5.2 Graphing Polynomial Functions

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1 Name Class Date 5.2 Graphing Polnomial Functions Essential Question: How do ou sketch the graph of a polnomial function in intercept form? Eplore 1 Investigating the End Behavior of the Graphs of Simple Polnomial Functions Linear, quadratic, and cubic functions belong to a more general class of functions called polnomial functions, which are categorized b their degree. Linear functions are polnomial functions of degree 1, quadratic functions are polnomial functions of degree 2, and cubic functions are polnomial functions of degree 3. In general, a polnomial function of degree n has the standard form p() = a n n + a n-1 n a a 1 + a 0, where a n, a n-1,..., a 2, a 1, and a 0 are real numbers called the coefficients of the epressions a n n, a n-1 n - 1,..., a 2 2, a 1, and a 0, which are the terms of the polnomial function. (Note that the constant term, a 0, appears to have no power of associated with it, but since 0 = 1, ou can write a 0 as a 0 0 and treat a 0 as the coefficient of the term.) A polnomial function of degree 4 is called a quartic function, while a polnomial function of degree 5 is called a quintic function. After degree 5, polnomial functions are generall referred to b their degree, as in a sith-degree polnomial function. Resource Locker A Use a graphing calculator to graph the polnomial functions ƒ() =, ƒ() = 2, ƒ() = 3, ƒ() = 4, ƒ() = 5, and ƒ() = 6. Then use the graph of each function to determine the function s domain, range, and end behavior. (Use interval notation for the domain and range.) Function Domain Range End Behavior f() = As +, f(). As -, f(). f() = 2 f() = 3 f() = 4 f() = 5 As +, f(). As -, f(). As +, f(). As -, f(). As +, f(). As -, f(). As +, f(). As -, f(). f() = 6 As +, f(). As -, f(). Module Lesson 2

2 B Use a graphing calculator to graph the polnomial functions ƒ() = -, ƒ() = - 2, ƒ() = - 3, ƒ() = - 4, ƒ() = - 5, and ƒ() = - 6. Then use the graph of each function to determine the function s domain, range, and end behavior. (Use interval notation for the domain and range.) Function Domain Range End Behavior f() = - f() = - 2 f() = - 3 f() = - 4 f() = - 5 f() = - 6 As +, f(). As -, f(). As +, f(). As -, f(). As +, f(). As -, f(). As +, f(). As -, f(). As +, f(). As -, f(). As +, f(). As -, f(). Reflect 1. How can ou generalize the results of this Eplore for ƒ() = n and ƒ() = - n where n is positive whole number? Module Lesson 2

3 Eplore 2 Investigating the -intercepts and Turning Points of the Graphs of Polnomial Functions The cubic function ƒ() = 3 has three factors, all of which happen to be. One or more of the s can be replaced with other linear factors in, such as - 2, without changing the fact that the function is cubic. In general, a polnomial function of the form p()= a( - 1 )( - 2 )...( - n )where a, 1, 2,..., and n are real numbers (that are not necessaril distinct) has degree n where n is the number of variable factors. The graph of p() = a( - 1 )( - 2 )...( - n )has 1, 2,..., and n as its -intercepts, which is wh the polnomial is said to be in intercept form. Since the graph of p() intersects the -ais onl at its -intercepts, the graph must move awa from and then move back toward the -ais between each pair of successive -intercepts, which means that the graph has a turning point between those -intercepts. Also, instead of crossing the -ais at an -intercept, the graph can be tangent to the -ais, and the point of tangenc becomes a turning point because the graph must move toward the -ais and then awa from it near the point of tangenc. The -coordinate of each turning point is a maimum or minimum value of the function. A maimum or minimum value is called global or absolute if the function never takes on a value that is greater than the maimum or less than the minimum. On the other hand, a maimum or minimum value is called local or relative if the function does take on values that are greater than the maimum or less than the minimum somewhere outside an interval where the maimum or minimum value occurs. A Use a graphing calculator to graph the cubic functions ƒ() = 3, ƒ() = 2 ( - 2), and ƒ()= ( - 2)( + 2). Then use the graph of each function to answer the questions in the table. Function f () = 3 f () = 2 ( - 2) f () = ( - 2)( + 2) How man distinct factors does f() have? What are the graph s -intercepts? Is the graph tangent to the -ais or does it cross the -ais at each -intercept? How man turning points does the graph have? How man global maimum values? How man local? How man global minimum values? How man local? Module Lesson 2

