91Ó°ÊÓ

When graphing linear inequalities, the slope-intercept form is incredibly useful. This form is written as \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. The slope tells you how steep the line is, or how much \(y\) changes for a change in \(x\).
For example, a slope of 3 means that for every 1 unit you move to the right on the x-axis, the y-value increases by 3 units. The y-intercept is where the line crosses the y-axis. It gives you a starting point on the graph when \(x = 0\).

• If the slope is positive, the line goes upwards as it moves from left to right.
• If the slope is negative, the line goes downwards as it moves from left to right.

In our inequality \(y > 3x + 2\), the slope is 3 and the y-intercept is 2, so the line will have a positive slope and will cross the y-axis at 2.

The concept of the boundary line is essential when dealing with graphing inequalities. The boundary line is derived from the equation you get when switching an inequality sign for an equals sign. For \(y > 3x + 2\), the boundary line is \(y = 3x + 2\).
This line divides the plane into two distinct regions - one that satisfies the inequality and one that does not. For strict inequalities (\(>\) or \(<\)), as in our example, the line itself is not included in the solution set. We represent this on a graph with a dotted or dashed line.

• A solid line is used for \(≥\) or \(≤\) inequalities since points on the line satisfy the condition.
• A dotted line indicates that points on this line do not satisfy the inequality, which is why it's used here.

The boundary line acts as a visual guide for where the solution area starts.

Shading is the final step in graphing linear inequalities and helps illustrate the solution area—the set of all points that satisfy the inequality. In the inequality \(y > 3x + 2\), the solution area is above the boundary line because \(y\) values must be greater than the corresponding \(y\) values on the line.
To determine the shading direction, you can use a test point. A common test point is the origin \((0,0)\), as it is easy to calculate. Substitute it into the inequality to see if it makes the inequality true.

• If \((0,0)\) satisfies the inequality, you shade the region containing this point.
• If it doesn't, you shade the opposite region.

In our case, substituting \(0\) for both \(x\) and \(y\) gives \(0 > 2\), which is false, so we shade the area not containing \((0,0)\), which is above the line. This shading visually represents all solutions to the inequality.