CHAPTER 6 Eponential Functions
6.1 EXPLORING THE CHARACTERISTICS OF EXPONENTIAL FUNCTIONS Chapter 6
EXPONENTIAL FUNCTIONS An eponential function is a function that has an in the eponent. Standard form: y a( b) where a 0, b 0, and b 1 Eponential graphs look different than any of the other graphs we ve seen before. We need to learn their characteristics so we can spot them easily.
COMPLETE THE FOLLOWING TABLE OF VALUES AND SKETCH THE GRAPH OF EACH FUNCTION: f() 3 10-3 = 0.001 2 10-2 = 0.01 1 10-1 = 0.1 0 10 0 = 1 1 10 1 = 10 2 10 2 = 100 3 10 3 = 1000
COMPLETE THE FOLLOWING TABLE OF VALUES AND SKETCH THE GRAPH OF EACH FUNCTION: f() 3 2(5) -3 = 0.016 2 2(5) -2 = 0.08 1 2(5) -1 = 0.4 0 2(5) 0 = 2 1 2(5) 1 = 10 2 2(5) 2 = 50 3 2(5) 3 = 250
COMPLETE THE FOLLOWING TABLE OF VALUES AND SKETCH THE GRAPH OF EACH FUNCTION: f() 3 8 2 4 1 2 0 1 1 0.5 2 0.25 3 0.125 f() 3 512 2 128 1 32 0 8 1 2 2 0.5 3 0.125
WHAT ARE THE DIFFERENCES AND SIMILARITIES? Similarities End behaviour Q2 to Q1 No -intercepts One y-intercept Domain is all real numbers Range: y>0 Dif ferences The first two graphs are increasing the second two graphs are decreasing
y 2 f() 2.25 1.5 0 1 1 2 2 4 y 3 f() 2.111 1.333 0 1 1 3 2 9 y 4 f() 2.063 1.25 0 1 1 4 2 16
y 2 y 3 y 4 Increasing or decreasing Number of -intercepts the y-intercept the end behaviour the domain the range What do you notice as the base gets larger?
y 1 2 f() y 1 3 f() y 1 4 f() 20 18 16 2 4 1 2 0 1 1.5 2 9 1 3 0 1 1.333 2 16 1 4 0 1 1.25 14 12 10 8 6 4 2 2.25 2.111 2.063-5 -4-3 -2-1 1 2 3 4 5-2
y 1 2 y 1 3 y 1 4 Number of -intercepts the y-intercept the end behaviour the domain the range y 1 4 What do you notice as the base gets smaller? What conclusions can we reach regarding the value of the base b?
y 3 2 f() 2.75 1 1.5 0 3 1 6 2 12 ( ) y 4 2 f() 2 1 1 2 0 4 1 8 2 16 ( ) y 5( 2) f() 2 1.25 1 2.5 0 5 1 10 2 20 y
y 3 2 ( ) y 4 2 ( ) y 5( 2) Number of -intercepts the y-intercept the end behaviour y 1 4 the domain the range
What do you notice as the value of a changes? How is the value of a related to the y -intercept?
PG. 337, #1-3 Independent Practice
6.2 RELATING THE CHARACTERISTICS OF AN EXPONENTIAL FUNCTION TO ITS EQUATION Chapter 6
JUST TO RECAP! f () = a(b) Where a is the leading coefficien t a 0 a represents the y intercept And b is the base of the eponential function When b 1, our function is increasing When 0 b 1, our function is decreasing
What is the value of a? What affect does a have the graph? What is the value of b? What affect does b have on the graph? 6 y 1 1 2 2 8 1 1 3 y 8 ( 2 ) y ( 6 ) y y-intercept is (0, 8) y-intercept is (0, 1) y-intercept is (0, 1) y 3 ( 08. ) y-intercept is (0, 3) 2 6 ½ 0.8 Graph is increasing from Q2 to Q1 Graph is increasing from Q2 to Q1 Graph is decreasing from Q2 to Q1 Graph is decreasing from Q2 to Q1
FOR EXPONENTIAL EQUATIONS, WE SAID. WHY IS THIS THE CASE? b 1 Graph: y 2 1 0 1 2 2( 1) f() y So, why can we not have a base of 1?
WE ALSO SAID, b 0. WHY CAN WE NOT HAVE A NEGATIVE BASE? Graph: y 2( 3) y f() 2.222 1 -.667 0 2 1-6 2 18 So, why can we not have a negative base?
