Domain and Range

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Answer Key & Explanations

Domain and Range · Grade 9 · Worksheet 1

1. A drone delivery service uses a quadratic function to model the height of its drones during delivery flights. The height h(t) in meters above ground level is given by h(t) = -2t² + 16t + 20, where t is time in seconds after takeoff. What is the maximum height the drone reaches during its flight? Answer: 52 Solution: We are given the height function: h(t) = -2t^2 + 16t + 20. This is a quadratic function in the form h(t) = at^2 + bt + c, with a = -2, b = 16, c = 20.
   Full step-by-step solution

   We are given the height function: h(t) = -2t^2 + 16t + 20. This is a quadratic function in the form h(t) = at^2 + bt + c, with a = -2, b = 16, c = 20. Since a is negative (a = -2), the parabola opens downward, so the vertex of the parabola gives the maximum height. Step 1: Find the time t at which the maximum height occurs. The vertex of a parabola at^2 + bt + c occurs at t = -b / (2a). Here, b = 16, a = -2. t = -16 / (2 * -2) = -16 / (-4) = 16 / 4 = 4. So the maximum height occurs at t = 4 seconds. Step 2: Find the height at t = 4 by substituting into h(t). h(4) = -2*(4)^2 + 16*(4) + 20 = -2*(16) + 64 + 20 = -32 + 64 + 20 = 32 + 20 = 52. Step 3: Conclusion. The maximum height the drone reaches is 52 meters. Answer: 52

2. Sophia is analyzing the path of a drone delivering a package. The drone's height above the ground (in meters) is modeled by the quadratic function h(t) = -3t² + 18t + 9, where t is the time in seconds after launch. The drone can only release its package when it is at least 24 meters high to ensure a safe drop. The graph of this function is a parabola opening downward. Determine the time interval (in seconds) during which the drone is at or above the required height of 24 meters. Then, from the graph of the function, state the domain and range of h(t) for the entire flight from launch until it returns to the ground (height = 0). Answer: Time interval: 1 ≤ t ≤ 5 seconds; Domain: 0 ≤ t ≤ 6.46 seconds; Range: 0 ≤ h(t) ≤ 36 meters Solution: Set the height function equal to 24 to find the boundary times: -3t² + 18t + 9 = 24. Rearrange to standard form: -3t² + 18t - 15 = 0. Multiply both sides by -1: 3t² - 18t + 15 = 0.
   Full step-by-step solution

   Step 1: Set the height function equal to 24 to find the boundary times: -3t² + 18t + 9 = 24. Step 2: Rearrange to standard form: -3t² + 18t - 15 = 0. Multiply both sides by -1: 3t² - 18t + 15 = 0. Step 3: Divide by 3: t² - 6t + 5 = 0. Step 4: Factor the quadratic: (t - 1)(t - 5) = 0. So t = 1 second and t = 5 seconds. Step 5: Since the parabola opens downward (coefficient of t² is negative), the drone is above 24 meters between these two times. Thus, the time interval is 1 ≤ t ≤ 5 seconds. Step 6: Find the domain: The flight starts at t = 0. Find when the drone returns to ground (h(t) = 0): -3t² + 18t + 9 = 0. Multiply by -1: 3t² - 18t - 9 = 0. Divide by 3: t² - 6t - 3 = 0. Step 7: Use the quadratic formula: t = [6 ± sqrt(36 + 12)] / 2 = [6 ± sqrt(48)] / 2 = [6 ± 4√3] / 2 = 3 ± 2√3. Since √3 ≈ 1.732, the positive root is 3 + 2(1.732) ≈ 3 + 3.464 = 6.464 seconds. So domain: 0 ≤ t ≤ 6.46 seconds (approximately). Step 8: Find the maximum height (vertex). The vertex occurs at t = -b/(2a) = -18/(2*(-3)) = -18/(-6) = 3 seconds. Step 9: Substitute t = 3 into h(t): h(3) = -3(9) + 18(3) + 9 = -27 + 54 + 9 = 36 meters. Step 10: The range is from ground level (0 m) to the maximum height (36 m). So range: 0 ≤ h(t) ≤ 36 meters. Final answer: Time interval: 1 ≤ t ≤ 5 seconds; Domain: 0 ≤ t ≤ 6.46 seconds; Range: 0 ≤ h(t) ≤ 36 meters.

3. Emma is designing a custom skateboard ramp for a local park. The ramp's cross-section follows a parabolic path modeled by the function h(x) = -0.5x² + 3x + 1, where h represents the height in feet above ground level and x represents the horizontal distance in feet from the ramp's starting point. The city requires that the ramp must be at least 4 feet high to accommodate advanced tricks. Over what interval of horizontal distance (in feet) will the ramp meet this height requirement? Answer: 1 ≤ x ≤ 5 Solution: This involves solving quadratic inequalities, where we find the points where the function equals our threshold value, then determine which intervals satisfy the inequality.
   Full step-by-step solution

   When analyzing quadratic functions that model real-world scenarios like ramps or projectile paths, we often need to determine when the function's output meets certain conditions. This involves solving quadratic inequalities, where we find the points where the function equals our threshold value, then determine which intervals satisfy the inequality. The solution represents the domain values (input) that produce range values (output) meeting our requirements.

