Answer
minimum, $-1$
Work Step by Step
1) The graph of the quadratic function $f(x) = ax^2+bx+c$ is a parabola that opens:
(a) Upward when $a \gt 0$ and its vertex has a minimum value.
(b) Downward when $a \lt 0$ and its vertex has a maximum value.
2) The coordinates of the vertex of a quadratic function $f(x) = ax^2+bx+c$ are given by: $\displaystyle(\frac{-b}{2a}, f(-\frac{2}{a}))$
On comparing $f(x)=4x^2-8x+3$ with $f(x) = ax^2+bx+c$, we get: $a=4, b=-8,c=3; a \gt 0$. We can see that the given function shows a graph of a parabola that opens upward. So, its vertex has a minimum value.
Thus, the minimum value at $x$ can be expressed as: $x=\displaystyle \frac{-b}{2a}=\frac{-(-8)}{2(4)}=1$
Therefore, the minimum value is: $f(1)=(4)(1)^2-(8)(1)+3=-1$
