17.2 Rational Functions, Domain Restrictions, Holes & Asymptotes

17.2 Rational Functions, Domain Restrictions, Holes & Asymptotes

Structure and Domain of Rational Functions

A rational function is defined as the quotient of two polynomial functions: $R(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials and $Q(x)$ is not the zero polynomial ($Q(x) \not\equiv 0$). Because division by zero is undefined in real analysis, the domain of $R(x)$ consists of all real numbers except the zeros of the denominator: $\text{Domain}(R) = \{x \in \mathbb{R} \mid Q(x) \neq 0\}$

A critical procedural rule tested on the FTCE Mathematics 6-12 exam is that the domain must be established before canceling any common factors. Canceling common factors simplifies algebraic evaluation but obscures values of $x$ where the original function is undefined.

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Discontinuity Classification: Removable Holes vs. Vertical Asymptotes

To analyze the discontinuities of $R(x)$, factor both $P(x)$ and $Q(x)$ completely over the real numbers. Suppose $x = c$ is a zero of the denominator, so $Q(c) = 0$. The geometric nature of the discontinuity at $x = c$ depends on whether $(x - c)$ is also a factor of the numerator $P(x)$:

1. Removable Discontinuity (Hole in the Graph): If $(x - c)$ appears in the numerator to a power greater than or equal to its power in the denominator—that is, if $(x - c)$ cancels completely from the denominator—then $x = c$ is a removable discontinuity. The graph of $R(x)$ behaves identically to the reduced function $R_{\text{reduced}}(x)$ everywhere except at $x = c$, where a single open point (hole) exists. The exact coordinates of the hole are $(c, L)$, where $L$ is the two-sided limit evaluated in the reduced expression: $L = \lim_{x \to c} R(x) = R_{\text{reduced}}(c)$

2. Vertical Asymptote (Non-Removable Infinite Discontinuity): If after canceling all common factors, $(x - c)$ remains in the denominator to power $k \ge 1$, then the line $x = c$ is a vertical asymptote. As $x$ approaches $c$, the denominator approaches $0$ while the numerator approaches a non-zero finite real number, forcing the quotient magnitude $|R(x)| \to \infty$.

Local Behavior Near Vertical Asymptotes (Factor Multiplicity)

The multiplicity $k$ of the remaining factor $(x - c)^k$ in the reduced denominator governs how the curve approaches the vertical asymptote:

• Odd Multiplicity ($k = 1, 3, 5, \dots$): The factor $(x - c)^k$ changes algebraic sign as $x$ passes through $c$. Consequently, $R(x)$ approaches opposite infinities on either side: one side shoots to $+\infty$ while the other shoots to $-\infty$.
• Even Multiplicity ($k = 2, 4, 6, \dots$): Because $(x - c)^k > 0$ for all $x \neq c$, the factor does not change sign across $c$. Consequently, $R(x)$ approaches the same infinity on both sides: both sides shoot toward $+\infty$ or both sides shoot toward $-\infty$.

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Horizontal, Slant (Oblique), and Polynomial Asymptotes

While vertical asymptotes describe local singularities where the denominator vanishes, horizontal and slant asymptotes describe global end behavior as $x \to \pm\infty$. Let the degrees of the numerator and denominator be $n = \deg(P)$ and $m = \deg(Q)$, with leading coefficients $a_n$ and $b_m$: $R(x) = \frac{a_n x^n + a_{n-1}x^{n-1} + \dots + a_0}{b_m x^m + b_{m-1}x^{m-1} + \dots + b_0}$

The existence and equation of non-vertical asymptotes depend entirely on the degree comparison between $n$ and $m$:

1. $n < m$ (Bottom-Heavy Rational Function): The denominator grows faster than the numerator as $|x| \to \infty$. Therefore, $\lim_{x \to \pm\infty} R(x) = 0$. The line $y = 0$ ($x$-axis) is a horizontal asymptote.

2. $n = m$ (Balanced Degrees): The numerator and denominator grow at identical asymptotic rates. Dividing every term by $x^n$ shows that all lower-order terms vanish as $|x| \to \infty$, leaving: $\lim_{x \to \pm\infty} R(x) = \frac{a_n}{b_m}$ The line $y = \frac{a_n}{b_m}$ is a horizontal asymptote.

3. $n = m + 1$ (Numerator Degree Exceeds Denominator by Exactly 1): There is no horizontal asymptote. By polynomial long division, $R(x)$ can be decomposed into a linear quotient plus a strictly proper rational remainder: $R(x) = (mx + b) + \frac{R_{\text{rem}}(x)}{Q(x)}$ where $\deg(R_{\text{rem}}) < \deg(Q)$. Because $\lim_{x \to \pm\infty} \frac{R_{\text{rem}}(x)}{Q(x)} = 0$, the graph approaches the line $y = mx + b$, which is a slant (oblique) asymptote.

