Math 111 Rational Functions Worksheet | Schemes and Mind Maps ... - Free Printable

Problem Analysis

The task involves analyzing rational functions to determine their domain, intercepts, holes, vertical asymptotes, and horizontal asymptotes. Additionally, we need to construct specific rational functions based on given properties.

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Part 1: Analyze Each Rational Function

#### Function 1: \( f(x) = \frac{x+1}{x+2} \)
1. Domain: The function is undefined when the denominator is zero. So, \( x + 2 = 0 \implies x = -2 \). The domain is \( x \in \mathbb{R} \setminus \{-2\} \).
2. \( y \)-intercept: Set \( x = 0 \):
\[
f(0) = \frac{0+1}{0+2} = \frac{1}{2}.
\]
The \( y \)-intercept is \( \left( 0, \frac{1}{2} \right) \).
3. \( x \)-intercept: Set \( f(x) = 0 \):
\[
\frac{x+1}{x+2} = 0 \implies x + 1 = 0 \implies x = -1.
\]
The \( x \)-intercept is \( (-1, 0) \).
4. Holes: There are no common factors in the numerator and denominator, so there are no holes.
5. Vertical Asymptotes: The vertical asymptote occurs where the denominator is zero, i.e., \( x = -2 \).
6. Horizontal Asymptotes: Compare the degrees of the numerator and denominator. Both are degree 1. The horizontal asymptote is given by the ratio of the leading coefficients:
\[
y = \frac{1}{1} = 1.
\]

#### Function 2: \( g(x) = \frac{3(x-2)(x-5)}{(x-2)(x+1)} \)
1. Domain: The function is undefined when the denominator is zero. So, \( (x-2)(x+1) = 0 \implies x = 2 \) or \( x = -1 \). However, \( x = 2 \) is a hole (since it cancels with the numerator). The domain is \( x \in \mathbb{R} \setminus \{-1\} \).
2. \( y \)-intercept: Set \( x = 0 \):
\[
g(0) = \frac{3(0-2)(0-5)}{(0-2)(0+1)} = \frac{3(-2)(-5)}{(-2)(1)} = \frac{30}{-2} = -15.
\]
The \( y \)-intercept is \( (0, -15) \).
3. \( x \)-intercept: Set \( g(x) = 0 \):
\[
\frac{3(x-2)(x-5)}{(x-2)(x+1)} = 0 \implies 3(x-2)(x-5) = 0 \implies x = 2 \text{ or } x = 5.
\]
However, \( x = 2 \) is a hole, so the only \( x \)-intercept is \( (5, 0) \).
4. Holes: The factor \( x-2 \) cancels in the numerator and denominator, so there is a hole at \( x = 2 \).
5. Vertical Asymptotes: The vertical asymptote occurs where the reduced denominator is zero, i.e., \( x + 1 = 0 \implies x = -1 \).
6. Horizontal Asymptotes: After canceling the common factor, the function simplifies to \( g(x) = \frac{3(x-5)}{x+1} \). The degrees of the numerator and denominator are both 1. The horizontal asymptote is given by the ratio of the leading coefficients:
\[
y = \frac{3}{1} = 3.
\]

#### Function 3: \( h(x) = \frac{1}{x^2 - x - 1} \)
1. Domain: The function is undefined when the denominator is zero. Solve \( x^2 - x - 1 = 0 \):
\[
x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}.
\]
The domain is \( x \in \mathbb{R} \setminus \left\{ \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2} \right\} \).
2. \( y \)-intercept: Set \( x = 0 \):
\[
h(0) = \frac{1}{0^2 - 0 - 1} = \frac{1}{-1} = -1.
\]
The \( y \)-intercept is \( (0, -1) \).
3. \( x \)-intercept: Set \( h(x) = 0 \):
\[
\frac{1}{x^2 - x - 1} = 0.
\]
This is impossible since the numerator is 1. There are no \( x \)-intercepts.
4. Holes: There are no common factors in the numerator and denominator, so there are no holes.
5. Vertical Asymptotes: The vertical asymptotes occur where the denominator is zero, i.e., \( x = \frac{1 + \sqrt{5}}{2} \) and \( x = \frac{1 - \sqrt{5}}{2} \).
6. Horizontal Asymptotes: The degree of the numerator is 0, and the degree of the denominator is 2. The horizontal asymptote is \( y = 0 \).

