Introductions to Basic Transformations

Hi my name is Daniel and today we learned about basic functions!!

This function can be expressed as f (x) = x²
 f (x) can also be y.f (x) = y
In class we reviewed some general equations to remember. 
They are the Vertex form and the Standard form
Vertex form is expressed as: y = a(x – h)² + k  and the variables a, h, k each tell you what the function would look like before even graphing.

The sign on "a" tell you whether the function opens up or down. If a < 0 then the function will most likely open down. If a > 0 then the function will most likely open up.
If a increases, the shape narrows, if a decreases, the shape becomes wider.
h and k will be the coordinates of the vertex used to plot the graph.

Standard form is expressed as: y = ax² + bx + c This is a quadratic equation because it is to the power of 2 (x²)

What you need to understand when you are graphing any quadratic is that you are moving, flipping, stretching which are called transformations.
Take the vertex form for example.  y = a(x – h)² + k 
We can move it up or down by adding a constant to our k-value.

k > 0 makes it go up

k < 0 makes it go down

We call k a vertical translation.





We can move the function to the left or right adding a constant to our h-value.

h > 0 makes it go left

h < 0 makes it go right

We call h a horizontal translation.





We can flip it upside down if we multiply the whole function by -1.
  f(x) = -(x²)



We can stretch it in the y-direction by multiplying a constant to our a-value.

a > 1 stretches it

0 < a < 1 compresses it.

 f(x) = 0.35(x²)


That's all you need to know! Hopefully this expanded your knowledge on the principles of transformation.

I CHOOSE #7 :D TO BE THE NEXT BLOGGER

HOW TO BECOME A CHAMPION?

Poland dethrones Brazil after winning its second World Championship.

HOW TO BE SICK AT PRECAL (The Binomial Theorem)

Hello my name is Marc. I will be teaching you about The Binomial Theorem.

The Binomial Theorem is when you expand the powers of binomials.

You can easily expand them by using Pascal's Triangle shown here:

Every time the power of a binomial increases you go down the triangle.

The sum of the exponent in each term equals to the power of the binomial expansion. 

The X variable begins with the same value as the power of the binomial and decreases by one in each successive term. 

The Y variable appears in the second term of the expansion and increases by one until it matches the power of the binomial.

The numbers of terms in the expansion contain N + 1 terms in the expansion.

The coefficients of the binomials are the combinations of the power number beginning with nCo and ending at nCn and are symmetrical.

These are important rules you should remember:

1. A binomial expansion where the exponent is n will have a n + 1 terms when expanded.

2. A binomial expansion where the exponent is even will have an odd number of terms when expanded causing the expansion to have a middle term.

3. A binomial expansion where the exponent is odd will have an even number of terms when expanded causing the expansion to not have a middle term. 

4.If a pattern occurs like 5C1=5C4 and 4C0=4C4 and 3c1=3C2 you will not have to calculate all combinations.

I CHOOSE DANIEL :) ok bye.

DO YOU WANT TO BUILD YOUR IQ? (Combinations)

COMBINATIONS

Hello, my name is Lloyd, and I will be teaching you about Combinations.

There is a difference between Combinations and Permutations.

Permutation is when we select and order the elements.
FORMULA: nPr = n!(n-r)!


and a...


Combination is when the order does not matter and you can choose.

FORMULA: nCr = r!(n-r)! 

The n is the total and r is how many you are choosing.

YOU CANNOT USE THE DASH METHOD, USE FORMULA!

An example of a combination:

Example 3: A student has a penny, a nickel, a dime, a quarter, and a half dollar and wishes to leave a tip consisting of exactly 3 coins. How many different tips are possible?

In this question there are 5 coins to choose from, and the student needs to pick 3 out of the 5 so you have...

n = 5 

and

 r = 3
So if you plug this in the formula you will get:

5C3 = 5! / 3!(5-3)!
     
        = 5x4x3! / 3!2!

        = 20 / 2

        = 10

There is 10 different possible combinations to tip with and the order doesn't matter.

The words I remember Mr. Piatek telling us about how to find if it's a combination is SELECT and COMMITTEE. hehe.

Also note that if the n's are the same...

nCx = nCy
n=x + y

Example: Solve for x.
xC6 = xC9
x=6+9
x=15



And thats it! Hope this helped... kind of.

I CHOOSE #18 FOR THE NEXT PERSON TO BLOG. (MARC)

Permutations with Case Restrictions.

Hi everyone, it's Vienna! On Friday, we learned "Permutations with Case Restrictions". Now if the last lessons were quiet challenging, this topic might confuse you even more. In this lesson, there are more restrictions demonstrated.

