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).
Best Answer
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.