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TimeGoesTooFast · Oct 17, 2013

I was wondering if anyone could help find the oblique asymptote for:

f(x)=x^3/(1-x)^2

I got y=x but the answers said it was y=x+2 and then solved it without dividing highest power...

HeroicPandas · Oct 17, 2013

$f(x) = \frac{x^3}{(1-x)^2} \\\\= \frac{x^3}{x^2 - 2x + 1} \\\\= \frac{x(x^2 - 2x +1)+2x^2 - x}{x^2 - 2x + 1} \\\\ = x + \frac{x^2 - 2x+1 +x^2 + x -1}{x^2 - 2x + 1} \\\\ =x+1 + \frac{x^2 +x - 1}{x^2 - 2x +1} \\ \\ = x+1 + \frac{x^2 - 2x + 1 + 3x -2}{x^2 - 2x + 1} \\\\ = x+1 + 1 + \frac{3x-2}{x^2 - 2x + 1}$
^If you don't understand, forget it

Long dividing would be the best way to find oblique assymptotes for this question (very quick - try it)

$x^2 - 2x + 1 |\overline{x^3 + 0x^2 + 0x + 0}$

EDIT: made a mistake

TimeGoesTooFast · Oct 17, 2013

HeroicPandas said:
$f(x) = \frac{x^3}{(1-x)^2} = \frac{x^3}{x^2 - 2x + 1} = \frac{x(x^2 - 2x +1)+2x^2 - x}{x^2 - 2x + 1} \\ \\ f(x) = x + \frac{x^2 - 2x+1 +x^2 + x -1}{x^2 - 2x + 1} = x+2 + \frac{x^2 +x - 1}{x^2 - 2x +1}$

^If you don't understand, forget it

Long dividing would be the best way to find oblique assymptotes for this question (very quick - try it)

$x^2 - 2x + 1 |\overline{x^3 + 0x^2 + 0x + 0}$

Haha, that was kind of like the answers. I was wondering, how you knew how to manipulate it like that?

Ah okay thanks :>)

HeroicPandas · Oct 17, 2013

TimeGoesTooFast said:
Haha, that was kind of like the answers. I was wondering, how you knew how to manipulate it like that?

Ah okay thanks :>)

Its an alternative way to long dividing - when the top degree of GREATER OR EQUAL TO bottom degree.
Begin with: $f(x) = \frac{x^3}{x^2 - 2x + 1}$

Write 'bottom on bottom'

$f(x) = \frac{x^2 - 2x + 1}{x^2 - 2x + 1}$

Do stuff to make it look like your true f(x) or else you've changed the question

$f(x) = \frac{x(x^2 - 2x + 1) + 2x^2 - x}{x^2 - 2x + 1}$

Simplfiy,

$f(x) = x + \frac{2x^2 - x}{x^2 - 2x + 1}$

Then keep doing it

Is this what you're asking?

EDIT: in bold

TimeGoesTooFast · Oct 17, 2013

Oh right lol. yep thanks again haha~

HeroicPandas · Oct 17, 2013

I realised I made a mistake lol (check again, I edited my post)

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