For example: How many 3-digit numbers greater than 300 can you make using the digits 1,2,3,4,5 and 6? No digits are repeated. (note: I will be using this question throughout the explanation)


In order to solve the question you must understand when to use and what cases mean. What Case restriction means is that there are different possibilities that you may use that fits in the question.

In the question above, it is asking for a number/numbers that are greater than 300; Well from 1 to 6, it acts as if it's in the hundreds. Like 1 is equivalent to 100 (depending on the question given). In this case, the number that is greater than 300 can be 3. But that is not it, there is also 4,5,6 that is greater than 300.

Cases:
300 and above > 300
400 > 300
500 > 300
600 >300


First Step: For questions like this, it is highly common you use the dash method just because there will be plenty of restrictions. On the example, it instructs you to create 3-digit numbers, that means that N=3. Since there are 4 numbers that are greater than 300, that means you must create 4 columns of the dash method (note: column is preferably called "cases" in this topic.)

First case with 3 dashes _____ x _____ x _____
Second case _____ x _____ x _____
Third Case _____ x _____ x _____
Fourth Case _____ x _____ x _____


Second Step: Plug in the numbers that are greater in the first dash. The reasoning to having 1 for the dash is because there is only one number given to you. If there are 3 2's, then you put 2.

3->    1     x           x _____
4->     1     x           x          
5->     1     x           x          
6->     1     x           x          


Third Step: In order to fill in the 2nd and 3rd dash, you need to know how many is left. It is indicated that "No repetitions are allowed". Since there are 6 options of numbers and we used one number already, there are five left. That means that the 5 is located in the 2nd dash while there will be 4 left in the third dash.

3->    1     x     5     x     4    
4->     1     x     5     x     4    
5->     1     x     5     x     4    
6->     1     x     5     x     4    


 Fourth Step: You calculate the total for each dash and it will give you the answer of 20. But that is only for 1 Case. You have to add all of the four cases and the final result is 80.

3->    1     x     5     x     4     = 20

4->     1     x     5     x     4     = 20
                                             +
5->     1     x     5     x     4     = 20

6->     1     x      5     x     4     = 20
                                                 80

In some questions such as this, you are able to do a case in one-shot. If there are for possibilities, that means that you can include that on the first dash. you would have four instead of one.

3,4,5,6->     5     x     5     x     4     = 80


When I said "some" I meant these type of questions "How many 4-digit odd numbers can you make using the digits 1 to 7 if he numbers must be less than 6000. No digits are repeated." For this example, you must know how to contribute your clues into the question; You have n=4, you must indicate the odd numbers in a dash, and the you will have 5 cases because 1,2,3,4,5 are less than 6000. It may sound complicated but it will click once you see a solution.

1->     1     x     5     x     4     x     3    <-3,5,7            = 60
2->     1     x     5     x     4     x     3    <-1,3,5,7         = 80
3->     1     x     5     x     4     x     3    <-1,5,7            = 60
4->     1     x     5     x     4     x     3    <-1,3,5,7         = 80

5->     1     x     5     x     4     x     3    <-1,3,7            = 60
6->     1      x     5     x     4    x     4    <-1,3,5,7         = 80

                                                                                    340

Do you see the pattern?. When you look at the first case, 1 is not included in the 4th dash because it is already being used and it is an odd number. When you look at the second dash, two is an even number so all the odd numbers are being shown on the 4th dash. As for the 5 and 4, that's how many numbers are left. Do you kind of get it?


Everyone deserves a break from all this lesson. For that, here is a little treat and enjoy! (And yes I got permission from Lloyd).

Do You Wanna Build A Snowman - Lloyd Umali
https://www.youtube.com/watch?v=b7ke-xEscQs



                                                                                                                                      Vienna Jaime

PERMUTATIONS PART 2

Yoooo what is up! Its your boy Jairo a.k.a joshy a.k.a Beyonce's long lost son!  Today's lesson was a continuation on permutation, Permutation Part Two!

We learned that permutation is a arrangement of the elements in an ordered list. When arranging items in specific order we can either use the the permutation formula OR the dash method. The permutation formula can only be used if the question doesn't allow repetition and the dash method will work on basically all situations.

*Remember, You always assume no repetition are allowed otherwise stated*

- Repetition means we are allowed to pick the same item more than once

  Restriction are when the questions indicates to place an item in specific location 

Examples

      -There are 15 horses in a horse show competition. The top three winning horses receive money. How many possible money winning orders are there for a competition with 15 horses?

-Using permutation formula


                                                      15P3 = 15! / (15-3)! =2730

                                                                        or

 -Using dash method

                                                         15 x 14 x 13 = 2730





      -There are 3 brothers and 3 sisters being seated in the bus for a field trip. How many arrangements are possible if they are seated alternately?


