Graphing Exponential Growth Functions An exponential function involves the expression b x where the base b is a positive number other than 1. In this lesson you will study exponential functions for which b > 1. To see the basic shape of the graph of an exponential function such as f(x) = 2 x, you can make a table of values and plot points, as shown below. x f(x) = 2 x Notice the end behavior of the graph. As x +, f(x) +, which means that the graph moves up to the right. –3 2 –3 = 1 8 –2 2 –2 = 1 4 As x –, f(x) 0, which means that the graph has the line y = 0 as an asymptote. 2 –1 = 1 2 –1 2 0 = 1 An asymptote is a line that a graph approaches as you move away from the origin. 1 2 1 = 2 2 2 2 = 4 3 2 3 = 8

Graphing Exponential Growth Functions ACTIVITY Developing Concepts INVESTIGATING GRAPHS OF EXPONENTIAL FUNCTIONS Graph y = • 2 x and y = 3 • 2 x. Compare the graphs with the graph of y = 2 x. 1 3 1 Graph y = – • 2 x and y = –5 • 2 x. Compare the graphs with the graph of y = 2 x. 1 5 2 3 Describe the effect of a on the graph of y = a • 2 x when a is positive and when a is negative.

Graphing Exponential Growth Functions In the previous slide you may have observed the following about the graph y = a • 2 x: The graph passes through the point (0, a). That is, the y-intercept is a. The x-axis is an asymptote of the graph. The domain is all real numbers. The range is y > 0 if a > 0 and y < 0 if a < 0. The characteristics of the graph of y = a • 2 x listed above are true of the graph of y = a b x. If a > 0 and b > 1, the function y = a b x is an exponential growth function.

Graphing Exponential Functions of the Form y = ab x Graph the function. y = • 3 x 1 2 SOLUTION Plot (0, ) and (1, ). Then, from left to right, draw a curve that begins just above the x-axis, passes through the two points, and moves up to the right. 3 2 1

( ) Graphing Exponential Functions of the Form y = ab x Graph the function. 1 2 ( ) y = 3 2 x – y = • 3 x SOLUTION SOLUTION Plot (0, –1) and (1, ). Then, from left to right, draw a curve that begins just below the x-axis, passes through the two points, and moves down to the right. 3 2 – Plot (0, ) and (1, ). Then, from left to right, draw a curve that begins just above the x-axis, passes through the two points, and moves up to the right. 1 2 3 2

Graphing Exponential Functions of the Form y = ab x Graph the function. y = • 3 x 1 2 SOLUTION Plot (0, ) and (1, ). Then, from left to right, draw a curve that begins just above the x-axis, passes through the two points, and moves up to the right. 3 2 1

( ) Graphing Exponential Functions of the Form y = ab x Graph the function. 1 2 ( ) y = 3 2 x – y = • 3 x SOLUTION SOLUTION Plot (0, –1) and (1, ). Then, from left to right, draw a curve that begins just below the x-axis, passes through the two points, and moves down to the right. 3 2 – Plot (0, ) and (1, ). Then, from left to right, draw a curve that begins just above the x-axis, passes through the two points, and moves up to the right. 3 2 1

Graphing a General Exponential Function To graph a general exponential function, begin by sketching the graph of y = ab x. Then translate the graph horizontally by h units and vertically by k units. y = ab x – h + k Graph y = 3 • 2 x – 1 – 4. State the domain and range. SOLUTION Begin by lightly sketching the graph of y = 3 • 2 x, which passes through (0, 3) and (1, 6). Then translate the graph 1 unit to the right and 4 units down. Notice that the graph passes through (1, –1) and (2, 2). The graph’s asymptote is the line y = – 4. The domain is all real numbers, and the range is y > – 4.

Using Exponential Growth Models When a real-life quantity increases by a fixed percent each year (or other time period), the amount y of quantity after t years can be modeled by this equation: y = a(1 + r) t In this model, a is the initial amount and r is the percent increase expressed as a decimal. The quantity 1 + r is called the growth factor.

Modeling Exponential Growth INTERNET HOSTS In January, 1993, there were about 1,313,000 Internet hosts. During the next five years, the number of hosts increased by about 100% per year. Write a model giving the number h (in millions) of hosts t years after 1993. About how many hosts were there in 1996? SOLUTION The initial amount is a = 1.313 and the percent increase is r = 1. So, the exponential growth model is: h = a(1 + r) t Write exponential growth model. = 1.313(1 + 1) t Substitute for a and r. = 1.313 • 2 t Simplify. Using this model, you can estimate the number of hosts in 1996 (t = 3) to be h = 1.313 • 2 3  10.5 million.

