Model Comparison

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Answer Key & Explanations

Model Comparison · Grade 11 · Worksheet 1

1. Emma is analyzing the growth of three different functions over time. On a coordinate grid, she plots the graphs of y = 5x, y = 5x², and y = 5(2^x) for x ≥ 0. For x = 0, all three functions start at (0,0) or (0,5). Looking at the graphs, which function will eventually have the greatest y-value for very large values of x, and which will have the smallest? Explain how the shapes of the graphs compare. Answer: Exponential (y = 5(2^x)) grows fastest; linear (y = 5x) grows slowest for large x. Solution: Evaluate each function at a small x, say x = 1. y = 5(1) = 5; y = 5(1)^2 = 5; y = 5(2)^1 = 10. At x = 1, the exponential is already ahead.
   Full step-by-step solution

   Step 1: Evaluate each function at a small x, say x = 1. y = 5(1) = 5; y = 5(1)^2 = 5; y = 5(2)^1 = 10. At x = 1, the exponential is already ahead. Step 2: Evaluate at x = 5. y = 5(5) = 25; y = 5(25) = 125; y = 5(32) = 160. The exponential leads, quadratic is second. Step 3: Evaluate at x = 10. y = 5(10) = 50; y = 5(100) = 500; y = 5(1024) = 5120. The exponential is now far ahead. Step 4: For very large x (e.g., x = 20): y = 5(20) = 100; y = 5(400) = 2000; y = 5(1,048,576) = 5,242,880. The exponential dwarfs the others. Step 5: Compare growth rates: Linear has constant slope 5. Quadratic has slope 10x, which increases linearly. Exponential has slope 5(2^x)ln(2), which grows exponentially. Since exponential growth multiplies by a constant factor each unit increase, it eventually surpasses any polynomial growth. The answer: For large x, the exponential function y = 5(2^x) grows fastest, and the linear function y = 5x grows slowest.

2. Compare f(x) = 9x + 14, g(x) = x^2 + 11, and h(x) = 4^x for large x. Which function grows fastest? Answer: h(x) = 4^x Solution: Analyze f(x) = 9x + 14 (linear). For large x, the growth rate is constant: each increase of 1 in x adds 9 to the output. Analyze g(x) = x^2 + 11 (quadratic).
   Full step-by-step solution

   Step 1: Analyze f(x) = 9x + 14 (linear). For large x, the growth rate is constant: each increase of 1 in x adds 9 to the output. Step 2: Analyze g(x) = x^2 + 11 (quadratic). For large x, the output grows proportionally to x^2. For example, when x = 10, g(10) = 111; when x = 20, g(20) = 411; the increase from x=10 to x=20 is 300. Step 3: Analyze h(x) = 4^x (exponential). For large x, each increase of 1 in x multiplies the output by 4. For example, h(10) = 1,048,576; h(11) = 4,194,304; the increase is huge. Step 4: Compare. For x = 10: f(10) = 104, g(10) = 111, h(10) = 1,048,576. For x = 20: f(20) = 194, g(20) = 411, h(20) = 4^20 = 1,099,511,627,776. Exponential growth far exceeds polynomial growth for large x. Therefore, h(x) = 4^x grows fastest.

3. Aroha is comparing three functions: f(x) = 9x + 11, g(x) = 3x² + 5, and h(x) = 7^x. For large x, which function grows fastest? Answer: h(x) = 7^x Solution: Identify the function types. f(x) = 9x + 11 is linear (degree 1). g(x) = 3x² + 5 is quadratic (degree 2).
   Full step-by-step solution

   Step 1: Identify the function types. f(x) = 9x + 11 is linear (degree 1). g(x) = 3x² + 5 is quadratic (degree 2). h(x) = 7^x is exponential (base 7 > 1). Step 2: Compare linear and quadratic. For large x, the quadratic term 3x² grows much faster than the linear term 9x. So g(x) > f(x) for sufficiently large x. Step 3: Compare quadratic and exponential. For large x, exponential functions with base > 1 always outgrow any polynomial function. Since 7^x grows by multiplying by 7 each time x increases by 1, while 3x² only grows by adding roughly 6x + 3 each step, the exponential function eventually far exceeds the quadratic. Step 4: Conclusion. For large x, h(x) = 7^x grows fastest. The answer is h(x) = 7^x.

4. Sophia is a civil engineer comparing three different models for the growth of traffic volume on a new highway over time. Model L predicts linear growth: V(t) = 7200 + 240t vehicles per day, where t is years after opening. Model Q predicts quadratic growth: V(t) = 30t² + 7200 vehicles per day. Model E predicts exponential growth: V(t) = 7200(1.12)^t vehicles per day. After 12 years, which model predicts the highest traffic volume? Answer: E Solution: Calculate Model L (linear) at t = 12. V(12) = 7200 + 240(12) = 7200 + 2880 = 10,080 vehicles per day. Calculate Model Q (quadratic) at t = 12.
   Full step-by-step solution

   Step 1: Calculate Model L (linear) at t = 12. V(12) = 7200 + 240(12) = 7200 + 2880 = 10,080 vehicles per day. Step 2: Calculate Model Q (quadratic) at t = 12. V(12) = 30(12)² + 7200 = 30(144) + 7200 = 4320 + 7200 = 11,520 vehicles per day. Step 3: Calculate Model E (exponential) at t = 12. V(12) = 7200(1.12)^12. First compute (1.12)^12: 1.12^2 = 1.2544 1.12^4 = (1.2544)^2 = 1.5735 1.12^8 = (1.5735)^2 = 2.4759 1.12^12 = 1.12^8 * 1.12^4 = 2.4759 * 1.5735 = 3.8958 (rounded) Then V(12) = 7200 * 3.8958 = 28,050 vehicles per day (rounded). Step 4: Compare the results. Model L: 10,080 Model Q: 11,520 Model E: 28,050 Model E predicts the highest traffic volume after 12 years. The answer is E.

