hw section 3.1 answer key word - hw section 3.1 answer key.docx created date 9/9/2017 5:10:52 pm

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Homework Section 3.1 MAT121 SECTION 3.1 HOMEWORK ANSWER KEY 10, 18, 26, 30, 35, 42 10. = โˆ’ + + a. The parabola opens down. b. The vertex is โˆ’2, 4 . c. The x-intercepts are 0,0 and (โˆ’4, 0). โˆ’ + 2 ! + 4 = 0 โˆ’ + 2 ! = โˆ’4 + 2 ! = 4 + 2 = 4 = โˆ’2 ยฑ 2 = 0, โˆ’4 d. The y-intercept is (0,0). 0 = โˆ’ 0 + 2 ! + 4 = 0 e. Sketch the graph. f. The axis of symmetry is = โˆ’2. g. The function has a maximum value of 4. h. Domain: (โˆ’โˆž, โˆž) Range: (โˆ’ โˆž, 4] 18. = + + a. Vertex Form: + 4 ! โˆ’ 9 ( ! + 8 + 16) + 7 โˆ’ 16 b. The vertex is โˆ’4, โˆ’9 . c. The x-intercepts are โˆ’1,0 and (โˆ’7, 0). + 1 + 7 = 0 d. The y-intercept is (0,7). e. Sketch the graph. f. The axis of symmetry is = โˆ’4 g. The function has a minimum value of โˆ’9. h. Domain: (โˆ’โˆž, โˆž) Range: [โˆ’9, โˆž)

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Page 1: HW Section 3.1 Answer Key Word - HW Section 3.1 Answer Key.docx Created Date 9/9/2017 5:10:52 PM

Homework Section 3.1

MAT121 SECTION 3.1

HOMEWORK ANSWER KEY

10, 18, 26, 30, 35, 42

10. ๐’ˆ ๐’™ = โˆ’ ๐’™+ ๐Ÿ ๐Ÿ + ๐Ÿ’

a. The parabola opens down.

b. The vertex is โˆ’2, 4 .

c. The x-intercepts are 0,0 and (โˆ’4, 0).

โˆ’ ๐‘ฅ + 2 ! + 4 = 0

โˆ’ ๐‘ฅ + 2 ! = โˆ’4

๐‘ฅ + 2 ! = 4

๐‘ฅ + 2 = 4

๐‘ฅ = โˆ’2ยฑ 2

๐‘ฅ = 0,โˆ’4

d. The y-intercept is (0,0).

๐‘” 0 = โˆ’ 0+ 2 ! + 4 = 0

e. Sketch the graph.

f. The axis of symmetry is ๐‘ฅ = โˆ’2.

g. The function has a maximum value of 4.

h. Domain: (โˆ’โˆž,โˆž) Range: (โˆ’ โˆž, 4]

18. ๐’ˆ ๐’™ = ๐’™๐Ÿ + ๐Ÿ–๐’™+ ๐Ÿ•

a. Vertex Form: ๐‘ฅ + 4 ! โˆ’ 9

(๐‘ฅ! + 8๐‘ฅ + 16)+ 7โˆ’ 16

b. The vertex is โˆ’4,โˆ’9 .

c. The x-intercepts are โˆ’1,0 and (โˆ’7, 0).

๐‘ฅ + 1 ๐‘ฅ + 7 = 0

d. The y-intercept is (0,7).

e. Sketch the graph.

f. The axis of symmetry is ๐‘ฅ = โˆ’4

g. The function has a minimum value of โˆ’9.

h. Domain: (โˆ’โˆž,โˆž) Range: [โˆ’9,โˆž)

Page 2: HW Section 3.1 Answer Key Word - HW Section 3.1 Answer Key.docx Created Date 9/9/2017 5:10:52 PM

Homework Section 3.1

26. ๐’‹ ๐’• = โˆ’ ๐Ÿ๐Ÿ’๐’•๐Ÿ + ๐Ÿ๐ŸŽ๐’•โˆ’ ๐Ÿ“

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ!!!""# = โˆ’๐‘2๐‘Ž =

โˆ’10

2 โˆ’ 14=โˆ’10

โˆ’ 12= 20

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ!!!""# = ๐‘— 20 = โˆ’14 20 ! + 10 20 โˆ’ 5 = โˆ’

14 400 + 200โˆ’ 5 = 95

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ = (20, 95)

30. ๐’Œ ๐’™ = ๐Ÿ๐’™๐Ÿ โˆ’ ๐Ÿ๐ŸŽ๐’™โˆ’ ๐Ÿ“

a. The parabola opens up.

b. The vertex is 2.5,โˆ’17.5 .

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ!!!""# = โˆ’๐‘2๐‘Ž =

โˆ’ โˆ’102 2 =

52 = 2.5

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ!!!""# = ๐‘˜52 = 2

52

!

โˆ’ 1052 โˆ’ 5 = โˆ’17.5

c. The x-intercepts are 5.46, 0 and (โˆ’.46, 0).

Via DQUAD: ๐‘ฅ = !!ยฑ !"

!

d. The y-intercept is (0,โˆ’5).

๐‘˜ 0 = 2 0 ! โˆ’ 10 0 โˆ’ 5 = โˆ’5

e. Sketch the graph.

f. The axis of symmetry is ๐‘ฅ = 2.5.

g. The function has a minimum value of -17.5.

h. Domain: (โˆ’โˆž,โˆž) Range: [โˆ’17.5, โˆž)

35. ๐’‰ ๐’™ = โˆ’๐ŸŽ.๐ŸŽ๐Ÿ๐Ÿ”๐’™๐Ÿ + ๐ŸŽ.๐Ÿ“๐Ÿ•๐Ÿ”๐’™+ ๐Ÿ‘

a. The maximum height occurs at the vertex. The maximum height is 11.1 meters.

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ!!!""# = โˆ’๐‘2๐‘Ž =

โˆ’0.5762 โˆ’0.026 = 11.1

b. The maximum height is 6.2 meters.

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ!!!""# = โ„Ž 11.1 = โˆ’0.026 11.1 + 0.576 11.1 + 3 = 6.2

c. The firefighter is 14 meters from the house.

โˆ’0.026๐‘ฅ! + 0.576๐‘ฅ + 3 = 6 simplified to โˆ’0.026๐‘ฅ! + 0.576๐‘ฅ โˆ’ 3 = 0

Via DQUAD: ๐‘ฅ = 8.4, 13.8

Page 3: HW Section 3.1 Answer Key Word - HW Section 3.1 Answer Key.docx Created Date 9/9/2017 5:10:52 PM

Homework Section 3.1

42.

a. ๐‘ƒ๐‘’๐‘Ÿ๐‘–๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ = 3๐‘ฅ + 4๐‘ฆ = 120

๐ด๐‘Ÿ๐‘’๐‘Ž = ๐‘ฅ๐‘ฆ

๐‘ฆ = โˆ’34 ๐‘ฅ + 30

๐ด๐‘Ÿ๐‘’๐‘Ž = ๐‘ฅ(โˆ’34 ๐‘ฅ + 30)

๐ด๐‘Ÿ๐‘’๐‘Ž = โˆ’34 ๐‘ฅ

! + 30๐‘ฅ

๐‘‰๐‘’๐‘Ÿ๐‘ก๐‘’๐‘ฅ!!!""# = โˆ’๐‘2๐‘Ž =

โˆ’30

2 โˆ’ 34=โˆ’30

โˆ’ 32= 20

๐‘ฆ = โˆ’34 20 + 30 = โˆ’15+ 30 = 15

The dimensions of an individual coop are 20๐‘“๐‘ก ร—15๐‘“๐‘ก.

b. The maximum area of an individual coop is 300๐‘“๐‘ก!.

๐ด๐‘Ÿ๐‘’๐‘Ž = ๐‘ฅ๐‘ฆ = 20 15 = 300