3.2 Properties of Rational Functions

Vanessa Coffelt

3.2 Properties of 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 ratio of polynomials.

Definition 3.1

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

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

where $p$ and $q$ are polynomial functions.^[1]

3.2.1 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.

Definition 3.2

A Laurent monomial function is either a monomial function (see Definition 2.4) or a function of the form $f(x) = \dfrac{a}{x^n} = ax^{-n}$ for $n \in \mathbb{N}$ .

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 $r(0)$ is not defined owing to the presence of $x$ in the denominator. That is, the domain of $r$ is $\{ x \in \mathbb{R} \, | \, x \neq 0\}$ or, using interval notation, $(-\infty, 0) \cup (0, \infty)$ . Of course excluding $0$ from the domain of $r$ serves only to pique our curiosity about the behavior of $r(x)$ when $x \approx 0$ . 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 on the left where it is apparent that for $x \approx 0$ , $r(x)$ is becoming unbounded.

As we investigate the end behavior of $r$ , we find that as $x \rightarrow \pm \infty$ , $r(x) \approx 0$ . Again, number sense agrees here with the data, because as the denominator of $\frac{1}{x}$ becomes unbounded, the value of the fraction should diminish. 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 $1 = 0$ 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 $c \neq 0$ be a real number. Then $\frac{1}{c}$ is defined and, moreover, $r \left(\frac{1}{c} \right) = \frac{1}{(1/c)} = c$ . This shows $c$ is in the range of $r$ . Hence, the range of $r$ is $\{ y \in \mathbb{R} \, | \, y \neq 0\}$ or, using interval notation, $(-\infty, 0) \cup (0, \infty)$ .

$\begin{array}{ccc} \begin{array}{|r||c|} \hline x & r(x) = \frac{1}{x} \\ \hline -0.01& -100 \\ \hline -0.001 & -1000 \\ \hline -0.0001 & -10000 \\ \hline -0.00001 & -100000 \\ \hline 0 & \text{undefined} \\ \hline 0.00001 & 100000 \\ \hline 0.0001 & 10000 \\ \hline 0.001 & 1000 \\ \hline 0.01 & 100 \\ \hline \end{array} & \hspace{1in} & \begin{array}{|r||c|} \hline x & r(x) = \frac{1}{x} \\ \hline -100000 & -0.00001 \\ \hline -10000 & -0.0001 \\ \hline -1000 & -0.001 \\ \hline -100 & -0.01 \\ \hline 0 & \text{undefined} \\ \hline 100 & 0.01 \\ \hline 1000 & 0.001 \\ \hline 10000 & 0.0001 \\ \hline 100000 & 0.00001 \\ \hline \end{array} \end{array}$

A graph with a portion of the curve in the third quadrant and another portion of the graph in the first quadrant. The ends of each portion approach the axes, but never cross them. The points (-1,-1) and (1,1) were marked on the curves. — Graph of Laurent Monomial Function

In order to more precisely describe the behavior near $0$ , we say `as $x$ approaches 0 from the left,’ written as $x \rightarrow 0^{-}$ , the function values $r(x) \rightarrow -\infty$ . By `from the left’ we mean we are considering $x$ -values slightly to the left of $0$ on the number line, such as $x = -0.001$ and $x = -0.0001$ in the table above. If we think of these numbers as all being $x$ -values where $x =$ `0 $-$ a little bit’, then the ` $-$ ‘ in the notation ` $x \rightarrow 0^{-}$ ‘ makes better sense. The notation to describe the $r(x)$ values, $r(x) \rightarrow -\infty$ , is used here in the same manner as it was in Section 2.2. That is, as $x \rightarrow 0^{-}$ , the values $r(x)$ are becoming unbounded in the negative direction.

Similarly, we say `as $x$ approaches $0$ from the right,’ that is as $x \rightarrow 0^{+}$ , $r(x) \rightarrow \infty$ . Here `from the right’ means we are using $x$ values slightly to the right of $0$ on the number line: numbers such as $x =0.001$ which could be described as ` $0 +$ a little bit.’ For these values of $x$ , the values of $r(x)$ become unbounded (in the positive direction) so we write $r(x) \rightarrow \infty$ here.

We can also use this notation to describe the end behavior, but here the numerical roles are reversed. We see as $x \rightarrow -\infty$ , $r(x) \rightarrow 0^{-}$ and as $x \rightarrow \infty$ , $r(x) \rightarrow 0^{+}$ .

The way we describe what is happening graphically is to say the line $x = 0$ is a vertical asymptote of the graph of $y = r(x)$ an the line $y = 0$ is a horizontal asymptote of the graph of $y = r(x)$ . Roughly speaking, asymptotes are lines which approximate functions as either the inputs or outputs become unbounded. While defined more precisely using the language of Calculus, we will do our best to formally define vertical and horizontal asymptotes below.

Definition 3.3

The line $x=c$ is called a vertical asymptote of the graph of a function $y=f(x)$ if as $x \rightarrow c^{-}$ or as $x \rightarrow c^{+}$ , either $f(x) \rightarrow \infty$ or $f(x) \rightarrow -\infty$ .

Definition 3.4

The line $y=c$ is called a horizontal asymptote of the graph of a function $y=f(x)$ if as $x \rightarrow -\infty$ or as $x \rightarrow \infty$ , $f(x) \rightarrow c$ .

Note that in Definition3.4, we write $f(x) \rightarrow c$ (not $f(x) \rightarrow c^{+}$ or $f(x) \rightarrow c^{-}$ ) because we are unconcerned from which direction the values $f(x)$ approach the value $c$ , just as long as they do so. As we shall see, the graphs of rational functions may, in fact, cross their horizontal asymptotes. If this happens, however, it does so only a finite number of times (at least in this chapter), and so for each choice of $x \rightarrow -\infty$ and $x \rightarrow \infty$ , $f(x)$ will approach $c$ from either below (in the case $f(x) \rightarrow c^{-}$ ) or above (in the case $f(x) \rightarrow c^{+}$ .) We leave $f(x) \rightarrow c$ generic in our definition, however, to allow this concept to apply to less tame specimens in the Precalculus zoo, one that cross horizontal asymptotes an infinite number of times.^[3]

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 2.2 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 $(1,1)$ are common to all three graphs as are the asymptotes $x = 0$ and $y = 0$ . As the $n$ increases, the graphs become steeper for $|x| < 1$ and flatten out more quickly for $|x|>1$ . Both the domain and range in each case appears to be $(-\infty, 0) \cup (0, \infty)$ . Indeed, owing to the $x$ in the denominator of $f(x) = \frac{1}{x^n}$ , $f(0)$ , and only $f(0)$ , is undefined. Hence the domain is $(-\infty, 0) \cup (0, \infty)$ . 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 $c \neq 0$ . This means $f\left( \sqrt[n]{\frac{1}{c}} \right) = c$ for every nonzero real number $c$ . If $c = 0$ , we are in the same situation as before: $\frac{1}{x^n} = 0$ has no real solution. This establishes the range is $(-\infty, 0) \cup (0, \infty)$ . Finally, each of the graphs appear to be symmetric about the origin. Indeed, for odd $n$ , $f(-x) = (-x)^{-n} = (-1)^{-n} x^{-n} = -x^{-n} = -f(x)$ , proving every member of this function family is odd.

