Day 116 – Combining Functions

Lesson: Comparing Linear and Exponential Functions A quantity is growing linearly when the same positive number is added to a quantity each time period. A quantity is growing exponentially when the same number greater than 1 is multiplied by a quantity each time period. Given an unlimited amount of time, exponential growth functions will always be larger than linear functions.

Lesson: Comparing Linear and Exponential Functions So, if the goal is to maximize a quantity and you have a choice between linear and exponential growth and time is unlimited, then you should choose exponential growth regardless of the starting quantity and regardless of the growth per time period.

Lesson: Comparing Linear and Exponential Functions When time is limited or if both functions are the same type, determine the value of each function at the end of the time interval to determine which is larger.

Example 1 The data in table A represents the value, in dollars, of a smartphone x months after purchase. The data in table B represents the number of minutes students spend texting based on grade level.

Example 1 Table A Table B Which set of data is more appropriately modeled with a linear function? An exponential function? x (months after purchase) 1 2 3 4 5 y (value in dollars) 475 386 309 248 198 x (grade level) 8 9 10 11 12 y (number of minutes texting) 32 50 78 99

Answer Data modeled by a linear function will grow by equal differences over equal intervals. There is often some slight variation in real-life data, the difference may not be exactly the same each time period but will be similar enough to indicate a linear trend. Determine the difference in y-values between equal x- value intervals.

Answer Table A: 386 − 475= −89 309 − 386= −77 248 − 309= −61 198 − 248= −50 Table B: 32 − 10 = 22 50 − 32 = 18 78 − 50 = 28 99 − 78 = 21 Since the differences in Table B are nearly identical, it is best modeled by a linear function.

Answer Data that is modeled by an exponential function grows by equal factors over equal intervals. There is often some slight variation in real-life data, the quotient may not be exactly the same each time period but will be similar enough to indicate an exponential. Determine the quotient of y-values between equal x-value intervals.

Answer Table A: 386/475 = 0.81 309/386 = 0.80 248/309 = 0.80 198/248 = 0.80 Table B: 32/10 = 3.2 50/32 = 1.56 78/50 = 1.56 99/78 = 1.30 Although two factors in table B are the same, the other ratios are not the same at all. Table A is better modeled by an exponential function.

Example 2 You have two options for your job salary. Option A: Salary starts at $50000. Salary grows by $1000 per year. Option B: Salary starts at $40000. Salary grows by 3% per year. Which option will yield the higher salary in the long run?

Answer For salary option A, the same number (you $1000 raise) is added per year. It is a linear function. For salary option B, your 3% raise results in a growth factor of 1.03, so the same greater than 1 is multiplied each time period. Since the time is unlimited, eventually option B will result in a higher salary.

Example 3 You have two options for your job salary. Option A: Salary starts at $50,000. Salary grows by $1000 per year Option B: Salary starts at $30,000. Salary grows by 3% per year. Which option will yield the higher salary after 20 years?

Answer These are the same salary situation from Example 1, so it has been determined that option A is linear and option B is exponential. Equations must be written for each salary 𝑡 years, then substitute 20 for 𝑡 in each equation.

Answer Option A: Use the formula 𝑦=𝑚𝑥+𝑏, where 𝑚=1000 and 𝑏=50,000. 𝑎 𝑡 =1000𝑡+50,000 𝑎 20 =1000 20 +50,000 =70,000

Answer Option B: Use the formula 𝑦=𝑎∙ 𝑏 𝑥 . The initial amount 𝑎 is $30,000. The growth factor 𝑏 is 1+3%=1+0.03. The equation 𝑦=30,000 1.03 𝑡 models your salary after 𝑡 years. Substitute 20 for 𝑡. 𝑏 𝑡 =30,000∙ 1.03 𝑡 𝑏 20 =30,000∙ 1.03 20 =54,183 After 20 years, Option A has a higher salary.

Example 4 You have two options for your job salary. Option A: Salary starts at $50000. Salary grows by 20.5% per year. Option B. Salary starts at $45000. Salary grows by 3% per year. Which option will yield the higher salary after 10 years?

Answer For both options, the salary is growing by the same percentage each time period, so both are modeled by exponential functions. Write equations for each situation and use them to determine the salary at 10 years.

Answer Option A: Use the formula 𝑦=𝑎∙ 𝑏 𝑥 . The initial amount a is $50,000. The growth factor 𝑏 is 1+2.5%=1+0.025=1.025. The equation 𝑦=50,000 1.025 𝑡 models your salary after 𝑡 years. Substitute 10 for 𝑡. 𝑏 𝑡 =50,000∙ 1.025 𝑡 𝑏 10 =50,000∙ 1.025 10 =64,004.23 The salary is approximately $64,004.

Answer Option B: Use the formula 𝑦=𝑎∙ 𝑏 𝑥 . The initial amount 𝑎 is $45,000. The growth factor 𝑏 is 1+3%=1+0.03=1.03. The equation 𝑦=45,000 1.03 𝑡 your salary after 𝑡 years. Substitute 10 for 𝑡. 𝑏 𝑡 =45,000∙ 1.03 𝑡 𝑏 10 =45,000∙(1.03)^10 =60,476.24 The salary is approximately $60,476. After 10 years, Option A yields the higher salary.