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Math  /  Algebra

Questionhich table of values represents a linear function?
A \begin{tabular}{|c|c|} \hlinexx & yy \\ \hline-2 & 1 \\ \hline-1 & 3 \\ \hline 0 & 5 \\ \hline 1 & 7 \\ \hline \end{tabular}
C \begin{tabular}{|c|c|} \hlinexx & yy \\ \hline-3 & 7 \\ \hline 0 & 6 \\ \hline 3 & 4 \\ \hline 9 & 0 \\ \hline \end{tabular}
B \begin{tabular}{|c|c|} \hlinexx & yy \\ \hline 1 & 3 \\ \hline 3 & 5 \\ \hline 5 & 7 \\ \hline 7 & 8 \\ \hline \end{tabular}
D \begin{tabular}{|c|c|} \hlinexx & yy \\ \hline-2 & 4 \\ \hline 1 & 1 \\ \hline 4 & -2 \\ \hline 7 & -4 \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. A linear function has a constant rate of change, or slope, between any two points.
2. We need to determine which table of values represents a linear function.

STEP 2

1. Calculate the rate of change for each table.
2. Determine if the rate of change is constant for each table.

STEP 3

For Table A, calculate the rate of change between successive points: - Between (−2,1)(-2, 1) and (−1,3)(-1, 3): 3−1−1−(−2)=21=2\frac{3 - 1}{-1 - (-2)} = \frac{2}{1} = 2 - Between (−1,3)(-1, 3) and (0,5) (0, 5): 5−30−(−1)=21=2\frac{5 - 3}{0 - (-1)} = \frac{2}{1} = 2 - Between (0,5) (0, 5) and (1,7) (1, 7): 7−51−0=21=2\frac{7 - 5}{1 - 0} = \frac{2}{1} = 2

STEP 4

For Table A, the rate of change is constant at 2, indicating a linear function.

STEP 5

For Table B, calculate the rate of change between successive points: - Between (1,3) (1, 3) and (3,5) (3, 5): 5−33−1=22=1\frac{5 - 3}{3 - 1} = \frac{2}{2} = 1 - Between (3,5) (3, 5) and (5,7) (5, 7): 7−55−3=22=1\frac{7 - 5}{5 - 3} = \frac{2}{2} = 1 - Between (5,7) (5, 7) and (7,8) (7, 8): 8−77−5=12\frac{8 - 7}{7 - 5} = \frac{1}{2}

STEP 6

For Table B, the rate of change is not constant, indicating it is not a linear function.

STEP 7

For Table C, calculate the rate of change between successive points: - Between (−3,7)(-3, 7) and (0,6) (0, 6): 6−70−(−3)=−13\frac{6 - 7}{0 - (-3)} = \frac{-1}{3} - Between (0,6) (0, 6) and (3,4) (3, 4): 4−63−0=−23\frac{4 - 6}{3 - 0} = \frac{-2}{3} - Between (3,4) (3, 4) and (9,0) (9, 0): 0−49−3=−46=−23\frac{0 - 4}{9 - 3} = \frac{-4}{6} = \frac{-2}{3}

STEP 8

For Table C, the rate of change is not constant, indicating it is not a linear function.

STEP 9

For Table D, calculate the rate of change between successive points: - Between (−2,4)(-2, 4) and (1,1) (1, 1): 1−41−(−2)=−33=−1\frac{1 - 4}{1 - (-2)} = \frac{-3}{3} = -1 - Between (1,1) (1, 1) and (4,−2) (4, -2): −2−14−1=−33=−1\frac{-2 - 1}{4 - 1} = \frac{-3}{3} = -1 - Between (4,−2) (4, -2) and (7,−4) (7, -4): −4−(−2)7−4=−23\frac{-4 - (-2)}{7 - 4} = \frac{-2}{3}

STEP 10

For Table D, the rate of change is not constant, indicating it is not a linear function.
The table that represents a linear function is Table A.
Table A \boxed{\text{Table A}}

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