If the parabola is rotated, then the general equation has at least an $x^{2}$ term, an $xy$ term, a $y^{2}$ term and either an $x$ term, a $y$ term, or both. Rotate it to be vertical, thus removing the $xy$ and $y^{2}$ terms.
From the general equation of a rotated parabola we can work out the focus and directrix. (See Rotation) The vertex is half way between these on a line through the focus that is perpendicular to the directrix.
From a rotated parabola in parametric form, take the vector derivative. Find its magnitude. Set the derivative of the magnitude to zero and solve for $t$. Use that $t$ in the original equation to get the vertex. Note: Just because a parbola has a form such as $$\left(\begin{array}{c} x\\ y \end{array}\right)=P+\left(\begin{array}{c} at+c\\ bt^{2}+d \end{array}\right)$$ does not mean that $P$ is the vertex! To have a separate vertex point, the form must be $$\left(\begin{array}{c} x\\ y \end{array}\right)=V+\left(\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right)\left(\begin{array}{c} 2pt\\ pt^{2} \end{array}\right) \tag{1} \label{1}$$ where $p$ is a constant. In $\eqref{1}$ if $\theta=0$, we get $$\left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{c} V_{x}\\ V_{y} \end{array}\right)+\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)\left(\begin{array}{c} 2pt\\ pt^{2} \end{array}\right)$$ which is a vertical open-up parabola. If $\theta$ increases to $\pi/4$, we get a counter-clockwise $45^{\circ}$ rotation into the second quadrant. Clearly, the angle, $\theta$, comes from the angle made by a line parallel to the directrix as it crosses the x-axis.