Graphing Rational Functions: Asymptotes, Holes and Zeros

Quick answer

A rational function is one polynomial divided by another. Factor both, and the graph's features can be read off: a factor that zeroes only the top gives a zero, one that zeroes only the bottom gives a vertical asymptote, and one that zeroes both gives a hole. Far from the origin, the leading terms take over and decide the horizontal asymptote.

What you'll learn

Find the zeros, holes and vertical asymptotes of a factored rational function
Explain why a graph approaches a vertical asymptote
Find the horizontal asymptote by comparing degrees

A fraction of two polynomials

A rational function divides one polynomial by another:

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

The rational expressions lesson simplified these as algebra. Graphed, they show features no polynomial has: a line the curve hugs without touching, and sometimes a single missing point.

h(x)
y = 2
x = 1

The graph of h(x) = (2x + 1)/(x − 1)

Four features describe the whole graph:

Feature	For $h$	Where it comes from
zero	$x = -\tfrac{1}{2}$	the top is zero
vertical asymptote	$x = 1$	the bottom is zero
horizontal asymptote	$y = 2$	the leading terms, $\tfrac{2x}{x}$
$y$ -intercept	$-1$	$h(0) = \tfrac{1}{-1}$

Why the graph shoots off near a vertical asymptote

Take a function with two factors in the denominator:

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

At $x = 3$ the denominator is zero and the numerator is $2$ . Close to $3$ , the numerator stays near $2$ while the denominator gets tiny, and dividing by something tiny makes the result huge.

$x$	$3.1$	$3.01$	$3.001$
$f(x)$	$4.12$	$40.12$	$400.12$

Each step ten times closer makes the output about ten times larger. There is no limit to how large it gets, so the graph climbs forever beside the line $x = 3$ without reaching it. That line is a vertical asymptote.

From the left the denominator is tiny and negative, so the values plunge instead: $f(2.99) \approx -39.88$ . The sign of each factor decides which way each side goes.

f(x)
x = −2
x = 3

The graph of f(x) = (x − 1)/((x − 3)(x + 2))

A vertical asymptote needs both conditions: the denominator is zero, and the numerator is not. When the numerator is zero too, something else happens.

Holes

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

The factor $x - 2$ zeroes both the top and the bottom. For every $x$ except $2$ it cancels, and $g(x) = x + 1$ . At $x = 2$ the original expression is $\tfrac{0}{0}$ , which has no value.

So the graph is the line $y = x + 1$ with one point missing, at $(2, 3)$ . That gap is a hole, drawn as an open circle.

g(x)

The graph of g(x) = (x² − x − 2)/(x − 2)

End behavior and horizontal asymptotes

Far from the origin, only the highest-degree terms matter, because they dwarf everything else. Compare the degrees of the top and bottom:

Degrees	Horizontal asymptote	Example
bottom higher	$y = 0$	$\tfrac{x - 1}{(x - 3)(x + 2)}$
equal	ratio of leading coefficients	$\tfrac{2x + 1}{x - 1} \to y = 2$
top higher	none	$\tfrac{x^2 + 1}{x - 1}$

For $h$ , the leading terms are $\tfrac{2x}{x} = 2$ . At $x = 1000$ the function is about $2.003$ , and at $x = -1000$ about $1.997$ — closing in on $2$ from both sides.

A graph can cross its horizontal asymptote closer in. The asymptote only describes where the graph heads far away. A vertical asymptote, by contrast, is never crossed, since the function has no value there.

Worked examples

Common mistakes

Practice problems

Find the zero, vertical asymptote and horizontal asymptote of $f(x) = \tfrac{x - 6}{x + 2}$ .

Answer

Zero $x = 6$ ; vertical asymptote $x = -2$ ; horizontal asymptote $y = 1$ .

Full solution

The top is zero at $6$ , the bottom at $-2$ , and the degrees are equal with leading coefficients $1$ and $1$ .
Find the vertical asymptotes of $f(x) = \tfrac{3}{x^2 - 9}$ .

Answer

$x = 3$ and $x = -3$

Full solution

$x^2 - 9 = (x - 3)(x + 3)$ is zero at $3$ and $-3$ , and the numerator $3$ is never zero.
Describe the graph of $g(x) = \tfrac{x^2 - 4}{x - 2}$ .

Answer

The line $y = x + 2$ with a hole at $(2, 4)$ .

