chapter 3: polynomial functions


Chapter 3: Polynomial Functions

3.1 Complex Numbers3.2 Quadratic Functions and Graphs3.3 Quadratic Equations and Inequalities3.4 Further Applications of Quadratic Functions and

Models3.5 Higher Degree Polynomial Functions and Graphs3.6 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


3.4 Applications of Quadratic Functions and Models

Example A 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) xxxA

xxxA1202)(

)2120()(2

Since x represents length, x > 0. Also, 120 – 2x > 0, or x < 60. Putting these restrictions together gives 0 < x < 60.


3.4 Applications: Area of a Rectangular Region

(c)

(d) Maximum value occurs at the vertex.

Figure 38 pg 3-60a

.120 ,2 where,1202)( 2 baxxxA

30)2(2

1202

a

bx

feet square 1800)30(120)30(2)30( 2 A


3.4 Finding the Volume of a Box

Example A 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.



(a) Using the drawing, we have Volume = length width height,

(b) Dimensions must be positive, so3x – 10 > 0 and x – 10 > 0, or

Both conditions are satisfied whenx > 10, so the theoretical domainis (10,).

(c)

50020015)5)(10)(103()(

2 xx

xxxV

.10and3

10 xx

17or3

11)17)(5515(0

935200150500200151435

2

2

xx

xxxxxx

Only 17 satisfies x > 10. The original dimensions : 17 inches by 3(17) = 51 inches.



(d) Set y1 = 15x2 – 200x + 500, y2 = 2000, and y3 = 3000. 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.


3.4 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.

0)10)(50(050040

9001204400804)302()202(

2

222

222

ssss

ssssssss

10or50 ss


3.4 A Problem Requiring the Pythagorean Theorem

Graphing Calculator Solution Replace s with x and find the x-intercepts of the graph of y1 – y2 = 0 where y1 = x2 +(2x + 20)2 and y2 = (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


3.4 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 1900.(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 1900 1920 1940 1960 1980 2000 2020 2040

% 65 and older, y 4.1 4.7 6.8 9.3 11.3 12.4 16.5 20.6



(a) L1 = x-list and L2 = 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.

1.40008418.)(0008418.600,191.46.20

1.4)0140(6.20

2

2

xxfaa

a



(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

chapter 3: polynomial functions

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