# Week 15 : Adding and Subtracting Rational Expressions with Monomial Denominators  The strategies for adding and subtracting rational numbers can be used to add and subtarct rational expressions : write the expressions with common denominator.

For the example : $\frac {2x}{7}\div\frac{5}{3x}$

So the non-permissible value is x = 0

A common denominator is $21x$ $\frac {2x}{7}\div\frac{5}{3x}$

= $\frac {6x^2+35}{21x}$ # Week 12 : Solving each liner-quadratic system algebraically # Week 11 : Solving quadratic inequalities in one variable

This week we started with new chapter: Solving quadratic inequalities in one variable. I think it same with quadratic function. So we can factor to solve it. So now let’s start with example.

Example :

2x-8>0
2x>8
x>4

4 is boundary points # Week 10 : Review Unit Test

As I think, in all four chapters I have a clear grasp of the essential knowledge that only chapter 1: Arithmetic and Geometric Sequence. I need to review important knowledge to prepare for Unit Test.  # Week 9 : Equivalent forms

This week we learned about equivalent forms.

So In short this chapter.

We learned 3 form to write quadratic function :

First one : General form : $y=ax^2+bx+c$

Second one : Standard form : $y=a(x-q)^2+p$

And the last one : Factored form : $y=a(x-x1)(x-x2)$

Cause factored form is new one we just learned so i will introduce little bit about it. And beside how can we change from the general from to factored or standard from. I will show you right now. # 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. 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 First : Perfect Square Trinomials

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

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!!!!!