## Overview of the unit

In this unit, we learned how to solve the equations that have X squared in it through different methods.

To get started into the unit, we first learned about what parabola is and how it works.

The Quadratic unit is divided into three parts:

Firstly,

Graphing Parabola from:

- Vertex form

- Factored form

Secondly,

Expanding and simplifying:

- expanding ends in standard form

- Factor: standard to vertex form

Lastly,

Completing the square: (Standard to Vertex form)

- Isolate X from vertex form to solve

- Quadratics formula: find X- intercept from standard form

- Solve a standard form equation

- Axis of Symmetry: Vertex from standard form

We then applied our understanding to solve a variety of word problems, where we needed to write the word problem to number problem. We also learned to check our answers by substituting the X- intercept into the real equation and we also talked about what it means when an equation have no solution and two solution.

## Parabola:

-Parabolas can open up and down

-The zero of a parabola is where the graph crosses the x - axis

-"Zeroes" can also be called "X - intercepts" or "roots"

-The axis of symmetry divides the parabola into two equal halves

-The vertex of the parabola is the point where the axis of symmetry and the parabola meet. It is the point where the parabola is at its maximum or minimum value

-The optimal value is the value of the Y co-ordinate of the vertex

-The Y- intercept of a parabola is where the graph crosses the y - axis

## Key Features of Quadratic Relations

y=x^2 is a simple equation, it basically tells that y is equal to x-squared.

To find out if the equation is a linear or a quadratic relationship you have to find the first and second differences.

- Linear relationships have equal first differences

- Quadratic relationships have equal second differences

## Graphing Parabola from Vertex form:

Vertex form: y=a(x-h)2+k

• (H) is the x- int for the vertex but with the opposite sign and (K) is the Y- int for the vertex
• Adding a number to x^2 after the square makes the vertex rise. (Vertical translation)
• Subtracting a number from x^2 after the square makes the vertex move down.
• Their is a step pattern for graphing an equation from vertex form: Over 1 up 1, over 2 up 4. This pattern does not change by adding or subtracting a number from x^2
• This pattern stays the same for the equations that do not have (A) value
• It is different for complex equations, such as the equations that have (A) value, you take the a value and times it by the step pattern.
• (H) also tells where the parabola is going left or right, if its + then going to the left, if - to the right. (Horizontal translations)
• (K )tells whether parabola is going to placing up or down, if - then down, if + then up. (Vertical translations)
• (A) affects the vertical stretch

## Y=(x+2)^2+5

Graphing from simple vertex form: Y=(x+2)^2+5

vertex: (-2,5) 2 to the left because it is negative and 5 units up because it is positive from the middle point of the graph.

To graph this:

1. Plot the vertex on the graph

2. Use the step pattern Over 1 up 1, over 2 up 4 to find the two points (x-int)

In this case you can use the exact same step pattern because the (A) Value is not given.

## Graphing a equation with (A)

Y=3(x+4)^2-10

Vertex: (-4,-10) 4 left, 10 down

To graph this:

1. Plot the vertex on the graph.
2. This time you will have to use a different step pattern to graph the equation,

## Finding equation of a parabola given the vertex

vertex (-2,1) x,y (0,-7) into y=a(x-h)2+k

To solve this:

1. you will have to substitute the vertex into the equation
2. Substitute the intercepts into the equation
3. Simplify the bracket
4. Subtract 8 from both sides, to get rid from one sides
5. Simplify the left side of the equation
6. Solve the bracket on the right side
7. Divide 4 from both sides
8. Now substitute a into the equation that already has h and k value in it.

y = a(x+2)^2 +1

-7= a(0+2)^2 +1

-7 = a(2)^2 +1

-7-1=a(2)^2+1-1

-7-1=a(2)^2

-8 = a(2)^2

-8/4 = a(4)/4

-2 = a

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

To check: Substitute -2 into the equation that has vertex and both intercepts in it.

Solution in the pictures...

