290 Chapter 1 Functions and Graphs

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1 90 Chapter Functions and Graphs 7. Studies show that teting while driving is as risk as driving with a 0.08 blood alcohol level, the standard for drunk driving. The bar graph shows the number of fatalities in the United States involving distracted driving from 004 through 008. Although the distracted categor involves such activities as talking on cellphones, conversing with passengers, and eating, eperts at the National Highwa Traffic Safet Administration claim that teting while driving is the clearest menace because it requires looking awa from the road. Number of Deaths Involving Distracted Driving Number of Highwa Fatalities in the United States Involving Distracted Driving (4, 5870) Year Number of Deaths Involving Distracted Driving Source: National Highwa Traffic Safet Administration (, 457) 0 4 Years after 004 a. Shown to the right of the bar graph is a scatter plot with a line passing through two of the data points. Use the two points whose coordinates are shown b the voice balloons to write the point-slope form of an equation that models the number of highwa fatalities involving distracted driving,, in the United States ears after 004. b. Write the equation from part (a) in slope-intercept form. Use function notation. c. In 00, surves showed overwhelming public support to ban teting while driving, although at that time onl 9 states and Washington, D.C., outlawed the practice. Without additional laws that penalize teting drivers, use the linear function ou obtained from part (b) to project the number of fatalities in the United States in 04 involving distracted driving. 8. Find the average rate of change of f() = - 5 from = 6 to = 0. - if Ú 9. If g() = b, find g(-) and g( 7 ). - if 6 In Eercises 0, find the domain of each function. 0. f() = f() = If f() = - 4 and g() =, find (f g)() and the domain of f g.. Epress h() = ( + ) 7 as a composition of two functions f and g so that h() = (f g)(). 4. Find the length and the midpoint of the line segment whose endpoints are (, -) and (5, ). 5. In 980, the winning time for women in the Olmpic 500-meter speed skating event was 4.78 seconds. The average rate of decrease in the winning time has been about 0.9 second per ear. a. Epress the winning time, T, in this event as a function of the number of ears after 980,. b. According to the function, when was winning time 5.7 seconds? 6. The annual ield per walnut tree is fairl constant at 50 pounds per tree when the number of trees per acre is 0 or fewer. For each additional tree over 0, the annual ield per tree for all trees on the acre decreases b.5 pounds due to overcrowding. a. Epress the ield per tree, Y, in pounds, as a function of the number of walnut trees per acre,. b. Epress the total ield for an acre, T, in pounds, as a function of the number of walnut trees per acre,. 7. You have 600 ards of fencing to enclose a rectangular field. Epress the area of the field, A, as a function of one of its dimensions,. 8. A closed rectangular bo with a square base has a volume of 8000 cubic centimeters. Epress the surface area of the bo, A, as a function of the length of a side of its square base,.

2 CHAPTER We are surrounded b evidence that the world is profoundl mathematical. After turning a somersault, a diver s path can be modeled b a quadratic function, f() = a + b + c, as can the path of a football tossed from quarterback to receiver or the path of a fl ipped coin. Even if ou throw an object directl upward, although its path is straight and vertical, its changing height over time is described b a quadratic function. And tailgaters beware: Whether ou re driving a car, a motorccle, or a truck on dr or wet roads, an arra of quadratic functions that model our required stopping distances at various speeds are available to help ou become a safer driver. HERE S WHERE YOU LL FIND THESE APPLICATIONS: The quadratic functions surrounding our long histor of throwing things appear throughout the chapter, including Eample 5 in Section. and Eample 5 in Section.7. Tailgaters should pa close attention to the Section.7 opener, Eercises and 9 9 in Eercise Set.7, and Eercise 9 in the Chapter Review Eercises. 9

3 9 Chapter Polnomial and Rational Functions SECTION. Comple Numbers Objectives Add and subtract comple numbers. Multipl comple numbers. Divide comple numbers. Perform operations with square roots of negative numbers. Solve quadratic equations with comple imaginar solutions. W ho is this kid warning us about our eeballs turning black if we attempt to find the square root of -9? Don t believe what ou hear on the street. Although square roots of negative numbers are not real numbers, the do pla a significant role in algebra. In this section, we move beond the real numbers and discuss square roots with negative radicands. The Imaginar Unit i In this section, we will stud equations whose solutions ma involve the square roots of negative numbers. Because the square of a real number is never negative, there is no real number such that = -. To provide a setting in which such equations have solutions, mathematicians have invented an epanded sstem Roz Chast/The New Yorker Collection/Cartoonbank of numbers, the comple numbers. The imaginar number i, defined to be a solution of the equation = -, is the basis of this new number sstem. The Imaginar Unit i The imaginar unit i is defined as i = -, where i = -. Using the imaginar unit i, we can epress the square root of an negative number as a real multiple of i. For eample, -5 = - 5 = i5 = 5i. We can check this result b squaring 5i and obtaining -5. ( 5i) = 5 i = 5(-) = -5 A new sstem of numbers, called comple numbers, is based on adding multiples of i, such as 5i, to real numbers. Real numbers a + bi with b = 0 Comple numbers a + bi Imaginar numbers a + bi with b 0 FIGURE. The comple number sstem Comple Numbers and Imaginar Numbers The set of all numbers in the form a + bi, with real numbers a and b, and i, the imaginar unit, is called the set of comple numbers. The real number a is called the real part and the real number b is called the imaginar part of the comple number a + bi. If b 0, then the comple number is called an imaginar number (Figure. ). An imaginar number in the form bi is called a pure imaginar number.

4 Section. Comple Numbers 9 Here are some eamples of comple numbers. Each number can be written in the form a + bi. 4+6i i=0+i =+0i a, the real part, is 4. b, the imaginar part, is 6. a, the real part, is 0. b, the imaginar part, is. a, the real part, is. b, the imaginar part, is 0. Can ou see that b, the imaginar part, is not zero in the first two comple numbers? Because b 0, these comple numbers are imaginar numbers. Furthermore, the imaginar number i is a pure imaginar number. B contrast, the imaginar part of the comple number on the right is zero. This comple number is not an imaginar number. The number, or + 0i, is a real number. A comple number is said to be simplified if it is epressed in the standard form a + bi. If b contains a radical, we usuall write i before the radical. For eample, we write 7 + i5 rather than 7 + 5i, which could easil be confused with 7 + 5i. Epressed in standard form, two comple numbers are equal if and onl if their real parts are equal and their imaginar parts are equal. Equalit of Comple Numbers a + bi = c + di if and onl if a = c and b = d. Add and subtract comple numbers. Operations with Comple Numbers The form of a comple number a + bi is like the binomial a + b. Consequentl, we can add, subtract, and multipl comple numbers using the same methods we used for binomials, remembering that i = -. Adding and Subtracting Comple Numbers. ( a + bi) + (c + di) = (a + c) + (b + d)i In words, this sas that ou add comple numbers b adding their real parts, adding their imaginar parts, and epressing the sum as a comple number.. ( a + bi) - (c + di) = (a - c) + (b - d)i In words, this sas that ou subtract comple numbers b subtracting their real parts, subtracting their imaginar parts, and epressing the difference as a comple number. EXAMPLE Adding and Subtracting Comple Numbers GREAT QUESTION! Are operations with comple numbers similar to operations with polnomials? Yes. The following eamples, using the same integers as in Eample, show how operations with comple numbers are just like operations with polnomials. a. (5 - ) + (7 + 4) = - 7 b. (-5 + ) - (- - 6) = = Perform the indicated operations, writing the result in standard form: a. ( 5 - i) + (7 + 4i) b. ( -5 + i) - (- - 6i). SOLUTION a. ( 5 - i) + (7 + 4i) = 5 - i i Remove the parentheses. = i + 4i Group real and imaginar terms. = (5 + 7) + (- + 4)i Add real parts and add imaginar parts. = - 7i Simplif. b. ( -5 + i) - (- - 6i) = -5 + i + + 6i Remove the parentheses. Change signs of real and imaginar parts in the comple number being subtracted. = i + 6i Group real and imaginar terms. = (-5 + ) + ( + 6)i Add real parts and add imaginar parts. = 6 + 7i Simplif.

5 94 Chapter Polnomial and Rational Functions Check Point Perform the indicated operations, writing the result in standard form: a. ( 5 - i) + ( + i) b. ( + 6i) - ( - i). Multiplication of comple numbers is performed the same wa as multiplication of polnomials, using the distributive propert and the FOIL method. After completing the multiplication, we replace an occurrences of i with -. This idea is illustrated in the net eample. Multipl comple numbers. EXAMPLE Multipling Comple Numbers Find the products: a. 4i( - 5i) b. ( 7 - i)(- - 5i). SOLUTION a. 4i( - 5i) = 4i # - 4i # 5i Distribute 4i throughout the parentheses. = i - 0i Multipl. = i - 0(-) Replace i with. = 0 + i Simplif to i + 0 and write in standard form. b. ( 7 - i)(- - 5i) F O I L = 4-5i+6i+5i = 4-5i+6i+5( ) = 4-5-5i+6i = 9-9i Use the FOIL method. i = Group real and imaginar terms. Combine real and imaginar terms. Check Point Find the products: a. 7i( - 9i) b. ( 5 + 4i)(6-7i). Divide comple numbers. Comple Conjugates and Division It is possible to multipl imaginar numbers and obtain a real number. This occurs when we multipl a + bi and a - bi. F O I L (a+bi)(a-bi)=a -abi+abi-b i =a -b ( ) =a +b Use the FOIL method. i = - Notice that this product eliminates i. For the comple number a + bi, we define its comple conjugate to be a - bi. The multiplication of comple conjugates results in a real number. Conjugate of a Comple Number The comple conjugate of the number a + bi is a - bi, and the comple conjugate of a - bi is a + bi. The multiplication of comple conjugates gives a real number. ( a + bi)(a - bi) = a + b ( a - bi)(a + bi) = a + b

6 Section. Comple Numbers 95 Comple conjugates are used to divide comple numbers. The goal of the division procedure is to obtain a real number in the denominator. This real number becomes the denominator of a and b in the quotient a + bi. B multipling the numerator and the denominator of the division b the comple conjugate of the denominator, ou will obtain this real number in the denominator. EXAMPLE Using Comple Conjugates to Divide Comple Numbers Divide and epress the result in standard form: SOLUTION 7 + 4i - 5i. The comple conjugate of the denominator, - 5i, is + 5i. Multiplication of both the numerator and the denominator b + 5i will eliminate i from the denominator while maintaining the value of the epression i - 5i (7 + 4i) ( + 5i) = # ( - 5i) ( + 5i) Multipl the numerator and the denominator b the comple conjugate of the denominator. F O I L 4+5i+8i+0i = +5 Use the FOIL method in the numerator and (a - bi)(a + bi) = a + b in the denominator. = 4 + 4i + 0(-) 9 In the numerator, combine imaginar terms and replace i with. In the denominator, + 5 = = 9. = i 9 Combine real terms in the numerator: 4 + 0( ) = 4-0 = 6. = i Epress the answer in standard form. 9 Observe that the quotient is epressed in the standard form a + bi, with a = and b = 4 9. Check Point Divide and epress the result in standard form: 5 + 4i 4 - i. Perform operations with square roots of negative numbers. Roots of Negative Numbers The square of 4i and the square of -4i both result in -6: ( 4 i) = 6i = 6(-) = -6 (-4i) = 6i = 6(-) = -6. Consequentl, in the comple number sstem - 6 has two square roots, namel, 4i and -4i. We call 4i the principal square root of -6. Principal Square Root of a Negative Number For an positive real number b, the principal square root of the negative number -b is defined b -b = ib.

7 96 Chapter Polnomial and Rational Functions Consider the multiplication problem 5i # i = 0i = 0(-) = -0. This problem can also be given in terms of principal square roots of negative numbers: -5 # -4. Because the product rule for radicals onl applies to real numbers, multipling radicands is incorrect. When performing operations with square roots of negative numbers, begin b epressing all square roots in terms of i. Then perform the indicated operation. Correct: Incorrect: -5 # -4 = i5 # i4-5 # -4 = (-5)(-4) = 5i # i = 00 = 0i = 0(-) = -0 = 0 EXAMPLE 4 Operations Involving Square Roots of Negative Numbers Perform the indicated operations and write the result in standard form: a b c. SOLUTION Begin b epressing all square roots of negative numbers in terms of i. a = i8 - i8 = i9 # - i4 # = i - i = i (A + B) = A + A B + B b. A + 5B =A +i5b =( ) +( )Ai5B+Ai5B =-i5+5i =-i5+5( ) = 4-i5 c = = -5 + i i 5 = i 5 = i b = ib 50 = 5 # = 5 Write the comple number in standard form. Simplif. Check Point 4 Perform the indicated operations and write the result in standard form: a b c..

8 Section. Comple Numbers 97 Solve quadratic equations with comple imaginar solutions. GREAT QUESTION! Where can I review quadratic equations and how to solve them? Read Section P.7, beginning on page 9. Quadratic Equations with Comple Imaginar Solutions We have seen that a quadratic equation can be epressed in the general form a + b + c = 0, a 0. All quadratic equations can be solved b the quadratic formula: = -b { b - 4ac. a Recall that the quantit b - 4ac, which appears under the radical sign in the quadratic formula, is called the discriminant. If the discriminant is negative, a quadratic equation has no real solutions. However, quadratic equations with negative discriminants do have two solutions. These solutions are imaginar numbers that are comple conjugates. EXAMPLE 5 A Quadratic Equation with Imaginar Solutions Solve using the quadratic formula: = 0. SOLUTION The given equation is in general form. Begin b identifing the values for a, b, and c. -+4=0 a = b = c = 4 = -b { b - 4ac a Use the quadratic formula. = -(-) { (-) - 4()(4) () = { = { = { i 6 { i = 6 = { i = { i Substitute the values for a, b, and c: a =, b = -, and c = 4. -(-) = and (-) = (-)(-) = 4. Subtract under the radical. Because the number under the radical sign is negative, the solutions will not be real numbers. -44 = 4()(-) = i Factor from the numerator. Divide numerator and denominator b. The solutions are comple conjugates, and the solution set is Write the comple numbers in standard form. b + i, - i r or b { i r. Check Point 5 Solve using the quadratic formula: - + = 0.

9 98 Chapter Polnomial and Rational Functions CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true.. The imaginar unit i is defined as i =, where i =.. The set of all numbers in the form a + bi is called the set of numbers. If b 0, then the number is also called a/an number. If b = 0, then the number is also called a/an number.. -9i + i = 4. 0i - (-4i) = 5. Consider the following multiplication problem: ( + i)(6-5i). Using the FOIL method, the product of the first terms is, the product of the outside terms is, and EXERCISE SET. Practice Eercises In Eercises 8, add or subtract as indicated and write the result in standard form.. ( 7 + i) + ( - 4i). ( - + 6i) + (4 - i). ( + i) - (5-7i) 4. ( i) - (-9 - i) (-5 + 4i) - (- - i) (-9 + i) - (-7 - i) 7. 8 i - (4-9i) 8. 5 i - ( - i) In Eercises 9 0, find each product and write the result in standard form. 9. -i(7i - 5 ) 0. -8i(i - 7 ). ( i)( + i). ( -4-8i)( + i). ( 7-5i)(- - i) 4. ( 8-4i)(- + 9i) 5. ( + 5i)( - 5i) 6. ( + 7i)( - 7i) 7. ( -5 + i)(-5 - i) 8. ( -7 - i)(-7 + i) 9. ( + i) 0. ( 5 - i) In Eercises 8, divide and epress the result in standard form... - i 4 + i i 5i i - i 8i -6i i + i 7. + i + i i 4 + i In Eercises 9 44, perform the indicated operations and write the result in standard form the product of the inside terms is of the last terms in terms of i is simplifies to. 6. The conjugate of - 9i is. 7. The division 7 + 4i - 5i. The product, which is performed b multipling the numerator and denominator b = 0 = 4 # 5 = 9. = -4 { 4-4 # # 5 # simplifies to = In Eercises 45 50, solve each quadratic equation using the quadratic formula. Epress solutions in standard form = = = = = = 4-6 Practice Plus In Eercises 5 56, perform the indicated operation(s) and write the result in standard form. 5. ( - i)( - i) - ( - i)( + i) 5. ( 8 + 9i)( - i) - ( - i)( + i) 5. ( + i) - ( - i) 54. ( 4 - i) - ( + i) Evaluate - + for = + i. 58. Evaluate for = - i.

