Slant asymptote of a function in x and y

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After looking for the asymptotes of the function:

$y^3+2y^2-x^2*y+y-x+4=0$

I found the answers y=0, y=-x-1 and y=x+1. This is almost exact: the last one should actually be y=x-1.

To find the result, I substituted x=my+c in the equation, which yielded $m=\pm 1$ and $c=\mp 1$, when c should be only -1. I can't find where I made a mistake. Any idea?

Benox

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Let's look for asymptote :

$y=a\cdot x+b$

substitute into original equation:

$x^3\cdot a\cdot (a^2-1)+x^2\cdot (3a^2\cdot b+2\cdot a^2-b)+x\cdot ...+\ const$

so from $a\cdot (a^2-1)=0$, $a=0,1,-1$ use $3a^2\cdot b+2\cdot a^2-b$ to find corresponding b values $b=0,-1,-1$