### By: Bryan Boodhoo

1. Explaining the basics

2. First and Second Differences

3. Parts of Vertex Form

4. Finding Equations from a graph

5. Graphing

6. Graphing in Factored Form

1. Common Factoring

2. Simple Trinomial Factoring

3. Complex Trinomial Factoring

4. Perfect Square Factoring

5. Difference of Squares

1. Completing the Square

3. The Discriminant

WORD PROBLEMS:

OTHER STUFF (MY VIDEO, REFLECTIONS, ASSESSMENTS, ETC.)

1. My Video

2. Assessment

3. Connecting Topics Together

4. Reflection

5. Other Videos to help (Not mine)

## 1. Quadratics 1: Explaining the basics

In the Quadratics 1 unit, it is mainly focused on understanding how parabolas are graphed and how to graph these parabolas when given the "Vertex" and "Factored" forms.

The parts of a parabola are clearly shown in the picture to the right. As you can see, a parabola may seem like a curved slope with a dip in the middle. Not to confuse you, there are similar principles between a slope and a parabola such as the x and y intercepts.

## 2. First and Second Differences

If given a table with x and y values, there is a way to tell if there is a Linear or Quadratic relationship. In order to find this out, we use First and Second Differences. If the first differences are all not the same, but the second differences are, then there is a Quadratic relation. If the first differences are all the same, then there is a Linear Relation ## 3. Parts of a Simple Vertex Form Equation: y=a(x-h)^2+k

The equation of y=a(x-h)^2+k is the most simple equation to explain how a quadratic equation in vertex form works.

a= The vertical stretch (how tight or loose the arms of the parabola will be)(If negative, parabola will open downwards)(You will use this number to find the step pattern when graphing a an equation)

h= The horizontal shift (how far the parabola moves from side to side)(If negative, vertex and axis of symmetry will be on right side of y-axis; if positive, vertex and axis of symmetry will be on left side of y-axis)

k= The vertical shift (how high the parabola moves up and down)

The vertex will be the coordinates that h and k will take place of and will be in the (x,y) form.

## 4. How to Determine the Equation from a Graph Recall the first diagram of a parabola that you looked at and see if you can find any of those parts on this graph, then write down those coordinates for those parts.

Using this knowledge, we can see that the vertex of this parabola is (1,5)

Since the parabola opens downward, we now know that "a" will be negative.

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

y=a(x-1)^2+5 (Subbed in vertex)

The last step now, is substituting a final point that the parabola is passing through. To do this, we simply find 2 coordinates and substitute them in place of x and y.

The point that we will be using is (0,2)

2=a(0-1)^2+5

2=a(-1)^2+5

2-5=a

-3=a

After finding the value of "a" the final equation is y=-3(x-1)^2+5

## 5. Graphing a Quadratic Equation

In order to do this, you will need to learn how to find the step pattern as well as the vertex.

Graphing y=(x+4)^2+3 will be fairly simple due to the fact that there is no altering of the step pattern involved. (Will be shown after this) One thing that we have to remember is that the basic step pattern when there is no "a" value will always be 1 to the side, 1 up, then, 2 to the side, 4 up. We must also remember that the step pattern will be applied to both sides of the vertex. From this equation, we can already tell that the vertex is shown as (-4,3) but will be drawn as (4,3). Now that we have this point all we have to do now is use the regular step pattern that we just learned and apply it to this vertex. The Final graphed parabola is shown above with the correct vertex and the correct step pattern.

## 6. Graphing in Factored Form

Lets start this off with a simple example.

Graph: y=(x+5)(x+1)

The equation we have here is written in a form called "Factored Form." It may seem different than the usual vertex form but it is not that hard to graph.

To start this off we must find the x-intercepts which are 5 and 1.

After this, we have to find the axis of symmetry. To find this we must get the 2 x-intercepts, add them together, then divide it by 2. From this we will get -3

Since we have found the AOS, this will also be our x value, now all we have to do is find the y value.

The final thing we have to do is find the vertex. To do this we must substitute the axis of symmetry for the x values

It will look like this when the AOS is subbed in: (Now we solve for y)

y=[(-3)+5][(-3)+1]

y=(2)(-2)

y=-4

There we have it:

x- intercepts are 5 and 1

Vertex is (-3,-4) What is factoring? Well, you may recall that you have previously learned to add, subtract, multiply, or divide a set of numbers in order to get one final number.

For example, 3x2=6

But, factoring is the exact opposite.

We need to start from the number 6, and find what numbers can be multiplied to get this.

The Quadratics 2 unit will be focused on all the different types of factoring.

## 1. Common Factoring

To common factor, you must find the GCF between all of the terms.

Factor 3x-12

To factor this, we must first find the GCF, which is 3. We know this is 3 because it is the lowest number that can be divided by all of the terms here.

Next we have to pull the GCF out and put the other terms in brackets

It will looks like this:

3(3x-12)

Then, we have to divide everything inside of the brackets by the GCF

3(3x-12)

3

After that, we are left with the final expression, which is:

3(x-4)

## 2. Simple Trinomial Factoring

The next type of factoring that we will learn is "Simple Trinomial Factoring."

The reason why it is given this name is because there are 3 terms and there is no coefficient in front of the x^2

Factor: x^2+5x+6

First we must look to see if there is a common factor in the expression, in this case there is none so we can skip that step.

Next, we must separate the x^2 into 2 brackets.

(x+ )(x+ )

Then, we must find two numbers that when multiplied, equal to 6 and also when added, equal to 5.

In this case the two numbers are 3 and 2 so we can input them into the brackets.

(x+2)(x+3)

There it is! We have our final answer.

Check (Fast way): In order to check that we have the right answer we must go from factored to standard form.

To do this we multiply the x terms (x^2)

Next we add the two terms 2 and 3 and add an x after it (5x)

Finally we multiply the 2 same terms to get the final number

This is how we end up where we started at x^2+5x+6

## More thorough way (Rainbow Method)(Same as above) ## 3. Complex Trinomial Factoring

Complex trinomials are similar to simple trinomials, its just that complex trinomials have a coefficient greater than 1.

Lets factor the expression: 6x^2+29x+35

Since there is a coefficient greater than 1 infront of the first term, we must find out 2 numbers that can be multiplied to get 6 and then separate them in the brackets.

(3x+ )(2x+ )

Next, we have to find 2 numbers that when multiplied, equal to 35.

This is the most important step and it will include trial and error due to the guess and check method.

The numbers that we will list are: 7 & 5 and 1 & 35

Then we input these numbers

(2x+5)(3x+7)

After this, we have to use the outside inside method to get the number "29"

In the picture below we will have to multiply the numbers 2 & 7 and the numbers 5 & 3, then add them too see if we get 29

And we do! So the final expression will be (2x+5)(3x+7) ## 4. Perfect Square Factoring

For this type of factoring, we are looking at how the expression has become a perfect square. A good example that we are going to solve is the expression a^2+2ab+b^2

The whole idea of perfect squaring is that every part of the expression has a perfect square root.

a=square root of a^2

ab=square root of 2ab

b=square root of b^2

Just like simple trinomial factoring, we must separate the first term in 2 brackets.

(a+ )(a+ )

Then we must find 2 terms that equal to b^2

In this case they are b and b

(a+b)(a+b)

We get the the second term "2ab" from multiplying the two terms by 2

Our final expression will be (a+b)^2 because it is in fact, a perfect square since (a+b) can be multiplied by itself in order to get a^2+2ab+b^2

## 5. Difference of Squares

To factor Difference of Squares equation, we must first know that there has two be a positive and negative term which makes the name "Difference."

Lets factor the expression 16x^2 - 25y^2

As we can see, the two squares for the two terms are (4x)^2 - (5y)^2

Just like complex trinomial factoring, we have to separate the first term by finding 2 numbers that equal to 16.

(4x+ )(4x- )

One of the brackets have to have a positive and one has to have a negative sign.

We then take the square of 25 and input it into each bracket.

(4x+5y)(4x-5y)

In the picture below it shows how to check if you got the right answer In the Quadratics 3 unit, the first topic we will focus on is completing the Square. You may recall the different forms of expressions and equations which include Vertex form, Factored form, and Standard form. Well, for completing the square we will be working on converting the Standard form to Vertex form. For the second and third topics, we will focus on using the quadratic formula and the discriminant to find and analyze the x-intercepts.

## 1. Completing the Square

For completing the Square, you will be taught how to convert an equation from standard form to vertex form. This technique will come in handy when you cannot graph a standard form equation.

