Question 762675
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The *[tex \LARGE x]-coordinate of the vertex of *[tex \LARGE \psi(x)\ =\ ax^2\ +\ bx\ +\ c] is given by *[tex \LARGE x_v\ =\ \frac{-b}{2a}]


The *[tex \LARGE y]-coordinate of the vertex of *[tex \LARGE \psi(x)\ =\ ax^2\ +\ bx\ +\ c], namely *[tex \LARGE y_v], is given by the value of the function at *[tex \LARGE x_v], namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v\ =\ \psi(x_v)\ =\ a(x_v)^2\ +\ b(x_v)\ +\ c].


Now that you have the coordinates of the vertex, plot *[tex \LARGE \left(x_v,\,y_v\right)].


As it turns out, *[tex \LARGE x_v] is an integer in this problem, so pick another value for *[tex \LARGE x] that is 1 larger than *[tex \LARGE x_v].  Evaluate the function at this value:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \psi(x_v\,+\,1)\ =\ a(x_v\,+\,1)^2\ +\ b(x_v\,+\,1)\ +\ c].


Then plot *[tex \LARGE \left(x_v\,+\,1,\,\psi(x_v\,+\,1)\right)].  Use the property of symmetry to determine that if *[tex \LARGE \left(x_v\,+\,1,\,\psi(x_v\,+\,1)\right)] is a point on your graph, *[tex \LARGE \left(x_v\,-\,1,\,\psi(x_v\,+\,1)\right)] must also be on the graph. 


Calculate  *[tex \LARGE x_v\ +\ 2] and repeat the above process for your other two points, giving you 5 points in all including the vertex.


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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