Quadratics

How could we defend our castle?

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.

Basic Quadratic Terminology

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.

Vertex Form

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.

Step Pattern

Factored Form

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)

Standard Form

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

Factoring

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.
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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.