Posted by Hannah on .
The value of an automobile was $15,780 in 1994 and depreciated to $5400 in 1999. Let y be the
value of the automobile in the year x, where x = 0 represents 1994. Write a linear equation in
slopeintercept form that models the value y of the automobile, x years after 1994.
Would the answer be y = 5,400 x + 15,780?

Math(Please check) 
Christina,
The yintercept is correct, but 5400 would not be the slope of the line. Remember that slope represents the rate of change, specifically the change in y divided by the change in x. In this case, y is the value of the automobile and x is the year. If that doesn't help, I can clarify it further.

Math(Please check) 
MathMate,
As Christina pointed out, the yintercept (value at year 0) is correct. However, the formula does not work for year 5, equivalent to 1999, because the depreciation of 5400 was accumulated over 5 years.
You only need a change in the slope to correct the equation. 
Math(Please check) 
Hannah,
Another answer choice was y=2076x + 15,780. Is this correct? If it is I do not understand how they go 2076. Thank you for your help!

Math(Please check) 
MathMate,
I do not understand how 2076 came about.
Since the value depreciated $5400 in 5 years, I would expect the slope to be $5400/5 = $1080 per year.
Does that make sense to you?