Question 678982
{{{f(x)=(x-2)^2}}}


{{{f(x)=x^2-4x+4}}}


Take the {{{derivative}}} f' of the function and set it equal to {{{0}}}. 

You need that because the derivative at a constant point is {{{0}}} plus, at that point, the function can change from increasing to decreasing and vice versa.

{{{f}}}'{{{(x) = 2x-4}}}

 {{{2x-4=0}}}

{{{2x= 4}}}

{{{x=2}}}, this means that when {{{x = 2}}}, the slope at {{{x =2}}} is {{{0}}}.

You plug in a number from either side of that number, x= 2 into f'(x) because by doing this you find out the slope of the line.

So I will pick, {{{x =-3}}} and {{{x =3}}} for simplicity's sake.

{{{2(-3)-4 = -6-4=-10 }}}.....=>..slope is negative
{{{2(3)-4= 6-4=2}}} .....=>..slope is positive


so, the function is decreasing for any value of {{{x}}} from ({{{-infinity}}},{{{2}}}),

increasing for any value of {{{x}}} from ({{{2}}},{{{infinity}}})



{{{ graph( 600, 600, -6, 10, -10, 10,x^2-4x+4) }}}