Question

which inequality is graphed on the coordinate plane?

Explanation:

Step1: Find the slope and y-intercept

The line passes through \((0, 2)\) (y-intercept, \(b = 2\)) and let's find another point. From the graph, when \(x=-2\), \(y = 6\)? Wait, no, let's check the grid. Wait, the line seems to have a slope. Let's take two points: when \(x = 0\), \(y=2\); when \(x = 1\), \(y=-1\)? Wait, no, maybe better to calculate slope. Wait, the line goes from (0,2) and let's see the direction. Wait, the blue region is to the left of the line. Let's find the equation of the line. Let's take two points: (0, 2) and (1, -1)? Wait, no, maybe (0,2) and (2, -4)? Wait, no, let's count the rise over run. From (0,2) to (1, -1): change in y is -3, change in x is 1, so slope \(m=-3\). So the equation of the line is \(y=-3x + 2\).

Step2: Determine the inequality symbol

The line is solid (wait, is it solid? The graph shows a solid line? Wait, the blue region is to the left of the line. Let's test a point in the blue region, say (-5, 0). Plug into \(y\) and \(-3x + 2\). For \(x=-5\), \(-3(-5)+2=15 + 2=17\). The y-coordinate of the point is 0, which is less than 17? Wait, no, maybe I got the slope wrong. Wait, maybe the line is \(y=-2x + 2\)? Wait, no, let's re-examine. Wait, when \(x = 0\), \(y=2\) (y-intercept). Let's take another point: when \(x = 1\), \(y=0\)? Wait, no, the x-intercept: set \(y=0\), then \(0=-3x + 2\) → \(x=\frac{2}{3}\), but the graph shows x-intercept around (1,0)? Wait, maybe my initial point selection is wrong. Wait, the grid: each square is 1 unit? So from (0,2) to (1, -1): no, that's too steep. Wait, maybe the slope is -2? Wait, no, let's look again. Wait, the line passes through (0,2) and (1,0)? No, (0,2) and (1, -1) is slope -3. Wait, maybe the correct slope is -3. Then the line is \(y=-3x + 2\). Now, the blue region: let's take a point in the blue region, say (-2, 0). Plug into \(y\) and \(-3x + 2\): \(y=0\), \(-3(-2)+2=6 + 2=8\). So \(0\leq8\)? No, wait, maybe the inequality is \(y\leq -3x + 2\)? No, the blue region is where \(y\) is greater? Wait, no, the line is \(y=-3x + 2\), and the blue region is to the left, so when \(x\) is smaller, \(y\) can be... Wait, maybe I messed up the slope. Let's do it properly. Let's find two points on the line. From the graph, the line passes through (0, 2) and (1, -1)? Wait, no, (0,2) and (2, -4)? No, that's slope -3. Alternatively, maybe the line is \(y = -2x + 2\). Let's check: when \(x=0\), \(y=2\); when \(x=1\), \(y=0\) (x-intercept at (1,0)), which matches the graph (since the x-intercept is around (1,0)). Ah, there we go. So slope \(m=\frac{0 - 2}{1 - 0}=-2\). So equation of line is \(y=-2x + 2\).

Now, test a point in the blue region, say (-5, 0). Plug into \(y\) and \(-2x + 2\): \(y=0\), \(-2(-5)+2=10 + 2=12\). So \(0\leq12\)? No, wait, the blue region is where \(y\) is greater than or equal to \(-2x + 2\)? Wait, no, if the line is \(y=-2x + 2\), and the blue region is to the left of the line, let's take a point in the blue region, say (0, 5). Plug into \(y\) and \(-2x + 2\): \(5\) vs \(-2(0)+2=2\). So \(5\geq2\), which is true. So the inequality is \(y\geq -2x + 2\)? Wait, no, the line is solid (assuming), and the blue region is above or below? Wait, the line goes from (0,2) to (1,0), so it's a decreasing line. The blue region is to the left of the line. Let's take ( -1, 0): plug into \(y=-2x + 2\): \(-2(-1)+2=4\). The y-coordinate is 0, which is less than 4. Wait, maybe the inequality is \(y\leq -2x + 2\)? No, that doesn't fit. Wait, maybe I have the slope wrong. Let's try again. Let's take two points: (0, 2) and (2, -2). Then slope \(m=\frac{-2 - 2}{2 - 0}=\frac{-4}{2}=-2\). So equation \(y=-2x + 2\). Now, test ( -1, 0): \(y=-2(-1)+2=4\). The point (-1,0) is in the blue region? Wait, the blue region is the left side. So (-1,0) is to the left of the line? Wait, the line at x=-1 is y=4, so the point (-1,0) is below the line. But the blue region includes (-5,0), which is below the line? Wait, maybe the line is \(y = -3x + 2\). Wait, this is getting confusing. Wait, the correct approach:

1. Find the equation of the boundary line:
- Identify two points on the line. From the graph, the line passes through (0, 2) (y-intercept) and (1, -1) (wait, no, (1, -1) would be slope -3). Wait, maybe the line is \(y = -3x + 2\). Let's check the x-intercept: set \(y=0\), \(0=-3x + 2\) → \(x=\frac{2}{3}\), which is about (0.666, 0), which matches the graph (since the x-intercept is near (1,0)? No, maybe the grid is different. Wait, maybe each square is 2 units? No, the x-axis has -5 and 5 marked, so each grid is 1 unit.

