Asymptote: Functions and Their Modeling - Mathematics (Grade 11)

What is an Asymptote?

Have you ever noticed a function graph that approaches a line but never touches it? Well, that line is called an asymptote!

An asymptote is a straight line that is approached by a function graph when its variable value approaches infinity or approaches a certain value. Imagine like you're walking towards a wall but never actually touching it, that's the concept of an asymptote.

Types of Asymptotes

There are three types of asymptotes you need to know:

Vertical Asymptote

A vertical asymptote is a vertical line that the graph approaches when the function value approaches positive or negative infinity.

Definition: The line $x = a$ is a vertical asymptote if:

When $x$ approaches $a$ from the left, $f(x) \to \pm\infty$
When $x$ approaches $a$ from the right, $f(x) \to \pm\infty$

How to find: For rational functions, vertical asymptotes occur when $\text{denominator} = 0$ and $\text{numerator} \neq 0$ , or when $Q(x) = 0$ and $P(x) \neq 0$ .

Horizontal Asymptote

A horizontal asymptote is a horizontal line that the graph approaches when $x$ approaches positive or negative infinity.

Definition: The line $y = b$ is a horizontal asymptote if:

$\lim_{x \to \infty} f(x) = b$
$\lim_{x \to -\infty} f(x) = b$

Oblique Asymptote (Oblique)

An oblique asymptote is a slanted line that the graph approaches when $x$ approaches infinity.

Definition: The line $y = mx + c$ is an oblique asymptote if:

\lim_{x \to \pm\infty} [f(x) - (mx + c)] = 0

Asymptotes in Rational Functions

Let's focus on rational functions $f(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials.

Finding Vertical Asymptotes

Steps:

Find the value of $x$ that makes $Q(x) = 0$
Check if $P(x) \neq 0$ at that value
If yes, then there is a vertical asymptote at $x = a$

Example: Determine the vertical asymptote of $f(x) = \frac{x + 3}{x - 2}$

Solution:

Denominator is zero when: $x - 2 = 0$ , so $x = 2$
When $x = 2$ , numerator is $2 + 3 = 5 \neq 0$
Therefore, vertical asymptote: $x = 2$

Let's look at the function behavior around the vertical asymptote:

$x$	$f(x) = \frac{x + 3}{x - 2}$	Description
$1.9$	$\frac{1.9 + 3}{1.9 - 2} = \frac{4.9}{-0.1} = -49$	Approaches $-\infty$
$1.99$	$\frac{4.99}{-0.01} = -499$	Getting more negative
$2.01$	$\frac{5.01}{0.01} = 501$	Approaches $+\infty$
$2.1$	$\frac{5.1}{0.1} = 51$	Getting more positive

Graph of

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

with Vertical Asymptote

Notice how the graph approaches the vertical line

x = 2

without ever touching it.

Finding Horizontal Asymptotes

Rules for rational functions:

Let the degree of numerator is $m$ and degree of denominator = $n$

If $m < n$ : Horizontal asymptote is $y = 0$
If $m = n$ : Horizontal asymptote is $y = \frac{a}{b}$ (ratio of leading coefficients)
If $m > n$ : No horizontal asymptote (but there might be an oblique asymptote)

Example: Determine the horizontal asymptote of:

$f(x) = \frac{2x + 1}{x^2 - 4}$
Solution:

The numerator degree is $= 1$ and the denominator degree is $= 2$ . Since $1 < 2$ , the horizontal asymptote is $y = 0$ .
$g(x) = \frac{3x^2 - 1}{2x^2 + 5}$
Solution:

The numerator degree is $= 2$ and the denominator degree is $= 2$ . Since the degrees are equal, the horizontal asymptote is $y = \frac{3}{2}$ .

Let's see how the function approaches the horizontal asymptote:

$x$	$g(x) = \frac{3x^2 - 1}{2x^2 + 5}$	Approaches
$10$	$\frac{3(100) - 1}{2(100) + 5} = \frac{299}{205} \approx 1.459$	$1.5$
$100$	$\frac{29999}{20005} \approx 1.4997$	$1.5$
$1000$	$\frac{2999999}{2000005} \approx 1.49997$	$1.5$

Graph of

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

with Horizontal Asymptote

The graph approaches

y = 1.5

when

x \to \pm\infty

Finding Oblique Asymptotes

Oblique asymptotes appear when $\text{degree of numerator} = \text{degree of denominator} + 1$ .

How to find: Perform polynomial division.

