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- anonymous

Find all asymptotes and intercepts:
y = 2ln(x-3)
How do I find the asymptotes and intercepts? I want to understand the reasoning.

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- anonymous

- jamiebookeater

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- anonymous

ln is your clue

- anonymous

I know it should be shifted by 3 to the right and the answer in the back of the book shows that as a vertical asymptote (x = 3), and I know multiplying by 2 will make the function larger. I just don't understand why x = 3 becomes the vertical asymptote and not the x intercept.

- anonymous

Because ln is always greater than 0

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- amistre64

ln(x) never equals 0 or less than 0

- amistre64

ln(x-3) is undefined at 3 and below

- anonymous

because you can't take ln(0) ?

- amistre64

you can try, but it like going to newyork to vist the grand canyon; when you get there, it aint gonna happen lol

- anonymous

So how could we write limit notation for the function?

- amistre64

lim[x->3] ln(x-3) = -inf if anything

- amistre64

that might have to be adjusted to say 'from the right'

- anonymous

This was actually a question from a graphing exercise. Trying to understand all of this so I can graph these correctly.

- anonymous

Thanks for the help!

- amistre64

yw :)

- anonymous

Amistre, you did so well on this and that triangle problem earlier, please take a look at this one, one guy is trying to find the answer, and I have tried everything http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4dd992160ada8b0bef2a41c7

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