Question

1. which of these tables represents a linear function? a. input, x 0 1 4 9 16 output, y 0 1 2 3 4 b. input, x -2 -1 0 1 2 output, y 4 1 0 1 4 c. input, x -2 -1 0 1 2 output, y -15 -9 -3 3 9 d. input, x -2 -1 0 1 2 output, y 8 1 0 -1 -8 permission is granted to reproduce for classroom use.

Explanation:

Step1: Recall linear function property

A linear function has a constant rate of change, i.e., the difference in \( y \)-values (output) divided by the difference in \( x \)-values (input) is constant (slope \( m=\frac{\Delta y}{\Delta x} \) is constant).

Step2: Analyze Option A

For A: \( x \) values: \( 0,1,4,9,16 \) (differences: \( 1 - 0 = 1 \), \( 4 - 1 = 3 \), \( 9 - 4 = 5 \), \( 16 - 9 = 7 \) – not constant). \( y \) values: \( 0,1,2,3,4 \) (differences: \( 1,1,1,1 \)). But since \( \Delta x \) is not constant, we check \( \frac{\Delta y}{\Delta x} \). From \( x=0 \) to \( x=1 \): \( \frac{1 - 0}{1 - 0}=1 \); \( x=1 \) to \( x=4 \): \( \frac{2 - 1}{4 - 1}=\frac{1}{3} \) – not constant. So A is not linear.

Step3: Analyze Option B

For B: \( x \) values: \( -2,-1,0,1,2 \) (differences: \( 1 \) each). \( y \) values: \( 4,1,0,1,4 \). Differences: \( 1 - 4 = -3 \), \( 0 - 1 = -1 \), \( 1 - 0 = 1 \), \( 4 - 1 = 3 \) – not constant. So B is not linear (it's quadratic, \( y = x^2 \)).

Step4: Analyze Option C

For C: \( x \) values: \( -2,-1,0,1,2 \) (differences: \( 1 \) each). \( y \) values: \( -15,-9,-3,3,9 \). Differences: \( -9 - (-15)=6 \), \( -3 - (-9)=6 \), \( 3 - (-3)=6 \), \( 9 - 3 = 6 \) – constant difference (slope \( m = \frac{6}{1}=6 \)). So constant rate of change.

Step5: Analyze Option D

For D: \( x \) values: \( -2,-1,0,1,2 \) (differences: \( 1 \) each). \( y \) values: \( 8,1,0,-1,-8 \). Differences: \( 1 - 8 = -7 \), \( 0 - 1 = -1 \), \( -1 - 0 = -1 \), \( -8 - (-1)= -7 \) – not constant. So D is not linear.

Answer:

C. Input, x: -2, -1, 0, 1, 2; Output, y: -15, -9, -3, 3, 9