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Difficulty Level: At Grade Created by: CK-12

Andrea invested money into a mutual fund. Every three months she received a bank statement indicating the growth of her investment. Andrea recorded the following data from her bank statements:

\begin{align*}&\text{Time} \ (months) \ \ 0 \qquad \ 3 \qquad \quad 6 \qquad \qquad 9 \qquad \quad 12 \qquad \quad \ 15\\ &\text{Value (\)} \qquad \ 1200 \quad 1224 \quad 1248.48 \quad 1273.45 \quad 1298.92 \quad 1324.90\end{align*}

i) How much was the initial investment?

ii) What was the rate of interest the bank applied to her investment? Express this as a percent.

iii) What is the exponential function that best models the value of Andrea’s investment?

iv) What will be the mutual fund’s value in two years?

v) Can you use technology to determine the length of time it will take for the mutual fund to double in value?

### Guidance

You have learned of exponential functions of the form \begin{align*}y=ab^x\end{align*} in which the \begin{align*}x\end{align*}-value increased by increments of one. However, it often occurs that data is collected using other intervals such as every three months, every two hours, every fifteen minutes, and so forth. You still want to obtain the exponential function to represent such a situation, but the function must be modified to accommodate this interval change.

The exponential function will now be expressed in the form \begin{align*}y=a(b)^{\frac{x}{c}}\end{align*} where:

• \begin{align*}a\end{align*}’ represents the initial value (y-intercept)
• \begin{align*}b\end{align*}’ represents the common ratio (the rate of growth or decay)
• \begin{align*}c\end{align*}’ represents the increment in value of \begin{align*}x\end{align*}.

With exponential functions of the form \begin{align*}y=ab^x\end{align*}, the horizontal asymptote was the \begin{align*}x\end{align*}-axis. A horizontal asymptote is a horizontal line that a function keeps getting closer and closer to. Look at the graph of the function \begin{align*}y=2^x\end{align*}. The growth curve approaches the \begin{align*}x\end{align*}-axis, but it will never touch the axis. The equation of the horizontal asymptote is \begin{align*}\boxed{y=0}\end{align*}.

Now look at the graph of the function \begin{align*}y=2^x+1\end{align*}. The growth curve approaches the line \begin{align*}y=1\end{align*} but it will never touch the line. The equation of the horizontal asymptote is \begin{align*}\boxed{y=1}\end{align*}.

For exponential functions of the form \begin{align*}y=a(b)^{\frac{x}{c}}+d\end{align*}:

• \begin{align*}a+d\end{align*}’ represents the initial value (y-intercept)
• \begin{align*}b\end{align*}’ represents the common ratio (the rate of growth or decay)
• \begin{align*}c\end{align*}’ represents the increment in value of \begin{align*}x\end{align*}.
• \begin{align*}y=d\end{align*}’ represents the horizontal asymptote.

#### Example A

Write an exponential function of the form \begin{align*}y=a(b)^{\frac{x}{c}}\end{align*} to describe the table of values.

\begin{align*}&X \qquad 0 \qquad 2 \qquad 4 \qquad 6 \quad \ 8 \quad \ \ 10\\ &Y \qquad 3 \qquad 6 \quad \ \ 12 \quad \ \ 24 \quad 48 \quad \ 96\end{align*}

Solution:

The initial value ‘\begin{align*}a\end{align*}’ is 3. The common ratio is:

\begin{align*}& \boxed{r=\frac{t_{n+1}}{t_n}=\frac{6}{3}=2} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{12}{6}=2} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{24}{12}=2}\\ &\boxed{r=\frac{t_{n+1}}{t_n}=\frac{48}{24}=2} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{96}{48}=2}\end{align*}

The increment in the \begin{align*}x\end{align*}-value is 2. The exponential function to describe the table of values is \begin{align*}\boxed{y=3(2)^{\frac{x}{2}}}\end{align*}.

#### Example B

A radioactive isotope decays exponentially over time according to the equation \begin{align*}A(t)=42\left(\frac{1}{2}\right)^{\frac{t}{20}}\end{align*}, where \begin{align*}A(t)\end{align*} is the amount of the isotope present at time \begin{align*}t\end{align*}, in days.

i) What is the half-life of the isotope?

ii) Explain in words what the equation represents.

iii) How much isotope will remain after 35 days?

