# Quadratic Relationships

### By: Saarukha Rajhkumar

## What Is Quadratics?

*Vertex Form=a(x-h)^2+k**Factored Form=a(x-r)(x-s)**Standard Form=ax^2+bx+c*

This flyer will help you better understand the quadratic equations, with lessons, videos and examples.

Down below, are some real life examples of parabolas we see in our lives

## All About Quadratic Relations

A* quadratic relation* is :

- A relation that has an equation (y=ax^2+bx+c)
- Where a, b and c are real numbers
- Where a ≠ 0

While linear equations are a straight line, a quadratic relation is a curve. Here are some new vocabulary words for this unit and a labelled diagram of a parabola.

## Vertex Form

## Learning Goals

**I am able to graph quadratic functions giving critical information and transformations where appropriate.**

**I am able to gather information from the vertex form of a quadratic (direction of opening, axis of symmetry, vertex, step pattern)**

**By given a set of transformations from the parent parabola, I am able to graph and state the equation of the parabola in vertex form.**

## Summary

*a(x-h)^2+k *

- Vertex (h,k)
- Axis of Symmetry (x=h)
- Optimal Value (y=k)
- a- direction of opening and compression/stretch
- h- horizontal translation
- k- vertical translation
- To find the y-intercepts sub x=0
- To find the x-intercepts set y=0
- Step Pattern

## Investigate Non-Linear Relations

__What are linear relationships?__

A linear relationship is a trend in the data that can be modelled by a straight line that shows a steady rate of increase or decrease.

__What are non-linear relationships?__

A non-linear relationship (quadratic) is a trend in the data that can be modelled by a curve. This curve is called a parabola.

Below, is an example of a linear and a non-linear relationship

## Quadratic Relations

__First and Second Differences__

In order to determine if an equation is linear or quadratic we must look at the first and second differences. First differences are the differences in the y-values chart.

__How to Calculate First and Second Differences__

In order to calculate the first differences in the y-values chart you should use the equation (y2-y1) where you subtract the top number from the bottom number. If the first differences are constant/same that means the relation is linear. Suppose the first differences aren't the same/constant you have to take the same steps as the first differences but this time you're subtracting the new numbers gotten from calculating the first differences rather than the existing y-values. If the second differences are same/constant that means the relation is quadratic. In some cases if you do the first differences and the second differences and you do not get constant/same numbers that means the relation is neither linear or quadratic.

## Write the equation in vertex form given transformations

__What Is Vertex Form?__

__Transformations of Vertex Form:__

*The "a" value*= Determines the stretch of the parabola, however if the "a" value is a fraction or a decimal it will compress the parabola. If the a value is a whole number then the parabola will get stretched. Lastly, if the a value has a negative sign in front of it then it will be reflected in the x-axis.*The "h" value*= Determines the horizontal translation to the left or the right of the parabola. If the number is negative it moves to the right side (positive) and if the number is positive it moves to the left side (negative)*The "k" value*=Determines the vertical translation, meaning how many units the parabola will move up or down

Here is an example:

- opens up (minimum at 7)
- vertically stretched by a factor of 15
- translated 5 units down
- translated 7 units right
- vertex: (-5,7)

## Graphing in Vertex Form

__Mapping Notation__

x | y= x² ← This is the notation for the base graph, y=x²

-2| y= (-2)²= 4 Using this mapping notation, you can plot the points on to the graph,

-1| y= (-1)²= 1 connect the lines and create a parabola.

0 | y= (0)²= 0

1 | y= (1)²= 1

2 | y= (2)²= 4

__Mapping Formula__

By using the mapping formula, you can go from the graph of y=x² to y= a (x-h)²+ k...

(x , y) → ( x+h, ay+k )

Example: y= (x-4)² +2 [a=1, h=4, k=2] (x,y) → (x+4, y+2)

__Steps to Graph__

1.Write mapping formula (x,y) → (x+h, ay+k)

2. Complete table of values for y=x²

3. Determine transformed "key" points (x+h)

4. Sketch base function, then new graph

Example: y= 2(x-3)² +1 [a=2, h=3, k=1]

(x,y) → (x+3, 2y+1)

x | y= x² → x | y= 2(x-3)²+ 1

-2 | 4 > (-2+3) > 1 | 2(4)+1 = 9

-1 | 1 > (-1+3) > 2 | 2 (1)+1 = 3

0 | 0 > (0+3) > 3 | 2 (0)+1 = 1

1 | 1 > (1+3) > 4 | 2 (1)+1 = 3

2 | 4 > (2+3) > 5 | 2 (4)+1 = 9

__Step Pattern__

When using step pattern to graph the parabola, multiply (a) by 1,3,5, etc.

