[Math] Polynomial Functions

I am having a hard time dealing with polynomial functions.

I am asked to state the end behaviours of f(x)=(x-3)(x+2)(x+5)as well as sketch the graph. How do I figure out the end behaviours of this graph?

I also need to determine the equation of the function with zeroes +/- 1, -2, and a y-intercept of -6. The answer section says the function is f(x)=3(x-1)(x+1)(x+2). How do I find this out? I see no connection with what I'm given to what the answer is.

And then there's one question which asks me to determine the remainder when x^3 + 2x^2 - 6x + 1 is divided by x + 2 without dividing. Again, how is this done?

If anyone can help, I would be really, really grateful.

I am asked to state the end behaviours of f(x)=(x-3)(x+2)(x+5)as well as sketch the graph. How do I figure out the end behaviours of this graph?

The polynomial is of degree 3 and has a positive leading coefficient (1). This can be determined be multiplying the terms of highest degree in each of the binomial multiples (x, x, and x) to arrive at 1*x^3.
From this we determine that the function has the same end-behavior as x^3. I'm not sure what terminology you'll want to use here, probably something along the lines of "the left side goes down, the right side goes up".

What would change if you had f(x)=(-x+3)(x+2)(x+5)?

I also need to determine the equation of the function with zeroes +/- 1, -2, and a y-intercept of -6. The answer section says the function is f(x)=3(x-1)(x+1)(x+2). How do I find this out? I see no connection with what I'm given to what the answer is.

With a zero at -1, the function is divisible by (x-(-1) ) or (x+1)
With a zero at +1, the function is divisible by (x-(+1) ) or (x-1)
With a zero at -2, the function is divisible by (x-(-2) ) or (x+2)

Thus the function is given as:

f(x) = (a)*(x+1)*(x+2)*(x-1)

But when x=0, f(x)=-6. Sub in these values:

-6 = (a)(0+1)(0+2)(0-1)
-6 = a*-2
3 = a

So finally:

f(x) = 3*(x+1)*(x+2)*(x-1)

What would change if the y-intercept was +8?

And then there's one question which asks me to determine the remainder when x^3 + 2x^2 - 6x + 1 is divided by x + 2 without dividing. Again, how is this done?

Solve the function for x = -2 (the point at which x + 2 = 0):

f(-2) = (-2)^3 + 2(-2)^2 - 6(-2) + 1
f(-2) = -8 +8 +12 +1
f(-2) = 13

You might try to do the long division by hand, and think about why this method would work as you do.

Originally Posted by Spearhead

I am having a hard time dealing with polynomial functions.

I am asked to state the end behaviours of f(x)=(x-3)(x+2)(x+5)as well as sketch the graph. How do I figure out the end behaviours of this graph?

Well, you can already place the points of the zeroes (3, -2, and -5) and you can find the y-intercept by plugging in 0 for x which gives you -30.

Before you even graph it, you know it will behave like a cubic because there are 3 x's, or if you multiply it all together an x^3. So you should already know the vague shape of the graph. Since there isn't a -5 or something in front of the equation, thus leaving the leading coefficient at a positive 1 (f(x)=1(x-3)(x+2)(x+5)), we know that it will be a positive.

This means, in the case of odd degrees, it starts at the "bottom" of the graph heading "up" and ends with the line heading "up" to the opposite corner. If x were to an even degree, both the left and right "ends" of the graph would be going in the same direction, the "ends" pointed "down" if the leading coefficient of x was negative and "up" if it were positive.

I would guess the rough graph would look something like this:
http://artpad.art.com/?krj6d5qd0co
(Sorry for the terrible illustration)
So the end behavior would be...positive.

Originally Posted by Spearhead

I also need to determine the equation of the function with zeroes +/- 1, -2, and a y-intercept of -6. The answer section says the function is f(x)=3(x-1)(x+1)(x+2). How do I find this out? I see no connection with what I'm given to what the answer is.

Well, the zeroes give you the pieces that are within the parentheses (with the opposite sign ie. -2 becomes x+2).

You have to remember that when x=0, y needs to equal -6. So, plug 0 in for x, we'll use z as the number/operation needed to make this true of this equation:

y=z(x-1)(x+1)(x+2)
-6= z(0-1)(0+1)(0+2)
-6=z*-1*1*2
-6=-2z
3=z

Thus, the equation you are trying to produce is y=3(x-1)(x+1)(x+2)

To check it, you could plug 0 in for the x's again just to be sure you get -6, or plug +/-1 or -2 in to see if you get 0. You do, so it's right.

Originally Posted by Spearhead

And then there's one question which asks me to determine the remainder when x^3 + 2x^2 - 6x + 1 is divided by x + 2 without dividing. Again, how is this done?

If anyone can help, I would be really, really grateful.

Without dividing? Not even synthetic division? Darn.

Alright, here's where I admit that I don't remember how to do this and turn you to a website that might help.

As far as I can gather, you are dividing f(x) by (x-a). Without dividing, you can determine that the remainder based on what f(a) is. So, plug "a" in (remembering that in (x-a), a is opposite of what it is due to the - sign).

Remainder=f(a)
Remainder=f(-2)=(-2)^3 + 2(-2)^2 - 6(-2) + 1
Remainder=f(-2)=(-8) + 8 +12 +1
Remainder=f(-2)=13
Remainder=13


I hope that helps a little bit.

EDIT: Oh, PIE beat me to it. -_-;

Thanks so much, you guys. The test today went more than smoothly because of your help.