Week 14 : Equivalent Rational Expression

This week we started with new chapter about rational expressions and equations.

So we will start with example so i can show you more clear about knowledge of it.

Week 13 : Reciprocal Linear Functions

This week we learned how to graphing recipropals of Quadratic Function

So let’s see how to graph it.

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.

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

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!!!!!

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.