Write Rational Functions: Practice Problems with Solutions
Learn how to write rational functions based on their key characteristics, including vertical and horizontal asymptotes, slant asymptotes, x-intercepts, zeros, and holes. Step-by-step solutions help you master each type of problem.
Table of Contents
Review of Rational Function Characteristics
A rational function has the form \( f(x) = \dfrac{P(x)}{Q(x)} \) where \( P \) and \( Q \) are polynomials. Key characteristics:
- Vertical asymptote at \( x = c \): \( Q(c) = 0 \) and \( P(c) \neq 0 \); the denominator contains a factor \( (x - c) \) that does not cancel.
- Hole at \( x = a \): Both \( P(a) = 0 \) and \( Q(a) = 0 \); the factor \( (x - a) \) cancels.
- Zero / x-intercept at \( x = b \): \( P(b) = 0 \) and \( Q(b) \neq 0 \).
- Horizontal asymptote \( y = L \): Numerator and denominator degrees are equal, and \( L = \dfrac{\text{leading coefficient of } P}{\text{leading coefficient of } Q} \).
- Slant asymptote \( y = mx + b \): Numerator degree is exactly one more than denominator degree; perform polynomial long division to find \( mx + b \).
Problem 1: Rational Function with Vertical and Horizontal Asymptotes
Write a rational function \( f \) that has a vertical asymptote at \( x = 2 \), a horizontal asymptote \( y = 3 \), and a zero at \( x = -5 \).
View Step-by-Step Solution
Since \( f \) has a vertical asymptote at \( x = 2 \), the denominator contains the factor \( (x - 2) \). Hence \( f \) has the form:
\[ f(x) = \frac{g(x)}{x - 2} \]The numerator \( g(x) \) must be of the same degree as the denominator to ensure a horizontal asymptote, and it must contain the factor \( (x + 5) \) since \( f \) has a zero at \( x = -5 \). To make the horizontal asymptote \( y = 3 \), the ratio of the leading coefficients must be 3. Therefore:
Verification: As \( x \to \infty \), \( f(x) \to 3/1 = 3 \) ✓. At \( x = -5 \), the numerator is zero while the denominator is \(-7 \neq 0\) ✓. At \( x = 2 \), the denominator is zero while the numerator is \( 21 \neq 0 \) ✓.
Problem 2: Rational Function with Two Vertical Asymptotes and No x-Intercept
Write a rational function \( g \) with vertical asymptotes at \( x = 3 \) and \( x = -3 \), a horizontal asymptote at \( y = -4 \), and with no x-intercept.
View Step-by-Step Solution
Since \( g \) has vertical asymptotes at \( x = 3 \) and \( x = -3 \), the denominator contains the product \( (x - 3)(x + 3) \). Hence:
\[ g(x) = \frac{h(x)}{(x - 3)(x + 3)} \]To achieve the horizontal asymptote \( y = -4 \), the numerator \( h(x) \) must be of the same degree as the denominator (degree 2) with a leading coefficient of \(-4\). To have no real zeros (no x-intercepts), the quadratic in the numerator must have no real roots. We can use a sum of squares:
Verification: The numerator \( -4x^2 - 6 \) is always negative (no real zeros) ✓. The horizontal asymptote is \( y = -4/1 = -4 \) ✓. Both \( x = 3 \) and \( x = -3 \) make the denominator zero and the numerator nonzero ✓.
Problem 3: Rational Function with Hole and Vertical Asymptote
Write a rational function \( h \) with a hole at \( x = 5 \), a vertical asymptote at \( x = -1 \), a horizontal asymptote at \( y = 2 \), and an x-intercept at \( x = 2 \).
View Step-by-Step Solution
Since \( h \) has a hole at \( x = 5 \), both numerator and denominator must share a common factor \( (x - 5) \).
The vertical asymptote at \( x = -1 \) gives a factor \( (x + 1) \) in the denominator, and the x-intercept at \( x = 2 \) gives a factor \( (x - 2) \) in the numerator.
To satisfy the horizontal asymptote \( y = 2 \), the numerator and denominator must have the same degree, and the ratio of leading coefficients must be 2. Multiplying the numerator by 2 gives:
Simplified (excluding \( x = 5 \)): \( h(x) = \dfrac{2(x - 2)}{x + 1} \).
