Edward purchased his home for $89,000. for the first three years after he moved in, the real estate market was very lively, so his property value grew by 4.8% every year. for the next five years, the market slowed down somewhat, and his property value grew by 2.6% every year. how much had the value of edward’s home increased after eight years, to the nearest hundred dollars?

Purchased Price: $89,000
1st to 3rd year: 4.8% or .048 (4.8/100)
4th to 8th year: 2.6% or .026

1st year: $89,000 x .048 = $4,272.00 
$89,000 + $4,272.00 = $93,272.00
Edward's Home Value 1st year, end - $93,272.00
2nd year: $93, 272 + 4,477.06 = $97,749.06
3rd year: $97,749.06 + $4,691.95 = $102,441.01

4th year to 8th year, same solution but change .048 to .026.

4th year: $102,441.01 x .026 = $2,663.47
$102,441.01 + 2,663.47  = $105,104.48
Edward's Home Value 4th year, end - $105,104.48

5th year: $105,104.48 + $2,732.72 = $107, 837.20
6th year: $107,837.20 + $2,803.77 = $110,641
7th year: $110,641 + $2,876.67 = $113,517.67
8th year: $113,517.67 + $2,951.46 = $116,496.13

ANSWER: $116,500

Abit confused about what you mean, but i think you might be referring to slope intercept form: y = mx + bin that case b, or the y intercept is already given.(0,-1)find the slope by using the formulay2 - y1/x2 - x1the final slope will be 4/2, equivalent to 2.now your equation should look like thisy = 4/2x -1sorry if this didn’t you out!