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ln graphs transformations.

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?
Reply 1
Avatar for Zacken
Zacken
22
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⁡xf(x) = \ln x then:

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

2) applying transformation 2 h(x)=g(x+4)=ln⁡(−2(x+4))=ln⁡(−2x−8)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)g(x) = f(-2x) = \ln(-2x) then

2) h(x)=g(x−2)=ln⁡(−2(x−2))=ln⁡(−2x+4)=ln⁡(4−2x)h(x) = g(x-2) = \ln (-2(x-2)) = \ln (-2x +4) = \ln (4-2x).
Reply 2
Avatar for SaadKaleem
SaadKaleem
OP
8
Original post by Zacken
If we call f(x)=ln⁡xf(x) = \ln x then:

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

2) applying transformation 2 h(x)=g(x+4)=ln⁡(−2(x+4))=ln⁡(−2x−8)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)g(x) = f(-2x) = \ln(-2x) then

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



Thank you so much! :smile: PRSOM
Reply 3
Avatar for James_mc
James_mc
2
Original post by SaadKaleem
Thank you so much! :smile: PRSOM


Just remember when you do a translation in x you replace all (x) with (x-a) so any coefficients affect the translation also :smile:
Reply 4
Avatar for SaadKaleem
SaadKaleem
OP
8
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 :smile:


Thank you aswell! :smile:
Reply 5
Avatar for Zacken
Zacken
22
Original post by SaadKaleem
Thank you so much! :smile: PRSOM


You're very welcome! :smile:
Reply 6
Avatar for SaadKaleem
SaadKaleem
OP
8
Original post by Zacken
You're very welcome! :smile:


Just noticed this....



Edexcel.... seriously? lol..
Reply 7
Avatar for Zacken
Zacken
22
Original post by SaadKaleem
Just noticed this....

Edexcel.... seriously? lol..


Hahaha. xD

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