oblique asymptotes (1 Viewer)

[h3ll h0und]

hey guys
umm..i noe how to find vertical and horizontal asymptotes but how do u find oblique asymptotes?
e.g....oblique asymptote for y=x - (1/x) is y=x (got this example from cambridge)

thanx in advance ... soz for my noobness

 

[nicko1990]

you gotta add or subtract graphs, kinda like superimposition of waves in physics. then from the added or subtracted graph you can see shape and infer the asymptote

 

[h3ll h0und]

so u just observe and infer? isnt their like any way apart from that? like wen finding the horizontal asymptote u use lim x->infinite

 

[unwanted80]

no, y=x is an asymptote, because the graph equation is in the simplest form already, so there is no need to simplify it further (long division).

 

[study-freak]

It's nothing difficult.

You first need to use division transformation to break fractions down to pronumeral 1+number/(pronumeral 2) format.

Then the next step would be to use lim(x->infinity) number/pronumeral 2, which gives you 0.
Then you are left with the pronumeral 1, and y=pronumeral 1 is the oblique asymptote.

Sorry, I can't explain this well.

I'll illustrate this with ur example.
y=x - (1/x) is y=x since lim (x->infinity)1/x=0 and you are left with x.

Was it comprehensible? lol My explanation doesn't seem so nicely worded.

 

[bored.of.u]

hey guys
umm..i noe how to find vertical and horizontal asymptotes but how do u find oblique asymptotes?
e.g....oblique asymptote for y=x - (1/x) is y=x (got this example from cambridge)

thanx in advance ... soz for my noobness

best way is to use lim where x-> infinity
therefore lim x->infinity y=x-(1/x)/(x/x)
lim x->infinity y=x-(0/1)
therefore lim x->infinity y=x

although i have only begun doing oblique asymptotes so i may be wrong...hope this helped hell hound ;D

 

[Timothy.Siu]

yeah...as x-->infinity y--->x

 

[kaz1]

It's similar to horizontal assymptotes.

 

[random-1005]

hey guys
umm..i noe how to find vertical and horizontal asymptotes but how do u find oblique asymptotes?
e.g....oblique asymptote for y=x - (1/x) is y=x (got this example from cambridge)

thanx in advance ... soz for my noobness

it is the same process for finding horizontal asymtotes, divide by highest power x in the denimantor and take the lim as x->infinity

 

[h3ll h0und]

oh ok....i get it now....so its just like finding horizontal asymptotes and in the end ur left with y=x so that's the oblique asymptote? thanx guys