Respuesta :

Answer:


Step-by-step explanation:

The question is faulty. It works once it gets to + 1, +2,  +3

The other x values should be 0 and - 1 instead of -1 and -3.

Here's the formula I get.

f(x) = 2*3^(2x + 1)

f(0) = 2*3^(1)

f(0)= 6

f(-1) = 2*3^(-2 + 1)

f(-1) = 2*3^(-1)

f(-1) = 2/3

f(-2) = 2*3^(-2*2 + 1)

f(-2) = 2*3^(-4 +1)

f(-2) = 2*3^(-3)

f(-2) = 2/27

Since my answers do not agree with the given answers, I can do nothing more to help you. I will pass this along to someone I know who will see through it.

Answer:

The base of the exponential is 3, which is the factor by which output increases when input increases by 1.

Step-by-step explanation:

You are looking for "b" in ...

... f(x) = a·b^x

Filling in values from the table, you have

... 2/3 = a·b^(-3)

... 6 = a·b^(-1)

Dividing the latter by the former gives ...

... 6/(2/3) = b^(-1)/b^(-3)

... 9 = b^2

... 9^(1/2) = b = 3 . . . . . the value you're looking for

_____

In case you're interested, "a" can be found from any of the table values. For example,

... 6 = a·3^(-1)

... 6·3 = a = 18

Then ...

... f(x) = 18·3^x

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