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    Hi, for the equation y=ln(4-x) why doesnt the graph cross the negative x axis since +4 shifts the graph to the left. I first reflected the graph over the y axis. I found that you have to changethe equation to this : y=ln-(x-4) ... Why ?thanks
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    (Original post by coconut64)
    Hi, for the equation y=ln(4-x) why doesnt the graph cross the negative x axis since +4 shifts the graph to the left. I first reflected the graph over the y axis. I found that you have to changethe equation to this : y=ln-(x-4) ... Why ?thanks
    It's going to cross the x-axis at the solution of ln(4-x)=0.

    Also the transformations that take y=lnx to y=ln(4-x) are

    reflection in the y-axis
    translation to the right by 4
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    (Original post by RichE)
    It's going to cross the x-axis at the solution of ln(4-x)=0.

    Also the transformations that take y=lnx to y=ln(4-x) are

    reflection in the y-axis
    translation to the right by 4
    My question is why is the translation to the right not to the left as its +4 thanks
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    (Original post by coconut64)
    Hi, for the equation y=ln(4-x) why doesnt the graph cross the negative x axis since +4 shifts the graph to the left. I first reflected the graph over the y axis. I found that you have to changethe equation to this : y=ln-(x-4) ... Why ?thanks
    I asked this question last year,

    (Original post by notnek)
    Let's try that.

    Start with f(x) = \ln x

    "Reflect the graph in the y axis" - this is a transformation f(x)\rightarrow f(-x)

    f(-x) = \ln (-x)
    Call this new function g(x) then "shift the graph 4 units in the negative x direction"

    is a transformation g(x)\rightarrow g(x+4).g(x) = \ln (-x)

    g(x+4) = \ln -(x+4) = \ln (-x -4)

    Can you see your mistake?
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    (Original post by NotNotBatman)
    I asked this question last year,
    The last part is the bit i dont get. How does it go from g(x+4) to g-(x+4) thanks
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    (Original post by coconut64)
    My question is why is the translation to the right not to the left as its +4 thanks
    reflection in the y-axis takes it to y=ln(-x)

    Translation to the right by 4 takes it to y = ln(-(x-4)) = ln(4-x)
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    (Original post by coconut64)
    The last part is the bit i dont get. How does it go from g(x+4) to g-(x+4) thanks
    It's g[-(x+4)] = g(-x-4)

    When you multiply by -1, you have to do it to the whole quantity in the bracket.
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    (Original post by NotNotBatman)
    It's g[-(x+4)] = g(-x-4)

    When you multiply by -1, you have to do it to the whole quantity in the bracket.
    Why dovyou multiply it by -1?
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    (Original post by coconut64)
    Why dovyou multiply it by -1?
    After the reflection you shifted the graph 4 units to the left, so replace any x with an x+4, so f(-x) becomes f(-(x+4)) which is wrong.
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    You can do the transformations in which ever order you're most comfortable with but depending on which one you do first the second transformation may be different in each case, you can either see it as a translation by vector (-4 0) followed by a reflection in the y axis OR you can say it is a reflection in the y axis followed by a translation by vector (4 0). Notice that the translations in each case are not the same.
    Either of these are correct.
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    (Original post by coconut64)
    My question is why is the translation to the right not to the left as its +4 thanks
    You start with this:



    Let's start with f(x) = \ln x. If you want to do f(x+ 4) then you get f(x+4) = \ln (4 + x) which shifts it to the left. But \ln (4+x) \neq \ln (4-x) (unsurprisingly).

    So to get f(x) = \ln (4-x) you need to do two things:

    1. f(-x) = \ln (-x) which reflects the graph in the y-axis. This gets you this:



    This is your new graph. Let's call it g(x) = \ln (-x). The graph of it is as above. Now you are going to do g(x - 4) = \ln (-(x-4)) = \ln (-x + 4) = \ln (4-x). This will shift the graph to the right by 4 units.

    The reason why g(x -4) = \ln (4-x) is because you need to replace the the x in g(x) by x-4.

    That is, pretend you have g(x) = \ln (-(x)). Then you need to replace (x) with (x-4).

    This gets you f(x-4) = \ln (-(x-4)) = \ln (-x + 4).

    If you had \ln x being transformed to \ln 2x then a shift, say f(x) = \ln 2x and you wanted to shift it 3 units to the left you'd do \ln (2(x + 3)) = \ln (2x + 6). Not \ln (2x + 3). Capiche?
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    (Original post by Zacken)
    You start with this:



    Let's start with f(x) = \ln x. If you want to do f(x+ 4) then you get f(x+4) = \ln (4 + x) which shifts it to the left. But \ln (4+x) \neq \ln (4-x) (unsurprisingly).

    So to get f(x) = \ln (4-x) you need to do two things:

    1. f(-x) = \ln (-x) which reflects the graph in the y-axis. This gets you this:



    This is your new graph. Let's call it g(x) = \ln (-x). The graph of it is as above. Now you are going to do g(x - 4) = \ln (-(x-4)) = \ln (-x + 4) = \ln (4-x). This will shift the graph to the right by 4 units.

    The reason why g(x -4) = \ln (4-x) is because you need to replace the the x in g(x) by x-4.

    That is, pretend you have g(x) = \ln (-(x)). Then you need to replace (x) with (x-4).

    This gets you f(x-4) = \ln (-(x-4)) = \ln (-x + 4).

    If you had \ln x being transformed to \ln 2x then a shift, say f(x) = \ln 2x and you wanted to shift it 3 units to the left you'd do \ln (2(x + 3)) = \ln (2x + 6). Not \ln (2x + 3). Capiche?
    Ah, okay that does ring a bell. I think I get it now, I will just do more practice to make it stick in my head. Thanks for your very detailed answer. Also, thanks for helping me with mechanics previously, I managed to get 93 ums!
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    Thanks guys I get it now, after a long time.
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    (Original post by coconut64)
    Ah, okay that does ring a bell. I think I get it now, I will just do more practice to make it stick in my head. Thanks for your very detailed answer. Also, thanks for helping me with mechanics previously, I managed to get 93 ums!
    Basically, the best thing to do is to always stick a bracket around (x) in the function when you see it, or at least do so whilst learning about it - you'll get used to it after .

    Like shifting \sqrt{2x} by 5 units. Write it as \sqrt{2(x)} then do the shift as \sqrt{2(x +5)}.

    (And congratulations on your mechanics! 93 UMS is fantastic. Glad I helped)
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    (Original post by Zacken)
    Basically, the best thing to do is to always stick a bracket around (x) in the function when you see it, or at least do so whilst learning about it - you'll get used to it after .

    Like shifting \sqrt{2x} by 5 units. Write it as \sqrt{2(x)} then do the shift as \sqrt{2(x +5)}.

    (And congratulations on your mechanics! 93 UMS is fantastic. Glad I helped)
    Will do, thanks again.
 
 
 
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