Problem 98 - Precalculus with Limits

Question

In Exercises 95-98, use a system of equations to find the quadratic function \(f(x) = ax^2 + bx + c\) that satisfies the equations. Solve the system using matrices. \(f(-2) = -3\), \(f(1) = -3\), \(f(2) = -11\)

Step-by-Step Solution

Answer

The quadratic function that satisfies the equations is \(f(x) = -0.75x^2 - 2.75x + 1\).

Step 1: Form equations

Substituting -2, 1 and 2 for x, we get \n1) \(a(-2^2) + b(-2) + c = -3\) -> \(4a - 2b + c = -3\) \n2) \(a(1^2) + b(1) + c = -3\) -> \(a + b + c = -3\) \n3) \(a(2^2) + b(2) + c = -11\) -> \(4a + 2b + c = -11\)

Step 2: Form the Matrix

Write the system as a matrix: \[\begin{align*}4 & -2 & 1 & -3 \1 & 1 & 1 & -3 \4 & 2 & 1 & -11 \\end{align*}\]

Step 3: Solve the system

Perform Gaussian elimination to find the reduced row echelon form: \[\begin{align*}1 & -0.5 & 0.25 & -0.75 \0 & 1 & 0.75 & -2.75 \0 & 0 & 1 & 1 \\end{align*}\]

Step 4: Interpret the solution

The solution to the system represents the coefficients a, b, and c of the function f(x) = ax^2 + bx + c: a = -0.75b = -2.75,c = 1

Key Concepts

Understanding Systems of Equations

A system of equations is a collection of two or more equations with the same set of variables. In our exercise, we are given three equations that come from the quadratic function \(f(x) = ax^2 + bx + c\). These equations capture specific values of the function and their corresponding known outputs.

To systematically solve for the coefficients \(a\), \(b\), and \(c\), we set up equations based on the given function values:

• \(4a - 2b + c = -3\)
• \(a + b + c = -3\)
• \(4a + 2b + c = -11\)

Each equation represents a line in a coordinate system, and solving the system means finding the point where all these lines intersect.

Solving systems of linear equations like this is fundamental in practices ranging from engineering to economics, as it helps to determine unknown variables based on given conditions. By using algebraic methods alongside valuable tools like matrices (which we will explore next), we can efficiently find these solutions.

The Role of Matrix Algebra

Matrix algebra offers a concise way to represent and solve systems of linear equations. Matrices are essentially rectangular arrays of numbers that can encapsulate multiple equations easily.

In our problem, we transform the system of equations into a matrix:
\[\begin{bmatrix}4 & -2 & 1 \1 & 1 & 1 \4 & 2 & 1\end{bmatrix}\begin{bmatrix} a \ b \ c \end{bmatrix} =\begin{bmatrix} -3 \ -3 \ -11 \end{bmatrix}\]This matrix equation represents the same system but makes it easier to apply various solving techniques, like Gaussian elimination.

One of the benefits of using matrices is the ability to perform operations like addition, multiplication, and elimination systematically, which simplifies the work involved in finding the values of \(a\), \(b\), and \(c\). By expressing our system in matrix form, we open the door to efficient computational techniques that can handle both simple and complex systems seamlessly.

What is Gaussian Elimination?

Gaussian elimination is an algorithmic process used to solve systems of linear equations. It systematically applies operations to transform the system’s matrix into what is known as "row-echelon form", or ideally "reduced row-echelon form", where solutions can be easily identified.

The steps typically include:

• Swapping rows if needed to position the equation with the largest coefficients at the top.
• Creating zeros below the pivot (leading entry) by scaling and subtracting rows.
• Finally, solving for the variables by back substitution once the matrix is in its simplest form.

In our example, Gaussian elimination helped us extract the coefficients from our system of equations efficiently, transforming our matrix into reduced row-echelon form: \[\begin{bmatrix}1 & -0.5 & 0.25 \0 & 1 & 0.75 \0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a \ b \ c \end{bmatrix} =\begin{bmatrix} -0.75 \ -2.75 \ 1 \end{bmatrix}\]The transformed matrix then directly provides the solutions: \(a = -0.75\), \(b = -2.75\), and \(c = 1\). Such processes underscore the power of matrix and algebra techniques in simplifying complex mathematical problems and finding solutions quickly.