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QuestionWhich table shows a linear function? A: (-3,-5), (-1,-2), (1,1), (3,5); B: (-4,3), (-1,1), (2,-1), (8,-5); C: (1,-5), (3,-2), (5,2), (7,6); D: (-3,3), (-1,0), (2,-3), (5,-6).

Studdy Solution

STEP 1

Assumptions1. A linear function has a constant rate of change (slope). . The rate of change can be calculated by the formulaRate of change=change in ychange in x=y−y1x−x1\text{Rate of change} = \frac{\text{change in } y}{\text{change in } x} = \frac{y - y1}{x - x1}3. If the rate of change is the same between all consecutive pairs of points, then the function is linear.

STEP 2

Let's calculate the rate of change for each table. Start with table A.
For the first two points in table A, the rate of change is−2−(−5)−1−(−)\frac{-2 - (-5)}{-1 - (-)}

STEP 3

implify the above expression to get the rate of change for the first two points in table A.
−2−(−5)−1−(−3)=32\frac{-2 - (-5)}{-1 - (-3)} = \frac{3}{2}

STEP 4

Calculate the rate of change for the second and third points in table A.
1−(−2)1−(−1)\frac{1 - (-2)}{1 - (-1)}

STEP 5

implify the above expression to get the rate of change for the second and third points in table A.
1−(−2)1−(−1)=32\frac{1 - (-2)}{1 - (-1)} = \frac{3}{2}

STEP 6

Calculate the rate of change for the third and fourth points in table A.
5−13−1\frac{5 -1}{3 -1}

STEP 7

implify the above expression to get the rate of change for the third and fourth points in table A.
5−13−1=42=2\frac{5 -1}{3 -1} = \frac{4}{2} =2

STEP 8

Notice that the rate of change is not constant in table A, so table A does not represent a linear function.

STEP 9

Repeat steps2-7 for tables B, C, and D.

STEP 10

For table B, calculate the rate of change for the first two points.
−3−−(−4)\frac{ -3}{- - (-4)}

STEP 11

implify the above expression to get the rate of change for the first two points in table B.
−3−−(−4)=−3\frac{ -3}{- - (-4)} = \frac{-}{3}

STEP 12

Calculate the rate of change for the second and third points in table B.
−−2−(−)\frac{- -}{2 - (-)}

STEP 13

implify the above expression to get the rate of change for the second and third points in table B.
−−2−(−)=−23\frac{- -}{2 - (-)} = \frac{-2}{3}

STEP 14

Calculate the rate of change for the third and fourth points in table B.
−−(−)8−2\frac{- - (-)}{8 -2}

STEP 15

implify the above expression to get the rate of change for the third and fourth points in table B.
−5−(−)8−2=−4=−23\frac{-5 - (-)}{8 -2} = \frac{-4}{} = -\frac{2}{3}

STEP 16

Notice that the rate of change is constant in table B, so table B does represent a linear function.

STEP 17

For table C and D, repeat steps2-7.

STEP 18

For table C, the rate of change is not constant, so table C does not represent a linear function.

STEP 19

For table D, the rate of change is not constant, so table D does not represent a linear function.
The table of values that represents a linear function is table B.

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