Question

svlc algebra 1a - standard (15260)
introduction to linear functions
which table represents a linear function?

first table (top left):

x	y
2	6
3	12
4	24

second table (top right):

x	y
2	5
3	9
4	14

third table (bottom left):

x	y
2	-4
3	-2
4	0

fourth table (bottom right):

x	y
2	-5
3	-7
4	-9

Explanation:

Step1: Recall linear function property

A linear function has a constant rate of change (slope), meaning the difference in \( y \)-values (\( \Delta y \)) divided by the difference in \( x \)-values (\( \Delta x \)) is constant. For \( x \) increasing by 1 (since \( \Delta x = 1 - 0 = 1 \) between consecutive \( x \)-values), \( \Delta y \) should be constant.

Step2: Analyze first table

\( x: 1, 2, 3, 4 \); \( y: 3, 6, 12, 24 \)
\( \Delta y_1 = 6 - 3 = 3 \)
\( \Delta y_2 = 12 - 6 = 6 \)
\( \Delta y_3 = 24 - 12 = 12 \)
Not constant, so not linear.

Step3: Analyze second table

\( x: 1, 2, 3, 4 \); \( y: 2, 5, 9, 14 \)
\( \Delta y_1 = 5 - 2 = 3 \)
\( \Delta y_2 = 9 - 5 = 4 \)
\( \Delta y_3 = 14 - 9 = 5 \)
Not constant, so not linear.

Step4: Analyze third table

\( x: 1, 2, 3, 4 \); \( y: -2, -4, -2, 0 \)
\( \Delta y_1 = -4 - (-2) = -2 \)
\( \Delta y_2 = -2 - (-4) = 2 \)
\( \Delta y_3 = 0 - (-2) = 2 \)
Not constant, so not linear.

Step5: Analyze fourth table

\( x: 1, 2, 3, 4 \); \( y: -3, -5, -7, -9 \)
\( \Delta y_1 = -5 - (-3) = -2 \)
\( \Delta y_2 = -7 - (-5) = -2 \)
\( \Delta y_3 = -9 - (-7) = -2 \)
Constant \( \Delta y = -2 \), so linear.

Answer:

The fourth table (with \( x: 1, 2, 3, 4 \) and \( y: -3, -5, -7, -9 \)) represents a linear function.