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Graphs of Polynomial Functions

By the end of this section, you will be able to:

  • 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

The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown below.

Year20062007200820092010201120122013
Revenues52.452.851.249.548.648.648.747.1

The revenue can be modeled by the polynomial function

R(t)=0.037t4+1.414t319.777t2+118.696t205.332R(t)=-0.037t^{4}+1.414t^{3}-19.777t^{2}+118.696t-205.332

where RR represents the revenue in millions of dollars and tt represents the year, with t=6t=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. The figures below show a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.

Example. Which of the four graphs below represents a polynomial function?

Solution. The graphs of ff and hh are graphs of polynomial functions. They are smooth and continuous.

The graphs of gg and kk are graphs of functions that are not polynomials. The graph of function gg has a sharp corner. The graph of function kk is not continuous.

Q&A. 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 ff is a polynomial function, the values of xx for which f(x)=0f(x)=0 are called zeros of ff. 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 xx-intercepts because at the xx-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 and trinomial factoring.
  • The polynomial is given in factored form.
  • Technology is used to determine the intercepts.

How To: given a polynomial function ff, find the xx-intercepts by factoring.

  1. Set f(x)=0f(x)=0.
  2. If the polynomial function is not given in factored form:
    1. Factor out any common monomial factors.
    2. Factor any factorable binomials or trinomials.
  3. Set each factor equal to zero and solve to find the xx-intercepts.

Example. Find the xx-intercepts of f(x)=x63x4+2x2f(x)=x^{6}-3x^{4}+2x^{2}.

Solution. We can attempt to factor this polynomial to find solutions for f(x)=0f(x)=0.

x63x4+2x2=0Factor out the greatest common factor.x2(x43x2+2)=0Factor the trinomial.x2(x21)(x22)=0 \begin{array}{lrcl} & x^{6}-3x^{4}+2x^{2} &=& 0 \\[4pt] \text{Factor out the greatest common factor.} & x^{2}(x^{4}-3x^{2}+2) &=& 0 \\[4pt] \text{Factor the trinomial.} & x^{2}(x^{2}-1)(x^{2}-2) &=& 0 \end{array}

Set each factor equal to zero.

x2=0orx21=0orx22=0x=0x2=1x2=2x=±1x=±2 \begin{array}{rclcrclcrcl} x^{2} &=& 0 & \text{or} & x^{2}-1 &=& 0 & \text{or} & x^{2}-2 &=& 0 \\[4pt] x &=& 0 & & x^{2} &=& 1 & & x^{2} &=& 2 \\[4pt] & & & & x &=& \pm 1 & & x &=& \pm\sqrt{2} \end{array}

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

Example. Find the xx-intercepts of f(x)=x35x2x+5f(x)=x^{3}-5x^{2}-x+5.

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

x35x2x+5=0Factor by grouping.x2(x5)(x5)=0Factor out the common factor.(x21)(x5)=0Factor the difference of squares.(x+1)(x1)(x5)=0 \begin{array}{lrcl} & x^{3}-5x^{2}-x+5 &=& 0 \\[4pt] \text{Factor by grouping.} & x^{2}(x-5)-(x-5) &=& 0 \\[4pt] \text{Factor out the common factor.} & (x^{2}-1)(x-5) &=& 0 \\[4pt] \text{Factor the difference of squares.} & (x+1)(x-1)(x-5) &=& 0 \end{array} x+1=0orx1=0orx5=0x=1x=1x=5 \begin{array}{rclcrclcrcl} x+1 &=& 0 & \text{or} & x-1 &=& 0 & \text{or} & x-5 &=& 0 \\[4pt] x &=& -1 & & x &=& 1 & & x &=& 5 \end{array}

There are three xx-intercepts: (1,0)(-1,0), (1,0)(1,0), and (5,0)(5,0).

Example. Find the yy- and xx-intercepts of g(x)=(x2)2(2x+3)g(x)=(x-2)^{2}(2x+3).

Solution. The yy-intercept can be found by evaluating g(0)g(0).

g(0)=(02)2(2(0)+3)=12 \begin{array}{lrcl} g(0) &=& (0-2)^{2}(2(0)+3) \\[4pt] &=& 12 \end{array}

So the yy-intercept is (0,12)(0,12).

The xx-intercepts can be found by solving g(x)=0g(x)=0.

(x2)2(2x+3)=0(x-2)^{2}(2x+3)=0(x2)2=0(2x+3)=0x2=0orx=32x=2 \begin{array}{rclcccrcl} (x-2)^{2} &=& 0 & & & & (2x+3) &=& 0 \\[4pt] x-2 &=& 0 & \text{or} & & & x &=& -\tfrac{3}{2} \\[4pt] x &=& 2 \end{array}

So the xx-intercepts are (2,0)(2,0) and (32,0)\left(-\tfrac{3}{2},0\right).

Analysis. We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial, shown below.

Example. Find the xx-intercepts of h(x)=x3+4x2+x6h(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, shown below, it appears that there are xx-intercepts at x=3x=-3, 2-2, and 11.

We can check whether these are correct by substituting these values for xx and verifying that h(3)=h(2)=h(1)=0h(-3)=h(-2)=h(1)=0.

h(3)=(3)3+4(3)2+(3)6=27+3636=0h(2)=(2)3+4(2)2+(2)6=8+1626=0h(1)=(1)3+4(1)2+(1)6=1+4+16=0 \begin{array}{rclclcl} h(-3) &=& (-3)^{3}+4(-3)^{2}+(-3)-6 &=& -27+36-3-6 &=& 0 \\[4pt] h(-2) &=& (-2)^{3}+4(-2)^{2}+(-2)-6 &=& -8+16-2-6 &=& 0 \\[4pt] h(1) &=& (1)^{3}+4(1)^{2}+(1)-6 &=& 1+4+1-6 &=& 0 \end{array}

Each xx-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.

h(x)=x3+4x2+x6=(x+3)(x+2)(x1) \begin{array}{lrcl} h(x) &=& x^{3}+4x^{2}+x-6 \\[4pt] &=& (x+3)(x+2)(x-1) \end{array}

Find the x-intercepts off(x)=x419x2+30xf(x)=x^4-19x^2+30x. Enter all of them, separated by commas.

Identifying zeros and their multiplicities

Graphs behave differently at various xx-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

f(x)=(x+3)(x2)2(x+1)3f(x)=(x+3)(x-2)^{2}(x+1)^{3}

Notice in the figure below that the behavior of the function at each of the xx-intercepts is different.

The xx-intercept 3-3 is the solution of equation (x+3)=0(x+3)=0. The graph passes directly through the xx-intercept at x=3x=-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 xx-intercept 22 is the repeated solution of equation (x2)2=0(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.

(x2)2=(x2)(x2)(x-2)^{2}=(x-2)(x-2)

The factor is repeated, that is, the factor (x2)(x-2) 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=2x=2, has multiplicity 2 because the factor (x2)(x-2) occurs twice.

The xx-intercept 1-1 is the repeated solution of factor (x+1)3=0(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(x)=x3f(x)=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 xx-axis. For zeros with odd multiplicities, the graphs cross or intersect the xx-axis. The three figures below show examples of graphs of polynomial functions with multiplicity 1, 2, and 3.

Single zero.

Zero with multiplicity 2.

Zero with multiplicity 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 xx-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 xx-axis.

Graphical behavior of polynomials at xx-intercepts. If a polynomial contains a factor of the form (xh)p(x-h)^{p}, the behavior near the xx-intercept hh is determined by the power pp. We say that x=hx=h is a zero of multiplicity pp.

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

The sum of the multiplicities is the degree of the polynomial function.

How To: given a graph of a polynomial function of degree nn, identify the zeros and their multiplicities.

  1. If the graph crosses the xx-axis and appears almost linear at the intercept, it is a single zero.
  2. If the graph touches the xx-axis and bounces off of the axis, it is a zero with even multiplicity.
  3. If the graph crosses the xx-axis at a zero, it is a zero with odd multiplicity.
  4. The sum of the multiplicities is nn.

Example. Use the graph of the degree-6 function below to identify its zeros and their possible multiplicities.

Solution. The polynomial function is of degree nn. The sum of the multiplicities must be nn.

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

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

The last zero occurs at x=4x=4. The graph crosses the xx-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.

Source note. OpenStax’s current source calls this “the function of degree 9”, but its own printed answer gives multiplicities 33, 22, and 22, which sum to 77, and the figure’s description calls it degree 66 — three different degrees for one graph. The printed edition of the book, and every reading consistent with the answer and the drawn curve, says 77, so this page says 77; the mathematics is unchanged.

The graph below shows a degree-7 polynomial function. What are its zeros and their multiplicities?

Determining end behavior

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

f(x)=anxn+an1xn1++a1x+a0f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}

will either ultimately rise or fall as xx increases without bound and will either rise or fall as xx 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 100-100 or 1,000-1{,}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, anxna_{n}x^{n}, is an even power function, as xx increases or decreases without bound, f(x)f(x) increases without bound. When the leading term is an odd power function, as xx decreases without bound, f(x)f(x) also decreases without bound; as xx increases without bound, f(x)f(x) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. The four cases below summarize every combination.

Even degree, positive leading coefficient (an>0a_n>0).

End behavior: as xx\to\infty, f(x)f(x)\to\infty; as xx\to-\infty, f(x)f(x)\to\infty.

Odd degree, positive leading coefficient (an>0a_n>0).

End behavior: as xx\to\infty, f(x)f(x)\to\infty; as xx\to-\infty, f(x)f(x)\to-\infty.

Even degree, negative leading coefficient (an<0a_n<0).

End behavior: as xx\to\infty, f(x)f(x)\to-\infty; as xx\to-\infty, f(x)f(x)\to-\infty.

Odd degree, negative leading coefficient (an<0a_n<0).

End behavior: as xx\to\infty, f(x)f(x)\to-\infty; as xx\to-\infty, f(x)f(x)\to\infty.

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)=x4x34x2+4xf(x)=x^{4}-x^{3}-4x^{2}+4x below. The graph has three turning points.

This function ff 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.

Interpreting turning points. A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).

A polynomial of degree nn will have at most n1n-1 turning points.

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

  1. f(x)=x3+4x53x2+1f(x)=-x^{3}+4x^{5}-3x^{2}+1
  2. f(x)=(x1)2(1+2x2)f(x)=-(x-1)^{2}(1+2x^{2})

Solution.

  1. First, rewrite the polynomial function in descending order: f(x)=4x5x33x2+1f(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 51=45-1=4.
  2. First, identify the leading term of the polynomial function if the function were expanded: an=(x2)(2x2)=2x4a_n=-(x^2)(2x^2)=-2x^4. Then identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is 41=34-1=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.

How To: given a polynomial function, sketch the graph.

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

Example. Sketch a graph of f(x)=2(x+3)2(x5)f(x)=-2(x+3)^{2}(x-5).

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

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

f(0)=2(0+3)2(05)=29(5)=90 \begin{array}{lrcl} f(0) &=& -2(0+3)^{2}(0-5) \\[4pt] &=& -2\cdot 9\cdot(-5) \\[4pt] &=& 90 \end{array}

The yy-intercept is (0,90)(0,90).

Additionally, we can see that the leading term, if this polynomial were multiplied out, would be 2x3-2x^{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. The figure below shows only that end behavior, before the rest of the graph is filled in.

To sketch this, we consider that:

  • As xx\to-\infty the function f(x)f(x)\to\infty, so we know the graph starts in the second quadrant and is decreasing toward the xx-axis.
  • Since f(x)=2(x+3)2(x5)f(-x)=-2(-x+3)^{2}(-x-5) is not equal to f(x)f(x), the graph does not display symmetry.
  • At (3,0)(-3,0), the graph bounces off of the xx-axis, so the function must start increasing. At (0,90)(0,90), the graph crosses the yy-axis at the yy-intercept.
  • 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 (5,0)(5,0).
  • As xx\to\infty the function f(x)f(x)\to-\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 complete graph for the polynomial function, shown below, and verify that the resulting graph looks like our sketch.

A Try It sketchesf(x)=14x(x1)4(x+3)3f(x)=\tfrac{1}{4}x(x-1)^4(x+3)^3. This function has three x-intercepts. Find them, separated by commas.

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 xx-axis, we can confirm that there is a zero between them. Consider a polynomial function ff whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers aa and bb in the domain of ff, if a<ba<b and f(a)f(b)f(a)\ne f(b), then the function ff takes on every value between f(a)f(a) and f(b)f(b).

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 ff at x=ax=a lies above the xx-axis and another point at x=bx=b lies below the xx-axis, there must exist a third point between x=ax=a and x=bx=b where the graph crosses the xx-axis. Call this point (c,f(c))(c,f(c)). This means that we are assured there is a solution cc where f(c)=0f(c)=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 xx-axis. The figure below shows that there is a zero between aa and bb.

Use of the Intermediate Value Theorem. Let ff be a polynomial function over an interval [a,b][a,b]. Bolzano’s Theorem (a corollary to the Intermediate Value Theorem) states that if f(a)f(a) and f(b)f(b) have opposite signs, then there exists at least one value cc between aa and bb for which f(c)=0f(c)=0.

Example. Show that the function f(x)=x35x2+3x+6f(x)=x^{3}-5x^{2}+3x+6 has at least two real zeros between x=1x=1 and x=4x=4.

Solution. As a start, evaluate f(x)f(x) at the integer values x=1,2,3x=1,2,3, and 44. See the table below.

xx1234
f(x)f(x)50−32

We see that one zero occurs at x=2x=2. Also, since f(3)f(3) is negative and f(4)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=1x=1 and x=4x=4.

Analysis. We can also see on the graph of the function below that there are two real zeros between x=1x=1 and x=4x=4.

Which pair of values confirms, by the Intermediate Value Theorem, that f(x)=7x^5-9x^4-x^2 has at least one real zero between x=1 and x=2?

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 xx-intercept where each factor is equal to zero, we can form a function that will pass through a set of xx-intercepts by introducing a corresponding set of factors.

Factored form of polynomials. If a polynomial of degree pp that can be factored into strictly linear factors has horizontal intercepts at x=x1,x2,,xnx=x_{1},x_{2},\dots,x_{n}, then the polynomial can be written in the factored form

f(x)=a(xx1)p1(xx2)p2(xxn)pnf(x)=a(x-x_{1})^{p_{1}}(x-x_{2})^{p_{2}}\cdots(x-x_{n})^{p_{n}}

where the powers pip_{i} on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor aa can be determined given a value of the function other than the xx-intercept.

How To: given a graph of a polynomial function, write a formula for the function.

  1. Identify the xx-intercepts of the graph to find the factors of the polynomial.
  2. Examine the behavior of the graph at the xx-intercepts to determine the multiplicity of each factor.
  3. Find the polynomial of least degree containing all the factors found in the previous step.
  4. Use any other point on the graph (the yy-intercept may be easiest) to determine the stretch factor.

Example. Write a formula for the polynomial function shown below.

Solution. This graph has three xx-intercepts: x=3x=-3, 22, and 55. The yy-intercept is located at (0,2)(0,-2). At x=3x=-3 and x=5x=5, the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At x=2x=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)(x2)2(x5)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 yy-intercept (0,2)(0,-2), to solve for aa.

f(0)=a(0+3)(02)2(05)2=a(0+3)(02)2(05)2=60aa=130 \begin{array}{lrcl} f(0) &=& a(0+3)(0-2)^{2}(0-5) \\[4pt] -2 &=& a(0+3)(0-2)^{2}(0-5) \\[4pt] -2 &=& -60a \\[4pt] a &=& \tfrac{1}{30} \end{array}

The graphed polynomial appears to represent the function f(x)=130(x+3)(x2)2(x5)f(x)=\tfrac{1}{30}(x+3)(x-2)^{2}(x-5).

Write a formula, in factored form, for the polynomial function shown in the figure above.

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.

Local and global extrema. A local maximum or local minimum at x=ax=a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around x=ax=a. If a function has a local maximum at aa, then f(a)f(x)f(a)\ge f(x) for all xx in an open interval around x=ax=a. If a function has a local minimum at aa, then f(a)f(x)f(a)\le f(x) for all xx in an open interval around x=ax=a.

