## anonymous 5 years ago what is an oblique asymptote?

1. anonymous

It's when you have an asymptote that's not parallel to the x- or y-axis.

2. anonymous

3. anonymous

The blue line's the oblique asymptote.

4. anonymous

The other asymptote, x = 0 (i.e. the y-axis) is not oblique.

5. anonymous

so when would you have one?

6. anonymous

I'm just looking for a function/./.

7. anonymous

You know, instead of me repeating everything, there's some useful information here: http://en.wikipedia.org/wiki/Asymptote

8. anonymous

You basically have an oblique asymptote if,$\lim_{x \rightarrow \infty}[f(x)-(mx+b)]=0$where f is your function and mx+b is a straight line.

9. anonymous

That's the definition.

10. anonymous

Are you okay with that or do you need to run through an example?

11. anonymous

may i please have an example. thankz

12. anonymous

my teacher said that we know there is an OA when the question is a fraction and the power of the numerator is greater than the power of the denominator or something like that

13. anonymous

but i was a bit confused about that

14. anonymous

Okay. You could have a function like$y=x+\frac{1}{x}= \frac{x^2+1}{x}$

15. anonymous

From the definition, we check for the slope of the OA:$m:=\lim_{x \rightarrow \pm \infty}\frac{f(x)}{x}=\lim_{x \rightarrow \pm \infty}\frac{(x+\frac{1}{x})}{x}=\lim_{x \rightarrow \pm \infty}(1+\frac{1}{x^2})=1$

16. anonymous

17. anonymous

Now you need to check for your y-intercept:

18. anonymous

$n:=\lim_{x \rightarrow \pm \infty}(f(x)-mx)=\lim_{x \rightarrow \pm \infty}((x+\frac{1}{x})-x)=\lim_{x \rightarrow \pm \infty}(\frac{1}{x})=0$

19. anonymous

(should read 0 at end). So the equation of your OA is$y=x$

20. anonymous

Both the limits have to exist for an OA.

21. anonymous

in your working out how did you get lim[1+(1/x^2)]

22. anonymous

how did you conclude that y=x?

23. anonymous

Because$\frac{x+\frac{1}{x}}{x}=\frac{x}{x}+ \frac{\frac{1}{x}}{x}=1+\frac{1}{x} \times \frac{1}{x}=1+\frac{1}{x^2}$

24. anonymous

y=x because m=1 and n = 0.

25. anonymous

The process is all about finding a line, y=mx+n (if it exists) that will be an OA.

26. anonymous

oh ok i think i should go and do some more practice egs thanks for writing all the working out! you make a brilliant teacher!

27. anonymous

you're welcome :)

28. anonymous

oh umm do you know any websites that have practice questions on differentiation curve sketching?

29. anonymous

Not off the top of my head. I can have a quick look.

30. anonymous

Is this what you're looking for? http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html

31. anonymous

Just do a Google search for "curve sketching problems".

32. anonymous

ahhh thank you very much for the site

33. anonymous

np