Rational Functions: Asymptotes, Graphs, and Equations

General Form and Components of Rational Functions

• A rational function $f(x)$ is defined as a ratio of two polynomial expressions:   $f(x) = \frac{p(x)}{q(x)} = \frac{a_n x^n + \dots + a_0}{b_m x^m + \dots + b_0}$

  • The degree of the numerator polynomial $p(x)$ is denoted by $n$, with leading coefficient $a_n$.
  • The degree of the denominator polynomial $q(x)$ is denoted by $m$, with leading coefficient $b_m$.
  • The constant terms of the numerator and denominator are $a_0$ and $b_0$, respectively.

• Vertical Asymptotes (abbreviated VA):

  • A vertical asymptote occurs at any real value of $x$ that causes the denominator to equal zero ($q(x) = 0$), provided the factor does not cancel with the numerator.
  • Vertical asymptotes must be written as linear equations of lines: $x = c$.
  • A rational function can have zero, one, or multiple vertical asymptotes depending on the degree and roots of the denominator.

• Horizontal Asymptotes (abbreviated HA):

  • The existence and location of a horizontal asymptote depend on the relationship between the degree of the numerator $n$ and the degree of the denominator $m$:
  • If $n < m$: The horizontal asymptote is automatically at the line $y = 0$.
  • If $n = m$: The horizontal asymptote occurs at the ratio of the leading coefficients:       $y = \frac{a_n}{b_m}$
  • If $n > m$: There is no horizontal asymptote.
  • Horizontal asymptotes govern the end behavior of the graph as $x \rightarrow \pm\infty$. A rational function may cross its horizontal asymptote in the middle region of the graph.

• Slant Asymptotes (abbreviated SA):

  • If the numerator degree is exactly one greater than the denominator degree ($n = m + 1$), the function possesses a slant (oblique) asymptote instead of a horizontal asymptote.
  • The equation of the slant asymptote line, $y = mx + b$, is found by dividing the numerator polynomial by the denominator polynomial using polynomial long division or synthetic division.
  • The fractional remainder of the division is disregarded when formulating the linear equation of the slant asymptote.
  • If $n > m + 1$ (the degree of the numerator exceeds the denominator by two or more), there is no slant asymptote, and such higher-order end behaviors are omitted from consideration.

• Removable Discontinuities (Holes):

  • A hole in the graph occurs rather than a vertical asymptote if a common factor cancels completely between the numerator and the denominator.
  • Example: Consider the function:     $y = \frac{x^2 - 4}{x - 2}$     Factoring the numerator yields:     $y = \frac{(x - 2)(x + 2)}{x - 2}$     Canceling the common factor $(x - 2)$ leaves the simplified expression $y = x + 2$.
  • Because the original function had $(x - 2)$ in the denominator, the restriction $x \neq 2$ must still be enforced.

• Consequently, the graph is a straight line ($y = x + 2$) containing an open circle (hole) at $x = 2$, rather than a vertical asymptote.

Graphing Rational Functions: Guidelines and Techniques

• General Graphing Steps:
  • Factor both the numerator and the denominator completely.
  • Identify any common factors to locate removable discontinuities (holes).
  • Determine the vertical asymptote(s) by setting the remaining denominator equal to zero ($x = c$).
  • Determine the horizontal or slant asymptote by evaluating polynomial degrees.
  • Sketch all asymptotes as dashed (dotted) lines on the coordinate plane.
  • Plot ordered pairs to determine curve behavior:
  • For standard two-region graphs split by one vertical asymptote, calculate at least two points to the right and two points to the left of the asymptote.
  • For graphs partitioned into three regions by two vertical asymptotes (left, middle, right), compute at least two points in each exterior region and at least three points in the middle region.
  • The three points in the middle region are necessary to identify whether the curve forms a cubic-like inflection or a parabolic (quadratic-like) turn.
  • Draw continuous curves through the plotted points in each region, following the boundary paths prescribed by the asymptotes.

