Graphs of Polynomial Functions

Recognizing Characteristics of Graphs of Polynomial Functions

Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.

Example 1

Recognizing Polynomial Functions

Which of the graphs in Figure 2 represents a polynomial function?

Solution

The graphs of $f$ and $h$ are graphs of polynomial functions. They are smooth and continuous.

The graphs of $g$ and $k$ are graphs of functions that are not polynomials. The graph of function $g$ has a sharp corner. The graph of function $k$ is not continuous.

Using Factoring to Find Zeros of Polynomial Functions

Recall that if $f$ is a polynomial function, the values of $x$ for which $f\left(x\right)=0$ are called zeros of $f.$ If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros.

We can use this method to find $x\text{-}$ intercepts because at the $x\text{-}$ intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases in this section:

1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring.
2. The polynomial is given in factored form.
3. Technology is used to determine the intercepts.

Example 2

Finding the x-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts of $f(x)=x^6-3x^4+2x^2.$

Solution

We can attempt to factor this polynomial to find solutions for $f\left(x\right)=0.$

$$\begin{array}{ll}\begin{array}{l}\\ x^6-3x^4+2x^2=0\end{array}\hfill & \begin{array}{l}\begin{array}{l}\\ \text{Factor out the greatest}\end{array}\hfill \\ \text{common factor}.\hfill \end{array}\hfill \\ x^2(x^4-3x^2+2)=0\hfill & \text{Factor the trinomial}.\hfill \\ x^2(x^2-1)(x^2-2)=0\hfill & \text{Set each factor equal to zero}.\hfill \end{array}$$

$$\begin{array}{lllllllll}\begin{array}{l}\\ \end{array}\hfill & \begin{array}{l}\\ \end{array}\hfill & \begin{array}{l}\\ \end{array}\hfill & \begin{array}{l}\\ \end{array}\hfill & \begin{array}{l}\\ (x^2-1)=0\end{array}\hfill & \begin{array}{l}\\ \end{array}\hfill & \begin{array}{l}\\ \end{array}\hfill & \begin{array}{l}\\ \end{array}\hfill & \begin{array}{l}\\ (x^2-2)=0\end{array}\hfill \\ x^2=0\hfill & \hfill & \text{or}\hfill & \hfill & x^2=1\hfill & \hfill & \text{or}\hfill & \hfill & x^2=2\hfill \\ x=0\hfill & \hfill & \hfill & \hfill & x=\pm 1\hfill & \hfill & \hfill & \hfill & x=\pm \sqrt{2}\hfill \end{array}$$

This gives us five $x\text{-}$ intercepts: $(0,0),(1,0),(-1,0),(\sqrt{2},0),$ and $(-\sqrt{2},0).$ See Figure 3. We can see that this is an even function.

Example 3

Finding the x-Intercepts of a Polynomial Function by Factoring

Find the $x\text{-}$ intercepts of $f(x)=x^3-5x^2-x+5.$

Solution

Find solutions for $f(x)=0$ by factoring.

$$\begin{array}{ll}\begin{array}{l}\hfill \\ x^3-5x^2-x+5=0\hfill \end{array}\hfill & \begin{array}{l}\hfill \\ \text{Factor by grouping}.\hfill \end{array}\hfill \\ x^2(x-5)-(x-5)=0\hfill & \text{Factor out the common factor}.\hfill \\ (x^2-1)(x-5)=0\hfill & \text{Factor the difference of squares}.\hfill \\ (x+1)(x-1)(x-5)=0\hfill & \text{Set each factor equal to zero}.\hfill \end{array}$$

$$\begin{array}{lllll}x+1=0\hfill & \text{or}\hfill & x-1=0\hfill & \text{or}\hfill & x-5=0\hfill \\ x=-1\hfill & \hfill & x=1\hfill & \hfill & x=5\hfill \end{array}$$

There are three $x\text{-}$ intercepts: $(-1,0),(1,0),$ and $\left(5,0\right).$ See Figure 4.

Example 4

Finding the y- and x-Intercepts of a Polynomial in Factored Form

Find the $y\text{-}$ and x-intercepts of $g(x)={(x-2)}^2(2x+3).$

Solution

The y-intercept can be found by evaluating $g\left(0\right).$

$$\begin{array}{c}g(0)={(0-2)}^2(2(0)+3)\\ =12\end{array}$$

So the y-intercept is $(0,12).$

The x-intercepts can be found by solving $g\left(x\right)=0.$

$${(x-2)}^2(2x+3)=0$$

$$\begin{array}{lllll}{(x-2)}^2=0\hfill & \hfill & \hfill & \hfill & (2x+3)=0\hfill \\ x-2=0\hfill & \hfill & \text{or}\hfill & \hfill & x=-\frac{3}{2}\hfill \\ x=2\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$$

So the $x\text{-}$ intercepts are $(2,0)$ and $\left(-\frac{3}{2},0\right).$

Analysis

We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5.

