Answer
Since for $y=0$, $\dfrac{dx}{dy}$ comes out to be zero.
if $\dfrac{dx}{dy}=0$, then $y=0$.
Since the y-coordinates of the points which intersect the x-axis are zero.
Thus $\dfrac{dx}{dy}$ will also be zero at these points.
Hence, the tangent line is vertical at the points where the curve intersects the x-axis.
Yes, the plot also confirms this conclusion.
Work Step by Step
Firstly differentiate $y^2 = x^3 − 4x$ with respect to $y$ using the chain rule.
We get, $2y=3x^2\dfrac{dx}{dy}-4\dfrac{dx}{dy}$
Now take $\dfrac{dx}{dy}$ common.
We get, $(3x^2-4)\dfrac{dx}{dy}=2y$
$\implies \dfrac{dx}{dy}=\dfrac{2y}{3x^2-4}$
If $\dfrac{dx}{dy}=0$, then $\dfrac{2y}{3x^2-4}=0$.
Which gives $2y=0$ or $y=0$.
Thus, if $\dfrac{dx}{dy}=0$, then $y=0$.
Hence proved.
Since the y-coordinates of the points which intersect the x-axis are zero.
Thus $\dfrac{dx}{dy}$ will also be zero at these points.
Since $\dfrac{dx}{dy}$ is zero only when the tangent line is vertical.
Hence, the tangent line is vertical at the points where the curve intersects the x-axis.
Yes, the plot also confirms this conclusion.
