![\"Escherichia](\"http:\/\/cnx.org\/resources\/f4bed7000de1219b5e7633366abef4a76617da73\/CNX_Precalc_Figure_04_00_001.jpg\")
Figure 1.<\/b> Electron micrograph of E.Coli bacteria (credit: \u201cMattosaurus,\u201d Wikimedia Commons)[\/caption]\r\nExponential Functions<\/h2>\r\n
In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data.<\/p>\r\n\r\n
Learning Outcomes<\/h3>\r\n\r\n \t- Evaluate exponential functions.<\/li>\r\n \t
- Find the equation of an exponential function.<\/li>\r\n \t
- Use compound interest formulas.<\/li>\r\n \t
- Evaluate exponential functions with base e.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.[footnote]http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.[\/footnote] If this rate continues, the population of India will exceed China\u2019s population by the year 2031. When populations grow rapidly, we often say that the growth is \"exponential,\" meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth <\/em>has a very specific meaning. In this section, we will take a look at exponential functions<\/em>, which model this kind of rapid growth.<\/p>\r\n\r\nEvaluate Exponential Functions<\/h2>\r\n
Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b<\/em>\u00a0to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:<\/p>\r\n\r\n\r\n \t- Let b\u00a0<\/em>= \u20139 and [latex]x=\\frac{1}{2}[\/latex]. Then [latex]f\\left(x\\right)=f\\left(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt{-9}[\/latex], which is not a real number.<\/li>\r\n<\/ul>\r\n
Why do we limit the base to positive values other than 1? Because base 1\u00a0results in the constant function. Observe what happens if the base is\u00a01:<\/p>\r\n\r\n
\r\n \t- Let b\u00a0<\/em>= 1. Then [latex]f\\left(x\\right)={1}^{x}=1[\/latex] for any value of x<\/em>.<\/li>\r\n<\/ul>\r\n
To evaluate an exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute x<\/em>\u00a0with the given value, and calculate the resulting power. For example:<\/p>\r\nLet [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n
[latex]\\begin{align}f\\left(x\\right)& ={2}^{x} \\\\ f\\left(3\\right)& ={2}^{3} && \\text{Substitute }x=3. \\\\ & =8 && \\text{Evaluate the power.} \\end{align}[\/latex]<\/div>\r\nTo evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:<\/p>\r\n
Let [latex]f\\left(x\\right)=30{\\left(2\\right)}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n
[latex]\\begin{align}f\\left(x\\right)& =30{\\left(2\\right)}^{x} \\\\ f\\left(3\\right)& =30{\\left(2\\right)}^{3} && \\text{Substitute }x=3. \\\\ & =30\\left(8\\right) && \\text{Simplify the power first.} \\\\ & =240 && \\text{Multiply.} \\end{align}[\/latex]<\/div>\r\nNote that if the order of operations were not followed, the result would be incorrect:<\/p>\r\n
[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/p>\r\n\r\n
\r\n\r\n\r\nExample 1: Evaluating Exponential Functions<\/h3>\r\n
Let [latex]f\\left(x\\right)=5{\\left(3\\right)}^{x+1}[\/latex]. Evaluate [latex]f\\left(2\\right)[\/latex] without using a calculator.<\/p>\r\n[reveal-answer q=\"847875\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847875\"]\r\n
Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\r\n
[latex]\\begin{align}f\\left(x\\right) & =5{\\left(3\\right)}^{x+1} \\\\ f\\left(2\\right) & =5{\\left(3\\right)}^{2+1} && \\text{Substitute }x=2. \\\\ & =5{\\left(3\\right)}^{3} && \\text{Add the exponents}. \\\\ & =5\\left(27\\right) && \\text{Simplify the power}. \\\\ & =135 && \\text{Multiply}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\nTry It<\/h3>\r\n
Let [latex]f\\left(x\\right)=8{\\left(1.2\\right)}^{x - 5}[\/latex]. Evaluate [latex]f\\left(3\\right)[\/latex] using a calculator. Round to four decimal places.<\/p>\r\n[reveal-answer q=\"292533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"292533\"]\r\n\r\n5.5556\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
Because the output of exponential functions increases very rapidly, the term \"exponential growth\" is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.<\/p>\r\n\r\n
\r\nA General Note: Exponential Growth<\/h3>\r\n
A function that models exponential growth<\/strong> grows by a rate proportional to the amount present. For any real number x<\/em>\u00a0and any positive real numbers a\u00a0<\/em>and b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\r\n\r\n[latex]\\text{ }f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\r\nwhere<\/p>\r\n\r\n
