the value of a car decreases as shown in the table below.\n| years after purchase | value ($) |\n| ---- |…

the value of a car decreases as shown in the table below.\n| years after purchase | value ($) |\n| ---- | ---- |\n| 0 | 25,000 |\n| 1 | 18,000 |\n| 2 | 14,000 |\n| 3 | 10,500 |\n| 4 | 8,000 |\n| 5 | 6,000 |\nwhich statements are true? check all that apply.\n□ the function that best represents the data is f(x)=24,512(0.755)^x.\n□ the function that best represents the data is f(x)=554x^2 - 5,439x + 24,600.\n□ the function decreases indefinitely.\n□ it is reasonable to interpolate to the value of the car at 4.5 years.\n□ it is reasonable to extrapolate to 40 years.

the value of a car decreases as shown in the table below.\n| years after purchase | value ($) |\n| ---- | ---- |\n| 0 | 25,000 |\n| 1 | 18,000 |\n| 2 | 14,000 |\n| 3 | 10,500 |\n| 4 | 8,000 |\n| 5 | 6,000 |\nwhich statements are true? check all that apply.\n□ the function that best represents the data is f(x)=24,512(0.755)^x.\n□ the function that best represents the data is f(x)=554x^2 - 5,439x + 24,600.\n□ the function decreases indefinitely.\n□ it is reasonable to interpolate to the value of the car at 4.5 years.\n□ it is reasonable to extrapolate to 40 years.

Answer

Explanation:

Step1: Check exponential - function fit

For an exponential function $y = a(b)^x$, when $x = 0$, $y=a$. For $f(x)=24512(0.755)^x$, when $x = 0$, $f(0)=24512(0.755)^0=24512$. For $x = 1$, $f(1)=24512\times0.755 = 24512\times\frac{755}{1000}=18506.56$. The data shows a pattern of exponential - decay.

Step2: Check quadratic - function fit

For the quadratic function $f(x)=554x^{2}-5439x + 24600$, when $x = 0$, $f(0)=24600$; when $x = 1$, $f(1)=554-5439 + 24600=19715$. It does not fit the data well.

Step3: Analyze the behavior of the function

The exponential function $y = a(b)^x$ with $0\lt b\lt1$ decreases indefinitely as $x$ increases.

Step4: Consider interpolation

Interpolation is finding a value within the range of the known data points. Since $4.5$ is between $4$ and $5$, it is reasonable to interpolate to find the value of the car at $4.5$ years.

Step5: Consider extrapolation

Extrapolating to 40 years is not reasonable because the car will reach a value of zero or be scrapped long before 40 years, and the model may not hold up over such a long - term period.

Answer:

The function that best represents the data is $f(x)=24512(0.755)^x$; The function decreases indefinitely; It is reasonable to interpolate to the value of the car at 4.5 years.