We need to either prove or refute the claim $6(n^2) + 2n = O(n^2)$

Question

I have no idea how we continue with the formal proof for this...I referred to the big O definition but do not understand how tight or exact we should make the upper bound.

Answer 1

You need to find constants $M,N$ st $6n^2+2n<Mn^2$ for all $n>N$. Well, for $n>1$ we have $2n<2n^2$, so $6n^2+2n<8n^2$. Hence $M=8,N=1$ suffice. You are not looking for the smallest possible $M,N$. Any values will do.

Incidentally, strictly you should say $6n^2+2n=O(n^2)$ as $n\to\infty$, but it is farily obvious that we are looking at large $n$ in this case.