Chapter 3.2 Rational Functions

In this section, we focus on rational functions. We will learn how rational functions form a rigid collection and how we use this information to sketch rational functions.

Sketching Reciprocals of Polynomials

Rational functions are quotients of polynomial functions and inherit their rigidity from the rigidity of polynomial functions.

Reduced Form

A rational function is written in simplest form (or reduced form) if it is written as a quotient of polynomials that do not have any common zeros.

It is important that we work with rational functions in reduced form. Make sure you understand what it means for a rational function to be in simplest form with this next example.

Example 1

Put a check by each rational function below that is given in simplest form:

$\dfrac{(x+3)^5(x-1)}{(x+3)(x+2)^2}$;
$\dfrac{x(2x-10)}{x(x+7)(x-5)}$;
$\dfrac{2x+1}{x}$;
$\dfrac{4(x-5)}{2}$;
$\dfrac{3(x+4)^2(x-4)}{(x+1)(x-1)(x-4)^3}$.

Not in simplest form; $x=-3$ is a common zero.
Not in simplest form: the numerator can be rewritten as $2x(x-5)$, so both the numerator and denominator have $x=5$ as a common zero.
It is in simplest form because there are no common zeros.
It is in simplest form because there are no common zeros.
Not in simplest form; $x=4$ is a common zero.

Just like with polynomials, understanding the zeros is crucial. However, because we are dividing by a polynomial, it is possible a rational function is undefined at specific real numbers. At this places, the rational function exhibits interesting behavior.

To articulate this, we need the following words.

Zero, Pole, Local Behavior

Take $f$ to be a rational function given in simplest form by the quotient \[f(x) = \frac{p(x)}{q(x)}.\]

A zero of $f$ of order $n$ is a zero of $p$ of order $n$.

A pole of $f$ of order $n$ is a zero of $q$ of order $n$

The zeros and poles of $f$ give the local behavior of $f$ and so $f$ inherits rigidity from $p$ and $q$ in this way.

Understand the difference between zeros and poles with this example.

Example 2

List all zeros and poles of $f$ with their orders, where \[f(x) = \dfrac{(x+3)(x-2)^3}{(x+2)^3(x-1)(x+5)^2}.\]

The zero $x=-3$ is order $1$ and the zero $x=2$ is order $3$.

The pole $x=-2$ is order $3$, the pole $x=1$ is order $1$, and the pole $x=-5$ is order $2$.

Inversion of the $y$-axis facilitates sketching reciprocals of polynomials that are written in factored form.

Sketch the polynomial in the denominator;
Invert the $y$-axis.

The transformation principle used to simplify the problem is: Use the $y$-axis inversion to go between a polynomial, which we already know how to sketch, and its reciprocal.

Here is another word we will use to describe poles.

Vertical Asymptote

If $a$ is a zero of the denominator of $f$, then the line \[x = a\] is a vertical asymptote of $f$.

The function $f$ appears to “follow” this line but cannot cross it because $f$ is not defined at $a$.

In this example, we will use the pole to sketch this simple rational function.

Example 3

Take $f$ to be given by \[f(x) = \dfrac{1}{x+2}.\] List the poles of $f$ and their respective orders. List the vertical asymptotes of $f$, sketch the denominator of $f$, and then use $y$-axis inversion to sketch $f$.

Pole of $f$ is $x=-2$ of order $1$. Therefore, $x=-2$ is a vertical asymptote. The denominator $x+2$ looks like this.

So by using $y$-axis inversion, $f$ will look like this

In this more complicated example, we will first graph what the rational function’s denominator looks like and then use $y$-axis inversion to provide a sketch of the rational function.

Example 4

Take $f$ to be given by \[f(x) = \dfrac{1}{(x+1)^2(x-2)^3(x-4)}.\] List the poles of $f$ and their respective orders. List the vertical asymptotes of $f$, sketch the denominator of $f$, and then use $y$-axis inversion to sketch $f$.

Poles of $f$ are $x=-1$ of order $2$, $x=2$ of order $3$, and $x=4$ of order $4$. Therefore, $x=-1$, $x=2$, and $x=4$ are vertical asymptote. The denominator $(x+1)^2(x-2)^3(x-4)$ looks like this:

The asymptotic behavior was $x^6$ and the zeros are $x=-1$ of order $2$, $x=2$ of order 3, and $x=4$ of order 1.

