Rational Expressions & Functions

A rational expression is a ratio of two polynomials, $\dfrac{P(x)}{Q(x)}$. It is undefined wherever the denominator equals zero --- those inputs are the excluded values. The domain is every real number except the excluded values.

To simplify: factor the numerator and denominator completely, then cancel any common factors. Always list excluded values from the original denominator.

Multiply: factor everything, cancel any factor in a numerator against a matching factor in a denominator, then write what remains.

Divide: multiply by the reciprocal of the second fraction --- “keep, change, flip” --- then proceed as a multiplication.

You can only combine fractions over a common denominator. Find the LCD: factor each denominator, then take each distinct factor to its highest power that appears. Rewrite each fraction with the LCD, combine the numerators, then simplify.

A complex fraction has fractions in its numerator, denominator, or both. The fastest method: find the LCD of all the little fractions and multiply the top and bottom by that LCD. This clears every small denominator at once.

To solve (not simplify) a rational equation, multiply every term by the LCD to clear all fractions, then solve the resulting polynomial equation. Because multiplying by a variable expression can introduce false roots, you must check each answer.

For $f(x)=\dfrac{P(x)}{Q(x)}$ in lowest terms:

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• Holes: a factor that cancels from top and bottom gives a hole at that $x$-value.
• Vertical asymptotes (VA): set the remaining denominator equal to $0$.
• Horizontal asymptote (HA): compare degrees of $P$ and $Q$.
• Intercepts: $x$-intercepts where $P(x)=0$; $y$-intercept at $f(0)$.

Every variation problem hides a constant $k$. Use one data point to find $k$, then answer the question.

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• Direct: $y=kx$ (as $x$ grows, $y$ grows proportionally).
• Inverse: $y=\dfrac{k}{x}$ (as $x$ grows, $y$ shrinks).
• Joint: $y=kxz$ ($y$ varies with the product of two quantities).

If one worker finishes a job in $a$ hours, their rate is $\dfrac{1}{a}$ job per hour. When several work together, rates add. For a shared time $t$:

For mixture problems, track the amount of pure substance and set the concentration equal to your target.

Both come from a zero of the original denominator, but they behave very differently. Factor top and bottom completely and compare the multiplicity of each root:

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• Vertical asymptote at $x=a$: the factor $(x-a)$ survives in the denominator of the reduced form. The function blows up ($\to\pm\infty$) there.
• Hole at $x=a$: the factor $(x-a)$ cancels completely. If it appears to power $p$ on top and $q$ on the bottom, the outcome depends on which is larger.

Compare powers: with $(x-a)^p$ on top and $(x-a)^q$ on the bottom, the net factor is $(x-a)^{p-q}$. If $p\ge q$ the point is a hole (removable); if $p<q$ it is a vertical asymptote of order $q-p$.

When $\deg P = \deg Q + 1$ (numerator exactly one degree higher), there is no horizontal asymptote --- instead the graph follows a slant line $y=mx+b$. Find it by polynomial long division:

As $x\to\pm\infty$ the remainder term $\to 0$, so the curve hugs the line $y=mx+b$. The remainder is what you discard for the asymptote --- but its sign tells you whether the curve sits above or below the line.

The reverse of adding fractions: split one complicated rational expression into a sum of simple ones. First make sure the fraction is proper ($\deg P<\deg Q$); if not, divide first. Then factor $Q$ and assign an unknown numerator to each factor type:

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• Distinct linear $(x-a)$: term $\dfrac{A}{x-a}$.
• Repeated linear $(x-a)^k$: one term per power, $\dfrac{A_1}{x-a}+\dfrac{A_2}{(x-a)^2}+\dots+\dfrac{A_k}{(x-a)^k}$.
• Irreducible quadratic $(x^2+bx+c)$: term $\dfrac{Bx+C}{x^2+bx+c}$ (a linear numerator).

Clear denominators and solve for the constants --- either by substituting convenient $x$-values or by matching coefficients.

Partial fractions turn many “impossible” sums into telescoping sums, where adjacent terms cancel in a cascade. The key identity is

Summing it, every interior term is cancelled by its neighbour, leaving only the first and last:

Many problems become easy once you let $s=x+\dfrac1x$. Squaring and cubing generate the other symmetric powers:

So knowing a single value $s=x+\dfrac1x$ instantly gives $x^2+\dfrac1{x^2}$ and $x^3+\dfrac1{x^3}$ without ever solving for $x$.

A continued fraction nests division inside division:

Evaluate a finite one by working from the bottom up. A periodic infinite continued fraction satisfies a self-referential equation: if $y$ equals the whole expression and the pattern repeats, substitute $y$ back into itself and solve the resulting (usually quadratic) equation --- keeping the positive root.