## How to tell if it is a Quadratic Function?

In order to find out if function you are dealing with, you must find the first and second differences. If the second differences are all the SAME, then the function is quadratic.
Since all the second differences are the same, it is a quadratic function.

## What are the forms of Quadratics?

There are three forms of quadratic equations:

Vertex Form

• Axis of Symmetry
• Optimal value
• Transformations (Translation, Vertical, or Horizontal, Vertical Stretch, and Reflection)
• X-Intercepts of Zeros (r and s)
• Step Pattern

Factored Form

• Zeros or X-Intercepts (r and s)
• Axis of Symmetry (x=(r+s)/2)
• Optimal Value (sub in)

Standard Form

• Axis of Symmetry (-b/2a)
• Optimal Value (sub in)
• Completing the Square to turn it into Vertex Form
• Factoring to turn into factored form
• Common Factor
• Simple trinomial
• Complex Trinomial
• Perfect Squares
• Difference of Squares

## Axis of Symmetry

As we know, in vertex form, (x, y)= (h, k).

To find the axis of symmetry in the vertex form the formula would be: y=(x-h)² where x=h.

For example, in y= -2(x+3)² -6, you would make it so x-h, and since x-3 (where x is 0) is -3, your axis of symmetry is -3. Simple!

## Optimal Value

When solving for the optimal value, you must remember that y= k. Optimal Value is your highest or lowest point, depending on the direction of opening. Essentially, the optimal value is the same as your vertex. In order to find the vertex, you need to change it to factored form and (h, k) would be your vertex right there.

## Transformations (Translation vertical or horizontal, vertical stretch, reflection)

When graphing by transformations, it is always important to remember that the base graph parabola is always y= x². When you multiplying the x by a single number, like 2, you will get y=2x². That means that when you graph the equation, you will have a vertical stretch of 2.
The 'k' value in the equation will shift the graph, either up or down. For example, if the 'k' value were negative 5, and our equation was y= 2x²-5, then the parabola would have a vertical shift of -5, making it go down 5 units.
Next, we will talk about horizontal shifts. If we were to change y= x² to y= x²-2, the horizontal shift would be 3 units to the right, because we changed the 'h' value.
For vertical compression, you must have an 'a' value that is not a whole number. For example, if we changed y= (x-3)² to y= (-1/2)(x-3)², your parabola would have a vertical compression of 2. If the 'a' value is negative, then it will change the direction of opening, which is what we will get into next.
A reflection in a quadratic equation is fairly simple. If your 'a' value is negative, that means that you will have a reflection from the regular y= x² parabola. For example, y= x² would have a direction of opening that goes up, while for y= -x² would have the same parabola, but with a direction of opening that goes down.

## X-Intercepts or Zeros (Sub y=0 and Solve)

To find the X-Intercept of a quadratic equation in vertex form, the easiest way is to sub 0 in for 'y'. It would look like this, for example:

y= (x-1)²-2

0= (x-1)²-2

2= (x-1)²

√2= (x-1)

1.41= x-1

1.41+1= x

x=2.41

(2.41, 0).

Simple!

## Step Pattern

The step pattern shows how to graph a parabola, as in it shows the points. It is the 'x' value squared. The basic step pattern, as it shows, gives you the parabola for the base graph, and the vertex for this parabola is always (0,0).
Here is a video on how to do the step pattern. I do not own this video at all. All content belong to InspireMath Tutorials. If you would like the URL, please right click on the video and get the URL.

## Zeros or X-Intercepts (r and s)

To find the Zeros, or the X-Intercepts of a quadratic equation, you will just need to know what the 'r' and 's' values are, and find the opposite form. For example, we will look at the equation y= a(x-3)(x+8). The X-Intercepts in this scenerio would be the oppostie of the 'r' and 's' values, which are -3 and +8. The opposite of -3 is +3, and the opposite of +8 would be -8, so those are you X-Intercepts. You could also make the two brackets equal 0, separate them, and solve for the two X-Intercepts.

