which is the graph of the linear inequality \\(\\frac{1}{2}x - 2y > -6\\)?

Explanation:

Step1: Rewrite the inequality

Start with (\frac{1}{2}x - 2y > -6). Subtract (\frac{1}{2}x) from both sides: (-2y > -\frac{1}{2}x - 6). Divide by (-2) (remember to reverse the inequality sign): (y < \frac{1}{4}x + 3).

Step2: Analyze the boundary line

The boundary line is (y = \frac{1}{4}x + 3), which has a slope of (\frac{1}{4}) and a y - intercept of (3). Since the inequality is (y < \frac{1}{4}x + 3), the boundary line should be dashed (because the inequality is strict, (>) or (<), not (\geq) or (\leq)).

Step3: Determine the shaded region

We test a point not on the line, say ((0,0)). Plug into (y < \frac{1}{4}x + 3): (0 < 0 + 3), which is true. So we shade the region that includes ((0,0)), which is below the line (y=\frac{1}{4}x + 3).

Now, looking at the graphs:

• The first graph: Check the boundary line (should be dashed) and shading. If the line is solid, it's wrong.
• The second graph: Boundary line is dashed, slope and y - intercept match, and shading is below the line (includes ((0,0)))? Wait, let's re - check. Wait, when we have (y < \frac{1}{4}x+3), the region below the line. Let's check the slope: from (y=\frac{1}{4}x + 3), when (x = 0), (y = 3); when (x = 4), (y=\frac{1}{4}(4)+3=4). So the line goes through ((0,3)) and ((4,4)). The second graph (the one in the middle - top, second from left) has a dashed line, correct slope, and shading below the line (since ((0,0)) is in the shaded region as (0<3)). The other graphs: one has a solid line (wrong, since inequality is strict), one has shading above (wrong, since (y <) the line), and the last one: let's check. Wait, the correct graph should have a dashed line, slope (\frac{1}{4}), y - intercept (3), and shading below. The second graph (the middle - top, second from left) matches. Wait, maybe the fourth graph? Wait, no. Wait, let's re - check the steps.

Wait, when we rewrite (\frac{1}{2}x-2y > - 6) as (y <\frac{1}{4}x + 3). The y - intercept is (3), so the line crosses the y - axis at (y = 3). The slope is (\frac{1}{4}), so for every 4 units right, it goes up 1 unit. Now, the dashed line (since (>) in the original inequality, when we manipulated to (y <), the boundary is dashed). The region below the line (since (y <)). The second graph (the one with the dashed line, passing through ((0,3)) approximately, and shading below) is the correct one. Wait, the fourth graph (the bottom - right) has a dashed line? Wait, the user's image: let's assume the second graph (the middle - top, second from left) is the correct one. Wait, maybe the second graph (the one with the dashed line, and shading above? No, wait ((0,0)) is in the region below. Wait, maybe I made a mistake. Wait, original inequality: (\frac{1}{2}x-2y > - 6). Let's test ((0,0)) in original: (\frac{1}{2}(0)-2(0)=0 > - 6), which is true. So the region that satisfies the original inequality is where ((0,0)) is. In the original inequality (\frac{1}{2}x-2y > - 6), we can also think in terms of the original inequality. Let's rewrite the original inequality as (\frac{1}{2}x-2y+6>0). At ((0,0)), (0 - 0+6 = 6>0), so ((0,0)) is in the solution set.

Now, the boundary line: (\frac{1}{2}x-2y=-6) or (x - 4y=-12) (multiplying both sides by 2). Solving for (y): (y=\frac{1}{4}x + 3), as before. So the line is dashed. The slope is positive ((\frac{1}{4})), y - intercept 3. The graph with dashed line, positive slope, y - intercept 3, and shading including ((0,0)) (below the line) is the correct one. Looking at the options, the second graph (the middle - top, second from left) should be the correct one. Wait, maybe the fourth graph? Wait, no. Let's check the slope again. The line (y = \frac{1}{4}x+3) has a gentle slope. The second graph (the one with the dashed line, going through near ((0,3)) and ((4,4))) with shading below (since ((0,0)) is in the shaded area) is correct.

Answer:

The second graph (the middle - top, second from the left)