Theorem: Given any parametric parabola, we can take the derivative of it, which will yield a vector, tangent to the parabola at any point $t$. When the magnitude of that vector is minimized, by setting its derivative to zero, then the $t$ that is found can be substituted into the original parabola to give its vertex.
We will not prove the theorem as that involves too much differential geometry. However, it only requires a tiny bit of derivative calculus to apply the theorem beneficially.
Example: Given a parabola parametrized as
$$\mathbf{r}=\left(\begin{array}{c}
x\\
y
\end{array}\right)=\left(\begin{array}{c}
3t^{2}+20t-3\\
-t^{2}-t-1
\end{array}\right),\,(-10\le t\le3)$$
find its vertex. Answer: Let's sketch the parabola (fig 1: panel a) to see what we are working with.
Figure 1: Sketch of the parabola in example 1. Panel (a) is a sketch of the parabola. Panel (b) is scaled in to show the region of the vertex. A point $P(t)$ is located at three different values of $t$, along with the vector $\mathbf{r'}(t)$. We see that the magnitude of $\mathbf{r'}(t)$ decreases as $P(t)$ approaches the vertex. Panel (c) shows the parabola with its vertex located using the $t$ value (-61/20).
According to the theorem, the derivative of the parametric equations will yield a vector. Let's name the parabola $\mathbf{r.}$ Then its derivative will be
$\mathbf{r'}$.
$$\mathbf{r'}=\left(\begin{array}{c}
6t+20\\
-2t-1
\end{array}\right)$$
Now let's make another sketch (fig. 1 b), where we locate a point, $P$ on the graph and then translate vector $\mathbf{r'}$ onto $P$. That way we can verify that it at least looks like a tangent. As the point gets close to the vertex, the length of vector $\mathbf{r'}(t)$ gets shorter. Again, according to the theorem, we want to find the $t$ that makes vector $\mathbf{r'}(t)$ as small as possible. To do that, we minimize the magnitude of $\mathbf{r'}$. The magnitude of $\mathbf{r'}$ is
$$\Vert\mathbf{r'}(t)\Vert=\sqrt{(6t+20)^{2}+(-2t-1)^{2}}.$$
Expanding $\Vert\mathbf{r'}(t)\Vert$, we get
$$\sqrt{40\;t^{2}+244\;t+401}, \tag{1} \label{1}$$
which is the length of the vector at any $t$. To minimize it, we need to take its derivative and set it to zero.
$$\frac{d\Vert\mathbf{r'}(t)\Vert}{dt}=\frac{40\;t+122}{\sqrt{40\;t^{2}+244\;t+401}}=0$$
Now solving for $t$,
$$40t=-122$$
$$t=-\frac{61}{20}.$$
Putting that value of $t$ into $\mathbf{r}$ will get the vertex.
$$\left(\begin{array}{c}
3t^{2}+20t-3\\
-t^{2}-t-1
\end{array}\right)=\left(\begin{array}{c}
3\left(-\frac{61}{20}\right)^{2}+20\left(-\frac{61}{20}\right)-3\\
-\left(-\frac{61}{20}\right)^{2}-\left(-\frac{61}{20}\right)-1
\end{array}\right)=\left(\begin{array}{c}
36.0925\\
-7.2525
\end{array}\right)$$
Rotation Angle of a Parametric Parabola
The angle of the directrix for a parabola is given by
$$\lim_{t\rightarrow\infty}\frac{y(t)}{x(t)}.$$
In a vertical, open-up position, the $\lim_{t\rightarrow\infty}\frac{y(t)}{x(t)}=\infty$, so the rotation angle is $\frac{\pi}{2}$ rotated already. Therefore, we have
$$\lim_{t\rightarrow\infty}\frac{y(t)}{x(t)}=\tan\left(\theta+\frac{\pi}{2}\right)=-\cot\theta.$$
Example: Find the rotation angle of a parabola given by $$\left(\begin{array}{c}
x\\
y
\end{array}\right)=\left(\begin{array}{c}
t^{2}-2t+1\\
2t^{2}+8t-1
\end{array}\right).$$ Figure 2: The original rotated equation is in black. The vertical version is in blue. $\theta$ is calculated by setting the $\lim_{t\rightarrow\infty}\frac{y(t)}{x(t)}=-\cot(\theta)$. The straight line is derived from the direction vector of $\theta,\mathbf{u}=(\cos\theta,\sin\theta)$. To get the vertical parabola, we used a counter-rotation matrix multiplied times the original parametric equation.Answer: This is very straight-forward. Just apply the formula.
$$\lim_{t\rightarrow\infty}\frac{2t^{2}+8t-1}{t^{2}-2t+1}.$$
This limit gives $\infty/\infty$ and thus we can use L'Hopital.
$$\lim_{t\rightarrow\infty}\frac{dy}{dx}=\frac{4}{2}=2.$$
Therefore
$$-\cot(\theta)=2$$
$$\theta=\tan^{-1}\left(-\frac{1}{2}\right)\approx0.46365$$
This value is perfect for rotating the parametric parabola to be vertical.
$$\text{Vertical Parabola}=\left(\begin{array}{cc}
\cos(-\theta) & -\sin(-\theta)\\
\sin(-\theta) & \cos(-\theta)
\end{array}\right)\left(\begin{array}{c}
t^{2}-2t+1\\
2t^{2}+8t-1
\end{array}\right)$$
