Question

in which table is y a non - linear function of x?
a.

x	y
-2	2.0
3	4.5
4	5.0

b.

x	y
-1	2
0	-1
4	-13

c.

x	y
-3	-5.5
-1	-2.5
5	6.5

d.

x	y
-4	12
-2	8
1	4

Explanation:

Step1: Recall linear function property

A linear function has a constant rate of change (slope), calculated as \(m = \frac{y_2 - y_1}{x_2 - x_1}\) for any two points \((x_1,y_1)\) and \((x_2,y_2)\). We'll check the slope between consecutive (or any) points in each table.

Step2: Check Table A

Points: \((-2, 2.0)\), \((3, 4.5)\), \((4, 5.0)\)
Slope between \((-2,2.0)\) and \((3,4.5)\): \(m_1=\frac{4.5 - 2.0}{3 - (-2)}=\frac{2.5}{5}=0.5\)
Slope between \((3,4.5)\) and \((4,5.0)\): \(m_2=\frac{5.0 - 4.5}{4 - 3}=\frac{0.5}{1}=0.5\)
Constant slope, so linear.

Step3: Check Table B

Points: \((-1, 2)\), \((0, -1)\), \((4, -13)\)
Slope between \((-1,2)\) and \((0,-1)\): \(m_1=\frac{-1 - 2}{0 - (-1)}=\frac{-3}{1}=-3\)
Slope between \((0,-1)\) and \((4,-13)\): \(m_2=\frac{-13 - (-1)}{4 - 0}=\frac{-12}{4}=-3\)
Constant slope, so linear.

Step4: Check Table C

Points: \((-3, -5.5)\), \((-1, -2.5)\), \((5, 6.5)\)
Slope between \((-3,-5.5)\) and \((-1,-2.5)\): \(m_1=\frac{-2.5 - (-5.5)}{-1 - (-3)}=\frac{3}{2}=1.5\)
Slope between \((-1,-2.5)\) and \((5,6.5)\): \(m_2=\frac{6.5 - (-2.5)}{5 - (-1)}=\frac{9}{6}=1.5\)
Constant slope, so linear.

Step5: Check Table D

Points: \((-4, 12)\), \((-2, 8)\), \((1, 4)\)
Slope between \((-4,12)\) and \((-2,8)\): \(m_1=\frac{8 - 12}{-2 - (-4)}=\frac{-4}{2}=-2\)
Slope between \((-2,8)\) and \((1,4)\): \(m_2=\frac{4 - 8}{1 - (-2)}=\frac{-4}{3}\approx -1.33\)
Slopes are not equal, so non - linear.

Answer:

D. The table with \(x\) values \(-4, -2, 1\) and \(y\) values \(12, 8, 4\) (i.e., Table D)