Rational Functions

Intuition

You divide all the time. Split a pizza of $12$ slices among $x$ friends and each gets $\frac{12}{x}$ slices. Drive $100$ miles at a speed of $x$ miles per hour and the trip takes $\frac{100}{x}$ hours. In each case the output is a division, and something interesting happens at the edges. Share the pizza among more and more friends and each portion shrinks toward nothing; share it among fewer and fewer and each portion grows huge. A rational function packages exactly this kind of rule, a polynomial divided by a polynomial. A rational function’s whole personality comes from one forbidden move: you can never divide by zero. That single rule punches gaps into the domain. Near those gaps the graph does something no polynomial ever does, racing off toward a vertical line it never reaches. Learning where the denominator vanishes, and what the curve does on each side of that gap, is the entire lesson.

Learning goals

Exclude the zeros of the denominator from the domain
Graph $\tfrac{1}{x}$ as a hyperbola with two branches
Tell a vertical asymptote from a hole by whether the factor cancels
Compare degrees to find the horizontal asymptote
Read asymptotes $x = h$ and $y = k$ from the shifted form
Clear denominators, then discard extraneous candidates

What a rational function is

A rational function is a ratio of two polynomials,

f(x) = \frac{p(x)}{q(x)},

where $q(x)$ is not the zero polynomial (otherwise you would be dividing by nothing). The word rational comes from ratio: just as a rational number is one integer over another, a rational function is one polynomial over another. Every polynomial you have already studied is a rational function in disguise, since you can write it over the denominator $1$ . But the new behavior appears the moment the denominator actually contains the variable.

The simplest genuine example, and the one the whole lesson is built on, is the reciprocal function

f(x) = \frac{1}{x}.

It sends $2$ to $\tfrac12$ , sends $\tfrac12$ to $2$ , and sends $-4$ to $-\tfrac14$ , turning each input into its reciprocal. A few more examples show the range of shapes the definition allows:

\frac{x + 1}{x - 3}, \qquad \frac{2x}{x^2 - 9}, \qquad \frac{x^2 - x - 6}{x^2 - 9}.

The rest of the lesson works out the two features that set rational functions apart from the polynomials of earlier chapters. Those two features are a domain with holes cut out of it, and a graph that runs off toward invisible guide-lines called asymptotes.

Finding the domain

A polynomial accepts every real number, but a rational function is choosier. Wherever the denominator $q(x)$ equals zero, the rule asks you to divide by zero, and that has no answer. So the domain of a rational function is every real number except the values that make the denominator zero. To find those forbidden inputs, set the denominator equal to zero and solve.

Why dividing by zero has no value#

Division is defined through multiplication. The quotient $\frac{a}{b}$ is the number that, multiplied by $b$ , gives back $a$ . So $\frac{a}{b} = c$ means exactly that $c \cdot b = a$ . Test what this demands when the divisor is zero.

Suppose first that $a \neq 0$ , and ask for $\frac{a}{0}$ . It would have to be a number $c$ with $c \cdot 0 = a$ . But anything times zero is zero,

c \cdot 0 = 0 \neq a,

so no such $c$ exists. There is simply no number that works.

Suppose instead that $a = 0$ , and ask for $\frac{0}{0}$ . Now the requirement $c \cdot 0 = 0$ is satisfied by every number $c$ , since anything times zero is zero. A quotient that could be any number at all pins down nothing, so it cannot be assigned a single value either. Either way, dividing by zero has no sensible answer, which is exactly why every input that makes the denominator zero is thrown out of the domain.

Worked example 1 Find the domain of three rational functions

For each one, set only the denominator to zero, since the numerator never restricts the domain.

For $f(x) = \dfrac{x + 5}{x - 3}$ , the denominator is zero when $x - 3 = 0$ , that is $x = 3$ . The domain is every real number except $3$ .

For $g(x) = \dfrac{2x - 1}{x^2 - x - 6}$ , factor the denominator to find its zeros:

x^2 - x - 6 = (x - 3)(x + 2) = 0 \quad \Longrightarrow \quad x = 3 \ \text{ or } \ x = -2.

The domain is every real number except $3$ and $-2$ .

For $h(x) = \dfrac{x + 1}{x^2 + 4}$ , set $x^2 + 4 = 0$ , which gives $x^2 = -4$ . No real number has a negative square, so the denominator is never zero, and the domain is all real numbers with nothing to exclude.

