Finding asymptotes (horizontal / vertical) and obliques Ok, well I would like to ask a few questions.
Let's say we have a function $y=\displaystyle\frac{2 x^2}{x+2}$.
So, the $D_f$ would be $\Bbb R \setminus\{-2\}$ right.
To find the asymptotes, for vertical asymptote, do we search it if there is no $D_f$ limitation (only $\Bbb R$?). And is -2 the vertical asymptote, the same as $D_f$?
How do we find horizontal asymptote, and the obliques?
 A: $$f(x) = \dfrac {2x^2}{(x + 2)}$$
The domain of the function is $\mathbb R \setminus \{-2\}$. Yes, you're correct that there is a vertical asymptote at $x = -2$. Note, as the $\lim_{x \to -2^+} f(x) = +\infty$, and $\lim_{x \to -2^{-}} f(x) = -\infty$
You can check for horizontal asymptotes by determining if/when $\lim_{x\to \pm \infty} f(x)= \pm \infty$ (if yes, exists, if no, then not).
To check for oblique asymptotes, check the limiting behavior of $\dfrac{f(x)}{x}$ as $x \to \pm \infty$. If it evaluates to $k$ then the slope of the line of the asymptote is $k$.
It helps in problems like this, as in all problems, to graph the function:

Note that the range of your function is $y\leq -16,\;\;y\geq 0$
Note also the vertical asymptote at $x = -2$
A: *

*Yes, only $x=-2$ is not allowed, so the domain of $f$ is indeed $\Bbb R\setminus\{-2\}$.

*Yes, the vertical asymptote is where the function wants to be $\pm\infty$ (in $y$ coordinate), so in this case it is at $x=-2$. But, this is not the same as $D_f$, rather its complement.

*For the horizontal asymptote (if any) check $\lim_{\pm\infty}f$ (so where $x$ wants to be $\pm\infty$).

*For the oblique asymptotes, their slope can be calculated by $\lim_{x\to\pm\infty}\frac{f(x)}{x}$.

