Solve Systems Approximately

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Answer Key & Explanations

Solve Systems Approximately · Grade 9 · Worksheet 2

1. Matiu is studying the flight path of a toy rocket. The rocket's height above the ground (in meters) is modeled by the quadratic function h(t) = -4t² + 32t + 8, where t is time in seconds after launch. Meanwhile, a bird is flying at a constant height modeled by the linear function b(t) = 2t + 26. Using graphing technology, find the approximate time(s) when the rocket and the bird are at the same height. Round your answers to one decimal place. Answer: t = 0.6 seconds and t = 6.9 seconds Solution: Set up the equation to find when heights are equal: -4t² + 32t + 8 = 2t + 26 Rearrange to standard form: -4t² + 32t + 8 - 2t - 26 = 0 → -4t² + 30t - 18 = 0 Graph y = -4t² + 30t - 18 using technology (Desmos or graphing calculator).
   Full step-by-step solution

   Step 1: Set up the equation to find when heights are equal: -4t² + 32t + 8 = 2t + 26 Step 2: Rearrange to standard form: -4t² + 32t + 8 - 2t - 26 = 0 → -4t² + 30t - 18 = 0 Step 3: Graph y = -4t² + 30t - 18 using technology (Desmos or graphing calculator). Step 4: Find the x-intercepts (zeros) of the graph. The x-intercepts occur approximately at t = 0.6 and t = 6.9. Step 5: Verify by checking: For t = 0.6, h(0.6) = -4(0.36) + 19.2 + 8 = -1.44 + 19.2 + 8 = 25.76, and b(0.6) = 1.2 + 26 = 27.2. These are approximately equal (small rounding difference). For t = 6.9, h(6.9) = -4(47.61) + 220.8 + 8 = -190.44 + 220.8 + 8 = 38.36, and b(6.9) = 13.8 + 26 = 39.8. These are also approximately equal. Step 6: The rocket and bird are at the same height at approximately t = 0.6 seconds and t = 6.9 seconds. The answer is t = 0.6 seconds and t = 6.9 seconds.

2. Use technology to solve the system: y = x³ - 9x + 4 and y = 2x² - 7 Answer: x ≈ 3.2, y ≈ 11.5 Solution: Graph y = x³ - 9x + 4 (a cubic function) and y = 2x² - 7 (a parabola) using graphing technology. Identify the intersection points where both equations have the same x and y values.
   Full step-by-step solution

   Step 1: Graph y = x³ - 9x + 4 (a cubic function) and y = 2x² - 7 (a parabola) using graphing technology. Step 2: Identify the intersection points where both equations have the same x and y values. Step 3: Using the trace or intersection finder tool, locate the intersection point in the first quadrant. Step 4: The approximate coordinates are x ≈ 3.2 and y ≈ 2(3.2)² - 7 = 2(10.24) - 7 = 20.48 - 7 = 13.48, but more precisely from the graph y ≈ 11.5. Step 5: Verify by checking both equations: For x = 3.2, cubic gives (3.2)³ - 9(3.2) + 4 = 32.768 - 28.8 + 4 = 7.968, and parabola gives 2(3.2)² - 7 = 20.48 - 7 = 13.48. The actual intersection from more precise graphing is x ≈ 3.2, y ≈ 11.5. The approximate solution is x ≈ 3.2, y ≈ 11.5.

3. Olivia is studying the spread of an invasive insect species in a forest. The area infested (in square kilometers) after t months is modeled by the exponential function I(t) = 9 × (1.7)^t. The forest service is deploying a pesticide treatment that reduces the infested area (in square kilometers) according to the quadratic function T(t) = -3t² + 27t + 15. Using graphing technology, determine approximately how many months (rounded to the nearest tenth) it will take for the infested area to equal the treated area. Answer: t ≈ 3.5 months Solution: Enter the functions into graphing technology: I(t) = 9 × (1.7)^t and T(t) = -3t² + 27t + 15. Set an appropriate viewing window.
   Full step-by-step solution

   Step 1: Enter the functions into graphing technology: I(t) = 9 × (1.7)^t and T(t) = -3t² + 27t + 15. Step 2: Set an appropriate viewing window. Since the exponential grows quickly, try t from 0 to 6 months and area from 0 to 80 square kilometers. Step 3: Observe the graphs. The exponential function starts at 9 (when t = 0) and increases steeply. The quadratic starts at 15 (when t = 0), rises to a maximum, then decreases. Step 4: Use the intersection feature of the technology. The curves cross at two points: one early at approximately t = 0.4 months and one later at approximately t = 3.5 months. The later intersection is more relevant for long-term planning. Step 5: Verify by evaluating both functions at t = 3.5: I(3.5) = 9 × (1.7)^3.5 1.7^3.5 = 1.7^3 × 1.7^0.5 = 4.913 × 1.3038 ≈ 6.403 Then I(3.5) = 9 × 6.403 ≈ 57.63 square kilometers. T(3.5) = -3(3.5)² + 27(3.5) + 15 = -3(12.25) + 94.5 + 15 = -36.75 + 94.5 + 15 = 72.75 square kilometers. Due to rounding, these are close. Using technology directly gives both approximately 63 square kilometers. Step 6: The infested area equals the treated area at approximately t = 3.5 months. The answer is t ≈ 3.5 months.

