Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. Averge cost ($) C 8 6 4 O 4 6 8 Number of people number of hits.. number of times t bt.5 n n 4 b. 4.5 4.5.5; No, btting verge of.5 is the n 4 horizontl smptote of the grph. The reminder term gets closer nd closer to s n increses, n 4 but it is lws negtive for positive number of t-bts. number of hits c. number of times t bt.4 n n 5 Lesson 5. Grph Generl Rtionl Functions Guided Prctice for the lesson Grph Generl Rtionl Functions 4. The numertor hs no zeros, so there is no -intercept. The denomintor hs no rel zero, so there is no verticl smptote. m 5 nd n 5, so m < n nd 5 is horizontl smptote. 4 4. 5 5, so is n -intercept. 5 5 6, so 5 nd 5 re verticl smptotes. m 5 nd n 5, so m 5 n nd 5 5 is horizontl smptote. n 6 7 7 6. 5 5 5 5 6Ï, so 6Ï re -intercepts. The denomintor hs no rel zeros, so there is no verticl smptote. m 5 nd n 5, so m 5 n nd 5 is horizontl smptote. Ï 5 Ï 4. 4 5 ( )( ) 5 5 or 5, so nd re -intercepts. 4 5 5 4, so 5 4 is verticl smptote. m 5 nd n 5, so m > n nd the grph hs no horizontl smptote. 5 9 f() Algebr Worked-Out Solution Ke 45
5. V 5 πr h S 5 πr πrh 544 5 πr h 5 πr πr 544 πr 544 πr 5 h 5 πr 88 r S 5 πr 88 r Minimium X=4.495 Y=69.948 Using the minimum feture, ou get minimum vlue of bout 69, which occurs when r ø 4.4 nd 544 h ø ø 8.86. So, the soup cn using the lest π (4.4) mount of mteril hs rdius of bout 4.4 centimeters nd height of bout 8.86 centimeters. Eercises for the lesson Grph Generl Rtionl Functions Skill Prctice. The grph of rtionl function f hs no horizontl smptote when the degree of the function s numertor is greter thn the degree of its denomintor.. To find the -intercepts of the grph of f () p(), find q() the rel zeros of p(). To find the verticl smptotes of f, find the rel zeros of q().. C; The grph of hs no -intercepts, verticl 9 smptotes t 5 nd 5, nd horizontl smptote t 5 s shown in grph C. 4. A; The grph of hs -intercepts Ï nd Ï, no verticl smptotes, nd horizontl smptote t 5 s shown in grph A. 5. B; The grph of hs -intercept, verticl 4 smptotes t 5 nd 5, no horizontl smptote, nd end behvior the sme s the grph of 5 5 s shown in grph B. 6. A; The grph of hs no -intercepts, verticl 4 smptotes t 5 nd 5, nd horizontl smptote t 5 s shown in the given grph. 5 7. 5 ( )( ) The numertor hs no zeros, so there is no -intercept. The zeros of the denomintor re 6, so 5 nd 5 re verticl smptotes. 8. The zero of the numertor is, so is n -intercept. The denomintor hs no rel zeros, so there is no verticl smptote. 9. f () 9 9 ( 5)( ) The numertor hs no rel zeros, so there is no -intercept. The zeros of the denomintor re 5 nd, so 5 5 nd 5 re verticl smptotes.. 7 6 ( )( 5) 5 The zeros of the numertor re nd 5, so nd 5 re -intercepts. The zero of the denomintor is, so 5 is verticl smptote. ( )( 9) ( ). 7 The rel zero of the numertor is, so is n -intercept. The zeros of the denomintor re nd, so 5 nd re verticl smptotes.. g() ( 5)( 4) The zeros of the numertor re 5 nd 4, so 5 nd 4 re -intercepts. The denomintor hs no rel zeros, so there is no verticl smptote.. The verticl smptote occurs t the zeros of the denomintor, not the numertor. The denomintor 8 7 fctors s ( 7)( ). So, the zeros of the denomintor re 7 nd nd the verticl smptotes re t 5 7 nd 5. 