Question

Chad invests $12,000 at a 5% annual interest rate, compounded annually. Write a function A(t) that finds the amount Chad has in his account after t years. Explain what the y-intercept represents. Describe the end behavior of the graph of A(t) as t increases. A. A(t) = 12000(1.05)^t; The y-intercept represents the initial investment. The graph increases exponentially as t increases. B. A(t) = 12000(0.05)^t; The y-intercept represents the initial investment. The graph decreases exponentially as t increases. C. A(t) = 12000(1.05)^t; The y-intercept represents the annual interest rate. The graph increases linearly as t increases. D. A(t) = 12000(0.05)^t; The y-intercept represents the annual interest rate. The graph decreases linearly as t increases.

Answer 1

The function A(t) = 12000(1.05)^t accurately represents Chad's investment over time, with the y-intercept being his initial investment of $12000 and a trend of increasing value as time goes on due to the 5% annual interest rate.

Step-by-step explanation:

The correct function for Chad's investment growth over time is A(t) = 12000(1.05)^t. In the equation, 't' represents time in years, '1.05' represents the annual interest rate (since 5% interest is equivalent to 1.05 in the function) and '$12000' is the initial investment. The y-intercept of the graph represents the initial investment amount of $12000.

As for the end behavior of the graph, as 't' (time) increases, the value of A(t) also increases. This is because the function is an example of exponential growth, where the growth rate (the 5% interest) is constant.

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