Question

which table shows exponential decay?

first table:

x	y
2	8
3	4
4	2

second table:

x	y
2	12
3	8
4	4

third table (partially shown):

x	y
2	42	(partially visible)

Explanation:

Step1: Recall exponential decay property

Exponential decay has a constant ratio between consecutive \( y \)-values (when \( x \) increases by 1). For a function \( y = ab^x \), the common ratio \( r=\frac{y_{n + 1}}{y_n}\) should be between 0 and 1 (since it's decay).

Step2: Analyze first table

For the first table:

• When \( x = 1 \), \( y = 16 \); \( x = 2 \), \( y = 8 \). Ratio: \(\frac{8}{16}=\frac{1}{2}\)
• \( x = 2 \), \( y = 8 \); \( x = 3 \), \( y = 4 \). Ratio: \(\frac{4}{8}=\frac{1}{2}\)
• \( x = 3 \), \( y = 4 \); \( x = 4 \), \( y = 2 \). Ratio: \(\frac{2}{4}=\frac{1}{2}\)

The ratio is constant (\(\frac{1}{2}\)) and between 0 and 1.

Step3: Analyze second table

For the second table:

• \( x = 1 \), \( y = 16 \); \( x = 2 \), \( y = 12 \). Ratio: \(\frac{12}{16}=\frac{3}{4}\)
• \( x = 2 \), \( y = 12 \); \( x = 3 \), \( y = 8 \). Ratio: \(\frac{8}{12}=\frac{2}{3}\)
• \( x = 3 \), \( y = 8 \); \( x = 4 \), \( y = 4 \). Ratio: \(\frac{4}{8}=\frac{1}{2}\)

The ratios are not constant, so it's linear or other, not exponential decay.

(Assuming the third table has a typo or is incomplete, but from given, first table has constant ratio for decay.)

Answer:

The first table (with \( x \): 1,2,3,4 and \( y \): 16,8,4,2) shows exponential decay.