Vertices: (.8, 5.), (.37, 3.563), (.6, 0.980), (5.373, 6.66), (.8, 7.88), (.95,.) Graph the equation for an value of P (the second graph shows the circle with P 5) and imagine increasing the value of P, enlarging the circle, until the last point within the feasible region is hit. This will show that the verte on the far right provides the maimum value. The maimum value of P occurs at (.37, 3.563), where P 7 7 3.879.. The matri a b c d 0 _ d ab bc 0 _ c ad bc 0 _ b ad _ a ad bc which means the inverse is 0 reduces to _ d b ad bc c a, where ad bc is nonzero. Because the determinant is defined as det ad bc, this is the same result as in Take Another Look Activit. CHAPTER 7 REFRESHING YOUR SKILLS POLYNOMIAL EXPRESSIONS. a. 8 6 b. 3 0.8.5 c. 3a a 6a a 3 3a a 6a a 3 3a 9a 7 d. 3 5 6 3 5 6 8 8. a. 6 9 3 3 3 3 9 b. 9 8 7 7 6 8 c. 6 6 d. 3 7 5 3 3 3 6 3. a. 5 5 5 5 5 b. 8 3 3 c. 6 6 6 6 36. a. ( 5)( 5) 0 5 b. ( )( 3) 5 c. ( 6)( 6) 36; 36 might be described as the difference of two squares because and 36 6 6. LESSON 7. SUPPORT EXAMPLES. th degree. Use the finite differences method to see that the fourth set of differences is constant. D 0 0 3 60 3 73 68 360 5 60 660 6 6.. D D 0 3 5 7 7 3 9 3 5 3 3 6 7 D D 3 D 8 08 9 300 36 60 8 08 6 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
The D values are constant, so it is a nd-degree polnomial. Start with a b c. Use three data points to get three equations and solve for the constants a, b, and c. a b c Using the data point (, ). a b c 7 Using the data point (, 7). 9a 3b c Using the data point (3, ). Solve this sstem to get a, b, and c. The equation is. EXERCISES. a. 3. The term with greatest eponent is 3. b.. The term with greatest eponent is 7. c. 7. The term with greatest eponent is 7. d. 5. The term with greatest eponent is 9 5.. a. Polnomial; 3rd degree; 5_ 9 3 3.5 3 b. Not a polnomial because p p has a negative eponent c. Not a polnomial because 3 3 has a non-integer eponent d. Polnomial; nd degree; alread in general form 3. a. No; {.,.6,.8, 0., 3.} b. No; {0.007, 0.006, 0.008, 0.00} c. No; {50, 50, 50}. 3. Use the finite differences method to see that the third set of differences is constant. 0 6 6 6 8 9 0 98 5 D D D 3 6 60 8 880 56 76 88 3 576 70 5. a. D {, 3,, 5, 6}, D {,,, }; nd degree n 3 5 6 s 3 6 0 5 D 3 5 6 6. a. d. s 0.5n 0.5n; s 78 when n. Because the function is quadratic, start with s an bn c. Use three data points to get three equations and solve for the constants a, b, and c. a b c a b c 3 9a 3b c 6 Using the data point (, ). Using the data point (, 3). Using the data point (3, 6). Solve this sstem to get a 0.5, b 0.5, and c 0. The equation is s 0.5n 0.5n. Now substitute for n, and solve for s: s 0.5() 0.5() 7 6 78. e. The pennies can be arranged to form triangles, where each row of a triangle has one more penn than the previous row. Using the polnomial s 0.5n 0.5n found in 5d, if ou form a triangle with n rows, it will contain s pennies. Laers 3 5 6 Blocks 5 30 55 9 b. _ 3 3 6. Use the finite differences method to find that the minimum degree of the polnomial is 3. 3 5 6 5 30 55 9 D 9 6 5 36 D 5 7 9 D 3 The data can be modeled b a 3rd-degree polnomial, so start with a 3 b c d. Use four data points to get four equations and solve for the constants a, b, c, and d. a b c d 8a b c d 5 7a 9b 3c d 6a 6b c d 30 Using the data point (, ). Using the data point (, 5). Using the data point (3, ). Using the data point (, 30). Solve this sstem to get a _ 3, b _, c _ 6, and d 0. The fastest wa to solve the sstem is with a reduced row-echelon matri. D b. The polnomial is nd degree, and the D values are constant. c. points. You have to find the finite differences twice, so ou need at least four data points to calculate two D values that can be compared. The equation is _ 3 3 6. Discovering Advanced Algebra Solutions Manual CHAPTER 7 7 00 Ke Curriculum Press
c. 0 blocks. Substitute 8 for and solve for : _ 3 (8)3 _ (8) _ 5 6 (8) _ 3 3 _ 3 0. d. laers. Substitute integer values for until equals 650. Because 8 results in 0 in 6c, start with 9. When, _ 3 ()3 _ () _ 6 () 576 7 650. 7. a. i. D {5., 5.3,.5,.3,., 33.9}; D {9.8, 9.8, 9.8, 9.8, 9.8} ii. D {59., 9.3, 39.5, 9.7, 9.9, 0.}; D {9.8, 9.8, 9.8, 9.8, 9.8} b. Both are nd-degree polnomials because the second differences are constant in each case. c. i. h.9t 0t 80. The function is quadratic, so start with h at bt c. Use three data points to get three equations and solve for the constants a, b, and c. c 80 9a 3b c 95.9 36a 6b c 3.6 Using the data point (0, 80). Using the data point (3, 95.9). Using the data point (6, 3.6). Solve this sstem to get a.9, b 0, and c 80. The equation is h.9t 0t 80. Check that the point (3, 95.9) satisfies this equation. 95.9?.9(3) 0(3) 80 95.9 95.9 ii. h.9t 6t. The function is quadratic, so start with h at bt c. Use three data points to get three equations and solve for the constants a, b, and c. c Using the data point (0, ). 9a 3b c 5.9 Using the data point (3, 5.9). 36a 6b c.6 Using the data point (6,.6). Solve this sstem to get a.9, b 6, and c. The equation is h.9t 6t. Check that the point (3, 5.9) satisfies this equation. 5.9?.9(3) 6(3) 5.9 5.9 8. a. D {0.07, 0.06, 0.0, 0.0, 0.0, 0.0, 0, 0, 0, 0.0, 0.0, 0.0, 0.03, 0.05, 0.07}. The graph is approimatel parabolic. b. D {0.0, 0.0, 0.0, 0.0, 0, 0.0, 0, 0, 0.0, 0, 0.0, 0.0, 0.0, 0.0}. These points are approimatel linear. D 3 {0.0, 0, 0.0, 0.0, 0.0, 0.0, 0, 0.0, 0.0, 0.0, 0, 0.0, 0}. There is no pattern to these numbers. c. 3rd-degree polnomial; the third differences are nearl constant and show no pattern. d. Possible answer: Let represent And s age in ears, and let represent his height in meters; 0.09 3 0.993.3.88. Start with the function a 3 b c d and use four data points, such as (9.5,.), (0.75,.3), (,.36), and (3.5,.5). Then solve the sstem to find values for a, b, c, and d, as in Eercise 7c. Note: When using this method, it is best to use four data points that are spread throughout the domain, rather than just using, for eample, the first four points. The model appears to be reasonable for about 8.5. 9. The data are represented b the quadratic function D {6, 0,, 8,, 6}; D {,,,, }. The second differences are constant, so a quadratic function epresses the relationship. Let represent the energ level, and let represent the maimum number of electrons. Start with the equation a b c, and use three data points to find the constants a, b, and c. a b c 6a b c 3 9a 7b c 98 Using the data point (, ). Using the data point (, 3). Using the data point (7, 98). Solve this sstem to get a, b 0, and c 0, so. 0. a. This is the image of the graph of translated horizontall units. 5 5 8 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
