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# ln graphs transformations.

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1. 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?
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 then:

1) Applying transformation 1 gives us then

2) applying transformation 2 .

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

1) then

2) .
3. (Original post by Zacken)
If we call then:

1) Applying transformation 1 gives us then

2) applying transformation 2 .

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

1) then

2) .

Thank you so much! PRSOM
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
5. (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!
Thank you so much! PRSOM
You're very welcome!
7. (Original post by Zacken)
You're very welcome!
Just noticed this....

Edexcel.... seriously? lol..
Just noticed this....

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

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