What are the Domain and Asymptotes of the Equation x²y + xy² = 2?

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Discussion Overview

The discussion centers on determining the domain and asymptotes of the equation x²y + xy² = 2. Participants explore various mathematical approaches to isolate y and analyze the behavior of the function as x approaches certain values, including zero and infinity.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the domain is (−∞, −2] ∪ (0, ∞).
  • Others mention asymptotes at x = 0, y = 0, and an oblique asymptote at y = -x.
  • A participant isolates y by rewriting the equation as y² + xy - 2/x = 0 and discusses the solutions derived from this form.
  • There are calculations regarding the limits of y as x approaches 0 and infinity, with some participants noting that the limit does not exist as x tends to 0 but approaches -∞ as x tends to infinity.
  • Some participants express uncertainty about the behavior of y for large negative x, with discussions suggesting it approaches -x and 0.
  • One participant raises the question of when x² + 8/x < 0, indicating that for -2 < x < 0, there are no real values of y that satisfy the equation.
  • It is noted that the function has three branches in the x-y plane, with specific asymptotic behavior in different quadrants.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the domain and asymptotic behavior of the function, and the discussion remains unresolved with differing interpretations of the limits and branches of the function.

Contextual Notes

Limitations include unresolved mathematical steps regarding the behavior of y as x approaches certain values and the implications of the derived conditions on the domain.

leprofece
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¿domain and asinptotestotes of x^2y+xy^2 = 2?

Answer (inf -2]U(0, inf +)
asinptotes AV x = 0 AH y = 0
Oblique x-y = 0

I can isolate y or i mean solving for y
 
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leprofece said:
¿domain and asinptotestotes of x^2y+xy^2 = 2?

Answer (inf -2]U(0, inf +)
asinptotes AV x = 0 AH y = 0
Oblique x-y = 0

I can isolate y or i mean solving for y

Yes, You can!... writing the equation as...

$\displaystyle y^{2} + x\ y - \frac{2}{x} =0\ (1)$

... You arrive to the soltutions...

$\displaystyle y = \frac{- x \pm \sqrt{x^{2} + \frac{8}{x}}}{2}\ (2)$

... and now You have to find the behavior of y if x tends to 0 or to infinity...

Kind regards

$\chi$ $\sigma$
 
calculating limit as x tends to 0 it doesnot exist but in \infty tends to -\infty
equating to 0 i got x = 2 /cubic root of 3
is that I have to do?
 
For 'large positive x' is...

$\displaystyle \frac{- x + \sqrt{x^{2} + \frac{8}{x}}}{2} \sim 0\ (1)$

... and...

$\displaystyle \frac{- x - \sqrt{x^{2} + \frac{8}{x}}}{2} \sim - x\ (2)$

What does it happen for 'large negative x'?...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
For 'large positive x' is...

$\displaystyle \frac{- x + \sqrt{x^{2} + \frac{8}{x}}}{2} \sim 0\ (1)$

... and...

$\displaystyle \frac{- x - \sqrt{x^{2} + \frac{8}{x}}}{2} \sim - x\ (2)$

What does it happen for 'large negative x'?...

Kind regards

$\chi$ $\sigma$

I got 0 and - infnity
 
chisigma said:
For 'large positive x' is...

$\displaystyle \frac{- x + \sqrt{x^{2} + \frac{8}{x}}}{2} \sim 0\ (1)$

... and...

$\displaystyle \frac{- x - \sqrt{x^{2} + \frac{8}{x}}}{2} \sim - x\ (2)$

What does it happen for 'large negative x'?...

Kind regards

$\chi$ $\sigma$

For 'large negative x' is...

$\displaystyle \frac{- x + \sqrt{x^{2} + \frac{8}{x}}}{2} \sim - x\ (1)$

... and...

$\displaystyle \frac{- x - \sqrt{x^{2} + \frac{8}{x}}}{2} \sim - 0\ (2)$

The asimptotes are y= 0, y = -x and for symmetry x=0...

Kind regards

$\chi$ $\sigma$
 
There is one more thing that must be investigated:

when is $x^2 + \dfrac{8}{x} < 0$?
 
Deveno said:
There is one more thing that must be investigated:

when is $x^2 + \dfrac{8}{x} < 0$?

It is clear that for -2 <x <0, there is no real value of y which satisfies the equation of the original post. The function defined in an implicit manner has three branches on the x, y plane: in the first quadrant with asymptotes x = 0 and y = 0, in the second quadrant with asymptotes y = 0 and y = - x, in the fourth quadrant with asymptotes x = 0 and y =-x ...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
It is clear that for -2 <x <0, there is no real value of y which satisfies the equation of the original post. The function defined in an implicit manner has three branches on the x, y plane: in the first quadrant with asymptotes x = 0 and y = 0, in the second quadrant with asymptotes y = 0 and y = - x, in the fourth quadrant with asymptotes x = 0 and y =-x ...

Kind regards

$\chi$ $\sigma$

Of course this would be clear to you, I just wanted to make sure the poster was aware of that fact (goes to domain of definition).
 

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