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majorbino
Aug 15, 2002, 01:14 PM
Help! Can anyone give an example of that? Thanks in advance!

whyte
Aug 15, 2002, 02:32 PM
Aysmptotes lang ang natatandaan ko.

Diba they're "lines" that approaches another closer and closer BUT will never touch until infinity ........

whyte
Aug 15, 2002, 02:35 PM
para akong tutor nito ...... :) :)


ONE (http://cne.gmu.edu/modules/dau/calculus/limits/limits8_bdy.html)

TWO (http://www.ping.be/math/asym.htm)

IMHOTEP
Aug 15, 2002, 04:27 PM
Dapat TC7 to the rescue!!!

juliuservin
Aug 15, 2002, 05:24 PM
OBLIQUE ASYMPTOTES
- are not Vertical nor Horizontal

y = mx + b is an oblique asymptote:

lim as x approaches + infinity [f(x) - (mx + b)] = 0

lim as x approaches - infinity [f(x) - (mx + b)] = 0

Example:

Find the oblique asymptotes of the average cost function:

c(x) = 1600 + .25x^2 x>0

c(x) = 1600/x + 0.25x

lim as x approaches + infinity [(1600/x + 0.25x) - (0.25x)] = 0

THEREFORE:

y = 0.25x is the oblique asympotote!!!!!!!!!!!!

melie
Aug 16, 2002, 04:08 AM
oblique asymptotes? test ko yan mamya. midterms... agh, wish me luck!

anyway, kapag *** highest degree ng numerator ng rational function ay mas malaki (by one degree) sa highest degree ng denominator, magkakaroon ng oblique asymptote. :) bale divide mo *** numerator by the denominator, tapos *** linear equation na sagot ang oblique asmptote. :)

yehey! tama ba ako? :)

agent1611
Aug 16, 2002, 05:21 AM
oblique asymptotes: are asymptotes whose directrices doesn't run parallel to either the x or y axes, instead their directrices are in lay man 'slanting' or having their own coordinates in the the plane.

dijkstra
Aug 16, 2002, 12:06 PM
Originally posted by juliuservin
OBLIQUE ASYMPTOTES
- are not Vertical nor Horizontal

y = mx + b is an oblique asymptote:

lim as x approaches + infinity [f(x) - (mx + b)] = 0

lim as x approaches - infinity [f(x) - (mx + b)] = 0
ok na sana ito but the example is wrong.




Example:

Find the oblique asymptotes of the average cost function:

c(x) = 1600 + .25x^2 x>0

c(x) = 1600/x + 0.25x

lim as x approaches + infinity [(1600/x + 0.25x) - (0.25x)] = 0

THEREFORE:

y = 0.25x is the oblique asympotote!!!!!!!!!!!!

"c(x) = 1600 + .25x^2" (x>0) is an equation of a parabola (2nd degree exponent). as far as we know, a parabola does not have an asymptote but latus rectum, directrix, focus and vertex.

the solution is obviously flawed.

look at the way she/he solved for the asymptote. the side of 1600 + .25x^2 is devided by x, but c(x) remains the same. deviding only one side by x already changes the equation.

(1) c(x) = 1600 + .25x^2 (x>0)

is not the same not the same as

(2) c(x) = 1600/x + .25x (x>0).

(1) does not have an asymptote, while (2) has. but, the derivation of (2) from (1) is flawed. the asymptote y = .25x is not the asymptote of (1) but of (2).

therefore, the conclusion of the above example is invalid. how can we get an asymptote of something that does not even have an asymptote? it does not make sense.

majorbino
Aug 16, 2002, 02:14 PM
Is it possible to find an oblique asymptote in an irrational function?

juliuservin
Aug 16, 2002, 03:37 PM
Originally posted by dijkstra

ok na sana ito but the example is wrong.





