Question 809284
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The general quadratic function is 


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


so if the graph of the desired quadratic function passes through (0,1), then the value of the function when the independent variable has a value of zero is 1, in other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(0)\ =\ a(0)^2\ +\ b(0)\ +\ c\ =\ 1]


which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0a\ +\ 0b\ +\ c\ =\ 1]


Using the coordinate values from the other two given points write two other three-variable linear equations in a, b, and c so that you create a 3X3 system.  Solve the system and you will have the three coefficients for your function.


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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