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.

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 ?? 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] (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 ??.)) 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 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 \(x \rightarrow -\infty \) and as \(x \rightarrow \infty \), \(r(x) \approx 0\). 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] (We’ll talk more about this end behavior shortly - stay tuned!)

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, \((-\infty , 0) \cup (0, \infty )\).

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

Similarly, we say ‘as \(x\) approaches \(0\) from the right,’ that is as \(x \rightarrow 0^{+}\), \(r(x) \rightarrow \infty \), or, more succinctly, \(\lim _{x \rightarrow 0^{+}} r(x) = \infty \). 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 \(x =0.001.\) These numbers could described as ‘\(0 + \text {a little bit}\),’ which justifies the ‘\(+\)’ in the notation ‘\(x \rightarrow 0^{+}\).’

In the GeoGebra interactive below, we can see what these limits mean geometrically.. Using the sliders, we can adjust values \(x \rightarrow 0^{+}\) and \(x \rightarrow 0^{-}\) and trace the corresponding points on the graph of \(y = r(x)\) to observe the unbounded behavior near \(x = 0\).

The way we describe what is happening graphically is to say the line \(x = 0\) is a vertical asymptote to the graph of \(y = r(x)\).

We can also use limit 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 \(x \rightarrow \infty \), \(r(x) \rightarrow 0^{+}\). When it comes to codifying these results using Calculus, we write \(\lim _{x \rightarrow -\infty } r(x) = 0\) and \(\lim _{x \rightarrow \infty } r(x) = 0\). Note that, unfortunately, we lose the directionality here on the limiting value - that is, we do not write \(\lim _{x \rightarrow -\infty } r(x) = 0^{-}\) or \(\lim _{x \rightarrow \infty } r(x) = 0^{+}\). Without getting too much into formal definitions, the reason is that if limiting values are finite, we express them as real numbers. ^[4] (We will, of course, offer more detail if the situation presents itself.) Period.

Once again, we have a GeoGebra interactive below to help us visualize what these limits mean graphically. We can adjust sliders to trace points on the graph of \(y = r(x)\) as \(x \rightarrow -\infty \) and \(x \rightarrow \infty \).

Here, we say the line \(y = 0\) 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.

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 ??, 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 \(|x|>1\).

To explain this graphical behavior, we consider the table of values below. Notice that as the exponent \(n\) in the denominator increases, the function values for \(|x|>1\) drop off considerably faster towards \(0\). Likewise, if \(|x|<1\), the higher the power in the denominator, the faster the function values grow (in absolute value.)

Both the domain and range for \(f(x) = \frac {1}{x^{n}}\) appear 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 \(c \neq 0\). Thus 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, 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.

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 \(y = 0\). Note here that both \(\lim _{x \rightarrow 0^{-}} f(x) = \infty \) and \(\lim _{x \rightarrow 0^{+}} f(x) = \infty \), so we may simply write \(\lim _{x \rightarrow 0} f(x) = \infty \).

As above, as \(n\) increases, the graphs of \(f(x) = \frac {1}{x^{n}}\) become steeper for \(|x|<1\) flatter for \(|x|>1\). This happens for the same reason as above and is demonstrated numerically in the table below.

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, the graphs of these functions are symmetric about the \(y\)-axis.

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

The proof of Theorem 1 is identical to the proof of Theorem ?? - just replace \(x^n\) with \(x^{-n}\). We nevertheless encourage the reader to work through the details ^[5] (We are, in fact, building to Theorem ?? in Section ??, 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.) and compare the results of this theorem with Theorems ??, ??, and ??.

We put Theorem 1 to good use in the following example.

In Example 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 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] (i.e., Laurent ‘Polynomials.’ This result is a combination of Theorems ?? in Section ?? and Theorem ?? in Section ?? .)

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. ^[7] (These functions arise in Differential Equations. The unfortunate name will make sense shortly.)

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 ?? in Section ??. 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 \(s\) 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 \(t_{0}\). We can define the average velocity as a function of any time \(t\) other than \(t_{0}\):

We must exclude \(t = t_{0}\) from the domain of \(\overline {v}\) in Definition 5 since, 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 explore this phenomenon in the next example.

Some notes about Example 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 \(\overline {v}(15)\) doesn’t exist, the fact that \(\lim _{t \rightarrow 15} \overline {v}(t) = -50\) means that we expect \(\overline {v}(15)\) to be \(-50\).This sentiment is exactly what a hole in the graph at \((15, -50)\) communicates.

