Explanation Given: The set of five points are, ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , ( 3 , 6 ) , ( − 1 , − 14 ) Approach: Consider the points ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , ( 3 , 6 ) , ( − 1 , − 14 ) . Substitute first two points in the equation of a line y = a x + b , to find the line that contains the first two points. Similarly, Substitute first three points in the equation of a quadratic y = a x 2 + b x + c , to find the quadratic that contains the first three points. Substitute first four points in the equation of a cubic y = a x 3 + b x 2 + c x + d , to find the cubic that contains the first four points. Substitute all five points in the equation of a bi-quadratic y = a x 4 + b x 3 + c x 2 + d x + e , to find the bi-quadratic that contains all the five points. Graph the points and functions in the same viewing rectangle on TI-83 plus calculator as follows- • Press on STAT and then, press on EDIT . • Enter the data. • Press Y= and enter the expressions. • Press the STAT PLOT and adjust the same. • Press WINDOW button to adjust the window settings. • Press GRAPH button to display the graph. Calculation: Consider the points ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , ( 3 , 6 ) , ( − 1 , − 14 ) . Substitute first two points ( 0 , 0 ) and ( 1 , 12 ) in the equation of a line y = a x + b , y = a x + b 0 = a ⋅ 0 + b b = 0 y = a x + b 12 = a ⋅ 1 + 0 a = 12 From the above equations, a = 12 b = 0 Equation of line passing through ( 0 , 0 ) and ( 1 , 12 ) is y = a x + b y = 12 x + 0 y = 12 x ( 1 ) Substitute first three points ( 0 , 0 ) , ( 1 , 12 ) , and ( 2 , 40 ) in the equation of a quadratic y = a x 2 + b x + c , y = a x 2 + b x + c 0 = a ⋅ 0 + b ⋅ 0 + c c = 0 y = a x 2 + b x + c 12 = a ⋅ 1 2 + b ⋅ 1 + 0 a + b = 12 ( 2 ) y = a x 2 + b x + c 40 = a ⋅ 2 2 + b ⋅ 2 + 0 2 a + b = 20 ( 3 ) Subtract equations ( 2 ) and ( 3 ) , 2 a + b − ( a + b ) = 20 − 12 a = 8 Substitute a = 8 in equation ( 2 ) , a + b = 12 8 + b = 12 b = 12 − 8 b = 4 Equation of quadratic passing through ( 0 , 0 ) , ( 1 , 12 ) , and ( 2 , 40 ) is y = a x 2 + b x + c y = 8 x 2 + 4 x + 0 y = 8 x 2 + 4 x ( 4 ) Substitute first four points ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , and ( 3 , 6 ) in the equation of a cubic y = a x 3 + b x 2 + c x + d , y = a x 3 + b x 2 + c x + d 0 = a ⋅ 0 + b ⋅ 0 + c ⋅ 0 + d d = 0 y = a x 3 + b x 2 + c x + d 12 = a ⋅ 1 3 + b ⋅ 1 2 + c ⋅ 1 + 0 a + b + c = 12 ( 5 ) y = a x 3 + b x 2 + c x + d 40 = 2 3 ⋅ a + 2 2 ⋅ b + 2 ⋅ c + 0 4 a + 2 b + c = 20 ( 6 ) y = a x 3 + b x 2 + c x + d 6 = a ⋅ 3 3 + b ⋅ 3 2 + c ⋅ 3 + 0 9 a + 3 b + c = 2 ( 7 ) Subtract equation ( 6 ) and equation ( 5 ) , 4 a + 2 b + c − ( a + b + c ) = 20 − 12 3 a + b = 8 ( 8 ) Subtract equation ( 7 ) and equation ( 6 ) , 9 a + 3 b + c − ( 4 a + 2 b + c ) = 2 − 20 5 a + b = − 18 ( 9 ) Subtract equation ( 9 ) and equation ( 8 ) , 5 a + b − ( 3 a + b ) = − 18 − 8 2 a = − 26 a = − 13 Substitute − 13 for a in equation ( 8 ) , 3 a + b = 8 3 ( − 13 ) + b = 8 b = 8 + 39 b = 47 Substitute − 13 for a , and 47 for b in equation ( 5 ) , a + b + c = 12 − 13 + 47 + c = 12 c = 12 − 34 c = − 22 Equation of cubic passing through ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , and ( 3 , 6 ) is y = a x 3 + b x 2 + c x + d y = − 13 x 3 + 47 x 2 − 22 x + 0 y = − 13 x 3 + 47 x 2 − 22 x ( 10 ) Substitute all five points ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , ( 3 , 6 ) , and ( − 1 , − 14 ) in the equation of a bi-quadratic y = a x 4 + b x 3 + c x 2 + d x + e ,

4th Edition

ISBN: 8220100548512

Author: Watson

Publisher: CENGAGE L

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Chapter 11.1, Problem 75E

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The line that contains the first two points, the quadratic that contains the first three points, the cubic that contains the first four points, and the fourth-degree polynomial that contains all five points and using a graphing device graph the points and functions in the same viewing rectangle.

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The line that contains the first two points, the quadratic that contains the first three points, the cubic that contains the first four points, and the fourth-degree polynomial that contains all five points and using a graphing device graph the points and functions in the same viewing rectangle.

Solution Summary: The author explains how to graph the points and functions in the same viewing rectangle using a graphing device.

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