2.5: Horizontal and Vertical Asymptotes
Rational functions abound in real life, we just don't always think of them that way. For example, suppose you are buying ice cream for you and your friends:
The ice cream shop has a deal going, if you buy one cone for $3, then additional cones are only $2.50 each. As you buy more and more cones (more and more lucky friends!), what is the average cost per cone?
This question can be solved with: \begin{align*}P(c)= (2.5c+3)/c\end{align*}
If you graph this function on a graphing calculator, it is interesting to note the horizontal asymptote at 2.5. What does this represent?
How might this process apply to movie tickets, where the first is $12, and each additional is $9? Does this graph also have an asymptote? What does it represent?
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James Sousa  Determining Vertical and Horizontal Asymptotes of Rational Functions
Guidance
Vertical and Horizontal Asymptotes
An asymptote is a line or curve to which a function's graph draws closer without touching it. Functions cannot cross a vertical asymptote, and they usually approach horizontal asymptotes in their end behavior (i.e. as \begin{align*}x\rightarrow\pm\infty\end{align*}
Looking at the graph of \begin{align*}f(x)=\frac{x+2}{(x1)(x+3)}\end{align*}
Finding a Vertical Asymptote
We can find vertical asymptotes by simply equating the denominator to zero and then solving for \begin{align*}x\end{align*}
\begin{align*}f(x)=\frac{P(x)}{Q(x)}\end{align*}
Then setting \begin{align*}Q(x)=0\end{align*}
So if
\begin{align*}f(x)=\frac{x+2}{(x1)(x+3)}\end{align*}
setting
\begin{align*}(x1)(x+3)=0\end{align*}
gives the vertical asymptotes at \begin{align*}x=1\end{align*}
Finding a Horizontal Asymptote
 Put the rational function in a standard form. That is, expand the numerator and denominator if they are written in a factored form.

Remove all terms except the terms that contain the largest exponents of \begin{align*}x\end{align*}
x in the numerator and the denominator. 
There are three possibilities:

If the degree of the numerator is smaller than the degree of the denominator, then the horizontal asymptote crosses the \begin{align*}y\end{align*}
y− axis at \begin{align*}y=0\end{align*}y=0 . That is, it is the \begin{align*}x\end{align*}x− axis itself.  If the degree of the denominator and the numerator are the same, then the horizontal asymptote equals to the ratio of the leading coefficients.
 If the degree of the numerator is larger than the degree of the denominator, then there is no horizontal asymptote.

If the degree of the numerator is smaller than the degree of the denominator, then the horizontal asymptote crosses the \begin{align*}y\end{align*}
Example A
Find the vertical and horizontal asymptotes of
\begin{align*}f(x)=\frac{2x^{3}2x^{2}+5}{3x^{3}81}\end{align*}
Solution: To find the vertical asymptote(s), set the denominator to zero and then solve for \begin{align*}x\end{align*}
\begin{align*}3x^{3}81 & = 0\\
3x^3 & = 81\\
x^3 & = 27\\
x & = \sqrt[3]{27}\\
x & = 3\end{align*}
Thus the graph has a vertical asymptote at \begin{align*}x=3\end{align*}
To find the horizontal asymptote, we follow the procedure above. Both the numerator and denominator are already written in standard form. Next, remove all terms except the largest exponents of \begin{align*}x\end{align*}
\begin{align*}\frac{2x^{3}}{3x^{3}}\end{align*}
Notice that the degree of the numerator and the denominator are the same and therefore the horizontal asymptote is the ratio of the coefficients,
\begin{align*}\frac{2x^{3}}{3x^{3}}=\frac{2}{3}\end{align*}
So the horizontal asymptote is at \begin{align*}y=\frac{2}{3}\end{align*}
Example B
Find the asymptotes of
\begin{align*}f(x)=\frac{3x2}{2x^{4}9}\end{align*}
Solution:
Remove all terms except the leading terms,
\begin{align*}\frac{3x}{2x^{4}}\end{align*}
Notice that the degree of the numerator is less than the degree of the denominator. Therefore, the horizontal asymptote is at \begin{align*}y=0\end{align*}
\begin{align*}2x^{4}9 & = 0\\
x^4 & = \frac{9}{4}\\
x & = \pm \frac{\sqrt{6}}{2}\end{align*}
Example C
What is the average price of movie tickets, where the first is $12, and each additional is $9, as more and more tickets are purchased?
Solution This question is very similar to the one in the introduction. It can be solved with: \begin{align*}P(t)= (9t+12)/t\end{align*}
Graphing this on a technological tool, we see a horizontal asymptote at 9, representing the price that each ticket approaches as more and more people (or tickets) share the additional cost of the first ticket. Note that if you were actually buying \begin{align*}\infty\end{align*}
Were you able to solve the question about the movie tickets, from the beginning of the lesson?
The first question asked the meaning of the horizontal asymptote at 2.5 on the graph of \begin{align*}(2.5c + 3)/c\end{align*} 

