\chapter{Polynomial and rational functions}
\section{Polynomial Functions}
\subsection{Definitions and terminology}
A \textit{monomial} (in the variable \(x\)) is a number times \(x^n\), where the
exponent \(n\) must be a whole number, and cannot be negative (it could be zero, in which
case \(x^0=1\) and the monomial is just a number).
\begin{example}
\[2x^3, \hspace{0.5cm} -x^2, \hspace{0.5cm} 2, \hspace{0.5cm} \dfrac{2}{3}x^4, \hspace{0.5cm} \dfrac{x}{\sqrt{2}}, \hspace{0.5cm} 2\pi \hspace{0.5cm}\] are all monomials.
\[ \dfrac{2}{x}, \hspace{0.5cm} 3x^{-1}, \hspace{0.5cm} \sqrt{x} \hspace{0.5cm}\] are not monomials (because of \(x\) in the
denominator, or with negative exponent, or inside a square root).
\end{example}
A \textit{polynomial} (in the variable \(x\)) is a sum of monomials.
\begin{example}
\[x^2-3x+1, \hspace{0.5cm} \pi x+ \sqrt{2}, \hspace{0.5cm} \dfrac{1}{2}x^3-\dfrac{2}{3}x^2, \hspace{0.5cm} \dfrac{x^2}{3}+\dfrac{x}{2}, \hspace{0.5cm} 3x^4 \hspace{0.5cm}\] are all polynomials.
\[ x+ \dfrac{1}{x}, \hspace{0.5cm} x^{-2}+1+x^2, \hspace{0.5cm} 1+x\sqrt{x} \hspace{0.5cm}\] are not polynomials (because of \(x\) in the
denominator, or with negative exponent, or inside a square root).
\end{example}
The various monomials making up a polynomial are also called \textit{terms}. A polynomial with
just two terms is called a \textit{binomial}. A polynomial with three terms is
called a \textit{trinomial}.
\begin{example}
\(2x+1\) and \(3x^4-x\) are binomials.
\(2x^2-x+3\) and \(x^3+x^2+5x\) are trinomials.
\end{example}
There are
several notions associated with polynomials: the degree, the coefficients, the leading coefficient,
the constant term. To understand what they are, look at the following example.
\begin{example}
Consider the polynomial \[p(x)=5x^3+6x^2-x+4.\]
\begin{itemize}
\item
This polynomial has four terms.
\item
The degree is 3.
\item The leading term is \(5x^3\).
\item
The coefficient of \(x^2\) is 6.
\item
The coefficient of \(x\) is \(-1\).
\item
The leading coefficient is 5.
\item
The constant term is 4.
\end{itemize}
\end{example}
So the \textit{degree} of a polynomial is the largest exponent of the variable.
In case there is no variable at all (so the polynomial is just a number), the degree is 0.
The term that contains the variable with the largest exponent is the \textit{leading term}.
The numbers that
multiply the variable in each term are the \textit{coefficients}. The \textit{leading coefficient}
is the coefficient of the leading term. The
\textit{constant term} is the term that does not contain the variable.
We will need to refer to the leading coefficient and the constant term quite often in
the rest of this chapter. So we are going to use the shorthand notation:
\[\begin{array}{|c|}
\hline
\mbox{LC\(=\)leading coefficient}, \hspace{1in} \mbox{CT\(=\)constant term}\\
\hline
\end{array}
\]
Make sure to distinguish between the numbers (or the coefficients) and the variable. Numbers
include \textbf{all} real numbers, not just the whole numbers. So \(2/3\), \(\sqrt{3}\), \(\pi\),
\(2\sqrt{5}-3\pi^2\) are all numbers.
In particular, when finding the degree,
do not confuse exponents for the variable with exponents that may appear in
some of the numbers.
\begin{example}\
\begin{enumerate}
\item
The polynomial
\(\displaystyle p(x)=4\pi^3 x^2 -\frac{x}{2} +2+\sqrt{3}\)
has degree 2 (not 3), the coefficient of \(x\) is \(-1/2\), the leading term
is \(4 \pi^3 x^2\), the leading coefficient
is \(4\pi^3\), and the constant term is \(2+\sqrt{3}\).
