Bertha is investing in a 3-year term investment account. So far, she has deposited 1,2 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 rate paid each year. Write a model polynomial, C(x), that represents the final amount of Bertha's investment account. Find the final amount of Bertha's investment account if the interest rate is 5.6 percent. Round the answer to two decimal places. 1. C(x)=□ 2. The final amount of Bertha's investment account if the interest rate is 5.6 percent is $□.

Answer 1

To create a polynomial model \( C(x) \) that represents the final amount of Bertha's investment account, we need to consider how much each deposit will grow over the 3-year term, depending on the timing of the deposits.

1. Investments and their growth:

   • For the first deposit of $1,200 made at the beginning of the first year, it will earn interest for 3 years: \[ 1200 \cdot x^3 \]
   • For the second deposit of $880 made at the beginning of the second year, it will earn interest for 2 years: \[ 880 \cdot x^2 \]
   • For the third deposit of $830 made at the beginning of the third year, it will earn interest for 1 year: \[ 830 \cdot x^1 \]

2. Combining the investments: The model polynomial \( C(x) \) can be expressed as: \[ C(x) = 1200x^3 + 880x^2 + 830x \]

3. Substituting the interest rate: If the interest rate \( r \) is 5.6%, then \( x = 1 + r = 1 + 0.056 = 1.056 \).

4. Calculating the final amount: We will substitute \( x = 1.056 \) into the polynomial \( C(x) \): \[ C(1.056) = 1200(1.056)^3 + 880(1.056)^2 + 830(1.056) \]

   First calculate \( (1.056)^3 \): \[ (1.056)^3 \approx 1.191016 \]

   Then calculate \( (1.056)^2 \): \[ (1.056)^2 \approx 1.118336 \]

   Now plug these values back into \( C(x) \): \[ C(1.056) = 1200(1.191016) + 880(1.118336) + 830(1.056) \] \[ C(1.056) \approx 1200 \cdot 1.191016 + 880 \cdot 1.118336 + 830 \cdot 1.056 \] \[ \approx 1429.2192 + 983.3288 + 875.88 \] \[ \approx 3288.428 \]

   Rounding to two decimal places: \[ \approx 3288.43 \]

Final Results:

1. The model polynomial \( C(x) \) is: \[ C(x) = 1200x^3 + 880x^2 + 830x \]
2. The final amount of Bertha's investment account if the interest rate is 5.6 percent is: \[ $3288.43 \]

Thus the complete answers are:

1. \( C(x) = 1200x^3 + 880x^2 + 830x \)
2. The final amount of Bertha's investment account if the interest rate is 5.6 percent is $3288.43.