Back to Basics -

Hey Guys! for this  (morning/afternoon/evening) I will tackled first a review regarding some previous stuffs from grade 11 about Quadratic Functions because I believe that before moving on another more complex topics we should go back to basics.  And I would also include about our previous lesson from yesterday the Transformation of Functions. Before moving on another lesson we should make sure that we are right on track so bare with me guys :)  

What my mom and dad used to say is that math is not about memorizing stuffs, but better yet its more of practicing stuffs.

They also told me that in life you really have to fail and to fail to become successful, because believe it or not failure builds character and will served as your stepping stone in life.

Study Plan 101:   

Remember this from way back from your grade 11 years... 

What are functions? 

- A function in math is the process of showing a relationship between the input and output of a problem. The number you put in is directly related to the number you get out.

f(x)= y

This means that for every  x value, there is only one y value

The most basic quadratic equation that you'll see  is x^2.

*note: A mind refresher for remembering stuffs.One way to graph a quadratic equation is using table of values. So this is the example of the table of values of the quadratic equation x^2. 

 

x	y
-2 -1  0  1  2	4 1 0 1 4

You can pick any number in the x-axis, and then plug them in your calculator solve it mentally. I chose this numbers for you to see how a parabola opens. Ex: your original equation is x^2. 

Always draw a nicely smooth curving line passing  neatly through the plotted points. And don't forget the arrow in the parabola

As for me this served as a guide for us to graph much complex and cooler quadratic equations. so keep this in mind and heart guys :)

* Second, you should know how to identify standard form from vertex form 

Graphing Quadratic Equations

The graphical representation of quadratic equations are based on the graph of a parabola. A parabola is an equation of the form y = a x² + bx + c. The most general parabola, shown at the right, has the equation y = x².
The coefficent, a, before the x² term determines the direction and the size of the parabola. For values of a > 0, the parabola opens upward while for values of a < 0, the parabola opens downwards. The graph at the right also shows the relationship between the value of a and the graph of the parabola.

The vertex is the maximum point for parabolas with a < 0 or minimum point for parabolas with a > 0. For parabolas of the form y = ax², the vertex is (0,0). The vertex of a parabola can be shifted however, and this change is reflected in the standard equation for parabolas. Given a parabola y=ax²+bx+c, we can find the x-coordinate of the vertex of the parabola using the formula x=-b/2a. The standard equation has the form y = a(x - h)² + k. The parabola y = ax² is shifted h units to the right and k units upwards, resulting in a parabola with vertex (h,k).

The standard form of a parabola's equation is generally expressed:

• y = ax ² + bx + c
  • The role of 'a'
    • If a> 0, the parabola opens upwards
    • if a< 0, it opens downwards.
  • The axis of symmetry
  The axis of symmetry is the line x = -b/2a
  
  

The vertex form of a parabola's equation is generally expressed as : 
y= a(x-h)²+k 

• (h,k) is the vertex as you can see in the picture below
• If a is positive then the parabola opens upwards like a regular "U".
• If a is negative, then the graph opens downwards like an upside down "U".

If |a| < 1, the graph of the parabola widens. This just means that the "U" shape of parabola stretches out sideways .

If |a| > 1, the graph of the graph becomes narrower(The effect is the opposite of |a| < 1).

You can see these trends when you look at how the curve y = ax2 moves as "a" changes:

As you can see, as the leading coefficient goes from very negative to slightly negative to zero (not really a quadratic) to slightly positive to very positive, the parabola goes from skinny upside-down to fat upside-down to a straight line (called a "degenerate" parabola) to a fat right-side-up to a skinny right-side-up.   

There is a simple, if slightly "dumb", way to remember the difference between right-side-up parabolas and upside-down parabolas:

positive quadratic y = x2	negative quadratic y = –x2

Graph the following parabola:

y = (x + 3)² + 4

The vertex is the point (-3,4) 

Axis of Symmetry x= -3

direction opens up because a is positive (+)

min value = y = 4

y-int= ( 0 + 3)^2 +4 = 13

 y = –(x–1)² + 1?

The vertex is the point (1,1) 

Axis of Symmetry x= 1

direction opens down because a is negative (-)

min value = y = 1

y-int= ( 1-1)^2 +1= 2

Example: Plot f(x) = 2x² - 12x + 16

First, let's note down:

• a = 2,
• b = -12, and
• c = 16

Now, what do we know?

