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

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The minimum value of a function f(x) is the value f(x) when f'(x) = 0 and f''(x) is positive.

The function we have is f(x) = 4x^2 - 8x + 1

f'(x) = 8x - 8

8x - 8 = 0

=> x = 1

f''(1) = 8 which is positive.

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

**The minimum value of the function is -3**

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).**