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Chapter 1: Problem 18

Each table describes a linear relationship. For each relationship, find the slope of the line and the \(y\) -intercept. Then write an equation for the relationship in the form \(y=m x+b .\) $$\begin{array}{|c|c|c|c|c|c|}\hline x & {-8} & {-3} & {3} & {5} & {10} \\\ \hline y & {26} & {11} & {-7} & {-13} & {-28} \\ \hline\end{array}$$

Short Answer

Expert verified
The slope of the line is m=-2, the y-intercept is b=10, and the equation is y=-2x+10.

Step by step solution

01

Identify two points to calculate the slope

Choose any two points from the table. Let's choose (-8, 26) and (10, -28).
02

Calculate the slope (m)

The formula for the slope is .
03

Calculate the slope using chosen points

Using the points (-8, 26) and (10, -28): .
04

Set up the equation using the slope

Using the slope obtained, start with the equation form: .
05

Solve for the y-intercept (b)

Substitute one point from the data table and the slope into the equation to solve for b. Substituting the point (-8, 26): .
06

Write the final equation of the line

Using the values of the slope and y-intercept, write the equation in the form .

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Key Concepts

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

Slope Calculation
To understand the linear equation in the form of \(y=mx+b\), we need to first identify the slope \(m\). The slope tells us how much the \(y\)-value changes for a one-unit change in the \(x\)-value. We calculate the slope as follows:

1. Select any two points from the table. For instance, let's use the points (-8, 26) and (10, -28).
2. Apply the slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
3. Plug in your chosen points: \( m = \frac{(-28) - 26}{10 - (-8)} = \frac{-54}{18} = -3 \)
Hence, the slope \(m\) is -3. This slope indicates a decrease of 3 units in \(y\) for every 1 unit increase in \(x\).
Y-intercept Identification
Next, we need to identify the y-intercept, denoted by \(b\). The y-intercept is the point where the line crosses the y-axis, which means it is the value of \(y\) when \(x=0\). However, we might not always have a point directly on the y-axis in our data table. So, we solve for \(b\) by substituting one of the given points and the slope into the equation form \(y = mx + b\):

1. Substitute \(x\), \(y\), and \(m\) into the equation. We'll use the point (-8, 26) and the slope \(m = -3\): \( 26 = (-3)(-8) + b \).
2. Simplify the equation to solve for \(b\): \( 26 = 24 + b \).
3. Subtract 24 from both sides: \( b = 2 \).

So, the y-intercept \(b\) is 2. This is the value of \(y\) when \(x\) is 0.
Equation of a Line
Now that we have the slope \(m\) and y-intercept \(b\), we can write the equation of the line. The general form of a linear equation is \(y = mx + b\):

1. Substitute the slope and y-intercept into the general form: \(y = -3x + 2\).

The final equation \(y = -3x + 2\) shows the relationship between \(x\) and \(y\) for the given data.

Understanding how to find the slope and y-intercept allows you to write equations for linear relationships with ease.
Utilize these equations to predict values and analyze trends accurately.

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Most popular questions from this chapter

In this exercise, you will apply what you have learned about writing equations for parallel lines. a. Write three equations whose graphs are parallel lines with positive slopes. Write the equations so that the graphs are equally spaced. b. Graph the lines, and verify that they are parallel. c. Write three equations whose graphs are parallel lines with negative slopes and are equally spaced. d. Graph the lines, and verify that they are parallel.

Each table describes a linear relationship. For each relationship, find the slope of the line and the \(y\) -intercept. Then write an equation for the relationship in the form \(y=m x+b .\) $$\begin{array}{|c|c|c|c|c|c|}\hline x & {2} & {4} & {6} & {8} & {10} \\\ \hline y & {8} & {12} & {16} & {20} & {24} \\ \hline\end{array}$$

\(\begin{array}{cl}{\text { Consider these four equations. }} & {} \\ {\text { i. } y=2 x-3} & {\text { i. } y=-\frac{1}{2} x-6} \\ {\text { iii. } y=\frac{2}{5} x+4} & {\text { iv. } y=-\frac{5}{2} x}\end{array}\) $$\begin{array}{l}{\text { a. Graph the four equations on one set of axes. Use the same scale }} \\ {\text { for each axis. Label the lines with the appropriate roman numerals. }} \\ {\text { b. What is the slope of each line? }} \\ {\text { c. What do you notice about the angle of intersection between }} \\ {\text { Lines i and ii? Between Lines iii and iv? }} \\ {\text { d. What is the relationship between the slopes of Lines i and i: }} \\ {\text { Between the slopes of Lines iii and iv? }}\end{array}$$ $$\begin{array}{l}{\text { e. Make a conjecture about the slopes of perpendicular lines. }} \\ {\text { f. Create two more lines with slopes that fit your conjecture. Are }} \\ {\text { they perpendicular? }} \\ {\text { 9. Write an equation for the line that passes through the point }(-1,4)} \\\ {\text { and is perpendicular to } y=\frac{1}{3} x+4 . \text { Check your answer by graph- }} \\ {\text { ing both lines on one set of axes. Use the same scale for each axis. }}\end{array}$$

Just as the \(y\) -intercept of a line is the \(y\) value at which the line crosses the \(y\) -axis, the \(x\) -intercept is the \(x\) value at which the line crosses the \(x\) -axis. In Exercises \(22-25,\) find an equation of a line with the given \(x\) -intercept and slope. \(x\) -intercept \(-2,\) slope \(-\frac{1}{2}\)

The table shows x and y values for a particular relationship. $$\begin{array}{|c|c|c|c|c|}\hline x & {6} & {3} & {1} & {2.5} \\ \hline y & {7} & {1} & {-3} & {0} \\ \hline\end{array}$$ $$\begin{array}{l}{\text { a. Graph the ordered pairs }(x, y) . \text { Make each axis scale from }-10} \\ {\text { to } 10 .} \\ {\text { b. Could the points represent a linear relationship? If so, write an }} \\ {\text { equation for the line. }} \\ {\text { c. From your graph, predict the } y \text { value for an } x \text { value of }-2 . \text { Check }} \\ {\text { your answer by substituting it into the equation. }}\end{array}$$ $$ \begin{array}{l}{\text { d. From your graph, find the } x \text { value for a } y \text { value of }-2 . \text { Check your }} \\ {\text { answer by substituting it into the equation. }} \\ {\text { e. Use your equation to find the } y \text { value for each of these } x \text { values: }} \\ {0,-- 1,-1.5,-2.5 . \text { Check that the corresponding points all lie on }} \\\ {\text { the line. }}\end{array} $$
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