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## Lesson 6-2 Exponential Functions

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**Lesson 6-2 Exponential Functions**Objective: To identify and solve exponential functions**Exponential Functions**• Exponential Functions have the form: • A constant raised to a variable power. f(x) = bx where b >0 & b ≠ 1**Graphs of Exponential Functions**• Enter into calculator and graph: • f(x) = 2x • [2nd] [table] to see list of data points • Notice how quickly exponential functions grow.**Properties of Exponential Functions**• If bu = bv then u = v When b > 0 and b≠ 1 • The graph of f(x) = bx always passes through the points (0,1) and (1,b) • The graph of f(x) = bx is the reflection about the y-axis of the graph of f(x)= • The graph of f(x) = bx has the horizontal asymptote y = 0.**Properties of Exponential Functions**• The domain of f(x) = bx is the set of real numbers: the range is the set of positive numbers. • f(x) = bxis increasing if b > 1; f(x) = bxis decreasing if 0< b< 1. • f(x) = bxis a one-to-one function since it passes the horizontal line test.**Graph the exponential Function:**• f(x) = 4x**Solving Exponential Equations**• 310 = 35x • 10 = 5x • x = 2 • 27 = (x-1)7 if x > 1 • 2 = x-1 • 3 = x • 33x = 9x-1 • 33x = 32(x-1) • 3x = 2x -2 • x = -2**Practice**• 28 = 2x+1 • 42x+1 = 411 • 8x+1 = 2**The Number e**• e is an irrational number • It is , as m gets larger and larger. • It is approximately 2.71828 • f(x)= ex is a natural exponential function • graph f(x)= ex on the calculator**Warm up**• Solve for x: • 1. 53 − 2x =5−x • 2. 32a=3−a • 3. 31 − 2x= 243**Applications of exponential functions**• Compound Interest • Continuous Compounding • Exponential Growth or decay (bacteria/ radiation half life)**Compound interest**• Compound interest means the each payment is calculated by including the interest previously earned on the investment.**formula**If you have a bank account whose principal = $1000, and your bank compounds the interest twice a year at an interest rate of 5%, how much money do you have in your account at the year's end?**Continous Compounding**• When n gets very large it approaches becoming continuous compounding. The formula is: • P = principal amount (initial investment)r = annual interest rate (as a decimal)t = number of yearsA = amount after time t**Example**• An amount of $2,340.00 is deposited in a bank paying an annual interest rate of 3.1%, compounded continuously. Find the balance after 3 years. • Solution • A = 2340 e(.031)(3) • A = 2568.06**Exponential Growth**• A = Pert...or... A = Pekt ...or... Q =ekt • ...or... Q = Q0ekt • k is the growth constant**Bacteria Growth**• In t hours the number of bacteria in a culture will grow to be approximately Q = Q0e2t where Q0 is the original number of bacteria. At 1 PM the culture has 50 bacteria. How many bacteria does it have at 4 PM? at noon? • Q = 50e2(3) Q = 50e2(-1) • Q = 50e6 Q = 50e-2 • Q = 20,248 Q = 7**Practice**• 1. If you start a bank account with $10,000 and your bank compounds the interest quarterly at an interest rate of 8%, how much money do you have at the year’s end ? (assume that you do not add or withdraw any money from the account) • 2. An amount of $1,240.00 is deposited in a bank paying an annual interest rate of 2.85 %, compounded continuously. Find the balance after 2½ years.**Solution**• 1. • 2. A = 1240e(.0285)(2.5) = $1,331.57**Exponential Decay**• An artifact originally had 12 grams of carbon-14 present. The decay model A = 12e-0.000121t describes the amount of carbon-14 present after t years. How many grams of carbon-14 will be present in this artifact after 10,000 years? • A = 12e-0.000121t • A = 12e-0.000121(10,000) • A = 12e-1.21 • A = 3.58