Explanation Concept used: A line that contain two point is written as, y = a x + b ........ (1) A quadratic curve that contains three points is written as, y = a x 2 + b x + c ........ (2) A cubic curve that contains four points is written as, y = a x 3 + b x 2 + c x + d …... (3) The fourth degree polynomial that contains five points is written as, y = a x 4 + b x 3 + c x 2 + d x + e ........ (4) Given information: The five points are ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , ( 3 , 6 ) , ( − 1 , − 14 ) Calculation: The first two points are ( 0 , 0 ) and ( 1 , 12 ) . Substitute the value of 0 for x, and 0 for y in equation (1) to obtain the value of b written as, y = a x + b 0 = a ⋅ ( 0 ) + b b = 0 Substitute the value of 1 for x, 12 for y and 0 for b in equation (1) to obtain the value of a written as, y = a x + b 12 = a ⋅ ( 1 ) + 0 a = 12 Here, the line that contain the first two point ( 0 , 0 ) and ( 1 , 12 ) is y = 12 x . To obtain graph of the line that contain first two point, first turn ON the calculator and press Y= button then insert the equation y = 12 x as shown in Figure 1. Figure 1 Now, press the WINDOW button and set the viewing window as shown in Figure 2. Figure 2 After this press the GRAPH button and the graph is obtained as shown in Figure 3. Figure 3 The quadratic that contains the first three points are ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) . Substitute the value of 0 for x, and 0 for y in equation (2) is written as, y = a x 2 + b x + c 0 = a ⋅ ( 0 ) + b ⋅ ( 0 ) + c c = 0 Substitute the value of 1 for x, 12 for y and 0 for c in equation (2) is written as, y = a x 2 + b x + c 12 = a ⋅ 1 2 + b ⋅ 1 + 0 a + b = 12 Substitute the value of 2 for x, 40 for y and 0 for c in equation (2) is written as, y = a x 2 + b x + c 40 = a ⋅ 2 2 + b ⋅ 2 + 0 4 a + 2 b = 40 Multiply by 2 in equation a + b = 12 and subtract from 4 a + 2 b = 40 to obtain the value of a is written as, ( 4 a + 2 b − 40 ) − 2 ( a + b − 12 ) = 0 4 a + 2 b − 40 − 2 a − 2 b + 24 = 0 2 a = 16 a = 8 Substitute the value of 8 for a in equation a + b = 12 to obtain the value of b is written as, a + b = 12 8 + b = 12 b = 12 − 8 b = 4 Substitute the value of 8 for a, 4 for b and 0 for c in equation y = a x 2 + b x + c to obtain the quadratic. y = a x 2 + b x + c y = 8 x 2 + 4 x + 0 y = 8 x 2 + 4 x Here, the quadratic that contain first three point ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) is y = 8 x 2 + 4 x To obtain graph of the quadratic that contain first three point, first turn ON the calculator and press Y= button then insert the equation y = 8 x 2 + 4 x as shown in Figure 1. Figure 4 Now, press the WINDOW button and set the viewing window as shown in Figure 2. Figure 5 After this press the GRAPH button and the graph is obtained as shown in Figure 3. Figure 6 The cubic that contains the first four points are ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , ( 3 , 6 ) . Substitute the value of 0 for x, and 0 for y in equation (3) is written as, y = a x 3 + b x 2 + c x + d 0 = a ⋅ ( 0 ) + b ⋅ ( 0 ) + c ⋅ ( 0 ) + d d = 0 Substitute the value of 1 for x, 12 for y and 0 for d in equation (3) is written as, y = a x 3 + b x 2 + c x + d 12 = a ⋅ 1 3 + b ⋅ 1 2 + c ⋅ 1 + 0 a + b + c = 12 Substitute the value of 2 for x, 40 for y and 0 for d in equation (3) is written as, y = a x 3 + b x 2 + c x + d 40 = a ⋅ 2 3 + b ⋅ 2 2 + c ⋅ 2 + 0 8 a + 4 b + 2 c = 40 4 a + 2 b + c = 20 Substitute the value of 3 for x, 6 for y and 0 for d in equation (3) is written as, y = a x 3 + b x 2 + c x + d 6 = a ⋅ 3 3 + b ⋅ 3 2 + c ⋅ 3 + 0 27 a + 9 b + 3 c = 6 9 a + 3 b + c = 2 Subtract equation a + b + c = 12 from 4 a + 2 b + 2 c = 20 to obtain the equation as, ( 4 a + 2 b + c − 20 ) − ( a + b + c − 12 ) = 0 4 a + 2 b + c − 20 − a − b − c + 12 = 0 3 a + b − 8 = 0 3 a + b = 8 Subtract equation 4 a + 2 b + c = 20 from 9 a + 3 b + c = 2 to obtain the equation as, ( 9 a + 3 b + c − 2 ) − ( 4 a + 2 b + c − 20 ) = 0 9 a + 3 b + c − 2 − 4 a − 2 b − c + 20 = 0 5 a + b + 18 = 0 5 a + b = − 18 Now, subtract 3 a + b = 8 from 5 a + b = − 18 to obtain the value of a is written as, 5 a + b + 18 − ( 3 a + b − 8 ) = 0 5 a + b + 18 − 3 a − b + 8 = 0 2 a + 26 = 0 2 a = − 26 Further solve the above equation. 2 a = − 26 a = − 26 2 a = − 13 Substitute the value of − 13 for a in equation 3 a + b = 8 to obtain the value of b is written as, 3 a + b = 8 3 ⋅ ( − 13 ) + b = 8 − 39 + b = 8 b = 47 Substitute the value of − 13 for a and 47 for b in equation a + b + c = 12 to obtain the value of c is written as, a + b + c = 12 − 13 + 47 + c = 12 35 + c = 12 c = − 22 Substitute the value of − 13 for a, 47 for b, − 22 for c and 0 for d in equation y = a x 3 + b x 2 + c x + d to obtain the cubic that contain three points. 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 Here, the cubic that contain first four point ( 0 , 0 ) , ( 1 , 12 ) , ( 2 , 40 ) , ( 3 , 6 ) is y = − 13 x 3 + 47 x 2 − 22 x To obtain graph of the cubic that contain first four point, first turn ON the calculator and press Y= button then insert the equation y = − 13 x 3 + 47 x 2 − 22 x as shown in Figure 1

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 6.1, Problem 75E

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The line that contain the first two points, the quadratic curve that contains the first three points, the cubic curve that contains the first four points, and the fourth-degree polynomial that contain all five points.

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Chapter 6.1, Problem 74E

Chapter 6.2, Problem 1E

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To graph: The line that contain the first two points, the quadratic curve that contains the first three points, the cubic curve that contains the first four points, and the fourth-degree polynomial that contain all five points.

Solution Summary: The author explains how the lines contain the first two points, the quadratic curve, and the fourth-degree polynomial.