Let $\;f:\mathbb R \rightarrow \mathbb R\;$ be a function such that $\;f(0)=0\;$. Denote $\;f^{λ}( \bullet) \equiv f(\bullet-λ)\;$ for some $\;λ \in \mathbb R\;$. Then, it's easy to see that $\;f^{λ}(λ)=0\;$
Now if $\;f:\mathbb R \rightarrow \mathbb R^n\;$ be a function which satisfies $\;f(0)=(0, \dots ,0)\;$, could I claim the same as the above for its translation? Furthermore, what will be the graph of the translation of $\;f\;$ ?
A function $f:\mathbb{R}\to\mathbb{R}^n$ may be viewed as vector $(f_1,\ldots,f_n)$ of $n$ functions, where $f_i:\mathbb{R}\to\mathbb{R}$.
If $\lambda\in \mathbb{R}$ then we may define $f^\lambda:\mathbb{R}\to\mathbb{R}^n$ by $f^\lambda=(f^\lambda_1,\ldots,f^\lambda_n)$, where $f^\lambda_i$, for $i=1,\ldots,n$, is defined as in the statement of the question.
Now $f(0)=0\in \mathbb{R}^n$ implies $f^i(0)=0\in \mathbb{R}$. Thus $f^\lambda_i(\lambda)=0\in \mathbb{R}$ and so we may conclude that $f^\lambda(\lambda)=0\in \mathbb{R}^n$.
For $i=1,\ldots,n$, the graph of $f^i_\lambda$, as a subset of $\mathbb{R}^2$, will be a translation of the graph of $f^i$.