So far in this unit, I have had only one thing that is troubling me and that is writing the equations on piecewise notation. I understand how to write the equation by the thing that is confusing for me is writing what is greater than or less than or equal to.

Down below you will see an example of Piecewise notation:

Graph y=|-2x+5| and write in piecewise notation

To find how this graph looks in piecewise notation we must first find the reciprocal of the equation. To find that I must multiply everything by -1.

f(x)= {-2x+5 { 2x-5

Now we must find if it is greater than or less than or equal to a number. To find this we need to look at the graph, first out x intercept or critical point is on 2.5 so our first so our first value will be x>=2.5. And for the bottom one we will get x < 2.5

So our final piecewise notaion will look something like this:

f(x)= {-2x+5 , x>=2.5 { 2x-5 , x < 2.5

]]>$latex x^2=y – 2x – 8

The first step is to choose a system that will be your base for the equations. I Chose y=2x+5

Next, organize the other equation to the y=mx +b format

Next, you must put it into your base equations

Then you let algebra take over and solve for x

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Linear inequalities will give us a straight line on a graph. When we are given inequalities we are usually given a y value that is greater than or equal to another value, that consists of an x variable. We can also be given a greater than/less than or equal to symbol. This greater than or less than help us decide which side of the line contains the real answers, which we shade in. We also need to determine if the line is broken or solid. The way we can determine that it is greater/less than (>,<) then it is a broken line. If it is greater/less than or equal to .

If we are given the equation of 3x + 7 < y. Just from looking at this equation we can determine the y-intercept which is 7. We can also determine the slope which is 3/1, which means it goes up 3 and over 1 each time.

(Insert Picture Here)

To know whether or not which side of the line contains the real answer, we must perform a test. We will insert the point (0,0) into the equation if it comes out there (left side equals to the right side). If we insert 0 then we will get that 7<0 which is false so we would shade in the right side of the line.

(Insert Picture Here)

The last and final thing that I need to mention is that you need to remember that one must remember to switch the signs if the coefficient of x is negative.

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- where (p,q) is the vertex coordinate
- where c represents the y-intercept
- Where x1 and x2 represent the x-intercepts

**Example:**

**Step 1: Find Vertex Form**

The first step we see how the vertex has transformed from (0,0). In this case, the p-value is -3 and the q value is -8. If we insert these numbers into the standard form we will be left with:

Now we just need to find the “a” value. According to the principle equation, the graph should go up by1 then by 3 and then by 5, etc. In this case, we see that the graph goes up by 2 and over one this means that the a value is congruent to

**Step 1: Convert To General Form**

Convert it from vertex form to general form. It is quite simple all you have to do is expand your equation and simplify again. Below you will see the process.

**Now you have the General Form of the equation which gives you your y-intercept**

**Step 3: Find Factored Form**

The reason to use factored for is to find x-intercepts to verify the position of your graph. Its really quite simple all one has to do is factor the equation in from the general form.

Take out the 2 from the equation:

Now you must factor what is in the prackets to create two sets of brackets like so:

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The Are 3 possible equations you can use:

The first formula adds q to the parent function which was listed first, this decides the Y-Intercept and so the parent either goes up or down when it is implemented.

The Second one changes the X-Intercept and changes the parabola location from side to side if p is being subtracted from the parabola the intercept moves right and the opposite happens when added.

The third function decides how wide or how skinny a parabola is, the higher the number the skinnier the shape. But if the number is smaller parabola will get wider. If ‘a’ is negative it reflects just like any function.

Below I will show examples for each of the 3 functions you can use. Please take into consideration how the parabola transforms.

As you can see the blue line is the principle function but when the 2 is added the line transforms 2 spaces up

As you can see parabola moves over to the right 4 spaces because the q value was negative.

and , in this example there are two variations. The 2 is represented in green and the decimal is represented by the color blue.

As you can see the smaller number is the wider the parabola gets (blue line) but as the value of “a” gets larger the parabola gets skinnier.

]]>- x > 0 (Then it has 2 solutions)
- x = 0 (Then it has 1 solution)
- x < 0 (Then it has no solutions)

When given the example

From the equation, we see a= 6, b = 10 and c= -1

Plugging these values into the discriminant, we get:

$latex b2−4ac $

$latex 10^2−4(6)(−1) $

= 124

This is a positive number, so the quadratic has two solutions.

This makes sense if we think about the corresponding graph.

Notice how it crosses the

-axis at two points.In other words, there are two solutions that have a

-value of 0, so there must be two solutions

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- Factoring the equation
- Completing the square
- Quadratic Formula

The areas that I found an still find the most difficult is solving a quadratic equation using the completion of squares method. I find this the hardest out of the three because there is almost no pattern and you tend to do many things to the formula at once. To use any of these three methods you must have the equations equal to zero making it quadratic. The Equation does not always need to have a degree of 2 it can be also solved to the degree of 1. To check your answers plug your (x) value back into the formula and make sure the left side equals the right.

Below you can find examples of the 3 methods of solving for a quadratic formula.

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For Example: x2 + 7x + 12

If you go through the factors of 12 you must find two number that adds to the second value in the expression(7)

1 * 12

2 * 6

3 * 4

4 * 3

Using the factors highlighted in red input them into your expression because 3 + 4 equals to 12

(x + 3) (x +4)

We were also taught what to do if the x2 value is greater than one. Ex: 2×2 + x – 6

To solve this we were introduced to the box method.

- Multiply 6 and 2x and find its factor. (12x) 3 and 4
- Fill in the first value in the top right corner
- Fill in the constant in the bottom right
- Fill in the two factors in the remaining corners
- Lastly find the common factor between the values from up to down and left to right

Claculations:

- 2×2 and 4x – 2x
- -3x and -6 – -3
- 2×2 and -3x – x
- 4x and -6 – 2

Final Answer: (2x – 3) (x + 2)

]]>Still, in this unit, we have not used a single formula because of the fact that throughout this unit we are not allowed to use a calculator.

Below you will see examples of all adding, subtracting, multiplying and dividing of radicals.

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