10 Section. Comple Numbers Evaluate 60. Evaluate Application Eercises for = i. for = 4i. Comple numbers are used in electronics to describe the current in an electric circuit. Ohm s law relates the current in a circuit, I, in amperes, the voltage of the circuit, E, in volts, and the resistance of the circuit, R, in ohms, b the formula E = IR. Use this formula to solve Eercises Find E, the voltage of a circuit, if I = (4-5i) amperes and R = ( + 7i) ohms. 6. Find E, the voltage of a circuit, if I = ( - i) amperes and R = ( + 5i) ohms. 6. The mathematician Girolamo Cardano is credited with the first use (in 545) of negative square roots in solving the now-famous problem, Find two numbers whose sum is 0 and whose product is 40. Show that the comple numbers 5 + i5 and 5 - i5 satisf the conditions of the problem. (Cardano did not use the smbolism i5 or even -5. He wrote R.m 5 for -5, meaning radi minus 5. He regarded the numbers 5 + R.m 5 and 5 - R.m 5 as fictitious or ghost numbers, and considered the problem manifestl impossible. But in a mathematicall adventurous spirit, he eclaimed, Nevertheless, we will operate. ) Writing in Mathematics 64. What is i? 65. Eplain how to add comple numbers. Provide an eample with our eplanation. 66. Eplain how to multipl comple numbers and give an eample. 67. What is the comple conjugate of + i? What happens when ou multipl this comple number b its comple conjugate? 68. Eplain how to divide comple numbers. Provide an eample with our eplanation. 69. Eplain each of the three jokes in the cartoon on page A stand-up comedian uses algebra in some jokes, including one about a telephone recording that announces You have just reached an imaginar number. Please multipl b i and dial again. Eplain the joke. Eplain the error in Eercises = -5 = i5 = 5i = -9 # -9 = 8 = 9 Critical Thinking Eercises Make Sense? In Eercises 7 76, determine whether each statement makes sense or does not make sense, and eplain our reasoning. 7. The humor in the cartoon at the top of the net column is based on the fact that rational and real have different meanings in mathematics and in everda speech. 007 GJ Caulkins 74. The word imaginar in imaginar numbers tells me that these numbers are undefined. 75. B writing the imaginar number 5 i, I can immediatel see that 5 is the constant and i is the variable. 76. When I add or subtract comple numbers, I am basicall combining like terms. In Eercises 77 80, determine whether each statement is true or false. If the statement is false, make the necessar change(s) to produce a true statement. 77. Some irrational numbers are not comple numbers. 78. ( + 7i)( - 7i) is an imaginar number i 5 + i = In the comple number sstem, + (the sum of two squares) can be factored as ( + i)( - i). In Eercises 8 8, perform the indicated operations and write the result in standard form ( + i)( - i) + i + i + - i - i 8 + i Preview Eercises Eercises will help ou prepare for the material covered in the net section. In Eercises 84 85, solve each quadratic equation b the method of our choice = -( - ) = Use the graph of f() = to graph g() = ( + ) +.

11 00 Chapter Polnomial and Rational Functions SECTION. Quadratic Functions Objectives Recognize characteristics of parabolas. Graph parabolas. Determine a quadratic function s minimum or maimum value. Solve problems involving a quadratic function s minimum or maimum value. Recognize characteristics of parabolas. Man sports involve objects that are thrown, kicked, or hit, and then proceed with no additional force of their own. Such objects are called projectiles. Paths of projectiles, as well as their heights over time, can be modeled b quadratic functions. We have seen that a quadratic function is an function of the form f() = a + b + c, where a, b, and c are real numbers, with a 0. A quadratic function is a polnomial function whose greatest eponent is. In this section, ou will learn to use graphs of quadratic functions to gain a visual understanding of the algebra that describes football, baseball, basketball, the shot put, and other projectile sports. Graphs of Quadratic Functions The graph of an quadratic function is called a parabola. Parabolas are shaped like bowls or inverted bowls, as shown in Figure.. If the coefficient of (the value of a in a + b + c ) is positive, the parabola opens upward. If the coefficient of is negative, the parabola opens downward. The verte (or turning point) of the parabola is the lowest point on the graph when it opens upward and the highest point on the graph when it opens downward. Ais of smmetr Verte (maimum point) f() = a + b + c (a > 0) f() = a + b + c (a < 0) Verte (minimum point) Ais of smmetr a > 0: Parabola opens upward. a < 0: Parabola opens downward. FIGURE. Characteristics of graphs of quadratic functions Look at the unusual image of the word mirror shown here. The artist, Scott Kim, has created the image so that the two halves of the whole are mirror images of each other. A parabola shares this kind of smmetr, in which a vertical line through the verte divides the figure in half. Parabolas are smmetric with respect to this line, called the ais of smmetr. If a parabola is folded along its ais of smmetr, the two halves match eactl.

12 Section. Quadratic Functions 0 Graph parabolas. Graphing Quadratic Functions in Standard Form In our earlier work with transformations, we applied a series of transformations to the graph of f() =. The graph of this function is a parabola. The verte for this parabola is (0, 0). In Figure.(a), the graph of f() = a for a 7 0 is shown in black; it opens upward. In Figure.(b), the graph of f() = a for a 6 0 is shown in black; it opens downward. Transformations of f() = a Ais of smmetr: = h Ais of smmetr: = h g() = a( h) + k Verte: (0, 0) Verte: (h, k) g() = a( h) + k f() = a f() = a Verte: (h, k) Verte: (0, 0) FIGURE.(a) a 7 0: Parabola opens upward. FIGURE.(b) a 6 0: Parabola opens downward. Figure.(a) and.(b) also show the graph of g() = a( - h) + k in blue. Compare these graphs to those of f() = a. Observe that h determines a horizontal shift and k determines a vertical shift of the graph of f() = a : g()=a(-h) +k. If h > 0, the graph of f() = a is shifted h units to the right. If k > 0, the graph of = a( h) is shifted k units up. Consequentl, the verte (0, 0) on the black graph of f() = a moves to the point ( h, k) on the blue graph of g() = a( - h) + k. The ais of smmetr is the vertical line whose equation is = h. The form of the epression for g is convenient because it immediatel identifies the verte of the parabola as (h, k). This is the standard form of a quadratic function. The Standard Form of a Quadratic Function The quadratic function f() = a( - h) + k, a 0 is in standard form. The graph of f is a parabola whose verte is the point (h, k). The parabola is smmetric with respect to the line = h. If a 7 0, the parabola opens upward; if a 6 0, the parabola opens downward.

13 0 Chapter Polnomial and Rational Functions The sign of a in f() = a( - h) + k determines whether the parabola opens upward or downward. Furthermore, if a is small, the parabola opens more flatl than if a is large. Here is a general procedure for graphing parabolas whose equations are in standard form: Graphing Quadratic Functions with Equations in Standard Form To graph f() = a( - h) + k,. Determine whether the parabola opens upward or downward. If a 7 0, it opens upward. If a 6 0, it opens downward.. Determine the verte of the parabola. The verte is (h, k).. Find b solving f() = 0. The function s real zeros are 4. Find b computing f(0). 5. Plot the intercepts, the verte, and additional points as necessar. Connect these points with a smooth curve that is shaped like a bowl or an inverted bowl. In the graphs that follow, we will show each ais of smmetr as a dashed vertical line. Because this vertical line passes through the verte, (h, k), its equation is = h. The line is dashed because it is not part of the parabola. EXAMPLE Graphing a Quadratic Function in Standard Form Graph the quadratic function f() = -( - ) + 8. SOLUTION We can graph this function b following the steps in the preceding bo. We begin b identifing values for a, h, and k. Standard form f()=a(-h) +k intercept: intercept: FIGURE.4 The graph of f() = -( - ) + 8 Verte: (, 8) -intercept: 5 Ais of smmetr: = a = h = k = 8 Given function f()= (-) +8 Step Determine how the parabola opens. Note that a, the coefficient of, is -. Thus, a 6 0; this negative value tells us that the parabola opens downward. Step Find the verte. The verte of the parabola is at (h, k). Because h = and k = 8, the parabola has its verte at (, 8). Step Find b solving f() 0. Replace f() with 0 in f() = -( - ) = -( - ) + 8 setting f() equal to 0. ( - ) = 8 Solve for. Add ( - ) to both sides of the equation. ( - ) = 4 Divide both sides b. - = 4 or - = - 4 Appl the square root propert. - = - = - 4 = = 5 = Add to both sides in each equation. are and 5. The parabola passes through (, 0) and (5, 0). Step 4 Find b computing f(0). Replace with 0 in f() = -( - ) + 8. f(0) = -(0 - ) + 8 = -(-) + 8 = -(9) + 8 = -0 is -0. The parabola passes through (0, -0). Step 5 Graph the parabola. With a verte at (, at 5 and, and ter cep t at -0, the graph of f is shown in Figure.4. The ais of smmetr is the vertical line whose equation is =.

14 Section. Quadratic Functions 0 Check Point Graph the quadratic function f() = -( - ) + 4. EXAMPLE Graphing a Quadratic Function in Standard Form Graph the quadratic function f() = ( + ) +. SOLUTION We begin b finding values for a, h, and k. f()=a(-h) +k f()=(+) + Standard form of quadratic function Given function f()=(-( )) + a = h = k = Ais of smmetr: = Verte: (, ) FIGURE.5 The graph of f() = ( + ) + -intercept: 0 Step Determine how the parabola opens. Note that a, the coefficient of, is. Thus, a 7 0; this positive value tells us that the parabola opens upward. Step Find the verte. The verte of the parabola is at (h, k). Because h = - and k =, the parabola has its verte at (-, ). Step Find b solving f() 0. Replace f() with 0 in f() = ( + ) +. Because the verte is at (-, ), which lies above and the parabola opens upward, it appears that this parabola has We can verif this observation algebraicall. 0 = ( + ) + Find setting f() equal to 0. - = ( + ) Solve for. Subtract from both sides. + = - or + = - - Appl the square root propert. + = i + = -i - = i = - + i = - - i The solutions are - { i. Because this equation has no real solutions, the parabola has Step 4 Find b computing f(0). Replace with 0 in f() = ( + ) +. f(0) = (0 + ) + = + = 9 + = 0 is 0. The parabola passes through (0, 0). Step 5 Graph the parabola. With a verte at (-, ), and at 0, the graph of f is shown in Figure.5. The ais of smmetr is the vertical line whose equation is = -. GREAT QUESTION! I m confused about finding h from the equation f() a( h) k. Can ou help me out? It s eas to make a sign error when finding h, of the verte. In f() = a( - h) + k, h is the number that follows the subtraction sign. f()= (-) +8 The number after the subtraction is : h =. f()=(+) + =(-( )) + The number after the subtraction is : h =.

15 04 Chapter Polnomial and Rational Functions Check Point Graph the quadratic function f() = ( - ) +. Graphing Quadratic Functions in the Form f () a b c Quadratic functions are frequentl epressed in the form f() = a + b + c. How can we identif the verte of a parabola whose equation is in this form? Completing the square provides the answer to this question. f() = a + b + c = aa + b a b + c Factor out a from a + b. =aa b b b + + b+c-aa b a 4a 4a Complete the square b adding the square of half the coefficient of. B completing the square, we added b a. To avoid changing the 4a function s equation, we must subtract this term. = aa + b a b + c - b 4a Write the trinomial as the square of a binomial and simplif the constant term. Compare this form of the equation with a quadratic function s standard form. Standard form f()=a(-h) +k b h = a k = c b 4a Equation under discussion f()=aa- a bb +cb a b 4a The important part of this observation is that h, of the verte, is - b a. can be found b evaluating the function at - b a. The Verte of a Parabola Whose Equation Is f () a b c Consider the parabola defined b the quadratic function f() = a + b + c. The parabola s verte is a- b a, f a- b a bb. is - b a. is found b substituting into the parabola s equation and evaluating the function at this value of. We can appl our five-step procedure to graph parabolas in the form f() = a + b + c.

16 Section. Quadratic Functions 05 Graphing Quadratic Functions with Equations in the Form f () a b c To graph f() = a + b + c,. Determine whether the parabola opens upward or downward. If a 7 0, it opens upward. If a 6 0, it opens downward.. Determine the verte of the parabola. The verte is a- b a, f a- b a bb.. Find b solving f() = 0. The real solutions of a + b + c = 0 are 4. Find b computing f(0). Because f(0) = c (the constant term in the function s equation), is c and the parabola passes through (0, c). 5. Plot the intercepts, the verte, and additional points as necessar. Connect these points with a smooth curve. EXAMPLE Graphing a Quadratic Function in the Form f() a b c Graph the quadratic function f() = Use the graph to identif the function s domain and its range. SOLUTION Step Determine how the parabola opens. Note that a, the coefficient of, is -. Thus, a 6 0; this negative value tells us that the parabola opens downward. Step Find the verte. We know that of the verte is = - b a. We identif a, b, and c in f() = a + b + c. f()= -+ a = b = c = Substitute the values of a and b into the equation for = - b a = - - (-) = - a - - b = -. of the verte is -. We substitute - for in the equation of the function, f() = - - +, to find The verte is at (-, ). f(-) = -(-) - (-) + = =. Step Find b solving f() 0. Replace f() with 0 in f() = We obtain 0 = This equation cannot be solved b factoring. We will use the quadratic formula to solve it. -+=0 a = b = c = b_b -4ac = = ( )_( ) -4( )() a ( ) To locate the -intercepts, we need decimal approimations. +8 Thus, there is no need to simplif =.4 or the radical form of the solutions. -8 = 0.4 = _4-( 4) are approimatel -.4 and 0.4. The parabola passes through ( -.4, 0) and (0.4, 0).

17 06 Chapter Polnomial and Rational Functions Step 4 Find b computing f(0). Replace with 0 in f() = f(0) = -0 - # 0 + = is, which is the constant term in the function s equation. The parabola passes through (0, ). Step 5 Graph the parabola. With a verte at (-, at approimatel -.4 and 0.4, and at, the graph of f is shown in Figure.6(a). The ais of smmetr is the vertical line whose equation is = -. Verte: (, ) -intercept: intercept: -intercept: Domain: Inputs on -ais include all real numbers Range: Outputs on -ais fall at or below. Ais of smmetr: = FIGURE.6(a) The graph of f() = FIGURE.6(b) Determining the domain and range of f() = GREAT QUESTION! Are there rules to find domains and ranges of quadratic functions? Yes. The domain of an quadratic function includes all real numbers. If the verte is the graph s highest point, the range includes all real numbers at or below of the verte. If the verte is the graph s lowest point, the range includes all real numbers at or above of the verte. Now we are read to determine the domain and range of f() = We can use the parabola, shown again in Figure.6(b), to do so. To find the domain, look for all the inputs on that correspond to points on the graph. As the graph widens and continues to fall at both ends, can ou see that these inputs include all real numbers? Domain of f is { is a real number} or (-, ). To find the range, look for all the outputs on that correspond to points on the graph. Figure.6(b) shows that the parabola s verte, (-, ), is the highest point on the graph. Because of the verte is, outputs on fall at or below. Ran ge o f f is { } or (-, ]. Check Point Graph the quadratic function f() = Use the graph to identif the function s domain and its range. Determine a quadratic function s minimum or maimum value. Minimum and Maimum Values of Quadratic Functions Consider the quadratic function f() = a + b + c. If a 7 0, the parabola opens upward and the verte is its lowest point. If a 6 0, the parabola opens downward and the verte is its highest point. of the verte is - b. Thus, we a can find the minimum or maimum value of f b evaluating the quadratic function at = - b a.

