2.2 Properties of Polynomial Functions and Their Graphs

Vanessa Coffelt

2.2 Properties of Polynomial Functions and Their Graphs

2.2.1 Monomial Function

Definition 2.4

A monomial function is a function of the form

$f(x) = b \qquad \text{or} \qquad f(x) = a x^{n},$

where $a$ and $b$ are real numbers, $a \neq 0$ and $n \in \mathbb{N}$ .

The domain of a monomial function is $(-\infty, \infty)$ .

Monomial functions, by definition, contain the constant functions along with a two parameter family of functions, $f(x) = ax^n$ . We use $x$ as the default independent variable here with $a$ and $n$ as parameters. From Section 0.1.1, we recall that the set $\mathbb{N} = \{ 1, 2, 3, \ldots \}$ is the set of natural numbers, so examples of monomial functions include $f(x) = 2x = 2x^{1}$ , $g(t) = -0.1 t^2$ , and $H(s) = \sqrt{2} \, s^{117}$ . Note that the function $f(x) = x^0$ is not a monomial function. Even though $x^0 = 1$ for all nonzero values of $x$ , $0^{0}$ is undefined,^[1] and hence $f(x) = x^0$ does not have a domain of $(-\infty, \infty)$ .^[2]

We begin our study of the graphs of polynomial functions by studying graphs of monomial functions. Starting with $f(x) = x^n$ where $n$ is even, we investigate the cases $n = 2$ , $4$ and $6$ . Numerically, we see that if $-1 < x < 1$ , $x^n$ becomes much smaller as $n$ increases whereas if $x<-1$ or $x>1$ , $x^n$ becomes much larger as $n$ increases. These trends manifest themselves geometrically as the graph `flattening’ for $|x|<1$ and `narrowing’ for $|x| > 1$ as $n$ increases.^[3]

$\begin{array}{|r||c|c|c|} \hline x & x^2 & x^4 & x^6 \\ \hline -2 & 4& 16& 64 \\ \hline -1 & 1 & 1& 1\\ \hline -0.5 & 0.25 & 0.0625& 0.015625 \\ \hline 0 & 0 & 0 & 0 \\ \hline 0.5 & 0.25 & 0.0625 & 0.015625 \\ \hline 1& 1 & 1& 1 \\ \hline 2 & 4 & 16 & 64 \\ \hline \end{array}$

Three graphs side by side. The first graph is a parabola opening up. The graph is labeled f(x) =x^2. The second graph is looks like a parabola that is slightly flatter at the vertex. The graph is labeled f(x)=x^4. The last graph is again similar to a parabola with an even flatter bottom. The graph is labeled f(x) = x^6. — Graphs of Even Degree Monomials

From the graphs, it appears as if the range of each of these functions is $[0, \infty)$ . When $n$ is even, $x^n \geq 0$ for all $x$ so the range of $f(x) = x^{n}$ is contained in $[0, \infty)$ . To show that the range of $f$ is all of $[0, \infty)$ , we note that the equation $x^n = c$ for $c \geq 0$ has (at least) one solution for every even integer $n$ , namely $x = \sqrt[n]{c}$ . (See Section 0.2 for a review of this notation.) Hence, $f(\sqrt[n]{c}) = (\sqrt[n]{c})^n = c$ which shows that every non-negative real number is in the range of $f$ .^[4]

Another item worthy of note is the symmetry about the line $x =0$ a.k.a the $y$ -axis. (See Definition ?? for a review of this concept.) With $n$ being even, $f(-x) = (-x)^n = x^n = f(x)$ . At the level of points, we have that for all $x$ , $(-x, f(-x)) = (-x,f(x))$ . Hence for every point $(x, f(x))$ on the graph of $f$ , the point symmetric about the $y$ -axis, $(-x, f(x))$ is on the graph, too. We give this sort of symmetry a name honoring its roots here with even-powered monomial functions:

Definition 2.5

A function $f$ is said to be even if $f(-x) = f(x)$ for all $x$ in the domain of $f$ .
NOTE: A function $f$ is even if and only if the graph of $y = f(x)$ is symmetric about the $y$ -axis.

An investigation of the odd powered monomial functions ( $n \geq 3$ ) yields similar results with the major difference being that when a negative number is raised to an odd natural number power the result is still negative. Numerically we see that for $|x| > 1$ the values of $|x^n|$ increase as $n$ increases and for $|x| <1$ the values of $|x^n|$ get closer to $0$ as $n$ increases. This translates graphically into a flattening behavior on the interval $(-1, 1)$ and a narrowing elsewhere. The graphs are shown below.

The range of these functions appear to be all real numbers, $(-\infty, \infty)$ which is algebraically sound as the equation $x^n = c$ has a solution for every real number,^[5] namely $x = \sqrt[n]{c}$ . Hence, for every real number $c$ , choose $x = \sqrt[n]{c}$ so that $f(x) = f(\sqrt[n]{c}) = (\sqrt[n]{c})^n =c$ . This shows that every real number is in the range of $f$ .

$\begin{array}{|r||c|c|c|} \hline x & x^3 & x^5 & x^7 \\ \hline -2 & -8& -32& -128 \\ \hline -1 & -1 & -1& -1\\ \hline -0.5 & 0.125 & -0.03125& -0.0078125 \\ \hline 0 & 0 & 0 & 0 \\ \hline 0.5 & 0.125 & 0.03125 & 0.0078125 \\ \hline 1& 1 & 1& 1 \\ \hline 2 & 8 & 32 & 128 \\ \hline \end{array}$

Three graphs side by side. The first graph is a increases from left to right, and crosses the x-axis at the point (0,0). The graph is labeled f(x) =x^3. The second graph is looks very similar to the first, but slightly flatter near (0,0). The graph is labeled f(x)=x^5. The last graph is again similar to a first graph, but even flatter near (0,0). The graph is labeled f(x) = x^7. — Graphs of Odd Degree Monomials

Here, as a result of $n$ being odd, $f(-x) = (-x)^n = -x^n = -f(x)$ . This means that whenever $(x, f(x))$ is on the graph, so is the point symmetric about the origin, $(-x, -f(x))$ . (Again, see Definition 1.1.) We generalize this property below. Not surprisingly, we name it in honor of its odd powered heritage:

Definition 2.6

A function $f$ is said to be odd if $f(-x) = -f(x)$ for all $x$ in the domain of $f$ .
NOTE: A function $f$ is odd if and only if the graph of $y = f(x)$ is symmetric about the origin.

The most important thing to take from the discussion above is the basic shape and common points on the graphs of $y = x^n$ for each of the families when $n$ even and $n$ is odd. While symmetry is nice and should be noted when present, even and odd symmetry are comparatively rare. The point of Definitions 2.5 and 2.6 is to give us the vocabulary to point out the symmetry when appropriate.

Moving on, we take a cue from Theorem 1.4 and prove the following.

Theorem 2.2

For real numbers $a$ , $h$ and $k$ with $a \neq 0$ , the graph of $F(x) = a(x-h)^n+k$ can be obtained from the graph of $f(x) = x^n$ by performing the following operations, in sequence:

add $h$ to the $x$ -coordinates of each 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 $y = (x-h)^n$ .
multiply the $y$ -coordinates of each 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 the $y$ -coordinates of each 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 $y = a(x-h)^n+k$

Proof. Our goal is to start with the graph of $f(x) = x^n$ and build it up to the graph of $F(x) = a(x-h)^n+k$ . We begin by examining $F_{1}(x) = (x-h)^n$ . The graph of $f(x) = x^n$ can be described as the set of points $\{ (c, c^n) \, | \, c \in \mathbb{R} \}$ .^[6] Likewise, the graph of $F_{1}$ can be described as the set of points $\{(x, (x-h)^n) \, | \, x \in \mathbb{R} \}$ . If we re-label $c =x-h$ so that $x = c+h$ , then as $x$ varies through all real numbers so does $c$ .^[7] Hence, we can describe the graph of $F_{1}$ as $\{ (c+h, c^n) \, | \, c \in \mathbb{R} \}$ . This means that we can obtain the graph of $F_{1}$ from the graph of $f$ by adding $h$ to each of the $x$ -coordinates of the points on the graph of $f$ and that establishes the first step of the theorem.