\item
The polynomial \(f(x)=2\pi^2\) has degree 0. The leading term is \(2\pi^2\).
The leading coefficient is \(2\pi^2\). The constant
term is \(2\pi^2\).
\end{enumerate}
\end{example}
A \textit{polynomial function} is just a function whose formula is given by a polynomial.
In Chapter 1, we defined the zero of a function \(f(x)\) to be a solution of the equation \(f(x)=0\).
When the function is actually a polynomial function, a zero is often also called a \textit{root}. So in the rest of this
module we will often use the word ``root'' instead of ``zero''.
\subsection{Graphs of polynomial functions}
In view of what we discussed in Units 2.1 and 2.2, we can now say that a quadratic function (whose graph is a parabola) is given by a polynomial of degree \(2\).
The vertex of a parabola is an example of a \textit{turning point}, that is, a point where the
graph turns from increasing to decreasing, or from decreasing to increasing.
\[
\begin{array}{cc}
\img{U2_4F1.jpg}{}{8em}{}& \hspace{1in} \img{U2_4F2.jpg}{}{8em}{}
\end{array}
\]
Recall that a quadratic function has no more than two \(x\)-intercepts (it could have one, or none).
And the shape is always basically the same: every quadratic function is just the
parent function \(y=x^2\)
after some shifting, reflection or distortion.
If we go up one degree, we find the polynomials of degree 3, also called \textit{cubic functions}.
The simplest one is the basic parent \(y=x^3\), whose graph we recall here below:
\[
\begin{array}{c}
\img{U2_4F3.jpg}{}{6em}{}
\end{array}
\]
Of course there are many more cubic functions: any polynomial of degree 3 will be one. But this
time it is no longer true that every cubic function will be the basic parent \(y=x^3\)
after some simple transformation. That happens only for the polynomials of degree 2 (the
quadratic functions).
So we can expect that there will be various
shapes for the graph of a cubic functions, some of which are shown in the picture below. Remember that
LC means leading coefficient.
\[
\begin{array}{cccc}
\img{U2_4F4.png}{}{8em}{} & \hspace{0cm}\img{U2_4F5.png}{}{8em}{}
\hspace{2ex}
\img{U2_4F6.png}{}{8em}{} & \hspace{0cm} \img{U2_4F7.png}{}{8em}{}\\
& \mbox{Polynomial functions of degree 3} & &
\end{array}
\]
Note a few important features of these graphs:
\begin{itemize}
\item
There are at most two turning points (and there could be less than that).
\item
There are at most three \(x\)-intercepts (and there could be less than that).
\item
Some \(x\)-intercepts cross the \(x\)-axis (we call them \textit{crossing} intercepts), and
some "bounce" off the \(x\)-axis (we call them \textit{bouncing}).
\item
If the LC is positive, the leftmost side and the rightmost side of the graph are both increasing.
We call this \textit{end behavior like \(x^3\)}, and picture it as:
\(\img{UU.png}{-1em}{}{2em}\).
\item
If the LC is negative, the leftmost side and the rightmost side of the graph are both decreasing.
We call this \textit{end behavior like \(-x^3\)}, and picture it as:
\(\img{DD.png}{-1em}{}{2em}\).
\end{itemize}
Let's now go up one more degree, to the polynomials of degree 4. As you can expect,
there are even more possible shapes, and some are shown in the picture below. The turning points
are marked by a black dot.
\[
\begin{array}{cccc}
\img{U2_4F8.png}{}{8em}{} & \hspace{0cm}\img{U2_4F9.png}{}{8em}{}
\hspace{2ex}
\img{U2_4F10.png}{}{8em}{} & \hspace{0cm} \img{U2_4F11.png}{}{8em}{}\\
& \mbox{Polynomial functions of degree 4} & &
\end{array}
\]
We note the following points:
\begin{itemize}
\item
There are at most three turning points (and there could be less than that).
\item
There are at most four \(x\)-intercepts (and there could be less than that).
\item
If the LC is positive, the leftmost side of the graph is decreasing and the rightmost side is
increasing. We call this \textit{end behavior like \(x^2\)},
and picture it as: \(\img{DU.png}{-1em}{}{2em}\). This is the same end behavior of \(y=x^2\).
