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    Can someone explain what happens to ln/e graphs when there is a modulus in the equasion, eg: y= ln l4x-12l
    Thanks!
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    I'm guessing you do exactly the same thing with them as with all of the other graphs. In this case, reflect in the x axis.
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    (Original post by unamed)
    I'm guessing you do exactly the same thing with them as with all of the other graphs. In this case, reflect in the x axis.
    Wouldn't that be a reflection in the y axis, because the modulus isn't around the whole function?
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    (Original post by definite_maybe)
    Wouldn't that be a reflection in the y axis, because the modulus isn't around the whole function?
    yes, it would. I should really look at the question more closely. Thank you for that!
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    (Original post by definite_maybe)
    Wouldn't that be a reflection in the y axis, because the modulus isn't around the whole function?
    The graph of lnx will be a reflection of e^{x} in the line y = x
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    (Original post by unamed)
    yes, it would. I should really look at the question more closely. Thank you for that!
    I'm guessing you're also doing core 3 tomorrow? I posted two questions in a maths thread called Integration, which I'm really stuck with from June 2007 paper. Any help?
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    (Original post by gottastudy)
    The graph of lnx will be a reflection of e^{x} in the line y = x
    but they're talking about modulus stuff, not that. :s

    edit:
    Also, it'd reflect in x=3, not in x=0/the y-axis.
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    (Original post by definite_maybe)
    I'm guessing you're also doing core 3 tomorrow? I posted two questions in a maths thread called Integration, which I'm really stuck with from June 2007 paper. Any help?
    yep. I'm doing edexcel, we haven't done integration yet - I'm sorry!
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    (Original post by definite_maybe)
    I'm guessing you're also doing core 3 tomorrow? I posted two questions in a maths thread called Integration, which I'm really stuck with from June 2007 paper. Any help?
    Thanks everyone, I'll muddle through... Ye I'm doing C3 tomorrow, not looking forward to it. I'll see if I can be of any help with the ones you've posted
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    (Original post by gingersnap_08)
    Thanks everyone, I'll muddle through... Ye I'm doing C3 tomorrow, not looking forward to it. I'll see if I can be of any help with the ones you've posted
    I might be able to help you out with integration.. can't find your questions, though.
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    (Original post by phen)
    I might be able to help you out with integration.. can't find your questions, though.
    http://www.thestudentroom.co.uk/show....php?t=1156798

    It's this one. OCR Core 3 June 2007. Thank you
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    (Original post by phen)
    but they're talking about modulus stuff, not that. :s

    edit:
    Also, it'd reflect in x=3, not in x=0/the y-axis.
    sorry, this is what I was confused about. it said something like that in the answers, (i cant remember exactly) and there seemed to be a new asymtote which the graph was reflected in
    could you explain how you got that answer please? thank youuu
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    (Original post by gingersnap_08)
    sorry, this is what I was confused about. it said something like that in the answers, (i cant remember exactly) and there seemed to be a new asymtote which the graph was reflected in
    could you explain how you got that answer please? thank youuu
    ..a new asymptote?
    Maybe you're comparing this situation to a 'normal' situation like 'ln|x|'. In that situation, there'd be a reflection in the horizontal asymptote of x=0.
    The horizontal asymptote is through the point where the stuff that's ln'd (lol pardon my terminology) equals zero, e.g. ln|x-4| has a horizontal asymptote for x-4=0 => x=4.

    edit:
    I just noticed I quoted the wrong person in my earlier post. :P Whoops. That's why I couldn't find the questions.
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    (Original post by gingersnap_08)
    Can someone explain what happens to ln/e graphs when there is a modulus in the equasion, eg: y= ln l4x-12l
    Thanks!
    Modulus means absolute value i.e. it forces what ever is in it to be non-negative, for graph like this you need to split your equation into separate cases to make sure whatever is in the modulus sign is non-negative

    With y= ln |4x-12|
    • 4x-12<0 if x<3, so for x < 3: y = ln (12-4x)
    • 4x-12>0 if x>3, so for x > 3: y = ln (4x-12)


    If you want to save on thinking about separate cases you could, for more simple equations like this note that y = |x| is a even function (symmetric in x=0) and use that to let you think about symmetry i.e. in this example y=ln |4x-12| = ln|4|+ln|x-3| must be symmetric in x=3.
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    Thank you!
 
 
 
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