How to find $n^2 - 2n - 3$ is $\Omega(n^2)$?

Question

How can I show that $n^2 - 2n - 3$ is $\Omega(n^2)$?

Answer 1

Since $2n+3$ is positive for any choice of $n$, $$n^2-2n-3\leq n^2$$ for any $n\geq 1$.

Now, $$-\frac 1 3 n^2\leq -3$$ and $$-\frac 1 3 n^2\leq -2n$$ when $n\geq 5$. Thus $$n^2-\frac 23n^2=\frac 13n^2\leq n^2-2n-3$$ when $n\geq 5$. We can then write $$\frac 13n^2\leq n^2-2n-3\leq n^2$$ for $n\geq 5$