# Quadratic Relationships

### Garthi Gobinathan

## Quadratic Relations Significance to Reality

Quadratics can be presented in a wide range of forms, however, being able to apply and understand different forms and information is my objective for you to comprehend this topic. Use this resource as a guide to lessons and to acquire a further understanding of any desired subject matter.

## Table of Contents

**Introduction to Quadratics**

- Key features of Quadratic relations
- Introduction to parabolas
- Ways to represent a Quadratic relation

**Types Of Equations**

- Vertex form
- Factored form
- Standard form

**Vertex form**

- Terminology
- Graphing Using Vertex Form
- Connection to Step Pattern
- Finding Equation and Analyzing Vertex Form

**Factored Form **

- Terminology
- Identifying Additional Information
- Example of Graphing in Factored Form and Its Relations
- Solving by Factored Form
- Algebraic Tiles
- Expanding and Simplifying

**Standard Form**

- Determining The Zeroes with The Quadratic Formula
- The Discriminant
- Determining Axis of Symmetry
- Determining Optimal Value
- Converting to Vertex Form (Completing the Square)
- Factoring from Standard Form

- Common Factoring
- Binomial Common Factoring
- Simple trinomial
- Complex Trinomial
- Special Cases
- Perfect Squares
- Difference of Squares

**Word Problems **

- Optimization
- Motion
- Revenue
- Geometric

**Reflection**

**Resources **

## Introduction to Quadratics

## Key Features of Quadratic Relations

**Vertex**

Definition : The maximum or minimum point on the graph. This point also determines the direction of opening of the parabola, whether being up or down.

Labeling it : (x,y)

**Minimum or Maximum Value (Optimal Value)**

Definition : The lowest or highest coordinate on the parabola.

Labeling it : y=

**Axis of Symmetry **

Labeling it : x=

**Y- Intercepts**

Definition : The coordinates that the parabola meets the y- axis on the graph.

Labeling it : y=

**X- Intercepts (Zeroes)**

Definition : The coordinates that the parabola meets the x- axis on the graph.

Labeling it : x=

## Introduction to Parabolas

**Basic Principles of Parabolas**

- Parabolas can open in the direction of up or down.
The zero of a parabola is where the graph crosses the x-axis.

Zeroes can also be called "x-intercepts".

Axis of symmetry divides the parabola into two halves .

The lowest or highest coordinate on the parabola is the optimal value.

The coordinate where the axis of symmetry and the optimal value meet is the vertex, the maximum or minimum point on the graph.

The y- intercepts is the coordinates where the parabola meets the y- axis.

## Ways to Represent a Parabola

**Tables of Value**

In quadratic relationships, when comparing the differences on a table of values, the second difference must be consistent throughout. If the difference is consistent on the first difference, it indicates it is a linear relationship.

**Graphing**

To represent a quadratic relationship on a graph, it is in a form of a parabola.

**Equation**

For basics, when there is an exponent of two in the equation, it indicates it is a quadratic relationship. However there are other types of equations to represent quadratic relationships, vertex form, factored form and standard form.

## Types of Equations

## Vertex Form

The vertex form is most simplified as y=x^2. To expand beyond this basic knowledge, there are also a more complicated form containing all components of vertical translation, horizontal translation and a stretch/compression factor, this equation is formatted as y=a(x-h)^2+k.

## Terminology

**What Does 'k' Represent?**

**k**, the 'k' represents the vertical translation. This determines whether the vertex will be shifting up or down on the y-axis, also basically determining the optimal value of the graph.

**What Does 'h' Represent?**

In the equation of y=a(x-**h**)^2+k, the 'h' represents the horizontal translation. This determines whether the graph's vertex will be shifting right or left,also basically determining the axis of symmetry of the parabola.

However, keep in mind that if the 'h' is negative, then it is identified to be moving right, if it is positive, it will be moving left.

Example

For instance, if h= -12, that would mean it would shift +12.

**What Does 'a' Represent?**

In the equation of y=**a**(x-h)^2+k, the 'a' represents the stretch/ compression factor of the parabola. This also determined the direction of opening of the graph, either up or down.

## Graphing using Vertex Form

**Example Question :*** Graph **y=2(x-3)^2+5*

To graph a quadratic relationship in vertex form, such as y=2(x-3)^2+5, you must use all the information given and be able to understand and comprehend how to apply it into a graph.

1. First identify the 'y' coordinate of the vertex using the 'k' value within the equation. In this case, five is the 'y' value, meaning the graph is moving five units up.

2. Secondly, to figure out the 'x' coordinate of the parabolas vertex, you use the 'h' value and switch signs to identify the correct coordinate. For this instance, the 'h' value is negative three, which means it is positive three when you graph. Therefore the 'x' value is 3 units right of the vertex.

