6.1 Exponential growth and decay

What does “a” and b” do?

a is the starting value

b is the multiplier

Identify growth vs decay function ***

growth / b is found as y at x = 1

decay / b is y at the x = -1

Uses in real life

Finances, depreciation (house, car etc), population

Compound interest problem ***

A = P(1 + r/n)nt

A = Final amount

P = Starting value / Principal amount

r = Rate as decimal (3% → 0.03)

n = # of compounds (1 → annual, 2 → semi-annual, 4 → quarterly, 12 → monthly)

t = Time in years

For example, if you invest $1,000 at an annual interest rate of 5% compounded annually for 3 years, you would substitute the values into the formula as follows:

A = 1000(1 + 0.05/1)1×3
A = 1000(1 + 0.05)3
A = 1000(1.05)3
A = 1000 * 1.157625
A ≈ 1157.63

Match an exponential function to a graph

Growth Functions: a > 0, b > 1

  • Increases overtime, J shape

  • Domain: all real (-infinity, +infinity)

  • Range: y > 0

  • Asymptope: y = 0

  • y-int: (0,a)

Decay Functions: a > 0, 0 < b < 1

  • Decreases overtime, reverse J

  • Domain: all real (-infinity, +infinity)

  • Range: y > 0

  • Asymptope: y = 0

  • y-int: (0,a)

For each graph below, determine what the value of “b’” is.

b in this function is the multiplier, meaning the increase is b. seeing.

b is the also the point at x 1

Real life problem ***

The value of a car y (in thousands of dollars) can be approximated by the model y = 25(0.85)t, where t is the number of years since the car was new.

a. Tell whether the model represents exponential growth or exponential decay.

Decay → (1 - r) = (1 - 0.85) = 0.15 or 15%

b. Identify the annual percent increase or decrease in the value ofthe car.

15% decrease

c. Estimate when the value of the car will be $8000.

7 years (no clue why)

You deposit $9000 in an account that pays 1.46% annual interest. Find the balance

after 3 years when the interest is compounded quarterly.

A = 9000(1 + 0.0146/4)4 × 3