Rational Functions Graphing Worksheet: Identify holes, vertical and horizontal asymptotes, and sketch graphs for four given rational functions.

To solve the problem, we need to analyze each rational function step by step to identify holes, vertical asymptotes, and horizontal asymptotes. Let's go through each function one by one.

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Function 1: \( f(x) = \frac{1}{3x + 3} \)

#### Step 1: Simplify the function
The function is already in its simplest form:
\[ f(x) = \frac{1}{3(x + 1)} \]

#### Step 2: Identify holes
Holes occur where both the numerator and the denominator are zero. Here, the numerator is 1, which is never zero. Therefore, there are no holes.
\[ \text{Holes: None} \]

#### Step 3: Identify vertical asymptotes
Vertical asymptotes occur where the denominator is zero (and the numerator is not zero). Set the denominator equal to zero:
\[ 3(x + 1) = 0 \]
\[ x + 1 = 0 \]
\[ x = -1 \]
So, there is a vertical asymptote at \( x = -1 \).
\[ \text{Vertical asymptotes: } x = -1 \]

#### Step 4: Identify horizontal asymptotes
For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator:
- Degree of numerator: 0 (constant)
- Degree of denominator: 1 (linear)
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).
\[ \text{Horizontal asymptotes: } y = 0 \]

#### Final Answer for Function 1
\[ \boxed{\text{Holes: None, Vertical asymptotes: } x = -1, \text{ Horizontal asymptotes: } y = 0} \]

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Function 2: \( f(x) = \frac{x^3 - 16x}{-3x^2 + 3x + 18} \)

#### Step 1: Factor the numerator and denominator
- Numerator: \( x^3 - 16x \)
\[ x^3 - 16x = x(x^2 - 16) = x(x - 4)(x + 4) \]
- Denominator: \( -3x^2 + 3x + 18 \)
\[ -3x^2 + 3x + 18 = -3(x^2 - x - 6) = -3(x - 3)(x + 2) \]
So, the function becomes:
\[ f(x) = \frac{x(x - 4)(x + 4)}{-3(x - 3)(x + 2)} \]

#### Step 2: Identify holes
Holes occur where both the numerator and the denominator are zero. There are no common factors between the numerator and the denominator, so there are no holes.
\[ \text{Holes: None} \]

#### Step 3: Identify vertical asymptotes
Vertical asymptotes occur where the denominator is zero (and the numerator is not zero). Set the denominator equal to zero:
\[ -3(x - 3)(x + 2) = 0 \]
\[ x - 3 = 0 \quad \text{or} \quad x + 2 = 0 \]
\[ x = 3 \quad \text{or} \quad x = -2 \]
So, there are vertical asymptotes at \( x = 3 \) and \( x = -2 \).
\[ \text{Vertical asymptotes: } x = 3, x = -2 \]

#### Step 4: Identify horizontal asymptotes
For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator:
- Degree of numerator: 3
- Degree of denominator: 2
Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, there is an oblique (slant) asymptote, but we are only asked for horizontal asymptotes.
\[ \text{Horizontal asymptotes: None} \]

#### Final Answer for Function 2
\[ \boxed{\text{Holes: None, Vertical asymptotes: } x = 3, x = -2, \text{ Horizontal asymptotes: None}} \]

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Function 3: \( f(x) = \frac{x^3 - x}{x^3 + 2x^2 - 3x} \)

#### Step 1: Factor the numerator and denominator
- Numerator: \( x^3 - x \)
\[ x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1) \]
- Denominator: \( x^3 + 2x^2 - 3x \)
\[ x^3 + 2x^2 - 3x = x(x^2 + 2x - 3) = x(x + 3)(x - 1) \]
So, the function becomes:
\[ f(x) = \frac{x(x - 1)(x + 1)}{x(x + 3)(x - 1)} \]

#### Step 2: Identify holes
Holes occur where both the numerator and the denominator are zero. The common factors are \( x \) and \( x - 1 \). These factors create holes at \( x = 0 \) and \( x = 1 \).
\[ \text{Holes: } x = 0, x = 1 \]

