Graphs of Rational Functions

Learning goals

Exclude the zeros of the original denominator
Define an asymptote as a line the curve stays near
Tell a hole from a vertical asymptote by whether the factor survives
Compare degrees for the horizontal asymptote
Divide to find an oblique asymptote when the degree exceeds by one
Cross a horizontal asymptote freely, but never a vertical one

The simplest rational function, and what its graph does

A rational function is one polynomial divided by another,

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

where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial. Because division by zero has no value, the domain of $f$ is every real number except the zeros of $Q$ .

Start with the smallest example there is, $f(x) = \dfrac{1}{x}$ , whose only excluded input is $x = 0$ . A short table shows what happens as $x$ creeps toward the forbidden value from the right:

$x$	$1$	$0.5$	$0.1$	$0.01$	$0.001$
$\frac{1}{x}$	$1$	$2$	$10$	$100$	$1000$

The output has no ceiling. Name any height you like, say a million, and $x = \tfrac{1}{2{,}000{,}000}$ already beats it. Travel the other way instead, out toward large $x$ , and the outputs shrink toward zero without ever reaching it: $f(1000) = 0.001$ , and $f(1{,}000{,}000) = 0.000001$ . Negative inputs behave the same way with the signs flipped.

The graph of y = 1/x. The two dashed lines are its asymptotes, and here they are the axes themselves: the curve climbs the y-axis without ever touching it, and flattens toward the x-axis without ever reaching it.

Both dashed lines in that picture are asymptotes. Here is the honest definition, and it is the one we will use all lesson:

A line is an asymptote of a curve when the distance from the curve to that line becomes and stays smaller than any distance you care to name. That happens once you travel far enough out along the curve.

“Smaller than any distance you care to name” is doing real work in that definition. That phrase is exactly the one a first course in calculus makes precise, with limits. We are not going to fake that machinery here. We do not need it: everything below is ordinary algebra, and it produces the right lines and the right picture. Naming the edge of your tools is not an apology, it is how you know what you have proved.

The definition covers all three kinds at once. The $y$ -axis is an asymptote of $y = \frac 1x$ because the points of the curve sit at horizontal distance $|x|$ from the line $x = 0$ . That horizontal distance can be made as small as you please while the curve runs off the top of the page. The $x$ -axis is an asymptote because the vertical distance $\left|\frac 1x\right|$ can be made as small as you please by taking $x$ far enough out.

Holes and vertical asymptotes

Every excluded input comes from a zero of the original denominator, and there are exactly two things a graph can do at one. Compare these two functions, which are undefined at the same place, $x = 2$ :

g(x) = \frac{x^2 - 4}{x - 2}, \qquad h(x) = \frac{1}{x - 2}.

For $g$ , factor and cancel. Since $x^2 - 4 = (x-2)(x+2)$ ,

g(x) = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad \text{for every } x \neq 2 .

That cancellation is the move from the simplifying lesson, and its warning applies here in full. Cancelling preserves the value at every input where the original was defined, but it does not restore the input that was thrown away. So $g$ agrees with the line $y = x+2$ everywhere except at $x=2$ , where $g$ has no value at all. Its graph is that line with one point removed: a hole.

For $h$ nothing cancels, and the outputs behave like the table from $\frac 1x$ . As $x$ approaches $2$ the denominator becomes tiny while the numerator stays $1$ , so the values blow up. The graph tears apart along the line $x = 2$ , which is a vertical asymptote.

Two functions, both undefined at x = 2, with completely different graphs. On the left the factor (x - 2) cancels, so the curve is an ordinary line with a single point punched out. On the right the factor survives in the denominator, so the values blow up along the dashed line.

So the same excluded input can produce a missing dot or a cliff, and the deciding question is whether the offending factor cancels. Here is why, and the argument is worth reading closely because it also tells you exactly how much cancelling is enough.

Why a factor that cancels leaves a hole, and one that survives forces a vertical asymptote#

Let $f(x) = \frac{P(x)}{Q(x)}$ and let $a$ be a zero of $Q$ . Pull out the largest power of $(x-a)$ that divides each polynomial, writing

P(x) = (x-a)^m\, p(x), \qquad Q(x) = (x-a)^n\, q(x),

where $n \ge 1$ because $a$ is a zero of $Q$ , where $m \ge 0$ , and where $p(a) \neq 0$ and $q(a) \neq 0$ . Those last two facts are exactly what “largest power” buys us. If $p(a)$ were zero then the factor theorem would hand us another copy of $(x-a)$ inside $p$ , and we would not have pulled out the largest power after all.

