Section 5.6: Rational Functions and Their Graphs

Definition and Examples of Rational Functions

• Definition of a Rational Function: A rational function is defined as a function that can be expressed as the quotient of two polynomial functions. It is written in the form:     $f(x) = \frac{p(x)}{q(x)}$

  • In this formula, $p(x)$ and $q(x)$ are polynomial functions.

  • A critical constraint is that the denominator polynomial must not be zero: $q(x) \neq 0$.

• Examples of Rational Functions:

  • $f(x) = \frac{3x}{x^2+5}$

  • $g(x) = \frac{x^3+7x^2-3x-4}{4x^3-6x+9}$

  • $y = - \frac{7}{(x+4)(x-8)}$

  • In each example, both the numerator and the denominator are polynomials.

Domain of a Rational Function

• General Rule: The domain of a rational function consists of the set of all real numbers except those values that make the denominator equal to $0$.

• Procedural Steps:

  1. Set the denominator expression to be "not equal to $0$".

  2. Solve for $x$.

  3. Exclude these values from the set of all real numbers.

• Example 1a: $f(x) = \frac{5x}{x^2-4}$

  • Denominator condition: $x^2 - 4 \neq 0$

  • $x^2 \neq 4 \implies x \neq -2, 2$

  • Domain in interval notation: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

• Example 1b: $g(x) = \frac{x^2+3x+2}{x^2+9}$

  • Denominator condition: $x^2 + 9 \neq 0$

  • $x^2 \neq -9$. Since the square of a real number is always non-negative, $x^2 + 9$ is always positive and never zero. There are no real solutions to $x^2 = -9$.

  • Domain: $(-\infty, \infty)$

• Example 1c: $r(x) = \frac{2x+3}{4(x-1)(x+3)}$

  • Denominator condition: $4(x-1)(x+3) \neq 0$

  • Factors solved: $x - 1 \neq 0 \implies x \neq 1$ and $x + 3 \neq 0 \implies x \neq -3$

  • Domain: $(-\infty, -3) \cup (-3, 1) \cup (1, \infty)$

• Example 1d: $y = \frac{(x+6)(x+9)}{(x+6)(x-4)}$

  • Denominator condition: $(x+6)(x-4) \neq 0 \implies x \neq -6, 4$

  • Domain: $(-\infty, -6) \cup (-6, 4) \cup (4, \infty)$

  • CRITICAL NOTE: Always find the domain BEFORE canceling out any common factors from the numerator and denominator.

• Practice Exercises (Think-Pair-Share 1):

  • a) $y = -\frac{2x(x-3)}{x^2(x+6)}$

  • b) $h(x) = \frac{6x^3-5x}{x^2+2x-8}$

Removable Discontinuities (Holes)

• Definition: A removable discontinuity occurs at $x = a$ if $(x - a)$ is a common factor in both the numerator and the denominator, provided the multiplicity of the factor in the numerator is greater than or equal to its multiplicity in the denominator.

• Visual Representation: On a graph, a removable discontinuity appears as a "hole" at the point where $x = a$.

• Procedure to Find Holes:

  1. Factorize both the numerator and the denominator completely.

  2. Identify common factors.

  3. Set the common factor equal to zero and solve for $x$.

• Example 2a: $f(x) = \frac{3x(x-5)}{-x(2x+3)}$

  • Common factor: $x$

  • Hole equation: $x = 0$. The graph has a hole at $x = 0$.

  • Domain Calculation (before canceling): Denominator $-x(2x+3)=0 \implies x=0, x=-\frac{3}{2}$.

  • Domain: $(-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, 0) \cup (0, \infty)$

• Example 2b: $t(x) = \frac{x^2-6x+5}{x^2-1}$

  • Factored form: $t(x) = \frac{(x-5)(x-1)}{(x+1)(x-1)}$

  • Common factor: $(x-1)$. Hole at $x - 1 = 0 \implies x = 1$.

Finding X- and Y-Intercepts

• Order of Operations: Factorize and reduce the function by canceling common factors BEFORE finding intercepts.

• X-Intercepts:

  • Set the numerator of the reduced function equal to $0$ and solve for $x$.