\r\n \t- a<\/em>\u00a0is the initial or starting value of the function.<\/li>\r\n \t
- b<\/em>\u00a0is the growth factor or growth multiplier per unit x<\/em>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
In more general terms, we have an exponential function<\/em>, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let\u2019s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex].<\/p>\r\nA few years of growth for these companies are illustrated below.<\/p>\r\n\r\n
\r\n\r\n\r\nYear,\u00a0x<\/em><\/th>\r\nStores, Company A<\/th>\r\n Stores, Company B<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n\r\n0<\/td>\r\n 100 + 50(0) = 100<\/td>\r\n 100(1 + 0.5)0<\/sup> = 100<\/td>\r\n<\/tr>\r\n\r\n1<\/td>\r\n 100 + 50(1) = 150<\/td>\r\n 100(1 + 0.5)1<\/sup> = 150<\/td>\r\n<\/tr>\r\n\r\n2<\/td>\r\n 100 + 50(2) = 200<\/td>\r\n 100(1 + 0.5)2<\/sup> = 225<\/td>\r\n<\/tr>\r\n\r\n3<\/td>\r\n 100 + 50(3) = 250<\/td>\r\n 100(1 + 0.5)3<\/sup> =\u00a0337.5<\/td>\r\n<\/tr>\r\n\r\nx<\/em><\/td>\r\nA<\/em>(x<\/em>) = 100 + 50x<\/td>\r\nB<\/em>(x<\/em>) = 100(1 + 0.5)x<\/em><\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe graphs comparing the number of stores for each company over a five-year period are shown in below.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010809\/CNX_Precalc_Figure_04_01_0012.jpg\")
Figure 2.<\/b> The graph shows the numbers of stores Companies A and B opened over a five-year period.[\/caption]\r\nNotice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.<\/p>\r\n
Now we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the base<\/em>, and x<\/em>\u00a0is called the exponent<\/em>.<\/p>\r\n\r\n\r\n\r\n\r\nExample 2: Evaluating a Real-World Exponential Model<\/h3>\r\n
At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?<\/p>\r\n[reveal-answer q=\"807378\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"807378\"]\r\nTo estimate the population in 2031, we evaluate the models for t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\r\n[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\r\n
There will be about 1.549 billion people in India in the year 2031.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\nTry It<\/h3>\r\n
The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex], where t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 2?<\/p>\r\n[reveal-answer q=\"740298\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"740298\"]\r\n\r\nAbout 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\u00a0Find the Equation of an Exponential Function<\/h2>\r\n\r\nIn the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a<\/em>\u00a0and b<\/em>, and evaluate the function.<\/p>\r\n\r\n\r\nHow To: Given two data points, write an exponential model.<\/h3>\r\n\r\n \t- If one of the data points has the form [latex]\\left(0,a\\right)[\/latex], then a<\/em>\u00a0is the initial value. Using a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex], and solve for b<\/em>.<\/li>\r\n \t
- If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex]. Solve the resulting system of two equations in two unknowns to find a<\/em>\u00a0and b<\/em>.<\/li>\r\n \t
- Using the a<\/em>\u00a0and b<\/em>\u00a0found in the steps above, write the exponential function in the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n\r\n\r\n\r\n
Example 3: Writing an Exponential Model When the Initial Value Is Known<\/h3>\r\n
In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N<\/em>(t<\/em>) representing the population N<\/em>\u00a0of deer over time t<\/em>.<\/p>\r\n[reveal-answer q=\"518916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"518916\"]\r\nWe let our independent variable t<\/em>\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a\u00a0<\/em>= 80. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find b<\/em>:<\/p>\r\n[latex]\\begin{align}N\\left(t\\right)& =80{b}^{t} \\\\ 180& =80{b}^{6} && \\text{Substitute using point }\\left(6, 180\\right). \\\\ \\frac{9}{4} & ={b}^{6} && \\text{Divide and write in lowest terms}. \\\\ b & ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}} && \\text{Isolate }b\\text{ using properties of exponents}. \\\\ b & \\approx 1.1447 && \\text{Round to 4 decimal places}. \\end{align}[\/latex]<\/p>\r\n
NOTE:<\/strong> Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.<\/em><\/p>\r\nThe exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)<\/p>\r\nWe can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex], and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].\r\n\r\n ![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010810\/CNX_Precalc_Figure_04_01_0022.jpg\")
<\/span>\r\nFigure 3.\u00a0<\/strong>Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], t<\/em>\u00a0years after 2006<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\nTry It<\/h3>\r\n
A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N<\/em>\u00a0of wolves over time t<\/em>.<\/p>\r\n[reveal-answer q=\"901847\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"901847\"]\r\n\r\n[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n\r\n\r\nExample 4: Writing an Exponential Model When the Initial Value is Not Known<\/h3>\r\n
Find an exponential function that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"935128\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"935128\"]\r\n
Because we don\u2019t have the initial value, we substitute both points into an equation of the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], and then solve the system for a<\/em>\u00a0and b<\/em>.<\/p>\r\n\r\n\r\n \t- Substituting [latex]\\left(-2,6\\right)[\/latex] gives [latex]6=a{b}^{-2}[\/latex]<\/li>\r\n \t
- Substituting [latex]\\left(2,1\\right)[\/latex] gives [latex]1=a{b}^{2}[\/latex]<\/li>\r\n<\/ul>\r\n
Use the first equation to solve for a<\/em>\u00a0in terms of b<\/em>:<\/p>\r\n[latex]\\begin{align}&6=ab^{-2}\\\\&\\frac{6}{b^{-2}}=a && \\text{Divide.} \\\\&a=6b^2 && \\text{Use properties of exponents to rewrite the denominator.}\\end{align}[\/latex]<\/span><\/p>\r\nSubstitute a<\/em>\u00a0in the second equation, and solve for b<\/em>:<\/p>\r\n[latex]\\begin{align}&1=ab^2 \\\\ &1=6b^2b^2=6b^4 && \\text{Substitute } a=6b^2. \\\\ &b=\\left(\\frac{1}{6}\\right)^\\frac{1}{4} && \\text{Use properties of exponents to isolate }b. \\\\ & b\\approx0.6389 && \\text{Round to 4 decimal places}. \\end{align}[\/latex]\r\n<\/span><\/p>\r\nUse the value of b<\/em>\u00a0in the first equation to solve for the value of a<\/em>:<\/p>\r\n[latex]a=6b^2\\approx6(0.6389)^2\\approx2.4492[\/latex]<\/span><\/p>\r\nThus, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex].<\/p>\r\nWe can graph our model to check our work. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(-2,\\text{ 6}\\right)[\/latex] and [latex]\\left(2,\\text{ 1}\\right)[\/latex]. The graph is an example of an exponential decay<\/strong> function.\r\n\r\n ![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010811\/CNX_Precalc_Figure_04_01_0032.jpg\")
<\/span>\r\nFigure 4.\u00a0<\/strong>The graph of [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex] models exponential decay.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\nTry It<\/h3>\r\n
Given the two points [latex]\\left(1,3\\right)[\/latex] and [latex]\\left(2,4.5\\right)[\/latex], find the equation of the exponential function that passes through these two points.<\/p>\r\n[reveal-answer q=\"15174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"15174\"]\r\n\r\n[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
\r\nTry It<\/h3>\r\n[ohm_question hide_question_numbers=1]174904[\/ohm_question]\r\n\r\n<\/div>\r\n\r\nQ & A<\/h3>\r\n
Do two points always determine a unique exponential function?<\/strong><\/p>\r\nYes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x<\/em>, which in many real world cases involves time.<\/em><\/p>\r\n\r\n<\/div>\r\n\r\nHow To: Given the graph of an exponential function, write its equation.<\/h3>\r\n\r\n \t- First, identify two points on the graph. Choose the y<\/em>-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.<\/li>\r\n \t
- If one of the data points is the y-<\/em>intercept [latex]\\left(0,a\\right)[\/latex] , then a<\/em>\u00a0is the initial value. Using a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex], and solve for b<\/em>.<\/li>\r\n \t
- If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex]. Solve the resulting system of two equations in two unknowns to find a<\/em>\u00a0and b<\/em>.<\/li>\r\n \t
- Write the exponential function, [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n\r\n\r\n\r\n
Example 5: Writing an Exponential Function Given Its Graph<\/h3>\r\n
Find an equation for the exponential function graphed in Figure 5.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010811\/CNX_Precalc_Figure_04_01_0042.jpg\")
