Question

which figure below is the graph of $6x - 2y \geq 24$? a graph, b graph, c graph, d graph

Explanation:

Step1: Rewrite the inequality

Start with \(6x - 2y \geq 24\). Solve for \(y\) to get the slope - intercept form (\(y = mx + b\)).
Subtract \(6x\) from both sides: \(-2y\geq - 6x + 24\).
Divide both sides by \(-2\). Remember that when dividing an inequality by a negative number, the direction of the inequality sign changes. So we have \(y\leq3x - 12\).

Step2: Analyze the boundary line

The boundary line is \(y = 3x-12\). The slope \(m = 3\) (positive, so the line should be increasing) and the \(y\) - intercept \(b=- 12\). To find the \(x\) - intercept, set \(y = 0\):
\(0=3x - 12\), then \(3x=12\), so \(x = 4\). So the line passes through \((4,0)\) and \((0,-12)\).

Step3: Analyze the inequality symbol and shading

Since the inequality is \(y\leq3x - 12\), the shading should be below the line (because \(y\) values are less than or equal to the line's \(y\) values). Also, the boundary line should be solid (because of the \(\geq\) in the original inequality, which we transformed, and the \(\leq\) in the slope - intercept form still indicates a solid line for the boundary).

Now let's analyze the options:

• Option A: The line has a positive slope, and the shading is on the side that would correspond to \(y\leq3x - 12\) (below the line with positive slope). The line passes through points that are consistent with \(y = 3x-12\) (e.g., when \(x = 4\), \(y = 0\); when \(x = 0\), \(y=-12\) which is in the general area of the graph).
• Option B: The line has a negative slope, which does not match the slope of \(3\) for \(y = 3x - 12\), so B is incorrect.
• Option C: The line has a negative slope, which does not match the slope of \(3\) for \(y = 3x - 12\), so C is incorrect.
• Option D: The line has a negative slope (or a slope that is not \(3\)) and the shading direction is wrong, so D is incorrect.

Answer:

A