Next, we consider the graph of $F_{2}(x) = a(x-h)^n$ as compared to the graph of $F_{1}(x) = (x-h)^n$ . The graph of $F_{1}$ is the set of points $\{ (x, (x-h)^n \, | \, x \in \mathbb{R} \}$ while the graph of $F_{2}$ is the set of points $\{ (x, a(x-h)^n) \, | \, x \in \mathbb{R} \}$ . The only difference between the points $(x, (x-h)^n)$ and $(x, a(x-h)^n)$ is that the $y$ -coordinate in the latter is $a$ times the $y$ -coordinate of the former.

In other words, to produce the graph of $F_{2}$ from the graph of $F_{1}$ , we take the $y$ -coordinate of each point on the graph of $F_{1}$ and multiply it by $a$ to get the corresponding point on the graph of $F_{2}$ . If $a>0$ , all we are doing is scaling the $y$ -axis by $a$ . If $a<0$ , then, in addition to scaling the $y$ -axis, we are also reflecting each point across the $x$ -axis. In either case, we have established the second step of the theorem.

Last, we compare the graph of $F(x) = a(x-h)^n + k$ to that of $F_{2}(x) = a(x-h)^n$ . Once again, we view the graphs as sets of points in the plane. The graph of $F_{2}$ is $\{ (x, a(x-h)^n) \, | \, x \in \mathbb{R} \}$ and the graph of $F$ is $\{ (x, a(x-h)^n+k) \, | \, x \in \mathbb{R} \}$ . Looking at the corresponding points, $(x, a(x-h)^n)$ and $(x, a(x-h)^n+k)$ , we see that we can obtain all of the points on the graph of $F$ by adding $k$ to each of the $y$ -coordinates to points on the graph of $F_{2}$ . This is equivalent to shifting every point vertically by $k$ units which establishes the third and final step in the theorem. $\qed$

This argument should sound familiar. The proof we presented above is more-or-less the same argument we presented after the proof of Theorem 1.4 in Section 1.4 but with ` $| \cdot |$ ‘ replaced by ` $(\cdot)^n$ .’ Also note that using $n =2$ in Theorem 2.2 establishes Theorem 2.1 in Section 2.1.

We now use Theorem 2.2 to graph two different “transformed” monomial functions. To provide the reader an opportunity to compare and contrast the graphical behaviors exhibited in the case when $n$ is even versus when $n$ is odd, we graph one of each case.

Example 2.2.1

Example 2.2.1.1

Use Theorem 2.2 to graph the following functions. Label at least three points on each graph. State the domain and range using interval notation.

$f(x) = -2(x+1)^4+3$

Solution:

Graph $f(x) = -2(x+1)^{4} + 3$ .

For $f(x) = -2(x+1)^4+3 = -2 (x-(-1))^4+3$ , we identify $n = 4$ , $a = -2$ , $h = -1$ , and $k = 3$ . Thus to graph $f$ , we start with $y = x^4$ and perform the following steps, in sequence, tracking the points $(-1,1)$ , $(0,0)$ and $(1,1)$ through each step:

Step 1: add $-1$ to the $x$ -coordinates of each of the points on the graph of $y=x^4$ :

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Step 2: multiply the $y$ -coordinates of each of the points on the graph of $f_1(x) = (x+1)^4$ by $-2$ :

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Step 3: add $3$ to the $y$ -coordinates of each of the points on the graph of $f_2 (x)=-2(x+1)^{4}$ :

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The domain here is $(-\infty, \infty)$ , while the range is $(-\infty, 3]$ .

Example 2.2.1.2

Use Theorem 2.2 to graph the following functions. Label at least three points on each graph. State the domain and range using interval notation.

$g(t) = \dfrac{(2t-1)^3}{5}$

Solution:

Graph $g(t) = \dfrac{(2t-1)^{3}}{5}$ .

To use Theorem 2.2 to graph $g(t) = \dfrac{(2t-1)^3}{5}$ , we must first rewrite the expression for $g(t)$ :

$\begin{array}{rcl} g(t) &=& \dfrac{(2t-1)^3}{5} \\[8pt] &=& \frac{1}{5} \left( 2 \left(t - \frac{1}{2} \right) \right)^3 \\[8pt] &=& \frac{1}{5} (2)^3 \left( t - \frac{1}{2} \right)^3 \\[8pt] &=& \frac{8}{5} \left( t - \frac{1}{2} \right)^3 \end{array}$

We identify $n = 3$ , $h = \frac{1}{2}$ and $a = \frac{8}{5}$ . Hence, we start with the graph of $y = t^3$ and perform the following steps, in sequence, tracking the points $(-1,-1)$ , $(0,0)$ and $(1,1)$ through each step:

Step 1: add $\frac{1}{2}$ to each of the $t$ -coordinates of each of the points on the graph of $y=t^3$ :

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Step 2: multiply each of the $y$ -coordinates of the graph of $f_{1} (t) = \left(t - \frac{1}{2}\right)^3$ by $\frac{8}{5}$ .

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Both the domain and range of $g$ is $(-\infty, \infty)$ .

Example 2.2.1 demonstrates two big ideas in mathematics: first, resolving a complex problem into smaller, simpler steps, and, second, the value of changing form.

Next we wish to focus on the so-called end behavior presented in each case.^[8] The end behavior of a function is a way to describe what is happening to the outputs from a function as the inputs approach the `ends’ of the domain. Due to the fact that the domain of monomial functions is $(-\infty, \infty)$ , we are looking to see what these functions do as their inputs `approach’ $\pm \infty$ . The best we can do is sample inputs and outputs and infer general behavior from these observations.\footnote{and let Calculus students prove our claims.} The good news is we’ve wrestled with this concept before. Indeed, every time we add `arrows’ to the graph of a function, we’ve indicated its end behavior. Let’s revisit the graph of $f(x) = x^2$ using the table below.

A table and a graph side by side. The table has two columns the first row are the titles for the columns, x and f(x)=x^2. The remaining rows are values of x and the corresponding values of f(x); -1000 and 1,000,000, -100 and 10,000, -10 and 100, 0 and 0, 10 and 100, 100 and 10,000, and 1000 and 1,000,000. The graph is of f(x) = x^2 with the ends of the x-axis and the end of f(x) highlighted with the infinity notation. — General End Behavior of an Even Function

As $x$ takes on smaller and smaller values,^[9], we see $f(x)$ takes on larger and larger positive values. The smaller $x$ we use, the larger the $f(x)$ becomes, seemingly without bound.^[10] We codify this behavior by writing as $x \rightarrow -\infty$ , $f(x) \rightarrow \infty$ . Graphically, the farther to the left we travel on the $x$ -axis, the farther up the $y$ -axis the function values travel. This is why we use an `arrow’ on the graph in Quadrant II heading upwards to the left. Similarly, we write as $x \rightarrow \infty$ , $f(x) \rightarrow \infty$ because as the $x$ values increase, so do the $f(x)$ values – seemingly without bound. Graphically we indicate this by an arrow on the graph in Quadrant I heading upwards to the right. This behavior holds for all functions $f(x) = x^n$ where $n \geq 2$ is even.

Repeating this investigation for $f(x) = x^3$ , we find as $x \rightarrow -\infty$ , $f(x)$ becomes unbounded in the negative direction, so we write $f(x) \rightarrow -\infty$ . As $x \rightarrow \infty$ , $f(x)$ becomes unbounded in the positive direction, so we write $f(x) \rightarrow \infty$ . This trend holds for all functions $f(x) = x^n$ where $n$ is odd.

A table and a graph side by side. The table has two columns the first row are the titles for the columns, x and f(x)=x^3. The remaining rows are values of x and the corresponding values of f(x); -1000 and -1,000,000,000, -100 and -1,000,000, -10 and -1000, 0 and 0, 10 and 1000, 100 and 1,000,000, and 1000 and 1,000,000,000. The graph is of f(x) = x^3 with the ends of the x-axis and the end of f(x) highlighted with the infinity notation. — General End Behavior of an Odd Function

Theorem 2.3 summarizes the end behavior of monomial functions. The results are a consequence of Theorem 2.2 in that the end behavior of a function of the form $y = ax^n$ only differs from that of $y = x^n$ if there is a reflection, that is, if $a<0$ .