• a is positive, so it is an "upwards" graph ("U" shaped)
• a is 2, so it is a little "squashed" compared to the x² graph

Next, let's calculate h:

h = -b/2a = -(-12)/(2x2) = 3

And next we can calculate k (using h=3):

k = f(3) = 2(3)² - 12·3 + 16 = 18-36+16 = -2

So now we can plot the graph (with real understanding!):

We also know: the vertex is (3,-2), and the axis is x=3

Just like Transformations in Geometry, you can move and resize the graphs of functions

You can move it up or down by adding a constant to the y-value:

g(x) = x² + C

Note: if you want to move the line down, just use a negative value for C.

• C > 0 moves it up
• C < 0 moves it down

 

You can move it left or right by adding a constant to the x-value:

g(x) = (x+C)²

Adding C moves the function to the left (the negative direction).

Why? Well imagine you are going to inherit a fortune when your age=25. If you change that to(age+4) = 25 then you would get it when you are 21. Adding 4 made it happen earlier.

• C > 0 moves it left
• C < 0 moves it right

An easy way to remember what happens to the graph when you add a constant:

add to y: go high
add to x: go left

BUT you must add C wherever x appears in the function (you are substituting x+C for x).

Example: the function v(x) = x³ - x² + 4x

Move C spaces to the left: w(x) = (x+C)³ - (x+C)² + 4(x+C)

You can stretch or compress it in the y-direction by multiplying the whole function by a constant.

g(x) = 0.35(x²)

• C > 1 stretches it
• 0 < C < 1 compresses it

 

You can stretch or compress it in the x-direction by multiplying x (wherever it appears) by a constant.

g(x) = (2x)²

• C > 1 compresses it
• 0 < C < 1 stretches it

Note that (unlike for the y-direction), bigger values cause more compression.

 

You can flip it upside down by multiplying the whole function by -1:

g(x) = -(x²)

This is also called reflection about the x-axis (the axis where y=0)

You can combine a negative value with a scaling.

Example: multiplying by -2 will flip it upside down AND stretch it in the y-direction.

 

You can flip it left-right by multiplying the x-value by -1:

g(x) = (-x)²

It really does flip it left and right! But you can't see it, because x² is symmetrical about the y-axis. So here is another example using √(x):

g(x) = √(-x)

This is also called reflection about the y-axis (the axis where x=0)

Summary

y = f(x) + C	C > 0 moves it up C < 0 moves it down
y = f(x + C)	C > 0 moves it left C < 0 moves it right
y = C·f(x)	C > 1 stretches it in the y-direction 0 < C < 1 compresses it
y = f(Cx)	C > 1 compresses it in the x-direction 0 < C < 1 stretches it
y = -f(x)	Reflects it about x-axis
y = f(-x)	Reflects it about y-axis

Examples

Example: the function g(x) = 1/x

Move 2 spaces up: h(x) = 1/x + 2

Move 3 spaces down: h(x) = 1/x - 3

Move 4 spaces to the right: h(x) = 1/(x-4) (play with the graph)

Move 5 spaces to the left: h(x) = 1/(x+5)

Stretch it by 2 in the y-direction: h(x) = 2/x

Compress it by 3 in the x-direction: h(x) = 1/(3x)

Flip it upside down: h(x) = -1/x

 Case 1: If k > 1, then  g(x) is a vertical expansion of scale factor k ( Stretch )

 Case 2: If 0 < k < 1, then  g(x) is a vertical compression of scale factor k (Shrink)

k > 1	0<k < 1

(Vertical shift) 

Note that:

   If k > 0, then  f(x) moves up by k units,

and If k < 0, then f(x) moves down by k units.

k = 3	k = -3

       Note:

          If k > 0; The translation is to the LEFT,     

        and if k < 0; The translation is to the RIGHT 

k > 0	k < 0

Note also that 

Sources:

Google: https://www.google.ca/

Mathisfun.com

shswisdompbworks

purplemath

Binomial Theorem

Hey guys!!! It's Vivian and I'm going to be blogging about binomial theorem today. Fun stuff!!!!!

So.. In class we learnt how to expand and simplify using the binomial expansion:

Example 1:
 (3x-y)^5

Okay, first off, we start off by listing our values.
Our a = 3xb = -y -> don't forget the negative sign!n value = 5 because our expression is to the power of 5.

Using the formula, we're going to expand the expression now. I wrote it out because it was really hard typing out all the power signs.

In the first step, we expanded out the equation (3x-y)^5 following the formula. We did this by plugging in our values into the equation.

 If you can't remember the formula, you can always remember that all the powers should equal to whatever the value of n is. In this case n is equal to 5. If you notice in the first term 5C0(3x)^5(-y)^0 the powers are 5 and 0. 5+0 = 5. Your terms 5C0, 5C1... etc should keep adding up until you reach 5. Which is what you should get in the end -> 5C5.

