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Which table of values represents a linear function?

A
xx yy
minus, 4−4 minus, 8−8
minus, 2−2 minus, 5−5
11 minus, 2−2
44 11
B
xx yy
minus, 8−8 77
minus, 4−4 22
00 minus, 3−3
44 minus, 8−8
C
xx yy
minus, 3−3 66
00 44
33 33
66 11
D
xx yy
minus, 3−3 77
00 33
33 00
55 minus, 2−2
Answer
Multiple Choice Answers
AA
BB
CC
DD

1 answer

To determine which table of values represents a linear function, we need to check if the change in \(y\) with respect to the change in \(x\) (the slope) is constant for all points in each table.

Let's analyze each table.

Table A:

\( (x, y) \)

  • \( (-4, -8) \)
  • \( (-2, -5) \)
  • \( (1, -2) \)
  • \( (4, 11) \)

Calculating the slopes:

  1. From \((-4, -8)\) to \((-2, -5)\): \[ \text{slope} = \frac{-5 - (-8)}{-2 - (-4)} = \frac{3}{2} \]

  2. From \((-2, -5)\) to \( (1, -2)\): \[ \text{slope} = \frac{-2 - (-5)}{1 - (-2)} = \frac{3}{3} = 1 \]

  3. From \( (1, -2)\) to \( (4, 11)\): \[ \text{slope} = \frac{11 - (-2)}{4 - 1} = \frac{13}{3} \]

The slopes are not constant, so Table A is not linear.

Table B:

\( (x, y) \)

  • \( (-8, 7) \)
  • \( (-4, 2) \)
  • \( (0, -3) \)
  • \( (4, -8) \)

Calculating the slopes:

  1. From \((-8, 7)\) to \((-4, 2)\): \[ \text{slope} = \frac{2 - 7}{-4 - (-8)} = \frac{-5}{4} \]

  2. From \((-4, 2)\) to \( (0, -3)\): \[ \text{slope} = \frac{-3 - 2}{0 - (-4)} = \frac{-5}{4} \]

  3. From \( (0, -3)\) to \( (4, -8)\): \[ \text{slope} = \frac{-8 - (-3)}{4 - 0} = \frac{-5}{4} \]

The slope is constant at \(-\frac{5}{4}\), so Table B is linear.

Table C:

\( (x, y) \)

  • \( (-3, 6) \)
  • \( (0, 4) \)
  • \( (3, 3) \)
  • \( (6, 1) \)

Calculating the slopes:

  1. From \((-3, 6)\) to \( (0, 4)\): \[ \text{slope} = \frac{4 - 6}{0 - (-3)} = \frac{-2}{3} \]

  2. From \( (0, 4)\) to \( (3, 3)\): \[ \text{slope} = \frac{3 - 4}{3 - 0} = \frac{-1}{3} \]

  3. From \( (3, 3)\) to \( (6, 1)\): \[ \text{slope} = \frac{1 - 3}{6 - 3} = \frac{-2}{3} \]

The slopes are not constant, so Table C is not linear.

Table D:

\( (x, y) \)

  • \( (-3, 7) \)
  • \( (0, 3) \)
  • \( (3, 0) \)
  • \( (5, -2) \)

Calculating the slopes:

  1. From \((-3, 7)\) to \( (0, 3)\): \[ \text{slope} = \frac{3 - 7}{0 - (-3)} = \frac{-4}{3} \]

  2. From \( (0, 3)\) to \( (3, 0)\): \[ \text{slope} = \frac{0 - 3}{3 - 0} = \frac{-3}{3} = -1 \]

  3. From \( (3, 0)\) to \( (5, -2)\): \[ \text{slope} = \frac{-2 - 0}{5 - 3} = \frac{-2}{2} = -1 \]

The slopes are not constant between all points, so Table D is not linear.

Conclusion:

The only table that represents a linear function is Table B.