What is the minimum value of $f(x) = (2x - 3)^2 + 5$?

Explanation

1. 01

   A square is never negative, so $(2x - 3)^2 \ge 0$ and $f(x) \ge 5$.

2. 02

   The square is 0 at $x = \frac{3}{2}$, where $f = 5$: the minimum value is 5. (There is no maximum: $f$ grows without limit.)