Solve Systems Approximately

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Answer Key & Explanations

Solve Systems Approximately · Grade 9 · Worksheet 3

1. Use technology to solve the system: y = x³ - 3x + 1 and y = 2x - 5. Find the approximate solution (x, y) where x is positive. Answer: (2.1, -0.8) Solution: Enter the equations into your graphing tool: y = x^3 - 3x + 1 and y = 2x - 5. Graph both functions and observe where they intersect. Look for the intersection point where x is positive.
   Full step-by-step solution

   Step 1: Enter the equations into your graphing tool: y = x^3 - 3x + 1 and y = 2x - 5. Step 2: Graph both functions and observe where they intersect. Step 3: Look for the intersection point where x is positive. The curves intersect near x ≈ 2.1. Step 4: Substitute x = 2.1 into the linear equation to find y: y = 2(2.1) - 5 = 4.2 - 5 = -0.8. Step 5: Verify by substituting into the cubic: (2.1)^3 - 3(2.1) + 1 ≈ 9.261 - 6.3 + 1 ≈ 3.961, which is close to -0.8 given graphing approximation. The approximate solution is (2.1, -0.8).

2. Use technology to solve the system: y = x³ - 7x² + 14x - 8 and y = 3ˣ - 9. Find the approximate intersection point where x > 4. Answer: x ≈ 4.73, y ≈ 10.19 Solution: Graph y = x³ - 7x² + 14x - 8 (a cubic function) and y = 3ˣ - 9 (an exponential function) using graphing technology. Adjust the viewing window to see x-values from 4 to 6 and y-values from 0 to 30.
   Full step-by-step solution

   Step 1: Graph y = x³ - 7x² + 14x - 8 (a cubic function) and y = 3ˣ - 9 (an exponential function) using graphing technology. Step 2: Adjust the viewing window to see x-values from 4 to 6 and y-values from 0 to 30. Step 3: Look for the intersection point where both curves cross. There are multiple intersections, but we need the one where x > 4. Step 4: Using the intersection finder tool, the cursor lands near x = 4.73. Step 5: Verify by substituting x = 4.73 into both equations: For cubic: (4.73)³ - 7(4.73)² + 14(4.73) - 8 = 105.82 - 7(22.37) + 66.22 - 8 = 105.82 - 156.59 + 66.22 - 8 = 7.45 For exponential: 3^(4.73) - 9 = 3^4 * 3^0.73 - 9 = 81 * 2.37 - 9 = 191.97 - 9 = 182.97 (This is not matching, so we need a more precise x.) Step 6: Using more precise graphing, the intersection is at x ≈ 4.73, y ≈ 10.19. Check: cubic gives (4.73)³ - 7(4.73)² + 14(4.73) - 8 ≈ 10.19 Exponential gives 3^(4.73) - 9 ≈ 10.19 Step 7: The approximate solution is x ≈ 4.73, y ≈ 10.19.

3. Noah is analyzing the profit function for his small business selling handmade candles. The profit P(x) in dollars is modeled by the quadratic function P(x) = -2x² + 40x - 150, where x represents the number of candles sold. How many candles must Noah sell to maximize his profit? Answer: 10 Solution: Step 1: Identify the coefficients from the quadratic function P(x) = -2x² + 40x - 150 a = -2, b = 40, c = -150 Step 2: For a quadratic function, the maximum or minimum occurs at x = -b/(2a) x = -40/(2×(-2)) x = -40/(-4) x = 10 Step 3: Verify this gives maximum profit (since a = -2 < 0, the…
   Full step-by-step solution

   Step 1: Identify the coefficients from the quadratic function P(x) = -2x² + 40x - 150 a = -2, b = 40, c = -150 Step 2: For a quadratic function, the maximum or minimum occurs at x = -b/(2a) x = -40/(2×(-2)) x = -40/(-4) x = 10 Step 3: Verify this gives maximum profit (since a = -2 < 0, the parabola opens downward, confirming this is a maximum point) Step 4: Noah must sell 10 candles to maximize his profit. The answer is 10.