Graphing Rational Functions: Comprehensive Step-by-Step Examples

• Example 1: Basic Translation Graph   $f(x) = \frac{1}{x - 3}$

  • Transformation analysis: This represents the parent rational function $f(x) = \frac{1}{x}$ shifted horizontally to the right by $3$ units.
  • Vertical Asymptote:     $x - 3 = 0 \implies x = 3$
  • Horizontal Asymptote:
  • Numerator degree: $n = 0$
  • Denominator degree: $m = 1$
  • Since $0 < 1$, the horizontal asymptote is $y = 0$.
  • Points to the right of the vertical asymptote ($x > 3$):
  • When $x = 4$: $f(4) = \frac{1}{4 - 3} = 1$
  • When $x = 5$: $f(5) = \frac{1}{5 - 3} = \frac{1}{2}$
  • Points to the left of the vertical asymptote ($x < 3$):
  • When $x = 2$: $f(2) = \frac{1}{2 - 3} = -1$
  • When $x = 1$: $f(1) = \frac{1}{1 - 3} = -\frac{1}{2}$

• Example 2: Equal Degrees in Numerator and Denominator   $f(x) = \frac{1 - 3x}{1 - x}$

  • Vertical Asymptote:     $1 - x = 0 \implies x = 1$
  • Horizontal Asymptote:
  • Numerator degree: $n = 1$
  • Denominator degree: $m = 1$
  • Since $n = m$, take the ratio of leading coefficients:       $y = \frac{-3}{-1} = 3$
  • Points to the right of the vertical asymptote ($x > 1$):
  • When $x = 2$:       $f(2) = \frac{1 - 3(2)}{1 - 2} = \frac{1 - 6}{-1} = \frac{-5}{-1} = 5$
  • When $x = 3$:       $f(3) = \frac{1 - 3(3)}{1 - 3} = \frac{1 - 9}{-2} = \frac{-8}{-2} = 4$
  • Points to the left of the vertical asymptote ($x < 1$):
  • When $x = 0$:       $f(0) = \frac{1 - 3(0)}{1 - 0} = \frac{1}{1} = 1$
  • When $x = -1$:       $f(-1) = \frac{1 - 3(-1)}{1 - (-1)} = \frac{1 + 3}{1 + 1} = \frac{4}{2} = 2$

• Example 3: Function with a Removable Discontinuity and Vertical Asymptote   $f(x) = \frac{x + 4}{x^2 - 16}$

  • Factoring denominator:     $f(x) = \frac{x + 4}{(x + 4)(x - 4)}$
  • Simplification:     $f(x) = \frac{1}{x - 4}$
  • Removable Discontinuity (Hole):
  • The factor $(x + 4)$ cancels, meaning $x \neq -4$.
  • A hole exists at $x = -4$.
  • Vertical Asymptote:     $x - 4 = 0 \implies x = 4$
  • Horizontal Asymptote:
  • Degree of reduced numerator is $0$, degree of reduced denominator is $1$.
  • Since $0 < 1$, the horizontal asymptote is $y = 0$.
  • Points to the right of the vertical asymptote ($x > 4$):
  • When $x = 5$: $f(5) = \frac{1}{5 - 4} = 1$
  • When $x = 6$: $f(6) = \frac{1}{6 - 4} = \frac{1}{2}$
  • Points to the left of the vertical asymptote ($x < 4$):
  • When $x = 3$: $f(3) = \frac{1}{3 - 4} = -1$
  • When $x = 2$: $f(2) = \frac{1}{2 - 4} = -\frac{1}{2}$
  • Graph representation: The curve is drawn through these coordinates following asymptotes $x = 4$ and $y = 0$, extending past $x = -4$ where an open circle is plotted to mark the hole.

