Hank started a savings account in June of 2002. On June 2006 he had $3800. On June 2014 he had 8600. If hanks saving is modeled by a linear function what was his initial deposit

If the savings is done by a linear function, then first solve the slope of the function:

m = (y2 - y1) / (x2 - x1)

m = (8600 - 3800) / (2014 - 2006)

m = $ 600 per year

using (8600, 2014)

y = mx + b

8600 = (600) (2014 - 2002) + b

b = $ 1400 is the initial deposit