<s>[INST] Is my proof that $\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n)$ correct? Prove that: $$\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n) $$
$$\frac{n^3-3n^2+2n}{n^2-4n}≥\frac{n^3-3n^2+2n-17}{n^2-4n+22} $$
now we choose$$ N≥6$$
thus
$$\frac{n^3-3n^2+3n^2-18n+2n}{n^2-4n}≥\frac{n^3-3n^2+2n}{n^2-4n}$$
and
$$\frac{n^3-16n}{n^2-4n}=\frac{n^2-16}{n-4}=\frac{(n-4)(n+4)}{n-4}=n+4$$
and if we pick $$c=2$$
we see that $$2n≥n+4$$
for all $$ N≥6$$
and thus
$$\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n) $$
 [/INST] First a small notation error, you define $N$ but don't use it. I think by $N\ge6$, you mean $n\ge N$ for $N=6$.
To prove $a_n = O(n)$ you need to prove $|a_n| \le Cn$ for  $n>N$, but you have proven $a_n\le Cn$. You should also have a separate check that for $n\ge6$ (or any larger number) that $a_n\ge 0$. Then $a_n=|a_n|$.  Also worth mentioning $n^2-4n+22$ has no integer roots.
Alternate proof:
$$ \left|\frac{n^3-3n^2+2n-17}{n^2-4n+22}\right|\le \frac{n^3+3n^3+2n^3+17n^3}{n^2|1-\frac4n+\frac{22}n|} = \frac{23n}{|1-\frac4n+\frac{22}n|}$$
for $n>100$ (for example), $\frac 4n<\frac14$ and $\frac{22}n<1/4$ so $1-\frac4n +\frac{22}n  > \frac12$. Thus
$$\frac{n^3-3n^2+2n-17}{n^2-4n+22} \le 46n$$
so $\frac{n^3-3n^2+2n-17}{n^2-4n+22}= O(n)$.
My constant is much worse and I proved it holds for $n$ much larger only, but these are things allowed by the $O(n)$ notation.
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