Question 806230
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Create the function *[tex \LARGE x\ +\ y] which is to say *[tex \LARGE \phi(x)\ =\ x\ +\ 1\ -\ x^2]


Now use whatever means you have at your disposal to find the *[tex \Large x] coordinate of the vertex of *[tex \LARGE -x^2\ +\ x\ +\ 1].  That gives you the *[tex \Large x] coordinate of the desired point.  Then evaluate *[tex \LARGE y\ =\ 1\ -\ x^2] at this *[tex \Large x] value to get the desired *[tex \Large y]-coordinate.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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