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You are about to travel to a

mystical land…

That many have heard of,

That many have heard of,

but few have ever seen.

Welcome to

OZThe

Land of

End Behaviors

of a function

Define a polynomi

al function

Factored form of polynomial

from x-intercepts

Identify zeros of a function

Determine number of real and non-real

solutions

Long division of

polynomialsSynthetic division to find zeros

Click a bubble to explore….

Polynomial Functions in…

Welcome to Munckinland Dorothy. Would you like to

learn about the end behaviors of functions?

The first thing that you need to know is that the

degree of a function determines which direction the end behaviors go. The Lollipop Guild will

explain.

So Dorothy. There are two different ways end behaviors can react in a polynomial

function. It can either look like a rainbow, like the beautiful ones we have in Oz, or like a funny sideways “s” thing. Ya’ understand?

I think so. You wouldn’t by chance have any pictures

would you?

Eh. Pictures. The gal wants pictures. Don’t worry, we got

some right here.

Lookie here. If we had an equation that had an even number as a degree, it’d look like this thing here. As you can tell, the end

behaviors are:X → - ∞ Y→∞

andX → ∞ Y→ ∞

But if the equation were reflected over the x- axis and the arch went the other way, both Y end behaviors would be -∞

And this is the funny “s” thing that equations with an odd degree look like when they’re graphed. Kind of weird, huh?

This equation’s end behaviors are: X→-∞ Y→-∞X→∞ Y→∞

If it were reflected over the x-axis, they would be:X→-∞ Y→∞X→∞ Y→-∞

Do you understand darling?

Yes. I think I do. Thank you Glinda.

End Behaviors

of a function

Define a polynomi

al function

Factored form of polynomial

from x-intercepts

Identify zeros of a function

Determine number of real and non-real

solutions

Long division of

polynomialsSynthetic division to find zeros

Click a bubble to explore….

Polynomial Functions in…

Scarecrow? Are you there?

Over here Dorothy.

Oh there you are! I have a question, do you know

anything about using long division with polynomial

functions?

Do I know anything? Haven’t you heard? I have a brain now!

Of course I know how to use long division with polynomial

functions. Watch this:

Pay attention to the leading x of the divisor and x² of the dividend. If you were to divide x² by x, what

would you have? X. Put that on top.

There. Now multiply that x that you just put on top by x+1, which turns into x² + 1x. Put that under

the x² and 9x of the dividend.

Perfect! You can now subtract x² +1x from x²-9x.

Fantastic. Now you can carry the -10 down from the dividend.

Now the process starts all over again. Take the -10x and divide it by the x of the divisor. It turns

into -10, so put that on top.

So now what you want to do is take the -10 that you just put on top and multiply it by the x+1. It turns into -10x-10. Put that under the other -10x-

10. Now subtract just like you did before.

And presto! Your work is done.

Oh Scarecrow, thank you! That makes it so

much clearer.

No problem Dorothy, I’d do anything to help. You are the one who helped

me get this brain. But now you better be off. You have many other things

to learn…

End Behaviors

of a function

Define a polynomi

al function

Factored form of polynomial

from x-intercepts

Identify zeros of a function

Determine number of real and non-real

solutions

Long division of

polynomialsSynthetic division to find zeros

Click a bubble to explore….

Polynomial Functions in…

Tin man? Tin man? I’ve heard you can teach me

synthetic division.

Synthetic division? I can help you with that. It’s quite easy.

Let’s look at the equation x²+5x+6 and try to divide it by x-1. First take all of the coefficients

and draw them inside a boxed section.

x² + 5x+6

. Just like that.

Now take the 1 form x-1 and plant it outside of the box.

Take the first coefficient and slide it down the division symbol.

Then multiply the one you just carried down by the planted one, and put the product under the next

coefficient, which would be 5.

Add 5 and 1 to get 6, and place that under the division bar.

Excellent! Now the process repeats itself. Multiply the carry down by one and place it under the next

coefficient.

Finally, add 6 and 6. The answer, if other than 0, is the remainder. 12 is the remainder here.

I had no idea it was that simple. That was incredibly helpful, thank you.

Why, you’re welcome! Have fun on your other adventures!