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

a. The parabola opens down.

b. The vertex is βˆ’2, 4 .

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

βˆ’ π‘₯ + 2 ! + 4 = 0

βˆ’ π‘₯ + 2 ! = βˆ’4

π‘₯ + 2 ! = 4

π‘₯ + 2 = 4

π‘₯ = βˆ’2Β± 2

π‘₯ = 0,βˆ’4

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

𝑔 0 = βˆ’ 0+ 2 ! + 4 = 0

e. Sketch the graph.

f. The axis of symmetry is π‘₯ = βˆ’2.

g. The function has a maximum value of 4.

h. Domain: (βˆ’βˆž,∞) Range: (βˆ’ ∞, 4]

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

a. Vertex Form: π‘₯ + 4 ! βˆ’ 9

(π‘₯! + 8π‘₯ + 16)+ 7βˆ’ 16

b. The vertex is βˆ’4,βˆ’9 .

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

π‘₯ + 1 π‘₯ + 7 = 0

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

e. Sketch the graph.

f. The axis of symmetry is π‘₯ = βˆ’4

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

h. Domain: (βˆ’βˆž,∞) Range: [βˆ’9,∞)

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. 𝒋 𝒕 = βˆ’ πŸπŸ’π’•πŸ + πŸπŸŽπ’•βˆ’ πŸ“

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯!!!""# = βˆ’π‘2π‘Ž =

βˆ’10

2 βˆ’ 14=βˆ’10

βˆ’ 12= 20

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯!!!""# = 𝑗 20 = βˆ’14 20 ! + 10 20 βˆ’ 5 = βˆ’

14 400 + 200βˆ’ 5 = 95

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯ = (20, 95)

30. π’Œ 𝒙 = πŸπ’™πŸ βˆ’ πŸπŸŽπ’™βˆ’ πŸ“

a. The parabola opens up.

b. The vertex is 2.5,βˆ’17.5 .

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯!!!""# = βˆ’π‘2π‘Ž =

βˆ’ βˆ’102 2 =

52 = 2.5

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯!!!""# = π‘˜52 = 2

52

!

βˆ’ 1052 βˆ’ 5 = βˆ’17.5

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

Via DQUAD: π‘₯ = !!Β± !"

!

d. The y-intercept is (0,βˆ’5).

π‘˜ 0 = 2 0 ! βˆ’ 10 0 βˆ’ 5 = βˆ’5

e. Sketch the graph.

f. The axis of symmetry is π‘₯ = 2.5.

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

h. Domain: (βˆ’βˆž,∞) Range: [βˆ’17.5, ∞)

35. 𝒉 𝒙 = βˆ’πŸŽ.πŸŽπŸπŸ”π’™πŸ + 𝟎.πŸ“πŸ•πŸ”π’™+ πŸ‘

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

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯!!!""# = βˆ’π‘2π‘Ž =

βˆ’0.5762 βˆ’0.026 = 11.1

b. The maximum height is 6.2 meters.

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯!!!""# = β„Ž 11.1 = βˆ’0.026 11.1 + 0.576 11.1 + 3 = 6.2

c. The firefighter is 14 meters from the house.

βˆ’0.026π‘₯! + 0.576π‘₯ + 3 = 6 simplified to βˆ’0.026π‘₯! + 0.576π‘₯ βˆ’ 3 = 0

Via DQUAD: π‘₯ = 8.4, 13.8

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Homework Section 3.1

42.

a. π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 3π‘₯ + 4𝑦 = 120

π΄π‘Ÿπ‘’π‘Ž = π‘₯𝑦

𝑦 = βˆ’34 π‘₯ + 30

π΄π‘Ÿπ‘’π‘Ž = π‘₯(βˆ’34 π‘₯ + 30)

π΄π‘Ÿπ‘’π‘Ž = βˆ’34 π‘₯

! + 30π‘₯

π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯!!!""# = βˆ’π‘2π‘Ž =

βˆ’30

2 βˆ’ 34=βˆ’30

βˆ’ 32= 20

𝑦 = βˆ’34 20 + 30 = βˆ’15+ 30 = 15

The dimensions of an individual coop are 20𝑓𝑑 Γ—15𝑓𝑑.

b. The maximum area of an individual coop is 300𝑓𝑑!.

π΄π‘Ÿπ‘’π‘Ž = π‘₯𝑦 = 20 15 = 300