Check your understanding

What is the domain of $f(x) = \dfrac{x + 2}{x^2 - x - 12}$ ?

A all

x

except

x = 4

and

x = -3

B all

x

except

x = -4

and

x = 3

C all

x

except

x = -2

D all real numbers

The reciprocal function and its hyperbola

To see what a rational graph looks like, plot the reciprocal $f(x) = \frac{1}{x}$ point by point, keeping in mind that $x = 0$ is barred from the domain:

$x$	$-2$	$-1$	$-\tfrac12$	$\tfrac12$	$1$	$2$
$y = \tfrac{1}{x}$	$-\tfrac12$	$-1$	$-2$	$2$	$1$	$\tfrac12$

The points fall into two separate pieces, called branches. For positive inputs both coordinates are positive, so that branch lives in the upper-right quadrant. For negative inputs both coordinates are negative, so the other branch sits in the lower-left. The two branches never join, because to pass from one to the other the curve would have to cross $x = 0$ , and there is no point there. This two-branch curve is a hyperbola.

Look at what happens at the edges, described in plain words rather than symbols. As $x$ grows very large in size, say $x = 100$ or $x = 1000$ , the reciprocal $\frac{1}{x}$ becomes very small, $0.01$ then $0.001$ , creeping toward zero without ever equalling it. So far out to the left and right the curve snuggles down against the x-axis. The horizontal line the graph approaches this way is a horizontal asymptote, and here it is the line $y = 0$ .

Now let $x$ get very close to $0$ , say $x = 0.01$ . Then $\frac{1}{x} = 100$ , and at $x = 0.001$ it is $1000$ : the closer $x$ creeps to zero, the more violently $\frac{1}{x}$ shoots up (or down, for small negative $x$ ). The curve rises along the y-axis without bound, hugging it ever more tightly but never landing on it, since $x = 0$ is not allowed. The vertical line the graph races along like this is a vertical asymptote, and here it is the line $x = 0$ .

The graph of y equals 1 over x is a hyperbola with two branches. For large x in either direction the curve flattens toward the x-axis, so y equals 0 is a horizontal asymptote; as x approaches 0 the curve shoots up or down along the y-axis, so x equals 0 is a vertical asymptote. The curve approaches both axes but never touches them, because x equals 0 is barred and 1 over x is never 0.

An asymptote is any straight line that the graph approaches more and more closely as you follow it outward. It is not part of the graph; it is a guide-line that describes where the curve is headed. Every rational function has this kind of edge behavior, and reading it off the formula is the goal of the next two sections.

Vertical asymptotes and holes

The reciprocal blew up at $x = 0$ because that is where its denominator vanished. The same thing happens in general, but with one twist: not every zero of the denominator produces an asymptote. What matters is whether the troublesome factor also lives in the numerator.

Why a leftover factor blows the graph up, and a canceling one does not#

Suppose the denominator is zero at $x = a$ , so it has a factor of $x - a$ . Two cases split apart.

In the first case $x - a$ divides the denominator but not the numerator. Then as $x$ moves very close to $a$ , the factor $x - a$ becomes a tiny number near zero. So the whole denominator is tiny, while the numerator, with no factor of $x - a$ to shrink it, settles on some nonzero value $N$ . A fixed nonzero number divided by an ever-smaller number produces an ever-larger result: dividing by $0.001$ multiplies the size by a thousand, dividing by $0.000001$ by a million. So the output grows without bound in size, positive on one side and negative on the other, and the curve races along the vertical line

x = a,

which is therefore a vertical asymptote.

In the second case $x - a$ divides both, and cancelling clears every copy of it from the denominator. After canceling, the function agrees, at every input except $x = a$ itself, with a simpler rational function that has a perfectly finite value at $a$ . Nothing blows up, because the shrinking denominator is matched by a shrinking numerator. The only leftover damage is that $x = a$ is still forbidden from the original domain, so the graph is that simpler curve with exactly one point removed. That missing point is a hole. The rule is clean, once you cancel as far as it will go and look at what is left. In that fully canceled form, a factor still sitting in the denominator gives an asymptote, and one with no copy left there gives a hole.

Worked example 2 Find the vertical asymptotes of $f(x) = \dfrac{x + 4}{(x - 1)(x + 2)}$

The denominator is zero at $x = 1$ and $x = -2$ . Check whether either factor also cancels with the numerator $x + 4$ .

