Question

graph the linear inequality below.
$y \geq -\frac{1}{3}x - 2$
show your work here
draw

Explanation:

Step1: Identify the boundary line

The inequality is \( y \geq -\frac{1}{3}x - 2 \). First, graph the boundary line \( y = -\frac{1}{3}x - 2 \). This is a linear equation in slope - intercept form (\( y=mx + b \)), where the slope \( m=-\frac{1}{3} \) and the y - intercept \( b=- 2 \). To graph the line, start by plotting the y - intercept at \( (0,-2) \). Then, use the slope to find another point. The slope \( -\frac{1}{3} \) means for every 3 units we move to the right (increase in x by 3), we move down 1 unit (decrease in y by 1). So from \( (0,-2) \), moving 3 units right to \( x = 3 \) and 1 unit down to \( y=-3 \), we get the point \( (3,-3) \). Since the inequality is \( \geq \), the boundary line should be a solid line (because the inequality includes the equal sign).

Step2: Determine the region to shade

To determine which side of the line to shade, we can use a test point. A common test point is \( (0,0) \) (as long as it is not on the boundary line). Substitute \( x = 0 \) and \( y = 0 \) into the inequality \( y\geq-\frac{1}{3}x - 2 \):
\( 0\geq-\frac{1}{3}(0)-2 \)
\( 0\geq - 2 \), which is a true statement. So we shade the region that contains the point \( (0,0) \), which is the region above the line \( y = -\frac{1}{3}x - 2 \).

Answer:

To graph \( y\geq-\frac{1}{3}x - 2 \):

1. Draw the solid line \( y = -\frac{1}{3}x - 2 \) (using the y - intercept \( (0,-2) \) and slope \( -\frac{1}{3} \) to plot additional points).
2. Shade the region above the line (the region that contains the test point \( (0,0) \) as it satisfies the inequality).