what is the relative minimum of y= x^2-4x=4 occurs at?

Question

use first derivative test

Answer 1

y = x^2 - 4x + 4
y' = 2x - 4

Setting y' to zero, we have:

2x - 4 = 0
2x = 4
x = 4 / 2
x = 2

When x < 2, y' < 0. This means y is decreasing for x < 2.
When x > 2, y' > 0. This means y is increasing for x > 2.

Hence, by 1st derivative test, y has a relative minimum at x = 2.