Problem 113 - Introductory Algebra for College Students

Question

Graph: \(y>-\frac{2}{3} x+1 .\) (Section 3.6, Example 3)

Step-by-Step Solution

Answer

The graph of the inequality \(y > -\frac{2}{3}x + 1\) is a line with a slope of -2/3 and y-intercept at 1. The region above the line is shaded to represent the values that satisfy the inequality.

Step 1: Plot the Linear Function

Take the linear equation \(y = -\frac{2}{3}x + 1\). It can be seen that it is in the slope-intercept form (y = mx + b), where m represents the slope of the line, and b is the y-intercept. Here, slope m = -\frac{2}{3} and y-intercept b = 1. Plot the y-intercept (0,1) and use the slope to find the next point. Since slope is rise over run, from y-intercept, go down 2 units (rise is negative) and right 3 units (run is positive) to get the next point. Draw a solid line to represent this equation, and make sure to write the equation of the line next to it.

Step 2: Shade the suitable region

The inequality is \(y > -\frac{2}{3}x + 1\), indicating that y-values are greater than the line formed. Consequently, to show this, shade the region above the line.

Step 3: Checking the solution

A check for any solution to an inequality could be done by selecting any arbitrary point from the shaded region. If this point when plugged into the initial inequality fulfills the inequality condition, it confirms the solution is correct.

Key Concepts

Slope-Intercept Form

Understanding the slope-intercept form of a linear equation is fundamental when graphing linear functions. This form is given by the format: \( y = mx + b \), where \( m \) represents the slope of the line, and \( b \) specifies the y-intercept.

The slope \( m \) is a measure of how steep the line is. It is calculated as the change in the y-coordinate, often referred to as the 'rise', over the change in the x-coordinate, referred to as the 'run'. In the equation \( y = -\frac{2}{3}x + 1 \), the slope is \( -\frac{2}{3} \), indicating the line falls 2 units for every 3 units it moves to the right.

The y-intercept \( b \) is the point where the line crosses the y-axis, also, it is the value of \( y \) when \( x \) is zero. For the same equation, the y-intercept is 1, depicted as the point (0,1) on the graph. Having these two critical pieces of information allows us to plot the initial point and set the direction of the line accurately.

Plotting Y-Intercept

Plotting the y-intercept is a crucial step in graphing a linear equation. The y-intercept is where the line crosses the y-axis. This point has an x-coordinate of 0.

To plot the y-intercept, locate the y-axis on a coordinate plane, which is the vertical axis. Then, move along the y-axis to the value of the y-intercept. For example, with our equation \( y = -\frac{2}{3}x + 1 \), the y-intercept is 1, so we would mark a point at (0,1) on the graph.

Understanding Through Visualization

Imagine the y-axis as a ladder you're ascending or descending. The y-intercept tells us at which rung of the ladder to start our journey. If it's positive, we climb up. If it's negative, we step down. For the exercise equation, you'd climb to the first rung, marking a clear starting point to guide the direction of the line using the slope.

Shading Number Line

After graphing the line for the inequality, it's essential to indicate which side of the line contains the solutions to the inequality. This is done by shading the appropriate side of the graph.

For the inequality \( y > -\frac{2}{3}x + 1 \), we shade above the line because the inequality suggests that the y-values are greater than the corresponding y-values on the line. In practical terms, pick any point not on the line as a test point. If the chosen test point satisfies the inequality when you substitute its coordinates into the original inequality, then the region that includes this test point is the solution area.

Why Shading Matters

Shading gives a visual representation of all possible solutions. It is a quick way to see an infinite number of points that satisfy the linear inequality. Without shading, it would be challenging to identify the area that represents the solutions from a mere glance at the graph.