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5 3 Practice Polynomial Functions

Learning Objectives

In this section, yous will:

  • Recognize characteristics of graphs of polynomial functions.
  • Use factoring to find zeros of polynomial functions.
  • Identify zeros and their multiplicities.
  • Decide 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 cablevision company from 2006 through 2013 is shown in Table i.

Year 2006 2007 2008 2009 2010 2011 2012 2013
Revenues 52.4 52.8 51.2 49.5 48.6 48.six 48.7 47.one

Table 1

The revenue can be modeled by the polynomial role

R ( t ) = 0.037 t 4 + 1.414 t 3 xix.777 t 2 + 118.696 t 205.332 R ( t ) = 0.037 t iv + ane.414 t 3 19.777 t 2 + 118.696 t 205.332

where R R represents the revenue in millions of dollars and t t represents the year, with t = 6 t = six 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, tin can be answered by examining the graph of the polynomial function. We take already explored the local behavior of quadratics, a special case of polynomials. In this department nosotros will explore the local behavior of polynomials in full general.

Recognizing Characteristics of Graphs of Polynomial Functions

Polynomial functions of degree 2 or more have graphs that do not take 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 role that is not a polynomial.

Graph of f(x)=x^3-0.01x.

Effigy i

Case 1

Recognizing Polynomial Functions

Which of the graphs in Figure ii represents a polynomial role?

Two graphs in which one has a polynomial function and the other has a function closely resembling a polynomial but is not.

Figure 2

Q&A

Do all polynomial functions accept as their domain all real numbers?

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

Using Factoring to Detect Zeros of Polynomial Functions

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

Nosotros tin use this method to find x - x - intercepts considering at the x - ten - intercepts we find the input values when the output value is goose egg. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively uncomplicated quadratic formula, the corresponding formulas for cubic and 4th-degree polynomials are not simple enough to remember, and formulas do not exist for full general higher-degree polynomials. Consequently, we will limit ourselves to iii cases:

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

How To

Given a polynomial role f , f , discover the ten-intercepts by factoring.

  1. Set f ( 10 ) = 0. f ( x ) = 0.
  2. If the polynomial part is not given in factored form:
    1. Factor out whatever common monomial factors.
    2. Factor any factorable binomials or trinomials.
  3. Set each cistron equal to zero and solve to find the x - 10 - intercepts.

Example 2

Finding the ten-Intercepts of a Polynomial Function past Factoring

Find the x-intercepts of f ( x ) = x vi 3 x 4 + 2 x ii . f ( x ) = 10 6 3 x four + 2 10 2 .

Case 3

Finding the ten-Intercepts of a Polynomial Office by Factoring

Find the x-intercepts of f ( x ) = ten 3 5 10 ii 10 + 5. f ( x ) = x 3 5 10 2 ten + five.

Example iv

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

Find the y- and x-intercepts of g ( x ) = ( x 2 ) ii ( 2 ten + iii ) . one thousand ( x ) = ( 10 2 ) two ( 2 10 + 3 ) .

Analysis

We tin ever bank check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Effigy 5.

Graph of f(x)=x^3-5x^2-x+5 with its three intercepts (-1, 0), (1, 0), and (5, 0).

Figure 5

Example 5

Finding the 10-Intercepts of a Polynomial Role Using a Graph

Discover the x-intercepts of h ( ten ) = 10 three + 4 x 2 + x 6. h ( ten ) = 10 3 + 4 10 2 + x 6.

Try It #1

Find the y- and ten-intercepts of the function f ( x ) = x 4 nineteen x 2 + 30 x . f ( x ) = x 4 nineteen 10 two + 30 x .

Identifying Zeros and Their Multiplicities

Graphs behave differently at various 10-intercepts. Sometimes, the graph volition cantankerous over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and "bounce" off.

Suppose, for case, we graph the function shown.

f ( x ) = ( x + 3 ) ( x ii ) two ( ten + ane ) 3 f ( ten ) = ( 10 + 3 ) ( x ii ) 2 ( ten + ane ) 3

Notice in Figure seven that the behavior of the function at each of the x-intercepts is different.