$\begin{array}{|r||c|c|c|} \hline x & \frac{1}{x} = x^{-1} & \frac{1}{x^3} = x^{-3} & \frac{1}{x^5} = x^{-5} \\ \hline -10 & -0.1 &-0.001& -0.00001 \\ \hline -1 & -1 & -1& -1\\ \hline -0.1 & -10 & -1000& -100000 \\ \hline 0 & \text{undefined} & \text{undefined} & \text{undefined} \\ \hline 0.1 & 10 & 1000& 100000 \\ \hline 1 & 1 & 1& 1\\ \hline 10 & 0.1 & 0.001& 0.00001 \\ \hline \end{array}$

Three graphs side by side. The first graph is a in the first and third quadrants and has a horizontal asymptote at y=0 and vertical asymptote at x=0. The graph is labeled f(x) =x^-1. The second graph is looks very similar to the first, but closer to the x-axis and further from the y -axis. The graph is labeled f(x)=x^-3. The last graph is again similar to a first graph, but even closer to the x-axis and further from the y-axis. The graph is labeled f(x) = x^-5. — Graphs of Laurent Monomials with Odd Negative Powers of x

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 $(-1,1)$ and $(1,1)$ , and asymptotes $x = 0$ and $y = 0$ . The same remarks about the steepness for $|x|<1$ and the flattening for $|x|>1$ also apply. For the same reasons as given above, the domain of each of these functions is $(-\infty, 0) \cup (0, \infty)$ . When it comes to the range, the fact $n$ is even tells us there are solutions to $\frac{1}{x^n} = c$ only if $c>0$ . It follows that the range is $(0, \infty)$ 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, all of the graphs of these functions is symmetric about the $y$ -axis.

$\begin{array}{|r||c|c|c|} \hline x & \frac{1}{x^2} = x^{-2} & \frac{1}{x^4} = x^{-4} & \frac{1}{x^6} = x^{-6} \\[2pt] \hline -10 &0.01 &0.0001& 1 \times 10^{-6} \\ \hline -1 & 1 & 1& 1\\ \hline -0.1 & 100 & 10000 & 1 \times 10^{6} \\ \hline 0 & \text{undefined} & \text{undefined} & \text{undefined} \\ \hline 0.1 & 100 & 10000 & 1 \times 10^{6} \\ \hline 1& 1 & 1& 1 \\ \hline 10 &0.01 &0.0001& 1 \times 10^{-6} \\ \hline \end{array}$

Three graphs side by side. The first graph is in the first and second quadrants and has a horizontal asymptote at y=0 and vertical asymptote at x=0. The graph is labeled f(x) =x^-2. The second graph is looks very similar to the first, but closer to the x-axis and further from the y -axis. The graph is labeled f(x)=x^-4. The last graph is again similar to a first graph, but even closer to the x-axis and further from the y-axis. The graph is labeled f(x) = x^-6. — Graphs of Laurent Monomials with Even Negative Powers of x

Not surprisingly, we have an analog to Theorem 2.2 for this family of Laurent monomial functions.

Theorem 3.1

For real numbers $a$ , $h$ , and $k$ with $a \neq 0$ , 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 $h > 0$ or left if $h < 0$ .
NOTE: This transforms the graph of $y = x^{-n}$ to $y = (x-h)^{-n}$ .
The vertical asymptote moves from $x=0$ 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 $a < 0$ .
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 $k > 0$ or down if $k< 0$ .
NOTE: This transforms the graph of $y = a(x-h)^{-n}$ to $y = a(x-h)^{-n}+k$ .
The horizontal asymptote moves from $y=0$ to $y=k$ .

The proof of Theorem 3.1 is identical to the proof of Theorem 2.2 – just replace $x^n$ with $x^{-n}$ . We nevertheless encourage the reader to work through the details^[4] and compare the results of this theorem with Theorems 1.4, 2.1, and 2.2.

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

Example 3.2.1

Example 3.2.1.1

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

$f(x) = (2x-3)^{-2}$

Solution:

Graph $f(x)=(2x-3)^{-2}$ .

In order to use Theorem 3.1, we first must put $f(x) = (2x-3)^{-2}$ into the form prescribed by the theorem. To that end, we factor:

$\begin{array}{rcl} f(x) &=& \left(2 \left[x - \frac{3}{2} \right] \right)^{-2} \\[6pt] &=& 2^{-2} \left(x - \frac{3}{2} \right)^{-2} \\[6pt] &=& \frac{1}{4} \left(x - \frac{3}{2} \right)^{-2} \end{array}$

We identify $n=2$ , $a=\frac{1}{4}$ and $h = \frac{3}{2}$ (and $k =0$ .) Per the theorem, we begin with the graph of $y = x^{-2}$ and track the two points $(-1,1)$ and $(1,1)$ along with the vertical and horizontal asymptotes $x = 0$ and $y=0$ , 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 $x = 0$ to $x = \frac{3}{2}$ (which we represent by a dashed line.)

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Step 2: multiply each of the $y$ -coordinates of each of the points on the graph of $f_1 = \left(x - \frac{3}{2} \right)^{-2}$ by $\frac{1}{4}$ .

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As we did not shift the graph vertically, the horizontal asymptote remains $y = 0$ . 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 $\left(-\infty, \frac{3}{2} \right) \cup \left(\frac{3}{2}, \infty \right)$ and the range of $f$ is $(-\infty, 0) \cup (0, \infty)$ .

Example 3.2.1.2

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

$g(t) = \dfrac{2t-1}{t+1}$

Solution:

Graph $g(t) = \dfrac{2t-1}{t+1}$ .

Using long division, we get

$\begin{array}{rcl} g(t) &=& \frac{2t-1}{t+1} \\[10pt] &=& - \frac{3}{t+1} + 2 \\[10pt] &=& \frac{-3}{(t-(-1))^{1}} + 2 \end{array}$

so we identify $n = 1$ , $a = -3$ , $h = -1$ , and $k = 2$ .

We start with the graph of $y = \frac{1}{t}$ with points $(-1,-1)$ , $(1,1)$ and asymptotes $t = 0$ and $y =0$ and track these through each of the steps.

Step 1: Add $-1$ 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 $t=0$ to $t = -1$ .

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Step 2: multiply each of the $y$ -coordinates of each of the points on the graph of $g_1 = \frac{1}{t+1}$ by $-3$ .

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Step 3: add $2$ to each of the $y$ -coordinates of each of the points on the graph of $g_2 = \frac{-3}{t+1}$ . This moves the horizontal asymptote from $y = 0$ to $y = 2$ .

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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 $(-\infty, 2) \cup (2, \infty)$ .

In Example 3.2.1, 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.

3.2.2 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.^[5]

Example 3.2.2

Example 3.2.2.1

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 $P(t) = \frac{100}{(5-t)^{2}}$ , $0 \leq t < 5$ .

Compute and interpret $P(0)$ .

Solution:

Compute and interpret $P(0)$ .

Substituting $t=0$ gives $P(0) = \frac{100}{(5-0)^2} = 4$ .

