## Background Info

Name

Quadratic is a Latin term. It is originated from quadratus, the past participle of quadrare which means "to make square."

Who invented the Quadratic Formula ???

Euclid, a Greek Mathematician, created an intellectual concept during 300 BC. Both Pythagoras and Euclid created a well based procedure to find multiple solutions of the quadratic equation.

1. Parabola

A symmetrical open plane cure (a U-Shape).

2. Vertex

The highest point/tip

Format: The X coordinate comes first. The Y coordinate comes second. (X,Y)

3. Axis of Symmetry (AOS)

A line of symmetry for a graph. The axis of symmetry is considered as a mirror to split images on either side.

4. Optimal Value (Maxima/Minima)

The largest and smallest value/point on the parabola. (y=___)

5. X-Intercept

The point which the graph crosses the x-axis. (x,0)

6. Y-Intercept

The point which the graph graph crosses the y-axis. (0,y) Another way of saying X-Intercept is Zeros.

## Axis of Symmetry

In the vertex, the ''x" coordinate will forever be your "Axis of Symmetry", because that is where the vertical line can be formed to split a parabola into 2 identical equal parts.

Vertex Form Equation: y = a (x-h)^2 + k

Vertex: (h,k)

Axis of Symmetry: x = h

Small Note: The Axis of Symmetry is a vertical line, so it has to be written: x = __

## Optimal Value (Maxima/Minima)

In the vertex, the "y" coordinate will forever be your Optimal Value, because that is where your horizontal line can be formed to identify the parabola's highest or lowest value/point.

Vertex Form Equation: y = a(x-h)^2 + k

Vertex: (h,k)

Optimal Value: y = k

Small Note: The Optimal Value is a horizontal line, so it has to be written: y = __

## Transformations

There are 4 types of Transformations to occur in a parabola in Vertex Form.

1. Vertical Translation

2. Horizontal Translation

3. Vertical Stretch

4. Reflection

## X-Intercepts/Zeros & Y-Intercept

X-Intercept/Zeros

To find the X-intercept/ Zeros, you have to understand that the coordinate is (x,0). Since we know this, to find the X-Intercept, the Y value in the equation of the parabola must be 0 to solve for the X value.

y = a (x-h)^2 + k

Y-Intercept

To find the Y-intercept, you have to understand that the coordinate is (0,y). Since we know this, to find the Y-Intercept, the X value in the equation of the parabola must be 0 to solve for the Y value.

## X-Intercept/Zeros

Since the equation y = a (x-r) (x-s) is already factored, it is easy to find the X-Intercept right away.

For Instant:

y = (x + 13) y = (x-6)

0 = x + 13 0 = x -6

-13 = x 6 = x

## Axis of Symmetry

In the Factored Form Equation we are already been provided 2 X-Intercepts. You can easily determine the Axis of Symmetry by finding by finding the average of both of them. By accomplishing that, you can find the middle of the X-Intercept.

Formula:

y = a (x-r) (x-s)

Final Formula:

x = r+s divided by 2

For Instant:

y = (x-r) (x-s)

x = (x-6) (x-8)

x = 6 + 8 divided by 2

x = 7

## Optimal Value

To determine the Optimal Value (Maxima/Minima) or Y Value in the vertex, simply add/ substitute the X Value from the results you've received from solving the Axis of Symmetry. Substitute the X Value into the original factored form equation.

For Instant:

y = (x-r) (x-s)

x = (x-6) (x-8)

x = 6 + 8 divided by 2

x = 7

Substitute

y = (7-6) (7-8)

y = (1) (-1)

y = -1

Conclusion

The Optimal Value is y = -1. Also the vertex is (7,-1)

## Zeroes

If we are being provided a standard form equation, it is certainly possible into converting the equation to Factored Form. However, there is a risk which it might not always work. In order to not come across these types of problem, it is more efficient to use the Quadratic Formula.

Formula:

ax^2 + bx + c

Converted Formula:

b + - square root b^2 - 4(a)(c) divided by 2(a)

For Instant:

Figure out the following X-Intercepts for the Standard Form Equation 3^2 + 12x + 6

Procedures;

1st Step: Substitute the Standard Equation into the Quadratic Formula

-12 +- square root (12)^2 - 4(3)(6) divided by 2(3)

2nd Step: Continue solving it

12 +- square root 144 - 72 divided by 6 // -12 +- square root 72 divided by 6 // -12 +-

8.49 divided by 6 //

Final Step: @ the end you will have two different X-Intercepts

1st X-Intercept:-0.585 // 2nd X-Intercept: -20.49

## Axis of Symmetry

To find out the AOS in a Standard Form Equation, you need to follow the rule/formula of

-b/2(a)

For Instant:

When finding the AOS for the given Stand Form Equation y = 4x^2 + 12x + 8, you need:

1st: Make the value of B negative (y = 4x^2 + 12x + 8)

Turn 12 into -12

2nd: You must divide the new negative B value by 2(a)

-12 divided by 2(3)

3rd: Finally, you solve everything

-12 divided by 6 = -2

In Conclusion, the AOS of this equation is (-2)

## Optimal Value (Maxima/Minima)

In a Standard Form Equation, to figure out the Optimal Value, you need to plug in the results you've received in the AOS by using the formula provided above this text -b divided by 2(a). Finally you need to substitute that value in for x and solve the entire equation.

For Instant:

Using the value we've received for x, substitute that in and solve.

(4(-2)^2 + 12(-2) + 8)

8^2 - 24 + 8

64 - 24 +8

48

In Conclusion, your Optimal Value is 48.

## Completing the square to turn to vertex form

What you have to do is very simple. All you have to do is take the Vertex Formula(y = a(x-h)^2 + k) and subtract it by ax^2 + bx +c

## Types of Factoring Equations

1. Common Factoring

2. Simple Trinomial

3. Complex Trinomial

4. Perfect Square

5. Differences of Squares

Simple Trinomials
Complex Trinomials

## Reflection & Assessment

In this unit, I've learned an incredible amount of knowledge about Quadratics. For example, I've learned new terminologies, graph a parabola, find x // y intercept, factor as many equations, etc. I kinda have a grasp with the application problems. The area that i most struggle with. Thanks to Mr. Anusic I believe some what i have improved. Without his help I wouldn't understand this unit. I struggled with the different types of transformations in a parabola.To sum up everything i've thoroughly enjoyed this unit and have learned alot.
Even tho I've received a fairly well mark I would like to acknowledge Mr Anusic to help me understand perfect square's and how to make your own steps to explain specific things in communication category. I improved on my guess and check ability and how to figure out perfect squares.