y ab  x

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Exponential Growth and Decay Formula:

Initial Starting Value # of times it grows or decays

y

ab x

Growth/Decay Factor

Exponential Growth

y

ab x b

1

Growth Factor: b = 1 + Rate of Growth

$10 is invested in a savings account where is grows 5% per year.

y

x y

10(1.05)

x

What is the y –intercept?

Exponential Decay

y

ab x

0

 

1

Decay Factor: b = 1 - Rate of Growth

10 grams of a particular liquid decays at a rate of 75% per day.

11

y y

 

10(0.25)

x x

11 10 10 9 9 8 8 7 7 6 6

Practice: Monthly benefits for Social Security in May 1992 were $23,307 million. Since then, benefits have increased about 5.4% per year.

a) Write an exponential function to model the growth of monthly Social Security benefits paid each year. (use millions

in your answer!)

y = 23,307(1+0.054) x  y = 23,307(1.054) x b) If benefits continue to grow at this rate, when will the monthly Social Security benefits reach $50,000 million?

50,000 = 23,307(1.054) x 1)Graph y = 23,307(1.054) x and y = 50,000 2) Solve 2.14527 = 1.054

x through guess and check

In 1984, funds for the Emergency Food Assistance program were about $1,075 million. Since 1984, this fund has decreased about 19% per year.

a) Write an exponential function to model this situation.

Y= 1,075(1 - 0.19) x  y = 1,075(0.81) x There is 81% of the fund LEFT each year b) Estimate the funds available for the Emergency Food Assistance program this year.

Y = 1075(0.81) 24  6.839 million Or graph the equation and TRACE with x = 24

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