## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 10 - 10.1 - Matrices and Systems of Equations - 10.1 Exercises - Page 712: 90

#### Answer

$f(x) = -2x^{2} - 2x + 1$

#### Work Step by Step

We must first solve for the functions in terms of a, b, and c: f(-2) = 4a - 2b + c = -3 f(1) = a + b + c = -3 f(2) = 4a + 2b + c = -11 We can then form the matrix, and use Gaussian elimination and back-substitution to solve the matrix: $\begin{bmatrix} 4 & -2 & 1 & |-3\\ 1 & 1 & 1 & |-3\\ 4 & 2 & 1 & |-11\\ \end{bmatrix}$ ~ $\begin{bmatrix} 4 & -2 & 1 & |-3\\ 0 & -6 & -3 & |9\\ 0 & -4 & 0 & |8\\ \end{bmatrix}$ ~ $\begin{bmatrix} 4 & -2 & 1 & |-3\\ 0 & 2 & 1 & |-3\\ 0 & -2 & 0 & |4\\ \end{bmatrix}$ ~ $\begin{bmatrix} 4 & -2 & 1 & |-3\\ 0 & 2 & 1 & |-3\\ 0 & 0 & 1 & |1\\ \end{bmatrix}$ C: c = 1 B: 2b + 1 = -3 2b = -4 b = -2 A: 4a - 2(-2) + 1 = -3 4a + 4 + 1 = -3 4a + 5 = -3 4a = -8 a = -2 Using the solution above, the quadratic function is: $f(x) = -2x^{2} - 2x + 1$

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