In another exercise is given: Find the parabola which is a displacement of $y = 2x^2 - 3x + 4$ which passes though the point $(2, -1)$ and has $x = 1$ as its symmetry axis.
I've reduced the based equation to the form: $y = 2(x - \frac{3}{4})^2 + \frac{23}{8}$, so the vertex is $(\frac{3}{4}, \frac{23}{8})$.
As the displaced equation has its symmetry axes on $x = 1$ it's known that its vertex is $(1, x)$ what can I do more to find the $y$ of the displaced equation ?
Thanks in advance.
EDIT: Corrected the typo