What is a rational function?

A quotient of two polynomials, f(x) = P(x)/Q(x), where the denominator must not be zero. A polynomial is the special case in which the denominator is a non-zero constant. The homographic function is the simplest non-trivial case: a quotient of two linear expressions.

How do you find the domain of a rational function?

Set the denominator to zero and exclude the numbers you find. For f(x) = (x + 1)/(x² − 9) solve x² − 9 = 0, giving x = 3 and x = −3, so the domain is every real number except those two. The numerator plays no part in this at all.

What is the difference between a pole and a hole?

Whether the numerator vanishes too. If the denominator has a zero and the numerator does not, you get a vertical asymptote — the values run off to infinity. If both vanish at the same point and the factor cancels, only a hole is left: the graph looks normal but one point is missing from it, as in f(x) = (x² − 4)/(x − 2) at x = 2.

How do you find the horizontal asymptote?

Compare the degrees of numerator and denominator. When the numerator has the lower degree, the asymptote is y = 0. When the degrees are equal, it is the ratio of the leading coefficients — for (3x² + 1)/(2x² − 5) that is y = 3/2. When the numerator has the higher degree, there is no horizontal asymptote.

When does a rational function have an oblique asymptote?

When the numerator has exactly one degree more than the denominator. Polynomial division then leaves a linear quotient plus a remainder that dies away for large arguments, and that quotient is the equation of the asymptote. An oblique and a horizontal asymptote never occur together — it is one role played by one line.

Can a graph cross its own asymptote?

The horizontal and oblique ones yes, the vertical one never. The function f(x) = (x² − 2x)/(x² + 1) has the horizontal asymptote y = 1 and crosses it at x = −0.5. An asymptote describes the behaviour of the curve at infinity, not a ban on meeting it at a finite distance. The vertical one cannot be crossed, because the function has no value there.

How many vertical asymptotes can a rational function have?

As many as the denominator has zeros that do not cancel against the numerator. The function 1/(x² − 4) has two, at x = −2 and x = 2, and its graph falls into three branches. The function 1/(x² + 1) has none, because its denominator never vanishes.

Why look for asymptotes at all?

Because they sketch the graph without computing dozens of points. Three lines — verticals at the poles and one horizontal or oblique — set the frame the curve has to fit into; add the zeros and the sign, and the shape is done. It is the same information that the limit of a function later states precisely.

Advanced level

Rational functions and asymptotes

A quotient of two polynomials, a few points thrown out of the domain, and three kinds of line the curve heads for. See when a zero of the denominator makes an asymptote and when it only makes a hole, how comparing degrees gives the horizontal asymptote, and where the oblique one comes from.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Inverse proportion and the homographic functionThe faster you drive, the shorter the drive — and the product of speed and time stays put. See how inverse proportion differs from direct, what the hyperbola y = a/x looks like, where its two asymptotes come from, and how to read them off a quotient of two linear expressions.
Rational expressions and equationsA fraction with polynomials where the numbers used to be — and the same four operations as on ordinary fractions, only preceded by factoring. The domain and the excluded numbers, cancelling, multiplying and dividing, adding and subtracting, and rational equations in product form.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Unit cost in manufacturing
Setting up a production line costs 40,000 zloty and each item costs 12, so the cost per item is k(n) = (40000 + 12n)/n. At 500 items that is 92 zloty, at 5000 it is 20, at 50,000 it is 12.80. The horizontal asymptote y = 12 says where the curve stops: it will never drop below the material cost, however far production grows.
Drug concentration in the blood
A pharmacologist models the concentration as C(t) = 5t/(t² + 4) mg/l. After one hour that is 1 mg/l, after two it peaks at 1.25 mg/l, after eight it is only 0.59 mg/l. The numerator has the lower degree, so the horizontal asymptote is y = 0: the drug eventually clears, and the curve says how fast.
Focal length of a lens
The lens equation 1/f = 1/x + 1/y rearranges to y = fx/(x − f) — a rational function with a pole at x = f. For a lens of focal length 5 cm and an object 8 cm away, the image forms at 13.3 cm. The closer the object comes to the focus, the further the image runs: at 5.1 cm it is already 255 cm away, and that is the physical content of a vertical asymptote.
Average speed there and back
A courier drives 60 km out at 80 km/h and returns at speed v. The average speed over the whole trip is not the plain average but v_avg = 160v/(80 + v). Returning at 40 km/h gives 53.3 km/h, not 60. The horizontal asymptote y = 160 marks the limit of the absurd: even an infinitely fast return cannot push the average past 160 km/h, because the outward leg already took 45 minutes.