4 B Use a graphing calculator to graph the quartic functions ƒ() = 4, ƒ() = 3 ( - 2), ƒ() = 2 ( - 2)( + 2), and ƒ()= ( - 2)( + 2)( + 3).Then use the graph of each function to answer the questions in the table. Function f () = 4 f () = 3 ( - 2) How man distinct factors? What are the -intercepts? Tangent to or cross the -ais at -intercepts? How man turning points? How man global maimum values? How man local? f () = 2 ( - 2) ( + 2) f () = ( - 2) ( + 2) ( + 3) How man global minimum values? How man local? Reflect 2. What determines how man -intercepts the graph of a polnomial function in intercept form has? 3. What determines whether the graph of a polnomial function in intercept form crosses the -ais or is tangent to it at an -intercept? 4. Suppose ou introduced a factor of -1 into each of the quartic functions in Step B. (For instance, ƒ() = 4 becomes ƒ() = - 4.) How would our answers to the questions about the functions and their graphs change? Module Lesson 2

5 Eplain 1 Sketching the Graph of Polnomial Functions in Intercept Form Given a polnomial function in intercept form, ou can sketch the function s graph b using the end behavior, the -intercepts, and the sign of the function values on intervals determined b the -intercepts. The sign of the function values tells ou whether the graph is above or below the -ais on a particular interval. You can find the sign of the function values b determining the sign of each factor and recognizing what the sign of the product of those factors is. Eample 1 Sketch the graph of the polnomial function. A ƒ () = ( + 2) ( 3) Identif the end behavior. For the function p () = a ( 1 ) ( 2 )... ( n ), the end behavior is determined b whether the degree n is even or odd and whether the constant factor a is positive or negative. For the given function f(), the degree is 3 and the constant factor a, which is 1, is positive, so ƒ() has the following end behavior: As +, ƒ() +. As -, ƒ() -. Identif the graph s -intercepts, and then use the sign of ƒ() on intervals determined b the -intercepts to find where the graph is above the -ais and where it s below the -ais. The -intercepts are = 0, = -2, and = 3. These three -intercepts divide the -ais into four intervals: < -2, -2 < < 0, 0 < < 3, and > 3. Interval Sign of the Constant Factor Sign of Sign of + 2 Sign of - 3 Sign of f () = ( +2) ( - 3) < < < < < > So, the graph of ƒ () is above the -ais on the intervals -2 < < 0 and > 3, and it s below the -ais on the intervals < -2 and 0 < < 3. Sketch the graph. While ou should be precise about where the graph crosses the -ais, ou do not need to be precise about the -coordinates of points on the graph that aren t on the -ais. Your sketch should simpl show where the graph lies above the -ais and where it lies below the -ais Module Lesson 2

6 B ƒ() = ( 4)( 1)( + 1)( + 2) Identif the end behavior. As +, ƒ(). As -, ƒ(). Identif the graph s -intercepts, and then use the sign of ƒ() on intervals determined b the -intercepts to find where the graph is above the -ais and where it s below the -ais. The -intercepts are =, =, =, =. Interval Sign of the Constant Factor Sign of - 4 Sign of - 1 Sign of + 1 Sign of + 2 Sign of f() = -( - 4) ( - 1) ( + 1) ( + 2) < < < < < < < > So, the graph of ƒ()is above the -ais on the intervals < < and < <, and it s below the -ais on the intervals <, < <, and >. Sketch the graph. Module Lesson 2

7 Your Turn Sketch the graph of the polnomial function. 5. ƒ () = - 2 ( - 4) Interval Sign of the Constant Factor Sign of 2 Sign of - 4 Sign of f () = 2 ( - 4) Eplain 2 Modeling with a Polnomial Function You can use cubic functions to model real-world situations. For eample, ou find the volume of a bo (a rectangular prism) b multipling the length, width, and height. If each dimension of the bo is given in terms of, then the volume is a cubic function of. Eample 2 To create an open-top bo out of a sheet of cardboard that is 9 inches long and 5 inches wide, ou make a square flap of side length inches in each corner b cutting along one of the flap s sides and folding along the other side. (In the first diagram, a solid line segment in the interior of the rectangle indicates a cut, while a dashed line segment indicates a fold.) After ou fold up the four sides of the bo (see the second diagram), ou glue each flap to the side it overlaps. To the nearest tenth, find the value of that maimizes the volume of the bo. 9 in. 5 in. Module Lesson 2

8 Analze Information Identif the important information. A square flap of side length inches is made in each corner of a rectangular sheet of cardboard. The sheet of cardboard measures 9 inches b 5 inches. Formulate a Plan Find the dimensions of the bo once the flaps have been made and the sides have been folded up. Create a volume function for the bo, graph the function on a graphing calculator, and use the graph to find the value of that maimizes the volume. Solve 1. Write epressions for the dimensions of the bo. Length of bo: 9 - Width of bo: 5 - Height of bo: 2. Write the volume function and determine its domain. V () = (9 - ) (5 - ) Because the length, width, and height of the bo must all be positive, the volume function s domain is determined b the following three constraints: 9-2 > 0, or < 5-2 > 0, or < > 0 Taken together, these constraints give a domain of 0 < <. 3. Use a graphing calculator to graph the volume function on its domain. Adjust the viewing window so ou can see the maimum. From the graphing calculator s CALC menu, select 4: maimum to locate the point where the maimum value occurs. So, V () 21.0 when, which means that the bo has a maimum volume of about 21 cubic inches when square flaps with a side length of 1 inch are made in the corners of the sheet of cardboard. Module Lesson 2

9 Justif and Evaluate Making square flaps with a side length of 1 inch means that the bo will be 7 inches long, 3 inches wide, and 1 inch high, so the volume is 21 cubic inches. As a check on this result, consider making square flaps with a side length of 0.9 inch and 1.1 inches: V (0.9) = (9-1.8) (5-1.8) (0.9) = V (1.1) = (9-2.2) (5-2.2) (1.1) = Both volumes are slightl less than 21 cubic inches, which suggests that 21 cubic inches is the maimum volume. Reflect 6. Discussion Although the volume function has three constraints on its domain, the domain involves onl two of them. Wh? Your Turn 7. To create an open-top bo out of a sheet of cardboard that is 25 inches long and 13 inches wide, ou make a square flap of side length inches in each corner b 13 in. cutting along one of the flap s sides and folding along the other. (In the diagram, a solid line segment in the interior of the rectangle indicates a cut, while a dashed line segment indicates a fold.) Once ou fold up the four sides of the 25 in. bo, ou glue each flap to the side it overlaps. To the nearest tenth, find the value of that maimizes the volume of the bo. Elaborate 8. Compare and contrast the domain, range, and end behavior of ƒ () = n when n is even and when n is odd. 9. Essential Question Check-In For a polnomial function in intercept form, wh is the constant factor important when graphing the function? Module Lesson 2

10 Evaluate: Homework and Practice Use a graphing calculator to graph the polnomial function. Then use the graph to determine the function s domain, range, and end behavior. (Use interval notation for the domain and range.) Online Homework Hints and Help Etra Practice 1. ƒ () = 7 2. ƒ () = ƒ () = ƒ () = - 8 Use a graphing calculator to graph the function. Then use the graph to determine the number of turning points and the number and tpe (global or local) of an maimum or minimum values. 5. ƒ () = ( + 1) ( + 3) 6. ƒ () = ( + 1) 2 ( - 1) ( - 2) 2 7. ƒ () = -( - 2) 3 8. ƒ () = -( - 1) ( + 2) Module Lesson 2