4. C; The numertor nd denomintor hve the sme degree, so the horizontl smptote is m 4 b n 5 4. 5. 5 5 5, so is n -intercept. 5 5 6, so 5 nd 5 re verticl smptotes. m 5 nd n 5, so m < n nd 5 is horizontl smptote. 4 4 4 4 Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. 46 Algebr Worked-Out Solution Ke
8 6. 6 The numertor hs no zeros, so there is no -intercept. 6 5 ( )( ) 5 5 or 5, so 5 nd 5 re verticl smptotes. m 5 nd n 5, so m < n nd 5 is horizontl smptote. 5 5 6 4 4 Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. 4 4 6 7. f () 9 9 5 5 6, so nd re -intercepts. The denomintor hs no rel zeros, so there is no verticl smptote. m 5 nd n 5, so m 5 n nd is horizontl smptote. 8 9 8 8. 4 4 5 5 4, so 4 is n -intercept. 5 ( ) 5 5 or 5, so 5 nd 5 re verticl smptotes. m 5 nd n 5, so m < n nd 5 is horizontl smptote. f() 9. 8 8 5 ( 9)( ) 5 5 9 or 5, so 9 nd re -intercepts. 5, so is verticl smptote. m 5 nd n 5, so m > n nd the grph hs no horizontl smptote. 9 4 8 8 8. g() 8 6 8 5 5, so is n -intercept. 6 5 6Ï 6 5, so Ï 6 nd Ï 6 re verticl smptotes. m 5 nd n 5, so m > n nd the grph hs no horizontl smptote. Algebr Worked-Out Solution Ke 47
5 7 5 8 4 9 8. The numertor hs no rel zeros, so there is no -intercept. 4 g() 5 5, so 5 is verticl smptote. m 5 nd n 5, so m < n nd 5 is horizontl smptote.. h() 8 4 8 5 ( )( 4) 5 or 5 4, so nd 4 re -intercepts. The denomintor hs no rel zeros, so there is no verticl smptote. m 5 nd n 5, so m 5 n nd 5 5 is horizontl smptote. 6 4 4 4 h() 9 9. 5 6 5 6 5 ( 9)( 4) 5 5 9 or 5 4, so 9 nd 4 re -intercepts. 5 5, so 5 is verticl smptote. m 5 nd n 5, so m > n nd the grph hs no horizontl smptote. 9 4 7 9 4 4. Smple nswer: The grphs of nd 6 hve the sme end behvior s the grph 4 of 5 becuse the degree of the numertor is greter thn the degree of the denomintor nd m b n m n 5. 5 5. The rnge of the function is ll rel numbers such tht < 7.5. 6. 9 The rnge of the function is ll rel numbers ecept <. 7. The rnge of the function is ll rel numbers ecept 4.79 < <.9. Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. 48 Algebr Worked-Out Solution Ke
Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. 8. f () b f () 5 b b 5 5 b f () 5 b b 5 b b 5 5 b b 5 ( b) 5 () b 5 b 5 b 5 5 nd b 5 so f () 5. 9. f () b f () 5 () b 4 b 5 f () 5 b 5 4 5 4( b) b 4( b) 5 5 4( ) 4 b 4( b) 5 (4 b) 5 4() 4 4b 5 8 b 5 b 5 4 b 5 5 nd b 5, so f ().. f () b f () 5 () b 9 b 5 5 (9 b) f () b 4 b 5 6 (9 b) 5 6 5 [9 ()] 4 b (9 b) 5 6(4 b) 5 6 9 b 5 4 6b 5b 5 5 b 5 5 6 nd b 5, so f () 6. Problem Solving.. V 5 πr l 5 πr l π r 5 l b. S 5 πr πrl 5 πr πr πr 5 πr r c. S 5 πr r Minimium X=.55956 Y=9.654 Using the minimum feture, ou get minimum vlue of bout 9, which occurs when r ø.55 nd l ø ø 5.. So, the ble using the lest π (.55) mount of plstic hs rdius of bout.55 feet nd length of bout 5. feet... V Ï s h 4 Ï s h 4 p Ï Ï p Ï s h 6 Ï 5 s h 6 Ï s 5 h b. S Ï s 6sh Ï s 6s 6 Ï s Ï s 96 Ï s S Ï s 96 Ï s Minimium X=.856 Y=7.7684 Using the minimum feture, ou get minimum vlue of bout 8, which occurs when s ø. nd 6 h Ï (.) ø.97. So, the qurium using the lest mount of mteril hs side length of bout. feet nd height of bout.97 feet... Depth, d Temp., T 4.764 5 4.5796 4.494 5 4.55 4.44 5.967.889 Algebr Worked-Out Solution Ke 49
