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Log transformations question

  1. Oct 15, 2009 #1
    Here is what the graph looks like on a graphing calculator (notice the equation at the top):

    http://img62.imageshack.us/img62/8898/graphingcalc.jpg [Broken]


    Here is what my graph looks like:

    http://img53.imageshack.us/img53/7475/lastscanc.jpg [Broken]

    I dont understand why the asymptote is shifted 6 units left instead of 3 units left, And the point on my graph dont seem quite right.

    i have the points (-1, -3)
    Yet the graphing calculator has (-1, -3.79)

    Note : where i wrote original points im referring to y = log x


    I dont see what i hsve done wrong, can someone help me.

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 15, 2009 #2

    Mark44

    Staff: Mentor

    Your first step should have taken care of the compressions/expansions, so you should have looked at y = 2ln(1/2*x) first. The 1/2*x inside the parentheses causes an expansion away from the vertical axis. After that, take care of reflections, and finally translations.
     
  4. Oct 15, 2009 #3
    thanks will try that
     
    Last edited: Oct 15, 2009
  5. Oct 15, 2009 #4

    Mark44

    Staff: Mentor

    Notice that y = -2log(.5x + 3) + 3 = -2log(.5(x + 6)) - 3. This means that
    1. The domain of this function is {x | x > -6}
    2. The graph of y = -2log(.5x) has been translated to the left by 6 units, and down by 3 units.
     
  6. Oct 15, 2009 #5
    You hit the nail on the head my friend, I forgot to factor Stupid me!!!
     
  7. Oct 15, 2009 #6
    ill try again and post back my result
     
  8. Oct 15, 2009 #7
    Got it thank you very much, forgot to factor.

    Here it is:
    http://img62.imageshack.us/img62/2599/lastscanjy.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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