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    What is the graph transformation f(1-x)? I'm trying to graph f(x)=2ln(1-x), and I got stretch in y axis factor 2, shift left 1 unit and reflect in y axis. However, the answers say it is a shift right one unit. I checked on an online graph maker and they also show a shift right. Can anyone explain why the 1 causes a shift to the right and not the left? I thought it had something to do with the fact it has been reflected but was not sure...
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    Actually I thought about it, I think it depends on the order in which you do the transformations.... Does anyone know what the correct order of doing transformations is? Like how we learn BODMAS, is there a kind of sequence that can be used to correctly do graph transformations?
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    (Original post by Uni12345678)
    What is the graph transformation f(1-x)? I'm trying to graph f(x)=2ln(1-x), and I got stretch in y axis factor 2, shift left 1 unit and reflect in y axis. However, the answers say it is a shift right one unit. I checked on an online graph maker and they also show a shift right. Can anyone explain why the 1 causes a shift to the right and not the left? I thought it had something to do with the fact it has been reflected but was not sure...
    Depends on the order, but there is no stretch in f(1-x) from f(x)

    Firstly it would be reflected in x=0 (the y-axis) - f(x) \rightarrow f(-x)

    Then it would be shifted by vector [1,0] - f(-x) \rightarrow f(-(x-1))=f(1-x)
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    (Original post by RDKGames)
    Depends on the order, but there is no stretch in f(1-x) from f(x)

    Firstly it would be reflected in x=0 (the y-axis) - f(x) \rightarrow f(-x)

    Then it would be shifted by vector [1,0] - f(-x) \rightarrow f(-(x-1))=f(1-x)
    Oh thank you! I get it! So when considering a transformation that involves moving along the x axis, the form must be including a positive x value as you have shown, yes?
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    (Original post by Uni12345678)
    Oh thank you! I get it! So when considering a transformation that involves moving along the x axis, the form must be including a positive x value as you have shown, yes?
    What do you mean?

    Translating y=f(x) by vector [a,b] would make it y-b=f(x-a)
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    (Original post by RDKGames)
    What do you mean?

    Translating f(x) by vector [a,0] would make it f(x-a)
    So when we learn the rules like f(x-1) is shift to right etc... The x must be positive for the transformation to be valid; if x is negative we can't just reflect and then move by 1 unit to the right because of -1, it must be to the right because the x is negative so we factorise to change it to. Positive by taking out negative 1 and that would make the transformation go to the left as it would create a positive 1...

    Okay it makes sense to me, I'm sorry it's complicated ahhaha
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    For transforming  f(x) to  f(ax+b) there are always two different sets of transformations.
    The first is a translation by vector  \begin{pmatrix} -b \\ 0 \end{pmatrix} followed by a stretch parallel to the x axis SF 1/a.
    It is also a stretch parallel to x axis SF 1/a followed by a translation by vector  \begin{pmatrix} -\frac{b}{a} \\ 0 \end{pmatrix} .
    These are the only order they can be in. Of course if a is negative then there will be a reflection in the y axis.
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    (Original post by Uni12345678)
    So when we learn the rules like f(x-1) is shift to right etc... The x must be positive for the transformation to be valid; if x is negative we can't just reflect and then move by 1 unit to the right because of -1, it must be to the right because the x is negative so we factorise to change it to. Positive by taking out negative 1 and that would make the transformation go to the left as it would create a positive 1...

    Okay it makes sense to me, I'm sorry it's complicated ahhaha
    The best advice I can give you is to put brackets around the x, and do any transformations concerning it inside that bracket. I will redo the first example with brackets to clear up any confusion and hopefully make it clearer:

    Reflection in the y-axis: f[(x)] \rightarrow f[(-x)]=f[-x]

    Translate by vector [1,0]: f[-(x)] \rightarrow f[-(x-1)]=f[1-x]

    Then for the banter we can choose to translate it by another vector [-8,0]: f[1-(x)] \rightarrow f[1-(x+8)]=f[-7-x]

    And then stretch it in along the x-axis by scale factor 1/2: f[-7-(x)] \rightarrow f[-7-(2x)]=f[-7-2x]
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    (Original post by RDKGames)
    The best advice I can give you is to put brackets around the x, and do any transformations concerning it inside that bracket. I will redo the first example with brackets to clear up any confusion and hopefully make it clearer:

    Reflection in the y-axis: f[(x)] \rightarrow f[(-x)]=f[-x]

    Translate by vector [1,0]: f[-(x)] \rightarrow f[-(x-1)]=f[1-x]

    Then for the banter we can choose to translate it by another vector [-8,0]: f[1-(x)] \rightarrow f[1-(x+8)]=f[-7-x]

    And then stretch it in along the x-axis by scale factor 1/2: f[-7-(x)] \rightarrow f[-7-(2x)]=f[-7-2x]
    Thank you! You've been very helpful, also your banter is brilliant 😂
 
 
 
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