y=2(x2−6x+m)
m=(b2)2
m=(−62)2
m=9
y=2(x2−6x+9−9)
y=2(x2−6x+9)−18
y=2(x−3)2−18
x=2(y−3)2−18
x+18=2(y−3)2
x+182=(y−3)2
±√12x+9=y−3
±√12x+9+3=y
Thus, ƒ−1(x)=±√12x+9+3. Don't forget the f−1(x) notation; I've been docked marks for this before.
Hopefully this helps!