For every $x$ other than $a$ we may cancel, and there are two outcomes.

Suppose first that $m \ge n$ , so the numerator carries at least as many copies of $(x-a)$ as the denominator. Cancelling all $n$ of the denominator’s copies leaves

f(x) = \frac{(x-a)^{m-n}\, p(x)}{q(x)} \qquad (x \neq a),

and the exponent $m-n$ is now zero or positive. The right-hand side is a perfectly ordinary expression at $x=a$ as well, because its denominator $q(a)$ is not zero, and it takes a finite value there. That value is $\frac{p(a)}{q(a)}$ when $m = n$ , since the factor $(x-a)^{m-n}$ is then absent altogether, and it is $0$ when $m > n$ , since the factor is present and vanishes at $a$ . So $f$ agrees, at every input where $f$ is defined, with a function whose graph sails straight through $x=a$ at an ordinary height. The original $f$ is still undefined at $a$ , since the original denominator vanishes there, so the graph of $f$ is that unremarkable curve with the single point above $x=a$ removed. That is a hole.

Suppose instead that $m < n$ , so the denominator keeps at least one copy of $(x-a)$ after cancelling. Write $k = n - m \ge 1$ ; then

f(x) = \frac{p(x)}{(x-a)^{k}\, q(x)} \qquad (x \neq a).

Look at the two pieces separately. The ratio $\frac{p(x)}{q(x)}$ is built from polynomials, whose graphs are unbroken curves, so a small change in $x$ only makes a small change in $p(x)$ and in $q(x)$ . Since $\frac{p(a)}{q(a)}$ is a nonzero number, that steadiness lets us pick a small interval around $a$ on which $\left|\frac{p(x)}{q(x)}\right|$ stays above some fixed positive number $c$ . The other piece, $(x-a)^{k}$ , we control completely: it is as small as we like once $x$ is close enough to $a$ . Putting the pieces together, on that interval

|f(x)| \;\ge\; \frac{c}{|x-a|^{k}} .

Now name any height $H$ , however large. The quantity $|x-a|^{k}$ is a product of $k$ copies of a number we can shrink at will. So we can bring $x$ close enough to $a$ to force $|x-a|^{k}$ below $\frac{c}{H}$ , and the inequality above then gives $|f(x)| \ge H$ . The curve therefore leaves every horizontal band you draw, no matter how tall. Yet the curve’s points sit at horizontal distance $|x-a|$ from the line $x=a$ , a distance we just made as small as we pleased. The line $x=a$ is a vertical asymptote.

Read the two cases again and notice what the dividing line actually is. It is not “does $(x-a)$ appear in the numerator”, it is does the numerator carry at least as many copies of $(x-a)$ as the denominator. The factor must cancel completely. A single $(x-1)$ upstairs against $(x-1)^2$ downstairs cancels one copy and leaves one behind, and one copy left in a denominator is all it takes.

Worked example 1 Find the domain, holes, and vertical asymptotes of $f(x)=\frac{x^2-9}{x^2-2x-3}$

Factor the top and the bottom completely:

f(x) = \frac{(x-3)(x+3)}{(x-3)(x+1)} .

The domain comes from the original denominator, whose zeros are $3$ and $-1$ , so the domain is every real number except $x = 3$ and $x = -1$ .

Now ask which of those two factors cancels. The factor $(x-3)$ appears once on top and once on the bottom, so it cancels completely:

f(x) = \frac{x+3}{x+1} \qquad (x \neq 3,\; x \neq -1).

Nothing is left of $(x-3)$ in the denominator, so $x=3$ gives a hole. Its height is the value of the simplified expression there:

\frac{3+3}{3+1} = \frac{6}{4} = \frac{3}{2},

so the hole sits at $\left(3, \tfrac 32\right)$ . The factor $(x+1)$ never cancelled, so $x = -1$ is a vertical asymptote.

Worked example 2 Classify the excluded inputs of $f(x)=\frac{x^2+3x-4}{(x-1)^2(x+2)}$

The numerator factors as $x^2 + 3x - 4 = (x+4)(x-1)$ , so

f(x) = \frac{(x+4)(x-1)}{(x-1)^2 (x+2)} .

