### Have Math in your Path!

• What is a Parabola?
• Parabolas in Real Life

• Investigating Vertex Form
• Factored Form
• Multiplying Binomials
• Common Factors and Simple Grouping
• Simple Trinomials
• Complex Trinomials
• Perfect Squares
• Difference of squares
• Recap
• Completing the square
• Standard form in algebra tiles
• Recap
• Connections
• Reflection

This unit is all about using formulas to solve for the information needed to graph parabolas (ex. x-intercepts, vertex, axis of symmetry, optimal value, etc.). The word quadratic comes from the root “quadratus” which is Latin for “to make square”. It deals with equations involving the variable x raised to the power of two.

## WHAT IS A PARABOLA?

A symmetrical open plane curve formed by the intersection of a cone with a plane parallel to its side. The path of a projectile under the influence of gravity ideally follows a curve of this shape. You will know that it is a quadratic equation if it has x².

## PARABOLAS IN REAL LIFE

Parabolas in real life

The graph of a quadratic relation is called a parabola. The Parabola has some important features......

1. Vertex: The point where the axis of symmetry and the parabola meet, it is the point where the parabola is at its maximum value (if the direction of opening is down) or minimum value (if the direction of opening is up) and it is the point where the graph changes directions. It is labelled as (x,y).
2. Optimal Value/ Minimum or Maximum Value: The value of the y co-ordinate of the vertex. It can be the lowest or highest values/ points. It is labelled as "y".
3. Axis of Symmetry: A vertical line that goes through the vertex and divides the parabola into two equal halves. It is labelled as "x".
4. Y- Intercept: Where the graph crosses the y-axis. It can be labelled as (0,y).
5. X- Intercept: Where the graph crosses the x-axis. It can also be called as "zeros" or "roots". It can be labelled as (x,0).

Big Idea – need more than a line, some things go up and then down, they have x- intercepts, a vertex, axis of symmetry and an optimal value USING TABLE OF VALUES, TECHNOLOGY AND FIRST DIFFERENCES TO ANALYZE QUADRATICS

To accurately graph any quadratic relation, you need to know the "key" points of the basic quadratic relation (y= x²): the simplest parabolic curve with no transformations applied. You can use a table of values to identify the "key" points for y= x² (the basic function of a parabola)...... You can determine if an equation is linear or quadratic by the use of a table of values.....

• If the relationship is linear, the first differences on the chart are constant
• If the relationship is quadratic, the second differences on the chart are constant, but the first differences are not But, you can also use a graph......

You can also determine if an equation is linear or quadratic by plotting the points on a graph...

• If the relationship is linear, the points on the graph lie along a line
• If the relationship is quadratic, the points on the graph will not lie along a line, but will form a curve

The standard parabola: y = x²
• is upright
• has a minimum turning point at (#, #)
• has an axis of symmetry : x = #
• has y-intercept at (0, #)
• has x-intercept at (#, 0)

WORD PROBLEM

1. A baseball is tossed into the air and follows the path h = -2t²+ 6t , where t is the time, in seconds, and h is the height of the baseball, in metres.

a) Sketch the path of the baseball. b) What is the maximum height of the baseball?
This question is asking for the optimal value (also known as the y value of the vertex). By analyzing the graph, you come to the conclusion that the maximum height of the baseball is 4.5m.

c) At what time will the baseball reach its maximum height?

This question is asking for the value of then x value in the vertex. By analyzing the graph, you come to the conclusion that the baseball reaches its maximum height of 4.5m at 1.5 seconds.

## Investigating Vertex Form The standard form of the equation of a quadratic function: y= ax² + bx + c can go through a variety of transformations to form y= a (x-h)² + k.