Example: Determine the oblique asymptote of $f(x) = \frac{x^2 + 2x - 1}{x - 1}$

Solution: Using polynomial division:

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

When $x \to \pm\infty$ , the term $\frac{2}{x - 1} \to 0$

Therefore, oblique asymptote: $y = x + 3$

Graph of

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

with Oblique Asymptote

The graph approaches the line

y = x + 3

when

x \to \pm\infty

Drawing Graphs with Asymptotes

Asymptotes are very helpful in drawing function graphs. Here are the steps:

Determine all asymptotes (vertical, horizontal, or oblique)
Draw asymptotes with dashed lines
Find intercepts with the axes
Determine some additional points
Draw the curve that approaches the asymptotes

Complete Example: Draw the graph of $f(x) = \frac{x + 1}{x - 2}$

Step $1$ : Find asymptotes

Vertical asymptote: $x = 2$ ( $\text{denominator} = 0$ )
Horizontal asymptote: $y = 1$ (same degree, coefficient ratio $= 1/1$ )

Step $2$ : Intercepts

$y$ -axis: $f(0) = \frac{0 + 1}{0 - 2} = -\frac{1}{2}$
$x$ -axis: $0 = \frac{x + 1}{x - 2}$ , so $x = -1$

Step $3$ : Behavior around asymptotes

When $x \to 2^-$ : $f(x) \to -\infty$
When $x \to 2^+$ : $f(x) \to +\infty$
When $x \to \pm\infty$ : $f(x) \to 1$

Step $4$ : Value table to help with drawing

$x$	$f(x) = \frac{x + 1}{x - 2}$	Description
$-3$	$\frac{-3 + 1}{-3 - 2} = \frac{-2}{-5} = 0.4$	Point in quadrant I
$-1$	$\frac{-1 + 1}{-1 - 2} = \frac{0}{-3} = 0$	$x$ -axis intercept
$0$	$\frac{0 + 1}{0 - 2} = \frac{1}{-2} = -0.5$	$y$ -axis intercept
$1$	$\frac{1 + 1}{1 - 2} = \frac{2}{-1} = -2$	Approaching vertical asymptote
$3$	$\frac{3 + 1}{3 - 2} = \frac{4}{1} = 4$	Right of asymptote
$5$	$\frac{5 + 1}{5 - 2} = \frac{6}{3} = 2$	Approaching horizontal asymptote

Complete Graph of

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

Graph with vertical asymptote

x = 2

and horizontal asymptote

y = 1

Practice Problems

Determine all asymptotes of $f(x) = \frac{2x^2 - 3x + 1}{x - 3}$
Determine the asymptotes of $g(x) = \frac{x^2 - 4}{x^2 - 9}$
The average cost function of a product is $C(x) = \frac{500 + 3x}{x}$ . Determine the minimum cost per unit that can be achieved.
Draw a sketch of the graph $h(x) = \frac{x}{x^2 - 1}$ complete with its asymptotes.

Answer Key

Answer $1$ :

$\text{degree of numerator }(2) = \text{degree of denominator }(1) + 1$
There is an oblique asymptote. By division: $f(x) = 2x + 3 + \frac{10}{x - 3}$
Vertical asymptote: $x = 3$
Oblique asymptote: $y = 2x + 3$

Answer $2$ :

Vertical asymptote: $x^2 - 9 = 0$ , so $x = 3$ and $x = -3$
But when $x = 2$ , $\text{numerator} = 0$ , so $x = 2$ is not an asymptote
When $x = -2$ , $\text{numerator} = 0$ , so $x = -2$ is not an asymptote
Horizontal asymptote: $y = 1$ (same degree, ratio $= 1/1$ )

Answer $3$ :

C(x) = \frac{500 + 3x}{x} = \frac{500}{x} + 3

When $x \to \infty$ , $\frac{500}{x} \to 0$ So minimum cost per unit $= 3$

Answer $4$ :

Vertical asymptotes: $x = 1$ and $x = -1$
Horizontal asymptote: $y = 0$ (degree of numerator is less than degree of denominator)
The graph has three separate parts due to two vertical asymptotes

Value table for $h(x) = \frac{x}{x^2 - 1}$ :

$x$	$h(x)$	Description
$-2$	$\frac{-2}{4-1} = -\frac{2}{3}$	Left part
$-0.5$	$\frac{-0.5}{0.25-1} = \frac{2}{3}$	Middle part
$0$	$\frac{0}{0-1} = 0$	Intercept
$0.5$	$\frac{0.5}{0.25-1} = -\frac{2}{3}$	Middle part
$2$	$\frac{2}{4-1} = \frac{2}{3}$	Right part

Graph of

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

with Two Vertical Asymptotes

Graph with vertical asymptotes at

x = -1

and

x = 1

, and horizontal asymptote

y = 0

What is an Asymptote?