Figure 5<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"446390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"446390\"]\r\n
We can choose the y<\/em>-intercept of the graph, [latex]\\left(0,3\\right)[\/latex], as our first point. This gives us the initial value, [latex]a=3[\/latex]. Next, choose a point on the curve some distance away from [latex]\\left(0,3\\right)[\/latex] that has integer coordinates. One such point is [latex]\\left(2,12\\right)[\/latex].<\/p>\r\n[latex]\\begin{align}y&=a{b}^{x} && \\text{Write the general form of an exponential equation}. \\\\ y&=3{b}^{x} && \\text{Substitute the initial value 3 for }a. \\\\ 12&=3{b}^{2} && \\text{Substitute in 12 for }y\\text{ and 2 for }x. \\\\ 4&={b}^{2} && \\text{Divide by 3}. \\\\ b&=\\pm 2 && \\text{Take the square root}.\\end{align}[\/latex]<\/p>\r\n
Because we restrict ourselves to positive values of b<\/em>, we will use b\u00a0<\/em>= 2. Substitute a<\/em>\u00a0and b<\/em>\u00a0into the standard form to yield the equation [latex]f\\left(x\\right)=3{\\left(2\\right)}^{x}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\nTry It<\/h3>\r\n
Find an equation for the exponential function graphed in Figure 6.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010811\/CNX_Precalc_Figure_04_01_0052.jpg\")
Figure 6<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"137067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"137067\"]\r\n\r\n[latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}[\/latex]. Answers may vary due to round-off error. The answer should be very close to [latex]1.4142{\\left(1.4142\\right)}^{x}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\nTry It<\/h3>\r\n[ohm_question hide_question_numbers=1]174253[\/ohm_question]\r\n\r\n<\/div>\r\n\r\nHow To: Given two points on the curve of an exponential function, use a graphing calculator to find the equation.<\/h3>\r\n\r\n \t- Press [STAT].<\/strong><\/li>\r\n \t
- Clear any existing entries in columns L1<\/strong> or L2.<\/strong><\/li>\r\n \t
- In L1<\/strong>, enter the x<\/em>-coordinates given.<\/li>\r\n \t
- In L2<\/strong>, enter the corresponding y<\/em>-coordinates.<\/li>\r\n \t
- Press [STAT] <\/strong>again. Cursor right to CALC<\/strong>, scroll down to ExpReg (Exponential Regression)<\/strong>, and press [ENTER].<\/strong><\/li>\r\n \t
- The screen displays the values of a<\/em> and b<\/em> in the exponential equation [latex]y=a\\cdot {b}^{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n\r\n\r\n\r\n
Example 6: Using a Graphing Calculator to Find an Exponential Function<\/h3>\r\n
Use a graphing calculator to find the exponential equation that includes the points [latex]\\left(2,24.8\\right)[\/latex] and [latex]\\left(5,198.4\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"455667\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455667\"]\r\n
Follow the guidelines above. First press [STAT]<\/strong>, [EDIT]<\/strong>, [1: Edit\u2026], <\/strong>and clear the lists L1<\/strong> and L2<\/strong>. Next, in the L1<\/strong> column, enter the x<\/em>-coordinates, 2 and 5. Do the same in the L2<\/strong> column for the y<\/em>-coordinates, 24.8 and 198.4.<\/p>\r\nNow press [STAT]<\/strong>, [CALC]<\/strong>, [0: ExpReg] <\/strong>and press [ENTER]<\/strong>. The values a\u00a0<\/em>= 6.2 and b\u00a0<\/em>= 2 will be displayed. The exponential equation is [latex]y=6.2\\cdot {2}^{x}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\nTry It<\/h3>\r\n
Use a graphing calculator to find the exponential function that includes the points (3, 75.98) and (6, 481.07).<\/p>\r\n[reveal-answer q=\"799504\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"799504\"]\r\n\r\n[latex]y\\approx 12\\cdot {1.85}^{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n
\u00a0Use Compound Interest Formulas<\/h2>\r\n
Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest<\/strong>. The term compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\r\nThe annual percentage rate (APR)<\/strong> of an account, also called the nominal rate<\/strong>, is the yearly interest rate earned by an investment account. The term\u00a0nominal<\/em>\u00a0is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\r\nWe can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t<\/em>, principal P<\/em>, APR