On a graph, the X-Intercepts would be here:

Those are where the X-Intercepts would be located, and next, we will be looking at how to find the axis of Symmetry (AOS) for the same equation.

## Axis of Symmetry (x=(r+s)/2)

In order to find the AOS, you will need to get the X-Intercepts you found, add them, and then divide them by two. For the equation y= a(x-3)(x+8), the X-Intercepts are +3, and -8, which we found before. You get those two numbers, add them, and divide by two, like this:

AOS= (3+(-8))/2)

AOS= ((-5))/2)

AOS= -2.5

On a graph, it would look like this:

Word Problem:

A ball is thrown from child one to child two, from left to right, where child one is left. The formula for the ball being thrown in the formula of y= a(x-8)(x+1). Where is child one located? Where is child two located? What is the distance between them?

In order to solve this equation, we need to look at two things. First, we must make the brackets equal 0. x-8=0 and x+1=0.

x-8+8=0+8

x=8

(8,0)

x+1=0

x+1-1=0

x=-1

(-1,0)

Since we know that the ball is thrown from left to right and that child 1 is left, child 1 is located at (-1,0) and child 2 is left at (8,0). The distance between them is 9 units, because no unit of measurement is given.

So that is how you find the Axis of Symmetry for a quadratic equation in factored form. So with factored form, we covered, how to find the Zeros and how to find the X-Intercept, but now we need to find the optimal value.

## Optimal Value

In order to find the optimal value of the parabola, you need to sub in the axis of symmetry in for the 'x' value. So far the equation we were working on, you would have an equation of y= a(-2.5-3)(-2.5+8). 'a' is 1, so then y= 1(-5.5)(5.5). So then y= (-5.5)(5.5) which is -30.25, so the optimal value, or vertex, is located at (-2.5, -30.25).
Word Problem:

A child throws a ball over a river. The ball is thrown in the formula of y= a(5+4)(5+3). The 'x' value is already given here, so what is the highest point the ball reaches?

Since the 'x' value is given, we just need to solve for 'y'.

y= a(5+4)(5+3)

y= 9+8

y=17

(5, 17)

The optimal value, or the highest point the ball is thrown, is at (5, 17).

## Standard Form

When solving using the quadratic formula, there is two things needed. First, you must have your equation in standard form. To start, we will use the equation y= 2x²-2x+124. Next, you must make 'y' equal 0. So you have 0= 2x²-2x+24. Next, you must know the quadratic formula, which is x= -b+/- √b²-4(a)(c)/2(a). Next, you must know what a, b, and c are equal to.

a=2

b=-2

c=24

Now all you need to do is substitute it into the formula, and solve!

x= -(-2)+/- √2²-4(-2)(24)/2(2)

x+ 2+/- √4+192/4

x= 2+/- √196/4

There are two ways to go from here.

x=198/4

x=49.5

(49.5, 0)

x=2-196/4

x= -194/4

x= -48.5

(-48.5, 0)

Those are your X-Intercepts for your parabola. You still need to find you axis of symmetry though, which is next.

## Axis Of Symmetry (-b/2a)

To find the AOS of an equation in standard form, you will need to first know your 'b' and 'a' values. From the past equation y= 2x²-2x+24, you know your 'b' is -2 and 'a' is 2. So you just substitute those values into the formula given, and you are done.

AOS= (-b/2a)

AOS= -(-2)/2(2)

AOS= 2/4

AOS= 0.5

(x, 0.5)

## Optimal Value (sub in)

For this, you just need to sub in the AOS into the 'x' value. So the equation y= 2x²-2x+24 is y= 2(0.5)²-2(0.5)+24.

Then y= 2(0.25)-1+24

y= 0.5-1+24

y= 23.5

23.5 is your maximum point in this scenario!

