Let $V=\mathbb R[t]_{\le 3}$ and $\cal A=\{p\in V \mid p(-1)=4 \land p(1)=6\}$. Show or disprove that $\cal A$ is a vector space. My approach: Let $p(x)=5+x$ and $q(x)=5+x^3$. Then $\cal A$ is not closed under addition, since $p,q\in \cal A, $ but $(p+q)(1)=p(1)+q(1)=(5+1)+(5+1)=12 \ne 6$. Thus $p+q\notin \cal A$.
I guess this attempt is okay, or?
Now I am looking for a polynom $f\in V$ and a vector subspace $\cal U$, such that $\cal A=f+U:=\{f+u\mid u\in U\}$. I do not know how to handle this here. I thought to take $f=p=5+x$, but this does not work I guess. Some help is appreciated!