Hailey purchased a new car in 2000 for $29,600. The value of the car has been depreciating exponentially at a constant rate. If the value of the car was $12,000 in the year 2007, then what would be the predicted value of the car in the year 2012, to the nearest dollar?

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Explanation:

x = number of years since the year 2000

eg: x = 1 represents the year 2001

y = value of the car in dollars

The point (x,y) = (0,29600) is on the curve and it's the y intercept. This is when Hailey bought the car with a value of $29600.

Another point on this curve is (7,12000) to indicate seven years have passed and the car is now worth $12000.

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Since the y intercept is 29600, this is the value of 'a' in the equation

y = a*b^x

Meaning we can update it to

y = 29600*b^x

Plug in the coordinates of the other point so we can solve for b

y = 29600*b^x

12000 = 29600*b^7

12000/29600 = b^7

b^7 = 12000/29600

b^7 = 0.405405

b = (0.405405)^(1/7)

b = 0.878991 approximately

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We found that

So we go from

y = a*b^x

to

y = 29600*0.878991^x

To verify this, plugging x = 0 should lead to y = 29600

Also, plugging in x = 7 should lead to y = 12000 or very close to it.

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The last step is to plug in x = 12 to find the car's value in 2012, i.e. x = 12 years after the year 2000.

So,

y = 29600*0.878991^x

y = 29600*0.878991^12

y = 6296.58182216087

y = 6297

The predicted car's value in 2007 is about $6297

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