The equation given is y≥(x–3)^2–1. Plotting points of x as 0,1,–1 and 3 we see the corresponding values as –1,3,15,–1. Remember that y≥, so Q1,2,4 will be satisfied here. However there is no point which will yield both -ve numbers for x and y coordinates so Q3 won't be satisfied here.

Please edit, correct and let me know if there are other efficient ways to do this.

which point u feel to be correct whereas I copied this as like the manhattan 5lb done. If u have still a problem though please have a look at that book name m said earlier. thanks

y ≥ (x−3)^2 - 1 same as parabola equation y = a(x-h)^2 + k

thus a=1 mean upper ward parabola & (h,k)= (3, -1) is vertex , if plot it on graph then this parabola does not intersect in 3rd quadrant thus C