Exploring the Domain and Asymptotes of x^2y+xy^2=2

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In summary, the given equation has a domain of (inf -2]U(0, inf +) and has asymptotes at x = 0 and y = 0. It also has an oblique asymptote at x-y = 0. When solving for y, it is important to consider the behavior of y as x tends to 0 or infinity, where it tends to -infinity. The function has three branches on the x-y plane, with asymptotes at x = 0 and y = 0 in the first quadrant, y = -x in the second quadrant, and x = 0 and y = -x in the fourth quadrant. It is also important to note that for -2 < x <
  • #1
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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  • #2
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

$\chi$ $\sigma$
 
  • #3
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?
 
  • #4
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

$\chi$ $\sigma$
 
  • #5
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
 
  • #6
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

$\chi$ $\sigma$
 
  • #7
There is one more thing that must be investigated:

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

$\chi$ $\sigma$
 
  • #9
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).
 

FAQ: Exploring the Domain and Asymptotes of x^2y+xy^2=2

What is the domain of the given function?

The domain of the function x^2y+xy^2=2 is all real numbers except for the values of x and y that make the function undefined, which are when the denominator of the fraction becomes zero. In this case, the denominator would be zero if either x or y is equal to zero. Therefore, the domain of this function is all real numbers except for x=0 and y=0.

How do I find the vertical and horizontal asymptotes of this function?

To find the vertical asymptotes of the given function, we need to determine the values of x and y that make the denominator of the fraction become zero. These values are the vertical asymptotes. In this case, the vertical asymptotes are x=0 and y=0. To find the horizontal asymptotes, we need to analyze the behavior of the function as x and y approach infinity. If the function approaches a constant value as x and y get larger and larger, then that constant is the horizontal asymptote. If the function approaches infinity or negative infinity, then there is no horizontal asymptote.

Can the function have any oblique asymptotes?

No, the given function x^2y+xy^2=2 does not have any oblique asymptotes. Oblique asymptotes occur when the degree of the numerator is one higher than the degree of the denominator. In this case, both the numerator and denominator have a degree of two, so there are no oblique asymptotes.

How do I graph this function?

To graph this function, you can first plot the points that satisfy the equation, keeping in mind the restrictions of the domain. Then, you can draw a smooth curve through these points to get an idea of the shape of the function. Additionally, you can use technology such as a graphing calculator or computer software to plot the function accurately.

Can this function have any roots or zeros?

Yes, this function can have roots or zeros. Roots or zeros occur when the function equals zero. In this case, the function can have multiple roots or zeros, which can be found by solving the equation x^2y+xy^2=2 for x or y. These roots or zeros can also be seen on the graph of the function as the points where the function intersects the x and y axes.

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