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.