GIVEN y a ( b ) where a 0, b 0, and b 1 Why is the range always y>0? Because mathematically, the graph will never touch the -ais and our graph will always be above the -ais. NOTE! This horizontal line that your eponential function approaches but never mathematically reaches is called an ASYMPTOTE.
EXAMPLES 1. Complete the following table: -intercept y-intercept Is the graph increasing or decreasing? How do you know? Domain Range End Behaviour y 1 y 11 3 2 ( 7) 8 y ( 15. ) y 3 7 2
2. EXAMPLES
EXAMPLES 3. Complete the following table using y 8 2 3 y 8 2 3 y-intercept is 1 graph has one -intercept Range: {y/y>0} Domain: {/>8} This is a decreasing eponential function. True False Why I Think So
EXAMPLES 4. Do the following tables represent eponential functions? Why or why not? a. -2-1 0 1 2 y 10 30 90 270 810 b. -2-1 0 1 2 y 100 20 4.8.16
PG. 346-351, #1, 3, 4, 6, 9, 11, 13, 14. Independent practice
EXAMPLE Predict the number of -intercepts, the y-intercept, the end behaviour, the domain, the range, and whether this function is increasing or decreasing: Use the equation of the function to make your predictions. Verify your predictions by creating a table of values and graphing the function. a = 9 b = 2/3 a > 0, and 0 < b < 1 This means that a is positive, and b is between 0 and 1. 0 -intercepts End behavior, since it s decreasing (b is smaller than 1), is going from quadrant II to quadrant I. Domain: { E R} Range: {y y > 0, y E R} y-intercept is at = 0: y = 9(2/3) 0 y = 9(1) y = 9
EXAMPLE Predict the number of -intercepts, the y-intercept, the end behaviour, the domain, and the range of the following function: Use the equation of the function to make your predictions. Verify your predictions by creating a table of values and graphing the function. What is the value of e? e = 2.718 So the equation can be written: y = 1(2.718) So: a = 1 b = 2.718 -intercepts: None Eponential functions don t have -intercepts. End behaviour: It is increasing, so it starts in quadrant II and goes to quadrant I. If b > 1, then the function is increasing. The y-intercept is at = 0, so it s at y = e 0 = 1.
6.3 SOLVING EXPONENTIAL EQUATIONS Chapter 6
When solving a polynomial equation, as in the first eample, we must find a value(s) for such that when the value of is multiplied by itself, the answer is 64. The second eample is an eponential equation. In this case, the value of represents how many times the base 2 is multiplied by itself.
ALGEBRAICALLY, EXPONENTIAL FUNCTIONS CAN BE SOLVED IN THREE WAYS: Write both sides of the equation with a common base (sometimes this will already be done for us). When both sides cannot be written, we will use guess and check. In the net chapter, we will use logarithms.
USING PRIME FACTORIZATION, REWRITE EACH OF THE FOLLOWING NUMBERS AS POWERS OF 2. 4 32 128 a. b. c.
WRITE EACH OF THE FOLLOWING AS POWERS WITH LOWEST BASE POSSIBLE: 27 125 36 a. b. c.
SOME NUMBERS CAN BE WRITTEN WITH MORE THAN ONE BASE: 16 81 64 a. b. c.
SOLVING EXPONENTIAL FUNCTIONS WHEN THE BASES ARE THE SAME If we have two powers that are equal to each other and their bases are the same, then their eponents must be equal. 8 3 7 2 2 5 5 a. b. c. 7 2 7 5 4
WHEN THE BASES ARE NOT THE SAME BUT BOTH SIDES OF THE EQUATION CAN BE WRITTEN WITH A COMMON BASE. 4 128 16 2 32 a. b. c. 4 9 5 3 27 8
IT IS NOT ALWAYS POSSIBLE TO WRITE BOTH SIDES WITH A COMMON BASE. WE THEN USE GUESS AND CHECK: 2 5 5 20 a. b. c. 3 100
WE WILL ALSO BE REWRITING NUMBERS AS POWERS WITH NEGATIVE AND FRACTIONAL EXPONENTS: Remember! n 1 n n m n m
WRITE EACH NUMBER AS A POWER WITH NEGATIVE EXPONENTS: 1 25 1 1000 a. b. c. 1 64
WRITE EACH NUMBER AS A POWER WITH FRACTIONAL EXPONENTS: 5 3 7 a. 4 3 b. c. 2 3
2 7 SOLVE FOR X: 1 27 9 3 4 2 25 5 a. b. c. 8 1
8 2 d. 3 4 e. 16 21 1 2 3
SOMETIMES, WE MAY HAVE TO REARRANGE THE EQUATION SO THAT THERE IS ONLY ONE POWER ON EITHER SIDE OF THE EQUATION: Solve for and verify your answer by substitution: a. 3(9) 5 27
SOMETIMES, WE MAY HAVE TO REARRANGE THE EQUATION SO THAT THERE IS ONLY ONE POWER ON EITHER SIDE OF THE EQUATION: b. First, we want to isolate the eponent. What s another way to write this? Now, we take just the eponents. Check: We need to write 64 in the form of a power of 4. 4 to the power of what is 64? 4 3 = 64
PG. 361-365, #2, 4, 6, AND 7 (ALGEBRAICALLY) AND #8 (BY GUESS AND CHECK NOT GRAPHING TECHNOLOGY). Independent practice
EXPONENTIAL FUNCTIONS CAN ALSO BE SOLVED GRAPHICALLY: 1. Solve for : Algebraically, we write both sides with a common base: 2 3 3 3 9 Graphically we determine where the two separate graphs intersect: y y 3 9 Then cancel the common bases and solve for : 2 They intersect at: 2
2. USE THE GRAPH TO DETERMINE THE SOLUTION FOR 3 1 9, THEN VERIFY ALGEBRAICALLY. y = 9 y Both equations intersect at 1 3 1 9 3 1 1 9 3 2 9 9 9
WHEN WE ARE UNABLE TO WRITE BOTH SIDES OF THE EQUATION WITH A COMMON BASE, WE USED GUESS AND CHECK. SIMILAR TO THE EXAMPLES ABOVE, WE CAN NOW USE GRAPHS TO APPROXIMATE OUR SOLUTION. 3. Solve for : a. b. 5 7 1.21 5 16 y 1.72
SOMETIMES, IT CAN BE EASIER TO USE A GRAPH THAN TO SOLVE ALGEBRAICALLY. 4. Use the graph to determine the solution for 7 2, 1 then verify using substitution. y 0.55 7 0.55 2.916 2 0.551 2.928
NOTE! In the net chapter, we will use logarithms to solve these types of equations. Most will find logarithms easier than guess and check.
APPLICATIONS OF EXPONENTIAL FUNCTIONS Eponential functions arise when a quantity changes by the same factor for each unit of time. Many real world phenomena can be modelled by functions that describe how things grow or decay as time passes. For eample: A mass of radioactive substance decreases by ½ every 462 years A population doubles every year A bank account increases by 0.1% each month
HALF-LIFE PROBLEMS In a half-life problem, the amount of a substance decreases by ½ every fied number of years. The formula is: Where: A 0 t h A(t) A( t) A0 represents the initial amount of the substance present represents the time represents the half-life ( the time it takes for the substance to decrease by ½ ) is the amount of substance present at time t 1 2 t h
EXAMPLES 1. The radioactive element cobalt-60 can be used to treat cancer patients. Radioactive elements decay into other elements in a predictable way over time. The percent of cobalt-60 left in a sample can be modeled by the half -life function: where A 0 t represents the initial amount of cobalt-60, 100%, and represents time in years after the initial amount. A( t) A 0 1 2 t 5.3
A) HOW LONG DOES IT TAKE FOR A SAMPLE OF COBALT-60 TO REDUCE TO ½ OF ITS INITIAL AMOUNT? A( t) A 0 1 2 t 5.3 50 100 1 2 1 1 2 2 t 1 5.3 t 5.3 t 5.3 t 5.3 How does this answer relate to the equation? We found the h or half-life of the function.
