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

Feb 13, 2017

See explanation.

#### Explanation:

[x^2] forms the sequence {n^2), with the limit $\infty$, as $x \texttt{o} \pm \infty$.

$y = {x}^{2} \left[{x}^{2}\right] \left(1 + \frac{1}{x}\right) \left(1 - \frac{2}{x}\right) \left(1 - \frac{3}{\left[{x}^{2}\right]}\right) \to \infty$, as $x \to \pm \infty$.

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

ends.

For example,

$y = \left(x + 1\right) \left(x - 2\right) \left(- 1\right) , x \in \left[\sqrt{2} , \sqrt{3}\right)$, giving

${\left(x - \frac{1}{2}\right)}^{2} = - \left(y - \frac{9}{4}\right)$

The size of a typical parabola, $a = \frac{1}{4 \left(\left[{n}^{2}\right] - 3\right)} \to 0$, as x to oo#.

The center $\to \left(\frac{1}{2} , - \infty\right)$, on the common axis $x = \frac{1}{2}$.

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.