India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.[footnote]http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.[\/footnote] If this rate continues, the population of India will exceed China\u2019s population by the year 2031. When populations grow rapidly, we often say that the growth is \"exponential,\" meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth <\/em>has a very specific meaning. In this section, we will take a look at exponential functions<\/em>, which model this kind of rapid growth.<\/p>\r\n\r\n Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b<\/em>\u00a0to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:<\/p>\r\n\r\n Why do we limit the base to positive values other than 1? Because base 1\u00a0results in the constant function. Observe what happens if the base is\u00a01:<\/p>\r\n\r\n To evaluate an exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute x<\/em>\u00a0with the given value, and calculate the resulting power. For example:<\/p>\r\n Let [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:<\/p>\r\n Let [latex]f\\left(x\\right)=30{\\left(2\\right)}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n Note that if the order of operations were not followed, the result would be incorrect:<\/p>\r\n [latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/p>\r\n\r\n Let [latex]f\\left(x\\right)=5{\\left(3\\right)}^{x+1}[\/latex]. Evaluate [latex]f\\left(2\\right)[\/latex] without using a calculator.<\/p>\r\n[reveal-answer q=\"847875\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847875\"]\r\n Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\r\n [latex]\\begin{align}f\\left(x\\right) & =5{\\left(3\\right)}^{x+1} \\\\ f\\left(2\\right) & =5{\\left(3\\right)}^{2+1} && \\text{Substitute }x=2. \\\\ & =5{\\left(3\\right)}^{3} && \\text{Add the exponents}. \\\\ & =5\\left(27\\right) && \\text{Simplify the power}. \\\\ & =135 && \\text{Multiply}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n Let [latex]f\\left(x\\right)=8{\\left(1.2\\right)}^{x - 5}[\/latex]. Evaluate [latex]f\\left(3\\right)[\/latex] using a calculator. Round to four decimal places.<\/p>\r\n[reveal-answer q=\"292533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"292533\"]\r\n\r\n5.5556\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n Because the output of exponential functions increases very rapidly, the term \"exponential growth\" is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.<\/p>\r\n\r\n A function that models exponential growth<\/strong> grows by a rate proportional to the amount present. For any real number x<\/em>\u00a0and any positive real numbers a\u00a0<\/em>and b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\r\n\r\n where<\/p>\r\n\r\n In more general terms, we have an exponential function<\/em>, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let\u2019s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex].<\/p>\r\n A few years of growth for these companies are illustrated below.<\/p>\r\n\r\n Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.<\/p>\r\n Now we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the base<\/em>, and x<\/em>\u00a0is called the exponent<\/em>.<\/p>\r\n\r\n At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?<\/p>\r\n[reveal-answer q=\"807378\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"807378\"]\r\n To estimate the population in 2031, we evaluate the models for t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\r\n [latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\r\n There will be about 1.549 billion people in India in the year 2031.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex], where t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 2?<\/p>\r\n[reveal-answer q=\"740298\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"740298\"]\r\n\r\nAbout 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a<\/em>\u00a0and b<\/em>, and evaluate the function.<\/p>\r\n\r\n In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N<\/em>(t<\/em>) representing the population N<\/em>\u00a0of deer over time t<\/em>.<\/p>\r\n[reveal-answer q=\"518916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"518916\"]\r\n We let our independent variable t<\/em>\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a\u00a0<\/em>= 80. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find b<\/em>:<\/p>\r\n [latex]\\begin{align}N\\left(t\\right)& =80{b}^{t} \\\\ 180& =80{b}^{6} && \\text{Substitute using point }\\left(6, 180\\right). \\\\ \\frac{9}{4} & ={b}^{6} && \\text{Divide and write in lowest terms}. \\\\ b & ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}} && \\text{Isolate }b\\text{ using properties of exponents}. \\\\ b & \\approx 1.1447 && \\text{Round to 4 decimal places}. \\end{align}[\/latex]<\/p>\r\n NOTE:<\/strong> Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.<\/em><\/p>\r\n The exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)<\/p>\r\nWe can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex], and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].\r\n\r\n Figure 3.\u00a0<\/strong>Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], t<\/em>\u00a0years after 2006<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N<\/em>\u00a0of wolves over time t<\/em>.<\/p>\r\n[reveal-answer q=\"901847\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"901847\"]\r\n\r\n[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n Find an exponential function that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"935128\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"935128\"]\r\n Because we don\u2019t have the initial value, we substitute both points into an equation of the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], and then solve the system for a<\/em>\u00a0and b<\/em>.<\/p>\r\n\r\n Use the first equation to solve for a<\/em>\u00a0in terms of b<\/em>:<\/p>\r\n [latex]\\begin{align}&6=ab^{-2}\\\\&\\frac{6}{b^{-2}}=a && \\text{Divide.} \\\\&a=6b^2 && \\text{Use properties of exponents to rewrite the denominator.}\\end{align}[\/latex]<\/span><\/p>\r\n Substitute a<\/em>\u00a0in the second equation, and solve for b<\/em>:<\/p>\r\n [latex]\\begin{align}&1=ab^2 \\\\ &1=6b^2b^2=6b^4 && \\text{Substitute } a=6b^2. \\\\ &b=\\left(\\frac{1}{6}\\right)^\\frac{1}{4} && \\text{Use properties of exponents to isolate }b. \\\\ & b\\approx0.6389 && \\text{Round to 4 decimal places}. \\end{align}[\/latex]\r\n<\/span><\/p>\r\n Use the value of b<\/em>\u00a0in the first equation to solve for the value of a<\/em>:<\/p>\r\n [latex]a=6b^2\\approx6(0.6389)^2\\approx2.4492[\/latex]<\/span><\/p>\r\n Thus, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex].<\/p>\r\nWe can graph our model to check our work. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(-2,\\text{ 6}\\right)[\/latex] and [latex]\\left(2,\\text{ 1}\\right)[\/latex]. The graph is an example of an exponential decay<\/strong> function.\r\n\r\n Figure 4.\u00a0<\/strong>The graph of [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex] models exponential decay.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n Given the two points [latex]\\left(1,3\\right)[\/latex] and [latex]\\left(2,4.5\\right)[\/latex], find the equation of the exponential function that passes through these two points.<\/p>\r\n[reveal-answer q=\"15174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"15174\"]\r\n\r\n[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n Do two points always determine a unique exponential function?<\/strong><\/p>\r\n Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x<\/em>, which in many real world cases involves time.<\/em><\/p>\r\n\r\n<\/div>\r\n Find an equation for the exponential function graphed in Figure 5.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"] We can choose the y<\/em>-intercept of the graph, [latex]\\left(0,3\\right)[\/latex], as our first point. This gives us the initial value, [latex]a=3[\/latex]. Next, choose a point on the curve some distance away from [latex]\\left(0,3\\right)[\/latex] that has integer coordinates. One such point is [latex]\\left(2,12\\right)[\/latex].<\/p>\r\n [latex]\\begin{align}y&=a{b}^{x} && \\text{Write the general form of an exponential equation}. \\\\ y&=3{b}^{x} && \\text{Substitute the initial value 3 for }a. \\\\ 12&=3{b}^{2} && \\text{Substitute in 12 for }y\\text{ and 2 for }x. \\\\ 4&={b}^{2} && \\text{Divide by 3}. \\\\ b&=\\pm 2 && \\text{Take the square root}.\\end{align}[\/latex]<\/p>\r\n Because we restrict ourselves to positive values of b<\/em>, we will use b\u00a0<\/em>= 2. Substitute a<\/em>\u00a0and b<\/em>\u00a0into the standard form to yield the equation [latex]f\\left(x\\right)=3{\\left(2\\right)}^{x}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n Find an equation for the exponential function graphed in Figure 6.