In the third edition of Introduction to Algorithms, the authors state:
... when $a$ > 0, any linear function $an+b$ is $O(n^2)$, which is easily verified by taking $c = a + |b|$ and $n_0 = max(1, -b/a).$
[In the above, $c$ and $n_0$ come from the standard definition of big-oh:
$f(n)$ is in $O(g(n))$ if there exist positive constants $c$ and $n_0$ such that $0 \le f(n) \le cg(n)$ for all $n \ge n_0$.]
I have no trouble accepting that $an+b \in O(n^2)$, but I'm questioning the motivation/origin of the weird choice of $n_0$.
If we assume that $ n \ge 1$, then, since $a > 0$, we have $an^2 \ge an$ and since $n^2 \ge 1$ we have $|b|n^2 \ge |b| \ge b$, implying that $(a+|b|)n^2 \ge an+b$. So it seems that $n_0$ could simply have been taken as $1$. Or am I messing up somewhere?