Question

write a recursive rule for the exponential function. (see example 3) 28. y = 19(4)^t

Explanation:

Step1: Recall recursive rule for exponential

For an exponential function \( y = a(b)^t \), the recursive rule is \( a_1 = a \), \( a_{t + 1}=b\cdot a_t \) (assuming \( t \) starts at 1, and \( a_t \) represents the value at time \( t \)).
Here, \( a = 19 \), \( b = 4 \).

Step2: Define initial term and recursive relation

• Initial term: When \( t = 1 \), \( y(1)=19(4)^1 = 19\times4=76 \)? Wait, no. Wait, actually, for the function \( y = 19(4)^t \), when \( t = 0 \), \( y(0)=19 \). But in recursive terms, usually we define \( a_1 \) as the first term (when \( t = 1 \), \( y(1)=19\times4 = 76 \)) and then \( a_{t+1}=4\cdot a_t \). Wait, let's correct. The general form of a recursive rule for an exponential function \( y = a\cdot r^t \) (where \( r \) is the common ratio) is:

Let \( a_1 \) be the value when \( t = 1 \), so \( a_1=a\cdot r \), and then \( a_{n + 1}=r\cdot a_n \) for \( n\geq1 \). Alternatively, if we take \( t \) starting at 0, then \( a_0=a \), and \( a_{t+1}=r\cdot a_t \).

Given \( y = 19(4)^t \), let's take \( t \) as a non - negative integer (starting at 0). Then:

• Initial condition: When \( t = 0 \), \( y(0)=19(4)^0=19 \). Let \( a_t=y(t) \). Then \( a_0 = 19 \).
• Recursive relation: \( a_{t + 1}=4\cdot a_t \), because \( y(t + 1)=19(4)^{t+1}=4\times19(4)^t = 4\cdot y(t) \), so \( a_{t+1}=4a_t \).

Alternatively, if we take \( t \) starting at 1:

• When \( t = 1 \), \( y(1)=19(4)^1 = 76 \), so \( a_1 = 76 \).
• And \( a_{t+1}=4\cdot a_t \), since \( y(t + 1)=19(4)^{t + 1}=4\times19(4)^t=4\cdot y(t) \).

The standard recursive rule for an exponential function \( y = a(b)^t \) (with \( t\in\mathbb{N}_0=\{0,1,2,\cdots\} \)) is:
\( a_0 = 19 \) (the initial term when \( t = 0 \)) and \( a_{t+1}=4\cdot a_t \) for \( t\geq0 \).
Or if we consider \( t \) starting at 1, \( a_1=19\times4 = 76 \) and \( a_{t + 1}=4\cdot a_t \) for \( t\geq1 \).

Let's verify. If \( a_0 = 19 \), then \( a_1=4\times a_0=4\times19 = 76 \), and \( y(1)=19(4)^1 = 76 \). \( a_2=4\times a_1=4\times76 = 304 \), and \( y(2)=19(4)^2=19\times16 = 304 \). So it works.

So the recursive rule is:
Initial term: \( a_0 = 19 \) (or \( a_1=76 \)) and the recursive formula \( a_{n+1}=4a_n \) (where \( n \) is the index, corresponding to \( t=n \) or \( t=n - 1 \) depending on the initial term).

Answer:

The recursive rule for the exponential function \( y = 19(4)^t \) is \( a_0=19 \) and \( a_{t + 1}=4a_t \) for \( t\geq0 \) (or \( a_1 = 76 \) and \( a_{n+1}=4a_n \) for \( n\geq1 \)). A common way to write it is \( \boldsymbol{a_1 = 76} \) (if we start at \( t = 1 \)) and \( \boldsymbol{a_{n+1}=4a_n} \) for \( n\geq1 \), or \( \boldsymbol{a_0 = 19} \) and \( \boldsymbol{a_{t+1}=4a_t} \) for \( t\geq0 \).