"c(x) = 1600 + .25x^2" (x>0) is an equation of a parabola (2nd degree exponent). as far as we know, a parabola does not have an asymptote but latus rectum, directrix, focus and vertex.

the solution is obviously flawed.

look at the way she/he solved for the asymptote. the side of 1600 + .25x^2 is devided by x, but c(x) remains the same. deviding only one side by x already changes the equation.

(1) c(x) = 1600 + .25x^2 (x>0)

is not the same not the same as

(2) c(x) = 1600/x + .25x (x>0).

(1) does not have an asymptote, while (2) has. but, the derivation of (2) from (1) is flawed. the asymptote y = .25x is not the asymptote of (1) but of (2).

therefore, the conclusion of the above example is invalid. how can we get an asymptote of something that does not even have an asymptote? it does not make sense.

---------------------------------------------------------------------------------
Eqn 1 - cost function
Eqn 2 - average cost function

kala ko clear na sa inyo ibig sabihin ng average cost function when i said : Find the ... average cost function ....

c(x) - cost function
c(x)/x - average cost function

(1) c(x) = 1600 + .25x^2
(2) c(x)/x = 1600/x + .25x

nakalimutan lang i type /x sa c(x) ... but that does not mean the example is flawed.

THIS IS THE MOST IMPORTANT PART:

y = 0.25x is the oblique asympotote because ... it makes the eqn equal to 0 as x approaches infinity (check mo uli un given definition sa taas)

I believe a CALCULUS solution is what majorbino looks for ...

http://www.ping.be/math/asym.gif

ayan sya oblique, hinde parabola

Check this example shown by whyte ...

http://www.ping.be/math/asym.htm#Example-:

am+dg

dijkstra
Aug 17, 2002, 12:43 PM
Originally posted by juliuservin


---------------------------------------------------------------------------------
Eqn 1 - cost function
Eqn 2 - average cost function

kala ko clear na sa inyo ibig sabihin ng average cost function when i said : Find the ... average cost function ....

c(x) - cost function
c(x)/x - average cost function

(1) c(x) = 1600 + .25x^2
(2) c(x)/x = 1600/x + .25x

nakalimutan lang i type /x sa c(x) ... but that does not mean the example is flawed.

THIS IS THE MOST IMPORTANT PART:

y = 0.25x is the oblique asympotote because ... it makes the eqn equal to 0 as x approaches infinity (check mo uli un given definition sa taas)

I believe a CALCULUS solution is what majorbino looks for ...

http://www.ping.be/math/asym.gif

ayan sya oblique, hinde parabola

Check this example shown by whyte ...

http://www.ping.be/math/asym.htm#Example-:

am+dg

c(x) = 1600 + .25x^2 does not have have an asymptote because this cost function is a parabola.

on the other hand, c(x)/x = 1600/x + .25x, an average cost function, has an oblique asymptote.

clear.

ad majorem+dei gloriam
basta ikaw lord

are you gonna take financial engineering?

juliuservin
Aug 18, 2002, 05:20 PM
Originally posted by dijkstra


c(x) = 1600 + .25x^2 does not have have an asymptote because this cost function is a parabola.

on the other hand, c(x)/x = 1600/x + .25x, an average cost function, has an oblique asymptote.

clear.

ad majorem+dei gloriam
basta ikaw lord

are you gonna take financial engineering?

No ...

MS Computer Science

whattaman
Aug 21, 2002, 08:20 AM
consult Mr Leithold's TC7. hindi ka matuto ng husto kugn dito ka magdedpend!

gud luck dude!

agent1611
Aug 21, 2002, 08:59 AM
naaliw ako sa inyo!:D talagang presenting it 'the way you want it'

ang saya nyong dalawa!:D

gusto n'yo para mas malinaw yung book ni pytel 3rd edition?:lol: :bounce::evil_lol:

topak_fos
Aug 21, 2002, 01:16 PM
... hay salamat, buti technical course ang kinuwa ko, 1yr lang tapos na.. la pang Math.... :redgrin: