SECTION 5.9 Building Exponential, Logarithmic, and Logistic Models from Data 367 Figure 60 y 5 abx, 0 , b , 1, a . 0 Exponential y x y 5 abx, a . 0, b . 1 Exponential y x y 5 a 1b In x, a . 0, b , 0 Logarithmic y x y 5 a 1b In x, a . 0, b . 0 Logarithmic y x y x 1 1ae2bx c y 5 , a . 0, b . 0, c . 0 Logistic Figure 61 TI-84 Plus CE 0 21 40,000 7 The correlation coefficient r will appear only if the model can be written as a linear expression. As it turns out, r will appear for the linear, power, exponential, and logarithmic models, since these models can be written as a linear expression. Remember, the closer r is to 1, the better the fit. 1 Build an Exponential Model from Data We saw in Section 5.7 that money earning compound interest grows exponentially, and we saw in Section 5.8 that growth and decay models also can behave exponentially. The next example shows how data can lead to an exponential model. Fitting an Exponential Function to Data Mariah deposited $20,000 into a well-diversified mutual fund 6 years ago. The data in Table 9 represent the value of the account each year for the last 7 years. (a) Using a graphing utility, draw a scatter plot with year as the independent variable. (b) Using a graphing utility, build an exponential model from the data. (c) Express the function found in part (b) in the form = A A e . kt 0 (d) Graph the exponential function found in part (b) or (c) on the scatter plot. (e) Using the solution to part (b) or (c), predict the value of the account after 10 years. (f) Interpret the value of k found in part (c). EXAMPLE 1 Solution (a) Enter the data into the graphing utility and draw the scatter plot as shown in Figure 61 on a TI-84 Plus CE. (b) A graphing utility fits the data in Table 9 to an exponential model of the form = y abx using the EXPonential REGression option. Figure 62 shows that ( ) = = y ab 19,820.43 1.085568 x x on a TI-84 Plus CE. Notice that = r 0.999, which is close to 1, indicating a good fit. (c) To express = y abx in the form = A A e , kt 0 where = x t and = y A, proceed as follows: = ab A e x kt 0 If = = x t 0, then = a A .0 This leads to ( ) = = = b e b e b e x kt x k t k x t = Because ( ) = = y ab 19,820.43 1.085568 , x x this means that = a 19,820.43 and = b 1.085568. = = = = a A b e 19,820.43 and 1.085568 k 0 To find k, rewrite = e 1.085568 k as a logarithm to obtain ( ) = ≈ k ln 1.085568 0.08210 As a result, = = A A e e 19,820.43 . kt t 0 0.08210 Year, x Account Value, y 0 20,000 1 21,516 2 23,355 3 24,885 4 27,484 5 30,053 6 32,622 Table 9 Figure 62 Exponential model using a TI-84 Plus CE (continued)
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