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A consumer buys a used car for $19300 with monthly payments of $380, an annual interest rate of 3.79%. Find the number of payments. Note: You should use a Graph in an Appropriate Window. List the window settings and draw the graph in your notebooks. Use the Intersect feature on your graphing calculator and write down an answer rounded to 3 decimal places. Give a whole number answer in this box in months for how many payments would be needed.

User Pedro Vaz
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the total for the car is $19300, with an interest of 3.79%. that means that the total amount the person will end up paying includes that interest:

I = P * r * t

Total Accrued value = P (1+r*t)

Total accrued value = 19300 (1+0.0379*t)

where we have used 0.0379 to represent the 3.79% interest in decimal form.

Notice that the total value paid is going to be $380 (the monthly payment) times 12 (for a full year of payments) times "t" (the number of years the person would be making payments:

380 * 12 * t = 19300 (1 +.0379 * t )

4560 t = 19300 (1 +.0379 * t)

We can try to solve this equation as suggested by your teacher via a graphing calculator. Allowme to plot the two expressions on each side of the equal sign, and find where they intersect. Give me a little time to create the image, please.

Where the line in blue is the expression as a function of time: for the monthly payments: 4580 * t

and the line in orange is the accrued value (total of the car price plus the interest accumulated over "t" number of years: 19300 (1+ 0.0379 t)

We found the crossing of the two lines at a point very close to 5 years: 5.041 years (rounded to three decimals as required).

Now we calculate the number of monthly payments multiplying 5.041 years times 12 (for the 12 months in every year) and get:

total number of payments is: 60.492 . Now this number rounded to the nearest whole number (in math terms) would be 60, but if we consider that the total will not be completely covered in 60 payments, if the 60 answer fails, please try 61.

A consumer buys a used car for $19300 with monthly payments of $380, an annual interest-example-1
User Wloescher
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