b. This is the image of the graph of translated verticall units. c. This is the image of the graph of translated horizontall units and verticall unit.. a..5 7 3 30 _ 30.5 b. 3 or ( ) 3 ( ) 8 ( ) or 3 or log 6 c. _.77 log 5 35 8 5 6 log 5 log 6 log 6 _.77 log 5. a. 3 3 6 9 3 3 3 b. ( 3)(3 ) 6 9 3 6 3 c. 5 5 3 3 5 Use the fact that 8 5 5 3 5(3). d. ( 3)( 5) 5 5 3. 3 9 _ 6 97 _ 6 The line connecting points A and B has slope 3 () 3 _. Use the point-slope equation with (3, 3) to get _ ( 3) 3 _ 3_. The feasible region is above the solid line, so the inequalit is _ 3_. The line connecting points A and C has slope 7 3 5 (3) _. Use the point-slope equation with (5, 7) to get _ ( 5) 7 _ 9_. The feasible region is below the solid line, so the inequalit is _ 9_. The line connecting points B and C has slope 7 () 5 6. Use the point-slope equation with (5, 7) to get 6 ( 5) 7 97 6 6. The feasible region is below the solid line, so the inequalit is 97 6 6.. ( 3)( )( ) 7 ( ) ( ) 7( ) ( ) 3 7 3 9 6 EXTENSIONS A. Results will var depending on the initial velocit at which the pillow is tossed upward, but the model should still be quadratic. Specificall, the model will be in the form.9 v 0 s 0, where v 0 is the initial velocit and s 0 is the initial height. The model for tossing the pillow upward has an -term, whereas the model for dropping it does not. B. See the solution to Take Another Look Activit on page 77. C. The finite differences never become constant for e, because it is not a polnomial function. See Calculator Note 7B. Label the first spreadsheet column in. Enter {,, 3,, 5, 6, 7, 8, 9, 0}. Label the second column out and define it as e (in). Then label the third column D and follow the Calculator Note to find the finite differences. Repeat this with the fourth and subsequent columns. You will find that the finite differences never become constant. LESSON 7. SUPPORT EXAMPLES. 0 and ; 8. 0, so 0, and 0, so. a. (3, 6) b. 6 ( 3) 6 ( 6 9) 6 8 Discovering Advanced Algebra Solutions Manual CHAPTER 7 9 00 Ke Curriculum Press
c. 3 d. Minimum b. 3 and 3. ( ) 8 ( ) 8 8 8 8 8 0 ( 5) ( )( 5) EXERCISES. a. Verte form b. Factored form c. Factored form and verte form d. None of these forms e. Factored form f. General form. In a c, identif the values of h and k for the equations in the form a( h) k. a. (, 3) b. (, ) c. (5, ) 3. In 3a c, identif the values of r and r for the equations in the form a r r. a. and c. and 5. Because the equations are equivalent, the graphs for each part should look like onl one parabola. a. ( ) 3 3 7 b. 0.5( ) 0.5 8 6 0.5 8 0.5 6 c. ( 5) 0 5 0 50 0 6 5. Because the equations are equivalent, the graphs for each part should look like onl one parabola. a. b. 0.5( )( 3) 0.5 6 0.5 0.5 3 c. ( )( 5) 7 0 0 6. a..5. Notice that the values in the table have smmetr about (.5, 9). b. (.5, 9); maimum. The line of smmetr passes through the verte, so (.5, 9) is the verte. It is a maimum because all other -values are less than 9. c. 3(.5) 9. The verte is (.5, 9), so start with the equation a(.5) 9 and substitute another data point to find a. For eample, substitute the coordinates (.5, 8) to get 8 a(.5.5) 9 7 a(3) 7 9a a 3 The equation is 3(.5) 9. 7. a. 0.5( h) 0.5 h h 0.5 h 0.5h 0.5 h 0.5h 50 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
b. a( ) a 8 6 a 8a 6a c. a( h) k a h h k a ah ah k d. 0.5( r)( ) 0.5 r r 0.5 0.5r r e. a( )( ) a 8 a a 8a f. a( r)( s) a r s rs a (r s) rs a a(r s) ars 8. a. a(.)( 0.8). Substitute the -intercepts,. and 0.8, for r and r in the form a r r. b..8(.)( 0.8). Substitute. for and.59 for and solve for a..59 a(..)(. 0.8).59 a(3.6)(0.) a _.59.8. c. 0.8,.608. First, find the mean of the -intercepts to get the -coordinate of the verte.. 0.8 _ 0.8. Then use the equation of the graph to find the -coordinate..8(0.8.)(0.8 0.8).608 d..8( 0.8).608. Substitute the values of h and k from the verte, h 0.8 and k.608, and the value of a from 8b in the form a( h) k. 9. In 9a c, use the graph to identif the - and -intercepts. a. ( )( ). The -intercepts are and, so start with the equation a( )( ). Use the -intercept point, (0, ), to solve for a. a(0 )(0 ) a a The equation is ( )( ). b. 0.5( )( 3). The -intercepts are and 3, so start with the equation a( )( 3). Use the -intercept point, (0, 3), to solve for a. 3 a(0 )(0 3) 3 6a a 0.5 The equation is 0.5( )( 3). c. _ 3 ( )( )( 3). The -intercepts are,, and 3, so start with the equation a( )( )( 3). Use the -intercept point, (0, ), to solve for a. a(0 )(0 )(0 3) 6a a 3 The equation is _ 3 ( )( )( 3). 0. a. Use the projection that the sales decrease b 5 packs per da with each $0.0 increase in selling price to get the number sold. Multipl the selling price b the number sold to get the revenue. Selling price ($).00.0.0.30.0 Number sold 00 95 90 85 80 Revenue ($) 00 09.50 8 5.50 3 b. D {9.5, 8.5, 7.5, 6.5}; D {,, } Number sold 00 95 90 85 80 Revenue ($) 00 D 09.50 8 5.50 3 9.5 8.5 7.5 6.5 D c. 50 300. The second differences are constant, so a quadratic function can model the data. Start with the equation a b c and use three data points to solve for a, b, and c..00a.00b c 00 Using the data point (.00, 00)..8a.0b c 8 Using the data point (.0, 8). 5.76a.0b c 3 Using the data point (.0, 3). Solve this sstem to get a 50, b 300, and c 0, so the equation is 50 300. d. The maimum revenue is $50 at a selling price of $3. Graph the function and trace to find the maimum.. a. Width (m) 5 0 5 0 5 Length (m) 35 30 5 0 5 Area m 75 300 375 00 375 Discovering Advanced Algebra Solutions Manual CHAPTER 7 5 00 Ke Curriculum Press
