Condition (1):

Since, y=x(x-4) is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ y= x(x-4) is continuous on [0,4].

Condition (2):

Here, y’= (x-4)+x which exist in [0,4].

So, y= x(x-4) is differentiable on (0,4).

Condition (3):

Here, y(0)=0(0-4)=0

And y(4)= 4(4-4)=0

i.e. y(0)=y(4)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,4) such that y’(c)=0

i.e. (c-4)+c=0

i.e. 2c-4=0

i.e. c=2

Value of c=2ϵ(0,4)

So,y(c)=y(2)=2(2-4)=-4

By geometric interpretation, (2,-4) is a point on a curve y=x(x-4),where tangent is parallel to x-axis.