Since $a$ and $b$ are integers, this means $f(n)\geq 0$ for all integers $n$. Now, observe that for all $n\in \mathbb{Z}$ and $r\in [0, 1)$, $$ 0\leq (1-r)f(n) + rf(n+1) = (1-r)(n^2 + an + b) + r((n+1)^2 + a(n+1) + b) = f(n+r) + r - r^2 = f(n+r) + \frac{1}{4} - \left(r - \frac{1}{2}\right)^2 $$ Hence, $f(n+r)\geq \left(r - \frac{1}{2}\right)^2 - \frac{1}{4} \geq -\frac{1}{4}$ for all $n\in \mathbb{Z}$ and $r\in [0, 1)$. But every real number can be expressed as $n + r$ for some $n\in \mathbb{Z}$ and $r\in [0, 1)$ (uniquely, in fact). So, indeed $f(x)\geq -\frac{1}{4}$ for all $x\in \mathbb{R}$.