Due to the fact that $t$ represents the number of days after the bacteria are introduced into the environment, $t =0$ corresponds to the day the bacteria are introduced. $P(t)$ is measured in thousands, thus $P(t) = 4$ means $4000$ bacteria are initially introduced into the environment.

Example 3.2.2.2

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 $P(t) = \frac{100}{(5-t)^{2}}$ , $0 \leq t < 5$ .

When will the population reach 100,000?

Solution:

When will the population reach 100,000?

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 $P(t) = 100$ . 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 $t = 4$ or $t=6$ . Of these two solutions, only $t=4$ in our domain, so this is the solution we keep.

Hence, it takes 4 days for the population of bacteria to reach 100,000.

Example 3.2.2.3

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 $P(t) = \frac{100}{(5-t)^{2}}$ , $0 \leq t < 5$ .

Graph $y = P(t)$ .

Solution:

Graph $y = P(t) =\frac{100}{(5-t)^{2}}$ . \vskip 0.15em

After a slight re-write, we have

$\begin{array}{rcl} P(t) &=& \frac{100}{(5-t)^2} \\[8pt] &=& \frac{100}{[(-1)(t-5)]^2} \\[8pt] &=& \frac{100}{(t-5)^2} \end{array}$

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 $0 \leq t < 5$ to arrive at the graph of $y = P(t)$ below on the right.

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Example 3.2.2.4

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 $P(t) = \frac{100}{(5-t)^{2}}$ , $0 \leq t < 5$ .

Determine and interpret the behavior of $P$ as $t \rightarrow 5^{-}$ .

Solution:

Determine and interpret the behavior of $P$ as $t \rightarrow 5^{-}$ .

We see that as $t \rightarrow 5^{-}$ , $P(t) \rightarrow \infty$ .

This means that the population of bacteria is increasing without bound as we near 5 days, which cannot actually happen. For this reason, $t=5$ is called the `doomsday’ for this population. There is no way any environment can support infinitely many bacteria, so shortly before $t = 5$ days the environment would collapse.

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 1.3.1.2 in Section 1.3.1. In that example, the function $s(t) = -5t^2+100t$ , $0 \leq t \leq 20$ gives the height of a model rocket above the Moon’s surface, in feet, $t$ seconds after liftoff. The function $s$ an example of a position function as 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 $t_{0}$ . We can define the average velocity as a function of any time $t$ other than $t_{0}$ :

Definition 3.5

Suppose $s(t)$ gives the position of an object at time $t$ and $t_{0}$ is a fixed time in the domain of $s$ . The average velocity between time $t$ and time $t_{0}$ is given by

$\overline{v}(t) = \dfrac{\Delta [s(t)]}{\Delta t} = \dfrac{s(t) - s(t_{0})}{t - t_{0}},$

provided $t \neq t_{0}$ .

It is clear why we must exclude $t = t_{0}$ from the domain of $\overline{v}$ in Definition 3.5 because otherwise we would have a $0$ in the denominator. What is interesting in this case however, is that substituting $t = t_{0}$ 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 can nevertheless explore this function in the next example.

Example 3.2.3

Example 3.2.3.1

Let $s(t) = -5t^2+100t$ , $0 \leq t \leq 20$ give the height of a model rocket above the Moon’s surface, in feet, $t$ seconds after liftoff.

Identify and simplify an expression for the average velocity of the rocket between times $t$ and $15$ .

Solution:

Identify and simplify an expression for the average velocity of the rocket between times $t$ and $15$ .

Using Definition 3.5 with $t_{0} = 15$ , we get:

$\begin{array}{rclr} \overline{v}(t) & = & \dfrac{s(t) - s(15)}{t-15} \\[10pt] & = & \dfrac{(-5t^2+100t) - 375}{t - 15} \\[10pt] & = & \dfrac{-5(t^2-20t+75)}{t - 15} \\[10pt] & = & \dfrac{-5(t-15)(t-5)}{t-15} \\[10pt] & = & \dfrac{-5\cancel{(t-15)} (t-5)}{\cancel{(t-15)}} \\[10pt] & = & -5(t-5) = -5t + 25 & \; \; t \neq 15 \end{array}$

The domain of $s$ is $0 \leq t \leq 20$ , thus our final answer is $\overline{v}(t) = -5t+25$ , for $t \in [0, 15) \cup (15, 20]$ .

Example 3.2.3.2

Let $s(t) = -5t^2+100t$ , $0 \leq t \leq 20$ give the height of a model rocket above the Moon’s surface, in feet, $t$ seconds after liftoff.

Compute and interpret $\overline{v}(14)$ .

Solution:

Compute and interpret $\overline{v}(14)$ .

We find $\overline{v}(14) = -5(14)+25 = -45$ .

This means 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.

Example 3.2.3.3

Let $s(t) = -5t^2+100t$ , $0 \leq t \leq 20$ give the height of a model rocket above the Moon’s surface, in feet, $t$ seconds after liftoff.

Graph $y = \overline{v}(t)$ . Interpret the intercepts.

Solution:

Graph $y = \overline{v}(t)$ . Interpret the intercepts.

The graph of $\overline{v}(t)$ is a portion of the line $y= -5t+25$ . The domain of $s$ is $[0, 20]$ and $\overline{v}(t)$ is not defined when $t = 15$ , therefore our graph is the line segment starting at $(0, 25)$ and ending at $(20, 75)$ with a hole at $(15, 50)$ .

The $y$ -intercept is $(0,25)$ which means on average, the rocket is traveling $25$ feet per second upwards.^[6]

To get the $t$ -intercept, we set $\overline{v}(t) = -5t+25 = 0$ and obtain $t = 5$ .

Hence, $\overline{v}(5) = 0$ or the average velocity between times $t = 5$ and $t = 15$ is $0$ . As you may recall, this is due to the rocket being at the same altitude ( $375$ feet) at both times, hence, $\Delta [s(t)]$ and, hence $\overline{v}(t) = 0$ .

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Example 3.2.3.4

Let $s(t) = -5t^2+100t$ , $0 \leq t \leq 20$ give the height of a model rocket above the Moon’s surface, in feet, $t$ seconds after liftoff.

Interpret the behavior of $\overline{v}$ as $t \rightarrow 15$ .

Solution:

Interpret the behavior of $\overline{v}$ as $t \rightarrow 15$ . \vskip 0.15em

From the graph, we see as $t \rightarrow 15$ , $\overline{v}(t) \rightarrow -50$ . (This is also borne out in the numerically in the included tables.)

This means as we sample the average velocity between time $t_{0} = 15$ and and times closer and closer to $15$ , the average velocity approaches $-50$ . This value is how we define the instantaneous velocity – that is, at $t=15$ seconds, the rocket is falling at a rate of $50$ feet per second towards the surface of the Moon.

$\begin{array}{|r||c|} \hline t & \overline{v}(t) \\ \hline 14.9 & -49.5 \\ \hline 14.99 & -49.95 \\ \hline 14.999 & -49.995 \\ \hline 15 & \text{undefined} \\ \hline 15.001 & -50.005 \\ \hline 15.01 & -50.05 \\ \hline 15.1 & -50.5 \\ \hline \end{array}$

If nothing else, Example 3.2.3 shows us that just because a value is excluded from the domain of a rational function doesn’t mean there will be a vertical asymptote to the graph there. In this case, the factor $(t-15)$ divided out from the denominator, thereby effectively removing the threat of dividing by $0$ . It turns out, this situation generalizes to the theorem below.

