I have to prove that $f(x)=ax+b$ for $a>0$ belongs to $O(n^2)$
where $O(g)=\{f:\exists c>0, n_{0}\geq 0\mbox{ s.t. } 0\leq f(n)\leq cg(n)\}$.

BROOF

$an+b\leq cn^{2}$
$\frac{a}{n}+\frac{b}{n^{2}}\leq c$ and $n\neq 0$.
and I need to find for which values of $n$ left-hand side is positive so I can express $c$ in terms of $n$.
$\frac{a}{n}+\frac{b}{n^{2}}>0$
$\frac{an}{n^{2}}+\frac{b}{n^{2}}>0$
since $n\neq 0$ and $n^2>0$ for $n>0$ and given that $n \in \{0, 1, 2, ...\}$ the sign of the equation depends only on $an+b$.
$an+b>0$
$n>-\frac{b}{a}$
$n\geq \lceil -\frac{b}{a} \rceil$
which combined with $n>0$ ($n\geq 1$) gives the minimum $n$ for which the inequality holds.
$n_{0}=\mbox{max}\{1, \lceil -\frac{b}{a} \rceil\}$.
For $h(n)=\frac{a}{n}+\frac{b}{n^{2}}$ we have $h(n_{0})=c$ as smallest solution.

Is my proof correct?