Question

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of values represents a linear function?

x	y
-2	2
-1	-2
1	-6

b

x	y
-1	1
3	-2
7	-5

x	y
0	2
2	1
4	-1

d

x	y
2	1
4	-1
6	-3

Explanation:

Step1: Recall linear function slope formula

For a linear function, the slope \( m=\frac{y_2 - y_1}{x_2 - x_1} \) should be constant between any two points \((x_1,y_1)\) and \((x_2,y_2)\).

Step2: Check Option (let's assume the first table is A, second B, third C, fourth D. Wait, the given tables: Let's label them as A (left top), B (right top), C (left bottom), D (right bottom). Wait, the user's image: first left table (A), right top (B), left bottom (C), right bottom (D). Let's check each:

Check Table A:

Points: \((-4,6)\), \((-2,2)\), \((-1,-2)\), \((1,-6)\)
Slope between \((-4,6)\) and \((-2,2)\): \( m_1=\frac{2 - 6}{-2 - (-4)}=\frac{-4}{2}=-2 \)
Slope between \((-2,2)\) and \((-1,-2)\): \( m_2=\frac{-2 - 2}{-1 - (-2)}=\frac{-4}{1}=-4 \)
\( m_1 eq m_2 \), so not linear.

Check Table B:

Points: \((-5,4)\), \((-1,1)\), \((3,-2)\), \((7,-5)\)
Slope between \((-5,4)\) and \((-1,1)\): \( m_1=\frac{1 - 4}{-1 - (-5)}=\frac{-3}{4}=-0.75 \)
Slope between \((-1,1)\) and \((3,-2)\): \( m_2=\frac{-2 - 1}{3 - (-1)}=\frac{-3}{4}=-0.75 \)
Slope between \((3,-2)\) and \((7,-5)\): \( m_3=\frac{-5 - (-2)}{7 - 3}=\frac{-3}{4}=-0.75 \)
Wait, but wait, let's check Table C and D. Wait, maybe I mislabeled. Wait the left bottom table (C) has points \((-2,3)\), \((0,2)\), \((2,1)\), \((4,-1)\)? No, left bottom: \((-2,3)\), \((0,2)\), \((2,1)\), \((4,-1)\)? Wait no, left bottom table: \( x=-2,y=3 \); \( x=0,y=2 \); \( x=2,y=1 \); \( x=4,y=-1 \). Let's check slope:

Check Table C:

Slope between \((-2,3)\) and \((0,2)\): \( m_1=\frac{2 - 3}{0 - (-2)}=\frac{-1}{2}=-0.5 \)
Slope between \((0,2)\) and \((2,1)\): \( m_2=\frac{1 - 2}{2 - 0}=\frac{-1}{2}=-0.5 \)
Slope between \((2,1)\) and \((4,-1)\): \( m_3=\frac{-1 - 1}{4 - 2}=\frac{-2}{2}=-1 \). Not constant.

Check Table D (right bottom):

Points: \((0,2)\), \((2,1)\), \((4,-1)\), \((6,-3)\)
Slope between \((0,2)\) and \((2,1)\): \( m_1=\frac{1 - 2}{2 - 0}=\frac{-1}{2}=-0.5 \)
Slope between \((2,1)\) and \((4,-1)\): \( m_2=\frac{-1 - 1}{4 - 2}=\frac{-2}{2}=-1 \). Not constant.

