Exponential Functions Chapter 6 Exponential Functions

6.1 – exploring the characteristics of exponential functions Chapter 6

Exponential functions What are exponents? An exponential function is a function that has an x in the exponent. Standard form: Exponential 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.

Exponential functions Can we make a table of values? x f(x) –3 10-3 = 0.001 –2 10-2 = 0.01 –1 10-1 = 0.1 100 = 1 1 101 = 10 2 102 = 100 3 103 = 1000

Exponential functions f(x) –3 2(5)-3 = 0.016 –2 2(5)-2 = 0.08 –1 2(5)-1 = 0.4 2(5)0 = 2 1 2(5)1 = 10 2 2(5)2 = 50 3 2(5)3 = 250 What are the differences and similarities?

Exponential functions f(x) –3 8 –2 4 –1 2 1 0.5 0.25 3 0.125 x f(x) –3 512 –2 128 –1 32 8 1 2 0.5 3 0.125

Consider the graphs we’ve seen What are the: number of x-intercepts the y-intercept the end behaviour the domain the range

Consider the graphs we’ve seen What are the: number of x-intercepts the y-intercept the end behaviour the domain the range

Pg. 337, #1-3

6.2 – Relating the characteristics of an exponential function to its equation Chapter 6

Exponential functions

example y = 1(2.718)x e = 2.718… So: a = 1 Predict the number of x-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. x-intercepts: None What is the value of e? Exponential functions don’t have x-intercepts. e = 2.718… End behaviour: It is increasing, so it starts in quadrant II and goes to quadrant I. So the equation can be written: y = 1(2.718)x If b > 1, then the function is increasing. So: a = 1 b = 2.718 The y-intercept is at x = 0, so it’s at y = e0 = 1.

example a > 0, and 0 < b < 1 a = 9 b = 2/3 0 x-intercepts Predict the number of x-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. 0 x-intercepts End behavior, since it’s decreasing (b is smaller than 1), is going from quadrant II to quadrant I. Domain: {x | x E R} Range: {y | y > 0, y E R} y-intercept is at x = 0: y = 9(2/3)0 y = 9(1) y = 9 a = 9 b = 2/3 a > 0, and 0 < b < 1 This means that a is positive, and b is between 0 and 1.

example

Pg. 346-351, #1, 3, 4, 6, 9, 11, 13, 14. Independent practice

6.3 – solving exponential equations Chapter 6

Exponent laws review

example Solve each equation, and verify your answer by substitution. a) b) c) a) First, we want to isolate the exponent. What’s another way to write this? Now, we take just the exponents. Check: We need to write 64 in the form of a power of 4.  4 to the power of what is 64?  43 = 64

example Solve each equation, and verify your answer by substitution. a) b) c) b) How can we write 125 as a power of 5? Now we take just the exponents. How can we write a square root as a power? Check:

example Solve each equation, and verify your answer by substitution. a) b) c) How can I write 9 as a power of 3? Check:

example There is 50% visibility at 5 m below the surface. 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% visibility at the surface but only 25% visibility 10 m below the surface. The team determined that the visibility for the dive could be modelled by the following half-life exponential function: A0 represents the percentage of light at the surface of the water; x represents the depth in metres; h represents the depth, in metres, where there is half the original visibility; and A(x) represents the percentage of light at the depth of x metres. At what depth will the visibility be half the visibility at the surface? Which variable are we looking for? What percentage of light is at the surface? A0 = 100 x = 10 A(x) = 25 There is 50% visibility at 5 m below the surface.

example Y1 = 2^(x + 1) Y2 = 5^(x – 1) x = 2.5 Solve the following exponential equation: Round your answer to one decimal place. Using your graphing calculator. Make each side of the equal sign one equation in your Y=. 2nd TRACE 5: INTERSECT Y1 = 2^(x + 1) Y2 = 5^(x – 1) ENTER THREE TIMES The “x =“ on the bottom left is your answer. x = 2.5

Pg. 361-365, #2, 3, 8, 10, 13, 16 Independent practice

6.4 – modelling data using exponential functions Chapter 6

example The population of Canada from 1871 to 1971 is shown in the table below. In the third column, the values have been rounded.

example Using graphing technology, create a graphical model and an algebraic exponential model for the data. Assuming that the population growth continued at the same rate to 2011, estimate the population in 2011. Round your answer to the nearest million. After entering in the data, we use STAT  CALC  0: ExpReg It gives us: a = 2.6685… b = 1.2398… y = 2.67(1.24)x If 1971 was the 10th interval we looked at, what interval is 2011?  It is the 14th interval  x = 14 In 2011, the population will be approximately 54 million people. y = 2.67(1.24)14 y = 54.3

example Sonja did an experiment 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.

example x = 33 y = 78.7(0.97)x y = 51ºC Y1 = 30 Y2 = 78.6812(0.97152)x Construct a scatter plot to display the data. Determine the equation of the exponential regression function that models Sonja’s data. Estimate the temperature of the water 15 min after the experiment began. Round your answer to the nearest degree. Estimate when the water reached a temperature of 30ºC. Round your answer to the nearest minute. Our calculator gives us: a = 78.6812… b = 0.97152… c) Let y = 30, and solve for x. When you are dealing with decimals like this, you should definitely use your calculator. y = 78.7(0.97)x 30 = 78.6812(0.97152)x b) Solve for y when x = 15 y = 78.6812(0.97152)15 y = 51.009 Y1 = 30 Y2 = 78.6812(0.97152)x x = 33 Find the intercept. y = 51ºC

Pg. 377-384, #1, 2, 4, 6, 9, 10, 13, 15 Independent practice

6.5 – financial applications involving exponential functions Chapter 6

Compound interest Compound interest is the interest earned on both the original amount that was invested and any interest that has accumulated over time. The future value is the amount that the investment will be worth after a certain amount of time. The compound interest formula: where: A(n) = future value P = principal i = interest rate per compounding period n = number of compounding periods The principal is the original amount of money that is invested or borrowed. The compounding period is the time over which interest is calculated and paid on an investment or loan.

example Brittany invested $2500 in an account that pays 3.5%/a compounded monthly. The following table gives the value of her investment for the first five months. Use exponential regression to determine the compound interest function that models this situation. Explain how the values in the table were determined. How long, in months, will it take for Brittany’s investment to grow to $3000. The interest rate is 3.5% per year, compounded monthly. So what is the interest rate per month?  0.035/12 = 0.002… P = 2500.00 i = 0.002

example Brittany invested $2500 in an account that pays 3.5%/a compounded monthly. The following table gives the value of her investment for the first five months. Use exponential regression to determine the compound interest function that models this situation. Explain how the values in the table were determined. How long, in months, will it take for Brittany’s investment to grow to $3000. c) Let A(n) = 3000 3000 = 2500(1.002)n Use your calculator. Y1 = 3000 Y2 = 2500(1.002)n x = 62.604 b) If we fill in different values for n, we find: A(1) = 2500(1.002)1 = 2507.29 A(2) = 2500(1.002)2 = 2514.60 A(3) = 2500(1.002)3 = 2521.94 It will take Brittany’s investment 63 months to grow to $3000.

example x = 11.3 200 = 2000(0.8)n A(n) = 2000(0.8)n Gina bought a camera for her studio 2 years ago. Her accountant told her that, starting from the beginning of the second year, she can claim a depreciation rate of 20% for the camera as a business expense. At the beginning of the second year, Gina’s camera was worth $2000. How long, in years from the time of the purchase, will it take for the camera to be worth only $200? Explain how the exponential regression function that models this situation relates to the depreciation of the camera. First let’s create our equation: What’s the principal? P = 2000 What’s our i? i = –0.2 A(n) = 2000(0.8)n What does the $200 represent?  Let A(n) = 200 x = 11.3 200 = 2000(0.8)n It will take 12 years for the camera to be worth only $200. Use your calculator. A(n) = 2000(1 – 0.2)n

example She will pay $230.77 in interest. Jessica borrowed $7500 from a bank to buy new equipment for her band. The bank is charging an interest rate of 3.6%/a compounded monthly. Jessica’s monthly loan payment is $400. Determine how long, to the nearest month, it will take Jessica to pay off the loan. The loan manager gave Jessica the following equation so she could determine how long it would take to pay off her loan (1.003)-n = 0.94375 where n represents the number of months How much interest will Jessica pay on her loan? b) How much will Jessica pay over those 19.326 months?  She pays $400 a month 400 x 19.326 = 7730.77 7730.77 – 7500.00 = 230.77 Solve for n, using your calculator. Y1 = (1.003)-n Y2 = 0.94375 x = 19.326 It will take her 20 months to pay of this loan. She will pay $230.77 in interest.

Pg. 395-399, #2, 3, 6, 8, 10, 11, 15, 18 Independent practice