Let’s solve this step-by-step. We are working with the function:

$f(x) = \frac{x - 7}{x^2 - 49}$

First, we’ll simplify and analyze the function, identify the intercepts and asymptotes, and find the function values at the given points: $x = -10, -9, 6, 8,$ and $9$.

---

Step 1: Simplify the function

The denominator $x^2 - 49$ can be factored:

$x^2 - 49 = (x - 7)(x + 7)$

Thus, the function becomes:

$f(x) = \frac{x - 7}{(x - 7)(x + 7)}$

For all $x \neq 7$, we can cancel the factor $x - 7$, giving:

$f(x) = \frac{1}{x + 7}, \quad \text{for } x \neq 7.$

Step 2: Identify key features

• Vertical asymptote: The vertical asymptote occurs when the denominator is 0. That happens when $x + 7 = 0$, i.e., at:

  $x = -7.$

• Hole in the graph: There's a removable discontinuity (hole) at $x = 7$, where the factor $x - 7$ was canceled out.

• x-intercept: Set the numerator $1 = 0$. However, this is impossible, so there are no x-intercepts.

• y-intercept: Evaluate $f(0)$.

  $f(0) = \frac{1}{0 + 7} = \frac{1}{7}.$

---

Step 3: Evaluate the function at the given points

1. At $x = -10$:

   $f(-10) = \frac{1}{-10 + 7} = \frac{1}{-3} = -\frac{1}{3}.$

2. At $x = -9$:

   $f(-9) = \frac{1}{-9 + 7} = \frac{1}{-2} = -\frac{1}{2}.$

3. At $x = 6$:

   $f(6) = \frac{1}{6 + 7} = \frac{1}{13}.$

4. At $x = 8$:

   $f(8) = \frac{1}{8 + 7} = \frac{1}{15}.$

5. At $x = 9$:

   $f(9) = \frac{1}{9 + 7} = \frac{1}{16}.$

---

Final Answer:

• $f(-10) = -\frac{1}{3}$
• $f(-9) = -\frac{1}{2}$
• $f(6) = \frac{1}{13}$
• $f(8) = \frac{1}{15}$
• $f(9) = \frac{1}{16}$

---

Would you like further clarification or a graph of this function to visualize it?

---

Related Questions:

1. What happens to $f(x)$ as $x \to -7^+$ and $x \to -7^-$?
2. What are horizontal asymptotes, if any, of the function?
3. How do you determine if a discontinuity is removable or non-removable?
4. What is the behavior of the function as $x \to \infty$?
5. How would the function change if we modified the numerator?

Tip: Always factor the denominator when analyzing rational functions—it simplifies detecting asymptotes and holes.