Rational functions

Ask the tutor about Rational functions Open the graphing workbench

A polynomial is a finite sum with a nonzero leading term, smooth on the whole line. Form the quotient of two such sums. Where is the graph still smooth, where does it blow up, and far away, who still decides the height?

Start with $y = b / x$

Freeze a positive constant $b > 0$ and look at

$y = \frac{b}{x}, x \in R, x \neq = 0.$

Drag $b$ . The green curve is an equilateral hyperbola: the coordinate axes are its asymptotes, and the blue marks sit at the vertices $S_{\pm} = (\pm b, \pm b)$ .

Drag b. The green curve is y = b/x for x ≠ 0. The axes are asymptotes; the blue marks are the vertices (±√b, ±√b). Far left and far right the height returns to 0; at x = 0 the graph has a pole.

As $x \to \pm \infty$ the height returns to $0$ . As $x$ approaches $0$ from the right the height goes to $+ \infty$ ; from the left it goes to $- \infty$ . So $x = 0$ is a pole — the denominator vanishes while the numerator stays off zero.

Shift the center: linear fractional maps

Keep numerator and denominator both degree one. Let $a, b, c, d$ be real with $c \neq = 0$ and determinant $Δ := a d - b c \neq = 0$ . Then

$y = \frac{a x + b}{c x + d}, x \in R, x \neq = - \frac{d}{c} .$

The excluded vertical line is again a pole. A translation $x = u - d / c$ and $y = w + a / c$ rewrites the same graph as

$w = - \frac{Δ}{c ^{2} u} .$

So after the shift, the picture is the same shape as $b / x$ , only the constant $- Δ/ c^{2}$ may change sign. The old origin lands at the new center $P (- d / c, a / c)$ .

Drag Δ. The green curve is y = −Δ/x for x ≠ 0 — the book’s shift shape w = −Δ/(c² u) with c = 1 — so the center sits at the origin of the dashed axes. The sign of Δ flips which pair of opposite quadrants the branches occupy.

Drag $Δ$ . When $Δ < 0$ the branches occupy one pair of opposite quadrants relative to the dashed axes through $P$ ; when $Δ > 0$ they occupy the other pair. The hypothesis $Δ \neq = 0$ is what keeps the rewritten coefficient off zero — otherwise the “fraction” would have collapsed to a constant.

Higher odd or even power in the denominator

Replace the first power by $n = 1, 2, \dots$ and keep $b > 0$ :

$y = \frac{b}{x ^{n}}, x \neq = 0.$

Drag n. The green curve is y = 1/x^n for x ≠ 0 (here b = 1). When n is even both arms rise the same way and the graph is even; when n is odd the arms take opposite signs and the graph is odd.

Drag $n$ (here $b = 1$ ). When $n$ is even the graph is an even function: both arms rise the same way as $x \to 0$ , and the picture is symmetric across the y-axis. When $n$ is odd the graph is odd: the arms take opposite signs, like $b / x$ . In every case the axes remain asymptotes.

Quadratic denominator — smooth or two poles

Now let the denominator be quadratic. First freeze $d > 0$ and compare two formulas whose denominators have no real root:

$y = \frac{1}{x ^{2} + d ^{2}}, y = \frac{x}{x ^{2} + d ^{2}}, x \in R .$

Drag d. The green curve is y = 1/(x² + d²). The denominator never zero, so the graph stays smooth on the whole line. The blue mark is the peak (0, 1/d²); widen d and the bell flattens.

Drag $d$ on the first. The bell is even; its peak is the blue mark at height $1/ d^{2}$ . Widen $d$ and the peak drops. The second formula (not drawn here) is odd: it crosses the origin and carries a local max and a local min at $x = \pm d$ . Because $x^{2} + d^{2} > 0$ for every real $x$ , both graphs are continuous and infinitely differentiable on the whole line — no poles.

Switch to two distinct real roots $x_{-} < x_{+}$ :

$y = \frac{1}{( x - x _{+} ) ( x - x _{-} )} .$

Partial fractions split the same function as

$y = \frac{1}{x _{+} - x _{-}} (\frac{1}{x - x _{+}} - \frac{1}{x - x _{-}}) .$

Drag the gap g. The green curve is y = 1/((x − g)(x + g)) with poles at ±g (orange dashed lines). Between the poles the graph stays negative; far left and far right the height returns to 0.

Drag the gap so the poles sit at $\pm g$ . Between them the graph stays negative; outside, it stays positive and returns to $0$ at infinity. Approaching either pole, the height escapes to $\pm \infty$ with opposite one-sided signs on the two sides of that pole. So both $x_{\pm}$ are poles.

Canceling a common factor can hide a hole. The formula

$y = \frac{x - 1}{x ^{2} - 1}$

is undefined at $x = 1$ , yet $x^{2} - 1 = (x - 1) (x + 1)$ rewrites it as $y = 1/ (x + 1)$ for every $x \neq = - 1$ . After the cancelation, $x = 1$ is only an apparent singularity; the true pole that remains is $x = - 1$ .

The same trichotomy appears for a general linear-over-quadratic

$y = \frac{a x + b}{x ^{2} + 2 c x + d}$

with $a^{2} + b^{2} \neq = 0$ . Let $D := c^{2} - d$ . Always $lim_{x \to \pm \infty} f (x) = 0$ . If $D > 0$ the denominator has two distinct real roots and $f$ splits into partial fractions with poles at both. If $D = 0$ there is a repeated real root and a single pole whose two sides both go to $+ \infty$ or both to $- \infty$ according to the sign of $a x_{+} + b$ . If $D < 0$ the quadratic stays positive for every real $x$ , so $f$ is continuous and infinitely differentiable on $R$ .

Name the general quotient

A real rational function is a quotient of two polynomials,

$y = f (x) = \frac{a _{n} x ^{n} + \dots + a _{1} x + a _{0}}{b _{m} x ^{m} + \dots + b _{1} x + b _{0}},$

with the denominator not the zero polynomial. Write $c := a_{n} / b_{m}$ . Far away the height matches the single power that survives after canceling the leading terms:

$x \to \pm \infty lim f (x) = x \to \pm \infty lim c x^{n - m} .$

When $c > 0$ , the limit as $x \to + \infty$ is $c$ if $n = m$ , $+ \infty$ if $n > m$ , and $0$ if $n < m$ ; as $x \to - \infty$ the same list holds except that $n > m$ follows the parity of $n - m$ (even $\to + \infty$ , odd $\to - \infty$ ). When $c < 0$ , swap each $\pm \infty$ for the opposite. That is the same leading-term logic as for polynomials, only the surviving degree is now the difference $n - m$ .

What this page put in place

A rational function is a quotient of polynomials. The model case $b / x$ is an equilateral hyperbola with a pole at the origin; every nondegenerate linear fractional map is that same shape after a shift of center; higher powers $b / x^{n}$ keep the axes as asymptotes and pick even or odd symmetry from $n$ ; a quadratic denominator is either pole-free and smooth, or it plants one or two real poles (or an apparent singularity after cancelation); and far away the height follows $c x^{n - m}$ .