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JenniferSmart1

  • 2 years ago

how do I graph \[y=-3cos(3x)\]? I know that 3 is the amplitude I know that the graph will be upside down. I know that the period is \[\frac{2\pi}{3}\]

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    • 2 years ago
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    Copy the image and begin by marking your amplitude, the highest and lowest point of the wave.|dw:1360290026921:dw|

  2. JenniferSmart1
    • 2 years ago
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    |dw:1360290074055:dw|

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    • 2 years ago
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    now label \[(\frac{2\pi}{3},f(\frac{2\pi}{3}))\]this point is the end of your period.

  4. JenniferSmart1
    • 2 years ago
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    |dw:1360290246963:dw|

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    • 2 years ago
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    \[f(x)=-3cos(3x)\]\[f(\frac{3\pi}{2})=-3cos(3(\frac{3\pi}{2}))=-3cos(9\pi/2)\]

  6. JenniferSmart1
    • 2 years ago
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    it's \[\frac{2\pi}{3}\] for the period

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    • 2 years ago
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    @JenniferSmart1 My fault. You evaluated it correctly, I would have gotten 0 for f(3pi/2) while f(2pi/3) is -3. Ok, so now can you finish the graph from here? So far you have done an excellent job.

  8. JenniferSmart1
    • 2 years ago
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    |dw:1360290496676:dw|

  9. JenniferSmart1
    • 2 years ago
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    ?

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    • 2 years ago
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    Could we label the highest point (x, f(x)) if we were asked?

  11. JenniferSmart1
    • 2 years ago
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    Let's see \[( \frac{\pi}{3} , 3)\] because \[\frac{2\pi}{3\cdot2}\]

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    • 2 years ago
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    Ok! I think your graph seems pretty complete, well done. Here is the actual graph for reference http://www.wolframalpha.com/input/?i=-3cos%283x%29

  13. JenniferSmart1
    • 2 years ago
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    I'm guessing because its half way between 2pi/3

  14. JenniferSmart1
    • 2 years ago
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    Thanks!

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