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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).

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Don't you mean $\ln x + 2$, or maybe $\ln(2+x)$? –  Javier Badia 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
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2 Answers

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.

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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 Badia 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 Badia Aug 21 '12 at 20:52
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This is a straight line, so no asymptotes. y-intercept is at ln(2)

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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
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