Again, another math question! (more detail below)

An investment is worth $2922 in 1995. By 1999 it has grown to $3454. Let y be the value of the investment in the year x, where x=0 represents 1995. Write a linear equation that represent the value of the investment, y, to the year x.

In the four years between 1995 and 1999, the investment grew $3,454 - $2,922, or $532, which is $133 per year, so:

y = $2,922 + $133x

y = 2922(1+rx)
We have to find r.
3454 = 2922(1+4r)
r = 0.0455
The linear relation is
y = 2922(1+0.0455x)
y = 133x + 2922

slope (m)= 3454-2922/ 1999-1995= 133

y-2922=133(x-1995)
y=133x+1060