abstract algebra – Plane quartics and morphisms

Suppose that $C$ is a non-hyperelliptic genus 3 curve defined over $mathbb{Q}$.
It is well known that non-hyperelliptic genus 3 curves are plane quartics, and these are also trigonal. So we have a morphism $C mapsto mathbb{P}^{1}$ of degree $3$, and the way in which one usually constructs this map is choosing a point $P$ in $C$ and projecting from $P$ to $mathbb{P}^{1} subseteq mathbb{P}^{2}.$

I am now wondering whether one can deduce that the degree 3 map $C mapsto mathbb{P}^{1}$ can always be defined over $mathbb{Q}$. If we have a rational point $P$ on $C$, then I guess you can always find a rational map $C mapsto mathbb{P}^{1}$ defined over $mathbb{Q}$ by the above construction. Moreover, if $C$ admits a model of the form $ax^4+ay^4+ text{lower order terms} =0$ for $a$ some fourth power, then I guess this is doable too.

I am now wondering whether one can always ensure that there’s a degree $3$ map which is defined over $mathbb{Q}$ if $C$ is defined over $mathbb{Q}$.