Question

4 multiple choice 1 point in the graph $y \leq -x$, which quadrant is completely shaded? \bigcirc quadrant i \bigcirc quadrant ii \bigcirc quadrant iii \bigcirc quadrant iv

Explanation:

Step1: Recall Quadrant Properties

In quadrants:

• Quadrant I: \( x>0, y>0 \)
• Quadrant II: \( x<0, y>0 \)
• Quadrant III: \( x<0, y<0 \)
• Quadrant IV: \( x>0, y<0 \)

Step2: Analyze \( y \leq -x \) in Each Quadrant

• Quadrant I: \( x>0, y>0 \). Then \( -x < 0 \), so \( y>0 \) and \( -x < 0 \), so \( y \leq -x \) (positive ≤ negative) is false.
• Quadrant II: \( x<0, y>0 \). Then \( -x > 0 \) (since \( x \) is negative, \( -x \) is positive). So \( y>0 \) and \( -x > 0 \), but \( y \leq -x \) would require \( y \leq \) positive value, but not all points here satisfy (e.g., \( x=-1, y=2 \): \( 2 \leq 1 \)? No).
• Quadrant III: \( x<0, y<0 \). Then \( -x > 0 \) (negative \( x \) → \( -x \) positive). Wait, no: \( x<0 \) → \( -x = |x| > 0 \), but \( y<0 \). Wait, correction: \( x<0, y<0 \), so \( -x = \text{positive} \), but \( y \) is negative. Wait, no—wait, \( y \leq -x \): if \( x<0 \), then \( -x = \text{positive} \), and \( y<0 \), so negative \( \leq \) positive: always true? Wait, no, let's take \( x=-2, y=-3 \): \( -3 \leq -(-2) = 2 \): yes. \( x=-1, y=-1 \): \( -1 \leq 1 \): yes. All points in Quadrant III: \( x<0, y<0 \), so \( -x = \text{positive} \), and \( y \) is negative, so negative \( \leq \) positive: always true. Wait, no—wait, maybe I messed up. Wait, \( y \leq -x \): if \( x<0 \), \( -x = |x| \), so \( y \leq |x| \). But in Quadrant III, \( y<0 \) and \( |x|>0 \), so \( y \) (negative) is always ≤ \( |x| \) (positive). So all points in Quadrant III satisfy \( y \leq -x \).
• Quadrant IV: \( x>0, y<0 \). Then \( -x < 0 \) (since \( x>0 \), \( -x \) is negative). So \( y<0 \) and \( -x < 0 \), so \( y \leq -x \) would require negative \( \leq \) negative, but not all points: e.g., \( x=1, y=-2 \): \( -2 \leq -1 \): yes. \( x=1, y=-0.5 \): \( -0.5 \leq -1 \)? No. So not all points satisfy.

Wait, correction: Let's re-express \( y \leq -x \) as \( y + x \leq 0 \).

• Quadrant I: \( x>0, y>0 \) → \( x + y > 0 \) → \( y + x \leq 0 \) false.
• Quadrant II: \( x<0, y>0 \) → \( x + y \) can be positive (e.g., \( x=-1, y=2 \): \( 1 > 0 \)) or negative (e.g., \( x=-3, y=1 \): \( -2 \leq 0 \)). So not all points.
• Quadrant III: \( x<0, y<0 \) → \( x + y < 0 \) (sum of two negatives) → \( x + y \leq 0 \) (since \( < 0 \) implies \( \leq 0 \)) → always true.
• Quadrant IV: \( x>0, y<0 \) → \( x + y \) can be positive (e.g., \( x=2, y=-1 \): \( 1 > 0 \)) or negative (e.g., \( x=1, y=-2 \): \( -1 \leq 0 \)). So not all points.

Thus, Quadrant III is completely shaded.

Answer:

quadrant III