I'm tutoring a freshman in college, and having to re-remember my college pre algebra classes from 20 years ago.

If I need to determine the end behavior for

$f(x)=x^6 + x^5... $

I'm good with that, but how do I quickly determine the end behavior if I'm given the similar formula but in a different form such as

$f(x) = 9x^5 (x-3)^3 (x+26)^8 (-x + 1)^2$

I have found a billion examples with a standard form polynomial, but hard to find the type above, but maybe I'm just calling it the wrong thing in google.

If he is trying to find this on a test, then multiplying all this out seems like it will take too long and waste valuable minutes, so I was taking a short cut of "dropping" down all the x values with their exponent, but not sure how to determine the positive/negative value.

$f(x) = 9x^5 (x-3)^3 (x+26)^8 (-x + 1)^2$

$x^5 \cdot x^3 \cdot x^8 \cdot x^2$

$5 + 3 + 8 + 2 = 18$

So above I know I would be $x^{18}$ but I can't figure out if it is positive or negative.

Help me obi-wan(s)... you (all) are my only hope..

  • $\begingroup$ What is "end behaviour"? $\endgroup$ – Lord Shark the Unknown Oct 26 '17 at 17:44
  • $\begingroup$ @LordSharktheUnknown, it's what some precalc and college algebra courses use to describe $\displaystyle\lim_{x \to \pm\infty}$ $\endgroup$ – tilper Oct 26 '17 at 17:46

Shortcut: Drop anything that doesn't contribute to the leading term when expanding. Being able to recognize what exactly contributes to that is a skill that takes some practice.

Your example: $f(x) = 9x^5(x-3)^3(x+26)^8(-x + 1)^2$

Imagine expanding each factor and taking the leading term of each factor. Leading terms include the coefficients! For example we'll want to keep the $9$ from the $9x^5$ factor. In this case it won't actually matter but it's a good habit to get into since sometimes it does matter (like if the coefficient were negative) and it's just easier to do it the same way every time.

In $9x^5$ there's nothing to expand, so you get $9x^5$.

From $(x-3)^3$ you get $x^3$.

From $(x+26)^8$ you get $x^8$.

From $(-x+1)^2$ you get $x^2$.

Now multiply these leading terms together: $9x^5 \cdot x^3 \cdot x^8 \cdot x^2 = 9x^{18}$

The coefficient is $9 > 0$.

So it's a positive coefficient with an even exponent.

Other example with minus signs we need to carefully track:

$g(x) = -2x^4(3x + 1)^2(-2x+1)^3$

In $-2x^4$ there's nothing to expand, so we get $-2x^4$.

From $(3x+1)^2$ we get $(3x)^2 = 9x^2$.

From $(-2x+1)^3$ we get $(-2x)^3 = -8x^3$.

Multiply leading terms: $-2x^4 \cdot 9x^2 \cdot (-8x^3) = 144x^9$

  • $\begingroup$ thank you. That is exactly what I was looking for! $\endgroup$ – Cade Thacker Oct 26 '17 at 19:00

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