Question

which inequality is graphed on the coordinate plane? a. $y\geq -3x + 2$ b. $y\leq -3x + 2$ c. $y > -3x + 2$ d. $y < -3x + 2$

Explanation:

Step1: Determine the equation of the boundary line

The boundary line intersects the y - axis at (0, 2) and has a slope. Let's calculate the slope. We can also see from the options that the equation of the line is in the form \(y = mx + b\), where \(b = 2\). To find the slope, we can use two points. Let's take the y - intercept (0, 2) and another point. From the graph, when \(x = 1\), \(y=-1\) (since the line goes down 3 units for 1 unit to the right, slope \(m=- 3\)). So the equation of the boundary line is \(y=-3x + 2\).

Step2: Determine if the line is solid or dashed

The boundary line in the graph is solid (since the shaded region includes the line), so the inequality will be either \(\geq\) or \(\leq\), not \(>\) or \(<\). So we can eliminate options C and D.

Step3: Determine the direction of the inequality

To find which side is shaded, we can test a point. Let's take the origin (0,0). Plugging into the inequality \(y\) compared to \(-3x + 2\). For the origin, \(y = 0\) and \(-3x+2=2\). Now, we check the shaded region. The shaded region includes the origin? Wait, no, looking at the graph, the shaded region is to the left of the line. Wait, let's take a point in the shaded region, say (-5, 0). Plug into \(y\) and \(-3x + 2\). \(y = 0\), \(-3(-5)+2=15 + 2=17\). Now, check the inequality. If we consider option A: \(y\geq-3x + 2\), for (-5,0), \(0\geq17\) is false. Wait, maybe I made a mistake. Wait, let's re - examine the graph. The line has a slope of - 3 (since for \(x\) increasing by 1, \(y\) decreases by 3). The y - intercept is (0,2). Now, the shaded region: let's take a point in the shaded area, say (0,0). Wait, no, (0,0) is in the shaded region? Wait, the graph shows that the shaded region is on the side where, when we plug in (0,0) into \(y\) and \(-3x + 2\), \(y = 0\), \(-3x+2 = 2\). If the line is solid and the shaded region is where \(y\) is greater than or equal to \(-3x + 2\)? Wait, no, let's think again. Wait, the slope is - 3, so the line goes from (0,2) down to (1, - 1), (2, - 4), etc. The shaded region: let's take a point like (-1, 5). Plug into \(y=-3x + 2\), when \(x=-1\), \(y=-3(-1)+2=5\). So the point (-1,5) is on the line? Wait, no, (-1,5): \(y = 5\), \(-3(-1)+2=5\), so it's on the line. Wait, maybe the shaded region is above the line? Wait, no, the graph's shaded area: looking at the y - axis, the line crosses the y - axis at (0,2). The shaded region is to the left of the line. Wait, let's use the correct method. The general form for a linear inequality: if the line is \(y=mx + b\), and the line is solid, then we test a point. Let's take (0,0). For the line \(y=-3x + 2\), at (0,0), \(0\) vs \(-3(0)+2 = 2\). Now, the shaded region: if the shaded region includes (0,0), then we check if \(0\leq - 3(0)+2\) (which is \(0\leq2\), true) or \(0\geq - 3(0)+2\) (which is \(0\geq2\), false). Wait, but the line is solid. Wait, maybe I misread the graph. Wait, the options: A is \(y\geq-3x + 2\), B is \(y\leq-3x + 2\). Let's check the slope and the y - intercept again. The line is \(y=-3x + 2\), solid line. Now, the shaded region: let's take a point in the shaded area, say (0,0). For option B: \(y\leq-3x + 2\), \(0\leq2\), which is true. Wait, but earlier I thought the shaded region was not including (0,0), but maybe I was wrong. Wait, the graph: the line goes from (0,2) down to (1, - 1), (2, - 4), etc. The shaded region: looking at the x - axis, when \(x = 0\), the shaded region is below the line? No, the y - axis: at \(x = 0\), the line is at \(y = 2\), and the shaded region is above or below? Wait, the graph's shaded area: the left - hand side of the line. Let's take \(x=-1\), then \(y=-3(-1)+2=5\). So the point (-1,0): \(y = 0\), and \(-3(-1)+2=5\). For option A: \(y\geq-3x + 2\), \(0\geq5\) is false. For option B: \(y\leq-3x + 2\), \(0\leq5\) is true. Wait, but the line is solid. Wait, maybe I made a mistake in the point selection. Wait, the correct way: the slope is - 3, y - intercept 2. The line is solid, so inequality is \(\geq\) or \(\leq\). Now, the shaded region: let's see the direction of the inequality. If the line is \(y=-3x + 