To solve this, let's examine the function and its characteristics step-by-step.

The function given is: $f(x) = \frac{x - 3}{x^2 - 9}$

Step 1: Find the x-intercepts and vertical asymptotes

1. Factor the denominator: $x^2 - 9 = (x + 3)(x - 3)$

   This means the function simplifies to: $f(x) = \frac{x - 3}{(x + 3)(x - 3)}$

   Since $x - 3$ cancels out, the function becomes: $f(x) = \frac{1}{x + 3} \quad \text{for } x \neq 3$

2. Identify the vertical asymptote:

   • The denominator $x + 3 = 0$ when $x = -3$, so there is a vertical asymptote at $x = -3$.

3. Find the x-intercepts:

   • Since $x - 3$ canceled out, there are no x-intercepts for this function.

Step 2: Evaluate the function at given x-values

Now that we have simplified the function to $f(x) = \frac{1}{x + 3}$, let’s evaluate it at each requested value of $x$:

1. At $x = -6$: $f(-6) = \frac{1}{-6 + 3} = \frac{1}{-3} = -\frac{1}{3}$

2. At $x = -5$: $f(-5) = \frac{1}{-5 + 3} = \frac{1}{-2} = -\frac{1}{2}$

3. At $x = 2$: $f(2) = \frac{1}{2 + 3} = \frac{1}{5}$

4. At $x = 4$: $f(4) = \frac{1}{4 + 3} = \frac{1}{7}$

5. At $x = 5$: $f(5) = \frac{1}{5 + 3} = \frac{1}{8}$

Summary of values

The values of $f(x)$ at the given points are:

• $f(-6) = -\frac{1}{3}$
• $f(-5) = -\frac{1}{2}$
• $f(2) = \frac{1}{5}$
• $f(4) = \frac{1}{7}$
• $f(5) = \frac{1}{8}$

Would you like a graph of this function or further explanation on any part of the solution?

Further Questions to Consider

1. What happens to the function as $x$ approaches the vertical asymptote at $x = -3$?
2. How does the function behave as $x \to \infty$ or $x \to -\infty$?
3. Are there any horizontal asymptotes in this function?
4. How would additional transformations, such as shifting or reflecting, affect the function's graph?
5. What would be the result if the function were redefined to avoid canceling $x - 3$?

Tip

Remember, vertical asymptotes occur where the function's denominator equals zero, resulting in undefined values for $f(x)$ at those points.