Worksheet for analyzing rational functions with four examples to determine asymptotes, intercepts, domain, range, and holes.

To solve the problem, we need to analyze each rational function step by step. Let's go through each function one by one.

Function 1: \( f(x) = \frac{-3x + 12}{x^2 - 3x - 4} \)

#### a) Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero and the numerator is nonzero. Solve:
\[ x^2 - 3x - 4 = 0 \]
Factorize:
\[ (x - 4)(x + 1) = 0 \]
So, \( x = 4 \) and \( x = -1 \).

Check if these values make the numerator zero:
- For \( x = 4 \): \( -3(4) + 12 = -12 + 12 = 0 \). This is a hole, not an asymptote.
- For \( x = -1 \): \( -3(-1) + 12 = 3 + 12 = 15 \neq 0 \). This is a vertical asymptote.

Thus, the vertical asymptote is \( x = -1 \).

#### b) Horizontal Asymptotes
The degree of the numerator is 1, and the degree of the denominator is 2. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).

#### c) Holes (Points of Discontinuity)
A hole occurs where both the numerator and the denominator are zero. From part (a), we found that \( x = 4 \) makes both the numerator and the denominator zero. So, there is a hole at \( x = 4 \).

#### d) X-intercept(s)
X-intercepts occur where the numerator is zero and the denominator is nonzero. Solve:
\[ -3x + 12 = 0 \]
\[ x = 4 \]
However, \( x = 4 \) is a hole, not an x-intercept. Therefore, there are no x-intercepts.

#### e) Y-intercept
The y-intercept occurs when \( x = 0 \):
\[ f(0) = \frac{-3(0) + 12}{0^2 - 3(0) - 4} = \frac{12}{-4} = -3 \]
So, the y-intercept is \( (0, -3) \).

#### f) Domain
The domain is all real numbers except where the denominator is zero. From part (a), the denominator is zero at \( x = -1 \) and \( x = 4 \). Therefore, the domain is:
\[ (-\infty, -1) \cup (-1, 4) \cup (4, \infty) \]

#### g) Range
The range is all real numbers except the value of the function at the hole and the horizontal asymptote. Since the horizontal asymptote is \( y = 0 \), and there is a hole at \( x = 4 \), we need to check the behavior around the asymptotes and the hole. The range is:
\[ (-\infty, 0) \cup (0, \infty) \]

Final Answer for Function 1:

\[
\boxed{
\begin{array}{l}
\text{a) Vertical Asymptote: } x = -1 \\
\text{b) Horizontal Asymptote: } y = 0 \\
\text{c) Hole: } x = 4 \\
\text{d) X-intercept: None} \\
\text{e) Y-intercept: } (0, -3) \\
\text{f) Domain: } (-\infty, -1) \cup (-1, 4) \cup (4, \infty) \\
\text{g) Range: } (-\infty, 0) \cup (0, \infty)
\end{array}
}
\]

Function 2: \( f(x) = \frac{x^2 - 3x}{2x^2 + 2x - 12} \)

#### a) Vertical Asymptotes
Solve:
\[ 2x^2 + 2x - 12 = 0 \]
Factorize:
\[ 2(x^2 + x - 6) = 0 \]
\[ 2(x + 3)(x - 2) = 0 \]
So, \( x = -3 \) and \( x = 2 \).

Check if these values make the numerator zero:
- For \( x = -3 \): \( (-3)^2 - 3(-3) = 9 + 9 = 18 \neq 0 \). This is a vertical asymptote.
- For \( x = 2 \): \( 2^2 - 3(2) = 4 - 6 = -2 \neq 0 \). This is a vertical asymptote.

Thus, the vertical asymptotes are \( x = -3 \) and \( x = 2 \).

#### b) Horizontal Asymptotes
The degree of the numerator is 2, and the degree of the denominator is 2. The horizontal asymptote is the ratio of the leading coefficients:
\[ y = \frac{1}{2} \]

#### c) Holes (Points of Discontinuity)
Factorize the numerator and the denominator:
\[ f(x) = \frac{x(x - 3)}{2(x + 3)(x - 2)} \]
There are no common factors, so there are no holes.

#### d) X-intercept(s)
X-intercepts occur where the numerator is zero and the denominator is nonzero. Solve:
\[ x(x - 3) = 0 \]
So, \( x = 0 \) and \( x = 3 \).

