Asymptotes

Asymptotes are imaginary lines that a curve approaches as it moves towards infinity. The distance between the curve and asymptote is about 0 as the curve moves to postive infinity (∞) or negative infinity (-∞).

Asymptotes come in 3 types:

  • Horizontal Asymptotes
  • Vertical Asymptotes
  • Oblique Asymptotes

A Horizontal Asymptotes (or HA) is an imaginary horizontal line expressed in the form \(y = k\). A function approaches a HA as \(x\) moves towards positive or negative infinity and the curve approaches a constant \(y\)-value, \(b\). In order to find the HA, we can compare the degrees in the numerator and denominator of a rational function:

  • If they have the same degree, then divide the coefficients of the largest degree terms to find the HA
  • If the denominator has a larger degree than the numerator, then the HA is located at \(y=0\) (the \(x\)-axis)
  • If the numerator has a larger degree than the denominator, then there is no HA

Additionally, a function can cross a HA.


A Vertical Asymptote (or VA) is an imaginary vertical line expressed in the form \(x = k\). A function approaches a VA when \(x\) approaches a constant \(y\)-value, \(c\) from left or right and the curve moves towards positive or negative infinity. In order to find the VA, we can set the denominator of a rational function equal to \(0\) and solve for \(x\). Additionally, a function cannot cross a VA.


An Oblique Asymptote (or OA) are slanted imaginary lines expressed in the form \(y = mx + b\) where \(m \neq 0\). The function approaches an OA as \(x\) approaches positive or negative infinity. An OA only occurs when the highest degree of a numerator is exactly 1 greater than that of the denominator. As a result, a function with an OA can never have a HA and vice versa. In order to find the OA, we can use Long Division.


Example

For the function \(f(x) = \cfrac{2x+4}{3x-6}\):

  1. Find the Asymptotes.
  2. Sketch what you can.

i. First, we can determine the Vertical Asymptote:

\(\text{VA}: 0 = 3x - 6\)

\(\text{VA}: 3x = 6\)

\(\text{VA}: \cfrac{\cancel{3}x}{\cancel{3}} = \cfrac{6}{3}\)

\(\text{VA}: x = 2\)

Next, we can determine that a Horizontal Asymptote exists since the numerator and denominator have the same degree (1). We can now determine what the HA is:

\(\text{HA}: y = \cfrac{2}{3}\)

Therefore, we can determine that the Vertical Asymptote is located at \(x = 2\) and the Horizontal Asymptote is located at \(y = \cfrac{2}{3}\).


ii. In order to draw an accurate sketch, we can first determine the intercepts and intervals of change.

We can start by determining the \(x\)-intercept:

\(\text{x-int}: 0 = 2x + 4\)

\(\text{x-int}: 2x = -4\)

\(\text{x-int}: x = -2\)

Next, we can determine the \(y\)-intercept:

\(\text{y-int}: y = \cfrac{2(0)+4}{3(0)-6}\)

\(\text{y-int}: y =\cfrac{4}{-6}\)

\(\text{y-int}: y = -\cfrac{2}{3}\)

In order to determine where the function falls above and below the \(x\)-axis, we can create a table outlining the relationship between the \(x\)-intercept/VA and the respective intervals:

Interval Test \(x\)-value \(f(x)\) Conclusion
\(\textcolor{green}{(-∞,-2)}\) \(-3\) \(f(-3)= 2/15\) \(f\) is Above \(x\)-axis
\(\textcolor{red}{(-2, 2)}\) \(0\) \(f(0)= -2/3\) \(f\) is Below \(x\)-axis
\(\textcolor{green}{(2, ∞)}\) \(3\) \(f(3)= 10/3\) \(f\) is Above \(x\)-axis

Based on this information, we can sketch our graph:

Graph of Reciprocal function with Vertical Asymptote at x=2 and Horizontal Asymptote at y=2/3.

For the equation \(y = \cfrac{1+x^2}{2-x}\):

  1. Find the Asymptotes.
  2. Sketch what you can.

Try these questions: