Question 1032348
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There are actually three ways to go about this.


First, you can simply use the fact that changing the variable from *[tex \Large x] to *[tex \Large x\ -\ a] translates the graph *[tex \Large a] units (to the right if *[tex \Large a\ >\ 0], left otherwise).  I.e. your new vertex is (4, -1)


Second, you can substitute *[tex \Large x\ -\ 2] for *[tex \Large x] in the definition of *[tex \Large f(x)] and then simplify.  Then calculate the *[tex \Large x]-coordinate of the new vertex using *[tex \Large x_v\ =\ -\frac{b}{2a}] and finally calculate *[tex \Large f(x_v)] to get the coordinates of the new vertex.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ -\ 4x\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ f(x\ -\ 2)\ =\ (x\ -\ 2)^2\ -\ 4(x\ -\ 2)\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ x^2\ -\ 8x\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v\ =\ -\frac{-8}{2}\ =\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(4)\ =\ (4)^2\ -\ 8(4)\ +\ 3\ =\ -1]


New vertex:  (4,-1)


Third, you can complete the square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ -\ 4x\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ (x^2\ -\ 4x\ +\ 4)\ +\ 3\ -\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ (x\ -\ 2)^2\ -\ 1]


So, by inspection, the original vertex is, indeed, (2,-1)


Now, define *[tex \Large g(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ f(x\ -\ 2)\ =\ ((x\ -\ 2)\ -2)^2\ -\ 1\ =\ (x\ -\ 4)^2\ -\ 1)]


And, again by inspection, the new vertex is (4, -1)


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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