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AdamK

  • one year ago

http://imgur.com/RPAkgxB #4. I've tried plotting points and I just get a straight line. Not sure if that's right or not. Here are what I think the transformations are: Right 1, stretch of 3, up 1

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  1. jim_thompson5910
    • one year ago
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    Were you able to figure out the domain and range of the transformed graph?

  2. anonymous
    • one year ago
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    can someone help me i realy need help

  3. AdamK
    • one year ago
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    No I can't find out the domain or range because none of the points I've plotted correspond to it

  4. jim_thompson5910
    • one year ago
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    What is the left most point on the given graph?

  5. AdamK
    • one year ago
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    (-1, 1)

  6. jim_thompson5910
    • one year ago
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    yes, so the left half of the domain of the original function starts with x = -1 the question is: which x value, when plugged into (1/3)*(x-1), gives a result of -1? ie what is the solution to (1/3)*(x-1) = -1 ?

  7. AdamK
    • one year ago
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    -2

  8. jim_thompson5910
    • one year ago
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    good

  9. jim_thompson5910
    • one year ago
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    this means that the left most part of the domain in the transformed function is x = -2

  10. jim_thompson5910
    • one year ago
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    if you plugged in something smaller than -2 (say -3), then (1/3)*(x-1) produces a number that is smaller than -1 (which is outside the domain of f(x))

  11. jim_thompson5910
    • one year ago
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    the goal is to stay in the domain of f(x) because we can't use x values that aren't defined for f(x)

  12. jim_thompson5910
    • one year ago
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    does that make sense?

  13. AdamK
    • one year ago
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    Not really. Why would x-values need to be defined for the original function when it's being transformed?

  14. jim_thompson5910
    • one year ago
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    notice how we have f[ `(1/3)*(x-1)` ] basically f[ T ] where T = (1/3)*(x-1)

  15. jim_thompson5910
    • one year ago
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    is it possible to have f[ T ] when T is say, T = 5 ?

  16. AdamK
    • one year ago
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    Oh, ok

  17. jim_thompson5910
    • one year ago
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    I guess in a way, you have to think backwards

  18. jim_thompson5910
    • one year ago
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    The lowest you can go is T = -1 So you solve for x in T = (1/3)*(x-1) and like you said, you'll get x = -2 So f[ (1/3)*(x-1) ] has the lower part of the domain be x = -2

  19. AdamK
    • one year ago
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    My final answer is "a."

  20. AdamK
    • one year ago
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    Hopefully I don't have to graph anything

  21. jim_thompson5910
    • one year ago
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    You don't. They just want the domain and range of the transformed function.

  22. AdamK
    • one year ago
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    Ok good

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