Theorem 3.2 Location of Vertical Asymptotes and Holes^[7]

Suppose $r$ is a rational function which can be written as $r(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ have no common zeros.^[8] Let $c$ be a real number which is not in the domain of $r$ .

If $q(c) \neq 0$ , then the graph of $y=r(x)$ has a hole at $\left(c, \frac{p(c)}{q(c)}\right)$
If $q(c) = 0$ , then the line $x=c$ is a vertical asymptote of the graph of $y=r(x)$ .

In English, Theorem 3.2 says that if $x=c$ is not in the domain of $r$ but, when we simplify $r(x)$ , it no longer makes the denominator $0$ , then we have a hole at $x=c$ . Otherwise, the line $x=c$ is a vertical asymptote of the graph of $y=r(x)$ . Like many properties of rational functions, we owe Theorem 3.2 to Calculus, but that won’t stop us from putting Theorem 3.2 to good use in the following example.

Example 3.2.4

Example 3.2.4.1

For each function below:

$f(x) = \dfrac{2x}{x^2-3}$

determine the values excluded from the domain.
determine whether each excluded value corresponds to a vertical asymptote or hole of the graph.
verify your answers with a rough sketch of a graph.
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.

Solution:

Determine the vertical behavior of $f(x) = \dfrac{2x}{x^2-3}$ .

To use Theorem 3.2, we first find all of the real numbers which aren’t in the domain of $f$ . To do so, we solve $x^2 - 3 = 0$ and get $x = \pm \sqrt{3}$ .

Because the expression $f(x)$ is in lowest terms (can you see why?), there is no division of like factors possible, and we conclude that the lines $x = -\sqrt{3}$ and $x=\sqrt{3}$ are vertical asymptotes to the graph of $y=f(x)$ . The graph verifies this claim, and from the graph, we see that as $x \rightarrow -\sqrt{3}^{\, -}$ , $f(x) \rightarrow -\infty$ , as $x\rightarrow -\sqrt{3}^{\, +}$ , $f(x) \rightarrow \infty$ , as $x \rightarrow \sqrt{3}^{\, -}$ , $f(x) \rightarrow -\infty$ , and finally as $x\rightarrow \sqrt{3}^{\, +}$ , $f(x) \rightarrow \infty$ .

As a side note, the graph of $f$ appears to be symmetric about the origin. Sure enough, we find: $f(-x) = \frac{2(-x)}{(-x)^2-3} = -\frac{2x}{x^2-3} = -f(x)$ , proving $f$ is odd.

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Example 3.2.4.2

For each function below:

$g(t) = \dfrac{t^2-t-6}{t^2-9}$

determine the values excluded from the domain.
determine whether each excluded value corresponds to a vertical asymptote or hole of the graph.
verify your answers with a rough sketch of a graph.
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.

Solution:

Determine the vertical behavior of $g(t) = \dfrac{t^2-t-6}{t^2-9}$ .

As above, we find the values excluded from the domain of $g$ by setting the denominator equal to $0$ . Solving $t^2 - 9 = 0$ gives $t = \pm 3$ .

In lowest terms

$\begin{array}{rcl} g(t) &=& \frac{t^2-t-6}{t^2-9} \\ &=& \frac{(t-3)(t+2)}{(t-3)(t+3)} \\ &=& \frac{t+2}{t+3} \end{array}$

Because $t=-3$ continues to be a zero of the denominator in the reduced formula, we know the line $t=-3$ is a vertical asymptote to the graph of $y=g(t)$ .

AS $t=3$ does not produce a ` $0$ ‘ in the denominator of the reduced formula, we have a hole at $t=3$ .

To find the $y$ -coordinate of the hole, we substitute $t=3$ into the reduced formula: $\frac{t+2}{t+3} = \frac{3+2}{3+3} = \frac{5}{6}$ so the hole is at $\left(3, \frac{5}{6}\right)$ .

Graphing $g$ we can definitely see the vertical asymptote $t=-3$ : as $t \rightarrow -3^{-}$ , $g(t) \rightarrow \infty$ and as $t \rightarrow -3^{+}$ , $g(t) \rightarrow -\infty$ . Near $t=3$ , the graph seems to have no interruptions (but we know $g$ is undefined at $t=3$ .) As $g$ appears to be increasing on $(-3, \infty)$ , we write as $t\rightarrow 3^{-}$ , $g(t) \rightarrow \frac{5}{6}^{-}$ , and as $t \rightarrow 3^{+}$ , $g(t) \rightarrow \frac{5}{6}^{+}$ .

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Example 3.2.4.3

For each function below:

$h(t) = \dfrac{t^2-t-6}{t^2+9}$

determine the values excluded from the domain.
determine whether each excluded value corresponds to a vertical asymptote or hole of the graph.
verify your answers with a rough sketch of a graph.
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.

Solution:

Determine the vertical behavior of $h(t) = \dfrac{t^2-t-6}{t^2+9}$ .

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.

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Example 3.2.4.4

For each function below:

$r(t) = \dfrac{t^2-t-6}{t^2+6t+9}$

determine the values excluded from the domain.
determine whether each excluded value corresponds to a vertical asymptote or hole of the graph.
verify your answers with a rough sketch of a graph.
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.

Solution:

Determine the vertical behavior of $r(t) = \dfrac{t^2-t-6}{t^2+6t+9}$ .

Setting the denominator of $r(t)$ to zero gives the equation $t^2+6t+9 = 0$ . We get the (repeated!) solution $t=-3$ .

Simplifying, we see

$\begin{array}{rcl} r(t) &=& \frac{t^2-t-6}{t^2+6t+9}\\ &=& \frac{(t-3)(t+3)}{(t+3)^2}\\ &=& \frac{t-3}{t+3} \end{array}$

Because $t=-3$ continues to produce a $0$ in the denominator of the reduced function, we know $t=-3$ is a vertical asymptote to the graph. The calculator bears this out, and, moreover, we see that as $t \rightarrow -3^{-}$ , $r(t) \rightarrow \infty$ and as $t \rightarrow -3^{+}$ , $r(t) \rightarrow -\infty$ .

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3.2.3 End Behavior

Now that we’ve thoroughly discussed behavior near values excluded from the domains of rational functions, 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.^[9]

Example 3.2.4

Example 3.2.5.1

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

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

Compute and interpret $N(0)$ .

Solution:

Compute and interpret $N(0)$ .

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

As $t$ represents the number of months since the beginning of the semester, $t=0$ 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.

Example 3.2.5.2

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

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

How long will it take until $300$ students will have had the flu?

Solution:

How long will it take until $300$ students will have had the flu?

We set $N(t) = \frac{1500t + 50}{3t+1} = 300$ and solve.

Clearing denominators gives $1500t + 50 = 300(3t+1)$ 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.

Example 3.2.5.3

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

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

Use Theorem 3.1 to graph $y = N(t)$ .

