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-