## Why do we use quadratics?

Quadratic equations, unlike linear equations, allow someone to see where a point will be if the pattern is not perfectly straight (linear). For example, throwing a ball into the sky will not have it go at the same velocity and at the same angle infinitely; it is bound to lose speed and fall to the ground. A method to find if an equation is linear or quadratic is using second differences, which I will teach in an upcoming topic!

## Overview of the Topics to be Covered

What is a Parabola?

- Vertex form

- Standard form

- Factored form

Vertex form

- Axis of symmetry

- Optimal Value

- Transformations

• Translations
• Stretched/Compressed
• Reflection
- x-intercepts/zeroes

- Step pattern

Standard form

- Zeroes

- Axis of symmetry

- Optimal value

- Completing the squares to convert to vertex form

- Factoring to turn to factored form

• Common
• Simple trinomial
• Complex trinomial
• Perfect squares
• Difference of squares

Factored Form

• Zeros or X-Intercepts
• Axis of Symmetry
• Optimal Value

Word Problems

-Sam's Room word problem

Reflection on the Unit

- Assessment reflections

## What is a Parabola?

A parabola is the result of graphing a quadratic equation. Unlike a linear equation, it is curved. A parabola is like a linear "line" in the sense that it gives information from a point on the graph. For example, graph the path of a soccer ball that has been kicked. If you find a certain height, you can be able to find the time that the soccer ball was at that height, and vice-versa!

A parabola has a few terms that make up the parabola, these include:

The vertex

This is the highest point of the parabola, it is made up of the axis of symmetry and the optimal value.

x-intercepts

This one is straight forward; it is the points of the parabola that are on the x-axis, where they "intercept". They are also called zeroes.

Axis of symmetry

This is the very middle of the parabola, it can be found by adding the x-intercepts and dividing them by 2. It is an x value.

Optimal Value

This is the highest point on the y-axis, it makes up the vertex with the axis of symmetry.

-VERTEX FORM

-STANDARD FORM

-FACTORED FORM

## Axis of Symmetry

Finding the axis of symmetry of a parabola is the easiest in a vertex form equation. Vertex form was given the name it has because the vertex is in the equation; (h,k). If the equation was y=3(x-6)² +7, then the vertex would be (6,7). All this talk about the vertex is because the axis of symmetry is the x in (x,y). So the axis of symmetry in the equation would be 6. Don't be fooled by the (x-6)², i wrote 6 instead of -6 because you must make the sign (+,-) the opposite.

## Optimal Value

The optimal value, much like the axis of symmetry, is also a part of the vertex. It is the y in (x,y). Because the vertex can be easily found in a vertex form equation the optimal value should be very easy to find, and it is! The optimal value in the equation y=3(x-6)² +7 is 7. Unlike the axis of symmetry, no changing of signs in needed.

## Translations

A translation is the movement of a parabola either up, down, left or right. The h in

y=a(x-h)² -k is what allows the parabola to move left or right. If the h is +3, then you move left by 3 units, if the h is -7, then you move right by 7 units.

## Stretched/Compressed

A parabola can be either vertically stretched or vertically compressed using the "a" value in

y=a(x-h)² +k. If the "a" value is positive that means it's being vertically stretched by the amount of "a". If the "a" value is negative that means the parabola is being vertically compressed by the amount of "a". Compressing or stretching would make the parabola wider or thinner.

## Reflection

A reflection to a parabola is simply making a "reflection" of the parabola along the x-axis. Reflections on the x-axis change whether the parabola is facing up or down. The sign alongside the x value dictate whether the parabola is facing up or down; a -x would mean the parabola is facing down, a +x value would mean the parabola is facing up.

## X-intercepts/Zeroes

To find the x-intercept(s) of a parabola which has an equation in vertex form, the simple way to find them is to make y=0 and solve for x. You then do + or - a number to get both x-intercepts. If there is only one intercept then when doing + or - then both answers will be the same.

For example let's use this equation:

y= 4(x-4)² - 20

0=4(x-4)² -20

20/4=4(x-4)²/4

5=(x-4)²

√ 5=(x-4)²

±2.24=x-4

±2.24+4=x

The x-intercepts would be

x=6.24

and

x=1.76

## step pattern

Here is a small video I made with Harjas and Shivam to explain step patterns and how you can find other points of a parabola with only one point and the equation itself!
Step patterns part 2

## Standard form

As you can see in the picture above, standard form is y=ax² + bx + c. Standard form is useful when someone wants to use complete the squares to factor a difficult question.

