1. 11-1 Investments and Polynomials Big Idea When amounts are invested periodically and earn interest from the time of investment, the total value can be represented by a polynomial. Goal See how situations in which money is periodically invested in a savings account can lead to polynomials, and do the requisite calculations.

2. A relative gives you $5,000 to putaway for college. 1. You put the money in an account that pays an annual yield of 4% the first year. How much will you have at the end of the year? 2. The account pays an annual yield of 4% during the second year. How much will you have after the second year? 3. The interest rate changes to an annual yield of 4.25% during the third year. How much will there be in the account at the end of three years? 1.5000(1+4%) 5000*1.04 $5200 2. 5200*1.04 $5408 or 5000*1.04² $5408 3. 5408(1+4.25%) 5408*1.0425 $5637.84

3. 4. As a New Year’s resolution, Bert has decided to deposit $100 in a savings account every January 2nd. The account yields 3% interest annually. How much will his savings be worth when he makes his fourth deposit? 100(1.03)³ + 100(1.03)² + 100(1.03) +100 109.2727 + 106.09 + 103 + 100  $418.36 Why isn’t it just 100(1.03) like the exponential growth? 4 100(1.03) means that you deposited only 100 for the entire 4 years not 100 each year. 4

4. 5. Janice has a savings account that has a scale factor of x. She makes deposits at regular yearly intervals. The first year she deposits $800, the 2nd year $300, the 3rd year $450, and the 4th year $775. What is her balance immediately after the 4th deposit? 800x³ + 300x² + 450x +775

5. 6. Which is more advantageous, to invest, $50 per year for 4 yrs. Or to invest $100 in the 1 st year and $100 in the 4 th year? In both instances, the money earns 3% interest a year. Option 1 50(1.03)³ + 50(1.03)² + 50(1.03) + 50 54.63635 + 53.045 + 51.5 + 50 $209.18 Option 2 100(1.03)³ + 100 = 109.2727 + 100 = $209.27 The 2 nd is slightly better than the first.

6. 7.Simplify the following expressions. a.(4r² + 10r – 6) + (7r² - 15r – 6) 4r² + 7r² + 10r - 15r – 6 – 6 11r² - 5r -12 b. (4r² + 10r – 6) - (7r² - 15r – 6) 4r² + 10r – 6 - 7r² + 15r + 6 4r² - 7r² + 10r + 15r – 6 + 6 -3r² + 25r c. Solve (4r² + 10r – 6) - (7r² - 15r – 6) = 0 -3r² + 25r = 0 r = -25 + √ 25² - 4(-3)0 / 2(-3) r = -25 + √25² / 2(-3) r = -25 + 25 / -6 r = 0/-6 = 0 r = -50/-6 = 8⅓