Vertical Tangent to a Curve

Question

Consider the curve $x^2 y = x^2 + x y^2 - x$. Determine the smallest value for $x$ such that the curve has vertical tangents. After implicitly differentiating and setting the denominator equal to zero, one finds that there exists two possible vertical curves occurring at $x = 0$ and $x = 2$. Now my question is do we include $x = 0$ as a possible value of $x$ for which there is a vertical tangent or not? Since when you divide both sides of equation by $x^2$, the curve is undefined at $x=0$.

Answer 1

You can rearrange $x^2y=x^2+xy^2-x$ as $x(x-y-1)(y-1)=0$ and hence the set of points on the "curve" $x^2y=x^2+xy^2-x$ is simply the union of the lines $x=0$, $y=x+1$ and $y=1$. Hence there will be no vertical tangents