Solution:

Use Theorem 3.1 to graph $y = N(t)$ .

To graph $y = N(t)$ , we first use long division to rewrite $N(t) = \frac{-450}{3t+1} + 500$ . From there, we get

$\begin{array}{rcl} N(t) &=& -\frac{450}{3t+1} + 500 \\[6pt] &=& \frac{-450}{3\left(t + \frac{1}{3}\right)} + 500 \\[6pt] &=& \frac{-150}{t+\frac{1}{3}} + 500 \end{array}$

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 $t \geq 0$ , we obtain the graph below on the right.

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Example 3.2.5.4

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

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

Determine and interpret the behavior of $N$ as $t \rightarrow \infty$ .

Solution:

Determine and interpret the behavior of $N$ as $t \rightarrow \infty$ .

From the graph, we see as $t \rightarrow \infty$ , $N(t) \rightarrow 500$ . (More specifically, $500^{-}$ .) 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 $y = N(t)$ 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 2.4 from Section 2.2.

The end behavior of the numerator of $N(t) = \frac{1500t + 50}{3t+1}$ is determined by its leading term, $1500t$ , and the end behavior of the denominator is likewise determined by its leading term, $3t$ . Hence, as $t \rightarrow \pm \infty$ ,

$\begin{array}{rcl} N(t) &=& \dfrac{1500t + 50}{3t+1} \\[6pt] &\approx & \dfrac{1500t}{3t} \\[6pt] &=& 500 \end{array}$

Hence as $t \rightarrow \pm \infty$ , $y = N(t) \rightarrow 500$ , producing the horizontal asymptote $y = 500$ .

This same reasoning can be used in general to argue the following theorem.

Theorem 3.3 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 $y=\frac{a}{b}$ is the^[10] horizontal asymptote of the graph of $y=r(x)$ .
If the degree of $p(x)$ is less than the degree of $q(x)$ , then $y=0$ 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 $x \rightarrow \pm \infty$ , Theorem 2.4 gives $r(x) \approx \frac{ax^n}{bx^m}$ , 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 $r(x) \approx \frac{ax^n}{bx^n} = \frac{a}{b}$ , 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, $r(x) \approx \frac{ax^n}{bx^m} = \frac{a}{bx^{m-n}} \rightarrow 0$ . As $x \rightarrow \pm \infty$ , $r(x)$ is more or less a fraction with a constant numerator, $a$ , but a denominator which is unbounded. Hence, $r(x) \rightarrow 0$ producing the horizontal asymptote $y = 0$ .

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 $r(x) \approx \frac{ax^n}{bx^m} = \frac{ax^{n-m}}{ b}$ , which is a monomial function from Section 2.2. As such, $r$ becomes unbounded as $x \rightarrow \pm \infty$ .

Note that in the two cases which produce horizontal asymptotes, the behavior of $r$ is identical as $x \rightarrow \infty$ and $x \rightarrow -\infty$ . Hence, if the graph of a rational function has a horizontal asymptote, there is only one.^[11]

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

Example 3.2.6

Example 3.2.6.1

For each function below:

$F(s) = \dfrac{5s}{s^2+1}$

use Theorem 2.4 to analytically determine the horizontal asymptotes of the graph, if any.
check your answers using Theorem 3.3 and a graph.
describe the end behavior of the graph using proper notation.
investigate any apparent symmetry of the graph about the $y$ -axis or origin.

Solution:

Determine the horizontal behavior of $F(s) = \dfrac{5s}{s^2+1}$ .

Using Theorem 2.4, we get as $s \pm \infty$ ,

$\begin{array}{rcl} F(s) &=& \frac{5s}{s^2+1}\\ [6pt] &\approx& \frac{5s}{s^2} \\[6pt] &=& \frac{5}{s} \end{array}$

Hence, as $s \rightarrow \infty$ , $F(s) \rightarrow 0$ , so $y = 0$ is a horizontal asymptote to the graph of $y = F(s)$ .

To check, we note that the degree of the numerator of $F(s)$ , $1$ , is less than the degree of the denominator, $2$ , so Theorem 3.3 gives $y=0$ is the horizontal asymptote. Graphically, we see as $s \rightarrow \pm \infty$ , $F(s) \rightarrow 0$ . More specifically, as $s \rightarrow -\infty$ , $F(s) \rightarrow 0^{-}$ and as $s \rightarrow \infty$ , $F(s) \rightarrow 0^{+}$ .

As a side note, the graph of $F$ appears to be symmetric about the origin. Indeed, $F(-s) = \frac{5(-s)}{(-s)^2+1} = -\frac{5s}{s^2+1}$ proving $F$ is odd.

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Example 3.2.6.2

For each function below:

$g(x) = \dfrac{x^2-4}{x+1}$

use Theorem 2.4 to analytically determine the horizontal asymptotes of the graph, if any.
check your answers using Theorem 3.3 and a graph.
describe the end behavior of the graph using proper notation.
investigate any apparent symmetry of the graph about the $y$ -axis or origin.

Solution:

Determine the horizontal behavior of $g(x) = \dfrac{x^2-4}{x+1}$ . \vskip 0.15em

As $x \rightarrow \pm \infty$ ,

$\begin{array}{rcl} g(x) &=& \frac{x^2-4}{x+1} \\[6pt] &\approx & \frac{x^2}{x} \\[6pt] &=& x \end{array}$

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 as the degree of the numerator of $g(x)$ is $2$ which is greater than the degree of the denominator, $1$ . By, there is no horizontal asymptote. From the graph, we see that the graph of $y=g(x)$ doesn’t appear to level off to a constant value, so there is no horizontal asymptote.^[12]

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Example 3.2.6.3

For each function below:

$h(t) = \dfrac{6t^3-3t+1}{5-2t^3}$

use Theorem 2.4 to analytically determine the horizontal asymptotes of the graph, if any.
check your answers using Theorem 3.3 and a graph.
describe the end behavior of the graph using proper notation.
investigate any apparent symmetry of the graph about the $y$ -axis or origin.

Solution:

Determine the horizontal behavior of $h(t) = \dfrac{6t^3-3t+1}{5-2t^3}$ .

We have

$\begin{array}{rcl} h(t) &=& \frac{6t^3-3t+1}{5-2t^3}\\[6pt] &\approx & \frac{6t^3}{-2t^3} \\[6pt] &=& -3\end{array}$

as $t \rightarrow \pm \infty$ , indicating a horizontal asymptote $y = -3$ .

Sure enough, the degrees of the numerator and denominator of $h(t)$ are both three, so 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 $t \rightarrow -\infty$ , $h(t) \rightarrow -3^{+}$ , and as $t \rightarrow \infty$ , $h(t) \rightarrow -3^{-}$ .

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Example 3.2.6.4

For each function below:

$r(x) = 2 - \dfrac{3x^2}{1-x^2}$

use Theorem 2.4 to analytically determine the horizontal asymptotes of the graph, if any.
check your answers using Theorem 3.3 and a graph.
describe the end behavior of the graph using proper notation.
investigate any apparent symmetry of the graph about the $y$ -axis or origin.

Solution:

Determine the horizontal behavior of $r(x) = 2 - \dfrac{3x^2}{1-x^2}$ .

