Question
Graph Points: 3 500, 3 220, 2 962.4, 2 725.4, 2 507.38, 2 306.78.
The S&P stock Index began the week at 3500, but fell by an average of 8% each day. Write an exponential function that models the data.
y= ___
The S&P stock Index began the week at 3500, but fell by an average of 8% each day. Write an exponential function that models the data.
y= ___
Answers
GPT-4o mini
To model the data given with an exponential function, we can use the general form of an exponential decay function:
\[ y = a \cdot e^{-kt} \]
where:
- \( y \) is the value at time \( t \),
- \( a \) is the initial amount (the starting value),
- \( k \) is the decay constant,
- \( t \) is time in days,
- \( e \) is the base of the natural logarithm.
In this case, we know the following:
- The initial value \( a = 3500 \) (the stock index at the beginning of the week).
- The stock falls by an average of 8% each day.
To express the daily decay rate in the exponential form, we need to calculate the decay constant \( k \). If the stock falls by 8% each day, then at the end of the first day, the value of \( y \) will be:
\[ y = 3500 \cdot (1 - 0.08) = 3500 \cdot 0.92 \]
For the second day, it will be:
\[ y = 3500 \cdot 0.92^2 \]
This gives us a general form where the decay is multiplicative:
General form for day \( t \):
\[ y = 3500 \cdot (0.92)^t \]
Now, converting the decay form into exponential form:
Since \( 0.92 \) can also be expressed as \( e^{-k} \), where \( k \) would need to be determined based on \( 0.92 = e^{-k} \).
Calculating \( k \):
\[
0.92 = e^{-k} \\
\ln(0.92) = -k \\
k = -\ln(0.92) \approx 0.0834
\]
Putting this back into our function gives:
\[
y = 3500 \cdot e^{-0.0834t}
\]
So the final exponential function that models the data is:
\[
y = 3500 \cdot e^{-0.0834t}
\]
Thus, the exponential function \( y \) can be written as:
\[
y = 3500 \cdot (0.92)^t
\]
You can use either form, but \( y = 3500 \cdot (0.92)^t \) may be simpler for practical calculations.
\[ y = a \cdot e^{-kt} \]
where:
- \( y \) is the value at time \( t \),
- \( a \) is the initial amount (the starting value),
- \( k \) is the decay constant,
- \( t \) is time in days,
- \( e \) is the base of the natural logarithm.
In this case, we know the following:
- The initial value \( a = 3500 \) (the stock index at the beginning of the week).
- The stock falls by an average of 8% each day.
To express the daily decay rate in the exponential form, we need to calculate the decay constant \( k \). If the stock falls by 8% each day, then at the end of the first day, the value of \( y \) will be:
\[ y = 3500 \cdot (1 - 0.08) = 3500 \cdot 0.92 \]
For the second day, it will be:
\[ y = 3500 \cdot 0.92^2 \]
This gives us a general form where the decay is multiplicative:
General form for day \( t \):
\[ y = 3500 \cdot (0.92)^t \]
Now, converting the decay form into exponential form:
Since \( 0.92 \) can also be expressed as \( e^{-k} \), where \( k \) would need to be determined based on \( 0.92 = e^{-k} \).
Calculating \( k \):
\[
0.92 = e^{-k} \\
\ln(0.92) = -k \\
k = -\ln(0.92) \approx 0.0834
\]
Putting this back into our function gives:
\[
y = 3500 \cdot e^{-0.0834t}
\]
So the final exponential function that models the data is:
\[
y = 3500 \cdot e^{-0.0834t}
\]
Thus, the exponential function \( y \) can be written as:
\[
y = 3500 \cdot (0.92)^t
\]
You can use either form, but \( y = 3500 \cdot (0.92)^t \) may be simpler for practical calculations.