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    For example take f(x) = ln(4-2x) and we have to sketch this:

    Can we not apply transformations to this eg: 1) ln(-2x) and then 2) ln(-2x+4)

    ln(-2x) graph is this according to Wolfram:


    if you apply f(x+4) to it, it goes to (-4.5,0)

    However, the actual graph of ln(4-2x) is quite different, with (1.5,0) x intercept.

    Am I applying the transformations wrong? I know that, you could easily substitute x=0 and y=0 and sketch it, however this is something which confused me.

    By the way, I noticed that f(-2x+4) could be rewritten as f(-2(x-2)), does this have to do something with it? and if so Why?
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    (Original post by SaadKaleem)
    For example take f(x) = ln(4-2x) and we have to sketch this:

    Can we not apply transformations to this eg: 1) ln(-2x) and then 2) ln(-2x+4)
    If we call f(x) = \ln x then:

    1) Applying transformation 1 gives us g(x) = f(-2x) = \ln (-2x) then

    2) applying transformation 2 h(x) = g(x + 4) = \ln (-2(x+4)) = \ln (-2x - 8).

    What you want to do if you want to do it in the order you suggested, then you need to apply:

    1) g(x) = f(-2x) = \ln(-2x) then

    2) h(x) = g(x-2) = \ln (-2(x-2)) = \ln (-2x +4) = \ln (4-2x).
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    (Original post by Zacken)
    If we call f(x) = \ln x then:

    1) Applying transformation 1 gives us g(x) = f(-2x) = \ln (-2x) then

    2) applying transformation 2 h(x) = g(x + 4) = \ln (-2(x+4)) = \ln (-2x - 8).

    What you want to do if you want to do it in the order you suggested, then you need to apply:

    1) g(x) = f(-2x) = \ln(-2x) then

    2) h(x) = g(x-2) = \ln (-2(x-2)) = \ln (-2x +4) = \ln (4-2x).

    Thank you so much! PRSOM
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    (Original post by SaadKaleem)
    Thank you so much! PRSOM
    Just remember when you do a translation in x you replace all (x) with (x-a) so any coefficients affect the translation also
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    (Original post by James_mc)
    Just remember when you do a translation in x you replace all (x) with (x-a) so any coefficients affect the translation also
    Thank you aswell!
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    (Original post by SaadKaleem)
    Thank you so much! PRSOM
    You're very welcome!
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    (Original post by Zacken)
    You're very welcome!
    Just noticed this....



    Edexcel.... seriously? lol..
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    (Original post by SaadKaleem)
    Just noticed this....

    Edexcel.... seriously? lol..
    Hahaha. xD
 
 
 
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