Example 5

Finding the x-Intercepts of a Polynomial Function Using a Graph

Find the $x\text{-}$ intercepts of $h(x)=x^3+4x^2+x-6.$

Solution

This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown in Figure 6, it appears that there are x-intercepts at $x=\mathrm{-3},\mathrm{-2},$ and $1.$

We can check whether these are correct by substituting these values for $x$ and verifying that

$$h(-3)=h(-2)=h(1)=0.$$

Since $h(x)=x^3+4x^2+x-6,$ we have:

$$\begin{array}{l}h(-3)={(-3)}^3+4{(-3)}^2+(-3)-6=-27+36-3-6=0\hfill \\ h(-2)={(-2)}^3+4{(-2)}^2+(-2)-6=-8+16-2-6=0\hfill \\ h(1)={(1)}^3+4{(1)}^2+(1)-6=1+4+1-6=0\hfill \end{array}$$

Each $x\text{-}$ intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.

$$\begin{array}{l}h(x)=x^3+4x^2+x-6\hfill \\ =(x+3)(x+2)(x-1)\hfill \end{array}$$

Identifying Zeros and Their Multiplicities

Graphs behave differently at various $x\text{-}$ intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and bounce off.

Suppose, for example, we graph the function

Notice in Figure 7 that the behavior of the function at each of the $x\text{-}$ intercepts is different.

The $x\text{-}$ intercept $\mathrm{-3}$ is the solution of equation $(x+3)=0.$ The graph passes directly through the $x\text{-}$ intercept at $x=\mathrm{-3.}$ The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.

The $x\text{-}$ intercept $2$ is the repeated solution of equation ${(x-2)}^2=0.$ The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.

The factor is repeated, that is, the factor $\left(x-2\right)$ appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, $x=2,$ has multiplicity 2 because the factor $\left(x-2\right)$ occurs twice.

The $x\text{-}$ intercept $-1$ is the repeated solution of factor ${(x+1)}^3=0.$ The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function $f\left(x\right)=x^3.$ We call this a triple zero, or a zero with multiplicity 3.

For zeros with even multiplicities, the graphs touch or are tangent to the $x\text{-}$ axis. For zeros with odd multiplicities, the graphs cross or intersect the $x\text{-}$ axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3.

For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the $x\text{-}$ axis.

For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the $x\text{-}$ axis.

Example 6

Identifying Zeros and Their Multiplicities

Use the graph of the function of degree 6 in Figure 9 to identify the zeros of the function and their possible multiplicities.

Solution

The polynomial function is of degree $n.$ The sum of the multiplicities must be $n.$

Starting from the left, the first zero occurs at $x=\mathrm{-3.}$ The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of $\mathrm{-3}$ has multiplicity $2.$

The next zero occurs at $x=\mathrm{-1.}$ The graph looks almost linear at this point. This is a single zero of multiplicity 1.

The last zero occurs at $x=4.$ The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6.

Determining End Behavior

As we have already learned, the behavior of a graph of a polynomial function of the form

will either ultimately rise or fall as $x$ increases without bound and will either rise or fall as $x$ decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000.

Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, $a_n{x}^n,$ is an even power function, as $x$ increases or decreases without bound, $f(x)$ increases without bound. When the leading term is an odd power function, as $x$ decreases without bound, $f(x)$ also decreases without bound; as $x$ increases without bound, $f(x)$ also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 11 summarizes all four cases.

Understanding the Relationship between Degree and Turning Points

In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function $f(x)=x^4-x^3-4x^2+4x$ in Figure 12. The graph has three turning points.

This function $f$ is a 4^th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.

Example 7

Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function

Find the maximum number of turning points of each polynomial function.

1. ⓐ $f(x)=-x^3+4x^5-3x^2+1$
2. ⓑ $f(x)=-{\left(x-1\right)}^2\left(1+2x^2\right)$

Solution

1. ⓐ $f(x)=-x{}^3+4x^5-3x^2+1$

   First, rewrite the polynomial function in descending order: $f(x)=4x^5-x^3-3x^2+1$

   Identify the degree of the polynomial function. This polynomial function is of degree 5.

   The maximum number of turning points is $5-1=4.$

2. ⓑ $f(x)=-{\left(x-1\right)}^2\left(1+2x^2\right)$

First, identify the leading term of the polynomial function if the function were expanded.

Then, identify the degree of the polynomial function. This polynomial function is of degree 4.

The maximum number of turning points is $4-1=3.$

Graphing Polynomial Functions

We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.

Example 8

Sketching the Graph of a Polynomial Function

Sketch a graph of $f(x)=\mathrm{-2}{(x+3)}^2(x-5).$

Solution

This graph has two $x\text{-}$ intercepts. At $x=\mathrm{-3},$ the factor is squared, indicating a multiplicity of 2. The graph will bounce at this $x\text{-}$ intercept. At $x=5,$ the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.