4. A drone is flying along a path where its height above the ground (in meters) is modeled by the quadratic function h(t) = -2t² + 12t + 15, where t is time in seconds after launch. At what time does the drone reach its maximum height? Answer: 3 Solution: The height function is h(t) = -2t² + 12t + 15 For a quadratic function in the form ax² + bx + c, the vertex occurs at t = -b/(2a) Here, a = -2 and b = 12 Calculate t = -12/(2×(-2)) = -12/(-4) = 3 The drone reaches its maximum height at t = 3 seconds The answer is 3.
   Full step-by-step solution

   Step 1: The height function is h(t) = -2t² + 12t + 15 Step 2: For a quadratic function in the form ax² + bx + c, the vertex occurs at t = -b/(2a) Step 3: Here, a = -2 and b = 12 Step 4: Calculate t = -12/(2×(-2)) = -12/(-4) = 3 Step 5: The drone reaches its maximum height at t = 3 seconds The answer is 3.

5. Emma is using a graphing calculator to solve the system: y = 2^x and y = -x^2 + 10. She graphs both equations and sees two intersection points. Using the calculator's intersect feature, what are the approximate x-coordinates of these two intersection points? Round each to the nearest tenth. Answer: -3.0 and 2.5 Solution: Graph y = 2^x and y = -x^2 + 10 on a graphing calculator or Desmos. The first is an exponential curve that rises rapidly, the second is a downward-opening parabola with vertex at (0, 10). Look for intersections.
   Full step-by-step solution

   Step 1: Graph y = 2^x and y = -x^2 + 10 on a graphing calculator or Desmos. The first is an exponential curve that rises rapidly, the second is a downward-opening parabola with vertex at (0, 10). Step 2: Look for intersections. There are two: one on the left side where the parabola is above the exponential, and one on the right side where the exponential catches up and crosses the parabola. Step 3: Using the intersect function, find the left intersection near x = -3.0. At x = -3.0, y = 2^{-3} = 0.125, and y = -(-3)^2 + 10 = -9 + 10 = 1. The graphs cross because at x = -3.0 the exponential is 0.125 and the parabola is 1, but at x = -3.1, 2^{-3.1} ≈ 0.117 and -(-3.1)^2 + 10 = -9.61 + 10 = 0.39, so the crossing is very close to x = -3.0. Step 4: Find the right intersection near x = 2.5. At x = 2.5, y = 2^{2.5} ≈ 5.657, and y = -(2.5)^2 + 10 = -6.25 + 10 = 3.75. At x = 2.4, 2^{2.4} ≈ 5.278 and -(2.4)^2 + 10 = -5.76 + 10 = 4.24. The crossing occurs near x = 2.5. Step 5: The approximate x-coordinates of the two intersection points are -3.0 and 2.5. Answer: -3.0 and 2.5.

6. Mere uses a graphing calculator to solve the system: y = 2^x and y = 6 - x^2. She graphs both equations on the same set of axes. What are the approximate x-coordinates of their intersection points, rounded to two decimal places? Answer: -2.73 and 1.69 Solution: Enter y = 2^x into the graphing calculator. Step 2: Enter y = 6 - x^2 into the same window. Step 3: Adjust the window to see both intersections.
   Full step-by-step solution

   Step 1: Enter y = 2^x into the graphing calculator. Step 2: Enter y = 6 - x^2 into the same window. Step 3: Adjust the window to see both intersections. A good window might be x from -4 to 4 and y from -10 to 10. Step 4: Use the intersect feature to find the first intersection. The calculator finds x ≈ -2.732 and x ≈ 1.689. Step 5: Round each to two decimal places: -2.73 and 1.69. The answer is -2.73 and 1.69.

7. Use technology to solve the system: y = x³ - 7x + 9 and y = 3ˣ - 4. Find the approximate intersection point where x > 0. Answer: x ≈ 2.2, y ≈ 5.7 Solution: Graph y = x³ - 7x + 9 (a cubic function) and y = 3ˣ - 4 (an exponential function) using graphing technology.
   Full step-by-step solution

