Bertha is investing in a 3 year term investment account. So far she has deposited $1,200 at the beginning of the first year and $880 at the beginning of the second year. She is planning to deposit another $830 at the beginning of the third year.

Use x = 1 + r, where r is the interest paid each year. Write a model polynomial C(x) that shows the final amount of Bertha’s investment account.

Find the final amount of Berthas investment account if the interest rate is 5.6 percent. Round the as we to two decimal places.

1 answer

To model Bertha's investment account using the given deposits and an interest rate, we first need to recognize how each deposit compounds over time.

Deposits and their compounding:

  1. The first deposit of $1,200 is made at the beginning of year 1, so it will be in the account for 3 years.
  2. The second deposit of $880 is made at the beginning of year 2, so it will be in the account for 2 years.
  3. The third deposit of $830 is made at the beginning of year 3, so it will be in the account for 1 year.

Let's combine these into a polynomial model \( C(x) \) where \( x = 1 + r \).

Using \( r = 0.056 \) (5.6 percent), we can calculate \( x \): \[ x = 1 + r = 1 + 0.056 = 1.056 \]

Construct the polynomial: The amount of each deposit at the end of the 3 years is as follows:

  1. For the first deposit of $1,200: \[ 1200 \cdot x^3 \]
  2. For the second deposit of $880: \[ 880 \cdot x^2 \]
  3. For the third deposit of $830: \[ 830 \cdot x^1 \]

Combining these, we have the polynomial: \[ C(x) = 1200x^3 + 880x^2 + 830x \]

Now plug in \( x = 1.056 \) and calculate \( C(1.056) \):

  1. Calculate \( x^3 \): \[ (1.056)^3 \approx 1.191016 \]
  2. Calculate \( x^2 \): \[ (1.056)^2 \approx 1.115936 \]
  3. Calculate \( x \): \[ (1.056)^1 = 1.056 \]

Now substituting back into the polynomial: \[ C(1.056) = 1200 \cdot 1.191016 + 880 \cdot 1.115936 + 830 \cdot 1.056 \] Calculating each component:

  1. \( 1200 \cdot 1.191016 \approx 1429.2192 \)
  2. \( 880 \cdot 1.115936 \approx 981.9128 \)
  3. \( 830 \cdot 1.056 \approx 875.88 \)

Finally, sum these amounts: \[ C(1.056) \approx 1429.2192 + 981.9128 + 875.88 \approx 3287.012 \]

Final Calculation: Rounding this to two decimal places, the final amount of Bertha’s investment account after 3 years is approximately: \[ \boxed{3287.01} \]