A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at aa, then f(a)f(x)f(a)\ge f(x) for all xx. If a function has a global minimum at aa, then f(a)f(x)f(a)\le f(x) for all xx.

We can see the difference between local and global extrema in the figure below.

Q&A. Do all polynomial functions have a global minimum or maximum?

No. Only polynomial functions of even degree have a global minimum or maximum. For example, f(x)=xf(x)=x has neither a global maximum nor a global minimum.

Example. 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 the one below, labeling the width of the cut-out squares with a variable, ww.

Notice that after a square is cut out from each end, it leaves a (142w)(14-2w) cm by (202w)(20-2w) cm rectangle for the base of the box, and the box will be ww cm tall. This gives the volume

V(w)=(202w)(142w)w=280w68w2+4w3 \begin{array}{lrcl} V(w) &=& (20-2w)(14-2w)w \\[4pt] &=& 280w-68w^{2}+4w^{3} \end{array}

Notice, since the factors are ww, 202w20-2w, and 142w14-2w, the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values ww may take on are greater than zero or less than 7. This means we will restrict the domain of this function to 0<w<70<w<7. Using technology to sketch the graph of V(w)V(w) on this reasonable domain, we get a graph like the one below. We can use this graph to estimate the maximum value for the volume, restricted to values for ww 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, [0,7][0,7]. 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 the figure below.

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.

Using the figure, what are the approximate local minimum and local maximum values of f(x)=-0.2(x-2)^3(x+1)^2(x-4) on [-1,4]? Enter the local minimum value first, then the local maximum value, separated by a comma.

Key concepts

  • Polynomial functions of degree 2 or more are smooth, continuous functions.
  • To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero.
  • Another way to find the xx-intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the xx-axis.
  • The multiplicity of a zero determines how the graph behaves at the xx-intercepts.
  • The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity.
  • The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity.
  • The end behavior of a polynomial function depends on the leading term.
  • The graph of a polynomial function changes direction at its turning points.
  • A polynomial function of degree nn has at most n1n-1 turning points.
  • To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most n1n-1 turning points.
  • Graphing a polynomial function helps to estimate local and global extrema.
  • The Intermediate Value Theorem tells us that if f(a)f(a) and f(b)f(b) have opposite signs, then there exists at least one value cc between aa and bb for which f(c)=0f(c)=0.

Key terms

global maximum — the highest turning point on a graph; f(a)f(a) where f(a)f(x)f(a)\ge f(x) for all xx. global minimum — the lowest turning point on a graph; f(a)f(a) where f(a)f(x)f(a)\le f(x) for all xx. Intermediate Value Theorem — for two numbers aa and bb in the domain of ff, if a<ba<b and f(a)f(b)f(a)\ne f(b), then the function ff takes on every value between f(a)f(a) and f(b)f(b); specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the xx-axis. multiplicity — the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form (xh)p(x-h)^{p}, x=hx=h is a zero of multiplicity pp.

Practice

Recognize characteristics of graphs of polynomial functions

For a polynomial function f, which statement correctly relates its x-intercepts to its zeros?

What is the domain of f(x)=x^3+x^2-20x?

Use factoring to find zeros of polynomial functions

Find the zeros of C(t)=3(t+2)(t-3)(t+5). Enter all of them, separated by commas.

Find the zeros of f(x)=x^3+x^2-4x-4. Enter all of them, separated by commas.

Identify zeros and their multiplicities

What is the multiplicity of the zero x=-3/2 for f(x)=x^2(2x+3)^5(x-4)^2?

Which statement correctly describes the zeros of f(x)=x^2(x^2+4x+4)?

If the graph of a polynomial function touches the x-axis at a zero and then changes direction without crossing, what must be true of that zero’s multiplicity?

Determine end behavior

As x approaches infinity, what happens to g(x)=(x+4)(x-1)^2?

As x approaches negative infinity, what happens to n(x)=-3x(x+2)(x-4)?

Understand the relationship between degree and turning points

What is the maximum number of turning points on the graph of f(x)=x^5-5x^3+4x?

What is the maximum number of turning points on the graph of f(x)=2x^3-x^2-8x+4?

