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AlgebraGrades 9–123 min read

Rational Functions

Rational functions divide polynomials and may have holes, asymptotes, intercepts, and separate branches.

Cheat sheet
Factoring a rational function reveals where its graph is missing, where it grows without bound, and where it crosses the axes.

Definition and domain

A rational function has form

$$ f(x)=\frac{P(x)}{Q(x)},qquad Q(x)\ne0, $$

where $P$ and $Q$ are polynomials. The domain excludes every real zero of the original denominator.

These exclusions can create vertical asymptotes or removable discontinuities called holes.

Holes versus vertical asymptotes

Factor numerator and denominator. If a denominator factor cancels, its zero usually creates a hole. If a denominator factor remains, its zero creates a vertical asymptote unless other unusual cancellation has changed the structure.

A hole has a finite missing coordinate found by substituting the excluded input into the simplified expression. Near a vertical asymptote, function values grow without bound in magnitude.

Worked example: full feature analysis

Horizontal asymptotes

Compare numerator degree $n$ with denominator degree $m$:

  • if $n<m$, horizontal asymptote $y=0$;
  • if $n=m$, horizontal asymptote equals the ratio of leading coefficients;
  • if $n>m$, there is no horizontal asymptote from this rule.

A graph may cross a horizontal asymptote. It describes end behaviour, not a forbidden line.

Polynomial and oblique asymptotes

If the numerator degree is exactly one more than the denominator degree, polynomial division produces a linear or oblique asymptote. More generally, division can give a higher-degree polynomial asymptote.

For

$$ \frac{x^2+1}{x-1}=x+1+\frac2{x-1}, $$

the remainder term approaches $0$ as $|x|$ grows, so the oblique asymptote is $y=x+1$.

Intercepts

An $x$-intercept occurs where the simplified numerator equals zero and the denominator is nonzero. A factor cancelled from the numerator does not create an intercept; it creates a hole.

The $y$-intercept is $f(0)$ if $0$ belongs to the domain.

Reciprocal parent function

The parent $y=1/x$ has asymptotes $x=0$ and $y=0$. The transformation

$$ y=\frac{a}{x-h}+k $$

has vertical asymptote $x=h$, horizontal asymptote $y=k$, and centre $(h,k)$. The sign of $a$ determines which opposite pair of regions contains the branches.

Sign and interval behaviour

Zeros and vertical asymptotes divide the domain into intervals. A sign chart determines where the function is positive or negative. Near a vertical asymptote, checking each side separately reveals whether the graph approaches $+\infty$ or $-\infty$.

One-sided behaviour matters because the two sides may differ.

Solving intersections

To find where a rational function meets another function, set the formulas equal, note restrictions, clear denominators, and solve. Any candidate excluded from the original domain must be rejected.

A graph can verify the number and approximate location of intersections.

Common mistakes

Calling every excluded value a vertical asymptote. Cancelled factors create holes.

Using the original numerator for intercepts after cancellation. A cancelled zero is missing, not an intercept.

Treating a horizontal asymptote as uncrossable. It describes end behaviour.

Forgetting the hole's $y$-coordinate. Substitute into the simplified formula.

Graphing across a vertical asymptote as one connected curve. The domain is split into branches.

Quick self-check

  • What are the original domain restrictions?
  • Which factors cancel and which remain below?
  • Where are holes and vertical asymptotes?
  • What end behaviour follows from degree or division?
  • Which intercepts survive the restrictions?
  • Does the graph's sign and branch behaviour match test values?
Check your understanding

Try it yourself

Hints are part of learning. Open one whenever it makes the next step feel possible.

1 practice question
Question 1Distinguish a hole and asymptote · Standard

For f(x) = (x² − 1)/(x² − x − 2), which description is correct?

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