# Week 14 -Rational Expressions and Equations

This week we learned how to make a function simply

For example:

$\frac{x^2-9}{x^2+8x+15}$

we need to factor first

so $\frac{(x+3)(x-3)}{(x+3)(x+5)}$

we can see the same thing from the top and the bottom, so we can get rid of them

so it is (x+3)

After that we can get $\frac{(x-3)}{(x+5)}$

Next step we have to make sure the range of values of “x” and what number can’t it become,

First we can see this formula:$\frac{(x+3)(x-3)}{(x+3)(x+5)}$ for values of bottom can’t be “0”

“x “can’t be -3 and -5

second we can see this formula :$\frac{(x-3)}{(x+5)}$ ,in the bottom x can’t be  -5

WARNING : you have to check the all function , even the first one ,one special is start by diving by a fraction , and then you have to multiply by the inverse of that fraction , and you have to look at the denominator of that fraction to make sure that the unknow numbers  before and after reciprocal and which number can’t be.(“0″ CAN’T BE THE BOTTOM”)

This week we learned reciprocal quadratic functions graph

So when we get the function, we have to graph the (parent function)quadratic formula first ,and then we have to find the invariant points ,which is when y=1 . y=-1 ,the points that on the parent function graph ,we can see this picture don’t have find the vertical asymptotes ,so there is no vertical asymptotes, because the x=0 is the asymptotes already so we don’t have to draw another one ,just have the horizontal asymptotes y=0 (in this term we just earn y=0)

First we also we have draw the parent formula, and there is no invariant point.we can find the reciprocal of number in the end (2) to find a point that the reciprocal of the quadratic function touch in the y line(x=0)

Ok so we draw the parent function again , find the x-intercept to find the vertical asymptotes (two) ,and there have the four invariant numbers ,and for the third graph line that function of the reciprocal of the quadratic function ,you also can find the point that reciprocal of the end number (four to be one over four:0.25) that where it touch at the y line(x=0)

# Week 12-Solving Absolute value Equations

This we learn how to solve the solution of the absolute value Equation

First we learned how to use the graph to solve

Ex

this question is asking that what’s the x solution are by looking the graph

we can see the equation

|-3x+3|=9

The y intercept is 3

So there two possible inside the absolute

First:-3x+3=9

-3x=6

x=-2

Second:-(-3x+3)=9

3x-3=9

3x=12

x=4

And because of when the y=9 touch the linear the x=-2,4

so the solution are -2,4.

EX:Another example is about the x square

Formula:$4+2x=|x^2 +8x+12|$

First possible

x is positive

(x+2)(x+4)=0

x=-2,x=-4

Second posible

x is negative

(x+2)(x+8)=0

x=-2,x=-8

Because absolute can’t be the negative.so when x=-4 ,x=-8

4+2x is negative , so they can’t be the solution of this,

so the solution is x=-2

# Week 11-Graphing Linear inequalities in two Variables

PART 1:If you want to slove the graph is in which part of, you can use the test point

Ex:

3x-2y>(=)-16

Is (-3, 4) can be in this part?

so 3(-3)-2(4)>(=)-16

-9-8>(=)-16

-17>(=)-16

From there we can see this point can be the formula used

so if this point is outside, so you can draw the part inside(can be use), so if this point is inside, you can draw outside(can be use).

WARN:And don’t forgot if the formula have (=) the line should be the soild line, if there is no (
=),so the line should be the broken line!

PART 2:SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY

How much solution can linear system have?

There possible:1 , countless ,no

And we leaned the possible of two quadratic functions

one, two and none

two solution is the max

same with the linear function with the quadratic function

(use the desmos if you can’t see the solutions(the point or points that they meet)

# Week 10-solving Quadratic inequalities in one variable

The formula that we learned

$ax^2 +bx+c<0$ $ax^2 +bx+c>0$ $ax^2 +bx+c<(=)0$ $ax^2 +bx+c>(=)0$

where a, b, and c are constants and a not =0

Ex:

The solution of the quadratic inequality

$x^2 -2x-3>0$ is the value of x for which y>0 ; that is ,the values of x for which the graph is above the x -axis.Visualize the shadow of the graph on the x-axis.

