Quadratic Applications

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Quadratic Applications · Grade 9 · Worksheet 2

1. Mason is designing a water fountain that shoots a stream of water into the air. The height of the water stream above the fountain's nozzle, in meters, is given by the quadratic function h(t) = -5t² + 18t + 2, where t is the time in seconds after the water leaves the nozzle. On a coordinate grid, the path of the water forms a parabola. What is the maximum height, in meters, that the water stream reaches above the nozzle? Answer: 18.2 Solution: Identify the coefficients in the quadratic function h(t) = -5t² + 18t + 2. Here, a = -5, b = 18, and c = 2. The vertex of a parabola given by h(t) = at² + bt + c occurs at t = -b / (2a).
   Full step-by-step solution

   Step 1: Identify the coefficients in the quadratic function h(t) = -5t² + 18t + 2. Here, a = -5, b = 18, and c = 2. Step 2: The vertex of a parabola given by h(t) = at² + bt + c occurs at t = -b / (2a). This gives the time at which the maximum height is reached. Step 3: Substitute a = -5 and b = 18 into the formula: t = -18 / (2 * -5) = -18 / -10 = 1.8 seconds. Step 4: To find the maximum height, substitute t = 1.8 into the original function: h(1.8) = -5(1.8)² + 18(1.8) + 2. Step 5: Calculate (1.8)² = 3.24. Step 6: Multiply: -5 * 3.24 = -16.2. Step 7: Multiply: 18 * 1.8 = 32.4. Step 8: Add the terms: h(1.8) = -16.2 + 32.4 + 2 = 18.2. The maximum height of the water stream is 18.2 meters above the nozzle.

2. A parabolic arch bridge has its vertex at (0, 20) and spans a horizontal distance of 40 meters, with the endpoints at (-20, 0) and (20, 0). The arch follows a quadratic function in the form y = ax² + bx + c. What is the height of the arch 10 meters from the center? Answer: 15 Solution: The vertex is at (0, 20). y = a(x - h)² + k, where (h, k) is the vertex. y = a(x - 0)² + 20 y = a x² + 20 The arch passes through (20, 0).
   Full step-by-step solution

   Let's solve step-by-step. --- **Step 1: Identify the vertex and general form** The vertex is at (0, 20). For a parabola, vertex form is: y = a(x - h)² + k, where (h, k) is the vertex. So: y = a(x - 0)² + 20 y = a x² + 20 --- **Step 2: Use a point to find 'a'** The arch passes through (20, 0). Substitute x = 20, y = 0: 0 = a(20)² + 20 0 = 400a + 20 400a = -20 a = -20 / 400 a = -1/20 --- **Step 3: Write the equation** y = (-1/20) x² + 20 --- **Step 4: Find height 10 meters from center** Center is at x = 0, so 10 meters from center means x = 10 or x = -10. Let's take x = 10: y = (-1/20)(10)² + 20 y = (-1/20)(100) + 20 y = -5 + 20 y = 15 --- **Step 5: Conclusion** The height of the arch 10 meters from the center is **15 meters**. --- **Final answer:** 15

3. Emma kicks a soccer ball whose height is given by h(t) = -5t² + 15t + 10. What is the maximum height of the ball? Answer: 21.25 Solution: The height function is h(t) = -5t² + 15t + 10. This is a downward-opening parabola since a = -5 < 0, so it has a maximum at its vertex. Find the time t when maximum height occurs using t = -b/(2a).
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 15t + 10. This is a downward-opening parabola since a = -5 < 0, so it has a maximum at its vertex. Step 2: Find the time t when maximum height occurs using t = -b/(2a). Here a = -5, b = 15, so t = -15/(2×(-5)) = -15/(-10) = 1.5 seconds. Step 3: Substitute t = 1.5 into the height function to find the maximum height: h(1.5) = -5(1.5)² + 15(1.5) + 10 = -5(2.25) + 22.5 + 10 = -11.25 + 22.5 + 10 = 21.25. Step 4: The maximum height is 21.25 units.

