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?
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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
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?
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 $\displaystyle ln(x^2 + 1)$ is $\displaystyle x^2 + 1$. -Dan
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