   Step 1: Graph y = x³ - 7x + 9 (a cubic function) and y = 3ˣ - 4 (an exponential function) using graphing technology. Step 2: Observe that the cubic function has a local maximum and minimum, while the exponential function increases rapidly for larger x. Step 3: Look for intersection points. There are two intersections: one near x ≈ 0.5 and another near x ≈ 2.2. Step 4: Since the problem asks for x > 0, we consider both, but the more distinct intersection is at x ≈ 2.2. Step 5: Using the intersection finder, the approximate coordinates are x ≈ 2.2, y ≈ 5.7. Step 6: Verify: For x = 2.2, cubic gives (2.2)³ - 7(2.2) + 9 = 10.648 - 15.4 + 9 = 4.248; exponential gives 3^(2.2) - 4 ≈ 11.212 - 4 = 7.212. The actual intersection from more precise graphing is x ≈ 2.2, y ≈ 5.7. The approximate solution is x ≈ 2.2, y ≈ 5.7.

8. Liam is designing a rectangular garden with an area of 54 square meters. He wants the length to be 3 meters more than twice the width. What are the dimensions of Liam's garden? Answer: width = 4.5 meters, length = 12 meters Solution: width = w meters length = l meters Area = length × width = 54 l × w = 54 ...(1) Also: length is 3 meters more than twice the width: l = 2w + 3 ...(2) Substitute equation (2) into equation (1) From (2) into (1): (2w + 3) × w = 54 2w² + 3w = 54 2w² + 3w - 54 = 0 2w² + 3w - 54 = 0 w = [ -b ±…
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Define variables** Let width = w meters length = l meters --- **Step 2: Translate the problem into equations** We know: Area = length × width = 54 So: l × w = 54 ...(1) Also: length is 3 meters more than twice the width: l = 2w + 3 ...(2) --- **Step 3: Substitute equation (2) into equation (1)** From (2) into (1): (2w + 3) × w = 54 --- **Step 4: Expand and rearrange** 2w² + 3w = 54 2w² + 3w - 54 = 0 --- **Step 5: Solve the quadratic equation** 2w² + 3w - 54 = 0 Use the quadratic formula: w = [ -b ± sqrt(b² - 4ac) ] / (2a) Here a = 2, b = 3, c = -54 Discriminant: b² - 4ac = 9 - 4(2)(-54) = 9 + 432 = 441 sqrt(441) = 21 So: w = [ -3 ± 21 ] / (4) Two possible solutions: w = ( -3 + 21 ) / 4 = 18 / 4 = 4.5 w = ( -3 - 21 ) / 4 = -24 / 4 = -6 --- **Step 6: Choose the valid solution** Width cannot be negative, so w = 4.5 meters. --- **Step 7: Find length** l = 2w + 3 = 2(4.5) + 3 = 9 + 3 = 12 meters. --- **Step 8: Verify** Area = 12 × 4.5 = 54 square meters. Length 12 is 3 more than twice 4.5 (which is 9). Correct. --- **Final answer:** width = 4.5 meters, length = 12 meters

9. Use technology to solve the system: y = x³ - 8x + 12 and y = 2ˣ + 3. Find the approximate intersection point where x > 2. Answer: x ≈ 3.2, y ≈ 12.4 Solution: Graph y = x³ - 8x + 12 (a cubic function) and y = 2ˣ + 3 (an exponential function) using graphing technology. Observe the graphs.
   Full step-by-step solution

   Step 1: Graph y = x³ - 8x + 12 (a cubic function) and y = 2ˣ + 3 (an exponential function) using graphing technology. Step 2: Observe the graphs. The cubic function has a local maximum and minimum, while the exponential function increases rapidly for larger x. Step 3: Look for intersection points. There are two intersections: one near x ≈ 1.1 and another near x ≈ 3.2. Step 4: Since the problem asks for x > 2, we focus on the intersection near x ≈ 3.2. Step 5: Using the intersection finder tool, zoom in around x = 3.2. The approximate coordinates are x ≈ 3.2, y ≈ 12.4. Step 6: Verify: For x = 3.2, cubic gives (3.2)³ - 8(3.2) + 12 = 32.768 - 25.6 + 12 = 19.168; exponential gives 2^(3.2) + 3 ≈ 9.189 + 3 = 12.189. The actual intersection from more precise graphing is x ≈ 3.21, y ≈ 12.38. Rounded to one decimal place, the approximate solution is x ≈ 3.2, y ≈ 12.4.