Just to elaborate on Michael Hardy's answer a little bit...
Suppose we have a rational function written like $f(x)=\frac{p(x)}{q(x)}$. Performing the long division $p(x)\div q(x)$ gives you a different way of writing the same function, $$f(x)=a(x)+\frac{r(x)}{q(x)},$$ as a polynomial plus a rational function whose numerator's degree is smaller than its denominator's degree. (To be more specific, $a(x)$ has degree equal to the degree of the original numerator $p(x)$ minus the degree of the denominator $q(x)$. The degree of $r(x)$ is smaller than that of $q(x)$ because $r(x)$ is the remainder of the long division.)
As you already understand, $\frac{r(x)}{q(x)}$ tends to zero as $x$ tends to infinity, because the degree of the numerator is smaller than the degree of its denominator. That means that as $x\rightarrow\infty$, the function $f(x)$ tends towards the polynomial $a(x)$.
If $p(x)$ has degree exactly one more than that of $q(x)$, then $a(x)$ has degree one: it's a line. We call this line a slant or oblique asymptote of $f(x)$. If the degree of $p(x)$ is more than one higher than that of $q(x)$, then $a(x)$ is a quadratic, or a cubic, or whatever, but it's not a line. So $f(x)$ doesn't have a slant asymptote in this direction. (We still might say something like $f(x)$ asymptotically approaches a(x).)
Hopefully this explains why asymptotes only occur when the degree of the numerator is exactly one more than that of the denominator. It also might give you a hint for how you can find slant asymptotes of functions that aren't rational: if you can rewrite your function as a line plus something that goes to zero, you've got yourself an asymptote!
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.
Best Answer
I assume that's $y=-{2\over x+3}-1$. There's a vertical asymptote where the function is undefined, and the function is undefined where it involves division by zero. For the horizontal asymptote, you have to ask yourself, what happens when $x$ grows without bound? when $x$ gets very large negative?
EDIT: Reading OP's comments and revised question, I think maybe this approach will be more what's wanted.
I take it you know that for $y=1/x$ the horizontal asymptote is $y=0$ and the vertical asymptote is $x=0$, and they meet at the point $(0,0)$. $y=1/(x+3)$ is the same graph but shifted 3 units to the left; that doesn't change the horizontal asymptote, but it shifts the vertical asymptote to $x=-3$. $y=-1/(x+3)$ flips (or, reflects) the graph in the $x$-axis, but has no effect on the asymptotes, which remain $y=0$ and $x=-3$. $y=-2/(x+3)$ stretches the graph, but again has no effect on the asymptotes. Finally, $y={-2\over x+3}-1$ lowers the graph by 1, so it changes the horizontal asymptote to $y=-1$, while not affecting the vertical asymptote.
So, the new asymptotes are $y=-1$ and $x=-3$, which meet at the point $(-3,-1)$, which is what you are calling the middle point.