Possible Transformations Include:

1. To graph y= ax² ,stretch or compress the graph of y = x² vertically by a factor of "a"

• if a < 0 (so if it's negative), the parabola is reflected in the x-axis
• If a > 1 or a < -1, then the graph is stretched vertically (narrows)
• If -1 < a < 0 or 0 < a < 1, then the graph is compressed vertically (widens) 2. To graph y= x²+ k, translate the graph of y = x² vertically k units

• if k > 0 (so if it's positive), then the graph is translated k units upward
• if k < 0 (so if it's negative), then the graph is translated k units downward 3. To graph y= (x- h)², translate the graph of y = x² horizontally h units

• if h > 0 (so if it's negative), then the graph is translated h units to the right
• if h < 0 (so if it's positive), then the graph is translated h units to the left  VERTEX FORM WITH STEP PATTERN

The step pattern is a pattern that identifies your next points in each parabola.

The original step pattern is "Over 1 up 1 and Over 2 up 4"

The value of how many steps up is the square of the value of the steps over.

This pattern shows you that when the point moves 1 over from the vertex, it then moves 1 up, and this pattern continues.

But in quadratic relations if there is an "a" value the step pattern changes.

If there is an "a" value then the squared number in the pattern needs to be multiplied by the "a" value.

Example: y= -2 (x-3)² + 5

Vertex: (3,5)

The step pattern would now be "Over 1 up -2 and Over 2 up -8"  WORD PROBLEM INVOLVING VERTEX FORM

1. The area available for a swimming pool in a backyard is a square with side length 20m. The square swimming pool is to be placed in the centre of the area. If the side length, in metres, of the swimming pool is x, then the area of the backyard remaining is given by the relation A = -x² +400.

a) Graph the relation. b) What are the intercepts? What do they represent?

By analyzing the graph, you come to the conclusion that the y- intercept is 400. This represents the area of the backyard if there is no swimming pool. You also come to the conclusion that the x-intercept is 20. This represents the side length of the swimming pool, in metres, if the pool completely fills the backyard.

FINDING EQUATIONS AND ANALYSING VERTEX FORM

Here are the steps to figuring out the equation when given the vertex and another point on the graph:

1. Sub the coordinates from the vertex into the equation.

• remember: y= a (x-h)² +k
• vertex= (h,k)
• h= opposite value (positive= negative and negative= positive)
2. Sub the coordinates from a point on the graph into the equation.
• sub the y value into the y of the equation
• sub the x value into the x of the equation
3. Now, solve and isolate for the a value since you have all the other values.

4. Write the equation.

Example: Analysing the given information from an equation to graph

Example 1: Write an equation for a parabola with vertex (3, -4) and passing through (2, -7).

y = a(x - h)² + k

The question provides the vertex, (3, -4). Thus (h, k) = (3, -4).

y = a(x - 3)² - 4

(2, -7) is on the quadratic. Thus (x, y) = (2, -7). Substitute these values into the equation: Therefore, the equation of the quadratic is y = -3(x - 3)² - 4

HOW TO COLLECT INFORMATION THAT CAN BE USED FOR GRAPHING IN VERTEX FORM

By using the vertex form, you can easily gather the information needed to graph the parabola (ex. you will find the vertex, direction of opening, axis of symmetry, etc.)

Example 1: y= (x+3)²

Vertex: (-3,0)

Axis of symmetry: X=-3

Stretch or compression factor relative to y=x²: None

Direction of opening: Upward

Value of x: X=-3

Value of y: Y=0

Example 2: y=(x+2)²+5

Vertex: (-2,5)

Axis of symmetry: X=-2

Stretch or compression factor relative to y=x²: None

Direction of opening: Upward

Value of x: X= -2

Value of y: Y=5

Example 3: y=3(x+7)²-2

Vertex: (-7,-2)

Axis of symmetry: X=-7

Stretch or compression factor relative to y=x²: 3

Direction of opening: Upward

Value of x: X= -7

Value of y: Y=-2

REAL LIFE SITUATION INVOLVING VERTEX FORM #1

The path of a football is modelled by the relation h= -0.25 (d-12)² +36, where d is the horizontal distance, in metres, after it was kicked, and h is the height, in metres, above the ground.

a) Sketch the path of the football. b) What is the maximum height of the football?

