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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!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    work backwards, determine your amp, period , phase shift, vertical/ horizontal shift

  3. anonymous
    • one year ago
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    can someone just show me how to solve the first one so that I can do the second one

  4. anonymous
    • one year ago
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    hey isn't that from con nexus

  5. anonymous
    • one year ago
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    what

  6. anonymous
    • one year ago
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    connections academy

  7. anonymous
    • one year ago
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    no its from my trig class in Texas

  8. anonymous
    • one year ago
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    ._. hmm bai

  9. anonymous
    • one year ago
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    can u help me or not?

  10. anonymous
    • one year ago
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    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
    • one year ago
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    @Nnesha you helped me before on these problems, can u help me one last time

  12. anonymous
    • one year ago
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    oh im not looking to graph, im just trying to find a function out of a graph

  13. anonymous
    • one year ago
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    @jim_thompson5910

  14. anonymous
    • one year ago
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    Please help @Vocaloid

  15. jim_thompson5910
    • one year ago
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    #3 has an asymptote, and its range is the set of all real numbers, so the function is either tan or cot

  16. anonymous
    • one year ago
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    ok

  17. jim_thompson5910
    • one year ago
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    let's assume it's tan(x) what is the value of tan(x) when x = pi/4

  18. anonymous
    • one year ago
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    its 1

  19. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    is this making sense?

  22. anonymous
    • one year ago
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    i see, so it has to be cotangent

  23. jim_thompson5910
    • one year ago
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    what is the value of cot(x) when x = pi/4 ?

  24. anonymous
    • one year ago
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    the answer is 1, but if I put a 4 behind cotangent: 4cot(pi/4) then it is 4

  25. jim_thompson5910
    • one year ago
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    good

  26. anonymous
    • one year ago
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    and it works with the rest as well

  27. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    yes 4*cot(x) = 0 when x = pi/2 and 4*cot(x) = -4 when x = 3pi/4

  29. anonymous
    • one year ago
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    ok so is thar it then, the function for this graphy is just y=4cot(x)

  30. anonymous
    • one year ago
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    that is it then*

  31. jim_thompson5910
    • one year ago
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    correct

  32. anonymous
    • one year ago
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    is that how you properly write it down:\[y=4\cot (x)\]

  33. jim_thompson5910
    • one year ago
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    yes

  34. jim_thompson5910
    • one year ago
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    or f(x) = 4*cot(x)

  35. anonymous
    • one year ago
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    how about the second one, that is the really confusing one

  36. jim_thompson5910
    • one year ago
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    what's the vertical distance from the very peak of one mountain to the very bottom of the valley? |dw:1438045445818:dw|

  37. anonymous
    • one year ago
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    1

  38. jim_thompson5910
    • one year ago
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    hint: look at the points (-pi/6, 10) and (5pi/6, -10) specifically the y coordinates

  39. anonymous
    • one year ago
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    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
    • one year ago
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    I knew the answer. I wanted to get you thinking on how to solve this problem

  41. anonymous
    • one year ago
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    oh

  42. anonymous
    • one year ago
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    sorry im stuck and i dont know where you were going with in the hint

  43. jim_thompson5910
    • one year ago
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    what is the change from y = 10 to y = -10

  44. anonymous
    • one year ago
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    20

  45. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    look at the graph and tell me what the two peak points are

  47. anonymous
    • one year ago
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    (-1,2) and (4,2)

  48. jim_thompson5910
    • one year ago
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    how are you getting them? look at the attachment. They give you the points

  49. anonymous
    • one year ago
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    arent they the two highest points

  50. jim_thompson5910
    • one year ago
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    they list the points below the graph

  51. jim_thompson5910
    • one year ago
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    2 of them are peak points

  52. jim_thompson5910
    • one year ago
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    yeah they are the highest points

  53. anonymous
    • one year ago
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    \[(4\pi/3,0) and (11\pi/6,10)\]

  54. jim_thompson5910
    • one year ago
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    there's a point higher than (4pi/3, 0)

  55. jim_thompson5910
    • one year ago
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    (11pi/6, 10) is one of the peak points

  56. anonymous
    • one year ago
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    (-pI/6,-10)

  57. jim_thompson5910
    • one year ago
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    yes, now subtract the x coordinates

  58. jim_thompson5910
    • one year ago
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    (11pi/6) - (-pi/6) = ???

  59. anonymous
    • one year ago
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    2pi

  60. jim_thompson5910
    • one year ago
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    so that is the period. Every 2pi units, the function repeats itself

  61. anonymous
    • one year ago
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    I see

  62. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    are we sure its a cosine function, could it not be something else

  64. jim_thompson5910
    • one year ago
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    either that or sine

  65. jim_thompson5910
    • one year ago
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    cosine is easier to work with though

  66. anonymous
    • one year ago
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    ok

  67. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    im confused as to how to write the equation out

  69. jim_thompson5910
    • one year ago
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    If T = 2pi/B is the period and we know T = 2pi is the period, then what is the value of B?

  70. anonymous
    • one year ago
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    1

  71. anonymous
    • one year ago
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    right?

  72. jim_thompson5910
    • one year ago
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    yes B = 1

  73. anonymous
    • one year ago
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    y=10cos(x+2pi)

  74. jim_thompson5910
    • one year ago
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    A = 10 B = 1 C = unknown to find C, solve C/B = -pi/6

  75. anonymous
    • one year ago
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    c=-pi/6

  76. anonymous
    • one year ago
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    ?

  77. jim_thompson5910
    • one year ago
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    yes, so the function is \[\Large f(x) = 10\cos\left(x + \frac{\pi}{6}\right)\]

  78. anonymous
    • one year ago
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    uhhhh, Im graphing this equation, but it looks nothing like the graph given

  79. anonymous
    • one year ago
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    @jim_thompson5910

  80. jim_thompson5910
    • one year ago
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    what are you using to graph?

  81. anonymous
    • one year ago
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    www.demos.com

  82. jim_thompson5910
    • one year ago
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    you mean this? https://www.desmos.com/calculator

  83. anonymous
    • one year ago
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    yep

  84. jim_thompson5910
    • one year ago
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    and you typed in 10*cos(x+pi/6) ?

  85. anonymous
    • one year ago
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    yes

  86. jim_thompson5910
    • one year ago
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    are you in radian mode?

  87. anonymous
    • one year ago
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    yes

  88. jim_thompson5910
    • one year ago
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    this is what I get

  89. anonymous
    • one year ago
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    oh wait it is working now

  90. anonymous
    • one year ago
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    if its not a problem can I ask you a way simpler question, that will only take 5 minutes

  91. anonymous
    • one year ago
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    how to we graph this function: \[y=-3\sec(\pi-2x)+5\]

  92. jim_thompson5910
    • one year ago
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    by using \[\Large \sec(x) = \frac{1}{\cos(x)}\]

  93. jim_thompson5910
    • one year ago
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    or desmos should be able to handle secant functions

  94. anonymous
    • one year ago
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    yeah, but i'm not getting a proper graph, im just getting something like this:

  95. jim_thompson5910
    • one year ago
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    that's what secant graphs look like. They have asymptotes and those U shapes

  96. anonymous
    • one year ago
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    oh ok thanks

  97. jim_thompson5910
    • one year ago
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    yw

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