Full solution

$\tfrac{(x - 2)(x + 2)}{x - 2} = x + 2$ for every $x \ne 2$ . At $x = 2$ the function is undefined, so the point $(2, 4)$ is missing.
Find the horizontal asymptote of $f(x) = \tfrac{5x^2 + 1}{2x^2 - 3}$ .

Answer

$y = \tfrac{5}{2}$

Full solution

The degrees are equal, so divide the leading coefficients: $\tfrac{5}{2}$ .
Does $f(x) = \tfrac{x + 1}{x^2 + 1}$ have any vertical asymptotes? What is its horizontal asymptote?

Answer

No vertical asymptotes; horizontal asymptote $y = 0$ .

Full solution

$x^2 + 1$ is at least $1$ for every $x$ , so the denominator is never zero.

The bottom has the higher degree, so the graph approaches $y = 0$ .
Find the $y$ -intercept of $h(x) = \tfrac{2x + 1}{x - 1}$ .

Answer

$-1$

Full solution

$h(0) = \tfrac{1}{-1} = -1$ .
For $f(x) = \tfrac{1}{x - 2}$ , find $f(2.01)$ and $f(1.99)$ . What do they show?

Answer

$100$ and $-100$ : the graph rises on the right of $x = 2$ and falls on the left.

Full solution

$\tfrac{1}{0.01} = 100$ and $\tfrac{1}{-0.01} = -100$ .

The denominator’s sign changes at $2$ , so the two branches head in opposite directions beside the asymptote.
For $f(x) = \tfrac{x - 1}{(x - 3)(x + 2)}$ , find the zero and the $y$ -intercept.

Answer

Zero at $x = 1$ ; $y$ -intercept $\tfrac{1}{6}$ .

Full solution

The top is zero at $1$ , where the bottom is not zero.

$f(0) = \tfrac{-1}{(-3)(2)} = \tfrac{1}{6}$ .
Explain why $x = 2$ is a hole in $\tfrac{x^2 - 4}{x - 2}$ rather than a vertical asymptote.

Answer

Both the top and bottom are zero there, and the common factor cancels.

Full solution

A vertical asymptote needs a tiny denominator divided into a numerator that stays away from zero, which makes the output huge.

Here the numerator shrinks at the same rate, because $x - 2$ is a factor of both. After canceling, the values near $2$ approach $4$ , a finite number, so the graph passes smoothly through the gap and only one point is missing.
Luis says $\tfrac{x^2 - 9}{x - 3}$ has a vertical asymptote at $x = 3$ , because the denominator is zero there. Find his error.

Hint

Factor the numerator.

Answer

The factor $x - 3$ cancels, so there is a hole at $(3, 6)$ , not an asymptote.

Full solution

$\tfrac{(x - 3)(x + 3)}{x - 3} = x + 3$ for $x \ne 3$ .

Near $x = 3$ the values approach $6$ instead of growing without bound. The graph is the line $y = x + 3$ with the point $(3, 6)$ missing.

A zero denominator makes a vertical asymptote only when the numerator is not also zero.

Frequently asked questions

How do I find the vertical asymptotes of a rational function?

Factor the numerator and denominator and cancel any common factors. Each remaining factor of the denominator that equals zero at x = a gives a vertical asymptote x = a.

What is the difference between a hole and a vertical asymptote?

A hole comes from a factor that cancels, so the graph is only missing one point. An asymptote comes from a factor left in the denominator, and the graph runs off toward infinity there.

How do I find the horizontal asymptote?

Compare degrees. If the bottom has the higher degree, it is y = 0. If the degrees are equal, it is the ratio of the leading coefficients. If the top is higher, there is none.

Can a graph cross a horizontal asymptote?

Yes. A horizontal asymptote describes behavior far from the origin, and the graph may cross it closer in. A vertical asymptote is never crossed.

Where are the zeros of a rational function?

Where the numerator is zero and the denominator is not, after canceling common factors.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSF.IF.C.7dInterpreting Functions(+) Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

A fraction of two polynomials

Why the graph shoots off near a vertical asymptote

Holes

End behavior and horizontal asymptotes

Worked examples

Common mistakes

Practice problems

Frequently asked questions

How do I find the vertical asymptotes of a rational function?

What is the difference between a hole and a vertical asymptote?

How do I find the horizontal asymptote?

Can a graph cross a horizontal asymptote?

Where are the zeros of a rational function?

What to learn next

Standards alignment

More Algebra 2 lessons