Graph of h(x)=x^3+4x^2+x-6.

Figure seven Identifying the beliefs of the graph at an x-intercept by examining the multiplicity of the zero.

The x-intercept x = −iii x = −3 is the solution of equation ( x + 3 ) = 0. ( x + 3 ) = 0. The graph passes directly through the x-intercept at x = −3. 10 = −3. The cistron is linear (has a degree of 1), and then the behavior most the intercept is like that of a line—it passes straight through the intercept. We call this a single nix considering the zero corresponds to a single factor of the function.

The x-intercept x = 2 10 = two is the repeated solution of equation ( x 2 ) 2 = 0. ( x 2 ) two = 0. The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree ii), so the beliefs about the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.

( x 2 ) 2 = ( x 2 ) ( x 2 ) ( ten 2 ) 2 = ( x two ) ( x 2 )

The factor is repeated, that is, the factor ( ten two ) ( x two ) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is chosen the multiplicity. The goose egg associated with this factor, x = 2 , x = ii , has multiplicity two because the gene ( x 2 ) ( x 2 ) occurs twice.

The x-intercept x = 1 ten = one is the repeated solution of factor ( x + ane ) 3 = 0. ( x + 1 ) 3 = 0. The graph passes through the axis at the intercept, simply flattens out a bit outset. This cistron is cubic (degree 3), so the behavior about the intercept is like that of a cubic—with the same Southward-shape near the intercept as the toolkit function f ( 10 ) = ten 3 . f ( 10 ) = ten 3 . We call this a triple zero, or a nothing with multiplicity three.

For zeros with even multiplicities, the graphs affect or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity one, 2, and 3.

Graph of f(x)=(x+3)(x-2)^2(x+1)^3.

Figure viii

For higher even powers, such as iv, half-dozen, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph volition appear flatter equally it approaches and leaves the x-axis.

For higher odd powers, such as five, 7, and 9, the graph will all the same cantankerous through the horizontal axis, just for each increasing odd power, the graph will announced flatter equally information technology approaches and leaves the ten-axis.

Graphical Behavior of Polynomials at x-Intercepts

If a polynomial contains a cistron of the form ( x h ) p , ( x h ) p , the beliefs near the x - x - intercept h h is determined by the power p . p . Nosotros say that x = h x = h is a naught of multiplicity p . p .

The graph of a polynomial function volition touch the ten-axis at zeros with even multiplicities. The graph will cross the x-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 part of degree n , n , place the zeros and their multiplicities.

  1. If the graph crosses the x-axis and appears virtually linear at the intercept, information technology is a single nil.
  2. If the graph touches the ten-centrality and bounces off of the centrality, it is a naught with even multiplicity.
  3. If the graph crosses the 10-axis at a nada, it is a zero with odd multiplicity.
  4. The sum of the multiplicities is n . n .

Case half-dozen

Identifying Zeros and Their Multiplicities

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

Three graphs showing three different polynomial functions with multiplicity 1, 2, and 3.

Figure 9

Try It #2

Use the graph of the function of degree ix in Figure 10 to identify the zeros of the function and their multiplicities.

Graph of an even-degree polynomial with degree 6.

Figure 10

Determining End Behavior

Every bit nosotros have already learned, the beliefs of a graph of a polynomial function of the form

f ( x ) = a n x n + a northward one x northward 1 + ... + a 1 x + a 0 f ( x ) = a northward x northward + a northward i x north 1 + ... + a ane x + a 0

will either ultimately rise or fall equally ten x increases without spring and will either rising or fall as x 10 decreases without bound. This is because for very large inputs, say 100 or ane,000, the leading term dominates the size of the output. The aforementioned is true for very small inputs, say –100 or –1,000.

Recall that nosotros call this behavior the terminate behavior of a function. Every bit we pointed out when discussing quadratic equations, when the leading term of a polynomial role, a n 10 n , a northward x n , is an fifty-fifty ability function, as x x increases or decreases without leap, f ( x ) f ( x ) increases without leap. When the leading term is an odd power function, as x x decreases without bound, f ( x ) f ( x ) also decreases without spring; as x 10 increases without bound, f ( ten ) f ( x ) also increases without jump. If the leading term is negative, it will modify the direction of the end beliefs. Figure 11 summarizes all four cases.

Graph of a polynomial function with degree 5.

Figure eleven

Agreement the Relationship between Degree and Turning Points

In addition to the end beliefs, recall that nosotros can analyze a polynomial office'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 ( 10 ) = 10 4 ten three 4 x 2 + 4 x f ( x ) = x four ten iii four x 2 + four x in Figure 12. The graph has three turning points.

Graph of an odd-degree polynomial with a negative leading coefficient. Note that as x goes to positive infinity, f(x) goes to negative infinity, and as x goes to negative infinity, f(x) goes to positive infinity.

Figure 12

This function f f is a 4th caste polynomial office and has three turning points. The maximum number of turning points of a polynomial function is ever i less than the degree of the function.

Interpreting Turning Points

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

A polynomial of degree n n volition have at most n 1 n ane turning points.

Example 7

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

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

  1. f ( x ) = 10 3 + iv x v 3 x 2 + 1 f ( ten ) = 10 three + four ten 5 3 10 ii + ane
  2. f ( x ) = ( x ane ) 2 ( 1 + 2 x 2 ) f ( ten ) = ( x ane ) 2 ( ane + 2 x 2 )

Graphing Polynomial Functions

We can use what we accept learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Permit usa put this all together and wait at the steps required to graph polynomial functions.

How To

Given a polynomial function, sketch the graph.

  1. Notice the intercepts.
  2. Check for symmetry. If the office is an even function, its graph is symmetrical virtually the y - y - centrality, 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 ( 10 ) = f ( x ) . f ( ten ) = f ( x ) .
  3. Use the multiplicities of the zeros to determine the beliefs of the polynomial at the x - x - intercepts.
  4. Determine the terminate beliefs by examining the leading term.
  5. Use the end behavior and the beliefs at the intercepts to sketch a graph.
  6. Ensure that the number of turning points does non exceed one less than the degree of the polynomial.
  7. Optionally, use technology to check the graph.

Example 8

Sketching the Graph of a Polynomial Function

Sketch a graph of f ( x ) = −2 ( ten + three ) two ( x five ) . f ( x ) = −2 ( 10 + iii ) two ( x v ) .

Try Information technology #iii

Sketch a graph of f ( x ) = one 4 x ( x i ) 4 ( x + three ) 3 . f ( x ) = 1 4 x ( 10 1 ) 4 ( x + three ) three .

Using the Intermediate Value Theorem

In some situations, nosotros may know 2 points on a graph but non the zeros. If those two points are on contrary sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial role f f whose graph is shine and continuous. The Intermediate Value Theorem states that for two numbers a a and b b in the domain of f , f , if a < b a < b and f ( a ) f ( b ) , f ( a ) f ( b ) , then the function f f takes on every value between f ( a ) f ( a ) and f ( b ) . f ( b ) . (While the theorem is intuitive, the proof is actually quite complicated and requires higher mathematics.) 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 f at x = a x = a lies above the x - x - centrality and another bespeak at 10 = b x = b lies below the x - 10 - axis, at that place must exist a third betoken between x = a x = a and x = b x = b where the graph crosses the 10 - x - axis. Phone call this point ( c , f ( c ) ) . ( c , f ( c ) ) . This ways that nosotros are bodacious in that location is a solution c c where f ( c ) = 0. f ( c ) = 0.

In other words, the Intermediate Value Theorem tells us that when a polynomial part changes from a negative value to a positive value, the function must cross the x - ten - centrality. Figure 17 shows that there is a zero betwixt a a and b . b .

Graph of an odd-degree polynomial function that shows a point f(a) that's negative, f(b) that's positive, and f(c) that's 0.

Figure 17 Using the Intermediate Value Theorem to show there exists a zilch.

Intermediate Value Theorem

Let f f be a polynomial function. The Intermediate Value Theorem states that if f ( a ) f ( a ) and f ( b ) f ( b ) take contrary signs, then there exists at least one value c c between a a and b b for which f ( c ) = 0. f ( c ) = 0.

Instance 9

Using the Intermediate Value Theorem

Show that the function f ( x ) = x iii five x 2 + 3 x + six f ( x ) = x iii five x two + iii x + six has at to the lowest degree two real zeros between x = one x = 1 and x = four. x = iv.

Analysis

We can besides come across on the graph of the function in Figure 18 that at that place are 2 real zeros betwixt x = ane x = 1 and ten = 4. ten = four.

Graph of f(x)=x^3-5x^2+3x+6 and shows, by the Intermediate Value Theorem, that there exists two zeros since f(1)=5  and f(4)=2 are positive and f(3) = -3 is negative.

Figure 18

Try It #4

Prove that the function f ( x ) = 7 x 5 9 10 4 x 2 f ( x ) = vii 10 5 nine x 4 x 2 has at to the lowest degree i real null between x = one x = 1 and x = 2. x = 2.

Writing Formulas for Polynomial Functions

Now that nosotros know how to discover zeros of polynomial functions, we can use them to write formulas based on graphs. Considering a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, nosotros can class a function that volition pass through a gear up of x-intercepts by introducing a respective set of factors.

Factored Form of Polynomials

If a polynomial of everyman degree p p has horizontal intercepts at x = ten one , ten 2 , , 10 due north , x = x 1 , x 2 , , x north , then the polynomial can be written in the factored form: f ( x ) = a ( 10 x ane ) p 1 ( x ten ii ) p 2 ( ten x northward ) p n f ( 10 ) = a ( x x 1 ) p ane ( x x 2 ) p ii ( x x due north ) p n where the powers p i p i on each cistron can be determined by the behavior of the graph at the respective intercept, and the stretch cistron a a can be adamant given a value of the function other than the 10-intercept.

How To

Given a graph of a polynomial part, write a formula for the office.

  1. Identify the x-intercepts of the graph to find the factors of the polynomial.
  2. Examine the behavior of the graph at the x-intercepts to make up one's mind the multiplicity of each factor.
  3. Notice the polynomial of least degree containing all the factors plant in the previous step.
  4. Use whatever other signal on the graph (the y-intercept may be easiest) to decide the stretch factor.

Example 10

Writing a Formula for a Polynomial Function from the Graph

Write a formula for the polynomial role shown in Figure nineteen.

Graph of a positive even-degree polynomial with zeros at x=-3, 2, 5 and y=-2.

Figure xix

Try Information technology #5

Given the graph shown in Figure xx, write a formula for the role shown.

Graph of a negative even-degree polynomial with zeros at x=-1, 2, 4 and y=-4.

Figure xx

Using Local and Global Extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the office past finding the vertex. For full general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur tin still exist 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 bespeak is the highest or lowest point on the entire graph. In these cases, nosotros say that the turning bespeak is a global maximum or a global minimum. These are too referred to every bit the absolute maximum and absolute minimum values of the function.

Local and Global Extrema

A local maximum or local minimum at x = a ten = a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open up interval around x = a . ten = a . If a function has a local maximum at a , a , then f ( a ) f ( x ) f ( a ) f ( x ) for all ten 10 in an open up interval effectually ten = a . x = a . If a function has a local minimum at a , a , and so f ( a ) f ( ten ) f ( a ) f ( x ) for all x x in an open interval around x = a . 10 = a .

A global maximum or global minimum is the output at the highest or everyman point of the function. If a function has a global maximum at a , a , then f ( a ) f ( 10 ) f ( a ) f ( x ) for all x . x . If a role has a global minimum at a , a , then f ( a ) f ( x ) f ( a ) f ( x ) for all ten . 10 .

Nosotros tin run into the difference between local and global extrema in Figure 21.

Graph of an even-degree polynomial that denotes the local maximum and minimum and the global maximum.

Figure 21

Q&A

Practise 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 ) = x f ( x ) = x has neither a global maximum nor a global minimum.

Case 11

Using Local Extrema to Solve Applications

An open up-summit box is to exist constructed by cutting out squares from each corner of a 14 cm by 20 cm canvas of plastic and and so folding up the sides. Notice the size of squares that should be cutting out to maximize the volume enclosed by the box.

