Algebra 2B PAP Workbook Page 162 Answers€¦ · Algebra 2B PAP Workbook Page 167 . disadvantage is if the buyer would resell the car, they would lose more money compared to the exponential

Algebra 2B PAP Workbook Page 162

Answers Section 5.1

1. 25 2. -16384 3. 6427

4. 1125

5. 52.139 10× 6. 1.6 × 105 7. 2.0 × 10−3 8. 6

2xy

9. 7

62mn

− 10. 8 545c d 11. 61

w 12.

6

3 327rq s

13. 18b 14. 1 15. 14 29

x y 16. 12

1z

17. 2 22n nx y 18. 4 4n nx y+ − 19. 4

1a b

20. 2 2

2 2

a ba b+

21. 8 1x + 22. 25 n 23. 39b aba

24. 3

10x 25. 9 98 53 12x x

x− 26.

2 23 24

b a a ba− 27. 23 4a a− 28. 13302 (4 )a a−

29. 2 30. ½ 31. 25/9 32. ¼ 33. -8 34. 1 12 2 ( )a b a b− 35.

1 12 2 (1 ) a b ab

−−

Section 6.1

1. 633/5 2. (−25)4/3 3. 1247/6 4. ( )43 57− 5. ( )313 6. ( )58 204 7. 9 8. 64 9. 16

10. 216 11. 1125

12. 2 13. 18

− 14. 159,049

15. 13,125

− 16. 186.014 17. 47.287

18. 132.257 19. 0.144 20. 3.339 21. 5.819 or 0.181 22. −3.208 23. 2± 24. ∅ 25. 0.580 26. 6.191 cm 27. 1.809 cm 28.a. -2 b. 2 c. 2 d. 2 n na a≠ when n is

even and a is negative 29. a. 2.3% b. 5.1% ± 2.06 30. 3

3 3 3. 272

ia − ±

=

35 5 3. 125

2ib

− ±=

c. cube roots of 1 are 1, 1 3

2i− ± ; cube roots of 8 are 2,

2 2 3 1 32

i i− ±= − ± ; cube roots of 64 are 4, 4 4 3 2 2 3

2i i− ±

= − ± d. cube roots of 3a

are a , 32

a ai− ± ;

Section 6.2

1. 4400 5 2. 33/2 3. 76/5 4. 4 5. 24 6 6. 63

7. 4 349x x 8. 7035

9. 4 1331

10. 33 24

11. 2 12. 4 4

2

x xy

13. 3 3x 14. 4 1 5

3

8x yz

15. 6 8 312 729x y z

xy 16.

4

2 28y xyz

x z

17. 23 120x 18. 42 3 19. 2 233x z xyz 20. 8 x 21. 3 50 25

5x x+ 22. 24x z xyz

23. 5 24. 8 8y x 25. 6 33 2ab ab


Section 6.3 1. 2x2 + x − x1/2 − 4 2. 6x2 + 2x1/2 − 5 3. 4x2 + x − 5x1/2 + 1 4. 2x2 − x + 7x1/2 − 6

5. 2x2 − 4x1/2 + 5 6. −4x2 + x − 3x1/2 + 1 7. 5 31 3

33xx

− − 8. x5/2 + x1/2 9. −3x1/6

10. 7 3 1 3

3x x− − 11.

5 2 1 2x xx+ 12.

5 6

3x− 13.

1 22 xx− ; positive real numbers

14. 1 23

1x −

; all real numbers greater than 1 15. 22 7 5

9x x− + ; all real numbers

16. ( )1 22

1

2x x− ; all real numbers < 0 and <> 1

2 17.

22 13

x x− − ; all real numbers

18. 1 4x ; positive real numbers Section 6.4 1. 1 1 1(0) 3, ( 1) 0, (1) 4, (0) 1, (3) 0, and (4) 1f f f f f f− − −= − = = = − = =

1 1 1

-1

1 1 12. ( ( )) ( ) 1 1 1 , 1; ( ( )) ( 1)111

1 1 1 1, 0 Therefore, the inverse of f(x)= is f ( ) 1.1 1 x-11 1

f f x f x x x f f x fx x

x

x x xx

x x

− − −= = + = − + = ≠ = + =−

−

= = ≠ = ++ −

1 1

1

2 34 32 3 4(2 3) 3( 4) 8 12 3 12 543. ( ( )) ( ) , 4; 2 34 2( 4) (2 3) 2 8 2 3 524

4 32 34 3 2(4 3) 3(2 ) 8 6 6 3 52g( ( )) ( ) ,4 32 (4 3) 4(2 ) 4 3 8 4 542

xx x x x x xxg g x g x xxx x x x x

xx

x x x x x xxg x g xxx x x x xx

− −

−

+−+ + − + + − −+= = = = = = ≠ −++ + − + + − −−+

−+− − + − − + −−= = = = = =

−− − + − − + −+−

-1

2

2 3 4 3Therefore, the inverse of g(x)= is g ( ) .4 2

x

x xxx x

≠

+ −=

+ −

4. ( )1 2 21 1( ) (3 1)3 3

h h x x x x− = + − = = 5. 1 1 3( ) , : ( , ); : ( , )2 2

f x x D R− = − + −∞ ∞ −∞ ∞

6. 1 5 20( ) , : ( , ); : ( , )3 3

f x x D R− = − −∞ ∞ −∞ ∞ 7. 1( ) 3 , 0; 3f x x x y− = + ≥ ≥

8. 1( ) 3 , 0; 3 f x x x y− = − ≥ ≤ 9. 1 21 5 5( ) , 0;2 2 2

f x x x y− = + ≤ ≥

10. 1 71( ) 32 ; : ( , ); : ( , )2

f x x D R− = −∞ ∞ −∞ ∞ 11. 1 4 1( ) , ; 03 1 3

f x x yx

− = > >+

12. 1 2( ) 4 , 2 0;0 2f x x x y− = − − ≤ ≤ ≤ ≤ 13. 1( ) 2 ; 0; 2f x x x y− = − ≥ ≥ −

3 x


14. 1( ) 2 2 ; 2; 2f x x x y− = − + ≥ − ≤ 15. 1 2( 1)2, 1; ( ) , 1, 21

xx y f x x yx

− − +≠ − ≠ = ≠ ≠ −

−

16. 13 3 3 4 3 3, ; ( ) , ,2 2 2 3 2 2

xx y f x x yx

− +≠ ≠ = ≠ ≠

− 17. k = 3 18. 4

7k = − 19. 1

2k = −

20. 21.

