Let $Q(x)$ be a polynomial in $x$ of order $n$. The Taylor polynomial of $Q(x)$ of order $n$ developed around $x=b$ (denoted by $P_{n,b}(x)$ ) corresponds to $Q(x)$ written in $(x-b)$. This can be proved using the error formula: $$R_{n}(x)=Q(x)-P_{n,b}(x)=\frac{Q^{n+1}(t)}{(n+1)!}(x-b)^{n+1} = 0$$ Since $Q^{n+1}(x)=0$ for all $x$.
Is there any way to visualize (maybe graphically) what this transformation represents?
Consider the polynomials as vectors. Then $Q(x)$ is the polynomial using the canonical basis i.e. the powers of $x$ and $P(x)$ is the polynomial after a change of basis. The new basis would be the powers of $(x-b)$.