example 3 Loan Balance

Chapter 1.3

A business property is purchased with a promise to pay off a $60,000 loan plus the $16,500 interest on this loan by making 60 monthly payments of $1275. The amount of money, y, remaining to be paid on $76,500 (the loan plus interest) is reduced by $1275 each month. Although the amount of money remaining to be paid changes every month, it can be modeled by the linear function

a. Find the x-intercept and the y-intercept of the graph of this linear function.

b. Interpret the intercepts in the context of this problem situation.

c. How should x and y be limited in this model so that they make sense in the application?

d. Use the intercepts and the results of part (c) to sketch the graph of the given equation.

76,500 1275y x

where x is the number of monthly payments made. We recognize that only integer values of x from 0 to 60 apply to this application.

2009 PBLPathways

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

a. Find the x-intercept and the y-intercept of the graph of this linear function.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

76,500 1275

0 76,500 1275

1275 76,500

76,500

127560

y x

x

x

x

x

x-intercepts: set y = 0

76,500 1275

76,500 1275(0)

76,500

y x

y

y

y-intercept: set x = 0

(60, 0)

(0, 76500)

2009 PBLPathways

b. Interpret the intercepts in the context of this problem situation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

(60, 0)

(0, 76500)

When x = 60, the values of y = 0.

“The loan is paid off in 60 months”.

When x = 0, the values of y = 76,500.

“A total of $76,500 must be repaid”.

2009 PBLPathways

b. Interpret the intercepts in the context of this problem situation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

(60, 0)

(0, 76500)

When x = 60, the values of y = 0.

“The loan is paid off in 60 months”.

When x = 0, the values of y = 76,500.

“A total of $76,500 must be repaid”.

2009 PBLPathways

b. Interpret the intercepts in the context of this problem situation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

(60, 0)

(0, 76500)

When x = 60, the values of y = 0.

“The loan is paid off in 60 months”.

When x = 0, the values of y = 76,500.

“A total of $76,500 must be repaid”.

2009 PBLPathways

b. Interpret the intercepts in the context of this problem situation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

(60, 0)

(0, 76500)

When x = 60, the values of y = 0.

“The loan is paid off in 60 months”.

When x = 0, the values of y = 76,500.

“A total of $76,500 must be repaid”.

2009 PBLPathways

b. Interpret the intercepts in the context of this problem situation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

(60, 0)

(0, 76500)

When x = 60, the values of y = 0.

“The loan is paid off in 60 months”.

When x = 0, the values of y = 76,500.

“A total of $76,500 must be repaid”.

2009 PBLPathways

c. How should x and y be limited in this model so that they make sense in the application?

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

Number of monthly payments made

Money remaining to be paid

Payments most be integers from 0 to 60.

Money remaining to paid is from 0 to $76,500.

2009 PBLPathways

c. How should x and y be limited in this model so that they make sense in the application?

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

Number of monthly payments made

Money remaining to be paid

Payments most be integers from 0 to 60.

Money remaining to paid is from 0 to $76,500.

2009 PBLPathways

c. How should x and y be limited in this model so that they make sense in the application?

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

Number of monthly payments made

Money remaining to be paid

Payments must be integers from 0 to 60.

Money remaining to paid is from 0 to $76,500.

2009 PBLPathways

c. How should x and y be limited in this model so that they make sense in the application?

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

Number of monthly payments made

Money remaining to be paid

Payments must be integers from 0 to 60.

Money remaining to be paid is from 0 to $76,500.

2009 PBLPathways

d. Use the intercepts and the results of part (c) to sketch the graph of the given equation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

2009 PBLPathways

d. Use the intercepts and the results of part (c) to sketch the graph of the given equation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

x

y

2009 PBLPathways

d. Use the intercepts and the results of part (c) to sketch the graph of the given equation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

The vertical axis has to include values from 0 to 76,500.

x

y

2009 PBLPathways

d. Use the intercepts and the results of part (c) to sketch the graph of the given equation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

The horizontal axis has to include values from 0 to 60.

x

y

2009 PBLPathways

d. Use the intercepts and the results of part (c) to sketch the graph of the given equation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

x

y

2009 PBLPathways

d. Use the intercepts and the results of part (c) to sketch the graph of the given equation.

76,500 1275y x

Money remaining to be paid

Number of monthly payments made

x

y