Math Problem Statement

What is the equation of the line represented by the table with points (-2, 0), (1, 1), (4, 2), and (7, 3)?

Solution

The image shows a table of values that describe a linear relationship between xx and yy:

\hline x & -2 & 1 & 4 & 7 \\ \hline y & 0 & 1 & 2 & 3 \\ \hline \end{array}$$ We can find the equation of the line represented by this table using the following steps: 1. **Identify two points**: - From the table, we can choose any two points. Let's take $$(-2, 0)$$ and $$(1, 1)$$. 2. **Find the slope**: - The formula for the slope $$m$$ between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Using points $$(-2, 0)$$ and $$(1, 1)$$: $$m = \frac{1 - 0}{1 - (-2)} = \frac{1}{3}$$ 3. **Use the point-slope form to find the equation**: - The point-slope form is: $$y - y_1 = m(x - x_1)$$ Using the point $$(1, 1)$$ and $$m = \frac{1}{3}$$: $$y - 1 = \frac{1}{3}(x - 1)$$ Simplifying this equation: $$y - 1 = \frac{1}{3}x - \frac{1}{3}$$ $$y = \frac{1}{3}x + \frac{2}{3}$$ Thus, the equation of the line is: $$y = \frac{1}{3}x + \frac{2}{3}$$ Would you like further explanation, or do you have any questions? --- Here are 5 follow-up questions: 1. How can you verify if all points in the table fit the equation $$y = \frac{1}{3}x + \frac{2}{3}$$? 2. How would you graph the line based on this equation and table? 3. What would the slope represent in a real-world context for a similar problem? 4. Can the line equation be written in standard form, and how would you do that? 5. What happens to the equation if we change one of the points? **Tip:** When finding the slope, ensure the correct order of subtraction in the formula $$\frac{y_2 - y_1}{x_2 - x_1}$$.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Equations
Slope
Point-Slope Form

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Point-slope form: y - y1 = m(x - x1)
Equation of a line: y = mx + b

Theorems

Slope-Intercept Theorem

Suitable Grade Level

Grades 8-10