18 Section. Quadratic Functions 07 Minimum and Maimum: Quadratic Functions Consider the quadratic function f() = a + b + c.. If a 7 0, then f has a minimum that occurs at = - b. This minimum a value is f a- b a b.. If a 6 0, then f has a maimum that occurs at = - b. This maimum a value is f a- b a b. In each case, the value of gives the location of the minimum or maimum value. The value of, or f a- b b, gives that minimum or maimum value. a EXAMPLE 4 Obtaining Information about a Quadratic Function from Its Equation Consider the quadratic function f() = a. Determine, without graphing, whether the function has a minimum value or a maimum value. b. Find the minimum or maimum value and determine where it occurs. c. Identif the function s domain and its range. SOLUTION We begin b identifing a, b, and c in the function s equation: f()= +6-. a = b = 6 c = a. Because a 6 0, the function has a maimum value. b. The maimum value occurs at = - b a = - 6 (-) = - 6 = -(-) =. -6 The maimum value occurs at = and the maimum value of f() = is Range is (, 0]. FIGURE.7 [ 6, 6, ] b [ 50, 0, 0] f() = - # + 6 # - = = -0. We see that the maimum is -0 at =. c. Like all quadratic functions, the domain is (-, ). Because the function s maimum value is -0, the range includes all real numbers at or below -0. The range is (-, -0]. We can use the graph of f() = to visualize the results of Eample 4. Figure.7 shows the graph in a [-6, 6, ] b [-50, 0, 0] viewing rectangle. The maimum function feature verifies that the function s maimum is -0 a t =. Notice that gives the location of the maimum and gives the maimum value. Notice, too, that the maimum value is -0 and not the ordered pair (, -0). Check Point 4 Repeat parts (a) through (c) of Eample 4 using the quadratic function f() =

19 08 Chapter Polnomial and Rational Functions Solve problems involving a quadratic function s minimum or maimum value. Applications of Quadratic Functions Man applied problems involve finding the maimum or minimum value of a quadratic function, as well as where this value occurs. Height of the Punted Football (feet) EXAMPLE 5 The Parabolic Path of a Punted Football SOLUTION Figure.8 shows that when a football is kicked, the nearest defensive plaer is 6 feet from the point of impact with the kicker s foot. The height of the punted football, f(), in feet, can be modeled b f() = , 4 where is the ball s horizontal distance, in feet, from the point of impact with the kicker s foot. a. What is the maimum height of the punt and how far from the point of impact does this occur? Distance from the Point of Impact (feet) b. How far must the nearest defensive plaer, who is 6 feet FIGURE.8 from the kicker s point of impact, reach to block the punt? c. If the ball is not blocked b the defensive plaer, how far down the field will it go before hitting the ground? d. Graph the function that models the football s parabolic path. a. We begin b identifing the numbers a, b, and c in the function s equation. f()= a = 0.0 b =.8 c = Because a 6 0, the function has a maimum that occurs at = - b a. = - b a = -.8 (-0.0) = -(-59) = 59 This means that the maimum height of the punt occurs 59 feet from the kicker s point of impact. The maimum height of the punt is f(59) = -0.0(59) +.8(59) + = 6.8, or 6.8 feet. b. Figure.8 shows that the defensive plaer is 6 feet from the kicker s point of impact. To block the punt, he must touch the football along its parabolic path. This means that we must find the height of the ball 6 feet from the kicker. Replace with 6 in the given function, f() = f(6) = -0.0(6) +.8(6) + = = 8.7 The defensive plaer must reach 8.7 feet above the ground to block the punt. c. Assuming that the ball is not blocked b the defensive plaer, we are interested in how far down the field it will go before hitting the ground. We are looking for the ball s horizontal distance,, when its height above the ground, f(), is 0 feet. To find replace f() with 0 in f() = We obtain 0 = , or = 0. The equation cannot be solved b factoring. We will use the quadratic formula to solve it.

20 Section. Quadratic Functions =0 a = 0.0 b =.8 c = The equation for determining the ball s maimum horizontal distance b b -4ac = = a Use a calculator to evaluate the radicand..8 (.8) -4( 0.0)() = ( 0.0) = or = Reject this value. We are interested in the football s height corresponding to horizontal distances from its point of impact onward, or 0. Use a calculator and round to the nearest tenth. If the football is not blocked b the defensive plaer, it will go approimatel 9.7 feet down the field before hitting the ground. d. In terms of graphing the model for the football s parabolic path, f() = , we have alread determined the verte and the ter cep t. verte: (59, 6.8) -intercept: 9.7 The ball s maimum height, 6.8 feet, occurs at a horizontal distance of 59 feet. The ball s maimum horizontal distance is approimatel 9.7 feet. Figure.8 indicates that is, meaning that the ball is kicked from a height of feet. Let s verif this value b replacing with 0 in f() = f(0) = -0.0 # # 0 + = = Using the verte, (59, 6.8), 9.7, and the graph of the equation that models the football s parabolic path is shown infigure.9. The graph is shown onl for Ú 0, indicating horizontal distances that begin at the football s impact with the kicker s foot and end with the ball hitting the ground. 40 Verte: (59, 6.8) The ball s maimum height is 6.8 feet. Football s Height (feet) f() = intercept: The ball is kicked from a height of feet Football s Horizontal Distance (feet) -intercept: 9.7 The ball hits the ground after traveling a maimum horizontal distance of 9.7 feet. FIGURE.9 The parabolic path of a punted football

21 0 Chapter Polnomial and Rational Functions Check Point 5 An archer s arrow follows a parabolic path. The height of the arrow, f(), in feet, can be modeled b f() = , where is the arrow s horizontal distance, in feet. a. What is the maimum height of the arrow and how far from its release does this occur? b. Find the horizontal distance the arrow travels before it hits the ground. Round to the nearest foot. c. Graph the function that models the arrow s parabolic path. Quadratic functions can also be used to model verbal conditions. Once we have obtained a quadratic function, we can then use of the verte to determine its maimum or minimum value. Here is a step-b-step strateg for solving these kinds of problems: Strateg for Solving Problems Involving Maimizing or Minimizing Quadratic Functions. Read the problem carefull and decide which quantit is to be maimized or minimized.. Use the conditions of the problem to epress the quantit as a function in one variable.. Rewrite the function in the form f() = a + b + c. 4. Calculate - b a. If a 7 0, f has a minimum at = - b. This minimum a value is f a- b a b. If a 6 0, f has a maimum at = - b. This maimum a value is f a- b a b. 5. Answer the question posed in the problem. EXAMPLE 6 Minimizing a Product Among all pairs of numbers whose difference is 0, find a pair whose product is as small as possible. What is the minimum product? SOLUTION Step Decide what must be maimized or minimized. We must minimize the product of two numbers. Calling the numbers and, and calling the product P, we must minimize P =. Step Epress this quantit as a function in one variable. In the formula P =, P is epressed in terms of two variables, and. However, because the difference of the numbers is 0, we can write - = 0. We can solve this equation for in terms of (or vice versa), substitute the result into P =, and obtain P as a function of one variable. - = Subtract from both sides of - = 0. = - 0 Multipl both sides of the equation b - and solve for. Now we substitute - 0 for in P =. P = = ( - 0) Because P is now a function of, we can write P() = ( - 0)

22 Section. Quadratic Functions Step Write the function in the form f() a b c. We appl the distributive propert to obtain P()=(-0)= -0. a = b = 0 TECHNOLOGY Numeric Connections The TABLE feature of a graphing utilit can be used to verif our work in Eample 6. Enter = 0, the function for the product, when one of the numbers is. The product is a minimum, 5, when one of the numbers is 5. Step 4 Calculate b. If a + 0, the function has a minimum at this value. The a voice balloons show that a = and b = -0. = - b a = - -0 () = -(-5) = 5 This means that the product, P, of two numbers whose difference is 0 is a minimum when one of the numbers,, is 5. Step 5 Answer the question posed b the problem. The problem asks for the two numbers and the minimum product. We found that one of the numbers,, is 5. Now we must find the second number,. = - 0 = 5-0 = -5 The number pair whose difference is 0 and whose product is as small as possible is 5, -5. The minimum product is 5(-5), or -5. Check Point 6 Among all pairs of numbers whose difference is 8, find a pair whose product is as small as possible. What is the minimum product? EXAMPLE 7 Maimizing Area You have 00 ards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maimize the enclosed area. What is the maimum area? SOLUTION Step Decide what must be maimized or minimized. We must maimize area. What we do not know are the rectangle s dimensions, and. Step Epress this quantit as a function in one variable. Because we must maimize the area of a rectangle, we have A =. We need to transform this into a function in which A is represented b one variable. Because ou have 00 ards of fencing, the perimeter of the rectangle is 00 ards. This means that + = 00. We can solve this equation for in terms of, substitute the result into A =, and obtain A as a function in one variable. We begin b solving for. = 00 - Subtract from both sides of + = = Divide both sides b. = 50 - Divide each term in the numerator b. 50 FIGURE.0 What value of will maimize the rectangle s area? Now we substitute 50 - for in A =. A = = (50 - ) The rectangle and its dimensions are illustrated in Figure.0. Because A is now a function of, we can write A() = (50 - ). This function models the area, A(), of an rectangle whose perimeter is 00 ards in terms of one of its dimensions,.

23 Chapter Polnomial and Rational Functions Step Write the function in the form f() a b c. We appl the distributive propert to obtain A()=(50-)=50- = +50. a = b = 50 TECHNOLOGY Graphic Connections The graph of the area function A() = (50 - ) was obtained with a graphing utilit using a [0, 50, ] b [0, 700, 5] viewing rectangle. The maimum function feature verifies that a maimum area of 65 square ards occurs when one of the dimensions is 5 ards. Step 4 Calculate b. If a * 0, the function has a maimum at this value. The a voice balloons show that a = - and b = 50. = - b a = - 50 (-) = 5 This means that the area, A(), of a rectangle with perimeter 00 ards is a maimum when one of the rectangle s dimensions,, is 5 ards. 50 FIGURE.0 (repeated) Step 5 Answer the question posed b the problem. We found that = 5. Figure.0 shows that the rectangle s other dimension is 50 - = 50-5 = 5. The dimensions of the rectangle that maimize the enclosed area are 5 ards b 5 ards. The rectangle that gives the maimum area is actuall a square with an area of 5 ards # 5 ards, or 65 square ards. Blitzer Bonus Check Point 7 You have 0 feet of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maimize the enclosed area. What is the maimum area? The abilit to epress a quantit to be maimized or minimized as a function in one variable plas a critical role in solving ma-min problems. In calculus, ou will learn a technique for maimizing or minimizing all functions, not onl quadratic functions. Addressing Stress Parabolicall Moderate stress, high performance Stress levels can help or hinder performance. The parabola in Figure. serves as a model that shows people under both low stress and high stress perform worse than their moderate-stress counterparts. Performance Low stress, low performance Level of Stress High stress, low performance FIGURE. Source: Herbert Benson, Your Maimum Mind, Random House, 987. CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true..the quadratic function f() = a( - h) + k, a 0, is in form. The graph of f is called a/an whose verte is the point. The graph opens upward if a and opens downward if a.. Consider the quadratic function f() = a + b + c, a 0. If a 7 0, then f has a minimum that occurs at =. This minimum value is. If a 6 0, then f has a maimum that occurs at =. This maimum value is.. True or false: The graph of f() = ( - ) + opens upward. 4. True or false: The graph of f() = ( + 5) + has its verte at (5, ).

24 Section. Quadratic Functions 5.True or false: The -coordinate of the verte of f() = is f(). 6. The difference between two numbers is 8. If one number is represented b, the other number can be epressed as. The product of the numbers, P(), epressed in the form P() = a + b + c, is P() =. 7. The perimeter of a rectangle is 80 feet. If the length of the rectangle is represented b, its width can be epressed as. The area of the rectangle, A(), epressed in the form A() = a + b + c, is A() =. EXERCISE SET. Practice Eercises In Eercises 4, the graph of a quadratic function is given. Write the function s equation, selecting from the following options... f() = ( + ) - g() = ( + ) + h() = ( - ) + j() = ( - ) In Eercises 5 8, the graph of a quadratic function is given. Write the function s equation, selecting from the following options. f() = + + g() = - + h() = - j() = In Eercises 9 6, find the coordinates of the verte for the parabola defined b the given quadratic function. 9. f() = ( - ) + 0. f() = -( - ) +. f() = -( + ) + 5. f() = -( + 4) - 8. f() = f() = f() = f() = In Eercises 7 8, use the verte and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola s ais of smmetr. Use the graph to determine the function s domain and range. 7. f() = ( - 4) - 8. f() = ( - ) - 9. f() = ( - ) + 0. f() = ( - ) +. - = ( - ). - = ( - ). f() = ( + ) - 4. f() = f() = 4 - ( - ) 6. f() = - ( - ) 7. f() = f() = f() = f() = f() = - +. f() = f() = f() = f() = f() = f() = f() = In Eercises 9 44, an equation of a quadratic function is given. a. Determine, without graphing, whether the function has a minimum value or a maimum value. b. Find the minimum or maimum value and determine where it occurs. c.identif the function s domain and its range. 9. f() = f() = f() = f() = f() = f() = 6-6

25 4 Chapter Polnomial and Rational Functions Practice Plus In Eercises 45 48, give the domain and the range of each quadratic function whose graph is described. 45. The verte is (-, -) and the parabola opens up. 46. The verte is (-, -4) and the parabola opens down. 47. Maimum = -6 at = Minimum = 8 at = -6 In Eercises 49 5, write an equation in standard form of the parabola that has the same shape as the graph of f() =, but with the given point as the verte. 49. (5, ) 50. (7, 4) 5. ( -0, -5 ) 5. ( -8, -6 ) In Eercises 5 56, write an equation in standard form of the parabola that has the same shape as the graph of f() = or g() = -, but with the given maimum or minimum. 5. Maimum = 4 at = Maimum = -7 at = Minimum = 0 at = 56. Minimum = 0 at = 9 Application Eercises An athlete whose event is the shot put releases the shot with the same initial velocit but at different angles. The figure shows the parabolic paths for shots released at angles of 5 and 65. Eercises are based on the functions that model the parabolic paths. Shot Put s Height (feet) Maimum height Maimum height f() = Shot released at 5 g() = Shot released at 65 Distance of throw or maimum horizontal distance Shot Put s Horizontal Distance (feet) 57. When the shot whose path is shown b the blue graph is released at an angle of 5, its height, f(), in feet, can be modeled b f() = , where is the shot s horizontal distance, in feet, from its point of release. Use this model to solve parts (a) through (c) and verif our answers using the blue graph. a. What is the maimum height of the shot and how far from its point of release does this occur? b. What is the shot s maimum horizontal distance, to the nearest tenth of a foot, or the distance of the throw? c. From what height was the shot released? When the shot whose path is shown b the red graph in the previous column is released at an angle of 65, its height, g(), in feet, can be modeled b g() = , where is the shot s horizontal distance, in feet, from its point of release. Use this model to solve parts (a) through (c) and verif our answers using the red graph. a. What is the maimum height, to the nearest tenth of a foot, of the shot and how far from its point of release does this occur? b. What is the shot s maimum horizontal distance, to the nearest tenth of a foot, or the distance of the throw? c. From what height was the shot released? 59. A ball is thrown upward and outward from a height of 6 feet. The height of the ball, f(), in feet, can be modeled b f() = , where is the ball s horizontal distance, in feet, from where it was thrown. a. What is the maimum height of the ball and how far from where it was thrown does this occur? b. How far does the ball travel horizontall before hitting the ground? Round to the nearest tenth of a foot. c. Graph the function that models the ball s parabolic path. 60. A ball is thrown upward and outward from a height of 6 feet. The height of the ball, f(), in feet, can be modeled b f() = , where is the ball s horizontal distance, in feet, from where it was thrown. a. What is the maimum height of the ball and how far from where it was thrown does this occur? b. How far does the ball travel horizontall before hitting the ground? Round to the nearest tenth of a foot. c. Graph the function that models the ball s parabolic path. 6. Among all pairs of numbers whose sum is 6, find a pair whose product is as large as possible. What is the maimum product? 6. Among all pairs of numbers whose sum is 0, find a pair whose product is as large as possible. What is the maimum product? 6. Among all pairs of numbers whose difference is 6, find a pair whose product is as small as possible. What is the minimum product? 64. Among all pairs of numbers whose difference is 4, find a pair whose product is as small as possible. What is the minimum product? 65. You have 600 feet of fencing to enclose a rectangular plot that borders on a river. If ou do not fence the side along the river, find the length and width of the plot that will maimize the area. What is the largest area that can be enclosed? 600 River

