## Introduction

Quadratic Relations are often seen in our daily lives, but we don't notice them. Examples of quadratics in the real world are the McDonald's arches, roller coasters such as the behemoth and the bottom of the Eiffel tower. Quadratics are graphed by using parabolas. Quadratic relations have constant second differences rather than constant first differences like linear relations. Parabolas consist of x-intercepts, y-intercepts, axis of symmetry, optimal value, zeroes and a vertex.

2. Analyzing Quadratics Using Table of Values

3. Explanation and transformation of Vertex Form Equation

4.Graphing by using Vertex Form Equations

5. Graphing Factored Form

6. Word Problems

7. Multiplying Binomials

8. Common Factoring and Simple Grouping

9. Perfect Square Trinomials and Difference of Perfect Squares

10. Factoring Simple Trinomials

11. Factoring Complex Trinomials

12. Word Problems

13. Completing the Square

16. Word Problems

18. Reflection

• X-intercepts: It is the point that touches the x-axis.
• Y-intercepts: It is the point that touches the y-axis.
• Axis of symmetry: It is a line that goes down the middle of the parabola which divides it into 2 equal halves.
• Optimal value: It is the value of the y co-ordinate of the vertex.
• Maximum/Minimum Value: It is also known as the optimal value. It is the highest (maximum) or lowest (minimum) point on the parabola.
• Zeroes: It is the point that crosses the x-axis. It can also be called "x-intercepts" or "roots".
• Vertex: It is the maximum or minimum point on the graph.
The parabola either opens upwards or downwards. You can figure out the direction of the opening by looking at the equation which will be explained later on. If the direction of opening for the parabola is down then it will have a maximum value and if the direction of opening is up then it will have a minimum value.

## Analyzing Quadratics Using Table of Values

Remember that in linear systems, the first differences (the difference between the y values in a table of value) are always the same. However, notice that in the table of values located to the right, the first differences are not the same. Instead, the second differences are the same. Therefore, if the second differences are the same but the first differences are different the relation is quadratic.

## Explanation and transformation of Vertex Form Equation

The vertex form of a parabola's equation is a(x-h)^2+k.
• The "a" in the equation tells us if the parabola is vertically stretched or compressed and by what factor. It also tells us if the parabola opens upwards or downwards. If "a" is one or a decimal that is smaller than 1 or larger than -1 (e.g. 0.5-1/2 or -0.75-3/4), then the parabola is vertically compressed. If "a" is larger than 1 or smaller than -1, then the parabola is vertically compressed. Also, if "a" is positive then the parabola opens upwards and if it is negative then the parabola opens downwards.
• The "h" in the equation tells us if the parabola is translated to the right or left from the origin. The "h" is the x co-ordinate of the vertex but it would be the opposite sign (e.g. if h is -5 in the equation, it would actually be +5). If "h" is positive the graph is translated to the left by the "h" units. If "h" is negative, then the graph is translated to the right from the origin.
• The "k" in the equation tells us if the parabola is translated up or down from the origin. It is the y co-ordinate of the vertex. If "k" is positive then the parabola moves up from the origin and if it is negative, it moves down from the origin.

## Example of Vertex Form Equation

y=a(x-h)^2+k
y=2(x-1)^2+5

• The "a" in this equation is positive 2 which means that the graph opens upwards and it is vertically stretched by a factor of 2.
• The "h" in the equation is -1 which means that the graph is translated to the right from the origin by 1 unit. Because "h" is -1, the x co-ordinate of the vertex will be positive one.
• The "k" in the equation is 5 which means the parabola is translated 5 units upwards. This also means that the y co-ordinate of the vertex will be 5.
You can notice these things in the graph that is attached with this note.

## Example of Vertex Form Equation

y=a(x-h)^2+k
y=-3(x+4)^2-6

• The "a" in this equation is negative three which means that the graph opens downwards and it is vertically stretched by a factor of 3
• The "h" in the equation is 4 which means that the graph is translated to the left from the origin by 4 units. Because "h" is +4, the x co-ordinate of the vertex will be negative 4.
• The "k" in the equation is -6 which means the parabola is translated 6 units down. This also means that the y co-ordinate of the vertex will be -6.
You can notice these things in the graph that is attached with this note.

## Graphing by using Vertex Form Equations

Step 1: You can graph a parabola by using the vertex form equation [y=a(x-h)^2+k)]. You can figure out one point (the vertex) by just looking at the equation. You know that "h" is the x co-ordinate of the vertex but with opposite signs and that "k" is the y co-ordinate of the vertex so you can plot that point on the graph.
Step 2: The rule for graphing is one over one and two over four. Look at the images below this note to see how to do step 2. Then plot the points that you figured out while doing step 2. Remember that whatever you do to one side will be reflected on the other side to create an axis of symmetry.
Step 3: Now that you know the points, connect the points to create a parabola. You have created a parabola by using the vertex from equation.

## Graphing Factored Form

Step 1: The factored form of quadratics is y=a(x-r)(x-s) where "r" and "s" are the x-intercepts of the graph but with the opposite signs (e.g. if r is -5 and s is +2 in the equation, the x-intercepts would be +5 and -2).
Step 2: You can also figure out the the vertex for the parabola by using the equation. The x co-ordinate of the vertex which is also known as the axis of symmetry can be found by adding the r and s when it has been changed to the opposite signs and then dividing it by 2.
Step 3: The y co-ordinate of the vertex which is also known as the optimal value can be found by solving the equation y=a(x-r)(x-s) by subbing in the x's with the axis of symmetry which you found in step 2. Now you have found the vertex for the parabola.
Step 4: You can also find out the direction of opening by looking at the "a" in the equation. Check to see if "a" is positive and negative and you can connect the three points you found out through steps 1 to 3 according to the direction of opening.

## Example of graphing factored form

y=-2(x+1)(x+3)
Step 1:

In this equation r is +1 and s is +3 but remember that you have to change the signs. So r will now be -1 and s will now be -3. You can now plot (-1,0) and (-3,0) in the graph.
Step 2:
Now find out the axis of symmetry by adding r and s and dividing it by 2.
-1+(-3)= -4/2= -2. The axis of symmetry is -2.
Step 3: To find out the optimal value, sub in the value of the axis of symmetry in the x's in the equation. Remember that you do not change the signs of the r and s to solve for y.
y=-2(-2+1)(-2+3)
y=-2(-1)(1)
y=-2(-1)
y=2
The optimal value is 2. Now you have both the x and y co-ordinate for the vertex. The vertex is (-2,2). You can plot the point on the graph.
Step 4: Now check if the parabola opens upward or downward. If "a" is positive, it opens up and if it is negative it opens down.
In this equation "a" is -2 so it opens downwards. Now connect the three points that you found out through the steps 1-3.

## Multiplying Binomials

In this, you will learn how to get from factored form to standard form. To do this you have to expand and simplify.

## Example of multiplying binomials

Use distributive property to find the answer.
(3x+5)(x-2)
= (3x)(x)=3x^2
= (3x)(-2)=-6x
= (5)(x)=5x
= (5)(-2)=-10
= 3x^2-6x+5x-10
= 3x^2-x-10
Multiplying Binomials

## Common Factoring and Simple Grouping

Common Factoring:
You have to find the GCF to common factor.
For example: In the equation 3x+9, the GCF is 3. So you have to divide both terms by 3. You put the answer you get inside the bracket and you put the 3 (the GCF) outside the bracket. The factored form for that equation will 3(x+3).

Another example would be 14m+21n.
Step 1: Find the GCF which is 7 in this case.
Step 2: Divide the terms in the equation by the GCF. 14m/7=2m, 21n/7=3n
Step 3: Put the GCF outside the bracket and put the quotients that you found by dividing the terms in the bracket. 7(2m+3n)

Simple Grouping:
When there are four terms, you can factor by grouping. For example, 2gh+4kh+3gm+6km has four terms.
Step 1: In order to solve this, you have to put the first two terms in a separate bracket and you have to put the next two terms in a separate bracket. In this case it would be like (2gh+4kh)(3gm+6km).
Step 2: Now you can use common factoring. Find a GCF for the first bracket and find a GCF for the second bracket. A GCF for 2gh and 4kh would be 2h so you divide the 2 terms by it which would be g and 2k. A GCF for 3gm and 6km would be 3m so you divide both the terms by 3m which would be g and 2k.
Step 3: Now put the GCF outside the brackets and put the terms inside the brackets.
For this equation it would be 2h(g+2k)3m(g+2k). Notice that the terms in the brackets are the same which means that you put the two GCF's in the a different bracket and you put the terms in the bracket in a separate bracket. The answer would be (2h+3m)(g+2f).
Common Factor

## Perfect Square Trinomials and Difference of Perfect Squares

Perfect Square Trinomials:
Perfect square trinomials can result by squaring binomials.
a^2+2ab+b^2=(a+b)^2
or
a^2-2ab+b^2=(a-b)^2
In a perfect square trinomial, the first and last terms are perfect squares, and the middle term is twice the product of the square roots of the first and last terms.
For example,
(3x+4)^2 is a perfect square binomial
=3x^2+2(3x)(4)+4^2
=3x^2+24x+4^2
=9x^2+24x+16 is a perfect square trinomial since the first and last terms are perfect squares and the middle term is a twice the product of the square roots of the first and last terms.
Difference of Perfect Squares:
This is when two perfect squares are subtracted. 4x^2-25 is an example of difference of perfect square since both 4 and 25 are perfect squares. When you factor this it would be in the form (a+b)(a-b). (2x+5)(2x-5)

## Factoring Simple Trinomials

The equation for this is x^2+bx+c. Notice how there is no coefficient in front of x^2 which is why this is a simple trinomial. To factor this, you have to find 2 factors of c that equal to b when you add them. Then you just put them in two separate brackets with two x's since there is a x^2.

For example:

x^2+bx+c
=x^2+6x+5
5=(1+5=6)
=(x+1)(x+5)

Some equations will have a coefficient before the x^2 but they will also be considered as simple trinomials if all the terms are divisible by the coefficient.

For example:
3x^2+15x+18 (All terms are divisible by the first coefficient which is 3)
=3x^2/3+15x/3+18/3
=3(x^2+5x+6) (Now you can follow the steps you used for the above example)
6=(3+2=5)
=(x+3)(x+2)

## Factoring Complex Trinomials

In complex trinomials, there are coefficients infront of the x^2 but all the terms are not divisible by it. The equation is ax^2+bx+c. To factor this, you have to multiply the first and last term. Then you have to find to factors for the product that adds up to the middle the number.

For example:
ax^2+bx+c
6x^2+11x+3
Step 1: =(6x^2)(3)=18=(2+9=11) Multiply the first and last term which equals to 18 in this case. Find 2 factors of that that equal to the middle term when you add them.
Step 2: =6x^2+2x+9x+3 Substitute the 2 factors into the equation with the middle term.
Step 3: =(6x^2+2x)(9x+3) Now find the GCF for each bracket.
Step 4: =(6x^2+2x/2x)(9x+3/3) Divide the brackets by the GCF
Step 5: =2x(3x+1)3(3x+1) Notice that both the brackets are the same. This means that you could put the terms outside the brackets (GCF) in a separate bracket and you could put the terms in the bracket in one separate bracket.
Step 6: (2x+3)(3x+1)

## Completing the Square

Completing the square is used to go from standard form to vertex form.

Examples:
1.
• Step 1: y=x^2+8x+3 {Put a bracket around the first two terms}
• Step 2: =(x^2+8x)+3 {If there is a coefficient multiplied with x^2 then you divide both the terms in the bracket with the coefficient.}
• Step 3: {Now divide the second term in the bracket by two and then square that number.)
• Step 4: =(x^2+8x+16)-16+3
• Step 5: =(x^2+8x+16)-13 {Now factor the perfect square}
• Step 6: =(x+4)^2-13 {Now this is in vertex form}
2.
• Step 1: -4x^2-8x-1
• Step 2: (-4x^2-8x)-1
• Step 3: -4(x^2+2x)-1
• Step 4: -4(x^2+2x+1)-1-1
• Step 5: -4(x^2+2x+1)+4-1
• Step 6: -4(x^2+2x+1)+3
• Step 7: -4(x+1)^2+3

There are 2 ways to solve quadratic equations.

1. Isolating x from factored form

In order to solve by isolating x from factored form, you have to write 0=in front of the factored form. Then you have to isolate x by doing the opposite.

So for example: (2x+4)(x-6)

Step 1: 0=(2x+4)(x-6)

Step 2: 0-4=(2x+4-4)0+6=(x-6+6)

Step 3: -4/2=(2x/2) x=6

Step 4: x=-2 x=6

The solution for this is x=-2 and x=6.

The second way to solve quadratic equations is the quadratic formula. Look at the note (images) attached with and below this note to see how to use the quadratic formula and what discriminant is.

## Word Problems

1. Converting Standard to Vertex Form
2. Factoring Perfect Square Trinomials
3. Factoring Difference of Squares