The product you get is the rise and the run increases by one each step.

Example:Sketch a graph for y= 3 (x-2)²- 9

- Vertex: (2,-9)
- a=3
- Step1 1x 3 = 3 (3/1)
- Step2 3x 3 = 9 (9/2)
- Step3 5x
- A.O.S: x = 2
- y-intercepts: x=0

y= 3(0-2)²-9

= 3(-2)²-9

= 3(4) -9

= 12-9

= 3

(0,3)

## Finding Zeros and Y-intercepts

Example:

y= 2(x-5)²- 50

Y- int (x=0) | x- int (y=0)

y= 2(0-5)²-50 | 0= 2(x-5)² -50

= 2(-5)²- 50 | 50= 2(x-5)²

= 2(25)-50 | 2 = 2

= 50-50 |√25= √(x-5)²

= 0 |5+5=x -5+5=x

(0,0) | 10=x 0=x

(10,0) (0,0)

## Word Problem

__Objects in the Air__

** • What was the initial height?**

* Find the y- intercept (x=0) and solve for y (the height when time=0)*

** • How long was it in the air?**

* Find the x- intercept (y=0) and solve for x (the time when ball hit ground, height=0)*

** • When did it hit the ground?**

* The time when ball hit the ground (y=0) so the x-intercept*

** • What is the maximum height of it?**

* The y- value of the vertex is the maximum height (y=k)*

** • At what time did it reach the maximum height? **

T*he x- value of the vertex is the time when it reached maximum height (x=h)*

** • Height of it at 3 seconds?**

* Substitute 3 into the x-value(t) of the equation and solve for y(h)*

__Example:__

**At a baseball game, a fan throws a baseball back onto the field. This is modelled by the equation h=-5(t-2)² +45. (h= height in metres, t= time in seconds).**

**a) What was the initial height? (t=0) **

h=-5(0-2)² +45

h=-5(-2)²+45

h=-20+45

h=25

∴ The initial height was 25m.

**b) What was the height after 3 seconds?**

h=-5(3-2)²+45

h=-5(1)²+45

h=-5+45

h=40

∴ The height of the ball was 40m after 3 seconds.

**c) What was the maximum height?**

The maximum point is the y-value of the vertex, (2,45), so the max height is 45m.

**d) What time did the ball reach its maximum height?**

The time the ball reached its maximum height is the x-value of the vertex, (2,45), so the time when the ball reached its maximum height is at 2 seconds.

## Factored Form

## Summary

*a(x-r)(x-s)*

- Value of a- shape and direction of opening
- Values of r and s- x-intercepts
- Axis of Symmetry- x+ (r+s) divided by 2 use this x value and sub into the equation to find the optimal value
- to find the y-intercept set x=0 and solve for y
**Types of Factoring:****Greatest Common Factor****Simple factoring (a=1)****Complex factoring****Special case - Difference of squares****Special case – Perfect square**

## Multiplying Binomials

When multiplying binomials the basic rule is to expand a simplify. To remember the steps of expanding and simplifying we use the acronym F.O.I.L which stands for first terms, outside terms, inside terms and last terms.

*Example:*

(3y+2) (4y+1)

=12y^2+3y+8y+2

=12y^2+11y+2

__Special Cases__

*Example 1:*

(6x - 7) ^2

= (6x - 7) (6x - 7)

= 36x^2 - 42x - 42x + 49 [the middle terms will always be the same]

= 36x^2 - 84x + 49

*Example 2:*

(2x + 5) (2x - 5)

= 4x^2 - 10x + 10x - 25

= 4x^2 - 25

## Common Factoring

Common factoring is the opposite of expanding!