26 Section. Quadratic Functions You have 00 feet of fencing to enclose a rectangular plot that borders on a river. If ou do not fence the side along the river, find the length and width of the plot that will maimize the area. What is the largest area that can be enclosed? You have 50 ards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maimize the enclosed area. What is the maimum area? 68. You have 80 ards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maimize the enclosed area. What is the maimum area? 69. A rectangular plaground is to be fenced off and divided in two b another fence parallel to one side of the plaground. Si hundred feet of fencing is used. Find the dimensions of the plaground that maimize the total enclosed area. What is the maimum area? 70. A rectangular plaground is to be fenced off and divided in two b another fence parallel to one side of the plaground. Four hundred feet of fencing is used. Find the dimensions of the plaground that maimize the total enclosed area. What is the maimum area? 7. A rain gutter is made from sheets of aluminum that are 0 inches wide b turning up the edges to form right angles. Determine the depth of the gutter that will maimize its cross-sectional area and allow the greatest amount of water to flow. What is the maimum cross-sectional area? 7. A rain gutter is made from sheets of aluminum that are inches wide b turning up the edges to form right angles. Determine the depth of the gutter that will maimize its cross-sectional area and allow the greatest amount of water to flow. What is the maimum cross-sectional area? If ou have difficult obtaining the functions to be maimized in Eercises 7 76, read Eample in Section.0 on pages On a certain route, an airline carries 8000 passengers per month, each paing $50. A market surve indicates that for each $ increase in the ticket price, the airline will lose 00 passengers. Find the ticket price that will maimize the airline s monthl revenue for the route. What is the maimum monthl revenue? 74. A car rental agenc can rent ever one of its 00 cars at $0 per da. For each $ increase in rate, five fewer cars are rented. Find the rental amount that will maimize the agenc s dail revenue. What is the maimum dail revenue? 75. The annual ield per walnut tree is fairl constant at 60 pounds per tree when the number of trees per acre is 0 or fewer. For each additional tree over 0, the annual ield per tree for all trees on the acre decreases b pounds due to overcrowding. How man walnut trees should be planted per acre to maimize the annual ield for the acre? What is the maimum number of pounds of walnuts per acre? 76. The annual ield per cherr tree is fairl constant at 50 pounds per tree when the number of trees per acre is 0 or fewer. For each additional tree over 0, the annual ield per tree for all trees on the acre decreases b pound due to overcrowding. How man cherr trees should be planted per acre to maimize the annual ield for the acre? What is the maimum number of pounds of cherries per acre? Writing in Mathematics 77. What is a quadratic function? 78. What is a parabola? Describe its shape. 79. Eplain how to decide whether a parabola opens upward or downward. 80. Describe how to find a parabola s verte if its equation is epressed in standard form. Give an eample. 8. Describe how to find a parabola s verte if its equation is in the form f() = a + b + c. Use f() = as an eample. 8. A parabola that opens upward has its verte at (, ). Describe as much as ou can about the parabola based on this information. Include in our discussion the number (if an) for the parabola. Technolog Eercises 8. Use a graphing utilit to verif an five of our hand-drawn graphs in Eercises a. Use a graphing utilit to graph = in a standard viewing rectangle. What do ou observe? b. Find the coordinates of the verte for the given quadratic function. c. The answer to part (b) is (0.5, -0.5). Because the leading coefficient,, of the given function is positive, the verte is a minimum point on the graph. Use this fact to help find a viewing rectangle that will give a relativel complete picture of the parabola. With an ais of smmetr at = 0.5, the setting for should etend past this, so tr Xmin = 0 and Xma = 0. The setting for should include (and probabl go below) of the graph s so tr Ymin = -0. Eperiment with Yma until our utilit shows the parabola s major features. d. In general, eplain how knowing the coordinates of a parabola s verte can help determine a reasonable viewing rectangle on a graphing utilit for obtaining a complete picture of the parabola. In Eercises 85 88, find the verte for each parabola. Then determine a reasonable viewing rectangle on our graphing utilit and use it to graph the quadratic function. 85. = = = =

27 6 Chapter Polnomial and Rational Functions 89. The bar graph shows the ratings of American Idol from season (00) through season 9 (00). Average Number of Viewers (millions) American Idol: Each Season s Champion and Average Number of Viewers Season Kell Clarkson Season Ruben Studdard Season Fantasia Barrino Season 4 Carrie Underwood Season 5 Talor Hicks Season 6 Jordin Sparks Season 7 David Cook Season 8 Kris Allen Season 9 Lee DeWze Source : Nielsen a.let represent American Idol s season number and let represent the average number of viewers, in millions. Use a graphing utilit to draw a scatter plot of the data. Eplain wh a quadratic function is appropriate for modeling these data. b. Use the quadratic regression feature to find the quadratic function that best fits the data. Round all numbers to two decimal places. c. Use the model in part (b) to determine the season in which American Idol had the greatest number of viewers. Round to the nearest whole number. According to the model, how man millions of viewers were there in that season? Round to one decimal place. d. How do the results obtained from the model in part (c) compare with the data displaed b the graph? e. Use a graphing utilit to draw a scatter plot of the data and graph the quadratic function of best fit on the scatter plot. Can ou see wh projections based on the graph have the producers of American Idol looking for a shake-up? Critical Thinking Eercises Make Sense? In Eercises 90 9, determine whether each statement makes sense or does not make sense, and eplain our reasoning. 90. I must have made an error when graphing this parabola because its ais of smmetr is is. 9. I like to think of a parabola s verte as the point where it intersects its ais of smmetr. 9. I threw a baseball verticall upward and its path was a parabola. 9. Figure.8 on page 08 shows that a linear function provides a better description of the football s path than a quadratic function. In Eercises 94 97, determine whether each statement is true or false. If the statement is false, make the necessar change(s) to produce a true statement. 94. No quadratic functions have a range of (-, ). 95. The verte of the parabola described b f() = ( - 5) - is at (5, ). 96. The graph of f() = -( + 4) - 8 has and 97. The maimum value of for the quadratic function f() = is. In Eercises 98 99, find the ais of smmetr for each parabola whose equation is given. Use the ais of smmetr to find a second point on the parabola is the same as the given point. 98. f() = ( + ) - 5; (-, - ) 99. f() = ( - ) + ; ( 6, ) In Eercises 00 0, write the equation of each parabola in standard form. 00. Verte: (-, -4); The graph passes through the point (, 4). 0. Verte: (-, -); The graph passes through the point ( -, - ). 0. Find the point on the line whose equation is + - = 0 that is closest to the origin. Hint: Minimize the distance function b minimizing the epression under the square root. 0. A 00-room hotel can rent ever one of its rooms at $80 per room. For each $ increase in rent, three fewer rooms are rented. Each rented room costs the hotel $0 to service per da. How much should the hotel charge for each room to maimize its dail profit? What is the maimum dail profit?

28 Section. Polnomial Functions and Their Graphs A track and field area is to be constructed in the shape of a rectangle with semicircles at each end. The inside perimeter of the track is to be 440 ards. Find the dimensions of the rectangle that maimize the area of the rectangular portion of the field. Group Eercise 05. Each group member should consult an almanac, newspaper, magazine, or the Internet to find data that initiall increase and then decrease, or vice versa, and therefore can be modeled b a quadratic function. Group members should select the two sets of data that are most interesting and relevant. For each data set selected, a. Use the quadratic regression feature of a graphing utilit to find the quadratic function that best fits the data. b. Use the equation of the quadratic function to make a prediction from the data. What circumstances might affect the accurac of our prediction? c. Use the equation of the quadratic function to write and solve a problem involving maimizing or minimizing the function. Preview Eercises Eercises will help ou prepare for the material covered in the net section. 06. Factor: If f() = - - 5, find f() and f(). Then eplain wh the continuous graph of f must cross between and. 08. Determine whether f() = is even, odd, or neither. Describe the smmetr, if an, for the graph of f. SECTION. Polnomial Functions and Their Graphs Objectives Identif polnomial functions. Recognize characteristics of graphs of polnomial functions. Determine end behavior. Use factoring to fi nd zeros of polnomial functions. Identif zeros and their multiplicities. Use the Intermediate Value Theorem. Understand the relationship between degree and turning points. Graph polnomial functions. I n 980, U.S. doctors diagnosed 4 cases of a rare form of cancer, Kaposi s sarcoma, that involved skin lesions, pneumonia, and severe immunological deficiencies. All cases involved ga men ranging in age from 6 to 5. B the end of 008, approimatel. million Americans, straight and ga, male and female, old and oung, were infected with the HIV virus. Modeling AIDS-related data and making predictions about the epidemic s havoc is serious business. Figure. on the net page shows the number of AIDS cases diagnosed in the United States from 98 through 008. Basketball plaer Magic Johnson (959 ) tested positive for HIV in 99.

29 8 Chapter Polnomial and Rational Functions AIDS Cases Diagnosed in the United States, Number of Cases Diagnosed 80,000 70,000 60,000 50,000 40,000 0,000 0,000 0, ,657 79,879 7,086 69,984 60,57 6,4 49,546 49,79 45,669 4,499 4,5 4,7 4,6 4,7 4,4 4,9 40,907 6,6 5,695 5,96 7,5 9,05 9,404, Year FIGURE. Source: Department of Health and Human Services Identif polnomial functions. Changing circumstances and unforeseen events can result in models for AIDS-related data that are not particularl useful over long periods of time. For eample, the function f() = models the number of AIDS cases diagnosed in the United States ears after 98. The model was obtained using a portion of the data shown infigure., namel cases diagnosed from 98 through 99, inclusive. Figure. shows the graph of f from 98 through 99. This function is an eample of a polnomial function of degree. Cases Diagnosed 60, f() = Years after 98 [0, 8, ] b [0, 60,000, 5000] FIGURE. The graph of a function modeling the number of AIDS diagnoses from 98 through 99 Definition of a Polnomial Function Let n be a nonnegative integer and let a n, a n -, c, a, a, a 0 be real numbers, with a n 0. The function defined b f() = a n n + a n - n - + g + a + a + a 0 is called a polnomial function of degree n. The number a n, the coefficient of the variable to the highest power, is called the leading coefficient. Polnomial Functions f()= Polnomial function of degree 5 g()= 4 (-)(+) = 4 ( +-6) = Polnomial function of degree 6 Not Polnomial Functions F()= + +5 = + +5 The eponent on the variable is not an integer. G()= + +5 = + +5 The eponent on the variable is not a nonnegative integer.

30 Section. Polnomial Functions and Their Graphs 9 A constant function f() = c, where c 0, is a polnomial function of degree 0. A linear function f() = m + b, where m 0, is a polnomial function of degree. A quadratic function f() = a + b + c, where a 0, is a polnomial function of degree. In this section, we focus on polnomial functions of degree or higher. Recognize characteristics of graphs of polnomial functions. Smooth, Continuous Graphs Polnomial functions of degree or higher have graphs that are smooth and continuous. B smooth, we mean that the graphs contain onl rounded curves with no sharp corners. B continuous, we mean that the graphs have no breaks and can be drawn without lifting our pencil from the rectangular coordinate sstem. These ideas are illustrated infigure.4. Graphs of Polnomial Functions Not Graphs of Polnomial Functions Smooth rounded corner Smooth rounded corner Sharp corner Smooth rounded corner Smooth rounded corners Discontinuous; a break in the graph Sharp corner FIGURE.4 Recognizing graphs of polnomial functions Determine end behavior. End Behavior of Polnomial Functions Figure.5 shows the graph of the function Cases Diagnosed 85, Years after 98 [0,, ] b [ 0,000, 85,000, 5000] Graph falls to the right. FIGURE.5 B etending the viewing rectangle, we see that is eventuall negative and the function no longer models the number of AIDS cases. f() = , which models the number of U.S. AIDS diagnoses from 98 through 99. Look what happens to the graph when we etend the ear up through 005. B ear (004), the values of are negative and the function no longer models AIDS diagnoses. We ve added an arrow to the graph at the far right to emphasize that it continues to decrease without bound. It is this far-right end behavior of the graph that makes it inappropriate for modeling AIDS cases into the future. The behavior of the graph of a function to the far left or the far right is called its end behavior. Although the graph of a polnomial function ma have intervals where it increases or decreases, the graph will eventuall rise or fall without bound as it moves far to the left or far to the right. How can ou determine whether the graph of a polnomial function goes up or down at each end? The end behavior of a polnomial function f() = a n n + a n - n - + g + a + a 0 depends upon the leading term a n n, because when is large, the other terms are relativel insignificant in size. In particular, the sign of the leading coefficient, a n, and the degree, n, of the polnomial function reveal its end behavior. In terms of end behavior, onl the term of highest degree counts, as summarized b the Leading Coefficient Test.

31 0 Chapter Polnomial and Rational Functions The Leading Coefficient Test As increases or decreases without bound, the graph of the polnomial function f() = a n n + a n - n - + a n - n - + g + a + a 0 (a n 0) eventuall rises or falls. In particular, If the leading coefficient is positive, the graph falls to the left and rises to the right. (b, Q). For n odd:. For n even: If the leading coefficient is negative, the graph rises to the left and falls to the right. (a, R) If the leading coefficient is positive, the graph rises to the left and rises to the right. (a, Q) a n > 0 a n < 0 a n > 0 a n < 0 If the leading coefficient is negative, the graph falls to the left and falls to the right. (b, R) Rises right Rises left Rises right Rises left Falls right Falls left Falls left Falls right Odd degree; positive leading coefficient Odd degree; negative leading coefficient Even degree; positive leading coefficient Even degree; negative leading coefficient DISCOVERY Verif each of the four cases of the Leading Coefficient Test b using a graphing utilit to graph f() =, f() = -, f() =, and f() = -. GREAT QUESTION! What s the bottom line on the Leading Coefficient Test? Odd-degree polnomial functions have graphs with opposite behavior at each end. Even-degree polnomial functions have graphs with the same behavior at each end. Here s a table to help ou remember the details: Leading Term: a n n Odd n Even n a n 0 a n 0 b Q a R a Q b R Opposite behavior at each end Same behavior at each end EXAMPLE Using the Leading Coefficient Test Use the Leading Coefficient Test to determine the end behavior of the graph of f() = SOLUTION We begin b identifing the sign of the leading coefficient and the degree of the polnomial. f()= + -- The leading coefficient,, is positive. The degree of the polnomial,, is odd.

