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### 1. **Solve Rational Equations and Rational Inequalities:**

The problem is:

\[
\frac{2}{x - 3} - \frac{1}{x + 1} \geq \frac{5x}{(x - 3)(x + 1)}
\]

#### **Step 1: Find a common denominator**
The common denominator for all terms is \((x - 3)(x + 1)\).

#### **Step 2: Multiply every term by \((x - 3)(x + 1)\)**
This eliminates the fractions:

\[
2(x + 1) - 1(x - 3) \geq 5x
\]

#### **Step 3: Expand and simplify**
Expand and simplify both sides:

\[
2x + 2 - x + 3 \geq 5x
\]

Combine like terms:

\[
x + 5 \geq 5x
\]

#### **Step 4: Solve for \(x\)**
Isolate \(x\) on one side:

\[
5 \geq 4x
\]

Divide both sides by 4:

\[
\frac{5}{4} \geq x \quad \text{or} \quad x \leq \frac{5}{4}
\]

#### **Step 5: Consider the domain**
\(x \neq 3\) and \(x \neq -1\) because these values would make the original denominators zero. So the solution is:

\[
x \leq \frac{5}{4}, \quad x \neq 3, -1
\]

### 2. **Table of Values and Graph:**

The function provided is:

\[
f(x) = \frac{x^2 - 1}{x - 2}
\]

#### **Step 1: Simplify the function**
Factor the numerator:

\[
f(x) = \frac{(x - 1)(x + 1)}{x - 2}
\]

#### **Step 2: Fill in the table of values**

Choose some \(x\) values and calculate the corresponding \(f(x)\):

\[
\begin{array}{c|c}
x & f(x) \\
\hline
-2 & \frac{(-2)^2 - 1}{-2 - 2} = \frac{4 - 1}{-4} = -\frac{3}{4} \\
0 & \frac{0^2 - 1}{0 - 2} = \frac{-1}{-2} = \frac{1}{2} \\
1 & \frac{1^2 - 1}{1 - 2} = \frac{0}{-1} = 0 \\
3 & \frac{3^2 - 1}{3 - 2} = \frac{9 - 1}{1} = 8 \\
4 & \frac{4^2 - 1}{4 - 2} = \frac{16 - 1}{2} = \frac{15}{2} = 7.5 \\
\end{array}
\]

#### **Step 3: Plot the points**

You would plot the points \((-2, -\frac{3}{4})\), \((0, \frac{1}{2})\), \((1, 0)\), \((3, 8)\), and \((4, 7.5)\) on the graph.

### 3. **State the forms of the rational function:**

For the function given:

\[
f(x) = \frac{3x + 1}{2x - 5}
\]

#### **Step 1: Vertical Asymptote**
Set the denominator equal to zero:

\[
2x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{2}
\]

So, there is a vertical asymptote at \(x = \frac{5}{2}\).

#### **Step 2: Horizontal Asymptote**
Since the degrees of the numerator and denominator are the same, the horizontal asymptote is the ratio of the leading coefficients:

\[
y = \frac{3}{2}
\]

#### **Step 3: X-intercept and Y-intercept**
- **X-intercept**: Set the numerator equal to zero:

\[
3x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{3}
\]

- **Y-intercept**: Set \(x = 0\):

\[
f(0) = \frac{3(0) + 1}{2(0) - 5} = -\frac{1}{5}
\]

### Summary:
1. **Solution to the inequality**: \(x \leq \frac{5}{4}, x \neq 3, -1\)
2. **Table of Values**: Completed and points plotted.
3. **Rational function forms**:
   - Vertical Asymptote: \(x = \frac{5}{2}\)
   - Horizontal Asymptote: \(y = \frac{3}{2}\)
   - X-intercept: \(x = -\frac{1}{3}\)
   - Y-intercept: \(y = -\frac{1}{5}\)

Would you like further details or have any specific questions about these solutions?

#### Here are 5 related questions to explore further:
1. How would the inequality change if the right-hand side was a different expression?
2. What would happen to the graph of \(f(x)\) if we shifted the function vertically or horizontally?
3. How can we determine the behavior of \(f(x)\) as \(x\) approaches infinity?
4. What are the steps to verifying the solutions of rational inequalities?
5. How do vertical asymptotes affect the behavior of rational functions near those points?

#### Tip:
Always double-check for extraneous solutions when solving rational equations or inequalities, especially when the solution involves values that might make the original denominator zero.

Learn how to solve rational equations and inequalities step by step with detailed examples and solutions. This problem covers rational equations, inequalities, and function analysis, suitable for advanced high school or college students.

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