Rational Expressions and Functions: Chapter Review

Vocabulary and notation

Rational expression, rational function

A quotient $\frac{p(x)}{q(x)}$ of polynomials with $q$ not the zero polynomial; read as a function of $x$ it is a rational function. Every polynomial is one, over denominator $1$. $\frac{|x|+1}{x}$ is not, since $|x|$ is not a polynomial.

Excluded value (restriction)

A real zero of the ORIGINAL denominator. The expression has no value there, and the exclusion survives every later step.

Lowest terms

Numerator and denominator share no common non-constant factor. Constants are exempt.

Complex fraction

A fraction carrying a fraction in its numerator, its denominator, or both. Its main bar is a division sign.

Extraneous root

A root of the cleared polynomial equation that is not in the original equation's domain. It is rejected however correctly it was derived.

Asymptote

A line the curve draws arbitrarily close to without settling on it. Horizontal and oblique (slanted) ones are approached as $x$ runs far out; a vertical one is approached as $x$ nears a value the function cannot take.

Hole

One missing point on a curve that is otherwise unbroken there, drawn as an open circle.

Formulas and theorems

• Domain of a rational expression

  $$q(x) \neq 0$$

  Use when Factor the ORIGINAL denominator and set each factor to zero; those are the excluded values. Numerators may vanish freely. Some expressions exclude nothing: $x^2+9$ has no real zero.

• Cancellation law

  $$\frac{ac}{bc} = \frac{a}{b}$$

  Use when $b \neq 0$ and $c \neq 0$. It divides out common FACTORS, never common terms, so factor completely first. Cancelling preserves the value wherever the original was defined, but never removes a restriction and never widens the domain.

  e.g. $\frac{4x^2-1}{2x-1} = \frac{(2x-1)(2x+1)}{2x-1} = 2x+1$ for $x \neq \frac{1}{2}$.

• Opposite factors

  $$a - b = -(b - a)$$

  Use when Any $a$ and $b$. Two factors differing only in the order of subtraction are not equal, so a factor of $-1$ comes out rather than nothing.

  e.g. $\frac{5-x}{x-5} = -1$ for $x \neq 5$.

• Product

  $$\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$$

  Use when $B \neq 0$ and $D \neq 0$, and nothing more: a product hides no condition, and $A$ and $C$ may vanish freely. No common denominator is wanted.

• Quotient

  $$\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} = \frac{AD}{BC}$$

  Use when THREE conditions: $B \neq 0$, $D \neq 0$, and $C \neq 0$, since the divisor equals zero exactly where $C$ does. Only $A$ may vanish. Flipping trades a visible condition for a hidden one, so no written form shows all three.

• Sum and difference over one denominator

  $$\frac{A}{D} \pm \frac{B}{D} = \frac{A \pm B}{D}$$

  Use when $D \neq 0$. The minus sign owns ALL of $B$, so bracket the numerator being subtracted before distributing it.

• Building a fraction up

  $$\frac{C}{D} = \frac{CK}{DK}$$

  Use when $D \neq 0$ and $K \neq 0$. Numerator AND denominator, or the value changes. On the way to an LCD this adds no new restriction.

• Least common denominator (LCD)

  $$\text{LCD} = \prod_i f_i^{\,\max_j e_{ij}}$$

  Use when The $f_i$ are the distinct irreducible factors across all the denominators, factored completely first, and $e_{ij}$ is the power of $f_i$ in denominator $j$: take each factor to the highest power it reaches in any ONE denominator. Numeric coefficients take their ordinary least common multiple. The product of the denominators is a legal common denominator but forces a needless cancellation whenever they share a factor. The factored LCD is zero at exactly the excluded values.

  e.g. $x^2-1 = (x-1)(x+1)$ and $x^2+2x+1 = (x+1)^2$ give LCD $(x-1)(x+1)^2$.

• Hole or vertical asymptote

  $$\begin{gathered} m \ge n \Rightarrow \text{hole at } x=a \\ m < n \Rightarrow \text{vertical asymptote } x=a \end{gathered}$$

  Text description

  Side by side: a curve broken only by one open circle at a hole, and a curve split into two branches that run off to infinity along a dashed vertical asymptote.

  Use when $a$ is a zero of the ORIGINAL denominator $Q$, and $m$ and $n$ count the copies of $(x-a)$ in the fully factored $P$ and $Q$, so $n \ge 1$ and $m \ge 0$. Appearing in the numerator is not enough; the factor must cancel COMPLETELY. A hole's height is the reduced expression evaluated at $a$.

