I've proven it's in $O(n^2)$ by having $n = 1$ and $c > 13$ (namely $c = 14$). How do I prove it for big theta?
Prove $f(n) = 3n^2+10n$ is in $\Theta(n^2)$
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$\begingroup$
asymptotics
proof-writing
1 Answers
4
The inequality in the other direction is easier, since $n^2\lt 3n^2 +10n$ if $n\ge 1$.