Question 282616
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The standard form of a parabola is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx\ +\ c]


Your equation is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 3x^2\ +\ 0x\ +\ 0]


A parabola opens upward when the coefficient on *[tex \Large x^2] is positive and downward when it is negative.  If it is zero, you don't have a parabola at all.


The vertex of such a parabola is at the point


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-b}{2a},\,f\left(\frac{-b}{2a}\right)\right)]


The *[tex \Large x]-coordinate of your vertex is *[tex \Large \frac{0}{6}\ =\ 0], so the *[tex \Large y]-coordinate is 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 3(0)^2\ +\ 0x\ +\ 0\ =\ 0] 


And your vertex is at (0, 0)


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