\item
If the LC is negative, the leftmost side of the graph is increasing and
the rightmost side is decreasing.We call this \textit{end behavior like \(-x^2\)},
and picture it as: \(\img{UD.png}{-1em}{}{2em}\). This is the same end behavior as \(y=-x^2\).
\end{itemize}
Note the difference in the end behavior between degree 3 and degree 4.
This is because 3 is odd and 4 is even. And this pattern continues for larger degrees.
Every polynomial of odd degree has end behavior like degree 3, and every polynomial
of even degree has end behavior like degree 2 or 4.
But if we are given a formula for a polynomial function, such as
\(f(x)=x^4-x^3-12x^2\), how can we find out what shape it has? The key is to completely
factor the polynomial (if it is not already completely factored!), as we show in the next example.
\begin{example}
We discuss how to sketch the graph of \(f(x)=x^4-x^3-12x^2\).
\begin{itemize}
\item Completely factor (if necessary):
\[
\begin{array}{rcll}
x^4-x^3-12x^2 & = & x^2(x^2-x-12)\\[2ex]
&=& x^2(x-4)(x+3)\\[2ex]
\end{array}
\]
\item When we set the factored polynomial equal to zero, each factor gives us a root,
corresponding to an \(x\)-intercept.
\[
\begin{array}{rclll}
x^2(x+3)(x-4)&=& 0 & \\[2ex]
x^2=0 & \mbox{or} & x-4=0 & \mbox{or} & x+3=0\\[2ex]
x=0 & \mbox{or} & x=4 & \mbox{or} & x=-3
\end{array}
\]
So there are three roots: \(x=0\), \(x=4\) and \(x=-3\), corresponding to the \(x\)-intercepts \((0,0)\), \((4,0)\) and
\((-3,0)\).
\item
Look at the \textbf{exponent} of each the three factors. So the factor \(x^2\) has exponent \(2\), the
factor \((x-4)=(x-4)^1\) has exponent \(1\), and the factor \((x+3)=(x+3)^1\) has exponent \(1\). These exponents are called the
\textit{multiplicity} of the root given by that factor.
The type of the \(x\)-intercept depends on whether the multiplicity
of the corresponding factor is \textbf{even} or \textbf{odd}. Even multiplicity gives a bouncing intercept, and odd multiplicity
gives a crossing intercept.
So \((0,0)\) will be a bouncing intercept, because \(2\) is even,
while \((4,0)\) and \((-3,0)\) will be crossing intercepts, because \(1\) is odd, as described in the table below:
\[\begin{array}{c|c|c|c}
\mbox{Factor} & \mbox{Root} & \mbox{Multiplicity} & \mbox{Type of intercept}\\
\hline
x^2 &x=0 & 2 & \mbox{Bouncing}\\
\hline
(x-4)^1 &x=4 & 1 & \mbox{Crossing}\\
\hline
(x+3)^1 & x=-3 & 1 & \mbox{Crossing}\\
\hline
\end{array}
\]
\item Find the \(y\)-intercept (if not already found), and
a few extra points using the formula \(f(x)=x^2(x-4)(x+3)\). In this example \((0,0)\) is
both \(x\)- and \(y\)-intercept.
We use \(x=1\), \(x=-1\) and \(x=2\) as extra points. We find
\[f(1)=1^2(1+3)(1-4)=-12,\]
\[f(-1)=(-1)^2(-1+3)(-1-4)=-10,\] and
\[f(2)=2^2(2-4)(2+3)=-40.\]
Note that it's easier to use the factored form of the polynomial to compute a function value.
