Answer
See below.
Work Step by Step
Let's compare $f(x)=4x^2-4x$ to $f(x)=ax^2+bx+c$. We can see that a=4, b=-4, c=0. a>0, hence the graph opens up, hence its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-4}{2\cdot 4}=0.5.$ Hence the minimum value is $f(0.5)=4(0.5)^2-4(0.5)=-1.$
