Question

1. graphing linear inequalities

graph the following inequalities. shade the correct region.

1. $y \leq 2x + 1$

Explanation:

Step1: Graph the boundary line

The inequality is \( y \leq 2x + 1 \). First, graph the line \( y = 2x + 1 \). The slope \( m = 2 \) and the y - intercept \( b = 1 \). Plot the y - intercept at \( (0,1) \). Then, use the slope to find another point. From \( (0,1) \), move up 2 units and right 1 unit to get \( (1,3) \), or down 2 units and left 1 unit to get \( (-1, - 1) \). Since the inequality is \( \leq \), the line should be solid (because the points on the line are included in the solution set).

Step2: Determine the shading region

To determine which side of the line to shade, we can use a test point that is not on the line. A common test point is \( (0,0) \). Substitute \( x = 0 \) and \( y = 0 \) into the inequality \( y\leq2x + 1 \):
\( 0\leq2(0)+1 \)
\( 0\leq1 \), which is a true statement. So, the region that contains the point \( (0,0) \) should be shaded. This region is below the line \( y = 2x+1 \) (since the test point \( (0,0) \) is below the line \( y = 2x + 1 \) when \( x = 0 \)).

Answer:

To graph \( y\leq2x + 1 \):

1. Draw a solid line with slope 2 and y - intercept 1 (passing through points like \( (0,1) \), \( (1,3) \), \( (-1,-1) \) etc.).
2. Shade the region below the solid line (the region that includes the point \( (0,0) \)).