Answer
See below.
Work Step by Step
Let's compare $f(x)=4x^2-8x+3$ to $f(x)=ax^2+bx+c$. We can see that a=4, b=-8, c=3. a>0, hence the graph opens up, hence its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-8}{2\cdot 4}=1.$ Hence the minimum value is $f(1)=4(1)^2-8(1)+3=-1.$