Find $x $- coordinates of the vertical tangent lines of a function.

Question

The function $6+(x^3)y = x(y^2)$ has vertical tangent lines. How do you find the functions of these lines?

Answer 1

Taking the derivative of the function,we obtain

$$ \frac{dy}{dx} = \frac{3x^2-y^2}{2xy-x^3} $$ which the is slope of the tangent.

We know that when the tangent is vertical, its slope is $\infty$. Hence, inorder to obtain points where the slope is $\infty$, just equate the denominator to zero.
$$2xy-x^3=0$$ Now, you will get an expression for $y$ in terms of $x$. Substitute them in the first equation and obtain the co-ordinates of the required points. The equation of the tangents can easily be found.

Answer 2

Standard answer from calculus. If the vertical line is at $x_{0}$, the equation is: $x = x_{0}$.

A more straightforward example is $$y^{2} = x.$$ The tangent line is $x=0$.