Use the formula Perimeter 80 (length) (width), or 0 length width, to get the second row. Use the formula Area length(width) to get the third row. b. (0 ), or 0. Because Perimeter (length) (width), 80 (length) and length 0. Use the formula Area length(width) to obtain (0 ). c. A width of 0 m (and length of 0 m) maimizes the area at 00 m. Graph the function and trace to find the maimum, which is also the verte. d. 0 m and 0 m. The equation (0 ) is in factored form and shows that the zeros are 0 and 0.. a. The graph is parabolic. As the temperature increases, the rate of photosnthesis also increases until a maimum rate is reached; then the rate decreases. b. At approimatel 3 C, the rate of photosnthesis is maimized at 00%. The verte of the parabola is approimatel (3, 00). c. 0 C and 6 C. The -intercepts of the parabola are 0 and 6. d. 0.9 ( 6); 0.9( 3) 00. For the factored form, substitute the -intercepts to start with the equation a( 0)( 6). Then use the verte, (3, 00), to find a. 00 a(3)(3 6) 00 59a a _ 59 00 0.9 The function in factored form is 0.9 ( 6). For the verte form, substitute the verte to start with the equation a( 3) 00. Then use the same value of a as above, so the equation in verte form is 0.9( 3) 00. 3. a. 5 b. 5 5 3 5 5 5 3 3 5 c. 9 d. 9 6 7 7 7 7 9. a. 3 () 3 (5) 5 3 3 9 3 3 b. ( 5) 3( 5) 5 3 5 5 c. ( 7) 7( 7) 7 7 9 9 d. (3 )(3 ) 3 (3 ) (3 ) 9 3 3 9 6 5. a. ( 5)( ). The numbers that multipl to 0 and add to 3 are 5 and. 5 5 0 b. ( )( ). The numbers that multipl to 6 and add to 8 are and. 6 c. ( 5)( 5). The numbers that multipl to 5 and add to 0 are 5 and 5. 5 5 5 5 5 d. Answers will var. Possible answer: The factored form of 8 6 is ( )( ), or ( ). It is a perfect square trinomial because the two factors are identical; possible answer: 6 9,. e. Answers will var. Possible answer: 6, 00. 6. a. f () 3() 3 5() 6 3(8) 5() 0 b. f () 3() 3 5() 6 3() 5() 7 5 c. f (0) 3(0) 3 5(0) 0 6 6 d. f 3 3 5 6 3 8 5 _ 5 _ 8 6 3 8 5 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
e. f 3 3 3 3 5 3 6 3 3 _ 7 6 5 _ 6 9 _ 3 0 _ 9 3 3 7. a. Sometimes true; true for all 0 and false for 0 b. Sometimes true; true for all and false for c. Alwas true IMPROVING YOUR REASONING SKILLS Let S represent the sum of the first n terms. S n n; S n ; S 3 n n; S d n d n For the first sequence, start with a table of values, and use the finite differences method. n S D 3 5 3 6 0 5 3 5 D The second differences are constant, so the sequence can be modeled b a quadratic equation. Start with the equation S an bn c, and use the three points (, ), (, 3), and (3, 6) to find a, b, and c. a b c Using the point (, ). 9a 3b c 6 Using the point (3, 6). 5a 5b c 5 Using the point (5, 5). Solve this sstem to get a _, b _, and c 0. Use a similar process for the second and third sequences. For an arithmetic sequence, note that the terms are, d, d, 3d,..., (n )d. Another wa to find a formula for the sum of the first n terms of an arithmetic sequence is to write S first in one order and then in the reverse order of summation. S ( d ) ( d ) [ (n 3)d ] [ (n )d ] [ (n )d ] S [ (n )d ] [ (n )d ] [ (n 3)d ] ( d ) ( d ) Now add S to itself, using these two orders of summation: S [ (n )d ] [ (n )d ] [ (n )d ] [ (n )d ] [ (n )d ] [ (n )d ] Each term in the sum of S is the same, and there are n terms, so S n[ (n )d ] n[ (n )d ] S This is equivalent to S d_ n d_ n LESSON 7.3 SUPPORT EXAMPLES. a. 0 5; (5) 5 b. _ 36 6; 6 or 6. ( 9) 9 ( 7) 7 3. Verte is at 3_, _ 7 3 9 5 3 3 5 3 3 9_ 7 5 3 3_ 7_ EXERCISES. a. ( 5) b. 5 c. ( 3), or 3 d. ( ). a. 0 _ 0 00 b. _ 7 _ 9, or.5 c. 6. First, factor out the. 6 _ You need to add a inside the parentheses to complete the square. Everthing inside the parentheses is multiplied b, so ou actuall add 6. d. 3. First, factor out the 3. 3 6 3 You need to add a inside the parentheses to complete the square. Everthing inside the parentheses is multiplied b 3, so ou actuall add 3. 3. a. ( 0) 6. As in a, ou need to add 00 to 0 to complete the square. 0 00 9 00 ( 0) 6 Discovering Advanced Algebra Solutions Manual CHAPTER 7 53 00 Ke Curriculum Press
b. ( 3.5) 3.75. As in b, ou need to add.5 to 7 to complete the square. 7.5 6.5 ( 3.5) 3.75 c. 6( ) 3. As an alternative to completing the square, identif the values of a, b, and c in general form, and use the formulas for h and k to write the verte form. a 6, b, c 7 h _ a b _ k c b _ a 7 _ 576 3 The verte form is 6( ) 3. d. 5( 0.8) 3.. Identif the coefficients and then calculate h and k: a 5, b 8, c 0 h b _ a k c b _ a _ 8 0.8 0 0 _ 6 3. 0 Use the values of a, h, and k in the verte form: 5( 0.8) 3.. a. No rewriting necessar; a 3, b, c 5 b. ; a, b 0, c c. 6 3; a, b 6, c 3 d. 3; a, b 3, c 0 5. (, ) a, b 6, c 0 h _ a b _ 6 k c b _ a (6) 0 8 6. 7.5( 3.5) 5.93 h _ a b 7.3 (7.5) k c b _ a 3.5 (7.3) 9.7 _ 5.93 (7.5) Rewrite the equation in verte form: 7.5( 3.5) 5.93. Use a calculator table to verif that the functions are approimatel equivalent. 7. a. Let represent time in seconds, and let represent height in meters;.9(.)(.7), or.9 8. 5.333. Use the factored form because ou are given the zeros,. and.7. Start with a(.)(.7). The acceleration due to gravit is.9 m/s, so a.9. b. 8. m/s. This is the coefficient of the -term. c. 5.333 m. At 0 s, the arrow starts at the bottom of the well. Substitute 0 for, or identif the constant term as the -intercept. The height at time 0 s is 5.333 m, so the well is 5.333 m deep. 8. a. Possible table, where represents the sides perpendicular to the building: 5 0 5 0 5 30 35 350 600 750 800 750 600 350 80. If l is the length of the rectangle, then the amount of fencing needed to build a fence using the building as one side is l. You have 80 ft of fencing, so 80 l, or l 80. The area is l (80 ), or 80. b. 0 ft; 800 ft. Complete the square for the function 80. 0 0 00 00 0 00 800 0 800 The verte is (0, 800), indicating a maimum area of 800 ft when the width is 0 ft. 9. Let represent time in seconds, and let represent height in meters;.9 7. 50. The acceleration due to gravit is.9 m/s, the initial velocit is 7. m/s, and the initial height is 50 m. Substitute these values in the projectile motion function, a v 0 s 0. 0. a..9t 00t 5. Start with the equation.9t v 0 t s 0, and use two data points to find v 0 and s 0..9 v 0 s 0 0. Using the point (, 0.). 78. v 0 s 0 36.6 Using the point (, 36.6). Solve this sstem to get v 0 00 and s 0 5, so.9t 00t 5. b. s 0 5 m; v 0 00 m/s 00 c. 0. s; 535 m. h 0. and (.9) 00 k 5 535, so the verte is approimatel (0., 535). This indicates a maimum (.9) height of 535 m at 0. s. 5 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