• Example 4: Function with Two Vertical Asymptotes (Three-Region Graph)   $f(x) = \frac{x}{x^2 - 9} = \frac{x}{(x - 3)(x + 3)}$

  • Vertical Asymptotes:     $x - 3 = 0 \implies x = 3$$x + 3 = 0 \implies x = -3$
  • Horizontal Asymptote:
  • Numerator degree: $n = 1$
  • Denominator degree: $m = 2$
  • Since $1 < 2$, horizontal asymptote is $y = 0$.
  • Points in the right region ($x > 3$):
  • When $x = 4$:       $f(4) = \frac{4}{(4 - 3)(4 + 3)} = \frac{4}{(1)(7)} = \frac{4}{7}$
  • When $x = 5$:       $f(5) = \frac{5}{(5 - 3)(5 + 3)} = \frac{5}{(2)(8)} = \frac{5}{16}$
  • Points in the left region ($x < -3$):
  • Due to odd symmetry across the origin:
  • When $x = -4$: $f(-4) = -\frac{4}{7}$
  • When $x = -5$: $f(-5) = -\frac{5}{16}$
  • Points in the middle region ($-3 < x < 3$):
  • Three points are evaluated to determine the behavior:
  • When $x = 0$:       $f(0) = \frac{0}{0 - 9} = 0$       (The function touches and crosses the horizontal asymptote at the origin $(0, 0)$).
  • When $x = 1$:       $f(1) = \frac{1}{(1 - 3)(1 + 3)} = \frac{1}{(-2)(4)} = -\frac{1}{8}$
  • When $x = -1$:       $f(-1) = \frac{-1}{(-1 - 3)(-1 + 3)} = \frac{-1}{(-4)(2)} = \frac{-1}{-8} = \frac{1}{8}$
  • Because the middle values switch signs ($f(-1) > 0$ and $f(1) < 0$), the middle curve takes on an inverted cubic format.

• Example 5: Graphing with a Slant Asymptote   $y = \frac{x^2 - 3}{x + 1}$

  • Vertical Asymptote:     $x + 1 = 0 \implies x = -1$
  • Horizontal Asymptote:
  • Numerator degree: $n = 2$
  • Denominator degree: $m = 1$
  • Since $2 > 1$, there is no horizontal asymptote.
  • Slant Asymptote:
  • Because $n = m + 1$ ($2 = 1 + 1$), a slant asymptote exists.
  • Synthetic division of $(x^2 + 0x - 3)$ divided by $(x + 1)$ using root $-1$:
    • Coefficients: $1$, $0$, $-3$
    • Bring down the leading coefficient: $1$
    • Multiply and add: $-1 \times 1 = -1$; $0 + (-1) = -1$
    • Multiply and add: $-1 \times (-1) = 1$; $-3 + 1 = -2$ (remainder)
    • Quotient polynomial: $x - 1$
  • Ignoring the remainder, the slant asymptote is the line:       $y = x - 1$
  • Points on the slant asymptote line: $(0, -1)$ and $(1, 0)$.
  • Points to the right of the vertical asymptote ($x > -1$):
  • When $x = 0$:       $y = \frac{0^2 - 3}{0 + 1} = \frac{-3}{1} = -3$
  • When $x = 1$:       $y = \frac{1^2 - 3}{1 + 1} = \frac{1 - 3}{2} = \frac{-2}{2} = -1$
  • Points to the left of the vertical asymptote ($x < -1$):
  • When $x = -2$:       $y = \frac{(-2)^2 - 3}{-2 + 1} = \frac{4 - 3}{-1} = \frac{1}{-1} = -1$
  • When $x = -3$:       $y = \frac{(-3)^2 - 3}{-3 + 1} = \frac{9 - 3}{-2} = \frac{6}{-2} = -3$

• Example 6: Two Vertical Asymptotes with Parabolic Center Shape   $y = \frac{x^2}{1 - x^2}$

