Copyright © 2006 Pearson Education, Inc Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Quadratic Functions and Equations 8 Quadratic Functions and Equations 8.1 Quadratic Equations 8.2 The Quadratic Formula 8.3 Applications Involving Quadratic Equations 8.4 Studying Solutions of Quadratic Equations 8.5 Equations Reducible to Quadratic 8.6 Quadratic Functions and Their Graphs 8.7 More About Graphing Quadratic Functions 8.8 Problem Solving and Quadratic Functions 8.9 Polynomial and Rational Inequalities Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Problem Solving and Quadratic Functions 8.8 Problem Solving and Quadratic Functions Maximum and Minimum Problems Fitting Quadratic Functions to Data Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Maximum and Minimum Problems We have seen that for any quadratic function f, the value of f (x) at the vertex is either a maximum or a minimum. Thus problems in which a quantity must be maximized or minimized can be solved by finding the coordinates of the vertex, assuming the problem can be modeled with a quadratic function. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example A farmer has 200 ft of fence with which to form a rectangular pen on his farm. If an existing fence forms one side of the rectangle, what dimensions will maximize the size of the area? Solution Familiarize. We make a drawing and label it, letting w = the width of the rectangle, in feet and l = the length of the rectangle, in feet. Existing fence w l Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Recall that Area = lw and Perimeter = 2w + 2l Recall that Area = lw and Perimeter = 2w + 2l. Since the existing fence forms one length of the rectangle, the fence will comprise three sides. Thus 2w + l = 200. 2. Translate. We have two equations: One guarantees that all 200 ft of fence will be used; the other expresses area in terms of length and width. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Factoring and completing the square, we get 3. Carryout. We need to express A as a function of l or w but not both. To do so, we solve for l in the first equation to obtain l = 200 – 2w. Substituting for l in the second equation, we get a quadratic function: A = (200 – 2w)w = 200w – 2w2 Factoring and completing the square, we get A = –2(w – 50)2 + 5000. The maximum area, 5000 ft2, occurs when w = 50 ft and l = 200 – 2(50), or 100 ft. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

4. Check. The check is left to the student. 5. State. The dimensions for the largest rectangular area for the pen that can be enclosed is 50 ft by 100 ft. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Fitting Quadratic Functions to Data Whenever a certain quadratic function fits a situation, that function can be determined if three inputs and their outputs are known. Each of the given ordered pairs is called a data point. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Use the data points (0, 10.4), (3, 16.8), and (6, 12.6) to find a quadratic function that fits the data. Solution We are looking for a function of the form f (x) = ax2 + bx + c, given that f (0) = 10.4, f (3) = 16.8, and f (6) = 12.6. Thus a(0)2 + b(0) + c = 10.4 a(3)2 + b(3) + c = 16.8 a(6)2 + b(6) + c = 12.6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Simplifying we see that we need to solve the system c = 10.4 (1) 9a + 3b + c = 16.8 (2) 36a + 6b + c = 12.6 (3) Substituting c = 10.4 into equations (2) and (3) and solving the resulting system we get So we have Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley