Precalculus: Rational Functions, Asymptotes, and Rational Equations

Fundamental Concepts of Rational Functions

• Rational Function: A function $f(x)$ that can be expressed as the quotient of two polynomial functions:     $f(x) = \frac{a(x)}{b(x)} = \frac{a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0}{b_m x^m + b_{m-1} x^{m-1} + \dots + b_1 x + b_0}$     where $b(x) \neq 0$ and the denominator polynomial has a degree greater than $0$.

• Asymptote: A line that a curve approaches arbitrarily closely as it heads toward infinity or a point of discontinuity.

• Vertical Asymptote: A line $x = c$ where the value of $f(x)$ approaches positive or negative infinity as $x$ approaches $c$ from either the left or the right.

• Horizontal Asymptote: A line $y = c$ that indicates the end behavior of the graph as $x$ approaches positive infinity ($\infty$) or negative infinity ($-\infty$).

• Oblique (Slant) Asymptote: A non-horizontal, non-vertical line asymptote that describes the end behavior of a rational function when the degree of the numerator is exactly one greater than the degree of the denominator ($n = m + 1$).

• Holes (Removable Discontinuities): Points on the graph where a rational function is undefined because both the numerator and the denominator equal zero for a specific value of $x$ (i.e., they share a common factor).

The Reciprocal Parent Function

• Parent Function Equation:     $f(x) = \frac{1}{x}$

• Domain:     $\{x \mid x \neq 0, x \in \mathbb{R}\}$ or $(-\infty, 0) \cup (0, \infty)$

• Range:     $\{y \mid y \neq 0, y \in \mathbb{R}\}$ or $(-\infty, 0) \cup (0, \infty)$

• Vertical Asymptote:     $x = 0$

• Horizontal Asymptote:     $y = 0$

• Symmetry: Origin symmetry (odd function, since $f(-x) = -f(x)$).

Limit Definitions of Vertical and Horizontal Asymptotes

• Vertical Asymptote Definition:     The line $x = c$ is a vertical asymptote of the graph of $f(x)$ if:     $\lim_{x \to c^-} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to c^+} f(x) = \pm\infty$     Example: For $f(x) = \frac{5}{(x + 2)^2}$, as $x \to -2$, $f(x) \to \infty$. Thus, the vertical asymptote is $x = -2$

• Horizontal Asymptote Definition:     The line $y = c$ is a horizontal asymptote of the graph of $f(x)$ if:     $\lim_{x \to -\infty} f(x) = c \quad \text{or} \quad \lim_{x \to \infty} f(x) = c$     Example: For $f(x) = \frac{6x^2}{x^2 + 2}$, as $x \to \pm\infty$, $f(x) \to 6$. Thus, the horizontal asymptote is $y = 6$

Key Rules for Graphing Rational Functions

Given a rational function $f(x) = \frac{a(x)}{b(x)}$ in simplest form (where $a(x)$ and $b(x)$ have no common factors other than $\pm 1$):

• Vertical Asymptotes: Occur at the real zeros of the denominator $b(x)$. Set $b(x) = 0$ and solve for $x$.

• Horizontal Asymptotes: Determined by comparing the degree $n$ of numerator $a(x)$ to the degree $m$ of denominator $b(x)$:

  • Case 1 ($n < m$): The horizontal asymptote is $y = 0$ ($x$-axis).

  • Case 2 ($n = m$): The horizontal asymptote is the ratio of the leading coefficients:         $y = \frac{a_n}{b_m}$

  • Case 3 ($n > m$): There is no horizontal asymptote (the graph may have an oblique asymptote or polynomial end behavior).

• Intercepts:

  • $x$-intercepts: Occur at the real zeros of the numerator $a(x)$ where $b(x) \neq 0$. Set $a(x) = 0$ and solve for $x$.

  • $y$--intercept: Calculated by evaluating $f(0)$, provided $x = 0$ is in the domain.

Oblique Asymptotes

• Condition: An oblique asymptote exists if and only if the degree $n$ of the numerator is exactly one greater than the degree $m$ of the denominator ($n = m + 1$).

• Determination Method: Divide the numerator $a(x)$ by the denominator $b(x)$ using polynomial long division or synthetic division:     $f(x) = \frac{a(x)}{b(x)} = q(x) + \frac{r(x)}{b(x)}$     As $x \to \pm\infty$, the remainder term $\frac{r(x)}{b(x)} \to 0$.     The equation of the oblique asymptote is given by the linear quotient polynomial:     $y = q(x)$

• Example: For $f(x) = \frac{x^2}{x + 1}$, dividing $x^2$ by $x + 1$ gives $x - 1 + \frac{1}{x + 1}$. The oblique asymptote is $y = x - 1$.

Step-by-Step Examples: Analyzing and Graphing Rational Functions

• Example 1A: Finding Asymptotes and Domain

  • Function:         $f(x) = \frac{x}{x - 1}$

  • Domain: Denominator cannot equal zero ($x - 1 \neq 0 \implies x \neq 1$).         $D = \{x \mid x \neq 1, x \in \mathbb{R}\}$

  • Vertical Asymptote: Zeros of denominator give $x = 1$

  • Horizontal Asymptote: Degree of numerator ($n = 1$) equals degree of denominator ($m = 1$). Ratio of leading coefficients is $\frac{1}{1} = 1$.         $y = 1$

  • End Behavior Check:         $\lim_{x \to -\infty} f(x) = 1 \quad \text{and} \quad \lim_{x \to \infty} f(x) = 1$

• Example 1B: Function with No Vertical Asymptotes

  • Function:         $f(x) = \frac{4x^2}{2x^2 + 1}$

  • Domain: The denominator $2x^2 + 1 = 0$ has imaginary zeros ($x = \pm \frac{i\sqrt{2}}{2}$).         $D = \{x \mid x \in \mathbb{R}\}$

  • Vertical Asymptotes: None.

  • Horizontal Asymptote: Degrees are equal ($n = 2, m = 2$). Ratio of leading coefficients is $\frac{4}{2} = 2$.         $y = 2$

• Example 2A: Graphing Rational Functions ($n < m$)

  • Function:         $k(x) = \frac{7}{x + 5}$

  • Domain: $D = \{x \mid x \neq -5, x \in \mathbb{R}\}$

  • Vertical Asymptote: $x = -5$

  • Horizontal Asymptote: Degree of numerator ($0$) is less than degree of denominator ($1$).         $y = 0$

  • Intercepts: No $x$-intercept because numerator $7 \neq 0$. $y$-intercept at $k(0) = \frac{7}{5} = 1.4$.

  • Selected Plotting Points: $k(4) = \frac{7}{9}$, $k(-6) = -7$.

• Example 2B: Graphing Functions with Multiple Vertical Asymptotes

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

  • Domain: $D = \{x \mid x \neq -2, 2, x \in \mathbb{R}\}$

  • Vertical Asymptotes: Set denominator to zero $\implies x = -2$ and $x = 2$

  • Horizontal Asymptote: Degree numerator ($1$) < degree denominator ($2$) $\implies y = 0$

  • Intercepts: $x$-intercept at numerator zero ($x = -1$); $y$-intercept at $f(0) = \frac{0 + 1}{0 - 4} = -0.25$

  • Test Intervals: $(-\infty, -2)$, $(-2, -1)$, $(-1, 2)$, $(2, \infty)$.

    • $f(-3) = -0.4$

    • $f(1) = -0.67$

    • $f(3) = 0.8$

• Example 3: Graphing Rational Functions with Oblique Asymptotes ($n = m + 1$)

  • Function:         $f(x) = \frac{x^2 + x - 8}{x + 3}$

  • Domain: $D = \{x \mid x \neq -3, x \in \mathbb{R}\}$

  • Vertical Asymptote: $x = -3$

  • Horizontal Asymptote: None ($n > m$).

  • Oblique Asymptote: Perform polynomial division $\frac{x^2 + x - 8}{x + 3} = x - 2 - \frac{2}{x + 3}$.         $y = x - 2$

  • $x$-intercepts: Solve $x^2 + x - 8 = 0$ using quadratic formula:         $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-8)}}{2(1)} = \frac{-1 \pm \sqrt{33}}{2} \approx 2.37 \text{ and } -3.37$

  • $y$-intercept: $f(0) = -\frac{8}{3} \approx -2.67$

  • Additional Test Points: $f(-4) = -4$.

• Example 4: Graphing Functions with Common Factors (Holes)

  • Function:         $h(x) = \frac{x^2 - 9}{x^2 - x - 6} = \frac{(x - 3)(x + 3)}{(x - 3)(x + 2)}$

  • Domain: $D = \{x \mid x \neq -2, 3, x \in \mathbb{R}\}$

  • Simplified Form:         $\tilde{h}(x) = \frac{x + 3}{x + 2} \quad \text{for } x \neq 3$

  • Vertical Asymptote: $x = -2$ (zero of simplified denominator).

  • Horizontal Asymptote: Degrees equal ($2 = 2$) $\implies y = 1$

  • $x$-intercept: $x = -3$ (zero of simplified numerator).

  • $y$-intercept: $h(0) = \frac{-9}{-6} = 1.5$

  • Hole (Removable Discontinuity): Occurs at $x = 3$. Calculate $y$-value using simplified form:         $\tilde{h}(3) = \frac{3 + 3}{3 + 2} = \frac{6}{5} = 1.2 \implies \text{Hole at } \left(3, \frac{6}{5}\right)$

Solving Rational Equations

• General Strategy: Multiply every term on both sides of the equation by the Least Common Denominator (LCD) of all rational terms to clear fractions. Solve the resulting polynomial equation and check all solutions against domain restrictions to eliminate extraneous solutions.

