Question

graph the inequality on the axes below.
$y \leq \frac{1}{3}x + 1$

Explanation:

Step1: Graph the boundary line

The inequality is \( y \leq \frac{1}{3}x + 1 \). First, graph the line \( y=\frac{1}{3}x + 1 \). The slope \( m=\frac{1}{3} \) 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 1 unit and right 3 units to get \( (3,2) \) (since slope \(=\frac{\text{rise}}{\text{run}}=\frac{1}{3}\)). Draw a solid line (because the inequality is \( \leq \), so the boundary line is included) through these points.

Step2: Shade the region

To determine which side of the line to shade, pick a test point not on the line. The origin \( (0,0) \) is a good choice. Substitute \( x = 0 \) and \( y=0 \) into the inequality: \( 0\leq\frac{1}{3}(0)+1 \), which simplifies to \( 0\leq1 \). This is true. So, shade the region that includes the origin (the region below the line \( y = \frac{1}{3}x+1 \)).

Answer:

(Graph with a solid line \( y=\frac{1}{3}x + 1 \) and the region below the line shaded)