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Chapter 1: Problem 9

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

Short Answer

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

Step by step solution

01

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 \[ P(x) = ax^2 + bx + c. \]
02

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:\[ 16a - 4b + c = -3 \c = 1 \a + b + c = 4.5.\]
03

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:\[16a - 4(3.5 - a) = -4 \16a - 14 + 4a = -4 \20a = 10 \a = 0.5.\]
04

Find Remaining Coefficients

Substitute \( a = 0.5 \) back to find \( b \):\[b = 3.5 - 0.5 = 3.\]
05

Write the Polynomial

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A matrix \(A\) is given below. In Exercises 16 20, a matrix \(B\) is given. Give the row operation that transforms \(A\) into \(B\). $$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 2 & 3\end{array}\right]$$ $$B=\left[\begin{array}{lll}1 & 1 & 1 \\ 2 & 1 & 2 \\ 1 & 2 & 3\end{array}\right]$$

Find the solution of the given problem by: (a) creating an appropriate system of linear equations (b) forming the augmented matrix that corresponds to this system (c) putting the augmented matrix into reduced row echelon form (d) interpreting the reduced row echelon form of the matrix as a solution. A rescue mission has 85 sandwiches, 65 bags of chips and 210 cookies. They know from experience that men will eat 2 sandwiches, 1 bag of chips and 4 cookies; women will eat 1 sandwich, a bag of chips and 2 cookies; kids will eat half a sandwhich, a bag of chips and 3 cookies. If they want to use all their food up, how many men, women and kids can they feed?

Use Gaussian Elimination to put the given matrix into reduced row echelon form. $$\left[\begin{array}{lll}2 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 1 & 2\end{array}\right]$$

Convert the given system of linear equations into an augmented matrix. $$ \begin{aligned} x_{1}+3 x_{2}-4 x_{3}+5 x_{4} &=17 \\ -x_{1}+4 x_{3}+8 x_{4} &=1 \\ 2 x_{1}+3 x_{2}+4 x_{3}+5 x_{4} &=6 \end{aligned} $$

Find the solution to the given linear system. If the system has infinite solutions, give 2 particular solutions. $$ \begin{aligned} -x_{1}-x_{2}+x_{3}+x_{4} &=0 \\ -2 x_{1}-2 x_{2}+x_{3} &=-1 \end{aligned} $$
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