The original denominator is zero at $x = 1$ and $x = -2$ , so those two inputs are out of the domain. The factor $(x-1)$ does appear in the numerator, which is precisely the trap. Count copies: one on top, two on the bottom. Cancelling one from each leaves a copy behind,

f(x) = \frac{x+4}{(x-1)(x+2)} \qquad (x \neq 1,\; x \neq -2),

and a surviving $(x-1)$ in the denominator means the values still blow up near $x=1$ . So $x = 1$ is a vertical asymptote, not a hole, and so is $x = -2$ . This function has no holes at all. Appearing in the numerator is not enough; the factor has to cancel completely.

Check your understanding

The function $f(x)=\dfrac{(x-5)(x+1)}{(x-5)(x-2)}$ is undefined at $x=5$ and at $x=2$ . What does its graph do at each?

A vertical asymptotes at

x = 5

and

x = 2

B a hole at

x = 2

and a vertical asymptote at

x = 5

C a hole at

x = 5

and a vertical asymptote at

x = 2

D a hole at

x = -1

and a vertical asymptote at

x = 2

End behavior: what the graph does far out

Vertical asymptotes tell you what happens near the bad points. End behavior is the other question: what does the curve do as you travel far to the right or far to the left? There is a three-case rule for this, and it is worth deriving rather than memorizing, because the derivation is one line and the rule falls out of it.

Take

f(x) = \frac{3x^2 + 5x - 1}{2x^2 - 7}

and divide the numerator and the denominator by $x^2$ , the highest power of $x$ in the denominator. For $x \neq 0$ this changes nothing about the value of the fraction:

f(x) = \frac{3 + \dfrac{5}{x} - \dfrac{1}{x^2}}{\,2 - \dfrac{7}{x^2}\,} .

Now every term that is not a leading coefficient has an $x$ underneath it, and those terms are exactly the ones that die away when $x$ is huge. At $x = 1000$ the numerator is $3 + 0.005 - 0.000001$ and the denominator is $2 - 0.000007$ , so $f(1000) \approx \frac{3.005}{2.000} \approx 1.5025$ , already within three thousandths of $\frac 32$ . Push $x$ further out and the little terms shrink further. The curve has the horizontal asymptote $y = \frac 32$ , and notice where that number came from: the ratio of the leading coefficients, $\frac{3}{2}$ , and nothing else.

The same division works in general. Given $f(x) = \frac{P(x)}{Q(x)}$ with $\deg P = m$ and $\deg Q = n$ , divide top and bottom by $x^n$ . The denominator turns into its leading coefficient plus a pile of terms with powers of $x$ underneath them. For large $|x|$ the denominator is as close to that leading coefficient as you like. Everything then depends on what the numerator turned into, and there are three possibilities.

The bottom outgrows the top ( $m < n$ ). Dividing by $x^n$ leaves every numerator term with a power of $x$ underneath it, so the whole numerator dies away while the denominator settles on a nonzero number. For example

\frac{2x+1}{x^2+3} = \frac{\dfrac{2}{x} + \dfrac{1}{x^2}}{\,1 + \dfrac{3}{x^2}\,},

whose numerator shrinks to nothing and whose denominator heads for $1$ . The horizontal asymptote is $y = 0$ , the $x$ -axis.

The degrees tie ( $m = n$ ). This is the case we just did. Both leading terms survive the division and everything else dies away, so the horizontal asymptote is $y = \frac{a}{b}$ , where $a$ and $b$ are the leading coefficients of $P$ and $Q$ .

The top outgrows the bottom ( $m > n$ ). Dividing by $x^n$ leaves the numerator with a positive power of $x$ still in it, so the numerator grows without bound while the denominator settles down. For example

\frac{x^3+1}{x^2+1} = \frac{x + \dfrac{1}{x^2}}{\,1 + \dfrac{1}{x^2}\,},

which behaves like $x$ itself for large $x$ . There is no horizontal asymptote: no horizontal line can hold a curve that climbs forever. When the top outgrows the bottom by exactly one degree, as it does here, the curve is chased by a slanted line instead, with one exception. The next section pins that exception down. When it outgrows the bottom by two degrees or more, there is no straight-line asymptote at all; the curve grows like a polynomial of degree $m-n$ .

One small bonus, and it saves work later. Cancelling a common factor from the top and the bottom lowers both degrees by the same amount and divides both leading coefficients by the same number. So the cancelling changes neither the comparison $m$ against $n$ nor the ratio $\frac ab$ . You may read the end behavior off the original form or off the simplified form, and you will get the same answer.