Have you ever noticed a function graph that approaches a line but never touches it? Well, that line is called an asymptote!

Types of Asymptotes

There are three types of asymptotes you need to know:

Vertical Asymptote

A vertical asymptote is a vertical line that the graph approaches when the function value approaches positive or negative infinity.

Definition: The line $x = a$ is a vertical asymptote if:

When $x$ approaches $a$ from the left, $f(x) \to \pm\infty$
When $x$ approaches $a$ from the right, $f(x) \to \pm\infty$

How to find: For rational functions, vertical asymptotes occur when $\text{denominator} = 0$ and $\text{numerator} \neq 0$ , or when $Q(x) = 0$ and $P(x) \neq 0$ .

Horizontal Asymptote

A horizontal asymptote is a horizontal line that the graph approaches when $x$ approaches positive or negative infinity.

Definition: The line $y = b$ is a horizontal asymptote if:

$\lim_{x \to \infty} f(x) = b$
$\lim_{x \to -\infty} f(x) = b$

Oblique Asymptote (Oblique)

An oblique asymptote is a slanted line that the graph approaches when $x$ approaches infinity.

Definition: The line $y = mx + c$ is an oblique asymptote if:

\lim_{x \to \pm\infty} [f(x) - (mx + c)] = 0

Asymptotes in Rational Functions

Let's focus on rational functions $f(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials.

Finding Vertical Asymptotes

Steps:

Find the value of $x$ that makes $Q(x) = 0$
Check if $P(x) \neq 0$ at that value
If yes, then there is a vertical asymptote at $x = a$

Example: Determine the vertical asymptote of $f(x) = \frac{x + 3}{x - 2}$

Solution:

Denominator is zero when: $x - 2 = 0$ , so $x = 2$
When $x = 2$ , numerator is $2 + 3 = 5 \neq 0$
Therefore, vertical asymptote: $x = 2$

Let's look at the function behavior around the vertical asymptote:

$x$	$f(x) = \frac{x + 3}{x - 2}$	Description
$1.9$	$\frac{1.9 + 3}{1.9 - 2} = \frac{4.9}{-0.1} = -49$	Approaches $-\infty$
$1.99$	$\frac{4.99}{-0.01} = -499$	Getting more negative
$2.01$	$\frac{5.01}{0.01} = 501$	Approaches $+\infty$
$2.1$	$\frac{5.1}{0.1} = 51$	Getting more positive

Graph of

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

with Vertical Asymptote

Notice how the graph approaches the vertical line

x = 2

without ever touching it.

Finding Horizontal Asymptotes

Rules for rational functions:

Let the degree of numerator is $m$ and degree of denominator = $n$

If $m < n$ : Horizontal asymptote is $y = 0$
If $m = n$ : Horizontal asymptote is $y = \frac{a}{b}$ (ratio of leading coefficients)
If $m > n$ : No horizontal asymptote (but there might be an oblique asymptote)

Example: Determine the horizontal asymptote of:

$f(x) = \frac{2x + 1}{x^2 - 4}$
Solution:

The numerator degree is $= 1$ and the denominator degree is $= 2$ . Since $1 < 2$ , the horizontal asymptote is $y = 0$ .
$g(x) = \frac{3x^2 - 1}{2x^2 + 5}$
Solution:

The numerator degree is $= 2$ and the denominator degree is $= 2$ . Since the degrees are equal, the horizontal asymptote is $y = \frac{3}{2}$ .

Let's see how the function approaches the horizontal asymptote:

$x$	$g(x) = \frac{3x^2 - 1}{2x^2 + 5}$	Approaches
$10$	$\frac{3(100) - 1}{2(100) + 5} = \frac{299}{205} \approx 1.459$	$1.5$
$100$	$\frac{29999}{20005} \approx 1.4997$	$1.5$
$1000$	$\frac{2999999}{2000005} \approx 1.49997$	$1.5$

Graph of

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

with Horizontal Asymptote

The graph approaches

y = 1.5

when

x \to \pm\infty

Finding Oblique Asymptotes

Oblique asymptotes appear when $\text{degree of numerator} = \text{degree of denominator} + 1$ .

How to find: Perform polynomial division.

