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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.

enter image description here

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?

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    Are you sure you've typed in the correct function? Please edit – 2017-02-26
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    Sorry - there was a typo. Fixed it @DavidQuinn. – 2017-02-26
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    https://www.wolframalpha.com/input/?i=draw+x+%3D+t%5E3-3t%5E2%2B2,++y+%3D+3t%5E3%2Bt%5E2-4t ... draw it a bit bigger – 2017-02-26
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    Are you perhaps looking at points on the graph where $x=0$ and $x=2$ instead of where $t=0,2$? – 2017-02-26

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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 $(-2,20)$

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    Where is the vertical tangent at $0$? – 2017-02-26
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    At $t=0$ we have the point $(2,0)$ – 2017-02-26
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    Ah got it, thanks! – 2017-02-26