Supposing I have an $n^{th}$-order polynomial curve $$y = \sum_{i=0}^n c_ix^i$$ and an $n^{th}$-order polynomial surface $$z = \sum_{i,j\in\mathbb{Z}^+\!,\ i+j=n} c_{ij}x^iy^j.$$ Now suppose that in each case, I want to transform the graphs of the functions by first rotating each graph with a rotation matrix $\mathbf{R}$ (planar rotation about the origin in $\mathbb{R}^2$, and rotation about some arbitrary axis through the origin in $\mathbb{R}^3$) and then translating by a displacement vector $\mathbf{d}$. How can I determine the implicit functions $C(x,y)$ and $S(x,y,z)$ such that the transformed curve and surface are defined by the zero level-sets $C(x,y)=0$ and $S(x,y,z)=0$, respectively?
$\textbf{Edit}$: if it's not convenient to do this for the general case of an $n^{th}$-order polynomial, I am particularly interested in the cases when $n=2$ and $n=4$.
Expand $$\begin{pmatrix}x\\y\end{pmatrix}=R\begin{pmatrix}u\\v\end{pmatrix}+t$$
and plug in
$$y -\sum_{i=0}^n c_ix^i=0.$$
Now you have an implicit equation in $u,v$.
Similar in 3D.