Question

1. the personnel expenses of a small city were $8.5 million in 2014. over the previous years, the personnel expenses increased at a rate of 3.2% annually.

a. use the given percentage growth rate to determine the growth factor.
b. assuming that this rate continues, write a function to express personnel costs c(t) in millions of dollars, where t represents years since 2014.
c. use your function to approximate the personnel costs in the year 2020.

1. the average tuition at american private universities in 2000 was $16,072. in 2005, it was $21,235. assuming the tuition costs are increasing at a constant percentage rate, create an exponential function t(n) that models the average tuition, t, as a function of the number of years since 2000, n. round to three decimal places as needed.
2. based on data from 2005 to 2012, the population of clarksville, tennessee can be approximated by the function: p(t)=113.4(1.036)^t where t represents the number of years since 2005 and p(t) represents the population in thousands.

a. does this function show exponential growth or decay? how can you tell?
b. state the coordinates of the vertical intercept and describe its practical meaning in this situation.
c. state the change factor and interpret its practical meaning (by determining the percentage rate at which the population is changing annually).

Explanation:

Step1: Find growth factor for personnel expenses

The growth - rate formula for a percentage $r$ is $b = 1+\frac{r}{100}$. Given $r = 3.2$, then $b=1+\frac{3.2}{100}=1.032$.

Step2: Write personnel cost function

The general form of an exponential - growth function is $C(t)=C_0b^t$, where $C_0$ is the initial amount and $b$ is the growth factor. Here, $C_0 = 8.5$ and $b = 1.032$, so $C(t)=8.5(1.032)^t$.

Step3: Calculate personnel costs in 2020

Since $t$ represents years since 2014, for 2020, $t = 2020 - 2014=6$. Then $C(6)=8.5(1.032)^6$. Using a calculator, $(1.032)^6\approx1.2056$, and $C(6)=8.5\times1.2056\approx10.248$ million dollars.

Step4: Find exponential function for tuition

The general form of an exponential function is $T(n)=T_0b^n$. We know $T_0 = 16072$ (tuition in 2000). For $n = 2005 - 2000 = 5$, $T(5)=21235$. Substituting into $T(n)=T_0b^n$ gives $21235 = 16072b^5$. Then $b^5=\frac{21235}{16072}\approx1.3212$. So, $b=\sqrt[5]{1.3212}\approx1.056$. The function is $T(n)=16072(1.056)^n$.

Step5: Analyze population function for Clarksville

• a. Determine growth or decay

The general form of an exponential function is $P(t)=P_0b^t$. In the function $P(t)=113.4(1.036)^t$, since $b = 1.036>1$, it shows exponential growth.

• b. Find vertical intercept

For an exponential function $y = ab^x$, the vertical intercept is found when $x = 0$. When $t = 0$ in $P(t)=113.4(1.036)^t$, $P(0)=113.4$. The coordinates of the vertical intercept are $(0,113.4)$. In 2005 (when $t = 0$), the population of Clarksville was 113400 people (since $P(t)$ is in thousands).

• c. Find change factor and percentage rate

The change factor is $b = 1.036$. Using the formula $b = 1+\frac{r}{100}$, where $r$ is the percentage rate of change. Solving $1.036=1+\frac{r}{100}$ gives $r = 3.6\%$. The population of Clarksville increases by $3.6\%$ each year.

Answer:

1.

• a. Growth factor: $1.032$
• b. Function: $C(t)=8.5(1.032)^t$
• c. Personnel costs in 2020: $\approx10.248$ million dollars

1. Exponential function for tuition: $T(n)=16072(1.056)^n$

3.

• a. Exponential growth because the base of the exponential function $b = 1.036>1$.
• b. Vertical intercept: $(0,113.4)$. In 2005, the population was 113400 people.
• c. Change factor: $1.036$. The population increases by $3.6\%$ each year.