Given any second degree equation in $x$ and $y$,
$ax^2+by^2+2gx+2fy+2hxy+c=0$
is it possible to find out the centre and/or the axis of the conic section it represents?
What information can I gather simply by looking at the curve?
For example one thing my teacher taught me was that if $h=0$ the axis of the curve(be it parabola,hyperbola,ellipse etc) is parallel to the coordinate axes.
I am looking for some other information that the equation can offer.
For obtaining the center (if there is one: if it is a parabola, there is no center), compute the partial derivatives (ask your teacher if necessary) of $f$ with respect to $x$, then to $y$, and write that they are zero: you will get a system of 2 linear equations with two unknowns, the solution of which is the coordinates of the center.
Let us take the example of
$$f(x,y)=4x^2 + y^2 + 6x y - 4 x=0$$
(plotted below: it's a hyperbola). The partial derivatives are:
$$\begin{cases}\frac{\partial f}{\partial x}&=&-4+8x+6y\\\frac{\partial f}{\partial y}&=&6x+2y\end{cases}$$
Equating these expressions to zero, we obtain $x=-2/5, y=6/5$ that can be verified to be the coordinates of the center of the hyperbola.
Furthermore, if the asymptotes are desired, we consider points at infinity that are obtained in the following way : we eliminate in (1) all terms which aren't of second degree. It remains
$$F(x,y):=y^2+6xy+4x^2=0\tag{2}$$
No, we look for what happens along lines $y=mx$ that we substitute in (2) giving the equation:
$$x^2(m^2+6m+4)=0$$
This is possible with the two slopes
$$m=-3\pm \sqrt{5}\tag{3}$$
As the asymptotes must pass through the center, their equations will be
$$y-6/5=m(x+2/5)$$
with $m$ given by (3).