Analyzing and Sketching Function: xy^2-x^2y+x+y=2 | Help with Homework Problem

[Taylor_1989]

Homework Statement

Hi guys I have been stuck on this for the better part of a day today and was wondering if anyone could help. I really can't seem to know even where to start.

I have to Analyse and sketch the function: $$xy^2-x^2y+x+y=2$$

I have not got a lot in the attempt solution because I am very lost. Can someone please give me an idea on how to tackle these types of problems.

Homework Equations

The Attempt at a Solution

1. $$x=0, y=2, y=0, x=2$$#

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[andrewkirk]

If you think of $x$ as a constant what you have is a quadratic equation in $y$, which you can solve, and then plot the $(x,y)$ pairs that are the solutions of that equation. By doing that for a number of different values of $x$ you can identify enough points to sketch to form an idea of the shape of the function.

 

[Taylor_1989]

This is the equation, that I got this morning, $$y=\frac{\left(x^2-1\right)\pm\sqrt{\left(1-x^2\right)^2-4x\left(x-2\right)}}{2x}$$. But this is where I was getting stuck because I could not find values for the quartic >0. This is why I thought this method I was doing was wrong. The values I got for x^4-6x+8x+1=0 was x=-1 and x=-0.544.

 

[Mark44]

This is the equation, that I got this morning, $$y=\frac{\left(x^2-1\right)\pm\sqrt{\left(1-x^2\right)^2-4x\left(x-2\right)}}{2x}$$. But this is where I was getting stuck because I could not find values for the quartic >0. This is why I thought this method I was doing was wrong. The values I got for x^4-6x+8x+1=0 was x=-1 and x=-0.544.

Your solution for y looks fine, as far as you took it, but when you simplify the radical you should get $x^4 - 6x^2 + 8x + 1$, which is different from what you show. Also, -1 is not a root of the equation ##$x^4 - 6x^2 + 8x + 1 = 0$.

Use andrewkirk's suggestion of putting in values of x to get y values. Do this to get a number of points to get an idea of what the graph looks like.

 

[haruspex]

On a whim, I tried the rotation x+y=u, x-y=v. This gave uv = -2±√(4+8v+v⁴). Marginally simpler? It makes an asymptote easy to find.

 

[Taylor_1989]

Thanks @Mark44 and @haruspex for the help. @haruspex at this moment in time I have not come across rotation only in matrices so ur method has sparked interest. Could you either expand on this on maybe point me to a website to read up of this method myself?

 

[haruspex]

Thanks @Mark44 and @haruspex for the help. @haruspex at this moment in time I have not come across rotation only in matrices so ur method has sparked interest. Could you either expand on this on maybe point me to a website to read up of this method myself?

It's just a change of coordinates. The u and v axes are at 45 degrees to the x and y axes. Actually it is not a pure rotation, there is also an expansion by a factor of √2. To make it a pure rotation you would have to use u=(x+y)/√2 etc. but for the purposes of sketching the graph you can fix that up later.
Here is another trick to try...
Figure out some forbidden regions. For what range of x does y have no solutions, and vice versa? Likewise u and v?