Step-by-step explanation:
We have been given a table, which represents the projected value of two different houses for three years.
Part A:
[tex]\text{Increase in value of house 1 after one year}=294,580-286,000[/tex]
[tex]\text{Increase in value of house 1 after one year}=8580[/tex]
[tex]\text{Increase in value of house 1 after two years}=303,417.40-294,580[/tex]
[tex]\text{Increase in value of house 1 after two years}=8837.4[/tex]
We can see from our given table that the value of house 1 is not increasing at a constant rate, while a linear function has a constant rate of change, therefore, an exponential function can be used to describe the value of the house 1 after a fixed number of years.
[tex]\text{Increase in value of house 2 after one year}=295,000-286,000[/tex]
[tex]\text{Increase in value of house 2 after one year}=9,000[/tex]
[tex]\text{Increase in value of house 2 after two years}=304,000-295,000[/tex]
[tex]\text{Increase in value of house 2 after two years}=9,000[/tex]
We can see from our given table that the value of house 2 is increasing at a constant rat that is $9,000 per year. Since a linear function has a constant rate of change, therefore, a linear function can be used to describe the value of the house 2 after a fixed number of years.
Part B:
Let x be the number of years after Dominique bought the house 1.
Since value of house 1 is increasing exponentially, so let us find increase percent of value of house 1.
[tex]\text{Increase }\%=\frac{\text{Final value-Initial value}}{\text{Initial value}}\times 100[/tex]
[tex]\text{Increase }\%=\frac{294,580-286,000}{286,000}\times 100[/tex]
[tex]\text{Increase }\%=\frac{8580}{286,000}\times 100[/tex]
[tex]\text{Increase }\%=0.03\times 100[/tex]
[tex]\text{Increase }\%=3[/tex]
[tex]\text{Increase }\%=\frac{303,417.40-294,580}{294,580}\times 100[/tex]
[tex]\text{Increase }\%=\frac{8837.4}{294,580}\times 100[/tex]
[tex]\text{Increase }\%=0.03\times 100[/tex]
[tex]\text{Increase }\%=3[/tex]
Therefore, the growth rate of house 1's value is 3%.
Since we know that an exponential function is in form: [tex]y=a*b^x[/tex], where,
a = Initial value,
b = For growth b is in form (1+r), where, r is rate in decimal form.
[tex]3\%=\frac{3}{100}=0.03[/tex]
Upon substituting our values in exponential function form we will get,
[tex]f(x)=286,000(1+0.03)^x[/tex], where, f(x) represents the value of the house 1, in dollars, after x years.
Therefore, the function [tex]f(x)=286,000(1.03)^x[/tex] represents the value of house 1 after x years.
Let x be the number of years after Dominique bought the house 2.
We can see that when Dominique bought house 2 it has a value of $286,000. This means that at x equals 0 value of house will be $286,000 and it will be our y-intercept.
Since value of house 2 is increasing 9000 per year, therefore, slope of our line be 9000.
Upon substituting these values in slope-intercept form of equation [tex](y=mx+b)[/tex] we will get,
[tex]f(x)=9000x+286,000[/tex], where, f(x) represents the value of the house 2, in dollars, after x years.
Therefore, the function [tex]f(x)=9000x+286,000[/tex] represents the value of house 2 after x years.
Part C:
Since values in exponential function increases faster than linear function, so the value of house 1 will be greater than value of house 2.
Let us find the value of house 1 and house 2 by substituting x=25 in our both functions.
[tex]f(25)=286,000(1.03)^{25}[/tex]
[tex]f(25)=286,000*2.0937779296542148[/tex]
[tex]f(25)=598820.48788[/tex]
We can see that value of house 1 after 25 years will be approx $598,820.48.
[tex]f(25)=9000*25+286,000[/tex]
[tex]f(25)=225,000+286,000[/tex]
[tex]f(25)=511,000[/tex]
We can see that value of house 2 after 25 years will be approx $511,000.
Since $511,000 is less than $598820.48, therefore, value of house 1 is greater than value of house 2.