At $x = 1$ the numerator is $1 + 4 = 5$ , and at $x = -2$ it is $-2 + 4 = 2$ . Neither is zero, so neither factor cancels, and both survive in the denominator.

x = 1 \quad \text{and} \quad x = -2

are therefore vertical asymptotes. Near $x = 1$ , for instance, the numerator stays close to $5$ while the denominator shrinks toward zero. So the fraction grows without bound, and the curve shoots off along the line $x = 1$ .

Worked example 3 Locate the hole and the asymptote of $f(x) = \dfrac{x^2 - x - 6}{x^2 - 9}$

Factor the top and bottom completely and see what cancels:

f(x) = \frac{x^2 - x - 6}{x^2 - 9} = \frac{(x - 3)(x + 2)}{(x - 3)(x + 3)}.

The factor $x - 3$ appears on both the top and the bottom, so it cancels, leaving

f(x) = \frac{x + 2}{x + 3} \qquad (x \neq 3).

Because $x - 3$ cancelled, $x = 3$ gives a hole, not an asymptote. Its height is the value of the reduced form there,

\frac{3 + 2}{3 + 3} = \frac{5}{6},

so the hole sits at $\left(3, \tfrac{5}{6}\right)$ . The other factor, $x + 3$ , does not cancel, so $x = -3$ is a genuine vertical asymptote. Both $3$ and $-3$ are still excluded from the domain, but for very different graphical reasons. The hole at $x = 3$ is a single missing point, and the asymptote at $x = -3$ is a line the curve flees along.

A hole, not an asymptote. Since (x squared minus 4) over (x minus 2) factors as (x minus 2)(x plus 2) over (x minus 2), the factor x minus 2 cancels and the graph is just the line y equals x plus 2, but with the single point at x equals 2 removed. The open circle at (2, 4) marks that hole. Nothing blows up, because the shrinking denominator is cancelled by a shrinking numerator.

Check your understanding

The graph of $f(x) = \dfrac{x - 4}{x^2 - 16}$ has a hole at which point?

(4, 0)

\left(4, \tfrac{1}{8}\right)

\left(-4, \tfrac{1}{8}\right)

D no hole; there is a vertical asymptote at

x = 4

Horizontal asymptotes and end behavior

Vertical asymptotes describe what the graph does near a forbidden input. The horizontal asymptote describes the opposite edge: what height the graph settles toward as $x$ runs far out to the left and right. The whole question turns on a single idea you can check by hand: when $x$ is huge, a polynomial is ruled by its highest-power term.

Take $x = 1000$ in $x^2 + 5x$ . The square term is $x^2 = 1{,}000{,}000$ , while $5x = 5{,}000$ is a thousand times smaller and barely nudges the total. The larger $x$ becomes, the more completely the top-power term dwarfs everything else. So for the purpose of the far-out behavior, you may compare a rational function’s top and bottom by their leading terms alone. Let $n$ be the degree of the numerator and $m$ the degree of the denominator. Three cases result.

If $n < m$ , the denominator’s top power outgrows the numerator’s, so you are dividing a slower-growing quantity by a faster-growing one, and the ratio is squeezed toward zero. The horizontal asymptote is

y = 0.

If $n = m$ , the top and bottom grow at the same rate, and each is dominated by its leading term. Writing the leading coefficients as $a$ (top) and $b$ (bottom), the ratio settles toward

y = \frac{a}{b}.

If $n > m$ , the numerator grows faster than the denominator, so the fraction grows without bound in size and never levels off. There is no horizontal asymptote. In the special case where the numerator is exactly one degree higher, the graph instead lines up far out with a slanted line called a slant asymptote. You find that line by dividing the polynomials, but the main fact to remember is that a top-heavy fraction has no horizontal asymptote at all.

Worked example 4 Find the horizontal asymptote in each degree case

Compare the degree of the top with the degree of the bottom.

For $f(x) = \dfrac{3x + 2}{x^2 - 1}$ , the top has degree $1$ and the bottom degree $2$ , so $n < m$ . The bottom wins for large $x$ , dragging the ratio toward zero:

y = 0.

For $g(x) = \dfrac{4x^2 - x}{2x^2 + 7}$ , the degrees are equal at $2$ , so the leading terms decide. The leading coefficients are $4$ on top and $2$ on the bottom:

y = \frac{4}{2} = 2.

You can sanity-check this by plugging in $x = 100$ : the top is about $4(10000) = 40000$ and the bottom about $2(10000) = 20000$ , a ratio near $2$ , exactly as predicted.

For $h(x) = \dfrac{x^3 + 1}{x^2 - 5}$ , the top has the higher degree, $3$ against $2$ , so $n > m$ . The fraction grows without bound and there is no horizontal asymptote.

Check your understanding

What is the horizontal asymptote of $f(x) = \dfrac{6x^2 + 1}{3x^2 - x}$ ?

y = 2

y = 0

y = \tfrac{1}{2}

D no horizontal asymptote

x-intercepts and shifting the hyperbola

A fraction equals zero only when its numerator equals zero (and its denominator does not). So to find where a rational graph crosses the x-axis, set the numerator to zero and solve. Then discard any solution that also makes the denominator zero, since that input is not even in the domain. For $f(x) = \dfrac{x - 3}{x + 1}$ the numerator is zero at $x = 3$ , and since the denominator there is $3 + 1 = 4 \neq 0$ , the x-intercept is $(3, 0)$ .

Reading a transformed reciprocal is just as direct. Every function of the form

f(x) = \frac{a}{x - h} + k

is the basic hyperbola $\frac{1}{x}$ shifted and scaled, using the same moves you applied to other graphs. The $x - h$ in the denominator shifts the graph right by $h$ , which drags the vertical asymptote from $x = 0$ over to $x = h$ . The $+k$ on the outside lifts the whole graph up by $k$ , carrying the horizontal asymptote from $y = 0$ up to $y = k$ . The factor $a$ stretches the branches, and a negative $a$ flips them across the horizontal asymptote. So the two asymptotes cross at the point $(h, k)$ , the new center of the hyperbola.

Shifting the hyperbola. The rule y equals 1 over (x minus 2), plus 1 takes the reciprocal graph and slides it right 2 and up 1. The vertical asymptote moves from x equals 0 to the dashed line x equals 2, and the horizontal asymptote moves from y equals 0 to the dashed line y equals 1. The two asymptotes cross at the new center (2, 1), and the two branches hug them just as before.

Worked example 5 Analyze $f(x) = \dfrac{2}{x + 1} - 3$

Match the rule to the form $f(x) = \dfrac{a}{x - h} + k$ . Writing $x + 1$ as $x - (-1)$ gives $h = -1$ , and the constant is $k = -3$ , with $a = 2$ .

The vertical asymptote sits where the denominator is zero, at $x = -1$ , and the horizontal asymptote is the added constant, $y = -3$ . The center of the hyperbola is where they cross, $(-1, -3)$ .

Now read off the intercepts. The y-intercept comes from $x = 0$ :

f(0) = \frac{2}{0 + 1} - 3 = 2 - 3 = -1,

so the graph crosses the y-axis at $(0, -1)$ . The x-intercept comes from setting $f(x) = 0$ :

\frac{2}{x + 1} = 3 \quad \Longrightarrow \quad 2 = 3(x + 1) \quad \Longrightarrow \quad x = -\frac{1}{3},

so the graph crosses the x-axis at $\left(-\tfrac{1}{3}, 0\right)$ . With the two asymptotes and these two crossings, the shape is completely pinned down.

Set that example on the figure below and watch it assemble itself: $a = 2$ , then step the inside shift down to $h = -1$ , then the outside shift down to $k = -3$ . The two dashed lines are the asymptotes, and they carry the graph with them.

Two things are worth doing deliberately once it is there. First, watch the expression while you move the inside shift. At $h = -1$ the rule reads $x + 1$ underneath, and the graph has gone left; the sign you see and the direction it travels disagree, every time. That disagreement is the whole reason the worked example rewrote $x + 1$ as $x - (-1)$ before reading off $h$ . Second, drive $a$ down through zero. Negative values flip both branches across the horizontal asymptote, and at $a = 0$ there is no hyperbola left at all: the rule collapses to the flat line $y = k$ . That line is still missing its point at $x = h$ , because a division by zero stays undefined no matter what you multiply it by.

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

Stretch a Inside h Outside k

y = 1/x. Its centre sits at (0, 0), not shifted at all. One branch lies above and to the right of it and the other below and to the left. It never reaches x = 0 or y = 0: those are its asymptotes.

A hyperbola on a coordinate plane, drawn with the two dashed lines it approaches but never reaches. The inside shift carries the vertical one sideways, the outside shift carries the horizontal one up and down, and the coefficient stretches the two branches or flips them over.