Step 1 - Find the GCF

Step 2 - Write the solution with brackets

*Examples:*

2x + 20 GCF = 2

= 2 (x+10)

14x^3 - 7x GCF = 7x

=7x (2x^2 - 1)

21c^4d^3 - 28c^2d^5 + 7cd^3 GCF = 7cd^3

=7cd^3 (3c^3 - 4cd^2 +1)

## Different Ways of Factoring

__Simple Trinomials__

To factor the trinomial, we need to find:

-> 2 numbers that ADD to give us b

-> 2 numbers that MULTIPLY to give us c

This can be called the product-sum method.

x^2 - 7x -18 -9 x 2 = -18

=(x-9) (x+2) -9 + 2 = -7

In some cases, you may need to common factor first.

2x^2 + 14x + 24

=2(x^2 + 7x + 12) 4 x 3 = 12

=2(x+4) (x+3) 4 + 3 = 7

__Binomial Common Factoring__

8x (y-7) + 3 (y-7)

= (y-7) (8x+3)

This is when the binomials are the common factor.

__Factor By Grouping__

When there is no common factor - we can group terms together that have a common factor.

d^2 + 5d + 3d + 15

= (d^2 + 5d) + (3d + 15) <- group like terms together

= d (d+5) + 3(d+5) <- the brackets should be the same

= (d+3)(d+5) <- factor the binomial

__Complex Trinomial Common Factoring__

A complex trinomial is when there is a coffeicent great than 1, in front of the x^2 term.

2 methods to factor complex trinomials are:

-> Decomposition

-> Trial and Error

__ How to do decomposition__

-Step 1: Multiply the 1st and the last number to find the 5x^2 - 14x + 8 product. 5 x 8 = 40

-Step 2: Use the product/sum method with the product above.

-10 x -4 = 40

-10 x -4 = -14

Step 3: Rewrite the middle terms with the 2 factors. = 5x^2 - 10x - 4x + 8

Step 4: Factor by grouping = (5x^2 - 10x) - (4x + 8)

= 5x (x -2) -4 (x -2)

Step 5: Binomial common factoring = (x - 2) (5x - 4)

## Factoring Special Quadratics

__Difference of Squares__

1,4, 9, 16, 25, 36... are all perfect squares.

*Example*

16x^2 + 9

Square root the 16 and 9

(4x + 9)(4x - 9)

The last term in the first bracket is always positive, and in second bracket, it is always negative. Or vise-versa.

__Perfect Square Trinomials__

The trinomial that results from squaring a binomial is called a perfect square trinomial. They can be factored using the patterns from expanding binomials:

a^2 + 2ab + b^2 = (a+b)^2

a^2 - 2ab + b^2 = (a-b)^2

->The first and last terms are perfect squares

->The middle term is twice the product of the square root of the first term and the square root of the last term.

*Example*

x^2 + 18x + 81

=(x+9)^2

## Solving Quadratics

__Find the Zeroes__

- The x-intercepts are called roots, zeroes, or x-intercepts.
- To find the zeroes, the equation must be in factored form, and one side must always equal 0.

*Example 1:*

x^2 + 11x + 30 = 0 ->This is a simple trinomial

(x + 5)(x+6) = 0 -> 5 x 6 = 30, 5 + 6 = 11

x+5=0 x+6=0 ->to solve for x, you must set each bracket = 0

x=-5 x=-6

To check: Substitute both zeroes into the left and right side of the original equation.

*Example 2:*

-2b^2 = -13b + 21 ->Bring everything to one side to make it = 0

0 = 2b^2 - 13b + 21 ->Factor the complex trinomial -6 x -7 = 42, -6 + -7 = -13

0 = (2b^2 - 6b) - (7b + 21)

0 = 2b (b -3) -7 (b-3)

0 = (2b-7)(b-3) ->Set each bracket = 0

2b-7 = 0 b-3=0

b = 3.5 b=3

## Graphing

**Graph x^2-8x-15**

__Step 1:__ Put the equation into factored form

y = x^2 - 8x + 15

y= (x-3)(x-5)

__Step 2:__ Find the zeroes by setting each bracket equal to 0

x=3 x=5

Plot these x intercepts onto your graph

__Step 3:__ Find the axis of symmetry (x value of the vertex)

x = 3+5

Divide 8 by 2 to get 4

x= 4

__Step 4:__ Find the optimal value (y value of the vertex)

y = x^2 - 8x + 15

y= (4)^2 - 8(4) + 15

y= -1

The vertex is (4,-1)

__Step 5:__ Plot the vertex and connect the points

## Word Problem

**The area of a rectangle is 65 cm^2. The length is 2 less than three times the width. Find the length and the width.**

To solve:

First, you need to find binomials that represent the dimensions of the rectangles.