( ] [ )[ ) [ )

[ ) [ )( ] [ )

2 2

1 1

( ) 6 10 ( ) 6 10: ,3 : 3,

: 1, : 1,

( ) 3 1 ( ) 3 1: 1, : 1,

: ,3 : 3,

f x x x f x x xD D

R R

f x x f x xD D

R R

− −

= − = = − =

−∞ ∞

∞ ∞

= − − = + −

∞ ∞

−∞ ∞

( ] [ )[ ) [ )

[ ) [ )( ] [ )

1 1

( ) 2 ( ) 2: , 2 : 2,

: 0, : 0,

( ) 2 ( ) 2: 0, : 0,

: , 2 : 2,

f x x f x xD D

R R

f x x f x xD D

R R

− −

= − − = +

−∞ − − ∞

∞ ∞

= − − = −

∞ ∞

−∞ − − ∞

Section 6.5 1. 2.

domain: x ≥ −4, range: y ≥ −2 domain: x ≥ 2, range: y ≥ 3 3. 4.

domain: x ≥ 0, range: y ≤ 3 domain: x ≥ −2, range: y ≤ 2 5. 6.

domain and range: all real numbers domain and range: all real numbers 7. 8.

domain and range: all real numbers domain and range: all real numbers


9. ; ; domain and range: all real numbers

10. 11. 12. 13.

O

O

O

O

14. 15. 16. 17.

O

Section 6.6

1. 152

2. 21 3. −4 4. ±5 5. 39± 6. ± 3 7. −6 8. 27 9. 120 10. 169

11. 39781

12. 32 3 13. 1,000 14. −6 15. 9 16. 33 6 6+ 17. no solution

18. 23± 19. 3 20. 73

21. −5 22. 3 23. no solution 24. 8116

25. 0

26. 34

− 27. no solution 28. 2512

29. 12 in. 30. 24 in 31. 3 22

± 32. 1 5 52

±

33. 1 34. 30, ,15

35. a) k =2 b) k<2 c) k>2 36. 9 4 66 1.56615

− +≈

Section 7.1 1. 2. 3.

( ) ( ): , ; : 3,D R−∞ ∞ − ∞ ( ) ( ): , ; : 1,D R−∞ ∞ ∞ ( ) ( ): , ; : ,3D R−∞ ∞ −∞


4. 5. 6.

( ) ( ): , ; : 5,D R−∞ ∞ − ∞ ( ) ( ): , ; : , 2D R−∞ ∞ −∞ ( ) ( ): , ; : 1.5,D R−∞ ∞ − ∞

7. ( )64 8xy = 8. ( )3 4xy = 9. ( )0.5 5xy = 10. ( )3 3xy = 11a.$8,396.71

b.$8,719.45 c.$8,121.19 14a. 2.13% b. y = 12,941,197 (1.0213)t c. 19,726,093 Section 7.2 1. decay 2. growth 3. growth 4. decay 5. decay 6. growth 7. 8. 9.

( ) ( ): , ; : 2,D R−∞ ∞ − ∞ ( ) ( ): , ; : 3,D R−∞ ∞ − ∞ ( ) ( ): , ; : 1,D R−∞ ∞ ∞

10. 11. 12.

( ) ( ): , ; : , 2.5D R−∞ ∞ −∞ ( ) ( ): , ; : 3,D R−∞ ∞ − ∞ ( ) ( ): , ; : 5,D R−∞ ∞ ∞

13. a. V = −6000t + 30,000 b. V = 30,000(0.775)t c.

During the first 2 years, the exponential model represents a faster depreciation. d. Book value after 1 year: Linear model: $24,000Exponential model: $23,250Book value after 3 years: Linear model: $12,000Exponential model: $13,965e. According to the linear model, the car will have no value after 5 years. According to the exponential model, there will never be a time when the car will have no value because the x-axis is a horizontal asymptote of the graph of the model. f. Linear model: An advantage for the buyer is the older the car, the less the buyer would have to pay compared to the exponential model. A


disadvantage is if the buyer would resell the car, they would lose more money compared to the exponential model. An advantage for the seller is during the first two years, the seller would be able to sell the car for more money compared to the exponential model. A disadvantage is after 5 years the car is not worth anything and would then be difficult to sell. Exponential model: An advantage for the buyer is during the first two years, the buyer would have to pay less for the car compared to the linear model. A disadvantage is if the buyer buys the car after two years, they would pay more compared to the linear model. An advantage for the buyer is that after 5 years the car is still valuable compared to the linear model. A disadvantage is that if the car were sold during the first two years, the seller would get less money for it compared to the linear model. 14a. V = 175,000(0.82)t b. $24,053.41 c. after 5 yr 15. V= 1600(0.8)t Section 7.3

1. 104e 2. 49e

3. 1227e 4. 632 xe 5. 9 3

8

xe − 6. 332 2x xe e

11. 12. 13.

( ) ( ): , ; : 3,D R−∞ ∞ − ∞ ( ) ( ): , ; : 1,D R−∞ ∞ ∞ ( ) ( ): , ; : 2,D R−∞ ∞ ∞

14. 15. 16.

( ) ( ): , ; : 1,D R−∞ ∞ ∞ ( ) ( ): , ; : 4,D R−∞ ∞ − ∞ ( ) ( ): , ; : 3,D R−∞ ∞ − ∞

17a. exponential growth b. c. 3 d. 23 e. 21

18. 2 2 2

2 2[ ( )]2 4

x x x xe e e ef x− − + + +

= =

; 2 2 2

2 2[ ( )]2 4

x x x xe e e eg x− − − − +

= =

2 2 2 22 2 2 2 4[ ( )] [ ( )] 1

4 4 4

x x x xe e e ef x g x− −+ + − +

− = − = =


Exponential Application Problems 1a. 192(0.933)Th = b. 40° is about 11.983 which is ¼ of 48 c. 60 3; 80 0.75° = ° =

2a. 2259 /865237076(10) TP −= b. 52.5 mm of mercury 3. $6414.27 4. $4204.80 5. $156.83 6. $5913.70 7. $4429.41 8. $4931.94 9. 7.25% 10. 28.285 million 11. 8417056 12. 0.49 ml 13. 3.259 g 14. 1128.59 g 15. 13.6% Section 7.4

1. 41/2 = 2 2. 34 = 81 3. 31 64

4

− =

4. −3 5. 32

6. 52

7. 23

8. 14

−

9. 32

10. π 11. 3π 12. 2π 13. 1254x −

14. etk

15. 21000π 16. 1( ) 7xf x− =

17. 1 3( )4

xf x− = 18.

21 1( )

2

xf x

−− =

19. 1 2( ) xf x e− −= 20. 1 3( ) 1xf x e− += +

21. 1 1 1( )3 3

f x x− = +

22. 23. 24.

