1.9 - Rational Functions and Vertical Asymptotes

Overview

A vertical asymptote is a vertical line

x = a

that a graph shoots up or down along, because the denominator is getting close to 0 there. In this topic you'll find vertical asymptotes, use multiplicity to tell an asymptote from a hole, and describe what happens on each side with one-sided limits. On the AP exam you'll do this from equations, graphs, and tables.

Where vertical asymptotes come from

Dividing by a number close to 0 gives a huge result:

\frac{1}{0.001} = 1{,}000

and

\frac{1}{-0.001} = -1{,}000

. So when the denominator of a rational function gets close to 0 and the numerator doesn't, the outputs blow up.

Rule: if

a

is a real zero of the denominator but not a real zero of the numerator, the graph has a vertical asymptote at

x = a

Look at

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

near

x = 3

$x$	$2.9$	$2.99$	$2.999$	$3$	$3.001$	$3.01$	$3.1$
$f(x)$	$-10$	$-100$	$-1{,}000$	undefined	$1{,}000$	$100$	$10$

The closer

x

gets to 3, the bigger the outputs get in size. From the left they're hugely negative, and from the right they're hugely positive.

When the numerator and denominator share a zero

a

makes both the numerator and the denominator 0, compare multiplicities, meaning how many times the factor

(x - a)

appears in each:

More copies in the denominator: vertical asymptote at $x = a$ . After canceling, a factor of $(x - a)$ is still left in the denominator.

At least as many copies in the numerator: a hole at $x = a$ , not an asymptote (Topic 1.10).

For example,

\frac{x - 2}{(x - 2)^2}

has one copy of

(x - 2)

on top and two on the bottom. Canceling leaves

\frac{1}{x - 2}

, so there's a vertical asymptote at

x = 2

One-sided limits

Near a vertical asymptote

x = a

, the outputs increase or decrease without bound, and the direction can be different on each side. We describe each side with a one-sided limit:

Notation	Meaning
$\lim_{x \to a^-} r(x)$	the limit as $x$ approaches $a$ from the left (inputs less than $a$ )
$\lim_{x \to a^+} r(x)$	the limit as $x$ approaches $a$ from the right (inputs greater than $a$ )

At a vertical asymptote, each one-sided limit is

\infty

-\infty

. For

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

\lim_{x \to 3^-} f(x) = -\infty, \qquad \lim_{x \to 3^+} f(x) = \infty

Finding the sign on each side

You don't need a calculator. Picture an input just to one side of

a

and track only the sign of each factor. The factor that's going to 0 is tiny, and its sign depends on which side you're on. Every other factor is close to its value at

a

, so its sign is easy to read.

Multiplicity tells you the pattern

1.9 - Rational Functions and Vertical Asymptotes

Odd multiplicity in the denominator, like $(x - 1)$ or $(x - 1)^3$ : the factor changes sign at $a$ , so the two sides go in opposite directions.

Even multiplicity, like $(x - 1)^2$ : the factor is positive on both sides, so both sides go in the same direction.

Use the multiplicity that's left after canceling any common factors. Also, a graph never crosses a vertical asymptote, since

x = a

isn't in the domain. (A horizontal asymptote is different: a graph can cross it.)

Examples

Example 1: Two vertical asymptotes

Find the vertical asymptotes of

g(x) = \frac{x + 1}{x^2 - x - 6}

, and describe the behavior on each side with one-sided limits.

Step 1: Factor.

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

Step 2: The denominator is 0 at

x = 3

and

x = -2

. The numerator is 0 only at

x = -1

, so there are no shared zeros. Vertical asymptotes:

x = 3

and

x = -2

Step 3: Check signs near

x = -2

. The factor

x + 1

is about

-1

(negative) and

x - 3

is about

-5

(negative).

Just right of $-2$ , $x + 2$ is tiny and positive: $\frac{(-)}{(-)(+)}$ is positive, so $g(x) \to \infty$ .

Just left of $-2$ , $x + 2$ is tiny and negative: $\frac{(-)}{(-)(-)}$ is negative, so $g(x) \to -\infty$ .

