Training Camp (cont 7)

Work from y = ab^x. This scenario relates monetary growth to time. Therefore, the amount of the investment, A(t), is the dependent variable, y.

In financial applications, the independent variable, x, represents the number of compounding periods after t years. In this case, interest is compounded semi-annually. This means every year has two compounding periods; therefore, t years have 2t compounding periods.

For this problem, the starting value is the initial investment, $500. Therefore, the a-value in this equation is $500.

The annual rate of interest is 6% compounded semi-annually. This means the interest rate per compounding period is . Therefore, the growth rate of the exponential function is 0.03. In the equation, this becomes b = 1 + 0.03 = 1.03.

The exponential function that models the investment is A(t) = 500(1.03)^2t.

Finding the time necessary for the investment to double means find t when A(t) = 1000.

Solve 1000 = 500(1.03)^2t using the graphical method introduced in Lesson 6B.

Step 1: Turn off the STAT PLOT; press 2^nd, Y=, ENTER, right arrow, ENTER.

Step 2: Clear the functions; press Y= then, CLEAR for each function that needs deleting.

Step 3: Select values for the viewing window; press WINDOW. Use the keypad to type the values. Use up and down arrow keys to scroll through the list.

Step 4: Input both sides of the equation; press Y=. Use the keypad to type both expressions.

Step 5: Graph the expressions; press GRAPH.

Step 6: Go to CALCULATE menu; press 2^nd, TRACE.

Step 7: Find the intersection of Y₁ and Y₂; press 5, ENTER, ENTER, ENTER.

Rounded to the nearest whole year, the answer is x = 12.

The investment of $500 at 6% per annum compounded semi-annually doubles after approximately 12 years.