The y-intercept is found by evaluating $f(0).$

$$\begin{array}{l}\hfill \\ f(0)=-2{(0+3)}^2(0-5)\hfill \\ =-2\cdot 9\cdot (-5)\hfill \\ =90\hfill \end{array}$$

The $y\text{-}$ intercept is $(0,90).$

Additionally, we can see the leading term, if this polynomial were multiplied out, would be $-2x^3,$ so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 13.

To sketch this, we consider that:

• As $x\to -\infty$ the function $f(x)\to \infty ,$ so we know the graph starts in the second quadrant and is decreasing toward the $x\text{-}$ axis.
• Since $f\left(-x\right)=\mathrm{-2}{\left(-x+3\right)}^2\left(-x–5\right)$ is not equal to $f\left(x\right),$ the graph does not display symmetry.
• At $\left(-3,0\right),$ the graph bounces off of the $x\text{-}$ axis, so the function must start increasing.

  At $\left(0,90\right),$ the graph crosses the $y\text{-}$ axis at the $y\text{-}$ intercept. See Figure 14.

Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at $\left(5,0\right).$ See Figure 15.

As $x\to \infty$ the function $f(x)\to \mathrm{-\infty },$ so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.

Using technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like our sketch in Figure 15.

Using the Intermediate Value Theorem

In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial function $f$ whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers $a$ and $b$ in the domain of $f,$ if $a<b$ and $f\left(a\right)\ne f\left(b\right),$ then the function $f$ takes on every value between $f\left(a\right)$ and $f\left(b\right).$ We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function $f$ at $x=a$ lies above the $x\text{-}$ axis and another point at $x=b$ lies below the $x\text{-}$ axis, there must exist a third point between $x=a$ and $x=b$ where the graph crosses the $x\text{-}$ axis. Call this point $\left(c,f\left(c\right)\right).$ This means that we are assured there is a solution $c$ where $f\left(c\right)=0.$

In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the $x\text{-}$ axis. Figure 17 shows that there is a zero between $a$ and $b.$

Example 9

Using the Intermediate Value Theorem

Show that the function $f(x)=x^3-5x^2+3x+6$ has at least two real zeros between $x=1$ and $x=4.$

Solution

As a start, evaluate $f(x)$ at the integer values $x=1,2,3,\text{and}4.$ See Table 2.

Table 2 ..

$x$	1	2	3	4
$f(x)$	5	0	–3	2

We see that one zero occurs at $x=2.$ Also, since $f(3)$ is negative and $f(4)$ is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.

We have shown that there are at least two real zeros between $x=1$ and $x=4.$

Analysis

We can also see on the graph of the function in Figure 18 that there are two real zeros between $x=1$ and $x=4.$

Writing Formulas for Polynomial Functions

Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an $x\text{-}$ intercept where each factor is equal to zero, we can form a function that will pass through a set of $x\text{-}$ intercepts by introducing a corresponding set of factors.

Example 10

Writing a Formula for a Polynomial Function from the Graph

Write a formula for the polynomial function shown in Figure 19.

Solution

This graph has three $x\text{-}$ intercepts: $x=\mathrm{-3},2,$ and $5.$ The $y\text{-}$ intercept is located at $(0,-2).$ At $x=\mathrm{-3}$ and $x=5,$ the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At $x=2,$ the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us

$$f(x)=a(x+3){(x-2)}^2(x-5)$$

To determine the stretch factor, we utilize another point on the graph. We will use the $y\text{-}$ intercept $(0,–2),$ to solve for $a.$

$$\begin{array}{l}f(0)=a(0+3){(0-2)}^2(0-5)\hfill \\ -2=a(0+3){(0-2)}^2(0-5)\hfill \\ -2=-60a\hfill \\ a=\frac{1}{30}\hfill \end{array}$$

The graphed polynomial appears to represent the function $f(x)=\frac{1}{30}(x+3){(x-2)}^2(x-5).$

Using Local and Global Extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function.

Example 11

Using Local Extrema to Solve Applications

An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.

Solution

We will start this problem by drawing a picture like that in Figure 22, labeling the width of the cut-out squares with a variable, $w.$

Notice that after a square is cut out from each end, it leaves a $\left(14-2w\right)$ cm by $\left(20-2w\right)$ cm rectangle for the base of the box, and the box will be $w$ cm tall. This gives the volume

$$\begin{array}{l}V(w)=(20-2w)(14-2w)w\hfill \\ =280w-68w^2+4w^3\hfill \end{array}$$

Notice, since the factors are $w,$ $20–2w$ and $14–2w,$ the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values $w$ may take on are greater than zero or less than 7. This means we will restrict the domain of this function to $0<w<7.$ Using technology to sketch the graph of $V\left(w\right)$ on this reasonable domain, we get a graph like that in Figure 23. We can use this graph to estimate the maximum value for the volume, restricted to values for $w$ that are reasonable for this problem—values from 0 to 7.

From this graph, we turn our focus to only the portion on the reasonable domain, $\left[0,7\right].$ We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 24.

From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side.