Question

which of the following linear inequalities matches the graph above?
a ( y > 2x - 1 )
b ( y < \frac{1}{3}x - 1 )
c ( y < 3x - 1 )
d ( y > \frac{1}{3}x - 1 )

Explanation:

Step1: Determine the slope of the dashed line

The line passes through the points \((0, -1)\) and \((3, 0)\). The slope \(m\) is calculated as \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0 - (-1)}{3 - 0}=\frac{1}{3}\).

Step2: Determine the inequality sign and the equation

The line is dashed, so the inequality is either \(y <\) or \(y >\). The shaded region is below the line (since for \(x = 0\), \(y=-1\) and the shaded area is below the line \(y=\frac{1}{3}x - 1\) when we check the direction of the inequality). Wait, no, let's re - check. Wait, when we look at the graph, the shaded region is above or below? Wait, the dashed line has a slope of \(\frac{1}{3}\) and y - intercept at \(- 1\) (since when \(x = 0\), \(y=-1\)). Let's take a test point, say \((0,0)\). Plug into the inequality. For option B: \(y<\frac{1}{3}x - 1\). Plug \(x = 0,y = 0\): \(0<\frac{1}{3}(0)-1\) which is \(0 < - 1\), false. For option D: \(y>\frac{1}{3}x - 1\). Plug \(x = 0,y = 0\): \(0>\frac{1}{3}(0)-1\) which is \(0>-1\), true. Wait, but wait the line is dashed, and the shaded region. Wait, maybe I made a mistake in slope. Wait, let's recalculate the slope. The two points on the dashed line: when \(x = 0\), \(y=-1\); when \(x = 3\), \(y = 0\). So slope \(m=\frac{0 - (-1)}{3-0}=\frac{1}{3}\). The equation of the line is \(y=\frac{1}{3}x-1\). Now, the shaded region: let's take a point in the shaded region, say \((0,0)\). Plug into the inequality. For option D: \(y>\frac{1}{3}x - 1\), \(0>\frac{1}{3}(0)-1\) (i.e., \(0 > - 1\)) which is true. For option B: \(y<\frac{1}{3}x - 1\), \(0<\frac{1}{3}(0)-1\) (i.e., \(0 < - 1\)) which is false. Wait, but the original option C is \(y < 3x-1\), slope 3, which is steeper. Option A has slope 2. So the correct slope is \(\frac{1}{3}\), and the inequality is \(y>\frac{1}{3}x - 1\) (option D)? Wait, no, maybe I messed up the direction of the shaded region. Wait, looking at the graph again, the dashed line goes from \((0,-1)\) to \((3,0)\), and the shaded region is below or above? Wait, the graph's shaded region: when \(x = 3\), the line is at \(y = 0\), and the shaded region is below? No, wait the blue shaded region. Wait, maybe the initial analysis of the slope was wrong. Wait, let's check the options again. Option B: \(y<\frac{1}{3}x - 1\), option D: \(y>\frac{1}{3}x - 1\). Let's take the point \((0,0)\) which is in the shaded region. For option B: \(0<\frac{1}{3}(0)-1\) → \(0 < - 1\) (false). For option D: \(0>\frac{1}{3}(0)-1\) → \(0>-1\) (true). So the correct inequality is \(y>\frac{1}{3}x - 1\), which is option D. Wait, but the original marked option is C? Wait, maybe I made a mistake in slope calculation. Wait, maybe the two points are \((0,-1)\) and \((- 3,0)\)? Let's check. If \(x=-3\), \(y = 0\) and \(x = 0\), \(y=-1\). Then slope \(m=\frac{0-(-1)}{-3 - 0}=\frac{1}{-3}=-\frac{1}{3}\)? No, that can't be. Wait, the graph's x - axis and y - axis: the positive x is to the right, positive y is up? Wait, the grid: the y - axis has negative values above zero? No, maybe the graph is drawn with a different orientation. Wait, maybe the y - axis is inverted. Wait, in the standard coordinate system, positive y is up. Let's re - examine the graph. The dashed line: when moving from left to right, does it go up or down? If the line goes from \((-3,0)\) to \((0,-1)\), then slope is \(\frac{-1 - 0}{0-(-3)}=-\frac{1}{3}\). But that doesn't match the options. Wait, the options have slopes \(\frac{1}{3}\), \(2\), \(3\), \(\frac{1}{3}\). Wait, maybe the correct slope is \(\frac{1}{3}\) and the inequality is \(y<\frac{1}{3}x - 1\) or \(y>\frac{1}{3}x - 1\). Wait, let's start over.

1. **Find the equation of the dashed line**:
- The general form of a linear equation is \(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept.
- The line passes through \((0,-1)\) (so \(b=-1\)) and \((3,0)\).
- The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0-(-1)}{3 - 0}=\frac{1}{3}\). So the equation of the line is \(y=\frac{1}{3}x-1\).
2. **Determine the inequality sign**:
- The line is dashed, so the inequality is either \(y>\frac{1}{3}x - 1\) or \(y<\frac{1}{3}x - 1\).
- We use a test point that is in the shaded region. Let's pick the origin \((0,0)\).
- For the inequality \(y>\frac{1}{3}x - 1\): Substitute \(x = 0\) and \(y = 0\) into the inequality: \(0>\frac{1}{3}(0)-1\), which simplifies to \(0>-1\). This is a true statement.
- For the inequality \(y<\frac{1}{3}x - 1\): Substitute \(x = 0\) and \(y = 0\) into the inequality: \(0<\frac{1}{3}(0)-1\), which simplifies to \(0 < - 1\). This is a false statement.

Answer:

D. \(y>\frac{1}{3}x - 1\)