Answer
The function has a minimum value of $-1$.
Work Step by Step
RECALL:
(1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens:
(i) upward when $a \gt0$ and has its vertex as its minimum; or
(ii) downward when $a\lt0$ and has its vertex as its maximum.
(2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$
Compare $f(x)=4x^2-4x$ to $f(x)=ax^2+bx+c$.
It can be seen that $a=4$, $b=-4$, $c=0$.
$a>0$, hence the graph opens up, hence it's vertex is a minimum.
The minimum value is at $x=-\frac{b}{2a}=-\frac{-4}{2\cdot 4}=0.5.$
Thus, the minimum value is $f(0.5)=4\cdot(0.5)^2-4\cdot(0.5)=-1.$
