Example 1
Completion requirements
| Example 1 |
Write an inequality to represent the graph shown.
Start by writing the equation of the boundary. One method is to use the slope-intercept form. The slope of the boundary is \(2\) and the \(y\)-intercept is \(5\), so the boundary is \(y = 2x + 5\).
The boundary is a dashed line, so the graph represents a strict inequality.
Use a test point from the solution region to determine the direction of the inequality symbol. The point \((–4, 2)\) is a point in the solution region.
Alternatively, the solution region is above the boundary, meaning the y-values are greater than the boundary, so the inequality must be \(y \gt 2x + 5\).
An inequality that corresponds to the graph is \(y \gt 2x + 5\).
The boundary is a dashed line, so the graph represents a strict inequality.
Use a test point from the solution region to determine the direction of the inequality symbol. The point \((–4, 2)\) is a point in the solution region.
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r} y \\ 2 \end{array}\] |
\(\begin{array}{l} 2x + 5 \\ 2\left( { - 4} \right) + 5 \\ - 3 \end{array}\) |
| \(\hspace{25pt}\)LS \(\gt \) RS | |
Alternatively, the solution region is above the boundary, meaning the y-values are greater than the boundary, so the inequality must be \(y \gt 2x + 5\).
An inequality that corresponds to the graph is \(y \gt 2x + 5\).