Rational Functions and Asymptotes

Rational Functions and Asymptotes — Moosa Academy

In a polynomial, $x$ only ever appears on top, so nothing unusual can happen. A rational function allows $x$ in the denominator too — and that one change produces the lines a curve approaches but never reaches.

Concept What a rational function is

f(x) = \frac{P(x)}{Q(x)}, \qquad Q(x) \neq 0

One polynomial divided by another, with the denominator never allowed to be zero.

Ordinary polynomials are the special case $Q(x) = 1$ , and constants the case where $P$ is a number as well. Both are rational functions — they simply never run into the difficulty, because their denominator can never vanish.

Theorem Vertical asymptotes: where the denominator is zero

Q(a) = 0

, then

x = a

is a vertical asymptote. The curve shoots off towards infinity there, coming ever closer to the line without ever meeting it.

f(x) = \dfrac{1}{x}

— denominator zero at

x = 0

, so one vertical asymptote

f(x) = \dfrac{1}{(x+2)(x-3)}

— zero at

x = -2

and

x = 3

, so two

The count is simply the number of roots of the denominator, so there is no limit to how many a function may have.

Theorem Horizontal asymptotes: compare the degrees

A horizontal asymptote describes what happens far out to the left and right. Which case applies depends only on the two degrees:

Denominator degree higher →

y = 0

Degrees equal →

y = \dfrac{a_n}{b_n}

, the ratio of the leading coefficients

Numerator degree higher → none at all

Unlike vertical asymptotes, there is at most one horizontal asymptote — never two. The three cases above are exhaustive, so checking the degrees settles the question immediately.

Example Both asymptotes for 1/x

Step 1 — the denominator is

x

, zero at

x = 0

⟹ vertical asymptote

x = 0

Step 2 — degrees: numerator 0, denominator 1

Step 3 — the denominator wins ⟹ horizontal asymptote

y = 0

⟹ the curve hugs both axes without ever touching them

This is the shape drawn above: two separate branches, each squeezed between the two dashed lines.

Example Equal degrees

Find the horizontal asymptote of $f(x) = \dfrac{2x^{2}+1}{3x^{2}+x}$ .

Step 1 — degrees: numerator 2, denominator 2 — equal

Step 2 — take the leading coefficients: 2 on top, 3 below

y = \frac{2}{3}

⟹ horizontal asymptote

y = \tfrac{2}{3}

Only the leading terms matter here. Far from the origin $2x^{2}$ and $3x^{2}$ dwarf everything else, so the fraction settles towards $\tfrac{2}{3}$ .

Summary

A rational function is P(x) ÷ Q(x), with Q(x) never zero.
Polynomials and constants are special cases where the denominator cannot vanish.
Every root of the denominator gives a vertical asymptote, so there may be many.
There is at most one horizontal asymptote, decided by comparing degrees.
Denominator higher → y = 0; equal → y = ratio of leading coefficients; numerator higher → none.
For 1/x both axes are asymptotes; for (2x²+1)/(3x²+x) the horizontal asymptote is y = 2/3.