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8-2 Properties of Exponential Functions

Learn about properties of exponential functions with varying constants. Explore graph translations, reflections, and asymptotes. Real-world examples and vocabulary included.

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8-2 Properties of Exponential Functions

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  1. 8-2 Properties of Exponential Functions Identifying the role of constants in y=abx Using e as a base

  2. Objectives Comparing Graphs The number e

  3. Vocabulary Parent Function: Stretch (|a| > 1) Shrink (0 < |a| < 1) Reflection: Vertical Translation: Horizontal Translation: Combined:

  4. Step 1: Make a table of values Step 2: Graph each function. xy = 3 · 2xy = –3 · 2x –3 –2 –1 0 3 –3 1 6 –6 2 12 –12 3 24 –24 3 8 3 8 – 3 4 3 4 – 1 2 1 2 1 1 – Graphing y = abx for 0<|a|<1 Graph y = 3 • 2x and y = –3 • 2x. Label the asymptote of each graph. y = 3 • 2xstretches y = 2x by a factor of 3. y = –3 • 2x reflects y = 3 • 2x in the x-axis.

  5. 1 2 1 2 1 2 Step 1: Graph y = 6. The horizontal asymptote is y = 0. x 1 2 x–3 Step 2:For y = 6 – 2, h = 3 and k = –2. So shift the graph of the present function 3 units right and 2 units down. Translating y = abx Graph y = 6 and y = 6 – 2. x x–3 The horizontal asymptote is y = –2.

  6. Relate: The amount of technetium-99m is an exponential function of the number of half-lives. The initial amount is 50 mg. The decay factor is . One half-life equals 6 h. 1 2 Define: Let y = the amount of technetium-99m. Let x = the number of hours elapsed. Then x = the number of half-lives. 1 6 Real-World Example Since a 100-mg supply of technetium-99m has a half-life of 6 hours, find the amount of technetium-99m that remains from a 50-mg supply after 25 hours.

  7. 1 6 1 2 x Write:y = 50 y = 50 Substitute 25 for x. 1 6 1 2 • 25 Continued (continued) 4.16 1 2 = 50 Simplify. 2.784 After 25 hours, about 2.784 mg of technetium-99m remains.

  8. Vocabulary The number e is a irrational and has an approximate value of 2.71828 Exponential functions with the base e are useful for describing continuous growth or decay models.

  9. Step 1: Graph y = ex. Step 2: Find y when x = 3. Evaluating e Graph y = ex. Evaluate e3 to four decimal places. The value of e3 is about 20.0855.

  10. Vocabulary Continuously Compounded Interest Formula amount in account rate A = Pert time principal

  11. 115.49 Simplify. Real-World Example Suppose you invest $100 at an annual interest rate of 4.8% compounded continuously. How much will you have in the account after 3 years? A = Pert = 100 • e0.048(3)Substitute 100 for P, 0.048 for r, and 3 for t. = 100 • e0.144Simplify. = 100 • (1.154884) Evaluate e0.144. You will have about $115.49 in the account after three years.

  12. Homework 8-2 p 442 #1,2,9,10,15,18,19,24,25

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