Question 94835: This is a question on my summer assignment that I've been working on for months!!! I've asked my friends for help, but they don't know how to solve it either!!! I hope you can help.
A business had a profit of $58,000 in 1989 and a profit of $74,000 in 1993. Find the rate of change. Write an equation that models the information given.
Rate of change=
Equation=
So far I have -
x/100 = 16,000/58,000
x = 1,600,000/58,000
Rate of Change = 27.5862% over 4 years or 6.8966% per year
Is that the correct answer for the rate of change?
My work for the equation is -
y2-y1/x2-x1
74,000-58,000/1993-1989
Equation = $16,000/4 years or $4000/1 year
I'm completely confused about the equation part! Would the slope equation be the correct equation for the answer? And if so, is this is correct solution?
Any insight or help that anyone can give me on this problem would be greatly appreciated!!! Thank-you VERY much!!!
Answer by tutorcecilia(2152)   (Show Source):
You can put this solution on YOUR website! First, it is necesaary to find the x- and y-values:
To find the x-values:
Let:
1989=year #0
1990=#1
1991=#2
1992-#3
1993=#4
.
The y-values are (58,000 when x=0; and 74,000 when x=4.
.
to find the rate of change use the fomula for the slope of a line:
Rate of change=Slope = m=
Pick two coordinate points (4; 74,000) and(0; 58,000) and plug them into the formula:
.
m= 
m=
m=$4000 as the rate of change.
What this means is that for every year, there is an additional profit of $4,000.
.
To find the equation, use the slope-intercept form of a line (y-mx+b):
y=mx+b
y=any y-value on the line
m=the sope of the line
x=any corresponding x-value on the line
b=the y-interception or point (0, b)
.
y=mx+b
58,000=(4,000)(0)+b [plug-in the values and solve for the b-term]
58,000=0+b
58,000=b
.
Plug-in only the value of the slope and the b-intercept to get the equation of the line:
y=4,000x + 58,000
.
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