# How to determine end behavior with my TI-83

So, I need to determine the end behavior of $P$ for this equation:

$$P(x) = −x^5 + 3x^2 + x.$$

If possible, how is this done?

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What do you mean by "end behavior" of a polynomial? – Per Manne Sep 30 '12 at 22:43
@jak: Taken care of... – Brandt Sep 30 '12 at 23:12
@myself: Nevermind...I see now that since P has an even degree and negative leading coefficient, its end behavior will look like this... y → - ∞ as x → ∞ and y → ∞ as x → - ∞ Reading is fundamental I suppose. – Brandt Sep 30 '12 at 23:13
@Brandt The end behavior you wrote down is correct for $P$, but it has odd degree, not even. – Austin Mohr Oct 2 '12 at 14:30

Your title asks how to determine the end behavior (also called limiting behavior) with a calculator, so I will speak to that.

To determine behavior as $x$ tends toward positive infinity, plug in a giant positive value for $x$. If the polynomial results in a positive (respectively, negative) value for that $x$, then the polynomial tends toward positive (respectively, negative) infinity.

To determine behavior as $x$ tends toward negative infinity, plug in a giant negative value for $x$. If the polynomial results in a positive (respectively, negative) value for that $x$, then the polynomial tends toward positive (respectively, negative) infinity.

Caveat

An important fact about polynomials: The maximum number of roots a polynomial can have is equal to the degree. For your example, this means at most five roots (possibly less).

To be absolutely sure you've determined your end behavior properly, you need to be sure the giant positive and negative $x$'s you picked is beyond (i.e. farther away from the origin than) all of the roots. In practice, however, plugging in something like $x = 10^{100}$ and $x = -10^{100}$ (or whatever your calculator can handle without exploding) will be well beyond the roots.

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The polynomial takes on all values of $x$. So you could just find all the roots and graph.

Standard methods won't get you anywhere because there is only one integer solution.

On WolframAlpha, I get a plot from $x \in [-1.5,2]$

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Did you mean these? $$\lim_{-\infty} P = +\infty\ \text{ and }\ \lim_{+\infty}P=-\infty$$ This is basically because $x^5$ is the leading summand and it is taken with negative coefficient ($-1$). For $x$ of large absolute value, since $\displaystyle\lim_{x\to\pm\infty}\frac{x^k}{x^5} = 0$ for $k<5$, so $$\lim_{\pm\infty} P = \lim_{x\to\pm\infty} -x^5(1-\frac{3x^2+x}{x^5}) = \lim_{x\to\pm\infty} -x^5$$

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