What would the "starting amount" be in these functions?

[bossbasslol]

Here are two functions: (they resemble growth and decay functions)

f(x) = 2*(1/8)^x and h(x) = -7*(1.567)^x-5

I heard that the first digit is the starting amount, such as 2 in f(x) and -7 in h(x).

This is proven wrong when I input h(0) into the second function. The reason I input 0 is because it is logically the starting amount of x, but this assumption immediately contradicts h(x)'s starting amount of "-7" if solved for x=0. The answer? -0.74.

Which claim should I follow? The "first-digit-is-the-starting-amount," claim or my "zero-is-the-logical-start-point," claim?

 

[Debsta]

Here are two functions: (they resemble growth and decay functions)

f(x) = 2*(1/8)^x and h(x) = -7*(1.567)^x-5

I heard that the first digit is the starting amount, such as 2 in f(x) and -7 in h(x).

This is proven wrong when I input h(0) into the second function. The reason I input 0 is because it is logically the starting amount of x, but this assumption immediately contradicts h(x)'s starting amount of "-7" if solved for x=0. The answer? -0.74.

Which claim should I follow? The "first-digit-is-the-starting-amount," claim or my "zero-is-the-logical-start-point," claim?

"The first digit is the starting amount" claim only works when the exponent is x (or a multiple of x) as in f(x)=2*(1/8)^x. When you substitute in x=0, you get 2*(1/8)^0 = 2*1 = 2.
The reason you get the starting digit is because (1/8)^0 =1 (anything raised to the power of 0 is equal to 1 (except 0 itself)) and so you are left with the 2.

But if the power is x-5 as in g(x), you won't get that 1 (when you substitute x=0) to leave you with "the first digit".

Don't learn a "rule" if you don't understand it. Always put in x=0 for "starting value" (ie if x is representing time).