Determine all values of $t$ for which the curve given parametrically by
$$x = t^3 - 3t^2 + 2, y = 3t^3 + t^2 - 4t$$
has a vertical tangent.
First, I found $\frac{dx}{dt}$, which is $3t^2 - 6t$ or $3t(t - 2)$. There would be a vertical tangent when $\frac{dx}{dt}$ is $0$, which happens when $t = 0, 2$. However, the graph of this parametric curve doesn't seem to support this.
I can see that there is a vertical tangent at $t = 2$, but I don't see there is one when $t = 0$. How do I resolve this discrepancy -- is my original analysis correct or is the graph correct?
If you render the graph on a larger scale you will see the vertical tangent at both points you have found. There is another such point at $(-2,20)$