CIRCULAR FUNCTIONS - DEGREE AND RADIANS MEASURE 2

Hi, My name is Kristine, I sit beside Maridel and Carl.

We started unit circle and I will be talking coterminal.
Coterminals are angles that share the same terminal side. There are two ways to get coterminal, you have positive and negative coterminals.

Lets learn how to get those coterminals.

POSITIVE COTERMINAL:
-To get postive coterminal you will have to add 360 degrees or 2 radians to the angle given until it is a positive number.

50° + 360° = 410°

 

 NEGATIVE COTERMINAL:
-To get negative coterminal you will have to subtract 360 degrees or 2 radians to the angle given until it is a negative number.

                                                               30° - 360° = -330°

YOU COULD ALSO DO A GENERAL FORM TO GET AN INFINITE AMOUNT OF COTERMINALS.

θ ± (360°)n or θ ±  2πn, where n is a natural number.


Thats all for today (: HAVE A GREAT DAY EVERYONE!

Polynomial Functions and Graphs

      Hello Mathematicians! Do you need help? ... No? I'm still going to blog :) haha.  This is Fatima scribing for the lesson on March 17.   Thanks to my table mates for shouting my number when they had to... seriously guys... thank you.

    I hope this will be helpful ^^ let's get started because there's 3 topics to cover.
(I think I'm supposed to put somewhere on this post that I don't own the pictures that I used but I'm not sure where, so I'll leave it here : I don't own any of the pictures used, I found them on google, please don't sue Mr. P.)

Analyzing graphs of polynomial functions

     Today we looked at polynomial functions and analyzed the following points:

• The least possible degree
• The sign of the leading coefficient
• The x-intercepts and factors of the function
• The intervals where the function is positive or negative

Here's an example... ( look i added numbers to the picture of the function)

Let's briefly go over how we get the least possible degree, and the sign of the leading coefficient. 

For the degree, we know it is one more than the number of bumps that we have or n+1. (for this example, it is 3+1=4) our degree is 4.

For the leading coefficient, our function ends in quadrants I and II. because we have an even exponent on our polynomial function, we know that this means we have a positive coefficient. ( if our coefficient was negative, it would be upside down)

In the same example, we can see that our x-intercepts are at: -10, -6, 4, and 10.
This means our factored form is : (x+10) (x+6) (x-4)(x-10)






Lastly, we need to know where our function is positive or negative. This is our domain, x, when our f(x), or y, is positive or negative.

I put circles on this one. hehe editing is so hard >.>

Here are the positive intervals (where f(x) is positive) :

     x<-10
     -6<x<4
     x>10

Here are the negative intervals (where f(x) is negative) :

     -10<x<-6
     4<x<10

phew are you tired? I am... there's more! Let's take a small snack break first...

 getting back to work...

Graphing Polynomial Functions using Transformations

     

Here is the basic forms of functions from x^1 to x^5.  The even exponentials are on the left side and the odd exponentials are on the right.

     The same rules apply as what we've learned from the previous unit. Do I have to go over it? Please check previous posts for help.

 
  our equation: y= a(b(x-c) ^3 +d 

Let's look at yet another example:

Here is a cubic function translated 2 units to both the left and right. It's similar to the translations we did right? okay! let's move on to the last topic... 

Is there a character limit on here?

Solving Problems Involving Polynomial Functions

     uh... I don't know if I'll be doing this correctly but here we go... 

it's time to put our smart glasses on..

 

Let's try the same example we have on our notes:

Bill is preparing to make an ice sculpture. He has a block of ice that is 3 ft wide, 4 ft high and 5 ft long. Bill wants to reduce the size of the block of ice by removing the same amount from each of the three dimensions. he wants to reduce the volume of the ice block to 24 ft^3.

a.) write a polynomial function

1. We know that volume = l*w*h (from our problem, we know that our length is 5, our height is 4, and out width is 3) so we have volume= 5*4*3
2. Next, Bill wants to subtract the same amount on each dimension. This means subtracting value x on length width and height.  volume= (5-x)(4-x)(3-x)
3. Lastly, we know that his final volume will be 24 ft^3 so,  24=(5-x)(4-x)(3-x)

 so the factored form of our polynomial function is 24=(x-5)(4-x)(3-x)

we now use FOIL to find our polynomial function...