#### e) Y-intercept
The y-intercept occurs when \( x = 0 \):
\[ f(0) = \frac{0^2 - 3(0)}{2(0)^2 + 2(0) - 12} = \frac{0}{-12} = 0 \]
So, the y-intercept is \( (0, 0) \).

#### f) Domain
The domain is all real numbers except where the denominator is zero. From part (a), the denominator is zero at \( x = -3 \) and \( x = 2 \). Therefore, the domain is:
\[ (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \]

#### g) Range
The range is all real numbers except the horizontal asymptote \( y = \frac{1}{2} \). Therefore, the range is:
\[ (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty) \]

Final Answer for Function 2:

\[
\boxed{
\begin{array}{l}
\text{a) Vertical Asymptotes: } x = -3, x = 2 \\
\text{b) Horizontal Asymptote: } y = \frac{1}{2} \\
\text{c) Holes: None} \\
\text{d) X-intercepts: } (0, 0), (3, 0) \\
\text{e) Y-intercept: } (0, 0) \\
\text{f) Domain: } (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \\
\text{g) Range: } (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)
\end{array}
}
\]

Function 3: \( f(x) = \frac{-2x^2 + 4x + 16}{x^2 - 5x + 4} \)

#### a) Vertical Asymptotes
Solve:
\[ x^2 - 5x + 4 = 0 \]
Factorize:
\[ (x - 4)(x - 1) = 0 \]
So, \( x = 4 \) and \( x = 1 \).

Check if these values make the numerator zero:
- For \( x = 4 \): \( -2(4)^2 + 4(4) + 16 = -32 + 16 + 16 = 0 \). This is a hole, not an asymptote.
- For \( x = 1 \): \( -2(1)^2 + 4(1) + 16 = -2 + 4 + 16 = 18 \neq 0 \). This is a vertical asymptote.

Thus, the vertical asymptote is \( x = 1 \).

#### b) Horizontal Asymptotes
The degree of the numerator is 2, and the degree of the denominator is 2. The horizontal asymptote is the ratio of the leading coefficients:
\[ y = \frac{-2}{1} = -2 \]

#### c) Holes (Points of Discontinuity)
Factorize the numerator and the denominator:
\[ f(x) = \frac{-2(x^2 - 2x - 8)}{(x - 4)(x - 1)} = \frac{-2(x - 4)(x + 2)}{(x - 4)(x - 1)} \]
There is a hole at \( x = 4 \).

#### d) X-intercept(s)
X-intercepts occur where the numerator is zero and the denominator is nonzero. Solve:
\[ -2(x - 4)(x + 2) = 0 \]
So, \( x = 4 \) and \( x = -2 \). However, \( x = 4 \) is a hole, so the only x-intercept is \( x = -2 \).

#### e) Y-intercept
The y-intercept occurs when \( x = 0 \):
\[ f(0) = \frac{-2(0)^2 + 4(0) + 16}{0^2 - 5(0) + 4} = \frac{16}{4} = 4 \]
So, the y-intercept is \( (0, 4) \).

#### f) Domain
The domain is all real numbers except where the denominator is zero. From part (a), the denominator is zero at \( x = 4 \) and \( x = 1 \). Therefore, the domain is:
\[ (-\infty, 1) \cup (1, 4) \cup (4, \infty) \]

#### g) Range
The range is all real numbers except the horizontal asymptote \( y = -2 \) and the value of the function at the hole. The range is:
\[ (-\infty, -2) \cup (-2, \infty) \]

Final Answer for Function 3:

\[
\boxed{
\begin{array}{l}
\text{a) Vertical Asymptote: } x = 1 \\
\text{b) Horizontal Asymptote: } y = -2 \\
\text{c) Hole: } x = 4 \\
\text{d) X-intercept: } (-2, 0) \\
\text{e) Y-intercept: } (0, 4) \\
\text{f) Domain: } (-\infty, 1) \cup (1, 4) \cup (4, \infty) \\
\text{g) Range: } (-\infty, -2) \cup (-2, \infty)
\end{array}
}
\]

Function 4: \( f(x) = \frac{x^2 + 7x + 12}{-2x^2 - 2x + 12} \)

#### a) Vertical Asymptotes
Solve:
\[ -2x^2 - 2x + 12 = 0 \]
Factorize:
\[ -2(x^2 + x - 6) = 0 \]
\[ -2(x + 3)(x - 2) = 0 \]
So, \( x = -3 \) and \( x = 2 \).