->Let w represent the width

->Let (3w-2) represent the length

Now, solve.

A= wl

A= w (3w-2) ->Expand and simplify

A= 3w^2 - 2w

65 = 3w^2 - 2w ->We know that A=65

0 = 3w^2 - 2w - 65 ->Make one side = 0

0 = 3w^2 - 15w + 13w -65 ->Factor the complex trinomial

0 = (3w^2 - 15w) + (13w-65) ->Factor by grouping

0 = 3w (w-5) + 13 (w-5)

0 = (w-5)(3w+13)

w-5 3w+13

w=5 w=-4.33

The width is 5cm. It cannot be -4.33 because the rectangle cannot have negative dimensions.

If w=5. Find the length

3(5)-2 = 13

Therefore the dimensions are 5x13.

## Standard Form

## Learning Goals

**I can find the number of zeros that a quadratic relationship has by calculating the discriminant.****I can start with Standard Form and solve for the roots of the equation by using the Quadratic Formula**

**I am able to solve application problems using standard form and discuss the values in context**

## Summary

*ax^2+bx+c*

- value of a - shape and direction of opening
- value of c - y-intercept
- find x-intercepts- use quadratic formula
- Find MAX/MIN-Complete the square to get vertex form

## Completing The Square

**Steps for if the coefficients of x² is one**

1. Take the middle number in the equation and then divide it by two and get that number and square it.

2. Put the new number into the equation. Remember that this number will always be added and subtracted.

3. Move the negative number outside.

4. Factor out the numbers in the brackets and add or subtract the numbers outside of the brackets.

**Steps for if the coefficients of x² is not one**

1. The same steps apply as if the coefficient is one but now you divide the first two terms with the coefficient of x².

2. After dividing the two terms by the coefficients of x² use the new middle term number to divide by two and square.

3. Proceed with the same steps as if the coefficient were one, but when you move the negative number outside of the bracket you have to multiply it with the coefficient of x²

## Quadratic Formula

**What is the Quadratic Formula? **

## The Discriminant

**What is the Discriminant?**

The discriminant is the number inside the square root of the quadratic formula (b²-4ac). It helps us tell how many x-intercepts the quadratic equation will have without having to use the whole quadratic formula. If the answer to the discriminant is less than zero (negative number) there will be no solution because you can not square root a negative number. If the answer to the discriminant is zero that means there will be one solution. If the answer is one or bigger that there will be two solutions.

## Word Problems

**Most Word Problems In Standard Form Ask For:**

- When the question is asking for the vertex - complete the square.
- To find the x-intercepts - use the quadratic formula.

**1. The height of a coin in the air t seconds after it is flipped can be modeled by**

**y = -16t^2 + 32t + 5, where t and y are measured in m. What is the maximum height that the coin reaches?**

**->To solve this, we need to complete the square because it is asking for the maximum height (y value of the vertex).**

y=(-16t^2 + 32t) + 5

y= -16(t^2 - 2) + 5

y= -16 (t^2 - 2 + 1) -1) +5 -> -16 multiplies with -1

y= -16 (t-1)^2 + 21

The vertex is (1,21). Therefore the maximum height of the coin is 21m.

## Reflection

__Assessment Reflection __

__Connections__

Throughout this unit I had made various connections within the various forms of a quadratic relation. From vertex form, to factored form and lastly the standard form I was able to connect major ideas to each unit and use steps to solve problems from each unit. For an example for standard form when you complete the square to determine the vertex for the equation you have to rearrange the equation into vertex form to determine the vertex by looking at the h and k values otherwise known as x and y. Also for each unit, just by graphing or even from the equation you can determine the axis of symmetry, a,x and y values and lastly the zeroes/x-intercepts. Lastly, all these different quadratic relations can relate to graphing because there are a variety of ways to solve the equation to determine its vertex and x-intercepts. For an example, if you want to find the x-intercepts in vertex from all you have to do is sub y=0 in the equation but when you are doing that in standard form you use the quadratic formula to determine the x-intercepts.

I hope to expand my knowledge more on the various forms of the quadratic relation and I hope to improve on on my skills even further in Grade 11.