 Solution

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$. If $B=(v_1,\dots ,v_n)$, its dual family satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Every functional $f$ obeysso the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a basis without any prior dimension argument.

For $p=a+bt+c{t}^2$,The basis dual to $(f_0,f_1,f_2)$ isas direct substitution gives $f_i(p_j)={\delta }_{ij}$. Hence the three functionals form a basis of $P_2^{\ast }$.