## Completing the Square (Standard to Vertex Form)

In this lesson, we will look at how to change an equation from standard form (y=ax²+bx+c) to vertex form (y=a(x-h)²-k). The reason we want to change it is because vertex form is better than standard form because it gives you the vertex, gives you the ability to solve and the ability to plot. Here is a video with great explanation on how to go from standard form to vertex form, which is completing the square:
I do not own this video, and this video belongs to V Anusic. If you would like the URL, right click on the video and get the URL.
3.14 Completing the square

## Factoring to turn to Factored Form

Common Factor:

Common factoring is when you divide the terms with one number, and isolating the 'a' value, and then turning it into factored form. We will look at how to common factor in a step by step method with an example.

The equation we are using first is y=8t²-24t-144. First, you isolate the 'a' and put the rest of the equation into brackets, so it would look like this: y= 8(t²-3t-18)

Next, you would have to find two numbers that add to -3 and when you multiply those two numbers, it must be -18.

After you find the factors, you realize that +3 and -6 work, because +3-6=-3, and +3*(-6)= -18.

Next, you remember that 4 can be divided by two, so you can split 4t² to 2t and 2t. You already have +3 and -6, so you just put it into factored form, and it looks like this:

y= 8(t+3)(t-6). You can check it by distributing the property.

8(t+3)(t-6)

=(8t+24)(t-6)

=8t²-48t+24t-144

=8t²-24t-144

Same equation as before! Remember that you can only common factor if there are common factors in the trinomial. If you don't have any common factors, you need to solve by the way you solve a simple trinomial, which we will look at next.

Simple Trinomial:

Simple trinomial's do not have any common factors in them. When you solve simple trinomials, you must first look for factors of the 'r' and 's' values. Next, you must do what you do when common factoring, and that is finding two numbers that add to the 'b' value, but those same numbers multiply to the 'c' value. Finally, you make sure that it is in factored form and that your 'x' value is squared. Here is a step by step example:

y= x²+6x+5

6 and 5 are the numbers we need to solve for. They cannot be common factored, so the best way to solve it is by doing 5+1 for the 'b' value and 5*1 for the 'c' value, and it works.

√x²= x+x.

Therefore, the factored form for the equation is y= (x+1)(x+5).

Remember, this would not work for standard equation with an 'a' value, so we will look at how to solve those next.

Complex Trinomial:

When solving a complex trinomial, you do the same thing as you do for solving simple trinomials, but only for the x²+bx+c part. But the 'a' value is a little more difficult. A simple way to do it is by using a little chart. You make a little 2 by 3 chart. The first thing you must do on the side is find two numbers that multiply that equals the 'a' value. We will use the expression 6x²+11x+3 for this. You put the two numbers in the first column.

Next, in the second column, you would find two numbers that multiply and equals the 'c' value. In this case, the 'c' value is 3.
Next, you need to fill in the column for the middle value. This one is a little tricky at first, but easy afterwards. You multiply the numbers opposite of each other, and you will get two numbers to fill in the rest of the chart. Those two numbers should add to give you the value of 'bx' and in this case, 11x.
Since you do not need to include 9 and 2 in the expression because you know that when you distribute the property, it will give you eleven, you just need to make the expression out of the first two column. The equation in this case is: y= (3x+1)(2x+3)

Check:

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

y= 6x²+9x+2x+3

y+ 6x²+11x+3

Same equation.

Here is a video of a word problem solved.

Sam's Room Word Problem
There are two easier ways, though to solve SOME equations. This involves square roots and squared equations, which is next in line.
Perfect Square:

When solving perfect squares, you must have a values that can be square rooted. All you need to do is square root the 'x' value (or 'ax'), and square root the 'c' value. However, the square root of 'c', as in the two numbers, must add and equal to 'b' value. Here is an example:

Next, we will look at difference of squares.
Difference of Squares:

Here is a video to show how to do special factoring, which in this case, would be difference of squares, as well as some more help with perfect squares.

3.10 Special factoring
This video does not belong to me. It belongs to V Anusic. For the video URL, please right click on the video and get the URL if you want it.
Here is an assessment I did, and it was a mini quiz. As you see, I struggled on the quiz, but it did help me understand factoring more. If you look at it, you can see the mistakes I made, and it will help you understand factoring as well.