Question 237858
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I think you want the equation for the line of symmetry for

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ (x\ -\ 2)(x\ -\ 2)]

Multiply the two binomials using FOIL.

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ -\ 4x\ + 4]

Now you have your equation in Standard Form, namely

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ f(x)\ =\ ax^2\ +\ bx\ +\ c]

The vertex of any such parabola is at the point

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-b}{2a},\,f\left(\frac{-b}{2a}\right)\right)]

Since the line of symmetry is a vertical line through the vertex, the equation of the line of symmetry must be:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b}{2a}]

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The set of parabolas with vertex at *[tex \Large \left(2,\,0\right)] is described by using the vertex form of the equation of a parabola.

The vertex form of a parabola with vertex at the point *[tex \Large \left(h,k\right)] is:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ = a(x\ -\ h)^2\ +\ k]

So your family of parabolas is:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ = a(x\ -\ 2)^2]

where *[tex \Large a\ \in\ \R,\ a\ \neq\ 0]

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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