f(x) = (x(x – 2))² = x² (x² - 4x + 4) = x4 – 4x³ + 4x²
f’(x) = 4x³ - 12x² + 8x
f’(x) = 4x (x² - 3x + 2)
= 4x (x - 2)(x - 1)
f’(x) = 0 ⇒ x = 0 or 1, 2
So, the tangent to curve f(x) is parallel to the x-axis if x = 0, x = 1 or x = 2.
Now points 0, 1 and 2 will divide the number line into 4 disjoint intervals (-∞, 0) (0, 1) (1, 2) (2, ∞)
Now in the interval (-∞, 0) and (1, 2) f’(x) < 0. So the function f(x) is strictly decreasing in these intervals.
f’(x) >0 in interval (0, 1) and (2, ∞)
So the function f(x) is strictly increasing in intervals (0, 1) and (2, ∞)
Tangent is parallel to X- axis if

dy

dx

= 0
Which gives us x = 0, 1, 2
Hence, x = 0, y = 0
X = 1, y = 1
x = 2, y = 0
Required points are (0, 0), (1, 1), (2, 0).