Example: Determine the oblique asymptote of $f(x) = \frac{x^2 + 2x - 1}{x - 1}$

Solution: Using polynomial division:

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

When $x \to \pm\infty$ , the term $\frac{2}{x - 1} \to 0$

Therefore, oblique asymptote: $y = x + 3$

Graph of

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

with Oblique Asymptote

The graph approaches the line

y = x + 3

when

x \to \pm\infty

Drawing Graphs with Asymptotes

Asymptotes are very helpful in drawing function graphs. Here are the steps:

Determine all asymptotes (vertical, horizontal, or oblique)
Draw asymptotes with dashed lines
Find intercepts with the axes
Determine some additional points
Draw the curve that approaches the asymptotes

Complete Example: Draw the graph of $f(x) = \frac{x + 1}{x - 2}$

Step $1$ : Find asymptotes

Vertical asymptote: $x = 2$ ( $\text{denominator} = 0$ )
Horizontal asymptote: $y = 1$ (same degree, coefficient ratio $= 1/1$ )

Step $2$ : Intercepts

$y$ -axis: $f(0) = \frac{0 + 1}{0 - 2} = -\frac{1}{2}$
$x$ -axis: $0 = \frac{x + 1}{x - 2}$ , so $x = -1$

Step $3$ : Behavior around asymptotes

When $x \to 2^-$ : $f(x) \to -\infty$
When $x \to 2^+$ : $f(x) \to +\infty$
When $x \to \pm\infty$ : $f(x) \to 1$

Step $4$ : Value table to help with drawing

$x$	$f(x) = \frac{x + 1}{x - 2}$	Description
$-3$	$\frac{-3 + 1}{-3 - 2} = \frac{-2}{-5} = 0.4$	Point in quadrant I
$-1$	$\frac{-1 + 1}{-1 - 2} = \frac{0}{-3} = 0$	$x$ -axis intercept
$0$	$\frac{0 + 1}{0 - 2} = \frac{1}{-2} = -0.5$	$y$ -axis intercept
$1$	$\frac{1 + 1}{1 - 2} = \frac{2}{-1} = -2$	Approaching vertical asymptote
$3$	$\frac{3 + 1}{3 - 2} = \frac{4}{1} = 4$	Right of asymptote
$5$	$\frac{5 + 1}{5 - 2} = \frac{6}{3} = 2$	Approaching horizontal asymptote

Complete Graph of

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

Graph with vertical asymptote

x = 2

and horizontal asymptote

y = 1

Practice Problems

Determine all asymptotes of $f(x) = \frac{2x^2 - 3x + 1}{x - 3}$
Determine the asymptotes of $g(x) = \frac{x^2 - 4}{x^2 - 9}$
The average cost function of a product is $C(x) = \frac{500 + 3x}{x}$ . Determine the minimum cost per unit that can be achieved.
Draw a sketch of the graph $h(x) = \frac{x}{x^2 - 1}$ complete with its asymptotes.

Answer Key

Answer $1$ :

$\text{degree of numerator }(2) = \text{degree of denominator }(1) + 1$
There is an oblique asymptote. By division: $f(x) = 2x + 3 + \frac{10}{x - 3}$
Vertical asymptote: $x = 3$
Oblique asymptote: $y = 2x + 3$

Answer $2$ :

Vertical asymptote: $x^2 - 9 = 0$ , so $x = 3$ and $x = -3$
But when $x = 2$ , $\text{numerator} = 0$ , so $x = 2$ is not an asymptote
When $x = -2$ , $\text{numerator} = 0$ , so $x = -2$ is not an asymptote
Horizontal asymptote: $y = 1$ (same degree, ratio $= 1/1$ )

Answer $3$ :

C(x) = \frac{500 + 3x}{x} = \frac{500}{x} + 3

When $x \to \infty$ , $\frac{500}{x} \to 0$ So minimum cost per unit $= 3$

Answer $4$ :

Vertical asymptotes: $x = 1$ and $x = -1$
Horizontal asymptote: $y = 0$ (degree of numerator is less than degree of denominator)
The graph has three separate parts due to two vertical asymptotes

Value table for $h(x) = \frac{x}{x^2 - 1}$ :

$x$	$h(x)$	Description
$-2$	$\frac{-2}{4-1} = -\frac{2}{3}$	Left part
$-0.5$	$\frac{-0.5}{0.25-1} = \frac{2}{3}$	Middle part
$0$	$\frac{0}{0-1} = 0$	Intercept
$0.5$	$\frac{0.5}{0.25-1} = -\frac{2}{3}$	Middle part
$2$	$\frac{2}{4-1} = \frac{2}{3}$	Right part

Graph of

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

with Two Vertical Asymptotes

Graph with vertical asymptotes at

x = -1

and

x = 1

, and horizontal asymptote

y = 0