Question 197247
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Expand the binomial, remove parentheses, and put the equation into the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2 + bx + c = 0]


The parabola will open upward if *[tex \LARGE a > 0] and downward if *[tex \LARGE a<0].


The coordinates of the vertex, *[tex \Large (x_v,y_v)], of a parabola described by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2 + bx + c = 0]


are given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v = \frac{-b}{2a}]


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


So calculate *[tex \LARGE x_v] then substitute that value for *[tex \LARGE x] in the function and calculate the value of the function for that value of the independent variable.


The equation of the axis of symmetry is *[tex \LARGE x = x_v], so just substitute the calculated value for *[tex \Large x_v]


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