( ) ( ): 2, ; : ,D R− ∞ −∞ ∞ ( ) ( ): 2, ; : ,D R− ∞ −∞ ∞ ( ) ( ): 1, ; : ,D R∞ −∞ ∞

25. log5 125 = 3 26. 27. ln 7.3890 = 2 28. ln 0.6065 = −0.5 29. 3 30. 3 31. 1 32. (0,1) (1, )x x x> − ≠ − ≥ ∞ Section 7.5

1. 3.332 2.1.659 3. 0.980 4. 3.738 5. 43log2

x + 6. log7 y + log7 z − log7 3

7. 2 log x + log y + log z 8. ln 7 + ln y − 2 ln x 9. 3 ln ln2

x y+ 10. 1log 2 log 2log2

x y+ −

11. log9 32 + 10 log9 x + 5log9 y +5 log9 z 12. 2 log6(x + 3) − log6 3 − 3 log6 y

13. ln (x2 − 1) − 4 ln x 14. 2 21ln 4 2ln ln(2 )

2y x y+ − + 15. 2

6 63 1log log ( 9)4 4

x x+ +

16. 35log3

17. 225log yx

18. 2 3

9logx y

19. 2

3

( 3)ln6( 2)

xx−+

20. 3

3

( 4)log

6x

x+

21. 2

2

3ln( 1)xx −

22. 1.869 23. −1.032 24. 2.665 25. 2

32

( 1) ( 8)ln( 1)

x xx x+ −

+

3 64 1

log 4 − =


26. ( )5 34log ( 2)( 2)x x x+ − 27.

2 6 12

2( 4) ( 1)108 log x x

x− −

+ 28. 3 3 3

49 6ln

125x y z

w v

29. –0.8 30. –0.5 31. 2 32. 16 33. 4 34. –2.5 35. 15. ln 2 ≈0.6931, ln 3 ≈ 1.0986, ln 4 ≈ 1.3862, ln 5 ≈ 1.6094, ln 6 ≈ 1.7917, ln 8 ≈ 2.0793, ln 9 ≈ 2.1972, ln 10 ≈ 2.3025, ln 12 ≈ 2.4848, ln 15 ≈ 2.7080, ln 16 ≈ 2.7724, ln 18 ≈ 2.8903, ln 20 ≈ 2.9956 Section 7.6

1) 2 2) 2,e e 3) 4 4) 72

5) 41 1 e− + + 6) 0 7) 14,4

8) ln(2 5)+

9) 152

10) 6 log38− + 11) 112

12) 2 ln 75 13) 21, e 14) 2 ln 5 2ln 3ln 5 3ln 3

+−

15) 1, 9, 5 34± 16) 1 17) 71,3

− 18) 16 19) 81 20) a) 0.567 b) 2.787

Logarithmic Application Problems 1. 7.727 year 2. 6.960 3a. ( )/ 32xy a= b. 4.755 hours c. 6 hours 4. 33830 years ago 5. 10,400 years 6. 209 days 7. no; 3561 years 8. 6028938764 years old 9. 8.711 min 10a. 18.616 b. it falls more slowly 11. 1:42 p.m. 12. 309.975 m 13. 100.7 = about 5

times more intense 14. 1396.683 volts 15. 7 16a. ( )1000 .06d

i = 16b. 9.821 m

17b. ( ) 0.034(1.947)tN t = 17c. 0.666( ) 0.03386 tN t e= 17e.1.847 17f. 6.2 hours 18a. ( ) 60(0.9919)td t = 18b. 126.85 kg 18c. 220 days 18d. y=0; no he will get closer and closer but never achieve his goal according to this model 19a. 20(1.0414)dp = 19b. blanks in order are 24.50, 36.74, 45, 55.11, 67.50, 82.67, 101.25, 124.01, 151.87 19c. (using 10 divisions) $3,263,569.80 19d. $3,263,569.80 Section 7.7

1. 1 (3)2

xy = 2. 1 (2)5

xy = 3. 3(4)xy = 4. 182

xy =

5. 25

3

xy =

6. 3 1

4 3

xy =

7. y = 37.069(6.848)x 8. y = 0.00188(1.327)x 9. y = 62.116(0.0302)x 10. y = 3x3

11. 212

y x= 12. y = 2x1.5 13. y = −4x2 14. 0.514

y x= 15. 2.512

y x= −

16. y = 12.936x3.3 18. y = 0.0138x1.05 19. y = 5.844(2.479)x 20. 0.846.2y x=

21. 1.5476(2.4257)xy = 22. 36y x= 23. 3(1.8)xy = 24. 0.9(4)xy =

25. 111 177220 220

y x= + ; prediction for 2020 = $6.86

Section 8.1 1. direct 2. neither 3. inverse 4. inverse 5. direct 6. neither 7. inverse

8. 8.64 ; 34.56yx

= 9. 15 ; 60yx

= 10. 5 ; 2.58

yx

= 11. 18 ; 540z xy=


12. 15 ; 2252

z xy= 13. 25 125;84 14

z xy= 14. 10 yr 15. 43725dp

= ; 3498 16. 4540.8

17a. 16560Vp

= b. c. 60 psi d. no, it will approach but never be 0

e. yes, that’s what inverse proportion means 18a. 2

4000sd

= b. 40 units c. 400,000 units

19a. intensity is inversely proportional to square of distance; 2

320Id

= b. 1.25 mr/hr; 3.2

mr/hr; 32,000 mr/hr c. 25.2982 meters 20a. 21150

b s= ; 730

r s= b. b(60) = 24 m;

r(60) = 14 m; b(120) = 96 m; r(120) = 28 m; b(180) = 216 m; r(180) = 42 m

c. 21 7150 30

t s s= + ; t(60) = 38 m; t(120) = 124 m; t(180) = 258 m d. 3.427 football fields

21a. 3 21.297d h= b. 458.530 cm; more than 10 times as big c. 83.243 m 22 a. 42500f pr= b. 163.84 cubic millimeters per second c. 244.14 units Section 8.2

1. x = 1; y = 5 2. x = −2; y = −3 3. 1 3;2 2

x y= = 4. 1 ; 24

x y= − = 5. x = 2; y = −1

6. 2 ; 53

x y= − =

7. 8. 9.

: 4; : 3D x R y≠ ≠ : 0; : 2D x R y≠ ≠ − 2: ; : 4

5D x R y≠ ≠

10. 11. 12.


: 1; : 1.5D x R y≠ − ≠ − : 0.25; : 1.25D x R y≠ − ≠ 2 4: ; :7 7

D x R y≠ ≠

13. 2 124

xyx−

=−

14. 6 143

xyx

− −=

+ 15. 0.75 12.5) ( ) ) 0 950

50xa C x b x

x+

= ≤ ≤+

c) Section 8.3

3 3 51. : 4, 4; int : 2, ; int : ; . : ; . : 4; f(x) does not5 16 2

74 have a removable discontinuity point; graph crosses its horizontal asymptote at = .7

D x x x y H A y V A x

x

≠ ≠ − − − − = = ±

−

32. : 3, 7; int : 3; int : ; . : 1; . : 7;f(x) has a7

3 removable discontinuity point (3, ); graph never crosses horizontal asymptote.5

D x x x y H A y V A x≠ ≠ − − − − = = −

33. : 2; int : 3; int : ; . : 0; . : None; f(x) has a 4

5 removable discontinuity point (2, ); graph crosses horizontal asymptote at (-3, 0).12

D x x y H A y V A≠ − − − − =

14. : 0; int : None; int : None; . : None; . : 0; Oblique Asy: 14

D x x y H A V A x y x≠ − − = = +

3 1 2 3 15. : ; int : , 2; int : ; . : None; . : ; Oblique Asy: 22 4 3 2 2

D x x y H A V A x y x≠ − − − = = −

6. : 0, 2, 1; int : 2; int : None; . : None; . : 0, 1; Oblique Asy: 1; removable discontinuity at (2,4)

D x x x x y H A V A x x y x≠ ≠ ≠ − − − − = = − = +

27. : 1; int : 0,4, 4; int : 0; . : None; . : 1; Oblique Asy:non-linear, 15D x x y H A V A x y x x≠ − − − = = + −

( )

5 58. : 1, ; int : 2.5; int : 1; . . : 0; . : 1, ; crosses 3 3

horizontal asymptote at 2.5,0

D x x x y H A y V A x x≠ ≠ − − − − − = = = −

−

59. : 6, 3; int : None; int : ; . . : 2; . : 6, 3; crosses 18

41 horizontal asymptote at ,26

D x x x y H A y V A x x≠ ≠ − − − − = = = −

−

510. : ( , ); int 1.58 ; int : ; . . : 1; . : None; does not cross 7

horizontal asymptote

D x y H A y V A−∞ ∞ − ≈ ± − − =

d) As the tank is filled, the rate at which the concentration of brine is increasing slows. The concentration of brine appears to approach 75%.


1.

O

2.

O

3.

O

4.

O

5.

O

6.

7. 8. 9.

10.

11. 2 20) ( ) 0.1 )3.169 cm )9.466 centsa C r r c dr

π= +

12. 2( 3)(2 54)) ( ) ) (0, ) ) 9 in. by 6 in. ) 96 inx xa A x b c dx

+ += ∞


Section 8.4

1. 72

xx+ 2.

2

2412

xx x

+− −

3. 2 3 9

3x x

x+ ++

4. 154

yx

5. 7 4

355

32x y

z 6. 2x

7. ( 5)( 4)3 ( 7)

x xx x+ −

+ 8. 2

2(2 3)( 2)x

x x++

9. ( 2)(3 4)(3 2)( 4)x xx x− −+ −

10. (2 5)( 4)3( 2)

x xx

+ −+

11. 1

xx +

12. 2 1

4 ( 2)xx x

++

13. ( 6)(2 1)( 2)

x xx x

−− −

14. ( 2)( 6)2

x xx

+ + 15. 21

xx +

16. (4 1)(5 6)( 2)(4 5)( 1)( 6)

x x xx x x+ + +

−+ + −

17. 35( 5)

4tt+ 18.

4

2( 1)nx

x + 19. ( 8)( 5)

( 1)

n n

n nx x

x x− −

+

20. a) 8π b) about 16,791.045 gal c) 30 ft by 15 ft by 5 ft

21. 1x h x

−− +

22. 11 1x h x+ + + +

23. 12 2x h x+ − + +

Section 8.5 1.2x(x + 3)(x − 3) 2. 2(x − 3)(x+ 4) 3. 3(x − 1)(x − 2)(x + 3) 4. (x + 1)(x − 3)(x + 5)(x + 6)

5. 4( 5)( 2)

xx x

++ −

6. 2 13

3( 3)( 2)x xx x− +− −

7. 2

33 2 3x x

x+ − 8. 6

1xx+

−+

9. 2 11 173 ( 2)

x xx x+ +

+

10. 22

( 1) ( 1)x

x x−

+ − 11.

27 22 93(2 1)( 3)

x xx x− −+ −

12. 2

27 13 1

2( 1)( 1)x x

x x x− +

− + + 13. 6

6x − 14. 2

5 1( 1)

xx

−−

15. 2( 3)( 5)

10 26x x

x x− +− −

16. 2

22 7

( 1)( 2)x xx x

− −+ −

17. 112x

− 18. 2

4 21t

t t+

−+

19. 36x xx+

20. (2 4)( 4)3

x x− + 21. System of eq. is 0

04

A B CB C

A

+ + = − + = − =

; solution is A = −4, B = 2, C = 2

22 a. domain: 1x ≠ , range: 0y ≠ b. 1( ( )) xf f xx−

= c. ( ( ( )))f f f x x= ; a line with

holes at (0, 0) and (1, 1) Section 8.6 1. 1 2. 6 3. 5 or 2 4. -8 5. -18 6. -2 7. no solution 8. 0 9. no solution 10. -9 or 3

11. -1 or -3 12. 2.5 or 0.5 13. 3 14. 6 or 12 15. no solution 16. 4( 1)( 2)( 3)( 4)

x xx x+ −+ +

Applications of Rational Functions

1a. 550 92 15 128.6715+ ⋅

= b. 550 92( ) nC nn+

= c. asymptotes 0n = and 92C =

d. The yearly expense of electricity continues no matter how many years the refrigerator works. The cost will never go below the $92, but the cost approaches $92.


e. Graphing the two functions 550 92( ) nC nn+

= and 21200 92( ) nC n

n+

= together or

reviewing a table of values will show the more expensive refrigerator remains more expensive annually although both approach $92 as n approaches infinity. 2a.

b. 13.76 micrograms which occurs at 18.2 minutes c. The graph shows that the concentration is 0 at time 0= . Within the first 15 minutes, the concentration rises sharply to the maximum at 18.2 minutes. After reaching the maximum, the kidneys begin cleansing the blood and rapidly remove the drug. d. There is only one asymptote in this rational function which is at 0C = . Oddly, this value actually exists in the range of the function at

0C = ; however, the graph tends toward 0C = as t − values increase to infinity. This characteristic gives the asymptotic behavior. e. Does not change asymptotes, but neg values of t make no sense in this problem

3a. The domains for both are 0x ≠ . 3b. asymptotes for both are 0x = and 5y = 3c. Both Tiffany and Adam are graphing the same function

4a. 30 mph 3.33 hours; 55 mph 1.82 hours; 65 mph 1.53 hours. 4b. 100tr

=

4c. Faster you travel, fewer hours on the road 5. If x is the number, then 1x

is the

reciprocal. Therefore the sum of a number and its reciprocal is represented by 1xx

+ .

Graph the function 1y xx

= + for positive x-values. The minimum occurs at the point (1,2).

Solving Rational Inequalities 1. 2x > − 2. ( , 5) [ 4.5, )−∞ − − ∞ 3. ( 4, 3) ( 2, )− − − ∞ 4. ( , 6) ( 4,4)−∞ − − 5. ( 3, 2) [2, )− − ∞ 6. ( 6, 2.4) (3, )− − ∞ 7. 2034 to 2044, non-inclusive 8. 250m > Section 9.1

1. (7,0) 2. (9, 3)− 3. 1 1(2 , 2 )m mx x y y− − 4. 7 7 5 3 13 5, ; , ; ,4 4 2 2 4 4

− − −

5. 3 9 3 1 3, ; 1, ; ,2 4 2 2 4

− − − − − −

6. 1 2 1 2 1 2 1 2 1 2 1 23 3 3 3, ; , ; ,4 4 2 2 4 4

x x y y x x y y x x y y+ + + + + +

7a. ( ) ( )2 2

2 22 1 2 12 1 2 1

13 3 3

x x y yd x x y y− − = + = − + −

b. 1 2 1 22 2,3 3

x x y y+ +


8. 5 42, ; 3,3 3

− −

9. 4 2, 2 ; , 13 3

− − − −

10a. 15 3 55= 10b.

2

3 3

1

m

m

− −

+

10d. The graph approaches y = 3. 11a. y x= 11b. 1y = 11c. 10 5( 4)3 2

y x= − +

12. 4 or 8y = − 13. 11 or 13x = − 14. 4 609y = ± 15. 17, 20 and 29

16. 2 triangles, vertices ( )2 3, 2± 17a. 50t 17b. 250 miles 17c. after 2 hours 17d. 50 mph

Parabolas - Sections 9.2 and 9.6 1. 2 16y x= 2. 2 12x y= − 3. 2 16x y= − 4. 2 14( 0.5)x y= − − 5. 2( 2) 8( 3)x y− = − + 6. 2( 2) 8( 4)y x+ = − 7. 2( 3) 4( 3)x y+ = − 8. 2( 2) 16( 1)y x+ = − − 9. 2 4.5y x= 10. (0,0); (0, 1); : 1V F d y− = 11. (0,0); (0.75,0); : 0.75V F d x = 12. (2, 3); (2, 4); : 2V F d y− − = − 13. ( 5, 3); ( 5.5, 3); : 4.5V F d x− − − − = − 14. (0, 2); ( 1, 2); : 1V F d x− = 15. ( 3, 2); ( 3, 1); : 3V F d y− − − − = −

16. 2( 1) 4( 3)y x− = + 17a. 2 49( 3) ( 4)6

x y− = − − 17b. 233,124

17c. 1624

y = 17d.

19324

17e. 19324

17f. They are the same 18. 5 feet 19. 22304 7238.23 ftπ ≈ 20. 4 2y x= −

21a. 2 640x y= − 21b. 8 feet 22a. 2x t= 22b.see table below 22c. yes t -3 -2 -1 0 1 2 3 x 9 4 1 0 1 4 9 y -3 -2 -1 0 1 2 3

23a. 2 20.5( 1) 1 0.5 0.5y t t t= − − = − − 23b. see table below 23c. yes t -3 -2 -1 0 1 2 3 x -3 -2 -1 0 1 2 3 y 7 3.5 1 -0.5 -1 -0.5 1

Circles - Section 9.3 and 9.6 1. 2 2( 2) ( 2) 5x y− + − = 2. 2 2( 3) ( 1) 9x y+ + − = 3. 2 2( 3) ( 1) 49x y+ + − =

4. 2 2( 4) ( 5) 25x y− + + = 5a. 22 2 ( 1)4

y x− = − 5b. 33 ( 5)4

y x− = +

6a. ( 2, 1); 5C r− − = 6b. int . 2 2 6; int . 1 21x y− = − ± − = − ± 6c. :[ 7,3]; :[ 4,6]D R− −

6d. not a function 7. 16 7(0, 3) and ,5 5

− − −

8a. 2( ) (4 ) 16A x x x= + − 8b. 2x =

9. 10. 11. 12.


13a. length = 2x; width = 22 2 36y x= − 13b. 24 36A x x= − 13c. 72 units2 14a. slope = -2/3; equation of line: y = (3/2)x 14b. (6, 9) 14c. 2 2 117x y+ =

15a. 5 16912 12

y x= − 15b. 212 169

5xxm

x− −

=−

15d. xm approaches a value of about 0.4;

this is almost equal to the slope 16. 2 22 41 1300

7 7 49x y + + − =

Ellipses - Section 9.4 and 9.6

1. 2.

3. 4.

5. 2 2

19 49x y

+ = 6. 2 2

1169 31x y

+ = 7. 2 2

136 24x y

+ = 8. 2 2

1169 81x y

+ =

9. 2 2

116 36x y

+ = 10. 2 2 2 2

2 22 2

4 4 41 or 19 194 4

x y x yh h h h

+ = + =

#11 #12 #13 #14 standard equation

11625

22=+

yx 19

)4(4

)3( 22=

−+

− yx

125100

22=+

yx 1425

22>+

yx

x–radius

5 2 10 5

y–radius

4 3 5 2

focal radius

3 5 35 21

center (0, 0) (3, 4) (0, 0) (0, 0) vertices (5, 0), (–5, 0),

(0, 4), (0, –4) (5, 4), (1, 4), (3, 1), (3, 7)

(10, 0), (–10, 0), (0, 5), (0, –5)

(5, 0), (–5, 0), (0, 2), (0, –2)

foci (3, 0), (–3, 0) (3, 54 ± ) ( 35± , 0) ( 21± , 0) x–int. (5, 0), (–5, 0) none (10, 0) (–10, 0) (5, 0) (–5, 0) y–int. (0, 4), (0, –4) none (0, 5) (0, –5) (0, 2) (0, –2)

vertices: (± 9/2, 0); co-vertices: (0, ± 5/3);

foci: 629( ,0)6±

vertices: ( 10,0)± co-vertices: (0, 2 2)± ; foci: ( 2,0)±

vertices: (0, ± 5/2); co-vertices: (± 4/3, 0);

foci: 161(0, )6±

vertices: (± 3, 0); co-vertices: (0, ± 1); foci: ( 2 2,0)±


#15 #16 #17 #18 standard equation 416

)3( 22=+

+ yx

19

)2(16

)1( 22=

−+

+ yx

1

9)2(

4)5( 22

=−

++ yx

14

)3(36

)6( 22

<+

++ yx

x–radius 4 4 2 6 y–radius 2 3 3 2 focal radius

32 7 5 24

center (–3, 0) (–1, 2) (–5, 2) (–6, –3) vertices (1, 0), (–7,

0), (–3, 2), (–3, –2)

(–5, 2), (3, 2), (–1, 5), (–1, –1)

(–3, 2), (–7, 2), (–5, 5), (–5, –1)

(0, –3), (–12, –3), (–6, –5), (–6, –1)

foci (–3 32± , 0)

( 71±− , 2) ( )52,5 ±− ( 246 ±− , –3)

x–int. (1, 0), (–7, 0)

±−0,

3

543

±−0,

3

5215 none

y–int (0,

4

7± )

±

4

1538,0

none (0, –3)

11. 12. 13.