Try It #6

Apply engineering science to find the maximum and minimum values on the interval [ −1 , iv ] [ −1 , four ] of the function f ( ten ) = 0.two ( x 2 ) 3 ( x + i ) ii ( ten four ) . f ( x ) = 0.2 ( x 2 ) iii ( x + 1 ) 2 ( x 4 ) .

5.3 Department Exercises

Verbal

1.

What is the difference between an x - ten - intercept and a zero of a polynomial part f ? f ?

two .

If a polynomial function of caste due north n has due north due north distinct zeros, what practise you know nearly the graph of the part?

3.

Explain how the Intermediate Value Theorem can assist u.s.a. in finding a cypher of a office.

4 .

Explain how the factored form of the polynomial helps us in graphing information technology.

5.

If the graph of a polynomial only touches the ten-centrality and and so changes direction, what tin nosotros conclude well-nigh the factored form of the polynomial?

Algebraic

For the following exercises, discover the ten - x - or t-intercepts of the polynomial functions.

six .

C ( t ) = 2 ( t four ) ( t + 1 ) ( t vi ) C ( t ) = 2 ( t 4 ) ( t + 1 ) ( t 6 )

seven.

C ( t ) = 3 ( t + 2 ) ( t iii ) ( t + 5 ) C ( t ) = 3 ( t + two ) ( t three ) ( t + 5 )

8 .

C ( t ) = 4 t ( t two ) ii ( t + 1 ) C ( t ) = four t ( t 2 ) 2 ( t + i )

9.

C ( t ) = two t ( t 3 ) ( t + i ) 2 C ( t ) = 2 t ( t 3 ) ( t + 1 ) 2

10 .

C ( t ) = 2 t 4 8 t 3 + 6 t 2 C ( t ) = 2 t four eight t 3 + 6 t 2

11.

C ( t ) = 4 t 4 + 12 t 3 40 t two C ( t ) = iv t 4 + 12 t 3 40 t 2

12 .

f ( x ) = x iv x two f ( x ) = 10 4 x 2

thirteen.

f ( x ) = x 3 + 10 2 twenty 10 f ( ten ) = x iii + x 2 20 10

xiv .

f ( x ) = x 3 + six 10 ii 7 ten f ( 10 ) = ten 3 + vi x ii seven x

15.

f ( x ) = 10 3 + ten 2 four x iv f ( x ) = 10 3 + ten ii 4 x 4

sixteen .

f ( ten ) = x 3 + 2 ten 2 9 x 18 f ( x ) = x 3 + 2 x 2 9 ten xviii

17.

f ( ten ) = ii x 3 10 2 viii x + 4 f ( x ) = 2 x three 10 2 eight x + 4

xviii .

f ( ten ) = x six seven 10 iii 8 f ( ten ) = ten 6 7 x 3 8

19.

f ( x ) = two 10 four + half-dozen 10 2 8 f ( x ) = 2 ten 4 + 6 x 2 8

20 .

f ( x ) = x 3 3 ten 2 10 + 3 f ( ten ) = 10 three 3 x 2 ten + iii

21.

f ( x ) = x six two 10 4 three ten 2 f ( x ) = x 6 2 x 4 iii 10 2

22 .

f ( x ) = x six 3 ten iv iv 10 ii f ( x ) = 10 half dozen 3 x four 4 x 2

23.

f ( ten ) = x v five 10 iii + 4 x f ( x ) = x 5 5 x 3 + 4 10

For the post-obit exercises, utilize the Intermediate Value Theorem to confirm that the given polynomial has at least 1 zero within the given interval.

24 .

f ( x ) = x three ix x , f ( x ) = x three 9 10 , between 10 = −4 x = −4 and x = −two. x = −2.

25.

f ( ten ) = x 3 9 x , f ( x ) = x 3 nine x , between x = 2 x = 2 and ten = 4. x = iv.

26 .

f ( ten ) = x 5 ii x , f ( x ) = 10 5 2 ten , between x = 1 x = 1 and x = two. x = ii.

27.

f ( x ) = x 4 + iv , f ( 10 ) = x iv + 4 , betwixt x = one x = 1 and x = three 10 = iii .

