Compute a canonical divisor Consider $C,C'$ two cubics in the plane with $C$ smooth. There are $9$ basepoint on the linear system generated by $C$ and $C'$ so if we blow them we get a map $X \to\mathbb P^1$, where $X$ is $\mathbb P^2$ blown-up at 9 points. Now is my question : how to compute $K_X$ ? I saw that $K_X = - C$. But I don't understand how to get it. Is the following argument correct ? $K_X = -c_1(X) = -c_1(f^*\mathbb P^2) = - f^*c_1(\mathbb P^2) = - f^*3H$ where $H$ is an hyperplane section. Now $3H \sim C$ and $f^*C$ it the strict transform of $C$. Thanks in advance for any remarks ! 

Here is a formula you should learn (say from Hartshorne). If $Y$ is a smooth surface, $p\in Y$ is a point and $\pi:X\to Y$ the blow up of $p$ and $E$ the exceptional divisor, then $K_X=\pi^*K_Y+E$. 
In your case, $K_X=f^*K_{\mathbb{P}^2}+\sum E_i$. Since $K_{\mathbb{P}^2}=-C$, we get $K_X=-f^*C+\sum E_i$ and thus $K_X$ is the negative of the proper transform of $C$. ($K_X=-C$ is not meaningful, since $C$ is not a curve on $X$).