The question is asking for the maximum height, which means that you need to find the optimal value (also known as the y-value of the vertex). In the form y= a (x-h) ² +k, k tells you the optimal value, which is the maximum height. Therefore, the maximum height of the football is 36m.

c) What is the horizontal distance when this occurs?

The question is asking for the horizontal distance when the football is at it's maximum height, which means that you need to find the x-value of the vertex. In the form

y= a (x-h) ² +k, h gives you the x-value of the vertex, but it needs to be changed to the opposite sign. Therefore the horizontal distance when the football is at it's maximum height is 12m.

d) What is the height of the football at a horizontal distance of 10m?

Since d represents the horizontal distance, we need to set the value of d in the equation h= -0.25 (d-12)² +36 to 10.

h= -0.25 (10-12)² +36

h= -0.25 (-2)² +36

h= -0.25 (4) +36

h= -1+36

h= 35

Therefore, the height of the football at a horizontal distance of 10m is 35m.

HOW TO FIND THE Y-INTERCEPT IN VERTEX FORM?

• Just to remind, the equation of a vertex form is y= a(x-h)² + k

• To find the y-intercept, first substitute y = 0.

• Then solve the equation step by step to find the y-value. Therefore the y-intercept is -24, and can be written as (0,-24).

HOW TO FIND THE X-INTERCEPTS (ZEROS) IN VERTEX FORM?

Here are the step-by-step instructions:

1. Set y = 0
2. Move k to the other side.
3. Divide both sides by "a"
4. Square root both sides
5. Move h to the other side.
6. You will get solutions, solve each of them REAL LIFE SITUATION INVOLVING VERTEX FORM #2

A football is punted into the air. Its height h, in metres, after t seconds is

h = -4.9(t - 2.4)² + 29

a) What was the height of the ball when it was kicked? Therefore, the ball was kicked from a height of 0.776 m.

b) What was the maximum height of the ball?

The maximum height is the optimal value of the parabola.

The k value in the equation represents the optimal value.

Therefore, the maximum height of the ball is 29 m.

c) When does the ball hit the ground?

Here we have to find the value of t when h = 0

Because we are looking for the time in seconds when the height is 0. Therefore, the ball hits the ground at approximately 4.83 seconds.

VIDEO: GRAPHING USING VERTEX FORM

Graphing Using Vertex Form

Big Idea – Vertex Form tells us about what the graph looks like

## Factored Form

• Factored Form Equation: y= a (x-r) (x-s)
• Zeros: x-intercepts= (r,0) (s,0) the r and s value are opposite values (positive= negative and negative= positive)
• Axis of Symmetry: a vertical line between the two zeros; gives you the x co-ordinate of the vertex
• formula to find the x co-ordinate of the vertex: x= r+s / 2
• optimal value: sub the x co-ordinate of the vertex into the equation to find the optimal value (the y co-ordinate of the vertex)
• direction of opening: up= a value is positive and down= a value is negative WHAT EACH VARIABLE REPRESENTS

-The zeros/x-intercepts: are the r and s value you find them by setting each (factor) equal to zero.

-The axis of symmetry: is the midpoint of the two zeroes

-The optimal value: is found by subbing the axis of symmetry value into the equation

Example 1: y=0.5(x+3)(x-9)

1) find the zeroes

y=0.5(x+3)(x-9)

0=0.5(x+3)(x-9)

x+3=0 and X-9=0

x=0-3 and x=0+9

x=-3 and x=9

2) find the axis of symmetry

X=-3+9/2

x=6/2

x=3

3) state the optimal value

y=0.5(x+3)(x-9)

y=0.5(3+3)( 3-9)

y=0.5(6)(-6)

y=-18

Example 2: y=2(x+1)(x+4)

1) find the zeroes

y=2(x+1)(x+4)

0=2(x+1)(x+4)

X+1=0 and X+4=0

X=0-1 x=0-4

x=-1 x=-4

2) find the axis of symmetry

X=-1+-4/2

x=-5/2

x=-2.5

3) state the optimal value

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

y=2(-2.5+1)(-2.5+4)

y=2(-1.5)(1.5)

y= -4.5

GRAPHING FROM FACTORED FORM

• You need 3 key points to graph your parabola from your equation in factored form.