If we apply Theorem 2.4 to the $\frac{3x^2}{1-x^2}$ term in the expression for $r(x)$ , we find

$\begin{array}{rcl} \frac{3x^2}{1-x^2} &\approx & \frac{3x^2}{-x^2} \\[6pt] &=& -3 \end{array}$

as $x \rightarrow \pm \infty$ . It seems reasonable to conclude, then, that

$\begin{array}{rcl} r(x) &=& 2 - \frac{3x^2}{1-x^2} \\[6pt] &\approx & 2 - (-3) \\[6pt] &=& 5 \end{array}$

so $y = 5$ is our horizontal asymptote.

In order to use Theorem 3.3 as stated, however, we need to rewrite the expression $r(x)$ with a single denominator: $r(x) = 2 - \frac{3x^2}{1-x^2} = \frac{2(1-x^2) - 3x^2}{1-x^2} = \frac{2-5x^2}{1-x^2}$ . Now we apply Theorem 3.3 and note the numerator and denominator have the same degree, we guarantee the horizontal asymptote is $y = \frac{-5}{-1} = 5$ . These calculations are borne out graphically where it appears as if as $x \rightarrow \pm \infty$ , $r(x) \rightarrow 5^{+}$ .

As a final note, the graph of $r$ appears to be symmetric about the $y$ -axis. We find $r(-x) = 2 - \frac{3(-x)^2}{1-(-x)^2} = 2 - \frac{3x^2}{1-x^2} = r(x)$ , proving $r$ is even.

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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 3.2.6.2. Performing long division,^[13] we get $g(x) = \frac{x^2-4}{x+1} = x-1 - \frac{3}{x+1}$ . Because the term $\frac{3}{x+1} \rightarrow 0$ as $x \rightarrow \pm \infty$ , 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 $x \rightarrow \pm \infty$ . We see this play out both numerically and graphically below. (As usual, the asymptote $y = x-1$ is denoted by a dashed line.)

$\begin{array}{ccc} \begin{array}{|r||c|c|} \hline x & g(x) & x-1 \\ \hline -10 & \approx -10.6667 & -11 \\ \hline -100 & \approx -100.9697 & -101 \\ \hline -1000 & \approx -1000.9970& -1001 \\ \hline -10000 & \approx -10000.9997 & -10001 \\ \hline \end{array} & \qquad \qquad & \begin{array}{|r||c|c|} \hline x & g(x) & x-1 \\ \hline 10 & \approx 8.7273 & 9 \\\hline 100 & \approx 98.9703 & 99 \\ \hline 1000 & \approx 998.9970 & 999 \\ \hline 10000 & \approx 9998.9997 & 9999 \\ \hline \end{array}\\ \end{array}$

A graph of a rational function with a numerator of x^2 - 4 and a denominator of x+1. The curve crosses the x-axis at x = -2 and x = 2. The graph crosses the y-axis at y = -4. The graph includes the slant asymptote (dashed line) y = x-1. The graph of the function approaches the dashed line as x goes to positive and negative infinity. — A Rational Function with a Slant Asymptote

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 $x \rightarrow \pm \infty$ , $g(x) \rightarrow x-1$ ‘ in order to have some notational consistency with what we have done earlier in this section when it comes to end behavior.^[14] In this case, we say the line $y=x-1$ is a slant asymptote^[15] of the graph of $y=g(x)$ . Informally, the graph of a rational function has a slant asymptote if, as $x \rightarrow \infty$ or as $x \rightarrow -\infty$ , the graph resembles a non-horizontal, or `slanted’ line. Formally, we define a slant asymptote as follows.

Definition 3.6

The line $y = mx+b$ where $m \neq 0$ is called a slant asymptote of the graph of a function $y=f(x)$ if as $x \rightarrow -\infty$ or as $x \rightarrow \infty$ , $f(x) \rightarrow mx+b$ .

A few remarks are in order. First, note that the stipulation $m \neq 0$ in Definition 3.6 is what makes the `slant’ asymptote `slanted’ as opposed to the case when $m=0$ in which case we’d have a horizontal asymptote. Secondly, while we have motivated what me mean intuitively by the notation ` $f(x) \rightarrow mx+b$ ,’ like so many ideas in this section, the formal definition requires Calculus. Another way to express this sentiment, however, is to rephrase ` $f(x) \rightarrow mx+b$ ‘ as ` $[f(x) - (mx+b)] \rightarrow 0$ .’ In other words, the graph of $y=f(x)$ has the slant asymptote $y = mx+b$ if and only if the graph of $y = f(x) - (mx+b)$ has a horizontal asymptote $y=0$ . If we wanted to, we could introduce the notations $f(x) \rightarrow (mx+b)^{+}$ to mean $[f(x)-(mx+b)] \rightarrow 0^{+}$ and $f(x) \rightarrow (mx+b)^{-}$ to mean $[f(x)-(mx+b)] \rightarrow 0^{-}$ , but these non-standard notations.

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 $x^2-4$ is $2$ , which is exactly one more than the degree if its denominator $x+1$ 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.

Theorem 3.4 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.^[16]

Example 3.2.7

Example 3.2.7.1

For each function below:

$f(x) = \dfrac{x^2-4x+2}{1-x}$

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

Solution:

Determine any slant asymptotes of $f(x) = \dfrac{x^2-4x+2}{1-x}$ .

The degree of the numerator is $2$ and the degree of the denominator is $1$ , so Theorem 3.4 guarantees a slant asymptote exists. To find it, we divide $1-x = -x+1$ into $x^2-4x+2$ and get a quotient of $-x+3$ , so our slant asymptote is $y=-x+3$ .

We confirm this graphically below.

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Example 3.2.7.2

For each function below:

$g(t) = \dfrac{t^2-4}{t-2}$

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

Solution:

Determine any slant asymptotes of $g(t) = \dfrac{t^2-4}{t-2}$ .

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,

$\begin{array}{rcl} g(t) &=& \frac{t^2-4}{t-2} \\[6pt] &=& \frac{(t+2)(t-2)}{(t-2)} \\[6pt] &=& \frac{(t+2) \cancel{(t-2)}}{\cancelto{1}{(t-2)}} \\[6pt] &=& t+2, \quad t \neq 2 \end{array}$

so we have that the slant asymptote, $y=t+2$ , is identical to the graph of $y=g(t)$ except at $t=2$ (where the latter has a `hole’ at $(2,4)$ .) While the word `asymptote’ has the connotation of `approaching but not equaling,’ Definitions 3.6 and 3.4 allow for these extreme cases.

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Example 3.2.7.3

For each function below:

$h(x) = \dfrac{x^3+1}{x^2-4}$

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

Solution:

Determine any slant asymptotes of $h(x) = \dfrac{x^3+1}{x^2-4}$ .

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 $\left(x^3+1 \right) \div \left(x^2-4\right)$ gives a quotient of just $x$ , so our slant asymptote is the line $y=x$ .

The graph 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 $x=1$ , 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.

