# Thread: How do you find the oblique asymptote for this specific rational function

1. ## How do you find the oblique asymptote for this specific rational function

The problem I'm trying to find the oblique asymptote for is: 8x^3+1/x^2-5x+6
I know the answers to this specific problem(vertical asymptotes at: x=2,x=3; oblique asymptote at:y=x+5) but I don't know how to arrive at the answer for the oblique asymptote for this specific problem. Can anyone help? I know about how to use long division to solve most of these types of problems, but this one I don't know how to achieve the correct answer for the oblique asymptote even with long division.

2. $\frac{8x^3+1}{x^2-5x+6}$

Use polynomial long division to change this into a form where the order of the numerator is less than the order of the denominator. It should be clearer then.

3. I understand that I should use polynomial long division, but even so I still don't arrive at the correct answer. Perhaps my calculations are wrong. If you've tried it already using long division and arrived at the oblique asymptote I have listed above, would you mind showing me your steps?

4. Check this for the method Polynomial Long Division

I get (did it rather quick)

$\frac{8x^3+1}{x^2-5x+6} = 8x+40+\frac{241}{x^2-5x+6}
$

5. Yea that's what I get when I work it out too. But my book is listing the answer for the oblique asymptote as y=x+5 and shows that I need to work it out as x^3-8/x^2-5x+6
Which I'm not understanding where x^3-8 comes from. Any ideas?

6. Originally Posted by micbal9388
I'm not understanding where x^3-8 comes from. Any ideas?
No I can't see how, maybe 8 has been taken out as a common factor. But not even.

$\frac{8x^3+1}{x^2-5x+6} = 8x+40+\frac{241}{x^2-5x+6} = 8(x+5)+\frac{241}{x^2-5x+6}$

7. Yea, thanks anyway. Ha, I've asked a bunch of people and no one knows. It's quite baffleing to me.

8. Yea I see what the problem is. My teacher created an incorrect answer sheet and confused the hell out of me. Thanks, you were right.