(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
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
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!