This table shows the value, v, of a car based on its age, a, assuming normal usage.

a. Find an equation that can be used to predict the value of the car, U, from the age of the car, a.

b. What is the average rate of change in the value of the car between years 7 and 8? Show your work or explain how you know.

c. At what age, in years, will the value of the car be closest to $5,000? Round your answer to the nearest year. Show your work or explain how you know.

  a. U(a) = 20000·0.9^a

  b. about -$957 dollars per year

  c. 13 years

Step-by-step explanation:

a) The age values are sequential, but the differences between "v" values are not constant. We suspect an exponential function. When we examine the ratios of adjacent table values, we find they are all 0.9, so the exponential function will be ...

  U = (initial value)·(0.9^(age))

  U(a) = 20000·0.9^a

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b) The average rate of change for the one year between age 7 and age 8 will be U(8) -U(7). A calculator evaluates that difference as -$956.59, about -$957.

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c) The car value drops by almost half in 6 years, so will drop by almost half of that in another 6 years. The car's age will be 12 or 13 years. A graph shows the age to be just over 13 years when the value is predicted to be $5000.

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A graphing calculator was used to evaluate the function and find the time associated with the given value. (See attached.) This is the "work" that was done to answer the questions.

The time to decrease to 1/4 of the original value can be found by solving ...

  5000 = 25000·0.9^a

The solution is ...

  a = log(.25)/log(0.9) ≈ 13.158 . . . . as found on the graph

Here, we have to examine the equation of the straight line which is denoted by: y = m x +c where "m" is the slope which represents the steepness and c is the y-intercept

Here, the two linear functions have the same slope "m" and the same y-intercept "c". When both these are the same, the two linear functions are representing the same straight line.

Therefore, Jeremy is correct in his argument.