### By: Saarukha Rajhkumar

Quadratics is all about parabolas. Parabolas are curves, meaning at one point -called vertex- the line turns. We use three different quadratic formulas to create, form and analyse these curves.

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

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

Relations that are non-linear are known to be quadratic

• 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.

## 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

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.

Linear, Quadratic or Neither using First and Second Differences Tables

## Write the equation in vertex form given transformations

What Is Vertex Form?

The transformed function y=a(x-h)² +k is the vertex form of the quadratic equation. The axis of symmetry (AOS) is represented by the h, and the vertex is represented by (h,k).

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:

Equation: y= 15 (x+5)²+ 7

• 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 3 = 15 (15/3)
• 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)

Graphing a parabola in vertex form | Quadratic equations | Algebra I | Khan Academy

## Finding Zeros and Y-intercepts

To find the zeros (x-intercepts) and y-intercepts, all you need to do is substitute 0 into x or y. While finding the x-intercepts, remember that when you square root the term on the left, there can be two possible square roots (1 positive and 1 negative) meaning there will be 2 x-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)

Finding x-intercepts (Vertex Form)

## 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?

The 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.

## Learning Goals

• I can factor polynomial expressions completely including GCFs, grouping, trinomials, difference of squares

• I am able to expand and simplify expressions in factored form

• I am able to solve for the x-intercepts and the vertex

## 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

Example 1: Multiplying a binomial by a binomial | Algebra I | Khan Academy

## 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 Trinomials by Trial and Error - Ex 2

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

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.

## 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.

Completing The Square-The Coefficient of x² is one
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

Completing The Square-The Coefficient of x² is not one

All quadratic equations in the form ax²+bx+c=0 can be solved using the quadratic formula. The quadratic formula is a direct way of calculating the roots (x-intercepts, the solution, the zeros). There could be 2,1, or 0 real solutions.

## 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.

The Discriminant

## 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.

Word Problem Geometric

## Reflection

Assessment Reflection

An assessment I chose to do a reflection on is the Quadratic Standard Form unit test. I chose this assessment to reflect on because this was my favourite unit as it was very straightforward and it was very easy to understand. The concepts were very easy to grasp on, as there were only two big ideas in this unit. The two ideas were completing the square where it helped you determine the vertex of the equation and the quadratic formula where it helped you find your x-intercepts of the equation. I had done well on this test by practicing and doing my homework. If I did not get something I go ask my classmates in class or I ask my teacher for help. By doing all this I have been able to understand the concepts fully. I also had done a big part at home by reviewing my notes and watching videos on these two major concepts so I can succeed on my TIPS assessments, quizzes and tests.

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.