Problem 64 - A Graphical Approach to College Algebra

Question

The values in the table are from a quadratic function \(f(x)=a x^{2}+b x+c .\) Find \(a, b,\) and \(c\). $$\begin{array}{c|c|c|c|c|c}x & -2 & -1 & 0 & 1 & 2 \\\\\hline f(x) & 2.9 & 1.26 & 0.56 & 0.8 & 1.98 \end{array}$$

Step-by-Step Solution

Answer

The coefficients are \(a = 0.47\), \(b = -0.23\), \(c = 0.56\).

Step 1: Write the General Form of a Quadratic Function

The general form of a quadratic function is given by \(f(x) = ax^2 + bx + c\). We need to determine the coefficients \(a\), \(b\), and \(c\) that satisfy this function for the given points.

Step 2: Set Up Equations Using Points

Use each pair of \((x, f(x))\) from the table to set up three equations.\\(-2: a(-2)^2 + b(-2) + c = 2.9 \, \Rightarrow\, 4a - 2b + c = 2.9\) \\(-1: a(-1)^2 + b(-1) + c = 1.26 \, \Rightarrow\, a - b + c = 1.26\) \\(0: a(0)^2 + b(0) + c = 0.56 \, \Rightarrow\, c = 0.56\)

Step 3: Solve for c

From the equation using \(x = 0\), we have \(c = 0.56\). This simplifies the system of equations.

Step 4: Substitute c and Solve the System

Substitute \(c = 0.56\) into the other two equations.\\(4a - 2b + 0.56 = 2.9\) \, leading to \, \(4a - 2b = 2.34\) \\(a - b + 0.56 = 1.26\) \, leading to \, \(a - b = 0.7\)

Step 5: Solve the System of Equations for a and b

Now solve the simultaneous equations: \\(4a - 2b = 2.34\) \\(a - b = 0.7\) \Multiply the second equation by 2 to facilitate elimination: \, \(2a - 2b = 1.4\) \Subtract this from the first equation: \, \(4a - 2b - 2a + 2b = 2.34 - 1.4\) \\(2a = 0.94\) \\(a = 0.47\)

Step 6: Find b

Substitute \(a = 0.47\) back into the equation \(a - b = 0.7\):\\(0.47 - b = 0.7\) \\(b = 0.47 - 0.7 = -0.23\)

Key Concepts

Finding Coefficients

When dealing with quadratic functions, the task of finding coefficients requires substituting the given values into the function formula. The general quadratic function is expressed as \(f(x) = ax^2 + bx + c\). By substituting different pairs of \(x\) and \(f(x)\) into this equation, we establish multiple equations. Each equation corresponds to a point on the curve defined by the quadratic function. For instance, if we plug in \(x = 0\), the equation simplifies to just \(c\), allowing us to easily determine its value from the corresponding \(f(x)\).
In our example, through substituting the given points from the table, we set up a series of equations:

• From \(x = -2\), we get \(4a - 2b + c = 2.9\).
• From \(x = -1\), the equation becomes \(a - b + c = 1.26\).
• From \(x = 0\), as stated, we simply find \(c = 0.56\).

By systematically solving these equations, we can determine the values of \(a\), \(b\), and \(c\).

Systems of Equations

The method of solving for coefficients often involves working with systems of equations. A system of equations is just a collection of two or more equations with a common set of variables. To solve the system, we use algebraic methods such as substitution or elimination. Here, elimination is particularly useful when dealing with equal numbers of unknowns and equations. We systematically eliminate one variable to find others.
In this problem, we substituted \(c = 0.56\) into the first two equations:

• \(4a - 2b + 0.56 = 2.9\), which simplifies to \(4a - 2b = 2.34\).
• \(a - b + 0.56 = 1.26\), simplifying to \(a - b = 0.7\).

Next, we multiply one of these equations by 2 to align the coefficients of one of the variables. This process enables easy subtraction to eliminate a variable, leaving us with a singular equation in one variable. Solving this provides a step into unraveling the entire system.

Algebraic Manipulations

Algebraic manipulations refer to the process of reshaping equations to isolate variables and simplify solutions. It’s crucial in making systems of equations more manageable. Here, we start by isolating one of the variables, typically \(c\), since one of our given \(x\) values, zero, simplifies its solving process. This newfound simplicity allows us to rearrange other equations to facilitate solving for \(a\) and \(b\).
Consider the equations we derived:

• \(2a - 2b = 1.4\)
• \(4a - 2b = 2.34\)

Starting with elimination, by subtracting \(2a - 2b = 1.4\) from \(4a - 2b = 2.34\), results in a "clean" equation:\(2a = 0.94\), thus \(a = 0.47\).
Substitution of \(a\'s\) value into another equation like \(a - b = 0.7\) helps us find \(b\):\(b = 0.47 - 0.7 = -0.23\).
This streamlined manipulation underscores the importance of breaking down the system into essential steps to unlock each variable value systematically.