Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Graph $f(x)=\ln x+2$ And find all intercepts and asymptotes.

I know exactly how the graph looks and I have a sketch in front of me. Now, for the intercepts, $x$-int=$-1$ and $y$-int$=-\infty+2$ which I'm guessing is just $-\infty$ because adding $2$ wouldn't make much of a difference on that level. So, I have my intercepts (correct me if they are wrong please) now for the asymptotes. For vertical: $\log(x)+2 \rightarrow \infty$ as $x\rightarrow 0$ thus making the vertical asymptote $0$. And for horizontal, it has to do with when $y \rightarrow\pm\infty$. But how do I find this. (The inly reason I know that the vertical asymptote is $0$ is because it's a $\log x$ function moved up $2$ units).

share|cite|improve this question
Don't you mean $\ln x + 2$, or maybe $\ln(2+x)$? – Javier Aug 21 '12 at 19:52
I completely switched those two around. It's corrected – Austin Broussard Aug 21 '12 at 19:53
If you are looking at $2+\ln(x)$ (I switched for clarity) then there is no $y$-intercept. The $x$-intercept is where $\ln(x)=-2$, which happens at $x=e^{-2}$. – André Nicolas Aug 21 '12 at 19:58
up vote 2 down vote accepted

It's not correct to say that the x-intercept is at $x=-1$. The x-intercept occurs when $f(x) = 0$, so we have to solve the following:

$$\begin{align} \ln x + 2 &= 0 \\ \ln x &= -2 \\ x &= e^{-2} \approx 0.13\end{align} $$

While your y-intercept is sort of correct, it doesn't make a lot of sense to say the the regular $\ln x $ funcion has a y-intercept at $y=-\infty$. Rather, you should say that the y-intercept is the value of $f(x)$ at $x=0$. But $f(x) = \ln x + 2$ is not defined at $x=0$, so this funcion doesn't have a y-intercept.

You mention at the end that you know this funcion is like $\ln x$ moved up two units. That should be all you need to know: It will still have the vertical asymptote at $x=0$, it will still go to infinity as $x$ goes to infinity, it will still have only one x-intercept. The shape is the same, but moved vertically.

share|cite|improve this answer
I did the wrong steps and got the wrong answer for the $x$-intercept. So my $y$-int is undefined? – Austin Broussard Aug 21 '12 at 20:02
And my real question is mainly about the horizontal asymptote. – Austin Broussard Aug 21 '12 at 20:02
@AustinBroussard: This is a minor terminology thing. Rather than undefined, I would say that there is no y-intercept. The functions $\ln x$ and $\ln x + 2$ are undefined at $x=0$, so there is no y-intercept. – Javier Aug 21 '12 at 20:04
Okay, I see what you mean now. How would I approach the horizontal asymptote? – Austin Broussard Aug 21 '12 at 20:05
@AustinBroussard: Yes, that's right. – Javier Aug 21 '12 at 20:52

This is a straight line, so no asymptotes. y-intercept is at ln(2)

share|cite|improve this answer
The problem is corrected. It is $f(x)=\ln x+2$ not $f(x)=\ln 2+x$ – Austin Broussard Aug 21 '12 at 19:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.