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# C3 modulus graphs watch

1. how would you draw the graph of y=lnl3x-6l XER where x doesnt= 2
and how do work out where the asymptote is ???
2. X doesn't equal 2, massive hint to where the asymptote is, no?
3. is it ln(3x-6) or ln(3x)-6?
4. Ok, think of the graph for ln x, call it f(x)...then ln |3x-6| becomes f(|3x-6|)...does that help?
5. I'm super confused with this as well. It would be really helpful if someone could tell me how you're meant to sketch graphs like ln(-x+1) -- I can never seem to get them right when they have that minus plus a constant. Thanks
6. (Original post by *MJ*)
X doesn't equal 2, massive hint to where the asymptote is, no?
ohhh i didnt know that meant thats where the asymptote will be thanks...how would you sketch it ??
7. (Original post by n_251)
Ok, think of the graph for ln x, call it f(x)...then ln |3x-6| becomes f(|3x-6|)...does that help?
hmm..a bit confused
8. (Original post by unamed)
is it ln(3x-6) or ln(3x)-6?
is it ln l3x-6l --> the lines mean that modulus sign
9. (Original post by klgyal)
is it ln l3x-6l --> the lines mean that modulus sign
I think I was asking so I knew where the lnx translates to..
10. (Original post by Dayxn2)
I'm super confused with this as well. It would be really helpful if someone could tell me how you're meant to sketch graphs like ln(-x+1) -- I can never seem to get them right when they have that minus plus a constant. Thanks
Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph is moved horizontally to the left by one and then reflected in y axis.

Hope this helps.
11. (Original post by JTeighty)
Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph reflected in y axis and moved horizontally to the left by one.

Hope this helps.

If so pls rep
how would you draw my graph please???
12. (Original post by klgyal)
how would you draw the graph of y=lnl3x-6l XER where x doesnt= 2
and how do work out where the asymptote is ???
It is a lnx graph with the x values modded and then a horizontal translation of 6 to the right and then a stretch of a third.

Asymptope will be x=2. Hence the question says x can't equal 2.
13. (Original post by JTeighty)
Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph reflected in y axis and moved horizontally to the left by one.

Hope this helps.

If so pls rep

Yeah, that's what I thought initially -- however, look:
http://www.wolframalpha.com/input/?i=y%3Dln%28-x%2B1%29

That should show the graph and you can see how it's not moved left, it's moved right!
14. (Original post by Dayxn2)
Yeah, that's what I thought initially -- however, look:
http://www.wolframalpha.com/input/?i=y%3Dln%28-x%2B1%29

That should show the graph and you can see how it's not moved left, it's moved right!
You are right. I made a mistake in my directions. First translate it TO THE LEFT and then reflect. So it goes one to the left then reflect.

When you have -x+1 and you want x, first minus one and then multiply by minus one.

Therefore i appologise and I was wrong. (but now I'm right)

(I have edited my previous post)
15. http://www.wolframalpha.com/input/?i=y%3Dln|%283x-6%29|

The graph the OP asked for.
16. (Original post by JTeighty)
It is a lnx graph with the x values modded and then a horizontal translation of 6 to the right and then a stretch of a third.

Asymptope will be x=2. Hence the question says x can't equal 2.

Pls rep if this helps
thanks ill rep you tomorrow--> already repped you today
17. (Original post by JTeighty)
You are right. I made a mistake in my directions. First translate it TO THE LEFT and then reflect. So it goes one to the left then reflect.

When you have -x+1 and you want x, first minus one and then multiply by minus one.

Therefore i appologise and I was wrong. (but now I'm right)

(I have edited my previous post)
Okay, so do you always do the adding/subtracting first, then the multiplication/division? That's not what I was taught... but hey, if it works!
18. (Original post by klgyal)
thanks ill rep you tomorrow--> already repped you today
Haha

Thanks

:-)
19. (Original post by JTeighty)
Think of the original graph. f(x)=lnx then use translations for f(-x+1). so the lnx graph is moved horizontally to the left by one and then reflected in y axis.

Hope this helps.

If so pls rep
i wrote that in another thread and i received three -ve reps saying that i was 'rep hunting'!!!
20. I would draw ln (3x-6) first. -->
-6 so the graph is shifted to the right by 6
3x so squish graph by x3
this should so you ur asymptote is now at 2 (which works if you make 3x-6 = 0)

then take the modulus of the function by reflecting eveything below the x axis up.

I think....

EDIT: seems this is wrong

You say that take ln|x|, (reflect lnx in y axis) then x 3 means squash by 3, and - 6 means shift six to the right.

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