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Use a system of equations to find the quadratic function  f(x) = ax2 + bx + c  that satisfies the equations. Solve the system using matrices. f(1)= -2, f(2)= -3, f(3)= -6

Use a system of equations to find the quadratic function  f(x) = ax2 + bx + c  that satisfies the equations. Solve the system using matrices. f(1)= -2, f(2)= -3, f(3)= -6

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.7: Solving Systems Of Linear And Quadratic Equations
Problem 23PPS
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Use a system of equations to find the quadratic function 

f(x) = ax2 + bx + c

 that satisfies the equations. Solve the system using matrices.

f(1)= -2, f(2)= -3, f(3)= -6

Step 1 The given quadratic function is f(x) = ax2 + bx + c. It is given that f(1) = -2, f(2) = -3, and f(3) Substitute x = 1, x = 2, and x = 3 into f(x) = ax2 + bx + c to form the system of linear equations. = -6. f(1) = a + b + c = -2 f(2) = 4a + 2b + C = -3 f(3) = 9a + 3b + c = -6 b + c = -2 4a + 2b + c = -3. 9a + 3b + c = -6 a + Therefore, the obtained system of equations is Write the associated augmented matrix for this system. 1 1 1: -2 4 2 1: -3 9 3 1: -6 To obtain the matrix in reduced row-echelon form, apply elementary row operations until every column that has a leading 1 has zeros in every position above and below its leading 1. Perform the operations (R2 - 4R1) on R2 and (R3 - 9R1) (R3 - 9R:) on R3. 1. -2 0 -2 |-3 -3 : 5 5 R2 - 4R1 R3 - 9R1 0 -6 : 12 -8 12
Transcribed Image Text:Step 1 The given quadratic function is f(x) = ax2 + bx + c. It is given that f(1) = -2, f(2) = -3, and f(3) Substitute x = 1, x = 2, and x = 3 into f(x) = ax2 + bx + c to form the system of linear equations. = -6. f(1) = a + b + c = -2 f(2) = 4a + 2b + C = -3 f(3) = 9a + 3b + c = -6 b + c = -2 4a + 2b + c = -3. 9a + 3b + c = -6 a + Therefore, the obtained system of equations is Write the associated augmented matrix for this system. 1 1 1: -2 4 2 1: -3 9 3 1: -6 To obtain the matrix in reduced row-echelon form, apply elementary row operations until every column that has a leading 1 has zeros in every position above and below its leading 1. Perform the operations (R2 - 4R1) on R2 and (R3 - 9R1) (R3 - 9R:) on R3. 1. -2 0 -2 |-3 -3 : 5 5 R2 - 4R1 R3 - 9R1 0 -6 : 12 -8 12
Perform the operation (-R2) on R2. 1. 1. 1 -2 3 : -5/2 2 -5/2 1 0 -6 -8 : 12 Perform the operation (R3 + 6R2) on R3. 1 1 1: -2 3 0 1 |-5/2 -5/2 R3 + 6R2 → 0 0 1 : |-3 -3 Step 3 (R2 -R3) Perform the operations R1 - R3) on R1 and R1 - R3 → 3R3 1 1 0 1 2. 0 0 1 -3 Perform the operation R1 R2 on R1. R1 - R2 1. -1 2 1 -3
Transcribed Image Text:Perform the operation (-R2) on R2. 1. 1. 1 -2 3 : -5/2 2 -5/2 1 0 -6 -8 : 12 Perform the operation (R3 + 6R2) on R3. 1 1 1: -2 3 0 1 |-5/2 -5/2 R3 + 6R2 → 0 0 1 : |-3 -3 Step 3 (R2 -R3) Perform the operations R1 - R3) on R1 and R1 - R3 → 3R3 1 1 0 1 2. 0 0 1 -3 Perform the operation R1 R2 on R1. R1 - R2 1. -1 2 1 -3
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