Understanding the Second Degree Polynomial Problem

The question asks us to find a specific second degree polynomial function, which we can call \(f(x)\). A second degree polynomial is also known as a quadratic function and has the general form:

\(f(x) = ax^2 + bx + c\)

where \(a\), \(b\), and \(c\) are constant coefficients. Since it's a second degree polynomial, the coefficient \(a\) cannot be zero.

We are provided with three specific values of the function at different points:

• \(f(0) = 5\)
• \(f(-1) = 10\)
• \(f(1) = 6\)

These three pieces of information will allow us to set up equations and solve for the three unknown coefficients \(a\), \(b\), and \(c\).

Using the Given Conditions to Find Coefficients

We will substitute each given condition into the general form of the second degree polynomial, \(f(x) = ax^2 + bx + c\), to create equations.

Condition 1: \(f(0) = 5\)

Let's substitute \(x = 0\) into the function:

\(f(0) = a(0)^2 + b(0) + c\)

We are given that \(f(0) = 5\), so:

\(5 = a(0) + b(0) + c\)

\(5 = 0 + 0 + c\)

\(c = 5\)

We have successfully found the value of the coefficient \(c\). Now we know that the polynomial has the form \(f(x) = ax^2 + bx + 5\).

Condition 2: \(f(-1) = 10\)

Now, substitute \(x = -1\) into the updated form of the function \(f(x) = ax^2 + bx + 5\):

\(f(-1) = a(-1)^2 + b(-1) + 5\)

We know that \(f(-1) = 10\), so:

\(10 = a(1) - b + 5\)

\(10 = a - b + 5\)

To simplify, subtract 5 from both sides of the equation:

\(10 - 5 = a - b\)

\(5 = a - b\)

This gives us our first linear equation involving \(a\) and \(b\):

Equation 1: \(a - b = 5\)

Condition 3: \(f(1) = 6\)

Finally, substitute \(x = 1\) into the function \(f(x) = ax^2 + bx + 5\):

\(f(1) = a(1)^2 + b(1) + 5\)

We are given that \(f(1) = 6\), so:

\(6 = a(1) + b + 5\)

\(6 = a + b + 5\)

To simplify, subtract 5 from both sides of the equation:

\(6 - 5 = a + b\)

\(1 = a + b\)

This gives us our second linear equation involving \(a\) and \(b\):

Equation 2: \(a + b = 1\)

Solving the System of Linear Equations

We now have a system of two linear equations with two variables, \(a\) and \(b\):

• Equation 1: \(a - b = 5\)
• Equation 2: \(a + b = 1\)

We can solve this system using the elimination method. Notice that the coefficients of \(b\) in the two equations are opposites (-1 and +1). If we add the two equations, the \(b\) terms will cancel out.

This simplifies to:

\(2a = 6\)

Divide both sides by 2 to find \(a\):

\(a = \frac{6}{2}\)

\(a = 3\)

Now that we have the value of \(a\), substitute \(a = 3\) into either Equation 1 or Equation 2 to solve for \(b\). Let's use Equation 2 (\(a + b = 1\)) as it is simpler:

\(3 + b = 1\)

Subtract 3 from both sides:

\(b = 1 - 3\)

\(b = -2\)

So, we have found all the coefficients: \(a = 3\), \(b = -2\), and \(c = 5\).

Constructing the Second Degree Polynomial

Substitute the values of \(a\), \(b\), and \(c\) back into the general form \(f(x) = ax^2 + bx + c\):

\(f(x) = 3x^2 + (-2)x + 5\)

\(f(x) = 3x^2 - 2x + 5\)

This is the second degree polynomial function that satisfies the given conditions.

Verification of the Polynomial Function

Let's check if the function \(f(x) = 3x^2 - 2x + 5\) satisfies the initial conditions:

• For \(x=0\): \(f(0) = 3(0)^2 - 2(0) + 5 = 0 - 0 + 5 = 5\). This matches the condition \(f(0)=5\).
• For \(x=-1\): \(f(-1) = 3(-1)^2 - 2(-1) + 5 = 3(1) - (-2) + 5 = 3 + 2 + 5 = 10\). This matches the condition \(f(-1)=10\).
• For \(x=1\): \(f(1) = 3(1)^2 - 2(1) + 5 = 3(1) - 2 + 5 = 3 - 2 + 5 = 1 + 5 = 6\). This matches the condition \(f(1)=6\).

The function \(f(x) = 3x^2 - 2x + 5\) correctly satisfies all three given conditions.

Revision Table: Finding Polynomials

Additional Information: Polynomial Interpolation Basics

This problem is a basic example of polynomial interpolation. Polynomial interpolation is the process of finding a polynomial that passes through a given set of points. If you have \(n+1\) distinct data points \((x_0, y_0), (x_1, y_1), \dots, (x_n, y_n)\), there is a unique polynomial of degree at most \(n\) that interpolates these points.

For a second degree polynomial (\(n=2\)), we need three points. The general form \(f(x) = ax^2 + bx + c\) has three coefficients (\(a, b, c\)). Each point gives us one linear equation in terms of these coefficients. Solving the resulting system of linear equations gives us the unique polynomial.

Other methods exist for polynomial interpolation, such as using Lagrange polynomials or Newton's form, but the method of setting up and solving a system of equations is fundamental and directly uses the definition of the polynomial.