Let $\mathbb K$ be a field, $X\subseteq \mathbb K^n$ a subset, and $\ f:\mathbb K^n\rightarrow \mathbb K^m$ an affine map. Using the standard notation $V$, $I$ for algebraic geometry, it should hold that $$V(I(f(X)))=f(V(I(X))),$$ i.e. that the Zariski closure $VI$ commutes with the affine transformation $f$. Why is that?
(To specify the notation more: for a set $Y$, we mean by $I(Y)$ the set of all polynomials vanishing on $Y$; and for a set of polynomials $J$, we mean by $V(J)$ the set of all zeros common to all polynomials in $J$.)
Note that I do NOT assume $f$ to be bijective, and that $X$ does not need to be an algebraic set.
The equality obviously doesn't hold for every polynomial mapping $\ f$, as that would mean that an image of Zariski closed set under polynomial mapping is always Zariski closed, which is not true (e.g. $f(x)=x^2:\mathbb R\rightarrow \mathbb R$ and $X=\mathbb R$).
EDIT: As pointed out by reuns, using the general topology fact that $\overline{f(X)}\supseteq f(\overline{X})$ for continuous maps, it is enough to show that $f(V(I(X)))$ is Zariski closed (i.e. an algebraic set).