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Math Help - translations of the graphs of sine and cosine functions

  1. #1
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    translations of the graphs of sine and cosine functions

    I have a test tmrw and just went over the information today so I am a little confused. Some of the graphs I get right but some of them I am getting the wrong X values.

    So if i have y=-4 sin 2(x-pi/2)
    so the Amplitude is 4
    and the period is pi
    but how in the world do i get the x axis?

    I've tried everything and i'll find one way works for one type but doesn't work for another. Is it really that way? I was told to multiply pi/2 by 1/4 2/4 and 3/4 but it doesnt come out correct. Does anyone know what to add or multiply by to get the x axis? Thanks
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  2. #2
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    Quote Originally Posted by KoolFlair View Post

    so the Amplitude is 4
    correct

    Quote Originally Posted by KoolFlair View Post

    and the period is pi
    correct


    Quote Originally Posted by KoolFlair View Post

    So if i have y=-4 sin 2(x-pi/2)
    but how in the world do i get the x axis?
    find the x axis intercepts by making y = 0

    y=-4 \sin 2(x-\frac{\pi}{2})

    0=-4 \sin 2(x-\frac{\pi}{2})

    0= \sin 2(x-\frac{\pi}{2})

    can you solve from here?
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  3. #3
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    Quote Originally Posted by KoolFlair View Post
    I have a test tmrw and just went over the information today so I am a little confused. Some of the graphs I get right but some of them I am getting the wrong X values.

    So if i have y=-4 sin 2(x-pi/2)
    so the Amplitude is 4
    and the period is pi
    but how in the world do i get the x axis?
    what do you mean by "getting the x-axis"?

    I've tried everything and i'll find one way works for one type but doesn't work for another. Is it really that way? I was told to multiply pi/2 by 1/4 2/4 and 3/4 but it doesnt come out correct. Does anyone know what to add or multiply by to get the x axis? Thanks
    ...
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  4. #4
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    im not sure how to finish it that way.

    I was told to make an inequality. So i got

    pi/2<x<9pi/2

    And pi/2 starts the x
    and
    9pi/2 ends as the 4th so i need to find the 3 in between. Do i add pi/2 to each one?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    ...

    like if you set up and x/y tree. I need to get the X then take the X and insert it into the cos x
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  6. #6
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    Quote Originally Posted by KoolFlair View Post
    im not sure how to finish it that way.

    I was told to make an inequality. So i got

    pi/2<x<9pi/2

    And pi/2 starts the x
    and
    9pi/2 ends as the 4th so i need to find the 3 in between. Do i add pi/2 to each one?
    The function

    y=-4 \sin 2(x-\frac{\pi}{2})

    has period of \pi with x-intercepts at 0+\frac{\pi}{2},\frac{\pi}{2}+\frac{\pi}{2} and \pi+\frac{\pi}{2}

    This gives all the solutions on [\frac{\pi}{2},\frac{3\pi}{2}]

    If you need solutions on [\frac{\pi}{2},\frac{9\pi}{2}]

    keep adding \pi to the solutions above
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  7. #7
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    Thanks pickslides
    Thats what i thought.


    But my solutions manuel says it should be

    Pi/2
    11pi/4
    5pi
    19pi/4
    9pi/2

    So how are they getting 11pi/4?
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  8. #8
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    You sure about that? I graphed the function and I'm getting x-intercepts at 0,\; \frac{\pi}{2},\; \pi,\; \frac{3\pi}{2},....

    If your original function is
    y = -4{\color{red}\cos} \left(2\left(x - \frac{\pi}{2}\right)\right)
    then the x-intercepts occur at
    \frac{\pi}{4},\; \frac{3\pi}{4},\; \frac{5\pi}{4},\; \frac{7\pi}{4},...
    but these answer still don't match yours.


    01
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