Answer
The points at which the tangent line is vertical are $(0,0)$ and $(1,0)$.
Work Step by Step
We apply implicit differentiation and solve for where $\frac{dy}{dx}$ is undefined to create a vertical tangent line:
$y^4 + y^2 = x(x-1)$
$4y^3 * \frac{dy}{dx} + 2y * \frac{dy}{dx} = 2x-1$
$\frac{dy}{dx} = \frac{2x-1}{4y^3+2y} = \frac{2x-1}{2y(2y^2+1)}$
$\frac{dy}{dx}$ is undefined when the denominator is $0$:
$2y(2y^2+1)=0$
$2y=0 \Rightarrow y=0$
$2y^2+1=0 \Rightarrow y^2 = -\frac{1}{2} \Rightarrow \text{No Solutions}$
For $y=0$, we must now solve for the x-coordinates:
$y^4 + y^2 = x(x-1)$
$0 = x(x-1) \Rightarrow x=0,1$
Thus, the points at which the tangent line is vertical are $(0,0)$ and $(1,0)$.