In step 2, what we have to do next is multiply the brackets through. I'm just going to do the first term just because it's really difficult typing everything out. First term: 5C0(3x)^5(-y)^0
First we start off with 5C0. 5C0 is simply just 1. You can either plug it into your calculator to find out the answer or you can use the formula: nCr = n!
                                                              ----
                                                            r! (n-r)!
5C0, our 5 = our n value, and 0 is our r value. Then we will get
   5!               5!            (5)(4)(3)(2)(1)
-------    =  ------ =    ------------------
0!(5-0)!       0!5!          (1)(5)(4)(3)(2)(1)

So our top is cancelled, and we're left with 1 at the bottom. 1/1 = 1. So then 1 is our answer for the 5C0. Still don't believe me? You can double check by plugging 5C0 into your calculator.

Next, we're trying to simplify the (3x)^5
(3x)^5 if we multiply that through (3)(3)(3)(3)(3) = 3^5 and (x)(x)(x)(x)(x) = x^5 |

Then we simplify (-y)^0
(-y)^0 = 1 because anything to the power of 0 is equal to 1.
Following that, you will continue to simplify the rest of the equation.
You should be left with what I have in step 2.

Then in your last step, you simplify what's remaining. Again, I'm only going to do the first term. (1)(3x)^5(1) = (1)(3)^5(x)^5(1) = 243x5. After you're done simplifying and multiplying everything, you should be left with the answer that I have boxed in.

Get it? Got it? Good. Next example.

Example 2) Find the 3rd term of (2x3+y2)^3
-> In this example we're looking for a specific term, and when we want to know a specific term we can use the formula:

tk+1 = nCka^n-k(b)^k

Again, we start off by listing our values.
a = 2x3
b = y2
n = 3 because our exponent is to the power of 3.
k = ? -> our k value is always 1 less then the term we're looking for. In this case, we're looking for the 3rd term, so our k value should be 3-1 = 2.

Since we have all the values we can simply just plug it into the formula now and solve!
tk+1 = nCka^n-k(b^k)
t2+1 = 3C2(2x3)^3-2(y2)^2
--> Now after we plug everything in we can just simplify the expression.
= (3)(2x3)^1(y4)
Now you just multiply and simplify the expression. You should get 6x3y4 as your answer.

Example 3) Find the coefficient of x2 in expansion of (x-3)^4
Notice how in this example it says find the COEFFICIENT, not find the TERM? Which means we have to do things a little differently as our first step this time.

It says find the coefficient of x2 in expansion of (x-3)^4.
Before we do anything, we have to find a pattern with x2 in the expression. We can do this by expanding out some of the terms.

First always list your terms!
a = x
b = -3
n = 4

(x-3)^4 = 4C0(x^4)(-3)^0 + 4C1(x)^3(-3)^1 + 4C2(x^2)(-3)^2
     -> x = x4 in the 1st term,  -> x3                   -> x2. Is this what we're looking for? Yes! = 3rd term.

After we find out the pattern, and we find that the 3rd term of the expression is = x2 we can plug it into this formula again:

tk+1 = nCka^n-k(b)^k
Values:
a = x
b = -3
n = 4
k = 2, this is because our 3rd term is = x2, so one less than 3 = 2.

t2+1 = 4C2(x)^4-2(-3)^2
=6(x^2)(9)
= 54x2
-> In this case our answer is going to be 54, because it's only asking for the COEFFICIENT, not the TERM.

                              Still having trouble? Check out the YouTube video above! ^^

         Well.. Our time is up everyone. Hope I helped and have a fabulous day everyone!!! :)

PAT'S BLOG :~)

Hey ya'll its Pat.
 
I will be blogging about some stuff that we learned in our pre-cal class this week. Mostly on Combinations and I will touch a little bit on Pascal's binomial theorem.
 
COMBINATIONS
 
first of all, what is the difference between PERMUTATION and COMBINATION?
 
we all learned that PERMUTATIONS are arrangements of a set of objects. The order matters and there are different formulas to solve for your permutation.
 

in COMBINATIONS however, the order does not matter. for combinations, we do not arrange the items. we just choose them.
 
the formula is.....
 
 
ncr where n = total and r = how many ou are choosing.
when you expand this formula it becomes.....          n!
                                                                        --------------
                                                                           r! (n-r)!

for example:

you have 10 teachers to pick from but you only need to choose 4 of them.
so if you plug it your formula.....

nCr = 10c4 = 210.

therefore, you have 210 combinations of teachers that you can pick. The order does not matter.

there are a few keywords that Mr. Piatek gave us for combinations:

CHOOSE
COMMITTEE
SELECT

Basically, we just have to remember that permutations are specific. Combinations are not specific.

moving on to
PASCAL'S BINOMIAL THEOREM

Yesterday in class, Mr Piatek showed us this pattern.
 

 and i foud this video that explains all the patterns that you find in the triangle!
 

....ok that's all im going to blog about for now. see you guys in class :~)

PreCal 40s

Hi guys this is Mac, blogging for the first time, woo! fun :)

I will be talking about what I have learned this week in our precal class. First of all I would like to thank Mr. Piatek for introducing me to this "blogging site" that I have never heard of but I'm just glad that I have. It'll be a very interesting semester with him :).

Permutations Part 1

FUNDAMENTAL COUNTING PRINCIPLE
I will introduce the fundamental counting principle with an example.

This counting principle is all about choices we might make given many possibilities.

Suppose most of your clothes are dirty and you are left with 2 pants and 3 shirts.

How many choices do you have or how many different ways can you dress?

Let's call the pants: pants #1 and pants #2

Let's call the shirts: shirt #1, shirt #2, and shirt #3

Then, a tree diagram as the one below can be used to show all the choices you can make

As you can see on the diagram, you can wear pants #1 with shirt # 1. That's one of your choices.

Count all the branches to see how many choices you have.

Since you have six branches, you have 6 choices.

However, notice that a quick multiplication of 2 × 3 will yield the same answer.

In general, if you have n choices for a first task and m choices for a second task, you have n × m choices for both tasks

In the example above, you have 2 choices for pants and 3 choices for shirts. Thus, you have 2 × 3 choices.

Another example:

You go a restaurant to get some breakfast. The menu says pancakes, waffles, or home fries. And for drink, coffee, juice, hot chocolate, and tea. How many different choices of food and drink do you have?

There 3 choices for food and 4 choices for drink.

Thus, you have a total of 3 × 4 = 12 choices.

Tips and advice:

Multiplication Principle: Key Words are "AND" "BOTH"
- when two events are dependent we multiply their results to find the total number of ways an event can occur.

Another example would be:
If there are 52 runners entered in a race, in how many ways can first AND (indicates that we have to multiply) second place be awarded?

therefore:     52        x      51           = 2652 ways
                 --------          ---------
               1st place       2nd place

So where does 51 come from? I am glad you asked :)

Well we have 52 runners in total and we can have any of these 52 individuals who can be awarded First, and since we can only have one first place winner, the other 51 individual can either place Second. And that is where 51 came from, and if you're not with me, Mr. Piatek can clarify the rest for you :)

Additive Principle: Key Words are "OR", "EITHER", "OPPOSITE"
- when two events are independent we add their results to find the total number of ways an event can occur.

An example would be:
Suppose that the executive of the Manitoba Association of Mathematics Teachers consists of three woman and two men. In how many ways can a president and a secretary be chosen if:

a) The president is to be female and the secretary male?

So here we have 3 Female and 2 Male:
3F = president
2M= secretary

therefore:    3        x        2     
              ---------           ----------   = 6 different ways
             President         Secretary

b) The president is to be male and the secretary female?

So now we have 2 Male and 3 Female:
2M= president
3F= secretary

therefore:   2         x         3
              --------           --------- = 6 different ways
             President       Secretary

c) The president and secretary are to be of the OPPOSITE (indicates that we have to add) sex?

So now it's asking for how many ways can the president and secretary are to be of the opposite sex, simple:

We have 3 females and 2 males, and we can name the females, Marjory, Lucy, and Gertrude and the males we can name them, John and Ash, just so it's easier for us to understand and not get mixed up.

 Since there are 6 different ways for them to become a president and secretary we write our own, let's say columns to show the 6 ways.

therefore:         1st column                                                      2nd column
             President(F)/ Secretary(M)                          President (M)/ Secretary(F)
                M                J                                                       J                  M
                M               A                                                       J                  L
                L                 J                                                       J                  G
                L                A                                                      A                 M
                G                J                                                       A                 L
                G               A                                                       A                 G
       -----------------------------                                        ----------------------------------
               = 6 ways                               +                                 = 6 ways

and 6+6 = 12

So now we add both of them and we get 12 different ways in total! :)

So far this is what I have in mind that I have learned over the week, and I hope you enjoyed browsing over my blog, and learning about Fundamental Counting Principles! Thank you!

Till next time ladies and gentlemen! Woo, Precal's fun and blogging! :)