B. Consider the following rational functions. Construct a table of values where x = - 2 -1, 0, 1, 2.

1. f(x) = 2/(x + 1)

2. f(x) = (3x)/(x + 3)

Let's create the tables of values for the given rational functions.

### 1. \( f(x) = \frac{2}{x + 1} \)

We will substitute \( x = -2, -1, 0, 1, 2 \) into the function:

| \( x \) | \( f(x) = \frac{2}{x + 1} \) |
|--------|-----------------------------|
|  -2    | \( f(-2) = \frac{2}{-2 + 1} = \frac{2}{-1} = -2 \) |
|  -1    | \( f(-1) = \frac{2}{-1 + 1} = \frac{2}{0} \) (undefined) |
|   0    | \( f(0) = \frac{2}{0 + 1} = \frac{2}{1} = 2 \) |
|   1    | \( f(1) = \frac{2}{1 + 1} = \frac{2}{2} = 1 \) |
|   2    | \( f(2) = \frac{2}{2 + 1} = \frac{2}{3} \approx 0.67 \) |

Notice that at \( x = -1 \), the function is undefined because the denominator becomes zero.

### 2. \( f(x) = \frac{3x}{x + 3} \)

Now, we substitute the same \( x \)-values into the second function:

| \( x \) | \( f(x) = \frac{3x}{x + 3} \) |
|--------|------------------------------|
|  -2    | \( f(-2) = \frac{3(-2)}{-2 + 3} = \frac{-6}{1} = -6 \) |
|  -1    | \( f(-1) = \frac{3(-1)}{-1 + 3} = \frac{-3}{2} = -1.5 \) |
|   0    | \( f(0) = \frac{3(0)}{0 + 3} = \frac{0}{3} = 0 \) |
|   1    | \( f(1) = \frac{3(1)}{1 + 3} = \frac{3}{4} = 0.75 \) |
|   2    | \( f(2) = \frac{3(2)}{2 + 3} = \frac{6}{5} = 1.2 \) |

### Summary:

1. For \( f(x) = \frac{2}{x + 1} \), the function is undefined at \( x = -1 \), and the other values are straightforward.
2. For \( f(x) = \frac{3x}{x + 3} \), the function is continuous, with various values for \( x \).

Let me know if you want more details or have questions.

Here are some questions that might expand your understanding:

1. How does the behavior of each function change as \( x \to -\infty \)?
2. What happens to each function as \( x \to \infty \)?
3. Can you determine the asymptotes of each function?
4. How would you sketch the graph of each function?
5. What does the undefined point at \( x = -1 \) in the first function signify?

**Tip:** Rational functions often have vertical asymptotes where the denominator becomes zero. Identifying these can help you better understand their behavior.

B. Consider the following rational functions. Construct a table of values where x = -2, -1, 0, 1, 2.

1. f(x) = 2/(x + 1)

2. f(x) = (3x)/(x + 3)

Explore how to construct a table of values for the rational functions f(x) = 2/(x + 1) and f(x) = (3x)/(x + 3) using x = -2, -1, 0, 1, 2. Learn about undefined points, function behavior, and key algebraic concepts suitable for Grades 9-11.

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