Theorem 2.3 End Behavior of Monomial Functions

Suppose $f(x) = a x^{n}$ where $a \neq 0$ is a real number and $n \in \mathbb{N}$ .

If $n$ is even:
If $n$ is odd:

2.2.2 Polynomial Functions

We are now in the position to discuss polynomial functions. Simply stated, polynomial functions are sums of monomial functions. The challenge becomes how to describe one of these beasts in general. Up until now, we have used distinct letters to indicate different parameters in our definitions of function families. In other words, we define constant functions as $f(x) = b$ , linear functions as $f(x) = mx+b$ , and quadratic functions as $f(x) = ax^2+bx+c$ . We even hinted at a function of the form $f(x) = ax^3+bx^2+cx+d$ . What happens if we wanted to describe a generic polynomial that required, say, 117 different parameters? Our work around is to use subscripted parameters, $a_{k}$ , that denote the coefficient of $x^{k}$ . For example, instead of writing a quadratic as $f(x) = ax^2+bx+c$ , we describe it as $f(x) = a_{2} x^2 + a_{1} x + a_{0}$ , where $a_{2}$ , $a_{1}$ , and $a_{0}$ are real numbers and $a_{2} \neq 0$ . As an added example, consider $f(x) = 4x^5 - 3x^2 + 2x - 5$ . We can re-write the formula for $f$ as $f(x)= 4x^5 + 0 x^{4} + 0 x^{3} + (-3)x^2 + 2 x + (-5).$ and identify $a_{5} = 4$ , $a_{4} = 0$ , $a_{3} = 0$ , $a_{2} = -3$ , $a_{1} = 2$ and $a_{0} = -5$ . This is the notation we use in the following definition.

Definition 2.7

A polynomial function is a function of the form

$f(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \ldots + a_{2} x^{2} + a_{1} x + a_{0},$

where $a_{0}$ , $a_{1}$ , $\ldots$ , $a_{n}$ are real numbers and $n \in \mathbb{N}.$

The domain of a polynomial function is $(-\infty, \infty).$

As usual, $x$ is used in Definition 2.7 as the independent variable with the $a_{k}$ each being a parameter. Even though we specify $n \in \mathbb{N}$ so $n \geq 1$ , the value of the $a_{k}$ are unrestricted. Hence, any constant function $f(x) = b$ can be written as $f(x) = 0 x + a_{0}$ , and so they are polynomials. Polynomials have an associated vocabulary, and hence, so do polynomial functions.

Definition 2.8

Given with and , we say
- The natural number $n$ is called the degree of the polynomial $f$ .
- The term $a_{n} x^{n}$ is called the leading term of the polynomial $f$ .
- The real number $a_{n}$ is called the leading coefficient of the polynomial $f$ .
- The real number $a_{0}$ is called the constant term of the polynomial $f$ .
If $f(x) = a_{0}$ , and $a_{0} \neq 0$ , we say $f$ has degree $0.$
If $f(x) = 0$ , we say $f$ has no degree.^[11]

Again, constant functions are split off in their own separate case Definition 2.8 because of the ambiguity of $0^0$ . (See the remarks following Definition 2.4.) A consequence of Definition 2.8 is that we can now think of nonzero constant functions as `zeroth’ degree polynomial functions, linear functions as `first’ degree polynomial functions, and quadratic functions as `second’ degree polynomial functions.

Example 2.2.2

Example 2.2.2.1

Determine the degree, leading term, leading coefficient and constant term of the following polynomial functions.

$f(x) = 4x^5 - 3x^2 + 2x - 5$

Solution:

Determine the degree, leading term, leading coefficient and constant term of $f(x) = 4x^5 - 3x^2 + 2x - 5$ .

There are no surprises with $f(x) = 4x^5 - 3x^2 + 2x - 5$ . It is written in the form of Definition 2.8, and we see that
the degree is $5$ ,
the leading term is $4x^5$ ,
the leading coefficient is $4$ , and
the constant term is $-5$ .

Example 2.2.2.2

Determine the degree, leading term, leading coefficient and constant term of the following polynomial functions.

$g(t) = 12t - t^3$

Solution:

Determine the degree, leading term, leading coefficient and constant term of $g(t) = 12t - t^3$ .

Two changes here: first, the independent variable is $t$ , not $x$ . Second, the form given in Definition 2.8 specifies the function be written in descending order of the powers of $x$ , or in this case, $t$ . To that end, we re-write $g(t) = 12t - t^3 = -t^3+12t,$ and see that
the degree of $g$ is $3$ ,
the leading term is $- t^3$ ,
the leading coefficient is $-1$ , and
the constant term is $0$ .

Example 2.2.2.3

Determine the degree, leading term, leading coefficient and constant term of the following polynomial functions.

$H(w) = \dfrac{4-w}{5}$

Solution:

Determine the degree, leading term, leading coefficient and constant term of $H(w) = \dfrac{4-w}{5}.$

We need to rewrite the formula for $H(w)$ so that it resembles the form given in Definition 2.8:

$\begin{array}{rcl} H(w) &=& \frac{4-w}{5} \\ &=& \frac{4}{5} - \frac{w}{5} \\ &=& -\frac{1}{5} w + \frac{4}{5} \end{array}$

We see
the degree of $H$ is $1$ ,
the leading term is $-\frac{1}{5} w$ ,
the leading coefficient is $-\frac{1}{5}$ , and
the constant term is $\frac{4}{5}.$

Example 2.2.2.4

Determine the degree, leading term, leading coefficient and constant term of the following polynomial functions.

$p(z) = (2z-1)^{3}(z-2)(3z+2)$

Solution:

Determine the degree, leading term, leading coefficient and constant term of $p(z) = (2z-1)^{3}(z-2)(3z+2).$

It may seem that we have some work ahead of us to get $p$ in the form of Definition 2.8. However, it is possible to glean the information requested about $p$ without multiplying out the entire expression $(2z-1)^{3}(z-2)(3z+2).$ The leading term of $p$ will be the term which has the highest power of $z$ . The way to get this term is to multiply the terms with the highest power of $z$ from each factor together – in other words, the leading term of $p(z)$ is the product of the leading terms of the factors of $p(z).$ Hence, the leading term of $p$ is $(2z)^3(z)(3z) = 24z^5 .$

This means that the degree of $p$ is $5$ and the leading coefficient is $24.$

As for the constant term, we can perform a similar operation. The constant term of $p$ is obtained by multiplying the constant terms from each of the factor: $(-1)^3(-2)(2) = 4.$

We now turn our attention to graphs of polynomial functions. As polynomial functions are sums of monomial functions, it stands to reason that some of the properties of those graphs carry over to more general polynomials. We first discuss end behavior. Consider $f(x) = x^3-75x+250$ . Below are two graphs of $f(x)$ (solid line) along with the graphs of its leading term, $y = x^3$ (dashed line.) On the left is a view `near’ the origin, while on the right is a `zoomed out’ view. Near the origin, the graphs have little in common, but as we look farther out, it becomes that the functions begin to look quite similar.

