How do you describe the end behavior of $y=(x+1)(x-2)([x^2]-3)$ ?

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Answer 1 · 2017-02-13T14:58:44

See explanation.

[x^2] forms the sequence {n^2), with the limit $oo$ , as $xtto +-oo$ .

$y = x^2 [x^2] (1+1/x)(1-2/x)(1-3/([x^2])) to oo$ , as $x to +-oo$ .

Piecewise, the graph is a series of arcs of parabolas, with holes at

ends.

For example,

$y= (x+1)(x-2)(-1), x in [sqrt2, sqrt3)$ , giving

$(x-1/2)^2=-(y-9/4)$

The size of a typical parabola, $a = 1/(4( [n^2]-3)) to 0$ , as x to oo#.

The center $to (1/2, -oo)$ , on the common axis $x = 1/2$ .

The arcs drift away from the common axis x = 1/2.

I have done some research for giving details. I would review, and if

necessary, revise my answer, later.

How do you describe the end behavior of y=(x+1)(x-2)([x^2]-3)y=(x+1)(x-2)([x^2]-3)?