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Example 3.2.7.4

For each function below:

$r(t) = 2t-1+\dfrac{4t^3}{1-t^2}$

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

Solution:

Determine any slant asymptotes of $r(t) = 2t-1+\dfrac{4t^3}{1-t^2}$ . \vskip 0.15em

For our last example, $r(t) = 2t-1+\frac{4t^3}{1-t^2}$ , 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 $\frac{4t^3}{1-t^2} = -4t + \frac{4t}{1-t^2}$ . We then get:

$\begin{array}{rcl} r(t) & = & 2t-1+\dfrac{4t^3}{1-t^2} \\ &= & 2t-1-4t+\dfrac{4t}{1-t^2} \\ & = & -2t-1 + \dfrac{4t}{1-t^2} \\ \end{array}$

As $t \rightarrow \pm \infty$ , Theorem 2.4 gives $\frac{4t}{1-t^2} \approx \frac{4t}{-t^2} = -\frac{4}{t} \rightarrow 0$ . Hence, as $t \rightarrow \pm \infty$ , $r(t) \rightarrow -2t-1$ , so $y = -2t-1$ is the slant asymptote to the graph as confirmed by the graph 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 $(0,-1)$ , as borne out by the computation $r(0) = -1$ . Hence $r$ cannot be odd. (Do you see why?)

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Our last example gives a real-world application of a slant asymptote. The problem features the concept of average profit. The average profit, denoted $\overline{P}(x)$ , 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 3.7

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

The average cost, $\overline{C}(x) = \dfrac{C(x)}{x}$ , $x > 0$ .
NOTE: The average cost is the cost per item produced.
The average profit, $\overline{P}(x) = \dfrac{P(x)}{x}$ , $x > 0$ .
NOTE:} The average profit is the profit per item sold.

You’ll explore average cost (and its relation to variable cost) in Exercise 37. For now, we refer the reader to to Example 2.1.3 in Section 2.1.

Example 3.2.8

Example 3.2.8.1

Recall the profit (in dollars) when $x$ PortaBoy game systems are produced and sold is given by $P(x) = -1.5x^2+170x-150$ , $0 \leq x \leq 166$ .

Determine and simplify an expression for the average profit, $\overline{P}(x)$ . What is the domain of $\overline{P}$ ?

Solution:

Determine and simplify an expression for the average profit, $\overline{P}(x)$ . What is the domain of $\overline{P}$ ?

We find

$\begin{array}{rcl} \overline{P}(x) &=& \frac{P(x)}{x} \\[6pt] &=& \frac{ -1.5x^2+170x-150}{x}\\[6pt] &=& -1.5x + 170 - \frac{150}{x} \end{array}$

While the domain of $P$ is $[0, 166]$ , $x \neq 0$ , thus the domain of $\overline{P}$ is $(0, 166]$ .

Example 3.2.8.2

Recall the profit (in dollars) when $x$ PortaBoy game systems are produced and sold is given by $P(x) = -1.5x^2+170x-150$ , $0 \leq x \leq 166$ .

Compute and interpret $\overline{P}(50)$ .

Solution:

Compute and interpret $\overline{P}(50)$ .

We find $\overline{P}(50) = -1.5(50)+170 - \frac{150}{50} = 92$ .

This means that when 50 PortaBoy systems are sold, the average profit is 92 dollars per system.

Example 3.2.8.3

Recall the profit (in dollars) when $x$ PortaBoy game systems are produced and sold is given by $P(x) = -1.5x^2+170x-150$ , $0 \leq x \leq 166$ .

Determine the slant asymptote to the graph of $y = \overline{P}(x)$ . Check your answer using a graph.

Solution:

Determine the slant asymptote to the graph of $y = \overline{P}(x)$ . Check your answer using a graph.

Technically, the graph of $y = \overline{P}(x)$ has no slant asymptote because the domain of the function is restricted to $(0, 166]$ .

That being said, if we were to let $x \rightarrow \infty$ , the term $\frac{150}{x} \rightarrow 0$ , so we’d have $\overline{P}(x) \rightarrow -1.5x + 170$ . This means the slant asymptote would be $y = -1.5x + 170$ .

We graph $y = \overline{P}(x)$ and $y = -1.5x+170$ .

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Example 3.2.8.4

Recall the profit (in dollars) when $x$ PortaBoy game systems are produced and sold is given by $P(x) = -1.5x^2+170x-150$ , $0 \leq x \leq 166$ .

Interpret the slope of the slant asymptote.

Solution:

Interpret the slope of the slant asymptote.

The slope of the slant asymptote $y = -1.5x+170$ is $-1.5$ . Because, ostensibly $\overline{P}(x) \approx -1.5 x + 170$ , this means that, as we sell more systems, the average profit is decreasing at about a rate of 1.50 dollars per system.

If the number 1.5 sounds familiar to this problem situation, it should. In Example 1.3.9 in Section 1.3.1, we determined the slope of the demand function to be $-1.5$ . In that situation, the $-1.5$ meant that in order to sell an additional system, the price had to drop by 1.50 dollars. 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.^[17]

3.2.4 Section Exercises

(Review of Long Division):^[18] In Exercises 1 – 6, use polynomial long division to perform the indicated division. Write the polynomial in the form $p(x) = d(x)q(x) + r(x)$ .

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

In Exercises 7 – 10, 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.

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

In Exercises 11 – 12, find a formula for each function below in the form $F(x) = \dfrac{a}{x-h}+k$ .

$x$ -intercept $(-1,0)$ , $y$ -intercept $\left(0, -\frac{1}{2} \right)$
$x$ -intercept $(3,0)$ , $y$ -intercept $\left(0, 3 \right)$

In Exercises 13 – 14, find a formula for each function below in the form $F(x) = \dfrac{a}{(x-h)^2}+k$ .

$x$ -intercepts $(-3,0)$ , $(-1,0)$ , $y$ -intercept $\left(0, 3 \right)$
$x$ -intercepts $(0,0)$ , $(1,0)$ , vertical asymptote: $x=\frac{1}{2}$

In Exercises 15 – 32, for the given rational function:

State the domain.
Identify any vertical asymptotes of the graph.
Identify any holes in the graph.
Determine the horizontal asymptote, if it exists.
Determine the slant asymptote, if it exists.
Graph the function and describe the behavior near the asymptotes.

