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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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As you have observed, a parabola with vertex $(h,k)$ and vertical axis of symmetry has equation $y=a(x-h)^2 + k$. The value $a$ determines the parabola's shape and direction. As a displacement of the given parabola, the target parabola must have $a=2$. (If the target were simply the same shape, but not necessarily a displacement, $a=-2$ would also be a possibility.) Can you proceed from there? –  Blue Jun 22 '12 at 21:31
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 $2(x-1)^2+c$. –  André Nicolas Jun 22 '12 at 21:33

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up vote 2 down vote accepted

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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Your equation for the original parabola is linear. –  Gerry Myerson Jun 23 '12 at 0:38
@GerryMyerson Thanks. –  talmid Jun 23 '12 at 1:01

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