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The History of Quadratic Functions

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Claire Nilsson

on 25 May 2015

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Transcript of The History of Quadratic Functions

The Origin of Quadratic Functions
2000 BC: to solve geometric problems pertaining to area, Babylonian mathematicians began to formulate the beginnings of the quadratic function
So how do we apply quadratic functions to real life?
Real World Application #1
You can use the quadratic formula to find when a thrown ball will hit the ground!
Real World Application #3:
We can use quadratic equations in the geometry of the creation of a field.
Real World Application #2:
The quadratic function is used in business to maximize profit. To do this you compare the price it takes to make the product and the price that you can sell it for.

The History and Impact of Quadratic Functions
By Claire Nilsson
400 BC: the Babylonians and Chinese both figured out to solve these problems by completing the square.
300 BC: Mathematicians Pythagoras (Italy) and Euclid (Egypt) found a way to solve quadratics by using geometry. Pythagoras discovered that ratios can be used to solve the quadratic equation. However, unlike Euclid, he didn't recognize that the ratios could be irrational.
Throughout history, quadratics have been an interest to mathematicians across the world.
700 AD: Indians finally had a way to write this equation, using the same decimal system that we use now. In India, mathematician Brahmagupta came up with the concept of 2 roots in the solution.
1100 AD: Indian mathematician Baskhara further refined the ideas of al-Khwarizmi and Brahmagupta.
1545 (Renaissance): An Italian mathematician, Girolamo Cardano, compiled the findings of preceding mathematicians like al-Khwarizmi and Euclid to create a form of the equation that allows for the existence for negative numbers.
800 AD: A Persian mathematician named al-Khwarizmi, used roots and squares of roots to write six equations. However, these equations rejected the idea of negative solutions.
Late 1500s: French mathematician François Viète came up with the symbolism that we use today. In 1637, René Descartes published La Géométrie, creating the modern form of the quadratic equation.
Well let's find out!
Works Cited
1. The quadratic equation is
ax^2+bx+c=0.

2. When you're using this equation to calculate the trajectory of a ball, you can change it to:
s(t) = –gt^2 + v0t + h0


This is the starting height when the ball is thrown
This is the velocity of the ball, or how fast it travels
This represents gravity
Height (s) at a specific time (t)
3. Then you plug in the numbers and factor to find the data that you're looking for!

Ex:
0 = t2 – 4t – 12
0 = (t – 6)(t + 2)
Production cost is $10
.
4. Solving for t gives you the distance of the throw. T couldn't be -2, so the answer is 6.
Here's an example:
Here's some variables for the equation!

q = quantity sold
s = selling price of the item
P = profit
"Quadratic Functions." HS-mathematics -. Web. 19 May 2015. http://hs-mathematics.wikispaces.com/Quadratic+Functions
"H2g2 - The History Behind The Quadratic Formula - Edited Entry." H2g2. Web. 19 May 2015. http://h2g2.com/edited_entry/A2982567
http://www.montereyinstitute.org/courses/Algebra1/COURSE_TEXT_RESOURCE/U10_L2_T1_text_container.html
"Everyday Examples of Situations to Apply Quadratic Equations | The Classroom | Synonym." The Classroom. Web. 19 May 2015. http://classroom.synonym.com/everyday-examples-situations-apply-quadratic-equations-10200.html
P = sq – 10q
q = -20s + 1200

Substitute
-20s + 1200
for q in profit formula!

P = s(-20s + 1200) – 10(-20s + 12000

P = -20s2 + 1200s + 200s – 12000

P = -20s2 + 1400s – 12000
Graph it! The vertex gives you the answer for maximum profit: $35
Here's an example:
If we want a field to have an area of 75 square feet and we want the width of the field to be 3 feet longer than the length. What are the dimensions of the field?
x = length of field
(x+3) = width of field

To find the area of the rectangle:
x (x+3) = 75
x^2 + 3x - 75 = 0

This is our quadratic formula!

The value of x can be found by using this formula:



Plug in a, b, and c and you get two solutions:
7.2892 and -10.2892

Since we're working with distance, the answer cant be positive, making the length of our field 7.2892 feet long.
Congrats!!!!!!
You did it,
...and you did it,
...and you did it! Good job guys!
Well that was easy, right? Let's do more!
The Chinese was greatly facilitated in the process by using the abacus.
Using geometry, this is what Euclid and Pythagoras came up with
Full transcript