B) WHAT PERCENT OF COBALT-60 WILL REMAIN IN A SAMPLE AFTER 10 YEARS? ROUND TO THE NEAREST PERCENT? A( t) A 1 A( t) 100 2 A( t) 0 1 2 27% t 5.3 10 5.3
C) DETERMINE HOW LONG IT WILL TAKE, TO THE NEAREST TENTH OF A YEAR, UNTIL ONLY 25% OF THE COBALT-60 REMAINS? A( t) A 0 25 100 1 2 1 2 t 5.3 t 5.3
THE HALF-LIFE OF A RADIOACTIVE ISOTOPE IS 30 HOURS. THE AMOUNT OF RADIOACTIVE ISOTOPE A(t), AT TIME t, CAN BE MODELED BY THE FUNCTION: A( t) A Determine algebraically how long it will take for a sample of 1792mg to decay to 56mg. 0 1 2 t 30
3.WHEN DIVING UNDERWATER, THE LIGHT DECREASES AS THE DEPTH OF THE DIVER INCREASES. ON A SUNNY DAY OFF THE COAST OF VANCOUVER ISLAND, A DIVING TEAM RECORDED 100% VISIBILIT Y AT THE SURFACE BUT ONLY 25% VISIBILIT Y 10M BELOW THE SURFACE. : The team determined that the visibility for the dive could be modeled by the following half-life formula: A( ) A0 1 2 where A 0 represents the percentage of light at the surface of the water, represents the depth in metres, h represents the depth in metres at which there is only half the original visibility, and A() represents the percentage of light at the depth of metres. h
AT WHAT DEPTH WILL THE VISIBILITY BE ONLY ½ OF WHAT IT WAS AT THE SURFACE? 1 25 100 2 10 h h 5 At 5m below the surface, you will only have 50% visibility
DOUBLING-TIME PROBLEMS IN A DOUBLING TIME PROBLEM, THE AMOUNT OF A SUBSTANCE DOUBLES EVERY FIXED NUMBER OF YEARS. THE FORMULA IS: Where: A(t) A 0 t d is the amount of substance present at time t represents the initial amount of the substance present represents the time t d A( t) A 2 0 represents the doubling time (the time it takes for the substance to double
NOTE! Some other applications might have an equation given with a number other than ½ or 2 as the base in the power. For eample, if a quantity triples, the base will be 3, if a quantity quadruples, the base will be 4.
EXAMPLES 1. The population of trout growing in a lake can be modeled by the function: t P( t) 200 25 where P(t) represents the number of trout and t represents the time in years after the initial count. How long will it take for there to be 6400 trout? 6400 200 t 25 t 25 years
2.SMALL RURAL WATER SYSTEMS ARE OFTEN CONTAMINATED WITH BACTERIA BY ANIMALS. SUPPOSE THAT A WATER TANK IS INFESTED WITH A COLONY OF 14000 E. COLI BACTERIA. IN THIS TANK, THE COLONY DOUBLES IN NUMBER EVERY 4 DAYS. The number of bacteria present in the tank after t days can be modeled by the function: A( t) 14000 t 24 Determine the value of t when A(t) equals 224,000. What does your answer mean in this contet? Eplain. 224000 14000 t 24 t 16 days This means that in 16 days, there will be 224000 E.coli bacteria in the water supply!
PG. 361-365, #1, 11, 12, 15, 16 Independent practice
JUST TO RECAP! Which is true of the table given below? (years) 0 3 6 9 12 y (amount) 10 20 40 80 160 Initial Amount Amount Growth (A) 10 doubles every three years (B) 10 triples every two years (C) 20 doubles every three years (D) 20 triples every two years
THE FUNCTION THAT MODELS THE DECAY OF CARBON -14 IS 1 5730 t At ( ) 100, WHERE A(T) IS THE NUMBER OF GRAMS 2 OF CARBON-14 PRESENT AT TIME T, IN YEARS. Which statement is true? (A) (B) (C) (D) The amount of carbon-14 doubles every 5730 years. There are 50 g of carbon-14 present initially. 14 g will be present after 50 years. 50 g of carbon-14 will be present after 5730 years.
REMEMBER! SOME OTHER APPLICATIONS MIGHT HAVE AN EQUATION GIVEN WITH A NUMBER OTHER THAN ½ OR 2 AS THE BASE IN THE POWER. Since 1950, the population of a certain community has tripled every 18 years. P( t) 624 t 318 a. If the initial population in 1950 was 624, how many people are there now in 2014? P(64) 3 P( 64) 624 18 31018 64
P( t) 624 t 318 b. When will the population reach 151 632 people? 151632 151632 624 624 624 624 t 318 t (3) 18 5 t t 18 90 243 3 5 t 318 t 318 In 2040, thepopulation will be about151632
P( t) 624 t 318 c. In what year will the population be approimately 200000 people? 200000 200000 624 320.5 t 318 624 624 624 t 318 t (3) 18 We cannot write of 3sowemust useguess and check toapproimateour 320.5as a power answer In about 94.5or 95 years That is, around 2045
6.4 MODELLING DATA USING EXPONENTIAL Chapter 6 FUNCTIONS
EXAMPLE The population of Canada from 1871 to 1971 is shown in the table below. In the third column, the values have been rounded.
a) Using graphing technology, create a graphical model and an algebraic eponential model for the data. b) Assuming that the population growth continued at the same rate to 2015, estimate the population in 2015. Round your answer to the nearest million. a) Enter in the data Use 0, 10, 20, 30, etc for the years where 1871 is year 0. Use STAT CALC 0: EpReg It gives us: EpReg y=a*b^ a = 2.6685 b = 1.0217 Write the equation of the eponential regression: y = 2.6685(1.0217) b) 2015 will be year 144. y = 2.6685(1.0217) y = 2.6685(1.0217) 144 y = 58.7 In 2015, the population will be approimately 59 million people.
