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?

Answer 1

First find the slope of the linear equation...

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

m=(10400-4200)/(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

The initial deposit was $1100.00.

My B-day is March 2007!

Answer 2

$1100

Explanation:

Set up 3 ordered pairs of year to money in the bank.

(2003,y);(2007,4200);(2015,10400)

Now make an equation in point slope form with the two known points.

y=mx+b

m=(10400-4200)/(2015-2007)=6200/8=

775

4200=775(2007)+b

-1551225=b

y=775x-1551225

now plug in 2003 into x, and get $1100