Answer
The function has a minimum value.
The minimum value is $4$.
Work Step by Step
Comparing $g(x)=8x^{2}-8x+6$ with $ax^{2}+bx+c$, we see that $a=8$ and $b=-8$.
As $a\gt0$, the parabola opens up and the function has a minimum value. The minimum value is the y-coordinate of the vertex.
The x-coordinate of the vertex is given by
$x=-\frac{b}{2a}=-\frac{-8}{2(8)}=\frac{1}{2}$
Evaluating the function at $x=\frac{1}{2}$, we get the y-coordinate of the vertex as
$g(\frac{1}{2})=8(\frac{1}{2})^{2}-8(\frac{1}{2})+6=4$
