In Big-O notation, what does it mean for T(n) to be upper bounded by something
Try it like this: "we can pick a constant $C$ such that, for sufficiently large $n$, $T(n)$ will always be less than $Cf(n)$".

Intuitively, this does indeed mean that lower-order terms and constant factors don't matter. Because lower-order terms stop mattering once $x$ gets sufficiently large, and constant factors can be cancelled out by an appropriate choice of $C$. If $T(n) = n^3 + 2019n^2 + 99999n + 10^{10}$, for example, that will still eventually be dominated by $Cn^3$, as long as $C &gt; 1$ and $n$ is large enough—the $n^3$ term in that expression will eventually outweigh everything else. So we say that $T(n) \in O(n^3)$. (Some people use an equals sign instead; the meaning is the same.)

The "sufficiently large $n$", by the way, is why it's called "asymptotic" complexity: we only care about what happens as $n$ goes toward infinity, not what happens for "small" values.