What is the vectorial equation for an $n$-degree polynomial in $2$D space? In the the two-dimensional Cartesian $xy$-plane, any non-vertical line is of the form $$y=m(x-h)+k\tag A$$ where $y$ varies with $x$, where $m$ is the slope of the line, and where $(h,k)\in\Bbb{R}^2$ is a fixed point on the line.
It is also possible to define lines (including vertical lines) by ‘tracing’ them out with a position vector: 
$$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} \\ \pmatrix{x\\y} = \pmatrix{h\\k} + \lambda\pmatrix{b_x\\b_y}\tag B$$
where $\mathbf r$ varies with $\lambda$ for all $\lambda\in(-\infty,\infty)$, where $(h,k)\in\Bbb{R}^2$ is also a point on the line, and where $\mathbf b$ is a “difference vector” such that $$\frac{b_y}{b_x}=m$$
The two loci are identical:
$$\bigcup_{x\in\Bbb{R}}\bigl\{ (x,y) : y = m(x-h)+k \bigr\}
=
\bigcup_{\lambda\in\Bbb{R}} \bigl\{ (x,y) : \mathbf{r}=\mathbf{a}+\lambda\mathbf{b}\bigr\}$$
It is certainly no coincidence that form $\text{(B)}$ so closely resembles the slope-intercept form of a line nor that $$\mathbf{b}=\pmatrix{\partial x/\partial x \\ \partial y/\partial x}$$

My question is: Given a polynomial $$P_n(x)=\sum_{i=0}^{n}c_ix^i$$ what is the general formula for a vector $\mathbf{r}(\lambda)$ such that
$$\bigcup_{x\in\Bbb{R}}\bigl\{ (x,y) : y = P_n(x) \bigr\}
=
\bigcup_{\lambda\in\Bbb{R}} \left\{ (x,y) : \mathbf{r}(\lambda) = \pmatrix{x\\y} \right\}\quad?$$
I have tried trying to reverse-engineer a formula using parametric representations of loci of the form $$\cases{x=f(t)\\y=g(t)}$$ but have had no luck.
 A: 
$$\pmatrix{x\\y} = \pmatrix{h\\k} + \lambda\pmatrix{b_x\\b_y}$$

The Taylor expansion of $\,P(x)= m(x-h)+k\,$ around $\,h\,$ is $\,P(x)=P(h)+\dfrac{P'(h)}{1!}(x-h)\,$, so the above can be thought of as $\displaystyle\,\pmatrix{x\\P(x)} = \pmatrix{h\\P(h)} + \lambda\pmatrix{1\\P'(h) \,/\, 1!}\,$ with $\,\lambda = x-h\,$.
For an $n^{th}$ degree polynomial, the equivalent Taylor expansion is:
$$
P(x)=P(h)+{\frac {P'(h)}{1!}}(x-h)+{\frac {P''(h)}{2!}}(x-h)^{2}+\cdots+{\frac {P^{(n)}(h)}{n!}}(x-h)^{n}
$$
Then the parametric representation with $\,\lambda = x-h\,$ could be written as:
$$
\pmatrix{x\\P(x)} = \pmatrix{h\\P(h)} + \lambda\pmatrix{1\\P'(h) \,/\, 1!} + \lambda^2\pmatrix{0\\P''(h) \,/\, 2!} + \ldots + \lambda^n\pmatrix{0\\P^{(n)}(h) \,/\, n!}
$$
Which is equivalent to $\,\mathbf{r} = \mathbf{b_0} + \lambda \mathbf{b_1}+ \lambda^2\mathbf{b_2}+\ldots+\lambda^n\mathbf{b_n}\,$.
A: As long as I took the time to write it up on my whiteboard, I thought I might as well just supplement @dxiv’s answer with this relationship:
$$
\newcommand{\b}[1]{ \begin{bmatrix} #1 \end{bmatrix} }
\bigcup_{x\in\Bbb R} \bigl\{ (x,y) : y=P(x) \bigr\}
=
\bigcup_{\lambda\in\Bbb R} \left\{ (x,y) : 
\pmatrix{x\\y} = 
\b{h\\P(h)} + 
\lambda \b{1\\P'(h)} + 
\sum_{i=2}^n \frac{\lambda^i}{i!} \b{0\\P^{(i)}(h)}
\right\}$$
or possibly even
$$\pmatrix{x\\y} = \sum_{i=0}^{\infty} \frac{\lambda^i}{i!} \b{h^{(i)}\\P^{(i)}(h)}$$
where $(i)$ loosely represents the $i$th derivative.
