Domain and Range

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Answer Key & Explanations

Domain and Range · Grade 9 · Worksheet 3

1. 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 above ground level and t represents time in seconds since launch. The drone's camera can only capture objects when it is at least 15 meters above the ground. During what time interval is the drone high enough to use its camera? Answer: 1 < t < 5 Solution: We are given the height function: h(t) = -2t^2 + 12t + 5. We want the time interval when h(t) >= 15.
   Full step-by-step solution

   We are given the height function: h(t) = -2t^2 + 12t + 5. We want the time interval when h(t) >= 15. Step 1: Set up the inequality -2t^2 + 12t + 5 >= 15 Step 2: Bring all terms to one side -2t^2 + 12t + 5 - 15 >= 0 -2t^2 + 12t - 10 >= 0 Step 3: Multiply through by -1 (reverse inequality sign) 2t^2 - 12t + 10 <= 0 Step 4: Divide by 2 to simplify t^2 - 6t + 5 <= 0 Step 5: Factor the quadratic (t - 1)(t - 5) <= 0 Step 6: Solve the inequality The parabola t^2 - 6t + 5 opens upward (coefficient of t^2 is positive), so it is <= 0 between the roots. Roots: t = 1 and t = 5. So inequality holds for 1 <= t <= 5. Step 7: Interpret the problem's wording The drone's camera works when it is at least 15 m above ground. At t = 1 and t = 5, h(t) = 15 exactly. If the camera requires strictly more than 15 m, then the interval is 1 < t < 5. If it requires at least 15 m, then it is 1 <= t <= 5. The correct answer given is 1 < t < 5, meaning the camera works only when strictly above 15 m. Step 8: Check h(t) at a point between 1 and 5 For example, t = 3: h(3) = -2*(9) + 12*3 + 5 = -18 + 36 + 5 = 23, which is > 15, so it works. Final answer: 1 < t < 5

2. f(x) = 2x² - 3x + 1, find f(-2) = ? Answer: 15 Solution: We are given the function f(x) = 2x² - 3x + 1 and asked to find f(-2). Substitute x = -2 into the function. f(-2) = 2*(-2)² - 3*(-2) + 1 Calculate (-2)².
   Full step-by-step solution

   We are given the function f(x) = 2x² - 3x + 1 and asked to find f(-2). Step 1: Substitute x = -2 into the function. f(-2) = 2*(-2)² - 3*(-2) + 1 Step 2: Calculate (-2)². (-2)² = (-2) * (-2) = 4 Step 3: Multiply 2 by 4. 2 * 4 = 8 Step 4: Multiply -3 by -2. -3 * (-2) = 6 Step 5: Now substitute these results back into the expression. f(-2) = 8 + 6 + 1 Step 6: Add the numbers. 8 + 6 = 14 14 + 1 = 15 Therefore, f(-2) = 15.

3. f(x) = 3x³ - 9x² + 5x - 7; f(3) = ? Answer: 8 Solution: Start with the function f(x) = 3x³ - 9x² + 5x - 7 Substitute x = 3 into the function: f(3) = 3(3)³ - 9(3)² + 5(3) - 7 Calculate the exponents first: (3)³ = 27 and (3)² = 9 Multiply: 3 × 27 = 81, -9 × 9 = -81, and 5 × 3 = 15 Rewrite the expression: f(3) = 81 - 81 + 15 - 7 Perform the operations…
   Full step-by-step solution

   Step 1: Start with the function f(x) = 3x³ - 9x² + 5x - 7 Step 2: Substitute x = 3 into the function: f(3) = 3(3)³ - 9(3)² + 5(3) - 7 Step 3: Calculate the exponents first: (3)³ = 27 and (3)² = 9 Step 4: Multiply: 3 × 27 = 81, -9 × 9 = -81, and 5 × 3 = 15 Step 5: Rewrite the expression: f(3) = 81 - 81 + 15 - 7 Step 6: Perform the operations from left to right: 81 - 81 = 0, then 0 + 15 = 15, then 15 - 7 = 8 The answer is 8.

4. Mere is analyzing the flight of a model rocket. The rocket's height above the ground, in meters, is modeled by the function h(t) = -4t² + 24t + 4, where t is the time in seconds after launch. She graphs this function to study its flight path. The graph is a parabola that opens downward. What is the domain and the range of the rocket's height function for the duration of its flight, from launch until it hits the ground? Express your answers using interval notation. Answer: Domain: [0, 6.1]; Range: [0, 40] Solution: The rocket's flight begins at launch, t = 0 seconds. The height at launch is h(0) = -4(0)² + 24(0) + 4 = 4 meters. The rocket hits the ground when h(t) = 0.
   Full step-by-step solution

   Step 1: The rocket's flight begins at launch, t = 0 seconds. The height at launch is h(0) = -4(0)² + 24(0) + 4 = 4 meters. Step 2: The rocket hits the ground when h(t) = 0. Solve -4t² + 24t + 4 = 0. Divide by -4: t² - 6t - 1 = 0. Use the quadratic formula: t = [6 ± sqrt(36 + 4)] / 2 = [6 ± sqrt(40)] / 2 = [6 ± 2*sqrt(10)] / 2 = 3 ± sqrt(10). sqrt(10) is approximately 3.16, so t = 3 + 3.16 = 6.16 seconds (positive root) or t = 3 - 3.16 = -0.16 seconds (negative, discard). The rocket hits the ground at t ≈ 6.16 seconds. Step 3: The domain (time from launch to ground) is all t from 0 to approximately 6.16. In interval notation: [0, 6.1] (rounded to one decimal place). Step 4: The range is the set of heights the rocket reaches. The maximum height occurs at the vertex. The vertex time is t = -b/(2a) = -24/(2*(-4)) = -24/(-8) = 3 seconds. Step 5: The maximum height is h(3) = -4(3)² + 24(3) + 4 = -36 + 72 + 4 = 40 meters. Step 6: The lowest height is 0 meters (ground), reached at t = 6.16 seconds. So the range of heights is from 0 to 40 meters. In interval notation: [0, 40]. Final answer: Domain: [0, 6.1]; Range: [0, 40]

