Quadratic Relations

By: Shivam Sharma

Table of Content

Vertex Form:

  • Axis of symmetry
  • Optimal value
  • Transformation
  • X-Intercepts or zeroes
  • Step pattern
Factored Form:
  • Zeroes or x-intercepts
  • Axis of symmetry
  • Optimal value
Standard Form:
  • 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.

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If the First Differences end up being identical as in the picture provided above, then the line will be linear. But if they are not identical, then you'll have to calculate the Second Differences. To calculate the Second Differences, you'll need to start at the bottom of the First Differences column and subtract the number by the number on top. Having Second Differences makes the line a quadratic relation.
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Step patterns part 2

Introduction to Parabolas

A parabola is the curved line in a graph formed by graphing a quadratic relation. Parabolas have terminology such as: x-intercepts (Zeros, roots), y-intercepts, Vertex, Axis of Symmetry, Optimal Value
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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

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Finding the x-intercepts


I will show how to find the x-intercepts for a quadratic relation in vertex form.


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.

3.6 Expanding
Common Factoring
3.7 Common Factoring
Simple Trinomials
3.8 Factoring Simple Trinomials
Complex Trinomials
3.9 Complex Trinomial Factoring
Perfect Squares & Difference of Squares
3.10 Special factoring

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:

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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.

Word Problems

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Self Reflection

Throughout the Quadratics unit, I've struggled a lot. I found it very challenging, especially word problems and also with knowing when to use which factoring method. However, after creating this website, I taught myself a lot about Quadratics independently and studied for hours until I've understood the concept. I still get confused about factoring and word problems but I definitely have gotten better at it. In the future, I should study more, ask teachers, family, and peers questions when confused, and stay more focused when being taught a lesson.
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