### BY: SARA AHMED

• Analyzing First & Second Differences
• Introduction to Parabolas

• Vertex Form
• Factored Form
• Standard Form

Vertex Form

• Key Features of Quadratic Relations
• Transformations
• Finding the x and

y intercepts

• Graphing Vertex Form

Factored Form

• Factors and Zeros
• Multiplying Binomials/ Expanding & Simplifying
• Common Factoring
• Factoring Simple Trinomials
• Factoring Complex Trinomials
• Special Cases
• Graphing Factored Form

Standard Form

• Maximum and Minimum Values/ Completing the Square
• Graphing using Standard Form
• Graphing using Standard Form & The Quadratic Formula

Solving Word Problems

• Vertex Form Problems
• Factored Form Problems
• Standard Form Problems

• Reflection
• Assessment

Quadratics are functions that can be described by an equation in the form

f(x) = a (x-h) ^2 + k. Quadratic equations form a curved line of data whereas linear function equations such as,y =mx+ b, form a straight line of data on a graph. The equation is a curved line of data because x is squared and this does not allow the line to stay linear, as it does not increase by the same amount each time. Quadratics are very useful in real life situations as well. For example, they can be used to track the path of a ball being kicked in the air or to measure the curve of a shape like a banana or the M in the McDonald's logo.

## How can you tell the difference between Linear and Quadratic Relations?

When the first difference is constant, the equation is linear
When the second difference is constant, the equation is quadratic

## INTRODUCTION TO ... PARABOLAS !!

When graphing a linear function, the end result is a straight line. When graphing a quadratic function the end result is a curved line with an opening either upwards or downwards called a parabola. The reason for this line being curved is due to the fact that in quadratic equations there is an x variable that is squared. This causes an increase in value starting off slow and later increasing rapidly. When throwing a football, it curves up into the air and falls down again following the path of a parabola.

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

FACTORED FORM : y = a(x - r)(x - s)

STANDARD FORM : y = ax^2 + bx + c

## Vertex Form

KEY FEATURES OF QUADRATICS IN THE VERTEX FORM

The Vertex Form provides key information that allows us to graph the parabola.

Key information includes:

• Vertex
• Axis Of Symmetry
• Optimal Value
• The Zeroes / Roots

## The Vertex: (x,y)

The vertex of the parabola is the tip of the curved line which is always either the highest or lowest point. The vertex is given in the equation as the h value and the k value (h = x, k = y). When putting h into the vertex, always switch the sign to the opposite of what is given in the equation. This is done because in the equation h is negative. The vertex is a combination of the axis of symmetry (x) and the optimal value (y).

The example below shows the vertex = (0, 3)

## Axis of Symmetry (AOS): x-h

The axis of symmetry (AOS) is also known as the vertical reflection point in the parabola which is located in the exact center. The x value in the vertex is the axis of symmetry.

The example below shows the AOS where x = 0

## Optimal Value: y=k

The optimal value is known as the maximum or minimum in a parabola. It is also known as the y value in the vertex that shows whether the parabola is opening upwards (min) or opening downwards (max). The reason why these values are called maximum or minimum points is because it can either be the largest or smallest y values in the parabola. The y in the vertex is also the optimal value.

The example below shows both maximum and minimum.

## The zeroes

The zeroes are the x- intercepts in a parabola where it passes through the x-axis

## How to find the Equation when only given the Vertex

Sub all given values into the equation , and set y=0 to solve for the remaining variable.

Step 4:SOOOOOLLLVEEEE!!
put it all together:

y= 0.05(6+8)^2+10

TRANSFORMATIONS

The transformed function f(x) = a(x-k) ^2 + k is the vertex form of the quadratic equation.

When graphing the vertex form there are 4 important things to look for in the given equation:

1. Vertical Stretch/Compression
2. Reflection in the x-axis
3. Horizontal Translation
4. Vertical Translation

1. Vertical Stretch/Compression

The vertical stretch/compression in a quadratic equation determines whether the parabola contains wide curve or a narrow curve. When a > 1 the parabola has a vertical stretch. When a negative sign is placed in front of it, the rule still applies however it is a < -1. When

0 < a > 1 the parabola has a vertical compression. Just like a stretch, when a negative sign is placed in front of it, the rule still applies but between 0 and -1.

2. Reflection in the x- axis

If the a value given in the quadratic equation has a negative sign in front of it the parabola is reflected into the x-axis meaning it will have a downwards opening. On the other hand, if the a value in the given equation has a positive sign in front of it the parabola will have an upwards opening.

When a has a negative in front of it:

Example : y = -1(x-1) ^2 +1

a = -1

When a does not have a negative in front of it

Example : y = 1(x-1)^2 + 1

a = 1

3. Horizontal Translation

The horizontal translation is the variable h in the vertex form. The h value determines where the x in the vertex lies. If the h value is written as a negative in the equation it will be placed on the graph as a positive, and if the h value is written as a positive in the equation it will be placed on the graph as a positive. When h is a negative value the translation on the graph is to the left, and when h is a positive value the translation on the graph is to the right.