  • The results are points of the form $(x, 0)$.

• Y-Intercept:

  • Substitute $x = 0$ into the original or reduced function, provided $x = 0$ is in the function's domain.

  • If $0$ makes the denominator zero, the function has no y-intercept.

• Example 3a: $f(x) = \frac{4x^2-7x}{2x(x-6)}$

  • Factored: $f(x) = \frac{x(4x-7)}{2x(x-6)}$

  • Reduced: $f(x) = \frac{4x-7}{2(x-6)}$

  • x-intercept: $4x - 7 = 0 \implies x = \frac{7}{4}$. Point: $(\frac{7}{4}, 0)$.

  • y-intercept: Since $x = 0$ was excluded from the domain (denominator was $2x(x-6)$), there is no y-intercept.

• Example 3b: $g(x) = \frac{x^2+x-12}{x^2+9x+20}$

  • Factored: $g(x) = \frac{(x+4)(x-3)}{(x+4)(x+5)}$

  • Reduced: $g(x) = \frac{x-3}{x+5}$

  • x-intercept: $x - 3 = 0 \implies x = 3$. Point: $(3, 0)$.

  • y-intercept: $g(0) = \frac{0-3}{0+5} = -\frac{3}{5}$. Point: $(0, -\frac{3}{5})$.

Arrow Notation

• Symbolic Summary:

  • $x \rightarrow a^-$: $x$ approaches $a$ from the left (x < a).

  • $x \rightarrow a^+$: $x$ approaches $a$ from the right (x > a).

  • $x \rightarrow \infty$: $x$ increases without bound (positive infinity).

  • $x \rightarrow -\infty$: $x$ decreases without bound (negative infinity).

  • $f(x) \rightarrow \infty$: Output increases without bound.

  • $f(x) \rightarrow -\infty$: Output decreases without bound.

  • $f(x) \rightarrow a$: Output approaches a constant value $a$.

• Behavior of $f(x) = \frac{1}{x}$ Example:

  • As $x \rightarrow -\infty, f(x) \rightarrow 0$

  • As $x \rightarrow \infty, f(x) \rightarrow 0$

  • As $x \rightarrow 0^-, f(x) \rightarrow -\infty$

  • As $x \rightarrow 0^+, f(x) \rightarrow \infty$

Vertical Asymptotes

• Definition: A vertical line $x = a$ is a vertical asymptote if the output values approach infinity or negative infinity as the input values approach $a$:

  • If $x \rightarrow a^-$ or $x \rightarrow a^+$, then $f(x) \rightarrow \pm \infty$.

• Procedure to Find Vertical Asymptotes:

  1. Factor the numerator and denominator.

  2. Reduce the function by canceling common factors.

  3. Set the remaining denominator equal to zero. The solutions are the equations of the vertical asymptotes.

• Example 4a: $f(x) = \frac{-5}{3x-1}$

  • Cannot be reduced further. Denominator zero: $3x - 1 = 0 \implies x = \frac{1}{3}$.

  • Arrow notation for this graph: As $x \rightarrow \frac{1}{3}^-, f(x) \rightarrow \infty$ and as $x \rightarrow \frac{1}{3}^+, f(x) \rightarrow -\infty$.

• Example 4b: $g(x) = \frac{x^2-5x+4}{(x^2-1)(x+4)}$

  • Factored: $g(x) = \frac{(x-1)(x-4)}{(x-1)(x+1)(x+4)}$

  • Reduced: $g(x) = \frac{x-4}{(x+1)(x+4)}$

  • Hole: At $x = 1$ (common factor).

  • Vertical Asymptotes: From remaining denominator, $x + 1 = 0 \implies x = -1$ and $x + 4 = 0 \implies x = -4$.

• Example 4c: $h(x) = \frac{x^2+7x+10}{x+5} = \frac{(x+2)(x+5)}{x+5} = x+2$

  • Hole exists at $x = -5$.

  • Since no $x$ remains in the denominator after reduction, there are no vertical asymptotes.

