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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblema 39
Capitolo 7, Problema 39

Find the quadratic function f(x) = ax² + bx + c for which ƒ( − 2) = −4, ƒ(1) = 2, and f(2) = 0.

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Start by writing the general form of the quadratic function: $f(x) = ax^{2} + bx + c$.
Use the given points to create a system of equations by substituting each \(x\) and \(f(x)\) value into the quadratic function:
For \(f(-2) = -4\), substitute \(x = -2\) and \(f(x) = -4\) to get: \(a(-2)^{2} + b(-2) + c = -4\).
For \(f(1) = 2\), substitute \(x = 1\) and \(f(x) = 2\) to get: \(a(1)^{2} + b(1) + c = 2\).
For \(f(2) = 0\), substitute \(x = 2\) and \(f(x) = 0\) to get: \(a(2)^{2} + b(2) + c = 0\).

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Quadratic Functions

A quadratic function is a polynomial of degree two, generally written as f(x) = ax² + bx + c, where a, b, and c are constants and a ≠ 0. Its graph is a parabola, and the coefficients determine its shape and position.
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Function Evaluation

Function evaluation involves substituting a specific input value into the function to find the corresponding output. For example, f(−2) means replacing x with −2 in the quadratic expression and calculating the result.
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Solving Systems of Equations

To find the coefficients a, b, and c, you set up equations based on the given function values and solve the resulting system of linear equations. This process often uses substitution or elimination methods to find the unknowns.
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a. Write each linear system as a matrix equation in the form AX = B. b. Solve the system using the inverse that is given for the coefficient matrix.

{x−y+z=82y−z=−72x+3y=1The inverse of [1−1102−1230] is [33−1−2−21−4−52].\(\begin{cases}\) x - y + z = 8 \\ 2y - z = -7 \\ 2x + 3y = 1 \(\end{cases}\) \\ \(\text{The inverse of }\) \(\begin{bmatrix}\) 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \(\end{bmatrix}\) \(\text{ is }\) \(\begin{bmatrix}\) 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \(\end{bmatrix}\)\(\text{.}\)

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In Exercises 37–44, use Cramer's Rule to solve each system. {4x−5y−6z=−1x−2y−5z=−122x−y=7\(\begin{cases}\) 4x - 5y - 6z = -1 \\ x - 2y - 5z = -12 \\ 2x - y = 7 \(\end{cases}\)

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In Exercises 39–42, find A^(-1) Check that AA^-1 = I and A^(-1)A = I

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Perform the indicated matrix operations given that A, B and C are defined as follows. If an operation is not defined, state the reason.

A=[40−3501],B=[51−2−2],C=[1−1−11]A=\(\begin{bmatrix}\)4 & 0\\ -3 & 5\\ 0 & 1\(\end{bmatrix}\),B=\(\begin{bmatrix}\)5 & 1\\ -2 & -2\(\end{bmatrix}\),C=\(\begin{bmatrix}\)1 & -1\\ -1 & 1\(\end{bmatrix}\)

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Find the cubic function f(x) = ax³ + bx² + cx + d for which ƒ( − 1) = 0, ƒ(1) = 2, ƒ(2) = 3, and ƒ(3) = 12.

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a. Write each linear system as a matrix equation in the form AX = B. b. Solve the system using the inverse that is given for the coefficient matrix.

{w−x+2y=−3x−y+z=4−w+x−y+2z=2−x+y−2z=−4The inverse of [1−12001−11−11−120−11−2] is [00−1−1141312120−10−1]\(\begin{cases}\) w - x + 2y \(\quad\)\(\quad\) = -3 \\ \(\quad\[\quad\) x - y + z = 4 \\ -w + x - y + 2z = 2 \\ \(\quad\]\quad\) -x + y - 2z = -4 \(\end{cases}\) \\ \(\text{The inverse of }\) \(\begin{bmatrix}\) 1 & -1 & 2 & 0 \\ 0 & 1 & -1 & 1 \\ -1 & 1 & -1 & 2 \\ 0 & -1 & 1 & -2 \(\end{bmatrix}\) \(\text{ is }\) \(\begin{bmatrix}\) 0 & 0 & -1 & -1 \\ 1 & 4 & 1 & 3 \\ 1 & 2 & 1 & 2 \\ 0 & -1 & 0 & -1 \(\end{bmatrix}\)

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