Vocabulary
Rational Function: A function that is a ratio of two polynomials. In other words, where the input value x, is defined in the numerator and denominator of a fraction.
Vertical Asymptote: A vertical line marking a specific value toward which the graph of a function may approach, but will never reach. Vertical asymptotes may be identified by setting the denominator of the rational function equal to zero and solving for x.
Horizontal Asymptote: A horizontal line marking a specific value toward which the graph of a function may approach, but will generally not cross. Horizontal asymptotes exist when the degree of the numerator of a rational function is less than or equal to the denominator.
Guided Practice
1) Evaluate the horizontal asymptote of the rational function:
\begin{align*}g(x)=\frac{2x^{4}9}{3x2}\end{align*}
2) Graph
\begin{align*}f(x)=\frac{x^{2}x2}{x1}\end{align*}
3) Graph
\begin{align*}g(x)=\frac{2x^{2}+1}{2x^{2}3x}\end{align*}
4) Graph
\begin{align*}T(x)=\frac{2x+1}{x1}\end{align*}
Solutions
1) Remove all terms except the leading terms of the numerator and denominator,

\begin{align*}\frac{2x^{4}}{3x}\end{align*}
2x43x  Here, the degree of the numerator is larger than the degree of the denominator. Thus there is no horizontal asymptote.
2) The vertical asymptote here is \begin{align*}x=1\end{align*}

\begin{align*}f(x)=\frac{x^{2}x2}{x1}=\frac{(x2)(x+1)}{x1}\end{align*}
f(x)=x2−x−2x−1=(x−2)(x+1)x−1 
Notice that the \begin{align*}x\end{align*}
x− intercepts are at \begin{align*}x=2\end{align*}x=2 and \begin{align*}x=1\end{align*}x=−1 . By polynomial division, we get 
\begin{align*}f(x)=x\frac{2}{x1}\end{align*}
f(x)=x−2x−1 
which indicates an oblique asymptote at \begin{align*}y=x\end{align*}
y=x . Plotting few additional points, we finally obtain the graph shown below.  {{Inline image source=Image:MA0273.jpgsize=250px}
3) The domain of \begin{align*}g\end{align*}
 \begin{align*}y=g(0)=\frac{1}{0}=\text{undefined}\end{align*}
 this tells us that there is no \begin{align*}y\end{align*}intercept. The \begin{align*}x\end{align*}intercept can be found by setting the numerator to zero,
 \begin{align*}2x^{2}+1 & = 0\\ 2x^2 & = 1\\ x^2 & = \frac{1}{2}\\ x & = \sqrt{\frac{1}{2}}\end{align*}
 Notice that this equation has no real solution and therefore, there is no \begin{align*}x\end{align*}intercept either.
 The vertical asymptote can be found by setting the denominator to zero,
 \begin{align*}2x^{2}3x & = 0\\ x(2x3) & = 0\end{align*}
 The two solutions are \begin{align*}x=0\end{align*} and \begin{align*}x=\frac{3}{2}\end{align*}, and these are the vertical asymptotes.
 Finally, the horizontal asymptote is found by analyzing the leading terms:
 \begin{align*}\frac{2x{}^{2}+1}{2x^{2}3x}\to\frac{2x^{2}}{2x^{2}}=1\end{align*}
 That is, \begin{align*}y=1\end{align*} is a horizontal asymptote. Again after substituting in some points, we can sketch the graph of \begin{align*}g(x)\end{align*} below.
4) Note that the domain of \begin{align*}T\end{align*} is the set of all real numbers except 1, that is \begin{align*}x\ne 1\end{align*}. This tells us that the line \begin{align*}x=1\end{align*} is a vertical asymptote of \begin{align*}T(x)\end{align*}. To graph \begin{align*}T\end{align*}, there are four important items that you need to find: The \begin{align*}y\end{align*}intercept, the \begin{align*}x\end{align*}intercept, the vertical asymptote and the horizontal asymptote.
 The \begin{align*}y\end{align*}intercept can be found by finding \begin{align*}y=T(0)\end{align*}.
 \begin{align*}y=T(0)=\frac{2(0)+1}{01}=1\end{align*}
 Thus the \begin{align*}y\end{align*}intercept is at point (0, 1).
 The \begin{align*}x\end{align*}intercept can be found by setting \begin{align*}y=T(x)=0\end{align*}.
 \begin{align*}\frac{2x+1}{x1}=0\end{align*}
 Note that a fraction \begin{align*}\frac{a}{b}=0\end{align*} if and only if \begin{align*}a=0\end{align*}, so we can set the numerator of \begin{align*}T9\end{align*} solve
 \begin{align*}2x+1 & = 0\\ x & = \frac{1}{2}\end{align*}
 Notice that we could have just set the numerator to zero and found the \begin{align*}x\end{align*}intercept. In general, set \begin{align*}P(x)=0\end{align*} to find the \begin{align*}x\end{align*}intercept for any rational function. BUT, you must make sure that the \begin{align*}x\end{align*}value you find is still defined for the function. If both \begin{align*}P(x)=0\end{align*} and \begin{align*}Q(x)=0\end{align*} for some value of \begin{align*}x\end{align*} then the graph has a hole in it.
 Next, the vertical asymptote. Set \begin{align*}Q(x)=0\end{align*}:
 \begin{align*}x1 & = 0\\ x & = 1\end{align*}
 ...and the horizontal asymptote:
 \begin{align*}\frac{2x+1}{x1}\to\frac{2x}{x}=2\end{align*}
 Therefore, the vertical asymptote is at \begin{align*}x=1\end{align*} and the horizontal asymptote is at \begin{align*}y=2\end{align*}. From this information, we can make the graph shown below.
Practice
 What is the definition of an asymptote?
 Which asymptote direction can a function not cross?
Identify the points of discontinuity, holes, vertical asymptotes, xintercepts and horizontal asymptote of each.
 \begin{align*}f(x) = \frac{1}{3x^2 + 3x  18}\end{align*}
 \begin{align*}f(x) = \frac{x  2}{x  4}\end{align*}
 \begin{align*}f(x) = \frac{x^3  x^2  6x}{3x^2  3x + 18}\end{align*}
 \begin{align*}f(x) = \frac{x^2 + x  6}{4x^2  16x  12}\end{align*}
Identify the points of discontinuity, holes, vertical asymptotes, and horizontal asymptote of each. Then sketch the graph.
 \begin{align*}f(x) = \frac{4}{x^2  3x}\end{align*}
 \begin{align*}f(x) = \frac{x  4}{4x  16}\end{align*}
 \begin{align*}f(x) = \frac{x + 4}{2x  6}\end{align*}
 \begin{align*}f(x) = \frac{x^3  9x}{3x^2  6x  9}\end{align*}
 \begin{align*}f(x) = \frac{3x^2  12x}{x^2  2x  3}\end{align*}
 \begin{align*}f(x) = \frac{x^3  16x}{4x^2 +4x + 24}\end{align*}
 \begin{align*}f(x) = \frac{x^2 + 2x}{4x + 8}\end{align*}
 \begin{align*}f(x) = \frac{x + 2}{2x + 6}\end{align*}
 \begin{align*}f(x) = \frac{2x^2 + 10x + 12}{x^2 + 3x + 2}\end{align*}
 \begin{align*}f(x) = \frac{3}{x  2}\end{align*}
The graph of a rational function is shown below:
Use the Graph to answer the following questions:
 as \begin{align*}x \to \infty\end{align*} what does \begin{align*}y \end{align*} approach?
 as \begin{align*}x \to \infty\end{align*} what does \begin{align*}y \end{align*} approach?
 as \begin{align*}x \to 2^+\end{align*} what does \begin{align*}y \end{align*} approach?
 as \begin{align*}x \to 4^\end{align*} what does \begin{align*}y \end{align*} approach?
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Term  Definition 

Asymptotes  An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions). 
Hole  A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero. 
Horizontal Asymptote  A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote. 
Intercept  The intercepts of a curve are the locations where the curve intersects the and axes. An intercept is a point at which the curve intersects the axis. A intercept is a point at which the curve intersects the axis. 
Intercepts  The intercepts of a curve are the locations where the curve intersects the and axes. An intercept is a point at which the curve intersects the axis. A intercept is a point at which the curve intersects the axis. 
Oblique Asymptote  An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division. 
Oblique Asymptotes  An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division. 
points of discontinuity  The points of discontinuity for a function are the input values of the function where the function is discontinuous. 
Rational Function  A rational function is any function that can be written as the ratio of two polynomial functions. 
Vertical Asymptote  A vertical asymptote is a vertical line marking a specific value toward which the graph of a function may approach, but will never reach. 
Image Attributions
Here you will learn about horizontal and vertical asymptotes and how to find and use them with the graphs of rational functions.