a hole at \( x = 5 \).

Would you like more details on any of the steps? 

Here are some related questions:
1. How do you find vertical asymptotes of a rational function?
2. What are the differences between a hole and a vertical asymptote in a function?
3. How would the graph of \( f(x) = \frac{-3x + 15}{5x - 25} \) look?
4. Can holes affect the domain of a function?
5. How do you find the domain and range of rational functions?

**Tip:** Always factor both the numerator and the denominator first when working with rational functions to simplify and identify any potential holes or asymptotes.

Learn how to find holes in the rational function f(x) = (-3x + 15) / (5x - 25). Understand the concept of removable discontinuities, factorization techniques, and how to simplify rational functions. Suitable for advanced high school students.

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