EXAMPLE Sonja did an eperiment to determine the cooling curve of water. She placed the same volume of hot water in three identical cups. Then she recorded the temperature of the water in each cup as it cooled over time. Her data for three trials is given as follows.
A) CONSTRUCT A SCATTER PLOT TO DISPLAY THE DATA. DETERMINE THE EQUATION OF THE EXPONENTIAL REGRESSION FUNCTION THAT MODELS SONJA S DATA. y
a) Determine the equation of the eponential regression function that models Sonja s data. b) Estimate the temperature of the EXAMPLE water 15 min after the eperiment began. Round your answer to the nearest degree. c) Estimate when the water reached a temperature of 30ºC. Round your answer to the nearest minute. a) Our calculator gives us: EpReg y=a*b^ a = 78.6812 b = 0.97152 y = 78.6812(0.97152) b) Solve for y when = 15 y = 78.6812(0.97152) 15 y = 51.009 y = 51ºC c) Let y = 30, and solve for. We can use guess and check or if we have the graph, we can approimate the -value when y=30. 30 = 78.6812(0.97152) = 33
OR WE CAN READ THE GRAPH y = 33
PG. 377-384, #1, 2, 4, 6, 9, 10, 13, 15 Independent practice
6.5 FINANCIAL APPLICATIONS INVOLVING EXPONENTIAL FUNCTIONS Chapter 6
When you invest money, your money earns interest. That is, the amount of money that you will have after a period of time will actually be greater than the amount that you invested. The etra money, above the amount you invested, is called interest. The principal is the original amount of money that is invested or borrowed.
THERE ARE TWO TYPES OF INTEREST: Simple Interest: The amount of interest that you earn is calculated ONLY based on the amount of the money that you invested. That is, you only earn interest on the money that you invested. Compound Interest: Interest is earned on two different things: I. The amount of money that you invest, and II. The interest that you earn on money that you invested.
SIMPLE INTEREST Eample 1. Ann invests $1000 into a savings account that earns an annual simple interest of 5%. The following table shows the amount of money in the account after the first three years. Year (t) Total Amount at the End of the Year (A) 0 $1000 1 $1050 ($1000 + $1000(0.05)(1)) 2 $1100 ($1000 + $1000(0.05)(2)) 3 $1150 ($1000 + $1000(0.05)(3))
NOTICE THAT EACH YEAR, AN EXTRA 5% OF THE ORIGINAL AMOUNT GETS ADDED ON. Consider what this data would look like on a graph: Total Amount At End Of Year y Year (t) Notice that the data appears to be linear. This is because we are adding on the same amount of money each year (5% of the original amount).
THE EQUATION WHICH REPRESENTS SIMPLE INTEREST IS: A P( 1 rt) or A P Prt Where: A P r t represents the amount present represents the principal amount = the interest percentage divided by 100 represents the number of years
COMPOUND INTEREST Eample 2. Bob invests $1000 into a savings account that earns an annual compound interest of 5%. This means that the Principal Amount ($1000) earns interest each year, and that the interest earned also earns more interest! The following table shows the amount of money in the account after the first three years. Year (t) Amount of Annual Interest 0 $1000 Total Amount at the End of the Year (A) 1 $1000 0.05 = 50 $1050 (1000(1.05) 1 ) 2 $1050 0.05 = 52.50 $1102.50 (1000(1.05) 2 ) 3 $1102.50 0.05 = 55.13 $1157.63 (1000(1.05) 3 ) 4 $1157.63 0.05 = 57.88 $1215.51 (1000(1.05) 4 )
NOTICE THAT EACH YEAR, AMOUNT INCREASES BY AN EVEN GREATER AMOUNT THAN IT INCREASED BY IN THE PREVIOUS YEAR. We can see this from the graph: y An etra point was added in on this graph to show that the pattern here is not linear, but is eponential.