so we first have to factor this formula

$x^2 -2x-3=0$

(x-3)(x+1)=0

ok, so we can get the value of x for equal to zero

x=3,x=-1

and because  y>0

so the part of y>0 are

-1>x>3

# Week 9-Modelling and Solving Problems with Quadratic Functions

This week we learned three formula form

General Form :$y=ax^2+bx+c$

Standard Form(Vertex form:Ms.Burton’s Favorite):$y=a(x-p)^2 +q$

X point form(x intercept form):y=(x-x_1)(x-x_2)

Ex:the rectangular have the area is to be bounded by 120 m of fencing(determine max area)

so we can get two formula

2L+2W=120

L*W=mix

so we can use the first formula get this form

L+W=60

L=60-W

SO PUT THIS FORMULA INTO THE SECOND FORMULA

(60-W)*W=MIX

We can get

$-(w-30)^2 +900=MIX$

So the vertex is (30,900):the 30 m is the w , and the $900 m^2$ is the MIX area

so we can use the formula  L=60-W ,to get the L=30

so it is a square .the special rectangular

# week 8-Analyzing Quadratic Functions of form

A quadratic function is any function that can be written in the form $y=ax^2+bx+c$, where a, b ,and c=R and a didn’t =0.

This is called the general form of the equation of a quadratic function.

The graph of ever quadratic function is a curve called a parabola.

The vertex of a parabola is its highest or lowest point . the vertex may be a minimum point or a maximum point.

The X-intercept is mean the points that when the parabola touch the horizontal line and the Y-intercept is mean the points that when the parabola touch the vertical line.

Analyzing quadratic functions of the form:

$y=x^2$ is the parents function,

The vertex is (0,0)， X-intercept is (0,0),Y-Intercept is (0,0)

We learned a new one:$Y =x^2+R$

This R is a positive number, if this number change to the more and more big, it will make the parents function goes up (+R),and the vertex goes up too.

The second one is :$y=x^2-R$

This R is a positive number,if this number change to the more and more small, it will make the parents function goes down(-R),and the vertex goes down too.

The third one is $y=ax^2$ , |a|>0

when |a|>1, the parabola will be more and more skinnier with the ‘a’number goes more and more big,

when |a|<1,the parabola will be more and more compression with the number goes more and more small.

WARN: If the “a” is a negative number the parabola opens down, if the “a “is a positive number , the parabola opens up

The fourth function is $y=(x+R)^2$

The R is a positive number, so the size of this number is the distance to the left the parents function

The fifth function is $y=(x-R)^2$

The R is a negative number, the size of this number is the distance to the right the parents function

THANK YOU FOR READING, HOPE YOU LEARN MANY FROM MY BLOG POST!

# Week 7-review

This week we did the review and skill check for the test

The important thing that we learned was discriminant:

$b^2-4ac=0$

There are one solution(2 equal solutions)

$b^2-4ac>0$

There are two distinct(unequal)solutions

$latex b^2-4ac<0$

There are no roots.

look the picture

# Week 6——Developing and Applying the Quadratic Formula

use the way that we learned to make ax²+bx+c=0 change  to

$\frac{-b+-\sqrt{b^2-4ac}}{2a}$

Ex:

x^2+9x -2=0

use the formula that we just make

a=1，b=9，c=-2

$\frac{-9+-\sqrt{9^2-4\cdot1\cdot-2}}{2\cdot1}$ $\frac{-9+-\sqrt{81+8}}{2}$ $\frac{-9+-\sqrt{89}}{2}$

finished

so we can use this formula to solve the question easily and correctly

### Aside

Add the square root of the same coefficient

ex: 1-3√5x=-3-2√5x

-3√5x+2√5x=-3-1

-√5x=-4

Than use the way that square both sides of this to get rid of the square root

5x=16

x=16/5

Review

$\frac{2\sqrt{6}}{\sqrt{7}+\sqrt{5}}$

[Than multiply both the top and the bottom by a number , which is goodfor removing the square root of the denominator(($\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})$=3-2=1)]

=$\frac{2\sqrt{6}(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}$

=$\frac{2\sqrt{42}-2\sqrt{30}}{7-5}$

=$\frac{2\sqrt{42}-2\sqrt{30}}{2}$

=$\sqrt{42}-\sqrt{30}$