Solutions:

i) The half life of the isotope is the increment in the \begin{align*}x\end{align*}-value which is ‘\begin{align*}c\end{align*}’. The half life is 20 days.

ii) The equation represents an isotope that has a half-life of 20 days. This means that over a period of 20 days the amount of the isotope remaining will be one-half of what it was originally.

\begin{align*}A(t)=42\left(\frac{1}{2}\right)^{\frac{t}{20}}\end{align*}

The equation represents the amount \begin{align*}(A)\end{align*} of a decaying isotope which remains at any time \begin{align*}x\end{align*}, in days. There are 42 units initially present and its half-life is 20 days.

iii) \begin{align*}A(t)&=42\left(\frac{1}{2}\right)^{\frac{t}{20}}\\ A(t)&=42\left(\frac{1}{2}\right)^{\frac{35}{20}}=12.5 \ units\end{align*}

#### Example C

Brian bought a cup of coffee from the cafeteria and placed it on the ground while recording its change in temperature (the temperature that day was \begin{align*}0^{\circ} C\end{align*}). The temperature of the coffee was recorded every 3 minutes, and the results are shown in the following table:

\begin{align*}&\text{Time } (min) \qquad 0 \qquad \ 3 \qquad \ 6 \qquad \ 9 \quad \ \ 12 \quad \ \ 15\\ &\text{Temp }(0^{\circ} C) \quad 90.0 \quad 81.0 \quad 72.9 \quad 65.6 \quad 59.0 \quad 53.1\end{align*}

i) Write an exponential function to describe the temperature of the coffee after ‘\begin{align*}t\end{align*}’ minutes.

ii) Determine the temperature of the coffee after 30 minutes.

iii) Determine the temperature of the coffee after I hour.

Solutions:

i) The initial value ‘\begin{align*}a\end{align*}’ is 90.

The common ratio is

\begin{align*}&\boxed{r=\frac{81}{90}=0.9} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{72.9}{81}=0.9} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{65.6}{72.9}=0.899=0.9}\\ &\boxed{r=\frac{t_{n+1}}{t_n}=\frac{59.0}{65.6}=0.899=0.9} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{53.1}{59}=0.9}\end{align*}

The increment in the \begin{align*}x\end{align*}-value is 3.

The exponential function to describe the table of values is \begin{align*}\boxed{y=90(0.9)^{\frac{x}{3}}}\end{align*}.

ii) Replace ‘\begin{align*}x\end{align*}’ in the exponent with ‘30’ and calculate the value of the function. \begin{align*}y=90(0.9)^{\frac{30}{3}}=31.4^{\circ} C\end{align*}

iii) One hour equals 60 minutes. Replace ‘\begin{align*}x\end{align*}’ in the exponent with ‘60’ and calculate the value of the function. \begin{align*}y=90(0.9)^{\frac{60}{3}}=10.9^{\circ} C\end{align*}

#### Example D

In the example above, the outdoor temperature was \begin{align*}0^{\circ} C\end{align*}. This temperature was chosen to make the value of the horizontal asymptote zero. More importantly, this enabled us to determine the value of the common ratio.

Now we will investigate the same problem again with the outdoor temperature now being \begin{align*}15^{\circ} C\end{align*}. What effect will this have on the value of the common ratio? How can the value of ‘\begin{align*}b\end{align*}’ be determined so that an exponential function can be created to model the data?

Brian bought a cup of coffee from the cafeteria and placed it on the ground while recording its change in temperature (the temperature that day was \begin{align*}15^{\circ} C\end{align*}). The temperature of the coffee was recorded every 3 minutes, and the results are shown in the following table:

\begin{align*}&\text{Time } (min) \qquad \ 0 \qquad \ \ 3 \qquad \ 6 \qquad 9 \qquad 12 \quad \ \ 15\\ &\text{Temp }(0^{\circ} C) \quad 105.0 \quad 96.0 \quad 87.9 \quad 80.6 \quad 74.0 \quad 68.1\end{align*}

i) Determine the rate at which the coffee is cooling.

ii) Write an exponential function to describe the temperature of the coffee after ‘\begin{align*}t\end{align*}’ minutes.

iii) Determine the temperature of the coffee after 30 minutes.

iv) Determine the temperature of the coffee after I hour.

Solutions:

i) We will first determine the common ratio by using the values in the given table.

\begin{align*}&\boxed{r=\frac{96}{105}=0.914} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{87.9}{96}=0.916} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{80.6}{87.9}=0.917}\\ & \boxed{r=\frac{t_{n+1}}{t_n}=\frac{74.0}{80.6}=0.918} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{68.1}{74.0}=0.920}\end{align*}

To eliminate the difficulty with identifying the common ratio, simply subtract the value of the horizontal asymptote (15) from each of the \begin{align*}y\end{align*}-values.

The common ratio is 0.9. Therefore, the temperature of the coffee is decreasing by 10% every 3 minutes.

ii) The value of the ‘\begin{align*}a\end{align*}’ is the initial temperature less the outdoor temperature. \begin{align*}a=105.0^{\circ} C - 15.0^{\circ} C = 90^{\circ} C\end{align*}

iii) The initial value ‘\begin{align*}a\end{align*}’ is 90. The common ratio ‘\begin{align*}b\end{align*}’ is 0.9. The increment in the \begin{align*}x\end{align*}-values is 3. Therefore, \begin{align*}c = 3\end{align*}. The coolest temperature that will be reached by the cooling coffee is that of the outdoor temperature. Therefore \begin{align*}d = 15\end{align*}. The exponential function that models the problem is \begin{align*}\boxed{y=90(0.9)^{\frac{x}{3}}+15}\end{align*}

iv) One hour equals 60 minutes. Replace ‘\begin{align*}x\end{align*}’ in the exponent with ‘60’ and calculate the value of the function. \begin{align*}\boxed{y=90(0.9)^{\frac{60}{3}}+15=25.9^{\circ} C}\end{align*}

#### Concept Problem Revisited

Andrea invested money into a mutual fund. Every three months she received a bank statement indicating the growth of her investment. Andrea recorded the following data from her bank statements:

\begin{align*}&\text{Time} \ (months) \ \ 0 \qquad \ 3 \qquad \quad 6 \qquad \qquad 9 \qquad \quad 12 \qquad \quad \ 15\\ &\text{Value (\)} \qquad \ 1200 \quad 1224 \quad 1248.48 \quad 1273.45 \quad 1298.92 \quad 1324.90\end{align*}

i) How much was the initial investment?

The initial investment was \$1200.

ii) What was the rate of interest the bank applied to her investment? Express this as a percent.

The rate of interest that the bank applied to her investment was

\begin{align*}& r=\frac{t_{n+1}}{t_n} && r=\frac{t_{n+1}}{t_n} && r=\frac{t_{n+1}}{t_n} && r=\frac{t_{n+1}}{t_n} && r=\frac{t_{n+1}}{t_n}\\ & r=\frac{1224}{1200} && r=\frac{1248.48}{1224} && r=\frac{1273.45}{1248.48} && r=\frac{1298.92}{1273.45} && r=\frac{1324.90}{1298.92}\\ & \boxed{r=1.02} && \boxed{r=1.02} && \boxed{r=1.02} && \boxed{r=1.02} && \boxed{r=1.02} \end{align*}

The common ratio is 1.02. Therefore the interest rate was 2%.

iii) What is the exponential function that best models the value of Andrea’s investment?

The increment of the \begin{align*}x\end{align*}-value is 3. Therefore ‘\begin{align*}c\end{align*}\begin{align*}= 3\end{align*}. The exponential function to model Andrea’s investment is \begin{align*}\boxed{y=1200(1.02)^{\frac{x}{3}}}\end{align*}

iv) What will be the mutual fund’s value in two years?

The value of ‘\begin{align*}c\end{align*}’ is 3. This represents the every 3 months that the bank applies the interest to the investment. Therefore, the value of ‘\begin{align*}x\end{align*}’ must be in months. There are 24 months in two years. The value of Andrea’s investment in two years time will be:

\begin{align*}\boxed{y=1200(1.02)^{\frac{24}{3}}=\ 1405.99}\end{align*}

v) Use technology to determine the length of time it will take for the mutual fund to double in value?

Press \begin{align*}y=\end{align*} and enter the equations \begin{align*}y=1200(1.02)^{\frac{x}{3}}\end{align*} and \begin{align*}y=2400\end{align*} as shown below.

Graph the equations

Determine the intersection point of the growth curve and the horizontal line by using \begin{align*}\boxed{2^{nd} \ \text{TRACE}} \ \boxed{5} \ \boxed{\text{ENTER}} \ \boxed{\text{ENTER}} \ \boxed{\text{ENTER}}\end{align*}

The intersection point is \begin{align*}(105.00837, 2400)\end{align*}. It will take 105 months (8 years and 9 months) for Andrea’s investment to double in value.

### Vocabulary

Horizontal Asymptote
A horizontal asymptote is a horizontal line that is approached but NOT touched by a curve as its distance from the starting point increases. This is true for both a growth curve and a decay curve.
Common Ratio
The common ratio is the constant that exists between successive terms and is determined by applying the formula \begin{align*}r=\frac{t_{n+1}}{t_n}\end{align*}. In an exponential function of the form \begin{align*}y=b^x\end{align*}, ‘\begin{align*}b\end{align*}’ represents the common ratio.
Decay Curve
A decay curve is the name given to the graph of an exponential function in which the common ratio is such that \begin{align*}0 . The graph is decreasing since the value of the function falls as the value of ‘\begin{align*}x\end{align*}’ increases. The following shows a decay curve:
Exponential Function
An exponential function is a function \begin{align*}(y)\end{align*} of the form \begin{align*}y=b^x\end{align*}, where ‘\begin{align*}b\end{align*}’ is the common ratio and ‘\begin{align*}x\end{align*}’ is an exponent that represents the variable.
Growth Curve
A growth curve is the name given to the graph of an exponential function in which the common ratio is such that \begin{align*}b>1\end{align*}. The graph is increasing since the value of the function rises as the value of ‘\begin{align*}x\end{align*}’ increases. The following shows a growth curve:

### Guided Practice

1. A bucket of tar is heated outside to repair a roof that is leaking. Once the leak has been repaired, the remaining tar is left in the bucket to cool down to use again at a later time. Its change in temperature, \begin{align*}T\end{align*}, measured in \begin{align*}^{\circ} C\end{align*}, with respect to time, \begin{align*}t\end{align*}, in minutes, can be modeled with the function \begin{align*}\boxed{T=95(0.72)^{\frac{t}{15}}+30}\end{align*}.

i) What is the initial temperature of the tar?
ii) What is the outdoor temperature?
iii) At what rate is the temperature of the tar changing over time?
iv) Determine the temperature of the tar after 30 minutes.
v) Determine the temperature of the tar after 2 hours.
vi) Draw a sketch to show the change in temperature of the tar over time.

2. Roberto’s car radiator is always overheating. He decided to monitor its change in temperature. The outdoor temperature was \begin{align*}73.4^{\circ} F\end{align*}, so he did not mind doing the task at hand. Roberto recorded the temperature that was determined every five minutes. The following table is the recorded temperatures.

\begin{align*}&\text{Time } (min) \qquad 0 \qquad \quad 5 \qquad \quad 10 \qquad \ 15 \qquad \ 20 \qquad \ 25\\ &\text{Temp } (0^{\circ} F) \quad 224.6 \quad 197.42 \quad 175.1 \quad 156.7 \quad 141.8 \quad 129.4\end{align*}
i) Determine the rate at which the radiator is cooling.
ii) Determine the value of ‘\begin{align*}a\end{align*}’ in \begin{align*}y=a(b)^{\frac{x}{c}}+d\end{align*} for this problem.
iii) What is the equation of the horizontal asymptote?
iv) Write an exponential function to indicate the temperature of the radiator after ‘\begin{align*}t\end{align*}’ minutes.
v) What was the temperature of the radiator after one hour?
vi) Draw a sketch of the radiator’s change in temperature over time.

3. A deadly bacteria is threatening the small town of Norman. The bacteria are doubling every 3 hours. If there were initially 250 spores, how many will be present in 12 hours?

1. i) The initial temperature is the temperature ‘\begin{align*}a\end{align*}’ plus the outdoor temperature of \begin{align*}30^{\circ} C\end{align*}.

The exponential function is of the form \begin{align*}y=a(b)^{\frac{x}{c}}+d\end{align*}. The initial temperature is \begin{align*}\boxed{a+d=95^{\circ} C + 30^{\circ} C=125^{\circ} C}\end{align*}.
ii) The outdoor temperature is the horizontal asymptote (‘\begin{align*}d\end{align*}’) of the function. The outdoor temperature is \begin{align*}\boxed{30^{\circ} C}\end{align*}.
iii) The common ratio ‘\begin{align*}b\end{align*}’ represents the rate expressed as a decimal that the tar is maintaining. Therefore the rate at which it is cooling down is \begin{align*}100-.72=0.28\\ 0.28 \times 100 \%=28 \%\end{align*}. The temperature of the tar is dropping 28% every 15 minutes.
iv) The temperature of the tar after 30 minutes is determined by replacing ‘\begin{align*}t\end{align*}’ in the exponent with 30 and then performing the calculation.
\begin{align*}T &=95(0.72)^{\frac{t}{15}}+30\\ T &=95(0.72)^{\frac{30}{15}}+30\\ T &=79.2^{\circ} C\end{align*}
v) There are 60 minutes in one hour and 120 minutes in two hours. The temperature of the tar after 2 hours is determined by replacing ‘\begin{align*}t\end{align*}’ in the exponent with 120 and then performing the calculation.
\begin{align*}T &=95(0.72)^{\frac{t}{15}}+30\\ T &=95(0.72)^{\frac{120}{15}}+30\\ T &=36.9^{\circ} C\end{align*}
vi) A sketch to represent \begin{align*}\boxed{T=95(0.72)^{\frac{t}{15}}+30}\end{align*} is shown below:

2. i) The rate by which the radiator is cooling must be determined by first subtracting the outdoor temperature of \begin{align*}73.4^{\circ} F\end{align*} from each of the temperature values.

\begin{align*}&\text{Time} \ (min) \qquad 0 \qquad \quad 5 \qquad \ \ 10 \qquad \ 15 \qquad \ 20 \qquad 25\\ &\text{Temp}(^{\circ} F) \qquad 224.6 \quad 197.4 \quad 175.1 \quad 156.7 \quad 141.8 \quad 129.4\\ &\text{Temp} - 73.4 \quad \ 151.2 \quad 124.0 \quad 101.7 \quad 83.3 \quad \ \ 68.4 \quad \ 56.0\end{align*}
Now, determine the common ratio using the final row of the table.
\begin{align*}& \boxed{r=\frac{t_{n+1}}{t_n}=\frac{124.0}{151.2}=0.820} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{101.7}{124.0}=0.820} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{83.3}{101.7}=0.819}\\ & \boxed{r=\frac{t_{n+1}}{t_n}=\frac{68.4}{83.3}=0.821} && \boxed{r=\frac{t_{n+1}}{t_n}=\frac{56.0}{68.4}=0.818}\end{align*}
Therefore the common ratio is 0.82. The rate at which the radiator is cooling is \begin{align*}\boxed{100 \% - 82 \% =18 \%}\end{align*} every five minutes.
ii) The value of ‘\begin{align*}a\end{align*}’ in \begin{align*}y=a(b)^{\frac{x}{c}}+d\end{align*} is the initial temperature given in the table less the outdoor temperature. \begin{align*}\boxed{a=224.6^{\circ} F-73.4^{\circ} F=151.2^{\circ} F}\end{align*}
iii) The equation of the horizontal asymptote is \begin{align*}y =\end{align*} the outdoor temperature or \begin{align*}\boxed{y=73.4}\end{align*}
iv) The exponential function that models the change in temperature of the radiator is:
\begin{align*}a&=151.2\\ b&=0.82\\ c&=5\\ d&=73.4\\ T&=151.2(0.82)^{\frac{t}{5}}+73.4\end{align*}
v) There are 60 minutes in one hour. Therefore \begin{align*}\boxed{T=151.2(0.82)^{\frac{60}{5}}+73.4=87.4^{\circ} F}\end{align*}
vi)

3. Write the exponential to model the number of spores produced over time.

\begin{align*}a&=250\\ b&=3\\ x&=12\\ y&=a(b)^x\\ y&=250(3)^{12}\end{align*}
Use the exponential function to determine the number of spores present after 12 hours.
\begin{align*}\boxed{y=250(3)^{12}=132860250 \ spores}\end{align*}

### Practice

Write an exponential function in the form \begin{align*}y=a(b)^{\frac{x}{c}}\end{align*} to model each table of values.

\begin{align*}&X \qquad 0 \qquad 5 \qquad 10 \qquad 15 \quad \ \ 20 \qquad \ 25\\ &Y \qquad 8 \qquad 4 \qquad \ 2 \qquad \ 1 \qquad 0.5 \qquad 0.25\end{align*}
\begin{align*}&X \qquad 0 \qquad 3 \quad \ \ 6 \qquad 9 \qquad \ 12 \qquad \ 15\\ &Y \qquad 2 \quad \ \ 10 \quad \ 50 \quad \ 250 \quad \ 1250 \quad \ 6250\end{align*}
\begin{align*}&X \quad -4 \qquad 0 \qquad 4 \qquad \ 8 \qquad \quad 12 \qquad \quad 16\\ &Y \qquad 6 \qquad 2.4 \quad 0.96 \quad 0.384 \quad 0.1536 \quad 0.06144\end{align*}
\begin{align*}&X \quad -0.6 \quad -0.3 \quad \ \ 0 \qquad 0.3 \qquad 0.6 \qquad \ \ 0.9\\ &Y \qquad \ 10 \qquad \ \ 12 \quad 14.4 \quad 17.28 \quad 20.736 \quad 24.8832\end{align*}
\begin{align*}&X \qquad 0 \qquad 0.1 \qquad 0.2 \qquad \ 0.3 \qquad \ 0.4 \qquad \ 0.5\\ &Y \qquad 5 \qquad \ 15 \qquad \ 45 \qquad \ 135 \qquad 405 \qquad 1215\end{align*}

A radioactive isotope decays at the rate indicated by the exponential function \begin{align*}A(t)=800\left(\frac{1}{2}\right)^{\frac{t}{1500}}\end{align*}, where ‘\begin{align*}t\end{align*}’ is the time in years and \begin{align*}A(t)\end{align*} is the amount of the isotope, in grams, remaining.

1. What is the initial mass of the isotope?
2. How long will it take for the isotope to be reduced to half of its original amount?
3. What will the mass of the isotope be after 4500 years?

For each of the following exponential functions state the equation of the horizontal asymptote, the \begin{align*}y\end{align*}-intercept, the range, and whether it is a growth or decay function.

1. \begin{align*}y=2^x+5\end{align*}
2. \begin{align*}y=2(3)^x\end{align*}
3. \begin{align*}y=6\left(\frac{1}{3}\right)^x+5\end{align*}
4. \begin{align*}y=4(0.4)^x+1.8\end{align*}
5. \begin{align*}y=12(1.25)^x\end{align*}

A hot cup of coffee cools exponentially with time as it sits on the teacher’s desk. Its change in temperature, \begin{align*}T\end{align*}, measured in \begin{align*}^{\circ}C\end{align*}, with respect to time, \begin{align*}t\end{align*}, in minutes, is modeled with the following function:

\begin{align*}T=82(0.6)^{\frac{t}{12}}+20\end{align*}

1. What is the initial temperature of the coffee?
2. What is the room temperature of the classroom?
3. At what rate is the temperature of the coffee changing over time?
4. What is the coffee’s temperature after 30 minutes?
5. What is the temperature of the coffee after 1 hour?

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Jan 16, 2013
Apr 29, 2014

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