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

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Chapter6: Linear Systems
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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
Transcribed Image Text: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) = -6.
Substitute x =
1, x = 2, and x = 3 into f(x) = ax2 + bx + c to form the system of linear equations.
f(1) = a + b + c = -2
f(2)
= 4a + 2b +c = -3
f(3)
= 9a + 3b + c = -6
a + b + C = -2
Therefore, the obtained system of equations is {4a + 2b + c = -3.
9a + 3b + C = -6
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) on R3.
1
-2
-2
-3
R2 - 4R1 →
R3 - 9R1
0 -6
-8
... ..
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) = -6. Substitute x = 1, x = 2, and x = 3 into f(x) = ax2 + bx + c to form the system of linear equations. f(1) = a + b + c = -2 f(2) = 4a + 2b +c = -3 f(3) = 9a + 3b + c = -6 a + b + C = -2 Therefore, the obtained system of equations is {4a + 2b + c = -3. 9a + 3b + C = -6 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) on R3. 1 -2 -2 -3 R2 - 4R1 → R3 - 9R1 0 -6 -8 ... ..
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