When you are graphing quadratics, it is the same as graphing linear equations but, quadratics have the curvy line, called a parabola. When you are graphing your points, it is best to graph three or more points. You are really going to need to point three or more points, because if there are less than three you will not have a correct graph, graphing more than three will insure that your graph will be correct. The biggest number that they say you have to graph will most likely not be able to be graphed because most of the graphs will not be big enough to graph that point. If you happen to somehow forget that the line has to be curved, having those extra points graphed will help remind you that the line will be curved. If you’re a value is positive, then the parabola will be a smile shape. If you’re a value is a negative, your parabola will be a sad face shape. In any and all functions, you have a trajectory, you start at a given spot and throw an object and measure the height and distance and out it on to a graph, the most common set up for a function like this is(ax^2+bx+c=0). Quadratic equation is a squared plus b squared = c squared. It's used to find the length of three sides of a triangle. The theory is the same as any other polynomial, and the Greeks found out that, this formula holds true, regardless of the different types or lengths of the sides. So, by using the same method you use to solve a typical polynomial, you can solve this equation as well. For the formula to actually work you must have your equation in this form, quadratic=0. The 2a in the bottom of the equation, is a 2a NOT just a 2. You also have to make sure that you do not drop the square root or the plus or minus in the middle of figuring out your problem. And the b^2 means b (b) not b (2). Do not try to take any shortcuts or slide by because your answer to the problem will be WRONG unless you take the problem and solve it step by step. The more mistakes, the more you will be wrong and you...

...Real World QuadraticFunctions
Read the following instructions in order to complete this assignment:
1. Solve problem 56 on pages 666-667 of Elementary and Intermediate Algebra.
2. Write a two to three page paper that is formatted in APA style and according to the Math Writing
Guide. Format your math work as shown in the example and be concise in your reasoning. In the
body of your essay, please make sure to include:
o An explanation of the basic shape and location of the graph and what it tells us about the
Profit function.
o Your solutions (it is a double question) to the above problem, making sure to include all
mathematical work for both problems.
o Discuss how and why this is important for managers to know, and explain what could
happen if the ideal conditions were not met.
The paper must be at least two pages in length and formatted according to APA style. Cite your resources
in text and on the reference page. For information regarding APA samples and tutorials, visit the Ashford
Writing Center, within the Learning Resources tab on the left navigation toolbar.
Carefully review the Grading Rubric for the criteria that will be used to evaluate your assignment.
Real World QuadraticFunctions
Solving real world quadratic problems is mandatory for business professionals
and managers. Profit functions routinely show up in their work tasks and these...

...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
--------------------------------------------------------------------------------
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...

...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...

...Section 2.1
Linear and QuadraticFunctions and Modeling
67
Chapter 2 Polynomial, Power, and Rational Functions
■ Section 2.1 Linear and QuadraticFunctions and Modeling
Exploration 1
1. –$2000 per year 2. The equation will have the form v(t)=mt+b. The value of the building after 0 year is v(0)=m(0)+b=b=50,000. The slope m is the rate of change, which is –2000 (dollars per year). So an equation for the value of the building (in dollars) as a function of the time (in years) is v(t)=–2000t+50,000. 3. v(0)=50,000 and v(16)=–2000(16)+50,000=18,000 dollars 4. The equation v(t)=39,000 becomes –2000t+50,000=39,000 –2000t=–11,000 t=5.5 years
6. (x-4)2=(x-4)(x-4)=x2-4x-4x+16 =x2-8x+16 7. 3(x-6)2=3(x-6)(x-6)=(3x-18)(x-6) =3x2-18x-18x+108=3x2-36x+108 8. –3(x+7)2=–3(x+7)(x+7) =(–3x-21)(x+7)=–3x2-21x-21x-147 =–3x2-42x-147 9. 2x2-4x+2=2(x2-2x+1)=2(x-1)(x-1) =2(x-1)2 10. 3x2+12x+12=3(x2+4x+4)=3(x+2)(x+2) =3(x+2)2
Section 2.1 Exercises
1. Not a polynomial function because of the exponent –5 2. Polynomial of degree 1 with leading coefficient 2 3. Polynomial of degree 5 with leading coefficient 2 4. Polynomial of degree 0 with leading coefficient 13 5. Not a polynomial function because of cube root 6. Polynomial of degree 2 with leading coefficient –5 5 5 5 18 7. m= so y-4= (x-2) ⇒ f(x)= x+ 7 7 7 7
y 5 (2, 4) x
Quick Review 2.1
1. y=8x+3.6 2. y=–1.8x-2 3...