Check your understanding

What is the horizontal asymptote of $f(x) = \dfrac{2x^3+7}{5x^2-x}$ ?

y = \frac{2}{5}

B there is no horizontal asymptote C

y = 0

y = 2

Oblique asymptotes, and why long division finds them

When the numerator’s degree is exactly one more than the denominator’s, the curve runs away from every horizontal line, and yet it is not wild. Far out, the curve hugs a slanted line. (Almost always. There is one exception, and we will corner it at the end of this section.) Finding that line needs no new tool. You already own it, from the polynomial division chapter.

Divide the polynomials. For

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

long division gives a quotient of $x+1$ and a remainder of $3$ : the first step takes $x(x-1) = x^2-x$ and leaves $x+2$ , and the second takes $1 \cdot (x-1) = x-1$ and leaves $3$ . Check it by multiplying back, which is the habit worth keeping: $(x+1)(x-1) + 3 = x^2 - 1 + 3 = x^2 + 2$ , the numerator we started with. (The remainder theorem agrees, since the numerator at $x=1$ is $1 + 2 = 3$ .) Written the way division always lets you write it,

f(x) = x + 1 + \frac{3}{x-1} .

Read that as a sentence. The function is a line, $y = x+1$ , plus a correction term $\frac{3}{x-1}$ . The correction is a proper fraction, because the remainder of a division always has smaller degree than the divisor. And a proper fraction is exactly the kind of thing that dies away far out. At $x = 101$ it is $\frac{3}{100} = 0.03$ ; at $x = 1001$ it is $0.003$ . So for large $x$ the curve sits a whisker above the line $y = x+1$ and settles onto it. That line is an oblique asymptote (also called a slant asymptote). The vertical distance from the curve to the line is precisely the correction term, so we can make that distance as small as we like by going far enough out. That is the definition, satisfied on the nose.

The graph of y = (x² + 2)/(x - 1), which long division rewrites as y = x + 1 + 3/(x - 1). The curve blows up along the vertical asymptote x = 1 and settles onto the slanted line y = x + 1 far out in both directions.

This is the chapter’s best payoff. Polynomial long division was introduced two chapters ago as a way to factor and to hunt for roots. It turns out to be the tool that reads the shape of a graph off its formula. The reason is that it splits a rational function into the part that matters far out (the quotient) and the part that vanishes far out (the remainder over the denominator).

And once you see it that way, the horizontal case is the same theorem. Divide $\frac{x-3}{x-2}$ and you get a quotient of $1$ with a remainder of $-1$ :

\frac{x-3}{x-2} = 1 - \frac{1}{x-2} .

A constant quotient is a horizontal asymptote, $y=1$ . A linear quotient is an oblique asymptote. Same division, same reasoning, and the degree comparison from the last section is just a shortcut for predicting which one you are about to get.

One condition is doing quiet work in all of this, and it is worth dragging into the light: the remainder has to be nonzero. The whole argument runs on the correction term $\frac{R(x)}{Q(x)}$ being small but present. If the division comes out exact, so that $R$ is zero, then $f$ is the quotient wherever it is defined, and there is nothing left for the curve to approach. Watch it happen:

\frac{3x^2+5x-2}{x+2} = \frac{(3x-1)(x+2)}{x+2} = 3x-1 \qquad (x \neq -2).

The numerator’s degree is one more than the denominator’s, exactly the case that is supposed to produce an oblique asymptote, and yet this graph has none. It does not approach the line $y=3x-1$ ; it is that line, with a hole punched at $(-2,-7)$ . So state the rule with its hypothesis attached. When the numerator’s degree is one more than the denominator’s and the division leaves a nonzero remainder, the quotient is an oblique asymptote. When the remainder is zero, factor and enjoy your line. This is the same exception you will meet again in a moment when we ask whether a graph can cross a horizontal asymptote.

Worked example 3 Find the asymptotes of $f(x)=\frac{2x^2-x-1}{x+1}$

First check for a hole, because you always check for a hole. The numerator factors as $2x^2 - x - 1 = (2x+1)(x-1)$ , which carries no factor of $(x+1)$ , so nothing cancels and $x = -1$ is a vertical asymptote.

The numerator has degree $2$ and the denominator degree $1$ , one more on top, so expect an oblique asymptote and go find it by long division. Dividing $2x^2 - x - 1$ by $x+1$ gives the quotient $2x-3$ with remainder $2$ , which checks out because $(2x-3)(x+1) + 2 = 2x^2 - x - 3 + 2 = 2x^2 - x - 1$ . Therefore

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

The oblique asymptote is $y = 2x-3$ . To see that the leftover term really is negligible far out, put $x = 101$ : the correction is $\frac{2}{102} \approx 0.02$ , so $f(101) \approx 199.02$ while the line gives $2(101) - 3 = 199$ . The curve is riding just above its asymptote, and the gap keeps closing.

Can a graph cross an asymptote?

Students routinely picture an asymptote as a fence the curve is forbidden to touch. That belief is half right, and the half that is wrong causes real errors.

A graph can never cross a vertical asymptote. This one needs no argument beyond the domain. If $x=a$ is a vertical asymptote then $a$ is not in the domain, so the graph has no point whose first coordinate is $a$ . There is nothing there to cross with.

A graph absolutely can cross a horizontal or an oblique asymptote, and often does. Look again at what the definition demands: the curve must approach the line far out. It says nothing whatever about the middle of the picture. A curve is free to cut straight through its horizontal asymptote near the origin and then come back and settle onto it.

Here is one doing exactly that. Take

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

The degrees tie, so the horizontal asymptote is $y = \frac 21 = 2$ . To find where the curve meets that line, set the function equal to $2$ and solve, which is a rational equation of the kind you solved last lesson:

\frac{2x^2+x-1}{x^2+1} = 2 \;\Longrightarrow\; 2x^2 + x - 1 = 2x^2 + 2 \;\Longrightarrow\; x = 3 .

Check it: $f(3) = \frac{18+3-1}{9+1} = \frac{20}{10} = 2$ . The curve crosses its horizontal asymptote at the point $(3,2)$ , and then goes on approaching it from above for the rest of the journey.

The graph of y = (2x² + x - 1)/(x² + 1) crosses its horizontal asymptote y = 2 at the point (3, 2), and still approaches that line far out in both directions. An asymptote is a promise about the far end of the curve, not a fence.

Why does one crossing not spoil the asymptote? Because solving $f(x) = c$ turns into the polynomial equation $P(x) - c\,Q(x) = 0$ , and there are only two ways that can go. Either $P - cQ$ is the zero polynomial, in which case $f$ equals $c$ at every input where it is defined. Then the graph does not approach that line, it lies on it (this is exactly what $\frac{x-6}{x-6} = 1$ does, a horizontal line with a hole). Or $P - cQ$ is a genuine nonzero polynomial, and then it has only finitely many roots, so the curve gets only finitely many chances to meet the line. Go far enough out and it has used them all up, after which it can only approach.

The same goes for oblique asymptotes, with one honest caveat. Since $f(x) - (\text{quotient})$ equals the remainder over the denominator, the curve meets its oblique asymptote exactly where the remainder is zero. When the denominator is linear the remainder is a constant, and a nonzero constant is never zero, so those curves never touch their slanted line. That is why the graph of $\frac{x^2+2}{x-1}$ stays strictly above $y = x+1$ on the right. But with a bigger denominator the remainder is free to have a root. Dividing $x^3+1$ by $x^2+1$ gives

\frac{x^3+1}{x^2+1} = x + \frac{1-x}{x^2+1},

whose remainder $1-x$ vanishes at $x=1$ , so that curve crosses its oblique asymptote $y=x$ at the point $(1,1)$ . Never say never.

Worked example 4 Does the graph of $f(x)=\frac{5x^2-3x}{2x^2+x-6}$ cross its horizontal asymptote?

Factor the denominator first, to be sure nothing cancels: $2x^2 + x - 6 = (2x-3)(x+2)$ , and the numerator is $x(5x-3)$ , which shares neither factor. So there are no holes, and the vertical asymptotes are $x = \frac 32$ and $x = -2$ .

The degrees tie at $2$ , so the horizontal asymptote is the ratio of the leading coefficients, $y = \frac 52$ . To find any crossing, set the function equal to that value and clear the fraction:

\frac{5x^2-3x}{2x^2+x-6} = \frac{5}{2} \;\Longrightarrow\; 2(5x^2 - 3x) = 5(2x^2 + x - 6).

Expand both sides and watch the quadratic terms destroy each other, which is exactly what always happens in this calculation when the degrees tie:

10x^2 - 6x = 10x^2 + 5x - 30 \;\Longrightarrow\; -11x = -30 \;\Longrightarrow\; x = \frac{30}{11}.

That value is in the domain (it is neither $\frac 32$ nor $-2$ ), so the crossing is real. The graph meets its horizontal asymptote at $\left(\frac{30}{11}, \frac 52\right)$ , roughly $(2.73,\, 2.5)$ , and approaches it from the other side afterwards.

Check your understanding

Which statement is true of the graph of a rational function?

A It can never cross any of its asymptotes. B It can cross a vertical asymptote but never a horizontal one. C It can cross a vertical asymptote, but only where the matching factor cancels. D It can cross a horizontal asymptote, but never a vertical one.

Putting it all together

Every tool is now on the table, so here is the whole sketch as one routine. Nothing in it is new.

Factor the numerator and the denominator completely.
Domain. Exclude every zero of the original denominator. Do this before you cancel anything.
Holes and vertical asymptotes. Cancel the common factors. An excluded input whose factor cancelled completely is a hole; anything still left in the denominator is a vertical asymptote.
Hole heights. Evaluate the simplified expression at each hole’s $x$ .
Intercepts. The $x$ -intercepts are the zeros of the simplified numerator that survive in the domain; the $y$ -intercept is $f(0)$ , if $0$ is in the domain.
End behavior. Compare the degrees for a horizontal asymptote, or divide for an oblique one.
Crossings. Solve $f(x) = c$ (or $f(x) = mx + b$ ) to see whether the curve meets that asymptote.
Sign chart. Mark the $x$ -intercepts and the vertical asymptotes on a number line and test one point in each interval. The sign can only change at those places, so one test point settles a whole interval, and the signs tell you which way each branch runs.

Worked example 5 Sketch $f(x)=\frac{x^2-x-6}{x^2-4}$ completely

Factor.

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

Domain. The original denominator is zero at $x = 2$ and $x = -2$ , so both are excluded.

Holes and vertical asymptotes. The factor $(x+2)$ appears once top and once bottom, so it cancels completely and $x = -2$ is a hole. The factor $(x-2)$ survives, so $x = 2$ is a vertical asymptote. After cancelling,

f(x) = \frac{x-3}{x-2} \qquad (x \neq 2,\; x \neq -2).

Hole height. Put $x = -2$ into the simplified expression:

\frac{-2-3}{-2-2} = \frac{-5}{-4} = \frac 54 ,

so the hole is at $\left(-2, \tfrac 54\right)$ .

Intercepts. The simplified numerator is zero at $x = 3$ , which is in the domain, so the $x$ -intercept is $(3,0)$ . For the $y$ -intercept, $f(0) = \frac{-3}{-2} = \frac 32$ , giving $\left(0, \tfrac 32\right)$ .

End behavior. The degrees tie ( $2$ and $2$ ) and both leading coefficients are $1$ , so the horizontal asymptote is $y = 1$ . Long division says the same thing in more detail:

f(x) = 1 - \frac{1}{x-2},

which even tells you which side the curve approaches from. For large positive $x$ the correction $-\frac{1}{x-2}$ is a small negative number, so the curve lies just below $y=1$ on the right. For large negative $x$ the correction is a small positive number, so the curve lies just above $y=1$ on the left.

Crossings. Setting $\frac{x-3}{x-2} = 1$ gives $x - 3 = x - 2$ , that is $-3 = -2$ , which is false. This graph never meets its horizontal asymptote. (The decomposition above already told us so: the correction $-\frac{1}{x-2}$ is never zero.)

Sign chart. The sign can change only at the zero $x=3$ and at the vertical asymptote $x=2$ . The hole at $x=-2$ changes nothing, since the factor that made it cancelled out of the formula. Test one point in each interval:

Interval	test value	$\dfrac{x-3}{x-2}$ at that value	the branch is
$x < 2$	$x=0$	$\frac{-3}{-2} = \frac 32 > 0$	above the $x$ -axis
$2 < x < 3$	$x=2.5$	$\frac{-0.5}{0.5} = -1 < 0$	below the $x$ -axis
$x > 3$	$x=4$	$\frac{1}{2} > 0$	above the $x$ -axis

Now assemble. On the left the curve comes in just above $y=1$ and passes through the hole’s neighbourhood, punching out the single point $\left(-2,\tfrac 54\right)$ . The curve then climbs through $\left(0, \tfrac 32\right)$ and, staying positive, rockets up the left side of $x=2$ . On the right of $x=2$ it starts far down (negative, per the sign chart), climbs through the intercept $(3,0)$ , and creeps back up toward $y=1$ from below.

