What if you had a function like \begin{align*}y=\frac{3x^2 - 2x + 1}{x + 2}\end{align*} ? How could you rewrite it to find its asymptotes? After completing this Concept, you'll be able to rewrite rational functions like this one using division.
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CK-12 Foundation: 1206S Rewriting Rational Functions Using Division
Guidance
In the last section we saw how to find vertical and horizontal asymptotes. Remember, the horizontal asymptote shows the value of @$\begin{align*}y\end{align*}@$ that the function approaches for large values of @$\begin{align*}x\end{align*}@$ . Let’s review the method for finding horizontal asymptotes and see how it’s related to polynomial division.
When it comes to finding asymptotes, there are basically four different types of rational functions.
Case 1: The polynomial in the numerator has a lower degree than the polynomial in the denominator.
Example A
Find the horizontal asymptote of @$\begin{align*}y=\frac{2}{x-1}\end{align*}@$ .
Solution:
We can’t reduce this fraction, and as @$\begin{align*}x\end{align*}@$ gets larger the denominator of the fraction gets much bigger than the numerator, so the whole fraction approaches zero.
The horizontal asymptote is @$\begin{align*}y = 0\end{align*}@$ .
Case 2: The polynomial in the numerator has the same degree as the polynomial in the denominator.
Example B
Find the horizontal asymptote of @$\begin{align*}y=\frac{3x+2}{x-1}\end{align*}@$ .
Solution:
In this case we can divide the two polynomials:
@$$\begin{align*}& \overset{\qquad \qquad \ 3}{x-1 \overline{ ) 3x+2 \;}}\\ & \qquad \underline{-3x+3}\\ & \qquad \qquad \quad 5\end{align*}@$$
So the expression can be written as @$\begin{align*}y=3+\frac{5}{x-1}\end{align*}@$ .
Because the denominator of the remainder is bigger than the numerator of the remainder, the remainder will approach zero for large values of @$\begin{align*}x\end{align*}@$ . Adding the 3 to that 0 means the whole expression will approach 3.
The horizontal asymptote is @$\begin{align*}y = 3\end{align*}@$ .
Case 3: The polynomial in the numerator has a degree that is one more than the polynomial in the denominator.
Example C
Find any asymptotes of @$\begin{align*}y=\frac{4x^2+3x+2}{x-1}\end{align*}@$ .
Solution:
We can do long division once again and rewrite the expression as @$\begin{align*}y=4x+7+\frac{9}{x-1}\end{align*}@$ . The fraction here approaches zero for large values of @$\begin{align*}x\end{align*}@$ , so the whole expression approaches @$\begin{align*}4x + 7\end{align*}@$ .
When the rational function approaches a straight line for large values of @$\begin{align*}x\end{align*}@$ , we say that the rational function has an oblique asymptote. In this case, then, the oblique asymptote is @$\begin{align*}y = 4x + 7\end{align*}@$ .
Case 4: The polynomial in the numerator has a degree that is two or more than the degree in the denominator.
Example D
Find any asymptotes of @$\begin{align*}y=\frac{x^3}{x-1}\end{align*}@$ .
This is actually the simplest case of all: the polynomial has no horizontal or oblique asymptotes.
Notice that a rational function will either have a horizontal asymptote, an oblique asymptote or neither kind. In other words, a function can’t have both; in fact, it can’t have more than one of either kind. On the other hand, a rational function can have any number of vertical asymptotes at the same time that it has horizontal or oblique asymptotes.
Watch this video for help with the Examples above.
CK-12 Foundation: Rewriting Rational Functions Using Division
Vocabulary
- When the rational function approaches a straight line for large values of @$\begin{align*}x\end{align*}@$ , we say that the rational function has an oblique asymptote.
Guided Practice
Find the horizontal or oblique asymptotes of the following rational functions.
a) @$\begin{align*}y=\frac{3x^2}{x^2+4}\end{align*}@$
b) @$\begin{align*}y=\frac{x-1}{3x^2-6}\end{align*}@$
c) @$\begin{align*}y=\frac{x^4+1}{x-5}\end{align*}@$
d) @$\begin{align*}y=\frac{x^3-3x^2+4x-1}{x^2-2}\end{align*}@$
Solution
a) When we simplify the function, we get @$\begin{align*}y=3-\frac{12}{x^2+4}\end{align*}@$ . There is a horizontal asymptote at @$\begin{align*}y = 3\end{align*}@$ .
b) We cannot divide the two polynomials. There is a horizontal asymptote at @$\begin{align*}y = 0\end{align*}@$ .
c) The power of the numerator is 3 more than the power of the denominator. There are no horizontal or oblique asymptotes.
d) When we simplify the function, we get @$\begin{align*}y=x-3+\frac{6x-7}{x^2-2}\end{align*}@$ . There is an oblique asymptote at @$\begin{align*}y = x - 3\end{align*}@$ .
Explore More
Find all asymptotes of the following rational functions:
- @$\begin{align*}\frac{x^2}{x-2}\end{align*}@$
- @$\begin{align*}\frac{1}{x+4}\end{align*}@$
- @$\begin{align*}\frac{x^2-1}{x^2+1}\end{align*}@$
- @$\begin{align*}\frac{x-4}{x^2-9}\end{align*}@$
- @$\begin{align*}\frac{x^2+2x+1}{4x-1}\end{align*}@$
- @$\begin{align*}\frac{x^3+1}{4x-1}\end{align*}@$
- @$\begin{align*}\frac{x-x^3}{x^2-6x-7}\end{align*}@$
- @$\begin{align*}\frac{x^4-2x}{8x+24}\end{align*}@$
Graph the following rational functions. Indicate all asymptotes on the graph:
- @$\begin{align*}\frac{x^2}{x+2}\end{align*}@$
- @$\begin{align*}\frac{x^3-1}{x^2-4}\end{align*}@$
- @$\begin{align*}\frac{x^2+1}{2x-4}\end{align*}@$
- @$\begin{align*}\frac{x-x^2}{3x+2}\end{align*}@$