An investment is worth $3518 in 1995. By 2000 it has grown to $5553. Let y be the value of the investment in the year x, where x = 0 represents 1995. Write a linear equation that models the value of the investment in the year x.

Question

An investment is worth $3518 in 1995. By 2000 it has grown to $5553. Let y be the value of the investment in the year x, where x = 0 represents 1995. Write a linear equation that models the value of the investment in the year x.
  A.  y = -407x + 7588 
  B.  y =1/407  x + 3518 
  C.  y = -407x + 3518 
  D.  y = 407x + 3518 
my guess is d but i am not sure and stuck on this question

Reiny · Answer

since the investment grows linearly,
you have two points,
(0, 3518) and (5,5553)
slope = (5553-3518)/(5-0) = 407
looking at the choices, the only one possible is D, since it has a slope of +407
making sure, it is correct ....
y = 407x + b
and since (0,3518) would be the y-intercept
D is the correct choice

Steve · Answer

The slope of the line is (5553-3518)/(2000-1995) = 407 That leaves only D as the choice.