Horizontal and Slant Asymptotes

• Concept: End behavior ($x \rightarrow \pm \infty$) of a rational function is determined by the ratio of the leading terms of the numerator and denominator.

• Determining Asymptotes by Degree (where $n$ = degree of numerator and $d$ = degree of denominator):

  • Case 1: n < d

    • There is a horizontal asymptote at the line $y = 0$ (the x-axis).

    • Example: $f(x) = \frac{4x+2}{x^2+4x-5}$.

  • Case 2: $n = d$

    • There is a horizontal asymptote at the line $y = \frac{a_n}{b_n}$, where $a_n$ is the leading coefficient of the numerator and $b_n$ is the leading coefficient of the denominator.

    • Example: $f(x) = \frac{3x^2+2}{x^2+4x-5} \implies y = \frac{3}{1} = 3$.

    • Example: $g(x) = \frac{-2x^3+x^2-9}{5x^3-6x+1} \implies y = -\frac{2}{5}$.

  • Case 3: n > d

    • No horizontal asymptote exists.

    • If $n = d + 1$, a slant (oblique) asymptote exists. Its equation is found by performing polynomial division; the quotient (ignoring remainder) is the equation of the slant asymptote.

    • Example: $f(x) = \frac{3x^2-2x+1}{x-1}$. Division yields a quotient of $3x+1$. Slant asymptote: $y = 3x+1$.

Graphing Rational Functions (Step-by-Step Process)

1. Factorize: Completely factor the numerator and denominator.

2. Domain: Identify all restrictions on $x$ (set denominator $\neq 0$).

3. Reduce: Cancel out common factors.

4. Holes: Set common factors equal to zero to find removable discontinuities.

5. Intercepts:

   • Find x-intercepts by setting the reduced numerator to zero.

   • Determine behavior (cross vs. bounce) based on multiplicity: Odd multiplicity crosses; even multiplicity bounces.

   • Find y-intercept by calculating $f(0)$.

6. Vertical Asymptotes: Set the reduced denominator to zero.

   • Multiplicity tells us behavior near the asymptote:

     • Even multiplicity factors in the denominator mean the graph goes toward the same infinity ($\infty$ or $-\infty$) on both sides of the asymptote.

     • Odd multiplicity factors mean the graph goes toward opposite infinities on either side.

7. Horizontal/Slant Asymptotes: Apply the degree-comparison rules.

8. Sketch: Combine all points, holes, and asymptotes to draw the curve.

Detailed Graphing Example

• Function: $f(x) = \frac{(x-2)^2(x+3)}{(x-1)^2(x^2+7x+12)}$

• Step 1 (Factoring): $f(x) = \frac{(x-2)^2(x+3)}{(x-1)^2(x+4)(x+3)}$

• Step 2 (Domain): $(x-1)^2(x+4)(x+3) \neq 0 \implies x \neq 1, -4, -3$

• Step 3 (Reduce): $f(x) = \frac{(x-2)^2}{(x-1)^2(x+4)}$

• Step 4 (Hole): Common factor $x+3$ implies a hole at $x = -3$.

• Step 5 (Intercepts):

  • X-intercept: $(x-2)^2 = 0 \implies x = 2$. Since the multiplicity is $2$, the graph bounces off the x-axis at $(2,0)$.

  • Y-intercept: $f(0) = \frac{(0-2)^2}{(0-1)^2(0+4)} = \frac{4}{1(4)} = 1$. Point: $(0,1)$.

• Step 6 (Vertical Asymptotes):

  • From $(x-1)^2(x+4) = 0$

  • $x = 1$: Multiplicity $2$. Same direction behavior. As $x \rightarrow 1^-, f(x) \rightarrow \infty$ and as $x \rightarrow 1^+, f(x) \rightarrow \infty$.

  • $x = -4$: Multiplicity $1$. Opposite direction behavior. As $x \rightarrow -4^-, f(x) \rightarrow -\infty$ and as $x \rightarrow -4^+, f(x) \rightarrow \infty$.

• Step 7 (Horizontal Asymptote): Master degree of numerator ($2$) vs master degree of denominator ($3$). Since 2 < 3, Case 1 applies: $y = 0$.