Question

which of the following tables could represent a linear function? select all that apply.

x	y
-6	-1
2	3
10	7

x	y
-2	-7
6	-5
14	9

x	y
0	4
5	3
10	1

x	y
8	10
14	7
20	4

Explanation:

Step1: Recall linear function rule

A linear function has a constant rate of change, i.e., the slope \( m=\frac{y_2 - y_1}{x_2 - x_1} \) is constant between any two points.

Step2: Analyze Orange Table (\(x,y\): \((-6,-1),(2,3),(10,7)\))

- Between \((-6,-1)\) and \((2,3)\): \( m_1=\frac{3 - (-1)}{2 - (-6)}=\frac{4}{8}=\frac{1}{2} \)
- Between \((2,3)\) and \((10,7)\): \( m_2=\frac{7 - 3}{10 - 2}=\frac{4}{8}=\frac{1}{2} \). Constant slope. Linear.

Step3: Analyze Green Table 1 (\(x,y\): \((-2,-7),(6,-5),(14,9)\))

- Between \((-2,-7)\) and \((6,-5)\): \( m_1=\frac{-5 - (-7)}{6 - (-2)}=\frac{2}{8}=\frac{1}{4} \)
- Between \((6,-5)\) and \((14,9)\): \( m_2=\frac{9 - (-5)}{14 - 6}=\frac{14}{8}=\frac{7}{4} \). Not constant. Not linear.

Step4: Analyze Green Table 2 (\(x,y\): \((0,4),(5,3),(10,1)\))

- Between \((0,4)\) and \((5,3)\): \( m_1=\frac{3 - 4}{5 - 0}=\frac{-1}{5} \)
- Between \((5,3)\) and \((10,1)\): \( m_2=\frac{1 - 3}{10 - 5}=\frac{-2}{5} \). Not constant. Not linear.

Step5: Analyze Blue Table (\(x,y\): \((8,10),(14,7),(20,4)\))

- Between \((8,10)\) and \((14,7)\): \( m_1=\frac{7 - 10}{14 - 8}=\frac{-3}{6}=-\frac{1}{2} \)
- Between \((14,7)\) and \((20,4)\): \( m_2=\frac{4 - 7}{20 - 14}=\frac{-3}{6}=-\frac{1}{2} \). Constant slope. Linear.

Wait, earlier mistake in Green Table 1: Wait, recalculate Green Table 1 (\((-2,-7),(6,-5),(14,9)\)):
Wait, \((-2,-7)\) to \((6,-5)\): \( \Delta y=-5 - (-7)=2 \), \( \Delta x=6 - (-2)=8 \), \( m=\frac{2}{8}=0.25 \)
\((6,-5)\) to \((14,9)\): \( \Delta y=9 - (-5)=14 \), \( \Delta x=14 - 6=8 \), \( m=\frac{14}{8}=1.75 \). Not constant. So Green Table 1 is not linear. Green Table 2: \((0,4)\) to \((5,3)\): \( \Delta y=-1 \), \( \Delta x=5 \), \( m=-0.2 \); \((5,3)\) to \((10,1)\): \( \Delta y=-2 \), \( \Delta x=5 \), \( m=-0.4 \). Not constant.

Wait, original Orange Table: \((-6,-1)\) to \((2,3)\): \( \Delta x=8 \), \( \Delta y=4 \), \( m=0.5 \); \((2,3)\) to \((10,7)\): \( \Delta x=8 \), \( \Delta y=4 \), \( m=0.5 \). Linear.

Blue Table: \((8,10)\) to \((14,7)\): \( \Delta x=6 \), \( \Delta y=-3 \), \( m=-0.5 \); \((14,7)\) to \((20,4)\): \( \Delta x=6 \), \( \Delta y=-3 \), \( m=-0.5 \). Linear.

Wait, maybe I misread the tables. Let's re-express all tables:

Table 1 (Orange):
x: -6, 2, 10; y: -1, 3, 7.
Differences: \( 2 - (-6)=8 \), \( 10 - 2=8 \) (constant \( \Delta x=8 \)). \( 3 - (-1)=4 \), \( 7 - 3=4 \) (constant \( \Delta y=4 \)). So linear (since \( \Delta y/\Delta x \) constant).

Table 2 (Green1):
x: -2, 6, 14; \( \Delta x=8, 8 \) (constant \( \Delta x=8 \)). y: -7, -5, 9. \( \Delta y=2, 14 \) (not constant). So not linear.

