Copyright © 2007 Pearson Education, Inc. Slide 3-1

Copyright © 2007 Pearson Education, Inc. Slide 3-2 Chapter 3: Polynomial Functions 3.1 Complex Numbers 3.2 Quadratic Functions and Graphs 3.3 Quadratic Equations and Inequalities 3.4 Further Applications of Quadratic Functions and Models 3.5 Higher Degree Polynomial Functions and Graphs 3.6 Topics in the Theory of Polynomial Functions (I) 3.7 Topics in the Theory of Polynomial Functions (II) 3.8 Polynomial Equations and Inequalities; Further Applications and Models

Copyright © 2007 Pearson Education, Inc. Slide Applications of Quadratic Functions and Models ExampleA farmer wishes to enclose a rectangular region. He has 120 feet of fencing, and plans to use one side of his barn as part of the enclosure. Let x represent the length of one side of the fencing. (a)Find a function A that represents the area of the region in terms of x. (b)What are the restrictions on x? (c)Graph the function in a viewing window that shows both x-intercepts and the vertex of the graph. (a)What is the maximum area the farmer can enclose? Solution (a)Area = width  length, so (b) Since x represents length, x > 0. Also, 120 – 2x > 0, or x < 60. Putting these restrictions together gives 0 < x < 60.

Copyright © 2007 Pearson Education, Inc. Slide Applications: Area of a Rectangular Region (c) (d) Maximum value occurs at the vertex. Figure 38 pg 3-60a

Copyright © 2007 Pearson Education, Inc. Slide Finding the Volume of a Box ExampleA machine produces rectangular sheets of metal satisfying the condition that the length is 3 times the width. Furthermore, equal size squares measuring 5 inches on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. (a)Determine a function V that expresses the volume of the box in terms of the width x of the original sheet of metal. (b)What restrictions must be placed on x? (c)If specifications call for the volume of such a box to be 1435 cubic inches, what should the dimensions of the original piece of metal be? (a)What dimensions of the original piece of metal will assure a volume greater than 2000 but less than 3000 cubic inches? Solve graphically.

Copyright © 2007 Pearson Education, Inc. Slide Finding the Volume of a Box (a)Using the drawing, we have Volume = length  width  height, (b)Dimensions must be positive, so 3x – 10 > 0 and x – 10 > 0, or Both conditions are satisfied when x > 10, so the theoretical domain is (10,  ). (c) Only 17 satisfies x > 10. The original dimensions : 17 inches by 3(17) = 51 inches.

Copyright © 2007 Pearson Education, Inc. Slide Finding the Volume of a Box (d)Set y 1 = 15x 2 – 200x + 500, y 2 = 2000, and y 3 = Points of intersection are approximately (18.7, 2000) and (21.2,3000) for x > 10. Therefore, the dimensions should be between 18.7 and 21.2 inches, with the corresponding length being 3(18.7)  56.1 and 3(21.2)  63.6 inches.

Copyright © 2007 Pearson Education, Inc. Slide A Problem Requiring the Pythagorean Theorem The longer leg of a right triangular lot is approximately 20 meters longer than twice the length of the shorter leg. The hypotenuse is approximately 10 meters longer than the length of the longer leg. Estimate the lengths of the sides of the triangular lot. Analytic Solution Let s = the length of the shorter leg in meters. Then 2s + 20 is the length of the longer leg, and (2s + 20) + 10 = 2s + 30 is the length of the hypotenuse. The approximate lengths of the sides are 50 m, 120 m, and 130 m.

Copyright © 2007 Pearson Education, Inc. Slide A Problem Requiring the Pythagorean Theorem Graphing Calculator Solution Replace s with x and find the x-intercepts of the graph of y 1 – y 2 = 0 where y 1 = x 2 + (2x + 20) 2 and y 2 = (2x + 30) 2. We use the x-intercept method because the y-values are very large. The x-intercept is 50, supporting the analytic solution. Figure 44 pg 3-64a

Copyright © 2007 Pearson Education, Inc. Slide Quadratic Models The percent of Americans 65 and older for selected years is shown in the table. (a)Plot data, letting x = 0 correspond to (b)Find a quadratic function, f (x) = a(x – h) 2 + k, that models the data by using the vertex (0,4.1). (c)Graph f in the same viewing window as the data. (d)Use the quadratic regression feature of a graphing calculator to determine the quadratic function g that provides the best fit. Year, x % 65 and older, y

Copyright © 2007 Pearson Education, Inc. Slide Quadratic Models (a)L 1 = x-list and L 2 = y-list (b)Substituting 0 for h and 4.1 for k, and choose the data point (2040,20.6) that corresponds to the values x = 140, y = f (140) = 20.6, we can solve for a.

Copyright © 2007 Pearson Education, Inc. Slide Quadratic Models (c) (d) This is a pretty good fit, especially for the later years. Note that choosing other second points would produce other models. Figure 46a pg 3-67a