Teacher Guide and Answer Key, Precalculus Activity

You are planning to sell a used 1988 automobile and want to establish an asking price that is competitive with that of other cars of the same make and model that are on the market. A review of newspaper advertisements for used cars yields the following data for 12 different cars of this make and model. You want to fit a least squares regression model to these data for use as a model in establishing the asking price for your car.

Production Year	1990	1991	1992	1987	1993	1991	1993	1985	1984	1982	1986	1979
Asking Price (in thousands of dollars)	6.0	7.7	8.8	3.4	9.8	8.4	8.9	1.5	1.6	1.4	2.0	1.0

Establish an asking price for the 1988 automobile using 4 models (linear, exponential, quadratic, and power). Let t = years since 1978 and p = price in thousands of dollars. Use your calculator's linear regression feature to determine a linear function in the form of:

a) p = mt + b using (t,p) data. p = .718997t-1.968552

p(10) = 5.2 Therefore, using the linear model, the predicted asking price for a 1988 automobile is $5200.

b) ln p = mt + b using the transformed data (t, ln p). Using exponential/logarithmic properties, re-express the equation as an exponential function in the form of p = ae^kt to determine the asking price.

ln p = .185021t - .492845

e^{ln p} = e ^{.185021t - .492845}

p = e ^.185021t ÷ e ^.492845

p = .610886 e ^.185021t

p(10) = 3.9 Therefore, using this exponential model, the predicted asking price for a 1988 automobile would be $3900.

 

c) sqrt(p) = mt + b using the transformed data (t, sqrt(p)). Using algebraic properties, re-express the equation in quadratic form, p = at² + bt + c, to determine the asking price of the automobile.

sqrt(p) = .175587t + .382519

sqrt(p)² =(.175587t + .382519)²

p = .030831t² +.134331t +.146321

p(10) = 4.6 Therefore, using this quadratic model, the predicted asking price for a 1988 automobile would be $4600.

d) ln p = m · ln t + b using the transformed data (ln t, ln p). Using exponential/logarithmic properties, re-express the equation as a power function, p = at^b, to determine the asking price of the automobile.

ln p = .953034 ln t - .679989

e^lnp = e^{.95034 ln t - .679989}

p = e ^.95034lnt ÷ e ^.679989

p = 0.506623•(t^.953034)

p(10) = 4.5 Therefore, using this power model, the predicted asking price of a 1988 automobile would be $4500.

e) Graph the 4 models with the data points. Which appears to be the best model? Explain why.