Quadratic Applications

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Answer Key & Explanations

Quadratic Applications · Grade 9 · Worksheet 3

1. A water fountain at a park shoots water into the air from a nozzle at ground level. The height of a water droplet, in meters, is given by the function h(d) = -0.05d² + 0.9d, where d is the horizontal distance from the nozzle in meters. Aroha notices that the stream of water forms a perfect parabolic arc. She wants to place a small cup at the highest point of the arc to catch a falling leaf. What is the maximum height of the water stream, and at what horizontal distance from the nozzle does this occur? Answer: Maximum height = 4.05 meters at d = 9 meters Solution: Identify the quadratic function: h(d) = -0.05d² + 0.9d. This is in standard form h(d) = ad² + bd + c, where a = -0.05, b = 0.9, and c = 0. The vertex of a parabola occurs at d = -b / (2a).
   Full step-by-step solution

   Step 1: Identify the quadratic function: h(d) = -0.05d² + 0.9d. This is in standard form h(d) = ad² + bd + c, where a = -0.05, b = 0.9, and c = 0. Step 2: The vertex of a parabola occurs at d = -b / (2a). Substitute the values: d = -0.9 / (2 * -0.05) = -0.9 / -0.1 = 9. Step 3: So the maximum height occurs at a horizontal distance of 9 meters from the nozzle. Step 4: Find the maximum height by substituting d = 9 into the function: h(9) = -0.05(9)² + 0.9(9) = -0.05(81) + 8.1 = -4.05 + 8.1 = 4.05. Step 5: Therefore, the maximum height of the water stream is 4.05 meters, occurring 9 meters from the nozzle. The answer is: Maximum height = 4.05 meters at d = 9 meters.

2. Mere kicks a soccer ball whose height is given by h(t) = -16t² + 48t + 4. Find the maximum height the ball reaches. Answer: 40 Solution: Identify the coefficients: a = -16, b = 48, c = 4. The maximum height occurs at the vertex. The time at the vertex is t = -b/(2a).
   Full step-by-step solution

   Step 1: Identify the coefficients: a = -16, b = 48, c = 4. Step 2: The maximum height occurs at the vertex. The time at the vertex is t = -b/(2a). Step 3: Calculate t = -48/(2 * -16) = -48/-32 = 1.5 seconds. Step 4: Substitute t = 1.5 into the height function: h(1.5) = -16(1.5)² + 48(1.5) + 4. Step 5: Calculate the exponent: (1.5)² = 2.25. Step 6: Multiply: -16 * 2.25 = -36, and 48 * 1.5 = 72. Step 7: Combine all terms: -36 + 72 + 4 = 40. The maximum height the ball reaches is 40 feet.

3. Aroha kicks a soccer ball whose height is given by h(t) = -5t² + 15t + 2. Find the maximum height of the ball. Answer: 13.25 Solution: The height function is h(t) = -5t² + 15t + 2. Since the coefficient of t² is negative, the parabola opens downward, so the vertex gives the maximum height. 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 + 2. Since the coefficient of t² is negative, the parabola opens downward, so the vertex gives the maximum height. 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: h(1.5) = -5(1.5)² + 15(1.5) + 2 Step 4: Calculate: -5(2.25) + 22.5 + 2 = -11.25 + 22.5 + 2 = 13.25 Step 5: The maximum height is 13.25 meters.

4. Mason throws a ball whose height is modeled by h(t) = -12t² + 72t + 7. Find the maximum height. Answer: 115 Solution: The height function is h(t) = -12t² + 72t + 7. This is a downward-opening parabola since a = -12 < 0. The maximum height occurs at the vertex.
   Full step-by-step solution

   Step 1: The height function is h(t) = -12t² + 72t + 7. This is a downward-opening parabola since a = -12 < 0. Step 2: The maximum height occurs at the vertex. For a quadratic in the form at² + bt + c, the t-coordinate of the vertex is t = -b/(2a). Step 3: Calculate t = -72/(2×(-12)) = -72/(-24) = 3 seconds. Step 4: Substitute t = 3 into the height function: h(3) = -12(3)² + 72(3) + 7 Step 5: Calculate: -12(9) + 72(3) + 7 = -108 + 216 + 7 Step 6: Simplify: -108 + 216 = 108, then 108 + 7 = 115 Step 7: The maximum height is 115 feet. The answer is 115.

5. A company's profit from selling handmade candles is modeled by the quadratic function P(x) = -2x² + 60x - 400, where x represents the number of candles sold and P(x) represents the profit in dollars. How many candles must the company sell to break even (reach zero profit)? Answer: 10 Solution: To break even, the profit must be zero, so set P(x) = 0: -2x² + 60x - 400 = 0 Divide the entire equation by -2 to simplify: x² - 30x + 200 = 0 Factor the quadratic equation: (x - 10)(x - 20) = 0 Set each factor equal to zero: x - 10 = 0 or x - 20 = 0 Solve for x: x = 10 or x = 20 Since the…
   Full step-by-step solution

   Step 1: To break even, the profit must be zero, so set P(x) = 0: -2x² + 60x - 400 = 0 Step 2: Divide the entire equation by -2 to simplify: x² - 30x + 200 = 0 Step 3: Factor the quadratic equation: (x - 10)(x - 20) = 0 Step 4: Set each factor equal to zero: x - 10 = 0 or x - 20 = 0 Step 5: Solve for x: x = 10 or x = 20 Step 6: Since the question asks for when the company first breaks even, the answer is the smaller value: x = 10 The company must sell 10 candles to break even.

6. Aroha throws a ball whose height is modeled by h(t) = -5t² + 15t + 2. Find the maximum height. Answer: 13.25 Solution: The height function is h(t) = -5t² + 15t + 2. This is a downward-opening parabola since a = -5 < 0, so it has a maximum at the vertex.
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 15t + 2. This is a downward-opening parabola since a = -5 < 0, so it has a maximum at the vertex. Step 2: The t-coordinate of the vertex is given by t = -b/(2a) = -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) + 2 Step 4: Calculate: -5(2.25) + 22.5 + 2 = -11.25 + 22.5 + 2 = 13.25 Step 5: The maximum height is 13.25 meters.

7. 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.

8. Matiu throws a ball whose height is modeled by h(t) = -5t² + 15t + 12. Find the maximum height reached by the ball. 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.
   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), where a = -5 and b = 15. Step 3: t = -15/(2×-5) = -15/-10 = 1.5 seconds Step 4: Substitute t = 1.5 into the height function to find the maximum height: h(1.5) = -5(1.5)² + 15(1.5) + 12 Step 5: Calculate: -5(2.25) + 22.5 + 12 = -11.25 + 22.5 + 12 = 23.25 Step 6: The maximum height reached by the ball is 23.25 meters.