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    how would you draw ln(3-x) - would you reflect ln(x) in the y-axis and then move it to the left by 3 or would you work out separetly where it cuts the x and y axis and then draw it
    please helpppp i soo confused what to do!!!!!!!!!
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    I'd go about it by getting the general graph in my head by doing what you said first, it's reflected in Y, then moved 3 to the RIGHT (asymptote at x=3). Then work out where it crosses the axis so you can draw it well
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    You can do either tbh.
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    (Original post by sheligalo)
    I'd go about it by getting the general graph in my head by doing what you said first, it's reflected in Y, then moved 3 to the RIGHT (asymptote at x=3). Then work out where it crosses the axis so you can draw it well
    why would you move it to the RIGHT? shouldn't it be to the LEFT
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    what paper is this????
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    (Original post by Remarqable M)
    what paper is this????
    its from my edexcel maths book
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    You definitely move it to the left along the x-axis, by -3.
    new asymtope will b at x=-3.
    it will cross y-axis at (0,ln3)
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    after reflection it moves 3 to the right
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    (Original post by infinity123321)
    after reflection it moves 3 to the right
    how how can it move to the right? the equation is f(x)=ln(-x+3)
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    your sketch has to be wrong
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    i used my graphical calculator and thats what it showed
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    (Original post by morebritishthanindian)
    your sketch has to be wrong
    oooh i seee what i got wrong anyways here is new sketch.
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    shouldnt it be :
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    The person who sketched their graph above (Now they deleted it) is wrong. lnx is a reflection of y=e^x in y=x not y=0.

    So the graph would be a translation of 3 to the left and then a reflection in x y axis.
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    It is the graph of y=ln(-x), but shifted along the x axis by +3.

    Here is a diagram, but where the graph just seems to stop at y=3, that's where the curve is tending to -infinity.
    http://tinyurl.com/yfhlwtb
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    (Original post by Remarqable M)
    oooh i seee what i got wrong anyways here is new sketch.
    :yep: thats right sorryi did e^x
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    (Original post by JTeighty)
    The person who sketched their graph above (Now they deleted it) is wrong. lnx is a reflection of y=e^x in y=x not y=0.

    So the graph would be a translation of 3 to the left and then a reflection in x y axis.
    i just realised it now! thank goodness i came on here
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    (Original post by Remarqable M)
    oooh i seee what i got wrong anyways here is new sketch.
    Correct
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    (Original post by JTeighty)
    Correct
    what about f(X)=-2^-x
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    (Original post by Remarqable M)
    what about f(X)=-2^-x
    Draw curve of y= 2^x (exponential curve passing through (0,1)

    Then reflect this in the y axis.
    Then reflect this in the x axis.
 
 
 
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