Two graphs side by side. Both graphs are demonstrating the behavior near x equals 5. Each graph contains a solid line and a dashed line. In the first, the graph has been zoomed in vertically. The dashed line is the graph of y=x^3 and the solid is the function f(x) near x equals 5. The second graph has the vertical axis zoomed out to the thousands. The dashed line is of y = x^3 and the solid line is the function f(x). — Comparing the Behavior of f(x) and y=x^3

This observation is borne out numerically as well. Based on the table below, as $x \rightarrow \pm \infty$ , it certainly appears as if $f(x) \approx g(x)$ . One way to think about what is happening numerically is that the leading term $x^3$ dominates the lower order terms $-75x$ and $250$ as $x \rightarrow \pm \infty$ . In other words, $x^3$ grows so much faster than $-75 x$ and $250$ that these `lower order terms’ don’t contribute anything of significance to the $x^3$ so $f(x) \approx x^3$ . Another way to see this is to rewrite $f(x)$ as^[12]

$\begin{array}{rcl} f(x) &=& x^3 - 75x + 250 \\[10pt] &=& x^3 \left(1 - \frac{75}{x^2} + \frac{250}{x^3} \right) \end{array}$

As $x \rightarrow \pm \infty$ , both $\frac{75}{x^2}$ and $\frac{250}{x^3}$ have constant numerators but denominators that are becoming unbounded. As such, both $\frac{75}{x^2}$ and $\frac{250}{x^3} \rightarrow 0$ . Therefore, as $x \rightarrow \pm \infty$ ,

$\begin{array}{rcl} f(x) &=& x^3 - 75x+250 \\[10pt] &=& x^3 \left(1 - \frac{75}{x^2} + \frac{250}{x^3} \right) \\[10pt] &\approx& x^3 (1 + 0 + 0) = x^3 \end{array}$

$\begin{array}{|r||c|c|c|c|c|c|} \hline x & f(x) = x^3 -75x+250 & x^3 & -75 x & 250 & \frac{75}{x^2} & \frac{250}{x^3} \\ \hline -1000 & \approx -1 \times 10^9 & -1 \times 10^9 &75000 & 250 & 7.5 \times 10^{-5} & -2.5 \times 10^{-7} \\ \hline -100 & \approx -9.9 \times 10^5 & -1 \times 10^6 & 7500 & 250 & 0.0075 & -2.5 \times 10^{-4} \\ \hline -10 & 0 & -1000 & 750 & 250 & 0.75 & -0.25\\ \hline 10 & 500 & 1000 & -750 & 250 & 0.75 & 0.25 \\ \hline 100 &\approx 9.9 \times 10^5 & 1 \times 10^6 & -7500 & 250 & 0.0075 & 2.5 \times 10^{-4} \\ \hline 1000 & \approx 1 \times 10^9 & 1 \times 10^9 & -75000 & 250 & 7.5 \times 10^{-5} & 2.5 \times 10^{-7} \\ \hline \end{array}$

Next, consider $g(x) = -0.01x^4 + 5x^2$ . Following the logic of the above example, we would expect the end behavior of $y=g(x)$ to mimic that of $y = -0.01 x^4$ . When we graph $y = g(x)$ (solid line) on the same set of axes as $y = -0.01x^4$ (dashed line), a view near the origin seems to suggest the exact opposite. However, zooming out reveals that the two graphs do share the same end behavior.^[13]

Two graphs side by side. Each graph has both a dashed curve and a solid curve. For the first the dashed is y=x^4 and the solid curve is the portion of g(x) near x equals 0. The second is a zoomed out version of both y=x^4 (dashed) and g(x) (solid). — Comparison of the General Behavior for y=x^4 and g(x)

Algebraically, for $x \rightarrow \pm \infty$ , even with the small coefficient of $-0.01$ , $-0.01x^4$ dominates the $5x^2$ term so $g(x) \approx -0.01 x^4$ . More precisely,

$\begin{array}{rcl} g(x) &=& -0.01x^4+5x^2 \\[10pt] &=& x^4 \left(-0.01 + \frac{5}{x^2} \right)\\[10pt] &\approx& x^4(-0.01 + 0) \\[10pt] &=& -0.01x^4 \end{array}$

The results of these last two examples are generalized below in Theorem 2.4.

Theorem 2.4 End Behavior for Polynomial Functions

The end behavior of polynomial function $f(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \ldots + a_{2} x^{2} + a_{1} x + a_{0}$ with $a_{n} \neq 0$ matches the end behavior of $y = a_{n} x^{n}$ .
That is, the end behavior of a polynomial function is determined by its leading term.

We argue Theorem 2.4 using an argument similar to ones used above. As $x \rightarrow \pm \infty$ ,

$\begin{array}{rcl} f(x) &=& x^{n} \left( a_{n} +\dfrac{a_{n-1}}{x}+ \ldots + \dfrac{a_{2}}{x^{n-2}} + \dfrac{a_{1}}{x^{n-1}}+\dfrac{a_{0}}{x^{n}}\right) \\[10pt] &\approx& x^n( a_{n} + 0 +\ldots 0) \\[8pt] &=& a_{n} x^n \end{array}$

If this argument looks a little fuzzy, it should. In Calculus, we have the tools necessary to more explicitly state what we mean by $\approx 0$ . For now, we’ll rely on number sense and algebraic intuition.^[14]

Now that we know how to determine the end behavior of polynomial functions, it’s time to investigate what happens `in between’ the ends. First and foremost, polynomial functions are continuous. Recall from Section 2.1 that, informally, graphs of continuous functions have no `breaks’ or `holes’ in them.^[15] Monomial functions are continuous (as far as we can tell) and polynomials are sums of monomial functions, so we conclude that polynomial functions are continuous as well. Moreover, the graphs of monomial functions, hence polynomial functions, are smooth. Once again, `smoothness’ is a concept defined precisely in Calculus, but for us, functions have no `corners’ or `sharp turns’. Below we find the graph of a function which is neither smooth nor continuous, and to its right we have a graph of a polynomial, for comparison. The function whose graph appears on the left fails to be continuous where it has a `break’ or `hole’ in the graph; everywhere else, the function is continuous. The function is continuous at the `corner’ and the `cusp’, but we consider these `sharp turns’, so these are places where the function fails to be smooth. Apart from these four places, the function is smooth and continuous. Polynomial functions are smooth and continuous everywhere, as exhibited in the graph on the right. The notion of smoothness is what tells us graphically that, for example, $f(x) = |x|$ , whose graph is the characteristic ` $\vee$ ‘ shape, cannot be a polynomial function, even though it is a piecewise-defined function comprised of polynomial functions. Knowing polynomial functions are continuous and smooth gives us an idea of how to `connect the dots’ when sketching the graph from points that we’re able to find analytically such as intercepts.

Two graphs side by side. The first graph includes a break, a corner, a cusp, and a hole in the graph. Each are labeled with the corresponding word. The graph is labeled Pathologies not found on graphs of polynomial functions. The second graph is a generic odd degree polynomial. The graph is labeled The graph of a polynomial function. — Properties that are not Polynomials and those that are Polynomials

Speaking of intercepts, we next focus our attention on the behavior of the graphs of polynomial functions near their zeros. Recall a zero $c$ of a function $f$ is a solution to $f(x) = 0$ . Geometrically, the zeros of a function are the $x$ -coordinates of the $x$ -intercepts of the graph of $y = f(x)$ .

Consider the polynomial function $f(x) = x^3 (x-2)^2 (x+1)$ .

To find the zeros of $f$ , we set $f(x) = x^3 (x-2)^2 (x+1) = 0$ . The expression $f(x)$ is already factored, so we set each factor equal to zero.^[16]

Solving $x^3 = 0$ gives $x = 0$ , $(x-2)^2 = 0$ gives $x = 2$ , and $x+1 = 0$ gives $x = -1$ . Hence, our zeros are $x = -1$ , $x = 0$ , and $x = 2$ .

Below, we graph $y = f(x)$ and observe the $x$ -intercepts $(-1,0)$ , $(0,0)$ and $(2,0)$ . We first note that the graph crosses through the $x$ -axis at $(-1,0)$ and $(0,0)$ , but the graph touches and rebounds at $(2,0)$ . Moreover, at $(-1,0)$ , the graph crosses through the axis is a fairly `linear’ fashion whereas there is a substantial amount of `flattening’ going on near $(0,0)$ . Our aim is to explain these observations and generalize them.

The graph of the function f(x). The graph opens upward and touches or crosses the graph at x equals -1, 0, and 2. — Graph of f(x)

First, let’s look at what’s happening with the formula $f(x) = x^3 (x-2)^2 (x+1)$ when $x \approx -1$ . We know the $x$ -intercept at $(-1,0)$ is due to the presence of the $(x+1)$ factor in the expression for $f(x)$ . So, in this sense, the factor $(x+1)$ is determining a major piece of the behavior of the graph near $x = -1$ . For that reason, we focus instead on the other two factors to see what contribution they make. We find when $x \approx -1$ , $x^3 \approx (-1)^3 = -1$ and $(x-2)^2 \approx (-1-2)^2 = 9$ .