#### Function 4: \( k(x) = \frac{x^2 - 2x - 1}{2x - 2} \)
1. Domain: The function is undefined when the denominator is zero. So, \( 2x - 2 = 0 \implies x = 1 \). The domain is \( x \in \mathbb{R} \setminus \{1\} \).
2. \( y \)-intercept: Set \( x = 0 \):
\[
k(0) = \frac{0^2 - 2(0) - 1}{2(0) - 2} = \frac{-1}{-2} = \frac{1}{2}.
\]
The \( y \)-intercept is \( \left( 0, \frac{1}{2} \right) \).
3. \( x \)-intercept: Set \( k(x) = 0 \):
\[
\frac{x^2 - 2x - 1}{2x - 2} = 0 \implies x^2 - 2x - 1 = 0.
\]
Solve the quadratic equation:
\[
x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}.
\]
The \( x \)-intercepts are \( (1 + \sqrt{2}, 0) \) and \( (1 - \sqrt{2}, 0) \).
4. Holes: There are no common factors in the numerator and denominator, so there are no holes.
5. Vertical Asymptotes: The vertical asymptote occurs where the denominator is zero, i.e., \( x = 1 \).
6. Horizontal Asymptotes: The degree of the numerator is 2, and the degree of the denominator is 1. Since the degree of the numerator is greater, there is no horizontal asymptote. Instead, there is an oblique asymptote, but we are not asked for that here.

#### Function 5: \( \ell(x) = \frac{2x - 1}{2x^2 - 7x + 3} \)
1. Domain: The function is undefined when the denominator is zero. Solve \( 2x^2 - 7x + 3 = 0 \):
\[
x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}.
\]
So, \( x = 3 \) or \( x = \frac{1}{2} \). The domain is \( x \in \mathbb{R} \setminus \left\{ 3, \frac{1}{2} \right\} \).
2. \( y \)-intercept: Set \( x = 0 \):
\[
\ell(0) = \frac{2(0) - 1}{2(0)^2 - 7(0) + 3} = \frac{-1}{3}.
\]
The \( y \)-intercept is \( \left( 0, -\frac{1}{3} \right) \).
3. \( x \)-intercept: Set \( \ell(x) = 0 \):
\[
\frac{2x - 1}{2x^2 - 7x + 3} = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}.
\]
However, \( x = \frac{1}{2} \) is a vertical asymptote, so there are no \( x \)-intercepts.
4. Holes: There are no common factors in the numerator and denominator, so there are no holes.
5. Vertical Asymptotes: The vertical asymptotes occur where the denominator is zero, i.e., \( x = 3 \) and \( x = \frac{1}{2} \).
6. Horizontal Asymptotes: The degree of the numerator is 1, and the degree of the denominator is 2. The horizontal asymptote is \( y = 0 \).

#### Function 6: \( m(x) = \frac{-3(x^2 + 2x + 1)}{4(x^2 - 1)} \)
1. Domain: The function is undefined when the denominator is zero. So, \( 4(x^2 - 1) = 0 \implies x^2 - 1 = 0 \implies x = \pm 1 \). The domain is \( x \in \mathbb{R} \setminus \{-1, 1\} \).
2. \( y \)-intercept: Set \( x = 0 \):
\[
m(0) = \frac{-3(0^2 + 2(0) + 1)}{4(0^2 - 1)} = \frac{-3}{-4} = \frac{3}{4}.
\]
The \( y \)-intercept is \( \left( 0, \frac{3}{4} \right) \).
3. \( x \)-intercept: Set \( m(x) = 0 \):
\[
\frac{-3(x^2 + 2x + 1)}{4(x^2 - 1)} = 0 \implies -3(x^2 + 2x + 1) = 0 \implies x^2 + 2x + 1 = 0 \implies (x+1)^2 = 0 \implies x = -1.
\]
However, \( x = -1 \) is a vertical asymptote, so there are no \( x \)-intercepts.
4. Holes: Factor the numerator and denominator:
\[
m(x) = \frac{-3(x+1)^2}{4(x-1)(x+1)} = \frac{-3(x+1)}{4(x-1)} \quad \text{(for } x \neq -1\text{)}.
\]
There is a hole at \( x = -1 \).
5. Vertical Asymptotes: The vertical asymptote occurs where the reduced denominator is zero, i.e., \( x = 1 \).
6. Horizontal Asymptotes: The degrees of the numerator and denominator are both 1. The horizontal asymptote is given by the ratio of the leading coefficients:
\[
y = \frac{-3}{4}.
\]

