Question

Maggy started a savings account in March of 2003. On March 2007, she had $4,200. On March 2015, she had $10,400. If Maggy's saving is modeled by a linear function, what was her initial deposit?

A) $1,100
Eliminate
B) $1,300
C) $1,500
D) $700

Answer 1

First we have to find the slope of the linear equation, slope=m=(y2-y1)/(x2-x1)

m=(10,400-4,200)/(2015-2007)=6200/8=775 so we so far have a line of:

s=775(y-2003)+b, using either original point we can solve for b, the y-intercept which is the initial value as well...I'll use point (2007, 4200) and we get:

4200=775(2007-2003)+b

4200=3100+b

b=$1100

s(y)=775(y-2003)+1100

Maggie initially deposited $1,100. Hope this helps!