Verification: The numerator has leading coefficient 2 and the denominator has leading coefficient 1, so the horizontal asymptote is \( y = 2 \) ✓. The common factor \( (x - 5) \) creates the hole at \( x = 5 \) ✓. \( x = 2 \) makes the numerator zero while the denominator is \( 3 \neq 0 \), giving an x-intercept ✓. \( x = -1 \) makes the denominator zero while the numerator is \(-6 \neq 0\), giving a vertical asymptote ✓.
Problem 4: Rational Function with Slant Asymptote
Write a rational function \( f \) with a slant asymptote \( y = x + 4 \), a vertical asymptote at \( x = 5 \), and one of the zeros at \( x = 2 \).
View Step-by-Step Solution
The graph of \( f \) has a slant asymptote \( y = x + 4 \) and a vertical asymptote at \( x = 5 \). Hence \( f(x) \) can be written in the form:
\[ f(x) = (x + 4) + \frac{a}{x - 5} \]where \( a \) is a constant determined from the zero at \( x = 2 \). Set \( f(2) = 0 \):
\[ f(2) = (2 + 4) + \frac{a}{2 - 5} = 0 \] \[ 6 + \frac{a}{-3} = 0 \implies \frac{a}{3} = 6 \implies a = 18 \]Substituting \( a \) back and finding a common denominator:
\[ f(x) = (x + 4) + \frac{18}{x - 5} = \frac{(x + 4)(x - 5) + 18}{x - 5} \] \[ f(x) = \frac{x^2 - x - 20 + 18}{x - 5} = \frac{x^2 - x - 2}{x - 5} \]Verification: Numerator degree 2, denominator degree 1 → slant asymptote exists. Dividing \( x^2 - x - 2 \) by \( x - 5 \) gives quotient \( x + 4 \) with remainder 18, so the slant asymptote is \( y = x + 4 \) ✓. At \( x = 2 \): \( 4 - 2 - 2 = 0 \) ✓. At \( x = 5 \): denominator is zero while numerator is \( 25 - 5 - 2 = 18 \neq 0 \) ✓.
Note: Factoring the numerator gives \( (x - 2)(x + 1) \), so the zeros are at \( x = 2 \) and \( x = -1 \).
Summary Table of Answers
| # | Given Characteristics | Function |
|---|---|---|
| 1 | VA \( x = 2 \), HA \( y = 3 \), zero at \( x = -5 \) | \( f(x) = \dfrac{3(x+5)}{x-2} \) |
| 2 | VA \( x = \pm 3 \), HA \( y = -4 \), no x-intercept | \( g(x) = \dfrac{-4x^2-6}{(x-3)(x+3)} \) |
| 3 | Hole at \( x = 5 \), VA \( x = -1 \), HA \( y = 2 \), zero at \( x = 2 \) | \( h(x) = \dfrac{2(x-5)(x-2)}{(x-5)(x+1)} \) |
| 4 | Slant \( y = x + 4 \), VA \( x = 5 \), zero at \( x = 2 \) | \( f(x) = \dfrac{x^2-x-2}{x-5} \) |
Frequently Asked Questions
How do you write a rational function given vertical and horizontal asymptotes?
Place factors corresponding to the vertical asymptotes in the denominator. To achieve the horizontal asymptote, ensure the numerator has the same degree as the denominator, and adjust the leading coefficient of the numerator so the ratio of leading coefficients equals the horizontal asymptote.
How do you include a slant asymptote in a rational function?
Write the function as the sum of the slant asymptote equation and a fractional term whose denominator contains the vertical asymptotes, such as \( f(x) = (mx + b) + \dfrac{a}{x - c} \), then solve for the constant \( a \) using a known point or zero.
What causes a hole in the graph of a rational function?
A hole occurs when the numerator and denominator share a common factor \( (x - a) \). After cancellation, the graph has a single missing point at \( x = a \) rather than an asymptote there.
How do you make a rational function with no x-intercepts?
Choose a numerator that has no real zeros, such as a sum of squares like \( x^2 + 1 \) or \( -4x^2 - 6 \). Since the numerator never equals zero for real \( x \), the function has no x-intercepts.
How do you determine the horizontal asymptote of a rational function?
Compare the degrees of the numerator and denominator. If the numerator degree is less than the denominator degree, the horizontal asymptote is \( y = 0 \). If the degrees are equal, it is \( y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} \). If the numerator degree is greater, there is no horizontal asymptote (there may be a slant asymptote instead).