Answer
minimum, $f(\frac{6}{5})=-\frac{16}{5}$
Work Step by Step
Given $f(x)=5x^2-12x+4$, we have $a=5\gt0, b=-12$, thus it has a minimum at $x=-\frac{b}{2a}=-\frac{-12}{2(5)}=\frac{6}{5}$ with $f(\frac{6}{5})=5(\frac{6}{5})^2-12(\frac{6}{5})+4=-\frac{16}{5}$
