We are asked to find a rational function that satisfies the following conditions:

• Vertical asymptotes: $x = -1$ and $x = 7$
• x-intercept: $(2, 0)$

Step 1: General form of a rational function

A rational function with vertical asymptotes at $x = -1$ and $x = 7$ would have a denominator of the form $(x + 1)(x - 7)$, because the denominator must be zero at $x = -1$ and $x = 7$.

Thus, the general form of the function can be written as: $f(x) = \frac{N(x)}{(x + 1)(x - 7)}$ where $N(x)$ is the numerator that we will determine next.

Step 2: Determining the numerator

We are also given that the function has an x-intercept at $(2, 0)$, meaning $f(2) = 0$. This implies that $x - 2$ must be a factor of the numerator $N(x)$. Therefore, we have: $N(x) = x - 2$

Step 3: Final function

Thus, the simplest rational function that satisfies the given conditions is: $f(x) = \frac{x - 2}{(x + 1)(x - 7)}$

Step 4: Match with options

Now, comparing this with the provided options:

• Option A: $f(x) = \frac{x + 2}{x^2 - 6x - 7}$ — incorrect, wrong numerator.
• Option B: $f(x) = \frac{x - 2}{x^2 + 6x + 7}$ — incorrect, wrong denominator.
• Option C: $f(x) = \frac{x - 2}{x^2 - 6x - 7}$ — correct, because $x^2 - 6x - 7$ factors into $(x + 1)(x - 7)$.
• Option D: $f(x) = \frac{x + 2}{x^2 - 6x + 7}$ — incorrect, wrong numerator and denominator.

The correct answer is Option C: $f(x) = \frac{x - 2}{x^2 - 6x - 7}$.

Would you like more details or have any questions?

---

Here are 5 related questions:

1. How do you determine the vertical asymptotes of a rational function?
2. How can you find the x-intercepts of a rational function?
3. What is the significance of factoring the denominator in rational functions?
4. How do horizontal asymptotes differ from vertical asymptotes in rational functions?
5. Can a rational function have more than one x-intercept?

Tip: Always check the factors of both the numerator and denominator to determine intercepts and asymptotes of rational functions.