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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: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 $d\ge 2$. Its Hessian curve is defined byEach 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 giveThis 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 $T_{p}C=\{y=0\}$, with $F_y(p)\ne 0$. The Euler theorem for homogeneous functions gives $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, whenceThe 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 $d\ge 3$ 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 $d\ge 3$. 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 $h^{\prime \prime }(x)$; 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 $d\ge 2$, rather than giving the meaningless negative value at $d=1$.

An inflection point of an algebraic plane curve is a smooth point where the tangent has intersection multiplicity at least three. For a homogeneous defining polynomial $P$ of degree $d$, its Hessian curve is defined by the determinant of the matrix of second partial derivatives, of degree $3(d-2)$. In characteristic zero, the smooth inflection points are exactly the smooth points of $C\cap \mathrm{Hess}(C)$.

For a local verification, move a smooth point to $[0:0:1]$ and its tangent to $y=0$, and scale so $P(x,y,1)=y+a{x}^2+bxy+c{y}^2+\cdots$. The tangent contact is at least three exactly when $a=0$. The homogeneous Euler identities give the Hessian matrix at that point as $\left(\begin{array}{ccc}2a& b& 0\\ b& 2c& d-1\\ 0& d-1& 0\end{array}\right)$, whose determinant is $-2a(d-1{)}^2$. This proves the criterion for $d>1$.

For the given cubic, the singularity equations are $x^2+\lambda yz=y^2+\lambda xz=z^2+\lambda xy=0$. No coordinate can vanish at a projective solution, since that would force all three to vanish. Multiplying the equations therefore gives $1=(-\lambda {)}^3$, so singularity requires ${\lambda }^3=-1$. Conversely, under that condition the point $[1:1:-1/\lambda ]$ satisfies all three equations, and the Euler identity places it on the cubic. Thus