Quadratic functions are used all the time, every day, all over the world. Even though right now, it doesn’t seem like this kind of math is ever going to creep back into our life. That is actually far from true. These math skill are crucial to have if one ever decides to do anything in engineering, or something like that. Those types of jobs are now becoming more and more popular, because the world is always going to need educated people who know how to construct or refurbish buildings and homes.

Sky scrapers are not easy types of building to build. Unlike children’s toy blocks, skyscrapers have to be able to withstand immense weather conditions, or else, like children’s blocks, the skyscraper would raze down to the ground. Skyscrapers call for an extensive amount of planning, and skill. First of all, each skyscraper is deigned much differently. They are different depending on climate, and usual weather conditions. For example, the structure of a building in Dubai is going to be much different from one in Alaska. Math is involved by using quadrics. Things that need to be considered are pressure vs. wind velocity. That means that the engineers need to take into consideration how much pressure is needed to withstand how much wind is going to be pushing the building.

Quadratic functions are actually used in NASA. NASA uses those functions to launch their spaceships into space. There is obviously a lot of planning and smarts that have to go into rocket science. The scientists have to take so much into consideration. Like not blowing something/someone up on accident, weight, height, velocity, wind..the list goes on and on. A way that quadratics relate to all of this is by using the parabola. It shows the rocket scientists the highest power term in the equation. An example of a quadratic that is a parabola is: x = c + b * x + a * x^2. That equation shows the distance that something will move vertically when rising against gravity. Obviously important for launching a...

...A quadratic equation is an equation that has a second-degree term and no higher terms. A second-degree term is a variable raised to the second power, like x2. When you graph a quadratic equation, you get a parabola, and the solutions to the quadratic equation represent where the parabola crosses the x-axis.
A quadratic equation can be written in the form:
quadratic equation,
where a, b, and c are numbers (a ≠0), and x is the variable. x is a solution (or a root) if it satisfies the equation ax2 + bx + c = 0.
Some examples of quadratic equations include:
3x2 + 9x - 2 = 0 6x2 + 11x = 7 4x2 = 13
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Solving a Quadratic Formula:
Some quadratic equations can be solved easily by factoring. Some simple-to-solve quadratic equations are:
x2 - 1 = 0
(x + 1)(x - 1) = 0
x = ±1
x2 - 5x + 6 = 0
(x - 2)(x - 3) = 0
x = 2, 3
Most second-degree equations are more difficult to solve, and cannot be solved by simple factoring. The quadratic formula is a general way of solving any quadratic equation:
quadratic formula
This formula gives two solutions, although the two solutions may be the same number. (When solving any polynomial equation of degree n, there are at most n solutions to that...

...Quadratic Equation:
Quadratic equations have many applications in the arts and sciences, business, economics, medicine and engineering. Quadratic Equation is a second-order polynomial equation in a single variable x.
A general quadratic equation is:
ax2 + bx + c = 0,
Where,
x is an unknown variable
a, b, and c are constants (Not equal to zero)
Special Forms:
* x² = n if n < 0, then x has no real value
* x² = n if n > 0, then x = ± n
* ax² + bx = 0 x = 0, x = -b/a
WAYS TO SOLVE QUADRATIC EQUATION
The ways through which quadratic equation can be solved are:
* Factorizing
* Completing the square
* Derivation of the quadratic formula
* Graphing for real roots
Quadratic Formula:
Completing the square can be used to derive a general formula for solving quadratic equations, the quadratic formula. The quadratic formula is in these two forms separately:
Steps to derive the quadratic formula:
All Quadratic Equations have the general form, aX² + bX + c = 0
The steps to derive quadratic formula are as follows:
Quadratic equations and functions are very important in business mathematics. Questions related to quadratic equations and functions cover a wide range of business...

...REVIEW FOR QUADRATICS TEST 1 ALG II CP1
I. Graphing from Vertex Form – Graph the following functions
(a) (b)
II. Graphing from Factored Form
(a) (b)
III. Graphing from Standard Form by Completing The Square – Graph the following functions by completing the square to get vertex form
(a) (b)
IV. Graphing from Standard Form using –b/2a – Graph the followingfunctions without completing the square
(a) (b)
V. Graphing from Standard Form by Factoring – Graph the following functions by factoring first
(a) (b)
VI. Determining the Equation from a graph or table – Determine the equation for each function. Don’t forget to solve for the a-value.
x
g(x)
-2
0
-1
15
0
24
3
15
4
0
VII. Converting from Vertex Form to Standard Form – Convert the following functions into Standard Form
(a) (b)
VIII. Converting from Factored Form to Standard Form – Convert the following functions into Standard Form
(a) (b)
IX. Converting from Standard Form to Factored Form – Factor the following functions to write them in Factored Form
(a) (b)
(c) (d)
X. Converting from Standard Form to Vertex Form by Completing The Square – Complete the square to write the following...

...The Zeros of a QuadraticFunction
MCR3U1 (Nature of the Roots)
MINDS ON...
The demand to create automotive parts is increasing.
BMW developed three different methods to develop these parts.
The profit function for each method is given below, where
y is the profit and x is the quantity of parts sold in thousands:
PROCESS A: P(x) = -0.5x2 + 3.2x –5.12
PROCESS B: P(x) = -0.5x2 + 4x – 5.12
PROCESS C: P(x) = -0.5x2 + 2.5x – 3.8
The graphs of the corresponding profit functions are shown below.
PROCESS A PROCESS B PROCESS C
Which process would you recommend? Explain.
LESSON 3 - 6 : The Zeros of a QuadraticFunction
MCR3U1 (Nature of the Roots)
Recall: y = ax2 + bx + c is a QuadraticFunction (a relation between x and y)
0 = ax2 + bx + c is a Quadratic Equation (let y = 0 to find the roots and ZEROS)
The roots of the quadratic equation ax2 + bx + c = 0 are , where the radicand, is called the DISCRIMINANT, D.
The value of the discriminant determines the number and nature the roots of a quadratic equation (that is, real/not, equal/distinct) and the number of x-intercepts.
Investigation
Complete the table below and observe the value of the discriminant, D, in each case.
Quadratic...