*This question uses restrictions therefore we can only use the dash method


                                3boys x 3girls x 2boys x 2girls x 1boy x 1girl = 36

                                3girls x 3boys x 2girls x 2boys x 1girl x 1boy = 36

                                                           36+36 = 72 ways


 Thats pretty much it! If you have any questions or didn't understand something feel free to ask in the comments and I might be able to help you or just google it lol

                                                                                                                  -Jairo Nato

Permutations Part One

Hey guys! It's Bea! Today, we learned more new things about Permutations and I'm going to be writing a recap of the lesson that we covered.

First things first, what is a Permutation? 

          - A permutation of a set of distinct objects is an arrangement of objects into different orders. To permute a set of objects means to rearrange them.

We learned a formula that expresses the number of permutations chosen from a set of objects and that is:

where n is the total number of things to choose from and r is the amount you choose out of it

So this means, that n always has to be greater than r because it wouldn't make sense if you're, for example, choosing 6 (r) out of 3 (n). It is not possible. So n ≥ r.

There are two types of permutations: 

• Permutations without repetition: Number of permutations of n-things, taken 'r' at a time but 
  nothing can be repeated. You can use either the formula or the dash method to calculate.
• Permutations with repetitions: Number of permutations of n-things, taken 'r' at a time when 
  each thing can be repeated r-times. You cannot use the formula to calculate permutations where 
  repetitions are allowed. To calculate, use the dash method ( _____ · _____ )

Example 1: How many 3 letter words composed from the 26 letters in the alphabet are possible if...

a.) No repetitions are allowed?

For this case, we can use the Permutation formula.

Our values are:

n= 26 

r= 3

26P3 = 26! / (26-3)!

         = 26! / 23!

         = 26·25·24·23! / 23!

         = 26·25·24

         = 15,600 words

b.) Repetitions are allowed?

We cannot use the formula here so we're going to use the dash method.

_______ · ________ · ________

1st letter    2nd letter     3rd letter

= 26 · 26 · 26

= 17, 576 words

That's it!I hope this helps you guys understand it better! See you all in class tomorrow! :)

Bea Clarissa Guan

factrial notation

Hiiiiiii.... everyone...its parul...

         today i am going to tell  u about factorial notation....

The symbon n! means to multiply all the positive integers from n consecutively down to 1 

n!=n(n-1)(n-2).........(3) (2) (1), where n is an that element which positive integers. 

exp. 1=

0!=1 it is always defined as 1 

exp.2=

         5!=5 (5-1)(5-2)(5-3)(5-4)(5-5) 

             5*4*3*2*1=120

exp.3 

        11!=  11*10*9*8*7*6!=  279n17600

exp.4=     

       n(n-5)!=24(n-6)!

           n(n-5) ( n-6)!=24 (n-6)!

cancel (n-6)!

           n(n-5)=24

           n*n-5n-24=0

         n*n-8n+3n-24=0

n(n-8)+3(n-8)=0

(n-8) (n+3)=0

n-8=0 or n+3=0

n=8  or n= (-3)

n=-3 reject it bcz n is an elemnt of negative integer....

    

  check ans... put the value of n=8 in eq.

      8(8-5)!=24(8-6)!

     8(3)!=24 (2)! 

     8*3*2*1=24*2*1

48=48

L.H.S=R.H.S

                     so its the easiest way to solve.........the factorial notation

          

Permutations

Hey everyone this is the one and only, Beyonce. Just kidding no it's Patrick.
Today's lesson was about Permutations.
Permutations is all about counting, without any real counting. Now what am I talking about? For example:

A cafe has a lunch special consisting of an egg or ham sandwich, milk, juice or coffee and a yogurt or pie for dessert. What are all the possible outcomes if you take one from each category?

It's easy to just list them all but there's a far more efficient way of doing things, using something called the "Fundemental Counting Principle," this principle involves reading word problems and translating them into a mathematical equation. Keywords such as "AND" or "BOTH" can translate into Multiplication while other words such as "OR," "EITHER" or "OPPOSITE" can translate to Addition. The question above we see that the word "OR" is present several times telling us to add the options for each category together:

2 Sandwiches, 3 Drinks and 2 Desserts.

The next step is to multiply each category to get the number of possible outcomes.

2 X 3 X 2 = 12

There are 12 possible meal combinations.

That's it! Make sure if you need any extra help or practice check out ther many examples in the booklet and ask for help! I hope this was helpful and good luck!

Patrick Eulalia

Welcome

Welcome to our blog. This space is designed for students of the Maples Collegiate, attending the Pre-Calculus 40S class, section D, with Mr.P. We are going to use this space to discuss our daily lessons, ask questions you didn't get a chance to ask in class, and to share your knowledge with other students. Most importantly we will use this blog to reflect on what we're learning.
Have a great semester.