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# If a new car is valued at \$21000 and 10 years later it is valued at \$3000, then what is the average rate of change of its value during those 10 years?

### 4 Answers by Expert Tutors

Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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"If a new car is valued at \$21000 and 10 years later it is valued at \$3000, then what is the average rate of change of its value during those 10 years?"

The Average Rate of Change of a function f(x) between x = x1 and x = x2 is defined as:

Δy/Δx = (f(x2) - f(x1))/(x2 - x1). [See any precalculus or calculus I text.]

For this question:

Δy/Δx = (3000 - 21000)/(10 - 0) = –18000/10 = –\$1800/year.

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Δy/Δx is the slope of the Secant Line between P1 and P2. If P2 is moved along the function graph closer and closer to P1, the secant line they define will approach the tangent line at P1. The slope of the tangent line is the slope of the function at P1 and the derivative of f at P1.
Eric Y. | SAT PrepSAT Prep
5.0 5.0 (5 lesson ratings) (5)
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Most likely, the car loses a percentage of its value every year.
If that is the case, then we should use the formula for exponential decay.

X = Xo(1-A)t

Xo is the original value
A is the percentage lost every year
t is the number of years
X is the new value

3000 = 21000(1-A)10
1/7 = (1-A)10
1/7(1/10) = (1-A)
1-A = .823
A  = 17.7% value lost per year

In order to visualize what is happening...
Initial value : 21000
After 1 year: 21000*.823 = 17283
After 2 years: 17283*.823 = 14223
...
After 10 years: 21000*.82310

Hmm, although I do think that the way the question is worded, this question is much simpler than I've made it out to be, and Eugene's answer is correct.

Since the problem asked for the "rate" of change, I would think this answer would work.

However, there's no way to know since we have no context of the problem.  (And his name is Stephen - he's from Eugene. ;-) LOL )
LOL! I don't mind being called Eugene, but thanks, Kay! 8-)
Stephen W. | Othello, Eltopia, Connell, Pasco, West Pasco - I am in your area!Othello, Eltopia, Connell, Pasco, West P...
4.9 4.9 (43 lesson ratings) (43)
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21,000 - 3,000 = 18,000

Divide by the number of years:

18,000 / 10 = Average rate of change: \$1,800

This is slope of the line, it indicates that price each year reduces by \$1800.
Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
4.7 4.7 (184 lesson ratings) (184)
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Hi Melissa,
There are several ways to answer this depending on how you interpret average rate of change. I'll answer it first using the simplest .

The  change in value =3000 -21000 =-18000 in 10 years so the average rate of change is -18000/10 =-1800 \$/year. Stated in words the car has lost \$1,800 in value per year.

You can also say that the percent change in value is 100*(3000-21000)/21000 =85.71% so the average percent decline in value is 8.57%/year.

Hope this helps, let me know if this is what you were looking for.

Jim