I'm supposed to figure out the Big-Oh notation of $7x^2$. Take a look at this.
Now this says:
Show that $7x^2$ is $O(x^3)$ When $x>7, 7x^2<x^3$,
So let $C=1$ and $k=7$, we see $7x^2$ is $O(x^3)$. Alternatively, when $x>1, 7x^2<7x^3$ and so that $C=7$ and
$k=1$, we have the relationship $7x^2$ is $O(x^3)$
By this logic shouldn't the Big Oh of $7x^2$ be: $$7x^2<8x^2$$ $$7x^2 \in O(x^2)$$ with C=8 and k=1? Since $7x^2$ is obviously less than $8x^2$ for each $k>=1$. Why do we need $x^3$?
At the same time the link says
Let $f(x)=a_nx^n+…+a_1x+a_0$, where a0, a1, …, an-1, an are all real numbers, then f(x) is $O(x^n)$
And doesn't the above mentioned rule specifically state that for a polynomial of degree n, the big oh will be $O(x^n)$?
What am I missing here?