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

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The discussion focuses on determining the domain and asymptotes of the equation x²y + xy² = 2. The domain is identified as (-∞, -2] ∪ (0, +∞), while the asymptotes are established as x = 0, y = 0, and the oblique asymptote y = -x. The behavior of y as x approaches 0 and infinity is analyzed, revealing that y tends to -∞ as x approaches infinity and does not exist as x approaches 0. The equation can be rewritten to isolate y, leading to solutions that further clarify the asymptotic behavior.

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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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