I have to compute the Zariski closure of the image of the following rational map: $f:{P}^2 \rightarrow \mathbb{P}^4$ $[x_0:x_1:x_2]\rightarrow [x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2]$
I have already proved that this map is a morphism from $U$ to $\mathbb{P}^4$, where $U=\mathbb{P}^2 \setminus \{[1:0:0]\}$ and that cannot be extended to a map defined on the whole $\mathbb{P}^2$.
Now, I don't know how to compute the Zariski closure of $f(U)$. I know the useful relation $\overline{S}=V(I(S))$ for a set $S$: does this mean that I have to compute the ideal of $f(U)$, or is there a better way?
Moreover, I am asked to say if this closure of $f(U)$ is irreducible and what is its dimension. I have already proved that the closure is irreducible if and only if $f(U)$ is irreducible, but I still do not know how to proceed.
Any help would be highly appreciated. Thanks.