Natural log graph transformation

I have a question that asks you to sketch the graph y=ln(4-x) I am able to get the correct answer by thinking about what will happen if I substituted numbers into the function. i.e. if x=4 then y=ln(4-4)=ln(0) which is impossible, therefore at x=4 there is an asymptote.

So after getting the answer correct this way I decided to try and do the same thing but working it out using the graph transformation rules. so the -x means that the graph y=ln(x) is reflected in the y axis, this is fine so far. But then the positive 4 would mean transforming this graph horizontally by a factor of -4 (as y=f(x+a) translates the graph horizontally by -a) (so moving it four spaces to the left, making the asymptote at x=-4) but this is not the case as it has been translated horizontally by a factor of +4 (it has moved four spaces to the right making the asymptote at x=4).

I'm guessing i've gone wrong somewhere so can someone please explain this to me as I wan't my understanding to be solid and be able to the question both ways (so as to check answers) thanks in advance :smile:

(edited 12 years ago)

Reply 1

MrGum

f(x+4) is a translation of f(x) by -4 in the x-direction as you say.

That is, if x is replaced with x+4. you get that translation.

but if f(x) is ln(-x), replacing x with x+4 gives ln(-(x+4)) which is not ln(-x+4).

What does x have to be replaced with to change ln(-x) into ln(-x+4) ?

(edited 12 years ago)

Reply 2

tdx

OP

6

ah right so if I was to think of ln(4-x) or ln(-x+4) in terms of different transformations of x I would need to set it out like ln(-(-4+x)) or ln(-(x-4)) therefore it is being translated by f(-x) which is a reflection in the y axis and also translated by f(x-4) which moves it along the x axis by +4? that correct? :biggrin:

Edit to answer your question x would need to be replaced with (-4+x) or (x-4)? :smile:

(edited 12 years ago)

Reply 3

MrGum

Basically yes, but I think the language and notation could be tidied up a bit (translation/transformation are not interchangeable and f(x) seems to be shifting in its meaning)

If f(x)=lnx

then f(-x)=ln(-x): so far we have a reflection in the y-axis.

Now if g(x)=ln(-x)

then ln(4-x)=g(x-4), so we get a translation of +4 in the x-direction.

The other way is to do the translation first

f(x)=ln(x), f(x+4)=ln(x+4): translation of -4 in the x-direction

Now if h(x)=ln(x+4), ln(-x+4)=h(-x), so we get a reflection in the y-axis.

So the overall transformation could be either "reflection in the y-axis followed by translation of +4 in x-direction" or "translation of -4 in x-direction followed by reflection in the y-axis". Note that swapping the order affects the transformations needed.

In this case it seems more natural to do it the first way, because it's more obvious to go from x to -x as a first step than to go to x+4, but either way works.

Reply 4

MrGum

Original post by tdx

Edit to answer your question x would need to be replaced with (-4+x) or (x-4)? :smile:

Yes!

Reply 5

tdx

OP

6

Original post by Mr Gum

Basically yes, but I think the language and notation could be tidied up a bit (translation/transformation are not interchangeable and f(x) seems to be shifting in its meaning)

If f(x)=lnx

then f(-x)=ln(-x): so far we have a reflection in the y-axis.

Now if g(x)=ln(-x)

then ln(4-x)=g(x-4), so we get a translation of +4 in the x-direction.

The other way is to do the translation first

f(x)=ln(x), f(x+4)=ln(x+4): translation of -4 in the x-direction

Now if h(x)=ln(x+4), ln(-x+4)=h(-x), so we get a reflection in the y-axis.

So the overall transformation could be either "reflection in the y-axis followed by translation of +4 in x-direction" or "translation of -4 in x-direction followed by reflection in the y-axis". Note that swapping the order affects the transformations needed.

In this case it seems more natural to do it the first way, because it's more obvious to go from x to -x as a first step than to go to x+4, but either way works.

ok, thanks :smile:

I get it now :biggrin:

and to tidy up my last post:

ah right so if I was to think of ln(4-x) or ln(-x+4) in terms of different transformations I would need to set it out like ln(-(-4+x)) or ln(-(x-4)) therefore it is being transformed by f(-x) which is a reflection in the y axis and also transformed by f(x-4) which translates it along the x axis by +4? that correct?

Edit: adding this I could also set it up as different functions with g(x)=ln(x) h(x)=x-4 and m(x)=-x if f(x)=ln(4-x) then f(x)=gmh(x)?

(edited 12 years ago)

Reply 6

MrGum

Original post by tdx

ok, thanks :smile:

I get it now :biggrin:

and to tidy up my last post:

ah right so if I was to think of ln(4-x) or ln(-x+4) in terms of different transformations I would need to set it out like ln(-(-4+x)) or ln(-(x-4)) therefore it is being transformed by f(-x) which is a reflection in the y axis and also transformed by f(x-4) which translates it along the x axis by +4? that correct?

Edit: adding this I could also set it up as different functions with g(x)=ln(x) h(x)=x-4 and m(x)=-x if f(x)=ln(4-x) then f(x)=gmh(x)?