• There are 4 main steps to graph your equation.

1. Find the x-intercepts of the relation.
2. Find the x-value of the vertex, which is the same as finding the axis of symmetry.
3. Find the y-value of the vertex, which is the same as finding the optimal value.
4. Now plot the 3 key points (two x-intercepts and the vertex) on the graph. Then connect the points and graph your parabola.

*Note that you might not always have two x-intercepts, there are cases where you have only 1 or NO x-intercept.

1. Finding the x-intercepts

• The example that we will be looking is :

y = (x+4)(x+2)

• Take your equation and set y equal to 0.

• You need to find to seperate solutions, therefore set each bracket equal to 0.

• Solve ecah bracket.

• You will get an answer that is x = # , write it as a point (X,0). 2. Finding the x-value of the vertex

The x-value of the vertex is the same as the axis of symmetry.

Use the formula (r+s)/2 to find the x-value.

You simply take the sum of the two x-intercepts that found before and divide it by 2. 3. Finding the y-value of the vertex

The y-value of the vertex is the same as the optimal value.

Take your original equation and replace the x with x-value of the vertex that we found.

Then solve for y 4. Graphing the equation using the key points

Equation : y = (x+4)(x+2)

Plot the 3 key points that we found through the above steps.

1. x-intercept #1 : (-4,0)

2. x-intercept #2 : ( 2,0)

3. Vertex : (-1,-9) REAL LIFE SITUATION INVOLVING FACTORED FORM

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 h is the height, in meters, above the ground.

a) Sketch a graph of the path of the rocket. b) How far from the wall is the rocket when it is launched?

When it comes to questions that ask about the distance that an object started at, it is referring to the x-intercept. The distance the object started at will be the smallest number between the two x-intercepts. Therefore, according to this equation, the rocket was launched 2m away from the wall.

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

When it comes to questions that ask about the distance that an object hits the ground at, it is referring to the x-intercept. The distance the object hits the ground at will be the larger number between the two x-intercepts. Therefore, according to this equation, the rocket lands on the ground 12m away from the wall.

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

This question is asking you to find the vertex. So, we will start off by solving for the axis of symmetry to get the distance and then substituting that value (the value of x) into the equation to get the height. Therefore, the rocket will reach a maximum height of 75m when it is 7m away from the wall.

CONVERTING FACTORED FORM INTO STANDARD FORM

To convert the equation of a quadratic relation from factored form to standard form,

we will expand, regroup, then simplify the factored form. VIDEO: GRAPHING USING FACTORED FORM
Graphing Using Factored Form

Big Idea – find x int by setting y=0, and vertex is between the two x-ints

## Multiplying Binomials

Factoring is an extremely important operation in mathematics. It is used in situations where a polynomial needs to be broken down unto its simplest parts. WORD PROBLEM INVOLVING MULTIPLYING BINOMIALS

1. A rectangle has width w centimetres and length 3 cm more than its width.

a) Draw a diagram of the rectangle. b) Express the area as a product.

The original formula for the area of a rectangle is a=(l)(w). Therefore, the formula to find the area is a= w(w+3).

c) Expand and simplify the area expression. Big Idea – can get from Factored form to Standard Form

## Common Factors and Simple Grouping

Common factoring should always be the first thing you look for when you have an equation in standard form. You should look to see if each of the terms in the equation has a similar component that each of the terms can easily be divided into. The common factor can either be a variable (x) or a coefficient (any number.)

