Understanding the "vertical shift" property of big Oh

Question

So I have difficulty understanding the big Oh property that says that if $\epsilon$ is some constant and $\epsilon < f$ on a neighborhood of infinity, then $\alpha + f = \mathcal{O}(f) $ . I have trouble understanding why that is true.

Answer 1

$$a< a C f \text{ for some constant } C>0$$

since we consider $f$ to be a positive function.

Thus $a+f\leq aCf+f \in O(f)$.