Question 623758
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Multiply the two binomials using FOIL until you have a quadratic function in standard form, i.e. that looks like:


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


Then calculate the *[tex \LARGE x]-coordinate of the vertex:  *[tex \LARGE x_v\ =\ \frac{-b}{2a}]


Then evaluate the function at that value of the independent variable, that is, calculate *[tex \LARGE f(x_v)].


By examination of the lead coefficient, determine whether the graph of *[tex \LARGE f] opens up (i.e. the vertex is a minimum) or opens down (i.e. the vertex is a maximum).


If the vertex is a minimum, then the value of *[tex \LARGE f(x_v)] is the low end of the range and the range increases without bound.  On the other hand, if the vertex is a maximum, then the value of *[tex \LARGE f(x_v}] is the top end of the range and the range decreases without bound.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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