• Example 6A: Basic Rational Equation

  • Equation:         $x - \frac{4}{x - 6} = 0$

  • LCD: $x - 6$ (Restriction: $x \neq 6$)

  • Multiply by LCD:         $x(x - 6) - \frac{4}{x - 6}(x - 6) = 0(x - 6)$         $x^2 - 6x - 4 = 0$

  • Apply Quadratic Formula:         $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-4)}}{2(1)} = \frac{6 \pm \sqrt{52}}{2} = \frac{6 \pm 2\sqrt{13}}{2} = 3 \pm \sqrt{13}$

  • Solutions: $x = 3 + \sqrt{13} \approx 6.61$ and $x = 3 - \sqrt{13} \approx -0.61$. Both are valid.

• Example 6B: Single Discontinuity Equation

  • Equation:         $\frac{9}{x - 5} + 3 = 0$

  • Multiply by $(x - 5)$:         $9 + 3(x - 5) = 0 \implies 9 + 3x - 15 = 0 \implies 3x - 6 = 0 \implies x = 2$

• Example 7A: Equation with Extraneous Solutions

  • Equation:         $x + \frac{x}{x - 1} = \frac{3x - 2}{x - 1}$

  • LCD: $x - 1$ (Restriction: $x \neq 1$)

  • Multiply by LCD:         $x(x - 1) + x = 3x - 2$         $x^2 - x + x = 3x - 2$         $x^2 - 3x + 2 = 0$

  • Factor:         $(x - 2)(x - 1) = 0 \implies x = 2 \quad \text{or} \quad x = 1$

  • Check Domain: $x = 1$ causes division by zero in the original equation and must be rejected as extraneous.

  • Final Solution: $x = 2$

• Example 7B: Equation with Quadratic LCD

  • Equation:         $\frac{x}{x + 2} + \frac{6}{x - 5} = \frac{14}{x^2 - 3x - 10}$

  • Factor Denominator: $x^2 - 3x - 10 = (x + 2)(x - 5)$. LCD is $(x + 2)(x - 5)$. (Restrictions: $x \neq -2, 5$)

  • Multiply by LCD:         $x(x - 5) + 6(x + 2) = 14$         $x^2 - 5x + 6x + 12 = 14$         $x^2 + x - 2 = 0$

  • Factor:         $(x + 2)(x - 1) = 0 \implies x = -2 \quad \text{or} \quad x = 1$

  • Check Domain: $x = -2$ is extraneous.

  • Final Solution: $x = 1$

Real-World Applications

• Example 8: Boat Rate in Water Current

  • Problem: The rate of a river current is $4\,\text{mi/h}$. A boat travels $6\,\text{miles}$ downstream (with current) and $6\,\text{miles}$ upstream (against current) in a total time of $2\,\text{hours}$. Find the speed $r$ of the boat in still water.

  • Relationship: $\text{Time} = \frac{\text{Distance}}{\text{Rate}}$

    • Rate downstream: $r + 4$

    • Rate upstream: $r - 4$

  • Equation:         $\frac{6}{r - 4} + \frac{6}{r + 4} = 2$

  • LCD: $(r - 4)(r + 4)$ (Restrictions: $r \neq \pm 4$)

  • Algebraic Solution:         $(r - 4)(r + 4)\frac{6}{r - 4} + (r - 4)(r + 4)\frac{6}{r + 4} = 2(r - 4)(r + 4)$         $6(r + 4) + 6(r - 4) = 2(r^2 - 16)$         $6r + 24 + 6r - 24 = 2r^2 - 32$         $12r = 2r^2 - 32$         $2r^2 - 12r - 32 = 0$         $r^2 - 6r - 16 = 0$         $(r - 8)(r + 2) = 0 \implies r = 8 \quad \text{or} \quad r = -2$

  • Conclusion: Speed cannot be negative, so $r = -2$ is discarded. The speed of the boat in still water is $8\,\text{mi/h}$.

Practice Problems and Solutions

• Problem 1: Asymptotes and Holes Analysis

  • Find vertical/horizontal asymptotes and holes for:         $k(x) = \frac{x^2 + x - 20}{x^2 + 7x + 10}$

  • Factored Form: $k(x) = \frac{(x + 5)(x - 4)}{(x + 5)(x + 2)}$

  • Domain: $x \neq -5, -2$

  • Vertical Asymptote: $x = -2$

  • Horizontal Asymptote: $y = 1$

  • Hole: At $x = -5$. Value: $\frac{-5 - 4}{-5 + 2} = \frac{-9}{-3} = 3 \implies (-5, 3)$

• Problem 2: Zeros of a Rational Function

  • Find all zeros for:         $f(x) = \frac{x^2 - 9}{x^2 - 2x - 15}$

  • Factored Form: $f(x) = \frac{(x - 3)(x + 3)}{(x - 5)(x + 3)}$

  • Domain restrictions: $x \neq 5, -3$

  • Simplified Form: $\frac{x - 3}{x - 5}$

  • Zero: Set simplified numerator to zero $\implies x = 3$ ($x = -3$ is a hole, not a zero).

• Problem 3: Zeros of High-Degree Polynomial

  • Find zeros for:         $f(x) = x^5 - 2x^4 - 4x^3 + 8x^2 - x - 2$

  • Zeros: $x = 1$ (multiplicity of 2), $x = -2$, $x = 1 \pm \sqrt{2}$