Explanation Given: The points are ( 2012 , 150 ) , ( 2013 , 180 ) , ( 2014 , 240 ) and ( 2015 , 360 ) . Calculation: The third degree polynomial function be, p ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 Substituting x = 2012 , 2013 , 2014 and 2015 into p ( x ) and equating the results to the respective y -values produces the system of linear equations in the variables a 0 , a 1 , a 2 and a 3 as shown below. p ( 2012 ) = a 0 + a 1 ( 2012 ) + a 2 ( 2012 ) 2 + a 3 ( 2012 ) 3 = a 0 + 2012 a 1 + 4048144 a 2 + 8144865728 a 3 = 150 p ( 2013 ) = a 0 + a 1 ( 2013 ) + a 2 ( 2013 ) 2 + a 3 ( 2013 ) 3 = a 0 + 2013 a 1 + 4052169 a 2 + 8157016197 a 3 = 180 p ( 2014 ) = a 0 + a 1 ( 2014 ) + a 2 ( 2014 ) 2 + a 3 ( 2014 ) 3 = a 0 + 2014 a 1 + 4056196 a 2 + 8169178744 a 3 = 240 p ( 2015 ) = a 0 + a 1 ( 2015 ) + a 2 ( 2015 ) 2 + a 3 ( 2015 ) 3 = a 0 + 2015 a 1 + 4060225 a 2 + 8181353375 a 3 = 360 The augmented matrix is, [ 1 2012 4048144 8144865728 150 1 2013 4052169 8157016197 180 1 2014 4056196 8169178744 240 1 2015 4060225 8181353375 360 ] Solving the system by using Gauss elimination with back substitution: − R 1 + R 2 → R 2 , − R 1 + R 3 → R 3 , − R 1 + R 4 → R 4 [ 1 2012 4048144 8144865728 150 0 1 4025 12150469 30 0 2 8052 24313016 90 0 3 12081 36487647 210 ] R 3 − 2 R 2 → R 3 , R 4 − 3 R 2 → R 4 [ 1 2012 4048144 8144865728 150 0 1 4025 12150469 30 0 0 2 12078 30 0 0 6 36240 120 ] 1 2 R 3 → R 3 [ 1 2012 4048144 8144865728 150 0 1 4025 12150469 30 0 0 1 6039 15 0 0 6 36240 120 ] R 4 − 6 R 3 → R 4 [ 1 2012 4048144 8144865728 150 0 1 4025 12150469 30 0 0 1 6039 15 0 0 0 6 30 ] 1 6 R 4 → R 4 [ 1 2012 4048144 8144865728 150 0 1 4025 12150469 30 0 0 1 6039 15 0 0 0 1 5 ] The above matrix is in row echelon form To determine b) To graph: The polynomial function showing the points ( 2012 , 150 ) , ( 2013 , 180 ) , ( 2014 , 240 ) and ( 2015 , 360 ) .

Elementary Linear Algebra - Text Only (Looseleaf)

8th Edition

ISBN: 9781305953208

Author: Larson

Publisher: Cengage Learning

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Chapter 1.3, Problem 10E

To determine

a)

To calculate:

The polynomial function whose graph passes through the points (2012,150),(2013,180),(2014,240) and (2015,360).

To determine

b)

To graph:

The polynomial function showing the points (2012,150),(2013,180),(2014,240) and (2015,360).

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Chapter 1.3, Problem 9E

Chapter 1.3, Problem 11E

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Chapter 1 Solutions

Elementary Linear Algebra - Text Only (Looseleaf)

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a) To calculate: The polynomial function whose graph passes through the points ( 2012 , 150 ) , ( 2013 , 180 ) , ( 2014 , 240 ) and ( 2015 , 360 ) .

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a) To calculate: The polynomial function whose graph passes through the points (2012,150),(2013,180),(2014,240) and (2015,360).

b) To graph: The polynomial function showing the points (2012,150),(2013,180),(2014,240) and (2015,360).

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Chapter 1 Solutions

a)

To calculate:

The polynomial function whose graph passes through the points (2012,150),(2013,180),(2014,240) and (2015,360).

b)

To graph:

The polynomial function showing the points (2012,150),(2013,180),(2014,240) and (2015,360).