Answer
See below
Work Step by Step
Given $V=P_2(R)$
We can write set $S$ as $S=\{p \in P_2(R):p(x)=ax^2+b\}$.
We can notice that $0x^2+0x+0=0 rightarrow 0\in S$. Hence, $S$ is nonempty (1)
Let $p(x)=ax^2+b$ and $q(x)=cx^2+d$
Obtain $p(x)+q(x)\\=ax^2+b+cx^2+d\\=x^2(a+c)+(b+d)\forall x \in (-\infty, \infty)$
Since $p(x)+q(x) \in S$, $p(x)+q(x)$ is closed under addition (2)
Let $p(x)=ax^2+b \in S$ and $c$ be a scalar
Obtain $kp(x)\\=k(ax^2+b)\\=ka(x^2)+kb$
Since $kp(x) \in S$, $kp(x)$ is closed under scalar multiplication (3)
From (1), (2), (3), $S$ is a subspace of $P_2(R)$