Step 4: Check signs near

x = 3

. The factor

x + 1

is about 4 and

x + 2

is about 5 (both positive).

Just right of 3, $x - 3$ is tiny and positive: $\frac{(+)}{(+)(+)}$ is positive, so $g(x) \to \infty$ .

Just left of 3, $x - 3$ is tiny and negative: $\frac{(+)}{(-)(+)}$ is negative, so $g(x) \to -\infty$ .

Answer:

\lim_{x \to -2^-} g(x) = -\infty, \quad \lim_{x \to -2^+} g(x) = \infty, \quad \lim_{x \to 3^-} g(x) = -\infty, \quad \lim_{x \to 3^+} g(x) = \infty

Both asymptotes come from factors with multiplicity 1, so each one has opposite directions on its two sides, just as the graph shows. The graph also crosses the

x

-axis at its zero,

x = -1

, and since the denominator has the greater degree,

y = 0

is a horizontal asymptote (Topic 1.7).

Example 2: Factor first (AP-style)

Let

h(x) = \frac{x^2 - x - 2}{x^3 - 3x^2 + 4}

. Find all vertical asymptotes of the graph of

h

, and write one-sided limits for each.

Step 1: Factor the numerator:

x^2 - x - 2 = (x - 2)(x + 1)

Step 2: Factor the denominator. Testing

x = -1

gives

-1 - 3 + 4 = 0

, so

(x + 1)

is a factor. Dividing it out:

x^3 - 3x^2 + 4 = (x + 1)(x^2 - 4x + 4) = (x + 1)(x - 2)^2

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

Step 3: Compare multiplicities at each zero of the denominator.

$x = 2$ : once in the numerator, twice in the denominator. The denominator has more, so it's a vertical asymptote.

$x = -1$ : once in each. Not more in the denominator, so it's a hole, not an asymptote.

Step 4: Cancel to see the behavior near

x = 2

. For

x \neq -1

and

x \neq 2

h(x) = \frac{1}{x - 2}

. Just left of 2 that's negative, and just right of 2 it's positive.

\lim_{x \to 2^-} h(x) = -\infty, \qquad \lim_{x \to 2^+} h(x) = \infty

Answer: The only vertical asymptote is

x = 2

. Even though the original denominator has

(x - 2)^2

, only one factor of

(x - 2)

is left after canceling, so the sides go in opposite directions. And even though

x = -1

makes the denominator 0, it's a hole at

\left(-1, -\frac{1}{3}\right)

, which you'll learn to locate in Topic 1.10.

Example 3: Reading a table

The table shows values of a rational function

r

near

x = 4

, where

r

is undefined. What do the values suggest about the graph near

x = 4

$x$	$3.9$	$3.99$	$3.999$	$4.001$	$4.01$	$4.1$
$r(x)$	$490$	$49{,}900$	$4{,}999{,}000$	$5{,}001{,}000$	$50{,}100$	$510$

Step 1: As

x

gets closer to 4 from either side, the outputs get larger and larger with no sign of stopping. That points to a vertical asymptote at

x = 4

Step 2: The outputs are positive on both sides.

\lim_{x \to 4^-} r(x) = \infty, \qquad \lim_{x \to 4^+} r(x) = \infty

Answer: The graph likely has a vertical asymptote at

x = 4

, rising on both sides. The same direction on both sides suggests an even multiplicity, like

(x - 4)^2

in the denominator. (These values come from

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

.) If the outputs had settled toward one number from both sides instead, that would point to a hole.

Common mistakes

Calling every zero of the denominator a vertical asymptote. Check the numerator first. In Example 2, $x = -1$ is a hole.

Deciding before factoring. Factor the numerator and denominator completely before you compare anything.

Using the multiplicity from before canceling. In Example 2, the denominator has $(x - 2)^2$ , but after canceling only one copy is left, so the sides go opposite ways.

Writing one two-sided limit when the sides disagree. If one side goes to $\infty$ and the other to $-\infty$ , write the two one-sided limits separately.

Drawing the graph across a vertical asymptote. The function isn't defined at $x = a$ .

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