Hence,

$\begin{array}{rcl} f(x) &=& x^3 (x-2)^2 (x+1) \\ &\approx& (-1)^3 (-1-2)^2 (x+1)\\ &=& -9(x+1) \end{array}$

Below on the left is a graph of $y = f(x)$ (the solid line) and the graph of $y = -9(x+1)$ (the dashed line.) Sure enough, these graphs approximate one another near $x = -1$ .

Likewise, let’s look near $x = 0$ . The $x$ -intercept $(0,0)$ is a result of the $x^3$ factor. For $x \approx 0$ , $(x-2)^2 \approx (0-2)^2 = 4$ and $(x+1) \approx (0+1) = 1$ , so

$\begin{array}{rcl} f(x) &=& x^3 (x-2)^2 (x+1)\\ &\approx& x^3 (-2)^2(1) \\ &=& 4x^3 \end{array}$

Below in the center picture, we have the graph of $y = f(x)$ (again, the solid line) and $y = 4x^3$ (the dashed line) near $x=0$ . Once again, the graphs verify our analysis.

Last, but not least, we analyze $f$ near $x = 2$ . Here, the intercept $(2,0)$ is due to the $(x-2)^2$ factor, so we look at the $x^3$ and $(x+1)$ factors. If $x \approx 2$ , $x^3 \approx (2)^3 = 8$ and $(x+1) \approx (2+1) = 3$ . Hence,

$\begin{array}{rcl} f(x) &=& x^3 (x-2)^2 (x+1) \\ &\approx& (2)^3 (x-2)^2 (2+1) \\ &=& 24(x-2)^2 \end{array}$

Sure enough, as evidenced below on the right, the graphs of $y = f(x)$ and $y = 24(x-2)^2$ .

Three graphs side by side by side. Each graph is an examination of f(x) near one of the roots. For each graph a portion of the curve at the root is drawn using a solid curve and a dashed line or curve is used to represent the approximate behavior. The first graph represents the behavior near x equals -1, the second at x equals 0, and the last at x eqals 2. — Behavior of f(x) near its intercepts

We generalize our observations in Theorem 2.5 below. Like many things we’ve seen in this text, a more precise statement and proof can be found in a course on Calculus.

Theorem 2.5

Suppose $f$ is a polynomial function and $f(x) = (x-c)^m q(x)$ where $m \in \mathbb{N}$ and $q(c) \neq 0$ . Then the the graph of $y = f(x)$ near $(c,0)$ resembles that of $y = q(c) (x-c)^m$ .

Let’s see how Theorem 2.5 applies to our findings regarding $f(x) = x^3 (x-2)^2 (x+1).$

For $c = -1$ , $(x-c) = (x-(-1)) = (x+1)$ . We rewrite $f(x) = x^3 (x-2)^2 (x+1) = (x-(-1))^1 \left[x^3(x-2)^2\right]$ and identify $m=1$ and $q(x) = x^3 (x-2)^2$ . We find $q(c) = q(-1) = (-1)^3(-1-2)^2 = -9$ so Theorem 2.5 says that near $(-1,0)$ , the graph of $y=f(x)$ resembles $y = q(-1)(x-(-1))^1 = -9(x+1)$ .

For $c=0$ , $(x-c) = (x-0) = x$ and we can rewrite $f(x) = x^3 (x-2)^2 (x+1) = (x-0)^3 \left[(x-2)^2 (x+1)\right] .$ We identify $m=3$ and $q(x) = (x-2)^2(x+1)$ . In this case $q(c) = q(0) = (0-2)^2(0+1) = 4$ , so Theorem 2.5 guarantees the graph of $y = f(x)$ near $x=0$ resembles $y = q(0)(x-0)^3 = 4x^3 .$

Lastly, for $c = 2$ , we see $f(x) = (x-2)^2 \left[x^3 (x+1)\right]$ and we identify $m = 2$ and $q(x) = x^3(x+1)$ . We find $q(2) = 2^3 (2+1)= 24$ , so Theorem 2.5 guarantees the graph of $y = f(x)$ resembles $y = 24(x-2)^2$ near $x=2$ .

As we already mentioned, the formal statement and proof of Theorem 2.5 require Calculus. For now, we can understand the theorem as follows. If we factor a polynomial function as $f(x) = (x-c)^m q(x)$ where $m \geq 1$ , then $x=c$ is a zero of $f$ , because $f(c) = (c-c)^m q(c) = 0 \cdot q(c) = 0$ . The stipulation that $q(c) \neq 0$ means that we have essentially factored the expression $f(x) = (x-c)^m q(x) = (\text{going to 0}) \cdot (\text{not going to 0})$ . Thinking back to Theorem 2.2, the graph $y = q(c) (x-c)^m$ has an $x$ -intercept at $(c,0)$ , a basic overall shape determined by the exponent $m$ , and end behavior determined by the sign of $q(c)$ . The fact that if $x=c$ is a zero then we are guaranteed we can factor $f(x) = (x-c)^m q(x)$ were $q(c) \neq 0$ and, moreover, such a factorization is unique (so that there’s only one value of $m$ possible for each zero) is a consequence of two theorems, Theorem 2.8 and The Factor Theorem, Theorem 2.10 which we’ll review in the next section. For now, we assume such a factorization is unique in order to define the following.

Definition 2.9

Suppose $f$ is a polynomial function and $m \in \mathbb{N}$ . If $f(x) = (x-c)^m q(x)$ where $q(c) \neq 0$ , we say $x=c$ is a zero of multiplicity $m$ .

So, for $f(x) = x^3 (x-2)^2 (x+1) = (x-0)^3(x-2)^2(x-(-1))^1$ , $x=0$ is a zero of multiplicity $3$ , $x=2$ is a zero of multiplicity $2$ , and $x =-1$ is a zero of multiplicity $1$ . Theorems 2.4 and 2.5 give us the following:

Theorem 2.6 The Role of Multiplicity

Suppose $f$ is a polynomial function and $x=c$ is a zero of multiplicity $m$ .

If $m$ is even, the graph of $y=f(x)$ touches and rebounds from the $x$ -axis at $(c,0)$ .
If $m$ is odd, the graph of $y=f(x)$ crosses through the $x$ -axis at $(c,0)$ .

Steps for Constructing a Sign Diagram for a Polynomial Function

Suppose $f$ is a polynomial function.

Compute the zeros of $f$ and place them on the number line with the number $0$ above them.
Choose a real number, called a test value, in each of the intervals determined in step 1.
Determine and record the sign of $f(x)$ for each test value in step 2.

Theorem 2.7

Suppose $f$ is a polynomial of degree $n \geq 1$ . Then $f$ has at most $n$ real zeros, counting multiplicities.

Connections Between Zeros, Factors and Graphs of Polynomial Functions

Suppose $p$ is a polynomial function of degree $n \geq 1$ . The following statements are equivalent:

The real number $c$ is a zero of $p$
$p(c) = 0$
$x = c$ is a solution to the polynomial equation $p(x) = 0$
$(x - c)$ is a factor of $p(x)$
The point $(c, 0)$ is an $x$ -intercept of the graph of $y = p(x)$

Our next example showcases how all of the above theories can assist in sketching relatively good graphs of polynomial functions without the assistance of technology.

Example 2.2.3

Example 2.2.3.1

Let $p(x) = (2x-1)(x+1)(1-x^4)$ . Compute all real zeros of $p$ and state their multiplicities.

Solution:

Compute all real zeros of $p(x) = (2x-1)(x+1)(1-x^4)$ and state their multiplicities.

To find the zeros of $p$ , we set $p(x) = (2x-1)(x+1)(1-x^4) = 0$ . The expression $p(x)$ is already (partially) factored, so we set each factor equal to $0$ and solve.

From $(2x-1) =0$ , we get $x=\frac{1}{2}$ ;
from $(x+1) = 0$ we get $x = -1$ ; and
from solving $1-x^4 =0$ we get $x = \pm 1$ .

