Respuesta :

Answer:

Table 2 is the right representation of exponential functio.

Step-by-step explanation:

We are given the following information in the question:

[tex]y = b^k,\\0<b<1[/tex]

If we look at table 2, then, the exponential function is of the form:

[tex]y(x) = \bigg(\displaystyle\frac{1}{3}\bigg)^x[/tex]

Putting different values of x, we have:

[tex]x = -3\\\\y(-3) = \bigg(\displaystyle\frac{1}{3}\bigg)^{-3} = 27\\\\x = -2\\\\y(-2) = \bigg(\displaystyle\frac{1}{3}\bigg)^{-3} = 9\\\\x = -1\\\\y(-1) = \bigg(\displaystyle\frac{1}{3}\bigg)^{-1} = 3\\\\x = -0\\\\y(0) = \bigg(\displaystyle\frac{1}{3}\bigg)^{0} = 1\\\\x = 1\\\\y(1) = \bigg(\displaystyle\frac{1}{3}\bigg)^{1} = \frac{1}{3}\\\\x = 2\\\\y(2) = \bigg(\displaystyle\frac{1}{3}\bigg)^{2} = \frac{1}{9}\\\\x = 3\\\\y(3) = \bigg(\displaystyle\frac{1}{3}\bigg)^{3} = \frac{1}{27}\\[/tex]

The exponential function given in the question is [tex]y=b^k[/tex] and the interval is given as [tex]0<b<1[/tex]. According to the obtained values one can conclude that: The table represents the exponential function [tex]y=b^k[/tex] is table-2.

Given information:

The exponential function [tex]y=b^k[/tex] where [tex]0<b<1[/tex]

If we see the table number 2,

Then exponential function is of the form [tex]y(x)=(1/3)^x[/tex]

putting the values of [tex]x[/tex] we have:

[tex]x=-3\\\\y(-3)=(1/3)^{-3}=27\\\\x=-2\\\\y(-2)=(1/3)^{-3}=9\\\\x=-1\\\\y(-1)=(1/3)^{-1}=3\\\\x=0\\\\y(0)=(1/3)^{0}=1\\\\x=1\\\\y(1)=(1/3)^{1}=(1/3)\\\\\\x=2\\\\y(2)=(1/3)^{2}=1/9\\\\x=3\\\\y(3)=(1/3)^{3}=1/27\\[/tex]

Hence, according to the obtained values one can conclude that:

The table represents the exponential function [tex]y=b^k[/tex] is table-2.

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