Question

1. at a local college, the enrollment can be modeled by the function ( s(t)=12.4(0.95)^{t} ) where ( s(t) ) represents the total enrollment (in thousands) and ( t ) is the number of years since 1994.

a. is the function increasing or decreasing? explain how you can determine this from the functions equation. ( downarrow ).95 < 1
b. from the equation, determine the vertical intercept. what practical meaning does this intercept have in this situation?
c. determine the growth or decay rate. what is the practical meaning of this rate in this situation?
d. according to the model, what was the enrollment at the college in the year 2000?
e. if the trend continues, in what year will the enrollment first dip below 7,500? use a table to obtain your answer.

Explanation:

Step1: Determine if function is increasing or decreasing

The general form of an exponential function is \( S(t)=a\cdot b^{t} \). If \( 0 < b< 1 \), the function is decreasing. Here \( a = 12.4 \) and \( b=0.95 \). Since \( 0.95<1 \), the function \( S(t)=12.4(0.95)^{t} \) is decreasing.

Step2: Find the vertical intercept

The vertical intercept occurs when \( t = 0 \). Substitute \( t = 0 \) into \( S(t)=12.4(0.95)^{t} \). Using the property \( a^{0}=1\) (\(a
eq0\)), we get \( S(0)=12.4(0.95)^{0}=12.4\times1 = 12.4 \). In this context, when \( t = 0 \) (the year 1994), the enrollment is \( 12.4\times1000=12400\) students.

Step3: Find the growth/decay rate

The general form of an exponential function is \( S(t)=a(1 + r)^{t}\) (for growth, \(r>0\)) or \( S(t)=a(1 - r)^{t}\) (for decay, \(0

Step4: Calculate enrollment in 2000

The year 2000 is \(t=2000 - 1994=6\) years after 1994. Substitute \(t = 6\) into \(S(t)=12.4(0.95)^{t}\). Then \(S(6)=12.4\times(0.95)^{6}\).

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The enrollment is approximately \(9.115\times1000 = 9115\) students.

Step5: Find the year enrollment dips below 7500

We want to find \(t\) when \(S(t)=12.4(0.95)^{t}<7.5\).

\(t\)	\(S(t)=12.4(0.95)^{t}\)
\(t = 7\)	\(12.4\times(0.95)^{7}=12.4\times0.95\times(0.95)^{6}\approx12.4\times0.95\times0.7350919\approx8.659\)
\(t = 8\)	\(12.4\times(0.95)^{8}\approx12.4\times0.6634204\approx8.226\)
\(t = 9\)	\(12.4\times(0.95)^{9}\approx12.4\times0.6302494\approx7.815\)
\(t = 10\)	\(12.4\times(0.95)^{10}\approx12.4\times0.5987369\approx7.424\)

Since \(t\) is the number of years since 1994, the year is \(1994 + 10=2004\)

Answer:

a. The function is decreasing. Since \(0.95<1\) in \(S(t)=12.4(0.95)^{t}\) (exponential decay form).
b. The vertical intercept is \((0,12.4)\). In 1994, the enrollment was 12400 students.
c. The decay rate is \(5\%\). Enrollment decreases by \(5\%\) each year.
d. Approximately 9115 students.
e. 2004.