3.1 Introduction to Rational Functions

If we add, subtract, or multiply polynomial functions, the result is another polynomial function. When we divide polynomial functions, however, we may not get a polynomial function. The result of dividing two polynomials is a rational function, so named because rational functions are ratios of polynomials.

Definition

A rational function is a function which is the ratio of polynomial functions. Said differently, $r$ is a rational function if it is of the form

$r (x) = \frac{p (x)}{q (x)},$

where $p$ and $q$ are polynomial functions.¹

Laurent Monomial Functions

As with polynomial functions, we begin our study of rational functions with what are, in some sense, the building blocks of rational functions, Laurent monomial functions.

Laurent monomial functions are named in honor of Pierre Alphonse Laurent and generalize the notion of `monomial function' from Chapter 2 to terms with negative exponents. Our study of these functions begins with an analysis of $r (x) = \frac{1}{x} = x^{- 1}$ , the reciprocal function. The first item worth noting is that $r ⁢ open parenthesis 0 close parenthesis$ is not defined owing to the presence of $x$ in the denominator. That is, the domain of $r$ is $open brace x ∈ ℝ vertical bar x not equal to 0 close brace$ or, using interval notation, $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . Of course excluding $0$ from the domain of $r$ serves only to pique our curiosity about the behavior of $r (x)$ when $x approximately equals 0$ . Thinking from a number sense perspective, the closer the denominator of $\frac{1}{x}$ is to $0$ , the larger the value of the fraction (in absolute value.)² So it stands to reason that as $x$ gets closer and closer to $0$ , the values for $r (x) = \frac{1}{x}$ should grow larger and larger (in absolute value.) This is borne out in the table below on the left where it is apparent that for $x approximately equals 0$ , $r (x)$ is becoming unbounded.

As we investigate the end behavior of $r$ , we find that as $x → minus ∞$ and as $x → ∞$ , $r ⁢ open parenthesis x close parenthesis approximately equals 0$ . Again, number sense agrees here with the data, since as the denominator of $\frac{1}{x}$ becomes unbounded, the value of the fraction should diminish.³ That being said, we could ask if the graph ever reaches the $x$ -axis. If we attempt to solve $y = r (x) = \frac{1}{x} = 0$ . we arrive at the contradiction $1 equals 0$ hence, $0$ is not in the range of $r$ . Every other real number besides $0$ is in the range of $r$ , however. To see this, let $c not equal to 0$ be a real number. Then $\frac{1}{c}$ is defined and, moreover, $r (\frac{1}{c}) = \frac{1}{(1 / c)} = c$ . This shows $c$ is in the range of $r$ . Hence, the range of $r$ is $open brace y ∈ ℝ vertical bar y not equal to 0 close brace$ or, using interval notation, $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ .

$\begin{matrix} x & r (x) = \frac{1}{x} \\ - 0.01 & - 100 \\ - 0.001 & - 1000 \\ - 0.0001 & - 10000 \\ - 0.00001 & - 100000 \\ 0 & undefined \\ 0.00001 & 100000 \\ 0.0001 & 10000 \\ 0.001 & 1000 \\ 0.01 & 100 \end{matrix}$

$\begin{matrix} x & r (x) = \frac{1}{x} \\ - 100000 & - 0.00001 \\ - 10000 & - 0.0001 \\ - 1000 & - 0.001 \\ - 100 & - 0.01 \\ 0 & undefined \\ 100 & 0.01 \\ 1000 & 0.001 \\ 10000 & 0.0001 \\ 100000 & 0.00001 \end{matrix}$

Coordinate-plane figure. — **Figure 3.1** $y = r (x) = \frac{1}{x}$

Like we did in Section 2.1, we'll borrow some notation from Calculus in order for us to codify the behavior as $x → 0$ . First off, note that the behavior of $r$ differs depending on which direction we approach $0$ . We describe the values $x less than 0$ but $x → 0$ (such as $x = - 0.01$ , $minus 0.001$ , etc.) as ` $x$ approaching $0$ from the left,' written as $x \to 0^{-}$ . If we think of these numbers as all being $x$ -values where $x = ‘ 0 - a little bit’$ , the the `' in the notation ` $x \to 0^{-}$ ' makes better sense. For these values, the function values $r ⁢ open parenthesis x close parenthesis → minus ∞$ . Using the limit notation introduced in Section 2.1, we'd write: $lim_{x \to 0^{-}} r (x) = - \infty$ .

Similarly, we say `as $x$ approaches $0$ from the right,' that is as $x \to 0^{+}$ , $r ⁢ open parenthesis x close parenthesis → ∞$ , or, more succinctly, $lim_{x \to 0^{+}} r (x) = \infty$ . As before, we understand `from the right' means we are using $x$ values slightly to the right of $0$ on the number line: numbers such as $x equals 0.001.$ These numbers could described as ` $0 plus a little bit$ ,' which justifies the ` $+$ ' in the notation ` $x \to 0^{+}$ .'

We can also use this notation to describe the end behavior, but here the numerical roles are reversed. We see as $x → minus ∞$ , $r (x) \to 0^{-}$ and as $x → ∞$ , $r (x) \to 0^{+}$ . When it comes to codifying these results using Calculus, we write $lim_{x \to - \infty} r (x) = 0$ and $lim x → ∞ r ⁢ open parenthesis x close parenthesis equals 0$ . Note that, unfortunately, we lose the directionality here on the limiting value - that is, we do not write $lim_{x \to - \infty} r (x) = 0^{-}$ or $lim_{x \to \infty} r (x) = 0^{+}$ . Without getting too much into formal definitions, the reason is that if limiting values are finite, we express them as real numbers.⁴ Period.

The way we describe what is happening graphically is to say the line $x equals 0$ is a vertical asymptote to the graph of $y = r (x)$ and the line $y equals 0$ is a horizontal asymptote to the graph of $y = r (x)$ . Roughly speaking, asymptotes are lines which approximate functions as either the inputs or outputs become unbounded.

Definition

The line $x = c$ is called a vertical asymptote of the graph of a function $y = f (x)$ if either of the limits $lim_{x \to c^{-}} f (x)$ or $lim_{x \to c^{+}} f (x)$ (or both) result in $∞$ or $minus ∞$ .

Definition

The line $y = c$ is called a horizontal asymptote of the graph of a function $y = f (x)$ either $lim_{x \to - \infty} f (x) = c$ or $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals c$ (or both).

The behaviors illustrated in the graph $r (x) = \frac{1}{x}$ are typical of functions of the form $f (x) = \frac{1}{x^{n}} = x^{- n}$ for natural numbers, $n$ . As with the monomial functions discussed in Section 2.1, the patterns that develop primarily depend on whether $n$ is odd or even. Having thoroughly discussed the graph of $y = \frac{1}{x} = x^{- 1}$ , we graph it along with $y = \frac{1}{x^{3}} = x^{- 3}$ and $y = \frac{1}{x^{5}} = x^{- 5}$ below. Note the points $(- 1, - 1)$ and $open parenthesis 1, 1 close parenthesis$ are common to all three graphs as are the asymptotes $x equals 0$ and $y equals 0$ . As the $n$ increases, the graphs become steeper for $vertical bar x vertical bar less than 1$ and flatten out more quickly for $vertical bar x vertical bar greater than 1$ . Both the domain and range in each case appears to be $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . Indeed, owing to the $x$ in the denominator of $f (x) = \frac{1}{x^{n}}$ , $f ⁢ open parenthesis 0 close parenthesis$ , and only $f ⁢ open parenthesis 0 close parenthesis$ , is undefined. Hence the domain is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . When thinking about the range, note the equation $f (x) = \frac{1}{x^{n}} = c$ has the solution $x = \sqrt[n]{\frac{1}{c}}$ as long as $c not equal to 0$ . Thus means $f (\sqrt[n]{\frac{1}{c}}) = c$ for every nonzero real number $c$ . If $c equals 0$ , we are in the same situation as before: $\frac{1}{x^{n}} = 0$ has no real solution. This establishes the range is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . Finally, each of the graphs appear to be symmetric about the origin. Indeed, since $n$ is odd, $f (- x) = {(- x)}^{- n} = {(- 1)}^{- n} x^{- n} = - x^{- n} = - f (x)$ , proving every member of this function family is odd.

$\begin{matrix} x & \frac{1}{x} = x^{- 1} & \frac{1}{x^{3}} = x^{- 3} & \frac{1}{x^{5}} = x^{- 5} \\ - 10 & - 0.1 & - 0.001 & - 0.00001 \\ - 1 & - 1 & - 1 & - 1 \\ - 0.1 & - 10 & - 1000 & - 100000 \\ 0 & undefined & undefined & undefined \\ 0.1 & 10 & 1000 & 100000 \\ 1 & 1 & 1 & 1 \\ 10 & 0.1 & 0.001 & 0.00001 \end{matrix}$

We repeat the same experiment with functions of the form $f (x) = \frac{1}{x^{n}} = x^{- n}$ where $n$ is even. $y = \frac{1}{x^{2}} = x^{- 2}$ , $y = \frac{1}{x^{4}} = x^{- 4}$ and $y = \frac{1}{x^{6}} = x^{- 6}$ . These graphs all share the points $open parenthesis minus 1, 1 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ , and asymptotes $x equals 0$ and $y equals 0$ . Note here that both $lim_{x \to 0^{-}} f (x) = \infty$ and $lim_{x \to 0^{+}} f (x) = \infty$ , so we may simply write $lim x → 0 f ⁢ open parenthesis x close parenthesis equals ∞$ .

The same remarks about the steepness for $vertical bar x vertical bar less than 1$ and the flattening for $vertical bar x vertical bar greater than 1$ also apply. For the same reasons as given above, the domain of each of these functions is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . When it comes to the range, the fact $n$ is even tells us there are solutions to $\frac{1}{x^{n}} = c$ only if $c greater than 0$ . It follows that the range is $open parenthesis 0, ∞ close parenthesis$ for each of these functions. Concerning symmetry, as $n$ is even, $f (- x) = {(- x)}^{- n} = {(- 1)}^{- n} x^{- n} = x^{- n} = f (x)$ , proving each member of this function family is even. Hence, the graphs of these functions are symmetric about the $y$ -axis.

$\begin{matrix} x & \frac{1}{x^{2}} = x^{- 2} & \frac{1}{x^{4}} = x^{- 4} & \frac{1}{x^{6}} = x^{- 6} \\ - 10 & 0.01 & 0.0001 & 1 \times 10^{- 6} \\ - 1 & 1 & 1 & 1 \\ - 0.1 & 100 & 10000 & 1 \times 10^{6} \\ 0 & undefined & undefined & undefined \\ 0.1 & 100 & 10000 & 1 \times 10^{6} \\ 1 & 1 & 1 & 1 \\ 10 & 0.01 & 0.0001 & 1 \times 10^{- 6} \end{matrix}$

Not surprisingly, we have an analog to Theorem 2.1 for this family of Laurent monomial functions.

Theorem

For real numbers $a$ , $h$ , and $k$ with $a not equal to 0$ , the graph of $F (x) = \frac{a}{{(x - h)}^{n}} + k = a {(x - h)}^{- n} + k$ can be obtained from the graph of $f (x) = \frac{1}{x^{n}} = x^{- n}$ by performing the following operations, in sequence:

add $h$ to each of the $x$ -coordinates of the points on the graph of $f$ . This results in a horizontal shift to the right if $h greater than 0$ or left if $h less than 0$ .
NOTE: This transforms the graph of $y = x^{- n}$ to $y = {(x - h)}^{- n}$ .
The vertical asymptote moves from $x equals 0$ to $x = h$ .
multiply the $y$ -coordinates of the points on the graph obtained in Step 1 by $a$ . This results in a vertical scaling, but may also include a reflection about the $x$ -axis if $a less than 0$ .
NOTE: This transforms the graph of $y = {(x - h)}^{- n}$ to $y = a {(x - h)}^{- n}$ .
add $k$ to each of the $y$ -coordinates of the points on the graph obtained in Step 2. This results in a vertical shift up if $k greater than 0$ or down if $k less than 0$ .
NOTE: This transforms the graph of $y = a {(x - h)}^{- n}$ to $y = a {(x - h)}^{- n} + k$ .
The horizontal asymptote moves from $y equals 0$ to $y = k$ .

The proof of Theorem 3.1 is identical to the proof of Theorem 2.1 - just replace $x^{n}$ with $x^{- n}$ . We nevertheless encourage the reader to work through the details⁵ and compare the results of this theorem with Theorems 1.2, 1.3, and 2.1.

We put Theorem 3.1 to good use in the following example.

Example 1

Use Theorem 3.1 to graph the following. Label at least two points and the asymptotes. State the domain and range using interval notation.

$f (x) = {(2 x - 3)}^{- 2}$
$g (t) = \frac{2 t - 1}{t + 1}$

In order to use Theorem 3.1, we first must put $f (x) = {(2 x - 3)}^{- 2}$ into the form prescribed by the theorem. To that end, we factor:
$f (x) = {(2 [x - \frac{3}{2}])}^{- 2} = 2^{- 2} {(x - \frac{3}{2})}^{- 2} = \frac{1}{4} {(x - \frac{3}{2})}^{- 2}$
We identify $n equals 2$ , $a = \frac{1}{4}$ and $h = \frac{3}{2}$ (and $k equals 0$ .) Per the theorem, we begin with the graph of $y = x^{- 2}$ and track the two points $open parenthesis minus 1, 1 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ along with the vertical and horizontal asymptotes $x equals 0$ and $y equals 0$ , respectively through each step.
Step 1: add $\frac{3}{2}$ to each of the $x$ -coordinates of each of the points on the graph of $y = x^{- 2}$ . This moves the vertical asymptote from $x equals 0$ to $x = \frac{3}{2}$ (which we represent by a dashed line.)
Figure 3.8 $y = x^{- 2}$
$\overset{add \frac{3}{2} to each x -coordinate}{\overset{}{\to}}$
Figure 3.9 $y = {(x - \frac{3}{2})}^{- 2}$
$(- 1, 1), VA: x = 0, (1, 1)$
$(\frac{1}{2}, 1), VA: x = \frac{3}{2}, (\frac{5}{2}, 1)$
Step 2: multiply each of the $y$ -coordinates of each of the points on the graph of $y = {(x - \frac{3}{2})}^{- 2}$ by $\frac{1}{4}$ .
Figure 3.10 $y = {(x - \frac{3}{2})}^{- 2}$
$\overset{multiply each y -coordinate by \frac{1}{4}}{\overset{}{\to}}$
Figure 3.11 $y = \frac{1}{4} {(x - \frac{3}{2})}^{- 2}$
$(\frac{1}{2}, 1), (\frac{5}{2}, 1)$
$(\frac{1}{2}, \frac{1}{4}), (\frac{5}{2}, \frac{1}{4})$
Since we did not shift the graph vertically, the horizontal asymptote remains $y equals 0$ . We can determine the domain and range of $f$ by tracking the changes to the domain and range of our progenitor function, $y = x^{- 2}$ . We get the domain and range of $f$ is $(- \infty, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$ and the range of $f$ is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ .
Using either long or synthetic division, we get
$g (t) = \frac{2 t - 1}{t + 1} = - \frac{3}{t + 1} + 2 = \frac{- 3}{{(t - (- 1))}^{1}} + 2$
so we identify $n equals 1$ , $a = - 3$ , $h = - 1$ , and $k equals 2$ . We start with the graph of $y = \frac{1}{t}$ with points $(- 1, - 1)$ , $open parenthesis 1, 1 close parenthesis$ and asymptotes $t equals 0$ and $y equals 0$ and track these through each of the steps.
Step 1: Add $minus 1$ to each of the $t$ -coordinates of each of the points on the graph of $y = \frac{1}{t}$ . This moves the vertical asymptote from $t equals 0$ to $t = - 1$ .
Figure 3.12 $y = \frac{1}{t}$
$→ add minus 1 to each t -coordinate$
Figure 3.13 $y = \frac{1}{t + 1}$
$(- 1, 1), VA: t = 0, (1, 1)$
$(- 2, 1), VA: t = - 1, (0, 1)$
Step 2: multiply each of the $y$ -coordinates of each of the points on the graph of $y = \frac{1}{t + 1}$ by $minus 3$ .
Figure 3.14 $y = \frac{1}{t + 1}$
$\overset{multiply each y -coordinate by - 3}{\overset{}{\to}}$
Figure 3.15 $y = \frac{- 3}{t + 1}$
$open parenthesis minus 2, 1 close parenthesis, open parenthesis 0, 1 close parenthesis$
$(- 2, 3), (0, - 3)$
Step 3: add $2$ to each of the $y$ -coordinates of each of the points on the graph of $y = \frac{- 3}{t + 1}$ . This moves the horizontal asymptote from $y equals 0$ to $y equals 2$ .
Figure 3.16 $y = \frac{- 3}{t + 1}$
$→ add 2 to each y -coordinate$
Figure 3.17 $y = \frac{- 3}{t + 1} + 2$
$(- 2, 3), (0, - 3), HA y = 0$
$(- 2, 5), (0, - 1), HA y = 2$
As above, we determine the domain and range of $g$ by tracking the changes in the domain and range of $y = \frac{1}{t}$ . We find the domain of $g$ is $(- \infty, - 1) \cup (- 1, \infty)$ and the range is $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ .

In Example 3.1.1, we once again see the benefit of changing the form of a function to make use of an important result. A natural question to ask is to what extent general rational functions can be rewritten to use Theorem 3.1. In the same way polynomial functions are sums of monomial functions, it turns out, allowing for non-real number coefficients, that every rational function can be written as a sum of (possibly shifted) Laurent monomial functions.⁶

Local Behavior near Excluded Values

We take time now to focus on behaviors of the graphs of rational functions near excluded values. We've already seen examples of one type of behavior: vertical asymptotes. Our next example gives us a physical interpretation of a vertical asymptote. This type of model arises from a family of equations cheerily named `doomsday' equations.⁷

Example 2

A mathematical model for the population $P (t)$ , in thousands, of a certain species of bacteria, $t$ days after it is introduced to an environment is given by $P (t) = \frac{100}{{(5 - t)}^{2}}$ , $0 less than or equal to t less than 5$ .

Find and interpret $P ⁢ open parenthesis 0 close parenthesis$ .
When will the population reach $100, 000$ ?
Graph $y = P (t)$ .
Find and interpret $lim_{t \to 5^{-}} P (t)$ .

Substituting $t equals 0$ gives $P (0) = \frac{100}{{(5 - 0)}^{2}} = 4$ . Since $t$ represents the number of days after the bacteria are introduced into the environment, $t equals 0$ corresponds to the day the bacteria are introduced. Since $P (t)$ is measured in thousands, $P ⁢ open parenthesis t close parenthesis equals 4$ means $4000$ bacteria are initially introduced into the environment.
To find when the population reaches $100, 000$ , we first need to remember that $P (t)$ is measured in thousands. In other words, $100, 000$ bacteria corresponds to $P ⁢ open parenthesis t close parenthesis equals 100$ . Hence, we need to solve $P (t) = \frac{100}{{(5 - t)}^{2}} = 100$ . Clearing denominators and dividing by $100$ gives ${(5 - t)}^{2} = 1$ , which, after extracting square roots, produces $t equals 4$ or $t equals 6$ . Of these two solutions, only $t equals 4$ in our domain, so this is the solution we keep. Hence, it takes $4$ days for the population of bacteria to reach $100, 000$ .
After a slight re-write, we have $P (t) = \frac{100}{{(5 - t)}^{2}} = \frac{100}{{[(- 1) (t - 5)]}^{2}} = \frac{100}{{(t - 5)}^{2}}$ . Using Theorem 3.1, we start with the graph of $y = \frac{1}{t^{2}}$ below on the left. After shifting the graph to the right $5$ units and stretching it vertically by a factor of $100$ (note, the graphs are not to scale!), we restrict the domain to $0 less than or equal to t less than 5$ to arrive at the graph of $y = P (t)$ below on the right.
From the graph, we see as $t \to 5^{-}$ , $P ⁢ open parenthesis t close parenthesis → ∞$ , so $lim_{t \to 5^{-}} P (t) = \infty$ . This means that the population of bacteria is increasing without bound as we near 5 days, which cannot actually happen. For this reason, $t equals 5$ is called the `doomsday' for this population. There is no way any environment can support infinitely many bacteria, so shortly before $t equals 5$ the environment would collapse.
Figure 3.18 $y = \frac{1}{t^{2}}$

Figure 3.19 $y = P (t)$
$(- 1, 1), VA: t = 0, (1, 1)$
$open parenthesis 4, 100 close parenthesis, VA: t equals 5$

Will all values excluded from the domain of a rational function produce vertical asymptotes in the graph? The short answer is `no.' There are milder interruptions that can occur - holes in the graph - which we explore in our next example.

To this end, we formalize the notion of average velocity - a concept we first encountered in Example 1.2.8 in Section 1.2. In that example, the function $s (t) = - 5 t^{2} + 100 t$ , $0 less than or equal to t less than or equal to 20$ gives the height of a model rocket above the Moon's surface, in feet, $t$ seconds after liftoff. The function $s$ is an example of a position function since it provides information about where the rocket is at time $t$ . In that example, we interpreted the average rate of change of $s$ over an interval as the average velocity of the rocket over that interval. The average velocity provides two pieces of information: the average speed of the rocket along with the rocket's direction.

Suppose we have a position function $s$ defined over an interval containing some fixed time $t sub 0$ . We can define the average velocity as a function of any time $t$ other than $t sub 0$ :

Definition

Let $s (t)$ be the position of an object at time $t$ and $t sub 0$ be a fixed time in the domain of $s$ . The average velocity between time $t$ and time $t sub 0$ for $t not equal to t sub 0$ is given by

$\bar{v} (t) = \frac{Δ [s (t)]}{Δ t} = \frac{s (t) - s (t_{0})}{t - t_{0}} .$

We must exclude $t equals t sub 0$ from the domain of $\bar{v}$ in Definition 3.5 since, otherwise, we would have a $0$ in the denominator. What is interesting in this case however, is that substituting $t equals t sub 0$ also produces $0$ in the numerator. (Do you see why?) While ` $\frac{0}{0}$ ' is undefined, it is more precisely called an `indeterminate form' and is studied extensively in Calculus. We explore this phenomenon in the next example.

Example 3

Let $s (t) = - 5 t^{2} + 100 t$ , $0 less than or equal to t less than or equal to 20$ give the height of a model rocket above the Moon's surface, in feet, $t$ seconds after liftoff.

Find and simplify an expression for the average velocity of the rocket between times $t$ and $15$ , $\bar{v} (t)$ .
Find and interpret $v bar ⁢ open parenthesis 14 close parenthesis$ .
Find and interpret $lim t → 15 v bar ⁢ open parenthesis t close parenthesis$ .
Graph $y = \bar{v} (t)$ . Interpret your answer to part graphically.

Using Definition 3.5 with $t sub 0 equals 15$ , we get:
$\begin{matrix} \bar{v} (t) & = & \frac{s (t) - s (15)}{t - 15}, & t \neq 15 \\ = & \frac{(- 5 t^{2} + 100 t) - 375}{t - 15} \\ = & \frac{- 5 (t^{2} - 20 t + 75)}{t - 15} \\ = & \frac{- 5 (t - 15) (t - 5)}{t - 15} \\ = & \frac{- 5 (t - 15) (t - 5)}{(t - 15)} \\ = & - 5 (t - 5) = - 5 t + 25, & t \neq 15 \end{matrix}$
Since the domain of $s$ is $0 less than or equal to t less than or equal to 20$ , our final answer is $\bar{v} (t) = - 5 t + 25$ , for $t ∈ open bracket 0, 15 close parenthesis union open parenthesis 15, 20 close bracket$ .
We find $\bar{v} (14) = - 5 (14) + 25 = - 45$ . This means that between $14$ and $15$ seconds after launch, the rocket was traveling, on average, a speed $45$ feet per second downwards, or falling back to the Moon's surface.
To find $lim t → 15 v bar ⁢ open parenthesis t close parenthesis$ , we need to analyze the outputs from $\bar{v} (t)$ as $t → 15$ . Using the table below on the left, it certainly appears to be the case that $lim_{t \to 15} \bar{v} (t) = - 50$ . Of course, we could be choosing the number $minus 50$ because of a bias towards nice, integer answers.⁸
However, we can argue a stronger case algebraically using the simplified formula $\bar{v} (t) = - 5 t + 25$ . As $t → 15$ , $- 5 t \to - 5 (15) = - 75$ so $- 5 t + 25 \to - 50$ . It stands to reason, then, that $lim_{t \to 15} \bar{v} (t) = - 50$ .
This means our average velocity approaches $minus 50$ feet per second as we sample times closer and closer to $t equals 15$ seconds after liftoff. Since we're pushing the $t$ -values to $t equals 15$ , instead of viewing the limit $minus 50$ as an average velocity calculated between two time values, we view $minus 50$ feet per second as the instantaneous velocity of the rocket at $t equals 15$ . That is, at $t equals 15$ seconds after liftoff, the rocket is traveling downwards at a rate of 50 feet per second.⁹
From part, we know $\bar{v} (t) = - 5 t + 25$ , for $t ∈ open bracket 0, 15 close parenthesis union open parenthesis 15, 20 close bracket$ Hence the graph of $\bar{v} (t)$ is a portion of the line $y = - 5 t + 25$ . Since $\bar{v} (t)$ is not defined when $t equals 15$ , our graph is the line segment starting at $open parenthesis 0, 25 close parenthesis$ and ending at $open parenthesis 20, 75 close parenthesis$ which skips over the point $open parenthesis 15, minus 50 close parenthesis$ , creating a hole in the graph.¹⁰
$\begin{matrix} t & \bar{v} (t) \\ 14.9 & - 49.5 \\ 14.99 & - 49.95 \\ 14.999 & - 49.995 \\ 15 & undefined \\ 15.001 & - 50.005 \\ 15.01 & - 50.05 \\ 15.1 & - 50.5 \end{matrix}$
Figure 3.20 $y = \bar{v} (t)$

Some notes about Example 3.1.3 are in order. First, excluded values from the domain of a rational function don't necessarily cause vertical asymptotes in the graph. Even though $v bar ⁢ open parenthesis 15 close parenthesis$ doesn't exist, the fact that $lim_{t \to 15} \bar{v} (t) = - 50$ means that we expect $v bar ⁢ open parenthesis 15 close parenthesis$ to be $minus 50$ .This sentiment is exactly what a hole in the graph at $open parenthesis 15, minus 50 close parenthesis$ communicates.

Second, in finding $lim_{t \to 15} \bar{v} (t) = - 50$ , we've taken some (more) steps into Calculus. Specifically, we used properties of the limit process that we've not yet formalized, let alone justified. The main idea is that the ` $\frac{0}{0}$ ' indeterminate form which occurs when we attempt to evaluate $v bar ⁢ open parenthesis 15 close parenthesis$ using the formula $\bar{v} (t) = \frac{- 5 t^{2} + 100 t - 375}{t - 15}$ is resolved when the factor $open parenthesis t minus 15 close parenthesis$ cancels from the denominator. We can algebraically reason what to expect out of the expression $\bar{v} (t) = - 5 t + 25$ as $t → 15$ because there is no longer any division by $0$ .

We will revisit these sorts of machinations later in the text in a bit more generality.¹¹ For now, we'll work to build some intuition with some classic hand-waving which we hope will do more good than harm.

Our next theorem generalizes our reasoning from this last example.

Theorem

Location of Vertical Asymptotes and Holes: ¹² Suppose $r$ is a rational function and $c$ is not in the domain of $r$ .

If $lim_{x \to c} r (x) = L$ where $L$ is a real number, then the graph of $r$ has a hole at $(c, L)$ .
Otherwise, the graph of $y = r (x)$ has a vertical asymptote $x = c$ .

Of course the first question to ask is how do we know if $lim_{x \to c} r (x)$ results in a real number, $L$ , and if so, how do we find $L$ ? It turns out that if the limit exists, then the same sort of algebraic cancellation which occurred Example 3.1.3 is guaranteed to happen.

Let's consider a generic rational function $r (x) = \frac{p (x)}{q (x)}$ where $p$ and $q$ are polynomial functions. The values ` $c$ ' excluded from the domain of $r$ are the zeros of $q$ : $q ⁢ open parenthesis c close parenthesis equals 0$ . We now have two cases to consider.

If $p ⁢ open parenthesis c close parenthesis not equal to 0$ , then as $x \to c$ , $r (x) = \frac{p (x)}{q (x)} \to \frac{p (c), a nonzero number}{0}$ which results in unbounded behavior (graphically, a vertical asymptote.)

If $p ⁢ open parenthesis c close parenthesis equals 0$ , then as $x \to c$ , $r (x) = \frac{p (x)}{q (x)} \to \frac{0}{0}$ , an indeterminate form. The Factor Theorem,¹³ guarantees both $p (x)$ and $q (x)$ contain factors of $(x - c)$ . This means we can simplify the expression $r (x)$ by cancelling common factors of $(x - c)$ . If all of the factors of $(x - c)$ in the denominator, $q (x)$ , cancel with factors in the numerator, $p (x)$ , then the division by $0$ is eliminated and we can proceed as in Example 3.1.3 to determine the limit.¹⁴ If some factors of $(x - c)$ remain in the denominator, then we're back to the first scenario and the graph will have a vertical asymptote.

We practice this methodology in the following example.

Example 4

For each function below:

determine the values excluded from the domain.
determine whether each excluded value corresponds to a vertical asymptote or hole in the graph.
verify your answers using a graphing utility.
describe the behavior of the graph near each excluded value using proper notation.
investigate any apparent symmetry of the graph about the $y$ -axis or origin.

$f (x) = \frac{2 x}{x^{2} - 3}$
$g (t) = \frac{t^{2} - t - 6}{t^{2} - 9}$
$h (t) = \frac{t^{2} - t - 6}{t^{2} + 9}$
$r (t) = \frac{t^{2} - t - 6}{t^{2} + 4 t + 4}$

We begin by finding the values excluded from the domain by setting the denominator equal to $0$ . Solving $x^{2} - 3 = 0$ , we get $x = \pm \sqrt{3}$ which factors the denominator as $(x - \sqrt{3}) (x + \sqrt{3})$ . Since $f (x)$ is in lowest terms (can you see why?), no cancellation occurs, hence, the lines $x = - \sqrt{3}$ and $x = \sqrt{3}$ are vertical asymptotes to the graph of $y = f (x)$ .
A graphing utility verifies this claim, and from the graph, we see that $lim_{x \to - {\sqrt{3}}^{-}} f (x) = - \infty$ , $lim_{x \to - {\sqrt{3}}^{+}} f (x) = \infty$ , $lim_{x \to {\sqrt{3}}^{-}} f (x) = - \infty$ , and $lim_{x \to {\sqrt{3}}^{+}} f (x) = \infty$ .
As a side note, the graph of $f$ appears to be symmetric about the origin. Sure enough, we find: $f (- x) = \frac{2 (- x)}{{(- x)}^{2} - 3} = - \frac{2 x}{x^{2} - 3} = - f (x)$ , proving $f$ is odd.
As above, we find the values excluded from the domain of $g$ finding the zeros of the denominator. Solving $t^{2} - 9 = 0$ gives $t = \pm 3$ . In this case, we can simplify the formula for $g (t)$ : $\frac{t^{2} - t - 6}{t^{2} - 9} = \frac{(t - 3) (t + 2)}{(t - 3) (t + 3)} = \frac{(t - 3) (t + 2)}{(t - 3) (t + 3)} = \frac{t + 2}{t + 3}$ . Hence, $g (t) = \frac{t + 2}{t + 3}$ provided $t not equal to 3$ .
Since the factor $open parenthesis t plus 3 close parenthesis$ , which corresponds to the zero $t = - 3$ , did not cancel from the denominator of $g (t)$ , we expect a vertical asymptote to the graph at $t = - 3$ . On the other hand, as $t → 3$ , $t plus 2 → 5$ and $t plus 3 → 6$ so $\frac{t + 2}{t + 3} \to \frac{3 + 2}{3 + 3} = \frac{5}{6}$ . This gives $lim t → 3 g ⁢ open parenthesis t close parenthesis equals$ $\frac{5}{6}$ . Hence we have a hole in the graph of $y = g (t)$ at $(3, \frac{5}{6})$ .
Graphing $g$ we can definitely see the vertical asymptote $t = - 3$ : as $lim_{t \to - 3^{-}} g (t) = \infty$ and $lim_{t \to - 3^{+}} g (t) = - \infty$ . Near $t equals 3$ , the graph seems to have no interruptions, but we know $g$ is undefined at $t equals 3$ . Depending on the graphing utility, we may or may not convince the display to show us the hole at $(3, \frac{5}{6})$ .
Figure 3.21
Figure 3.22
The graph of $y = f (x)$
The graph of $y = g (t)$
Setting the denominator of the expression for $h (t)$ to $0$ gives $t^{2} + 9 = 0$ , which has no real solutions. Accordingly, the graph of $y = h (t)$ (at least as much as we can discern from the technology) is devoid of both vertical asymptotes and holes. Using terms defined in Section 2.1, the function $h$ is both continuous and smooth.¹⁵
Setting the denominator of $r (t)$ to zero gives the equation $t^{2} + 4 t + 4 = 0$ . We get the (repeated!) solution $t = - 2$ . Simplifying, we get $\frac{t^{2} - t - 6}{t^{2} + 4 t + 4} = \frac{(t - 3) (t + 2)}{(t + 2) (t + 2)} = \frac{(t - 3) (t + 2)}{(t + 2) (t + 2)} = \frac{t - 3}{t + 2}$ . Since not all factors of $open parenthesis t plus 2 close parenthesis$ cancelled from the denominator, $t = - 2$ continues to produce a $0$ in the denominator. Hence $t = - 2$ is a vertical asymptote to the graph. A graphing utility bears this out. Specifically, $lim_{t \to - 2^{-}} r (t) = \infty$ and $lim_{t \to - 2^{+}} r (t) = - \infty$ .
Figure 3.23
Figure 3.24
The graph of $y = h (t)$
The graph of $y = r (t)$

End Behavior

Now that we've discussed behavior near values excluded from the domains of rational functions, let's focus our attention on end behavior. We have already seen one example of this in the form of horizontal asymptotes. Our next example of the section gives us a real-world application of a horizontal asymptote.¹⁶

Example 5

The number of students $N (t)$ at local college who have had the flu $t$ months after the semester begins can be modeled by:

$N (t) = \frac{1500 t + 50}{3 t + 1}, t \geq 0 .$

Find and interpret $N ⁢ open parenthesis 0 close parenthesis$ .
How long will it take until $300$ students will have had the flu?
Use Theorem 3.1 to graph $y = N (t)$ .
Find and interpret $lim t → ∞ N ⁢ open parenthesis t close parenthesis$ .

Substituting $t equals 0$ gives $N (0) = \frac{1500 (0) + 50}{1 + 3 (0)} = 50$ . Since $t$ represents the number of months since the beginning of the semester, $t equals 0$ describes the state of the flu outbreak at the beginning of the semester. Hence, at the beginning of the semester, $50$ students have had the flu.
We set $N (t) = \frac{1500 t + 50}{3 t + 1} = 300$ and solve. Clearing denominators gives $1500 t + 50 = 300 (3 t + 1)$ from which we get $t = \frac{5}{12}$ . This means it will take $\frac{5}{12}$ months, or about 13 days, for $300$ students to have had the flu.
To graph $y = N (t)$ , we first use long division to rewrite $N (t) = \frac{- 450}{3 t + 1} + 500$ . From there, we get
$N (t) = - \frac{450}{3 t + 1} + 500 = \frac{- 450}{3 (t + \frac{1}{3})} + 500 = \frac{- 150}{t + \frac{1}{3}} + 500$
Using Theorem 3.1, we start with the graph of $y = \frac{1}{t}$ below on the left and perform the following steps: shift the graph to the left by $\frac{1}{3}$ units, stretch the graph vertically by a factor of $150$ , reflect the graph across the $t$ -axis, and finally, shift the graph up $500$ units. As the domain of $N$ is $t greater than or equal to 0$ , we obtain the graph below on the right.
Figure 3.25 $y = \frac{1}{t}$

Figure 3.26 $y = N (t)$
$(- 1, 1), (1, 1), HA: y = 0$
$(\frac{2}{3}, 350), HA: y = 500$
Owing to the horizontal asymptote, $y equals 500$ , we have $lim t → ∞ N ⁢ open parenthesis t close parenthesis equals 500$ . (More specifically, as $t → ∞$ , $N (t) \to 500^{-}$ .) This means as time goes by, only a total of 500 students will have ever had the flu.

We determined the horizontal asymptote to the graph of $y = N (t)$ in Example 3.1.5 by rewriting $N (t)$ into a form compatible with Theorem 3.1, and while there is nothing wrong with this approach, it will simply not work for general rational functions which cannot be rewritten this way. To that end, we revisit this problem using Theorem 2.3 from Section 2.1. The end behavior of the numerator of $N (t) = \frac{1500 t + 50}{3 t + 1}$ is determined by its leading term, $1500 ⁢ t$ , and the end behavior of the denominator is likewise determined by its leading term, $3 ⁢ t$ . Hence, as $t → ∞$ :

$N (t) = \frac{1500 t + 50}{3 t + 1} \approx \frac{1500 t}{3 t} = 500 .$

Hence $lim t → ∞ N ⁢ open parenthesis t close parenthesis equals 500$ so $y equals 500$ is the horizontal asymptote. This same reasoning can be used in general to argue the following theorem.

Theorem

Location of Horizontal Asymptotes: Suppose $r$ is a rational function and $r (x) = \frac{p (x)}{q (x)}$ , where $p$ and $q$ are polynomial functions with leading coefficients $a$ and $b$ , respectively.

If the degree of $p (x)$ is the same as the degree of $q (x)$ , then $lim x → ∞ r ⁢ open parenthesis x close parenthesis equals$ $\frac{a}{b}$ so $y = \frac{a}{b}$ is the¹⁷ horizontal asymptote of the graph of $y = r (x)$ .
If the degree of $p (x)$ is less than the degree of $q (x)$ , then $lim x → ∞ r ⁢ open parenthesis x close parenthesis equals 0$ so $y equals 0$ (the $x$ -axis) is the horizontal asymptote of the graph of $y = r (x)$ .
If the degree of $p (x)$ is greater than the degree of $q (x)$ , then the graph of $y = r (x)$ has no horizontal asymptotes.

So see why Theorem 3.3 works, suppose $r (x) = \frac{p (x)}{q (x)}$ where $a$ is the leading coefficient of $p (x)$ and $b$ is the leading coefficient of $q (x)$ . As $x → minus ∞$ or $x → ∞$ , Theorem 2.3 gives $r (x) \approx \frac{a x^{n}}{b x^{m}}$ , where $n$ and $m$ are the degrees of $p (x)$ and $q (x)$ , respectively.

If the degree of $p (x)$ and the degree of $q (x)$ are the same, then $n = m$ so that $r (x) \approx \frac{a x^{n}}{b x^{n}} = \frac{a}{b}$ . Hence $lim x → minus ∞ r ⁢ open parenthesis x close parenthesis$ $= \frac{a}{b}$ and $lim x → ∞ r ⁢ open parenthesis x close parenthesis$ $= \frac{a}{b}$ which means $y = \frac{a}{b}$ is the horizontal asymptote in this case.

If the degree of $p (x)$ is less than the degree of $q (x)$ , then $n < m$ , so $m - n$ is a positive number, and hence, $r (x) \approx \frac{a x^{n}}{b x^{m}} = \frac{a}{b x^{m - n}} \to 0$ . As $x → minus ∞$ or $x → ∞$ , $r (x)$ is more or less a fraction with a constant numerator, $a$ , but a denominator which is unbounded. Hence, $lim_{x \to - \infty} r (x) = 0$ and $lim x → ∞ r ⁢ open parenthesis x close parenthesis equals 0$ producing the horizontal asymptote $y equals 0$ .

If the degree of $p (x)$ is greater than the degree of $q (x)$ , then $n > m$ , and hence $n - m$ is a positive number and $r (x) \approx \frac{a x^{n}}{b x^{m}} = \frac{a x^{n - m}}{b}$ , which is a monomial function from Section 2.1. As such, $r$ becomes unbounded as $x → minus ∞$ or $x → ∞$ .

Note that in the two cases which produce horizontal asymptotes, the behavior of $r$ is identical as $x → minus ∞$ and $x → ∞$ . Hence, if the graph of a rational function has a horizontal asymptote, there is only one.¹⁸

We put Theorem 3.3 to good use in the following example.

Example 6

For each function below:

use Theorem 2.3 to analytically determine the horizontal asymptotes to the graph, if any.
check your answers using Theorem 3.3 and a graphing utility.
describe the end behavior of the graph using proper notation.
investigate any apparent symmetry of the graph about the $y$ -axis or origin.

$F (s) = \frac{5 s}{s^{2} + 1}$
$g (x) = \frac{x^{2} - 4}{x + 1}$
$h (t) = \frac{6 t^{3} - 3 t + 1}{5 - 2 t^{3}}$
$r (x) = 2 - \frac{3 x^{2}}{1 - x^{2}}$

Using Theorem 2.3, we get as $s → minus ∞$ or $s → ∞$ , $F (s) = \frac{5 s}{s^{2} + 1} \approx \frac{5 s}{s^{2}} = \frac{5}{s}$ . Hence, $lim_{s \to - \infty} F (s) = 0$ and $lim s → ∞ F ⁢ open parenthesis s close parenthesis equals 0$ so $y equals 0$ is a horizontal asymptote to the graph.
Alternatively, to use Theorem 3.3 note the degree of the numerator of $F (s)$ , $1$ , is less than the degree of the denominator, $2$ , so $y equals 0$ as the horizontal asymptote using this approach as well.
Graphically, as $s → minus ∞$ or $s → ∞$ , the graph $y = F (s)$ approaches the $s$ -axis ( $y equals 0$ ). More specifically, as $s → minus ∞$ , $F (s) \to 0^{-}$ and as $s → ∞$ , $F (s) \to 0^{+}$ .
As a side note, the graph of $F$ appears to be symmetric about the origin. Indeed, $F (- s) = \frac{5 (- s)}{{(- s)}^{2} + 1} = - \frac{5 s}{s^{2} + 1}$ proving $F$ is odd.
As $x → minus ∞$ or $x → ∞$ , $g (x) = \frac{x^{2} - 4}{x + 1} \approx \frac{x^{2}}{x} = x$ , and while $y = x$ is a line, it is not a horizontal line. Hence, we conclude the graph of $y = g (x)$ has no horizontal asymptotes. Sure enough, Theorem 3.3 supports this since the degree of the numerator of $g (x)$ is $2$ which is greater than the degree of the denominator, $1$ . From the graph, we see that the graph of $y = g (x)$ doesn't appear to level off to a constant value, confirming there is no horizontal asymptote.¹⁹
Figure 3.27
Figure 3.28
The graph of $y = F (s)$
The graph of $y = g (x)$
As $t → minus ∞$ or $t → ∞$ , $h (t) = \frac{6 t^{3} - 3 t + 1}{5 - 2 t^{3}} \approx \frac{6 t^{3}}{- 2 t^{3}} = - 3$ . Hence, $lim_{t \to - \infty} h (t) = - 3$ and $lim_{t \to \infty} h (t) = - 3$ , indicating a horizontal asymptote $y = - 3$ . Sure enough, since the degrees of the numerator and denominator of $h (t)$ are both three, Theorem 3.3 tells us $y = \frac{6}{- 2} = - 3$ is the horizontal asymptote. We see from the graph of $y = h (t)$ that as $t → minus ∞$ , $h (t) \to - 3^{+}$ , and as $t → ∞$ , $h (t) \to - 3^{-}$ .
If we apply Theorem 2.3 to the term $\frac{3 x^{2}}{1 - x^{2}}$ in the expression for $r (x)$ , we find $\frac{3 x^{2}}{1 - x^{2}} \approx \frac{3 x^{2}}{- x^{2}} = - 3$ as $x → minus ∞$ or $x → ∞$ . It seems reasonable to conclude, then, that $lim_{x \to - \infty} r (x) = 2 - (- 3) = 5$ and likewise $lim_{x \to \infty} r (x) = 2 - (- 3) = 5$ so $y equals 5$ is our horizontal asymptote.
In order to double check this calculation using Theorem 3.3, however, we need to rewrite the expression $r (x)$ with a single denominator: $r (x) = 2 - \frac{3 x^{2}}{1 - x^{2}} = \frac{2 (1 - x^{2}) - 3 x^{2}}{1 - x^{2}} = \frac{2 - 5 x^{2}}{1 - x^{2}}$ . Now we apply Theorem 3.3 and note since the numerator and denominator have the same degree, we are guaranteed the horizontal asymptote is $y = \frac{- 5}{- 1} = 5$ .
Both calculations are borne out graphically below where it appears as if as $x → minus ∞$ or $x → ∞$ , $r (x) \to 5^{+}$ .
As a final note, the graph of $r$ appears to be symmetric about the $y$ axis. We find $r (- x) = 2 - \frac{3 {(- x)}^{2}}{1 - {(- x)}^{2}} = 2 - \frac{3 x^{2}}{1 - x^{2}} = r (x)$ , proving $r$ is even.
Figure 3.29
Figure 3.30
The graph of $y = h (t)$
The graph of $y = r (x)$

We close this section with a discussion of the third (and final!) kind of asymptote which can be associated with the graphs of rational functions. Let us return to the function $g (x) = \frac{x^{2} - 4}{x + 1}$ in Example 3.1.6. Performing long division,²⁰ we get $g (x) = \frac{x^{2} - 4}{x + 1} = x - 1 - \frac{3}{x + 1}$ . Since the term $\frac{3}{x + 1} \to 0$ as $x → minus ∞$ and as $x → ∞$ , it stands to reason that as $x$ becomes unbounded, the function values $g (x) = x - 1 - \frac{3}{x + 1} \approx x - 1$ . Geometrically, this means that the graph of $y = g (x)$ should resemble the line $y = x - 1$ as $x → minus ∞$ and $x → ∞$ . We see this play out both numerically and graphically below. (As usual, the asymptote $y = x - 1$ is denoted by a dashed line.)

Table 3.1

\begin{matrix} x & g (x) & x - 1 \\ - 10 & \approx - 10.6667 & - 11 \\ - 100 & \approx - 100.9697 & - 101 \\ - 1000 & \approx - 1000.9970 & - 1001 \\ - 10000 & \approx - 10000.9997 & - 10001 \end{matrix}

\begin{matrix} x & g (x) & x - 1 \\ 10 & \approx 8.7273 & 9 \\ 100 & \approx 98.9703 & 99 \\ 1000 & \approx 998.9970 & 999 \\ 10000 & \approx 9998.9997 & 9999 \end{matrix}

The way we symbolize the relationship between the end behavior of $y = g (x)$ with that of the line $y = x - 1$ is to write `as $x → minus ∞$ and $x → ∞$ , $g ⁢ open parenthesis x close parenthesis → x minus 1$ ' in order to have some notational consistency with what we have done earlier in this section when it comes to end behavior.²¹ In this case, we say the line $y = x - 1$ is a slant asymptote ²² to the graph of $y = g (x)$ . Informally, the graph of a rational function has a slant asymptote if, as $x → minus ∞$ or as $x → ∞$ , the graph resembles a non-horizontal, or `slanted' line. More formally, we define a slant asymptote as follows.

Definition

The line $y = m x + b$ where $m not equal to 0$ is called a slant asymptote of the graph of a function $y = f (x)$ if as $x → minus ∞$ or as $x → ∞$ , $f (x) \to m x + b$ .

A few remarks are in order. First, note that the stipulation $m not equal to 0$ in Definition 3.6 is what makes the `slant' asymptote `slanted' as opposed to the case when $m equals 0$ in which case we'd have a horizontal asymptote.

Secondly, while we have motivated what me mean intuitively by the notation ` $f (x) \to m x + b$ ,' like so many ideas in this section, the formal definition requires Calculus. Another way to express this sentiment, however, is to rephrase ` $f (x) \to m x + b$ ' as ` $[f (x) - (m x + b)] \to 0$ .' In other words, the graph of $y = f (x)$ has the slant asymptote $y = m x + b$ if and only if the graph of $y = f (x) - (m x + b)$ has a horizontal asymptote $y equals 0$ . This last sentiment can be encoded using limit notation as follows.

Definition

The line $y = m x + b$ where $m not equal to 0$ is called a slant asymptote of the graph of a function $y = f (x)$ if either $lim_{x \to - \infty} [f (x) - (m x + b)] = 0$ or $lim_{x \to \infty} [f (x) - (m x + b)] = 0$ (or both!).

Our next task is to determine the conditions under which the graph of a rational function has a slant asymptote, and if it does, how to find it. In the case of $g (x) = \frac{x^{2} - 4}{x + 1}$ , the degree of the numerator $x squared minus 4$ is $2$ , which is exactly one more than the degree if its denominator $x plus 1$ which is $1$ . This results in a linear quotient polynomial, and it is this quotient polynomial which is the slant asymptote. Generalizing this situation gives us the following theorem.²³

Theorem

Determination of Slant Asymptotes: Suppose $r$ is a rational function and $r (x) = \frac{p (x)}{q (x)}$ , where the degree of $p$ is exactly one more than the degree of $q$ . Then the graph of $y = r (x)$ has the slant asymptote $y = L (x)$ where $L (x)$ is the quotient obtained by dividing $p (x)$ by $q (x)$ .

In the same way that Theorem 3.3 gives us an easy way to see if the graph of a rational function $r (x) = \frac{p (x)}{q (x)}$ has a horizontal asymptote by comparing the degrees of the numerator and denominator, Theorem 3.4 gives us an easy way to check for slant asymptotes. Unlike Theorem 3.3, which gives us a quick way to find the horizontal asymptotes (if any exist), Theorem 3.4 gives us no such `short-cut'. If a slant asymptote exists, we have no recourse but to use long division to find it.²⁴

Example 7

For each of the following functions:

find the slant asymptote, if it exists.
verify your answer using a graphing utility.
investigate any apparent symmetry of the graph about the $y$ -axis or origin.

$f (x) = \frac{x^{2} - 4 x + 2}{1 - x}$
$g (t) = \frac{t^{2} - 4}{t - 2}$
$h (x) = \frac{x^{3} + 1}{x^{2} - 4}$
$r (t) = 2 t - 1 + \frac{4 t^{3}}{1 - t^{2}}$

The degree of the numerator is $2$ and the degree of the denominator is $1$ , so Theorem 3.4 guarantees us a slant asymptote. To find it, we divide $1 - x = - x + 1$ into $x^{2} - 4 x + 2$ and get a quotient of $- x + 3$ , so our slant asymptote is $y = - x + 3$ . We confirm this graphically below.
As with the previous example, the degree of the numerator $g (t) = \frac{t^{2} - 4}{t - 2}$ is $2$ and the degree of the denominator is $1$ , so Theorem 3.4 applies. In this case,
$g (t) = \frac{t^{2} - 4}{t - 2} = \frac{(t + 2) (t - 2)}{(t - 2)} = \frac{(t + 2) (t - 2)}{{(t - 2)}^{1}} = t + 2, t \neq 2$
so we have that the slant asymptote $y = t + 2$ is identical to the graph of $y = g (t)$ except at $t equals 2$ (where the latter has a `hole' at $open parenthesis 2, 4 close parenthesis$ .) While the word `asymptote' has the connotation of `approaching but not equaling,' Definitions 3.4 and 3.6 allow for these extreme cases.
Figure 3.32
Figure 3.33
The graph of $y = f (x)$
The graph of $y = g (t)$
For $h (x) = \frac{x^{3} + 1}{x^{2} - 4}$ , the degree of the numerator is $3$ and the degree of the denominator is $2$ so again, we are guaranteed the existence of a slant asymptote. The long division $(x^{3} + 1) \div (x^{2} - 4)$ gives a quotient of just $x$ , so our slant asymptote is the line $y = x$ . The graphing utility confirms this.
Note the graph of $h$ appears to be symmetric about the origin. We check $h (- x) = \frac{{(- x)}^{3} + 1}{{(- x)}^{2} - 4} = \frac{- x^{3} + 1}{x^{2} - 4} = - \frac{x^{3} - 1}{x^{2} - 4}$ . However, $- h (x) = - \frac{x^{3} + 1}{x^{2} - 4}$ , so it appears as if $h (- x) \neq - h (x)$ for all $x$ . Checking $x equals 1$ , we find $h (1) = - \frac{2}{3}$ but $h (- 1) = 0$ which shows the graph of $h$ , is in fact, not symmetric about the origin.
For our last example, $r (t) = 2 t - 1 + \frac{4 t^{3}}{1 - t^{2}}$ , the expression $r (t)$ is not in the form to apply Theorem 3.4 directly. We can, nevertheless, appeal to the spirit of the theorem and use long division to rewrite the term $\frac{4 t^{3}}{1 - t^{2}} = - 4 t + \frac{4 t}{1 - t^{2}}$ . We then get:
$\begin{matrix} r (t) & = & 2 t - 1 + \frac{4 t^{3}}{1 - t^{2}} \\ = & 2 t - 1 - 4 t + \frac{4 t}{1 - t^{2}} \\ = & - 2 t - 1 + \frac{4 t}{1 - t^{2}} \end{matrix}$
As $t → minus ∞$ or $x → ∞$ , Theorem 2.3 gives $\frac{4 t}{1 - t^{2}} \approx \frac{4 t}{- t^{2}} = - \frac{4}{t} \to 0$ . Hence, as $t → minus ∞$ or $t → ∞$ , $r (t) \to - 2 t - 1$ , so $y = - 2 t - 1$ is the slant asymptote to the graph as confirmed by the graphing utility below. From a distance, the graph of $r$ appears to be symmetric about the origin. However, if we look carefully, we see the $y$ -intercept is $open parenthesis 0, minus 1 close parenthesis$ , as borne out by the computation $r (0) = - 1$ . Hence $r$ cannot be odd.²⁵
Figure 3.34
Figure 3.35
The graph of $y = h (x)$
The graph of $y = r (t)$

Our last example gives a real-world application of a slant asymptote. The problem features the concept of average profit. The average profit, denoted $\bar{P} (x)$ , is the total profit, $P (x)$ , divided by the number of items sold, $x$ . In English, the average profit tells us the profit made per item sold. It, along with average cost, is defined below.

Definition

Let $C (x)$ and $P (x)$ represent the cost and profit to make and sell $x$ items, respectively.

The average cost, $\bar{C} (x) = \frac{C (x)}{x}$ , $x greater than 0$ .
NOTE: The average cost is the cost per item produced.
The average profit, $\bar{P} (x) = \frac{P (x)}{x}$ , $x greater than 0$ .
NOTE: The average profit is the profit per item sold.

You'll explore average cost (and its relation to variable cost) in Exercise. For now, we refer the reader to to Example 1.4.3 in Section 1.4.

Example 8

Recall the profit (in dollars) when $x$ PortaBoy game systems are produced and sold is given by $P (x) = - 1.5 x^{2} + 170 x - 150$ , $0 less than or equal to x less than or equal to 166$ .

Find and simplify an expression for the average profit, $\bar{P} (x)$ . What is the domain of $\bar{P}$ ?
Find and interpret $P bar ⁢ open parenthesis 50 close parenthesis$ .
Determine the slant asymptote to the graph of $y = \bar{P} (x)$ . Check your answer using a graphing utility.
Interpret the slope of the slant asymptote.

We find $\bar{P} (x) = \frac{P (x)}{x} = \frac{- 1.5 x^{2} + 170 x - 150}{x} = - 1.5 x + 170 - \frac{150}{x}$ . Since the domain of $P$ is $open bracket 0, 166 close bracket$ but $x not equal to 0$ , the domain of $\bar{P}$ is $open parenthesis 0, 166 close bracket$ .
We find $\bar{P} (50) = - 1.5 (50) + 170 - \frac{150}{50} = 92$ . This means that when $50$ PortaBoy systems are sold, the average profit is $$ 92$ per system.
Technically, the graph of $y = \bar{P} (x)$ has no slant asymptote since the domain of the function is restricted to $open parenthesis 0, 166 close bracket$ . That being said, if we were to let $x → ∞$ , the term $\frac{150}{x} \to 0$ , so we'd have $\bar{P} (x) \to - 1.5 x + 170$ . This means the slant asymptote would be $y = - 1.5 x + 170$ . We graph $y = \bar{P} (x)$ and $y = - 1.5 x + 170$ .
Figure 3.36
The slope of the slant asymptote $y = - 1.5 x + 170$ is $minus 1.5$ . Allowing for $x → ∞$ , $\bar{P} (x) \approx - 1.5 x + 170$ which means as we sell more systems, the average profit is decreasing at about a rate of $$ 1.50$ per system. If the number $1.5$ sounds familiar to this problem situation, it should. In Example 1.2.4 in Section 1.2, we determined the slope of the demand function to be $minus 1.5$ . In that situation, the $minus 1.5$ meant that in order to sell an additional system, the price had to drop by $$ 1.50$ . The fact the average profit is decreasing at more or less this same rate means the loss in profit per system can be attributed to the reduction in price needed to sell each additional system.²⁶

Exercises

(Review of Long Division):²⁷ In Exercises -, use polynomial long division to perform the indicated division. Write the polynomial in the form $p (x) = d (x) q (x) + r (x)$ .

$(4 x^{2} + 3 x - 1) \div (x - 3)$
$(2 x^{3} - x + 1) \div (x^{2} + x + 1)$
$(5 x^{4} - 3 x^{3} + 2 x^{2} - 1) \div (x^{2} + 4)$
$(- x^{5} + 7 x^{3} - x) \div (x^{3} - x^{2} + 1)$
$(9 x^{3} + 5) \div (2 x - 3)$
$(4 x^{2} - x - 23) \div (x^{2} - 1)$

In Exercises -, given the pair of functions $f$ and $F$ , sketch the graph of $y = F (x)$ by starting with the graph of $y = f (x)$ and using Theorem 3.1. Track at least two points and the asymptotes. State the domain and range using interval notation.

$f (x) = \frac{1}{x}$ , $F (x) = \frac{1}{x - 2} + 1$
$f (x) = \frac{1}{x}$ , $F (x) = \frac{2 x}{x + 1}$
$f (x) = x^{- 1}$ , $F (x) = 4 x {(2 x + 1)}^{- 1}$
$f (x) = x^{- 2}$ , $F (x) = - {(x - 1)}^{- 2} + 3$

In Exercises -, find a formula for each function below in the form $F (x) = \frac{a}{x - h} + k$ .

$y = F (x)$
Figure 3.37 $x$ -intercept $open parenthesis minus 1, 0 close parenthesis$ , $y$ -intercept $(0, - \frac{1}{2})$
$y = F (x)$
Figure 3.38 $x$ -intercept $open parenthesis 3, 0 close parenthesis$ , $y$ -intercept $open parenthesis 0, 3 close parenthesis$

In Exercises -, find a formula for each function below in the form $F (x) = \frac{a}{{(x - h)}^{2}} + k$ .

$y = F (x)$
Figure 3.39 $x$ -intercepts $open parenthesis minus 3, 0 close parenthesis$ , $open parenthesis minus 1, 0 close parenthesis$ , $y$ -intercept $open parenthesis 0, 3 close parenthesis$
$y = F (x)$
Figure 3.40 $x$ -intercepts $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 1, 0 close parenthesis$ , Vertical Asymptote: $x = \frac{1}{2}$

In Exercises -, for the given rational function:

State the domain.
Identify any vertical asymptotes of the graph.

Identify any holes in the graph.
Find the horizontal asymptote, if it exists.

Find the slant asymptote, if it exists.
Graph the function using a graphing utility and describe the behavior near the asymptotes.
$f (x) = \frac{x}{3 x - 6}$
$f (x) = \frac{3 + 7 x}{5 - 2 x}$
$f (x) = \frac{x}{x^{2} + x - 12}$
$g (t) = \frac{t}{t^{2} + 1}$
$g (t) = \frac{t + 7}{{(t + 3)}^{2}}$
$g (t) = \frac{t^{3} + 1}{t^{2} - 1}$
$r (z) = \frac{4 z}{z^{2} + 4}$
$r (z) = \frac{4 z}{z^{2} - 4}$
$r (z) = \frac{z^{2} - z - 12}{z^{2} + z - 6}$
$f (x) = \frac{3 x^{2} - 5 x - 2}{x^{2} - 9}$
$f (x) = \frac{x^{3} + 2 x^{2} + x}{x^{2} - x - 2}$
$f (x) = \frac{x^{3} - 3 x + 1}{x^{2} + 1}$
$g (t) = \frac{2 t^{2} + 5 t - 3}{3 t + 2}$
$g (t) = \frac{- t^{3} + 4 t}{t^{2} - 9}$
$g (t) = \frac{- 5 t^{4} - 3 t^{3} + t^{2} - 10}{t^{3} - 3 t^{2} + 3 t - 1}$
$r (z) = \frac{z^{3}}{1 - z}$
$r (z) = \frac{18 - 2 z^{2}}{z^{2} - 9}$
$r (z) = \frac{z^{3} - 4 z^{2} - 4 z - 5}{z^{2} + z + 1}$
The cost $C (p)$ in dollars to remove $p$ % of the invasive Ippizuti fish species from Sasquatch Pond is:
$C (p) = \frac{1770 p}{100 - p}, 0 \leq p < 100$
1. Find and interpret $C ⁢ open parenthesis 25 close parenthesis$ and $C ⁢ open parenthesis 95 close parenthesis$ .
2. What does the vertical asymptote at $x equals 100$ mean within the context of the problem?
3. What percentage of the Ippizuti fish can you remove for $40000?
In the scenario of Example 3.1.3, $s (t) = - 5 t^{2} + 100 t$ , $0 less than or equal to t less than or equal to 20$ gives the height of a model rocket above the Moon's surface, in feet, $t$ seconds after liftoff. For each of the times $t sub 0$ listed below, find and simplify a the formula for the average velocity $\bar{v} (t)$ between $t$ and $t sub 0$ (see Definition 3.5) and use $\bar{v} (t)$ to find and interpret the instantaneous velocity of the rocket at $t equals t sub 0$ (See Example 3.1.3).
1. $t sub 0 equals 5$
2. $t sub 0 equals 9$
3. $t sub 0 equals 10$
4. $t sub 0 equals 11$
The population of Sasquatch in Portage County $t$ years after the year 1803 is modeled by the function
$P (t) = \frac{150 t}{t + 15} .$
Find and interpret the horizontal asymptote of the graph of $y = P (t)$ and explain what it means.
The cost in dollars, $C (x)$ to make $x$ dOpi media players is $C (x) = 100 x + 2000$ , $x greater than or equal to 0$ . You may wish to review the concepts of fixed and variable costs introduced in Example 1.2.3 in Section 1.2.2.
1. Find a formula for the average cost $\bar{C} (x)$ .
2. Find and interpret $C bar ⁢ open parenthesis 1 close parenthesis$ and $C bar ⁢ open parenthesis 100 close parenthesis$ .
3. How many dOpis need to be produced so that the average cost per dOpi is $$ 200$ ?
4. Find and interpret $lim_{x \to 0^{+}} \bar{C} (x)$ .
5. Interpret the behavior of $\bar{C} (x)$ as $x → ∞$ .
This exercise explores the relationships between fixed cost, variable cost, and average cost. The reader is encouraged to revisit Example 1.2.3 in Section 1.2.2 as needed. Suppose the cost in dollars $C (x)$ to make $x$ items is given by $C (x) = m x + b$ where $m$ and $b$ are positive real numbers.
1. Show the fixed cost (the money spent even if no items are made) is $b$ .
2. Show the variable cost (the increase in cost per item made) is $m$ .
3. Find a formula for the average cost when making $x$ items, $\bar{C} (x)$ .
4. Show $\bar{C} (x) > m$ for all $x greater than 0$ and, moreover, $\bar{C} (x) \to m^{+}$ as $x → ∞$ .
5. Interpret $\bar{C} (x) \to m^{+}$ both geometrically and in terms of fixed, variable, and average costs.
Suppose the price-demand function for a particular product is given by $p (x) = m x + b$ where $x$ is the number of items made and sold for $p (x)$ dollars. Here, $m less than 0$ and $b greater than 0$ . If the cost (in dollars) to make $x$ of these products is also a linear function $C (x)$ , show that the graph of the average profit function $\bar{P} (x)$ has a slant asymptote with slope $m$ and interpret.

In Exercise in Section 2.1, we fit a few polynomial models to the following electric circuit data. The circuit was built with a variable resistor. For each of the following resistance values (measured in kilo-ohms, $k Ω$ ), the corresponding power to the load (measured in milliwatts, $m W$ ) is given below.²⁸

**Table 3.2**
Resistance: ( $k Ω$ )	1.012	2.199	3.275	4.676	6.805	9.975
Power: ( $m W$ )	1.063	1.496	1.610	1.613	1.505	1.314

Using some fundamental laws of circuit analysis mixed with a healthy dose of algebra, we can derive the actual formula relating power $P (x)$ to resistance $x$ :

$P (x) = \frac{25 x}{{(x + 3.9)}^{2}}, x \geq 0 .$

Graph the data along with the function $y = P (x)$ using a graphing utility.
Use a graphing utility to approximate the maximum power that can be delivered to the load. What is the corresponding resistance value?
Find and interpret the end behavior of $P (x)$ as $x → ∞$ .

Let $f (x) = \frac{a x^{2} - c}{x + 3}$ . Find values for $a$ and $c$ so the graph of $f$ has a hole at $open parenthesis minus 3, 12 close parenthesis$ .
Let $f (x) = \frac{a x^{n} - 4}{2 x^{2} + 1}$ .
1. Find values for $a$ and $n$ so the graph of $y = f (x)$ has the horizontal asymptote $y equals 3$ .
2. Find values for $a$ and $n$ so the graph of $y = f (x)$ has the slant asymptote $y equals 5 ⁢ x$ .
Suppose $p$ is a polynomial function and $a$ is a real number. Define $r (x) = \frac{p (x) - p (a)}{x - a}$ . Use the Factor Theorem, Theorem 2.8, to prove the graph of $y = r (x)$ has a hole at $x = a$ .
For each function $f (x)$ listed below, compute the average rate of change over the indicated interval.²⁹ What trends do you observe? How do your answers manifest themselves graphically? How do you results compare with those of Exercise in Section 2.1?
$\begin{matrix} f (x) & [0.9, 1.1] & [0.99, 1.01] & [0.999, 1.001] & [0.9999, 1.0001] \\ x^{- 1} \\ x^{- 2} \\ x^{- 3} \\ x^{- 4} \end{matrix}$
In his now famous 1919 dissertation The Learning Curve Equation, Louis Leon Thurstone presents a rational function which models the number of words a person can type in four minutes as a function of the number of pages of practice one has completed.³⁰ Using his original notation and original language, we have $Y = \frac{L (X + P)}{(X + P) + R}$ where $L$ is the predicted practice limit in terms of speed units, $X$ is pages written, $Y$ is writing speed in terms of words in four minutes, $P$ is equivalent previous practice in terms of pages and $R$ is the rate of learning. In Figure 5 of the paper, he graphs a scatter plot and the curve $Y = \frac{216 (X + 19)}{X + 148}$ . Discuss this equation with your classmates. How would you update the notation? Explain what the horizontal asymptote of the graph means. You should take some time to look at the original paper. Skip over the computations you don't understand yet and try to get a sense of the time and place in which the study was conducted.

Answers

$4 x^{2} + 3 x - 1 = (x - 3) (4 x + 15) + 44$
$2 x^{3} - x + 1 = (x^{2} + x + 1) (2 x - 2) + (- x + 3)$
$5 x^{4} - 3 x^{3} + 2 x^{2} - 1 = (x^{2} + 4) (5 x^{2} - 3 x - 18) + (12 x + 71)$
$- x^{5} + 7 x^{3} - x = (x^{3} - x^{2} + 1) (- x^{2} - x + 6) + (7 x^{2} - 6)$
$9 x^{3} + 5 = (2 x - 3) (\frac{9}{2} x^{2} + \frac{27}{4} x + \frac{81}{8}) + \frac{283}{8}$
$4 x^{2} - x - 23 = (x^{2} - 1) (4) + (- x - 19)$
$F (x) = \frac{1}{x - 2} + 1$ [10pt] Domain: $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ Range: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$ Vertical asymptote: $x equals 2$ Horizontal asymptote: $y equals 1$
Figure 3.41
$F (x) = \frac{2 x}{x + 1} = \frac{- 2}{x + 1} + 2$ [10pt] Domain: $(- \infty, - 1) \cup (- 1, \infty)$ Range: $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ Vertical asymptote: $x = - 1$ Horizontal asymptote: $y equals 2$
Figure 3.42
$F (x) = 4 x {(2 x + 1)}^{- 1} = \frac{4 x}{2 x + 1} = \frac{- 1}{x + \frac{1}{2}} + 2$ [10pt] Domain: $(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$ Range: $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ Vertical asymptote: $x = - \frac{1}{2}$ Horizontal asymptote: $y equals 2$
Figure 3.43
$F (x) = - {(x - 1)}^{- 2} + 3 = \frac{- 1}{{(x - 1)}^{2}} + 3$ Domain: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$ Range: $open parenthesis minus ∞, 3 close parenthesis union open parenthesis 3, ∞ close parenthesis$ Vertical asymptote: $x equals 1$ Horizontal asymptote: $y equals 3$
Figure 3.44
$F (x) = \frac{1}{x + 2} - 1$
$F (x) = \frac{- 2}{x - 1} + 1$
$F (x) = \frac{- 4}{{(x + 2)}^{2}} + 4$
$F (x) = \frac{1}{{(x - \frac{1}{2})}^{2}} - 4$
$f (x) = \frac{x}{3 x - 6}$ Domain: $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ Vertical asymptote: $x equals 2$ $lim_{x \to 2^{-}} f (x) = - \infty$ , $lim_{x \to 2^{+}} f (x) = \infty$ No holes in the graph Horizontal asymptote: $y = \frac{1}{3}$ $lim_{x \to - \infty} f (x) =$ $\frac{1}{3}$ More specifically: as $x \to - \infty, f (x) \to {\frac{1}{3}}^{-}$ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals$ $\frac{1}{3}$ More specifically: as $x \to \infty, f (x) \to {\frac{1}{3}}^{+}$
$f (x) = \frac{3 + 7 x}{5 - 2 x}$ Domain: $(- \infty, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$ Vertical asymptote: $x = \frac{5}{2}$ $lim_{x \to {\frac{5}{2}}^{-}} f (x) = \infty$ , $lim_{x \to {\frac{5}{2}}^{+}} f (x) = - \infty$ No holes in the graph Horizontal asymptote: $y = - \frac{7}{2}$ $lim_{x \to - \infty} f (x) =$ $- \frac{7}{2}$ More specifically: as $x \to - \infty, f (x) \to - {\frac{7}{2}}^{+}$ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals$ $- \frac{7}{2}$ More specifically: as $x \to \infty, f (x) \to - {\frac{7}{2}}^{-}$
$f (x) = \frac{x}{x^{2} + x - 12} = \frac{x}{(x + 4) (x - 3)}$ Domain: $(- \infty, - 4) \cup (- 4, 3) \cup (3, \infty)$ Vertical asymptotes: $x = - 4, x = 3$ $lim_{x \to - 4^{-}} f (x) = - \infty$ , $lim_{x \to - 4^{+}} f (x) = \infty$ $lim_{x \to 3^{-}} f (x) = - \infty$ , $lim_{x \to 3^{+}} f (x) = \infty$ No holes in the graph Horizontal asymptote: $y equals 0$ $lim_{x \to - \infty} f (x) = 0$ More specifically, as $x \to - \infty, f (x) \to 0^{-}$ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 0$ More specifically, as $x \to \infty, f (x) \to 0^{+}$
$g (t) = \frac{t}{t^{2} + 1}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$ No vertical asymptotes No holes in the graph Horizontal asymptote: $y equals 0$ $lim_{t \to - \infty} g (t) = 0$ More specifically, as $t \to - \infty, g (t) \to 0^{-}$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals 0$ More specifically, as $t \to \infty, g (t) \to 0^{+}$
$g (t) = \frac{t + 7}{{(t + 3)}^{2}}$ Domain: $(- \infty, - 3) \cup (- 3, \infty)$ Vertical asymptote: $t = - 3$ $lim_{t \to - 3} g (t) = \infty$ No holes in the graph Horizontal asymptote: $y equals 0$ $lim_{t \to - \infty} g (t) = 0$ ³¹More specifically, as $t \to - \infty, g (t) \to 0^{-}$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals 0$ More specifically, as $t \to \infty, g (t) \to 0^{+}$
$g (t) = \frac{t^{3} + 1}{t^{2} - 1} = \frac{t^{2} - t + 1}{t - 1}$ Domain: $(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$ Vertical asymptote: $t equals 1$ $lim_{t \to 1^{-}} g (t) = - \infty$ , $lim_{t \to 1^{+}} g (t) = \infty$ Hole at $(- 1, - \frac{3}{2})$ Slant asymptote: $y = t$ $lim_{t \to - \infty} g (t) = - \infty$ As $t → minus ∞$ , the graph is below $y = t$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ As $t → ∞$ , the graph is above $y = t$
$r (z) = \frac{4 z}{z^{2} + 4}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$ No vertical asymptotes No holes in the graph Horizontal asymptote: $y equals 0$ $lim_{z \to - \infty} r (z) = 0$ More specifically, as $z \to - \infty, r (z) \to 0^{-}$ $lim z → ∞ r ⁢ open parenthesis z close parenthesis equals 0$ More specifically, as $z \to \infty, r (z) \to 0^{+}$
$r (z) = \frac{4 z}{z^{2} - 4} = \frac{4 z}{(z + 2) (z - 2)}$ Domain: $(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$ Vertical asymptotes: $z = - 2, z = 2$ $lim_{z \to - 2^{-}} r (z) = - \infty$ , $lim_{z \to - 2^{+}} r (z) = \infty$ $lim_{z \to 2^{-}} r (z) = - \infty$ , $lim_{z \to 2^{+}} r (z) = \infty$ No holes in the graph Horizontal asymptote: $y equals 0$ $lim_{z \to - \infty} r (z) = 0$ More specifically, as $z \to - \infty, r (z) \to 0^{-}$ $lim z → ∞ r ⁢ open parenthesis z close parenthesis equals 0$ More specifically, as $z \to \infty, r (z) \to 0^{+}$
$r (z) = \frac{z^{2} - z - 12}{z^{2} + z - 6} = \frac{z - 4}{z - 2}$ Domain: $(- \infty, - 3) \cup (- 3, 2) \cup (2, \infty)$ Vertical asymptote: $z equals 2$ $lim_{z \to 2^{-}} r (z) = \infty$ , $lim_{z \to 2^{+}} r (z) = - \infty$ Hole at $(- 3, \frac{7}{5})$ Horizontal asymptote: $y equals 1$ $lim_{z \to - \infty} r (z) = 1$ More specifically, as $z \to - \infty, r (z) \to 1^{+}$ $lim z → ∞ r ⁢ open parenthesis z close parenthesis equals 1$ More specifically, as $z \to \infty, r (z) \to 1^{-}$
$f (x) = \frac{3 x^{2} - 5 x - 2}{x^{2} - 9} = \frac{(3 x + 1) (x - 2)}{(x + 3) (x - 3)}$ Domain: $(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$ Vertical asymptotes: $x = - 3, x = 3$ $lim_{x \to - 3^{-}} f (x) = \infty$ , $lim_{x \to - 3^{+}} f (x) = - \infty$ $lim_{x \to 3^{-}} f (x) = - \infty$ , $lim_{x \to 3^{+}} f (x) = \infty$ No holes in the graph Horizontal asymptote: $y equals 3$ $lim_{x \to - \infty} f (x) = 3$ More specifically, as $x \to - \infty, f (x) \to 3^{+}$ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 3$ More specifically, as $x \to \infty, f (x) \to 3^{-}$
$f (x) = \frac{x^{3} + 2 x^{2} + x}{x^{2} - x - 2} = \frac{x (x + 1)}{x - 2}$ Domain: $(- \infty, - 1) \cup (- 1, 2) \cup (2, \infty)$ Vertical asymptote: $x equals 2$ $lim_{x \to 2^{-}} f (x) = - \infty$ , $lim_{x \to 2^{+}} f (x) = \infty$ Hole at $open parenthesis minus 1, 0 close parenthesis$ Slant asymptote: $y = x + 3$ $lim_{x \to - \infty} f (x) = - \infty$ As $x → minus ∞$ , the graph is below $y = x + 3$ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals ∞$ As $x → ∞$ , the graph is above $y = x + 3$
$f (x) = \frac{x^{3} - 3 x + 1}{x^{2} + 1}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$ No vertical asymptotes No holes in the graph Slant asymptote: $y = x$ $lim_{x \to - \infty} f (x) = - \infty$ As $x → minus ∞$ , the graph is above $y = x$ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals ∞$ As $x → ∞$ , the graph is below $y = x$
$g (t) = \frac{2 t^{2} + 5 t - 3}{3 t + 2}$ Domain: $(- \infty, - \frac{2}{3}) \cup (- \frac{2}{3}, \infty)$ Vertical asymptote: $t = - \frac{2}{3}$ $lim_{t \to - {\frac{2}{3}}^{-}} g (t) = \infty$ , $lim_{t \to - {\frac{2}{3}}^{+}} g (t) = - \infty$ No holes in the graph Slant asymptote: $y = \frac{2}{3} t + \frac{11}{9}$ $lim_{t \to - \infty} g (t) = - \infty$ As $t → minus ∞$ , the graph is above $y = \frac{2}{3} t + \frac{11}{9}$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ As $t → ∞$ , the graph is below $y = \frac{2}{3} t + \frac{11}{9}$
$g (t) = \frac{- t^{3} + 4 t}{t^{2} - 9} = \frac{- t^{3} + 4 t}{(t - 3) (t + 3)}$ Domain: $(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$ Vertical asymptotes: $t = - 3$ , $t equals 3$ $lim_{t \to - 3^{-}} g (t) = \infty$ , $lim_{t \to - 3^{+}} g (t) = - \infty$ $lim_{t \to 3^{-}} g (t) = \infty$ , $lim_{t \to 3^{+}} g (t) = - \infty$ No holes in the graph Slant asymptote: $y = - t$ $lim_{t \to - \infty} g (t) = \infty$ As $t → minus ∞$ , the graph is above $y = - t$ $lim_{t \to \infty} g (t) = - \infty$ As $t → ∞$ , the graph is below $y = - t$
$g (t) = \frac{- 5 t^{4} - 3 t^{3} + t^{2} - 10}{t^{3} - 3 t^{2} + 3 t - 1} = \frac{- 5 t^{4} - 3 t^{3} + t^{2} - 10}{{(t - 1)}^{3}}$ Domain: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$ Vertical asymptotes: $t equals 1$ $lim_{t \to 1^{-}} g (t) = \infty$ , $lim_{t \to 1^{+}} g (t) = - \infty$ No holes in the graph Slant asymptote: $y = - 5 t - 18$ $lim_{t \to - \infty} g (t) = \infty$ As $t → minus ∞$ , the graph is above $y = - 5 t - 18$ $lim_{t \to \infty} g (t) = - \infty$ As $t → ∞$ , the graph is below $y = - 5 t - 18$
$r (z) = \frac{z^{3}}{1 - z}$ Domain: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$ Vertical asymptote: $z equals 1$ $lim_{z \to 1^{-}} r (z) = \infty$ $lim_{z \to 1^{+}} r (z) = - \infty$ No holes in the graph No horizontal or slant asymptote $lim_{z \to - \infty} r (z) = - \infty$ $lim_{z \to \infty} r (z) = - \infty$
$r (z) = \frac{18 - 2 z^{2}}{z^{2} - 9} = - 2$ Domain: $(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$ No vertical asymptotes Holes in the graph at $(- 3, - 2)$ and $open parenthesis 3, minus 2 close parenthesis$ Horizontal asymptote $y = - 2$ $lim_{z \to - \infty} r (z) = - 2$ $lim_{z \to \infty} r (z) = - 2$
$r (z) = \frac{z^{3} - 4 z^{2} - 4 z - 5}{z^{2} + z + 1} = z - 5$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$ No vertical asymptotes No holes in the graph Slant asymptote: $y = z - 5$ $lim_{z \to - \infty} r (z) = - \infty$ $lim z → ∞ r ⁢ open parenthesis z close parenthesis equals ∞$ $r (z) = z - 5$ everywhere.
1. $C ⁢ open parenthesis 25 close parenthesis equals 590$ means it costs $590 to remove 25% of the fish and and $C ⁢ open parenthesis 95 close parenthesis equals 33630$ means it would cost $33630 to remove 95% of the fish from the pond.
2. The vertical asymptote at $x equals 100$ means that as we try to remove 100% of the fish from the pond, the cost increases without bound; i.e., it's impossible to remove all of the fish.
3. For $40000 you could remove about 95.76% of the fish.
1. $\bar{v} (t) = \frac{s (t) - s (5)}{t - 5} = \frac{- 5 t^{2} + 100 t - 375}{t - 5} = - 5 t + 75$ , $t not equal to 5$ . The instantaneous velocity of the rocket when $t sub 0 equals 5$ is $- 5 (5) + 75 = 50$ meaning it is traveling $50$ feet per second upwards.
2. $\bar{v} (t) = \frac{s (t) - s (9)}{t - 9} = \frac{- 5 t^{2} + 100 t - 495}{t - 9} = - 5 t + 55$ , $t not equal to 9$ . The instantaneous velocity of the rocket when $t sub 0 equals 9$ is $- 5 (9) + 55 = 10$ , so the rocket has slowed to $10$ feet per second (but still heading up.)
3. $\bar{v} (t) = \frac{s (t) - s (10)}{t - 10} = \frac{- 5 t^{2} + 100 t - 495}{t - 10} = - 5 t + 50$ , $t not equal to 10$ . The instantaneous velocity of the rocket when $t sub 0 equals 10$ is $- 5 (10) + 50 = 0$ , so the rocket has momentarily stopped! In Example 1.2.8, we learned the rocket reaches its maximum height when $t equals 10$ seconds, which means the rocket must change direction from heading up to coming back down, so it makes sense that for this instant, its velocity is $0$ .
4. $\bar{v} (t) = \frac{s (t) - s (11)}{t - 11} = \frac{- 5 t^{2} + 100 t - 495}{t - 11} = - 5 t + 45$ , $t not equal to 11$ . The instantaneous velocity of the rocket when $t sub 0 equals 11$ is $- 5 (11) + 45 = - 10$ meaning the rocket has, indeed, changed direction and is heading downwards at a rate of $10$ feet per second. (Note the symmetry here between this answer and our answer when $t equals 9$ .)
The horizontal asymptote of the graph of $P (t) = \frac{150 t}{t + 15}$ is $y equals 150$ and it means that the model predicts the population of Sasquatch in Portage County will never exceed 150.
1. $\bar{C} (x) = \frac{100 x + 2000}{x} = 100 + \frac{2000}{x}$ , $x greater than 0$ .
2. $C bar ⁢ open parenthesis 1 close parenthesis equals 2100$ and $C bar ⁢ open parenthesis 100 close parenthesis equals 120$ . When just $1$ dOpi is produced, the cost per dOpi is $$ 2100$ , but when $100$ dOpis are produced, the cost per dOpi is $$ 120$ .
3. $C bar ⁢ open parenthesis x close parenthesis equals 200$ when $x equals 20$ . So to get the cost per dOpi to $$ 200$ , $20$ dOpis need to be produced.
4. We find $lim_{x \to 0^{+}} \bar{C} (x) = \infty$ . This means that as fewer and fewer dOpis are produced, the cost per dOpi becomes unbounded. In this situation, there is a fixed cost of $$ 2000$ ( $C ⁢ open parenthesis 0 close parenthesis equals 2000$ ), we are trying to spread that $$ 2000$ over fewer and fewer dOpis.
5. As $x → ∞$ , $\bar{C} (x) \to 100^{+}$ . This means that as more and more dOpis are produced, the cost per dOpi approaches $$ 100$ , but is always a little more than $$ 100$ . Since $$ 100$ is the variable cost per dOpi ( $C (x) = \underline{100} x + 2000$ ), it means that no matter how many dOpis are produced, the average cost per dOpi will always be a bit higher than the variable cost to produce a dOpi. As before, we can attribute this to the $$ 2000$ fixed cost, which factors into the average cost per dOpi no matter how many dOpis are produced.
1. The cost to make $0$ items is $C (0) = m (0) + b = b$ . Hence, so the fixed costs are $b$ .
2. $C (x) = m x + b$ is a linear function with slope $m greater than 0$ . Hence, the cost increases at a rate of $m$ dollars per item made. Hence, the variable cost is $m$ .
3. $\bar{C} (x) = \frac{C (x)}{x} = \frac{m x + b}{x} = m + \frac{b}{x}$ for $x greater than 0$ .
4. Since $b greater than 0$ , $\bar{C} (x) = m + \frac{b}{x} > m$ for $x greater than 0$ . As $x → ∞$ , $\frac{b}{x} \to 0$ so $\bar{C} (x) = m + \frac{b}{x} \to m$ .
5. Geometrically, the graph of $y = \bar{C} (x)$ has a horizontal asymptote $y = m$ , the variable cost. In terms of costs, as more items are produced, the affect of the fixed cost on the average cost, $\frac{b}{x}$ falls away so that the average cost per item approaches the variable cost to make each item.
If $p (x) = m x + b$ and $C (x)$ is linear, say $C (x) = r x + s$ , then we can compute the the profit function (in general) as: $P (x) = x p (x) - C (x) = x (m x + b) - (r x + s)$ which simplifies to $P (x) = m x^{2} + (b - r) x - s$ . Hence, the average profit $\bar{P} (x) = \frac{P (x)}{x} = \frac{m x^{2} + (b - r) x - s}{x} = m x + (b - r) - \frac{s}{x}$ . We see that as $x → ∞$ , $\frac{s}{x} \to 0$ so $\bar{P} (x) \approx m x + (b - r)$ . Hence, $y = m x + (b - r)$ is the slant asymptote to $y = \bar{P} (x)$ . This means that as more items are sold, the average profit is decreasing at approximately the same rate as the price function is decreasing, $m$ dollars per item. That is, to sell one additional item, we drop the price $p (x)$ by $m$ dollars which results in a drop in the average profit by approximately $m$ dollars.
1. Figure 3.45
2. The maximum power is approximately $1.603 ⁢ m ⁢ W$ which corresponds to $3.9 ⁢ k ⁢ Ω$ .
3. As $x \to \infty, P (x) \to 0^{+}$ which means as the resistance increases without bound, the power diminishes to zero.
$a = - 2$ and $c = - 18$ so $f (x) = \frac{- 2 x^{2} + 18}{x + 3}$ .
1. $a equals 6$ and $n equals 2$ so $f (x) = \frac{6 x^{2} - 4}{2 x^{2} + 1}$
2. $a equals 10$ and $n equals 3$ so $f (x) = \frac{10 x^{3} - 4}{2 x^{2} + 1}$ .
If we define $f (x) = p (x) - p (a)$ then $f$ is a polynomial function with $f (a) = p (a) - p (a) = 0$ . The Factor Theorem guarantees $(x - a)$ is a factor of $f (x)$ , that is, $f (x) = p (x) - p (a) = (x - a) q (x)$ for some polynomial $q (x)$ . Hence, $r (x) = \frac{p (x) - p (a)}{x - a} = \frac{(x - a) q (x)}{x - a} = q (x)$ so the graph of $y = r (x)$ is the same as the graph of the polynomial $y = q (x)$ except for a hole when $x = a$ .
The slope of the curves near $x equals 1$ matches the exponent on $x$ . This exactly what we saw in Exercise in Section 2.1.
$\begin{matrix} f (x) & [0.9, 1.1] & [0.99, 1.01] & [0.999, 1.001] & [0.9999, 1.0001] \\ x^{- 1} & - 1.0101 & - 1.0001 & \approx - 1 & \approx - 1 \\ x^{- 2} & - 2.0406 & - 2.0004 & \approx - 2 & \approx - 2 \\ x^{- 3} & - 3.1021 & - 3.0010 & \approx - 3 & \approx - 3 \\ x^{- 4} & - 4.2057 & - 4.0020 & \approx - 4 & \approx - 4 \end{matrix}$

According to this definition, all polynomial functions are also rational functions. (Take $q ⁢ open parenthesis x close parenthesis equals 1$ ). ↩
Technically speaking, $- 1 \times 10^{117}$ is a `small' number (since it is very far to the left on the number line.) However, it's absolute value, $1 \times 10^{117}$ is very large. If you read footnotes, you've seen this clarification before …(see Section 2.1.) ↩
We'll talk more about this end behavior shortly - stay tuned! ↩
We will, of course, offer more detail if the situation presents itself. ↩
We are, in fact, building to Theorem 5.11 in Section 5.4, so the more you see the forest for the trees, the better off you'll be when the time comes to generalize these moves to all functions. ↩
i.e., Laurent `Polynomials.' This result is a combination of Theorems 2.16 in Section 2.4 and Theorem 9.10 in Section 9.6. ↩
These functions arise in Differential Equations. The unfortunate name will make sense shortly. ↩
Or maybe the limit is actually $minus 49.999999865$ which, owing to calculation limitations of the graphing utility rounds to $minus 50$ . ↩
It is worth noting that Exercise from Section 1.2 has you arrive at this limit via a table. ↩
We've seen and discussed such holes in the graph as far back as Example 1.1.4 in Section 1.1 … ↩
See Section 6.1. ↩
Or, `How to tell your asymptote from a hole in the graph.' ↩
Theorem 2.8 in Section 2.2 ↩
Using the vocabulary from Section 2.1, the limit will exist if the multiplicity of $c$ as a zero for $p$ is greater than or equal to the multiplicity of $c$ as a zero for $q$ . ↩
We'll remind you more about continuous functions in Section 3.2 … ↩
Though the population below is more accurately modeled with the functions in Chapter 7, we approximate it (using Calculus, of course!) using a rational function. ↩
The use of the definite article will be justified momentarily. ↩
We will (first) encounter functions with more than one horizontal asymptote in Chapter 4.1. ↩
Sit tight! We'll revisit this function and its end behavior shortly. ↩
See the remarks following Theorem 3.3. ↩
Other notations include $g ⁢ open parenthesis x close parenthesis ≍ x minus 1$ or $g ⁢ open parenthesis x close parenthesis ∼ x minus 1$ . ↩
Also called an `oblique' asymptote in some, ostensibly higher class (and more expensive), texts. ↩
Once again, this theorem is brought to you courtesy of Theorem 2.2 and Calculus. ↩
That's OK, though. In the next section, we'll use long division to analyze end behavior and it's worth the effort! ↩
Do you see why? ↩
We generalize this result in Exercise . ↩
For more review, see Section A.8.2. ↩
The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set. ↩
See Definition 1.8 in Section 1.2.4 for a review of this concept, as needed. ↩
This paper, which is now in the public domain and can be found here , is from a bygone era when students at business schools took typing classes on manual typewriters. ↩
This is hard to see on the calculator, but trust me, the graph is below the $t$ -axis to the left of $t = - 7$ . ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.