WHY!! asmyptotes -_- (1 Viewer)

[Seraph]

Hey pplz im having some probs

check this out y = x/(x+1)

obviously there is an asymptote at x = -1

but where are the y asymptotes and why???
and is there a stat pt for this at (3,0.75) ?? (yea i need to sketch it).....

 

[sunny]

Think of it this way,

for y to equal 1, the numerator and denominator of the RHS have to be the same.

But they can never ever be the same since the numerator is x, and the denominator is x+1

Hence the RHS can never simplify to 1, thus you get a horizontal (y) asymptote at y=1.

 

[Seraph]

why does y have to equal 1?

ok hang on i just tried something , i threw in some random numbers into that equation and
yea y is trying to approach 1 , but how can we say that y = 1????

 

[KeypadSDM]

y = x/(x + 1)
xy + y = x
y = x - xy
y = x(1 - y)
x = y/(1 - y)

And thus, there must be asymptotes in both the y and x axis.

More importantly, it asymptotes at y = 1

 

[Seraph]

oh nice , is that all you have to do for any equation to get the y asymptote
just re arrange so x = ????

 

[KeypadSDM]

Exactomundo. It's one method of doing it.
The other is using limits.

As x -> oo
y -> oo/(oo + 1)
y -> 1/(1 + (1/oo))
y -> 1
Thus y = 1 is an asymptote

 

[Rahul]

i like the way of using limits, it always made more sense.

for y = x/(x+1), remember that you can never divide thru by 0. in the case where x=-1, you would have a situation where 1/0, which cannot be. so x cannot be -1

 

[KeypadSDM]

Originally posted by Rahul
i like the way of using limits, it always made more sense.

But the way of rearranging the equation, that's prefectly legit as well, right?

I found that it made much more algebraic sense, and was much easier to relate to others.

 

[...]

eh

go the algebra method

Key's first approach

simple and its easy to understand

sif infinity rahul

 

[Rahul]

hmm....i'll show you a trick i used.

well it made more practical sense to me. "as y got infinitely big, x would be nearly 1 or something"

all you do in sub in a large number, such as 100, for say x in y=x/(x+1). it would be a number close to 1, thus the asympotote is at 1. a bit shifty in its working, but all you will need to show is:
as x -> oo, y -> 1...etc.

 

[...]

but that would be very applicable when ur drawing grpahs to determine how the graph will "end"

 

[Rahul]

nah it works for asymptotes aswell. it helped for me for curve sketching in 3u/4u.

Originally posted by ...
but that would be very applicable when ur drawing grpahs to determine how the graph will "end"

hmm....actually i dont really know what you mean.

 

[...]

u never understand me


/me runs off crying

 

[KeypadSDM]

Originally posted by Rahul
hmm....actually i dont really know what you mean.

Lol, weak. Like where the 'ends' of the graph, i.e. at infinity, lie.

 

[...]

Originally posted by KeypadSDM
Lol, weak. Like where the 'ends' of the graph, i.e. at infinity, lie.

yay

as in will it increase as u said towards y-infinity or x-infinity


ur truly a maths god key

 

[Seraph]

Originally posted by KeypadSDM
But the way of rearranging the equation, that's prefectly legit as well, right?

I found that it made much more algebraic sense, and was much easier to relate to others.

yes if i can use this way it would be much more easier !!!

plus all y asmyptotes dont have to be as x --> oo do they?

 

[Seraph]

Originally posted by Rahul
hmm....i'll show you a trick i used.

well it made more practical sense to me. "as y got infinitely big, x would be nearly 1 or something"

all you do in sub in a large number, such as 100, for say x in y=x/(x+1). it would be a number close to 1, thus the asympotote is at 1. a bit shifty in its working, but all you will need to show is:
as x -> oo, y -> 1...etc.

thats kinda what i did
subbed big numbers in
and i coudl see that it was approaching 1 but i couldnt see the connection too much with the infinity oo

BTW
...
For saying arsenal sux!!!!!
guess whose on top of the table now

 

[Rahul]

i couldnt explain it in a post. but it is just the way i did it.

 

[Xayma]

Originally posted by Seraph
plus all y asmyptotes dont have to be as x --> oo do they?

No they dont. Depending on what value x cant equal will affect it as well for example 1/x you also have y asmyptotes as x--->0 or for 1/(x+2) you have y---> infinity as x-->-2

 

[KeypadSDM]

Originally posted by Xayma
No they dont. Depending on what value x cant equal will affect it as well for example 1/x you also have y asmyptotes as x--->0 or for 1/(x+2) you have y---> infinity as x-->-2

Isn't that an x asymptote? x = -2 is the asymptote?

Anyway, I'm pretty sure all asymptotes in y (or x) must be to infinity in x (or y), that's the very definition of an asymptote, the value the function will reach when infinity has been obtained.