Modeling Exponential Growth INTERNET HOSTS In January, 1993, there were about 1,313,000 Internet hosts. During the next five years, the number of hosts increased by about 100% per year. Graph the model. SOLUTION The graph passes through the points (0, 1.313) and (1, 2.626). It has the t-axis as an asymptote. To make an accurate graph, plot a few other points. Then draw a smooth curve through the points.

Modeling Exponential Growth INTERNET HOSTS In January, 1993, there were about 1,313,000 Internet hosts. During the next five years, the number of hosts increased by about 100% per year. Use the graph to estimate the year when there were 30 million hosts. SOLUTION Using the graph, you can estimate that the number of hosts was 30 million sometime during 1997 (t  4.5).

Modeling Exponential Growth In the previous example the annual percent increase was 100%. This translated into a growth factor of 2, which means that the number of Internet hosts doubled each year. People often confuse percent increase and growth factor, especially when a percent increase is 100% or more. For example, a percent increase of 200% means that a quantity tripled, because the growth factor is 1 + 2 = 3. When you hear or read reports of how a quantity has changed, be sure to pay attention to whether a percent increase or a growth factor is being discussed.

( ) Using Exponential Growth Models COMPOUND INTEREST Exponential growth functions are used in real-life situations involving compound interest. Compound interest is interest paid on the initial investment, called the principal, and on previously earned interest. (Interest paid only on the principal is called simple interest.) Although interest earned is expressed as an annual percent, the interest is usually compounded more frequently than once per year. Therefore, the formula y = a(1 + r) t must be modified for compound interest problems. COMPOUND INTEREST Consider an initial principal P deposited in an account that pays interest at an annual rate r (expressed as a decimal), compounded n times per year. The amount A in the account after t years can be modeled by this equation: A = P 1 + rn nt ( )

( ) ( ) Finding the Balance in an Account FINANCE You deposit $1000 in an account that pays 8% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency. annually SOLUTION With interest compounded annually, the balance at the end of 1 year is: A = P 1 + rn nt ( ) Write compound interest model. ( ) A = 1000 1 + 0.08 1 1 • 1 P = 1000, r = 0.08, n = 1, t = 1 = 1000(1.08)1 Simplify. = 1080 Use a calculator. The balance at the end of 1 year is $1080.

( ) ( ) Finding the Balance in an Account FINANCE You deposit $1000 in an account that pays 8% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency. quarterly SOLUTION With interest compounded quarterly, the balance at the end of 1 year is: A = P 1 + rn nt ( ) Write compound interest model. ( ) A = 1000 1 + 0.08 4 4 • 1 P = 1000, r = 0.08, n = 4, t = 1 = 1000(1.02)4 Simplify.  1082.43 Use a calculator. The balance at the end of 1 year is $1082.43.

( ) ( ) Finding the Balance in an Account FINANCE You deposit $1000 in an account that pays 8% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency. daily SOLUTION With interest compounded daily, the balance at the end of 1 year is: A = P 1 + rn nt ( ) Write compound interest model. ( ) A = 1000 1 + 0.08 365 365 • 1 P = 1000, r = 0.08, n = 365, t = 1 1000(1.000219) 365 Simplify.  1083.28 Use a calculator. The balance at the end of 1 year is $1083.28.

( ) Graphing Exponential Decay Functions In this lesson you will study exponential decay functions, which have the form ƒ(x) = a b x where a > 0 and 0 < b < 1. State whether f(x) is an exponential growth or exponential decay function. ( ) f(x) = 5 2 3 x SOLUTION Because 0 < b < 1, f is an exponential decay function.

( ) ( ) Graphing Exponential Decay Functions In this lesson you will study exponential decay functions, which have the form ƒ(x) = a b x where a > 0 and 0 < b < 1. State whether f(x) is an exponential growth or exponential decay function. ( ) f(x) = 5 2 3 x ( ) f(x) = 8 3 2 x SOLUTION SOLUTION Because 0 < b < 1, f is an exponential decay function. Because b > 1, f is an exponential growth function.

