Question 1160
      Since any denominator cannot be 0, 
      so x-3 cannot be 0 in y=G(x)= 1 +  2/(x-3)^2 and 3 should be
      excluded from the domain of G.
      Hence,the domain of G = R -{3} = (-oo,3) U(3,+oo)
      Next,we see that 2/(x-3)^2 > 0 for all  real x, so G(x) > 1+0 = 1.
      So, the range of G is (1, +oo)
      
      When x =0, G(0) = 1 + 2/9 = 11/9 is y-intercept.
      Since y=G(x) >0 for all x, there are no x-intercepts of y=G(x).   

      The denominator in G(x) is x - 3= 0, which is a vertical asymptote 
      of G.
      As x--> +oo, G(x) --> 1.So, a horizontal asymptote is y=1.