Example: y = -1 (x-1)^2 + 1

• The h value in the equation is -1, meaning on the graph it will be placed as +1
• This is a horizontal translation 1 unit right

The Example below shows how the horizontal translation effects the placing of a parabola on a graph.

4. Vertical Translation

The vertical translation is the k value in the vertex form. The k value determines where the y in the vertex lies. When the k is a positive the y value moves up and when it is a negative

the y value moves down. For example, if the k value is positive 7, the y value in the vertex would be positive 7 and this would be a vertical translation 7 units up.

Example: y = -1 (x-1)^2+1

This is a vertical translation 1 unit up

The Example below shows how the vertical translation effects the placing of a parabola on a graph.

EXAMPLE OF EVERYTHING : y = -2 (x -5)^2 - 10

• Vertical Stretch of -2
• Reflected onto the x-axis
• Horizontal translation 5 units right
• Vertical translation 10 units down

How to Find the Equation When Given Only the Transformations

When forming the equation after given the transformations, sub in all the numbers you know based on the variables the transformations represent as done below.

FINDING THE X-INTERCEPTS

When trying to find the x-intercepts of a parabola, simply set the y variable to 0, and then solve for x.

Finding x-intercepts (Vertex Form)
After getting to the answer at the end of this video, you must solve the remainder of the equation by following these steps:

## FINAL STEP

(-1.4, 0) , and (-4.6, 0)

FINDING THE Y-INTERCEPT

When solving for the y-intercept, you set the x variable to zero and solve.

GRAPHING VERTEX FORM

Graphing using the Step Pattern Notation

Graphing using the Mapping Notation Formula:

This formula is an accurate and simple way to graph any quadratic equation. This method can be used instead of the step pattern (notation), and it makes going from y=x^2 to y=a(x-h)^2+k simple to graph by using its formula: (x+k, ay +k) -----> (x, y)

For Example: y = 2 (x-5)^2 + 4

The Mapping Notation of this Quadratic equation would be (x + 5, 2y +4)

**Remember that h is always changes to the opposite sign when graphing**

## FACTORED FORM

Factored Form: y = a(x - r)(x - s)

FACTORS AND ZEROS

MULTIPLYING BINOMIALS/ EXPANDING & SIMPLIFYING

F.O.I.L

When multiplying binomials, the two variables in the first bracket are both multiplied with each variables.In the second bracket. This method is known as foil

. Foil stands for " First outside, inside last" this means that the first variable in the first brackets is multiplied to the two variables in the second brackets first, and then the second variable multiples next.

**DO NOT MULTIPLY THE TWO VARIABLES IN THE SAME BRACKET TOGETHER**

What Foil Looks Like :
Simplify

Add together the like terms after expanding (Foil) to great a smaller equation.

When two separate variables are multiplying the same bracket, they are added together to create their own bracket.

COMMON FACTORING

Common factoring is the opposite of expanding. Expanding involves multiplying while factoring involves dividing.

Different ways to Factor :

1. Finding the GCF
2. Factor by Grouping

Finding the GCF

When using this method, find the greatest common factor of the polynomial's terms. This will include the GCF of its coefficients and the GCF of its variables.

In this example, the GCF of the coefficient was 2 and the GCF of its variables was y^2

If there are two binomials that are exactly the same, think of the Binomial as one factor
Factor by Grouping

Group together the terms that have the same factor.

Then simplify as shown below

FACTORING SIMPLE TRINOMIALS

When given a quadratic equation in standard form, and use factoring to get Factored Form

x^2 + bx +c = (x+r)(x+s)

Standard ------> Factored

Where

• r +s =b
• rs = c

Some Steps to follow when Factoring Simple Trinomials:

1. Look at the signs of b and c

• If c is negative, one of r or s is negative
• If b and c are positive, both r and s are positive
• If b is negative and c is positive, both r and s are negative

2. Find the Product and Sum

• Find two numbers whose product is c
• Find two numbers whose sum is b

Example 1: if c is negative one of r or s is negative

Example 2: If c and b are both positive, r and s will both be positive
Example 3: If b is negative and c is positive, both r and s are negative

FACTORING COMPLEX TRINOMIALS

When factoring complex trinomials, we use similar steps as to when factoring simple trinomials.

Factoring Complex Trinomials

SPECIAL CASES

A polynomial of the form a^2x^2- b is a difference of squares and can be factored as (ax-b) (ax+b)

When factoring a special case you must look for numbers that are perfect squares. Another thing to look at is the Sign in between the terms. A perfect Square will never have a positive sign in between the

terms,

because this means that the two brackets (ax-b)(ax+b) are not in the correct form and have two of the negative symbols because:

(-)(-)= + <<< INCORRECT (not a difference of square)

(-)(+)= - <<< CORRECT

Some Common perfect Squares

1,4, 9, 16, 25, 36, 49, 61, 81, 100, 121, 144

Example of Special cases:

x^2- 4 ----> (x-2)(x+2)

6x^2 - 25y ----> (3x^2 - 5) (3x^2 +5)

