Given: 35(0.57^x) Let A=35, a=0.57, and the given equation be y. So, y=A(a^x) ··· Eq.1
1. Growth Factor: Since A>0, and 0<a<1, the limit of y,as x approaches -infinity is +infinity, e.g. 0.57^0=1,and 0.57^-100=approx. 2.6(10^24). And the limit of y,as x approaches +infinity, is 0 from above, e.g. 0.57^100=3.9/(10^25). So, y exponentialy decreases as x increases.
Therefore, the growth factor is: a=0.57 (0<a<1)
2. Rate of Growth: The derivative is the instantaneous rate of change of a function: variable y to variable x. To find the rate, we differentiate y with respect to x. Use a rule of derivative operation and an identity such as {fc(x)}'=cf'(x), and (a^x)'=(a^x)ln(a). From Eq.1, we have: y'=A(a^x)ln(a) Here, ln(a) is a constant, so the equation above is restated: y'=A·ln(a)·a^x ··· Eq.2
Plug the values of A and a into Eq.2. We have: y'=35(ln0.57)(0.57^x) Since ln0.57<0 (0.57<e) and 0.57^x >0, then y'<0
Therefore, the rate of growth is: y'=35(ln0.75)(0.57^x), and y'<0 in the interval: -infinity<x<+infinity.
3. Initial Value: From Eq.2, y''=A{(In(a))^2}a^x=35{(ln0.57)^2}(0.57^x) Here,(ln0.57)^2>0, so y''>0. Since y'<0 and y''>0, the graph of y concavely decreases downwards as x increases from -infinity to +infinity.
Therefore, 35(0.57^x) doesn't have the initial value, but the limits: +infinity and 0 from above. It has its place value for any x.