## Factored form of a Quadratic relation

y= a(x-r) (x-s)

Zeroes: are found by setting y equal to zero, so:

x-r=0 means x=r and x-s = 0 means x=s

Axis of symmetry (AOS): is the midpoint of the two zeroes, so: x= r+s / 2

Optimal value: is found by subbing the AOS value into an equation

## Y= -(x-2)(x+4)

1. Zeroes: find the zeroes by setting y equal to zero, so: x-2=0 means x=2 and x+4 = 0 means x=-4
2. Axis of symmetry (AOS): is the midpoint of the two zeroes, so: x= 2-4 / 2 = -1
3. Optimal value: is found by subbing the AOS value into an equation Y= -(-1-2)(-1+4)
4. Write the vertex (x,y), take the x value from AOS and y value from optimal value
5. Vertex: (-1,9)

## Y=2(x-1)(x-4)

1. Zeroes: find the zeroes by setting y equal to zero, so: x-1=0 means x=1 and x-4 = 0 means x= 4
2. Axis of symmetry (AOS): is the midpoint of the two zeroes, so: x= 1+4/ 2 = 2.5
3. Optimal value: is found by subbing the AOS value into an equation Y=2(2.5-1)(2.5-4)
4. Write the vertex (x,y), take the x value from AOS and y value from optimal value
5. Vertex: (2.5,-4.5)

## Y= -(x-2)(x+4)

Factored form equations are really easy to graph once you have solved an equation.

1. Find the zeroes by setting y equal to 0
2. Get the x-int from AOS and y-int from optimal value
3. Plot those points on the graph
4. Plot the zeroes on the graph

## Practice word problem

• Recall: length can be described as a distance from Ist x to 2nd X (Zeroes)
• Height is the y value of the vertex (First AOS to find x then optimal value to find y)

The Humber Bridge in Toronto can be represented by an equation H=-0.5(X-2)(X-14), where x represents the horizontal distance from the start of a bridge in meters, and H represents the height of the arches in meters.

a) What length of the bridge do the arches cover?

To find the length of the bridge You will have to find both x values

H=-0.5(X-2)(X-14)

1st x x-2 = 0

x-2+2 =0+2 Add 2 from both sides to get rid from one side

x=2

2nd x : X-14=0

X-14+14 =0+14

X= 14

Therefore, the two x values are: 2 and 14

To find the length subtract 2 from 14

The Arch covers 12 m of bridge

b) Show mathematically how high the arches are at their highest point.

Find AOS: 2+14/2 = 8 , X= 8

Optimal Value: H=-0.5(X-2)(X-14)

H=-0.5(8-2)(8-14) Substitute the x value

H=-0.5(6)(-6)

H=-0.5(-36)

H= 18

Therefore, the highest point that arches reach is 18 m

## Word Problem

Write a quadratic relation which has vertex below the x-axis and two x-intercepts.

For this type of questions, you have to think about all the things that we have learned in this unit.

1. Write an equation which has vertex below x-axis

y= 0.5(x-2)(x-10)

2. Find x-intercepts

1st x-2+2=0+2 2nd x-10+10=0+10

x=2 x=10

3. Find AOS by adding the x-intercepts and dividing them by 2

2+10/2

X= 6

4. Find Optimal value by subbing X value into the equation

y=0.5(6-2)(6-10)

y= 0.5(4)(-4)

y= 0.5(-16)

y=-8

5. Write the vertex using the AOS, x value and optimal's y value

Vertex: (6,-8)

3.6 Expanding

## Factoring

Factoring

• Opposite of expanding and simplifying
• looking for numbers that multiply to end up within the original expression
• Factor mean, find the factor of 4: 2*2, 4*1
Common factoring

• Things that divide
• if the term in an expression have any "Factor" in common, we can "factor them out"

## Factoring simple trinomials

x^2+17x +60

To factor this equation:

1. Look at the last number and find two number that are divides by it and adds to the second number
2. You take those number and write them in factored form (x+c) (x+s)
3. simplify the equation

x^2+17x +60

Possible number that times by 60 and adds to 17 is: (12,5)

(x+5) (x+12)

To check: x^2 +12x +5x +60

x^2+17x+60

## Factoring Complex Trinomials

To factor: y= 6x^2 +11x +3

you will need to make a chart,

1. two numbers that multiply to 6: 3*2
2. Two numbers that multiply to 3: 3*1
3. write those numbers on the chart, as shown in the picture
4. Cross multiply
5. The answer in the third column has to add up the middle coefficient in the equation
6. keep flipping the numbers within the same column until you get the right answer
7. Finally Write the equation in factored form (x+c) (x+s)