32 Falls left 5 4 Rises right FIGURE.6 The graph of f() = Section. Polnomial Functions and Their Graphs The degree of the function f is, which is odd. Odd-degree polnomial functions have graphs with opposite behavior at each end. The leading coefficient,, is positive. Thus, the graph falls to the left and rises to the right ( b, Q). The graph of f is shown in Figure.6. Check Point Use the Leading Coefficient Test to determine the end behavior of the graph of f() = 4-4. EXAMPLE Using the Leading Coefficient Test Use the Leading Coefficient Test to determine the end behavior of the graph of f() = -4 ( - ) ( + 5). SOLUTION Although the equation for f is in factored form, it is not necessar to multipl to determine the degree of the function. f()= 4 (-) (+5) Degree of this factor is. Degree of this factor is. Degree of this factor is. When multipling eponential epressions with the same base, we add the eponents. This means that the degree of f i s + +, or 6, which is even. Even-degree polnomial functions have graphs with the same behavior at each end. Without multipling out, ou can see that the leading coefficient is -4, which is negative. Thus, the graph of f falls to the left and falls to the right (b, R). Check Point Use the Leading Coefficient Test to determine the end behavior of the graph of f() = ( - )( + 5). EXAMPLE Using the Leading Coefficient Test Use end behavior to eplain wh f() = is onl an appropriate model for AIDS diagnoses for a limited time period. SOLUTION We begin b identifing the sign of the leading coefficient and the degree of the polnomial. f()= The leading coefficient, 49, is negative. The degree of the polnomial,, is odd. The degree of f is, which is odd. Odd-degree polnomial functions have graphs with opposite behavior at each end. The leading coefficient, -49, is negative. Thus, the graph rises to the left and falls to the right (a, R). The fact that the graph falls to the right indicates that at some point the number of AIDS diagnoses will be negative, an impossibilit. If a function has a graph that decreases without bound over time, it will not be capable of modeling nonnegative phenomena over long time periods. Model breakdown will eventuall occur.

33 Chapter Polnomial and Rational Functions Check Point The polnomial function f() = models the ratio of students to computers in U.S. public schools ears after 980. Use end behavior to determine whether this function could be an appropriate model for computers in the classroom well into the twent-first centur. Eplain our answer. If ou use a graphing utilit to graph a polnomial function, it is important to select a viewing rectangle that accuratel reveals the graph s end behavior. If the viewing rectangle, or window, is too small, it ma not accuratel show a complete graph with the appropriate end behavior. FIGURE.7 [ 8, 8, ] b [ 0, 0, ] EXAMPLE 4 Using the Leading Coefficient Test The graph of f() = was obtained with a graphing utilit using a [-8, 8, ] b [-0, 0, ] viewing rectangle. The graph is shown in Figure.7. Is this a complete graph that shows the end behavior of the function? SOLUTION We begin b identifing the sign of the leading coefficient and the degree of the polnomial. f()= The leading coefficient,, is negative. The degree of the polnomial, 4, is even. [ 0, 0, ] b [ 000, 750, 50] FIGURE.8 FIGURE.9 Use factoring to fi nd zeros of polnomial functions. The degree of f is 4, which is even. Even-degree polnomial functions have graphs with the same behavior at each end. The leading coefficient, -, is negative. Thus, the graph should fall to the left and fall to the right ( b, R). The graph in Figure.7 is falling to the left, but it is not falling to the right. Therefore, the graph is not complete enough to show end behavior. A more complete graph of the function is shown in a larger viewing rectangle infigure.8. Check Point 4The graph of f() = is shown in a standard viewing rectangle in Figure.9. Use the Leading Coefficient Test to determine whether this is a complete graph that shows the end behavior of the function. Eplain our answer. Zeros of Polnomial Functions If f is a polnomial function, then the values of for which f() is equal to 0 are called the zeros of f. These values of are the roots, or solutions, of the polnomial equation f() = 0. Each real root of the polnomial equation appears as of the graph of the polnomial function. EXAMPLE 5 Finding Zeros of a Polnomial Function Find all zeros of f() = SOLUTION B definition, the zeros are the values of for which f() is equal to 0. Thus, we set f() equal to 0: f() = = 0.

34 5 4 -intercept: -intercept: intercept: 5 FIGURE.0 TECHNOLOGY Section. Polnomial Functions and Their Graphs We solve the polnomial equation = 0 for as follows: = 0 This is the equation needed to find the function s zeros. ( + ) - ( + ) = 0 Factor from the first two terms and - from the last two terms. ( + )( - ) = 0 A common factor of + is factored from the epression. + = 0 or - = 0 Set each factor equal to 0. = - = Solve for. = { Remember that if = d, then = { d. The zeros of f are -, -, and. The graph of f in Figure.0 shows that each zero is The graph passes through the points (-, 0), (-, 0), and (, 0). Graphic and Numeric Connections A graphing utilit can be used to verif that -, -, and are the three real zeros of f() = Numeric Check Displa a table for the function. Enter = +. Graphic Check Displa a graph for the function. indicate that -, -, and are the real zeros.,, and are the real zeros. is equal to 0 when =, =, and =. -intercept: -intercept: -intercept: [ 6, 6, ] b [ 6, 6, ] The utilit s ZERO feature on the graph of f also verifies that -, -, and are the function s real zeros. Check Point 5Find all zeros of f() = intercept: 0 -intercept: FIGURE. The zeros of f() = , namel 0 and, are for the graph of f. EXAMPLE 6 Finding Zeros of a Polnomial Function Find all zeros of f() = SOLUTION We find the zeros of f b setting f() equal to 0 and solving the resulting equation = 0 We now have a polnomial equation = 0 Multipl both sides b -. This step is optional. ( ) = 0 Factor out. ( - ) = 0 Factor completel. = 0 or ( - ) = 0 Set each factor equal to 0. = 0 = Solve for. The zeros of f() = are 0 and. The graph of f, shown in Figure., at 0 and. The graph passes through the points (0, 0) and (, 0).

35 4 Chapter Polnomial and Rational Functions Check Point 6Find all zeros of f() = 4-4. GREAT QUESTION! Can zeros of polnomial functions alwas be found using one or more of the factoring techniques that were reviewed in Section P.5? No. You ll be learning additional strategies for finding zeros of polnomial functions in the net two sections of this chapter. Identif zeros and their multiplicities. Multiplicities of Zeros We can use the results of factoring to epress a polnomial as a product of factors. For instance, in Eample 6, we can use our factoring to epress the function s equation as follows: f()= = ( )= (-). The factor occurs twice: =. The factor ( ) occurs twice: ( ) = ( )( ). -intercept: 0 -intercept: FIGURE. (repeated) The graph of f() = Notice that each factor occurs twice. In factoring the equation for the polnomial function f, if the same factor - r occurs k times, but not k + times, we call r a zero with multiplicit k. For the polnomial function f() = - ( - ), 0 and are both zeros with multiplicit. Multiplicit provides another connection between zeros and graphs. The multiplicit of a zero tells us whether the graph of a polnomial function touches at the zero and turns around or if the graph crosses at the zero. For eample, look again at the graph of f() = in Figure.. Each zero, 0 and, is a zero with multiplicit. The graph of f touches, but does not cross, at each of these zeros of even multiplicit. B contrast, a graph crosses at zeros of odd multiplicit. Multiplicit If r is a zero of even multiplicit, then the graph touches and turns around at r. If r is a zero of odd multiplicit, then the graph crosses at r. Regardless of whether the multiplicit of a zero is even or odd, graphs tend to flatten out near zeros with multiplicit greater than one. GREAT QUESTION! If r is a zero of even multiplicit, how come the graph of f doesn t just cross at r? Because r is a zero of even multiplicit, the sign of f() does not change from one side of r to the other side of r. This means that the graph must turn around at r. On the other hand, if r is a zero of odd multiplicit, the sign of f() changes from one side of r to the other side. That s wh the graph crosses at r. If a polnomial function s equation is epressed as a product of linear factors, we can quickl identif zeros and their multiplicities. EXAMPLE 7 Finding Zeros and Their Multiplicities Find the zeros of f() = ( + )( - ) and give the multiplicit of each zero. State whether the graph crosses or touches and turns around at each zero. SOLUTION We find the zeros of f b setting f() equal to 0: ( + )( - ) = 0.

36 Section. Polnomial Functions and Their Graphs 5 Set each variable factor equal to 0. + = 0 = = 0 = is a zero of odd multiplicit. Graph crosses -ais. q(+) (-) =0 This eponent is. Thus, the multiplicit of is. This eponent is. Thus, the multiplicit of is. The zeros of f() = ( + )( - ) are -, with multiplicit, and, with multiplicit. Because the multiplicit of - is odd, the graph crosses at is a zero of even multiplicit. Graph touches -ais, flattens, and turns around. [,, ] b [ 0, 0, ] FIGURE. The graph of f() = ( + )( - ) this zero. Because the multiplicit of is even, the graph touches and turns around at this zero. These relationships are illustrated b the graph of f in Figure.. Check Point 7Find the zeros of f() = -4 + ( - 5) and give the multiplicit of each zero. State whether the graph crosses or touches and turns around at each zero. Use the Intermediate Value Theorem. (b, f(b)) f(b) > 0 The Intermediate Value Theorem The Intermediate Value Theorem tells us of the eistence of real zeros. The idea behind the theorem is illustrated infigure.. The figure shows that if (a, f(a)) lies below and (b, f(b)) lies above the smooth, continuous graph of the polnomial function f must cross at some value c between a and b. This value is a real zero for the function. These observations are summarized in the Intermediate Value Theorem. The Intermediate Value Theorem for Polnomial Functions f(c) = 0 a c b Let f be a polnomial function with real coefficients. If f(a) and f(b) have opposite signs, then there is at least one value of c between a and b for which f(c) = 0. Equivalentl, the equation f() = 0 has at least one real root between a and b. (a, f(a)) f(a) < 0 FIGURE. The graph must cross at some value between a and b. = 5 EXAMPLE 8 Using the Intermediate Value Theorem Show that the polnomial function f() = has a real zero between and. SOLUTION Let us evaluate f at and at. If f() and f() have opposite signs, then there is at least one real zero between and. Using f() = - - 5, we obtain f()= - -5=8-4-5= f() is negative. and f()= - -5=7-6-5=6. f() is positive. FIGURE.4 [,, ] b [ 0, 0, ] Because f() = - and f() = 6, the sign change shows that the polnomial function has a real zero between and. This zero is actuall irrational and is approimated using a graphing utilit s ZERO feature as in Figure.4.

37 6 Chapter Polnomial and Rational Functions Check Point 8 Show that the polnomial function f() = has a real zero between - and -. Understand the relationship between degree and turning points. Turning points: from increasing to decreasing f() = Turning points: from decreasing to increasing FIGURE.5 Graph with four turning points Graph polnomial functions. Turning Points of Polnomial Functions The graph of f() = is shown in Figure.5. The graph has four smooth turning points. At each turning point in Figure.5, the graph changes direction from increasing to decreasing or vice versa. The given equation has 5 as its greatest eponent and is therefore a polnomial function of degree 5. Notice that the graph has four turning points. In general, if f is a polnomial function of degree n, then the graph of f has at most n turning points. Figure.5 illustrates that of each turning point is either a relative maimum or a relative minimum of f. Without the aid of a graphing utilit or a knowledge of calculus, it is difficult and often impossible to locate turning points of polnomial functions with degrees greater than. If necessar, test values can be taken between to get a general idea of how high the graph rises or how low the graph falls. For the purpose of graphing in this section, a general estimate is sometimes appropriate and necessar. A Strateg for Graphing Polnomial Functions Here s a general strateg for graphing a polnomial function. A graphing utilit is a valuable complement, but not a necessar component, to this strateg. If ou are using a graphing utilit, some of the steps listed in the following bo will help ou to select a viewing rectangle that shows the important parts of the graph. GREAT QUESTION! When graphing a polnomial function, how do I determine the location of its turning points? Without calculus, it is often impossible to give the eact location of turning points. However, ou can obtain additional points satisfing the function to estimate how high the graph rises or how low it falls. To obtain these points, use values of between (and to the left and right of) e r c e pt s. Graphing a Polnomial Function f() = a n n + a n - n - + a n - n - + g + a + a 0, a n 0. Use the Leading Coefficient Test to determine the graph s end behavior.. b setting f() = 0 and solving the resulting polnomial equation. If there is at r as a result of ( - r) k in the complete factorization of f(), then a. If k is even, the graph touches at r and turns around. b. If k is odd, the graph crosses at r. c. If k 7, the graph flattens out near (r, 0).. Find b computing f(0). 4. Use smmetr, if applicable, to help draw the graph: smmetr: f(- ) = f() b. Origin smmetr: f(-) = -f(). 5. Use the fact that the maimum number of turning points of the graph is n -, where n is the degree of the polnomial function, to check whether it is drawn correctl. EXAMPLE 9 Graphing a Polnomial Function Graph: f() = SOLUTION Step Determine end behavior. Identif the sign of a n, the leading coefficient, and the degree, n, of the polnomial function.

38 Section. Polnomial Functions and Their Graphs 7 f()= Rises left Rises right The leading coefficient,, is positive. The degree of the polnomial function, 4, is even. Because the degree, 4, is even, the graph has the same behavior at each end. The leading coefficient,, is positive. Thus, the graph rises to the left and rises to the right. Step (zeros of the function) b setting f() = 0 Set f () equal to 0. ( - )( - ) = 0 Factor. ( + )( - )( + )( - ) = 0 Factor completel. ( + ) ( - ) = 0 Epress the factorization in a more compact form. ( + ) = 0 or ( - ) = 0 Set each factorization equal to 0. = - = Solve for. We see that - and are both repeated zeros with multiplicit. Because of the even multiplicit, the graph touches at - and and turns around. Furthermore, the graph tends to flatten out near these zeros with multiplicit greater than one. Rises left Rises right Step Find b computing f(0). We use f() = and compute f(0). f(0) = # 0 + = There is at, so the graph passes through (0, ). It appears that is a relative maimum, but we need more information to be certain FIGURE.6 The graph of f() = Step 4 Use possible smmetr to help draw the graph. Our partial graph smmetr. Let s verif this b finding f(-). f() = Replace with. f( )=( ) 4 -( ) += Because f(-) = f(), the graph of f is smmetric with respect to Figure.6 shows the graph of f() = Step 5 Use the fact that the maimum number of turning points of the graph is n to check whether it is drawn correctl. Because n = 4, the maimum number of turning points is 4 -, or. Because the graph in Figure.6 has three turning points, we have not violated the maimum number possible. Can ou see how this verifies that is indeed a relative maimum and (0, ) is a turning point? If the graph rose above on either side of = 0, it would have to rise above on the other side as well because of smmetr. This would require additional turning points to smoothl curve back to The graph alread has three turning points, which is the maimum number for a fourth-degree polnomial function.

39 8 Chapter Polnomial and Rational Functions Check Point 9Use the five-step strateg to graph f() = -. EXAMPLE 0 Graphing a Polnomial Function Graph: f() = -( - ) ( + ). SOLUTION Step Determine end behavior. Identif the sign of a n, the leading coefficient, and the degree, n, of the polnomial function. f()= (-) (+) The leading term is, or. The leading coefficient,, is negative. The degree of the polnomial function,, is odd. Because the degree,, is odd, the graph has opposite behavior at each end. The leading coefficient, -, is negative. Thus, the graph rises to the left and falls to the right. Rises left Falls right Step intercepts (zeros of the function) b setting f() 0. -( - ) ( + ) = 0 Set f() equal to 0. ( - ) = 0 or + = 0 Set each variable factor equal to 0. = = - Solve for. We see that the zeros are and -. The multiplicit of is even, so the graph touches at, flattens, and turns around. The multiplicit of - is odd, so the graph crosses at -. Rises left Falls right Step Find b computing f(0). We use f() = -( - ) ( + ) and compute f(0). f(0) = -(0 - ) (0 + ) = -()() = -4

40 Section. Polnomial Functions and Their Graphs 9 There is at -4, so the graph passes through (0, -4). We're not sure about the graph's behavior for between and 0 or between 0 and. Additional points will be helpful. Furthermore, we ma need to rescale the -ais. 4 Based on the note in the voice balloon, let s evaluate the function at -.5, -, -0.5, and 0.5, as well as at -,, and f() ( ) ( ) In order to accommodate these points, we ll scale from -50 to 50, with each tick mark representing 0 units. Step 4 Use possible smmetr to help draw the graph. Our partial graph illustrates that we have neither -ais smmetr nor origin smmetr. Using end behavior, intercepts, and the points from our table, Figure.7 shows the graph of f() = -( - ) ( + ). We cannot be sure about the eact location of this turning point. (Yes, our author is certain there's a relative minimum at : He took calculus!) FIGURE.7 The graph of f() = -( - ) ( + ) Step 5 Use the fact that the maimum number of turning points of the graph is n to check whether it is drawn correctl. The leading term is -, so n =. The maimum number of turning points is -, or. Because the graph in Figure.7 has two turning points, we have not violated the maimum number possible. Check Point 0 Use the five-step strateg to graph f() = ( + ) ( - ).

