# Find the slant asymptote?

• March 19th 2009, 02:45 PM
puzzledwithpolynomials
Find the slant asymptote?
(x^3-4x²+2x-5)/(x²+2)
• March 19th 2009, 03:05 PM
skeeter
Quote:

Originally Posted by puzzledwithpolynomials
Find the slant asymptote?

(x^3-4x²+2x-5)/(x²+2)

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
• March 21st 2009, 04:29 AM
SeanBE
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
• March 21st 2009, 10:58 AM
skeeter
Quote:

Originally Posted by SeanBE
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

quotient is ...

$x - 4 + \frac{3}{x^2+2}$

slant asymptote is $y = x - 4$
• March 21st 2009, 11:26 AM
Hush_Hush
A slant asymptote always has this form:

y=mx+n

Where :

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

and

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

In this particular case of yours you have :

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

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

So the fianl result will be :

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

That's all,

Have a nice day!