symplectic classes on rational surfaces.  Hi. I have a stupid question. 
Let $M$ be a blow-up of the complex projective plane at $k$ generic points. 
Then we can choose an orthoginal basis (with respect to the cup product) $H, E_1, \cdots, E_k$ of $H^2(M;\mathbb{Z})$ such that $H^2 = 1, E_i^2 = -1$ for each $i=1,\cdots,k$. Then my question is, 
For a given class $C = aH + b_1E_1 + \cdots b_kE_k \in H^2(M;\mathbb{Z})$, how can we check 
whether $C$ is a symplectic class or not? (I mean, how can we know there exists a symplectic form which represents $C$?). For example, if $k=2$, is there a symplectic form 
which represents a class $2H - E_1 - E_2$?
I'd really appriciate for your any comment. Thank you in advance. 
 A: This answer is rewritten and include more details
First of all I highly recommend you the article of Paul Biran From Symplectic Packing to Algebraic Geometry and Back available on the page  http://www.math.tau.ac.il/~biranp/Publications/Pubications.html , especially theorem 3.2.
Your question basically asks "what is the symplectic cone of $\mathbb CP^2$ blown up in a finite number of points?" This question was answered by Paul Biran (check theorem 3.2. from the above article), though the answer is not 100% explicit.
Also, it is known that the symplectic cone of $\mathbb CP^2$ blown up in up to $9$ points coincides with the Kahler cone if the points are chosen so that the resulting surface has only $-1$ curves (in particular it is Fano if the number of points is at most $8$). This permits one to answer your last question (that is done below). In fact Kahler cones of Fano surfaces rather classical objects and all basic questions about them can be answered.
I would like to add that from a certain conjecture from algebraic geometry - Harbourne-Hirschowitz conjecture, it follows that the Kahler cone of $\mathbb CP^2$ blown up in a very generic collection of points  coincides with its symplectic cone (Literately this conjecture says the following 
: any integral curve with negative self-intersection on the blow-up of $\mathbb CP^2$ 
at a set of points in very general position is a smooth rational curve with self-intersection $−1$. In order to deduce the statement that the symplectic cone coincides with the Kahler one you have to use SW theory). Habourine-Hirschwitz conjecture is open even for $\mathbb CP^2$ blown up in $10$ points, and the famous Nagata conjecture is a partial case of it.
Now let us answer the last bit of the question. The class $2H-E_1-E_2$ is not symplectic. The symplectic cone of $\mathbb CP^2$ blown up in $2$ points coincides with the Kahler cone of $\mathbb CP^2$ blown up in two points. And $H-E_1-E_2$ is a rational $-1$-curve on this surface, while $(2H-E_1-E_2)\cdot (H-E_1-E_2)=0$, so $2H-E_1-E_2$ is not ample, and hence not symplectic. 
