Asymptotes (1 Viewer)

[Premus]

i know how to find horizontal and vertical asymptotes for curves......
but do u know if there is an oblique asymptote?

eg: y = x^2 / ( 3 - x)

is there any special trick ?

Thanks

 

[smallcattle]

oblique asymptote???

never heard of such term

 

[Premus]

such as lines.... y = x

 

[tempco]

long divide the function.. you should get something like y = ax + b + (c/3-x)

y = ax + b <- thats the oblique assymptote, since when x -> +- infinity, c/3-x -> 0

 

[Dougie]

oblique asymptote???

never heard of such term

how could u not have heard of them!

 

[Premus]

awesome!
thanks

 

[smallcattle]

how could u not have heard of them!

i dont know, what topic is this under. parabola?

 

[jumb]

long divide the function.. you should get something like y = ax + b + (c/3-x)

y = ax + b <- thats the oblique assymptote, since when x -> +- infinity, c/3-x -> 0

Wow! I wish our teache told US that a fucking years ago.

 

[sub]

uhh, in 4u its graph sketching i dont know the 3u equivalent...sry

btw, u dont need to long divide...
for vertical asymptotes: denominator = 0
for horizontal/obliques: use x --> ±infinity straight off. then only consider the terms of the highest power in the fraction. from the example u get (x^2)/x which becomes x...therefore the oblique asymptote is y = x.
however, if u find this confusing...just long divide...no reason to get confused b4 the exam...this is just an alternative method hope that helps some goodluck tomoro

 

[jumb]

I guess you could just superimpose them and guestimate

 

[Premus]

but from the long division method...i get y = -x
could u explain ur method pls?

Thanks

 

[jumb]

i got y = -x - 3 as the asymptote

recheck your division

 

[Premus]

x^2 / ( 3-x ) = -x + 3x/ ( 3-x)

 

[danif]

heya, i have a hypothetical (dontcha just love those)

when finding oblique asymptotes - you divide by the highest power of x in the denominator correct? what if the highest power is in the numerator? what do you do then? i.e. so you dont get 1/x = 0 as lim--> 00 (that's infinity) e.g:

x^3+ 3x^2 + 2x - 5
-------------------------
4x^2 - 2x^2 + 9

N.B: that is not an actual example so if the answer isnt pretty then you know why =)


thanks =)

 

[becany]

i know how to find horizontal and vertical asymptotes for curves......
but do u know if there is an oblique asymptote?

eg: y = x^2 / ( 3 - x)

is there any special trick ?

Thanks

Ok for that example...is the oblique asymptote - x - 3?? now im confused...i thought i understood limits

 

[tennille]

If you divide y = x^2 / ( 3 - x) by the highest power ,ie x^2, you'll still get y = -x as the asymptote.

 

[underthebridge]

If you divide y = x^2 / ( 3 - x) by the highest power ,ie x^2, you'll still get y = -x as the asymptote.

wouldn't you have 1/(3/x^2 - 1/x) kinda hard to look at on the computer i know, but then wen x approaches infinty, the denominator approaches zero, which doesnt give u an asymptote. Therefore you have to divide by the HIGHEST POWER IN THE DENOMINATOR. btw, i worked out the oblique asymptote to be y = -x

 

[Xayma]

No what you do is you evaluate the highest power first, so you take the limit of 3/x² as x--> &infin; first. so it becomes 0 and you are left with 1/(-1/x) ie -x.

A similar thing is done with
lim.... sin x
x->&infin;.... x

 

[Sirius Black]

disagree

I got y=-x-3 for oblique asymptote

write y=x² / (3-x)
as y=(x²-9+9)/(3-x)
= (x²-9)/(3-x)+9/(3-x)
=-(x+3) + 9/(3-x)
as x--->infinity; y-----> -(x+3)
could anyone confirm my answer?

 

[tennille]

x^2/ x^2 = 1 (numerator).

3/x^2 = 0 (it is equal to 0 because as x approaches infinity, it will approach zero, like Xayma said)

x/x^2 = 1/x

Therefore, you get....

y= ____1____
0 - 1/x

which is y= -x