41 0 Chapter Polnomial and Rational Functions CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true.. The degree of the polnomial function f() = - ( - )( + 5) is. The leading coefficient is.. True or false: Some polnomial functions of degree or higher have breaks in their graphs.. The behavior of the graph of a polnomial function to the far left or the far right is called its behavior, which depends upon the term. 4.The graph of f() = to the left and to the right. 5.The graph of f() = - to the left and to the right. 6.The graph of f() = to the left and to the right. 7.The graph of f() = - to the left and to the right. 8. True or false: Odd-degree polnomial functions have graphs with opposite behavior at each end. 9. True or false: Even-degree polnomial functions have graphs with the same behavior at each end. 0. Ever real zero of a polnomial function appears as a/an of the graph..if r is a zero of even multiplicit, then the graph touches and at r. If r is a zero of odd multiplicit, then the graph at r..if f is a polnomial function and f(a) and f(b) have opposite signs, then there must be at least one value of c between a and b for which f(c) =. This result is called the Theorem.. If f is a polnomial function of degree n, then the graph of f has at most turning points. EXERCISE SET. Practice Eercises In Eercises 0, determine which functions are polnomial functions. For those that are, identif the degree.. f() = f() = g() = p g() = 67 + p h() = h() = f() = f() = f() = f() = + 7 In Eercises 4, identif which graphs are not those of polnomial functions... In Eercises 5 8, use the Leading Coefficient Test to determine the end behavior of the graph of the given polnomial function. Then use this end behavior to match the polnomial function with its graph. [ The graphs are labeled (a) through (d).] 5. f() = f() = f() = ( - ) 8. f() = a. b. c d In Eercises 9 4, use the Leading Coefficient Test to determine the end behavior of the graph of the polnomial function. 9. f() = f() = f() = f() = f() = f() =

42 Section. Polnomial Functions and Their Graphs In Eercises 5, find the zeros for each polnomial function and give the multiplicit for each zero. State whether the graph crosses or touches and turns around, at each zero. 5. f() = ( - 5)( + 4) 6. f() = ( + 5)( + ) 7. f() = 4( - )( + 6) 8. f() = - + ( - 4) 9. f() = f() = f() = f() = In Eercises 40, use the Intermediate Value Theorem to show that each polnomial has a real zero between the given integers.. f() = - - ; between and 4. f() = ; between 0 and 5. f() = ; between - and 0 6. f() = ; between and 7. f() = ; between - and - 8. f() = ; between and 9. f() = ; between - and f() = ; between and 58. f() = ( + ) ( + ) 59. f() = - ( - )( + ) 60. f() = - ( + )( - ) 6. f() = - ( - ) ( + 5 ) 6. f() = - ( - ) ( + ) 6. f() = ( - ) ( + 4)( - ) 64. f() = ( + )( + ) ( + 4 ) Practice Plus In Eercises 65 7, complete graphs of polnomial functions whose zeros are integers are shown. a. Find the zeros and state whether the multiplicit of each zero is even or odd. b. Write an equation, epressed as the product of factors, of a polnomial function that might have each graph. Use a leading coefficient of or -, and make the degree of f as small as possible. c. Use both the equation in part (b) and the graph to find the -intercept In Eercises 4 64, a.use the Leading Coefficient Test to determine the graph s end behavior. b.find the -intercepts. State whether the graph crosses or touches and turns around, at each intercept. c.find the -intercept. d.determine whether the graph smmetr, origin smmetr, or neither. e.if necessar, find a few additional points and graph the function. Use the maimum number of turning points to check whether it is drawn correctl. 4. f() = f() = f() = [ 5, 5, ] b [,, ] 68. [ 6, 6, ] b [ 40, 40, 0] 44. f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = [, 6, ] b [ 0, 0, ] [ 4, 4, ] b [ 40, 4, 4] [,, ] b [ 0, 0, ] [, 5, ] b [ 40, 4, 4] 54. f() = f() = -( - ) ( - 4 ) 56. f() = -( - 4) ( - 5 ) 57. f() = ( - ) ( + ) [,, ] b [ 5, 0, ] [,, ] b [ 5, 0, ]

43 Chapter Polnomial and Rational Functions Application Eercises Eperts fear that without conservation efforts, tigers could disappear from the wild b 0. Just one hundred ears ago, there were at least 00,000 wild tigers. B 00, the estimated world tiger population was 00. The bar graph shows the estimated world tiger population for selected ears from 970 through 00. Also shown is a polnomial function, with its graph, that models the data. Use this information to solve Eercises World Tiger Population 40,000 5,000 0,000 5,000 0,000 5,000 0, a. Find and interpret f(40). Identif this information as a point on the graph of f. b. Does f(40) overestimate or underestimate the actual data shown b the bar graph? B how much? c. Use the Leading Coefficient Test to determine the end behavior to the right for the graph of f. Will this function be useful in modeling the world tiger population if conservation efforts to save wild tigers fail? Eplain our answer. 74. a. Find and interpret f(0). Identif this information as a point on the graph of f. d. Does f(0) overestimate or underestimate the actual data shown b the bar graph? B how much? c. Use the Leading Coefficient Test to determine the end behavior to the right for the graph of f. Might this function be useful in modeling the world tiger population if conservation efforts to save wild tigers are successful? Eplain our answer. 75. During a diagnostic evaluation, a -ear-old woman eperienced a panic attack a few minutes after she had been asked to rela her whole bod. The graph shows the rapid increase in heart rate during the panic attack. Heart Rate (beats per minute) Estimated World Tiger Population 7, ,000 Source : World Wildlife Fund Heart Rate before and during a Panic Attack Baseline,500 Year 600 Panic Attack Relaation 00 Onset of Panic attack Time (minutes) Source: Davis and Palladino, Pscholog, Fifth Edition, Prentice Hall, 007. Vanishing Tigers World Tiger Population Graph of a Polnomial Model for the Data 40,000 5,000 0,000 5,000 0,000 5,000 0, f() = , Years after 970 a. For which time periods during the diagnostic evaluation was the woman s heart rate increasing? b. For which time periods during the diagnostic evaluation was the woman s heart rate decreasing? c. How man turning points (from increasing to decreasing or from decreasing to increasing) occurred for the woman s heart rate during the first minutes of the diagnostic evaluation? d. Suppose that a polnomial function is used to model the data displaed b the graph using (time during the evaluation, heart rate). Use the number of turning points to determine the degree of the polnomial function of best fit. e. For the model in part (d), should the leading coefficient of the polnomial function be positive or negative? Eplain our answer. f. Use the graph to estimate the woman s maimum heart rate during the first minutes of the diagnostic evaluation. After how man minutes did this occur? g. Use the graph to estimate the woman s minimum heart rate during the first minutes of the diagnostic evaluation. After how man minutes did this occur? 76. Volatilit at the Pump The graph shows the average price per gallon of gasoline in the United States in Januar for the period from 005 through 0. Average Price per Gallon $.5 $.00 $.75 $.50 $.5 $.00 $.75 $.50 Average Januar Gasoline Price in the United States Year Source : U.S. Energ Information Administration

44 Section. Polnomial Functions and Their Graphs a. For which ears was the average price per gallon in Januar increasing? b. For which ears was the average price per gallon in Januar decreasing? c. How man turning points (from increasing to decreasing or from decreasing to increasing) does the graph have for the period shown? d. Suppose that a polnomial function is used to model the data displaed b the graph using (number of ears after 005, average Januar price per gallon). Use the number of turning points to determine the degree of the polnomial function of best fit. e. For the model in part (d), should the leading coefficient of the polnomial function be positive or negative? Eplain our answer. f. Use the graph to estimate the maimum average Januar price per gallon. In which ear did this occur? g. Use the graph to estimate the minimum average Januar price per gallon. In which ear did this occur? Write a polnomial function that imitates the end behavior of each graph in Eercises The dashed portions of the graphs indicate that ou should focus onl on imitating the left and right behavior of the graph and can be fleible about what occurs between the left and right ends. Then use our graphing utilit to graph the polnomial function and verif that ou imitated the end behavior shown in the given graph Writing in Mathematics 77. What is a polnomial function? 78. What do we mean when we describe the graph of a polnomial function as smooth and continuous? 79. What is meant b the end behavior of a polnomial function? 80. Eplain how to use the Leading Coefficient Test to determine the end behavior of a polnomial function. 8. Wh is a third-degree polnomial function with a negative leading coefficient not appropriate for modeling nonnegative real-world phenomena over a long period of time? 8. What are the zeros of a polnomial function and how are the found? 8. Eplain the relationship between the multiplicit of a zero and whether or not the graph crosses or touches at that zero. 84. If f is a polnomial function, and f(a) and f(b) have opposite signs, what must occur between a and b? If f(a) and f(b) have the same sign, does it necessaril mean that this will not occur? Eplain our answer. 85. Eplain the relationship between the degree of a polnomial function and the number of turning points on its graph. 86. Can the graph of a polnomial function have Eplain. 87. Can the graph of a polnomial function have Eplain. 88. Describe a strateg for graphing a polnomial function. In our description, mention intercepts, the polnomial s degree, and turning points. Technolog Eercises 89. Use a graphing utilit to verif an five of the graphs that ou drew b hand in Eercises In Eercises 94 97, use a graphing utilit with a viewing rectangle large enough to show end behavior to graph each polnomial function. 94. f() = f() = f() = f() = In Eercises 98 99, use a graphing utilit to graph f and g in the same viewing rectangle. Then use the ZOOM OUT feature to show that f and g have identical end behavior. 98. f() = - 6 +, g() = 99. f() = , g() = - 4 Critical Thinking Eercises Make Sense? In Eercises 00 0, determine whether each statement makes sense or does not make sense, and eplain our reasoning. 00. When I m tring to determine end behavior, it s the coefficient of the leading term of a polnomial function that I should inspect. 0. I graphed f() = ( + ) ( - 4), and the graph touched and turned around at I m graphing a fourth-degree polnomial function with four turning points. 0. Although I have not et learned techniques for finding of f() = , I can easil determine In Eercises 04 07, determine whether each statement is true or false. If the statement is false, make the necessar change(s) to produce a true statement. 04. If f() = - + 4, then the graph of f falls to the left and falls to the right. 05. A mathematical model that is a polnomial of degree n whose leading term is a n n, n odd and a n 6 0, is ideall suited to describe phenomena that have positive values over unlimited periods of time.

45 4 Chapter Polnomial and Rational Functions 06. There is more than one third-degree polnomial function with the same 07. The graph of a function with origin smmetr can rise to the left and rise to the right. Use the descriptions in Eercises to write an equation of a polnomial function with the given characteristics. Use a graphing utilit to graph our function to see if ou are correct. If not, modif the function s equation and repeat this process. 08. Crosses is a t -4, 0, and ; lies above between -4 and 0; lies below between 0 and 09. Touches at 0 and crosses at ; lies below between 0 and Preview Eercises Eercises 0 will help ou prepare for the material covered in the net section. 0. Divide 77 b without using a calculator. Write the answer as quotient + remainder. divisor. Rewrite in descending powers of.. Use = + - to factor completel. SECTION.4 Dividing Polnomials; Remainder and Factor Theorems Objectives Use long division to divide polnomials. Use snthetic division to divide polnomials. Evaluate a polnomial using the Remainder Theorem. Use the Factor Theorem to solve a polnomial equation. Use long division to divide polnomials. A moth has moved into our closet. She appeared in our bedroom at night, but somehow her relativel stout bod escaped our clutches. Within a few weeks, swarms of moths in our tattered wardrobe suggest that Mama Moth was in the famil wa. There must be at least 00 critters nesting in ever crevice of our clothing. Two hundred plus moth-tkes from one female moth is this possible? Indeed it is. The number of eggs, f(), in a female moth is a function of her abdominal width,, in millimeters, modeled b f() = ,.5.5. Because there are 00 moths feasting on our favorite sweaters, Mama s abdominal width can be estimated b finding the solutions of the polnomial equation = 00. How can we solve such an equation? You might begin b subtracting 00 from both sides to obtain zero on one side. But then what? The factoring that we used in the previous section will not work in this situation. In Section.5, we will present techniques for solving certain kinds of polnomial equations. These techniques will further enhance our abilit to manipulate algebraicall the polnomial functions that model our world. Because these techniques are based on understanding polnomial division, in this section we look at two methods for dividing polnomials. (We ll return to Mama Moth s abdominal width in the Eercise Set.) Long Division of Polnomials and the Division Algorithm We begin b looking at division b a polnomial containing more than one term, such as Divisor has two terms and is a binomial. The polnomial dividend has three terms and is a trinomial.

46 Section.4 Dividing Polnomials; Remainder and Factor Theorems 5 When a divisor has more than one term, the four steps used to divide whole numbers divide, multipl, subtract, bring down the net term form the repetitive procedure for polnomial long division. EXAMPLE Long Division of Polnomials Divide b +. SOLUTION The following steps illustrate how polnomial division is ver similar to numerical division. 7( + ) = 7 + ( + ) = Change signs of the polnomial being subtracted T Remainder Arrange the terms of the dividend ( ) and the divisor ( + ) in descending powers of. Divide (the first term in the dividend) b (the first term in the divisor): =. Align like terms. Multipl each term in the divisor ( + ) b, aligning terms of the product under like terms in the dividend. Subtract + from + 0 b changing the sign of each term in the lower epression and adding. Bring down from the original dividend and add algebraicall to form a new dividend. Find the second term of the quotient. Divide the first term of 7 + b, the first term of the divisor: 7 = 7. Multipl the divisor ( + ) b 7, aligning under like terms in the new dividend. Then subtract to obtain the remainder of 0. The quotient is + 7. Because the remainder is 0, we can conclude that + is a factor of and = + 7.

47 6 Chapter Polnomial and Rational Functions Check Point Divide b + 9. Before considering additional eamples, let s summarize the general procedure for dividing one polnomial b another. Long Division of Polnomials. Arrange the terms of both the dividend and the divisor in descending powers of an variable.. Divide the first term in the dividend b the first term in the divisor. The result is the first term of the quotient.. Multipl ever term in the divisor b the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up. 4. Subtract the product from the dividend. 5. Bring down the net term in the original dividend and write it net to the remainder to form a new dividend. 6. Use this new epression as the dividend and repeat this process until the remainder can no longer be divided. This will occur when the degree of the remainder (the highest eponent on a variable in the remainder) is less than the degree of the divisor. In our net long division, we will obtain a nonzero remainder. EXAMPLE Long Division of Polnomials Divide b -. SOLUTION We begin b writing the dividend in descending powers of = ( ) = Change signs of the polnomial -5 being subtracted. 6 Divide: =. Multipl: (-)=6-4. Subtract 6-4 from 6 - and bring down 5. Now we divide b to obtain, multipl and the divisor, and subtract. ( ) = Change signs of the polnomial being subtracted Divide: =. Multipl: (-)= -. Subtract - from -5 and bring down 4.

