Question 724219
Assuming you meant y equals x squared (y=x^2):
To transform {{{y=x^2}}} into {{{y=-3(2x-4)^2}}} 
you need a horizontal translation by 2 units right,
because you replaced {{{z=x-2}}} by {{{x}}} and then by {{{2z=2(x-2)=2x-4}}}.
A little algebra to get a simplified formula for the same function could have helped:
{{{y=-3(2x-4)^2}}} --> {{{y=-3(2(x-2))^2}}} --> {{{y=-3(2^2(x-2)^2)}}} --> {{{y=-3(4(x-2)^2)}}} --> {{{y=-12(x-2)^2}}}
Then you would need a vertical stretch by 12, reflection on x-axis, and horizontal translation 2 units right.
 
NOTE: If you meant {{{y=sqrt(x)}}} and {{{y=-3sqrt(2x-4)}}}
the same thing applies.
{{{z(x)=(x-2)}}} puts {{{z}}} 2 units to the left by subtracting 4, but
in the compound function {{{f(g(x))}}}, to get the same y value you need to add 4 to x:
{{{f(z+2)=f(x-2+2)=f((x+2)-2)=f(x)=y}}}