Using this equation, you could also find the, Zeroes, Axis of symmetry, and Optimal value

Recall: y= a(x-r) (x-s)

Zeroes: are found by setting y equal to zero, so:

x-r=0 means x=r and x-s = 0 means x=s

Axis of symmetry (AOS): is the midpoint of the two zeroes, so: x= r+s / 2

Optimal value: is found by subbing the AOS value into an equation

y= (3x+1)(2x+3)

Zeroes: The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at least one of the factors must be zero, I'll set them eachequal to zero:

3x+1=0 To find x value of: 3x+1

3x+1-1 =0 -1 Subtract 1 from both sides

3x/3=-1/3 Divide 3 from both side

X= -1/3

To x value of: (2x+3)

2x+3-3=0-3 Subtract 3 from both sides

2x/2=-3/2 Divide 2 from both side

x = -3/2

AOS: -0.33 -1.5 / 2

-1.83 / 2

x= -0.9

Optimal value: 6x^2 +11x +3 Substitute the x value

6(-0.9)^2 + 11(-0.9)+3

6(0.84)- 9.9 + 3

5.04 - 9.9 + 3

y= -1.86

Vertex: (-0.9,-1.8)

## Multiplying Polynomials

(x+4)^2 to multiply this, you just write the same thing twice and then simplify it

(x+4)(x+4)

x^2 +4x +4x +16

## Word problem

a. Determine an equation for the area of the following shape:

x(x+2)+(x-1)(2x+4)

2. simplify the equation

x^2 +2x + 2x^2 +4x -2x -4

x^2+ 2x^2+2x+4x -2x -4

3x^2 +4x -4

b. If the area of the shape is 11m^2, determine the x.

3x^2 +4x -4= 11

3x^2 +4x -4-11= 11-11

3x^2 +4x -15 = 0

To find the x-intercepts you have to change the equation into factored form.

(3x-5) (x+3) =0

Now, find the x-intercepts

1st x: (3x-5) 2nd x: (x+3)

3x-5+5 =0+5 x+3-3=0-3

3x =5

x= 5/3

2nd x = -3

Therefore, the x value is 5/3 because it cannot be negative

## Completing the Square

Moving from standard form to vertex form

Y= ax^2+bx+c ==> y= (x-h)^2+k

Completing the Square - Solving Quadratic Equations

## Solving by Square rooting

Solving an equation by square rooting both sides, we can end up with two answers(+or-)

x^2 +1 =17

x^2 +1-1=17-1

x^2 (square root) = 16(square root)

x= + or - 4 because for any positive number there are two numbers you can square to get that number. For example there are two numbers you can square to get 16:4 and -4. In general, for some positive number we will call "x", the two numbers you can square to get x are called the square roots of x. One is the positive square root and the other is called the negative square root. The positive square root of 16 is 4 and the negative square root of 16 is -4.

## Example:

x^2 +4x -4 = 0 Keep all terms containing x on one side. Move the constant to the right.

x^2 +4x = 4

x^2 +4x + __ = 4 + __Get ready to create a perfect square on the left. Balance the equation.

x^2 +4x + _4_ = 4 + _4_

x^2 +4x + 4 = 8 Take half of the x-term coefficient and square it. Add this value to both sides.

(x+2)^2 = 8 Simplify and write the perfect square on the left.

x+2= +/- 8(Square root) Take the square root of both sides. Be sure to allow for both plus and minus.

x = -2 +/- 8(Square root) = -2 +/- 2 square root 2

x= -2 + 2 square root 2

x= -2 - 2 square root 2 Solve for x.

## Song to remember Quadratic formula

The solutions of some quadratic equations are not rational, and cannot be factored. For such equations, the most common method of solution is the quadratic formula. The quadratic formula can be used to solve ANY quadratic equation, even those that can be factored.

The equation must be set equal to zero before using the formula.

