Growth or Decay?
Is the following growth or decay: f(x) = 2x
Growth, b = 2, b > 1
The population in the town of Huntersville is presently 38,300. The town grows at an annual rate of 1.2%. Find the number population of the town after 9 years.
f(t) = 38300(1 + 0.012)t
f(9) = 38300(1.012)9
Approximately 42,640 people
Use y = 250(1.2)t, What is the initial value?
Initial Value: 250
Write an exponential growth function to model the situation. A population of 422,000 increases by 12% each year.
f(x) = a(1 + r)x
f(x) = 422000(1 + 0.12)x
f(x) = 422000(1.12)x
Simplify: 6tv0
6t
Is the following growth or decay: f(x)=100(0.5)x
Decay, b = 0.5, 0 < b < 1
$1,200 is invested at an annual rate of 3.2%. How much money will the account have after 12 years?
A = P(1 + r/n)(n)(t)
A = 1200(1 + 0.032/1)(1)(12)
$1751.21
Use y = 250(1.2)t, what is the growth factor?
Growth Factor: 1.2
The population of Baconburg started off at 20,000, and grew by 13% each year. Write an equation to find how long it took the population to grow to 43,000.
f(t) = a(1 + r)t
43000 = 20000(1 + 0.13)t
43000 = 20000(1.13)t
Simplify: 2(4)(x-2)
8 / x2
Is the following growth or decay: f(x)= 100(1.4)x
Growth, b = 1.4, b > 1
The population in the town of Deersburgh is presently 42,500. The town has been growing at a steady rate of 2.7%. What will the population of the town be in 5 years?
f(t) = 42500(1 + 0.027)t
f(5) = 42500(1.027)5
Approximately 48,555 people
Use y = 42500(1.37)t, what is the growth rate (percent)?
Growth Rate: r = |b-1| = |1.37-1| = 0.37 = 37%
The value of a car bought for $45,000 decreases in value each year. After 10 years the car is now worth $15,000. Write an equation to determine the rate at which the car depreciated.
f(t) = a(1 - r)t
15000 = 45000(1 - r)10
Simplify: 4(x-2)(g3)
4g3 / x2
Is the following growth or decay: f(x) = 7 (1.5)2x
Growth, b = 1.52 = 2.25, b > 1
The value of a car was $22,000 when it was purchased. They car depreciates at a rate of 19% per year. How much will the car be worth in 8 years?
f(t) = 22000(1 - 0.19)t
f(8) = 22000(0.81)8
About $4,076.64
Use y = 9.8e(1/2)t, what is the growth factor?
Growth Factor: e(1/2) = 1.65
The fish in The Magic Forest Lake were declining at an annual rate at 1.5%. Their current number is estimated at 2500. Write an equation to model the decreasing number of fish.
y = a(1 - r)x
y = 2500(1 - 0.015)x
y = 2500(.985)x
Simplify: ( (4-2)(x-5)(y-9)(z-3) )0
1
Is the following growth or decay:
f(x) = -350e(1/2)x
Growth, b = e(1/2) = 1.65, b > 1
The value of a stock when purchased is $10 a share. However, over the past 5 days the price went down at a constant rate of 4%. How much is the stock worth now?
f(t) = 10(1 - 0.04)t
f(5) = 10(0.096)5
About $8.15 a share.
y = 9.8(0.65)(3/4)t, what is the decay rate (percent)?
Decay Rate: b = 0.65(3/4) = 0.72
r = |b - 1| = |0.72 - 1| = 0.28 = 28%
There was $10,000 invested in a money market that pays 1.91% interest. After 5 years there is $11,000 in the account. Write an equation to find the compounding periods this bank is using to pay interest.
A = P(1+r/n)(n)(t)
11000 = 10000(1 + 0.0191/n)n(5)
( (x-2)(y2) ) / ( (b-4)(c4) )
( (b4)(y2) ) / ( (c4)(x2) )