Question 812501
Just a change in position from the reference graph of f(x).  Up 12 units and to the right by 3 units.  The function, g(x) is in standard form, h=3 and k=12.  Standard Form for vertical parabola is {{{y=a(x-h)^2+k}}}.


See carefully that while f(x) has a minimum point for vertex, now g(x) has a=-1, so now g(x) has its vertex as a MAXIMUM point, so parabola opens downward.


f(x)=x^2 shown in green, and g(x) in red:
{{{graph(400,400,-20,20,-20,20,-(x-3)^2+12,x^2)}}}