Let's say I take a pencil and start drawing a curve on $xy$ plane. The curve is continuous and for each value of $x$ there is only one corresponding value of $y$.
So question that interests me is - for any curve I draw, is there a polynomial that describes that curve?
On
Yes, there are several methods how to get this polynomial like Lagranges polynomial https://en.wikipedia.org/wiki/Lagrange_polynomial or newtons polynomial https://en.wikipedia.org/wiki/Newton_polynomial which is faster method.
On
Effectively you're asking "Is every continuous function (on some interval of real numbers) a polynomial?" The answer is a resounding "No", but in a sense that's difficult to make precise in a non-technical way. Loosely, polynomials represent a Very Thin Subset of the space of all continuous functions.
There are plenty of familiar functions that are not polynomials, such as quotients of polynomials having no common non-constant factor, the natural exponential function, the circular (trig) functions. Further, if you add any such function to a polynomial, you get another non-polynomial. (Why?) That already starts to suggest how "rare" polynomials are even among the functions familiar from (pre-)calculus.
As David Ullrich notes, every continuous function is approximated as closely as you like on a closed, bounded interval, in the sense that if you draw your continuous curve with a pencil whose lead has positive diameter (but diameter as small as you like), then over every closed, bounded interval, there is a polynomial whose graph lies within the region covered by your pencil. This isn't easy to prove, but requires nothing beyond undergraduate-level analysis.
If you meant a line, then yes, for each line you draw satisfying the property you mentioned there would be a polynomial of the form $ax+b$ that describes it.
And if you meant a general curve, then consider the following: Draw a continuous curve on $[0,1]$ satisfying $f(0)=f(1)=0$ and such that this curve has an additional root in some point $c\in(0,1)$. Repeat this process with all intervals of the form $[n,n+1]$, i.e. make it periodic. This graph is continuous and has infinitely many roots, but no polynomial can have that much roots.
What if one adds the assumption that the curve has only finitely many roots? Well draw a random continuous shape on a finite interval such that it crosses the $x$-axis in one instance, then at the 'sides' of this curve on this interval, draw two lines parallel to the $x$-axis so as to make it continuous over $\mathbb{R}$. Now a basic property of polynomials p with degree $n\gt0$ is that $\lim\limits_{|x|\to\infty}p(x)=\pm\infty$. Yet in this case the limits at $\pm\infty$ will be real numbers contradicting that property.
What if one adds the assumption that the curve goes to $\pm\infty$ at $\pm\infty$? Well, a polynomial $p$ can be written as $p(x)=\sum\limits_{k=0}^na_kx^k$ (and $p(0)=a_0$), thus $p''(x)=\sum\limits_{k=0}^{n-2}b_kx^k$. This last polynomial will have finitely many roots, thus finitely many changes in its sign. So $p$ will have finitely many changes in its convexity (at one interval it will be convex upward, and in another convex downward). And as we did previously, we can draw a curve such that it is
And hence it can't be represented by a polynomial because of the fourth property.
Polynomials are differentiable everywhere. You may pick up any continuous function that isn't differentiable at one or more points (e.g. $x\mapsto|x|$ isn't differentiable at $x=0$) and you get such curve.
These type of constructions generally rely on making a curve that violates a property of polynomials. But to be much more general, your question boils down to whether any continuous function is a polynomial, and the answer is that only a tiny set of continuous functions are a polynomial.