What is the minimum value of the function y=4x^2-8x+1?

To determine the extreme value of the function, we'll do
the first derivative test.

We'll determine the 1st
derivative of the function:

y' = 8x -
8

We'll cancel the
derivative:

8x - 8 = 0

We'll
move -8 to the right side:

8x = 8 => x =
1

The roots of the 1st derivative represent the critical
values of the function.

We'll calculate the minimum value
of y, replacing x by 1 in the expression of the
function:

f(1) = 4-8+1 =
-3

The minimum value of the given function is
reached at the point (1 ; -3).