48 Section.4 Dividing Polnomials; Remainder and Factor Theorems 7 Now we divide - b to obtain -, multipl - and the divisor, and subtract. ( ) = Change signs of the polnomial being subtracted Remainder Divide: =. Multipl: (-)= +. Subtract + from +4, leaving a remainder of. The quotient is + - and the remainder is. When there is a nonzero remainder, as in this eample, list the quotient, plus the remainder above the divisor. Thus, = Remainder above divisor The Division Algorithm Quotient An important propert of division can be illustrated b clearing fractions in the equation that concluded Eample. Multipling both sides of this equation b - results in the following equation: =(-)( +-)+. Dividend Divisor Quotient Remainder Polnomial long division is checked b multipling the divisor with the quotient and then adding the remainder. This should give the dividend. The process illustrates the Division Algorithm. If f() and d() are polnomials, with d() 0, and the degree of d() is less than or equal to the degree of f(), then there eist unique polnomials q() and r() such that f() = d() q() + r(). Dividend Divisor Quotient Remainder The remainder, r(), equals 0 or it is of degree less than the degree of d(). If r() = 0, we sa that d() divides evenl into f() and that d() and q() are factors of f(). Check Point Divide b -. Epress the result in the form quotient, plus remainder divided b divisor. If a power of is missing in either a dividend or a divisor, add that power of with a coefficient of 0 and then divide. In this wa, like terms will be aligned as ou carr out the long division. EXAMPLE Long Division of Polnomials Divide b -.

49 8 Chapter Polnomial and Rational Functions SOLUTION We write the dividend, , a s to keep all like terms aligned. Multipl ( ) = ( ) = ( ) = Remainder The division process is finished because the degree of 7-5, which is, is less than the degree of the divisor -, which is. The answer is = Check Point Divide b -. Use snthetic division to divide polnomials. Dividing Polnomials Using Snthetic Division We can use snthetic division to divide polnomials if the divisor is of the form - c. This method provides a quotient more quickl than long division. Let s compare the two methods showing divided b -. Long Division Snthetic Division Divisor c; c = Quotient Dividend Remainder Notice the relationship between the polnomials in the long division process and the numbers that appear in snthetic division. The divisor is. This is, or c, in c. These are the coefficients of the dividend These are the coefficients of the quotient This is the remainder.

50 Section.4 Dividing Polnomials; Remainder and Factor Theorems 9 Now let s look at the steps involved in snthetic division. Snthetic Division To divide a polnomial b - c:. Arrange the polnomial in descending powers, with a 0 coefficient for an missing term.. Write c for the divisor, - c. To the right, write the coefficients of the dividend.. Write the leading coefficient of the dividend on the bottom row. 4. Multipl c (in this case, ) times the value just written on the bottom row. Write the product in the net column in the second row. 5. Add the values in this new column, writing the sum in the bottom row. Eample T Bring down Multipl b : = Add Repeat this series of multiplications and additions until all columns are filled in Add. Multipl b : 7 = Add. Multipl b : 6 = Use the numbers in the last row to write the quotient, plus the remainder above the divisor. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in this row is the remainder. Written from the last row of the snthetic division EXAMPLE 4 Using Snthetic Division Use snthetic division to divide b +. SOLUTION The divisor must be in the form - c. Thus, we write + as - (-). This means that c = -. Writing a 0 coefficient for the in the dividend, we can epress the division as follows: - (-)

51 40 Chapter Polnomial and Rational Functions Now we are read to set up the problem so that we can use snthetic division. Use the coefficients of the dividend in descending powers of. - (-) The given division problem (repeated) This is c in ( ). 5 We begin the snthetic division process b bringing down 5. This is followed b a series of multiplications and additions Bring down T 5 4. Multipl: ( 0) Multipl 0 b.. Multipl: (5) Multipl 5 b.. Add: 0 ( 0) Add Add: Add. 6. Multipl: (6) Multipl 6 b. 7. Add: 8 ( 5) Add The numbers in the last row represent the coefficients of the quotient and the remainder. The degree of the first term of the quotient is one less than that of the dividend. Because the degree of the dividend, , is, the degree of the quotient is. This means that the 5 in the last row represents The quotient is The remainder is 44. Thus, Check Point 4 Use snthetic division to divide b +. Evaluate a polnomial using the Remainder Theorem. The Remainder Theorem Let s consider the Division Algorithm when the dividend, f(), is divided b - c. In this case, the remainder must be a constant because its degree is less than one, the degree of - c. f() = ( -c)q() + r Dividend Divisor Quotient The remainder, r, is a constant when dividing b c.

52 Section.4 Dividing Polnomials; Remainder and Factor Theorems 4 Now let s evaluate f at c. f(c) = (c - c)q(c) + r Find f (c) b letting = c in f () = ( - c)q() + r. This will give an epression for r. f(c) = 0 # q(c) + r c - c = 0 f(c) = r 0 # q(c) = 0 and 0 + r = r. What does this last equation mean? If a polnomial is divided b - c, the remainder is the value of the polnomial at c. This result is called the Remainder Theorem. The Remainder Theorem If the polnomial f() is divided b - c, then the remainder is f(c). Eample 5 shows how we can use the Remainder Theorem to evaluate a polnomial function at. Rather than substituting for, we divide the function b -. The remainder is f(). EXAMPLE 5 Using the Remainder Theorem to Evaluate a Polnomial Function Given f() = , use the Remainder Theorem to find f(). SOLUTION B the Remainder Theorem, if f() is divided b -, then the remainder is f(). We ll use snthetic division to divide Remainder The remainder, 5, is the value of f( ). Thus, f() = 5. We can verif that this is correct b evaluating f() directl. Using f() = , we obtain f() = - 4 # + 5 # + = = 5. Check Point 5 Given f() = , use the Remainder Theorem to find f(-4). Use the Factor Theorem to solve a polnomial equation. The Factor Theorem Let s look again at the Division Algorithm when the divisor is of the form - c. f() = ( -c)q() + r Dividend Divisor Quotient Constant remainder B the Remainder Theorem, the remainder r is f(c), so we can substitute f(c) for r: f() = ( - c)q() + f(c). Notice that if f(c) = 0, then f() = ( - c)q() so that - c is a factor of f(). This means that for the polnomial function f(), if f(c) = 0, then - c is a factor of f().

53 4 Chapter Polnomial and Rational Functions Let s reverse directions and see what happens if - c is a factor of f(). This means that f() = ( - c)q(). If we replace in f() = ( - c)q() with c, we obtain f(c) = (c - c)q(c) = 0 # q(c) = 0. Thus, if - c is a factor of f(), then f(c) = 0. We have proved a result known as the Factor Theorem. The Factor Theorem Let f() be a polnomial. a. If f(c) = 0, then - c is a factor of f(). b. If - c is a factor of f(), then f(c) = 0. The eample that follows shows how the Factor Theorem can be used to solve a polnomial equation. EXAMPLE 6 Using the Factor Theorem Solve the equation = 0 given that is a zero of f() = SOLUTION We are given that is a zero of f() = This means that f() = 0. Because f() = 0, the Factor Theorem tells us that - is a factor of f(). We ll use snthetic division to divide f() b -. TECHNOLOGY Graphic Connections Because the solution set of = 0 is 5-,, 6, this implies that the polnomial function f() = (or zeros) at -,, and. This is verified b the graph of f. -intercept: -intercept: -intercept: [ 0, 0, ] b [ 5, 5, ] The remainder, 0, verifies that is a factor of + 6. Now we can solve the polnomial equation = Equivalentl, = ( - )( + - ). This is the given equation. ( - )( + - ) = 0 Factor using the result from the snthetic division. ( - )( - )( + ) = 0 Factor the trinomial. - = 0 or - = 0 or + = 0 Set each factor equal to 0. = = = - Solve for. The solution set is 5-,, 6.

54 CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. Section.4 Dividing Polnomials; Remainder and Factor Theorems 4 Based on the Factor Theorem, the following statements are useful in solving polnomial equations:. If f() is divided b - c and the remainder is zero, then c is a zero of f and c is a root of the polnomial equation f() = 0.. If f() is divided b - c and the remainder is zero, then - c is a factor of f(). Check Point 6 Solve the equation = 0 given that - is a zero of f() = Consider the following long division problem: We begin the division process b rewriting the dividend as.. Consider the following long division problem: We begin the division process b dividing b. We obtain. We write this result above in the dividend.. In the following long division problem, the first step has been completed: The net step is to multipl and. We obtain. We write this result below. 4. In the following long division problem, the first two steps have been completed: The net step is to subtract from. We obtain. Then we bring down and form the new dividend. EXERCISE SET.4 Practice Eercises In Eercises 6, divide using long division. State the quotient, q(), and the remainder, r().. ( ), ( + 5 ). ( + - 0), ( - ). ( ), ( + ) 5. In the following long division problem, most of the steps have been completed: ? Completing the step designated b the question mark, we obtain. Thus, the quotient is and the remainder is. The answer to this long division problem is. 6. After performing polnomial long division, the answer ma be checked b multipling the b the, and then adding the. You should obtain the. 7. To divide b - 4 using snthetic division, the first step is to write. 8. To divide b + 5 using snthetic division, the first step is to write. 9. True or false: means = The Remainder Theorem states that if the polnomial f() is divided b - c, then the remainder is.. The Factor Theorem states that if f is a polnomial function and f(c) = 0, then is a factor of f(). 4. ( ), ( - ) 5. ( ), ( - ) 6. ( ), ( + 4 ) 7. ( + - 4), ( - ) 8. ( ), ( - )

55 44 Chapter Polnomial and Rational Functions In Eercises 7, divide using snthetic division. 7. ( + - 0), ( - ) 8. ( + - ), ( - ) 9. ( + 7-0), ( + 5 ) 45. Solve the equation = 0 given that - is a root. 46. Solve the equation = 0 given that - is a root. Practice Plus In Eercises 47 50, use the graph or the table to determine a solution of each equation. Use snthetic division to verif that this number is a solution of the equation. Then solve the polnomial equation = ( ), ( + ). ( ), ( - ). ( ), ( - ). ( ), ( - ) 4. ( ), ( - ) 5. ( ), (5 + ) 6. ( ), (6 + ) 0 4 = [0, 4, ] b [ 5, 5, 5] = In Eercises 40, use snthetic division and the Remainder Theorem to find the indicated function value.. f() = ; f( 4 ) 4. f() = ; f( ) 5. f() = ; f(- ) 6. f() = ; f(- ) 7. f() = ; f( ) 8. f() = ; f( ) 9. f() = ; f a- b 40. f() = ; f a- b 4. Use snthetic division to divide f() = b +. Use the result to find all zeros of f. 4. Use snthetic division to divide f() = b +. 4 = [ 4, 0, ] b [ 5, 5, 5] = 0 = = 0 = Use the result to find all zeros of f. 4. Solve the equation = 0 given that is a zero of f() = Solve the equation = 0 given that - is a zero of f() = Application Eercises 5. a. Use snthetic division to show that is a solution of the polnomial equation = 0.

56 Section.4 Dividing Polnomials; Remainder and Factor Theorems 45 b. Use the solution from part (a) to solve this problem. The number of eggs, f(), in a female moth is a function of her abdominal width,, in millimeters, modeled b f() = What is the abdominal width when there are eggs? 5. a. Use snthetic division to show that is a solution of the polnomial equation h + 4h - 7 = 0. b. Use the solution from part (a) to solve this problem. The width of a rectangular bo is twice the height and the length is 7 inches more than the height. If the volume is 7 cubic inches, find the dimensions of the bo. h h h + 7 In Eercises 5 54, write a polnomial that represents the length of each rectangle. 5. The area is square units. The width is + 0. units. 54. The width is + 4 units. The area is square units. During the 980s, the controversial economist Arthur Laffer promoted the idea that ta increases lead to a reduction in government revenue. Called suppl-side economics, the theor uses functions such as f() =, This function models the government ta revenue, f(), in tens of billions of dollars, in terms of the ta rate,. The graph of the function is shown. It illustrates ta revenue decreasing quite dramaticall as the ta rate increases. At a ta rate of (gasp) 00%, the government takes all our mone and no one has an incentive to work. With no income earned, zero dollars in ta revenue is generated. Government Ta Revenue (tens of billions of dollars) f() = At a 00% ta rate, $0 in ta revenue is generated Ta Rate Use function f and its graph to solve Eercises a.find and interpret f(0). Identif the solution as a point on the graph of the function. b. Rewrite the function b using long division to perform ( ), ( - 0). Then use this new form of the function to find f(0). Do ou obtain the same answer as ou did in part (a)? c. Is f a polnomial function? Eplain our answer. 56. a. Find and interpret f(40). Identif the solution as a point on the graph of the function. b. Rewrite the function b using long division to perform ( ), ( - 0). Then use this new form of the function to find f(40). Do ou obtain the same answer as ou did in part (a)? c. Is f a polnomial function? Eplain our answer. Writing in Mathematics 57. Eplain how to perform long division of polnomials. Use divided b - in our eplanation. 58. In our own words, state the Division Algorithm. 59. How can the Division Algorithm be used to check the quotient and remainder in a long division problem? 60. Eplain how to perform snthetic division. Use the division problem in Eercise 57 to support our eplanation. 6. State the Remainder Theorem. 6. Eplain how the Remainder Theorem can be used to find f(-6 ) if f() = What advantage is there to using the Remainder Theorem in this situation rather than evaluating f(-6) directl? 6. How can the Factor Theorem be used to determine if - is a factor of - - +? 64. If ou know that - is a zero of f() = , eplain how to solve the equation = 0. Technolog Eercise 65. For each equation that ou solved in Eercises 4 46, use a graphing utilit to graph the polnomial function defined b the left side of the equation. Use end behavior to obtain a complete graph. Then use the graph to verif our solutions. Critical Thinking Eercises Make Sense? In Eercises 66 69, determine whether each statement makes sense or does not make sense, and eplain our reasoning. 66. When performing the division ( 5 + ), ( + ), there s no need for me to follow all the steps involved in polnomial long division because I can work the problem in m head and see that the quotient must be Ever time I divide polnomials using snthetic division, I am using a highl condensed form of the long division procedure where omitting the variables and eponents does not involve the loss of an essential data.

57 46 Chapter Polnomial and Rational Functions 68. The onl nongraphic method that I have for evaluating a function at a given value is to substitute that value into the function s equation. 69. I found the zeros of function f, but I still need to find the solutions of the equation f() = 0. In Eercises 70 7, determine whether each statement is true or false. If the statement is false, make the necessar change(s) to produce a true statement. 70. If a trinomial in of degree 6 is divided b a trinomial in of degree, the degree of the quotient is. 7. Snthetic division can be used to find the quotient of and An problem that can be done b snthetic division can also be done b the method for long division of polnomials. 7. If a polnomial long-division problem results in a remainder that is a whole number, then the divisor is a factor of the dividend. 74. Find k so that 4 + is a factor of k. 75. When is divided b a polnomial, the quotient is - and the remainder is. Find the polnomial. 76. Find the quotient of n + and n Snthetic division is a process for dividing a polnomial b - c. The coefficient of in the divisor is. How might snthetic division be used if ou are dividing b - 4? 78. Use snthetic division to show that 5 is a solution of = 0. Then solve the polnomial equation. Preview Eercises Eercises 79 8 will help ou prepare for the material covered in the net section. 79. Solve: = Solve: = Let f() = a n ( ). If f() = -50, determine the value of a n. SECTION.5 Zeros of Polnomial Functions Objectives Use the Rational Zero Theorem to fi nd possible rational zeros. Find zeros of a polnomial function. Solve polnomial equations. Use the Linear Factorization Theorem to fi nd polnomials with given zeros. Use Descartes s Rule of Signs. You stole m formula! Tartaglia s Secret Formula for One Solution of m n = C B a n b + a m b + n - C B a n b + a m b - n P opularizers of mathematics are sharing bizarre stories that are giving math a secure place in popular culture. One episode, able to compete with the wildest fare served up b television talk shows and the tabloids, involves three Italian mathematicians and, of all things, zeros of polnomial functions. Tartaglia ( ), poor and starving, has found a formula that gives a root for a third-degree polnomial equation. Cardano (50 576) begs Tartaglia to reveal the secret formula, wheedling it from him with the promise he will find the impoverished Tartaglia a patron. Then Cardano publishes his famous work Ars Magna, in which he presents Tartaglia s formula as his own. Cardano uses his most talented student, Ferrari (5 565), who derived a formula for a root of a fourth-degree polnomial equation, to falsel accuse Tartaglia of plagiarism. The dispute becomes violent and Tartaglia is fortunate to escape alive.