5. Compare f(x) = 13x + 20, g(x) = 3x² + 11, h(x) = 4^x for large x. Which function grows fastest? Answer: h(x) = 4^x Solution: Analyze f(x) = 13x + 20 (linear function). As x increases, f(x) grows at a constant rate of 13 per unit increase in x. For large x, f(x) behaves like 13x.
   Full step-by-step solution

   Step 1: Analyze f(x) = 13x + 20 (linear function). As x increases, f(x) grows at a constant rate of 13 per unit increase in x. For large x, f(x) behaves like 13x. Step 2: Analyze g(x) = 3x² + 11 (quadratic function). As x increases, g(x) grows proportionally to x². For large x, g(x) behaves like 3x². Quadratic growth is faster than linear growth because x² grows faster than x. Step 3: Analyze h(x) = 4^x (exponential function with base 4 > 1). As x increases, h(x) multiplies by 4 for each unit increase in x. For large x, h(x) behaves like 4^x. Step 4: Compare growth rates. For small x, linear or quadratic might be larger, but for large x: - Linear: f(x) ≈ 13x - Quadratic: g(x) ≈ 3x² - Exponential: h(x) ≈ 4^x Exponential functions with base > 1 always grow faster than any polynomial function (including quadratic) for sufficiently large x. Therefore, h(x) = 4^x grows fastest. The answer is h(x) = 4^x.

6. Compare f(x) = 14x + 22, g(x) = 2x² + 18, h(x) = 6^x for large x. Which function grows fastest? Answer: h(x) = 6^x Solution: Identify the function types. f(x) = 14x + 22 is linear (degree 1). g(x) = 2x² + 18 is quadratic (degree 2).
   Full step-by-step solution

   Step 1: Identify the function types. f(x) = 14x + 22 is linear (degree 1). g(x) = 2x² + 18 is quadratic (degree 2). h(x) = 6^x is exponential (base 6 > 1). Step 2: Compare linear and quadratic. For large x, the quadratic term 2x² dominates the linear term 14x, so g(x) > f(x) for sufficiently large x. Step 3: Compare quadratic and exponential. For large x, an exponential function with base > 1 grows faster than any polynomial function. Since 6^x grows much faster than 2x² for large x, h(x) > g(x) for sufficiently large x. Step 4: Order the functions by growth rate for large x: f(x) < g(x) < h(x). Therefore, h(x) = 6^x grows fastest for large x.

7. A research team is studying bacterial growth in a lab culture. Initially, there are 500 bacteria. After 3 hours, the population reaches 4,000 bacteria. The team models this growth using the exponential function P(t) = P₀e^(kt), where P(t) is the population at time t hours. Determine the growth constant k (to four decimal places) and calculate the predicted population after 8 hours. Answer: k ≈ 0.6931, P(8) ≈ 128000 Solution: Write down the given information. P(t) = P₀ * e^(k t) Initial population P₀ = 500 bacteria. After t = 3 hours, P(3) = 4000 bacteria.
   Full step-by-step solution

   Step 1: Write down the given information. The exponential growth model is: P(t) = P₀ * e^(k t) Initial population P₀ = 500 bacteria. After t = 3 hours, P(3) = 4000 bacteria. Step 2: Substitute the known values into the equation to find k. P(3) = 4000 = 500 * e^(k * 3) Step 3: Solve for e^(3k). Divide both sides by 500: 4000 / 500 = e^(3k) 8 = e^(3k) Step 4: Take the natural logarithm (ln) of both sides to solve for the exponent. ln(8) = ln(e^(3k)) ln(8) = 3k Step 5: Solve for k. k = ln(8) / 3 Step 6: Calculate the numerical value of k. First, note that 8 = 2^3, so ln(8) = ln(2^3) = 3 * ln(2). We know ln(2) ≈ 0.693147. Therefore: k = (3 * ln(2)) / 3 k = ln(2) k ≈ 0.693147 Rounded to four decimal places: k ≈ 0.6931 Step 7: Use the model to predict the population after t = 8 hours. P(8) = P₀ * e^(k * 8) P(8) = 500 * e^(0.6931 * 8) Step 8: Calculate the exponent. First, calculate 0.6931 * 8 = 5.5448 Step 9: Calculate e^(5.5448). We can also calculate this exactly using k = ln(2). P(8) = 500 * e^(ln(2) * 8) P(8) = 500 * e^(ln(2^8)) P(8) = 500 * 2^8 Step 10: Calculate 2^8. 2^8 = 256 Step 11: Calculate the final population. P(8) = 500 * 256 P(8) = 128000 Therefore, the growth constant k is approximately 0.6931, and the predicted population after 8 hours is 128,000 bacteria.