4. A parabola is graphed on a coordinate plane with vertex at (2, -1) and passing through the point (4, 3). The parabola opens upward. What is the equation of this parabola in vertex form? Answer: y = (x - 2)² - 1 Solution: y = a(x - h)² + k where (h, k) is the vertex. The vertex is (2, -1), so h = 2 and k = -1. y = a(x - 2)² - 1 Use the given point (4, 3) to find 'a' The parabola passes through (4, 3), so when x = 4, y = 3.
   Full step-by-step solution

   Let's solve this step by step. --- **Step 1: Recall the vertex form of a parabola** The vertex form is: y = a(x - h)² + k where (h, k) is the vertex. --- **Step 2: Substitute the vertex into the equation** The vertex is (2, -1), so h = 2 and k = -1. Substitute h and k: y = a(x - 2)² - 1 --- **Step 3: Use the given point (4, 3) to find 'a'** The parabola passes through (4, 3), so when x = 4, y = 3. Substitute x = 4, y = 3 into the equation: 3 = a(4 - 2)² - 1 --- **Step 4: Simplify and solve for a** 3 = a(2)² - 1 3 = a(4) - 1 3 = 4a - 1 Add 1 to both sides: 3 + 1 = 4a 4 = 4a Divide both sides by 4: a = 1 --- **Step 5: Write the final equation** Substitute a = 1, h = 2, k = -1 into vertex form: y = 1(x - 2)² - 1 Simplify: y = (x - 2)² - 1 --- **Final answer:** y = (x - 2)² - 1

5. f(x) = 6x² - 8x + 4; f(2) = ? Answer: 12 Solution: Start with the function f(x) = 6x² - 8x + 4 Substitute x = 2 into the function: f(2) = 6(2)² - 8(2) + 4 Calculate the exponent first: (2)² = 4 Multiply: 6 × 4 = 24 and -8 × 2 = -16 Combine all terms: 24 - 16 + 4 Perform the operations from left to right: 24 - 16 = 8, then 8 + 4 = 12 The answer…
   Full step-by-step solution

   Step 1: Start with the function f(x) = 6x² - 8x + 4 Step 2: Substitute x = 2 into the function: f(2) = 6(2)² - 8(2) + 4 Step 3: Calculate the exponent first: (2)² = 4 Step 4: Multiply: 6 × 4 = 24 and -8 × 2 = -16 Step 5: Combine all terms: 24 - 16 + 4 Step 6: Perform the operations from left to right: 24 - 16 = 8, then 8 + 4 = 12 The answer is 12.

6. f(x) = 3x² - 8x + 5; f(4) = ? Answer: 21 Solution: Start with the function f(x) = 3x² - 8x + 5 Substitute x = 4 into the function: f(4) = 3(4)² - 8(4) + 5 Calculate the exponent first: (4)² = 16 Multiply: 3 × 16 = 48 and -8 × 4 = -32 Rewrite the expression: f(4) = 48 - 32 + 5 Perform the operations from left to right: 48 - 32 = 16, then 16 + 5 =…
   Full step-by-step solution

   Step 1: Start with the function f(x) = 3x² - 8x + 5 Step 2: Substitute x = 4 into the function: f(4) = 3(4)² - 8(4) + 5 Step 3: Calculate the exponent first: (4)² = 16 Step 4: Multiply: 3 × 16 = 48 and -8 × 4 = -32 Step 5: Rewrite the expression: f(4) = 48 - 32 + 5 Step 6: Perform the operations from left to right: 48 - 32 = 16, then 16 + 5 = 21 The answer is 21.

7. A drone is flying over a park following a parabolic path described by the function h(t) = -2t² + 12t + 5, where h represents the drone's height in meters and t represents time in seconds. What is the maximum height the drone reaches during its flight? Answer: 23 meters Solution: Identify the type of function. This is a quadratic function in the form h(t) = at² + bt + c, where a = -2, b = 12, and c = 5.
   Full step-by-step solution

   Let's find the maximum height of the drone using the function h(t) = -2t² + 12t + 5. Step 1: Identify the type of function. This is a quadratic function in the form h(t) = at² + bt + c, where a = -2, b = 12, and c = 5. Since a = -2 is negative, the parabola opens downward, meaning it has a maximum point at its vertex. Step 2: Find the time t at which the maximum height occurs. For a quadratic function, the vertex occurs at t = -b / (2a). Substitute b = 12 and a = -2: t = -12 / (2 * -2) t = -12 / (-4) t = 3 seconds. Step 3: Calculate the maximum height by substituting t = 3 into h(t). h(3) = -2*(3)² + 12*(3) + 5 First, compute (3)² = 9. So, -2 * 9 = -18. Next, 12 * 3 = 36. Now add them: h(3) = -18 + 36 + 5 h(3) = 18 + 5 h(3) = 23 meters. Step 4: State the conclusion. The maximum height the drone reaches is 23 meters. ANSWER: 23 meters