So the question asks me to find the x values of the vertical tangent of the equation $xy^2 - x^3y = 6$
After differentiating you get this equation: $y’ = \frac{3x^2y-y^2}{2xy-x^3}$
And then you set denominator to 0 to find the y, which is $\frac{x^2}{2}$
But then you plug that y equation back to the original equation to find the x-value. My question: why do we do that? What is the intuition behind this step?
I thank you in advance!
Vertical tangents occur when $y=\frac{x^2}{2}$. If you then plug that value in for $y$ back into the original equation, you will find all possible values of $x$ for which vertical tangents occur. Right?
So, when $y=\frac{x^2}{2}$, $xy^2-x^3y=6$ has a vertical tangent. Therefore, $x\big(\frac{x^2}{2}\big)^2-x^3\frac{x^2}{2}=6$ represents a vertical tangent for all $x$. Solve for $x$, you find the locations of your vertical tangents.