Quadratic Applications

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Quadratic Applications · Grade 9 · Worksheet 3

1. Matiu throws a ball whose height is given by h(t) = -16t² + 48t + 4. Find the maximum height. Answer: 40 Solution: The height function is h(t) = -16t² + 48t + 4 Find the time when maximum height occurs using t = -b/(2a) a = -16, b = 48, so t = -48/(2×-16) = -48/-32 = 1.5 seconds Substitute t = 1.5 into the height function h(1.5) = -16(1.5)² + 48(1.5) + 4 = -16(2.25) + 72 + 4 = -36 + 72 + 4 = 40 The maximum…
   Full step-by-step solution

   Step 1: The height function is h(t) = -16t² + 48t + 4 Step 2: Find the time when maximum height occurs using t = -b/(2a) Step 3: a = -16, b = 48, so t = -48/(2×-16) = -48/-32 = 1.5 seconds Step 4: Substitute t = 1.5 into the height function Step 5: h(1.5) = -16(1.5)² + 48(1.5) + 4 = -16(2.25) + 72 + 4 = -36 + 72 + 4 = 40 Step 6: The maximum height is 40 feet.

2. f(x) = 3x² - 12x + 5, f(2) = ? Answer: -7 Solution: Start with the function f(x) = 3x² - 12x + 5 Substitute x = 2 into the function: f(2) = 3(2)² - 12(2) + 5 Calculate the exponent first: (2)² = 4 Multiply: 3 × 4 = 12 and 12 × 2 = 24 Rewrite the expression: f(2) = 12 - 24 + 5 Perform the operations from left to right: 12 - 24 = -12, then -12 + 5…
   Full step-by-step solution

   Step 1: Start with the function f(x) = 3x² - 12x + 5 Step 2: Substitute x = 2 into the function: f(2) = 3(2)² - 12(2) + 5 Step 3: Calculate the exponent first: (2)² = 4 Step 4: Multiply: 3 × 4 = 12 and 12 × 2 = 24 Step 5: Rewrite the expression: f(2) = 12 - 24 + 5 Step 6: Perform the operations from left to right: 12 - 24 = -12, then -12 + 5 = -7 The answer is -7.

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. A parabolic arch bridge spans a river with its vertex at the highest point of the arch, 15 meters above the water. The arch is modeled by a quadratic function in vertex form: y = a(x - h)² + k, where (h, k) is the vertex. The arch meets the water at two points that are 50 meters apart horizontally. If the left support is at the origin (0, 0) on a coordinate plane, what is the equation of the parabola? Answer: y = -0.024x² + 1.2x Solution: Identify the vertex. The highest point is at the vertex, which is halfway between the two supports. Since the supports are 50 meters apart and the left support is at (0, 0), the vertex is at (25, 15).
   Full step-by-step solution

   Step 1: Identify the vertex. The highest point is at the vertex, which is halfway between the two supports. Since the supports are 50 meters apart and the left support is at (0, 0), the vertex is at (25, 15). Step 2: Write the vertex form equation: y = a(x - 25)² + 15. Step 3: Use a known point to find 'a'. The left support is at (0, 0), so substitute x = 0 and y = 0 into the equation: 0 = a(0 - 25)² + 15. Step 4: Simplify: 0 = a(625) + 15. Step 5: Solve for 'a': 625a = -15, so a = -15/625 = -0.024. Step 6: Write the final equation: y = -0.024(x - 25)² + 15. Step 7: Expand to standard form if needed: y = -0.024(x² - 50x + 625) + 15 = -0.024x² + 1.2x - 15 + 15 = -0.024x² + 1.2x. The equation of the parabola is y = -0.024x² + 1.2x.

5. 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

6. 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.

7. Isabella launches a water balloon from a slingshot at the top of a 15-meter tall building. The height of the water balloon above the ground is modeled by the quadratic function h(t) = -5t² + 20t + 15, where h is the height in meters and t is the time in seconds after launch. What is the maximum height the water balloon reaches above the ground? Answer: 35 Solution: The height function is h(t) = -5t² + 20t + 15. This is a downward-opening parabola because the coefficient of t² is negative (-5 < 0), so the vertex gives the maximum height.
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 20t + 15. Step 2: This is a downward-opening parabola because the coefficient of t² is negative (-5 < 0), so the vertex gives the maximum height. Step 3: Find the time at the vertex using the formula t = -b/(2a), where a = -5 and b = 20. Step 4: t = -20 / (2 × (-5)) = -20 / (-10) = 2 seconds. Step 5: Substitute t = 2 into the height function to find the maximum height: h(2) = -5(2)² + 20(2) + 15. Step 6: h(2) = -5(4) + 40 + 15 = -20 + 40 + 15 = 35. Step 7: The maximum height the water balloon reaches above the ground is 35 meters.

8. f(x) = x² - 4x + 3, f(5) = ? Answer: 8 Solution: Write the function: f(x) = x² - 4x + 3 Substitute x = 5 into the function: f(5) = (5)² - 4(5) + 3 Calculate the square: (5)² = 25 Calculate the multiplication: 4(5) = 20 Substitute these values: f(5) = 25 - 20 + 3 Perform the subtraction: 25 - 20 = 5 Perform the addition: 5 + 3 = 8 The answer is 8.
   Full step-by-step solution

   Step 1: Write the function: f(x) = x² - 4x + 3 Step 2: Substitute x = 5 into the function: f(5) = (5)² - 4(5) + 3 Step 3: Calculate the square: (5)² = 25 Step 4: Calculate the multiplication: 4(5) = 20 Step 5: Substitute these values: f(5) = 25 - 20 + 3 Step 6: Perform the subtraction: 25 - 20 = 5 Step 7: Perform the addition: 5 + 3 = 8 The answer is 8.