what is an oblique asymptote?

- anonymous

what is an oblique asymptote?

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- anonymous

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

- anonymous

##### 1 Attachment

- anonymous

The blue line's the oblique asymptote.

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- anonymous

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

- anonymous

so when would you have one?

- anonymous

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

- anonymous

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

- 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

That's the definition.

- anonymous

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

- anonymous

may i please have an example. thankz

- 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

but i was a bit confused about that

- anonymous

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

- 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

so that's your slope

- anonymous

Now you need to check for your y-intercept:

- 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

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

- anonymous

Both the limits have to exist for an OA.

- anonymous

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

- anonymous

how did you conclude that y=x?

- 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

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

- anonymous

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

- 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

you're welcome :)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ahhh thank you very much for the site

- anonymous

np

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