Exponential Growth and Decay Formula:
Initial Starting Value # of times it grows or decays y
 ab x
Growth/Decay Factor

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?

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