28 .

f ( x ) = −2 x 3 x , f ( x ) = −2 x 3 x , between 10 = –1 x = –one and x = one. x = 1.

29.

f ( ten ) = x 3 100 ten + 2 , f ( ten ) = 10 3 100 x + 2 , between x = 0.01 10 = 0.01 and x = 0.1 x = 0.i

For the following exercises, find the zeros and give the multiplicity of each.

30 .

f ( x ) = ( x + two ) three ( x 3 ) 2 f ( x ) = ( x + 2 ) 3 ( 10 3 ) 2

31.

f ( x ) = 10 ii ( 2 x + 3 ) 5 ( x four ) 2 f ( x ) = ten ii ( ii x + 3 ) 5 ( ten 4 ) 2

32 .

f ( x ) = x 3 ( x i ) 3 ( ten + 2 ) f ( 10 ) = x three ( x 1 ) 3 ( x + 2 )

33.

f ( x ) = x 2 ( x 2 + four x + 4 ) f ( ten ) = x 2 ( x ii + four 10 + 4 )

34 .

f ( x ) = ( 2 ten + i ) 3 ( ix 10 2 vi 10 + 1 ) f ( 10 ) = ( 2 x + 1 ) 3 ( 9 ten 2 6 ten + i )

35.

f ( ten ) = ( 3 x + 2 ) 5 ( x 2 10 10 + 25 ) f ( ten ) = ( 3 ten + two ) five ( 10 ii 10 10 + 25 )

36 .

f ( x ) = x ( 4 x 2 12 x + 9 ) ( x 2 + 8 x + 16 ) f ( x ) = 10 ( iv x two 12 x + 9 ) ( x ii + viii x + 16 )

37.

f ( ten ) = x 6 x five 2 ten iv f ( x ) = x 6 x v 2 10 four

38 .

f ( ten ) = iii ten iv + 6 x 3 + 3 x 2 f ( ten ) = 3 x 4 + 6 ten 3 + 3 10 two

39.

f ( x ) = iv x 5 12 x 4 + 9 x 3 f ( x ) = 4 x v 12 10 iv + 9 x iii

40 .

f ( x ) = 2 x 4 ( 10 iii 4 x two + iv x ) f ( 10 ) = 2 x 4 ( x 3 4 x 2 + 4 x )

41.

f ( x ) = iv ten four ( nine x iv 12 x 3 + 4 x 2 ) f ( x ) = iv x 4 ( 9 x 4 12 x 3 + 4 x 2 )

Graphical

For the following exercises, graph the polynomial functions. Annotation x - x - and y - y - intercepts, multiplicity, and end behavior.

42 .

f ( x ) = ( x + 3 ) 2 ( 10 2 ) f ( x ) = ( x + 3 ) 2 ( x two )

43.

m ( x ) = ( x + 4 ) ( x one ) ii g ( ten ) = ( x + four ) ( x 1 ) 2

44 .

h ( x ) = ( x i ) 3 ( x + three ) two h ( x ) = ( ten i ) iii ( ten + 3 ) 2

45.

k ( x ) = ( x 3 ) 3 ( x 2 ) ii k ( x ) = ( x 3 ) three ( 10 2 ) 2

46 .

1000 ( 10 ) = ii x ( x 1 ) ( x + 3 ) g ( x ) = 2 ten ( x 1 ) ( x + iii )

47.

northward ( 10 ) = iii x ( x + 2 ) ( x 4 ) northward ( x ) = 3 x ( x + ii ) ( 10 4 )

For the post-obit exercises, utilise the graphs to write the formula for a polynomial function of least degree.

48 .

Graph of a positive odd-degree polynomial with zeros at x=-2, 1, and 3.

50 .

Graph of a negative odd-degree polynomial with zeros at x=-1, and 2.

52 .

Graph of a negative even-degree polynomial with zeros at x=-3, -2, 3, and 4.

For the following exercises, apply the graph to place zeros and multiplicity.

54 .

Graph of a positive even-degree polynomial with zeros at x=-4, -2, and 3.

56 .

Graph of a negative odd-degree polynomial with zeros at x=-3, -2, and 1.

