# Math Help - about ln functions

For ln(x^2+1)

Here's what I have written:

Domain: all real numbers
x intercept: 0
Asymptote:

Not sure....I mean, I know that y will never be negative...is that an asymptote? Is there a vertical asymptote?

2. Originally Posted by theowne
For ln(x^2+1)

Here's what I have written:

Domain: all real numbers
x intercept: 0
Asymptote:

Not sure....I mean, I know that y will never be negative...is that an asymptote? Is there a vertical asymptote?
x = 0 is a vertical asymptote for y = ln(x). However, since your argument never gets to 0 it does not have one.

-Dan

3. However, since your argument never gets to 0 it does not have one.
What do you mean by this? Do you mean f(x)? Isn't f(x) = 0 when x=0?

4. Originally Posted by theowne
What do you mean by this? Do you mean f(x)? Isn't f(x) = 0 when x=0?
The argument is the quantity inside the function. So the argument of $ln(x^2 + 1)$ is $x^2 + 1$.

-Dan