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?

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.