Hence, the zeros are $x = -1$ , $x = \frac{1}{2}$ , and $x = 1$ .

In order to determine the multiplicities, we need to factor $p(x)$ as so we can identify the $m$ and $q(x)$ as described in Definition 2.9.

The zero $x = -1$ corresponds to the factor $(x+1)$ . Notice, however, that writing $p(x) = (x+1)^1 \left[(2x-1)(1-x^4)\right]$ with $m = 1$ and $q(x) = (2x-1)(1-x^4)$ does not satisfy Definition 2.9 because, $q(-1) = (2(-1)-1)(1-(-1)^4) = 0$ . Indeed, we can factor $(1-x^4) = (1-x^2)(1+x^2) = (1-x)(1+x)(x^2+1)$ so that

$\begin{array}{rcl} p(x) &=& (2x-1)(x+1)(1-x^4)\\ &=& (2x-1)(x+1)(1-x)(1+x)(x^2+1)\\ &=& (x+1)^2 \left[(2x-1)(1-x)(x^2+1) \right] \end{array}$

Identifying $q(x) = (2x-1)(1-x)(x^2+1)$ , we find $q(-1) = (2(-1)-1)(1-(-1))((-1)^2+1) = -12 \neq 0$ , which means the multiplicity of $x=-1$ is $m=2$ .

The zero $x = \frac{1}{2}$ came from the factor $(2x-1) = 2 (x-\frac{1}{2})$ , so we have

$\begin{array}{rcl} p(x) &=& (2x-1)(x+1)^2(1-x)(x^2+1) \\ &=& (x -\frac{1}{2})^{1} \left[2 (x+1)^2(1-x)(x^2+1) \right]. \end{array}$

If we identify $q(x) = 2 (x+1)^2(1-x)(x^2+1)$ , we find $q(\frac{1}{2}) = \frac{45}{16} \neq 0$ so multiplicity here is $m=1$ .

Last but not least, we turn our attention to our last zero, $x = 1$ , which we obtained from solving $1-x^4=0$ . However, from $p(x) = (2x-1)(x+1)^2(1-x)(x^2+1)$ , we see the zero $x=1$ corresponds to the factor $(1-x) = -(x-1)$ . We have

$\begin{array}{rcl} p(x) &=& (x-1)^{1}\left[- (2x-1)(x+1)^2(x^2+1)\right] . \end{array}$

Identifying $q(x) = - (2x-1)(x+1)^2(x^2+1)$ , we see $q(1) = -8$ , so the multiplicity $m = 1$ here as well.

Example 2.2.3.2

Let $p(x) = (2x-1)(x+1)(1-x^4)$ . Describe the behavior of the graph of $y = p(x)$ near each of the $x$ -intercepts.

Solution:

Describe the behavior of the graph of $y = p(x)$ near each of the $x$ -intercepts.

From Theorem 2.6, because the multiplicities of $x = \frac{1}{2}$ and $x = 1$ are both odd, we know the graph of $y = p(x)$ crosses through the $x$ -axis at $(\frac{1}{2}, 0)$ and $(1,0)$ . More specifically, due to the fact that the multiplicity for both of these zeros is $1$ , the graph will look locally linear at these points.

Based on our calculations above, near $x = \frac{1}{2}$ , the graph will resemble the increasing line $y = \frac{45}{16} (x - \frac{1}{2})$ , and near $x = 1$ , the graph will resemble the decreasing line $y = -8(x-1)$ .

As the multiplicity of $x = -1$ is even, we know the graph of $y = p(x)$ touches and rebounds at $(-1,0)$ . The multiplicity of $x=-1$ is $2$ , thus the rebound will look locally like a parabola. More specifically, the graph near $x = -1$ will resemble $y = -12(x+1)^2$ .

Example 2.2.3.3

Let $p(x) = (2x-1)(x+1)(1-x^4)$ . Determine the end behavior and $y$ -intercept of the graph of $y = p(x)$ .

Solution:

Determine the end behavior and $y$ -intercept of the graph of $y = p(x)$ .

Per Theorem 2.4, the end behavior of $y =p(x)$ , matches the end behavior of its leading term. As in Example 2.2.2, we multiply the leading terms from each factor together to obtain the leading term for $p(x)$ :

$\begin{array}{rcl} p(x) &=& (2x-1)(x+1)(1-x^4) \\ &=& (2x)(x)(-x^4) + \ldots \\ &=& -2x^6 + \ldots \end{array}$

The degree here, $6$ , is even and the leading coefficient $-2 <0$ , so we know as $x \rightarrow \pm \infty$ , $p(x) \rightarrow -\infty$ .

To find the $y$ -intercept, we determine $p(0) = (2(0)-1)(0+1)(1-0^4) = -1$ , hence, the $y$ -intercept is $(0,-1).$

Example 2.2.3.4

Let $p(x) = (2x-1)(x+1)(1-x^4)$ . Sketch $y = p(x)$ .

Solution:

Sketch $y = p(x)$ .

From the end behavior, $x \rightarrow -\infty$ , $p(x) \rightarrow -\infty$ , we start the graph in Quadrant III and head towards $(-1,0)$ .

At $(-1,0)$ , we `bounce’ off of the $x$ -axis and head towards the $y$ -intercept, $(0,-1)$ .

We then head towards $\left(\frac{1}{2}, 0 \right)$ and cross through the $x$ -axis there.

Finally, we head back to the $x$ -axis and cross through at $(1,0)$ .

Owing to the end behavior $x \rightarrow \infty$ , $p(x) \rightarrow -\infty$ , we exit the picture in Quadrant IV.

Remember polynomial functions are continuous and smooth, thus we have no holes or gaps in the graph, and all the `turns’ are rounded (no abrupt turns or corners.) We produce something resembling the next graph.

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A couple of remarks about Example 2.2.3 are in order. First, notice that the factor $(x^2+1)$ was more of a spectator in our discussion of the zeros of $p$ . Indeed, if we set $x^2+1 = 0$ , we have $x^2=-1$ which provides no real solutions.^[17] That being said, the factor $x^2+1$ does affect the shape of the graph. Next, when connecting up the graph from $(-1,0)$ to $(0,-1)$ to $\left(\frac{1}{2}, 0 \right)$ , there really is no way for us to know how low the graph goes, or where the lowest point is between $x = -1$ and $x = \frac{1}{2}$ unless we plot more points. Likewise, we have no idea how high the graph gets between $x = \frac{1}{2}$ and $x = 1$ . While there are ways to determine these points analytically, more often than not, finding them requires Calculus. As these points do play an important role in many applications, we’ll need to discuss them in this course and, when required, we’ll use technology to find them. For that reason, we have the following definition:

Definition 2.10

Suppose $f$ is a function with $f(a) = b$ .

We say $f$ has a local minimum at the point $(a,b)$ if and only if there is an open interval $I$ containing $a$ for which $f(a) \leq f(x)$ for all $x$ in $I$ . The value $f(a) = b$ is called `a local minimum value of $f$ .’ That is, $b$ is the minimum $f(x)$ value over an open interval containing $a$ . Graphically, no points `near’ a local minimum are lower than $(a,b)$ .
We say $f$ has a local maximum at the point $(a,b)$ if and only if there is an open interval $I$ containing $a$ for which $f(a) \geq f(x)$ for all $x$ in $I$ . The value $f(a) = b$ is called `a local maximum value of $f$ .’ That is, $b$ is the maximum $f(x)$ value over an open interval containing $a$ . Graphically, no points `near’ a local maximum are higher than $(a,b)$ .

Taken together, the local maximums and local minimums of a function, if they exist, are called the local extrema of the function.

Once again, the terminology used in Definition 2.10 blurs the line between the function $f$ and its outputs, $f(x)$ . Also, some textbooks use the terms `relative’ minimum and `relative’ maximum instead of the adjective `local.’ Lastly, note the definition of local extrema requires an open interval exist in the domain containing $a$ in order for $(a, f(a))$ to be a candidate for a local maximum or local minimum. We’ll have more to say about this in later chapters. If our open interval happens to be $(-\infty, \infty)$ , then our local extrema are the extrema of $f$ – we’ll see an example of this momentarily.