3. Use the basic information given so far and plot the vertex.

4. Finally, use the 'a' value to determine whether the parabola is moving upwards or downwards and the compression/ stretch factor. Moreover, if the 'a' value is negative, then it is facing downwards, however if it is positive, it is upwards. Also, to determine the compression/ stretch factor by the definite value. In this case, the parabola is facing upwards and the stretch factor is two.

To have a deeper understanding of graphing various compression/stretch factors, it is essential to understand the method of step patterning.

In this case, when you go over one, you go up one. however, when you go over two, you go up four. Further on, when you go over three, you get up nine. Basically, substitute the 'x' value with numbers in a sequential order to identify the relationship. Afterwards, connect the coordinates and you have successfully created the parabola for y=x^2. Ultimately, the substitute the 'x' value with your desired 'x' coordinate and solve the equation to determine the 'y' coordinate.

Connecting back to our original problem, we used our basic understanding of step patterning and have have substituted the 'x' coordinates and have plotted the coordinated to both sides of the parabola. The image of the parabola below represents the graph to y=2(x-3)^2+5.

## Finding Equation and Analyzing Vertex Form

**Exploring the affects of 'a' in y=ax^2**

## Factored Form

## Terminology

**What Does 'r' and 's' Represent?**

In factored form, 'r' and 's' are considered the x-intercepts or the zeroes, however when graphing the x-intercepts are opposites to their signs.

Example

For instance, in y=(x-7)(x-3), the x- intercepts are positive seven and positive three.

**What Does 'a' Represent?**

In relation to the vertex form, 'a' has control to the stretch/ compression factor of the parabola and the direction of opening of the graph, either up or down. Also in addition it can impact the optimal value of the parabola in factored form.

## Identifying Additional Information

**How Do I Determine the Axis of Symmetry?**

To determine the axis of symmetry, you can follow a simple formula of x=(r+s)/2. To simplify, you must add both of the x-intercepts to receive the x-coordinate to the axis of symmetry, however they must be added in opposite of their signs. Then divide the results by two to determine the axis of symmetry.

**How Do I Determine the Optimal Value?**

Using the information from the axis of symmetry, you substitute the 'x' value in y=a(x-r)(x-s) and solve the equation to determine the optimal value.

## Example of Graphing in Factored Form and Its Relations

**Question : Graph y=3(x-2)(x-8).**

**Find Axis of Symmetry. **

Determine x-intercepts.

x= 2, x=8

Sub x-intercepts in axis of symmetry formula.

x=(r+s)/2

x=(2+8)/2

x=10/2

x=5

Axis of symmetry is x=5

**Find Optimal Value**

Sub the axis of symmetry value in the original equation.

y=3(5-2)(5-8)

y=3(3)(-3)

y=3(-9)

*At this stage, we are multiplying the results so far with the 'a' value, for 'a' has the impact to change the overall orientation and stretch/compression value, hence adjusting the 'a' value would drastically change the optimal value.

y=-27

The optimal value is y=-27.

**Determine The Vertex.**

Use the values of the axis of symmetry and the optimal value as (x,y).

x=5 and y=-27

(5,-27)

**Graph Using Required Information **

Plot the vertex and the x-intercepts, then finally connect the dots. View image below to understand the final product as a visual representation.

## Solving by Factored Form

**Example**

If given to solve (x)(y)=0, then it means that either 'x' is zero or 'y' is zero, therefore recieving a product of zero.

**Question : Solve 0= x^2+5x+4 By Factoring**

**Convert Standard Form into Factored Form.**

0= x^2+5x+4

0=(x+4)(x+1)

**Set Each Variable To Zero To Determine Possible Outcomes.**

0=x+4, 0=x+1

**Isolate For 'x'.**

-4=x, -1=x

**Check By Substituting 'x's into the Original Equation. **

x=-4

0= x^2+5x+4

0= (-4)^2+5(-4)+4

0= 16-20+4

0=0

x=-1

0= x^2+5x+4

0= (-1)^2+5(-1)+4

0= 1-5+4

0=0

Therefore, since the right side equals to the left side it confirms that our answer is correct.

## Algebraic Tiles

quadrilateral dimensions, it helps to understand and comprehend the factored formula. The following video uses the concept of algebraic tiles to teach how expanding and factoring work.

## Expanding and Simplifying

**Example : Expand and Simplify (x+3)(x+4) using FOIL.**

**1. F**OIL **(Firsts)**

(**x**+3)(**x**+4)

x^2

**2.** F**O**IL **(Outer)**

(**x**+3)(x+**4**)

x^2+4x

**3.** FO**I**L **(Inner)**

(x+**3**)(**x**+4)

x^2+4x+3x

**4.** FOI**L (Lasts)**

(x+**3**)(x+**4**)

x^2+4x+3x+12

**5. Simplify **

x^2+4x+3x+12

x^2+7x+12

Therefore the expanded and simplified version to the original question is x^2+7x+12.

