(x-1)^3/x if i try to divide to get the asymptote i get x^2 is that right?

Question

anonymous · Answer

don't think you can divide that based on it's not like terms

anonymous · Answer

umm lol u use long division method

anonymous · Answer

oh..i thought u meant divide as in exponents

anonymous · Answer

sorry still waking up

anonymous · Answer

well (x-1)^3 expanded out is:
$$x^3-3x^2+3x-1$$
divided by x you get:
$$x^2-3x+3-\frac{1}{x}$$

anonymous · Answer

so a vertical asymptote would be x = 0
and as x gets really really really large, it will approach the graph of
$$x^2-3x+3$$

anonymous · Answer

no i dont think so http://www.coolmath.com/graphit/ gave me the answer im looking for

anonymous · Answer

what is the answer its giving?

anonymous · Answer

the asymptote is x^2

anonymous · Answer

it is n't x^2

anonymous · Answer

how are you seeing that?

anonymous · Answer

nvm you are right

anonymous · Answer

sorry for doubting ur awesomness

anonymous · Answer

youre not entirely wrong though i guess.  One could say that as x goes to positive and negative infinity, the (3x+3) part isnt going to contribute as much as the x^2 term.  So in a sense, it will be asymptotic to x^2.

anonymous · Answer

especially that +3, that isnt going to do anything with large values of x.