...Name: ________________________ Date:__________________
Year & Section: _________________ Teacher: _______________
Reviewer: Quadratic Equations
I. Multiple Choice: Choose the letter of the correct answer. Show your solution.
1. What are the values of x that satisfy the equation 3 – 27x2 = 0?
A. x = [pic]3 B. x = [pic] C. x = [pic] D. x = [pic]
2. What are the solutions of the equation 6x2 + 9x – 15 = 0?
A. 1, - 15 B. 1, [pic] C. – 1, - 5 D. 3, [pic]
3. For which equation is – 3 NOT a solution?
A. x2 – 2x – 15 = 0 C. 2x2 + 12x = - 18
B. x2 – 21 = 4x D. 9 + x2 = 0
4. What are the solutions of the equation (2x – 7)2 = 25?
A. 6, - 6 B. 6, 1 C. 6, -1 D. – 6, - 1
5. Find the sum of the solutions to the equation x2 + 2x – 15 = 0.
A. 8 B. – 8 C. 2 D. – 2
6. Find the product of the solutions to the equation x2 – 8x = 9.
A. 6 B. – 6 C. 9 D. – 9
7. Which equation has [pic]as a solution?
A. (2x – 5)(x + 1) = 0 C. (5x + 2)(x + 1) = 0
B. 5x – 2)(x + 1) D. (2x + 5)(x + 1)
8. The equation x2 – 3x + a = 0 has two roots. One root of the equation is 2. What is the other root?
A. – 2 B. – 1 C. 1 D. 3
9. What is the quadratic equation determine by the roots 3 and – 4 ?
A. x2 + x – 12 = 0 C. x2 + x + 12 = 0
B. x2 – x – 12 = 0 D. x2 – x + 12 = 0
10. One solution to...

...
Assignment 4: Real World QuadraticFunctions
MAT222: Intermediate Algebra
Professor Andrea Grych
Assignment 4: Real World QuadraticFunctions
Managers and business people use quadratic equations on a daily basis in order to find out how much of a profit can be made. The following problem is an example of that. On page number 666 of the textbook, problem number 56 (Dugopolski, 2012) states that in order to get maximum profits, a chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P = −25x2 + 300x. What number of clerks will maximize the profit, and what is the maximum possible profit?
In order to solve the first question of what number of clerks will maximize the profit, the function -25x + 300x = 0 must be solved in order to get the x-intercepts of the parabola.
-25x2 + 300x = 0 Divide both sides by -1 to get rid of the negative.
25x2 – 300x = 0 Factor the left side of the problem.
25x (x – 12) = 0 Now use the Zero Factor Property.
25x = 0 or x – 12 = 0 Solve each problem.
x = 0 or x = 12 This means that the parabola will cross the x-axis at 0 and 12.
Now that we know where the parabola will cross the x –axis at, the next question can be asked. What number of clerks will...

...329
Quadratic Equations
Chapter-15
Quadratic Equations
Important Definitions and Related Concepts
1. Quadratic Equation
If p(x) is a quadratic polynomial, then p(x) = 0 is called
a quadratic equation. The general formula of a quadratic
equation is ax 2 + bx + c = 0; where a, b, c are real
numbers and a 0. For example, x2 – 6x + 4 = 0 is a
quadratic equation.
2. Roots of aQuadratic Equation
Let p(x) = 0 be a quadratic equation, then the values of
x satisfying p(x) = 0 are called its roots or zeros.
For example, 25x2 – 30x + 9 = 0 is a quadratic equation.
3
And the value of x =
is the solution of the given
5
equation.
3
Since, if we put x =
in 25x2 – 30x + 9 = 0, we have,
5
2
3
3
LHS = 25 × – 30 ×
+ 9
5
5
= 9 – 18 + 9 = 0 = RHS
Finding the roots of a quadratic equation is known as
solving the quadratic equation.
5. Methods of Solving Quadratic Equation
( i ) By Factorization
This can be understood by the examples given
below:
2
Ex. 1: Solve: 25 x 30 x 9 0
Soln: 25x 2 30x 9 0 is equivalent to
5x 2 25x 3 32
0
5 x 32 0
3 3
3
,
or simply x
as the
5 5
5
required solution.
This gives x
Ex. 2: Find the solutions of the quadratic equation
x 2 6 x 5 0 and check the solutions.
2
Soln: The quadratic polynomial...

...
Real World QuadraticFunctions
MAT222: Intermediate Algebra
Argenia L. McCray
Professor: Eric Bienstock
October 27, 2014
QuadraticFunctions
This week we have been learning the many different quadraticfunctions. Throughout the world the quadraticfunctions are being used / or being implicated into their system of employment, business, and in all schools. To say that the quadraticfunction has limited/ or less of it many possibilities which is available to be used in solving our many problem today issues. I will used one today to help me to solve my prediction and to helping me make some much better decisions.
In this lesson we was given this problem to work out, “A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P=-25x2+300x. What number of clerks will maximize the profit, and what is the maximum possible profit?” In order to solving this problem it requires all the following steps listed below (Dugopolski, M. 2012).
One way to finding the x-intercept is to do the following
-25x2 + 300x = 0
25x2-300x = 0 we will now divide -1 to both sides to get rid of the negative in front of -25
25x (x-12) = 0 Then we will factored the left side of the problem of our equation
25x =...