I'm just starting to learn Big O Notation and I was trying to understand how this function would scale:


If the function was $n^2$, it would be quadratic, so O(n^2). However, the denominator gives me pause. Is this still a O(n^2) function? This is not homework. Thanks in advance for the help and clarification.


The denominator is just a constant so $O(\cdot)$ doesn't care about it. For all $n\geq 0$, $\tfrac14n(n-3)\leq 1\cdot n^2$ so $\tfrac14n(n-3)\in O(n^2)$.

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