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)= \frac{(2.5c+3)}{c}\end{align*}, where P(c)is price per cone based on number purchased, and c is the number of cones.
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?
Watch This
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*}, you will notice that it has two vertical asymptotes (the vertical dotted lines), one is at \begin{align*}x=1\end{align*} and the other is at \begin{align*}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*}. In other words, if
\begin{align*}f(x)=\frac{P(x)}{Q(x)}\end{align*}
Then setting \begin{align*}Q(x)=0\end{align*}, will give the vertical asymptote(s).
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*} and \begin{align*}x=3\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*} 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*}axis at \begin{align*}y=0\end{align*}. That is, it is the \begin{align*}x\end{align*}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.
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*} that are found in the numerator and the denominator:
\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*}, i.e., the \begin{align*}x\end{align*}axis plays the role of the horizontal asymptote. To find the vertical asymptote, set the denominator equal to zero and solve:
\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*}, where P(t)is price per ticket based on number purchased, and t is the number of tickets.
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*} tickets, the price per ticket would mathematically be $9 per ticket!
Concept question wrapup: 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*} (the cost per cone if the first is $3 and each additional is only $2.50). That asymptote represents the minimum cost per cone, as the number of cones purchased increases. The more cones you buy, the closer the cost of each approaches $2.50, because the additional $0.50 of the initial cone is "shared" among more purchases! 

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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*}
Answers:
1) Remove all terms except the leading terms of the numerator and denominator,
 \begin{align*}\frac{2x^{4}}{3x}\end{align*}
 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*}. Factoring the numerator and get
 \begin{align*}f(x)=\frac{x^{2}x2}{x1}=\frac{(x2)(x+1)}{x1}\end{align*}
 Notice that the \begin{align*}x\end{align*}intercepts are at \begin{align*}x=2\end{align*} and \begin{align*}x=1\end{align*}. By polynomial division, we get
 \begin{align*}f(x)=x\frac{2}{x1}\end{align*}
 which indicates an oblique asymptote at \begin{align*}y=x\end{align*}. Plotting few additional points, we finally obtain the graph shown below.
3) The domain of \begin{align*}g\end{align*} is the set of all real numbers except 0 and \begin{align*}\frac{3}{2}\end{align*}, that is \begin{align*}\left \{ xx\ne 0 \ \text{and} \ x\ne\frac{3}{2} \right \}\end{align*}. The \begin{align*}y\end{align*}intercept is
 \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.
Explore More
 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?