1. $f(x) = \dfrac{x}{3x - 6}$
2. $f(x) = \dfrac{3 + 7x}{5 - 2x}$
3. $f(x) = \dfrac{x}{x^{2} + x - 12}$
4. $g(t) = \dfrac{t}{t^{2} + 1}$
5. $g(t) = \dfrac{t + 7}{(t + 3)^{2}}$
6. $g(t) = \dfrac{t^{3} + 1}{t^{2} - 1}$
7. $r(z) = \dfrac{4z}{z^2+4}$
8. $r(z) = \dfrac{4z}{z^2-4}$
9. $r(z) = \dfrac{z^2-z-12}{z^2+z-6}$
10. $f(x) = \dfrac{3x^2-5x-2}{x^2-9}$
11. $f(x) = \dfrac{x^3+2x^2+x}{x^2-x-2}$
12. $f(x) = \dfrac{x^{3} - 3x + 1}{x^{2} + 1}$
13. $g(t) = \dfrac{2t^{2} + 5t - 3}{3t + 2}$
14. $g(t) = \dfrac{-t^{3} + 4t}{t^{2} - 9}$
15. $g(t) = \dfrac{-5t^{4} - 3t^{3} + t^{2} - 10}{t^{3} - 3t^{2} + 3t - 1}$
16. $r(z) = \dfrac{z^3}{1-z}$
17. $r(z) = \dfrac{18-2z^2}{z^2-9}$
18. $r(z) = \dfrac{z^3-4z^2-4z-5}{z^2+z+1}$
19. The cost in dollars to remove \% of the invasive Ippizuti fish species from Sasquatch Pond is:
  $C(p) = \frac{1770p}{100 - p}, \quad 0 \leq p < 100$
  1. Compute and interpret $C(25)$ and $C(95)$ .
  2. What does the vertical asymptote at $x = 100$ mean within the context of the problem?
  3. What percentage of the Ippizuti fish can you remove for 40000 dollars?
20. In the scenario of Example 3.2.3, , gives the height of a model rocket above the Moon’s surface, in feet, seconds after liftoff. For each of the times listed below, find and simplify a the formula for the average velocity between and (see Definition 3.5) and use to find and interpret the instantaneous velocity of the rocket at .
  1. $t_{0} = 5$
  2. $t_{0} = 9$
  3. $t_{0} = 10$
  4. $t_{0} = 11$
21. The population of Sasquatch in Portage County $t$ years after the year 1803 is modeled by the function
  $P(t) = \frac{150t}{t + 15}.$
  
  Find and interpret the horizontal asymptote of the graph of $y = P(t)$ and explain what it means.
22. The cost in dollars, to make dOpi media players is , . You may wish to review the concepts of fixed and variable costs introduced in Example 1.3.8 in Section 1.3.3.
  1. Write a formula for the average cost $\overline{C}(x)$ .
  2. Compute and interpret $\overline{C}(1)$ and $\overline{C}(100)$ .
  3. How many dOpis need to be produced so that the average cost per dOpi is 200 dollars?
  4. Interpret the behavior of $\overline{C}(x)$ as $x \rightarrow 0^{+}$ .
  5. Interpret the behavior of $\overline{C}(x)$ as $x \rightarrow \infty$ .
23. This exercise explores the relationships between fixed cost, variable cost, and average cost. The reader is encouraged to revisit Example 1.3.8 in Section 1.3.3 as needed. Suppose the cost in dollars to make items is given by where and 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. Write a formula for the average cost when making $x$ items, $\overline{C}(x)$ .
  4. Show $\overline{C}(x) > m$ for all $x>0$ and, moreover, $\overline{C}(x) \rightarrow m^{+}$ as $x \rightarrow \infty$ .
  5. Interpret $\overline{C}(x) \rightarrow m^{+}$ both geometrically and in terms of fixed, variable, and average costs.
24. Suppose the price-demand function for a particular product is given by $p(x) = mx + b$ where $x$ is the number of items made and sold for $p(x)$ dollars. Here, $m<0$ and $b>0$ . 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 $\overline{P}(x)$ has a slant asymptote with slope $m$ and interpret.
25. Electric circuits are built with a variable resistor. For each of the following resistance values (measured in kilo-ohms, ), the corresponding power to the load (measured in milliwatts, ) is given below.^[19]
  $\begin{array}{|l|r|r|r|r|r|r|} \hline \text{Resistance: }(k \Omega) & 1.012 & 2.199 & 3.275 & 4.676 & 6.805 & 9.975 \\ \hline \text{Power: }(mW) & 1.063 & 1.496 & 1.610 & 1.613 & 1.505 & 1.314 \\ \hline \end{array}$
  
  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 $x$ :
  
  $P(x) = \frac{25x}{(x + 3.9)^2}, \quad x \geq 0.$
  1. Use a graphing utility to approximate the maximum power that can be delivered to the load. What is the corresponding resistance value?
  2. State and interpret the end behavior of $P(x)$ as $x \rightarrow \infty$ .
26. Let $f(x) = \dfrac{ax^2-c}{x+3}$ . Find values for $a$ and $c$ so the graph of $f$ has a hole at $(-3, 12)$ .
27. Let .
  1. Determine values for $a$ and $n$ so the graph of $y = f(x)$ has the horizontal asymptote $y = 3$ .
  2. Determine values for $a$ and $n$ so the graph of $y=f(x)$ has the slant asymptote $y = 5x$ .
28. Suppose $p$ is a polynomial function and $a$ is a real number. Define $r(x)= \dfrac{p(x) - p(a)}{x-a}$ . Use the Factor Theorem, Theorem 2.10, to prove the graph of $y = r(x)$ has a hole at $x =a$ .
29. For each function $f(x)$ listed below, compute the average rate of change over the indicated interval.^[20] What trends do you observe? How do your answers manifest themselves graphically? How do you results compare with those of Exercise 51 in Section 2.2?
  $\begin{array}{|r||c|c|c|c|} \hline f(x) & [0.9, 1.1] & [0.99, 1.01] &[0.999, 1.001] & [0.9999, 1.0001] \\ \hline x^{-1} &&&& \\ \hline x^{-2} &&&& \\ \hline x^{-3} &&&& \\ \hline x^{-4} &&&& \\ \hline \end{array}$
30. In his now famous 1919 dissertation $\underline{The Learning Curve Equation}$ , 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.^[21] 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.

Section 3.2 Exercise Answers can be found in the Appendix … Coming soon

According to this definition, all polynomial functions are also rational functions. (Take $q(x) = 1$ ). ↵
Technically speaking, $-1 \times 10^{117}$ is a `small' number (because it is very far to the left on the number line.) However, it's absolute value, $1 \times 10^{117}$ is very large. ↵
See Exercise 23 in Section 7.3. ↵
We are, in fact, building on Theorem 1.12 in Section 1.6, 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. ↵
These functions arise in Differential Equations. The unfortunate name will make sense shortly. ↵
Note that the rocket has already started its descent at $t = 10$ seconds (see Example 1.3.1.2 in Section 1.3.1.) However, the rocket is still at a higher altitude at when $t =15$ than $t=0$ which produces a positive average velocity. ↵
Or, `How to tell your asymptote from a hole in the graph.' ↵
In other words, $r(x)$ is in lowest terms. ↵
Though the population below is more accurately modeled with the functions in Chapter 5, 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 $g(x) \asymp x-1$ or $g(x) \sim x-1$ . ↵
Also called an `oblique' asymptote in some, ostensibly higher class (and more expensive), texts. ↵
That's OK, though. In the next section, we'll use long division to analyze end behavior and it's worth the effort! ↵
We generalize this result in Exercise38. ↵
For more review, see Section 2.3. ↵
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 1.11 in Section 1.3.4 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. ↵

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Definition 3.1

3.2.1 Laurent Monomial Functions

Definition 3.2

Definition 3.3

Definition 3.4

Example 3.2.1

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3.2.2 Local Behavior Near Excluded Values

Example 3.2.2

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Definition 3.5

Example 3.2.3

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Example 3.2.4

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3.2.3 End Behavior

Example 3.2.4

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Example 3.2.6

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Definition 3.6

Example 3.2.7

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Definition 3.7

Example 3.2.8

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