= Solution
A <nonsingular point of an algebraic curve> $p$ is an <inflection point of an algebraic plane curve>, or <flex>, if its <projective tangent line> has contact of order at least three:
$$
I_p(C,T_pC)\geq3.
$$
Here the <projective tangent line> must not be a component of $C$. Ordinary tangency has contact order two.
Let $C=V(F)$, where $F$ is a <homogeneous polynomial> of degree $d\geq2$. Its <Hessian curve> is defined by
$$
H_F=\det\begin{pmatrix}
F_{xx}&F_{xy}&F_{xz}\\
F_{yx}&F_{yy}&F_{yz}\\
F_{zx}&F_{zy}&F_{zz}
\end{pmatrix}=0.
$$
Each entry has degree $d-2$, so a nonzero $H_F$ has degree $3(d-2)$. By the <Hessian criterion for a flex>, the <flexes> are exactly the points of $C\cap V(H_F)$. The stated no-common-component assumption and <Bézout's theorem> therefore give
$$
\boxed{\#\{\text{distinct flexes of }C\}\leq3d(d-2).}
$$
This counts distinct points; higher contact can use more than one unit of <intersection multiplicity> at a single point.
For completeness, the <Hessian criterion for a flex> follows by putting $p=[0:0:1]$ and $T_pC=\{y=0\}$, with $F_y(p)\ne0$. The <Euler theorem for homogeneous functions> gives $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, whence
$$
H_F(p)=-(d-1)^2F_y(p)^2F_{xx}(p).
$$
The restriction $F(x,0,1)$ already has vanishing constant and linear terms. Its quadratic term vanishes exactly when the <intersection multiplicity> with the tangent is at least three, proving the criterion.
For $d\geq3$ the bound is the actual <maximum number of flexes of a smooth plane curve>: a general nonsingular curve has only contact of order three, and the corresponding intersections with its <Hessian curve> are simple. One way to see genericity is to vary a point and a line through it, a three-dimensional family. Contact of order at least four imposes four independent linear conditions on the coefficients of a degree-$d$ equation for $d\geq3$. Thus the locus of equations with such contact is a proper exceptional locus. Locally, if the curve is $y=h(x)$, its <Hessian curve> restricted to $C$ is a nonvanishing factor times $h''(x)$; contact exactly three gives a simple zero.
A nonsingular <plane conic> has no <flexes>, agreeing with the formula at $d=2$. A line is excluded from this formulation: its tangent is itself and its <Hessian matrix> is zero, so the no-common-component hypothesis fails. The formula is therefore intended for $d\geq2$, rather than giving the meaningless negative value at $d=1$.
Back to article page