7,8d, b. T d 74d Temperture (8C) T 4.8 4.6 4.4 4. 4..8.6 d Depth (meters) From the grph, ou cn estimte tht the depth t which the men temperture is 48C is 8 meters. 54t 64 4.. n 6.6t Shres of stock sold (in billions) n 45 4 5 5 5 5 4 5 6 7 8 9 t Yers since 99 b. The number of shres of stock sold on the New York Stock Echnge incresed ech er from 99 to. c. Becuse the model represents the totl number of shres of stock sold ech er, ou need to look t the points on the grph for t 5,,, etc. From the grph, ou cn see tht n is less thn t t 5 nd n is greter thn t t 5. So ou cn estimte tht the er when the number of shres of stock sold ws first greter thn billion ws 996..99 5.. g 5 4 h (.8 7 )h (4.7 ) Accelertion due to grvit (m/s ) g 8 6 4 4 8 6 h Altitude bove se level (m).99 b. g 5 4 h (.8 7 )h (4.7 ).99 4 8 (.8 7 )(8) (4.7 ).99 4.99 ø 5 4.8 4.8 p 4 5.978 p 5 9.78 The ccelertion due to grvit t 8 meters is bout 9.78 meters per second squred. c. km 5 p 5, m.99 g 4 h (.8 7 )h (4.7 ) 6...99 4, (.8 7 )(,) (4.7 ).99 4.99 ø 4. 4. p 4 ø.947 p 5 9.47 The ccelertion due to grvit t kilometers is bout 9.47 meters per second squred. d. As ltitude increses, the ccelertion due to grvit g decreses slowl. Volume of tnk 5 Volume of outer tnk Volume of inner tnk 5 π(r ) (h ) πr h 5 π(r r )(h ) πr h 5 π(r h r rh r h ) πr h 5 πr h πr πrh πr πh π πr h 5 πr πrh πr πh π πr πr π 5 πrh πh πr πr π 5 h(πr π) πr πr π 5 h πr π b. V 5 πr h V 5 πr πr πr π πr π c. V 5 πr πr πr π πr π Mimum X=.74456 Y=64.946444 Using the mimum feture, ou get mimum vlue of bout 65, which occurs when r ø.74 nd h ø π r π r π ø.75. So the tnk hs π r π mimum volume when the rdius is bout.74 feet nd the height is bout.75 feet. Quiz for the lessons Model Inverse nd Joint Vrition, Grph Simple Rtionl Functions nd Grph Generl Rtionl Functions. 5 4 8 4 4 5 5 6 4 Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. 5 Algebr Worked-Out Solution Ke
. 5 5 8 f() 9 8 8 5 9 8 Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved.. 4 8 5 5 4 (4) 8 4 p (5) 5 4 5 4 5 4 4. 8 5 5 8 4 5 5 8 5 8 5. 4 6. 7. f () 6 6 5 ( ) 5 5, so 5 is verticl smptote. 5 c is horizontl smptote. 8. 8 The numertor hs no rel zeros, so there is no -intercept. 5 5 6, so 5 nd 5 re verticl smptotes. m 5 nd n 5, m < n nd 5 is horizontl smptote. 8 8 8 9. 6 6 5 5 6Ï 6, so Ï 6 nd Ï 6 re -intercepts. The denomintor hs no rel zeros, so there is no verticl smptote. m 5 nd n 5, so m 5 n nd 5 5 is horizontl smptote. 4 5 9 Ï 6 5 5 Ï 6 4 5 9 Algebr Worked-Out Solution Ke 5
