Question

Find the polynomial with the smallest degree that goes through the given points. $$(-4,-3),(0,1) \text { and }(1,4.5)$$

Short answer

The polynomial is \( P(x) = 0.5x^2 + 3x + 1 \).

Step-by-step solution

Determine Polynomial Form

To find a polynomial that passes through given points, assume the smallest-degree polynomial. For three points, start with a quadratic polynomial of the form \[ P(x) = ax^2 + bx + c. \]

Set Up the Equation System

Use the given points to create a system of equations. Substitute each point into the polynomial equation:\[ P(-4) = a(-4)^2 + b(-4) + c = -3 \P(0) = a(0)^2 + b(0) + c = 1 \P(1) = a(1)^2 + b(1) + c = 4.5.\] This simplifies to:\[ 16a - 4b + c = -3 \c = 1 \a + b + c = 4.5.\]

Solve the Equation System

Substitute \( c = 1 \) into the other equations:\[16a - 4b + 1 = -3 \a + b + 1 = 4.5.\]Rearrange these to:\[16a - 4b = -4 \a + b = 3.5.\]Solve these equations using substitution or elimination. Here, we notice:\[b = 3.5 - a.\]Substitute in the first:\[16a - 4(3.5 - a) = -4 \16a - 14 + 4a = -4 \20a = 10 \a = 0.5.\]

Find Remaining Coefficients

Substitute \( a = 0.5 \) back to find \( b \):\[b = 3.5 - 0.5 = 3.\]

Write the Polynomial

Substitute the values of \( a \), \( b \), and \( c \) back into the polynomial form:\[ P(x) = 0.5x^2 + 3x + 1. \]

Key concepts

Quadratic Polynomial

A quadratic polynomial is a polynomial of degree two, which means that its highest power of the variable is squared. The general form of a quadratic polynomial is given by: \[ P(x) = ax^2 + bx + c \] where:

• a, b, and c are constants with a not equal to zero.
• x is the variable.

In the context of polynomial interpolation, a quadratic polynomial is often used when you need a smooth curve that fits a set of three points closely. Using three points ensures that the quadratic polynomial is uniquely determined. Start by plugging the coordinates into the polynomial equation to reveal more about the values of a, b, and c. Each set of coordinates corresponds to a specific linear equation in terms of a, b, and c, which collectively form a system of equations.

System of Equations

A system of equations refers to a set of multiple equations that are solved together since they share common variables. In solving for the coefficients of a quadratic polynomial passing through given points, you create a system of linear equations by plugging the points into the polynomial expression. For example:

• Using point \((-4, -3)\), we have: \(16a - 4b + c = -3\).
• Using point \((0, 1)\), we simplify to \(c = 1\).
• Using point \((1, 4.5)\), the equation is \(a + b + c = 4.5\).

These equations form a system that can be solved by finding the values of . Solving a system of equations is critical for finding the exact coefficients needed to define the quadratic polynomial accurately.

Substitution Method

The substitution method is a technique used for solving systems of equations where one equation is solved for one variable, and then this expression is substituted into other equations. Here's how it generally works:

• First, isolate one variable in one of the simpler equations. For instance, in the equation \(c = 1\), \(c\) is already isolated.
• Substitute \(c = 1\) into the other equations: \(16a - 4b + 1 = -3\) and \(a + b + 1 = 4.5\).
• This substitution simplifies the system, leaving you with a reduced number of equations with fewer variables. Solve these step by step.

The substitution method simplifies complex relationships and reduces them to simpler forms, making systematic solution easier.

Elimination Method

The elimination method involves manipulating equations in a system to eliminate one of the variables, making it easier to solve the remaining equations. Here's a step-by-step idea of how it is applied:

• First, ensure the equations are lined up with variables and constants matched in order.
• Choose a variable to eliminate and manipulate the equations to make the coefficients of this variable equal in magnitude but opposite in sign across two equations.
• For example, from \(16a - 4b = -4\) and using \(b = 3.5 - a\) obtained from another equation, there's a simplified way of seeing \(20a = 10\).
• This leads to finding \(a = 0.5\) and allowing substitution into a simplified remaining equation to find the other variables.

The elimination method requires strategic steps to simplify calculations, turning multi-variable problems into linear equation solutions.