= Solution
Choose a point $p\in C$ and project from $p$ to a line not containing $p$. Every line through $p$ meets the quartic in four points counted with <intersection multiplicity> by <Bézout theorem>; removing the fixed intersection at $p$ leaves a generic fibre of three points. This <projection from a point on a plane curve> therefore defines a nonconstant morphism
$$
C\longrightarrow\mathbb P^1
$$
of degree three, so the <gonality> is at most three.
It cannot be one, because a degree-one morphism between <smooth projective curves> is an <isomorphism of algebraic varieties>, whereas $C$ has genus three and the <projective line> has genus zero. It cannot be two either: that would make $C$ a <hyperelliptic curve>. The <canonical map of a hyperelliptic curve> is not a <closed immersion>[closed embedding], but for a <smooth plane quartic> the <canonical map of a smooth plane curve> is exactly its given embedding $C\hookrightarrow\mathbb P^2$, since $K_C\cong\mathcal O_C(1)$. This contradiction proves
$$
\boxed{d=\operatorname{gon}(C)=3.}
$$
Back to article page
