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C3 maths exponential functions

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1. Can anyone help? I have drawn the function ln x and transformed it using the transformation rules. However, I have drawn ln (-x) and that doesn't seem to follow the transformation rules. Anyone know why?
Thanks
2. It does,y = ln(-x) is a reflection in the y axis of the curve y=ln(x)
3. (Original post by peterhw)
Can anyone help? I have drawn the function ln x and transformed it using the transformation rules. However, I have drawn ln (-x) and that doesn't seem to follow the transformation rules. Anyone know why?
Thanks
Depends on the transformation you're trying to perform. If it's a reflection in the y-axis (x=0), then well done you got it.
4. I should maybe ha e been clearer. If I plot ln(X+4) ln X moves left 4 but if I plot ln(-X+4) then ln(-X) moves right 4. Why is that?
5. (Original post by peterhw)
I should maybe ha e been clearer. If I plot ln(X+4) ln X moves left 4 but if I plot ln(-X+4) then ln(-X) moves right 4. Why is that?
It doesn't work like that.

is a translation by vector [-4,0] (so you got that bit correct)

is a translation by vector [-4,0] followed by a reflection in the y-axis.

Translation by vector [4,0] would be represented by which is not the same as
6. (Original post by peterhw)
I should maybe ha e been clearer. If I plot ln(X+4) ln X moves left 4 but if I plot ln(-X+4) then ln(-X) moves right 4. Why is that?
This gets asked a lot, look at post 11, credit goes to Zacken for this good explanation.
http://www.thestudentroom.co.uk/show...4#post67429014
7. (Original post by RDKGames)
It doesn't work like that.

is a translation by vector [-4,0] (so you got that bit correct)

is a translation by vector [-4,0] followed by a reflection in the y-axis.

Translation by vector [4,0] would be represented by which is not the same as
Thank you

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