. a. n p 00. Write the equation of the line that passes through the points (0, 60) and (5, 50). Obtain the second point, (5, 50), from the information that a $5 increase in price results in 0 fewer sales. The slope of the line is _ 50 60 5 0. Use the point-slope equation to get n (p 0) 60, or n p 00. b. R(p) p 00p. The revenue is the price per T-shirt times the number of T-shirts sold, R(p) pn p(p 00), or R(p) p 00p. c. Verte form: R(p) ( p 5),50. The verte is (5, 50). This means that the maimum revenue is $,50 when the price is $5. h 00 00 5 and k 0,50, so () () the verte form is R( p) ( p 5),50 and the verte is (5, 50). d. Between $5 and $35. Find the points where R( p),050, or p 00p,050. p 00p,050 0 p 50p 55 0 ( p 5)( p 35) 0 p 5 or p 35 An price between $5 and $35 will ield at least $,050 in revenue.. a. ( ) 3( ) 6 b. ( ) ( ) 3 3., 3, or _. B the zero-product propert, 0, 3 0, or 0, so, 3, or _.. a. ( 5) 7 3 ; ellipse. For the vertical stretch, replace with _ 3 to get the equation _ 3. For the translation, replace with ( 5) and with ( 7) to get the equation ( 5) 7 3. b. B A C 5 D 0 5 A(5, 0), B(6, 7), C (5, ), D(, 7); center: (5, 7) 5. a. Let represent the ear, and let represent the number of endangered species. b. Answers will var. The median-median line is ŷ 5.6 90,89. c. Answers will var. Using the equation from 5b, approimatel,5 species in 005; 3,78 species in 050. Substitute 005 for : ŷ 5.6(005) 90,89,9.. Substitute 050 for : ŷ 5.6(050) 90,89 3,73. d. The prediction of about 9 species is too high. Students might lower their prediction for 050, but there aren t ver much data on which to base a prediction? e. The more recent data appear to lie on a curve that is leveling off. Possible answers: a logarithmic function, a quadratic function. Answers ma var quite a bit, but students should justif their choice of model. LESSON 7. SUPPORT EXAMPLES. 5, so 0 5 and a, b 5, and c for the quadratic formula. 5 (5) _ ()() _ 5 5 () _ 5 0.09 or.79. 0 EXERCISES. a. 3 3 0 0; a 3, b 3, c 0 b. 5 3 0; a, b 5, c 3 c. 3 5 0; a 3, b 5, c d. 3 3 0; a 3, b 3, c Discovering Advanced Algebra Solutions Manual CHAPTER 7 55 00 Ke Curriculum Press
e. 5 50 0; a, b 5, c 50 ( ) ( ) ( )( ) 56 8 3 58 8 0 5 50. In a d, be sure to use parentheses as necessar. a. 0.0 b. 5.898 c. 0.3 d. 8.3 3. a. or 5. Use factoring and the zeroproduct propert. 6 5 0 ( )( 5) 0 or 5 Alternativel, ou can use the quadratic formula. b. or 9. Use factoring and the zeroproduct propert. 7 8 0 ( )( 9) 0 or 9 Alternativel, ou can use the quadratic formula. c. or.. Use the quadratic formula; a 5, b, c 7. _ 5 7 5 or.. a. ( )( 5) b. ( )( 9) c. 5( )(.) 5. a. _ and or 0.5 and. 7 7 ()() _ 7 _ 9 3 () _ 7 9 _ and b. 3 6 or 5.95 and 0.5505. 6 _ (6) ()(3) 6 36 6 6 6 _ 3 () 3 6 c. 3 or.73. 0 _ 0 _ ()(6) _ 8 () 3.73 d. No real solutions. 5 _ 5 ()() () 5 _ 5 3 5 7, and since the solution has an imaginar part, the equation has no real zeros. 6. a. 9 0 0. The quadratic formula shows that a, b 9, and c 0. _ b. 9 c. 9 and 9 7. In 7a c, substitute the -intercepts for r and r in the factored form, a r r. In each, a can be an nonzero value, or ou can leave it as a variable. a. a( 3)( 3) for a 0 b. a( ) _ 5, or a( )(5 ), for a 0. The second equation is found b multipling the factor _ 5 b the denominator of the root, 5. Note that _ 5 0 and 5 0 both have the solution _ 5. c. a r r for a 0 8. The solution includes the square root of 36, so there are no real solutions. The graph shows no -intercepts. Before using the quadratic function, evaluate b ac. If b ac 0, then there will be no real solutions. 9. The function can be an quadratic function for which b ac is negative, as shown in Eercise 8. Sample answer:. 0. The mean of the two solutions is _ b b ac _ b b ac a a b a _ a b The -coordinate of the verte, b a, is midwa between the two -intercepts. If there are no -intercepts, the roots are nonreal. In this situation, the -coordinate of the verte is midwa between the two comple solutions on the comple plane. 56 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
. a. 6.8 9. Time (min) Amount of water (L).5.5 38. 30.0 9.6 7. D 8. 0.. D The model appears to be quadratic because the second differences are constant. Start with the equation a b c, and use three data points to solve for the coefficients a, b, and c. a b c 38..5a.5b c 30.0 a b c 9.6 Using the point (, 38.). Using the point (.5, 30.0). Using the point (, 9.6). Solve this sstem to get a, b 6.8, and c 9., so the equation is 6.8 9.. b. 9. L. The plug was pulled at 0 min, so substitute 0 for : (0) 6.8(0) 9. 9.. Alternativel, identif the constant term, 9., as the -intercept. c..76 min. Use the quadratic formula to find the time when there is 0 L of water in the tub. 6.8 _ (6.8) ()(9.) ().6 or.76 Because represents time after the plug was pulled, the onl solution that makes sense is the positive one,.76. The tub empties in.76 min.. _ a a, a _ 5 a _ a a a Original equation. a a Multipl both sides b a (a ). a a 0 Subtract a from both sides. a _ 5 Use the quadratic formula. Because onl a positive value makes sense in the 5 contet of the rectangle, a. 3. a. 9 ( 7) or () 9 [ (7)]. Take the square root of the constant, _ 9 7, and then multipl b to get. This is the -coefficient of the perfect squares ( 7) ( 7). b. 0 5 ( 5). _ 0 5, so add 5 to 0 to complete the square. d. 8 8 ( ). It s easiest to first fill in the third and fourth blanks. When ou factor from 8, ou get in the third blank. The factor of carries over to the last epression, so put in the fourth blank.? 8? (?) Now look at the middle epression and determine what to add to make it a perfect square. Take the square root of,, and then double it. This gives in the second blank and in the fifth blank.? 8 ( ) Finall, distribute the through the middle epression to fill in the first blank with (), or 8. 8 8 ( ) You can follow the same procedure to find the other possible solution, (8) 8 () ().. a.. Switch and, and then solve for. ( ) b. _. Switch and, and then solve for. ( ) ( ) c. _ 6. Switch and, and then solve for. 5 5 ( ) 6 6 ( ) _ 6 _ 6 5. a. 5 ( 3)( 5) 5 6 5 5 Original equation. Use the distributive propert. Combine terms. c. 3 9_ 3_ 3_ 9_, so add 9_ to 3 to complete the square. Discovering Advanced Algebra Solutions Manual CHAPTER 7 57 00 Ke Curriculum Press
