Is there a simple explanation for why rational plane curves of degree $>2$ are singular? Its a well-known result that smooth projective plane curves of degree $d$ have genus $(d-1)(d-2)/2$, so in particular, smooth curves of degree $1$ and $2$ are genus 0, and those of higher degree have positive genus.  It seems like there should be a simple explanation for this in terms of the behavior of dimension 3 subspaces of degree $>2$ polynomials (something about them having too many zeros), using the standard criterion that the map determined by a linear system is an isomorphism if and only if contains functions that distinguish points, and with non-zero derivative at each point, but I have not managed to find it in any reference, or figure out what it is myself.

Do you have a simple explanation of this fact?

 A: There is nothing special about rational curves in some sense.
Proposition: Given a smooth projective curve $C$, a very ample line bundle $L$ on $C$, and a 3-dimensional subspace $V\subset \Gamma(C,L)$ which is base-point free, the map $\phi:C\to\mathbb{P}(V)$ is an embedding if and only if $V=\Gamma(C,L)$.
The reason is that that $\phi$ is obtained by projection of the embedding $\eta:C\to\mathbb{P}(\Gamma(C,L))$ from the linear projective subspace $\mathbb{P}(W)$ where $W=\Gamma(C,L)/V$. It follows that $\mathbb{P}(W)$ is of codimension 3 in $\mathbb{P}(\Gamma(C,L))$ and thus must intersect the secant variety $S(C)$ in $\mathbb{P}(V)$ (which has dimension 3).

The response to my comment above is as follows. The tangent variety $T(C)$ is a surface inside $S(C)$. For some curves $C$ in $\mathbb{P}(\Gamma(C,L))$ (such as the rational normal curve), it is possible to find $V$ such that $\mathbb{P}(W)$ meets $S(C)$ at exactly one point of $T(C)$.
A: There is an elementary argument based on Bezout count of solutions of an algebraic system. It is given on pages 102-103 of the book "A Treatise on Algebraic Plane Curves" by Coolidge.
A: After reading the reference suggested by Abdelmalek, I think I have a proof I’m happy with (which is basically the one from the book in more modern language);  I suspect this is also equivalent to the observation in terms of the secant variety from Kapil’s answer.  This is a little more advanced than I was looking for, but I’ve already given my lecture, so this is more for my own satisfaction at this point.
Given three homogeneous polynomials $f_1,f_2,f_3$ of degree $d$ on $\mathbb{P}^1$, we can look for point where this map fails to be injective (we have a node in the image) by looking at whether the section $g_{ij}([X_0:X_1],[Y_0:Y_1])=\frac{f_i(X_0,X_1) f_j(Y_0,Y_1)-f_j(X_0,X_1)f_i(Y_0,Y_1)}{X_0Y_1-X_1Y_0}$ of $\mathcal{O}(d-1)\boxtimes\mathcal{O}(d-1)$ on $\mathbb{P}^1\times \mathbb{P}^1$ vanishes for all $i,j$.  Of course, enough to check this for $i=1$ and $j=2,3$. The multi-Bézout theorem (equivalently, cup product) tells me that there should be $2(d-1)^2$ simultaneous zeros of $g_{12}$ and $g_{13}$ (up to multiplicity).  A simultaneous zero of these corresponds to a cusp if $[X_0:X_1]=[Y_0:Y_1]$ and to a node if $[X_0:X_1]\neq [Y_0:Y_1]$, except when $f_1(X_0,X_1)=f_1(Y_0,Y_1)=0$ and the intersection multiplicity is 1 (in which case, it’s a smooth point).  This final exceptional case accounts for $d^2-d$ points (up to multiplicity).  So, there are $(2(d-1)^2)-d(d-1)=(d-1)(d-2)$ intersection points left that give singularities.  The symmetry under swapping X’s and Y’s tells us that generically we have $(d-1)(d-2)/2$ nodes, as expected from the fact that a smooth curve of this degree has that genus.