Below we use a graphing utility to graph $y= p(x) = (2x-1)(x+1)(1-x^4)$ . We first consider the point $(-1,0)$ . Even though there are points on the graph of $y = p(x)$ that are higher than $(-1,0)$ , locally, $(-1,0)$ is the top of a hill. To satisfy Definition 2.10, we need to provide an open interval on which $p(-1) = 0$ is the largest, or maximum function value. Note the definition requires us to provide \textit{just one} open interval. One that works is the interval $(-1.5, -0.5)$ . We could use any smaller interval or go as large as $\left(-\infty, \frac{1}{2} \right)$ (can you see why?) Next we encounter a `low’ point at approximately $(-0.2353, -1.1211)$ . More specifically, for all $x$ in the interval, say, $(-0.5, 0)$ , $p(x) \geq -1.1211$ , Hence, we have a local minimum at $(-0.2353, -1.1211)$ . Lastly, at $(0.811, 0.639)$ , we are back to a high point. In fact, $0.639$ isn’t just a local maximum value, based on the graph, it is the maximum of $p$ . Here, we may choose the open interval $(-\infty, \infty)$ as the open interval required by Definition 2.10, because for all $x$ , $p(x) \leq 0.639$ . It is important to note that there is no minimum value of $p$ despite there being a local minimum value.^[18]

The graph of p(x) with the local minimum and local maximums labeled. The local minimum was at (-0.235, -1.121), while the local maximums were at (-1,0) and (0.811, 0.639). — Graph of p(x) using technology

We close this section with a classic application of a third degree polynomial function.

Example 2.2.4

Example 2.2.4.1

A box with no top is to be fashioned from a $10$ inch $\times$ $12$ inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let $x$ denote the length of the side of the square which is removed from each corner.

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Write an expression for $V(x)$ , the volume of the box produced by removing squares of edge length $x$ . Include an appropriate domain.

Solution:

Write an expression for $V(x)$ , the volume of the box produced by removing squares of edge length $x$ . Include an appropriate domain.

From Geometry, we know that Volume = width $\times$ height $\times$ depth. The key is to find each of these quantities in terms of $x$ . From the figure, we see that the height of the box is $x$ itself. The cardboard piece is initially $10$ inches wide. Removing squares with a side length of $x$ inches from each corner leaves $10-2x$ inches for the width.^[19] As for the depth, the cardboard is initially $12$ inches long, so after cutting out $x$ inches from each side, we would have $12-2x$ inches remaining. Hence, we get

$V(x) = x(10-2x)(12-2x)$

To find a suitable applied domain, we note that to make a box at all we need $x > 0$ . Also the shorter of the two dimensions of the cardboard is $10$ inches, and as we are removing $2x$ inches from this dimension, we also require $10 - 2x > 0$ or $x < 5$ . Hence, our applied domain is $0 < x < 5$ .

Example 2.2.4.2

A box with no top is to be fashioned from a $10$ inch $\times$ $12$ inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let $x$ denote the length of the side of the square which is removed from each corner.

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Use a graphing utility to help you determine the value of $x$ which produces the box with the largest volume. What is the largest volume? Round your answers to two decimal places.

Solution:

Use a graphing utility to help you determine the value of $x$ which produces the box with the largest volume. What is the largest volume? Round your answers to two decimal places.

Using a graph and technology, we identify a local maximum at approximately $(1.811, 96.771)$ . Because the domain of $V$ is restricted to the interval $(0,5)$ , the maximum of $V$ is here as well.

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This means the maximum volume attainable is approximately 96.77 cubic inches when we remove squares of approximately 1.81 inches per side.

Notice that there is a very slight, but important, difference between the function $V(x) = x(10-2x)(12-2x)$ , $0 < x < 5$ from Example 2.2.4 and the function $p(x) = x(10-2x)(12-2x)$ : their domains. The domain of $V$ is restricted to the interval $(0,5)$ while the domain of $p$ is $(-\infty, \infty)$ . Indeed, the function $V$ has a maximum of (approximately) 96.771 at (approximately) x = 1.811 whereas for the function $p$ , 96.771 is a local maximum value only. We leave it to the reader to verify that $V$ has neither a minimum nor a local minimum.

2.2.3 Section Exercises

In Exercises 1 – 6, 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 2.2. Track at least three points of your choice through the transformations. State the domain and range of $g$ .

$f(x) = x^3$ and $F(x) = (x + 2)^{3} + 1$
$f(x) = x^4$ and $F(x) = (x + 2)^{4} + 1$
$f(x) = x^4$ and $F(x) = 2 - 3(x - 1)^{4}$
$f(x) = x^5$ and $F(x) = -x^{5} - 3$
$f(x) = x^5$ and $F(x) = (x+1)^5+10$
$f(x) = x^6$ and $F(x) = 8-x^6$

In Exercises 7 – 8, find a formula for each function below in the form $F(x) = a(x-h)^3+k$ .

In Exercises 9 – 10, find a formula for each function below in the form $F(x) = a(x-h)^4+k$ .

In Exercises 11 – 20, find the degree, the leading term, the leading coefficient, the constant term and the end behavior of the given polynomial function.

$f(x) = 4-x-3x^2$
$g(x) = 3x^5 - 2x^2 + x + 1$
$q(r) = 1 - 16r^{4}$
$Z(b) = 42b - b^{3}$
$f(x) = \sqrt{3}x^{17} + 22.5x^{10} - \pi x^{7} + \frac{1}{3}$
$s(t) = -4.9t^{2} + v_{0}t + s_{0}$
$P(x) = (x - 1)(x - 2)(x - 3)(x - 4)$
$p(t) = -t^2(3 - 5t)(t^{2} + t + 4)$
$f(x) = -2x^3(x+1)(x+2)^2$
$G(t) = 4(t-2)^2\left(t+\frac{1}{2}\right)$

In Exercises 21 – 30, find the real zeros of the given polynomial and their corresponding multiplicities. Use this information along with end behavior to provide a rough sketch of the graph of the polynomial function.

$a(x) = x(x + 2)^{2}$
$g(t) = t(t + 2)^{3}$
$f(z) = -2(z-2)^2(z+1)$
$g(x) = (2x+1)^2(x-3)$
$F(t) = t^{3}(t+ 2)^{2}$
$P(z) = (z- 1)(z - 2)(z - 3)(z - 4)$
$Q(x) = (x + 5)^{2}(x - 3)^{4}$
$h(t) = t^2(t-2)^2(t+2)^2$
$H(z) = (3-z)(z^2+1)$
$Z(x) = x(42 - x^{2})$

In Exercises 31 – 45, determine analytically if the following functions are even, odd or neither.

$f(x) = 7x$
$g(t) = 7t + 2$
$p(z) = 7$
$F(s) = 3s^2 - 4$
$h(t) = 4-t^2$
$g(x) = x^2-x-6$
$f(x) = 2x^3 - x$
$p(z) = -z^5 + 2z^3 - z$
$G(t) = t^{6} - t^{4} + t^{2} + 9$
$G(s) = s(s^2 - 1)$
$f(x) = (x^2+1)(x-1)$
$H(t) = (t^2-1)(t^4+t^2+3)$
$g(t) = t(t-2)(t+2)$
$P(z) = (2z^{5} - 3z)(5z^3+z)$
$f(x) =0$
Suppose is a polynomial function written in the form of Definition 2.7.
1. If the nonzero terms of $p(x)$ consist of even powers of $x$ (or a constant), explain why $p$ is even.
2. If the nonzero terms of $p(x)$ consist of odd powers of $x$ , explain why $p$ is odd.
3. If $p(x)$ the nonzero terms of $p(x)$ contain at least one odd power of $x$ and one even power of $x$ (or a constant term), then $p$ is neither even nor odd.
Use the results of Exercise 46 to determine whether the following functions are even, odd, or neither.
1. $p(x) = 3x^4 + x^2 - 1$
2. $F(s) = s^3 - 14s$
3. $f(t) = 2t^5 - t^2 + 1$
4. $g(x) =x^3(x^2+1)$
Show $f(x) = |x|$ is an even function.
Rework Example 2.2.4 assuming the box is to be made from an 8.5 inch by 11 inch sheet of paper. Using scissors and tape, construct the box. Are you surprised?^[20]
For each function $f(x)$ listed below, compute the average rate of change over the indicated interval.^[21] What trends do you observe? How do your answers manifest themselves graphically?
$\begin{array}{|r||c|c|c|c|c|c|} \hline f(x) & [-0.1, 0] & [0, 0.1] &[0.9, 1] & [1, 1.1] & [1.9, 2] & [2, 2.1] \\ \hline 1 &&&&&& \\ \hline x &&&&&& \\ \hline x^2 &&&&&& \\ \hline x^3 &&&&&& \\ \hline x^4 &&&&&& \\ \hline x^5 &&&&&& \\ \hline \end{array}$
For each function $f(x)$ listed below, compute the average rate of change over the indicated interval.^[22] What trends do you observe? How do your answers manifest themselves graphically?
$\begin{array}{|r||c|c|c|c|} \hline f(x) & [0.9, 1.1] & [0.99, 1.01] &[0.999, 1.001] & [0.9999, 1.0001] \\ \hline 1 &&&& \\ \hline x &&&& \\ \hline x^2 &&&& \\ \hline x^3 &&&& \\ \hline x^4 &&&& \\ \hline x^5 &&&& \\ \hline \end{array}$

