PLEASE HELP ME WITH THIS....

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