Problem 62 - 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 \((1,2),(2,3),\) and \((-1,6) .(\) Hint: See Exercise 61.)

Step-by-Step Solution

Answer

The values are \(a = 1\), \(b = -2\), and \(c = 3\).

Step 1: Substitute Each Ordered Pair

To find the values of \(a, b,\) and \(c\), first substitute each ordered pair into the equation \(y = ax^2 + bx + c\). This will give us three equations.**For \((1, 2)\):**\[2 = a(1)^2 + b(1) + c\] which simplifies to: \[2 = a + b + c\]**For \((2, 3)\):**\[3 = a(2)^2 + b(2) + c\] which simplifies to: \[3 = 4a + 2b + c\]**For \((-1, 6)\):**\[6 = a(-1)^2 + b(-1) + c\] which simplifies to: \[6 = a - b + c\]

Step 2: Set Up the System of Equations

From Step 1, we have the following system of linear equations:1. \(a + b + c = 2\)2. \(4a + 2b + c = 3\)3. \(a - b + c = 6\)

Step 3: Eliminate One Variable

To eliminate \(c\), subtract the first equation from the second and the first equation from the third.**Equation 2 - Equation 1:**\[(4a + 2b + c) - (a + b + c) = 3 - 2\] results in \[3a + b = 1\]**Equation 3 - Equation 1:**\[(a - b + c) - (a + b + c) = 6 - 2\] results in \[-2b = 4\]From the last result, we simplify to find \(b\): \[b = -2\]

Step 4: Solve for Remaining Variables

Substitute \(b = -2\) into \(3a + b = 1\):\[3a - 2 = 1\]Add 2 to both sides:\[3a = 3\]Divide by 3:\[a = 1\]

Step 5: Find the Value of c

Substitute \(a = 1\) and \(b = -2\) back into the first equation: \(a + b + c = 2\).\[1 - 2 + c = 2\]Combine like terms:\[-1 + c = 2\]Add 1 to both sides to solve for \(c\):\[c = 3\]

Step 6: State the Solution

The values that satisfy the equation for the given ordered pairs are \(a = 1\), \(b = -2\), and \(c = 3\).

Key Concepts

Solving Systems of Equations

Solving systems of equations involves finding values for variables that satisfy all equations in the system simultaneously. In the context of quadratic equations, we often deal with a system formed by substituting ordered pairs into a quadratic expression, like in our original exercise.
Consider the quadratic equation \( y = ax^2 + bx + c \) and three ordered pairs: \((1, 2), (2, 3), (-1, 6)\). By substituting each pair into the equation, we generate a system of linear equations, one for each pair.
Let's take a look at this transformation. First, for \((1, 2)\), substituting gives \( 2 = a(1)^2 + b(1) + c \), resulting in the equation \( a + b + c = 2 \). Repeating this process for the other two pairs generates two more linear equations:
\(4a + 2b + c = 3\) and \(a - b + c = 6\).
This provides us an approach to break down a complex quadratic function into simpler, manageable systems. The goal now is to solve for the unknowns \(a\), \(b\), and \(c\).
This makes the task approachable by reducing it to basic algebraic manipulations.

Substitution Method

The substitution method is an effective strategy to solve systems of equations, especially linear ones. It involves expressing one variable in terms of another and substituting this expression into the other equations to simplify or reduce the number of variables.
In the provided solution, after setting up the system of equations, we use substitution by first targeting variable \(c\) for elimination. Using basic subtraction between the equations allows us to simplify two of the equations.

• By subtracting the first equation from the second, we eliminate \(c\), resulting in \(3a + b = 1\).
• Similarly, by subtracting the first from the third equation, we also eliminate \(c\), yielding \(-2b = 4\).

This is strategic because removing one variable reduces the complexity and the number of unknowns.
Once we determine \(b = -2\) from the equation \(-2b = 4\), we can directly substitute \(b\) back into another equation \(3a + b = 1\) to solve for \(a\).
Substitution is a powerful method as it simplifies solving systems of equations into a series of straightforward steps.

Ordered Pairs

In mathematics, an ordered pair is simply a pair of numbers used to represent a point in a plane. It is written in the form \((x, y)\) where \(x\) is the horizontal value, and \(y\) is the vertical value. Ordered pairs are vital in determining solutions in equations, particularly in graphing and solving problems involving coordinate systems.
To utilize ordered pairs in the context of quadratic equations like our exercise, these pairs are plugged into the function to set up systems of equations. Each pair represents a unique point on the parabola described by the quadratic equation.
In our exercise, the ordered pairs \((1, 2)\), \((2, 3)\), and \((-1, 6)\) provide specific points through which the parabola \(y = ax^2 + bx + c\) must pass. By forming separate equations from each pair, we can explore the relationships between \(x\), \(y\), and the unknown coefficients \(a\), \(b\), and \(c\).
Recognizing how ordered pairs convert into usable equations highlights the connection between geometric visualization and algebraic expression, aiding in solving broader mathematical problems.