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Math Help - X-intercepts of a graph

  1. #1
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    X-intercepts of a graph

    Hi all!

    Ive been having some trouble with the same types of questions over and over again. I cannot seem to get them right..

    I have trouble finding the x-intercepts, after a series of transformations to the original graph.

    Here is an example:

    y = 2-3cos 2(θ - π/2)

    I know that:
    - amplitude: 3
    - reflection about the x-axis
    - translation of π/2 units in the positive direction of the x-axis
    - dilation of factor 1/2 from the y-axis
    - translation of 2 units in the positive direction of the y-axis

    I understand how it has been transformed from the original cos graph, but i can't get the x-intercepts after all the transformations. How do i start it? Can anyone give me a good method??

    Anyones help would be greatly appreciated.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by steph_r View Post
    Hi all!

    Ive been having some trouble with the same types of questions over and over again. I cannot seem to get them right..

    I have trouble finding the x-intercepts, after a series of transformations to the original graph.

    Here is an example:

    y = 2-3cos 2(θ - π/2)

    I know that:
    - amplitude: 3
    - reflection about the x-axis
    - translation of π/2 units in the positive direction of the x-axis
    - dilation of factor 1/2 from the y-axis
    - translation of 2 units in the positive direction of the y-axis

    I understand how it has been transformed from the original cos graph, but i can't get the x-intercepts after all the transformations. How do i start it? Can anyone give me a good method??

    Anyones help would be greatly appreciated.
    For that example you could use cos(A+B) = cosAcosB - sinAsinB to reduce cos(\theta - \frac{\pi}{2}) to -sin(\theta) and solve that way.

    The phase shift means each point will cross the x axis \frac{\pi}{2} 'later' so your original intercepts would now be at k + \frac{\pi}{2} rather than at k

    Applying the dilation factor means each point is at half it's old value (it is "squashed" so to speak). This means your points which were at k+\frac{\pi}{2} would be at \frac{1}{2}(k+\frac{\pi}{2})
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  3. #3
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    Smile

    Quote Originally Posted by e^(i*pi) View Post
    For that example you could use cos(A+B) = cosAcosB - sinAsinB to reduce cos(\theta - \frac{\pi}{2}) to -sin(\theta) and solve that way.

    The phase shift means each point will cross the x axis \frac{\pi}{2} 'later' so your original intercepts would now be at k + \frac{\pi}{2} rather than at k

    Applying the dilation factor means each point is at half it's old value (it is "squashed" so to speak). This means your points which were at k+\frac{\pi}{2} would be at \frac{1}{2}(k+\frac{\pi}{2})
    Thank you i'll give that a try, and i hope if i have any questions u can help me out thank you
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by steph_r View Post
    Thank you i'll give that a try, and i hope if i have any questions u can help me out thank you
    All right, I'll do what I can before I turn into bed XD
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