4. Sophia launches a model rocket from a platform 7 meters above ground level. The height of the rocket, h(t), in meters, t seconds after launch is given by the quadratic function h(t) = -4.9t² + 49t + 7. On a graph of this function, the parabola opens downward and crosses the vertical axis at (0, 7). What is the maximum height, in meters, reached by the rocket? Answer: 129.5 Solution: Identify the coefficients of the quadratic function in the form h(t) = at² + bt + c. Here, a = -4.9, b = 49, and c = 7. The vertex of a parabola occurs at t = -b / (2a).
   Full step-by-step solution

   Step 1: Identify the coefficients of the quadratic function in the form h(t) = at² + bt + c. Here, a = -4.9, b = 49, and c = 7. Step 2: The vertex of a parabola occurs at t = -b / (2a). Substitute the values: t = -49 / (2 * -4.9) t = -49 / (-9.8) t = 5 seconds. Step 3: To find the maximum height, substitute t = 5 into the original equation: h(5) = -4.9(5)² + 49(5) + 7 h(5) = -4.9(25) + 245 + 7 h(5) = -122.5 + 245 + 7 h(5) = 122.5 + 7 h(5) = 129.5 meters. The maximum height reached by the rocket is 129.5 meters.

5. Sophia throws a ball whose height is modeled by h(t) = -16t² + 96t + 6. Find the maximum height of the ball. Answer: 150 Solution: The height function is h(t) = -16t² + 96t + 6. This is a downward-opening parabola (since a = -16 < 0), so its vertex gives the maximum height.
   Full step-by-step solution

   Step 1: The height function is h(t) = -16t² + 96t + 6. This is a downward-opening parabola (since a = -16 < 0), so its vertex gives the maximum height. Step 2: Find the time t at which the maximum height occurs using the vertex formula t = -b/(2a). Here, a = -16 and b = 96. Step 3: Calculate t = -96 / (2 * -16) = -96 / -32 = 3 seconds. Step 4: Substitute t = 3 back into the height function to find the maximum height: h(3) = -16(3)² + 96(3) + 6. Step 5: Calculate the exponent: 3² = 9. Step 6: Multiply: -16 * 9 = -144 and 96 * 3 = 288. Step 7: Combine all terms: -144 + 288 + 6 = 150. The maximum height of the ball is 150 feet.

6. Matiu kicks a soccer ball whose height is modeled by h(t) = -5t² + 15t + 12. What is the maximum height the ball reaches? Answer: 23.25 Solution: The height function is h(t) = -5t² + 15t + 12. Since the coefficient of t² is negative, the parabola opens downward, and the vertex gives the maximum height. The t-coordinate of the vertex is found using t = -b/(2a).
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 15t + 12. Since the coefficient of t² is negative, the parabola opens downward, and the vertex gives the maximum height. Step 2: The t-coordinate of the vertex is found using t = -b/(2a). Here, a = -5 and b = 15, so t = -15/(2×-5) = -15/-10 = 1.5 seconds. Step 3: Substitute t = 1.5 into the height function: h(1.5) = -5(1.5)² + 15(1.5) + 12 Step 4: Calculate (1.5)² = 2.25, so -5 × 2.25 = -11.25 Step 5: Calculate 15 × 1.5 = 22.5 Step 6: Combine all terms: -11.25 + 22.5 + 12 = 11.25 + 12 = 23.25 Step 7: The maximum height is 23.25 meters.

7. A rectangular garden is being designed against a long wall. A total of 88 feet of fencing is available to enclose the other three sides (the wall forms the fourth side). Let x represent the width of the garden (the sides perpendicular to the wall). Write a quadratic function for the area A of the garden in terms of x, and determine the maximum possible area of the garden. Answer: 968 square feet Solution: Let x = width (the two sides perpendicular to the wall). Let L = length (the side parallel to the wall). The fencing is used for the two widths and one length: 2x + L = 88.
   Full step-by-step solution

   Step 1: Let x = width (the two sides perpendicular to the wall). Let L = length (the side parallel to the wall). The fencing is used for the two widths and one length: 2x + L = 88. Step 2: Solve for L: L = 88 - 2x. Step 3: Area A = length * width = x * L = x(88 - 2x) = 88x - 2x^2. Step 4: This is a quadratic function A(x) = -2x^2 + 88x, which opens downward. The vertex gives the maximum area. Step 5: For a quadratic ax^2 + bx + c, the vertex x-coordinate is x = -b/(2a). Here a = -2, b = 88, so x = -88/(2 * -2) = -88/(-4) = 22. Step 6: The maximum area is A(22) = 88(22) - 2(22)^2 = 1936 - 2(484) = 1936 - 968 = 968. The maximum possible area of the garden is 968 square feet.