Question

1. name the domain and range for each relation. plot the points for each on the coordinate plane. which relation falls along a line?

relation a: (-4, -7), (-2, -3), (0, 1), (1, 3), (3, 7)
domain (x) range (y)

relation b: (0, 5), (0, -5), (5, 0), (-5, 0)
domain (x) range (y)

Explanation:

Step1: Analyze Relation A (Domain)

The domain of a relation is the set of all x - values. For Relation A: \((-4, -7), (-2, -3), (0, 1), (1, 3), (3, 7)\), the x - values are \(-4, -2, 0, 1, 3\). So the domain of Relation A is \(\{-4, -2, 0, 1, 3\}\).

Step2: Analyze Relation A (Range)

The range of a relation is the set of all y - values. For Relation A, the y - values are \(-7, -3, 1, 3, 7\). So the range of Relation A is \(\{-7, -3, 1, 3, 7\}\).

Step3: Analyze Relation B (Domain)

For Relation B: \((0, 5), (0, -5), (5, 0), (-5, 0)\), the x - values are \(0, 0, 5, -5\). Removing duplicates, the domain of Relation B is \(\{-5, 0, 5\}\).

Step4: Analyze Relation B (Range)

For Relation B, the y - values are \(5, -5, 0, 0\). Removing duplicates, the range of Relation B is \(\{-5, 0, 5\}\).

Step5: Check which relation is linear

To check if a relation falls along a line, we can check the slope between consecutive points (for Relation A).

• Slope between \((-4, -7)\) and \((-2, -3)\): \(m=\frac{-3 - (-7)}{-2-(-4)}=\frac{4}{2} = 2\)
• Slope between \((-2, -3)\) and \((0, 1)\): \(m=\frac{1-(-3)}{0 - (-2)}=\frac{4}{2}=2\)
• Slope between \((0, 1)\) and \((1, 3)\): \(m=\frac{3 - 1}{1-0}=\frac{2}{1}=2\)
• Slope between \((1, 3)\) and \((3, 7)\): \(m=\frac{7 - 3}{3 - 1}=\frac{4}{2}=2\)

Since the slope between consecutive points of Relation A is constant (\(m = 2\)), Relation A falls along a line. For Relation B, the points are \((0,5),(0, - 5),(5,0),(-5,0)\), which form a cross (intercepts of the axes), not a straight line.

Answer:

Relation A:

• Domain: \(\{-4, -2, 0, 1, 3\}\)
• Range: \(\{-7, -3, 1, 3, 7\}\)

Relation B:

• Domain: \(\{-5, 0, 5\}\)
• Range: \(\{-5, 0, 5\}\)

Linear Relation:

Relation A falls along a line.