Question

I could use some help with this problem for my algebra 2 class, I'm pretty lost, please give an explanation with your answer, much appreciated!

Question:
Mrs. Galacia is building a chicken farm in 2021 with an initial population of 3750 chickens. The farm grows at a rate of 2.15% annually.
Part (a) Use the exponential growth model to write an equation that estimates the population t years after 2021.
Part (b) Estimate the population of the town in 2036. (I was given an extra step here saying, "Must show plug in step before using calculator")

Answer 1

`y=3750 \cdot 1.0215^t`

(where
`y` is the population and
`t` is time in years after 2021)

5159

Explanation:

Part (a)

General form of an exponential function:
`y=ab^x`

where:

• 
  `a` is the y-intercept (or initial value)
• 
  `b` is the base (or growth factor) in decimal form
• 
  `x` is the independent variable

If
`b > 1` then it is an increasing function

If
`0 < b < 1` then it is a decreasing function

We are told that the initial population is 3750. Therefore,
`a=3750`

We are told that the farm grows at a rate of 2.15% annually. Therefore, if it grows then every year it is 100% + 2.15% = 102.15% of the previous year.

Convert the percentage into a decimal:

102.15% = 102.15/100 = 1.0215

Therefore,
`b=1.0215`

We are told that the independent variable is
`t` (in years).

Therefore, the equation is
`y=3750 \cdot 1.0215^t`

(where
`y` is the population and
`t` is time in years after 2021)

------------------------------------------------------------------------------------------

Part (b)

The year 2036 is 15 years after 2021. Therefore, substitute
`t = 15` into the equation and solve for
`y`:

`\implies y=3750 \cdot 1.0215^(15)`

`\implies y=5159.49068...`

Therefore, an estimate of the population of the town in 2036 is 5159.