Question 5013: f(x) = x^3-x^2-6x divided by x^2-3x+2
I want to find:
a) Any zeros ____
b) vertical asymptotes ___
c) horizontal asymptotes___
d) the domain of the function_______
e) the range of the function____
f) where f(x)is greater or = to 0: ________
this can be done by using a calculator but need steps to see how to do it.
Thanks
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! f(x) = (x^3-x^2-6x)
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(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)
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(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)
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(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
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