58 GREAT QUESTION! Section.5 Zeros of Polnomial Functions 47 The noise from this You Stole M Formula episode is quieted b the work of French mathematician Evariste Galois (8 8). Galois proved that there is no general formula for finding roots of polnomial equations of degree 5 or higher. There are, however, methods for finding roots. In this section, we stud methods for finding zeros of polnomial functions. We begin with a theorem that plas an important role in this process. When finding zeros of polnomial functions, what kinds of numbers can I get? Here s a quick eample that involves all possible kinds of zeros: f()=(+)(-)(+)(-)(-4+5i)(-4-5i). Zeros:,,,, 4-5i, 4+5i Rational zeros Irrational zeros Comple imaginar zeros Real zeros Nonreal zeros Use the Rational Zero Theorem to fi nd possible rational zeros. The Rational Zero Theorem The Rational Zero Theorem provides us with a tool that we can use to make a list of all possible rational zeros of a polnomial function. Equivalentl, the theorem gives all possible rational roots of a polnomial equation. Not ever number in the list will be a zero of the function, but ever rational zero of the polnomial function will appear somewhere in the list. The Rational Zero Theorem If f() = a n n + a n - n - + g + a + a 0 has integer coefficients and p p q (where is reduced to lowest terms) is a rational zero of f, then p is a factor of q the constant term, a 0, and q is a factor of the leading coefficient, a n. You can eplore the wh behind the Rational Zero Theorem in Eercise 9 of Eercise Set.5. For now, let s see if we can figure out what the theorem tells us about possible rational zeros. To use the theorem, list all the integers that are factors of the constant term, a 0. Then list all the integers that are factors of the leading coefficient, a n. Finall, list all possible rational zeros: Possible rational zeros = Factors of the constant term Factors of the leading coefficient. EXAMPLE Using the Rational Zero Theorem List all possible rational zeros of f() = SOLUTION The constant term is 4. We list all of its factors: {, {, {4. The leading coefficient is -. Its factors are {. Factors of the constant term, 4: {, {, {4 Factors of the leading coefficient, -: {

59 48 Chapter Polnomial and Rational Functions Because Possible rational zeros = Factors of the constant term Factors of the leading coefficient, we must take each number in the first row, {, {, {4, and divide b each number in the second row, {. Factors of 4,, 4 Possible rational zeros= = =,, 4 Factors of Divide ± b ±. Divide ± b ±. Divide ±4 b ±. GREAT QUESTION! What s the relationship among zeros, roots, The zeros of a function f are the roots, or solutions, of the equation f() = 0. Furthermore, the real zeros, or real roots, are of the graph of f. There are si possible rational zeros. The graph of f() = is shown in Figure.8. are - and. Thus, - and are the actual rational zeros. Check Point List all possible rational zeros of f() = is a rational zero. is a rational zero. FIGURE.8 The graph of f() = shows that - and are rational zeros. EXAMPLE Using the Rational Zero Theorem List all possible rational zeros of f() = SOLUTION The constant term is - and the leading coefficient is 5. Possible rational zeros = Factors of the constant term, - Factors of the leading coefficient, 5 = {, { {, {, {5, {5 =,,,,, 5, 5, 5 5 Divide ± and ± b ±. Divide ± and ± b ±. Divide ± and ± b ±5. Divide ± and ± b ±5. There are 6 possible rational zeros. The actual solution set of = 0 is 5-, -, 5 6, which contains three of the 6 possible zeros. Check Point List all possible rational zeros of f() = Find zeros of a polnomial function. How do we determine which (if an) of the possible rational zeros are rational zeros of the polnomial function? To find the first rational zero, we can use a trial-and-error process involving snthetic division: If f() is divided b - c and the remainder is zero, then c is a zero of f. After we identif the first rational zero, we use the result of the snthetic division to factor the original polnomial. Then we set each factor equal to zero to identif an additional rational zeros. EXAMPLE Finding Zeros of a Polnomial Function Find all zeros of f() =

60 SOLUTION We begin b listing all possible rational zeros. Possible rational zeros Section.5 Zeros of Polnomial Functions 49 Factors of the constant term, 6,,, 6 = = =,,, 6 Factors of the leading coefficient, Divide the eight numbers in the numerator b ±. Now we will use snthetic division to see if we can find a rational zero among the possible rational zeros {, {, {, {6. Keep in mind that if f() is divided b - c and the remainder is zero, then c is a zero of f. Let s start b testing. If is not a rational zero, then we will test other possible rational zeros. Possible rational zero Test. Coefficients of f() = The nonzero remainder shows that is not a zero. Possible rational zero Test. Coefficients of f() = The zero remainder shows that is a zero. The zero remainder tells us that is a zero of the polnomial function f() = Equivalentl, is a solution, or root, of the polnomial equation = 0. Thus, - is a factor of the polnomial. The first three numbers in the bottom row of the snthetic division on the right,, 4, and, give the coefficients of the other factor. This factor is = 0 Finding the zeros of f () = is the same as finding the roots of this equation. ( - )( ) = 0 Factor using the result from the snthetic division. ( - )( + )( + ) = 0 Factor completel. - = 0 or + = 0 or + = 0 Set each factor equal to zero. = = - = - Solve for. The solution set is {-, -, }. The zeros of f are -, -, and. Check Point Find all zeros of f() = Our work in Eample involved finding zeros of a third-degree polnomial function. The Rational Zero Theorem is a tool that allows us to rewrite such functions as products of two factors, one linear and one quadratic. Zeros of the quadratic factor are found b factoring, the quadratic formula, or the square root propert. EXAMPLE 4 Finding Zeros of a Polnomial Function Find all zeros of f() = SOLUTION We begin b listing all possible rational zeros. Possible rational zeros = Factors of the constant term, - Factors of the leading coefficient, = {, { { = {, {

61 50 Chapter Polnomial and Rational Functions Now we will use snthetic division to see if we can find a rational zero among the four possible rational zeros, {, {, of f() = Test. Test. Test. Test The zero remainder when testing - tells us that - is a zero of the polnomial function f() = To find all zeros of f, we proceed as follows: = 0 Finding the zeros of f is the same thing as finding the roots of f () = 0. ( + )( ) = 0 This result is from the last snthetic division, shown above. The first three numbers in the bottom row,, 4, and -, give the coefficients of the second factor. + = 0 or = 0 Set each factor equal to 0. = -. Solve the linear equation. We can use the quadratic formula to solve = 0. = -b { b - 4ac a = -4 { 4-4()(-) () = = -4 { 0-4 { 5 We use the quadratic formula because cannot be factored. Let a =, b = 4, and c = -. Multipl and subtract under the radical: 4-4()(-) = 6 - (-4) = = 0. 0 = 4 # 5 = 5 = - { 5 Divide the numerator and the denominator b. The solution set is 5-, - - 5, The zeros of f() = are -, - - 5, and Among these three real zeros, one zero is rational and two are irrational. Check Point 4 Find all zeros of f() = If the degree of a polnomial function or equation is 4 or higher, it is often necessar to find more than one linear factor b snthetic division. One wa to speed up the process of finding the first zero is to graph the function. is a zero. Solve polnomial equations. EXAMPLE 5 Solving a Polnomial Equation Solve: = 0. SOLUTION Recall that we refer to the zeros of a polnomial function and the roots of a polnomial equation. Because we are given an equation, we will use the word roots, rather than zeros, in the solution process. We begin b listing all possible rational roots.

62 Possible rational roots = Section.5 Zeros of Polnomial Functions 5 Factors of the constant term, 4 Factors of the leading coefficient, {, {, {, {4, {6, {8, {, {4 = { = {, {, {, {4, {6, {8, {, {4 Part of the graph of f() = is shown in Figure.9. Because is, we will test b snthetic division and show that it is a root of the given equation. Without the graph, the procedure would be to start the trial-and-error snthetic division with and proceed until a zero remainder is found, as we did in Eample 4. -intercept: FIGURE.9 The graph of f() = in a [-, 5, ] b [-, 0, ] viewing rectangle Careful! = The zero remainder indicates that is a root of = 0. Now we can rewrite the given equation in factored form = 0 This is the given equation. ( - )( ) = 0 This is the result obtained from the snthetic division. The first four numbers in the bottom row,,, -, and -, give the coefficients of the second factor. - = 0 or = 0 Set each factor equal to 0. We can use the same approach to look for rational roots of the polnomial equation = 0, listing all possible rational roots. Without the graph infigure.9, the procedure would be to start testing possible rational roots b trial-and-error snthetic division. However, take a second look at the graph in Figure.9. Because the graph turns around at, this means that is a root of even multiplicit. Thus, must also be a root of = 0, confirmed b the following snthetic division These are the coefficients of + = 0. The zero remainder indicates that is a root of + = 0. Now we can solve the original equation as follows: = 0 This is the given equation. ( - )( ) = 0 This factorization was obtained from the first snthetic division. ( - )( - )( ) = 0 This factorization was obtained from the second snthetic division. The first three numbers in the bottom row,, 4, and 6, give the coefficients of the third factor. - = 0 or - = 0 or = 0 Set each factor equal to 0. = =. Solve the linear equations.

63 5 Chapter Polnomial and Rational Functions -intercept: FIGURE.9 (repeated) The graph of f() = in a [-, 5, ] b [-, 0, ] viewing rectangle We can use the quadratic formula to solve = 0. = -b { b - 4ac a = -4 { 4-4()(6) () = = -4 { -8-4 { i = - { i Simplif. We use the quadratic formula because cannot be factored. Let a =, b = 4, and c = 6. Multipl and subtract under the radical: 4-4()(6) = 6-4 = = 4()(-) = i The solution set of the original equation, = 0, is 5, - - i, - + i6. Figure.9, repeated in the margin, shows that a graphing utilit does not reveal the two imaginar roots. In Eample 5, is a repeated root of the equation with multiplicit. Counting this multiple root separatel, the fourth-degree equation = 0 has four roots:,, - + i, and - - i. The equation and its roots illustrate two general properties: Properties of Roots of Polnomial Equations. If a polnomial equation is of degree n, then counting multiple roots separatel, the equation has n roots.. If a + bi is a root of a polnomial equation with real coefficients (b 0), then the imaginar number a - bi is also a root. Imaginar roots, if the eist, occur in conjugate pairs. Check Point 5 Solve: = 0. The Fundamental Theorem of Algebra The fact that a polnomial equation of degree n has n roots is a consequence of a theorem proved in 799 b a -ear-old student named Carl Friedrich Gauss in his doctoral dissertation. His result is called the Fundamental Theorem of Algebra. GREAT QUESTION! Do all polnomial equations have at least one imaginar root? No. As ou read the Fundamental Theorem of Algebra, don t confuse comple root with imaginar root and conclude that ever polnomial equation has at least one imaginar root. Recall that comple numbers, a + bi, include both real numbers ( b = 0) and imaginar numbers ( b 0 ). The Fundamental Theorem of Algebra If f() is a polnomial of degree n, where n Ú, then the equation f() = 0 has at least one comple root. Suppose, for eample, that f() = 0 represents a polnomial equation of degree n. B the Fundamental Theorem of Algebra, we know that this equation has at least one comple root; we ll call it c. B the Factor Theorem, we know that - c is a factor of f(). Therefore, we obtain ( - c )q () = 0 The degree of the polnomial q () is n -. - c = 0 or q () = 0. Set each factor equal to 0. If the degree of q () is at least, b the Fundamental Theorem of Algebra, the equation q () = 0 has at least one comple root. We ll call it c. The Factor Theorem gives us q () = 0 The degree of q () is n -. ( - c )q () = 0 The degree of q () is n -. - c = 0 or q () = 0. Set each factor equal to 0.

64 Section.5 Zeros of Polnomial Functions 5 Let s see what we have up to this point and then continue the process. f() = 0 This is the original polnomial equation of degree n. ( - c )q () = 0 This is the result from our first application of the Fundamental Theorem. ( - c )( - c )q () = 0 This is the result from our second application of the Fundamental Theorem. B continuing this process, we will obtain the product of n linear factors. Setting each of these linear factors equal to zero results in n comple roots. Thus, if f() is a polnomial of degree n, where n Ú, then f() = 0 has eactl n roots, where roots are counted according to their multiplicit. Use the Linear Factorization Theorem to fi nd polnomials with given zeros. The Linear Factorization Theorem In Eample 5, we found that = 0 has 5, - { i6 as a solution set, where is a repeated root with multiplicit. The polnomial can be factored over the comple nonreal numbers as follows: f()= These are the four zeros. =[-( +i)][-( -i)](-)(-). These are the linear factors. This fourth-degree polnomial has four linear factors. Just as an nth@degree polnomial equation has n roots, an nth@degree polnomial has n linear factors. This is formall stated as the Linear Factorization Theorem. The Linear Factorization Theorem If f() = a n n + a n - n - + g + a + a 0, where n Ú and a n 0, then f() = a n ( - c )( - c ) g ( - c n ), where c, c, c, c n are comple numbers (possibl real and not necessaril distinct). In words: An nth@degree polnomial can be epressed as the product of a nonzero constant and n linear factors, where each linear factor has a leading coefficient of. Man of our problems involving polnomial functions and polnomial equations dealt with the process of finding zeros and roots. The Linear Factorization Theorem enables us to reverse this process, finding a polnomial function when the zeros are given. EXAMPLE 6 Finding a Polnomial Function with Given Zeros Find a fourth-degree polnomial function f() with real coefficients that has -,, and i as zeros and such that f() = -50. SOLUTION Because i is a zero and the polnomial has real coefficients, the conjugate, -i, must also be a zero. We can now use the Linear Factorization Theorem. f() = a n ( - c )( - c )( - c )( - c 4 ) This is the linear factorization for a fourth-degree polnomial. = a n ( + )( - )( - i)( + i) Use the given zeros: c = -, c =,c = i, and, from above, c 4 = -i. = a n ( - 4)( + ) Multipl: ( - i)( + i) = - i = - (-) = +.

65 54 Chapter Polnomial and Rational Functions TECHNOLOGY Graphic Connections The graph of f() = , shown in a [-,, ] b [-00, 0, 0] viewing rectangle, verifies that - and are real zeros. B tracing along the curve, we can check that f() = -50. is a zero. is a zero. f() = a n ( ) Complete the multiplication for f() = a n ( - 4)( + ). f() = a n ( 4 - # - 4) = -50 To find a n, use the fact that f() = -50. a n ( ) = -50 Solve for a n. 50a n = -50 Simplif: = 50. a n = - Divide both sides b 50. Substituting - for a n in the formula for f(), we obtain f() = -( ). Equivalentl, f() = Check Point 6 Find a third-degree polnomial function f() with real coefficients that has - and i as zeros and such that f() = 8. Use Descartes s Rule of Signs. Descartes s Rule of Signs Because an nth@degree polnomial equation might have roots that are imaginar numbers, we should note that such an equation can have at most n real roots. Descartes s Rule of Signs provides even more specific information about the number of real zeros that a polnomial can have. The rule is based on considering variations in sign between consecutive coefficients. For eample, the function f() = has three sign changes: f()= sign change sign change sign change An equation can have as man true [positive] roots as it contains changes of sign, from plus to minus or from minus to plus. René Descartes ( ) in La Géométrie (67) GREAT QUESTION! Does Descartes s Rule of Signs include both rational and irrational zeros? Yes. The number of real zeros given b Descartes s Rule of Signs includes rational zeros from a list of possible rational zeros, as well as irrational zeros not on the list. It does not include an imaginar zeros. Descartes s Rule of Signs Let f() = a n n + a n - n - + g + a + a + a 0 be a polnomial with real coefficients.. The number of positive real zeros of f is either a. the same as the number of sign changes of f() or b. less than the number of sign changes of f() b a positive even integer. If f() has onl one variation in sign, then f has eactl one positive real zero.. The number of negative real zeros of f is either a. the same as the number of sign changes of f(-) or b. less than the number of sign changes of f(-) b a positive even integer. If f(-) has onl one variation in sign, then f has eactl one negative real zero.

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