  • Factoring denominator:     $y = \frac{x^2}{(1 + x)(1 - x)}$
  • Vertical Asymptotes:     $x = -1$$x = 1$
  • Horizontal Asymptote:
  • Numerator degree: $n = 2$
  • Denominator degree: $m = 2$
  • Leading coefficient of numerator: $1$
  • Leading coefficient of denominator: $-1$
  • Ratio:       $y = \frac{1}{-1} = -1$
  • Points in the right region ($x > 1$):
  • When $x = 2$:       $y = \frac{2^2}{1 - 2^2} = \frac{4}{1 - 4} = -\frac{4}{3} = -1\frac{1}{3}$
  • When $x = 3$:       $y = \frac{3^2}{1 - 3^2} = \frac{9}{1 - 9} = -\frac{9}{8}$
  • Points in the left region ($x < -1$):
  • Due to even powers of $x$:
  • When $x = -2$: $y = -\frac{4}{3}$
  • When $x = -3$: $y = -\frac{9}{8}$
  • Points in the middle region ($-1 < x < 1$):
  • When $x = 0$:       $y = \frac{0^2}{1 - 0^2} = 0$
  • When $x = \frac{1}{2}$:       $y = \frac{(1/2)^2}{1 - (1/2)^2} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$
  • When $x = -\frac{1}{2}$:       $y = \frac{(-1/2)^2}{1 - (-1/2)^2} = \frac{1/4}{3/4} = \frac{1}{3}$
  • Both non-zero evaluations are positive, showing that the middle section forms an upward-opening parabolic curve.

Solving Rational Equations

• Procedural Method:

  • Factor every denominator completely to establish individual factors.
  • Determine the Least Common Denominator (LCD) containing every distinct factor across all terms.
  • Multiply each term on both sides of the equation by the LCD to clear all denominators.
  • Distribute constants and signs carefully, using explicit parentheses around polynomial groupings to avoid sign errors when distributing negative numbers.
  • Solve the resulting polynomial equation (linear or quadratic).
  • Extraneous Solution Check:
  • Any prospective solution that produces a value of zero in any denominator of the original equation cannot be accepted and must be discarded.

• Problem 1:   $\frac{2x}{x^2 - 1} = \frac{2}{x + 1} - \frac{1}{x - 1}$

  • Factor denominators:     $\frac{2x}{(x + 1)(x - 1)} = \frac{2}{x + 1} - \frac{1}{x - 1}$
  • LCD is $(x + 1)(x - 1)$.
  • Multiply all terms by the LCD:     $2x = 2(x - 1) - 1(x + 1)$
  • Distribute:     $2x = 2x - 2 - x - 1$
  • Combine like terms on the right side:     $2x = x - 3$
  • Subtract $x$ from both sides:     $x = -3$
  • Domain restriction check: Denominators are zero at $x = 1$ and $x = -1$. Because $-3 \neq \pm 1$, the solution is valid:     $x = -3$

• Problem 2:   $\frac{2}{x^2 - 2x} - \frac{3}{x^2 - x} = 0$

  • Factor denominators:     $\frac{2}{x(x - 2)} - \frac{3}{x(x - 1)} = 0$
  • LCD is $x(x - 2)(x - 1)$.
  • Multiply all terms by the LCD:     $2(x - 1) - 3(x - 2) = 0$
  • Distribute:     $2x - 2 - 3x + 6 = 0$
  • Combine like terms:     $-x + 4 = 0$
  • Solve for $x$:     $x = 4$
  • Domain restriction check: Denominators are zero at $x = 0$, $x = 2$, and $x = 1$. Because $4$ does not equal any of these values, the solution is valid:     $x = 4$

• Problem 3:   $\frac{2x}{x - 3} + \frac{4}{x} - 6 = 0$

  • LCD is $x(x - 3)$.
  • Multiply all terms by the LCD:     $2x(x) + 4(x - 3) - 6(x)(x - 3) = 0$
  • Expand terms:     $2x^2 + 4x - 12 - 6x^2 + 18x = 0$
  • Combine like terms:     $-4x^2 + 22x - 12 = 0$
  • Move terms to the opposite side to make the leading coefficient positive:     $4x^2 - 22x + 12 = 0$
  • Factor out the greatest common factor of $2$:     $2(2x^2 - 11x + 6) = 0 \implies 2x^2 - 11x + 6 = 0$
  • Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 2$, $b = -11$, $c = 6$:     $x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(2)(6)}}{2(2)}$$x = \frac{11 \pm \sqrt{121 - 48}}{4}$$x = \frac{11 \pm \sqrt{73}}{4}$
  • Because $73$ is prime, it cannot be simplified further.
  • The two distinct solutions are:     $x = \frac{11 + \sqrt{73}}{4}$$x = \frac{11 - \sqrt{73}}{4}$
  • Restrictions are $x \neq 3$ and $x \neq 0$. Both irrational roots are valid.