Second, in finding \(\lim _{t \rightarrow 15} \overline {v}(t) = -50\), 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 \(\overline {v}(15)\) using the formula \(\overline {v}(t) = \frac {-5t^2+100t - 375}{t - 15}\) is resolved when the factor \((t-15)\) cancels from the denominator. We can algebraically reason what to expect out of the expression \(\overline {v}(t) = -5t+25\) as \(t \rightarrow 15\) 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] (See Section ??.) 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.

Of course the first question to ask is how do we know if \(\lim _{x \rightarrow c} r(x)\) 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 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 \(q\): \(q(c) =0\). We now have two cases to consider.

If \(p(c) \neq 0\), then as \(x \rightarrow c\), \(r(x) = \frac {p(x)}{q(x)} \rightarrow \frac {\text {$p(c)$, a nonzero number}}{0}\) which results in unbounded behavior (graphically, a vertical asymptote.)

If \(p(c) = 0\), then as \(x \rightarrow c\), \(r(x) = \frac {p(x)}{q(x)} \rightarrow \frac {0}{0}\), an indeterminate form. The Factor Theorem, ^[13] (Theorem ?? in Section ??) 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 to determine the limit. ^[14] (Using the vocabulary from Section ??, 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 \(q\).) 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.

3 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. ^[17] (Though the population below is more accurately modeled with the functions in Chapter ??, we approximate it (using Calculus, of course!) using a rational function.)

We determined the horizontal asymptote to the graph of \(y = N(t)\) in Example 5 by rewriting \(N(t)\) into a form compatible with Theorem 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 ?? from Section ??. 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 \(t \rightarrow \infty \):

Hence \(\lim _{t \rightarrow \infty } N(t) = 500\) so \(y = 500\) is the horizontal asymptote. This same reasoning can be used in general to argue the following theorem.

So see why Theorem 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 \(x \rightarrow -\infty \) or \(x \rightarrow \infty \), Theorem ?? 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 \(r(x) \approx \frac {ax^n}{bx^n} = \frac {a}{b}\). Hence \(\lim _{x \rightarrow -\infty } r(x) =\frac {a}{b}\) and \(\lim _{x \rightarrow \infty } r(x) =\frac {a}{b}\) 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 \(x \rightarrow -\infty \) or \(x \rightarrow \infty \), \(r(x)\) is more or less a fraction with a constant numerator, \(a\), but a denominator which is unbounded. Hence, \(\lim _{x \rightarrow -\infty } r(x) = 0\) and \(\lim _{x \rightarrow \infty } r(x) = 0\) 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 ??. As such, \(r\) becomes unbounded as \(x \rightarrow -\infty \) or \(x \rightarrow \infty \).

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. ^[19] (We will (first) encounter functions with more than one horizontal asymptote in Chapter ??.)

We put Theorem 3 to good use in the following example.

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 6. Performing long division, ^[21] (See the remarks following Theorem 3.) we get \(g(x) = \frac {x^2-4}{x+1} = x-1 - \frac {3}{x+1}\). Since the term \(\frac {3}{x+1} \rightarrow 0\) as \(x \rightarrow - \infty \) and as \(x \rightarrow \infty \) , it stands to reason that as \(x\) becomes unbounded, the function values \(g(x) = x-1 - \frac {3}{x+1} \approx x-1\). Numercially, we see this idea play out in the tables below.

Geometrically, this means that the graph of \(y=g(x)\) should resemble the line \(y = x-1\) as \(x \rightarrow -\infty \) and \(x \rightarrow \infty \). Our graph below provided by desmos confirms this.

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 \(x \rightarrow -\infty \) and \(x \rightarrow \infty \), \(g(x) \rightarrow x-1\)’ in order to have some notational consistency with what we have done earlier in this section when it comes to end behavior. ^[22] (Other notations include \(g(x) \asymp x-1\) or \(g(x) \sim x-1\).) In this case, we say the line \(y=x-1\) is a slant asymptote ^[23] (Also called an ‘oblique’ asymptote in some, ostensibly higher class (and more expensive), texts.) to 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. More formally, we define a slant asymptote as follows.

A few remarks are in order. First, note that the stipulation \(m \neq 0\) in Definition 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 \(y=0\). This last sentiment can be encoded using limit notation as follows.

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. ^[24] (Once again, this theorem is brought to you courtesy of Theorem ?? and Calculus.)

In the same way that Theorem 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 4 gives us an easy way to check for slant asymptotes. Unlike Theorem 3, which gives us a quick way to find the horizontal asymptotes (if any exist), Theorem 4 gives us no such ‘short-cut’. If a slant asymptote exists, we have no recourse but to use long division to find it. ^[25] (That’s OK, though. In the next section, we’ll use long division to analyze end behavior and it’s worth the effort!)

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.

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

4 Exercises