Common factoring:

Example 1: 8x + 6

Step #1: Find GCF ( what is the greatest # in common between 8 and 6)

GCF=2

Step #2: Write solution with brackets

8x/2 + 6/2 = 4x + 3

= 2(4x + 3)

Example 2: 12x³ - 6x²

step #1: GCF= 6x²

Step #2: 6x²(2x-1)

6x²(2x-1)

=12x³ - 6x²

Example 4: 2x(x+4) -5(x+4)

Step #1: GCF= (x+4)

Step #2: (x+4) (2x-5)

WORD PROBLEM INVOLVING COMMON FACTORS AND SIMPLE GROUPING

Write an expression, in factored form, for the shaded region. a=(5x)(9x)- (3y)(5y)

a= 45x²- 15y²

The gcf of 45 and 15 is 15.

a= 15 (3x²- y²)

Big Idea – Factoring will take us from Standard form back to Factored form

## Simple trinomials

Simple trinomials have 3 terms, and can be written in the form y = ax² + bx + c

Many polynomials such as x² + 7x + 12, can be written as the products of 2 binomials or the form (x+r) and (x+s).

Expanding:

x² + 7x + 12= (x+3) (x+4)

Factoring: When factorign a polynomial of the form ax² + bx + c (when a=1), we want to find:

1. Two numbers that ADD to give you the value of b
2. Two numbers that MULTIPLY to give you the value of c

Example: Factor x² - 8x + 12

= (x-6) (x-2)

-6 + -2= -8

-6 x -2= 12

## NOTE REAL LIFE SITUATION INVOLVING SIMPLE TRINOMIALS

The height of a rock thrown from a walkway over a lagoon can be approximated by the formula h= t² -8t+ 15, where t is time in seconds, and h is the height, in metres.

a) Write the formula in factored form. b)When will the rock hit the water?

In the form y= a (x-r) (x-s) the r and s values gives you the values of the x-intercepts but it needs to be changed to the opposite sign. The question asks for the distance at which the rock hits the water, so it is asking for the value of the second x-intercept. Therefore, the rock will hit the water in 5 seconds.

Big Idea – Factoring will take us from Standard form back to Factored form

## Complex Trinomials

Complex Trinomials follow the same rules as Simple Trinomials (see above), however the "a" term on a complex trinomial is not 1. That makes it a little trickier, and more often the method trial and error or guess and check will have to be used.

## Note   REAL LIFE SITUATION INVOLVING COMPLEX TRINOMIALS

The height of a ball thrown from the top of a ladder can be approximated by the formula h=-2t²+4t+48, where t is the time, in seconds and h is the height, in metres.

a) Write the formula in factored form. b) Determine when the ball will hit the ground.

In the form y= a (x-r) (x-s) the r and s values gives you the values of the x-intercepts but it needs to be changed to the opposite sign. The question asks for the distance at which the ball hits the ground, so it is asking for the value of the second x-intercept. Therefore, the ball will hit the ground in 6 seconds.

Big Idea – Factoring will take us from Standard form back to Factored form

## Perfect Squares

What is a perfect square?

Perfect Square Trinomial is the product of two binomials. But, both the binomials are same. When factoring some quadratics which gives identical factors, that quadratics are Perfect Square Trinomials.

In this section, you will learn to recognize trinomials of the form x² + 2ax + a² and factor them to (x+a)².

Example of a perfect square:

x² + 10x + 25

= (x + 5)(x + 5)

= (x + 5)²

Question 1: Factor the trinomial x² + 4x + 4
Solution:

Given x² + 4x + 4

The above trinomial can be written as x² + 2x + 2x + 4

Take x as common from first two terms,

x(x + 2) + 2x + 4

Take 2 as common from last two terms.

x(x + 2) + 2(x + 2)

(x + 2)(x + 2)

In this it is the product of two identical binomials

(x + 2)2 , which is a perfect square.

