# Quadratic Relations

### By: Shivam Sharma

## Table of Content

Vertex Form:

- Axis of symmetry
- Optimal value
- Transformation
- X-Intercepts or zeroes
- Step pattern

- Zeroes or x-intercepts
- Axis of symmetry
- Optimal value

- Zeroes
- Axis of symmetry
- Optimal value
- Completing the square to turn to vertex form
- Factoring to turn to factored form (Common factoring, Simple Trinomial, Complex Trinomial, Perfect Square, Difference of Squares

## Introduction to Quadratics

__First & Second Differences__A quadratic relation will always have first differences. You can calculate the first differences by subtracting the Y value by the closest minor Y value.

## Introduction to Parabolas

## Vertex Form: y=a(x-h)²+k

__How to find a equation with a given vertex__I will be explaining how to find an equation when given a vertex.

__Example 1:__

Step 1: Vertex: (2,5). From this we can tell that the x-intercept is 2 and the y-intercept is 5. Now we can plug the two numbers into the Vertex Form. y=a(x-h)²+k is the Vertex Form, and when we plug in the x and y intercept, we need to understand that h is x and k is y. So we would write y=a(x-2)²+5 (When we put in the x-intercept in the h, we have to multiply the number by -1 so we flip the positive/negative). When we graph the equation, we graph the x-intercept starting from 0 and move left/right according to whether the x-intercept is positive or negative. In this case, we will move 2 places to the right. Next we will graph the y-intercept starting from 0 and move up/down according to whether the y-intercept is positive or negative. In this case, we will move 5 places up.

Step 2: Now that we have plugged in the x and y intercept, we now substitute (-2,0) in the equation.

Y=a(x-2)²+5

Y=a(-2-2)²+5

Y=a(-4)²+5

Y=16a+5

__-5 = 16a__

16 16

__-5__ = a

16

Now sub in the a value that we just found into the equation to complete the vertex form. Y=-5/16(x-2)²+5

__Finding the x-intercepts__

__Example 1:__

Step 1: Y=2(x-5)²+10. First off, we have to put the k value to the other side. 10=2(x-5)².

Step 2: Divide the equation by the a value, which in this case is 2.

__10__ = __2(x-5)__²

2 2

5=(x-5)²

Step 3: To get rid of the ², square root the equation

5=(x-5)²

√5=√(x-5)²

2.24=x-5

Step 4: Bring -5 to the other side of the equation

2.24+5=x

7.24 = x

Step 5: To find the other x-intercept, take your Y and multiply it by -1, then do Step 4 again

-2.24+5=x

2.76=x

__Vertically Compressed/Stretched__If the a value is 1/-1, >1 or -1<, then the parabola will be vertically stretched (The parabola will be skinny). If the a value is between > -1 and < 1, then the parabola will be vertically compressed (The parabola will be fat).

## Factored Form: Y=a(x-r)(x-s)

__Expanding__Expanding is used when you want to change Factored Form equations into Standard Form. This video will thoroughly explain how to Expand.

__Common Factoring__

__Simple Trinomials__

__Complex Trinomials__

__Perfect Squares & Difference of Squares__## Standard Form: Y= ax²+bx+c

__Completing the Squares__

__Example 1: __Y=2x²+10x+4

Step 1: Put brackets on the first 2 terms

Y=(2x²+10x)+4

Step 2: Factor the terms in the bracket

Y=2(x²+5x)+4

Step 3: Get your b value, divide it by 2 and then square it. Place it in between bx and c

Y=2(x²+5x+6.25-6.25)+4

Step 4: Put the -6.25 outside the bracket and multiply it with the a value

Y=2(x²+5x+6.25)+(2)(-6.25)+4

Step 5: Solve whatever you can

Y=2(x²+5x+6.25)-12.5+4

Y=2(x²+5x+6.25)-8.5

Step 6: Divide the b value by 2 and get rid of "5x+6.25"

Y=2(x²+2.5x)-8.5

Step 7: Factor the ² out of the bracket

Y=2(x+2.5)²-8.5

__Quadratic Formula__

The Quadratic Formula is used to find an equation's x-intercepts when factoring it is not easy with other methods.

The Quadratic Formula is:

__Example 1:__0=2x²+6x+4

Step 1: Find the a, b, and c value

a = 2, b = 6, c = 4

Step 2: Plug in the a, b, and c values into the quadratic formula

x = (-b ± √b² -4ac) / 2a

x = [-6 ± √6² - 4(2)(4)] / 2(2)

Step 3: Solve the equation

x = [-6 ± √(6² - 4x2x4)] / 2(2)

x = [-6 ± √(36 - 32)] / 4

x = (-6± 4) / 4

x = (-6 + 4) / 4 x= (-6 - 4) / 4

x= -2 / 4 x= -10/4

x = -0.5 x= -2.5

__Discriminant__

The Discriminant of an equation quickly helps tell how many x-intercepts the parabola will have. The Discriminant is the b²-4ac part of the quadratic formula. When you plug in your values into the quadratic formula and solve it partially, you will find your discriminant. If b²-4ac = a positive number, then you will have 2 x-intercepts. If b²-4ac = 0, then you will have 1 x-intercept. Finally, if b²-4ac = a negative number, then you will not have any x-intercepts.