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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

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    There is$a$typo that may or may not be a problem, need $(x-3/4)^2$, not $(X^2-3/4)^2$. The current axis of symmetry is $x=3/4$, so you need to move to the right by $1/4$. New thing will be $2(x-1)^2+c$.2012-06-22

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Your equation for the original parabola should read $y = 2\left(x-\dfrac{3}{4}\right)^2+\dfrac{23}{8}$. The displaced parabola will have the equation $y = 2(x-x_v)^2 + y_v$, where $(x_v, y_v)$ is the vertex. Since the axis is $x=1$, $x_v$ is $1$. The point $(2,-1)$ is on the parabola, so we have $-1 = 2(2-1)^2 + y_v \Rightarrow y_v = -1 -2 = -3$. Then, the equation of the displaced parabola is $y = 2(x-1)^2 -3$.

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    Notice that the only thing we needed from the equation of the original parabola was the coefficient $2$ and the fact that its axis is parallel to the $y$ axis. We don't need to know the vertex of the original parabola, only the vertex of the new one.2018-08-19