If $f(x) = -3x^2 + 6x + 2$, prove that $f(x)$ is $O(x^2)$ I had a question on Big-O. 
If $f(x) = -3x^2 + 6x + 2$, prove that $f(x)$ is $O(x^2)$
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. 

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. 
