Identify all the asymptotes of the function $f(x,y)=x^2-2xy-2x-3y+1=0$ By degree of the equation, I know that there has to be 2 asymptotes. I have applied the general procedure to find the asymptotes as follows:

*

*Separate the terms of the equation in according to its degree. So I get $\phi_2=(x^2-2xy)$ and $\phi_1=(-2x-3y)$

*Then I put $x=1$ and $y=m$ in $\phi_2$ to get $m=1/2$

*Next I find $c= \frac{-\phi_1}{\phi_2'}$ to get $c=-7/4$

*Thus I get only one asymptote: $y=\frac{x}{2}-\frac{7}{4}$
But how do I get second asymptote? Wolfram

$x^2-2xy-2x-3y+1=0 \implies y = \dfrac{x^2-2x+1}{2x+3}$
This is a rational function and as the degree of the polynomial in the numerator is one higher than the one in denominator, the function will have an oblique asymptote and no horizontal asymptote.
You already found the oblique asymptote otherwise you can rewrite the function as (use polynomial long division method),
$\displaystyle y = \frac{(2x+3) \left(\frac{x}{2} - \frac{7}{4}\right) + \frac{25}{4}}{2x+3} = \left(\frac{x}{2} - \frac{7}{4}\right) + \frac{25/4}{2x+3}$
So the oblique asymptote is $\displaystyle y = \frac{x}{2} - \frac{7}{4}$
In addition to an oblique or horizontal asymptote, a rational function will have vertical asymptotes at the singularities of the function. Finding vertical asymptotes is simple as for that, you just need to make the denominator zero.
Hence $x = - \dfrac{3}{2}$ is a vertical asymptote.
