Posted by **Hannah** on Sunday, January 30, 2011 at 6:39pm.

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

slope-intercept 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**, Sunday, January 30, 2011 at 6:48pm
The y-intercept 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**, Sunday, January 30, 2011 at 6:58pm
As Christina pointed out, the y-intercept (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**, Sunday, January 30, 2011 at 9:28pm
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**, Monday, January 31, 2011 at 12:05am
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?

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