RECALL THE FORMAT FOR AN EXPONENTIAL FUNCTION IS: y a b ( ) Remember! y is the dependent variable, which in this case is A (the amount present). a is the initial amount, which in this case is $1000. b is the value by which the dependent variable gets multiplied by each year, which in this case is 1.05 (confirm this by looking at the table of values). is the independent variable, which in this case is the year.
PUTTING ALL THIS TOGETHER, WE GET THE FORMULA FOR COMPOUND INTEREST: where: A(n) = future value P = principal i = interest rate per compounding period n = number of compounding periods The future value is the amount that the investment will be worth after a certain amount of time. The compounding period is the time over which interest is calculated and paid on an investment or loan.
NOTICE THAT THE INTEREST RATE PER COMPOUNDING PERIOD IN THE PREVIOUS EXAMPLE WAS 5% PER YEAR. Thus, i 0. 05 1 0. 05
COMPOUNDING PERIODS CAN BE DAILY, WEEKLY, SEMI-MONTHLY, MONTHLY, QUARTERLY, SEMI- ANNUALLY, OR ANNUALLY. The table below shows how many times interest is paid, and the interest rate for each of those options. Compounding Period Daily Weekly Semi-Monthly Monthly Quarterly Semi-Annually Annually Number of Times Interest is Paid 365 times per year 52 times per year 24 times per year 12 times per year 4 times per year 2 times per year 1 time per year Interest Rate Per Compounding Period annual rate i 365 annual rate i 52 annual rate i 24 annual rate i 12 annual rate i 4 annual rate i 2 annual rate i 1
EXAMPLE 1. DEB INVESTS $1000 WHICH HAS A COMPOUND INTEREST RATE OF 6% COMPOUNDED SEMI - ANNUALLY. WRITE AN EXPONENTIAL EQUATION REPRESENTING THE SITUATION AND USE IT TO DETERMINE HOW MUCH MONEY SHE WILL HAVE IN 6 YEARS.
EXAMPLE 2. A PRINCIPAL AMOUNT OF $3000 WAS INVESTED FOR 5 YEARS IN AN ACCOUNT IN WHICH 8% COMPOUND INTEREST WAS COMPOUNDED QUARTERLY. HOW MUCH MONEY WAS IN THE ACCOUNT AT THE END OF 5 YEARS?
YOUR TURN 1. Consider the following statements: Emily invested $3000 for a term of 5 years with a simple interest rate of 4% per year. Zack invested $3000 for a term of 5 years with a compound interest rate of 4% per year compounded annually. Which investment results in the greatest value after 5 years?
2. $3000 WAS INVESTED AT 6% PER YEAR COMPOUNDED MONTHLY. WRITE THE EXPONENTIAL FUNCTION WHICH MODELS THIS SITUATION. WHAT WILL BE THE FUTURE VALUE AFTER 4 YEARS?
3. $2000 IS INVESTED AT 6% PER YEAR COMPOUNDED SEMI - ANNUALLY. CAROL DEFINED THE EXPONENTIAL FUNCTION AS n A 2000(1.03) WHERE IS THE # OF 6 MONTH PERIODS. n Is Carol s reasoning correct? Why or why not?
4. JOE INVESTS $5000 IN A HIGH -INTEREST SAVINGS BOND THAT HAS AN ANNUAL INTEREST RATE OF 9% COMPOUNDED MONTHLY. Write the eponential function in the form of which models this situation. A P( 1 i) n What will be the future value of the bond after 1 year?
5. $2000 IS INVESTED FOR 3 YEARS THAT HAS AN ANNUAL INTEREST RATE OF 9% COMPOUNDED MONTHLY. LUCAS SOLVED THE FOLLOWING EQUATION: 3 A 2000(1.0075) Identify the error that Lucas made. Correct the error and solve the problem.
6. $1000 IS INVESTED AT 8.2% PER YEAR FOR 5 YEARS. USING THE EQUATION, DETERMINE THE ACCOUNT BALANCE IF IT IS COMPOUNDED : a. annually. b. quarterly
6. $1000 IS INVESTED AT 8.2% PER YEAR FOR 5 YEARS. USING THE EQUATION, DETERMINE THE ACCOUNT BALANCE IF IT IS COMPOUNDED : c. monthly d. daily
WOULD IT MAKE MORE OF A DIFFERENCE IF THE INTEREST IS ACCRUED FOR 25 YEARS? EXPLAIN YOUR REASONING.
PG. 395-399, #2, 3, 6, 8, 10, 11, 15, 18 Independent practice