Question

this is the graph of a linear inequality. write the inequality in slope - intercept form.
write your answer with y first, followed by an inequality symbol. use integers, proper fractions, and improper fractions in simplest form.

Explanation:

Step1: Find the slope-intercept form of the line

The slope-intercept form of a line is \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. From the graph, the line passes through \( (0, 2) \) (so \( b = 2 \)) and \( (1, 0) \). The slope \( m \) is calculated as \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2}{1 - 0} = -2 \). So the equation of the line is \( y = -2x + 2 \).

Step2: Determine the inequality symbol

The shaded region is above the line (since the blue area is on the left/upper side of the line), and the line is solid (from the graph's appearance, though we can check the shading). Wait, actually, looking at the graph, the line is solid? Wait, no, let's check the shading. The shaded area is above the line? Wait, when \( x = 0 \), the line is at \( y = 2 \), and the shaded area includes \( y \) values greater than that? Wait, no, let's take a test point. Let's take \( (0, 3) \), which is in the shaded area. Plug into the line equation: \( 3 \) vs \( -2(0) + 2 = 2 \). So \( 3 > 2 \), so the inequality is \( y \geq -2x + 2 \)? Wait, no, wait the line: let's check another point. Wait, the line goes from \( (0, 2) \) to \( (1, 0) \), and the shaded area is to the left of the line. Wait, maybe I made a mistake. Wait, let's re-examine the graph. The blue shaded area: when \( x \) is negative, \( y \) is higher. Wait, the line has a slope of -2, y-intercept 2. Let's check the boundary: the line is solid (so the inequality includes equality). Now, take a point in the shaded region, say \( (-3, 0) \). Plug into \( y \) and \( -2x + 2 \): \( 0 \) vs \( -2(-3) + 2 = 6 + 2 = 8 \). Wait, that's not right. Wait, maybe I got the slope wrong. Wait, let's take two points: (0, 2) and (let's see, when x=1, y=0; x=2, y=-2; x=3, y=-4; so the slope is (0 - 2)/(1 - 0) = -2, correct. Wait, maybe the shaded area is above the line? Wait, when x=0, the line is at y=2, and the shaded area is above that? Wait, the graph shows the blue area on the left, which is when x is smaller, y is larger. Wait, maybe the correct test point is (0, 3): plug into y and -2x + 2: 3 vs 2, so 3 > 2, so the inequality is \( y \geq -2x + 2 \)? Wait, but the line is solid, so the inequality is \( y \geq -2x + 2 \)? Wait, no, maybe the line is dashed? Wait, the original graph: the line is solid? Wait, the problem says "this is the graph of a linear inequality". Let me check again. The line: from (0,2) to (1,0), and the shaded area is above the line? Wait, no, when x increases, y decreases. So the region above the line would be where y is greater than -2x + 2. Let's confirm with (0, 3): 3 > -2(0) + 2 = 2, so yes, that's in the shaded area. So the inequality is \( y \geq -2x + 2 \)? Wait, but maybe the line is dashed? Wait, the graph's line: looking at the image, the line is solid (the blue line is solid). So the inequality is \( y \geq -2x + 2 \)? Wait, no, wait the shaded area: in the graph, the blue area is on the left side of the line. Wait, maybe I messed up the direction. Let's use the slope-intercept form correctly. The general form: if the line is \( y = mx + b \), and the shaded region is above the line, then \( y \geq mx + b \) (if solid) or \( y > mx + b \) (if dashed). Since the line appears solid (from the graph), the inequality is \( y \geq -2x + 2 \)? Wait, but let's check the point (0, 2): it's on the line, so it should satisfy the inequality. \( 2 \geq -2(0) + 2 \) is true (2=2). So that works. So the inequality is \( y \geq -2x + 2 \)? Wait, but wait the graph: when x=0, the line is at y=2, and the shaded area is above that? Wait, the blue area in the graph: when x is negative, y is higher. So yes, the shaded area is where y is greater than or equal to -2x + 2.

Wait, but maybe I made a mistake in the slope. Wait, let's recalculate the slope. Two points: (0, 2) and (1, 0). Slope \( m = (0 - 2)/(1 - 0) = -2 \), correct. So the line is \( y = -2x + 2 \). The shaded region is above the line (since for x=0, shaded area includes y > 2), so the inequality is \( y \geq -2x + 2 \). Wait, but let's check another point: ( -1, 4). Plug into \( -2x + 2 \): \( -2(-1) + 2 = 2 + 2 = 4 \). So \( y = 4 \) is on the line, and ( -1, 5) would be \( 5 \geq 4 \), which is true. So yes, the inequality is \( y \geq -2x + 2 \). Wait, but the original graph: maybe the line is dashed? Wait, the problem's graph: the line is solid (the blue line is a solid boundary), so the inequality includes equality. So the slope-intercept form of the inequality is \( y \geq -2x + 2 \).

Wait, but wait, maybe I got the direction wrong. Let's check the shading again. The blue shaded area: when x is -3, y is, say, 0? No, wait the grid: each square is 1 unit. So (0, 2) is on the line, (1, 0), (2, -2), etc. The shaded area is to the left of the line (where x is smaller), so for a given x, y is larger than the line's y-value. So yes, \( y \geq -2x + 2 \).

Answer:

\( y \geq -2x + 2 \)