Check if these values make the numerator zero:
- For \( x = -3 \): \( (-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0 \). This is a hole, not an asymptote.
- For \( x = 2 \): \( 2^2 + 7(2) + 12 = 4 + 14 + 12 = 30 \neq 0 \). This is a vertical asymptote.

Thus, the vertical asymptote is \( x = 2 \).

#### b) Horizontal Asymptotes
The degree of the numerator is 2, and the degree of the denominator is 2. The horizontal asymptote is the ratio of the leading coefficients:
\[ y = \frac{1}{-2} = -\frac{1}{2} \]

#### c) Holes (Points of Discontinuity)
Factorize the numerator and the denominator:
\[ f(x) = \frac{(x + 3)(x + 4)}{-2(x + 3)(x - 2)} \]
There is a hole at \( x = -3 \).

#### d) X-intercept(s)
X-intercepts occur where the numerator is zero and the denominator is nonzero. Solve:
\[ (x + 3)(x + 4) = 0 \]
So, \( x = -3 \) and \( x = -4 \). However, \( x = -3 \) is a hole, so the only x-intercept is \( x = -4 \).

#### e) Y-intercept
The y-intercept occurs when \( x = 0 \):
\[ f(0) = \frac{0^2 + 7(0) + 12}{-2(0)^2 - 2(0) + 12} = \frac{12}{12} = 1 \]
So, the y-intercept is \( (0, 1) \).

#### f) Domain
The domain is all real numbers except where the denominator is zero. From part (a), the denominator is zero at \( x = -3 \) and \( x = 2 \). Therefore, the domain is:
\[ (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \]

#### g) Range
The range is all real numbers except the horizontal asymptote \( y = -\frac{1}{2} \) and the value of the function at the hole. The range is:
\[ (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty) \]

Final Answer for Function 4:

\[
\boxed{
\begin{array}{l}
\text{a) Vertical Asymptote: } x = 2 \\
\text{b) Horizontal Asymptote: } y = -\frac{1}{2} \\
\text{c) Hole: } x = -3 \\
\text{d) X-intercept: } (-4, 0) \\
\text{e) Y-intercept: } (0, 1) \\
\text{f) Domain: } (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \\
\text{g) Range: } (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)
\end{array}
}
\]

Final Answer for All Functions:

\[
\boxed{
\begin{array}{l}
\text{Function 1: } \begin{array}{l}
\text{a) Vertical Asymptote: } x = -1 \\
\text{b) Horizontal Asymptote: } y = 0 \\
\text{c) Hole: } x = 4 \\
\text{d) X-intercept: None} \\
\text{e) Y-intercept: } (0, -3) \\
\text{f) Domain: } (-\infty, -1) \cup (-1, 4) \cup (4, \infty) \\
\text{g) Range: } (-\infty, 0) \cup (0, \infty)
\end{array} \\
\text{Function 2: } \begin{array}{l}
\text{a) Vertical Asymptotes: } x = -3, x = 2 \\
\text{b) Horizontal Asymptote: } y = \frac{1}{2} \\
\text{c) Holes: None} \\
\text{d) X-intercepts: } (0, 0), (3, 0) \\
\text{e) Y-intercept: } (0, 0) \\
\text{f) Domain: } (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \\
\text{g) Range: } (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)
\end{array} \\
\text{Function 3: } \begin{array}{l}
\text{a) Vertical Asymptote: } x = 1 \\
\text{b) Horizontal Asymptote: } y = -2 \\
\text{c) Hole: } x = 4 \\
\text{d) X-intercept: } (-2, 0) \\
\text{e) Y-intercept: } (0, 4) \\
\text{f) Domain: } (-\infty, 1) \cup (1, 4) \cup (4, \infty) \\
\text{g) Range: } (-\infty, -2) \cup (-2, \infty)
\end{array} \\
\text{Function 4: } \begin{array}{l}
\text{a) Vertical Asymptote: } x = 2 \\
\text{b) Horizontal Asymptote: } y = -\frac{1}{2} \\
\text{c) Hole: } x = -3 \\
\text{d) X-intercept: } (-4, 0) \\
\text{e) Y-intercept: } (0, 1) \\
\text{f) Domain: } (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \\
\text{g) Range: } (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)
\end{array}
\end{array}
}
\]