Roberta invested $600 into a mutual fund that paid 4% interest each year compounded annually.
i) Complete a table showing the value of the mutual fund for the first five years.
ii) Write an exponential function of the form \begin{align*}y=a\cdot b^x\end{align*} to describe the value of the mutual fund.
iii) Use the exponential function to determine the value of the mutual fund in 15 years.
Watch This
Khan Academy Exponential Growth Functions
Khan Academy Exponential Decay Functions
Guidance
An exponential function is a function with a variable in the exponent. Two examples of exponential functions are shown below:
\begin{align*}\boxed{y=2^x}\end{align*}
\begin{align*}\boxed{y=\left(\frac{1}{2}\right)^x}\end{align*}
Here are some facts to notice about the functions and their graphs:
 The graph of \begin{align*}y=2^x\end{align*} is an increasing curve. It shows growth.
 Each yvalue of \begin{align*}y=2^x\end{align*} is 2 times the previous yvalue (for the integer values of x). For example, the points on the graph go from (0, 1) to (1, 2) to (2, 4). The next point would be (3, 8). The yvalues keep being multiplied by 2.
 The graph of \begin{align*}y=\left(\frac{1}{2}\right)^x\end{align*} is a decreasing curve. It shows decay.
 Each yvalue of \begin{align*}y=\left(\frac{1}{2}\right)^x\end{align*} is \begin{align*}\frac{1}{2}\end{align*} the value of the previous yvalue (for the integer values of x). For example, the points on the graph go from (0, 1) to \begin{align*}(1, \frac{1}{2})\end{align*} to \begin{align*}(2, \frac{1}{4})\end{align*}. The yvalues keep being multiplied by \begin{align*}\frac{1}{2}\end{align*}.
 Both graphs have a yintercept of 1. This is because anything to the zero power is equal to 1.
 The domain of each function is \begin{align*}D=\{x x \in R\}\end{align*}.
 The range for each function is \begin{align*}R=\{y  y>0, \ y \in R \}\end{align*}.
Based on the above observations, you can deduce that an exponential function of the form \begin{align*}y=ab^x\end{align*} where \begin{align*}b >0\end{align*} has the following properties:
Properties of an Exponential Function of the form \begin{align*}y=ab^x \ (b>0)\end{align*}
 ‘\begin{align*}b\end{align*}’ is the value of the common ratio. Within the function, as the xvalue increases by 1, the yvalue is multiplied by the common ratio.
 If \begin{align*}b > 1 \end{align*} then the curve will represent exponential growth.
 If \begin{align*}0 < b < 1\end{align*} then the curve will represent exponential decay.
 Every exponential function of the form \begin{align*}y=ab^x\end{align*} will pass through the point \begin{align*}(0, a)\end{align*}. \begin{align*}a\end{align*} will always be the yintercept of the function, or its value at time 0.
 Every exponential function of the form \begin{align*}y=ab^x\end{align*} will have the domain and range: \begin{align*}D=\{x x \in R \} \ and \ R=\{y  y>0, \ y \in R \}\end{align*}
Example A
For the following tables of values that represent exponential functions, determine the common ratio:
i)

 \begin{align*}& x \quad 0 \quad 1 \quad 2 \quad 3 \quad \ 4 \quad \cdots\\ & y \quad 1 \quad 2 \quad 4 \quad 8 \quad 16 \quad \cdots\end{align*}
ii)

 \begin{align*}& x \quad \ 0 \qquad 1 \quad \ 2 \qquad 3 \qquad \ 4 \quad \ \cdots\\ & y \quad 100 \quad 50 \quad 25 \quad 12.5 \quad 6.25 \quad \cdots\end{align*}
Solutions:
i) The common ratio is a constant that is determined by \begin{align*}r=\frac{t_{n+1}}{t_n}\end{align*}.
\begin{align*}& r=\frac{t_{n+1}}{t_n}=\frac{2}{1}=2\\ & r=\frac{t_{n+1}}{t_n}=\frac{4}{2}=2\\ & r=\frac{t_{n+1}}{t_n}=\frac{8}{4}=2\\ & r=\frac{t_{n+1}}{t_n}=\frac{16}{8}=2\\ & \boxed{\text{The common ratio is} \ 2.}\end{align*}
ii) The common ratio is a constant that is determined by \begin{align*}r=\frac{t_{n+1}}{t_n}\end{align*}.
\begin{align*}& r=\frac{t_{n+1}}{t_n}\\ & r=\frac{50}{100}=\frac{1}{2}\\ & r=\frac{25}{50}=\frac{1}{2}\\ & r=\frac{12.5}{25}=\frac{1}{2}\\ & r=\frac{6.25}{12}=\frac{1}{2}\\ & \boxed{\text{The common ratio is } \frac{1}{2}.}\end{align*}
Example B
Using the exponential function \begin{align*}\boxed{f(x)=3^x}\end{align*}, determine the value of each of the following:
i) \begin{align*}f(2)\end{align*}
ii) \begin{align*}f(3)\end{align*}
iii) \begin{align*}f(0)\end{align*}
iv) \begin{align*}f(4)\end{align*}
v) \begin{align*}f(2)\end{align*}
Solutions: \begin{align*}f(x)=3^x\end{align*} is another way to express \begin{align*}y=3^x\end{align*}. To determine the value of the function for the given values, replace the exponent with that value and evaluate the expression.
i)
\begin{align*}& f(x)=3^x\\ & f({\color{red}2})=3^{\color{red}2}\\ & f(2)={\color{red}9}\\ & \boxed{f(2)=9}\end{align*}
ii)
\begin{align*}& f(x)=3^x\\ & f({\color{red}3})=3^{\color{red}3}\\ & f(3)={\color{red}27}\\ & \boxed{f(3)=27}\end{align*}
iii)
\begin{align*}& f(x)=3^x\\ & f({\color{red}0})=3^{\color{red}0}\\ & f(0)={\color{red}1}\\ & \boxed{f(0)=1}\end{align*}
iv)
\begin{align*}& f(x)=3^x\\ & f({\color{red}4})=3^{\color{red}4}\\ & f(4)={\color{red}81}\\ & \boxed{f(4)=81}\end{align*}
v)
\begin{align*}& f(x)=3^x\\ & f({\color{red}2})=3^{\color{red}2}\\ & f(2)={\color{red}\frac{1}{3^2}}\\ & \boxed{f(2)=\frac{1}{9}}\end{align*}
Example C
On January 1, Juan invested $1.00 at his bank at a rate of 10% interest compounded daily.
i) Create a table of values for the first 8 days of the investment.
ii) What is the common ratio?
iii) Determine the equation of the function that would best represent Juan’s investment.
iv) How much money will Juan have in his account on January 31?
v) If Juan had originally invested $100 instead of $1.00 at 10%, what exponential equation would describe the investment. How much money would he have in his account on January 31?
Solution:
i)
\begin{align*}&1.00(.10)=0.10 && 1.10(.10)=0.11 && 1.21(.10)=0.12 && 1.33(.10)=0.13\\ &1.00+0.10=1.10 && 1.10+0.11=1.21 && 1.21+0.12=1.33 && 1.33+0.13=1.46\\ \\ &1.46(.10)=0.15 && 1.61(.10)=0.16 && 1.77(.10)=0.18 && 1.95(.10)=0.20\\ &1.46+0.15=1.61 && 1.61+0.16=1.77 && 1.77+0.18=1.95 && 1.95+0.20=2.15\end{align*}
\begin{align*}&\# \ \text{ of days} \qquad 0 \quad \ \ 1 \qquad 2 \qquad \ 3 \qquad \ 4 \qquad \ 5 \qquad \ 6 \qquad \ 7 \qquad \ 8\\ &\text{Money } (\$) \qquad 1 \quad 1.10 \quad 1.21 \quad 1.33 \quad 1.46 \quad 1.61 \quad 1.77 \quad 1.95 \quad 2.15\end{align*}
ii) The common ratio is a constant that is determined by \begin{align*}\frac{t_{n+1}}{t_n}\end{align*}. Therefore, the common ratio for this problem is \begin{align*}r=\frac{t_{n+1}}{t_n} \rightarrow \frac{1.10}{1}=1.10 \rightarrow \frac{1.21}{1.10}=1.10 \rightarrow \frac{1.33}{1.21}=1.10\end{align*}.
The common ratio is \begin{align*}\boxed{1.10}\end{align*}.
iii) The equation of the function to model Juan’s investment is \begin{align*}y=1.10^x\end{align*}
iv)
\begin{align*}y=1.10^x \rightarrow y=1.10^{31} \rightarrow \boxed{y=\$ 19.19}\end{align*}.
On January 31, Juan will have $19.19 in his account.
v)
\begin{align*}y&=100(1.10)^x\\ y&=100(1.10)^{31} \rightarrow y={\color{red}\$1919.43} \rightarrow \boxed{y=\$1919.43}.\end{align*}
On January 31, Juan would have $1919.43 in his account if he had invested $100 instead of $1.00.
Concept Problem Revisited
Roberta invested $600 into a mutual fund that paid 4% interest each year compounded annually.
i)
\begin{align*}& 600(.04)=24 && 624(.04)=24.96 && 648.96(.04)=25.96 \\ &600+24=624 && 624+24.96=648.96 && 648.96+25.96=674.92\\ \\ &674.92(.04)=27.00 && 701.92(.04)=28.08\\ &674.92+27.00=701.92 && 701.92+28.08=730.00\end{align*}
\begin{align*}&\text{Time} \ (years) \ \ 0 \qquad 1 \qquad \ \ 2 \qquad \quad \ 3 \qquad \quad \ 4 \qquad \quad \ 5\\ &\text{Value (\$)} \qquad 600 \quad 624 \quad 648.96 \quad 674.92 \quad 701.92 \quad 730.00\end{align*}
ii) The initial value is $600. The common ratio is 1.04 which represents the initial investment and the interest rate of 4%. The exponent is the time in years. The exponential function is \begin{align*}y=600(1.04)^x\end{align*} or \begin{align*}v=600(1.04)^t\end{align*}.
iii)
\begin{align*}& v=600(1.04)^t\\ & v=600(1.04)^{\color{red}15}\\ & v={\color{red}\$1080.57}\\ &\boxed{v=\$1080.57}\end{align*}
The value of the mutual fund in fifteen years will be $1080.57.
Guided Practice
1. The graph below shows the change in value of two comic books purchases in the year 2000. Both comics were expected to be good investments, but one of them did not perform as expected. Use the graphs to answer the questions.
 a) What was the purchase price of each comic book?
 b) Which comic book shows exponential growth?
 c) Which comic book shows exponential decay?
 d) In what year were both comic books equal in value?
 e) State the domain and range for each comic.
2. Paulette bought a Bobby Orr rookie card for $300. The value of the card appreciates (increases) by 30% each year.
 a) Complete a table of values to show the first five years of the investment.
 b) Determine the common ratio for the successive terms.
 c) Determine the equation of the exponential function that models this investment.
3. Due to the closure of the pulp and paper mill, the population of the small town is decreasing at a rate of 12% annually. If there are now 2400 people living in the town, what will the town's projected population be in eight years?
Answers:
1. a) The purchase price of each comic book is the \begin{align*}y\end{align*}intercept. The \begin{align*}y\end{align*}intercept is the initial value of the books. The Spiderman comic book cost $30.00 and the Superman comic book cost $5.00.
 b) The Superman comic book shows exponential growth.
 c) The Spiderman comic book shows exponential decay.
 d) In 2005 both comic books were equal in value. The graphs intersect at approximately (5, $12.50), where 5 represents five years after the books were purchased.
 e) The domain and range for each comic is \begin{align*}D=\{x x \in R\}\end{align*} and \begin{align*}R=\{yy>0, \ y \in R\}\end{align*}
2.
\begin{align*}&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\ &300+90=390 && 390+117=507 && 507+152.10=659.10\\ \\ &659.10(.30)=197.73 && 856.83(.30)=257.05\\ &659.10+197.73=856.83 && 856.83+257.05=1113.88\end{align*}
 a)
 \begin{align*}&\text{Time} \ (years) \ \ 0 \qquad 1 \qquad 2 \qquad \quad 3 \qquad \quad 4 \qquad \quad \ 5\\ &\text{Value (\$)} \qquad 300 \quad 390 \quad 507 \quad 659.10 \quad 856.83 \quad 1113.88\end{align*}
 b)
 \begin{align*}& r=\frac{t_{n+1}}{t_n}\\ & r=\frac{390}{300}=1.3\\ & r=\frac{507}{390}=1.3\\ & r=\frac{659.10}{507}=1.3\\ & r=\frac{856.83}{659.10}=1.3\\ & r=\frac{1113.88}{856.83}=1.3\end{align*}
 c) The exponential function that would model Paulette's investment is \begin{align*}\boxed{y=300(1.3)^x \ or \ v=300(1.3)^t}\end{align*}
3. The town's population is decreasing by 12% annually. The simplest way to use this in an exponential function is to use the percent of the population that still exists each year – 88%.
 Therefore, the exponential function would consist of the present population \begin{align*}(a)\end{align*}, the common ratio is 0.88 \begin{align*}(b)\end{align*} and the time in years would be the exponent \begin{align*}(x)\end{align*}. The function is \begin{align*}p=2400(0.88)^t\end{align*}
 The population in eight years would be
 \begin{align*}& p=2400(0.88)^t\\ & p=2400(0.88)^8\\ & p=863.123\\ & p \approx 863 \ people\end{align*}
Explore More
Brandon bought a car for $13,000. The value of the car depreciates by 20% each year.
 Complete a table of values to show the car’s values for the first five years.
 Determine the exponential function that would model the depreciation of Brandon’s car.
For each of the following exponential functions, identify the common ratio and the \begin{align*}y\end{align*}intercept, and tell if the function represents a growth or decay curve.
 \begin{align*}y=4(5)^x\end{align*}
 \begin{align*}y=13(2.3)^x\end{align*}
 \begin{align*}y=0.85(0.16)^x\end{align*}
 \begin{align*}y=1.6(0.5)^x\end{align*}
 \begin{align*}y=0.4(2.1)^x\end{align*}
Match each graph below with its corresponding equation:
 \begin{align*}y=2^x\end{align*}
 \begin{align*}y=3^x\end{align*}
 \begin{align*}y=2(3)^x\end{align*}
 \begin{align*}y=3(2)^x\end{align*}
 Do these graphs represent growth or decay?
Match each graph below with its corresponding equation:
 \begin{align*}y=0.5^x\end{align*}
 \begin{align*}y=0.2^x\end{align*}
 \begin{align*}y=2(0.5)^x\end{align*}
 \begin{align*}y=3(0.2)^x\end{align*}
 Do these graphs represent growth or decay?
 Jolene purchased a summer home for $120,000 in 2002. If the property has consistently increased in value by 11% each year, what will be the value of her summer home in 2012?