Question

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

Answer 1

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