     24=(5-x)(4-x)(3-x)

     24=(20-9x+x^2)(3-x)

     24= -x^3+12x-47x+60

b.) how much should he remove from each direction? (this question is asking what the value of x is, so we go back to our polynomial function and solve for x)

     24= -x^3+12x-47x+60

     x^3-12x+47x+60-24=0

    x^3-12x+47x-36=0    

here, we use integral factors to find our possible values of x. (Again, please refer to previous posts for integral factors)

p(1)= 1^3-12(1)^2+47(1)-36

      = 0

so one of our factors is (x-1).

next we use synthetic division to get factor out (x-1)... I don't know how to show that on here so let's fast forward to the factored form...

x^2-11x+36 (not factorable)

this function is not factorable; therefore, we disregarded the other roots and came to a conclusion that

x=1

Let's check:

(5-1)(4-1)(3-1)= 24

(4)(3)(2)= 24 

IT'S CORRECT! yaaas we're done! 

Alright mathematicians! I hope this blog was helpful in a way or another!  Time to study because we have a test soon!

-end of scribe-

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,

LONG DIVISION

SOLVING LONG DIVISIONS!!!

Hello everyone, my name is Shubham. The one who sits next to Vincent, the guy who picked me up today to scribe next. I told you to choose a number instead of choosing me, but you choose me. At that time i was going to kill you for choosing me, but never mind :)

 So today i am gonna discuss, what i have learned in Mr. P's class today. which is to solve Long Divisions. 

There are two methods of solving long divisions:-

1. The normal division method which is a bit time consuming.

2. Synthetic Division method which is quite easy and saves our time. 

The result of division of a polynomial in x, P(x), by a binomial of the form P(x) = Q(x)+ R 

                                                                                                                         x-a               x-a

                                                          

=> NORMAL DIVISION METHOD

Example: divide  8x²+9x+3x³ -2 by x+2

Steps:

  1. write the dividend and divisor polynomial in descending powers.

       3x³+8x²+3x

2. Divide the leading term of the dividend by first term.

       3x³  = 3x²

        x

3. Multiple the divisor by newly formed term of the quotient using the distributive law and subtract the result.

                                                           3x² + 2x -1                    

                                             x+2   ) 3x³+8x²+3x-2

                                                      -    3x³+6x²

                                                                   2x²+3x

                                                                   2x²+4x

                                                                          -x -2

                                                                          -x -2

                                                                                 0

  factors = (x+2)(3x-x)(x+1)

                             

=> SYNTHETIC DIVISION

    Example: Divide 3x⁴ - 5x² - 4x + 4  by x+2

     

   Solution: P(x)= 0  =>  x+3=0 =>  x= -2

                    x-a     

Now divide all the coefficients by 3. And if any coefficient of power of x is missing in the         followings the let that coefficient be ZERO

like in this example the coefficient of x³ is missing then let that coefficient of x³  be zero

                                                                             

                                      -2  ) 2  +0  -5  +4  +  4

                                           +      -4   8   -6    -4

                                               2  -4   3   -2     0

Factor => 2x³-4x² +3x -2

The key steps of synthetic division are as follows:-

1. Arrange the coefficient of f(x) in the order of descending power of x

2. After writing the divisor in the form x-a, use "a" to generate the 2nd and 3rd rows of number as follow

3. Bring down the 1st coefficient of the dividend and multiple by "a" 

4. Then add the product to the 2nd coefficient of dividend.Repeat the process until a product is added to the constant term of (x).

Well, those are the ways Mr. P taught us in class to solve these long divisions and I hope all the readers of this blog will like amd take help of it to solve the questions.:)