\item Determine the end behavior: the degree is even, and the LC is positive, so the
end behavior is like \(x^2\): \(\img{DU.png}{-0.5em}{}{2em} \)
\item
Summarize the information found in a table, and plot all the points. B means a bouncing intercept
and C a crossing intercept.
\[
\begin{array}{cc}
\begin{array}{c|c|l}
x & f(x) & \\
\hline
\mbox{End behavior} & \mbox{like }x^2 & \img{DU.png}{-0.5em}{}{2em} \\
0 & 0 & \mbox{Bouncing and \(y\)-intercept}\\[2ex]
-3 & 0 & \mbox{Crossing}\\[2ex]
4 & 0 & \mbox{Crossing}\\[2ex]
1 & -12 & \\[2ex]
-1 & -10 & \\[2ex]
2 & -40
\end{array}
& \img{U2_4F13.png}{-8em}{12em}{}
\end{array}
\]
\item Join the plotted points with a smooth curve (no sharp turns or corners!),
making sure the \(x\)-intercepts are of the right type, and the end behavior is like \(x^2\).
\[
\img{U2_4F12.png}{}{10em}{}
\]
\begin{center} Graph of \(f(x)=x^4-x^3-12x^2\)
\end{center}
\end{itemize}
\end{example}
We summarize the information needed to draw graphs of polynomial functions:
| Shape of graph for a polynomial function of degree \(n\) |
|---|
| There are at most \(n-1\) turning points |
| There are at most \(n\) \(x\)-intercepts |
| There are two types of \(x\)-intercepts: |
| \(\hspace{2ex} \bullet\) Crossing for odd multiplicity of the root, |
| \(\hspace{2ex} \bullet\) Bouncing for even multiplicity. |
| If \(n\) is odd, then the end behavior is: |
| \(\hspace{2ex} \bullet\) Like \(x^3\) if LC \(> 0\) |
| \(\hspace{2ex} \bullet\) Like \(-x^3\) if LC \(< 0\) |
| If \(n\) is even, then the end behavior is: |
| \(\hspace{2ex} \bullet\) Like \(x^2\) if LC \(> 0\) |
| \(\hspace{2ex} \bullet\) Like \(-x^2\) if LC\(< 0\). |
| Steps to draw the graph of a polynomial function |
|---|
| \(
\begin{array}{l}
\bullet \mbox{ Completely factor the polynomial}\\[1ex]
\bullet \mbox{ Find the roots, and their multiplicities}\\[1ex]
\bullet \mbox{ Determine the type of \(x\)-intercepts:}\\
\hspace{1ex} \mbox{Crossing for odd multiplicity, Bouncing for even} \\[1ex]
\bullet \mbox{ Find the \(y\)-intercept (if not already found)}\\[1ex]
\bullet \mbox{ Find a few extra points, near the \(x\)-intercepts}\\[1ex]
\bullet \mbox{ Determine the end behavior according to the table above}\\[1ex]
\bullet \mbox{ Plot all the points}\\[1ex]
\bullet \mbox{ Join the plotted points with a smooth curve without turns or corners}
\end{array}\) |
Problems
\problem
Decide if each of the following functions is a polynomial function. If it is, give the degree, the leading coefficient, and the constant term.
\begin{enumerate}
\item \(f(x)=4x^3-3x-5\)
\item \(g(x)=2-3x-4x^2\)
\item \(h(x)=2\sqrt{x}-3\)
\item \(k(x)=\pi^4x^3-2x^2+x+3-\pi\)
\item \(p(x)=x^5\)
\item \(q(x)=x+\displaystyle\frac{3}{x}\)
\item \(r(x) = -1\).
\end{enumerate}
\begin{sol}
\[\begin{array}{|c|c|c|c|c|}
\hline
\mbox{function} & \mbox{Polynomial?} & \mbox{degree} & \mbox{LC} & \mbox{CT}\\
\hline
f(x)=4x^3-3x-5 & \mbox{Yes} & 3 & 4 & -5 \\
\hline
g(x)=2-3x-4x^2 & \mbox{Yes} & 2 & -4 & 2\\
\hline
h(x)=2\sqrt{x}-3 & \mbox{No} & \mbox{N/A} & \mbox{N/A} & \mbox{N/A}\\
\hline
k(x)=\pi^4x^3-2x^2+x+3-\pi & \mbox{Yes} & 3 & \pi^4 & 3-\pi \\
\hline
\displaystyle p(x)=x^5 & \mbox{Yes} & 5 & 1 & 0 \\
\hline
q(x)=x+\displaystyle\frac{3}{x} & \mbox{No}& \mbox{N/A} & \mbox{N/A} & \mbox{N/A} \\[1ex]
\hline
r(x)=-1 & \mbox{Yes}& 0 & -1 & -1 \\
\hline
\end{array}
\]
\end{sol} \mproblem
Decide if each of the following functions is a polynomial function. If it is, give the degree, the leading coefficient, and the constant term.