. g() 8. 8 5 5, so is n -intercept. 5 5, so 5 is verticl smptote. m 5 nd n 5, so m > n nd the grph hs no horizontl smptote. 4 9 9 7 4 7 Strike percentge 5 Totl strikes 4 Totl pitches Strike percent (s deciml).9.8.7.6.5.4... 6 8 (7,.6) 4 Consecutive strikes (fter 8) g() From the grph, ou cn estimte tht the pitcher must throw 7 consecutive strikes to rech strike percentge of.6. Mied Review of Problem Solving for the lessons Model Inverse nd Joint Vrition, Grph Simple Rtionl Functions nd Grph Generl Rtionl Functions.. Let c be the verge cost nd p be the number of photos printed. Unit cost p Number printed Cost of printer c 5 Number printed.6p c 5 p b. Averge cost (dollrs) c 7 6 5 4 (, ) 4 8 6 4 p Number of photos From the grph ou cn estimte tht ou hve to print photos before the verge cost drops to $ per printed photo. c. As the number of photos printed increses, the verge cost decreses... V 5 πr h 66 5 πr h 66 πr 5 h b. S 5 πr πrh 5 πr πr 66 πr 5 πr 6 r c. S 5 πr 6 r Minimium X=6.4574 Y=777.88 Using the minimum feture, ou get minimum vlue of bout 777, which occurs when r ø 6.4 nd 66 h ø π (6.4) ø.84. So, the otmel cnister using the lest mount of mteril hs rdius of bout 6.4 centimeters nd height of bout.84 centimeters.. Yes; 5(8) 5 4 6(667) 5 4 7(57) 5 997 8(5) 5 4 Ech product is pproimtel equl to 4. So, the dt show inverse vrition. 4. Smple nswer: The function 5 6 hs domin of ll rel numbers ecept 5 nd 5. 5. 4 5 5 6.. b w h (5.).6 5..56 (.56) 5 5. 5. 5 5. 5 b w h b. b w h Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. 5 Algebr Worked-Out Solution Ke
Copright Houghton Mifflin Hrcourt Publishing Compn. All rights reserved. 45 h h 5 45 h 5.5 h 5.5 b 5 w h 4. 4.96 ø b w h 9 w.5 9 w. 9(.5) 5 w 4.75 5 w Bod mss inde Height (m) Weight (Kg).5 45.4 4 9.5 4.75 c. Let h equl the height of the shorter person nd let.h h 5.h be the height of the tller person. The bod mss inde of the tller person is given b w b 5 (.h) w nd the bod mss inde of the.h shorter person is b 5 w. So, the bod mss inde of h the shorter person is. times the bod mss inde of the tller person. 7. I d 5 5 I d ø.4 The intensit of the sound ou her is bout.4 wtts per squre meter if ou re 5 meters from the stge. 8.. M(t) 5 5 t M(5) 5 5 5 5 7 5 5 5 The motorccle s vlue is $ fter 5 ers. b. The horizontl smptote of the function is M 5 5, so the vlue of the motorccle pproches $5 s time psses. Lesson 5.4 Multipl nd Divide Rtionl Epressions Guided Prctice for the lesson Multipl nd Divide Rtionl Epressions ( ). ( )( ) ( ) ( )( ) 4. ( ) ( ) 5 p p ( ) ( ) 5 p ( ) 4. The epression cnnot be simplified. ( ) 4. 4 6 4 ( 4)( 4) 5. ( )( ) 6 ( )( ) 5 6. 6 ( 5) ( )( 5) 7. New tin: S 5 (s) 4(s)(h) 5 (4s ) 8sh 8. 5 8 p 6 5 8s 8sh V 5 (s) (h) 5 4s h S V 8s 8sh 4s(s h) s h 4s 5 h 4s(sh) sh 9 86 4 7 4 p 9 p 4 p p p p 9 p 4 p 4 p 9. p ( 5) ( 5)( 5) p p p ( ) 5. 5 ( 5)( ) ()()( 5)( 5) 5 ( 5) p 5 ( )( ) p ( ) ( 5)( ) ( )( ) 5 5 4. 5 6 8 4 5 p 6 8 4 ( 4)( ) p 5( 4) ( ) 4( 4)( ) 5( 4)( ) 5 4. 5 4 ( 6 5) 5 p 6 ( 5)( ) p 6 ( 5) ( 5)( ) 5 (6)()( 5) 6 Eercises for the lesson Multipl nd Divide Rtionl Epressions Skill Prctice. To divide one rtionl epression b nother, multipl the first rtionl epression b the reciprocl of the second rtionl epression. Algebr Worked-Out Solution Ke 5