( ) ( ) ( ) Graphing Exponential Decay Functions In this lesson you will study exponential decay functions, which have the form ƒ(x) = a b x where a > 0 and 0 < b < 1. State whether f(x) is an exponential growth or exponential decay function. ( ) f(x) = 5 2 3 x ( ) f(x) = 8 3 2 x f(x) = 10(3) –x SOLUTION SOLUTION SOLUTION Because 0 < b < 1, f is an exponential decay function. Because b > 1, f is an exponential growth function. Rewrite the function as: . Because 0 < b < 1, f is an exponential decay function. ( ) f(x) = 10 1 3 x

Recognizing and Graphing Exponential Growth and Decay To see the basic shape of the graph of an exponential decay function, you can make a table of values and plot points, as shown below. Notice the end behavior of the graph. As x –, f(x) +, which means that the graph moves up to the left. As x +, f(x) 0, which means that the graph has the line y = 0 as an asymptote. f(x) = 1 2 x ( ) = 8 ( ) 1 2 –3 –3 = 4 ( ) 1 2 –2 –2 = 2 ( ) 1 2 –1 –1 = 1 ( ) 1 2 ( ) 1 2 = 1 ( ) 1 2 = 4 2 ( ) 1 2 3 = 8 3

Recognizing and Graphing Exponential Growth and Decay To see the basic shape of the graph of an exponential decay function, you can make a table of values and plot points, as shown below. Notice the end behavior of the graph. As x –, f(x) +, which means that the graph moves up to the left. As x +, f(x) 0, which means that the graph has the line y = 0 as an asymptote. f(x) = 1 2 x ( ) = 8 ( ) 1 2 –3 –3 = 4 ( ) 1 2 –2 –2 = 2 ( ) 1 2 –1 –1 = 1 ( ) 1 2 Recall that in general the graph of an exponential function y = a b x passes through the point (0, a) and has the x-axis as an asymptote. The domain is all real numbers, and the range is y > 0 if a > 0 and y < 0 if a < 0. ( ) 1 2 = 1 ( ) 1 2 = 4 2 ( ) 1 2 3 = 8 3

( ) ( ) Graphing Exponential Functions of the Form y = ab x Graph the function. ( ) y = 3 1 4 x SOLUTION Plot (0, 3) and . Then, from right to left, draw a curve that begins just above the x-axis, passes through the two points, and moves up to the left. ( ) 3 4 1,

( ) ( ) Graphing Exponential Functions of the Form y = ab x Graph the function. ( ) y = –5 2 3 x SOLUTION Plot (0, –5) and . Then, from right to left, draw a curve that begins just below the x-axis, passes through the two points, and moves down to the left. 10 3 – 1, ( )

( ) ( ) ( ) ( ) Graphing a General Exponential Function Remember that to graph a general exponential function, begin by sketching the graph of y = a b x. Then translate the graph horizontally by h units and vertically by k units. y = a b x – h + k Graph . State the domain and range. ( ) y = –3 + 1 1 2 x + 2 SOLUTION Begin by lightly sketching the graph of , which passes through (0, –3) and . ( ) y = –3 1 2 x 1, 3 – ( ) Then translate the graph 2 units to the left and 1 unit up. Notice that the graph passes through (–2, –2) and . The graph’s asymptote is the line y = 1. The domain is all real numbers, and the range is y < 1. ( ) –1, 1 2 –

Using Exponential Decay Models When a real-life quantity decreases by a fixed percent each year (or other time period), the amount y of quantity after t years can be modeled by the equation: y = a(1 – r) t In this model, a is the initial amount and r is the percent decrease expressed as a decimal. The quantity 1 – r is called the decay factor.

Modeling Exponential Decay You buy a new car for $24,000. The value y of the car decreases by 16% each year. Write an exponential decay model for the value of the car. Use the model to estimate the value after 2 years. SOLUTION Let t be the number of years since you bought the car. The exponential decay model is: y = a(1 – r) t Write exponential decay model. = 24,000(1 – 0.16) t Substitute for a and r. = 24,000(0.84) t Simplify. When t = 2, the value is y = 24,000(0.84) 2  $16,934.

Modeling Exponential Decay You buy a new car for $24,000. The value y of the car decreases by 16% each year. Graph the model. SOLUTION The graph of the model is shown at the right. Notice that it passes through the points (0, 24,000) and (1, 20,160). The asymptote of the graph is the line y = 0.

Modeling Exponential Decay You buy a new car for $24,000. The value y of the car decreases by 16% each year. Use the graph to estimate when the car will have a value of $12,000. SOLUTION Using the graph, you can see that the value of the car will drop to $12,000 after about 4 years.

Modeling Exponential Decay In the previous example the percent decrease, 16%, tells you how much value the car loses from one year to the next. The decay factor, 0.84, tells you what fraction of the car’s value remains from one year to the next. The closer the percent decrease for some quantity is to 0%, the more the quantity is conserved or retained over time. The closer the percent decrease is to 100%, the more the quantity is used or lost over time.