\begin{enumerate}
\item \(f(x)=x^3-2x^2+4x+3\)
\item \(g(x)=3x-\sqrt{x}+5\)
\item \(h(x)=1-x-x^2\)
\item \(k(x)=5x^4\)
\item \(p(x)=4\sqrt{3}\)
\item \( \displaystyle q(x)= x+2+\frac{1}{x}\)
\item \(r(x)=2\pi^3x^2-3x+\pi+1\)
\end{enumerate}
\problem
Sketch the graph of the following functions and label all intercepts.
\begin{enumerate}
\item \(f(x)=x^4-4x^2\)
\item \(g(x)=-(x-2)^2(x+3)\)
\item \(h(x)=-(x+1)(x-2)(x+2)^2\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item \(f(x)=x^4-4x^2\)
\begin{itemize}
\item Factor the polynomial:
\[
\begin{array}{rcll}
x^4-4x^2 &=& x^2(x^2-4)\\[2ex]
& = & x^2(x-2)(x+2)
\end{array}
\]
\item Find all intercepts:
\[x=0, \hspace{1cm} x=2, \hspace{1cm} x=-2\]
\item Determine the type of \(x\)-intercepts:
\begin{center}
\(x=0\): even exponent, bouncing.
\(x=-2\): odd exponent, crossing.
\(x=2\): odd exponent, crossing.
\end{center}
\item The \(y\)- intercept is \((0,0)\). Find a few extra points:
\[f(x)=x^2(x-2)(x+2)\]
\[f(1)=(1)^2(1-2)(1+2)=-3,\]
\[f(-1)=(-1)^2(-1-2)(-1+2)=-3.\]
\item The degree is even and the LC is positive, so the end behavior
is like \(x^2\) \(\img{DU.png}{-0.5em}{}{2em} \)
\item
Gather all information and plot all the points,
together with the information about the type of intercept:
\[
\begin{array}{cc}
\begin{array}{c|c|l}
x & f(x) & \\
\hline
\mbox{End behavior} & \mbox{like } x^2 & \img{DU.png}{-0.5em}{}{2em} \\
0 & 0 & \mbox{Bouncing and \(y\)-intercept}\\[2ex]
-2 & 0 & \mbox{Crossing}\\[2ex]
2 & 0 & \mbox{Crossing}\\[2ex]
1 & -3 & \\[2ex]
-1 & -3 &
\end{array} & \img{U2_4F14.png}{-8em}{12em}{}
\end{array}
\]
\item Join the plotted points with a smooth curve.
Make sure the end behavior is like \(x^2\).
\[
\img{U2_4F15.png}{}{12em}{}
\]
\begin{center} Graph of \(f(x)=x^4-4x^2\)
\end{center}
\end{itemize}
\item \(g(x)=-(x-2)^2(x+3)\)
\begin{itemize}
\item The polynomial is already factored.
\item Find all \(x\)-intercepts:
\[x=2, \hspace{1cm} x=-3\]
\item Determine the type of \(x\)-intercept:
\begin{center}
\(x=2\): even exponent, bouncing.
\(x=-3\): odd exponent, crossing.
\end{center}
\item Find the \(y\)-intercept, and a few extra points:
\[g(x)=-(x-2)^2(x+3)\]
\[\mbox{\(y\)-intercept: } \hspace{0.3cm} g(0)=-(0-2)^2(0+3)=-12 \]
\[g(1)=-(1-2)^2(1+3)=-4,\]
\[g(-2)=-(-2-2)^2(-2+3)=-16,\]
\[g(3)=-(3-2)^2(3+3)=-6.\]
\item We need the degree and the leading coefficient. We could
expand \(-(x-2)^2(x+3)\). But we don't need to do that.
The degree will be the sum of the degrees of each factor.
\(-(x-2)^2\) has degree 2 and \((x+3)\) has degree 1, so the degree of \(g(x)\) will be 3.
