I have to count parameters for the intersection of a quadric and a cubic surface in $\mathbb{P}^3$, up to linear automorphisms of $\mathbb{P}^3$. I take account of the theorem according to which a not hyperelliptic algebraic curve $X$ of genus $4$ is such that the canonical map $\Phi_K$ embeds $X$ into $\mathbb{P}^3$ as a smooth curve of degree $6$ (by the vanishing of a quadratic and a cubic polynomial). So I have:
number of parameters $=$ dimension of degree $6$ polynomials in $\mathbb{P}^3$ - dimension of linear automorphisms of $\mathbb{P}^3$ - 1. But this count doesn't give $9$, as I expect. What's wrong?
There are $9$ parameters of quadrics in $\mathbb P^3$, and $15$ parameters of automorphisms of $\mathbb P^3$. Thus in your count, you may reduce to having a fixed quadric, which then has a projective automorphism group of dimension $6$.
There are $19$ parameters of cubic surfaces, but if we quotient out by the $6$ parameters in the automorphisms of our quadric, we are left with $13$ parameters. Note that $13 = 9 + 4$; the point is that a canonically embedded genus $4$ curve is contained in a single quadric (which after a projective transformation we may as well take to be our chosen quadric), but is contained in a $4$-parameter family of cubic surfaces.