Solve Systems Approximately

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Answer Key & Explanations

Solve Systems Approximately · Grade 9 · Worksheet 2

1. Emma is monitoring the height of a rocket launched from a platform. The rocket's height (in meters) above the ground is modeled by the quadratic function h(t) = -5t² + 40t + 10, where t is time in seconds after launch. A nearby drone's height is modeled by the linear function d(t) = 10t + 25. Using graphing technology, determine approximately how many seconds after launch the rocket and the drone are at the same height. Round your answer to the nearest tenth of a second. Answer: 0.6 Solution: To find when the rocket and drone are at the same height, we need to solve h(t) = d(t). Step 2: This means -5t² + 40t + 10 = 10t + 25. Step 3: Rearranging gives -5t² + 30t - 15 = 0, or divide by -5: t² - 6t + 3 = 0.
   Full step-by-step solution

   Step 1: To find when the rocket and drone are at the same height, we need to solve h(t) = d(t). Step 2: This means -5t² + 40t + 10 = 10t + 25. Step 3: Rearranging gives -5t² + 30t - 15 = 0, or divide by -5: t² - 6t + 3 = 0. Step 4: Using a graphing calculator or Desmos, graph y = -5t² + 40t + 10 and y = 10t + 25. Step 5: Find the intersection point(s). The two curves intersect at approximately t = 0.6 seconds (and also at t = 5.4 seconds, but the first intersection is at 0.6). Step 6: The rocket and drone are at the same height at approximately 0.6 seconds after launch. The answer is 0.6.

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. Emma is analyzing the profit function for her small business. The profit P(x) in dollars from selling x units of her product is modeled by the quadratic equation P(x) = -2x² + 120x - 1000. Using graphing technology, determine approximately how many units Emma needs to sell to break even (where profit equals zero). Round your answer to the nearest whole number. Answer: 10 Solution: Set up the equation for break-even point: P(x) = 0 Solve -2x² + 120x - 1000 = 0 Divide the entire equation by -2 to simplify: x² - 60x + 500 = 0 Use the quadratic formula: x = [60 ± sqrt(60² - 4×1×500)] / (2×1) Calculate the discriminant: 3600 - 2000 = 1600 Find the square root: sqrt(1600) = 40…
   Full step-by-step solution

   Step 1: Set up the equation for break-even point: P(x) = 0 Step 2: Solve -2x² + 120x - 1000 = 0 Step 3: Divide the entire equation by -2 to simplify: x² - 60x + 500 = 0 Step 4: Use the quadratic formula: x = [60 ± sqrt(60² - 4×1×500)] / (2×1) Step 5: Calculate the discriminant: 3600 - 2000 = 1600 Step 6: Find the square root: sqrt(1600) = 40 Step 7: Apply the quadratic formula: x = [60 ± 40] / 2 Step 8: Calculate the two solutions: x = (60 + 40)/2 = 100/2 = 50 and x = (60 - 40)/2 = 20/2 = 10 Step 9: The break-even points occur at x = 10 and x = 50 units. Since we're looking for when she first breaks even, the answer is 10 units. The answer is 10.

4. 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.

5. Use technology to solve the system: y = x³ - 6x² + 12x - 8 and y = 2x - 4. Find the intersection points approximately. Answer: (2, 0) and (4, 4) Solution: Enter y = x³ - 6x² + 12x - 8 and y = 2x - 4 into Desmos or a graphing calculator. Observe the graphs and identify intersection points. The curves intersect at approximately x = 2 and x = 4.
   Full step-by-step solution

   Step 1: Enter y = x³ - 6x² + 12x - 8 and y = 2x - 4 into Desmos or a graphing calculator. Step 2: Observe the graphs and identify intersection points. Step 3: The curves intersect at approximately x = 2 and x = 4. Step 4: For x = 2: y = 2(2) - 4 = 0, so point is (2, 0). Step 5: For x = 4: y = 2(4) - 4 = 4, so point is (4, 4). Step 6: Verify by checking if these points satisfy both equations. The approximate intersection points are (2, 0) and (4, 4).

6. A company models its monthly profit P (in thousands of dollars) with the quadratic function P(x) = -0.5x² + 12x - 18, where x represents the number of units sold (in hundreds). Using technology, find the number of units (to the nearest whole number) that maximizes the monthly profit. Answer: 1200 Solution: P(x) = -0.5x² + 12x - 18 where P is in thousands of dollars, and x is in hundreds of units sold. This is a quadratic function with a negative coefficient for x² (-0.5), so the graph is a downward-opening parabola.
   Full step-by-step solution

   We are given the profit function: P(x) = -0.5x² + 12x - 18 where P is in thousands of dollars, and x is in hundreds of units sold. --- **Step 1: Understand the problem** This is a quadratic function with a negative coefficient for x² (-0.5), so the graph is a downward-opening parabola. The maximum profit occurs at the vertex. --- **Step 2: Vertex formula** For a quadratic function ax² + bx + c, the x-coordinate of the vertex is: x = -b / (2a) Here: a = -0.5 b = 12 c = -18 So: x = -12 / (2 * -0.5) x = -12 / (-1) x = 12 --- **Step 3: Interpret x** x is in hundreds of units. So x = 12 means 12 hundreds of units = 1200 units. --- **Step 4: Verify with nearest whole number** The problem says "to the nearest whole number" — since x = 12 exactly, the number of units is 1200 exactly. --- **Step 5: Conclusion** The number of units that maximizes monthly profit is 1200. --- **Final answer:** 1200

7. A local electronics store is analyzing smartphone sales data. The number of phones sold per day follows a normal distribution with a mean of 75 phones and a standard deviation of 12 phones. The store manager wants to know approximately what percentage of days they sell between 63 and 87 phones. Use technology to find this probability. Answer: 68 Solution: Identify the mean (μ = 75) and standard deviation (σ = 12) Calculate how many standard deviations 63 and 87 are from the mean Lower bound: (63 - 75)/12 = -12/12 = -1 standard deviation Upper bound: (87 - 75)/12 = 12/12 = +1 standard deviation According to the empirical rule for normal…
   Full step-by-step solution

   Step 1: Identify the mean (μ = 75) and standard deviation (σ = 12) Step 2: Calculate how many standard deviations 63 and 87 are from the mean Lower bound: (63 - 75)/12 = -12/12 = -1 standard deviation Upper bound: (87 - 75)/12 = 12/12 = +1 standard deviation Step 3: According to the empirical rule for normal distributions, approximately 68% of data falls within 1 standard deviation of the mean Step 4: Therefore, the store sells between 63 and 87 phones on approximately 68% of days The answer is 68.