Wait, earlier Table B: Wait, let's recheck Table B points: \((-5,4)\), \((-1,1)\), \((3,-2)\), \((7,-5)\)
Slope between \((-5,4)\) and \((-1,1)\): \( \frac{1 - 4}{-1 - (-5)}=\frac{-3}{4}=-0.75 \)
Slope between \((-1,1)\) and \((3,-2)\): \( \frac{-2 - 1}{3 - (-1)}=\frac{-3}{4}=-0.75 \)
Slope between \((3,-2)\) and \((7,-5)\): \( \frac{-5 - (-2)}{7 - 3}=\frac{-3}{4}=-0.75 \). Wait, but maybe I made a mistake with Table C and D. Wait the left bottom table (C) is \((-2,3)\), \((0,2)\), \((2,1)\), \((4,-1)\)? No, looking at the image: left bottom table: \( x=-2,y=3 \); \( x=0,y=2 \); \( x=2,y=1 \); \( x=4,-1 \). Wait, the right bottom table (D) is \( x=0,y=2 \); \( x=2,y=1 \); \( x=4,-1 \); \( x=6,-3 \). Wait, maybe the correct one is Table B? Wait no, let's check Table C (left bottom) again. Wait, no, the user's problem: Let's check the table with \( x=-2,y=3 \); \( x=0,y=2 \); \( x=2,y=1 \); \( x=4,-1 \): no, slope changes. Wait the table with \( x=0,y=2 \); \( x=2,y=1 \); \( x=4,-1 \); \( x=6,-3 \): slope between (0,2) and (2,1) is -0.5, (2,1) and (4,-1) is -1, so no. Wait Table B: points (-5,4), (-1,1), (3,-2), (7,-5). The x - differences: from -5 to -1: +4; -1 to 3: +4; 3 to 7: +4. y - differences: 4 to 1: -3; 1 to -2: -3; -2 to -5: -3. So slope \( m=\frac{-3}{4} \) constant. Wait, but maybe the correct answer is Table B? Wait, no, let's check the other tables. Wait the first table (A): x from -4 to -2: +2, y from 6 to 2: -4 (slope -2); -2 to -1: +1, y from 2 to -2: -4 (slope -4). Not constant. Table C (left bottom): x from -2 to 0: +2, y from 3 to 2: -1 (slope -0.5); 0 to 2: +2, y from 2 to 1: -1 (slope -0.5); 2 to 4: +2, y from 1 to -1: -2 (slope -1). Not constant. Table D (right bottom): x from 0 to 2: +2, y from 2 to 1: -1 (slope -0.5); 2 to 4: +2, y from 1 to -1: -2 (slope -1); 4 to 6: +2, y from -1 to -3: -2 (slope -1). Wait, no, slope changes. Wait Table B: x increases by 4 each time (-5 to -1: 4, -1 to 3:4, 3 to7:4), y decreases by 3 each time (4 to1: -3, 1 to -2: -3, -2 to -5: -3). So slope is constant (\( m=\frac{-3}{4} \)), so it's linear. Wait, but maybe the correct table is the one with constant slope. Wait, maybe I mislabeled. Let's re-express:

Wait the four tables:

1. Top left (A): \( x: -4, -2, -1, 1 \); \( y:6,2,-2,-6 \)
2. Top right (B): \( x: -5, -1, 3, 7 \); \( y:4,1,-2,-5 \)
3. Bottom left (C): \( x: -2, 0, 2, 4 \); \( y:3,2,1,-1 \)
4. Bottom right (D): \( x:0,2,4,6 \); \( y:2,1,-1,-3 \)

Check Table B:

Between (-5,4) and (-1,1): \( m = (1 - 4)/(-1 - (-5)) = (-3)/4 = -3/4 \)

Between (-1,1) and (3,-2): \( m = (-2 - 1)/(3 - (-1)) = (-3)/4 = -3/4 \)

Between (3,-2) and (7,-5): \( m = (-5 - (-2))/(7 - 3) = (-3)/4 = -3/4 \)

So slope is constant. Now check Table C:

Between (-2,3) and (0,2): \( m=(2 - 3)/(0 - (-2)) = (-1)/2 = -1/2 \)

Between (0,2) and (2,1): \( m=(1 - 2)/(2 - 0) = (-1)/2 = -1/2 \)

Between (2,1) and (4,-1): \( m=(-1 - 1)/(4 - 2) = (-2)/2 = -1 \). Not constant.

Table D:

Between (0,2) and (2,1): \( m=(1 - 2)/(2 - 0) = -1/2 \)

Between (2,1) and (4,-1): \( m=(-1 - 1)/(4 - 2) = -2/2 = -1 \). Not constant.

Table A:

Between (-4,6) and (-2,2): \( m=(2 - 6)/(-2 - (-4)) = (-4)/2 = -2 \)

Between (-2,2) and (-1,-2): \( m=(-2 - 2)/(-1 - (-2)) = (-4)/1 = -4 \). Not constant.

So Table B has constant slope, so it's linear. Wait, but maybe the bottom left table (C) was misread. Wait no, the bottom left table's last y is -1, so slope changes. So the linear function is Table B (assuming the right top table is B). Wait, but maybe the user's labels: the problem says "which table of values represents a linear function?" So the correct one is the table with constant slope, which is Table B (the right top one) or maybe the bottom left? Wait no, our calculation shows Table B has constant slope. Wait, maybe I made a mistake with Table C. Wait Table C: x from -2 to 0: +2, y from 3 to 2: -1 (slope -0.5); 0 to 2: +2, y from 2 to 1: -1 (slope -0.5); 2 to 4: +2, y from 1 to -1: -2 (slope -1). So slope changes. Table B: x increases by 4, y decreases by 3, so slope constant. So the answer is the table with x: -5, -1, 3, 7 and y:4,1,-2,-5 (Table B). Wait, but maybe the options are labeled as B. So the correct table is the one with constant slope, which is the top right table (B).

Answer:

The table with \( x \) values \(-5, -1, 3, 7\) and \( y \) values \(4, 1, -2, -5\) (labeled as B) represents a linear function.