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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's when you have an asymptote that's not parallel to the x or yaxis.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The blue line's the oblique asymptote.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The other asymptote, x = 0 (i.e. the yaxis) is not oblique.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so when would you have one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm just looking for a function/./.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know, instead of me repeating everything, there's some useful information here: http://en.wikipedia.org/wiki/Asymptote

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0That's the definition.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you okay with that or do you need to run through an example?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0may i please have an example. thankz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my 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
 5 years ago
Best ResponseYou've already chosen the best response.0but i was a bit confused about that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. You could have a function like\[y=x+\frac{1}{x}= \frac{x^2+1}{x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From 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
 5 years ago
Best ResponseYou've already chosen the best response.0Now you need to check for your yintercept:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[n:=\lim_{x \rightarrow \pm \infty}(f(x)mx)=\lim_{x \rightarrow \pm \infty}((x+\frac{1}{x})x)=\lim_{x \rightarrow \pm \infty}(\frac{1}{x})=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(should read 0 at end). So the equation of your OA is\[y=x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Both the limits have to exist for an OA.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in your working out how did you get lim[1+(1/x^2)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you conclude that y=x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because\[\frac{x+\frac{1}{x}}{x}=\frac{x}{x}+ \frac{\frac{1}{x}}{x}=1+\frac{1}{x} \times \frac{1}{x}=1+\frac{1}{x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=x because m=1 and n = 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The process is all about finding a line, y=mx+n (if it exists) that will be an OA.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh umm do you know any websites that have practice questions on differentiation curve sketching?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not off the top of my head. I can have a quick look.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this what you're looking for? http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just do a Google search for "curve sketching problems".

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh thank you very much for the site
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