Question

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

Explanation:

Step1: Rewrite the inequality in slope - intercept form

We start with the inequality \(\frac{1}{2}x - 2y>-6\).
Subtract \(\frac{1}{2}x\) from both sides: \(- 2y>-\frac{1}{2}x - 6\).
Divide both sides by \(-2\). Remember that when we divide or multiply an inequality by a negative number, the direction of the inequality sign changes. So we get \(y <\frac{1}{4}x + 3\).

Step2: Analyze the boundary line

The boundary line of the inequality \(y <\frac{1}{4}x + 3\) is the line \(y=\frac{1}{4}x + 3\). Since the inequality is strict (\(y <\frac{1}{4}x+3\), not \(y\leq\frac{1}{4}x + 3\)), the boundary line should be dashed.

Step3: Analyze the region

We need to find the region where \(y\) is less than \(\frac{1}{4}x + 3\). Let's test the origin \((0,0)\) in the inequality \(y<\frac{1}{4}x + 3\). Substitute \(x = 0\) and \(y = 0\) into the inequality: \(0<\frac{1}{4}(0)+3\), which simplifies to \(0 < 3\), a true statement. So the region containing the origin should be shaded.

Now let's analyze the four graphs:
- For the boundary line: The line \(y=\frac{1}{4}x + 3\) has a slope of \(\frac{1}{4}\) and a y - intercept of \(3\). The first two graphs have a boundary line with a different slope (the slope of the line in the first two graphs seems to be \(\frac{1}{2}\) approximately, while our line has a slope of \(\frac{1}{4}\))? Wait, no, let's re - examine. Wait, maybe I made a mistake in slope calculation. Wait, let's re - do the inequality manipulation.

Wait, starting over: \(\frac{1}{2}x-2y>-6\)

Subtract \(\frac{1}{2}x\): \(-2y>-\frac{1}{2}x - 6\)

Divide by \(-2\): \(y <\frac{1}{4}x+3\). The slope is \(\frac{1}{4}\), y - intercept \(3\).

Now, the third and fourth graphs: The boundary line in the third and fourth graphs: let's check the y - intercept. When \(x = 0\), \(y\) - intercept of the line in the third and fourth graphs is \(2\)? Wait, no, maybe I misread the graphs. Wait, the original inequality: let's find the x and y intercepts of the boundary line \(y=\frac{1}{4}x + 3\). When \(y = 0\), \(0=\frac{1}{4}x+3\), \(\frac{1}{4}x=-3\), \(x=-12\). When \(x = 0\), \(y = 3\).

Wait, maybe the graphs are different. Wait, the first two graphs have a boundary line with y - intercept around 2 - 3? Wait, no, let's look at the four graphs:

First graph: dashed line, shaded above? No, wait, our inequality is \(y <\frac{1}{4}x + 3\), so shaded below the line.

Wait, the third and fourth graphs: the boundary line in the third graph is dashed, fourth is solid. Our inequality has a strict inequality, so boundary line is dashed. So we can eliminate the second and fourth graphs (since they have solid lines). Now between the first and third graphs.

Wait, let's test a point. Let's take \(x = 0\), \(y = 0\). In the inequality \(y<\frac{1}{4}x + 3\), \(0<3\), which is true. So the region containing \((0,0)\) should be shaded. In the first graph, the shaded region is above the line? No, wait, the first graph's shaded region is above the dashed line. The third graph's shaded region is below the dashed line. Wait, our inequality is \(y <\frac{1}{4}x+3\), so we shade below the line.

Wait, maybe I messed up the inequality direction. Wait, let's re - express the original inequality:

\(\frac{1}{2}x-2y>-6\)

\(-2y>-\frac{1}{2}x - 6\)

\(y <\frac{1}{4}x + 3\) (since dividing by negative flips the inequality). So \(y\) is less than the line, so we shade below the line.

Now, the third graph: dashed line, shaded below the line (since the shaded region is at the bottom, including the origin? Wait, the origin \((0,0)\) is in the shaded region of the third graph? Let's see, the third graph's shaded region is below the dashed line. The first graph's shaded region is above the dashed line.

