Week 8 : Properties of Quadratic Functions

This week we started with new lesson : Graphing Quadraction Functions about the determine the value of vertex, formula, what the graph looks like and how to draw the parabola.

So let’s start the things I studies in this week :

y=ax^2+bx+c : It is general form of quadratic
With a (coefficent) it will helps us know the graph will be big or small.

Also, we can know : If quadratic positive y=x^2 , the parabola will be go up (+) and contrast if quadratic negative y=-x^2 , the parabola will go down.

It will be like this in graphing:

Vertex : highest and lowest point (-1,4)

Axis of Symmetry : which the parabola is symmetric (-1) of above picture

x-intercepts : zero of function or we can determine it by Quadractic Formula

y-intercepts : it depends on c

Maximum point : when the graph opens down Because the intersection point between x and y is at the top

Minimum point : when the graph opens up so the intersection point between x and y is at the bottom

Pattern of Parent Function :

y=x^2 it will show stretch / compress : 1,3,5,7,9..

But if y=2x^2 the stretch / compress : 2,4,6,10..

Domain : the value of x and the complete set of possible values of the independent variable, make sure it is real number.

Range : the value of y

(x-p)^2 : depends on the value of p, It will move to right or left

Let’s star with example :

(x+7)^2 : when p is positive the vertex move to left
(x-7)^2 : when p is negative the verter move to right

Week 6 : Perfect square trinomials – The quadratic formula

Today, we will start with perfect square trinomials and The quadratic Formula

First : Perfect Square Trinomials
I will help you know more about that with first example :

We have already discussed perfect square trinomials:

(a+b)^2= a^2+2ab+b^2
(a-b)^2= a^2-2ab+b^2

We know : a^2 : Square of first term of binomial
2ab : twice the product of binomial’s first and last terms
b^2 : Square of last term of binomial

Let’s start with example :

Factor : x^2+12x+36

Like my way, i always try the last term the numbers multiply together to get it like this :

36×1
18×2
12×3
9×4
6×6

I will choose 6×6 because if they multiply together, i will get 36 and when they add together, i will get 12 like the exercise i gave above.

You can do it faster than my way with perfect square.

Answer: (x+6)(x+6) or (x+6)

Example 2 : 9x^2-6x+1

The leading coefficient is not 1 (x^2). Its 9 but Both 9x^2 and 1 are perfect squares, and 6x is twice the product of 3x and 1.

So we will know a = 3a and b = 1.

Then get the answer : (3x-1)^2 or (3x-1)(3x-1)

Quadratic Formula : We will use this formula when we can not use Factoring and Binomial with complicated exercise. This formula will help us get answer easier and faster.

x(1) = (-b+-(b^2-4ac)\div(2a)
And now let’s start :

2x^2+6x+9=0

We knew : a=2
b=6
c=9

We will apply this formula :

Week 5 : Factoring Polynomials

This week we start to learn the new unit. That’s factoring polynomial expressions
Let’s star with the first example :
x^2-4x-12

If the leading coefficient is 1 like this. The process to do this exercise is so easy. The only two numbers have sum is −4 and that multiply them to give −12 are −6 and 2. So if you wanna find the number have sum = -4 from -12 you can try by this way :
-12×1
-6×2
-4×3
-3×4

So right now we will chosse one into them. Which ones will help us get -4. That’s -6 and 2 ( -6+2=4 ).

So let’s try with another example :

x^2-10x+24

It has a leading coefficient of 1, find two numbers with a product of 24 and a sum of −10.

Same as above exercise we will try each number multiply to get 24

24×1
12×2
8×3
4×6

Which one will help us get 10. Thats last. 4+6

Then we will get x^2-4x-6x+24. So we got it!!!!

Example 3 : 2x^2+9x-5

This example diffirent with two example exercise cause the leading coefficient is not 1 (x^2) . We still find two numbers, and those numbers will still add up to 9.

2x^2+10x-x-5

RIght? Cause 10x-1=9x. Its same but i change it to be easy to a simpler equation

Then we will distribute them together then get :

2x(x+5)-1(x+5)

Between 2x^2+10x the same point is 2x then we will divide each number for 2x to get (x+5)

Same way with -1(x+5)

Finally we will make it simpler by the way, they have same (x+5). We will make the general figures between them.

(x+5)(2x-1)

Done!!!!!

Week 4 : Simplifying And Adding Radicals

Hi guys,Today i will tell you about the Simplifying and Adding radicals.

So let’s start with the first example :

Adding : 2\sqrt{11} + 8\sqrt{11}

So we can know : the two radicals inside the roots. they have same the number “11”. This means you can combine them but only combine the number before the roots. We will still keep the radicals inside roots

Let’s go to another example :

8\sqrt{5} + \sqrt{3} + 3\sqrt{5} + 10\sqrt{3}

So with this example we can not adding with diffirent radicals. First, I always sort the number have same radicals are next to each other. Then we will adding each number have the same radicals. We will get like this :

8\sqrt{5} + 3\sqrt{5} + \sqrt{3} + 10\sqrt{3}= 11\sqrt{5} + 11\sqrt{3}

Just remember, never do combine the number have diffirent radicals like
8\sqrt{5} + \sqrt{3} and then get 9\sqrt{8} . It is the big mistake if you combine it together.

This is not correct because \sqrt{5} and \sqrt{3} are not same radicals so we can not be added them together.

Week 3 : Absolute Value Of A Real Number – Simplyfying Radicals

The absolute value of a number is its distance from 0
Example : The absolute value of 3 or -3 is 3 because they have same distance from 0 and absolute value of 3.

The symbol for absolute value is a bar |∣vertical bar on each side of the number.

Example : The absolute value of 25, we can write it like this : |25|=25

Simplyfying Radicals :

A radical is a number that has a fraction as its exponent.

Example :

\sqrt{20}= \sqrt{4.5}
\sqrt{20}=2\sqrt{5}

\sqrt{20} : Intire
2\sqrt{5} : a mix of whole numbers and radicals

Another example : \sqrt{75}= \sqrt{25.3}
\sqrt{75}=5\sqrt(3)

Ex: \sqrt[3]{96}= \sqrt[3]{16.6}=2\sqrt[3]{12}

\sqrt[3]{-81}= \sqrt[3]{-27.3}=-3\sqrt[3]{3}

Week 2 : Geometric Sequence

Geometric Sequences
In a Geometric Sequence each term is found by multiplying the previous term by a constant.

Example : 2,6,18,54, 163,486,1458,…

This sequence has a factor of 2 between each number

Each term is found by multiplying the previous term by 3 ( except the first term : 2 )

Infinite Gemetric Series :

In Infinite Geoemtric Series, when : r>1 or r<-1 it will be converges Let's see my example, and see what happends : 1,1/3,1/9,1/27,.. We have: a = 1 (the first term) r = 1/3 (halves each time) And so use this formula : [latex]S_ \infty[/latex] [latex]S_ \infty[/latex]=1/1-1/3 [latex]S_ \infty[/latex]=2/3 (0,666666..)

Week1 : Arithmetic Sequences

t_n t_{50} S_n=\frac{n}{2}(t_1+t_n)

In an Arithmetic Sequence the difference between one term and the next is a constant.

In General we could write an arithmetic sequence like this:

{at, t+d, t+2d, t+3d, … }

where:

t is the first term, and
d is the difference between the terms (called the “common difference”)
Example : My Arithmetic Sequences : 11,2,-7,-16,-25,-34

Know :
– t = 11 (the first term)
– d = -9 (the “common difference” between terms)

We can write an Arithmetic Sequence as a formula :
t_n= t1+(n-1).d

Using the Arithmetic Sequence formula:

tn=t1+(n-1).d
tn=11+(n-1).(-9)
-34=11+9-9n
-34=20-9n
n=(6)

Example 2 : I want to determine t50
I know : t1= 11
n=50
n-1=50-1=49
Using the Arithmetic Sequence Rule :

t_{50}= t1+49.d
t_{50}= 11+49.(-9)
t_{50}= -430

To sum up the terms of this arithmetic sequence:

t1 + (t2+d) + (t3+2d) + (t4+3d) + …

Use this formula: S_n=\frac{n}{2}(t_1+t_n)

Example: Determine the sum of this Arithmetic Sequence : 11,2,-7,-16,-25,-34

The values of t,d and n are:

t = 11 (the first term)
d = -9 (the “common difference” between terms)
n (how many terms to add up)

S_n=\frac{n}{2}(t_1+t_n)
S_n=\frac{6}{2}(11+-34)
S_n=(3.-23)
S_n=(-69)