• Horizontal asymptote by degree

  $$\begin{gathered} m < n: y = 0 \\ m = n: y = \frac{a_m}{b_n} \\ m > n: \text{none} \end{gathered}$$

  Use when $m = \deg P$, $n = \deg Q$, and $a_m$, $b_n$ the LEADING coefficients, not the constant terms. Cancelling a common factor changes neither the comparison nor the ratio, so either form may be read. A graph may cross a horizontal asymptote.

  e.g. $\frac{4-x^2}{2x^2+3}$ ties at degree $2$, so $y = -\frac{1}{2}$.

• Oblique asymptote by long division

  $$f(x) = S(x) + \frac{R(x)}{Q(x)}$$

  Text description

  A curve running close to a dashed slanted line at both far ends, bowing away from it in the middle, and crossing it once.

  Use when $S$ and $R$ are the quotient and remainder of $P \div Q$, so $\deg R < \deg Q$. $S$ is an oblique asymptote exactly when $\deg P = \deg Q + 1$ AND $R$ is not the zero polynomial; an exact division means the graph IS that line, with a hole at each cancelled factor. Past one degree of excess there is no straight-line asymptote. The curve meets $S$ wherever $R(x) = 0$.

  e.g. $\frac{x^2-5}{x+3} = x - 3 + \frac{4}{x+3}$, so $y = x-3$.

Problem types, step by step

Exam traps

• Trap Reading the restrictions off the final answer, so $\frac{x^2-4}{x^2-2x} = \frac{x+2}{x}$ is reported with $x \neq 0$ alone.

  Fix The answer is only what survived cancelling, so it has already forgotten what cancelled. The original denominator $x(x-2)$ also forces $x \neq 2$, where the reduced form happily returns $2$.

• Trap Crossing out a symbol visible above and below, so $\frac{x+2}{x}$ is reported as $2$.

  Fix Only FACTORS cancel, and a factor multiplies all of its side. At $x=3$ that expression is $\frac{5}{3}$. Likewise $\frac{x^2-9}{x^2+9}$ is already in lowest terms: at $x=1$ it is $-\frac{4}{5}$.

• Trap In $\frac{x+9}{x-1} \div (x+4)$, collecting only $x \neq 1$, because that is the sole denominator on the page.

  Fix Dividing by an expression demands that expression be nonzero, so $x \neq -4$ as well. That condition comes from the division symbol alone and sits in no denominator anywhere.

• Trap Letting a subtraction reach only the first term, so $\frac{5x-2}{x^2-9} - \frac{3x+4}{x^2-9}$ gets the numerator $2x+2$.

  Fix $(5x-2)-(3x+4) = 2x-6$. Bracket, then distribute as its own step. The lost sign also destroys the factor $x-3$ that reduces the answer to $\frac{2}{x+3}$.

• Trap Calling $x=a$ a hole as soon as $(x-a)$ appears in the numerator, so $\frac{x-2}{(x-2)^2(x+1)}$ is said to have a hole at $x=2$.

  Fix Count copies: one against two cancels once and leaves $\frac{1}{(x-2)(x+1)}$, so $x=2$ is a VERTICAL ASYMPTOTE. Cancel as far as it goes, then read the reduced denominator.

• Trap Treating a horizontal asymptote as a fence the curve may not touch.

  Fix It describes end behaviour only. Solve $f(x)=c$ to find crossings: $\frac{x^2+2x}{x^2+4}=1$ gives $x=2$, so that curve cuts $y=1$ at $(2,1)$. Only a VERTICAL asymptote can never be crossed, its input being outside the domain.

• Trap Reporting $y = \frac{2}{5}$ as the horizontal asymptote of $\frac{2x^3+7}{5x^2-x}$, since those are the leading coefficients.

  Fix The ratio applies only when the degrees TIE. Here $3 > 2$, so no horizontal line catches the curve; an excess of exactly one degree gives a slanted asymptote instead, found by long division.

• Trap Certifying a candidate by substituting it into the CLEARED equation.

  Fix Every candidate satisfies that equation by construction, so the test proves nothing. Check it against the ORIGINAL excluded values. Substituting into the original only catches arithmetic slips.

• Trap Dividing both sides by an expression holding the variable, so $x(x-2) = 3(x-2)$ yields $x=3$ alone.

  Fix That step is not reversible either, and it LOSES the root $x=2$, where both sides are $0$. Move everything to one side and factor: $(x-2)(x-3)=0$.

• Trap Answering "every real number" when the variable cancels into a true statement.

  Fix $\frac{3x}{x-2} = 3 + \frac{6}{x-2}$ collapses to $3x = 3x$, so the solution set is the DOMAIN: every real number except $2$. An identity holds only where both sides exist.