For the following exercises, use the given data about the polynomial graph to write the equation.

57.

Degree 3. Zeros at ten = –2, ten = –two, x = 1, x = 1, and x = 3. ten = three. y-intercept at ( 0 , –4 ) . ( 0 , –4 ) .

58 .

Degree iii. Zeros at 10 = –5, x = –five, ten = –2 , x = –2 , and ten = 1. x = i. y-intercept at ( 0 , 6 ) ( 0 , vi )

59.

Caste five. Roots of multiplicity 2 at x = 3 x = three and 10 = i x = 1 , and a root of multiplicity 1 at x = –three. x = –3. y-intercept at ( 0 , 9 ) ( 0 , ix )

60 .

Degree four. Root of multiplicity ii at x = iv, x = 4, and a roots of multiplicity 1 at x = 1 x = one and x = –two. x = –2. y-intercept at ( 0 , 3 ) . ( 0 , 3 ) .

61.

Degree 5. Double zero at x = 1 , ten = 1 , and triple cypher at x = 3. x = 3. Passes through the point ( 2 , 15 ) . ( 2 , 15 ) .

62 .

Degree three. Zeros at ten = 4 , x = 4 , ten = 3 , x = iii , and x = 2. 10 = two. y-intercept at ( 0 , −24 ) . ( 0 , −24 ) .

63.

Degree 3. Zeros at x = −3 , x = −three , x = −2 x = −2 and x = i. x = 1. y-intercept at ( 0 , 12 ) . ( 0 , 12 ) .

64 .

Degree 5. Roots of multiplicity two at x = −iii ten = −3 and x = 2 10 = 2 and a root of multiplicity one at ten = −two. x = −ii.

y-intercept at ( 0 , 4 ) . ( 0 , 4 ) .

65.

Degree 4. Roots of multiplicity 2 at x = 1 two x = 1 two and roots of multiplicity 1 at x = 6 x = 6 and 10 = −two. x = −2.

y-intercept at ( 0, 18 ) . ( 0, 18 ) .

66 .

Double cypher at x = −3 10 = −3 and triple zero at x = 0. x = 0. Passes through the point ( 1 , 32 ) . ( 1 , 32 ) .

Technology

For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum.

67.

f ( 10 ) = ten 3 ten ane f ( ten ) = 10 three ten one

68 .

f ( x ) = 2 x 3 3 x 1 f ( 10 ) = 2 x 3 3 x ane

69.

f ( x ) = x four + x f ( x ) = 10 4 + x

70 .

f ( ten ) = x 4 + 3 x 2 f ( x ) = x iv + three x 2

71.

f ( 10 ) = x iv x iii + 1 f ( x ) = x 4 x 3 + 1

Extensions

For the following exercises, utilize the graphs to write a polynomial role of least degree.

72 .

Graph of a positive odd-degree polynomial with zeros at x=--2/3, and 4/3 and y=8.

74 .

Graph of a positive odd-degree polynomial with zeros at x=--300, and 100 and y=-90000.

Existent-World Applications

For the following exercises, write the polynomial function that models the given situation.

75.

A rectangle has a length of ten units and a width of 8 units. Squares of x x by ten 10 units are cutting out of each corner, and then the sides are folded up to create an open box. Express the volume of the box every bit a polynomial part in terms of x . ten .

76 .

Consider the same rectangle of the preceding problem. Squares of two 10 2 x by 2 x 2 x units are cutting out of each corner. Express the volume of the box as a polynomial in terms of ten . x .

77.

A foursquare has sides of 12 units. Squares x + 1 x + 1 by 10 + ane ten + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box equally a function in terms of x . x .

78 .

A cylinder has a radius of x + 2 ten + 2 units and a elevation of 3 units greater. Express the book of the cylinder as a polynomial role.

79.

A right circular cone has a radius of three x + 6 3 x + 6 and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is V = one iii π r 2 h 5 = 1 3 π r 2 h for radius r r and height h . h .

5 3 Practice Polynomial Functions,

Source: https://openstax.org/books/college-algebra/pages/5-3-graphs-of-polynomial-functions

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