The value of a car decreases at a constant rate. After 1 year the value of the car is $20,000. After 2 more yearsit is $14,000. Write an equation in slope-intercept that represents the value y (in dollars) of the car after x years.

it dropped 6K in 2 years. That's 3K per year.

So, it started at 3K more than 20K.

y = 23000 - 3000x

steve how did you get that

get what? They told you that the value dropped from 20K to 14K in two years.

If you can't see how that relates to the slope of the graph, then you seriously need to review your text or look at examples online.

Since it dropped 3K in one year, and a year after it was bought it was worth 20K, then that means that in the beginning, it was worth 3K more -- $23K.

So, at time x=0 (0 years after purchase), y= 23000

To write an equation in slope-intercept form, we need to express the equation in the form y = mx + b, where m is the slope and b is the y-intercept.

In this case, we know that the value of the car decreases at a constant rate. Let's consider the decrease in value:

The car's value decreases by $20,000 - $14,000 = $6,000 in just 2 years. Therefore, the car's value decreases by $6,000/2 = $3,000 per year.

Since the car's value decreases by $3,000 per year, we can say that the slope of the equation is -3,000.

Now, let's find the y-intercept. After 1 year, the value of the car is $20,000. This means that when x = 1 (1 year), y = $20,000.

Substituting these values into the slope-intercept form equation, we have:

y = mx + b
$20,000 = -3,000(1) + b
$20,000 = -3,000 + b

Now, rearrange the equation to solve for b:

$20,000 + 3,000 = b
$23,000 = b

Now we have the value of b, which is the y-intercept.

So, the equation that represents the value y (in dollars) of the car after x years is:

y = -3,000x + $23,000