**50 ! will give ! ** dominique is thinking about buying a house for $286,000. the table below shows the projected value of two different houses for three years. number of years 1 | 2 | 3 house 1 (value in dollars) 294,580 | 303,417.40 | 312,519.92 house 2 (value in dollars) 295,000 | 304,000 | 313,000 part a: what type of function, linear or exponential, can be used to describe the value of each of the houses after a fixed number of years? explain your answer. part b: write one function for each house to describe the value of the house f(x), in dollars, after x years. part c: dominique wants to purchase a house that would have the greatest value in 25 years. will there be any significant difference in the value of either house after 25 years? explain your answer, and show the value of each house after 25 years.

The values of two houses (in thousands of dollars)

A) the nature of the function, linear or exponential, that can be used to model the value after x years

B) the actual function f(x) that can be used in each case

C) f(25) for each house. Is there a significant difference?

A) The oddball numbers give you a clue immediately that the value of house 1 will be best modeled by an exponential function.

The value of house 2 is increasing steadily at 9,000 per year, so is modeled by a linear function.

B) The ratio of values from a given year to the year before for house 1 is

... 294.58/286 = 1.03

A check for other years reveals the same ratio, so the exponential function can be written for house 1 as

... f(x) = 286·1.03^x . . . . . value of house 1

In part A we determined the year-to-year difference in value for house 2 is 9,000. That is the slope of the linear function. Then (in thousands), that function is

... f(x) = 286 +9x . . . . . value of house 2

C) After 25 years, the house values are (in thousands of dollars)

The value of house 1 has more than doubled in the same time that the value of house 2 has increased by about 79%. This is a significant difference.

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An exponential function will always outperform a linear function over a long enough time period.