# Slant asymptote of a function in x and y

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

• Is there a reason you used $x=my+c$ rather than $y=mx+c$? It looks possible that there might have been a confusion between $x$ and $y$ , especially since $x=y+1$ is the same as $y=x-1$. – Henry Jun 17 '15 at 22:49
• Thanks Henry. It doesn't work with $$y=mx+x$$, because a $$m^3*y^3$$ comes up, which would mean $$m=0$$. I would have thought it should produce the same result; I must admit I have not quite fully digested the theory behind it (still looking for a good textbook explaining that). – benox Jun 18 '15 at 7:15
• Anyway your second hint was it: I was so convinced the mistake was in the more extended lines I didn't see I used y instead of x at the end of the calculation. – benox Jun 18 '15 at 7:27

$y=a\cdot x+b$
$x^3\cdot a\cdot (a^2-1)+x^2\cdot (3a^2\cdot b+2\cdot a^2-b)+x\cdot ...+\ const$
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 $b=0,-1,-1$