## Standard Form

## Determining The Zeroes with The Quadratic Formula

**1. Firstly, identify each of the variables in the equation of 'a', 'b' and 'c'.**

y=5x^2+6x+1

a= 5 b=6 c=1

**2. Substitute the values in the quadratic formula**

x=(-b+/-√b^2-4ac)/2a

*To clarify, the reason there is +/- in x=(-b**+/-**√b^2-4ac)/2a, is because there are two outcomes and each is to add and subtract to have two different x-intercepts. As we continue, I will have a subheading of 'Solution 1' and 'Solution 2' to indicate the two different x-intercepts.

Part 1

x=(-6+√6^2-4(5)(1))/2(5)

Part 2

x=(-6-√6^2-4(5)(1))/2(5)

**3. Solve each equation to determine x-intercepts**

Part 1

x=(-6+√6^2-4(5)(1))/2(5)

x=(-6+√36-20)/10

x=(-6+√16)/10

x=(-6+4)/10

x=-2/10

x=-0.2

Part 2

x=(-6-√6^2-4(5)(1))/2(5)

x=(-6-√36-20)/10

x=(-6-√16)/10

x=(-6-4)/10

x=-10/10

x=-1

**5. The x-intercepts or zeroes are x=-0.2 and x=-1. **

## The Discriminant

- If the discriminant value is negative, it indicates that there is no x-intercept. Therefore in graphed form, the parabola does not meet the x-axis.
- If the discriminant value is zero, it indicates that there is only one x-intercept. Therefore in graphed form, the parabola only meets the x-axis once,at the vertex.
- If the discriminant value is above zero, it indicates that there is only two x-intercept. Therefore in graphed form, the parabola only meets the x-axis twice.

## Determining Axis of Symmetry

**1. Identify the required values for the variables. **

y=5x^2+6x+1

a= 5 b=6 c=1

**2. Substitute the essential values in y=-b/2a.**

y=-b/2a

y=-6/2(5)

y=-6/10

y=-0.6

**3. The axis of symmetry is x=-0.6. **

## Determining The Optimal Value

**1. Sub in the 'x' value in the standard form equation.**

y=5x^2+6x+1

y=5(-0.6)^2+6(-0.6)+1

y=5(0.36)+(-3.6)+1

y=1.8+(-3.6)+1

y=-0.8

**2. The optimal value is y=-0.8.**

**Overall, the vertex is (-0.6, -0.8) and two x-intercepts of x=-0.2 and x=-1 **

The image below is a visual representation of an parabola based on the example equation, to prove my solution is correct.

## Converting Standard to Vertex Form (Completing the Square)

**Question : Complete the square of y=x^2-6x+4**

**1. Group the x-terms **

y=x^2-6x+4

y=(x^2-6)+4

**2. Divide the coefficient of the middle term by two, square it, then add and subtract that number within the bracket. **

y=(x^2-6x)+4

(6/2)^2 = 9

y=(x^2-6x+9-9)+4

**3. Remove the subtracted term from the brackets. **

y=(x^2-6x+9)-9+4

y=(x^2-6x+9)-5

**4. Factor the brackets (trinomial) as a perfect square trinomial. **

y=(x^2-6x+9)-5

**Therefore, the vertex form of the original equation is y=(x-3)^2-5.**

To convert from standard form to vertex form, we can easily use a method called 'completing the square'. However, this time, we will have an 'a' value larger than one , therefore we would need to common factor it out.

**Question : Complete the square to y=2x^2+12x+13.**

**1. Group the x-terms. **

y=2x^2+12x+13

y=(2x^2+12x)+13

**2. Common factor out the 'a' value and the 'b' value. However, do not factor out the 'x'. **

y=(2x^2+12x)+13

y=2(x^2+6x)+13

**3. ****Divide the coefficient of the middle term by two, square it, then add and subtract that number within the bracket.**

y=2(x^2+6x)+13

(6/2)^2=9

y=2(x^2+6x+9-9)+13

**4.**** Remove the subtracted term from the brackets. Also, multiply it by the 'a' value on the way out. **

y=2(x^2+6x+9-9)+13

y=2(x^2+6x+9)-18+13

y=2(x^2+6x+9)-5

**5. Write a perfect square trinomial as the square of a binomial.**

** **y=2(x+3)^2-5

The vertex form of the original equation is **y=2(x+3)^2-5. **

## Factoring from Standard Form

## Word Problems

## Reflection

In focus of my test results below, I had done average on knowledge and understanding portion of the assessment, however I did quite poorly on the application part. If possible, I would have tried to practice more to reinforce my understanding of the concepts. Also, the most recent I have received, I have got a level of 4+ on the assessment focussing on the different types of factoring. This proves my understanding of the concept through my ability to explain all seven types of situations. Also, in the word problem assessments, I have found it very simple to apply and communicate for it just takes real life situations and incorporates it with the basic knowledge we have learnt of quadratics. Apart from my strengths, I admit to having difficulties in the start of quadratics for it was a complete new concept from linear relations that was more difficult to grasp. However, through studying all the aspects and the application portions of it, I believe I have improved greatly at the end.