b. ( ) Original equation. Use the distributive propert. Combine terms. 6. a. 00. B the Pthagorean Theorem, 0. 00 00 The function is 00 because onl positive values of make sense in the contet of this problem. b. Approimatel 7.3 ft. Substitute 0 for in the equation for 6a: _ 00 0 7.3 ft. c. Approimatel 8.7 ft. Solve the equation 0 from 6a for : 00. Substitute 8 for : _ 00 8 8.7 ft. 7. a and k have length 5.083 ft, b and j have length 33.3 ft, c and i have length 8.75 ft, d and h have length 8.3 ft, e and g have length.083 ft, and f has length 0 ft. The total length of support cable needed for the portion of the bridge between two towers is 9.6 ft. Note: If ou consider that there is a support cable on each side of the bridge, our answer will be 58. _ 3 ft. Start b putting the bridge on a coordinate plane with the verte of the parabola at (0, 0) and the tips of the towers at the points (80, 75) and (80, 75). ( 80, 75) (80, 75) a b c d e g f h i j k The equation in verte form of this parabola is a because the verte is (0, 0). Substitute the coordinates of the point (80, 75) to find a: 75 a 80, so a _ 75 80 _ 3. 56 Hence the equation of the parabola is _ 3 56. The support cables break up the interval from the verte to one of the towers into si equal intervals, each of length 80 6 0 3 3.3. Now use the equation _ 3 56 to compute the lengths of the support cables. Distance Support cable from center (ft) length (ft) Support cable a 66.6 5.083 b 53.3 33.3 c 0 8.75 d 6.6 8.3 e 3.3.083 f 0 0 g 3.3.083 h 6.6 8.3 i 0 8.75 j 53.3 33.3 k 66.6 5.083 The total length of support cables needed between two towers is the sum of the lengths, 9.6 ft. EXTENSION See the Sketchpad demonstration Quadratic Functions for a pre-made sketch and guided instructions. You should find that as a changes, the verte follows a linear path, specificall b_ c. As b changes, the verte follows a parabolic path, specificall a c. As c changes, the verte follows a linear path, specificall b a. LESSON 7.5 SUPPORT EXAMPLES. (3 i )( 6i ) 8i 8i i 6i 6i. (3 i ) ( i ) ( i ) _ ( i ) 3 6i i i i i i _ 3 5i 5 5i _ 5 i 3. _ ()(5) _ 0 () i 9 EXERCISES. a. (5 i) (3 5i) (5 3) ( 5)i 8 i b. (6 i ) ( i) [6 ()] ( )i 7 c. ( 3i ) ( 5i) ( ) (3 5)i i d. (.35.7i ) (.9 3.3i ) (.35.9) (.7 3.3)i.56 0.6i 58 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
. a. (5 i )(3 5i ) 5 5i 3i 5i 5 i 5() 0 i b. 6 i c. 3i( 5i ) 6i 5i 6i 5() 5 6i d. (.35.7i )(.9 3.3i ).5385 7.80i 3.306i 8.997i.5385.08i 8.997().53.08i 3. To find the conjugate, negate the imaginar part. a. 5 i b. i c. 3i d..35.7i. A: 5 i, B: 3i, C: 3 i, D: 6, E: i, F:, G: i 5. a. Comple b. c. d. e. Real All real numbers are also comple. Rational Irrational No number is both rational and irrational. Comple Imaginar All imaginar numbers are also comple. Imaginar Real No number is both real and imaginar. Real Comple Imaginar All real numbers are comple and all imaginar numbers are comple, but no number is both real and imaginar. 6. 5 0. An product in the form ( a)( b) epands to (a b) ab. Letting a i and b i, note that a and b are comple conjugates and that a b and ab 5. So, [ ( i )][ ( i )] 5. 7. a. i 3 i i i i b. i i i c. i 5 i i i i d. i 0 i 5 i 5 i i i 8. i 7 i i, i 6 Imaginar i, i 5 i 3, i 7 i, i 8 Real Each multiplication b i rotates another 90 counterclockwise. The all correspond to, i,, or i. To find i 7, use the fact that i. i 7 i 6 i i i i i. In general, divide the eponent b and raise i to the power of the remainder. 9. 0..6i. Multipl the numerator and denominator b the comple conjugate of the denominator, i. 3i i _ i 8i i 5 0..6i 0. a. ; comple and real ± b. i; comple and imaginar i c. i ; comple. Using the quadratic formula, the solutions are _ () _ 6 _ 8 () _ i i d. _ i _ 3 ; comple. Rewrite the equation as 0. Using the quadratic formula, the solutions are _ 3 () i 3 i e. ; comple. Rewrite the equation as 0 3. Using the quadratic formula, the solutions are () ()(3) 6 () _ 8 i _ i _ i Discovering Advanced Algebra Solutions Manual CHAPTER 7 59 00 Ke Curriculum Press
. For a d, substitute the values of the zeros into the factored form r r, and then convert to general form. a. ( 3)( 5) 5 b. ( 3.5)( 3.5) 7.5 c. ( 5i )( 5i ) 5i 5 d. [ ( i )][ ( i )] ( i ) ( i ) ( i )( i ) ( i i ) i 5. 6 50. Start with a quadratic equation whose zeros are 3i and its comple conjugate, 3i. [ ( 3i )][ ( 3i )] ( 3i ) ( 3i ) ( 3i )( 3i ) 8 5 The function 8 5 has the correct roots, but its -intercept is 5, half of 50. Stretch it verticall b a factor of to get the desired function, 8 5 6 50. 3. a. 5 _ 3 i 0.83i or 5 _ 3 i 0.83i. Use the quadratic formula with a, b 0i, and c 9i 9. 0i 00i ()(9) 0i 36 (i ) 0i i 36 0 3 i 5 _ 3 i 0.83i or 0.83i b. i or i 3i 0. Use the quadratic formula with a, b 3i, and c. 3i 9i _ () 3i _ 9 8 () _ 3i i i or i c. The coefficients of the quadratic equations are nonreal.. a. b ac 0. The solutions are nonreal when b ac 0, or when b ac. In this case, when ou use the quadratic formula, ou are taking the square root of a negative number. b. b ac 0. The solutions are real when b ac 0, or when b ac. c. b ac 0. In the quadratic formula, the two solutions are equal when b ac 0, or when b ac, because the square root of zero is zero. 5. For 5a d, use recursion on our calculator to evaluate z 0 to z 5 and to verif the long-run value. a. 0, 0, 0, 0, 0, 0; remains constant at 0 b. 0, i, i, i, i, i; alternates between i and i c. 0, i, 3i, 7 7i, 97i, 907 93i; no recognizable pattern in the long run d. 0, 0. 0.i, 0. 0.8i, 0.66 0.3i, 0.877056 0.300838i, 0.6078 0.7778585i; approaches 0.063 0.7937653i in the long run 60 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