14. 15. 16.

17. 18.

.


19. 19

)12(5

)4( 22=

−+

+ yx 20. 11216

22=+

yx 21. 125

)7(4

)5( 22=

−+

+ yx

22. 198

22=+

yx 23. 19

)2(25

)1( 22=

−+

− yx 24.

636

36

49

2

2

2

==

=⋅

⋅⋅=⋅

rr

r

r

ππ

ππ

25a. (20 )A a aπ= − 25b. 2 2

1196 36x y

+ = 25c. see table below 26. 32

a 7 8 9 10 11 12 13

A 285.88 301.59 311.02 314.16 311.02 301.59 285.88

25c. Because the maximum area occurs when a = 10 and a + b = 20, b is also 10. So, the

equation becomes 2 2

2 21 100100 100x y x y+ = → + = which is the equation of a circle.

Hyperbolas - Lesson 9.5 and 9.6

31. (0,0); . : ; ( 5,0); ( 34,0); . :5

2. (0,0); . : ; (0, 4); (0, 2 5); . : 2

3. (0,0); . : ; (0, 3); (0, 10); . : 3

4. (0,0); . : ; ( 2,0); ( 6,0

C Tran axis x axis V F Asy y x

C Tran axis y axis V F Asy y x

C Tran axis y axis V F Asy y x

C Tran axis x axis V F

− ± ± = ±

− ± ± = ±

− ± ± = ±

− ± ± ); . : 2Asy y x= ±

1.

O

2.

O

3.

O

4.

O


2 2 2 2 2 2 2 2 2 2

2 2 22 2 2

2 2 2 2

( 4) ( 3)5. 1 6. 1 7. 1 8. 1 9. 116 20 4 5 36 9 9 16 4 12

( 4) ( 7) ( 4)10. ( 1) 1 11. ( 5) 1 12. ( 3) 18 3 8

( 1) ( 1) ( 1) 9( 1)13. 1 14. 14 9 4 16

15. ( 3,1); . : 1; ( 7,1),

y x x y y x x y y x

y y xx x y

x y y x

C Tran axis y V

+ +− = − = − = − = − =

− − +− − = − − = − − =

− + + −− = − =

− = −

22

5(1,1); ( 3 41,1); . : 1 ( 3)4

916. ( 1, 2); . : 1; ( 1, 11), ( 1,7); ( 1, 2 106); . : 2 ( 1)5

( 2)17. ( 1) 1; ( 1, 2); . : 2; ( 2, 2), (0, 2); ( 1 3, 2);2

(. : 2 2( 1) 18.

F Asy y x

C Tran axis x V F Asy y x

yx C Tran axis y V F

yAsy y x

− ± − = ± +

− − = − − − − − − ± + = ± +

++ − = − − = − − − − − ± −

+ = ± +2

21) ( 2) 1; ( 2,1); . : 2;4

( 2,3), ( 2, 1); ( 2,1 5); . : 1 2( 2)19. ) ( 5,0), (5,0) ) 2 2 4 8, 8 5 8 13

x C Tran axis x

V F Asy y xa Q R b PR PQ a then PR PQ

−− + = − = −

− − − − ± − = ± +

− − = = ⋅ = = + = + =20 21. 22. 23.

. 24. 25. 26.

#20 #21 #22 #23 standard equation

2 21

25 16

x y− =

2 21

9 16

x y+− =

2 2( 2) ( 3) 136 25

x y−

− −=

2 2( 3) ( 6) 136 64

x y− +

− +=

x–radius 5 3 6 6 y–radius 4 4 5 8 focal radius 41 5 61 10 center (0, 0) (0, 0) (2, 3) (3, –6)


vertices ( 5± , 0) ( )0, 4± (8, 3) and (–4, 3) (3, 2) and (3, –14) foci ( 41± , 0) ( )0, 5± (2 61± , 0) (3, 4) and (3, –16) transverse axis

horiz., length = 10

vert., length = 8

horiz., length = 12

vert., length = 16

conjugate axis

vert., length = 8

horiz., length = 6

vert. , length = 10 horiz. , length = 12

asymptotes 45

y x= ± 43

y x= ± ( )53 26

y x− = ± − ( )46 33

y x+ = ± −

x–intercepts

(5, 0), (–5, 0)

none 6 342 ,05

±

none

y–intercepts

none (0, 4), (0, –4) none ( )0, 6 3 10− ±

#24 #25 #26 standard equation

2 2( 1) ( 4) 116 1

x y−

− −=

2 2( 3) ( 2) 116 36

x y−

− +=

2 21

144 25x y

− =

x–radius 4 4 12 y–radius 1 6 5 focal radius 17 2 13 13 center (1, 4) (3, –2) (0, 0) vertices (5, 4) and (–3, 4) (7, –2) and (–1, –2) (12, 0) and (–12, 0) foci ( 5 17± , 4) ( 3 2 13± , 4) (13, 0) and (–13, 0) transverse axis

horiz., length = 8 horiz., length = 8 horiz., length = 24

conjugate axis

vert. , length = 2 vert. , length = 122 vert., length = 10

asymptotes ( )16 14

y x− = ± − ( )32 32

y x+ = ± − 512

y x= ±

x–intercepts

(2, 0) and (0, 0) 4 103 ,03

±

(12, 0) and (–12, 0)

y–intercepts

none none none

27. 2 2

19 7x y

− = 28. 2 2

116 4x y

− = 29. 2 2( 2) ( 1) 1

39 25x y

− ++ −

= 30. 2 2( 3) ( 3) 1

9 16x y

− +− −

=

31. 2 2( 2) ( 5) 1

36 64x y

−+ −

=

Conics Challenges and Extras 1. 2 4 ( )y h x h= − − , where h is the x-coordinate of the vertex and the focal radius


2. 2 2( 6) ( 2) 1

9 7x y− −

− = 3. if 8, then the graph would be the point (1, – 2); if greater than

8, there would be no graph since the equation would have no solution

4. 2 (3 3) 12x y+ = 5. 250,3

6. length = 4p

7a. Endpoints of latera recta:

9 9 9 9, 7 , , 7 ; , 7 ; , 74 4 4 4

− − − −

7b. Endpoints of latera recta:

4 4 4 4, 5 , , 5 ; , 5 ; , 53 3 3 3

− − − −

7c. Endpoints of latera recta:

3 5 3 5 3 5 3 52, , 2, ; 2, ; 2,5 5 5 5

− − − −

7d. First solve for y in the equation 2 2

2 2 1x ya b

+ = to obtain the positive solution

2

21 xy ba

= − . This represents half the length of the latus rectum. Because the focus

( ,0)c and the latus rectum directly above the focus have the same x–coordinate, substitute c

for x in the equation for y to get 2

21 cy ba

= ± − . Add the expressions underneath the radical

to get 2 2

2

a cy ba−

= ± . Then notice that for an ellipse, a2 − c2 = b2. So, the equation for y

simplifies to 2 2

2

b by ba a

= ± = ± . Therefore, the length of the latus rectum is 22b

a 8. The

eccentricity is cea

= . Solve this equation for c and then square each side to obtain 2 2 2e a c= . You know that for an ellipse, 2 2 2c a b= − . Substitute e2a2 for c2 and solve for b2

to obtain 2 2 2(1 )b a e= − . Now substitute this expression for b2 in the equation for an ellipse


to obtain 2 2

2 2 2 1(1 )

x ya a e

+ =−

. As e approaches 0 and a remains fixed, the ellipse approaches

the shape of a circle. 9a. apogee: about 405,508 km 9b. perigee: about 363,292 km 9c. The apogee A is A = a + c. The perigee P is P = a − c.

Substitute these values for A and P into ( ) 22

A P a c a c c ceA P a c a c a a− + − −

= = = =+ + + −

Lesson 9.7 1. (5, −9), (−3, 7) 2. (3, −1), (−1, 3) 3. no points of intersection 4. ( 10,2), ( 10,2)−

5. 9 4 6 9 4 6, , ,5 5 5 5

−

6. (1.5, 1.5), ( 6, 69)− − − 7. (), (), (), () 8. about 53.2 mi

9. yes; about 2.39 ft by 2.39 ft by 6.13 ft; or 4.35 ft by 4.35 ft by 1.85 ft 10. 4 5b = ± Conics – Systems of Inequalities See teacher for correct results Section 12.1

1 2 3 4 5 2 1 2 1 1 11. ) 0, , , , , ) 2, 1, , , , ) 1, 2, , 4, ,62 3 4 5 6 3 2 5 3 3 5

a b c− − −

2 11

1 12. ) ) ) 2 1 ) ( 1) ) ( 1)3

nn n

n n n n nn

ea a b a c a n d a n e an n

+−= = = − = − ⋅ = −

1

92

-11 1 1

1 1 1 1 13. ) 1,2,6,24,120,720 ) 1,1,2,3,5,8 4. ) 1 ...2 3 -1

1) ! 1 2! 3! ... ( -1)! ! 5. ) ) 6. ) 45 ) 39 ) 322

n

k

n n

nk k k

a b ak n n

b k n n a k b a b c

=

= = =

= + + + + +

= + + + + + −

∑

∑ ∑ ∑

Section 12.2

1 11. arithmetic 2. 3( 1) 5 (3 5) 3, 5, 3n ns s n n s d+ − = + + − + = = = .

13 13. 50 4. ) 96, 3 ) 96 ( 1)( 3) 3 99 na a a d b a n n= = = − = + − − = − + 21( ) 3 195) 5. 1305 6. 2360 7. 1

2 2 2n

nn a ac S n n x+

= = − + =

Section 12.3 2 3 15 15

2 3 15 15 15

2 12 1

3 3 3 3 1 1 3(1 3 ) 1 11. ... (3 3 3 ... 3 ) (1 3 ) (3 1)9 9 9 9 9 9 1 3 6 61 5 5 5 1 1 (1 5 ) 1 12. ... (1 5 5 ... 5 ) (1 5 ) (5 1)9 9 9 9 9 9 1 5 36 36

n nn n n

or

or−

−

−+ + + + = + + + + = ⋅ = − − −

−−

+ + + + = + + + + = ⋅ = − − −−

1615

15 15 16 16

4 (1 3 )4(1 3 ) 2 233. 2(1 3 ) 2(3 1) 4. (1 3 ) (3 1) 1 3 1 3 3 3

or or−−

= − − − = − − −− −


44 1 5

3 9 27 1 1 4 5 205. 1 ... 6. 7. 1, -43 3 14 16 64 7 31 ( ) 1 14 4 4

5 (1 5 ) 1 18. a) r = 5, 5 5 5 b) (1 5 ) (5 1)1 5 2500 2500

nn n n n

n na S or−

− − −

− + − + = = = =− − + −

−= ⋅ = = = − − −

−

Section 12.4

1. 2 2. 7556

3. 552

− 4. 10 5. doesn’t exist 6. 32

7. 3 8. doesn’t exist

9. –72 10. 162 11. doesn’t exist 12. 2 4 and 3x x< < ≠ 13. 3 1 and 2x x− < < − ≠ −

14. 1 52+ 15. 2 16. a) 11000(1.03)n

na −= b) 150(1.12)nnp −=

c) 1

11000(1.03)

50(1.12)

n

n

−

− ; Since the price of the shares are increasing faster than the investment

amount, the number of shares you are able to buy will decrease from the initial 20 shares each year. Section 12.5 1. 9, 37, 149, 597, 2389, 9557 2. –4, 0, 9, 25, 50, 86 3. 3, 3, 3, 2, 0, –4 4. 1 15, 7n na a a −= − = + 5. 1 17, 3n na a a −= = 6. ( )( )1 2 2 12, 3, n n na a a a a− −= − = = 7. 1 15, n na a na −= = 8. 0 1 0 2 1 3 22, ( ) 1, ( ) 8, ( ) 19,x x f x x f x x f x= = = − = = = = −

4 3( ) 62x f x= = ; the four iterates are 1,8, 19,62− − 9. 0 1 05, ( ) 2,x x f x= = = −

2 1 3 2 4 3( ) 12, ( ) 82, ( ) 6312x f x x f x x f x= = = = = = ; the four iterates are 2,12,82,6312−

10. 0 1 0 2 1 3 2 4 35 27 472, ( ) 3, ( ) , ( ) , ( )2 10 18

x x f x x f x x f x x f x= = = = = = = = = ; the four

iterates are 5 27 473, , ,2 10 18

11. 0 1 0 2 1 3 21 113, ( ) 2, ( ) , ( ) ,2 2

x x f x x f x x f x= = = = = = = −

4 3109( )22

x f x= = − ; the four iterates are 1 11 1092, , ,2 2 22− −

12a. 1 1 1 12000 4210; 0.0035 0.99653500 7n n n nw w w w w− − −= = − + = +

12b. 19 18 17 16 1520

4 4 4 4 4(210)(.9965) (.9965) (.9965) (.9965) (.9965) ...7 7 7 7 7

w = + + + + + + =

1819

0

4210(.9965) (.9965 ) 206.9887

n

n=

+ =

∑ 13. 3

1 4 3 42 144a a a−= = ⋅ =

14. 61 6 5 5

45 93125 6255

aa a −= = = =


15.

