1. **State the problem:** We need to determine which of the given tables represents a nonlinear function. 2. **Recall the definition:** A function is linear if the rate of change (difference in y divided by difference in x) is constant between all points. 3. **Check each table:** - Table A: Differences in x: $4-2=2$, $6-4=2$, $8-6=2$ Differences in y: $10-6=4$, $14-10=4$, $18-14=4$ Rate of change: $\frac{4}{2}=2$ for all intervals, so linear. - Table B: Differences in x: $3-2=1$, $4-3=1$, $5-4=1$ Differences in y: $10-6=4$, $14-10=4$, $18-14=4$ Rate of change: $\frac{4}{1}=4$ constant, so linear. - Table C: Differences in x: $4-2=2$, $6-4=2$, $8-6=2$ Differences in y: $2-6=-4$, $-2-2=-4$, $-6-(-2)=-4$ Rate of change: $\frac{-4}{2}=-2$ constant, so linear. - Table D: Differences in x: $3-2=1$, $4-3=1$, $5-4=1$ Differences in y: $2-8=-6$, $0-2=-2$, $-6-0=-6$ Rate of change: $\frac{-6}{1}=-6$, then $\frac{-2}{1}=-2$, then $\frac{-6}{1}=-6$ Rate of change is not constant, so nonlinear. 4. **Conclusion:** Table D represents a nonlinear function. **Final answer:** Table D is nonlinear.