After man recursions: _ 3 5 6. a. ; 3. or 0.. Use the quadratic formula with a, b 6, and b. c 3. 6 _ 6 _ (3) _ 6 5 _ 3 5 3. or 0. 3 5 _ 3_ 5, 3 5 _ (0, 3) The graph crosses the -ais at (0, 3) because the constant term is 3, which is also the -intercept. To find the verte, complete the square. 6 3 3 3 3 9 9 3 3 9 9 3 b. 8 (0, 7.5) 5 6 50 5, 3 30 (0, 0) (.5, 0) 6 8 0 c. (, ), (, ), (, ), (, ), (3, ), (, 3), (, 3), (3, 3), (, ), (, ), (3, ), (, ), (, 5), (, 5), (3, 5), (, 6) EXTENSIONS A. See the solutions to Take Another Look Activities and 5 on page 78. B. Sample answer: In electronics, the state of an element in a circuit is described b two real numbers: voltage and current. These two states can be combined with a comple number in the form voltage (current)i. Similarl, inductance and capacitance can be represented with a comple number. The laws of electricit, therefore, can be epressed using comple arithmetic. C. Results will var. LESSON 7.6 SUPPORT EXAMPLES. The -intercepts are at (, 0), (, 0), and (3, 0); the -intercept is at (0, ). The -intercepts occur where 0, where 0, and where 3 0. The -intercept occurs at ()()(3).. ( )( 5) 5 5 5 3 5 EXERCISES. a. -intercepts:.5, 6; -intercept:.5. To find the -intercepts, identif the values of r and r in factored form. r.5 and r 6. To find the -intercept, substitute 0 for and solve for. 0.5(0.5)(0 6).5 Check b graphing. 3 5 _ The verte is 3, 5 _. 7. a. Let represent the first integer, and let represent the second integer. 0 0 3 30 5 Discovering Advanced Algebra Solutions Manual CHAPTER 7 6 00 Ke Curriculum Press
b. -intercept: ; -intercept: 8. To find the -intercepts, identif the values of r and r in factored form. r r. To find the -intercept, substitute 0 for and solve for. 3(0 )(0 ) 8 Check b graphing. b. 0.5(.5)( 6). Start with the factored equation a(.5)( 6), using the -intercepts,.5 and 6. Then substitute the coordinates of the -intercept, (0,.5), to find a..5 a(0.5)(0 6) 9a a 0.5 The equation is 0.5(.5)( 6). c. -intercepts: 3,, 5; -intercept: 60. To find the -intercepts, identif r 3, r, and r 3 5 from the factored form. To find the -intercept, substitute 0 for and solve for. (0 3)(0 )(0 5) 60 Check b graphing. d. -intercepts: 3, 3; -intercept: 35. To find the -intercepts, identif r r 3 and r 3 3 from the factored form. To find the -intercept, substitute 0 for and solve for. 3. a. ( )( 6) 6 0 b. ( 3)( 3) 3 3 9 6 9 c. ( 8)( 8) 6 3 6 d. 3( )( )( 5) 3 ( 5) 3 3 5 0 3 3 5 60. a. 7.5,.5, 3.. Identif the values of r, r, and r 3 from the factored form. b. 50. Substitute 0 for and solve for :.5(0 7.5)(0.5)(0 3.) 50. c..5( 7.5)(.5)( 3.).5( 7.5) 0.7 8.5 3 0.7 8 7.5 5.5 60.5 3 8..75 60.5 3 0.5 6.875 50 d. 5(0 3)(0 3)(0 3) 35 Check b graphing. If the functions are equivalent, ou should see onl one curve.. a. ( )( ). Start with the factored equation a( )( ), using the -intercepts, and. Then substitute the coordinates of the verte, (3, ), to find a. 5. a. Approimatel.9 units; approimatel 0 cubic units. Graph the function (6 )(0 ), zoom in on the peak, and then trace the function to approimate the maimum volume. a(3 )(3 ) a a The equation is ( )( ). 6 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
b. 5 and approimatel.8. Graph the linear function 300 along with (6 )(0 ) and trace to approimate the intersections. g. ( 8i )( 8i ). The roots of the polnomial 6 are 8i, so it can be factored as ( 8i )( 8i ). h. 7 7. The roots of the polnomial 7 are 7, so it can be factored as 7 7. i. ( 3). Factor out the common factor,. 7. a. Sample answer: c. The function doesn t model volumes of real boes outside the interval 0 8. For 0, two dimensions of the bo are negative, so the product is positive. This means that this impossible bo has a possible volume. d. The function doesn t model volumes of real boes outside the interval 0 8. For 8 0, one of the dimensions of the bo would be negative, so the product is negative. 6. a. ( )( 0). First, factor out the common factor, : 88 80 0 ( )( 0). b. 6 5_ 3 _, or (3 5)( ). Use the quadratic formula: The zeros of the polnomial are 7 (7) 6 (5) 7 3 6 _, or 5_ 3 or _. Hence the polnomial factors as 6 5_ 3 _ 3 5_ 3 _ (3 5)( ). c. ( )( )( 5) 3 5 0 ( 5) ( 5) ( 5) Original equation. Factor out the common factor from 3 5, and factor out the common factor from 0. Factor out the common factor, ( 5). ( )( )( 5) Factor into ( )( ). d. ( )( 3)( ). Graph the function 3 6 38 on our calculator. It has -intercepts, 3, and. Use these values for r, r, and r 3 in the factored form along with the value of a from the leading coefficient, a. So the factored form is ( )( 3)( ). e. (a b)(a b), or (a b). This is the general form of a perfect square. f. ( 8)( 8). The roots of the polnomial 6 are 8, so it can be factored as ( 8)( 8). An parabola with its verte on the -ais has one real zero. b. Sample answer: An parabola that does not cross the -ais has no real zeros. c. Not possible. An parabola intersects the -ais at most twice. d. Sample answer: An cubic curve that crosses the -ais once has one real zero. e. Sample answer: An cubic curve that crosses the -ais once and then touches the -ais (has a double root) has two real zeros. f. Not possible. An cubic curve intersects the -ais at least once. 8. 0.3(.5)( )( ).7 a (.5)( )( ).7 a (0.5)(3)(6).7 9a 0.3 a Discovering Advanced Algebra Solutions Manual CHAPTER 7 63 00 Ke Curriculum Press
9. a. a ( 5)( 3)( 6) b. ( 5)( 3)( 6). Substitute the coordinates of the -intercept, (0, 80), and solve for a. 80 a (0 5)(0 3)(0 6) 80 90a a c. Replace with 00: ( 5)( 3)( 6) 00 d. Replace with : (( ) 5)(( ) 3)(( ) 6) ( 9)( )( ) 0. a. (T t), or T Tt t T T TT Tt t t Tt tt b. (T t), or T Tt t c. 0.70 t. Because T is the dominant gene, the people who have either TT or Tt constitute 70% of the population. Looking at the equation in 9b, this means T Tt 0.70, so 0.70 t. d. t 0.58. Use the equation from 0c to solve for t. t 0.70 t 0.30 t 0.30 0.58 Onl the positive root makes sense. e. Subtract t from to find T: T 0.58 0.5. f. T T 0.5 0.5 0.05, or about 0% of the population.. No. These points are collinear. An attempt to find a nd-degree polnomial function will give a 0.. 0.5( ) 3. Start with the verte form of the equation, a ( ) 3, and use the point (, ) to find a. a ( ) 3 9 36a a 0.5 So the equation is 0.5( ) 3. 3. a. 50. 7. b. _ 3 3.6 69 3 or 3 _ 3 The equation 3 has no real solution. c.. 0.. 6.3; 8.7 or 3.9 (.) 0.. 0.. 0. d. 3 6. a. f () 3 5. Write the function as ( 3 5), then switch and and solve for. ( 5) 3 3 5 3 5 So, f () 3_ 5. Check our answer: f () 3 ( 5) 6; f (6) 3_ (6) 5. b. g () 3 ( 6) 3. Write the function as 6 ( 3) 3, then switch and and solve for. 6 ( 3) 3 6 ( 3) 3 ( 6) 3 3 ( 6) 3 3 So, g () 3 ( 6) 3. Check our answer: g (5) 6 (5 3) 3 6 ; g () 3 ( 6) 3 3 8 5. c. h () log (7 ). Write the function as 7, then switch and and solve for. 7 7 7 log (7 ) So, h () log (7 ). Check our answer: h() 7 7 3; h (3) log (7 3) log log. 5. f ().5. The first differences are all, so the function is linear. Find an equation for the line through an two of the points. The slope of the line through the points (.,.5) and (.6, 5.5) is _ 5.5.5.6..5. Use point-slope form to find the equation: f ().5(.).5.5. 6 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