Also, the boundary line: the line \(y=\frac{1}{4}x + 3\) has a y - intercept of 3. Wait, the line in the third graph: when \(x = 0\), \(y = 2\)? No, maybe the graphs are labeled differently. Wait, maybe the correct graph is the third one? Wait, no, let's check the x - intercept. For the boundary line \(y=\frac{1}{4}x + 3\), when \(y = 0\), \(x=-12\). The third and fourth graphs have a boundary line that passes through \((- 8,0)\) and \((0,2)\), because the slope between \((-8,0)\) and \((0,2)\) is \(\frac{2 - 0}{0+8}=\frac{1}{4}\). Ah! There we go. So the line \(y=\frac{1}{4}x + 2\)? Wait, no, if \(x=-8\), \(y = 0\), then \(y=\frac{1}{4}(x + 8)=\frac{1}{4}x+2\). Wait, maybe I made a mistake in the inequality manipulation.

Wait, let's start over with the inequality \(\frac{1}{2}x-2y>-6\)

Let's find the x and y intercepts of the boundary line \(\frac{1}{2}x-2y=-6\) (the equality case).

For x - intercept, set \(y = 0\): \(\frac{1}{2}x=-6\), \(x=-12\). For y - intercept, set \(x = 0\): \(-2y=-6\), \(y = 3\). Wait, but the graphs have a line with x - intercept \(-8\) and y - intercept \(2\). So maybe there was a miscalculation.

Wait, maybe the original inequality is \(\frac{1}{2}x - y>-2\)? No, the user wrote \(\frac{1}{2}x-2y>-6\). Wait, let's re - calculate the intercepts for \(\frac{1}{2}x-2y=-6\):

When \(x = 0\): \(-2y=-6\), \(y = 3\)

When \(y = 0\): \(\frac{1}{2}x=-6\), \(x=-12\)

But the graphs provided have a line with x - intercept \(-8\) and y - intercept \(2\). So maybe the inequality is \(\frac{1}{2}x - y>-2\), but the user wrote \(\frac{1}{2}x-2y>-6\). Wait, \(\frac{1}{2}x-2y>-6\) is equivalent to \(x - 4y>-12\) (multiplying both sides by 2). The line \(x - 4y=-12\) (or \(y=\frac{1}{4}x + 3\)) has x - intercept \(-12\) and y - intercept \(3\).

But the graphs in the problem: the first two graphs have a line with x - intercept \(-8\) and y - intercept \(2\) (since when \(x=-8\), \(y = 0\); when \(x = 0\), \(y = 2\), slope \(\frac{2-0}{0 + 8}=\frac{1}{4}\)? Wait, \(\frac{2-0}{0-(-8)}=\frac{2}{8}=\frac{1}{4}\). Oh! Wait, the line in the first two graphs: when \(x=-8\), \(y = 0\); \(x = 0\), \(y = 2\). So the equation of the line is \(y=\frac{1}{4}x+2\). Wait, maybe there was a typo in the inequality, or maybe I misread the inequality.

Wait, the user's inequality is \(\frac{1}{2}x-2y>-6\). Let's rewrite it as \(2y<\frac{1}{2}x + 6\) (adding \(2y\) and \(6\) to both sides), then \(y<\frac{1}{4}x + 3\). But the line in the graphs has \(y\) - intercept \(2\). Wait, maybe the inequality is \(\frac{1}{2}x-2y>-4\). Then \( - 2y>-\frac{1}{2}x-4\), \(y<\frac{1}{4}x + 2\), which matches the line in the graphs (x - intercept \(-8\), y - intercept \(2\)).

Assuming that maybe there was a typo and the inequality is \(\frac{1}{2}x-2y>-4\) (or maybe I misread the original inequality as \(-6\) instead of \(-4\)), but let's proceed with the graphs.

The inequality \(y<\frac{1}{4}x + 2\) (assuming the correct inequality is \(\frac{1}{2}x-2y>-4\)):

- Boundary line: dashed (since strict inequality), so eliminate the second and fourth graphs (solid lines).

- Region: \(y<\frac{1}{4}x + 2\), so shade below the line. The first graph is shaded above the line, the third graph is shaded below the line.

So the correct graph is the third one (the one with dashed line and shaded below the line).

Answer:

The graph of the linear inequality \(\frac{1}{2}x - 2y>-6\) (assuming a possible typo in the intercept, but based on the graph's line characteristics) is the third graph (the one with the dashed boundary line and the shaded region below the line, containing the origin - like area). If we consider the strict inequality (dashed line) and the region below the line (since \(y<\frac{1}{4}x + 3\) or the adjusted \(y<\frac{1}{4}x + 2\) based on the graph's line), the correct graph is the third one (the bottom - left graph with dashed line and shaded below).