## zeroes

Finding zeroes in standard form is a little more complicated. It involves using the quadratic formula which is: x= -b+/- √b²-4(a)(c)/2(a)

Example:

To explain how zeroes are found using the quadratic formula, I will be using this equation: x² + 3x - 4

Firstly, we must find the a, b, and c parts of the equation, they are:

a = 1

b = 3

c = -4

Now you can simply substitute those into the quadratic equation:

x= -3+/- √3²-4(1)(-4)/2(1)

x= -3+/- √9+16/2

x= -3+√25/2

x= 1

x= -3-√25/2

x= -4

Therefore, the zeroes of the equation are x=1 and x=-4.

Next I will teach you how to find the axis of symmetry!

## axis of symmetry

After finding the zeroes, the axis of symmetry is very easy to find. You simply need to get both zeroes, which in our case is 1 and -4, and add them then divide them by 2.

= (1+(-4))/2

=-3/2

=-1.5

This is how the equation would look when graphed on Desmos. A link to Desmos is below.

## Optimal Value

After finding the axis of symmetry, finding the optimal value is easy. All you need to do is sub in the axis of symmetry in the original equation. It would replace x;

x² + 3x - 4

-1.5² + 3(-1.5) - 4

2.25 - 4.5 - 4

= -6.25

Therefore, the optimal value of the equation would be -6.25. The vertex if the equation would be (-1.5,-6.25).

## Completing the squares (factoring to vertex form)

Completing the squares is a handy tool used to turn a standard form equation into a vertex form equation, which allows the vertex to be easily spotted. It is also much easier to graph!

Below is a video by Mr. Anusic teaching how to use the 'completing the squares' technique to change a standard form equation into vertex form:

3.14 Completing the square

## Factoring standard form to factored form

There are numerous ways to get to factored form from standard form. They include:

• Common factoring
• Simple trinomial
• Complex trinomial
• Perfect squares
• Difference of squares

Below is a great video by Mr. Anusic that covers the different forms of factoring.

3.11 Factoring

## Zeros or X-Intercepts

Finding the x-intercepts in factored form is by far the easiest, it is in the equation. In the equation y=2(x-3)(x-4) the x-intercepts would be x=3 and x=4. It's basically making the brackets equal to 0, or just changing the sign of the number in the bracket!
The factored for equation for a parabola like this would be: y=a(x-1)(x-3) +3

## Axis of Symmetry

Axis of symmetry is also very easy to find in a factored form equation. You simply add both x-intercepts found and divide them by two! The equation would be something like this:

(x=(r+s)/2). For the parabola above, the axis of symmetry would be:

(AOS=(3+1)/2)

AOS=4/2

AOS=2

Therefore the axis of symmetry would be 2.

## Optimal Value

To find the axis of symmetry, you need to sub in the axis of symmetry found into the original factored form equation. It would look something like this with the previous equation:

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

y=1(-1) +3

y=-1 +3

y=2

Therefore the optimal value in this equation would be 2!

## Word Problem

Here is a word problem me and Makilthan made to solve. We simply expanded and common factored to find out the are of Sam's room!
Sam's Room Word Problem

## Reflection of assessment

Quadratics this year has been quite "up and down" for me. I did well on the first test, as you can see below, did bad on the second test, and I think I did quite well on the 3rd and final test. There is A LOT of things covered in quadratics so there is no room for slacking off. I personally was having trouble all the time and I don't think I would have even passed the tests if it wasn't for Ms. Johal putting up the review homework and overview of the test on edmodo. Who could also forget about Mr. Anusic's video's? I think the main thing that really brought me down from quadratics was because there were MANY steps. My memory isn't always the best and I would forget how to do a word problem i reviewed the day before. Overall quadratics is a hard and important unit in Grade 10 math, but if you study like I did (somewhat), you should come up with a good mark!
Above is an assessment i recently have done on quadratics, it was a mini test. As you can see, I did well on the quiz, but don't be fooled! I practiced my whole life up until this test and I have done hours of training to have come where I am today. If you want to be successful in quadratics, or math in general, the special, super magical, ingredient is to study. The wise words of Nick never fail.