Asymptote

Lines That a Graph Approaches

An asymptote is a line that a graph approaches in a particular limiting process. Near a vertical asymptote, the function grows without bound. Near a horizontal or oblique asymptote, the distance between the graph and the line tends to zero as the input grows in magnitude.

A graph may cross a horizontal or oblique asymptote at a finite input. The asymptote describes the value that the graph approaches far away. It does not act as a wall.

Types of Asymptotes

Asymptotes fall into three types. The type follows from how the graph approaches the line, so each one has its own identifying feature. The three types are named after the direction of the line the graph approaches.

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 at least one of these one-sided limits is infinite:

• 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: First cancel any common polynomial factors. In the simplified rational function, a vertical asymptote occurs when $Q(x) = 0$ while $P(x) \neq 0$. If the factor cancels, the graph has a hole at that input.

Horizontal Asymptote

A horizontal asymptote is a horizontal line that the graph approaches as $x$ tends toward positive infinity, negative infinity, or both.

Definition: The line $y = b$ is a horizontal asymptote in each direction where the corresponding limit holds:

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

Oblique Asymptote

An oblique asymptote is a slanted line that the graph approaches as $x$ tends toward positive infinity, negative infinity, or both.

Definition: The line $y = mx + c$ is an oblique asymptote in a direction where the difference tends to zero:

Asymptotes in Rational Functions

Consider rational functions $f(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials. For these functions the asymptotes follow directly from the zeros of the denominator and from the degrees of the two polynomials.

Finding Vertical Asymptotes

Steps:

1. Find the value of $x$ that makes $Q(x) = 0$
2. Check if $P(x) \neq 0$ at that value
3. 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: $x = 2$

The following values show the function's behavior around the vertical asymptote:

Finding Horizontal Asymptotes

Rules for rational functions:

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

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

Example: Determine the horizontal asymptote of:

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

   Solution:

   The numerator degree is $= 1$ and the denominator degree is $= 2$. Since $1 < 2$, the horizontal asymptote is $y = 0$.

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

   Solution:

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

The following values show how the function approaches the horizontal asymptote:

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:

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

Oblique asymptote: $y = x + 3$

Drawing Graphs with Asymptotes

Draw the asymptotes first. A vertical asymptote marks the input value approached as the magnitude of the function's output grows without bound. A horizontal or oblique asymptote describes the graph's end behavior as the magnitude of the input grows. Then use the following steps:

1. Determine all asymptotes (vertical, horizontal, or oblique)
2. Draw asymptotes with dashed lines
3. Find intercepts with the axes
4. Determine some additional points
5. 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

Practice Problems

Each problem asks for an asymptote of a given function, so find the term that drives the curve first, either a denominator that reaches zero or an exponential term that dominates.

1. Determine all asymptotes of $f(x) = \frac{2x^2 - 3x + 1}{x - 3}$

2. Determine the asymptotes of $g(x) = \frac{x^2 - 4}{x^2 - 9}$

3. The average cost function of a product is $C(x) = \frac{500 + 3x}{x}$. Determine the limiting cost per unit as production grows without bound.

4. Draw a sketch of the graph $h(x) = \frac{x}{x^2 - 1}$ complete with its asymptotes.

Worked Solutions

Problem 1:

The denominator is zero at $x = 3$, while the numerator there is $2(3)^2 - 3(3) + 1 = 10$. Because the numerator does not vanish, $x = 3$ is a vertical asymptote.

Polynomial division separates the long-term linear behavior from the remainder:

As $x \to \pm\infty$, the remainder $\frac{10}{x - 3} \to 0$. Therefore the oblique asymptote is $y = 2x + 3$.

Problem 2:

Factor both polynomials before classifying their zeros:

There is no common factor to cancel. The denominator zeros $x = -3$ and $x = 3$ are therefore vertical asymptotes. The numerator zeros $x = -2$ and $x = 2$ are $x$-intercepts. Since numerator and denominator have the same degree and both leading coefficients are $1$, the horizontal asymptote is $y = 1$.

Problem 3:

When $x \to \infty$, $\frac{500}{x} \to 0$. Therefore $C(x) \to 3$. The cost approaches $3$ per unit but does not equal that value for any finite positive $x$.

Problem 4:

Factor the denominator as $x^2 - 1 = (x - 1)(x + 1)$. Neither factor cancels with the numerator, so the vertical asymptotes are $x = -1$ and $x = 1$. Because the numerator degree is smaller than the denominator degree, $h(x) \to 0$ as $x \to \pm\infty$. Hence the horizontal asymptote is $y = 0$.

The asymptotes split the domain into three intervals. The sign of $h(x)$ is negative on $x < -1$, positive on $-1 < x < 0$, negative on $0 < x < 1$, and positive on $x > 1$. These signs, together with the intercept $(0,0)$, determine the placement of all three branches.

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