Level D

If the car is depreciating at a constant rate, this indicates that the function is linear. We can find the equation for the linear function by first finding the slope of the line. Let's do that using the slope formula. The coordinates we will use to do this are found in (year, value). Our coordinates are (0, 21500) and (3, 14120). The first coordinate says that before any time goes by at all, meaning when the car was brand new, its value was 21500. The second coordinate says that after 3 years, the value of the car was 14120.

$m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Filling in:

$m=\frac{14120-21500}{3-0}=\frac{-7380}{3}=-2460$

This is the slope. In the context of our problem, it means that for each year that goes by, the car's value goes down $2460.

Now we can use that slope and one of our coordinates to write the equation of the line. We can choose either one of the coordinates to write our equation. Let's use the first one, since it will be easier. Rest assured that regardless of which one you choose, you will get the same equation. We fill in the slope-inteercept form of a line, y = mx + b, where y is the y value from the coordinate, m is the slope we solved for, x is the x value from the same coordinate, and b is the y-intercept. We are solving for b.

21500 = -2460(0) + b and

21500 = 0 + b so

b = 21500. Our equation, then, is

y = -2460x + 21500

If we want to find the value of the car, y, in 5 years, x, we plug in 5 for x and solve for y.

y = -2460(5) + 21500 and

y = -12300 + 21500 so

y = $9200

That means that in 5 year the car will be worth only $9200 when it is 5 years old.

5 0

3 years ago

Determine the quadratic function f whose graph is given. The vertex is (1,-3) and y-intercept is -2