MAT 222 Algebra

Running head: QUADRATIC FUNCTIONS 1 Running header should use a shortened version of the title if the title is long. Page number is

located at right margin.

(full title; centered horizontally & vertically)

Real World Quadratic Functions

John Q. Student

MAT 222 Week 4 Assignment

Instructor’s Name

Date

QUADRATIC FUNCTIONS 2

Real World Quadratic Functions (title required on first line)

Quadratic functions are perhaps the best example of how math concepts can be combined into a single problem. To solve these, rules for order of operations, solving equations,

exponents, and radicals must be used. Because multiple variables are involved and affect the

outcome, quadratics are extension of functions as well.

The following example is similar to #56 on page 666 (Dugopolski, 2012) and is in the

form ax2 + bx + c = 0. The profit function P(x) = -12x2 + 600x where x is the number of clerks

working. The x-intercepts of the parabola can be found by solving -12x2 + 600x = 0.

-12x2 + 600x = 0 Divide both sides by -1.

12x2 – 600x = 0 Factor the left side.

12x(x – 50) = 0 Use Zero Factor Property

12x = 0 or x – 50 = 0 Solve each equation.

x = 0 or x = 50 The parabola will cross the x-axis at 0 and 50.

This quadratic function has a large a value which means the parabola will be narrow. It

also has a negative a value so the parabola will open downward. This means there will be a

maximum value of the graph at the vertex, which will happen at the x value of –b/(2a), where

b = 600 and a = -12 in this case.

What number of clerks will maximize the profit?

–b/(2a)

-600/2(-12) Notice the b value is now negative because of the negative in the formula.

-600/-24

25 25 clerks working will maximize profit.

What is the maximum possible profit when this many clerks are working?

QUADRATIC FUNCTIONS 3

P(x) = -12x2 + 600x Start with original profit function to find P(25).

P(25) = -12(25)2 + 600(25) Substitute 25 for all x’s in the problem.

P(25) = -12(625) + 15000 The exponent must be solved first.

P(25) = -7500 + 15000

P(25) = $7500 They can expect a possible $7500 profit when 25 clerks are

working.

Basically, the graph shows there will be no profit made when zero clerks are working or

when 50 clerks are working. The maximum profit will occur when 25 clerks are working. The

graph of this function is only relevant in the first quadrant because negative clerks cannot exist.

Conclusion paragraph would go here. Remember to include 4-5 sentences to make a

complete paragraph.

QUADRATIC FUNCTIONS 4

Reference Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY:

McGraw-Hill Publishing.

Use the word ‘Reference’ or ‘References’ as the title.

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