#### Function 7: \( s(x) = \frac{5x^2 - 20}{3x^2 + 12x + 12} \)
1. Domain: The function is undefined when the denominator is zero. So, \( 3x^2 + 12x + 12 = 0 \implies x^2 + 4x + 4 = 0 \implies (x+2)^2 = 0 \implies x = -2 \). The domain is \( x \in \mathbb{R} \setminus \{-2\} \).
2. \( y \)-intercept: Set \( x = 0 \):
\[
s(0) = \frac{5(0)^2 - 20}{3(0)^2 + 12(0) + 12} = \frac{-20}{12} = -\frac{5}{3}.
\]
The \( y \)-intercept is \( \left( 0, -\frac{5}{3} \right) \).
3. \( x \)-intercept: Set \( s(x) = 0 \):
\[
\frac{5x^2 - 20}{3x^2 + 12x + 12} = 0 \implies 5x^2 - 20 = 0 \implies x^2 = 4 \implies x = \pm 2.
\]
However, \( x = -2 \) is a vertical asymptote, so the only \( x \)-intercept is \( (2, 0) \).
4. Holes: Factor the numerator and denominator:
\[
s(x) = \frac{5(x^2 - 4)}{3(x+2)^2} = \frac{5(x-2)(x+2)}{3(x+2)^2} = \frac{5(x-2)}{3(x+2)} \quad \text{(for } x \neq -2\text{)}.
\]
There is a hole at \( x = -2 \).
5. Vertical Asymptotes: The vertical asymptote occurs where the reduced denominator is zero, i.e., \( x = -2 \).
6. Horizontal Asymptotes: The degrees of the numerator and denominator are both 1. The horizontal asymptote is given by the ratio of the leading coefficients:
\[
y = \frac{5}{3}.
\]

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Part 2: Construct Specific Rational Functions

#### Problem 1: Construct \( p(x) \) with Properties
1. \( p(3) = 0 \): The numerator must have a factor of \( x - 3 \).
2. \( p(2) \) is not defined: The denominator must have a factor of \( x - 2 \).
3. Horizontal asymptote at \( y = \frac{1}{3} \): The degrees of the numerator and denominator must be equal, and the ratio of the leading coefficients must be \( \frac{1}{3} \).

A possible function is:
\[
p(x) = \frac{x-3}{3(x-2)}.
\]

#### Problem 2: Construct \( q(x) \) with Properties
1. \( q(8) = 0 \): The numerator must have a factor of \( x - 8 \).
2. Hole at \( x = 4 \): The numerator and denominator must have a common factor of \( x - 4 \).
3. Horizontal asymptote at \( y = 0 \): The degree of the numerator must be less than the degree of the denominator.

A possible function is:
\[
q(x) = \frac{(x-8)(x-4)}{(x-4)(x^2 + 1)} = \frac{x-8}{x^2 + 1} \quad \text{(for } x \neq 4\text{)}.
\]

#### Problem 3: Construct \( R(x) \) with Properties
1. Zero at \( x = 2 \): The numerator must have a factor of \( x - 2 \).
2. Hole at \( x = 3 \): The numerator and denominator must have a common factor of \( x - 3 \).
3. Vertical asymptote at \( x = 4 \): The denominator must have a factor of \( x - 4 \).
4. Horizontal asymptote at \( y = -\frac{2}{3} \): The degrees of the numerator and denominator must be equal, and the ratio of the leading coefficients must be \( -\frac{2}{3} \).

A possible function is:
\[
R(x) = \frac{-2(x-2)(x-3)}{3(x-3)(x-4)} = \frac{-2(x-2)}{3(x-4)} \quad \text{(for } x \neq 3\text{)}.
\]

#### Problem 4: Construct a Rational Function with No Holes or Vertical Asymptotes and Horizontal Asymptote at \( y = 0 \)
1. No holes: The numerator and denominator should have no common factors.
2. No vertical asymptotes: The denominator should have no real roots.
3. Horizontal asymptote at \( y = 0 \): The degree of the numerator must be less than the degree of the denominator.

A possible function is:
\[
f(x) = \frac{1}{x^2 + 1}.
\]

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Final Answers

\[
\boxed{
\begin{aligned}
1. & \quad p(x) = \frac{x-3}{3(x-2)} \\
2. & \quad q(x) = \frac{x-8}{x^2 + 1} \\
3. & \quad R(x) = \frac{-2(x-2)}{3(x-4)} \\
4. & \quad f(x) = \frac{1}{x^2 + 1}
\end{aligned}
}
\]