(x^3-4x²+2x-5)/(x²+2)

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- Mar 19th 2009, 02:45 PMpuzzledwithpolynomialsFind the slant asymptote?
(x^3-4x²+2x-5)/(x²+2)

- Mar 19th 2009, 03:05 PMskeeter
do the long division ... the quotient will be a linear expression that is the equation of the slant asymptote (ignore the remainder).

here's a link with some examples ...

Slant, or Oblique, Asymptotes - Mar 21st 2009, 04:29 AMSeanBE
Did you figure out the answer? I learned about oblique/slant asymptotes a week ago and I thought I might try this problem out for practice. Would the answer by any chance be : x-4 R 4x+3 OR x-4 + (4x+3/x^2-2) Oblique asymptote being x-4 . Not sure If I am allowed to supply answers but I would just like to see if I am right.

Sean - Mar 21st 2009, 10:58 AMskeeter
- Mar 21st 2009, 11:26 AMHush_Hush
A slant asymptote always has this form:

y=mx+n

Where :

$\displaystyle m=\lim_{x\rightarrow\inf}\frac{f(x)}{x} $

and

$\displaystyle n=\lim_{x\rightarrow\inf}(f(x)-mx)$

In this particular case of yours you have :

$\displaystyle m=\lim_{x\rightarrow\inf}\frac{\frac{x^3-4x^2+2x-5}{x^2+2}}{x}=\lim_{x\rightarrow\inf}\frac{x^3-4x^2+2x-5}{x^3+2x} = 1 $

$\displaystyle n=\lim_{x\rightarrow\inf}(\frac{x^3-4x^2+2x-5}{x^2+2}-x)=\lim_{x\rightarrow\inf}\frac{-4x^2-5}{x^2+2}=-4$

So the fianl result will be :

$\displaystyle y=mx+n=1*x-4 =x-4$

That's all,

Have a nice day!