Quadratic Modeling

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Quadratic Modeling · Grade 10 · Worksheet 3

1. A company is designing a new smartphone case and wants to model the relationship between the selling price and weekly profit. Their market research shows that when they sell the case for $20, they make a weekly profit of $800, and when they sell it for $30, they make a weekly profit of $1200. The maximum profit occurs at a selling price of $35. Find the quadratic function P(x) = ax² + bx + c that models their weekly profit, where x is the selling price in dollars and P(x) is the profit in dollars. Answer: P(x) = -4x² + 280x - 4000 Solution: Quadratic functions are often used to model relationships where there's an optimal value, like maximum profit. The vertex form P(x) = a(x - h)² + k is particularly useful when you know the vertex coordinates (h,k), which represent the optimal point.
   Full step-by-step solution

   Quadratic functions are often used to model relationships where there's an optimal value, like maximum profit. The vertex form P(x) = a(x - h)² + k is particularly useful when you know the vertex coordinates (h,k), which represent the optimal point. You can then use additional data points to solve for the remaining coefficient.

2. Hana is designing a parabolic water fountain for a community garden. The water jet follows a parabolic path, with the nozzle located at ground level at the origin (0, 0). The water reaches its maximum height of 8 meters at a horizontal distance of 6 meters from the nozzle. The water then lands back on the ground at a point (12, 0). On a coordinate grid, each unit represents 1 meter. Hana wants to place a decorative ring at a height of 6 meters above the ground. How far horizontally from the nozzle will the water pass through this ring? Answer: 3 meters and 9 meters Solution: The parabola has vertex at (6, 8) and passes through (0, 0). Use vertex form: h(x) = a(x - 6)^2 + 8. Substitute (0, 0): 0 = a(0 - 6)^2 + 8 => 0 = 36a + 8 => 36a = -8 => a = -8/36 = -2/9.
   Full step-by-step solution

   Step 1: The parabola has vertex at (6, 8) and passes through (0, 0). Use vertex form: h(x) = a(x - 6)^2 + 8. Step 2: Substitute (0, 0): 0 = a(0 - 6)^2 + 8 => 0 = 36a + 8 => 36a = -8 => a = -8/36 = -2/9. Step 3: The quadratic model is h(x) = -2/9 (x - 6)^2 + 8. Step 4: Set h(x) = 6: 6 = -2/9 (x - 6)^2 + 8. Step 5: Subtract 8 from both sides: -2 = -2/9 (x - 6)^2. Step 6: Multiply both sides by -9/2: 9 = (x - 6)^2. Step 7: Take the square root: x - 6 = 3 or x - 6 = -3 => x = 9 or x = 3. Step 8: The water passes through the ring at horizontal distances of 3 meters and 9 meters from the nozzle. Final answer: 3 meters and 9 meters

3. Mere is designing a parabolic water fountain for a park. The fountain's water stream follows a parabolic path. The water spout is at ground level, and the stream reaches a maximum height of 4 meters at a horizontal distance of 6 meters from the spout. The stream lands back on the ground at a horizontal distance of 12 meters from the spout. Mere wants to know the height of the water stream at a horizontal distance of 4 meters from the spout. What is this height? Answer: 3.56 Solution: Set up the coordinate system. Place the water spout at (0,0) and the landing point at (12,0). The vertex is at (6,4).
   Full step-by-step solution

   Step 1: Set up the coordinate system. Place the water spout at (0,0) and the landing point at (12,0). The vertex is at (6,4). Step 2: Use the vertex form of a parabola: y = a(x - h)^2 + k, where (h,k) is the vertex. Here h = 6 and k = 4, so y = a(x - 6)^2 + 4. Step 3: Use the point (0,0) to find a. Substitute x = 0 and y = 0: 0 = a(0 - 6)^2 + 4 → 0 = a(36) + 4 → 36a = -4 → a = -4/36 = -1/9. Step 4: The equation is y = (-1/9)(x - 6)^2 + 4. Step 5: Find the height at x = 4: y = (-1/9)(4 - 6)^2 + 4 = (-1/9)(-2)^2 + 4 = (-1/9)(4) + 4 = -4/9 + 4 = -0.444... + 4 = 3.555... Step 6: Round to two decimal places: 3.56 meters. The height of the water stream at 4 meters from the spout is 3.56 meters.