Question 2: Factor the trinomial x² + 8x + 16
Solution:

Given x² + 8x + 16

The above trinomial can be written as x² + 4x + 4x + 16

Take x as common from first two terms,

x(x + 4) + 4x + 16

Take 4 as common from last two terms.

x(x + 4) + 4(x + 4)

(x + 4)(x + 4)

In this, it is the product of two identical binomials

(x + 4)2, which is a perfect square.  REAL LIFE SITUATION INVOLVING PERFECT SQUARES

The cost, in dollars, of operating an appliance per day is given by the formula

C= 2t² -24t +150, where t is the time, in months, the appliance is running. What is the minimum cost of running the appliance and when does this happen?

C= 2t² -24t +150

C= 2 (t² -12t) +150

C= 2 (t² -12t +36 -36) +150

C= 2 (t² -12t +36) -36 +150

C= 2 (t² -12t +36) -72+150

C= 2 (t² -12t +36) +78

C= 2 (t-6)² +78

In the form y= a (x-h)² +k, the h value gives you the x value of the vertex but you need to change it to the opposite sign and the k value gives you the y value of the vertex. Therefore, the minimum cost of running the appliance is \$78 and this occurs when the appliance runs for 6 months.

Big Idea – Factoring will take us from Standard form back to Factored form

## Difference of Squares

When working with a difference of squares, you must check to see that:

1. Each term is a squared term
2. There is a minus sign between the two terms
3. The two terms that are squared are separated by a subtraction sign like this: a² - b²
This is useful because it can be factored into (a+b)(a−b)  REAL LIFE SITUATION INVOLVING A DIFFERENCE OF SQUARES

A square has side length 8a. One dimension is increased by 3b and the other is decreased by 3b.

a) Find an algebraic expression for the resulting area. Expand and simplify.

Using the formula for the area of a square you will plug in the information provided to create a formula.

Given Information:

• Side length= 8a
• One side= +3b
• Another side= -3b

Therefore the equation is (8a+3b) (8a-3b).

a= (8a+3b) (8a-3b)

a= 64a²-9b²

b) Calculate the area of the square if a represents 4cm and b represents 2cm.

a= 64a²-9b²

a= 64 (4)²-9 (2)²

a=64 (16)-9 (4)

a= 1024- 36

a= 988

Therefore, the area of the square is 988cm².

Big Idea – Factoring will take us from Standard form back to Factored form

## RECAP ## COMPLETING THE SQUARE

Completing the square is a technique used to solve quadratic equations, graph quadratic functions, and evaluate integrals. This technique can be used when factoring a quadratic equation does not work or to find irrational and complex roots. You learn how to rewrite a quadratic relation of the form y= ax² +bx +c in the form y= a(x-h)² + k by completing the square.

## Note    ## Examples

REAL LIFE SITUATION INVOLVING COMPLETING THE SQUARES #1

1.The path of a ball is modelled by the equation y= -x² +2x +3, where x is the horizontal distance, in metres, from a fence and y is the height, in metres, above the ground.

a) What is the maximum height of the ball, and at what horizontal distance does it occur?

y= -x² +2x +3

y= -1(x² -2x) +3

y= -1(x² -2x +1- 1) +3

y= -1(x² -2x +1) -1+ 3

y= -1(x² -2x +1) +1+3

y= -1(x² -2x +1) +4

y= -1(x² -1)² +4

Therefore the maximum height of the ball is 4m and this occurs when the horizontal distance is at 1m.

REAL LIFE SITUATION INVOLVING COMPLETING THE SQUARES #2

2. Alex runs a snowboard rental business that charges \$12 per snowboard and averages 36 rentals per day. She discovers that for each \$0.50 decrease in price, her business rents out two additional snowboards per day. At what price can Alex maximize her revenue?

Let r= the total revenue, in dollars

Let x= the number of \$0.50 decreases in price

Revenue is the product of the price and the number rented.

r= (12 -0.5x) (36+ 2x)

r= (12 -0.5x) (36+ 2x)

r= 432+ 6x -x²

r= -x² +6x +432

r= -1(x² -6x) +432

r= -1(x² -6x +9 -9) +432

r= -1(x² -6x +9) -9 +432

r= -1(x² -6x +9) +9 +432

r= -1(x² -6x +9) +441

r= -1(x²-3)² +441

The relation reaches a maximum value of 441 when x=3.

