Recognize characteristics of graphs of polynomial functions. Use factoring to find zeros of polynomial functions. Identify zeros and their multiplicities. Determine end behavior. Understand the relationship between degree and turning points. Graph polynomial functions. Use the Intermediate Value Theorem.

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The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table \(\PageIndex{1}\).

Table \(\PageIndex{1}\) Year Revenues
2006 2007 2008 2009 2010 2011 2012 2013
52.4 52.8 51.2 49.5 48.6 48.6 48.7 47.1

The revenue can be modeled by the polynomial function

\

where \(R\) represents the revenue in millions of dollars and \(t\) represents the year, with \(t=6\)corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general.


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 \(\PageIndex{1}\) shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.

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Figure \(\PageIndex{1}\)


Example \(\PageIndex{1}\): Recognizing Polynomial Functions

Which of the graphs in Figure \(\PageIndex{2}\) represents a polynomial function?

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api/deki/files/876/QA.png?revision=1" /> Do all polynomial functions have as their domain all real numbers?

Yes. Any real number is a valid input for a polynomial function.


Using Factoring to Find Zeros of Polynomial Functions

Recall that if \(f\) is a polynomial function, the values of \(x\) for which \(f(x)=0\) 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. (Also, any value \(x=a\) that is a zero of a polynomial function yields a factor of the polynomial, of the form \(x-a)\).(

We can use this method to find x-intercepts because at the x-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:

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

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Example \(\PageIndex{3}\): Finding the x-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts of \(f(x)=x^3−5x^2−x+5\).

Solution

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

\<\begin{align} x^3−5x^2−x+5&=0 &\text{Factor by grouping.} \\ x^2(x−5)−(x−5)&=0 &\text{Factor out the common factor.} \\ (x^2−1)(x−5)&=0 &\text{Factor the difference of squares.} \\ (x+1)(x−1)(x−5)&=0 &\text{Set each factor equal to zero.} \end{align}\>

\<\begin{align} x+1&=0 & &\text{or} & x−1&=0 & &\text{or} & x−5&=0 \\ x&=−1 &&& x&=1 &&& x&=5\end{align}\>

There are three x-intercepts: \((−1,0)\), \((1,0)\), and \((5,0)\). See Figure \(\PageIndex{4}\).

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api/deki/files/869/try-it.png?revision=1" /> \(\PageIndex{1}\)

Use a graphing utility (like Desmos) to find the y-and x-intercepts of the function \(f(x)=x^4−19x^2+30x\).

Answer y-intercept \((0,0)\); x-intercepts \((0,0)\), \((–5,0)\), \((2,0)\), and \((3,0)\)
Identifying Zeros and Their Multiplicities

Graphs behave differently at various x-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 \(\PageIndex{7}\) that the behavior of the function at each of the x-intercepts is different.

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Figure \(\PageIndex{8}\): Three graphs showing three different 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-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-axis.


Graphical Behavior of Polynomials at x-Intercepts

If a polynomial contains a factor of the form \((x−h)^p\), the behavior near the x-intercept \(h\) is determined by the power \(p\). We say that \(x=h\) is a zero of multiplicity \(p\).

The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities.

The sum of the multiplicities is no greater than the degree of the polynomial function.




api/deki/files/869/try-it.png?revision=1" /> \(\PageIndex{2}\)

Use the graph of the function of degree 5 in Figure \(\PageIndex{10}\) to identify the zeros of the function and their multiplicities.

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api/deki/files/13581/CNX_Precalc_Figure_03_04_011abcdn.jpg?revision=1&size=bestfit&width=697&height=824" />

Figure \(\PageIndex{11}\).


api/deki/files/13580/CNX_Precalc_Figure_03_04_015.jpg?revision=1" />

Figure \(\PageIndex{12}\): Graph of \(f(x)=x^4-x^3-4x^2+4x\)

This function \(f\) is a 4th 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 \(\PageIndex{7}\): Finding the Maximum possible Number of Turning Points Using the Degree of a Polynomial Function

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

\(f(x)=−x^3+4x^5−3x^2+1\) \(f(x)=−(x−1)^2(1+2x^2)\)

Solution

a. \(f(x)=−x^3+4x^5−3x^2+1\)

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 possible number of turning points is \(\; 5−1=4\).

b. \(f(x)=−(x−1)^2(1+2x^2)\)

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

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

The maximum possible number of turning points is \(\; 4−1=3\).




api/deki/files/871/how-to.png?revision=1" />Given a polynomial function, sketch the graph.

Find the intercepts, if possible. Check for symmetry. If the function is an even function, its graph is symmetrical about the y-axis, that is, \(f(−x)=f(x)\). If a function is an odd function, its graph is symmetrical about the origin, that is, \(f(−x)=−f(x)\). Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts. Determine the end behavior by examining the leading term. Use the end behavior and the behavior at the intercepts to sketch a graph. Ensure that the number of turning points does not exceed one less than the degree of the polynomial. Optionally, use technology to check the graph.

Example \(\PageIndex{8}\): Sketching the Graph of a Polynomial Function

Sketch a graph of \(f(x)=−2(x+3)^2(x−5)\).

Solution

This graph has two x-intercepts. At \(x=−3\), the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-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{align} f(0)&=−2(0+3)^2(0−5) \\ &=−2⋅9⋅(−5) \\ &=90 \end{align}\>

The y-intercept is \((0,90)\).

Additionally, we can see the leading term, if this polynomial were multiplied out, would be \(−2x3\), 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 \(\PageIndex{13}\).

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Figure \(\PageIndex{13}\): Showing the distribution for the leading term.

To sketch this, we consider that:

As \(x{\rightarrow}−{\infty}\) the function \(f(x){\rightarrow}{\infty}\),so we know the graph starts in the second quadrant and is decreasing toward the x-axis. Since \(f(−x)=−2(−x+3)^2(−x–5)\) is not equal to \(f(x)\), the graph does not display symmetry. At \((−3,0)\), the graph bounces off of thex-axis, so the function must start increasing. At \((0,90)\), the graph crosses the y-axis at the y-intercept. See Figure \(\PageIndex{14}\).

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Figure \(\PageIndex{14}\): Graph of the end behavior and intercepts, \((-3, 0)\) and \((0, 90)\), for the function \(f(x)=-2(x+3)^2(x-5)\).

Somewhere before or 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 \((5,0)\). See Figure \(\PageIndex{15}\).

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Figure \(\PageIndex{16}\): The complete graph of the polynomial function \(f(x)=−2(x+3)^2(x−5)\).


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Figure \(\PageIndex{17}\): Graph of \(f(x)=\frac{1}{6}(x−1)^3(x+2)(x+3)\)


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 \(x\) 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 \(x=4\).

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Analysis

We can also see on the graph of the function in Figure \(\PageIndex{19}\) that there are two real zeros between \(x=1\) and \(x=4\).