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To solve this problem, let’s analyze the conditions and the answer choices:

### Given Conditions
1. **Vertical Asymptotes**: \( x = -7 \) and \( x = 3 \).
   - Vertical asymptotes typically occur where the denominator of a rational function is zero.
   - For asymptotes at \( x = -7 \) and \( x = 3 \), the denominator should include factors \( (x + 7) \) and \( (x - 3) \).

2. **Horizontal Asymptote**: \( y = \frac{6}{5} \).
   - Horizontal asymptotes in rational functions are determined by the degrees of the numerator and the denominator.
   - Since the horizontal asymptote is \( \frac{6}{5} \), the degrees of the numerator and denominator should be the same. The leading coefficients in the numerator and denominator should form the fraction \( \frac{6}{5} \).

3. **x-intercept**: \( (8, 0) \).
   - The x-intercept occurs where the function is zero, meaning the numerator must be zero at \( x = 8 \).
   - So, the numerator should have a factor of \( (x - 8) \).

### Analyzing Answer Choices
Now, let’s evaluate each option:

#### Option A: 
\[ f(x) = \frac{x - 8}{x^2 + 4x - 21} \]
   - Denominator factors as \( (x + 7)(x - 3) \), so this matches the vertical asymptotes condition.
   - The degrees of numerator and denominator are not the same (numerator is degree 1, denominator is degree 2), so it doesn’t have a horizontal asymptote \( y = \frac{6}{5} \).
   - This does not satisfy all conditions.

#### Option B: 
\[ f(x) = \frac{x^2 - 9x + 8}{x^2 + 4x - 21} \]
   - Denominator factors as \( (x + 7)(x - 3) \), matching the vertical asymptotes condition.
   - Both numerator and denominator are degree 2, which is correct for a horizontal asymptote.
   - However, for a horizontal asymptote \( y = \frac{6}{5} \), the leading coefficient ratio should be \( \frac{6}{5} \), not \( 1 \).
   - This does not satisfy all conditions.

#### Option C:
\[ f(x) = \frac{6x^2 - 96x + 384}{5x^2 + 20x - 105} \]
   - Denominator factors as \( (x + 7)(x - 3) \), matching the vertical asymptotes condition.
   - Both numerator and denominator are degree 2, allowing for a horizontal asymptote.
   - The leading coefficients are 6 in the numerator and 5 in the denominator, giving a horizontal asymptote \( y = \frac{6}{5} \).
   - When \( x = 8 \), the numerator becomes zero: \( 6(8)^2 - 96(8) + 384 = 0 \), so this gives an x-intercept at \( (8, 0) \).
   - **This option satisfies all conditions**.

#### Option D:
\[ f(x) = \frac{6x^2 - 66x + 144}{5x^2 + 20x - 105} \]
   - Denominator factors as \( (x + 7)(x - 3) \), matching the vertical asymptotes condition.
   - The degrees of the numerator and denominator are the same, allowing for a horizontal asymptote.
   - However, when evaluated, this does not provide the correct x-intercept at \( (8, 0) \).
   - This does not satisfy all conditions.

### Conclusion
The correct answer is:
**Option C: \( f(x) = \frac{6x^2 - 96x + 384}{5x^2 + 20x - 105} \)**

Would you like more details or have any questions?

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### Related Questions
1. How do vertical asymptotes affect the domain of a rational function?
2. How are horizontal asymptotes determined for rational functions of different degrees?
3. Why do rational functions have x-intercepts, and how are they found?
4. What conditions are necessary for a rational function to have no vertical asymptotes?
5. How does changing the leading coefficient affect the horizontal asymptote?

#### Tip
When solving rational function problems with multiple conditions, breaking down each condition separately can simplify identifying the correct answer.

Find a rational function that satisfies the given conditions: Vertical asymptotes at x = -7 and x = 3, horizontal asymptote y = 6/5, and x-intercept (8, 0).

Learn how to find a rational function that satisfies conditions on vertical and horizontal asymptotes, as well as an x-intercept. Explore properties of rational functions and asymptote calculations. Suitable for Grades 10-12.

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