There should be three price reductions of \$0.50 to maximize the revenue.

12- 0.5 (3)= 10.50

Therefore, a price of \$10.50 maximizes Alex's revenue.

Big Idea – getting from standard form to vertex form

## Note ## Example ## Word Problem ## Standard form in algebra tiles

You can also use algebra tiles, to convert the same equation into standard form.

Think of this like a room with dimensions (x+2) by (x+1). The area of the room is (x+2)(x+1). We are trying to determine another way of expressing the area. · To solve a quadratic equation from standard form, we use the Quadratic formula.

· We could find the x-intercepts of the equation using this formula.

· We use this when it is hard to solve an equation by factoring.

Ex: When there are decimals

· We also use quadratic formula in situations where the equation doesn't factor at all.

The quadratic formula is used to solve a very specific type of equation, called a quadratic equation. These equations are usually written in the following form: ## Examples  FINDING THE AXIS OF SYMMETRY

• Finding the axis of symmetry is the same as finding x-value of the vertex.

• You can use a simple formula to find the axis symmetry when the equation is it's standard form.

• You could also use the previous method that we learned but if you made a mistake in finding your x-intercepts then your vertex will also be wrong.

The formula is: EXAMPLE

y = 2x² +3x -5

a = 2

b = 3

axis of symmetry : x = -b / 2a

= -(3) / 2(2)

= -3 / 4

= -0.75 WORD PROBLEM

• Now let's look at a real life situation in which we could solve and equation given in standard form.

• Knowing how to find the axis of symmetry, optimal value, vertex and x-intercepts will help you answer the questions from a word problem. VIDEO: GRAPHING INVOLVING FACTORED FORM
Graphing Using Standard Form

## recap ## Connections

Throughout this unit I have notices that every topic that we learnt within the unit can be linked. For example, you can change vertex form to standard form and vice versa, factored form to standard form and vice versa, and vertex form to factored form. Now converting the equations into the 3 different forms can require either expanding, factoring, completing the square, using the quadratic formula, etc. In the process of these conversions you also will gather the necessary information to be able to graph the parabola of the equation (vertex, x-intercepts, axis of symmetry, etc.) A more specific example of this would be that through the equations we gather the information needed to graph the parabola, but in standard form you will not be able to gather any information by using the equation. So, you must convert it to either factor or vertex form to proceed with the procedure to graph the parabola. The quadratics unit is all linked together and knowing and understanding each topic and how they connect to each other will help you better understand the unit as a whole.

## REFLECTION

Throughout the unit, I have learned a lot about quadratics. When we first started this unit, I felt kind of overwhelmed with all the information coming at me. I immediately thought that quadratics was going to be hard and I was going to fail the whole unit. But once we started the unit bit by bit, (mini test by mini test), I started to understand it better and like it more. One of the reasons for why I ended up enjoying this unit is because I listened to my teacher (with some distraction here and there), took notes , did my homework and reviewed. Thus my mini test reflected on my studying and it shows how my hard work actually payed off. I also learnt how easy it is to fall behind and how hard it is to catch back up, and from that I learned I need to work on asking people for help instead of getting frustrated about it and not getting anything done. This explains what happened with my minitest. I didn't really understand well so instead of clarifying my doubts, I chose to try and figure it out by myself..... and judging from my mark it is obvious that idea didn't work out so well. But after I had some help and did more practice I became familiar with my mistakes and was able to correct them. For instance, for this question on one of my tests I got 1 mark out of 5, and this really upset me. But I didn't let that stop me from finding out how to solve this question the right way. I was on the right track but instead of trying to solve this question using a chart (which takes a lot of time), I should have just solved it simply, with the less time consuming method of solving with an equation (as shown below).  On the other hand, I have had some really good tests. This was one test in which I was confident with and I really enjoyed learning about this unit. This just proves that with hard work and determination, I can be successful in math!