Monday, December 2nd

Review for Final:

1.What is the variable(s) in the expression?

2.What is the constant in the expression?

WARM UP

Grade Check

Grades Left the Semester1.1 more quiz 2.1 more Warm up(Daily

grade)3. Exponential Test (Test

Grade) 4. Semester I Final (Final

grade) 5.3 Weekly Reviews-(daily

Grade)

Weekly Review 1. We will have three

weekly Reviews 2. Each will count as a

Daily Grade 3. They are eligible to

replace QUIZ grades

High School GPA A: 4.0B: 3.0 C. 2.0 D. 1.0 F-Receive no Credit. You will have to retake first semester all over again during semester II.

Tutoring Option 1: Ms. Evans

Tuesday and ThursdaysBefore School 6:40-7:00After School 2:10-2:30

Tutoring Option 2: Lunch

Tuesday and ThursdaysEither lunch

Go to room 400 FIRST, then to lunch

Tutoring Option 3: Algebra

Department*Check Schedule in Back

Discussion QuestionWhat’s the difference between exponential

growth and exponential decay equations?!

Growth & Decay in graph

Growth Decay

Growth & Decay in Equation

x

y

322

y

x

y

232Growth

Decay

Growth and Decay in Table

x -2 -1 0 1 2

y 2 4 8 16 32

x -2 -1 0 1 2

y 32 16 8 4 2

For compound interest • annually means “once per year” (n = 1).• quarterly means “4 times per year” (n =4).• monthly means “12 times per year” (n = 12).

Reading Math

Write a compound interest function to model each situation. Then find the balance after the given number of years.

$1200 invested at a rate of 2% compounded quarterly; 3 years.

Step 1 Write the compound interest function for this situation.

= 1200(1.005)4t

Write the formula.

Substitute 1200 for P, 0.02 for r, and 4 for n.

Simplify.

Example #1

Step 2 Find the balance after 3 years.

≈ 1274.01

Substitute 3 for t.A = 1200(1.005)4(3)

= 1200(1.005)12

Use a calculator and round to the nearest hundredth.

The balance after 3 years is $1,274.01.

Write a compound interest function to model each situation. Then find the balance after the given number of years.$15,000 invested at a rate of 4.8% compounded monthly; 2 years.

Step 1 Write the compound interest function for this situation.

Write the formula.

Substitute 15,000 for P, 0.048 for r, and 12 for n.

= 15,000(1.004)12t Simplify.

Example #2

Step 2 Find the balance after 2 years.

≈ 16,508.22

Substitute 2 for t.A = 15,000(1.004)12(2)

= 15,000(1.004)24 Use a calculator and round to the nearest hundredth.

The balance after 2 years is $16,508.22.

Write a compound interest function to model each situation. Then find the balance after the given number of years.$1200 invested at a rate of 3.5% compounded quarterly; 4 yearsStep 1 Write the compound interest function for this situation.

Write the formula.

Substitute 1,200 for P, 0.035 for r, and 4 for n.

= 1,200(1.00875)4t Simplify.

Example #3

Step 2 Find the balance after 4 years.

1379.49

Substitute 4 for t.A = 1200(1.00875)4(4)

= 1200(1.00875)16

Use a calculator and round to the nearest hundredth.

The balance after 4 years is $1,379.49.

1. The number of employees at a certain company is 1440 and is increasing at a rate of 1.5% per year. Write an exponential growth function to model this situation. Then find the number of employees in the company after 9 years.

y = 1440(1.015)t; 16462. $12,000 invested at a rate of 6% compounded

quarterly; 15 yearsA = 12,000(1 + .06/4)4t, =$29,318.64

Write a compound interest function to model each situation. Then find the balance

after the given number of years.

LESSON SUMMARY