#### Step 3: Identify vertical asymptotes
Vertical asymptotes occur where the denominator is zero (and the numerator is not zero). After canceling the common factors, the denominator is \( x + 3 \). Set it equal to zero:
\[ x + 3 = 0 \]
\[ x = -3 \]
So, there is a vertical asymptote at \( x = -3 \).
\[ \text{Vertical asymptotes: } x = -3 \]

#### Step 4: Identify horizontal asymptotes
For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator:
- Degree of numerator: 3 - 2 = 1 (after canceling common factors)
- Degree of denominator: 3 - 2 = 1 (after canceling common factors)
Since the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients:
\[ y = \frac{1}{1} = 1 \]
\[ \text{Horizontal asymptotes: } y = 1 \]

#### Final Answer for Function 3
\[ \boxed{\text{Holes: } x = 0, x = 1, \text{ Vertical asymptotes: } x = -3, \text{ Horizontal asymptotes: } y = 1} \]

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Function 4: \( f(x) = \frac{x^3 - 6x^2 + 8x}{-3x^2 + 9x - 6} \)

#### Step 1: Factor the numerator and denominator
- Numerator: \( x^3 - 6x^2 + 8x \)
\[ x^3 - 6x^2 + 8x = x(x^2 - 6x + 8) = x(x - 2)(x - 4) \]
- Denominator: \( -3x^2 + 9x - 6 \)
\[ -3x^2 + 9x - 6 = -3(x^2 - 3x + 2) = -3(x - 1)(x - 2) \]
So, the function becomes:
\[ f(x) = \frac{x(x - 2)(x - 4)}{-3(x - 1)(x - 2)} \]

#### Step 2: Identify holes
Holes occur where both the numerator and the denominator are zero. The common factor is \( x - 2 \). This creates a hole at \( x = 2 \).
\[ \text{Holes: } x = 2 \]

#### Step 3: Identify vertical asymptotes
Vertical asymptotes occur where the denominator is zero (and the numerator is not zero). After canceling the common factor, the denominator is \( -3(x - 1) \). Set it equal to zero:
\[ x - 1 = 0 \]
\[ x = 1 \]
So, there is a vertical asymptote at \( x = 1 \).
\[ \text{Vertical asymptotes: } x = 1 \]

#### Step 4: Identify horizontal asymptotes
For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator:
- Degree of numerator: 3 - 1 = 2 (after canceling common factors)
- Degree of denominator: 2 - 1 = 1 (after canceling common factors)
Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, there is an oblique (slant) asymptote, but we are only asked for horizontal asymptotes.
\[ \text{Horizontal asymptotes: None} \]

#### Final Answer for Function 4
\[ \boxed{\text{Holes: } x = 2, \text{ Vertical asymptotes: } x = 1, \text{ Horizontal asymptotes: None}} \]

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Summary of All Answers

1. \( f(x) = \frac{1}{3x + 3} \)
\[ \boxed{\text{Holes: None, Vertical asymptotes: } x = -1, \text{ Horizontal asymptotes: } y = 0} \]

2. \( f(x) = \frac{x^3 - 16x}{-3x^2 + 3x + 18} \)
\[ \boxed{\text{Holes: None, Vertical asymptotes: } x = 3, x = -2, \text{ Horizontal asymptotes: None}} \]

3. \( f(x) = \frac{x^3 - x}{x^3 + 2x^2 - 3x} \)
\[ \boxed{\text{Holes: } x = 0, x = 1, \text{ Vertical asymptotes: } x = -3, \text{ Horizontal asymptotes: } y = 1} \]

4. \( f(x) = \frac{x^3 - 6x^2 + 8x}{-3x^2 + 9x - 6} \)
\[ \boxed{\text{Holes: } x = 2, \text{ Vertical asymptotes: } x = 1, \text{ Horizontal asymptotes: None}} \]