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    (Original post by imasillynarb)
    Can anyone show me what the graph of ln|3x-6| looks like?
    this is how i drew it:
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    (Original post by crana)
    If you can imagine it this is how I drew it:

    Asymptote at x = 2

    line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

    then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

    rosie
    i got a dip at x=2

    coz dy/dx = 3/(3x-6) = 0 hence x is 2

    but i couldnt get a corresponding y value ... ill upload an image from my graphic calc

    it was wierd in a way ...
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    (Original post by mockel)
    this is how i drew it:
    spot on my friend ... thts wat i did ... but i made the curve come very close to x=2 so tht it seemed tht they were joined

    (Original post by infekt)
    i got a dip at x=2

    coz dy/dx = 3/(3x-6) = 0 hence x is 2

    but i couldnt get a corresponding y value ... ill upload an image from my graphic calc

    it was wierd in a way ...
    the only thing is that at x=2, y is undefined, so I don't think dy/dx is valid there if you see what I mean? Which is why you couldn't get a y value.

    Im no great shakes at maths but if there is no "turning point" as such then the dy/dx = 0 could be that at x =2 the gradient *is* 0, because there is no graph there?

    If that makes sense.

    I dont have a graphics calculator but I am fairly sure that it is undefined ie asymptote at x=2
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    (Original post by crana)
    (your way is nicer MB!)
    Thanks,

    MB
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    (Original post by BossLady)
    hey guys, was ques 5 the binomials right with [x^3-1/2x]^12? and then 6 was that iteration stuff (zzz) and 7 was that trap rule stuff?
    you know for the iteration, how did u show it was 1.70 or whatever, did u just choose 1.65 and 1.75, stick it in and show sign change ya/?
    Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?

    For the "independent of x" term in the binomial question, it's possible without expanding the whole expression. Since the first term has postive index (x^3) and the second term has negative index (-0.5x^-1), you need to find the stage in the binomial expansion that makes both x terms have the same index number, and then they will cancel each other, leaving only a number and no x-term. So you need the term when (-0.5x^-1) is to the power of 9, and therefore (x^3) will be to power of 3. (Total powers must equal 12, because that is n.)

    So you then get 12C9 . (x^3)^3 . (-0.5x^-1)^9
    the x terms cancel out and you're left with a fraction which cancels down to
    -55/128

    (Original post by satin)
    Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?
    That;s what I did (1.695 etc)
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    Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?

    spot on dude!

    btw, what was the one for.. I think it was the iteration one and what value of x0 is this not valid for?

    I think I put as it was something like root (blah blah /x+1) that it was when x=-1 so it is over 0 and hence undefinined..
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    (Original post by infekt)
    spot on my friend ... thts wat i did ... but i made the curve come very close to x=2 so tht it seemed tht they were joined
    yeah i did that in the exam, but my mouse control isn't very good
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    hey ppl, u no wat sum1 jus told me?? dat last nyt a copy of da p2 edexcel got stolen+sum yoke put it online sumwer askin 4 help wif ans bt den da site ppl contacted edexcel! so wat y'all tink will happen?? :confused: c if we afta do it again.... :mad:
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    could the graph go below the x axis??? i thought it would reflect in x=2 but i tohught that it wouldnt cross the axis
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    (Original post by mockel)
    yeah i did that in the exam, but my mouse control isn't very good
    What I mean is, anyone have some sort of graph sketching program and can like save the picture and post it for me?

    (Original post by lexazver203)
    could the graph go below the x axis??? i thought it would reflect in x=2 but i tohught that it wouldnt cross the axis
    Sure it can

    for example: x = 2.01
    y = ln |3x - 6| = ln (0.03) = -3.506..

    (Original post by imasillynarb)
    What I mean is, anyone have some sort of graph sketching program and can like save the picture and post it for me?
    Try
    http://www.coolmath.com/graphit/

    however it does seem to get a tad confused with modulus's, you can't really put them in.
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    (Original post by satin)
    Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?
    Yeah, thats what I did
 
 
 
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