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8.2 – Properties of Exponential Functions Review: what is an asymptote? “Walking halfway to the wall” An Asymptote is a line that a graph approaches as x or y increases in absolute value. In this example, the asymptote is the x axis. y 10(2) x Graphing y=abx when a<0 1 x 1 x • Ex: Graph y (2) and y (2) 2 2 Sketch your prediction of what the graph will look like Where is the asymptote? Translating y=abx • How does the equation change if we want to move both graphs up 4 units? Predictions? y y 1 x 1 (2) and y (2) x 2 2 1 x 1 (2) 4 and y (2) x 4 2 2 Question: where is the asymptote now? To move the graph up or down, add or subtract units at the end of the equations. No need to use inverses – if you want to go up, add; if you want to go down, subtract. Translating y=abx • How does the equation change if we want to move both graphs left 4 units? Predictions? 1 x 1 (2) and y (2) x 2 2 1 1 y (2) x 4 and y (2) x 4 2 2 y To move the graph left or right, add or subtract units to the exponent of the equation. Reminder: use the inverse of how you want the graph to move (e.g. x-4 will move to the right; x+4 will move to the left) Let’s try some • Graph each function as a translation of y=9(3)x a) y 9(3) x 1 b) y 9(3) 4 x Make a table of values for each Graph, from -3 to 3 c) y 9(3) x 4 1 y =9(3)x+1 y =9(3)x-4-1 y =9(3)x-4 “e = 2.718” What is base “e” ? e is an irrational number, approximately equal to 2.718. Exponential functions with a base of e are useful for describing continuous growth or decay. In the graph below, y = e is the asymptote to the graph. y=e Graphing ex • Using your graphing calculators, graph y=ex. Evaluate e4 to four decimal places. We now need to evaluate where x=4 2. Press 2nd, Calc and select 1 (value). Press enter 3. We are evaluating when x=4. Enter 4 for x and press enter. The value of e4 is about 54.59815 Your turn: evaluate e-3 0.0498 So, what is “e” good for??? Continuously Compounding Interest • A = Pert A = amount of money in the account P = principal (how much is deposited) r = annual rate of interest t = time (in years) Example: Continuously Compounded Interest Problem • You invest $1,050 at an annual interest rate of 5.5%, compounded continuously. How much will you have in the account after 5 years? •Start with: P=$1050, r=5.5% = 0.055, t=5 A = Pert Substitute in for p, r, and t 1050(e)0.055(5) 1050(e)0.275 Simplify they power 1050(1.316531) Evaluate e0.275 with your calculator A = $1382.36 Simplify Let’s try one: • Suppose you invest $1,300 at an annual interest rate of 4.3%, compounded continuously. How much will you have in the account after three years? Suppose you invest $1,300 at an annual interest rate of 4.3%, compounded continuously. How much will you have in the account after three years?