4. $n > m + 1$ (Numerator Degree Exceeds Denominator by More Than 1): The function has neither a horizontal nor a slant asymptote. Instead, polynomial division produces a non-linear quotient $Q(x)$ of degree $n - m$ that represents the polynomial end behavior (e.g., parabolic end behavior if $n = m + 2$).

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Intersections with Horizontal and Slant Asymptotes

A pervasive student misconception is that 'a curve can never cross an asymptote.' On the FTCE exam, distinguish sharply between vertical and non-vertical asymptotes:

• Vertical Asymptotes can NEVER be crossed, because the function is undefined at those input values ($x = c \notin \text{Domain}$). A graph intersecting a vertical line at a function value would violate the vertical line test.
• Horizontal and Slant Asymptotes CAN be intersected, sometimes multiple times or infinitely often (as with damped oscillatory functions). Non-vertical asymptotes describe boundary behavior at infinity ($x \to \pm\infty$), imposing no restriction on function values for finite $x$.

To find where a rational function intersects its horizontal or slant asymptote, set the function equal to the asymptote equation $R(x) = y_{\text{asymptote}}$ and solve the resulting algebraic equation for real solutions $x$.

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Asymptote Rules Decision Matrix

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Worked Exemplar: Slant Asymptote, Removable Discontinuity, and Asymptotic Behavior

Problem: Analyze the rational function $f(x) = \frac{2x^3 - 4x^2 - 6x}{x^2 - 1}$. Determine its domain, identify all discontinuities as holes or vertical asymptotes, find all horizontal or slant asymptotes, check for asymptote intersections, and state all intercepts.

1. Factor Completely and Establish Domain: $P(x) = 2x(x^2 - 2x - 3) = 2x(x - 3)(x + 1)$ $Q(x) = (x - 1)(x + 1)$ The denominator vanishes when $x = 1$ or $x = -1$. The domain is all real numbers except $1$ and $-1$: $\text{Domain}(f) = \{x \in \mathbb{R} \mid x \neq 1, x \neq -1\} = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$

2. Classify Discontinuities: The factor $(x + 1)$ appears in both numerator and denominator and cancels completely: $f_{\text{reduced}}(x) = \frac{2x(x - 3)}{x - 1} = \frac{2x^2 - 6x}{x - 1}, \quad x \neq -1$

   • Removable Discontinuity (Hole) at $x = -1$: Evaluate the limit in the reduced expression: $y_{\text{hole}} = \lim_{x \to -1} f(x) = \frac{2(-1)(-1 - 3)}{-1 - 1} = \frac{2(-1)(-4)}{-2} = \frac{8}{-2} = -4$ The graph has a removable hole at $(-1, -4)$.
   • Vertical Asymptote at $x = 1$: The non-removable denominator zero is $x = 1$ (multiplicity 1). The line $x = 1$ is a vertical asymptote. Examining one-sided limits: $\lim_{x \to 1^+} \frac{2x(x - 3)}{x - 1} = \frac{2(1)(-2)}{0^+} = \frac{-4}{0^+} = -\infty$ $\lim_{x \to 1^-} \frac{2x(x - 3)}{x - 1} = \frac{-4}{0^-} = +\infty$ Because the factor $(x - 1)^1$ has odd multiplicity, the graph approaches opposite infinities on either side.

3. Determine End Behavior (Slant Asymptote): The degree of the numerator is $n = 3$ and the denominator is $m = 2$. Since $n = m + 1$, there is no horizontal asymptote; instead, there is a slant asymptote. Dividing the reduced numerator $2x^2 - 6x$ by $x - 1$: $\frac{2x^2 - 6x}{x - 1} = 2x - 4 - \frac{4}{x - 1}$ As $x \to \pm\infty$, $-\frac{4}{x - 1} \to 0$. The slant asymptote is the line $y = 2x - 4$.

4. Check for Intersections with the Slant Asymptote: Set $f_{\text{reduced}}(x) = 2x - 4$: $2x - 4 - \frac{4}{x - 1} = 2x - 4 \implies -\frac{4}{x - 1} = 0$ Because $-4 \neq 0$, this equation has no solution. The graph never intersects its slant asymptote.

5. Intercepts:

   • $y$-intercept: $f(0) = \frac{2(0)(0 - 3)}{0 - 1} = 0 \implies (0, 0)$.
   • $x$-intercepts: set reduced numerator equal to zero: $2x(x - 3) = 0 \implies x = 0$ or $x = 3$. The $x$-intercepts are $(0, 0)$ and $(3, 0)$. Note that $x = -1$ was a zero of $P(x)$, but it is excluded from the domain as a hole and does not produce an $x$-intercept.