Polynomial Graphs: End Behavior

Hello Lgene here , I am going to discuss the "end behavior" of polynomial graphs.

When you're graphing (or looking at a graph of) polynomials, it can help to already have an idea of what basic polynomial shapes look like. One of the aspects of this is "end behavior", and it's pretty easy.

Even Degree

As you can see, even-degree polynomials are either "up" on both ends (entering and then leaving the graphing "box" through the "top") or "down" on both ends (entering and then leaving through the "bottom"), depending on whether the polynomial has, respectively, a positive or negative leading coefficient. 

Odd Degree

Odd-degree polynomials have ends that head off in opposite directions. If they start "down" (entering the graphing "box" through the "bottom") and go "up" (leaving the graphing "box" through the "top"), they're positive polynomials; if they start "up" and go "down", they're negative polynomials.

*Things to remember*

            -Even degree Polynomial's ends, end in same direction.

            -Odd degree Polynomial's ends ,ends in opposite directions.

And...

Graphing High Degree Polynomials(i was supposed to post this last Thursday, but i didn't have a chance,  because Mr.P didn't accept me on the blog.. peace :P)

When factoring polynomial, one approach is to find one of the roots (which we will call r)

Of the polynomial, then factor out (x-r).The resulting polynomial has a smaller degree, and we can repeat the process again and again until we have the final result.

Example: X³-6x²-x+30=0

If r is the root, then r must be factor of 30. Therefore possible values for r are±1; ±2; ±3; ±5; ±6; ±10; ±15; ±30.

We can test whether a possible root is an actual root by plugging the value in and seeing if the expression becomes 0; 2 is not a root since 2³ – 6(2)²  - 2  +  30 = 8 -24 -2 + 30 = 12, but -2 is a root since (-2)³ – 6(-2)²  - (-2) +30 = - 8 - 24 +2 + 30 =0.

Note that if the root is -2, then the factor is (x+2); specially, pay attention to the change of the sign!

Then we calculate x³ – 6x² - x + 30 / x+ 2, and get x² - 8x + 15, with remainder of zero.(take a look on how to do SYNTETHIC DIVISION at Shubham's blog)

Now we consider the polynomial x² – 8x + 15; if r is a root, then r must be a factor of 15. That narrows it down to ±1; ±3; ±5; ±15.

As it turns out,  3 is a root since (3)² – 8(3) + 15 =9 - 24 +15 = 0.

Therefore we now calculate x² - 8x + 15 / x-3, and get x-5.

Since we divided first by (x - 2), then by (x – 3), and finally got (x-5), we conclude that x³ – 6x² –x + 30 = (x+2) (x-3) (x-5).

Lastly,