LESSON 7.7 SUPPORT EXAMPLES. i. 3 6 ( 3)()[ ( i )][ ( i )] ( 3)()( i )( i ) 3 i i i i 3 3 3 3 6 6 3 6 EXERCISES. For ad, identif the -intercepts. a. 5, 3, and 7 b. 6, 3,, and 6 c. 5 and d. 5, 3,,, and 6. a. (0, 05) b. (0, 08) c. (0, 00) d. (0, 90) 3. a. 3, because it has 3 -intercepts b., because it has -intercepts c., because it has -intercepts d. 5, because it has 5 -intercepts. a. ( 5)( 3)( 7). First use the zeros to start with the factored form a ( 5)( 3)( 7), and then use the -intercept point, (0, 05), to find a. 05 a (0 5)(0 3)(0 7) c. 0( 5)( ). First use the zeros to start with the factored form a( 5)( ), and then use the -intercept, (0, 00), to find a. 00 a(0 5)(0 ) 00 0a a 0 d. 0.5( 5)( 3)( )( )( 6). First use the zeros to start with the factored form a( 5)( 3)( )( )( 6), and then use the -intercept, (0, 90), to find a. 90 a(0 5)(0 3)(0 )(0 )(0 6) 90 360a a 0.5 5. a. a( ) where a 0 b. a( ) where a 0 c. a( ) 3 where a 0; or a( ) r r where a 0, and r and r are nonreal comple conjugates 6. a. (See table below.) b. i. ii. 05 05a a b. 0.5( 6)( 3)( )( 6). First use the zeros to start with the factored form a( 6)( 3)( )( 6), and then use the -intercept, (0, 08), to find a. iii. 08 a(0 6)(0 3)(0 )(0 6) 08 6a a 0.5 Lesson 7.7, Eercise 6a Polnomial function Degree Roots i. ii. two at 3, one at 5, one at iii. one at 3, two at 5, one at iv. a( 3) ( 5)( ) 5 v. 5 one at 3, one at 5, three at vi. 6 one at 3, two at 5, three at Discovering Advanced Algebra Solutions Manual CHAPTER 7 65 00 Ke Curriculum Press
iv. b. ( )( 5)( ). Start with the factored form a( )( 5)( ). Use the -intercept point, (0, 60), to find a. 60 a ()(5)() 60 80a a v. vi. c. A factor raised to the power of results in an -intercept that crosses the -ais. A factor raised to the power of results in an -intercept that touches but does not cross the -ais. A factor raised to the power of 3 results in an -intercept that crosses the -ais in a curved fashion. It appears that if the power of the factor is odd, the graph crosses the -ais, whereas if the power of the factor is even, it touches but does not cross the -ais. 7. a. b. 5. There are onl four -intercepts, but 5 is at least a double root because the graph touches the -ais at (5, 0) but does not cross it. c. ( 5) ( )( ). Use the zeros 0, 5 (double root),, and to start with the factored form a ( 0)( 5) ( )( ). Then use the point (, 6) to find a. c. a _ 3 _ 5, or a (3 )(5 ) where a 0. Start with the factored form a (3 )(5 ). Use the -intercept point, (0, 0), to find a. 0 a (0)()() 0 0 In this case, the a-value is not uniquel defined. An nonzero value for a will give the same -intercepts. d. ( 5i )( 5i )( ) 3 ( ), or ( 5)( ) 3 ( ). Comple roots alwas come in conjugate pairs, so 5i is also a root. Start with the factored form a ( 5i )( 5i )( ) 3 ( ). Use the -intercept point, (0, 00), to find a. 00 a (5i )(5i )() 3 () 00 00i a 00 00a a 9. The leading coefficient is equal to the -intercept divided b the product of the zeros if the degree of the function is even. The leading coefficient is equal to the -intercept divided b times the product of the zeros if the degree of the function is odd. Note: This solution fails if f (0) 0, because the denominator (the product of the roots) is zero. 0. Graphs will be a basic W shape, pointed either up or down. The points of the W can var in relative length or be basicall the same. In the etreme case, or, the W flattens to a U. Sample graphs: 6 a ( 5) ( )( 0)( ) 6 6a a 8. a. ( )( 5)( ). Start with the factored form a( )( 5)( ). Use the -intercept point, (0, 80), to find a. 80 a ()(5)() 80 80a a 66 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
. a. i. ( 5) ( )( ). 5 is a double root because the graph onl touches the -ais at (5, 0). The other roots are and. The graph opens upward, so the leading coefficient must be. ii. ( 5) ( )( ). This is the image of graph i after a reflection across the -ais, so replace with. ( 5) ( )( ) ( 5) ( )( ) iii. ( 5) ( )( ). 5 and are double roots because the graph onl touches the -ais at those -values. The other root is. The -values increase as the -values get increasingl more positive, and the -values decrease as the -values get decreasingl more negative, so the leading coefficient must be. iv. ( 5)( ) 3 ( ). is a triple root because the graph curves a bit as it crosses the -ais at (, 0). The other roots are 5 and. The -values decrease as the -values get increasingl more positive, and the -values increase as the -values get decreasingl more negative, so the leading coefficient must be. b. i. 5, 5,, and. a. i. ii. 5, 5,, and iii. 5, 5,,, and iv. 5,,,, and ii. iii. 5 iv. 5 b. i. 3. The graph has two local minimums, at (5, 0) and about (0, 50), and one local maimum, at about (3, 6). ii. 3. The graph has one local minimum, at about (3, 6), and two local maimums, at (5, 0) and about (0, 50). iii.. The graph has two local minimums, at about (3.3, 70) and (, 0), and two local maimums, at (5, 0) and about (0.7, 70). iv.. The graph has one local minimum, at about (.3, 5), and one local maimum, at about (0.3, 5). c. The number of etreme values of a polnomial function of degree n is at most n. d. i. n ii. n iii. n iv. n 3. a. Sample answer: The graph should have at most five -intercepts and at most four etreme values. b. Sample answer: The graph should have at most si -intercepts and at most five etreme values. c. Sample answer: The graph should have at most seven -intercepts and at most si etreme values.. a. 3 or 5. Use the quadratic formula. _ 3 69 0 6 _ or 5 3 3 7 _ 6 _ 30 6 or _ 6 b. 3_ or _ 3. Use the quadratic formula. _ 7 7 _ 8 or _ 3 or 3 c. For a, the factors of the constant term are,, 5, 0, and the factors of the leading coefficient are, 3. For b, the factors of the constant term are, 3, and the factors of the leading coefficient are,, 3, 6. Each root is the quotient of a factor of the constant term and a factor of the leading coefficient. Discovering Advanced Algebra Solutions Manual CHAPTER 7 67 00 Ke Curriculum Press