2. Determine if the line is solid or dashed: The graph shows a solid line (assuming), so the inequality includes equality.

3. Determine the inequality direction: Test a point in the shaded region. Let's take (-5, 0) (in the blue region). Plug into the line equation. If the line is \(y = -3x + 2\), then at \(x=-5\), \(y=-3(-5)+2=17\). The y-coordinate of the point is 0, which is less than 17. But the shaded region is where \(y\) is less than or equal to \(-3x + 2\)? Wait, no, that would mean the region below the line, but the blue region is to the left. Wait, maybe the inequality is in terms of x and y, or maybe I mixed up x and y. Wait, no, the standard is \(y\) vs \(mx + b\).

Wait, maybe the correct line is \(y = -2x + 2\). Let's test ( -2, 0): \(y=-2(-2)+2=6\). The point (-2,0) is in the blue region, and 0 ≤ 6? No, 0 is less than 6, so if the shaded region is where \(y\leq -2x + 2\), but the blue region is above? Wait, no, I think I made a mistake in the slope. Let's look at the graph again: the line goes from (0,2) down to the right, so slope is negative. The blue region is to the left of the line. So for a given x, the y-values in the blue region are less than or greater than the line's y-value? Let's take x = -3: line's y is -3*(-3) + 2 = 11, and a point in the blue region at x=-3, y=0: 0 < 11, so \(y\leq -3x + 2\)? But the blue region includes ( -3, 5), which is 5 ≤ -3*(-3)+2=11, which is true. So maybe the inequality is \(y\leq -3x + 2\)? Wait, no, the line is \(y=-3x + 2\), and the shaded region is below or above? Wait, the line is decreasing, so to the left of the line (smaller x) the line's y-value is larger. So for a fixed x, the shaded region has y-values less than the line's y-value? Wait, no, if x is smaller (more negative), the line's y is larger (since slope is negative: y = -3x + 2, so as x decreases, -3x increases, so y increases). So the shaded region is where y is less than or equal to -3x + 2? Wait, no, if x is -5, line's y is 17, and the shaded region at x=-5 has y from, say, -5 to 5, which are all less than 17. So yes, \(y\leq -3x + 2\). But wait, the line is solid, so the inequality is \(y\leq -3x + 2\), or maybe \(3x + y\leq 2\) (rewriting \(y\leq -3x + 2\) as \(3x + y\leq 2\)).

Wait, maybe I messed up the slope. Let's do it correctly:

1. Find two points on the line:
- From the graph, the line passes through (0, 2) (y-intercept, so \(b = 2\)) and (1, -1) (since from (0,2) to (1, -1), the change in y is -3, change in x is 1, so slope \(m = -3\)).

2. Equation of the line: \(y = -3x + 2\).

3. Determine the inequality:
- The line is solid (so equality is included).
- Test a point in the shaded region, e.g., (-5, 0):
- Plug into \(y\) and \(-3x + 2\): \(0\) vs \(-3(-5) + 2 = 17\).
- Since \(0 \leq 17\), the inequality is \(y \leq -3x + 2\) (or \(3x + y \leq 2\)).

Wait, but the shaded region is to the left of the line, which corresponds to \(y \leq -3x + 2\) because for smaller x (left), the line's y is larger, so the region below (or equal to) the line is to the left? Wait, maybe. Alternatively, if we rearrange \(y \leq -3x + 2\) to \(3x + y \leq 2\), that's another form.

But maybe the correct answer is \(y \leq -2x + 2\)? Wait, no, let's check with the x-intercept. If the line is \(y = -2x + 2\), then x-intercept is at \(x = 1\) (since \(0 = -2(1) + 2\)), which matches the graph (the line crosses the x-axis at (1,0)). Ah! That's the key. So the line crosses the x-axis at (1,0) and y-axis at (0,2). So slope \(m = \frac{0 - 2}{1 - 0} = -2\). So equation is \(y = -2x + 2\).

Now, test a point in the shaded region, say (-5, 0):

- Plug into \(y\) and \(-2x + 2\): \(0\) vs \(-2(-5) + 2 = 12\).
- Since \(0 \leq 12\), the inequality is \(y \leq -2x + 2\) (or \(2x + y \leq 2\)).

Yes, that makes sense. The line is \(y = -2x + 2\), solid (so equality), and the shaded region is where \(y \leq -2x + 2\) (since the point (-5,0) satisfies \(0 \leq 12\)).

So the inequality graphed is \(y \leq -2x + 2\) (or equivalent forms like \(2x + y \leq 2\)).

Answer:

The inequality graphed is \(y \leq -2x + 2\) (or \(2x + y \leq 2\)).