All formulas

A rational function
$f(x) = \frac{P(x)}{Q(x)}, \qquad Q(x) \neq 0$
a quotient of two polynomials
The domain
$D_f = \{x \in \mathbb{R} : Q(x) \neq 0\}$
the zeros of the denominator are excluded
A pole
$x = x_0 \ \text{— a vertical asymptote}$
when Q(x_0)=0 and P(x_0) \neq 0
Horizontal asymptote
$\deg P < \deg Q \implies y = 0$
the denominator grows faster
Horizontal asymptote
$\deg P = \deg Q \implies y = \frac{a_P}{a_Q}$
the ratio of the leading coefficients
Oblique asymptote
$\deg P = \deg Q + 1 \implies y = ax + b$
the quotient of the polynomial division

The cost of making one item falls the more items you make — but it will never fall below the cost of the material. That floor is an asymptote, and the formula describing it is a rational function.

Definition and domain

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

where $P$ and $Q$ are polynomials. Everything peculiar about this family comes from one prohibition — you may not divide by zero — so the zeros of the denominator drop out of the domain:

D_f = \{x \in \mathbb{R} : Q(x) \neq 0\}

Find the domain of f(x) = (2x + 5)/(x² − 3x − 10).

A pole or a hole

A zero of the denominator does not always give an asymptote. What settles it is whether the numerator vanishes along with it.

case	what happens	example
$Q(x_0) = 0$ , $P(x_0) \neq 0$	a pole — vertical asymptote	$\dfrac{1}{x-2}$ at $x = 2$
$Q(x_0) = 0$ , $P(x_0) = 0$ , the factor cancels	a hole — one point missing	$\dfrac{x^2-4}{x-2}$ at $x = 2$

In the second case cancelling leaves $x + 2$ , an ordinary line — but the point $x = 2$ still does not belong to the domain. The graph looks like a line with one point punched out of it.

Cancelling a fraction does not restore an excluded argument. The domain is settled before simplifying, never after.

The vertical asymptote

At a pole the values grow or fall without bound. For $f(x) = \tfrac{1}{x-2}$ :

$x$	$2.1$	$2.01$	$2.001$	$1.999$	$1.99$
$f(x)$	$10$	$100$	$1000$	$-1000$	$-100$

From the right the curve runs upwards, from the left downwards — the typical behaviour at a simple pole. At a double pole, as in $\tfrac{1}{(x-2)^2}$ , both branches run the same way.

A vertical asymptote stands at every zero of the denominator that does not cancel against the numerator. So there may be two poles, or three, or none.

Two poles, three branches. The denominator x² − 4 vanishes at −2 and at 2, so the domain falls into three intervals and the graph into three separate pieces.

The horizontal asymptote

For very large arguments the value of the fraction is decided by the leading powers alone — the remaining terms grow too slowly to change anything. So the whole rule reduces to comparing degrees:

degrees	horizontal asymptote	example
$\deg P < \deg Q$	$y = 0$	$\dfrac{2x+1}{x^2-3} \to 0$
$\deg P = \deg Q$	$y = \dfrac{a_P}{a_Q}$	$\dfrac{3x^2+x}{2x^2-5} \to \dfrac{3}{2}$
$\deg P > \deg Q$	none	$\dfrac{x^3}{x+1}$ grows without bound

where $a_P$ and $a_Q$ are the leading coefficients.

A horizontal asymptote may be crossed. The function $f(x) = \tfrac{x^2 - 2x}{x^2+1}$ tends to $y = 1$ and crosses that line on the way — an asymptote speaks about behaviour at infinity, not about a ban at a finite distance. That distinguishes it from the asymptote of the hyperbola in the lesson on inverse proportion, which cannot be crossed.

The oblique asymptote

When the numerator has exactly one degree more, the graph at infinity comes to resemble not a horizontal line but a slanted one. You find it by dividing the polynomials.

f(x) = \frac{x^2 + 1}{x}

Divide term by term:

f(x) = \frac{x^2}{x} + \frac{1}{x} = x + \frac{1}{x}

The term $\tfrac{1}{x}$ tends to zero, so for large $|x|$ the function behaves like $y = x$ .

A vertical asymptote at zero and an oblique one at y = x. The further from the origin, the tighter the curve hugs the slanted line — the gap between them is exactly 1/x.

Find every asymptote of f(x) = (x² + 3x − 4)/(x − 1).

Find the asymptotes of f(x) = (2x² − x + 3)/(x + 1).

Sketching the graph in four steps

Domain — the zeros of the denominator.
Cancelling — is any of them a hole rather than a pole?
Asymptotes — verticals at the poles, plus a horizontal or an oblique one, never both.
Zeros and sign — wherever the numerator vanishes.

Four steps give the frame the curve has to fit into, without drawing a table of values. The same behaviour is later stated precisely by the limit of a function: a horizontal asymptote is a limit at infinity, and a vertical one an infinite limit at a point.

Practice

The set checks the three kinds of asymptote separately: the vertical one (a zero of the denominator that does not cancel), the horizontal one (comparing degrees) and the oblique one (polynomial division). The vertical and horizontal ones are typed as a number — the line is named in the prompt — and the oblique one as the equation of a line, e.g. x+3. The domain is not asked here: the algebra branch practises it, with rational expressions, where its rule is taught.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

Vertical asymptote x: f(x) = (x − 4)/(x − 3)

Common mistakes

Claiming a vertical asymptote without checking for cancelling — if the numerator vanishes alongside the denominator, you get a hole rather than a pole.
Fixing the domain after simplifying the fraction — cancelling does not restore an excluded argument.
Reading the horizontal asymptote off the constant terms — what counts are the coefficients of the leading powers.
Quoting a horizontal and an oblique asymptote at once — a function has at most one of them.
Believing a graph can never cross an asymptote — it can cross a horizontal or oblique one; the vertical one it cannot, because there is no value there.
Looking for an oblique asymptote when the degree gap exceeds one — then there is none at all, and the graph outgrows every line.

Formula card

Topic: Rational functions and asymptotes

A rational function
$f(x) = \frac{P(x)}{Q(x)}, \qquad Q(x) \neq 0$
a quotient of two polynomials
The domain
$D_f = \{x \in \mathbb{R} : Q(x) \neq 0\}$
the zeros of the denominator are excluded
A pole
$x = x_0 \ \text{— a vertical asymptote}$
when Q(x_0)=0 and P(x_0) \neq 0
Horizontal asymptote
$\deg P < \deg Q \implies y = 0$
the denominator grows faster
Horizontal asymptote
$\deg P = \deg Q \implies y = \frac{a_P}{a_Q}$
the ratio of the leading coefficients
Oblique asymptote
$\deg P = \deg Q + 1 \implies y = ax + b$
the quotient of the polynomial division

The function f(x) = (x² + 1)/x. A vertical asymptote stands at zero, and for large arguments the curve hugs the line y = x — the oblique asymptote, which is what dividing the numerator by the denominator leaves.

The function f(x) = 1/(x² − 4) has two poles, at x = −2 and x = 2, so its graph comes in three branches. The horizontal asymptote is y = 0, because the numerator has the lower degree.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min

Rational functions and asymptotes

Before you start

Where this is used

All formulas

Definition and domain

A pole or a hole

The vertical asymptote

The horizontal asymptote

The oblique asymptote

Sketching the graph in four steps

Practice

Practice

Common mistakes

Formula card

Frequently asked questions

What is a rational function?

How do you find the domain of a rational function?

What is the difference between a pole and a hole?

How do you find the horizontal asymptote?

When does a rational function have an oblique asymptote?

Can a graph cross its own asymptote?

How many vertical asymptotes can a rational function have?

Why look for asymptotes at all?

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Why a function's graph can never double back