anonymous
  • anonymous
what is an oblique asymptote?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
It's when you have an asymptote that's not parallel to the x- or y-axis.
anonymous
  • anonymous
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anonymous
  • anonymous
The blue line's the oblique asymptote.

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anonymous
  • anonymous
The other asymptote, x = 0 (i.e. the y-axis) is not oblique.
anonymous
  • anonymous
so when would you have one?
anonymous
  • anonymous
I'm just looking for a function/./.
anonymous
  • anonymous
You know, instead of me repeating everything, there's some useful information here: http://en.wikipedia.org/wiki/Asymptote
anonymous
  • 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.
anonymous
  • anonymous
That's the definition.
anonymous
  • anonymous
Are you okay with that or do you need to run through an example?
anonymous
  • anonymous
may i please have an example. thankz
anonymous
  • 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
anonymous
  • anonymous
but i was a bit confused about that
anonymous
  • anonymous
Okay. You could have a function like\[y=x+\frac{1}{x}= \frac{x^2+1}{x}\]
anonymous
  • anonymous
From the definition, we check for the slope of the OA:\[m:=\lim_{x \rightarrow \pm \infty}\frac{f(x)}{x}=\lim_{x \rightarrow \pm \infty}\frac{(x+\frac{1}{x})}{x}=\lim_{x \rightarrow \pm \infty}(1+\frac{1}{x^2})=1\]
anonymous
  • anonymous
so that's your slope
anonymous
  • anonymous
Now you need to check for your y-intercept:
anonymous
  • anonymous
\[n:=\lim_{x \rightarrow \pm \infty}(f(x)-mx)=\lim_{x \rightarrow \pm \infty}((x+\frac{1}{x})-x)=\lim_{x \rightarrow \pm \infty}(\frac{1}{x})=0\]
anonymous
  • anonymous
(should read 0 at end). So the equation of your OA is\[y=x\]
anonymous
  • anonymous
Both the limits have to exist for an OA.
anonymous
  • anonymous
in your working out how did you get lim[1+(1/x^2)]
anonymous
  • anonymous
how did you conclude that y=x?
anonymous
  • anonymous
Because\[\frac{x+\frac{1}{x}}{x}=\frac{x}{x}+ \frac{\frac{1}{x}}{x}=1+\frac{1}{x} \times \frac{1}{x}=1+\frac{1}{x^2}\]
anonymous
  • anonymous
y=x because m=1 and n = 0.
anonymous
  • anonymous
The process is all about finding a line, y=mx+n (if it exists) that will be an OA.
anonymous
  • 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!
anonymous
  • anonymous
you're welcome :)
anonymous
  • anonymous
oh umm do you know any websites that have practice questions on differentiation curve sketching?
anonymous
  • anonymous
Not off the top of my head. I can have a quick look.
anonymous
  • anonymous
Is this what you're looking for? http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
anonymous
  • anonymous
Just do a Google search for "curve sketching problems".
anonymous
  • anonymous
ahhh thank you very much for the site
anonymous
  • anonymous
np

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