1

2

3

4

11

5 11

1 1 31 1.52 1 21 3 2 17 1.4162 2 3 121 17 12 577 1.4142152 12 17 4081 577 408 665857 1.4142142 408 577 4708321 665857 470832 8.86731 10 1.4142142 470832 665857 6.27014 10

As approaches , approacn

a

a

a

a

a

n a

= ⋅ + = =

= ⋅ + = =

= ⋅ + = ≈

= ⋅ + = ≈

×= ⋅ + ≈ ≈

×∞ hes 2.

Area Under Curve – Now You Try LHR = 15750 un2; RHR = 27750 un2; MPR = 20875 un2; Trapezoid = 21500 un2 Area Estimation Exercises #1 1. 24 un2 2. 7.085 un2 3. 8.198 un2 4. 1.885 un2 5. 171.680 ft2 6. 33.279 grams 7a. total cost 7b. $250,000; under 76c. $437,500; over 7d. $334,375; slightly under 7e. $343,750; slightly over 8. 6,086.4 in2 Area Estimation Exercises #2 1a. 3.75 b. 5.75 c. 4.625 d. 4.75 2. 6.173 3b. total cars entered parking lot c. $1810 4. 50,000 barrels 5. 1900 cm3 6. 140/3 miles 7a. 113.54 ft b. 150.46 ft c. 135.6 ft d. 131.99 ft 8a. 255 b. 255 c. 258.6 d. 255 Position/Velocity/Acceleration Worksheet #1

1a.

20 0 360 3 8

( ) 60 540 8 915 135 9 11

10 140 11 14

t tt

v t t tt t

t t

≤ ≤ ≤ ≤= − + ≤ ≤− + ≤ ≤

− ≤ ≤

1b.

20 0 30 3 8

( ) 60 8 915 9 11

10 11 14

tt

a t ttt

≤ ≤ ≤ ≤= − ≤ ≤− ≤ ≤

≤ ≤

1c. [0,3];[9,11] 1d. [8,9];[11,14] 1e. 420 inches 1f. 75 inches 1g. 289 inches 1h. 495 inches 2a. –14 2b. at –14 cm 2c. t = 7 seconds; t = 1 second 2d. 18 cm 2e. –32 cm 2f. vertex (4, 18); t-int = 1 and 7; x(t)-intercept = –14 2g. [0, 4]; positive; positive; move to right 2h. [4, ]∞ ; negative; negative; move to left 2i. [0,1);(7, )∞ ; left of origin 2j. (1, 7); right of origin 2k. student graphs 2l. (4, )∞ ; moving left; yes 2m. (0, 4); moving right; yes 2n. t = 4; vertex of the position function; 18 cm

2o. cm4 secsec

− 2p. ( 4, )− ∞ ; it is speeding up 2q. (0, 4); slowing down

3a. student graph; 0 2.5t≤ ≤ 3b. speed is increasing downward 3c. – 40 ft/sec


3d. – 64 ft/sec 3e. 0 ft/sec 3f. ( ) 32v t t= − 3g. ft32 sec

sec

− 3h. –64 ft/sec; instantaneous

velocity at time t = 2 3i. 64 ft 3j. 100 ft 3k. 100 ft; yes it is the same 4a. 3 2( ) 3x t t t t= − − + 4b. student graphs 4c. t = 0.6453; t = 1.3 4d. change direction 4e. 8 in/sec 4f. 40 in/sec 4g. t = 1 4h. it is not moving (it is changing directions) 4i. –1.3312; 1.333; at t = 0.36 seconds the particle is moving left at 1.3312 in/sec; at t = 1.276 seconds the particle is moving right at 1.333 in/sec 4j. 47 inches

5a. 2 21 3 27( ) ln2 4 4

x t t t t= − + 5b. t e> 5c. 0 1t< <

5d. speed in increasing because velocity and acceleration are both positive 5e. 71.641 units Position/Velocity/Acceleration Handout #2 1. Left: (2, 3) and (5, 6); right: (0, 1); Still: (1, 2) and (3, 5) 2a. move up: (0, 1) and (5, 7); move down: (1, 5); speed up: (1, 2) and (5, 6); slow down: (0,1), (3, 5) and (6, 7) b. pos: (3, 6); neg: (0, 2); zero: (2, 3) and (7, 9) 3a. at 2 and 7 b. -3, 0, 3, -1.5 4a. 10 m b. 2 m/sec c. 5 m/sec d. 2 m/sec2 e. t = 1.5 f. -0.25 5a. 45 in/min b. -24 in/min2 c. t>1 6a. 26 ft/sec b. t>2 c. 0<t<2 7a. 0<t<4 b. t>4 c. up: 0<t<2 and t>4; down: 2<t<4 8a. up: 1<t<7/3 and t>11/3; down: 0<t<1 and 7/3<t<11/3 b. at t=1 and t = 11/3 c. at t = 1 and t = 11/3 d. t = 1 and t = 5 9. at t = 1, a = -6 and at t = 3, a = 6 10. at t = 1, speed = 0 m/sec, at t = 2, speed = 1 m/sec 11. 38 mph; 7 hours 12a. ( ) ln 3 3x t t t t= − + b. -2e + 3 c. 20 t e< < 13a. 3 b. 0<t<1 c. decreasing d. increasing e. up: 0<t<0.236 and t>1 down: at t = 1 14a. c = 3 b. 3 inches right of origin c. right; its increasing d. decreasing e. 1 and 5 f. 45 inches 15a. -33 b. increasing c. 4 d. 53; toward the origin

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Algebra 2B PAP Workbook Page 162 Answers€¦ · Algebra 2B PAP Workbook Page 167 . disadvantage is if the buyer would resell the car, they would lose more money compared to the exponential