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anonymous

  • 5 years ago

what is an oblique asymptote?

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  1. anonymous
    • 5 years ago
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    It's when you have an asymptote that's not parallel to the x- or y-axis.

  2. anonymous
    • 5 years ago
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  3. anonymous
    • 5 years ago
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    The blue line's the oblique asymptote.

  4. anonymous
    • 5 years ago
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    The other asymptote, x = 0 (i.e. the y-axis) is not oblique.

  5. anonymous
    • 5 years ago
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    so when would you have one?

  6. anonymous
    • 5 years ago
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    I'm just looking for a function/./.

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

  8. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    That's the definition.

  10. anonymous
    • 5 years ago
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    Are you okay with that or do you need to run through an example?

  11. anonymous
    • 5 years ago
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    may i please have an example. thankz

  12. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    but i was a bit confused about that

  14. anonymous
    • 5 years ago
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    Okay. You could have a function like\[y=x+\frac{1}{x}= \frac{x^2+1}{x}\]

  15. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    so that's your slope

  17. anonymous
    • 5 years ago
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    Now you need to check for your y-intercept:

  18. anonymous
    • 5 years ago
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    \[n:=\lim_{x \rightarrow \pm \infty}(f(x)-mx)=\lim_{x \rightarrow \pm \infty}((x+\frac{1}{x})-x)=\lim_{x \rightarrow \pm \infty}(\frac{1}{x})=0\]

  19. anonymous
    • 5 years ago
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    (should read 0 at end). So the equation of your OA is\[y=x\]

  20. anonymous
    • 5 years ago
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    Both the limits have to exist for an OA.

  21. anonymous
    • 5 years ago
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    in your working out how did you get lim[1+(1/x^2)]

  22. anonymous
    • 5 years ago
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    how did you conclude that y=x?

  23. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    y=x because m=1 and n = 0.

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

  26. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    you're welcome :)

  28. anonymous
    • 5 years ago
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    oh umm do you know any websites that have practice questions on differentiation curve sketching?

  29. anonymous
    • 5 years ago
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    Not off the top of my head. I can have a quick look.

  30. anonymous
    • 5 years ago
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    Is this what you're looking for? http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html

  31. anonymous
    • 5 years ago
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    Just do a Google search for "curve sketching problems".

  32. anonymous
    • 5 years ago
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    ahhh thank you very much for the site

  33. anonymous
    • 5 years ago
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    np

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