And the leading coefficient will be the product of the leading coefficient of each factor.
The LC of \(-(x-2)^2\) is \(-1\), and the LC of \((x-2)\) is 1, so the LC of \(g(x)\) is \(-1\). So
we have an odd degree with a negative LC, and the end behavior will be like \(-x^3\)
\(\img{DD.png}{-1em}{}{2em}\).
\item
Gather all information and plot all the points,
together with the information about the type of intercept:
\[
\begin{array}{cc}
\begin{array}{c|c|c}
x & g(x) & \\
\hline
\mbox{End behavior} & \mbox{like } -x^3 & \img{DD.png}{-1em}{}{2em}\\[1.5ex]
0 & -12 & \mbox{\(y\)-intercept}\\[2ex]
2 & 0 & \mbox{Bouncing}\\[2ex]
-3 & 0 & \mbox{Crossing}\\[2ex]
1 & -4 & \\[2ex]
-2 & -16 & \\[2ex]
3 & -6
\end{array} & \img{U2_4F16.png}{-8em}{12em}{}
\end{array}
\]
\item Join the plotted points with a smooth curve.
Make sure the end behavior is like \(-x^3\).
\[
\img{U2_4F17.png}{}{12em}{}
\]
\begin{center} Graph of \(g(x)=-(x-2)^2(x+3)\)
\end{center}
\end{itemize}
\item \(h(x)=-(x+1)(x-2)(x+2)^2\)
\begin{itemize}
\item The polynomial is already factored.
\item Find all \(x\)-intercepts:
\[x=-1, \hspace{1cm} x=2, \hspace{1cm} x=-2\]
\item Determine the type of \(x\)-intercept:
\begin{center}
\(x=-1\): odd exponent, crossing.
\(x=2\): odd exponent, crossing.
\(x=-2\): even exponent, bouncing.
\end{center}
\item Find the \(y\)-intercept, and a few extra points:
\[h(x)=-(x+1)(x-2)(x+2)^2\]
\[\mbox{\(y\)-intercept: } \hspace{0.3cm} h(0)=-(0+1)(0-2)(0+2)^2=8 \]
\[h(1)=-(1+1)(1-2)(1+2)^2=18,\]
\[h(-3)=-(-3+1)(-3-2)(-3+2)^2=-10.\]
\item The degrees of the factors \(-(x+1)\), \((x-2)\), \((x+2)^2\) are 1,1,2 and so the
degree of \(h(x)\) will be \(1+1+2=4\). The LC of \(-(x+1)\) is \(-1\), and the other two are 1,
so the LC of \(h(x)\) is \(-1\).
So we have an even degree with a negative LC, and the end behavior will be like \(-x^2\)
\(\img{UD.png}{-1em}{}{2em}\).
\item
Gather all information and plot all the points,
together with the information about the type of intercept:
\[
\begin{array}{cc}
\begin{array}{c|c|c}
x & h(x) & \\
\hline
\mbox{End behavior} & \mbox{like } -x^2 & \img{UD.png}{-1em}{}{2em} \\
0 & -8 & \mbox{\(y\)-intercept}\\[2ex]
-1 & 0 & \mbox{Crossing}\\[2ex]
2 & 0 & \mbox{Crossing}\\[2ex]
-2 & 0 & \mbox{Bouncing}\\[2ex]
1 & -18 & \\[2ex]
-3 & 10 &
\end{array}
& \img{U2_4F18.png}{-8em}{12em}{}
\end{array}
\]
\item Join the plotted points with a smooth curve.
Make sure the end behavior is like \(-x^2\).
\[
\img{U2_4F19.png}{}{12em}{}
\]
\begin{center} Graph of \(h(x)=-(x+1)(x-2)(x+2)^2\)
\end{center}
\end{itemize}
\end{enumerate}
\end{sol} \mproblem
Sketch the graph of the following functions, and label all intercepts.
\begin{enumerate}
\item \(f(x)=-x(x+2)(x-2)\)
\item \(g(x)=(x+1)^2(x-3)(x+4)\)
\item \(h(x)=x^4-9x^2\)
\end{enumerate}