Question

pretest: special linear relationships
a. $y \geq -3x + 2$
b. $y \leq -3x + 2$
c. $y > -3x + 2$
d. $y < -3x + 2$

Explanation:

Step1: Analyze the line's equation

First, find the slope and y - intercept of the boundary line. The line passes through \((0, 2)\) (y - intercept) and we can find another point. Let's take two points on the line: when \(x = 0\), \(y=2\); when \(y = - 4\), let's find \(x\). The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's use \((0,2)\) and \((2,-4)\) (by looking at the grid, from \(x = 0,y = 2\) to \(x=2,y=-4\), the change in \(y\) is \(- 6\) and change in \(x\) is \(2\), so slope \(m=\frac{-4 - 2}{2-0}=\frac{-6}{2}=-3\). So the equation of the line is \(y=-3x + 2\).

Step2: Determine the inequality sign (dashed or solid line)

The line in the graph is a solid line? Wait, looking at the graph, the boundary line is solid? Wait, no, wait the options have \(\leq,\geq,>,<\). Wait, the shaded region: let's test a point. Let's take the origin \((0,0)\). Plug into the inequality. Let's check each option:

- Option A: \(y\geq - 3x+2\). At \((0,0)\), \(0\geq2\)? No.
- Option B: \(y\leq - 3x + 2\). At \((0,0)\), \(0\leq2\)? Yes. But wait, the shaded region: wait the graph's shaded area. Wait, maybe I made a mistake. Wait, let's look at the direction of the shading. The line is \(y=-3x + 2\). Let's check the slope again. Wait, when \(x\) increases, \(y\) decreases (since slope is - 3). The shaded region: let's take a point in the shaded area, say \((-5,0)\). Plug into \(y < - 3x+2\): \(0 < - 3\times(-5)+2=15 + 2=17\), which is true. Plug into \(y > - 3x+2\): \(0>17\)? No. Wait, maybe the line is dashed? Wait, the options: A is \(y\geq - 3x + 2\), B is \(y\leq - 3x + 2\), C is \(y > - 3x + 2\), D is \(y < - 3x + 2\). Wait, the boundary line: if the line is solid, then the inequality is \(\leq\) or \(\geq\). If it's dashed, then \(>\) or \(<\). Wait, looking at the graph, the boundary line seems to be solid? Wait, no, maybe the original graph has a solid line. Wait, let's re - evaluate.

Wait, let's take the point \((0,0)\). Let's check option D: \(y < - 3x+2\). \(0 < 2\), which is true. Option B: \(y\leq - 3x + 2\), \(0\leq2\) is also true. Wait, maybe the line is dashed? Wait, the problem's graph: maybe the line is dashed. Wait, the options have \(>\) and \(<\) (dashed line) and \(\geq,\leq\) (solid line). Let's check the slope again. Let's find two points on the line. When \(x = 0\), \(y = 2\) (from the graph, the line crosses the y - axis at \(y = 2\)). When \(x = 1\), \(y=-3(1)+2=-1\). So the line passes through \((0,2)\) and \((1,-1)\). Now, the shaded region: let's take a point to the left of the line, say \((-1,0)\). Plug into \(y < - 3x+2\): \(0 < - 3\times(-1)+2=3 + 2=5\), true. Plug into \(y > - 3x+2\): \(0>5\)? No. Plug into \(y\leq - 3x + 2\): \(0\leq5\), true. Plug into \(y\geq - 3x + 2\): \(0\geq5\)? No. Now, check the line type. If the line is dashed, then it's \(>\) or \(<\); if solid, \(\leq\) or \(\geq\). Looking at the graph, the boundary line appears to be solid? Wait, no, maybe the original graph has a dashed line. Wait, the options: D is \(y < - 3x+2\). Let's check another point. Take \((0,3)\). Plug into D: \(3 < 2\)? No. Take \((0,-1)\): \(-1 < 2\), yes. Wait, the shaded region: in the graph, the shaded area is below the line? Wait, no, the line has a negative slope, so as \(x\) increases, \(y\) decreases. The shaded region is to the left of the line. Wait, maybe I messed up the direction. Let's use the slope - intercept form. The line is \(y=-3x + 2\). The inequality: if the shaded region is where \(y\) is less than \(-3x + 2\), then \(y < - 3x+2\). Let's check the point \((0,0)\): \(0 < 2\), which is true, and \((0,0)\) is in the shaded region (from the graph, the shaded area includes the origin? Wait, the graph's shaded area: the blue area. Let's see the grid: the origin \((0,0)\) is in the blue area. So \(0 < - 3\times0+2=2\), which is true. So the inequality is \(y < - 3x + 2\), which is option D.

Answer:

D. \(y < - 3x + 2\)