Solving a rational equation

To solve an equation that has variables in a denominator, clear the fractions by multiplying every term by a common denominator. Then solve the polynomial equation that remains, and finally throw out any candidate that makes an original denominator zero. That last step is not optional. Multiplying by an expression that could be zero can introduce a false solution, an extraneous one, exactly as squaring did for radical equations.

Worked example 6 Solve two rational equations, one of which has an extraneous solution

First solve $\dfrac{3}{x + 1} = \dfrac{2}{x - 4}$ . Multiply both sides by $(x + 1)(x - 4)$ to clear the denominators, which leaves a linear equation:

3(x - 4) = 2(x + 1) \quad \Longrightarrow \quad 3x - 12 = 2x + 2 \quad \Longrightarrow \quad x = 14.

The value $x = 14$ makes neither denominator zero, so it is genuine, and the only solution is $x = 14$ .

Now solve $\dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 3$ . Multiply every term by $x - 2$ :

x = 2 + 3(x - 2) \quad \Longrightarrow \quad x = 3x - 4 \quad \Longrightarrow \quad x = 2.

Before celebrating, check the candidate against the original denominators. Here $x = 2$ makes $x - 2 = 0$ , so it was never in the domain. The candidate is extraneous, and the equation has no solution. The multiplication step manufactured it, which is why every answer must be tested in the original equation.

Common mistakes

Letting the numerator restrict the domain. Only the denominator can force an exclusion. For $\dfrac{x - 5}{x + 2}$ the domain is every $x \neq -2$ ; the numerator’s zero at $x = 5$ is a perfectly good input (it is the x-intercept), not a forbidden one.
Calling every denominator zero a vertical asymptote. A zero of the denominator is a vertical asymptote only if its factor does not cancel with the numerator. If the factor cancels, that input is a hole instead. Always factor and cancel first.
Reading the horizontal asymptote off the constant terms. End behavior is set by the highest-power terms, not the constants. For $\dfrac{4x^2 - x}{2x^2 + 7}$ the asymptote is $y = \tfrac{4}{2} = 2$ , from the leading coefficients, not $y = \tfrac{0}{7}$ .
Expecting a horizontal asymptote when the top is heavier. If the numerator has the higher degree, there is no horizontal asymptote at all, because the fraction grows without bound as $x$ gets large.
Forgetting to check for extraneous solutions. Multiplying an equation by a variable expression can create a solution that breaks an original denominator. Every candidate must be tested, and any that makes a denominator zero is discarded.
Assuming a graph can never cross a horizontal asymptote. A graph never crosses a vertical asymptote, because the function is undefined there. A horizontal asymptote only describes the far-out behavior, so a curve can cross it at ordinary $x$ -values before settling toward it at the edges. The reciprocal $\dfrac{1}{x}$ happens to meet neither of its axes, but that is special to $\dfrac{1}{x}$ , not a rule for every rational function.

Key takeaways

A rational function $f(x) = \dfrac{p(x)}{q(x)}$ is one polynomial over another. Its domain is every real number except the zeros of the denominator, because dividing by zero has no value.
The reciprocal $\dfrac{1}{x}$ graphs as a hyperbola with two branches. Its vertical asymptote is $x = 0$ and its horizontal asymptote is $y = 0$ ; the curve approaches both without touching them.
A zero of the denominator whose factor does not cancel gives a vertical asymptote, where the output blows up. A zero whose factor cancels with the numerator gives a hole instead, a single missing point.
The horizontal asymptote comes from comparing degrees $n$ (top) and $m$ (bottom). When $n < m$ the asymptote is $y = 0$ ; when $n = m$ it is $y = \dfrac{a}{b}$ , the ratio of leading coefficients; and when $n > m$ there is no horizontal asymptote.
x-intercepts are the zeros of the numerator that stay in the domain. In the form $\dfrac{a}{x - h} + k$ , the asymptotes are $x = h$ and $y = k$ , crossing at the center $(h, k)$ .
To solve a rational equation, clear denominators, solve, and discard any extraneous candidate that makes an original denominator zero.

Rational Functions

Learning goals

What a rational function is

Finding the domain

The reciprocal function and its hyperbola

Vertical asymptotes and holes

Horizontal asymptotes and end behavior

x-intercepts and shifting the hyperbola

Solving a rational equation

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)