**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.

no one knows lol

step-by-step explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).