5. 3 5; 0 a 6 where a 0. Using the quadratic formula, the solutions to the equation a b c 0 are b _ b ac a _ b b ac a a This shows that if 3 5 is a root of a quadratic equation with rational coefficients, then the other root must be 3 5. An quadratic equation with these roots must be of the form 0 a 3 5 3 5 a 3 5 3 5 3 5 3 5 a { 3 5 3 5 9 5 5 5()} 0 a 6 6. a. Q (3) (3) (3) 0 9 6 0 3 b. Q 5 5 5 0 _ 5 5 c. Q 3 0 _ 5 3 3 0 8 6 0 36 8 d. Q ( 3i ) ( 3i ) ( 3i ) 0 6i 9i 6i 0 0 7. a. 5_, _ 3. Let matri [A] represent the coefficients, matri [X] represent the variables, and matri [B] represent the constant terms. 9 3 7 Find the inverse of matri [A] on our calculator. [A] _ 0 _ 3 0 _ 5 _ 5 Multipl [A] [B] to solve the sstem. _ 0 _ 5 _ 3 0 _ 5 7 0 5 0 _ 5 _ 5 3 The solution to the sstem is 5_ and _ 3. b. 5_, _ 3 Augmented matri R R R 0 9 3 9 5 7 0 3 5 R R R 0 R R 0 _ 5 R R 0 0 5 0 5 0 0 0 5 0 5 3 8. Approimatel 7.9 knots. Let l represent the blue whale s length in feet, and let S represent its speed in knots. Froude s Law states that S k _ l for some constant k. Use the point (75, 0) to find k. 0 k _ 75 k 5 0 3, so k 5 3 3 3. Thus the function relating a blue whale s length to its 3 speed is S _ 3 l, or S 3l 3. To find out how fast a 60-foot-long blue whale would swim, substitute 60 for l: S _ 3(60) 3 7.9. So a 60-ft blue whale would swim about 7.9 knots. EXTENSIONS A. See the solution to Take Another Look Activit on page 77. B. Answers will var. LESSON 7.8 SUPPORT EXAMPLE a. The other zeros are 0,, and. 7 3 6 () 3 7 6 3 7 6 3 0 ()( 3) ()( 3)( )( ) This ields zeros of 0, 3,, and. b. ()( 3)( ) EXERCISES. a. a 3 7 3. You first divide 3 3 b to get 3, and then ou divide 7 b to get 7, and finall divide 3 b to get 3. b. b 6 3. In the first step of the long division, ou multipl b the divisor, 3, to get 6 3.. a. 3 3 38 5 ( 5)3 7 3 b. 6 3 9 6 (3 ) 5 3 68 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press
3. a. a. Multipl the divisor,, b 3 to get. b. b. Add the column: 5 (3). c. c 7. The column must sum to : c 3, so c 7. d. d. The divisor, d, times the numbers in the bottom row must equal the net entr in the middle row. d, d 3, and d (), so d.. For ad, the dividend comes from the first line, the divisor comes from the upper left corner, and the quotient comes from the last line. a. 3 3 7 ( )3 b. 3 5 ( 3) 7 c. 3 8 7 6 (.5) d. 3 7 ( ) 3 5. 5, 5, 3,, 5, 5_, 3_, _. The numerator must be a factor of 5 and the denominator must be a factor of. 6. a. 7 3 b. P () ( )6 3 8. Use snthetic division to divide 6 5 3 7 5 b. It shows a remainder of. 6 5 7 5 6 8 6 8 c. P() ( ) 8 0. Use snthetic division to divide 3 0 6 b. It shows a remainder of 0. 0 6 6 8 0 7. a. (3i ) 3 (3i) 8(3i ) 9 5i 9 5i 9 0 Remember that i and i 3 i. b. 3i and. The conjugate pair ( 3i) and ( 3i) will be factors, so their product, 9, will also be a factor. Use long division to find the other factor. 9 ) 3 8 9 3 0 8 9 9 0 This shows 3 8 9 9( ), so the zeros are 3i and. 8. a.. There are four zeros because the function is a th-degree polnomial. b.,, 5, and. First look for rational zeros. Their denominators must be factors of the leading coefficient,, so an rational zeros are in fact integers that are factors of 0. The possibilities are,, 5, and 0. Tr dividing b one of these possibilities, such as. 3 3 0 7 0 7 0 0 The remainder is 0, so ( ) is a factor. Now tr dividing the quotient, 3 7 0, b another possibilit, such as. 7 0 0 6 5 0 The remainder is 0, so ( ) is also a factor. Now factor the quotient, 6 5 ( 5)( ). The zeros are,, 5, and. c. ( )( )( 5)( ) 9. a. The zeros are 0, 7, and. 3 5 () 5 ()( 7)( ) and this ields zeros of 0, 7, and. b. The zeros are 5, 6, and. 3 3 8 60 5 3 8 60 5 0 60 8 0 ( 5) 8 ( 5)( + 6)( + ) This ields zeros of 5, 6, and. 0. ( 3)( 5)( ), or ( 3)( 5) _. Use trial and error to find a zero from the list of possible rational zeros. Tr an integer first, such as 3. 3 3 3 5 6 7 5 9 5 0 Now tr 5. 5 9 5 0 5 0 These snthetic divisions show that 3 and 5 are zeros and that 3 3 3 5 ( 3)( 5)( ). Discovering Advanced Algebra Solutions Manual CHAPTER 7 69 00 Ke Curriculum Press
. For a d, first graph the function and approimate the -intercepts. If there are an nonreal roots (and ou ll know this because the graph has fewer -intercepts than the degree of the polnomial), use snthetic division to factor the polnomial. Finall, use the quadratic formula, or an other solution method, to find the zeros of an remaining quadratic factor. a.,, and i. The graph shows -intercepts at approimatel,, and. Now do snthetic division with these zeros to factor the polnomial. 0 0 6 6 6 6 3 6 8 0 3 6 8 8 8 0 8 0 8 0 0 So the factored form of the polnomial function is ( )( )( ). Find the zeros of the quadratic factor: 0, so, and i. The five zeros are,, and i. b. 7.0, 0.93, and 0.5. In graphing the function, it looks like one of the -intercepts is between 8 and 6, one is between and 0, and one is between 0 and. The three -intercepts are approimatel 7.0, 0.93, and 0.5. c. 6.605,.50, 7.556, and.669 0.7i. Note: While these zeros were found using unrounded values, the intermediate values shown in this solution are rounded approimations. In graphing the function, it looks like one of the -intercepts is between 6 and 8, one is between 0 and, and one is between 6 and 8. Multipling out 0.( ) 5 6( ) 3 ( ) will be cumbersome. Instead, notice that 0.( ) 5 6( ) 3 ( ) is a translation right units of the polnomial function 0. 5 6 3. The real zeros of 0. 5 6 3 are each less than the real zeros of 0.( ) 5 6( ) 3 ( ). 6.605 5.395.50 0.50 7.556 5.556 Use snthetic division to factor 0. 5 6 3. 5.395 0. 0 6 0.08 5.8 0.97 0.9 0..08 0.8 0.03 0.9 0 0.50 0..08 0.8 0.03 0.9 0.0 0.9 0.3 0.9 0. 0.98 0.67 0.37 0 5.556 0. 0.98 0.67 0.37. 0.7 0.37 0. 0.3 0.07 0 So, 0. 5 6 3 ( 5.395) ( 0.50)( 5.556)0. 0.3 0.05. Now use the quadratic formula to find the zeros of the quadratic factor. 0.3 _ (0.3) (0.)(0.07) (0.) 0.33 0.7i Now add to each of these zeros to account for the translation:.669 0.7i. The five zeros are approimatel 6.605,.50, 7.556, and.669 0.7i. d. 3.033,.63, and 0.3 0.8i. In graphing the function, it looks like one of the -intercepts is between 3.5 and.5 and that one is between and 3. 70 CHAPTER 7 Discovering Advanced Algebra Solutions Manual 00 Ke Curriculum Press