Question 583247
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First of all, if you are going to say that it is a function, write it like a function, namely:


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


Then re-write to expand the squared binomial:


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


Now, evaluate each of the factors for its sign if *[tex \Large x] is a relatively large positive number, say 100.


First factor is +
Second factor is -
Third factor is +
Fourth factor is +


Plus times Minus times Plus times Plus is Minus.  As *[tex \LARGE x] increases without bound, i.e. *[tex \LARGE x\ \right\ \infty], *[tex \LARGE y] decreases without bound, i.e. *[tex \LARGE y\ \right\ -\infty].


Repeat the process with a relatively large negative number, -100 should do nicely:


First factor is -
Second factor is +
Third factor is -
Fourth factor is -


Minus times Plus times Minus times Minus is Minus.  As *[tex \LARGE x] decreases without bound, i.e. *[tex \LARGE x\ \right\ -\infty], *[tex \LARGE y] decreases without bound, i.e. *[tex \LARGE y\ \right\ -\infty].



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