Table 3 (Green2):
x: 0, 5, 10; \( \Delta x=5, 5 \) (constant \( \Delta x=5 \)). y: 4, 3, 1. \( \Delta y=-1, -2 \) (not constant). So not linear.

Table 4 (Blue):
x: 8, 14, 20; \( \Delta x=6, 6 \) (constant \( \Delta x=6 \)). y: 10, 7, 4. \( \Delta y=-3, -3 \) (constant \( \Delta y=-3 \)). So linear.

Wait, but the original checkmarks: maybe the user's table has different values? Wait, maybe I made a mistake. Let's re-express the tables as given:

First table (orange):
x: -6, 2, 10; y: -1, 3, 7.

Second table (green1):
x: -2, 6, 14; y: -7, -5, 9.

Third table (green2):
x: 0, 5, 10; y: 4, 3, 1.

Fourth table (blue):
x: 8, 14, 20; y: 10, 7, 4.

Now, for a linear function, when \( \Delta x \) is constant (which it is in each table: 8, 8, 8 for first; 8, 8 for second; 5, 5 for third; 6, 6 for fourth), then \( \Delta y \) must be constant.

First table: \( \Delta y=3 - (-1)=4 \), \( 7 - 3=4 \). Constant. Linear.

Second table: \( \Delta y=-5 - (-7)=2 \), \( 9 - (-5)=14 \). Not constant. Not linear.

Third table: \( \Delta y=3 - 4=-1 \), \( 1 - 3=-2 \). Not constant. Not linear.

Fourth table: \( \Delta y=7 - 10=-3 \), \( 4 - 7=-3 \). Constant. Linear.

Wait, but the original image has checkmarks on all, but that's incorrect. So the correct tables are the first (orange) and fourth (blue), and wait, maybe I misread the second table's y-values. Wait, maybe the second table is (-2,-7), (6,-5), (14,9)? Wait, no, maybe the second table's y-values are different. Wait, maybe the user's table has a typo, but based on calculation:

First table (orange): linear.

Fourth table (blue): linear.

Wait, maybe the second table is (-2,-7), (6,-5), (14,3)? No, the given is 9. So perhaps the initial analysis was wrong. Wait, let's use slope formula for all pairs:

First table:
Points (-6,-1), (2,3), (10,7).

Slope between (-6,-1) and (2,3): (3 - (-1))/(2 - (-6))=4/8=0.5.

Slope between (2,3) and (10,7): (7 - 3)/(10 - 2)=4/8=0.5. Linear.

Second table:
Points (-2,-7), (6,-5), (14,9).

Slope between (-2,-7) and (6,-5): (-5 - (-7))/(6 - (-2))=2/8=0.25.

Slope between (6,-5) and (14,9): (9 - (-5))/(14 - 6)=14/8=1.75. Not linear.

Third table:
Points (0,4), (5,3), (10,1).

Slope between (0,4) and (5,3): (3 - 4)/(5 - 0)=-1/5=-0.2.

Slope between (5,3) and (10,1): (1 - 3)/(10 - 5)=-2/5=-0.4. Not linear.

Fourth table:
Points (8,10), (14,7), (20,4).

Slope between (8,10) and (14,7): (7 - 10)/(14 - 8)=-3/6=-0.5.

Slope between (14,7) and (20,4): (4 - 7)/(20 - 14)=-3/6=-0.5. Linear.

So the correct tables are the first (orange) and fourth (blue). But maybe the second table's y-value at x=14 is -3? No, the given is 9. So the answer should be the first and fourth tables. But the original checkmarks are on all, which is wrong. So based on calculation, the linear tables are the first (x: -6,2,10; y: -1,3,7) and the fourth (x:8,14,20; y:10,7,4).

Wait, maybe I made a mistake in the second table. Let's recheck:

Second table: x=-2, y=-7; x=6, y=-5; x=14, y=9.

Δx from -2 to 6: 8; Δy: -5 - (-7)=2.

Δx from 6 to 14: 8; Δy: 9 - (-5)=14. 2≠14, so not linear.

Third table: x=0, y=4; x=5, y=3; x=10, y=1.

Δx=5, Δy=3-4=-1; Δx=5, Δy=1-3=-2. -1≠-2, not linear.

So the correct tables are the first (orange) and fourth (blue).

Answer:

The first table (with \( x \) values -6, 2, 10 and \( y \) values -1, 3, 7) and the fourth table (with \( x \) values 8, 14, 20 and \( y \) values 10, 7, 4) could represent linear functions.