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    how would you draw the graph of y=lnl3x-6l XER where x doesnt= 2
    and how do work out where the asymptote is ???
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    X doesn't equal 2, massive hint to where the asymptote is, no? :rolleyes:
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    is it ln(3x-6) or ln(3x)-6?
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    Ok, think of the graph for ln x, call it f(x)...then ln |3x-6| becomes f(|3x-6|)...does that help?
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    I'm super confused with this as well. It would be really helpful if someone could tell me how you're meant to sketch graphs like ln(-x+1) -- I can never seem to get them right when they have that minus plus a constant. Thanks :love:
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    (Original post by *MJ*)
    X doesn't equal 2, massive hint to where the asymptote is, no? :rolleyes:
    ohhh i didnt know that meant thats where the asymptote will be thanks...how would you sketch it ??
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    (Original post by n_251)
    Ok, think of the graph for ln x, call it f(x)...then ln |3x-6| becomes f(|3x-6|)...does that help?
    hmm..a bit confused
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    (Original post by unamed)
    is it ln(3x-6) or ln(3x)-6?
    is it ln l3x-6l --> the lines mean that modulus sign
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    (Original post by klgyal)
    is it ln l3x-6l --> the lines mean that modulus sign
    I think I was asking so I knew where the lnx translates to..
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    (Original post by Dayxn2)
    I'm super confused with this as well. It would be really helpful if someone could tell me how you're meant to sketch graphs like ln(-x+1) -- I can never seem to get them right when they have that minus plus a constant. Thanks :love:
    Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph is moved horizontally to the left by one and then reflected in y axis.

    Hope this helps.
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    (Original post by JTeighty)
    Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph reflected in y axis and moved horizontally to the left by one.

    Hope this helps.

    If so pls rep
    how would you draw my graph please???
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    (Original post by klgyal)
    how would you draw the graph of y=lnl3x-6l XER where x doesnt= 2
    and how do work out where the asymptote is ???
    It is a lnx graph with the x values modded and then a horizontal translation of 6 to the right and then a stretch of a third.

    Asymptope will be x=2. Hence the question says x can't equal 2.
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    (Original post by JTeighty)
    Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph reflected in y axis and moved horizontally to the left by one.

    Hope this helps.

    If so pls rep

    Yeah, that's what I thought initially -- however, look:
    http://www.wolframalpha.com/input/?i=y%3Dln%28-x%2B1%29

    That should show the graph and you can see how it's not moved left, it's moved right! :confused:
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    (Original post by Dayxn2)
    Yeah, that's what I thought initially -- however, look:
    http://www.wolframalpha.com/input/?i=y%3Dln%28-x%2B1%29

    That should show the graph and you can see how it's not moved left, it's moved right! :confused:
    You are right. I made a mistake in my directions. First translate it TO THE LEFT and then reflect. So it goes one to the left then reflect.

    When you have -x+1 and you want x, first minus one and then multiply by minus one.

    Therefore i appologise and I was wrong. (but now I'm right)

    (I have edited my previous post)
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    http://www.wolframalpha.com/input/?i=y%3Dln|%283x-6%29|

    The graph the OP asked for.
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    (Original post by JTeighty)
    It is a lnx graph with the x values modded and then a horizontal translation of 6 to the right and then a stretch of a third.

    Asymptope will be x=2. Hence the question says x can't equal 2.

    Pls rep if this helps
    thanks ill rep you tomorrow--> already repped you today
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    (Original post by JTeighty)
    You are right. I made a mistake in my directions. First translate it TO THE LEFT and then reflect. So it goes one to the left then reflect.

    When you have -x+1 and you want x, first minus one and then multiply by minus one.

    Therefore i appologise and I was wrong. (but now I'm right)

    (I have edited my previous post)
    Okay, so do you always do the adding/subtracting first, then the multiplication/division? That's not what I was taught... but hey, if it works! :p:
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    (Original post by klgyal)
    thanks ill rep you tomorrow--> already repped you today
    Haha

    Thanks

    :-)
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    (Original post by JTeighty)
    Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph is moved horizontally to the left by one and then reflected in y axis.

    Hope this helps.

    If so pls rep
    i wrote that in another thread and i received three -ve reps saying that i was 'rep hunting'!!!
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    I would draw ln (3x-6) first. -->
    -6 so the graph is shifted to the right by 6
    3x so squish graph by x3
    this should so you ur asymptote is now at 2 (which works if you make 3x-6 = 0)

    then take the modulus of the function by reflecting eveything below the x axis up.

    I think....

    EDIT: seems this is wrong

    You say that take ln|x|, (reflect lnx in y axis) then x 3 means squash by 3, and - 6 means shift six to the right.
 
 
 
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