Graphing For n00bs (1 Viewer)

[velox]

how do i find the oblique asymtote for this graph y= (x^2)/(x+1)
Is it just y=0?

 

[velox]

oh crap i typed out the question wrong. It should be x^3 not x^2.

 

[velox]

but dont you have to do the lim as X approached infinity?

 

[velox]

u know how u get this x3=(x+1)(x2-x+1)-1 Cant u get rid of the (x+1) by cancelling out? So you just get x2-x+1-1?

 

[velox]

ok so u cant cancel out
But when u let x go to inifinity how do you get y = x^2 -x +1 ??

 

[velox]

but how do u get 1/(x+1) ?

 

[velox]

ok from the above equation, how do u get to 1/(x+1) ?

 

[velox]

did u split the fractions before u did that?

 

[CrashOveride]

ping, pong.

first ive seen tywebb get involved in more humbler tasks ^^

 

[Slidey]

y= (x^3)/(x+1)
y= (x^3+1)/(x+1) - 1/(x+1)
y= (x+1)(x^2-x+1)/(x+1) - 1/(x+1)
y= (x^2-x+1) - 1/(x+1)

That's how tywebb smanipulated the fraction. It's a skill you should acquire.

Here's a simpler example: (ax)/(x+1). To divide this fraction, we add a/(x+1) to get: (ax+a)/(x+1), which is a(x+1)/(x+1)=a. But since we added a/(x+1), we now take it away to get a - a/(x+1).

 

[velox]

yeah thats what i just got now. Thanks for the confirmation slide.

 

[velox]

if i have the graph y = ((X^3) -1)/(x-1) I need to find the oblique asymtote
I divided and got y= x^2 + x + 1
Then what do i do?

 

[mojako]

complete the square and graph the quadratic (parabola).
But dont forget its not defined at x=1
(denominator = 0)

 

[velox]

yeah thats what cambridge does. thanks

 

[SeDaTeD]

when x -> infinity, 3/(x+3) -> 0
so y will approach y = (x-1)

*edit - hmm, seems as if your question was removed by the time i posted this.

 

[velox]

i have this graph y = x/((x^(1/2)) +2) I divided it by the highest power of x and i get this (open the attachment). How can i simplify this so i can find the asymtotes?

 

[velox]

when x -> infinity, 3/(x+3) -> 0
so y will approach y = (x-1)

*edit - hmm, seems as if your question was removed by the time i posted this.

yeah, sorry, i should think more before i post up the questions

 

[queenie]

ok.. this is prob wrong, but anyway:

- here are no asymptotes (horizont or vert.)
- the curve only exists where x>0
- the curve gets pretty large as x gets large

so i get something like a y^2 =4ay parabola looking thing, above the x-axis, intersecting at 0.

there are no vert. asymp, but i cant seem to get the horixontal one... :S

 

[velox]

hmmm, ill post the answer tommorrow when i get it.

 

[SeDaTeD]

y = x/(sqrtx +2)
= x/( sqrtx + 2) . (2 - sqrtx)/(2 - sqrtx)
= (2x - xsqrtx)/(4 - x)
= (xsqrtx - 2x)/(x - 4)
= (xsqrtx - 4sqrtx + 4sqrtx - 2x + 8 - 8)/(x-4)
= sqrtx - 2 + 4(sqrtx - 2)/(x-4)

when x-> inf, 4(sqrtx - 2)/(x-4) -> 0
therefore y-> sqrtx - 2
an asymptote would be y = sqrtx - 2

the graph also passes through (0,0) and (4,1)