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] Use a graphing calculator to find the exponential equation that includes the points [latex]\\left(2,24.8\\right)[\/latex] and [latex]\\left(5,198.4\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"455667\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455667\"]\r\n Follow the guidelines above. First press [STAT]<\/strong>, [EDIT]<\/strong>, [1: Edit\u2026], <\/strong>and clear the lists L1<\/strong> and L2<\/strong>. Next, in the L1<\/strong> column, enter the x<\/em>-coordinates, 2 and 5. Do the same in the L2<\/strong> column for the y<\/em>-coordinates, 24.8 and 198.4.<\/p>\r\n Now press [STAT]<\/strong>, [CALC]<\/strong>, [0: ExpReg] <\/strong>and press [ENTER]<\/strong>. The values a\u00a0<\/em>= 6.2 and b\u00a0<\/em>= 2 will be displayed. The exponential equation is [latex]y=6.2\\cdot {2}^{x}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n Use a graphing calculator to find the exponential function that includes the points (3, 75.98) and (6, 481.07).<\/p>\r\n[reveal-answer q=\"799504\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"799504\"]\r\n\r\n[latex]y\\approx 12\\cdot {1.85}^{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest<\/strong>. The term compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\r\n The annual percentage rate (APR)<\/strong> of an account, also called the nominal rate<\/strong>, is the yearly interest rate earned by an investment account. The term\u00a0nominal<\/em>\u00a0is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\r\n We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t<\/em>, principal P<\/em>, APR Evaluate Exponential Functions<\/h2>\r\n
\r\n \t
\r\n \t
Example 1: Evaluating Exponential Functions<\/h3>\r\n
Try It<\/h3>\r\n
A General Note: Exponential Growth<\/h3>\r\n
\r\n \t
\r\n\r\n
\r\n Year,\u00a0x<\/em><\/th>\r\n Stores, Company A<\/th>\r\n Stores, Company B<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n\r\n 0<\/td>\r\n 100 + 50(0) = 100<\/td>\r\n 100(1 + 0.5)0<\/sup> = 100<\/td>\r\n<\/tr>\r\n \r\n 1<\/td>\r\n 100 + 50(1) = 150<\/td>\r\n 100(1 + 0.5)1<\/sup> = 150<\/td>\r\n<\/tr>\r\n \r\n 2<\/td>\r\n 100 + 50(2) = 200<\/td>\r\n 100(1 + 0.5)2<\/sup> = 225<\/td>\r\n<\/tr>\r\n \r\n 3<\/td>\r\n 100 + 50(3) = 250<\/td>\r\n 100(1 + 0.5)3<\/sup> =\u00a0337.5<\/td>\r\n<\/tr>\r\n \r\n x<\/em><\/td>\r\n A<\/em>(x<\/em>) = 100 + 50x<\/td>\r\n B<\/em>(x<\/em>) = 100(1 + 0.5)x<\/em><\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe graphs comparing the number of stores for each company over a five-year period are shown in below.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] Figure 2.<\/b> The graph shows the numbers of stores Companies A and B opened over a five-year period.[\/caption]\r\nExample 2: Evaluating a Real-World Exponential Model<\/h3>\r\n
Try It<\/h3>\r\n
\u00a0Find the Equation of an Exponential Function<\/h2>\r\n
How To: Given two data points, write an exponential model.<\/h3>\r\n
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Example 3: Writing an Exponential Model When the Initial Value Is Known<\/h3>\r\n
<\/span>\r\nTry It<\/h3>\r\n
Example 4: Writing an Exponential Model When the Initial Value is Not Known<\/h3>\r\n
\r\n \t
<\/span>\r\nTry It<\/h3>\r\n
Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174904[\/ohm_question]\r\n\r\n<\/div>\r\n
Q & A<\/h3>\r\n
How To: Given the graph of an exponential function, write its equation.<\/h3>\r\n
\r\n \t
Example 5: Writing an Exponential Function Given Its Graph<\/h3>\r\n
Figure 5<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"446390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"446390\"]\r\nTry It<\/h3>\r\n
Figure 6<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"137067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"137067\"]\r\n\r\n[latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}[\/latex]. Answers may vary due to round-off error. The answer should be very close to [latex]1.4142{\\left(1.4142\\right)}^{x}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTry It<\/h3>\r\n[ohm_question hide_question_numbers=1]174253[\/ohm_question]\r\n\r\n<\/div>\r\n
How To: Given two points on the curve of an exponential function, use a graphing calculator to find the equation.<\/h3>\r\n
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Example 6: Using a Graphing Calculator to Find an Exponential Function<\/h3>\r\n
Try It<\/h3>\r\n
\u00a0Use Compound Interest Formulas<\/h2>\r\n