So by using $y$-axis inversion, $f$ will look like this

Asymptotic Behavior

In the previous examples, we specialized to rational functions that were of the form

\[f(x)=\frac{1}{p(x)},\]

where $p$ was a polynomial. However, a rational function may have a more complicated numerator.

To be able to make sense of how this may affect the graph, we need to understand how this affects its asymptotic behavior.

Here are two words we will use to describe asymptotic behavior:

Horizontal Asymptote and Slant Asymptote

A rational function $f$ uniquely decomposes into a sum $p + r$ of a polynomial function $p$ and a proper rational function $r$.

For large values of $N$, $f\Big|_{(-\infty,-N)\cup(N,\infty)}$ looks like $p\Big|_{(-\infty,-N)\cup(N,\infty)}$.

The function $f$ is asymptotically equal to zero if $p$ is the zero polynomial.

If $p$ is a line, then $p$ is a horizontal asymptote of $f$ if $p$ is horizontal, otherwise it is a slant asymptote of $f$.

Let’s understand this definition and also revisit vertical asymptotes with this example.

Example 5

List, classify, and sketch all asymptotes of the rational function $f$, where \[f(x) = \dfrac{x-3}{(x-2)^2(x+1)}.\]

The vertical asymptotes are $x=2$ and $x=-1$, with order 2 and order 1 respectively.

The horizontal asymptote can be determine by writing $f$ as $p+r$ for some polynomial $p$ and proper rational function $r$. In this case, $f$ is already a proper rational function, so $\dfrac{x-3}{(x-2)^2(x+1)}=0+ \dfrac{x-3}{(x-2)^2(x+1)}$, where $p=0$. Thus the $y=0$ is a horizontal asymptote.

So the vertical asymptotes of $f$ will look like this

Again, understand how to find asymptotes of a rational function with this example.

Example 6

List, classify, and sketch all asymptotes of the rational function $f$, where \[f(x) = \dfrac{3(x-2)^2 - (x-2) + 5}{x-2}.\]

The vertical asymptote is $x=2$ which is order $1$.

To find the horizontal asymptote we need to write $f$ as the sum of a polynomial plus a proper rational function.

First, expand the numerator: $(x-2)^2 - (x-2) + 5=3 x^2 - 13 x + 19$. Now rewrite $x^2 - 13 x + 19$ as $Q(x)\cdot(x-2)+R(x)$. Since the degrees must match up, $Q(x)=ax+b$ and $R(x)=c$.

\[ \begin{align*} 3 x^2 - 13 x + 19&=(ax+b)(x-2)+c\\ &=ax^2-2ax+bx-2b+c\\ &=ax^2+(-2a+b)x+[-2b+c]\\ \end{align*} \]

Hence we need to solve \[ \begin{cases} 3&=a\\ -13&=-2a+b\\ 19&=-2b+c \end{cases}. \]

Hence we get $a=3$, $b=-7$, and $c=5$. Thus,

\[ 3 x^2 - 13 x + 19=(3x-7)(x-2)+5 \]

\[ \frac{3 x^2 - 13 x + 19}{x-2}=(3x-7)+\frac{5}{x-2}. \]

Thus the slant asymptote is $3x-7.$

The asymptotes look like this

Take $f$ to be a rational function that uniquely decomposes into a sum $p + r$ of a polynomial function $p$ and a proper rational function $r$.

It can be difficult to precisely determine $p$, but exact determination of $p$ is not usually necessary.

Asymptotically Equal To

The function $f$ is asymptotically equal to $Cx^n$ if the leading term of $p$ is $Cx^n$.

All we need to do is identify the leading term of the numerator and the denominator and then divide them.

Example 7

Find a $C$ and an $n$ so that $f$, $g$, and $h$ are asymptotically equal to $Cx^n$, where $f$, $g$, and $h$ are given by

$f(x) = \dfrac{x-1}{x+2}$;
$g(x) = \dfrac{2x^2-1}{x+3}$;
$h(x) = \dfrac{x^3}{3x-3}$.

The leading term of the numerator is $x$ and the leading term of the denominator is $x$. Thus $f$ is asymptotically equal to $\tfrac{x}{x}=1$

The leading term of the numerator is $2x^2$ and the leading term of the denominator is $x$. Thus $f$ is asymptotically equal to $\tfrac{2x^2}{x}=2x$

The leading term of the numerator is $x^3$ and the leading term of the denominator is $3x$. Thus $f$ is asymptotically equal to $\tfrac{x^3}{3x}=2x^2$

Sketching Rational Functions

Rational functions with linear denominators readily break into manageable pieces and so can be sketched more accurately than more general rational functions.

