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Chapter 1: Problem 7

Find the polynomial with the smallest degree that goes through the given points. $$(-2,14) \text { and }(3,4)$$

Short Answer

Expert verified
The polynomial is \( y = -2x + 10 \).

Step by step solution

01

Understand the Problem

We need to find a polynomial of the smallest degree that passes through the points (-2, 14) and (3, 4). A polynomial passing through two points is typically linear, which is a polynomial of degree 1.
02

Set Up the Polynomial Equation

The general form of a linear polynomial (degree 1) is:\[ y = mx + b \]where \( m \) is the slope and \( b \) is the y-intercept. We need to find \( m \) and \( b \).
03

Calculate the Slope

Use the two points to determine the slope \( m \) using the formula:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]Substituting the points (-2, 14) and (3, 4), we get:\[ m = \frac{4 - 14}{3 + 2} = \frac{-10}{5} = -2 \]
04

Find the Y-Intercept

Using the slope \( m = -2 \) and one of the given points, say (-2, 14), plug these into the equation \( y = mx + b \) to solve for \( b \):\[ 14 = -2(-2) + b \] \[ 14 = 4 + b \] \[ b = 10 \]
05

Write the Final Polynomial

Now that we have the slope \( m = -2 \) and the y-intercept \( b = 10 \), the equation of our polynomial is:\[ y = -2x + 10 \]

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Key Concepts

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

Polynomial Degree
The degree of a polynomial tells us how many solutions (or roots) it can have. It is equal to the highest power of the variable, often denoted as \( x \). In our problem, we're looking for the smallest polynomial that passes through two points, which means we need a polynomial with a low degree to keep it simple.

A polynomial of degree \( n \) can have up to \( n \) intersections with the x-axis, and it can have a maximum of \( n-1 \) turns. Since we're given only two points, a linear polynomial (degree 1) typically suffices to connect these points directly.
Linear Polynomial
A linear polynomial is the simplest type of polynomial equation. It is a first-degree polynomial, which implies that the highest power of \( x \) is 1. The general form of a linear polynomial is \( y = mx + b \).

- The term \( mx \) represents the slope of the line, indicating how steep the line is.- The term \( b \) represents the y-intercept. This is the point where the line crosses the y-axis.

Linear polynomials are a key concept in algebra because they represent straight lines. In many real-world applications, linear equations are used to model relationships between variables, where changes in one variable cause proportional changes in another.
Slope Calculation
The slope of a line is an important concept in linear algebra. It indicates how steep the line is. We calculate the slope \( m \) using the slope formula:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]This formula finds the change in \( y \) divided by the change in \( x \).

For example, given the points \((-2, 14)\) and \((3, 4)\), we can substitute into the formula:
  • \( y_2 = 4 \), \( y_1 = 14 \)
  • \( x_2 = 3 \), \( x_1 = -2 \)
Plugging these values in gives us:\[ m = \frac{4 - 14}{3 - (-2)} = \frac{-10}{5} = -2 \]

A negative slope indicates a line that decreases from left to right, which is what we have here.
Y-Intercept
The y-intercept of a line is the value of \( y \) where the line crosses the y-axis. In simple terms, it's where the line is when \( x \) equals 0.

Knowing the slope of the line, we can determine the y-intercept \( b \) by choosing one of our given points and using the line equation \( y = mx + b \). Suppose we use the point \((-2, 14)\) and our calculated slope \( m = -2 \):

Substitute these into the equation:
  • \( 14 = -2(-2) + b \)
  • \( 14 = 4 + b \)
  • Solve for \( b \) to get \( b = 10 \)
So, the y-intercept is 10. The point \((0, 10)\) is where the line crosses the y-axis.

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Most popular questions from this chapter

State for which values of \(k\) the given system will have exactly 1 solution, infinite solutions, or no solution. $$ \begin{array}{l} x_{1}+2 x_{2}=1 \\ x_{1}+k x_{2}=2 \end{array} $$

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

Use Gaussian Elimination to put the given matrix into reduced row echelon form. $$\left[\begin{array}{cccc}4 & 1 & 8 & 15 \\ 1 & 1 & 2 & 7 \\ 3 & 1 & 5 & 11\end{array}\right]$$

State for which values of \(k\) the given system will have exactly 1 solution, infinite solutions, or no solution. $$ \begin{array}{l} x_{1}+2 x_{2}=1 \\ x_{1}+k x_{2}=1 \end{array} $$

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