Plot points between and beyond each​ x-intercept and vertical asymptote. Find the value of the function at the given value of x.

x

negative 9

negative 8

5

7

8

f left parenthesis x right parenthesis equals StartFraction x minus 6 Over x squared minus 36 EndFraction

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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.

Plot points between and beyond each​ x-intercept and vertical asymptote. Find the value of the function at the given value of x: x = -9, -8, 5, 7, 8. The function is f(x) = (x - 6) / (x^2 - 36).

This tutorial explains how to plot points for the rational function f(x) = (x - 6) / (x^2 - 36) and identifies vertical asymptotes and values of the function at specific x-values, such as -9, -8, 5, 7, and 8. Suitable for high school students, the solution includes simplification, evaluation, and asymptote analysis.

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