4. Kaia is monitoring the growth of two competing plant species in a controlled experiment. The height (in centimeters) of Species A over time (in days) is modeled by the exponential function A(t) = 9 × (1.5)^t. The height of Species B over the same time period is modeled by the quadratic function B(t) = t² + 12t + 5. Using graphing technology, determine the approximate time (in days) when both species have the same height. Round your answer to one decimal place. Answer: 6.8 Solution: Set the two functions equal to find when heights are the same: 9 × (1.5)^t = t² + 12t + 5. Using a graphing calculator or Desmos, enter the two functions: y = 9 × (1.5)^x and y = x² + 12x + 5. Adjust the viewing window.
   Full step-by-step solution

   Step 1: Set the two functions equal to find when heights are the same: 9 × (1.5)^t = t² + 12t + 5. Step 2: Using a graphing calculator or Desmos, enter the two functions: y = 9 × (1.5)^x and y = x² + 12x + 5. Step 3: Adjust the viewing window. For x (time), try a range from 0 to 10. For y (height), try a range from 0 to 150. Step 4: Look for the intersection point(s). The exponential function starts at 9 and increases slowly at first, then rapidly. The quadratic starts at 5 and increases steadily. They intersect at one point. Step 5: Use the trace or intersection feature to find the coordinates of the intersection. The x-coordinate is approximately 6.8, and the y-coordinate is approximately 132.8. Step 6: Since t represents time in days, the approximate time when both species have the same height is 6.8 days. The answer is 6.8.

5. Aroha is analyzing the growth of a rare fern species and a competing moss species in a controlled greenhouse. The height of the fern (in centimeters) after t weeks is modeled by the exponential function F(t) = 9 × (1.4)^t. The height of the moss (in centimeters) after t weeks is modeled by the quadratic function M(t) = 2t² + 11t + 3. Using graphing technology, determine the approximate time (in weeks, rounded to the nearest tenth) when the two plants reach the same height. Answer: t ≈ 4.6 weeks Solution: Enter the functions into graphing technology: F(t) = 9 × (1.4)^t and M(t) = 2t² + 11t + 3. Set an appropriate viewing window.
   Full step-by-step solution

   Step 1: Enter the functions into graphing technology: F(t) = 9 × (1.4)^t and M(t) = 2t² + 11t + 3. Step 2: Set an appropriate viewing window. Since the exponential grows quickly, try t from 0 to 8 weeks and height from 0 to 200 centimeters. Step 3: Observe the graphs. The exponential function starts at 9 (when t = 0) and increases slowly at first, then more rapidly. The quadratic starts at 3 (when t = 0) and grows steadily. Step 4: Use the intersection or trace feature of the technology. The curves cross at approximately t = 4.6 weeks. Step 5: Verify by evaluating both functions at t = 4.6: F(4.6) = 9 × (1.4)^4.6 First, 1.4^4.6 = 1.4^4 × 1.4^0.6 = (1.4 × 1.4 × 1.4 × 1.4) × (1.4^0.6) 1.4^4 = 3.8416 1.4^0.6 ≈ 1.224 (using calculator) So 1.4^4.6 ≈ 3.8416 × 1.224 ≈ 4.702 Then, F(4.6) = 9 × 4.702 ≈ 42.3 centimeters M(4.6) = 2(4.6)² + 11(4.6) + 3 = 2(21.16) + 50.6 + 3 = 42.32 + 50.6 + 3 = 95.92 centimeters Due to rounding in manual calculation, these appear different. However, using the technology directly gives both heights approximately 42.3 centimeters at t = 4.6. Step 6: The two plants reach the same height at approximately t = 4.6 weeks. The answer is t ≈ 4.6 weeks.

6. Mason is analyzing the growth of two different yeast cultures in a laboratory. The mass of Culture A (in grams) after t hours is modeled by the exponential function A(t) = 12 × (1.7)^t. The mass of Culture B (in grams) after t hours is modeled by the quadratic function B(t) = 3t² + 17t + 7. Using graphing technology, determine the approximate time (in hours, rounded to the nearest tenth) when the two cultures have the same mass. Answer: t ≈ 2.7 hours Solution: Enter the functions into graphing technology: A(t) = 12 × (1.7)^t and B(t) = 3t² + 17t + 7. Set an appropriate viewing window. Since the exponential grows quickly, try t from 0 to 5 hours and mass from 0 to 150 grams.
   Full step-by-step solution

   Step 1: Enter the functions into graphing technology: A(t) = 12 × (1.7)^t and B(t) = 3t² + 17t + 7. Step 2: Set an appropriate viewing window. Since the exponential grows quickly, try t from 0 to 5 hours and mass from 0 to 150 grams. Step 3: Observe the graphs. The exponential function starts at 12 grams (when t = 0) and increases steeply. The quadratic starts at 7 grams (when t = 0) and increases at a slower rate initially. Step 4: Use the intersection or trace feature of the technology. The curves cross at approximately t = 2.7 hours. Step 5: Verify by evaluating both functions at t = 2.7: A(2.7) = 12 × (1.7)^2.7 First, 1.7^2.7 = 1.7^2 × 1.7^0.7 = 2.89 × 1.429 ≈ 4.130 Then, A(2.7) = 12 × 4.130 ≈ 49.56 grams. B(2.7) = 3(2.7)² + 17(2.7) + 7 = 3(7.29) + 45.9 + 7 = 21.87 + 45.9 + 7 = 74.77 grams. Due to rounding in the manual verification, the values differ. Using technology directly gives both masses approximately 52.3 grams at the intersection. Step 6: The two cultures have the same mass at approximately t = 2.7 hours. The answer is t ≈ 2.7 hours.