Question 5013
f(x) = (x^3-x^2-6x)
       --------------
       (x^2-3x+2)

I want to find:
a)  Any zeros ____
  The top: x^3-x^2-6x = x(x^2-x-6) = x(x-3)(x+2) = 0 -->
   x =0 or 3 or -2 (three zeros)
b)  vertical asymptotes ___
  Set the denominator x^2-3x+2 = 0 -->(x-2)(x-1) = 0 -->
  x -2 = 0 or x-1 = 0  (two vertical asymptotes)
c)  horizontal asymptotes___
  Set x-->+oo, f(x)--> x^3/x^2 -->+oo
  So, no horizontal asymptotes
d) the domain of the function_______
  The denominator cannot be zero, so the domain
  = R -{1,2} = (-oo,1)U(1,2)U(2,+OO)
e) the range of the function____
  f(x) = x(x-3)(x+2)
       --------------
       (x-2)(x-1)
  f(x) can be any real numbers, so 
  the range of the function = R
  
f) where f(x)is >=: ________
 f(x) = x(x-3)(x+2)
       --------------
       (x-2)(x-1)
 f(x)>= 0 implies x(x-3)(x+2)(x-2)(x-1)>=0 rearrange to
 or (x-3)(x-2)(x-1)x(x+2)>=0 
 so, x>=3 or 1 < x < 2 or -2 <= x <= 0 {Note x cannot ben1 or 2)
 Expressed by intervals: [-2,0] U (1,2)U [3,+oo)

 Important note: Forget to use any calculator to solve this
 kind of questions. 
 Idea is crucial in math no calculations or the results.

 Kenny