5. The function f(x) = -2(x - 3)^2 + 8 represents the height of a ball thrown upward in meters after x seconds. What is the maximum height the ball reaches? Answer: 8 Solution: f(x) = -2(x - 3)^2 + 8 This represents the height of a ball in meters after x seconds. f(x) = a(x - h)^2 + k where (h, k) is the vertex of the parabola. f(x) = -2(x - 3)^2 + 8 a = -2 h = 3 k = 8 So the vertex is (3, 8).
   Full step-by-step solution

   Let's solve the problem step by step. We are given: f(x) = -2(x - 3)^2 + 8 This represents the height of a ball in meters after x seconds. --- **Step 1: Identify the form of the function** The function is in vertex form: f(x) = a(x - h)^2 + k where (h, k) is the vertex of the parabola. --- **Step 2: Compare with vertex form** f(x) = -2(x - 3)^2 + 8 Here: a = -2 h = 3 k = 8 So the vertex is (3, 8). --- **Step 3: Determine if the vertex is maximum or minimum** Since a = -2 is negative, the parabola opens downward. That means the vertex is the maximum point of the function. --- **Step 4: Interpret the vertex** The vertex (3, 8) means: At x = 3 seconds, the height is f(3) = 8 meters. This is the maximum height. --- **Step 5: Conclusion** The maximum height the ball reaches is 8 meters. --- **Final Answer:** 8

6. The function f(x) = -0.5(x - 4)^2 + 18 models the height of a diver above the water in feet, where x is the horizontal distance from the diving board in feet. What is the maximum height the diver reaches above the water? Answer: 18 Solution: The function f(x) = -0.5(x - 4)^2 + 18 is in vertex form f(x) = a(x - h)^2 + k In vertex form, the vertex is at (h, k) = (4, 18) Since the coefficient a = -0.5 is negative, the parabola opens downward For a downward-opening parabola, the vertex represents the maximum point The y-coordinate of…
   Full step-by-step solution

   Step 1: The function f(x) = -0.5(x - 4)^2 + 18 is in vertex form f(x) = a(x - h)^2 + k Step 2: In vertex form, the vertex is at (h, k) = (4, 18) Step 3: Since the coefficient a = -0.5 is negative, the parabola opens downward Step 4: For a downward-opening parabola, the vertex represents the maximum point Step 5: The y-coordinate of the vertex (k = 18) gives the maximum height Step 6: Therefore, the maximum height the diver reaches is 18 feet The answer is 18.

7. Mason is designing a parabolic arch for a new skatepark. The height of the arch above the ground, in feet, is modeled by the function h(x) = -x² + 14x - 40, where x is the horizontal distance in feet from the left base of the arch. Sketch the graph of this function and determine the domain and range of the arch's height over its full span from one base to the other. Answer: Domain: [4, 10]; Range: [0, 9] Solution: Find where the arch meets the ground (h(x) = 0). Solve -x² + 14x - 40 = 0. Multiply by -1: x² - 14x + 40 = 0.
   Full step-by-step solution

   Step 1: Find where the arch meets the ground (h(x) = 0). Solve -x² + 14x - 40 = 0. Multiply by -1: x² - 14x + 40 = 0. Factor: (x - 4)(x - 10) = 0. So x = 4 and x = 10. These are the left and right bases. Step 2: The domain is the set of all x-values for which the arch exists above ground. Since the arch spans from x = 4 to x = 10, the domain is [4, 10]. Step 3: Find the vertex (maximum height). For h(x) = -x² + 14x - 40, a = -1, b = 14. Vertex x = -b/(2a) = -14/(2 * -1) = -14/-2 = 7. Step 4: Find the maximum height: h(7) = -(7)² + 14(7) - 40 = -49 + 98 - 40 = 9. Step 5: The arch height ranges from 0 at the bases to 9 feet at the peak. So the range is [0, 9]. Answer: Domain: [4, 10]; Range: [0, 9]

8. A right triangle is drawn on a coordinate plane with vertices at (0,0), (6,0), and (6,4). A circle is circumscribed around this triangle, passing through all three vertices. What is the equation of this circle in standard form? Answer: (x-3)^2+(y-2)^2=13 Solution: Identify that for a right triangle, the hypotenuse is the diameter of the circumscribed circle. The vertices (0,0) and (6,4) are the endpoints of the hypotenuse since the right angle is at (6,0).
   Full step-by-step solution

   Step 1: Identify that for a right triangle, the hypotenuse is the diameter of the circumscribed circle. Step 2: The vertices (0,0) and (6,4) are the endpoints of the hypotenuse since the right angle is at (6,0). Step 3: Find the center of the circle by calculating the midpoint of the hypotenuse: ((0+6)/2, (0+4)/2) = (3,2). Step 4: Calculate the radius by finding half the length of the hypotenuse: distance between (0,0) and (6,4) = sqrt((6-0)^2+(4-0)^2) = sqrt(36+16) = sqrt(52) = 2sqrt(13). Half of this is sqrt(13). Step 5: The radius squared is (sqrt(13))^2 = 13. Step 6: Write the equation in standard form: (x-3)^2+(y-2)^2=13.