Question 1038656
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In order for *[tex \LARGE a(x)] to be zero, either *[tex \LARGE x] has to be zero or *[tex \LARGE x\ +\ 2] has to be zero.  Since the *[tex \LARGE x\ +\ 2] factor is repeated twice (in other words, squared), the multiplicity of the zero *[tex \LARGE x\ =\ -2] is 2.


Regardless of the value of *[tex \LARGE x], *[tex \LARGE (x\ +\ 2)^2\ >\ 0].  Hence the sign of *[tex \LARGE a(x)] is dependent on the sign of *[tex \LARGE x].


So, as *[tex \LARGE x] gets large in the negative direction, *[tex \LARGE a(x)] also gets large in the negative direction.  As *[tex \LARGE x] gets large in the positive direction, *[tex \LARGE a(x)] gets large in the positive direction.


At zero, *[tex \LARGE a(x)] transitions from negative to positive going from left to right.


At -2, since the *[tex \LARGE x\ +\ 2] factor is squared, you approach zero from below, become tangent to the *[tex \LARGE x]-axis at *[tex \LARGE x\ =\ -2] and then decreases again.  Somewhere between -2 and 0, the function reverses direction again and starts to increase.


*[illustration 3rdDegreePolyFunctionGraphAnalysis_(2).jpg]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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