# Quadratics

### By: Bryan Boodhoo

## Table of Contents:

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

QUADRATICS 2:

1. Common Factoring

2. Simple Trinomial Factoring

3. Complex Trinomial Factoring

4. Perfect Square Factoring

5. Difference of Squares

QUADRATICS 3:

1. Completing the Square

2. Quadratic Formula

3. The Discriminant

WORD PROBLEMS:

1. Quadratics 1 word problem

2. Quadratics 2 word problem

3. Quadratics 3 word problem

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

*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

## 3. Parts of a Simple Vertex Form Equation: y=a(x-h)^2+k

**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

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

## Try these 3 examples

## 5. Graphing a Quadratic Equation

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.

## 6. Graphing in Factored Form

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)

## Quadratics 2: Factoring

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

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 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

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

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

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

## Quadratics 3: Completing the Square, Quadratic Formula, and the Discriminant

## 1. Completing the Square

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

## 2. The Quadratic Fomula

## Example

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.

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

## 3. The Discriminant

The Discriminant of the quadratic formula tells us how many solutions the formula has.

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

## Quadratics 1 word problem

At what horizontal distance does the ball land?

The ball lands at 50 metres.

## Quadratics 2 word problem

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)

## Quadratics 3 word problem

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