Let's solve the problem step by step.

Step 1: Simplifying the Function

The given function is: $f(x) = \frac{x - 6}{x^2 - 36}$

The denominator can be factored as: $x^2 - 36 = (x + 6)(x - 6)$

Thus, the function simplifies to: $f(x) = \frac{x - 6}{(x + 6)(x - 6)}$

Since $x - 6$ in the numerator and denominator cancel out (except when $x = 6$), we get: $f(x) = \frac{1}{x + 6}, \quad x \neq 6$

Step 2: Identifying Vertical Asymptotes and x-intercepts

• Vertical Asymptote: The vertical asymptote occurs where the denominator is zero. Therefore, $x + 6 = 0$ gives $x = -6$.
• x-intercept: There is no x-intercept for the simplified function $f(x) = \frac{1}{x + 6}$, because the numerator is always 1.

Step 3: Evaluating the Function at Given Values of $x$

Let's find $f(x)$ for each of the given values.

1. For $x = -9$: $f(-9) = \frac{1}{-9 + 6} = \frac{1}{-3} = -\frac{1}{3}$

2. For $x = -8$: $f(-8) = \frac{1}{-8 + 6} = \frac{1}{-2} = -\frac{1}{2}$

3. For $x = 5$: $f(5) = \frac{1}{5 + 6} = \frac{1}{11}$

4. For $x = 7$: $f(7) = \frac{1}{7 + 6} = \frac{1}{13}$

5. For $x = 8$: $f(8) = \frac{1}{8 + 6} = \frac{1}{14}$

Step 4: Plotting Points

To plot the function, note:

• There is a vertical asymptote at $x = -6$.
• The values calculated above can be plotted as points on the graph of the function.

Summary of Evaluated Values:

• $f(-9) = -\frac{1}{3}$
• $f(-8) = -\frac{1}{2}$
• $f(5) = \frac{1}{11}$
• $f(7) = \frac{1}{13}$
• $f(8) = \frac{1}{14}$

Would you like me to explain the graphing process in more detail, or do you have any specific questions?

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Related Questions:

1. How do you find horizontal asymptotes of a rational function?
2. What happens to the function near a vertical asymptote?
3. How do you determine the domain of a rational function?
4. What are the steps for graphing a rational function with holes?
5. Can rational functions have both horizontal and oblique asymptotes?

Tip:

Always factor the denominator of a rational function to identify potential vertical asymptotes and simplify the function.