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.

Question

Mathematics

Heidi

7 February, 20:08

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.

+1

Answers (1)

Know the Answer?

Madeleine Riggs · Answer 1

A) the first house is exponential, because the increase of value changes with each year

for the second house the change is always + 9,000 so it's linear

B)

house1:

f (x) = a*b^x

I:f (1) = a*b^1=294,580

a=294,580/b

II: f (2) = a*b^2=303,417.40

insert a=294,580/b into II:

294,580/b*b^2=303,417.40

294,580*b=303,417.40

b=303,417.40/294,580=1,03

-> calculate a

a=294,580/b

a=294,580/1,03

a=286,000

so house1: f (x) = 286000*1,03^x

house2:

I:f (1) = 295,000

II:f (2) = 304,000

f (x) = mx+d

difference is always 9000=m

-> calculate d with I

295000=9000*1+d

286000=d

house2: f (x) = 9000*x+286000

C)

house1:

f (25) = 286000*1,03^25=598820.487881

house2:

f (25) = 9000*25+286000=511000

so in 25 years house1 would have a significantly higher value (87820.48)