If $f(x) = -3x^2 + 6x + 2$, prove that $f(x)$ is $O(x^2)$

Question

I had a question on Big-O.

Generally my teacher gave us these strategies to follow: 1. Eliminate Negative Terms 2. Multiply to match highest order term.

So in this case Would I be getting rid of $-3x^2$ term and end up with $6x+2$?

So my $c = 8$ and $k = 1$.

Any help would be appreciated.

I wouldn't say you "eliminate" negative terms so much as you make them positive. e.g. $$-3x^2 \leq 3x^2$$ — The Chaz 2.0, Mar 27 '17 at 03:10

score 0 · Answer 1 · answered Mar 27 '17 at 03:13

Consider that by the triangle inequality we have, $$|f(x)| = |-3x^2+6x+2| \leq |-3x^2|+|6x|+2$$

then for $x>0$ we have that

$$|f(x)| \leq 3x^2+6x+2$$

and for any $x>1$ (because $x^2>x>1$), we clearly than have that

$$|f(x)|\leq 3x^2+6x^2+2x^2 = 11x^2 = 11\cdot|x^2|.$$

Since this holds for all $x>1$, we have that $f(x)$ is $O(x^2)$ by definition.

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