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asymptotes problem (1 Viewer)

ronaldinho

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how do i find the asymptotes for this question without polynomials

can it be done without knowing polynomails?

the back of the book has it as a parabola...

y = x^2 + 1/x^2
 
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Riviet

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As x approaches positive or negative infinity, 1/x2 approaches zero, since x2 approaches infinity. Therefore y approaches x2.
 

rocky1989

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whenever i have to figure out asymptotes...i find that the easiest way is just to first look at the denominator and find out when it equals 0 (the obvious ones), and then i just sub in x=999999999.....etc. then you are able to see where it is finishing up. so no matter how much bigger it gets, it is forever approaching y=1. hope that helps, its a pretty simple technique but it seems to work. so the 2 asymptotes will be x=0 and y=1.
 

rocky1989

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As x approaches positive or negative infinity, 1/x<sup>2</sup> approaches zero, since x<sup>2</sup> approaches infinity. Therefore y approaches x<sup>2</sup>.
nope
 

ronaldinho

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lol smart one ^^.. dats i i editted it now.. cos it had the denominator in a weird place :p

i was like wtf? how did he know where the denom. spose to be?

how come ur not subbing in infinity for the x^2 as well riviet?

wont it be y approaches infinity.. so no asymptote?
 
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rocky1989

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The curve is y= x<sup>2</sup> + (1/x<sup>2</sup>)

Quote his original post and see for yourself. ;)
righto...but how did you know what was originally written.
it didn't look like y=x^2 + 1/x^2 at all
 

ronaldinho

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where i went wrong?

horizontal asymptote

i think the def. for hor. asymp. is

if y = l as x approaches infinity then its a horizontal asymptote...

but when i sub in x = infinity i get y = infinity

vertical asymptote

i think defn. is y=infinity when x---> positive oor negative c (domain)

--------------------

when i do all this it doesnt work.. so how u know if its parabola.. thanks bye!
 

SoulSearcher

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Vertical asymptote is y = 0, because you cannot have x = 0 otherwise 1/x2 would not be defined.

For a horizontal asymptote, note the behaviour of the graph as it approaches x = 1 and x = -1. x = infinity won't work because the graph acts like a parabola as x approaches infinity, and thus the graph will approach infinity as x approaches infinity.
 

followme

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sketch y=x^2 and y=(1/x)^2 then add. there's only one asymp ie x=0 (vertical line)
 

conman

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ronaldinho said:
where i went wrong?

horizontal asymptote

i think the def. for hor. asymp. is

if y = l as x approaches infinity then its a horizontal asymptote...

but when i sub in x = infinity i get y = infinity

vertical asymptote

i think defn. is y=infinity when x---> positive oor negative c (domain)

--------------------

when i do all this it doesnt work.. so how u know if its parabola.. thanks bye!
Vertical asymptote
- when there is a border (check for domain)
- Let x goes to - infinity and + infinity if both results are equal to - infinity or -infinity so there is vertical asymptote.
* Only one type occurs if there is a border don't need to find lim

Horizontal asymptote:
- Let x goes - infinity and + infinity. Results are equal to a constant then there is a horizontal asymptote:

Slaint asymptote(when there is slaint asymptote don't need to check horizontal and vice versa).
- When the power on top is higher than the power at the bottom.
 

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