Function Fitting

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Answer Key & Explanations

Function Fitting · Grade 11 · Worksheet 1

1. The population of a certain bacterial culture grows according to the model P(t) = 1200e^(0.05t), where t is time in hours. After how many hours will the population reach 3000 bacteria? Round your answer to the nearest tenth of an hour. Answer: 18.3 Solution: P(t) = 1200 e^(0.05 t) We want to find t when P(t) = 3000. Set up the equation. 3000 = 1200 e^(0.05 t) Divide both sides by 1200 to isolate the exponential term.
   Full step-by-step solution

   We are given the population model: P(t) = 1200 e^(0.05 t) We want to find t when P(t) = 3000. Step 1: Set up the equation. 3000 = 1200 e^(0.05 t) Step 2: Divide both sides by 1200 to isolate the exponential term. 3000 / 1200 = e^(0.05 t) Simplify the fraction: 3000 ÷ 1200 = 30/12 = 5/2 = 2.5 So: 2.5 = e^(0.05 t) Step 3: Take the natural logarithm of both sides to solve for t. ln(2.5) = ln(e^(0.05 t)) Using the property ln(e^x) = x: ln(2.5) = 0.05 t Step 4: Solve for t. t = ln(2.5) / 0.05 Step 5: Compute ln(2.5). ln(2.5) ≈ 0.9162907319 (using a calculator) Step 6: Divide by 0.05. t ≈ 0.9162907319 / 0.05 Dividing by 0.05 is the same as multiplying by 20: t ≈ 0.9162907319 × 20 t ≈ 18.325814638 Step 7: Round to the nearest tenth of an hour. t ≈ 18.3 Final answer: 18.3 hours

2. Aroha records the following data: (1, 3), (2, 9), (3, 27), (4, 81), (5, 243). Which type of function best models this data: linear, quadratic, or exponential? Justify your answer and write the equation of the model. Answer: Exponential; y = 3^x Solution: Check for linearity by computing first differences: 9-3=6, 27-9=18, 81-27=54, 243-81=162. Not constant, so not linear. Check for quadratic by computing second differences: 18-6=12, 54-18=36, 162-54=108.
   Full step-by-step solution

   Step 1: Check for linearity by computing first differences: 9-3=6, 27-9=18, 81-27=54, 243-81=162. Not constant, so not linear. Step 2: Check for quadratic by computing second differences: 18-6=12, 54-18=36, 162-54=108. Not constant, so not quadratic. Step 3: Check for exponential by computing ratios: 9/3=3, 27/9=3, 81/27=3, 243/81=3. Constant ratio of 3, so exponential. Step 4: The general form is y = a * b^x. Using point (1,3): 3 = a * b^1. Using ratio b=3, so 3 = a * 3, thus a = 1. Step 5: The model is y = 1 * 3^x = 3^x. The answer is exponential; y = 3^x.

3. Liam is analyzing the growth of a bacterial culture in his biology lab. The culture starts with 200 bacteria and doubles every 3 hours. He needs to model the population P(t) after t hours using an exponential function. Write the function that models the bacterial population. Answer: P(t) = 200 * 2^(t/3) Solution: Identify the initial population. The problem says the culture starts with 200 bacteria. So when \( t = 0 \), \( P(0) = 200 \).
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Identify the initial population.** The problem says the culture starts with 200 bacteria. So when \( t = 0 \), \( P(0) = 200 \). --- **Step 2: Identify the growth pattern.** The population doubles every 3 hours. This means that every 3 hours, the population is multiplied by 2. --- **Step 3: General exponential form.** An exponential growth model is: \[ P(t) = P_0 \cdot a^{t} \] But here, the doubling time is 3 hours, so we adjust the exponent to match the doubling period. If it doubles every 3 hours, then after \( t \) hours, the number of doubling periods is \( t / 3 \). So: \[ P(t) = 200 \cdot 2^{t/3} \] --- **Step 4: Check the formula with an example.** At \( t = 0 \): \( P(0) = 200 \cdot 2^{0} = 200 \) ✅ At \( t = 3 \): \( P(3) = 200 \cdot 2^{3/3} = 200 \cdot 2^1 = 400 \) ✅ (doubled) At \( t = 6 \): \( P(6) = 200 \cdot 2^{6/3} = 200 \cdot 2^2 = 800 \) ✅ (doubled again) --- **Step 5: Final answer.** The function is: \[ P(t) = 200 \cdot 2^{t/3} \]

4. Ava records the following data: (1, 6), (2, 11), (3, 16), (4, 21), (5, 26). Determine which function type (linear, quadratic, or exponential) best fits this data, and write the equation of the best-fitting function. Answer: Linear; y = 5x + 1 Solution: Calculate first differences: 11-6 = 5, 16-11 = 5, 21-16 = 5, 26-21 = 5. The first differences are constant (5). Since first differences are constant, the data is linear.
   Full step-by-step solution

   Step 1: Calculate first differences: 11-6 = 5, 16-11 = 5, 21-16 = 5, 26-21 = 5. The first differences are constant (5). Step 2: Since first differences are constant, the data is linear. Step 3: The slope m = 5. Use point (1, 6): y = mx + b => 6 = 5(1) + b => b = 1. Step 4: Equation: y = 5x + 1. The answer is linear; y = 5x + 1.

5. Emma records the following data: (1, 7), (2, 21), (3, 63), (4, 189), (5, 567). Determine which function type (linear, quadratic, or exponential) best fits the data, and estimate the value when x = 6. Answer: 1701 Solution: Check for linear: differences between y-values: 21-7=14, 63-21=42, 189-63=126, 567-189=378. Not constant, so not linear. Check for quadratic: second differences: 42-14=28, 126-42=84, 378-126=252.
   Full step-by-step solution

   Step 1: Check for linear: differences between y-values: 21-7=14, 63-21=42, 189-63=126, 567-189=378. Not constant, so not linear. Step 2: Check for quadratic: second differences: 42-14=28, 126-42=84, 378-126=252. Not constant, so not quadratic. Step 3: Check for exponential: ratios of consecutive y-values: 21/7=3, 63/21=3, 189/63=3, 567/189=3. Constant ratio of 3, so exponential. Step 4: The function is y = 7 * 3^(x-1). For x=6: y = 7 * 3^(5) = 7 * 243 = 1701. The answer is 1701.

6. Liam is analyzing the growth of a bacteria culture in his biology lab. The culture starts with 200 bacteria and doubles every 3 hours. He needs to determine after how many hours the culture will reach 12,800 bacteria. Write your answer as a number. Answer: 18 Solution: The bacteria start at 200 and double every 3 hours. We want the time \( t \) (in hours) when the number of bacteria is 12,800.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Understand the problem** The bacteria start at 200 and double every 3 hours. We want the time \( t \) (in hours) when the number of bacteria is 12,800. --- **Step 2: Set up the exponential growth formula** The number of bacteria after \( t \) hours is: \[ N(t) = N_0 \times 2^{(t / T)} \] where \( N_0 = 200 \) (initial amount) \( T = 3 \) hours (doubling time) \( N(t) = 12800 \) (final amount) So: \[ 12800 = 200 \times 2^{(t / 3)} \] --- **Step 3: Isolate the exponential term** Divide both sides by 200: \[ 12800 / 200 = 2^{(t / 3)} \] \[ 64 = 2^{(t / 3)} \] --- **Step 4: Express 64 as a power of 2** We know \( 64 = 2^6 \), so: \[ 2^6 = 2^{(t / 3)} \] --- **Step 5: Equate the exponents** Since the bases are the same (base 2), we can set the exponents equal: \[ 6 = t / 3 \] --- **Step 6: Solve for \( t \)** Multiply both sides by 3: \[ t = 6 \times 3 = 18 \] --- **Step 7: Conclusion** The bacteria culture reaches 12,800 after **18 hours**. --- **Final answer:** 18

7. Ava is studying the cooling of a liquid in a chemistry lab. She records the temperature of the liquid every 10 minutes and obtains the following data: at t = 0 minutes, temperature = 96°C; at t = 10 minutes, temperature = 76°C; at t = 20 minutes, temperature = 61°C; at t = 30 minutes, temperature = 51°C; at t = 40 minutes, temperature = 46°C; at t = 50 minutes, temperature = 43°C. Which type of function (linear, quadratic, or exponential) best models this cooling data, and what is the approximate temperature at t = 60 minutes based on that model? Answer: Exponential; approximately 41°C Solution: Calculate first differences (temperature change each 10 minutes): 96-76 = 20, 76-61 = 15, 61-51 = 10, 51-46 = 5, 46-43 = 3. These differences (20, 15, 10, 5, 3) are not constant, so it is not linear.
   Full step-by-step solution

   Step 1: Calculate first differences (temperature change each 10 minutes): 96-76 = 20, 76-61 = 15, 61-51 = 10, 51-46 = 5, 46-43 = 3. These differences (20, 15, 10, 5, 3) are not constant, so it is not linear. Step 2: Calculate second differences: 15-20 = -5, 10-15 = -5, 5-10 = -5, 3-5 = -2. These are not constant (the last is -2 instead of -5), so it is not exactly quadratic. Step 3: Calculate ratios of consecutive temperatures: 76/96 = 0.7917, 61/76 = 0.8026, 51/61 = 0.8361, 46/51 = 0.9020, 43/46 = 0.9348. The ratios are not exactly constant, but the pattern of decreasing differences suggests an exponential decay approaching room temperature. The data closely fits an exponential model with a horizontal asymptote around 41°C. Step 4: Using the exponential decay pattern, the temperature decreases by a factor of roughly 0.8 each interval initially, then slows as it approaches 41°C. Extrapolating, at t = 60 minutes the temperature is approximately 41°C. The answer is exponential; approximately 41°C.

8. Aroha records the following data: (1, 5), (2, 11), (3, 19), (4, 29), (5, 41). Determine which type of function (linear, quadratic, or exponential) best fits the data, and write the equation of the best-fitting function. Answer: quadratic; y = x^2 + 3x + 1 Solution: Compute first differences: 11-5=6, 19-11=8, 29-19=10, 41-29=12. First differences are not constant (6,8,10,12), so not linear. Compute second differences: 8-6=2, 10-8=2, 12-10=2.
   Full step-by-step solution

   Step 1: Compute first differences: 11-5=6, 19-11=8, 29-19=10, 41-29=12. First differences are not constant (6,8,10,12), so not linear. Step 2: Compute second differences: 8-6=2, 10-8=2, 12-10=2. Second differences are constant (2), so the data is quadratic. Step 3: Assume quadratic form y = ax^2 + bx + c. Use three points: (1,5), (2,11), (3,19). For (1,5): a + b + c = 5 For (2,11): 4a + 2b + c = 11 For (3,19): 9a + 3b + c = 19 Step 4: Subtract first equation from second: (4a+2b+c) - (a+b+c) = 11-5 => 3a + b = 6 Subtract second from third: (9a+3b+c) - (4a+2b+c) = 19-11 => 5a + b = 8 Step 5: Subtract these two results: (5a+b) - (3a+b) = 8-6 => 2a = 2 => a = 1 Then 3(1)+b=6 => b = 3 Then from a+b+c=5: 1+3+c=5 => c = 1 Step 6: Equation is y = x^2 + 3x + 1. Check with (4,29): 16+12+1=29 correct. With (5,41): 25+15+1=41 correct. The answer is quadratic; y = x^2 + 3x + 1.