Question

does the graph of \(y=-\log(x+1)\) have aysmptotes?

Answer 1

Yes. x = -1 is its vertical asymptote.

Answer 2

which one of these doesn't have asymptotes then?
\[y={2\over x^2-1}\]\[y=e^{x-2}\]\[y={4x\over x^2+1}\]\[xy=1\]

Answer 3

It will be xy = 1.

Answer 4

but xy=1 is \(y={1\over x}\) and that has an asymptote at x=0...??

Answer 5

Ah, true! Sorry about that. y = 4x / (x^2 + 1) should be the answer. =)

Answer 6

Since x^2 can never come out to be a negative number, the function will never be undefined.

Answer 7

ohhhh....i thought it saidm x^2-1 not +1 ...thanks @Dreweed ! :)

Answer 8

No problem! I am glad to help. ^.^