11 Sketch the graph the polnomial function. 9. ƒ() = 2 ( - 2) Interval Sign of f()= 2 ( - 2) 10. ƒ() = -( + 1)( - 2) ( - 3) Interval Sign of f()= -( + 1) ( - 2) ( - 3) 11. ƒ() = ( + 2) 2 ( - 1) Interval Sign of f()= ( + 2) 2 ( - 1) Module Lesson 2

12 12. To create an open-top bo out of a sheet of cardboard that is 6 inches long and 3 inches wide, ou make a square flap of side length inches in each corner b cutting along one of the flap s sides and folding along the other. Once ou fold up the four sides of the bo, ou glue each flap to the side it overlaps. To the nearest tenth, find the value of that maimizes the volume of the bo. 6 in. 3 in. 13. The template shows how to create a bo from a square sheet of cardboard that has a side length of 36 inches. In the template, solid line segments indicate cuts, dashed line segments indicate folds, and graed rectangles indicate pieces removed. The vertical strip that is 2 inches wide on the left side of the template is a flap that will be glued to the side of the bo that it overlaps when the bo is folded up. The horizontal strips that are 2 inches wide at the top and bottom of the template are also flaps that will overlap to form the top and bottom of the bo when the bo is folded up. Write a volume function for the bo in terms of onl. (You will need to determine a relationship between and first.) Then, to the nearest tenth, find the dimensions of the bo with maimum volume. 2 in in. 36 in. Module Lesson 2

13 Write a cubic function in intercept form for the given graph, whose -intercepts are integers. Assume that the constant factor a is either 1 or Write a quartic function in intercept form for the given graph, whose -intercepts are integers. Assume that the constant factor a is either 1 or Module Lesson 2

14 18. Multiple Response Select all statements that appl to the graph of ƒ() = ( - 1) 2 ( + 2). A. The -intercepts are = 1 and = -2. B. The -intercepts are = -1 and = 2. C. The graph crosses the -ais at = 1 and is tangent to the -ais at = -2. D. The graph crosses the -ais at = -1 and is tangent to the -ais at = 2. E. The graph is tangent to the -ais at = 1 and crosses the -ais at = -2. F. The graph is tangent to the -ais at = -1 and crosses the -ais at = 2. G. A local minimum occurs on the interval -2 < < 1, and a local maimum occurs at = 1. H. A local maimum occurs on the interval -2 < < 1, and a local minimum occurs at = 1. I. A local minimum occurs on the interval -1 < < 2, and a local maimum occurs at = 2. J. A local maimum occurs on the interval -1 < < 2, and a local minimum occurs at = 2. H.O.T. Focus on Higher Order Thinking 19. Eplain the Error A student was asked to sketch the graph of the function ƒ() = 2 ( - 3). Describe what the student did wrong. Then sketch the correct graph. Module Lesson 2

15 20. Make a Prediction Knowing the characteristics of the graphs of cubic and quartic functions in intercept form, sketch the graph of the quintic function ƒ()= 2 ( + 2)( - 2) Represent Real-World Situations A rectangular piece of sheet metal is rolled and riveted to form a circular tube that is open at both ends, as shown. The sheet metal has a perimeter of 36 inches. Each of the two sides of the rectangle that form the two ends of the tube has a length of inches, and the tube has a circumference of - 1 inches because an overlap of 1 inch is needed for the rivets. Write a volume function for the tube in terms of. Then, to the nearest tenth, find the value of that maimizes the volume of the tube. - 1 Module Lesson 2

16 Lesson Performance Task The template shows how to create a bo with a lid from a sheet of card stock that is 10 inches wide and 24 inches long. In the template, solid line segments indicate cuts, and dashed line segments indicate folds. The square flaps, each with a side length of inches, are glued to the sides the overlap when the bo is folded up. The bo has a bottom and four upright sides. The lid, which is attached to one of the upright sides, has three upright sides of its own. Assume that the three sides of the lid can be tucked inside the bo when the lid is closed. 10 in. 24 in. a. Write a polnomial function that represents the volume of the bo, and state its domain. b. Use a graphing calculator to find the value of that will produce the bo with maimum volume. What are the dimensions of that bo? Module Lesson 2