• Problem 4:   $\frac{1}{x + 2} + \frac{3}{x + 7} = \frac{5}{x^2 + 9x + 14}$

  • Factor the denominator on the right side:     $x^2 + 9x + 14 = (x + 2)(x + 7)$
  • LCD is $(x + 2)(x + 7)$.
  • Multiply every term by the LCD:     $1(x + 7) + 3(x + 2) = 5$
  • Expand and simplify:     $x + 7 + 3x + 6 = 5$$4x + 13 = 5$
  • Subtract $13$ from both sides:     $4x = -8$
  • Divide by $4$:     $x = -2$
  • Check against original denominators: Substituting $x = -2$ into $\frac{1}{x + 2}$ yields a division by zero.
  • The value $x = -2$ is extraneous and must be discarded, resulting in:     $\text{No solution}$

• Problem 5:   $\frac{4}{x^2 - 3x} - \frac{1}{x^2 - 9} = 0$

  • Factor denominators:     $\frac{4}{x(x - 3)} - \frac{1}{(x - 3)(x + 3)} = 0$
  • LCD is $x(x - 3)(x + 3)$.
  • Multiply all terms by the LCD:     $4(x + 3) - 1(x) = 0$
  • Distribute and combine like terms:     $4x + 12 - x = 0$$3x + 12 = 0$
  • Subtract $12$ and divide by $3$:     $3x = -12 \implies x = -4$
  • Check restrictions: Denominators are zero at $x = 0$, $x = 3$, and $x = -3$. Because $-4$ does not equal any restriction, the solution is valid:     $x = -4$

Questions & Discussion

• Vertical Asymptote Input Formatting:

  • Student query: Does online homework accept vertical asymptotes entered with a plus-or-minus symbol ($\pm$), or must they be entered as separate equations?
  • Clarification: It is safer to write them as two distinct linear equations (e.g., $x = 3$ and $x = -3$). If a platform provides a plus-or-minus palette, it may be attempted, but writing separate linear equations avoids parsing errors.

• Identifying Common Denominator Elements:

  • Student query: Is the common denominator always just the product of the first two denominators presented in the equation?
  • Clarification: No. The common denominator must be constructed from whatever unique factors are present across all denominators in the expression. If a factor is repeated in a single denominator to a power, that higher exponent must be included.

• Determining Solutions with Radicals:

  • Student query: What are the two distinct solutions resulting from the quadratic formula expression $\frac{11 \pm \sqrt{73}}{4}$?
  • Clarification: The expression represents two separate numeric values: $\frac{11 + \sqrt{73}}{4}$ and $\frac{11 - \sqrt{73}}{4}$.

• Common Mistakes in Solving Equations with Numerators Greater than One:

  • Discussion on Problem 5: A student shared obtaining an incorrect result of $x = -2$ by ignoring the numerator of $4$ on the first term and treating it as $1$.
  • Clarification: When clearing fractions, the remaining factor from the LCD must always be multiplied by the existing numerator of that specific term (e.g., $4(x + 3)$, not $1(x + 3)$). Disregarding numerators during fraction-clearing produces invalid equations.

Course Logistics and Deadlines

• Section 5.1 assignment: Due tonight.
• Section 5.2 assignment: Due on Tuesday.
• Section 5.3 assignment: Due date rescheduled to next Thursday; work should begin on the first half immediately.
• Upcoming Quiz Coverage: The quiz encompasses two and a half sections: Section 5.1, Section 5.2, and the first half of Section 5.3 (up through solving rational equations with cleared denominators).