Graph polynomial functions

Write a formula, in factored form, for the degree-3 polynomial function with zeros at x=-2, x=1, and x=3, and y-intercept (0,-4).

Write a formula, in factored form, for the degree-3 polynomial function with zeros at x=-3, x=-2, and x=1, and y-intercept (0,12).

Write a formula, in factored form, for the degree-3 polynomial function shown in the figure above.

Use the Intermediate Value Theorem

Explain how the Intermediate Value Theorem can assist us in finding a zero of a function.

Which pair of values confirms, by the Intermediate Value Theorem, that f(x)=x^3-9x has at least one real zero between x=2 and x=4?

Which pair of values confirms, by the Intermediate Value Theorem, that f(x)=-x^4+4 has at least one real zero between x=1 and x=3?


This section is adapted from Precalculus 2e, Section 3.4: Graphs of Polynomial Functions by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated every graph as an accessible inline SVG generated from its exact polynomial coefficients (expanded by hand from each factored form) — the smooth/discontinuous comparison pair, the four-panel “which graph is a polynomial” figure, f(x)=x63x4+2x2f(x)=x^6-3x^4+2x^2, f(x)=x35x2x+5f(x)=x^3-5x^2-x+5, g(x)=(x2)2(2x+3)g(x)=(x-2)^2(2x+3), h(x)=x3+4x2+x6h(x)=x^3+4x^2+x-6, f(x)=(x+3)(x2)2(x+1)3f(x)=(x+3)(x-2)^2(x+1)^3, the single-zero/multiplicity-2/multiplicity-3 panels, a fitted degree-6 example with zeros 3-3 (mult. 2), 1-1 (mult. 1), 44 (mult. 3), a fitted degree-7 Try-It with zeros 5-5 (mult. 3), 1-1 (mult. 2), 33 (mult. 2), the four end-behavior corner-arc panels, f(x)=x4x34x2+4xf(x)=x^4-x^3-4x^2+4x with its turning points labeled, the end-behavior tails and complete graph of f(x)=2(x+3)2(x5)f(x)=-2(x+3)^2(x-5) (condensing the source’s four-step progressive sketch to a start and an end state, narrated in between), a fitted monotonic cubic for the generic Intermediate Value Theorem illustration, f(x)=x35x2+3x+6f(x)=x^3-5x^2+3x+6 with f(1)f(1), f(3)f(3), and f(4)f(4) labeled, f(x)=130(x+3)(x2)2(x5)f(x)=\tfrac{1}{30}(x+3)(x-2)^2(x-5), the Try-It graph for f(x)=18(x2)3(x+1)2(x4)f(x)=-\tfrac18(x-2)^3(x+1)^2(x-4) with its intercepts and yy-intercept labeled, a fitted degree-4 local/global extrema illustration, the corner-cutout box diagram, V(w)=280w68w2+4w3V(w)=280w-68w^2+4w^3 at full scale and zoomed near its maximum, and the labeled local extrema of f(x)=0.2(x2)3(x+1)2(x4)f(x)=-0.2(x-2)^3(x+1)^2(x-4); presented the revenue table and the Intermediate Value Theorem sign-check table (Table 2) as Markdown tables; omitted the decorative “How To” and “Q&A” icon glyphs, which carry no mathematics; omitted the Media box linking to an external openstax.org resource; converted every retained “Try It” into an interactive component, adapting the two “sketch a graph” prompts into a check of the resulting x-intercepts (since freehand sketching is not gradable here), the graph-reading multiplicity prompt and both Intermediate Value Theorem “explain” prompts into multiple choice among paraphrased statements (a categorical judgment is never a fill-in number), and the “find the maximum and minimum values” prompt into a single ordered fill-in (minimum then maximum) since the source asks for both values from one graph; and adapted 16 selected end-of-section exercises — an x-intercept/zero vocabulary question, a domain question, two zero-finding-by-factoring problems, a multiplicity-reading problem, a zeros-and-multiplicities multiple choice, an even-multiplicity concept check, two end-behavior judgments, two maximum-turning-points computations, three “write the formula from the given information or graph” problems, an Intermediate-Value-Theorem explanation, and two sign-change confirmations — into 17 interactive components in a closing Practice block, one group per objective.