A Regular point on a curve defined as the following
Def.1 (Singular and Regular points of planar curves)
Suppose that $S$ is a curve in $\mathbb{R}^2$ and $a\in S$. If, for every $r>0,S\cap B(a;r)$ is not a $C^1$ graph $~~~($of a function $(a,b)\to\mathbb{R})$, then $S$ is singular at $a$. If there exists some $r>0$ such that $S\cap B(a;r)$ is a $C^1$ graph $($of a function $(a,b)\to\mathbb{R})$, then $S$ is regular at $a$.
Also basicly, a curve is Regular if and only if every point on the curve is a regular point.
Now we consider a curve defined by a graph of some polynomial i.e. $$S=\{(x,y):y=a_1x+\dots+a_nx^n\}$$ Is it correct to say $S$ must be a Regular curve? My intuition is since polynomial functions are continuous everywhere, also all the partial derivative exists, so polynomials are continious differatiable, i.e. it's a $C^1$ function, this seems implies every point in $S$ has a neighbourhood intersect with $S$ gives a $C^1$ graph, which is the def of a regular point. However I'm not sure about my intuition. Please tell me if i'm wrong. If it's correct, how do I write a rigorous proof for this?
I'm assuming here that $C^1$ graph means the graph of $f(x,y) = 0$ for some $C^1$ function $f$? In that case, you've just got to show that $f(x,y) = p(x) - y$ is $C^1$ everywhere, then $S$ is a $C^1$ curve and, for any $r>0$, $S\cap B(a;r) \subset S$ will necessarily also be one.
This is easily shown by calculating the partials, assuming you take it as given that $x^n$ is differentiable everywhere and the linear combination of differentiable functions is differentiable.