strange choice of constant when showing $an + b = O(n^2)$ ("Introduction to Algorithms" book) 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 $a &gt; 0$, 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?
Because of the requirement $an+b&gt;0$. The definition requires $0 \leq f(n)$.