...While the ultimate goal is the same, to determine the value(s) that hold true for the equation, solving quadratic equations requires much more than simply isolating the variable, as is required in solving linear equations. This piece will outline the different types of quadratic equations, strategies for solving each type, as well as other methods of solutions such as Completing the Square and using the Quadratic Formula. Knowledge of factoring perfect square trinomials and simplifying radical expression are needed for this piece. Let’s take a look!
Standard Form of a Quadratic Equation
ax2+ bx+c=0
Where a, b, and c are integers
and a≥1
I. To solve an equation in the form ax2+c=k, for some value k. This is the simplest quadratic equation to solve, because the middle term is missing.
Strategy: To isolate the square term and then take the square root of both sides.
Ex. 1) Isolate the square term, divide both sides by 2
Take the square root of both sides
2x2=40
2x22= 40 2
x2 =20
Remember there are two possible solutions
x2= 20
Simplify radical; Solutions
x= ± 20
x=± 25
(Please refer to previous instructional materials Simplifying Radical Expressions )
II. To solve a quadratic equation arranged in the form ax2+ bx=0.
Strategy: To factor the binomial using the greatest common factor...

...
Real World QuadraticFunctions
Maximum profit. 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 find the point at which profit is maximized, I must find the critical points of the first derivative of the equation.
Coefficient of x^2 is negative, so the maximum value of P will be found at the parabola's vertex. I also know that it has to be symmetric, so the average of or mid-point between the roots of the function will give me the x value of the vertex.
The x-coordinate of the vertex is given by: x = b/2·a, and so the maximum value of P will be found at x = b/2·a
My equation is already in the standard form: P=ax^2+bx+c, p = -25x2 + 300x quadratic equation Therefore I can find the max profit by finding the value of x of the axis of symmetry and find the vertex with that: this is a quadratic equation with a negative coefficient of x^2, so I know that the max is on the axis of symmetry.
The formula for the axis of symmetry; x = -b/ (2a), in this equation a = -25, b = 300
X = - 300 / 2 · (-25) I will Simplify the equation
x = -300/ -50 Divide
x = 6
The basic shape of the graph in this equation of...

...MCR3U0: Unit 2 – Equivalent Expressions and
QuadraticFunctions
Radical Expressions
1) Express as a mixed radical in simplest form.
a)
c)
b)
e)
d)
f)
2) Simplify.
a)
b)
d)
e)
c)
f)
3) Simplify.
a)
b)
c)
d)
e)
f)
4) Simplify.
a)
d)
b)
e)
f)
c)
For questions 5 to 9, calculate the exact values and express your answers in simplest radical form.
5) Calculate the length of the diagonal of a square with side length 4 cm.
6) A square has an area of 450 cm2. Calculate the side length.
7) Determine the length of the diagonal of a rectangle with dimensions 3 cm
9 cm.
8) Determine the length of the line segment from A(-2, 7) to B(4, 1).
9) Calculate the perimeter and area of the triangle to the right.
10) If
and
, which is greatest,
or
?
11) Express each radical in simplest form.
a)
c)
b)
d)
12) Simplify
FMSS 2013
.
Page 1 of 16
Solutions
1a)
2a)
3a)
4a)
4e)
8)
11a)
1b)
1c)
1d)
2c) 32
2b)
3b)
3c)
4b)
1e)
2d)
3d)
9) Perimeter =
11b)
3e)
4c)
5)
4f)
1f)
2e)
cm
2f) -140
3f)
4d)
6)
units, Area = 12 square units 10)
11c)
11d) –
cm
7)
cm
12)
Polynomial Expressions
13) Expand and Simplify
a)
b)
c)
d)
e)
f)
14) Expand and Simplify
a)
b)
c)
d)
e)
f)
15) Expand and Simplify
a)
b)
d)
e)
16) Factor
a)
b)
c)
d)
e)
f)
17) Factor
a)
b)
c)
d)
e)
f)
18) Factor
a)
b)
c)
d)
e)
f)
19) Show that
and
are equivalent.
20) Show that
and
are not equivalent.
FMSS 2013
Page 2 of 16
21) a) Is
equivalent to
? Justify your...

...
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...