The finished sketch of y = (x² - x - 6)/(x² - 4). The open circle at (-2, 5/4) is the hole, the dashed lines are the vertical asymptote x = 2 and the horizontal asymptote y = 1, and the filled dots are the intercepts (0, 3/2) and (3, 0).

Look at what that picture is. Every feature on it is an algebraic fact you established without plotting a single extra point: the hole is the cancelled factor, and the vertical asymptote is the surviving one. The horizontal asymptote is the tie in degrees, and the sign chart chose the direction of each branch. The domain, tracked as a side condition since the first lesson of this chapter, is now the most visible thing on the page.

Common mistakes

Reading the domain off the simplified form. The excluded inputs come from the original denominator. In $\frac{(x-3)(x+2)}{(x-2)(x+2)}$ the value $x=-2$ is out of the domain forever, even though the simplified $\frac{x-3}{x-2}$ shows no sign of it. Cancelling changes the formula, not the function.
Calling every zero of the denominator a vertical asymptote. A zero of the denominator is an excluded input, and that is all you know until you check whether its factor cancels. If it cancels completely you get a hole, not an asymptote.
Accepting a partial cancellation. One $(x-1)$ upstairs against $(x-1)^2$ downstairs leaves a copy behind, so $x=1$ is still a vertical asymptote. Count the copies; the factor has to cancel completely.
Treating an asymptote as a fence. A graph may cross its horizontal or oblique asymptote, and you find where by solving $f(x)=c$ . Only a vertical asymptote can never be crossed, and that is because its $x$ is not in the domain at all.
Reading the horizontal asymptote off the wrong terms. It comes from the leading coefficients, not the constants. The asymptote of $\frac{3x+8}{2x+1}$ is $y = \frac 32$ , not $y = 8$ .
Expecting a horizontal asymptote when the top outgrows the bottom. If $\deg P > \deg Q$ there is none. You get an oblique asymptote only in the single case $\deg P = \deg Q + 1$ .
Drawing a hole as a solid dot, or forgetting it entirely. Mark it as an open circle. It is the one feature of the graph that a plotting program will usually hide from you.

Key takeaways

A rational function is $f(x) = \frac{P(x)}{Q(x)}$ with $P$ and $Q$ polynomials. Its domain excludes exactly the zeros of the original $Q$ .
A line is an asymptote when the distance from the curve to the line becomes and stays smaller than any distance you name, once you go far enough out. Making “arbitrarily close” precise is the business of calculus; the algebra here already finds the right lines.
At a zero $a$ of the denominator: if $(x-a)$ cancels completely, the graph has a hole at $x=a$ , whose height is the simplified expression evaluated there. If any copy survives in the denominator, $x=a$ is a vertical asymptote.
End behavior comes from dividing the top and the bottom by $x^{\deg Q}$ , which kills every term but the leading ones. With $\deg P = m$ and $\deg Q = n$ : if $m<n$ the asymptote is $y=0$ ; if $m=n$ it is $y = \frac ab$ , the ratio of leading coefficients; if $m>n$ there is no horizontal asymptote.
If $m = n+1$ and the division leaves a nonzero remainder, polynomial long division writes $f(x) = (\text{linear quotient}) + \frac{\text{remainder}}{Q(x)}$ . The correction term dies away far out, so the linear quotient is an oblique asymptote. If instead the division comes out exact, the graph is that line (with a hole at each cancelled factor) and there is no asymptote. Nothing is left for the curve to approach.
A graph can cross a horizontal or an oblique asymptote, but it can never cross a vertical one, because that input is not in the domain. Solve $f(x)=c$ to find where it crosses the horizontal or oblique one.
To sketch: factor, take the domain, sort the holes from the vertical asymptotes, then find the intercepts. Settle the end behavior, check for crossings, and run a sign chart between the interesting $x$ -values.

Graphs of Rational Functions

Learning goals

The simplest rational function, and what its graph does

Holes and vertical asymptotes

End behavior: what the graph does far out

Oblique asymptotes, and why long division finds them

Can a graph cross an asymptote?

Putting it all together

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)