Problem 61 - Algebra A Combined Function

Question

Find the values of \(a, b\), and \(c\) such that the equation \(y=a x^{2}+b x+c\) has ordered pair solutions \((1,6),(-1,-2),\) and \((0,-1) .\) To do so, substitute each ordered pair solution into the equation. Each time, the result is an equation in three unknowns: \(a, b,\) and \(c .\) Then solve the resulting system of three linear equations in three unknowns, \(a, b,\) and \(c\).

Step-by-Step Solution

Answer

Values are \(a = 3\), \(b = 4\), \(c = -1\).

Step 1: Substitute first ordered pair

Substitute the ordered pair \((1, 6)\) into the equation to get the first linear equation.\[\begin{align*}y &= ax^2 + bx + c \y &= 6, \quad x = 1 \6 &= a(1)^2 + b(1) + c \6 &= a + b + c \\end{align*}\]The first equation is:\[a + b + c = 6\]

Step 2: Substitute second ordered pair

Substitute the ordered pair \((-1, -2)\) into the equation to get the second linear equation.\[\begin{align*}y &= ax^2 + bx + c \y &= -2, \quad x = -1 \-2 &= a(-1)^2 + b(-1) + c \-2 &= a - b + c \\end{align*}\]The second equation is:\[a - b + c = -2\]

Step 3: Substitute third ordered pair

Substitute the ordered pair \((0, -1)\) into the equation to get the third linear equation.\[\begin{align*}y &= ax^2 + bx + c \y &= -1, \quad x = 0 \-1 &= a(0)^2 + b(0) + c \-1 &= c \\end{align*}\]The third equation is:\[c = -1\]

Step 4: Solving for c

From Step 3, we already have directly:\[c = -1\]This resolves the unknown \(c\).

Step 5: Substitute c into first and second equations

Use the value of \(c\) found in Step 4 and substitute it back into the first two equations.Substituting \(c = -1\) into the first equation:\[a + b - 1 = 6\]Thus,\[a + b = 7\]Substituting \(c = -1\) into the second equation:\[a - b - 1 = -2\]Thus,\[a - b = -1\]

Step 6: Solve the system of equations

Now solve the system of equations:1. \(a + b = 7\)2. \(a - b = -1\)Add the two equations to eliminate \(b\):\[\begin{align*}(a + b) + (a - b) &= 7 + (-1) \2a &= 6 \a &= 3 \\end{align*}\]Now substitute \(a = 3\) back into the first equation to find \(b\):\[3 + b = 7\]\[b = 4\]

Step 7: Conclusion and verification

Thus, the values are \(a = 3\), \(b = 4\), and \(c = -1\).Verify by substituting all values back into the original quadratic equation for each ordered pair.For \((1,6)\):\[6 = 3(1)^2 + 4(1) - 1 = 6\] \ For \((-1,-2)\):\[-2 = 3(-1)^2 + 4(-1) - 1 = -2\] \ For \((0,-1)\):\[-1 = 3(0)^2 + 4(0) - 1 = -1\]All equations check out.

Key Concepts

Systems of Linear Equations

A system of linear equations is simply a set of linear equations with multiple variables. These equations are solved to find the variable values that satisfy all equations simultaneously. In our exercise, we were tasked with finding values of the coefficients \(a\), \(b\), and \(c\) that satisfy three different ordered pairs, inserted into a quadratic equation.

When you have three unknowns, like in our task, you need three equations to determine each unknown distinctly. The reason for three equations is that each equation gives you different information about the variables. You worked with

• \(a + b + c = 6\)
• \(a - b + c = -2\)
• \(c = -1\)

Together, these equations form a system that helps us pinpoint the precise values for \(a\), \(b\), and \(c\). By solving these systems, you identify values that satisfy all equations—giving you coefficients that satisfy the given conditions.

Substitution Method

The substitution method is a technique to solve systems of equations. It's particularly useful when one of the equations is easy to solve for a single variable. After finding that variable, you substitute its value into the other equations to find the remaining unknowns.

In the example, we quickly found \(c = -1\) from the third equation \(c = -1\). Knowing this, we substituted \(-1\) for \(c\) back into the other two equations. Here's how it looked:

• \(a + b - 1 = 6\) simplifies to \(a + b = 7\)
• \(a - b - 1 = -2\) simplifies to \(a - b = -1\)

These steps helped in reducing the system to only two variables. Once simplified, you can solve such equations through additional methods like elimination—adding or subtracting equations to eliminate one of the remaining variables.

Ordered Pairs

Ordered pairs are fundamental in the context of equations and graphs. An ordered pair like \((x, y)\) represents a point on the Cartesian plane where \(x\) is the horizontal position, and \(y\) is the vertical position.

In our case, we were given specific ordered pairs: \((1, 6)\), \((-1, -2)\), and \((0, -1)\). Each ordered pair provides a unique condition for the quadratic equation. These points suggest that if \(x\) is plugged into the quadratic equation, \(y\) will be the result, helping us confirm if the equation is correctly structured.

Ordered pairs serve as checkpoints that verify if the equation satisfies particular conditions. They help `translate` algebraic expressions into numerical terms that are much easier to understand and work with.

Polynomial Functions

Polynomial functions are expressions composed of variables and coefficients. They include terms in the form of \(ax^n\) where \(n\) is a non-negative integer. Quadratic functions, a type of polynomial, have \(n = 2\), and are expressed in the form \(ax^2 + bx + c\).

In our problem, the polynomial is a quadratic function \(y = ax^2 + bx + c\). Using given conditions in the form of ordered pairs, we locate correct coefficients \(a\), \(b\), and \(c\) that match the given outputs. This polynomial determines a parabola when plotted on a graph.

Understanding polynomial functions is crucial since they model many real-world scenarios. From physics to finance, grasping how to manipulate and solve them becomes a valuable tool in the problem-solving toolbox.