Special Case: Trinomials

A polynomial of the form a^2x^2 +/- 2abx +b^2 is a perfect square trinomial and can be factored as (ax+b)^2

When factoring this you must follow these steps:

1. Square root a^2 and b^2
2. Check if 2ab is equal to the middle term
3. Plug the square rooted Numbers into (ax+b)^2

Example:

The Equation

GRAPHING PARABOLAS USING FACTORED FORM

• The zeros of a parabola are its x-intercepts
• The r and s variables are the x-intercepts (r,0) & (s,0)
• The a variable provides the shape and direction of the opening (within the parabola)
• To find the vertex, use the zeros to find the AOS, and substitute this x value in the equation to solve for y

When graphing the x-intercepts and vertex:

1. Factor the equation
2. Solve for r and s by making each factor equal to zero
3. Find the AOS
4. Substitute the x variable into the equation and solve for y to find the vertex
5. Put together your Vertex (AOS & y-intercept)
6. Using the Vertex and x-intercepts, graph the Parabola!

My Video Example:

## Standard Form

Standard Form: y = ax^2 + bx + c

COMPLETING THE SQUARE

When completing the square, we are going from the standard form to the vertex form.

Example 3: Completing the square | Quadratic equations | Algebra I | Khan Academy
EXAMPLE
The Discriminant

Will determine how many solutions there will be. (solutions= x-intercepts)

• When the Discriminant is a negative there are 0 solutions
• When the Discriminant is equal to 0 there is 1 solution
• When the Discriminant is a positive there are 2 solutions
An example of Solving for the x-intercepts
Therefore, the x intercepts= (0.28,0) (-6,0)
May 10, 2016

## GRAPHING PARABOLAS IN STANDARD FORM

Graphing Parabolas In Standard Form (Quadratic Functions)

## SOLVING WORD PROBLEMS

Words Problems Given in the Vertex Form:

Example: Flight Path of an Object Word Problem

The height of the ball is thrown following the path,

ℎ = −5 (t − 3)^2 + 46.5,

where h is the height in meters and t is

time

in seconds.

a) What is the maximum height of the ball?

h= 46.5 > the optimal value from the vertex

b) How long does it take the ball to reach its maximum height?

t= 3 > the x value from the vertex

c) What was the initial height of the ball when it was thrown?

To find the height, set t=0 and solve for h.

Therefore, the initial height of the ball when it was thrown was 1.5 metres.

d) How long was the ball in the air?

This question is asking when the ball lands on the ground.

Set h=0 and solve for t.

Therefore, the ball was in the air for 6.05 seconds.

e) What is the height of the ball at 1 second?

Set t=1 to solve for the height of the ball at 1 second

Word Problems Given in the Factored Form:

Example: Flight Path of an object Word Problem

The predicted flight path of a toy rocket used in a mathematics project is defined by the relation h= -3 (d-2)(d-12) , where d is the horizontal distance, in metres, from a wall, and his the height, in metres above the ground.

a) How far from the wall is the rocket when it is launched?

b) How far from the wall is the rocket when it lands on the ground?

when these questions are asked, find the x- intercepts (the zeros)

a) 2m

b) 12m

c) What is the maximum height of the rocket, and how far, horizontally, is it from the wall at that moment?

When this question is asked, find the AOS using the x-intercepts, and find the y-intercept by subbing in the x value into the original equation.

Word Problems Given in the Standard Form:

Example: Flight of an Object

The height of a rocket thrown from a walkway over a lagoon can be approximated by the formula h=5t^2+20t+60, where t is the time in seconds and h is the height in meters.

a) write the formula in factored form

b) when will the rock hit the water?

meaning: find the x-intercepts

MY REFLECTION ON THE QUADRATICS UNIT

The Quadratics Unit was long and difficult, however by the end of it I learned many techniques and steps to help me solve equations and word problems using the different forms: Standard, Vertex, and Factored form. In the first unit of Quadratics which was Quadratics in the Vertex Form, I was quite content with my work. I had found the first couple lessons in the unit to be simple and straight forward. These lessons included the introduction to parabolas, quadratic functions, how to write equations in vertex form given the transformations, and mapping notations. I have a clear understanding of how to graph parabolas, how to use the vertex form equations to determine what was the vertex of the parabola, and the AOS (Axis of Symmetry). Once the Easy parts were over i

struggled through the rest of the unit not being able to understand what was being done and why it was being done. The struggle mainly came from finding the x-intercepts. I did not understand why there was a negative and positive sign being

places in front of the square root symbol, and I did not understand the concept as it was new to me. As the 3 units went on i grew to understand the concept of the negative and positive sing both being placed at the same time. The second unit of Quadratics was simple, but there were many rules and steps to be followed that sometimes it could get confusing trying to memorize when to use what method and how to use it. It was easy to get 2 different types of equations mixed up. The Last unit of Quadratics was the easiest in my opinion. I did struggle through the unit though . After completing this website,

i've

noticed that i have become a lot better at quadratics and that its not so bad after all