4. Tane is designing a parabolic water fountain for a public park. The water jet will be launched from a nozzle at ground level, and the arch of the water will be modeled by a quadratic function. The fountain is designed so that the water reaches a maximum height of 9 meters above the ground, and this maximum occurs exactly 7 meters horizontally from the nozzle. Tane wants to know the horizontal distance from the nozzle where the water will first hit the ground again. Assuming the fountain operates symmetrically and the water's path follows a quadratic model, what is that horizontal distance in meters? Answer: 14 Solution: Set up the coordinate system. Place the nozzle at the origin (0, 0). The vertex (maximum) is at (7, 9).
   Full step-by-step solution

   Step 1: Set up the coordinate system. Place the nozzle at the origin (0, 0). The vertex (maximum) is at (7, 9). Since the parabola opens downward, use vertex form: h(x) = a(x - 7)^2 + 9, where h(x) is the height at horizontal distance x. Step 2: Use the point (0, 0) to find a. Substitute x = 0, h(0) = 0: 0 = a(0 - 7)^2 + 9 0 = a(49) + 9 -9 = 49a a = -9/49 Step 3: The quadratic model is h(x) = -9/49 (x - 7)^2 + 9. Step 4: Find where the water hits the ground, i.e., where h(x) = 0: 0 = -9/49 (x - 7)^2 + 9 9/49 (x - 7)^2 = 9 (x - 7)^2 = 9 * 49/9 = 49 x - 7 = ±7 x = 7 + 7 = 14 or x = 7 - 7 = 0 Step 5: x = 0 is the starting point (nozzle). The water hits the ground again at x = 14 meters. The answer is 14.

5. A company's profit P(x) = -2x² + 80x - 600, where x is units sold. Find the number of units that maximizes profit. Answer: 20 Solution: The profit function is P(x) = -2x² + 80x - 600, which is a quadratic function in standard form ax² + bx + c. For a quadratic function, the x-coordinate of the vertex is given by x = -b/(2a). Here, a = -2 and b = 80.
   Full step-by-step solution

   Step 1: The profit function is P(x) = -2x² + 80x - 600, which is a quadratic function in standard form ax² + bx + c. Step 2: For a quadratic function, the x-coordinate of the vertex is given by x = -b/(2a). Step 3: Here, a = -2 and b = 80. Step 4: Substitute into the formula: x = -80/(2×(-2)) = -80/(-4) = 20. Step 5: Since a = -2 is negative, the parabola opens downward, so the vertex represents a maximum point. Step 6: Therefore, the number of units that maximizes profit is 20.

6. A rectangular garden has a perimeter of 60 meters. If the area is modeled by A(w) = w(30 - w), find the maximum possible area of the garden. Answer: 225 Solution: The area function is A(w) = w(30 - w) = -w² + 30w This is a quadratic function opening downward (a = -1 < 0), so it has a maximum at the vertex The vertex occurs at w = -b/(2a) = -30/(2×-1) = -30/-2 = 15 Substitute w = 15 into the area function: A(15) = 15(30 - 15) = 15 × 15 = 225 The maximum…
   Full step-by-step solution

   Step 1: The area function is A(w) = w(30 - w) = -w² + 30w Step 2: This is a quadratic function opening downward (a = -1 < 0), so it has a maximum at the vertex Step 3: The vertex occurs at w = -b/(2a) = -30/(2×-1) = -30/-2 = 15 Step 4: Substitute w = 15 into the area function: A(15) = 15(30 - 15) = 15 × 15 = 225 Step 5: The maximum possible area is 225 square meters.