Question 326886
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The vertex for any parabola expressed in standard form:


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


is at *[tex \Large \left(\frac{-b}{2a},p\left(\frac{-b}{2a}\right)\right)]


In your case *[tex \Large a\ =\ 3], *[tex \Large b\ =\ -2], and *[tex \Large c\ =\ 0], so *[tex \Large \frac{-b}{2a}\ =\ \frac{-(-2)}{2(3)}\ =\ \frac{1}{3}]


so


is at *[tex \Large \left(\frac{1}{3},p\left(\frac{1}{3}\right)\right)\ =\ \left(\frac{1}{3},\ 3\left(\frac{1}{3}\right)^2\ -\ 2\left(\frac{1}{3}\right)\right)]


You can do your own arithmetic to calculate the value of your function.



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