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 \(f(x) = \cfrac{2x+4}{3x-6}\):
- Find the Asymptotes.
- 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 \(\boldsymbol{x = 2}\) and the Horizontal Asymptote is located at \(\boldsymbol{y = \cfrac{2}{3}}\).
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:
For the equation \(y = \cfrac{1+x^2}{2-x}\):
- Find the Asymptotes.
- Sketch what you can.
i. First, we can determine the Vertical Asymptote:
\(\text{VA}: 0 = 2 - x\)
\(\text{VA}: x = 2\)
Next, we can determine that an Oblique Asymptote exists since the numerator is exactly 1 degree greater than the denominator. We can now determine what the OA is using Long Division:
\(= \require{enclose}-x+2\enclose{longdiv}{x^2+1}\)
\(= \begin{array}{r}-x\\ -x+2\enclose{longdiv}{x^2+1}\end{array}\)
\(= \begin{array}{r}-x\\ -x+2\enclose{longdiv}{x^2+1}\\-(x^2-2x)\end{array}\)
\(= \begin{array}{r}-x\\ -x+2\enclose{longdiv}{x^2+1}\\-(x^2-2x)\\ \hline 2x+1\end{array}\)
\(= \begin{array}{r}-x-2\\ -x+2\enclose{longdiv}{x^2+1}\\-(x^2-2x)\\ \hline 2x+1\end{array}\)
\(= \begin{array}{r}-x-2\\ -x+2\enclose{longdiv}{x^2+1}\\-(x^2-2x)\\ \hline 2x+1 \\ -(2x-4)\end{array}\)
\(= \begin{array}{r}-x-2\\ -x+2\enclose{longdiv}{x^2+1}\\-(x^2-2x)\\ \hline 2x+1 \\ -(2x-4) \\ \hline 5\end{array}\)
Since the degree of the remainder \((5)\) is less than the degree of the divisor \((-x+2)\), we are done performing the long division.
Therefore, we can determine that the Vertical Asymptote is located \(x = 2\) and the formula of the Oblique Asymptote is \(y = -x-2\).
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:
\(0 = 1 + x^2\)
\(x^2 = -1\)
\(\sqrt{x^2} = \sqrt{-1}\)
\(x = \text{UNDEFINED}\)
Next, we can determine the \(y\)-intercept:
\(\text{y-int}: y = \cfrac{1+(0)^2}{2-0}\)
\(\text{y-int}: y = \cfrac{1}{2}\)
With the Oblique Asymptote, we can determine what \(x\)-value(s) it can't touch:
\(0 = -x-2\)
\(x = -2\)
Additionally, we can use the Oblique Asymptote to determine what \(y\)-value(s) it can't touch:
\(y = -(0) - 2\)
\(y = -2\)
Therefore, we can determine that the function can't touch \(x = -2\) or \(y = -2\).
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 | \(y(x)\) | Conclusion |
|---|---|---|---|
| \(\textcolor{green}{(-∞,2)}\) | \(1\) | \(y(1)= 2\) | \(y\) is Above \(x\)-axis |
| \(\textcolor{red}{(2, ∞)}\) | \(3\) | \(y(3)= -10\) | \(y\) is Below \(x\)-axis |
Based on this information, we can sketch our graph:
