## anonymous one year ago PLEASE CAN SOMEONE HELP ME WITH 2 TRIG QUESTIONS THAT I HAVE BEEN STUCK ON FOR ABOUT 2 DAYS AND NOBODY SEEMS TO HELP ME!

1. anonymous

2. anonymous

work backwards, determine your amp, period , phase shift, vertical/ horizontal shift

3. anonymous

can someone just show me how to solve the first one so that I can do the second one

4. anonymous

hey isn't that from con nexus

5. anonymous

what

6. anonymous

7. anonymous

no its from my trig class in Texas

8. anonymous

._. hmm bai

9. anonymous

can u help me or not?

10. anonymous

if that problem is really from where you say it is, can you give me a link so I cam find a solution

11. anonymous

@Nnesha you helped me before on these problems, can u help me one last time

12. anonymous

oh im not looking to graph, im just trying to find a function out of a graph

13. anonymous

@jim_thompson5910

14. anonymous

15. jim_thompson5910

#3 has an asymptote, and its range is the set of all real numbers, so the function is either tan or cot

16. anonymous

ok

17. jim_thompson5910

let's assume it's tan(x) what is the value of tan(x) when x = pi/4

18. anonymous

its 1

19. jim_thompson5910

so IF the point (pi/4, 1) was on the graph, then tan(x) might be the answer. But it's not

20. jim_thompson5910

something like 4*tan(x) would work but if we plugged in x = pi/2, then we'd get undefined. The function f(x) = tan(x) is undefined when x = pi/2. So it turns out that tan doesn't even work at all

21. jim_thompson5910

is this making sense?

22. anonymous

i see, so it has to be cotangent

23. jim_thompson5910

what is the value of cot(x) when x = pi/4 ?

24. anonymous

the answer is 1, but if I put a 4 behind cotangent: 4cot(pi/4) then it is 4

25. jim_thompson5910

good

26. anonymous

and it works with the rest as well

27. jim_thompson5910

so it appears that 4*cot(x) works. Check the other x coordinates to see if it holds up for the other two points

28. jim_thompson5910

yes 4*cot(x) = 0 when x = pi/2 and 4*cot(x) = -4 when x = 3pi/4

29. anonymous

ok so is thar it then, the function for this graphy is just y=4cot(x)

30. anonymous

that is it then*

31. jim_thompson5910

correct

32. anonymous

is that how you properly write it down:$y=4\cot (x)$

33. jim_thompson5910

yes

34. jim_thompson5910

or f(x) = 4*cot(x)

35. anonymous

how about the second one, that is the really confusing one

36. jim_thompson5910

what's the vertical distance from the very peak of one mountain to the very bottom of the valley? |dw:1438045445818:dw|

37. anonymous

1

38. jim_thompson5910

hint: look at the points (-pi/6, 10) and (5pi/6, -10) specifically the y coordinates

39. anonymous

ok i looked at them, just a question when you asked me the distance from the peak of the mountain to the bottom, did you already know the answer or were you asking me because the graph was unclear?

40. jim_thompson5910

I knew the answer. I wanted to get you thinking on how to solve this problem

41. anonymous

oh

42. anonymous

sorry im stuck and i dont know where you were going with in the hint

43. jim_thompson5910

what is the change from y = 10 to y = -10

44. anonymous

20

45. jim_thompson5910

cut that in half to get 20/2 = 10 the amplitude is 10 units. This is the distance from the center to the peak

46. jim_thompson5910

look at the graph and tell me what the two peak points are

47. anonymous

(-1,2) and (4,2)

48. jim_thompson5910

how are you getting them? look at the attachment. They give you the points

49. anonymous

arent they the two highest points

50. jim_thompson5910

they list the points below the graph

51. jim_thompson5910

2 of them are peak points

52. jim_thompson5910

yeah they are the highest points

53. anonymous

$(4\pi/3,0) and (11\pi/6,10)$

54. jim_thompson5910

there's a point higher than (4pi/3, 0)

55. jim_thompson5910

(11pi/6, 10) is one of the peak points

56. anonymous

(-pI/6,-10)

57. jim_thompson5910

yes, now subtract the x coordinates

58. jim_thompson5910

(11pi/6) - (-pi/6) = ???

59. anonymous

2pi

60. jim_thompson5910

so that is the period. Every 2pi units, the function repeats itself

61. anonymous

I see

62. jim_thompson5910

let's assume this is a cosine function when x = 0, cos(x) = 1 and 10*cos(x) = 10 however, the point (0,10) isn't on this function. Instead, it's shifted to the left pi/6 units. So we need to account for the phase shift phase shift = -pi/6 the negative tells us to go left

63. anonymous

are we sure its a cosine function, could it not be something else

64. jim_thompson5910

either that or sine

65. jim_thompson5910

cosine is easier to work with though

66. anonymous

ok

67. jim_thompson5910

A = amplitude = 10 T = period = 2pi P = phase shift = -pi/6 ------------------------------------------------------- The general cosine equation is y = A*cos(Bx - C) + D where |A| is the amplitude T = 2pi/B is the period C/B is the phase shift y = D is the midline. In this case, the midline is y = 0 so D = 0

68. anonymous

im confused as to how to write the equation out

69. jim_thompson5910

If T = 2pi/B is the period and we know T = 2pi is the period, then what is the value of B?

70. anonymous

1

71. anonymous

right?

72. jim_thompson5910

yes B = 1

73. anonymous

y=10cos(x+2pi)

74. jim_thompson5910

A = 10 B = 1 C = unknown to find C, solve C/B = -pi/6

75. anonymous

c=-pi/6

76. anonymous

?

77. jim_thompson5910

yes, so the function is $\Large f(x) = 10\cos\left(x + \frac{\pi}{6}\right)$

78. anonymous

uhhhh, Im graphing this equation, but it looks nothing like the graph given

79. anonymous

@jim_thompson5910

80. jim_thompson5910

what are you using to graph?

81. anonymous

www.demos.com

82. jim_thompson5910

you mean this? https://www.desmos.com/calculator

83. anonymous

yep

84. jim_thompson5910

and you typed in 10*cos(x+pi/6) ?

85. anonymous

yes

86. jim_thompson5910

87. anonymous

yes

88. jim_thompson5910

this is what I get

89. anonymous

oh wait it is working now

90. anonymous

if its not a problem can I ask you a way simpler question, that will only take 5 minutes

91. anonymous

how to we graph this function: $y=-3\sec(\pi-2x)+5$

92. jim_thompson5910

by using $\Large \sec(x) = \frac{1}{\cos(x)}$

93. jim_thompson5910

or desmos should be able to handle secant functions

94. anonymous

yeah, but i'm not getting a proper graph, im just getting something like this:

95. jim_thompson5910

that's what secant graphs look like. They have asymptotes and those U shapes

96. anonymous

oh ok thanks

97. jim_thompson5910

yw