Let's first find the value of y when x = 2 + c:
f(2 + c) = (2 + c)((2 + c) - 4)
f(2 + c) = (2 + c)(2 + c - 4)
f(2 + c) = (2 + c)(c - 2)
f(2 + c) = 2c + c^2 - 4 - 2c
f(2 + c) = c^2 - 4
Therefore, the coordinates for the point (2 + c, y) will be (2 + c, c^2 - 4).
Now, let's consider the point (c-2, y):
f(c - 2) = (c - 2)((c - 2) - 4)
f(c - 2) = (c - 2)(c - 2 - 4)
f(c - 2) = (c - 2)(c - 6)
f(c - 2) = c^2 - 8c + 12
Therefore, the coordinates for the point (c - 2, y) will be (c - 2, c^2 - 8c + 12).
Since the function f(x) = x(x-4) is symmetric around x = 2, we can see that the functions at points (2 + c, y) and (c - 2, y) are not equal. Therefore, the second statement " (c-2,y) " will not be on the graph of f(x).