2\), and the shaded region is where \(y\) is greater than or equal to the line? Wait, no, when \(x\) is negative, \(-3x\) is positive, so \(-3x + 2\) is large. Wait, maybe the initial point selection was wrong. Let's take (0,3). Plug into \(y=-3x + 2\), \(y = 3\), \(-3(0)+2=2\). So \(3\geq2\), which is true. And (0,3) is in the shaded region? Looking at the graph, the shaded region is above the line? Wait, the graph shows the shaded area on the left side of the line. Wait, maybe the correct answer is A. Wait, let's re - examine the options. The boundary line is solid, so the inequality is either \(\geq\) or \(\leq\). The slope is - 3, y - intercept 2. Now, the shaded region: if we take a point in the shaded area, say (-2, 0). For option A: \(y\geq-3x + 2\), \(0\geq-3(-2)+2=6 + 2=8\)? No, that's false. Wait, I'm confused. Wait, maybe the line is \(y=-3x + 2\), and the shaded region is where \(y\) is greater than or equal to the line. Wait, let's check the y - intercept. The line crosses the y - axis at (0,2). The shaded region includes the area above the line? Wait, the graph's blue area: looking at the y - axis, above \(y = 2\) on the left? No, the grid: each square is 1 unit. The line goes from (0,2) to (1, - 1), so it's a steeply decreasing line. The shaded region is to the left of the line. Let's use the formula for linear inequalities: the inequality \(y\geq mx + b\) shades above the line, \(y\leq mx + b\) shades below the line (when \(m\) is negative, "above" and "below" can be a bit tricky, but the rule is based on the test point). Let's take the origin (0,0). For the line \(y=-3x + 2\), at (0,0), \(y = 0\) and \(-3x+2 = 2\). Now, if we shade above the line, then \(y\geq-3x + 2\) would shade where \(y\) is greater than or equal to \(-3x + 2\). At (0,0), \(0\geq2\) is false, so the origin is not in the shaded region for \(y\geq-3x + 2\). But if we shade below the line, \(y\leq-3x + 2\), at (0,0), \(0\leq2\) is true, so the origin is in the shaded region. Wait, but looking at the graph, does the origin lie in the shaded region? The graph shows the shaded region on the left side of the line. The origin (0,0) is on the right side of the line? Wait, no, the line at \(x = 0\) is at \(y = 2\), so the origin (0,0) is below the line. Wait, maybe the graph's shaded region is below the line. So if the line is solid and the shaded region is below the line, the inequality is \(y\leq-3x + 2\)? But wait, let's check the slope again. Wait, the line has a slope of - 3, so for \(x\) increasing, \(y\) decreases. The y - intercept is (0,2). Now, let's take a point in the shaded area, say (-5, 5). Plug into \(y\leq-3x + 2\): \(5\leq-3(-5)+2=15 + 2 = 17\), which is true. Plug into \(y\geq-3x + 2\): \(5\geq17\), which is false. Wait, but the line is solid. Wait, maybe I made a mistake in the slope. Wait, the line goes from (0,2) to (1, - 1), so the change in \(y\) is \(-1 - 2=-3\), change in \(x\) is \(1-0 = 1\), so slope \(m=-3\), correct. The equation of the line is \(y=-3x + 2\), correct. Now, the boundary line is solid, so the inequality is either \(\geq\) or \(\leq\). Now, the shaded region: let's see the direction. If we consider the inequality \(y\geq-3x + 2\), it shades above the line (since for a line \(y=mx + b\), \(y\geq mx + b\) shades above when \(m\) is negative? Wait, no, the rule is: for \(y>mx + b\), shade above the line (if \(m\) is negative, "above" is towards the positive y - direction), and \(y<mx + b\) shades below. But when the line is solid, it's \(\geq\) or \(\leq\). Wait, let's take a point above the line, say (0,3). For \(y=-3x + 2\), at \(x = 0\), \(y = 2\). So (0,3) is above the line. Plug into \(y\geq-3x + 2\): \(3\geq2\), which is true. Is (0,3) in the shaded region? Looking at the graph, the shaded region is on the left side of the line, which includes (0,3). Now, take a point below the line, say (0,0). Plug into \(y\geq-3x + 2\): \(0\geq2\), which is false, so (0,0) is not in the shaded region. Which matches the graph, because the shaded region does not include (0,0). So the correct inequality is \(y\geq-3x + 2\), because the line is solid (so \(\geq\) or \(\leq\)) and the shaded region is above the line (where \(y\) is greater than or equal to \(-3x + 2\)).

Answer:

A. \(y\geq-3x + 2\)