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Exponential Growth and Decay. M & M Lab. Part 1- Growth. Part 2-Decay. What happened to the number of M&Ms?. What happened to the number of M&Ms?. Increased. Decreased. What was true about the values of a &b?. What was true about the values of a &b?. a is the starting number of M&Ms
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M & M Lab Part 1- Growth Part 2-Decay What happened to the number of M&Ms? What happened to the number of M&Ms? • Increased • Decreased What was true about the values of a &b? What was true about the values of a &b? • a is the starting number of M&Ms • b >1 • a is the starting number of M&Ms • b < 1
10.1– Exponential Funtions Ex. 1 Sketch the graph of y = 2x. Then state the functions domain & range. *Sketch of graph must havey-intercept & four other coordinate points. Use the “table” function on your TI-83!
Ex. 2 Sketch the graph of y = (⅞)x. Then state the functions domain & range.
Exponential Growth – for any equation in the form y = bx, b> 1 • Exponential Decay – for any equation in the form y = bx, b< 1 Ex. 3 Determine whether each exponential function represents exponential growth or decay. a. y = 7x growth b. y = (0.5)x decay c. y = 0.3(5)x growth d. y = 4-x = decay
Exponential Growth Model Instead of x, t is the variable. (1+r) is called the growth factor. a is the initial amount. y is the amount after t years.
Exponential GrowthCompound Interest A is the balance. P is the principal deposited in an account. r is the interest rate changed to a decimal (% 100). n is the number of times the interest is compounded in a year. Annually n = 1 Semiannually n = 2 Quarterly n = 4 Monthly n = 12
Compound Interest Problem You deposit $500 in an account that pays 3% interest. Find the balance after 2 years if the interest is compounded quarterly. A = ? P = $500
Exponential Decay Model y is the quantity after t years. t is the variable. a is the initial amount. (1-r) is the decay factor.
Exponential DecayDepreciation You buy a new car for $22,000. The value of the car decreases by 12.5%. What is the value of the car after 4 years. y = ? a = $22,000