Respuesta :
The option are missing in the question. The options are :
A. P = 2, a = 1
B. [tex]$P=\frac{1}{2} ; a =\frac{1}{3}$[/tex]
C. [tex]$P=\frac{1}{2} ; a =1$[/tex]
D. P = 2, a = 3
Solution :
The given function is [tex]$f(x)= Pa^x$[/tex]
So for the function to be an exponential growth, a should be a positive number and should be larger than 1. If it less than 1 or a fraction, then it is a decay. If the value of a is negative, then it would be between positive and negative alternately.
When the four option being substituted in the function, we get
A). It is a constant function since [tex]$2(1^x)=2$[/tex]
B). Here, the value of a is a fraction which is less than 1, so it is a decay function. [tex]$f(x)=\frac{1}{2}\left(\frac{1}{3}\right)^x$[/tex]
C). It is a constant function since the value of a is 1.
D). Here a = 3. So substituting, as the value of x increases by 1, the value of the function, f(x) increases by 3 times.
[tex]$f(x)=2(3)^x$[/tex]
Therefore, option (D). represents an exponential function.
Answer:
Its P = 1/5 ; a = 2
Step-by-step explanation:
I am doing the test; I got it right.