Decompose such functions into a sum of a polynomial and the reciprocal of a polynomial of degree 1.
Find the zeros by finding the zeros of the numerator of the original function.
Overlay sketches of the polynomial part of the quotient and the part that is a proper fraction separately to approximate how they add.
Sketch the quotient of linear functions by simply scaling and shifting the reciprocal of the identity function.

Here is an example.

Example 8

Sketch $f$ and its asymptotes, where \[f(x) = \dfrac{x-1}{x+2}.\]

The function $f$ can be rewritten as $1-\frac{3}{x+2}$, so it has a horizontal asymptote at $y=1$.

The function $f$ has a zero $x=1$, order 1

The function $f$ has a pole of $x=-2$ order 1.

The function therefore looks like this:

Let’s see how this rational function looks.

Example 9

Sketch $f$ and its asymptotes, where \[f(x) = \dfrac{x^2-4}{x+1}. \quad \text{Note: $= x-1 - \frac{3}{x+1}$}\]

The function $f$ can be rewritten as $x-1 - \frac{3}{x+1}$, so it has a slant asymptote at $y=x-1$.

The function $f$ has a zero $x=2$ and $x=-2$, both order 1

The function $f$ has a pole of $x=-1$ order 1.

The function therefore looks like this:

More general rational functions may still be written as a sum of a polynomial and a proper rational function.

Finding the quotient requires a more involved calculation and there is not a great payoff in making such a calculation since knowing the asymptotic behavior is usually sufficient for sketching the function.

Remember that determining the asymptotic behavior only involves finding the degree of the quotient.

Example 10

Write the rational function $f$ as the sum of a polynomial and a proper rational function, where \[\hspace{.5in} f(x) = \frac{(3x+5)^2(x+2)(x-1)}{(x+1)(x-3)}. \hspace{.5in} \text{Note: $= \frac{9x^4+39x^3+37x^2-35x-50}{x^2-2x-3}$}\] Find a $C$ and an $n$ so that $f$ is asymptotically equal to $Cx^n$.

The function can be rewritten like $f(x)=9x^2+57x+178+\frac{492x+484}{x^2-2x-3}.$

The function $f$ is asymptotically equal to $9x^2$.

In general, it is straightforward to roughly sketch the rational functions of the form \[f(x) = \frac{a(x-a_1) (x-a_2) \cdots (x-a_n)}{b(x-b_1) (x-b_2) \cdots (x-b_m)}\] by using the local and global behavior of $f$.

This is done by finding the zeros and their orders (local behavior), the poles and their orders (local behavior), and noticing that (global behavior) the function asymptotically behaves like \[g(x) = \frac{a}{b} x^{n-m}.\]

Warning: We may miss certain ``wiggles’’ in the shape of $f$, but we will have properly captured the properties of $f$ near zeros, all asymptotic properties of $f$, and the sign of $f$ at any point.

Example 11

Sketch $f$, where \[f(x) = \dfrac{x-1}{(x-6)(x+1)}.\]

The zero is $x=1$, order 1.

The pole is $x=6$ and $x=-1$, both order 1.

The function $f$ is asymptotically equal to $\frac{1}{x}$, so there is a horizontal asymptote at $y=0$.

The function looks like this:

In this next example, pay attention to how the information comes together.

Example 12

Sketch $f$, where \[f(x) = \dfrac{(x-3)^3}{(x-2)^2(x+1)}.\]

The zero is $x=3$, order 3.

The pole is $x=2$ order 2, and $x=-1$, order 1.

The function $f$ is asymptotically equal to $1$, so there is a horizontal asymptote at $y=1$.

The function looks like this:

In this final example, use the graphical information to extract as much as possible what features the rational function may have.

Example 13

Below is a sketch of a rational function, $Q$. List all zeros and poles of $Q$. For each zero, try to determine the order of the zero. For each pole, say whether it is an odd or even order pole. Say if $Q$ has a horizontal asymptote, a slant asymptote, or whether $Q$ is asymptotically equal to a polynomial of even or odd degree that is at least 2.

Zeros $x=-7$ (even order), $x=-1$ (odd order), $x=2$ (odd order) , $x=4$ (even order)

Pole $x=-5$ (odd order), $=1$ (even order),

No horizontal asymptote or slant asymptote. $Q$ is asymptotically equal to a polynomial of odd degree.

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