Lets complete the square for the equation: y=-2x^2+28x-15

y=-2x^2+28x-15

y=(-2x^2+28x)-15 (Separate the x terms)

y=-2(x^2-14x)-15 (Common factor the -2 as the "a" value)

y=-2(x^2-14x+49-49)-15 (Divide 14 by 2 then Square it, also subtract the same number)

y=-2((x-7)^2-49)-15 (Equation becomes a perfect square)

y=-2(x-7)^2+98-15 (Then expand using the -2)

y=-2(x-7)^2+83 (Add/Subtract the final values outside the brackets)

Final equation: y=-2(x-7)^2+83

We have looked at solving quadratic equations by factoring, but this method may not always work for every single type of equation. The quadratic formula can be used in cases for equations that cannot be factored. The reason as to why there is a plus/minus symbol is because there are two possible zeroes that can be found when you choose the adding or subtracting option. ## Example

Lets try and solve to find the zeroes for the equation, 2x^+9x+6=0

For the quadratic formula, the letters a,b, and c are represented in the equation you are going to solve by the first, second, and third terms. The first term will represent a, the second will represent b, and the third will of course represent c.

a=2

b=9

c=6

Then, we must next input the values of a,b, and c. We then continue to find the zeroes by solving this equation with the values that were just inserted As said before, the plus/minus sign means that it can be subtracted or added. So we will square 33, then add/subtract it from -9, and finally divide it by 4.

Finally we have the values for the two zeroes which will be 0.81 and 3.68.

## 3. The Discriminant

The number inside the square root of the quadratic formula is called the Discriminant.

The Discriminant of the quadratic formula tells us how many solutions the formula has. To simplify this, if the discriminant is:

Greater than 0, there are two solutions

Exactly 0, then there is only one solution

Less than 0, there is no real solution

Examples:

Find out how many solutions there are to the discriminant, (-4)^2-4(1)(7)

(-4)^2-4(1)(7)

-> 16-28

-> -12

No real solutions since the discriminant is less than 0

Find out how many solutions there are to the discriminant, (3)^2+4(5)(2)

(3)^2+4(5)(2)

-> 9+40

-> 49

2 solutions since the discriminant is more than 0

Find out how many solutions there are to the discriminant, (4)^2-4(2)(2)

(4)^2-4(2)(2)

-> 16-16

->0

No real solutions since the discriminant is 0

## WORD PROBLEMS

A couple of word problems with solutions showing you how to complete them from Quadratics 1, 2, and, 3.

The path of a soccer ball is represented by the equation h=-0.06d(d-50), where h represents height, in metres, of the soccer ball above the ground and d is the horizontal distance, in metres, of the soccer ball from the player.

At what horizontal distance does the ball land?

The ball lands at 50 metres.

Determine the length and width of a rectangle with an area of x^2+17x+60

For this problem we will have to do simple trinomial factoring.

So,

(x+12)(x+5)

The reason why this is the answer is because in simple trinomial factoring we have to split to first term. (x and x)

Then we have to find two numbers which can be added to get the second term. (12 and 17)

But, at the same time, these 2 numbers also have to multiply to get the final term. (60)

A ball is thrown into the air. Its height in metres after t seconds is given by the equation, h=-4.9(t-2.4)^2+29

What was the height of the ball when it was thrown?

sub t=0

h=-4.9(0-2.4)^2+29

=0.776

The ball was 0.776 metres high when it was thrown.

What was the maximum height of the ball, at what time?

The maximum height of the ball was 29 metres at 2.4 seconds

How high was the ball after 3 seconds?

sub t=3

h=-4.9(3-2.4)^2+29

=27.2 m

The ball was 27.2 metres high after 3 seconds

## 1. The Video I made - How to Graph a parabola in vertex form

How to graph a parabola Bryan Boodhoo

## 3. Connecting topics Together

If you look at it closely, everything in quadratics was somehow related to another topic. One of the easiest relations to see was how every type of equation from Quadratics 1,2, and 3 was possible to graph. Also, it was possible to convert the three types of equations (Vertex form, factored form, and standard form) from one to another form. For example, since I have successfully completed all of the quadratics units, it is possible for me to graph a standard form equation in different ways. I could do this by completing the square, factoring, or even using the quadratic formula. Overall, it is evident to see from my explanation as to how everything in quadratics is related in a special way.

## 4. Reflection

Overall, I found that Quadratics was a very interesting unit. In the beginning I wasn't very confident with it but later I put the pieces together and started to improve. I noticed that towards the factoring unit in Quadratics 2 is where I felt the most confident with my work. Also, with the word problems in the Quadratics 3 unit I didn't really understand it, but later I looked into it with more detail and went at it with a better approach. In the Quadratics unit we learned vertex form, factored form, and standard form. The whole Quadratics unit was basically like a step up from the Linear systems unit that we learned in grade 9. To conclude, I learned at lot of new things through trial and error.

## 5. Other Videos to help (Not mine)

3.3 More Graphing from Vertex Form
3.11 Factoring
Common Factors
3.8 Factoring Simple Trinomials
3.9 Complex Trinomial Factoring