• This method is especially useful if the quadratic equation is not factorable.
• Make sure that the quadratic equation you are trying to solve is set equal to 0 before plugging the quadratic equation's coefficients a, b, and c into the formula.
• The number inside the square root (b^2-4ac) of the quadratic formula is called discriminant, it tells us how many solutions a quadratic relation has.
• If b^2-4ac is greater than 0 then the equation will have two solution + and *
• If b^2-4ac is equal to 0 then the equation will have one solution which is going to be the x-int of the vertex
• if b^2-4ac is less than 0 then the equation will have no real solution

## Finding x- int using -b/2a

This way is easier to find the x - intercept than the quadratic formula

In this case you do not have to through the step of finding zeroes and then AOS

-b/2a takes you straight to the AOS

Example: y= 3x^2 -14x-5

-b/2a ==> -14)/ 2(3)

x = 14/6

x = 2.33 (AOS)

## Example 1:

A rectangle has a length which is six more than twice the width. The area is 230m^2. determine the dimensions of the rectangle.

First, you should try drawing the shape and labeling the sides.

Length: 2x+6

Width: x

Area: 240m^2

(2x+6)(x) = 240

## Example 2:

The sum of the squares of two consecutive integers is 61. Determine the two integers.

The formula is x^2 + (x+1)^2 = the total number

x^2 + (x+1)^2 = 61

## Example 3:

The flight of a ball is modelled by a equation h = -5t^2 +20t +25 where h represents the height of the ball in meters, and t represents the amount of time the ball has spent in the air in seconds.

a. Write the equation in factored form, determine the x-intercepts and vertex, and graph.

1. Write the equation in simplified form (y= a(x-r) (x-s)), by using the chart
2. h = -5t^2 +20t +25 ==> h= -5(t^2-4t-5)
3. You need 2 numbers that multiply by -5 add up to -4
4. After you have completed the chart, write the equation in factored form which in this case is h= -5(x-5)(x+1)

• X-intercepts: x-5=0

x-5+5=0+5

x= 5

X-intercept: x+1=0

x+1-1=0 -1

x=-1

• To find the Vertex first you need to find AOS: add both of the X-intercepts and divide them by 2 ==> 5-1/2 = 2. Therefore, x = 2
• Then you need to find the optimal value: sub in the x- intercept into the factored equation
h= -5(x-5)(x+1)

h= -5(2-5)(2+1) expand and simplify the equation

h= -5(-3)(3)

h= -5(-9)

h= -45

Therefore, the x value is the x-int of the vertex and h value is the y- int of the vertex

Vertex: (2,-45)

b. When does the ball hit the ground?

To find when the ball lands on the ground, you need to look at the x- int and the + one will be the answer.

The ball hits the ground when h=0 and at 5 seconds.

c. What is the highest the ball flies, and when does that happen?

To find the highest point you will need to look at the y - int, which in this case is 45m and it happens at 2 seconds.

Graph...

## Reflection

Overall, the quadratics unit had many parts to it and it showed many ways of solving an equation using different methods to get the same answer. We first learned what a parabola is and then used different methods to graph the parabola. we also learned how to graph from vertex and factored form. Then in the second part, we learned to simply and expand an equation and also how to change equations from standard form to vertex form. In the last part of the unit we learned how to find the axis of symmetry and the vertex using the quadratic formula.Therefore their are many methods to solve an equation and they are all related to each other. All the equations that we have to solve comes from standard form and we then use the standard form to change it into factored form and we also used the standard form to find the x- intercepts using different methods. Factoring and Quadratic formula are both used to find x-intercepts but through different methods. In the beginning we learned how to find the x-intercepts through factoring but then we learned different methods which made it easier for us because we then had more options and could use the one that we found easy. Overall, some parts of this unit were hard and some were easy for me to understand.

I liked how this unit was broken into three parts because it made it easy for us to understand. I really enjoyed learning this unit because it was interesting. As we learned more concepts it sometime became confusing when to use which method.

I could have definitely improved on some things such as application questions on the unit test and the quizzes. I did fairly well on the tests but I was not successful at explaining my answers clearly. In order to improve this, I could have practiced more and thought about all the different methods to find the correct answers.

This unit will also help me succeed in grade 11 and 12 math and also in real life situations.

This is the question that I did really poorly on and I could have definitely improved it by explaining my answer clearly. I could have also put more thought to my answer to make it better. But overall, I did well on my quizzes and tests but my weakness is explaining the application questions just like this one. But now, I will work harder on application and communication questions .