In Exercises 52 – 54, suppose the revenue $R$ , in \textit{thousands} of dollars, from producing and selling $x$ \textit{hundred} LCD TVs is given by $R(x) = -5x^3+35x^2+155x$ for $0 \leq x \leq 10.07$ .

Graph $y = R(x)$ and determine the number of TVs which should be sold to maximize revenue. What is the maximum revenue?
Assume the cost, in thousands of dollars, to produce $x$ hundred LCD TVs is given by the function $C(x) = 200x + 25$ for $x \geq 0$ . Find and simplify an expression for the profit function $P(x)$ .(Remember: Profit = Revenue – Cost.)
Graph $y = P(x)$ and determine the number of TVs which should be sold to maximize profit. What is the maximum profit?
While developing their newest game, Sasquatch Attack!, the makers of the PortaBoy (from Example 1.3.8) revised their cost function and now use $C(x) = .03x^{3} - 4.5x^{2} + 225x + 250$ , for $x \geq 0$ . As before, $C(x)$ is the cost to make $x$ PortaBoy Game Systems. Market research indicates that the demand function $p(x) = -1.5x + 250$ remains unchanged. Use a graphing utility to find the production level $x$ that maximizes the profit made by producing and selling $x$ PortaBoy game systems.
According to US Postal regulations, a rectangular shipping box must satisfy the following inequality: “Length + Girth 130 inches” for Parcel Post and “Length + Girth 108 inches” for other services. Let’s assume we have a closed rectangular box with a square face of side length as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, .
1. Assuming that we’ll be mailing a box via Parcel Post where Length + Girth $=$ 130 inches, express the length of the box in terms of $x$ and then express the volume $V$ of the box in terms of $x$ .
2. Find the dimensions of the box of maximum volume that can be shipped via Parcel Post.
3. Repeat parts 56a and 56b if the box is shipped using “other services”.
Below is a graph of a polynomial function . Answer the following questions about based on the graph provided.
1. Describe the end behavior of $y = p(x)$ .
2. List the real zeros of $p$ along with their respective multiplicities.
3. List the local minimums and local maximums of the graph of $y = p(x)$ .
4. What can be said about the degree of and leading coefficient $p(x)$ ?
5. It turns out that $p(x)$ is a seventh degree polynomial.^[23] How can this be?
Use the graph of $y= p(x) = (2x-1)(x+1)(1-x^4)$ prior to Example 2.2.4 to estimate the largest open interval containing $x = -0.235$ which satisfies the the criteria for `local minimum’ in Definition 2.10.
(This is a follow-up to Exercises 96 in Section 1.3.1 and 33 in Section 2.1.) The Lagrange Interpolate function for four points: , , , where , , , and are four distinct real numbers is given by the formula:
$\begin{array}{rcl}L(x) & = & y_{0} \dfrac{(x - x_{1}) (x - x_{2}) (x-x_{3})}{(x_{0} - x_{1})(x_{0} - x_{2})(x_{0} - x_{3})}+ y_{1} \dfrac{(x - x_{0}) (x - x_{2}) (x-x_{3})}{(x_{1} - x_{0})(x_{1} - x_{2})(x_{1} - x_{3})} \\ [15pt] && +y_{2} \dfrac{(x - x_{0}) (x - x_{1}) (x-x_{3})}{(x_{2} - x_{0})(x_{2} - x_{1})(x_{2} - x_{3})}+ y_{3} \dfrac{(x - x_{0}) (x - x_{1}) (x-x_{2})}{(x_{3} - x_{0})(x_{3} - x_{1})(x_{3} - x_{2})} \\ \end{array}$
1. Choose four points with different $x$ -values and construct the Lagrange Interpolate for those points. Verify each of the points lies on the polynomial.
2. Verify that, in general, $L(x_{0}) = y_{0}$ , $L(x_{1}) = y_{1}$ , $L(x_{2}) = y_{2}$ , and $L(x_{3}) = y_{3}$ .
3. Find $L(x)$ for the points $(-1,1)$ , $(0,0)$ , $(1,1)$ and $(2,4)$ . What happens?
4. Find $L(x)$ for the points $(-1,0)$ , $(0,1)$ , $(1,2)$ and $(2,3)$ . What happens?
5. Generalize the formula for $L(x)$ to five points. What’s the pattern?

Section 2.2 Exercise Answers can be found in the Appendix … Coming soon

More specifically, $0^0$ is an indeterminate form. These are studied extensively in Calculus. ↵
This is why we do not describe monomial functions as having the form $f(x) = ax^n$ for any whole number $n$ . See Section 0.1.1. ↵
Recall that $|x| < 1$ is equivalent to $-1<x<1$ and $|x|>1$ is equivalent to $x<-1$ or $x>1$ . Using absolute values allow us to describe these sets of real numbers more succinctly. ↵
This should sound familiar - see the comments regarding the range of $f(x) = x^2$ in Section 2.1. ↵
Do you see the importance of $n$ being odd here? ↵
We are using the dummy variable $c$ here instead of $x$ for reasons that will become apparent shortly. ↵
That is, for a fixed number $h$ every real number $c$ can be written as $x-h$ for some real number $x$ , and every real number $x$ can be written as $c + h$ for some real number $c$ . ↵
Sometimes called the `long run' behavior. ↵
said differently, negative values that are larger in absolute value ↵
That is, the $f(x)$ values grow larger than any positive number. They are `unbounded.' ↵
Some authors say $f(x) = 0$ has degree $-\infty$ for reasons not even we will go into. ↵
We are considering $x \rightarrow \pm \infty$ , thus we are not concerned with $x$ even being close to $0$ , so these fractions will all be defined. ↵
Or at least they appear to within the limits of the technology. ↵
Both of which, by the way, can lead one astray, so we must proceed cautiously. ↵
Again, the formal definition of `continuity' and properties of continuous functions are discussed in Calculus. ↵
in accordance with the Zero Product Property of the Real Numbers - see Section 0.1. ↵
The solutions are $x = \pm i$ - see Section 0.5.6. ↵
Some books use the adjectives `global' or `absolute' when describing the extreme values of a function to distinguish them from their local counterparts. ↵
There's no harm in taking an extra step here and making sure this makes sense. If we chopped out a $1$ inch square from each side, then the width would be $8$ inches, so chopping out $x$ inches would leave $10-2x$ inches. ↵
Consider decorating the box and presenting it to your instructor. If done well enough, maybe your instructor will issue you some bonus points. Or maybe not. ↵
See Definition 1.11 in Section 1.3.4 for a review of this concept, as needed. ↵
See Definition 1.11 in Section 1.3.4 for a review of this concept, as needed. ↵
to be exact, $p(x) = -0.1\left(x+1.5\right)^2\left(3x\right)\left(x-1\right)^3\left(x+5\right)$ . ↵