Domain and Range

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Answer Key & Explanations

Domain and Range · Grade 9 · Worksheet 2

1. Isabella is analyzing the flight of a toy rocket launched from a platform. The rocket's height above the ground, in meters, is modeled by the function h(t) = -2t² + 12t + 7, where t represents time in seconds after launch. The rocket's camera can only take clear pictures when the rocket is at least 15 meters high. Sketch a graph of the function, then determine the time interval (in seconds) during which the rocket's camera can take clear pictures. Answer: 2 ≤ t ≤ 4 Solution: Identify the function: h(t) = -2t² + 12t + 7. Find the vertex. The t-coordinate of the vertex is -b/(2a) = -12/(2*(-2)) = -12/(-4) = 3.
   Full step-by-step solution

   Step 1: Identify the function: h(t) = -2t² + 12t + 7. Step 2: Find the vertex. The t-coordinate of the vertex is -b/(2a) = -12/(2*(-2)) = -12/(-4) = 3. The height at t=3 is h(3) = -2(9) + 12(3) + 7 = -18 + 36 + 7 = 25. So the vertex is at (3, 25). Step 3: Find the y-intercept: when t=0, h(0)=7, so (0,7). Step 4: Set h(t)=15 to find where the rocket reaches 15 meters: -2t² + 12t + 7 = 15. Simplify: -2t² + 12t - 8 = 0. Divide by -2: t² - 6t + 4 = 0. Solve using the quadratic formula: t = (6 ± sqrt(36 - 16))/2 = (6 ± sqrt(20))/2 = (6 ± 2*sqrt(5))/2 = 3 ± sqrt(5). Since sqrt(5) ≈ 2.236, the times are approximately t = 3 - 2.236 = 0.764 and t = 3 + 2.236 = 5.236. However, these are the times when h(t)=15, but the rocket is above 15 between them. Step 5: Check: The parabola opens downward (a=-2 < 0), so the rocket is above 15 meters between the two times when it equals 15. So the interval is 3 - sqrt(5) ≤ t ≤ 3 + sqrt(5). But wait, we need exact values: t² - 6t + 4 = 0 gives t = 3 ± sqrt(5). But sqrt(5) ≈ 2.236, so t ≈ 0.764 and t ≈ 5.236. However, the rocket only launches at t=0 and lands at t when h(t)=0. Solve h(t)=0: -2t² + 12t + 7 = 0 => 2t² - 12t - 7 = 0 => t = (12 ± sqrt(144 + 56))/4 = (12 ± sqrt(200))/4 = (12 ± 10*sqrt(2))/4 = 3 ± (5*sqrt(2))/2 ≈ 3 ± 3.535. So landing at about t=6.535. The interval from t=0.764 to t=5.236 is within the flight. Step 6: The exact interval where h(t) ≥ 15 is from t = 3 - sqrt(5) to t = 3 + sqrt(5). But since the problem likely expects a simple interval, and the graph shows symmetry, the answer is 2 ≤ t ≤ 4 when rounded? Let's re-check: 3 - sqrt(5) ≈ 0.764, 3 + sqrt(5) ≈ 5.236. So the correct interval is approximately 0.764 ≤ t ≤ 5.236. But the problem asks for the exact interval in simplest radical form. So the exact answer is 3 - sqrt(5) ≤ t ≤ 3 + sqrt(5). However, the instruction says answer should be a plain number, so I'll use the exact interval. The time interval is 3 - sqrt(5) ≤ t ≤ 3 + sqrt(5).

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

   Let's solve this step by step. --- **Step 1: Recall vertex form of a quadratic** The vertex form of a quadratic function is: y = a(x - h)² + k where (h, k) is the vertex. --- **Step 2: Substitute the vertex into the equation** We are told the vertex is (2, -3), so h = 2 and k = -3. Substitute: y = a(x - 2)² - 3 --- **Step 3: Use the given point to find 'a'** The parabola passes through (4, 1). Substitute x = 4 and y = 1 into the equation: 1 = a(4 - 2)² - 3 --- **Step 4: Simplify and solve for a** 1 = a(2)² - 3 1 = a(4) - 3 1 + 3 = 4a 4 = 4a a = 1 --- **Step 5: Write the final equation** Substitute a = 1, h = 2, k = -3 into vertex form: y = 1(x - 2)² - 3 We usually write it without the 1: y = (x - 2)² - 3 --- **Final answer:** y = (x - 2)² - 3

3. Emma is designing a rectangular garden with a fixed perimeter of 60 meters. She wants to model the area of the garden as a function of its width. If the width is represented by x meters, what is the maximum possible area the garden can have? Answer: 225 Solution: Let the width be x meters and the length be L meters. The perimeter is 60 meters, so 2x + 2L = 60. Simplify to get x + L = 30, so L = 30 - x.
   Full step-by-step solution

   Step 1: Let the width be x meters and the length be L meters. Step 2: The perimeter is 60 meters, so 2x + 2L = 60. Step 3: Simplify to get x + L = 30, so L = 30 - x. Step 4: The area A = length × width = x(30 - x) = 30x - x². Step 5: This is a quadratic function A(x) = -x² + 30x, which opens downward. Step 6: The maximum occurs at the vertex x = -b/(2a) = -30/(2×(-1)) = 15. Step 7: Substitute x = 15 into the area function: A = 15(30 - 15) = 15 × 15 = 225. Step 8: The maximum possible area is 225 square meters.

4. f(x) = 2x³ - 11x² + 12x + 9; f(4) = ? Answer: 9 Solution: Start with the function f(x) = 2x³ - 11x² + 12x + 9 Substitute x = 4 into the function: f(4) = 2(4)³ - 11(4)² + 12(4) + 9 Calculate the exponents first: (4)³ = 64 and (4)² = 16 Multiply: 2 × 64 = 128, -11 × 16 = -176, and 12 × 4 = 48 Rewrite the expression: f(4) = 128 - 176 + 48 + 9 Perform the…
   Full step-by-step solution

   Step 1: Start with the function f(x) = 2x³ - 11x² + 12x + 9 Step 2: Substitute x = 4 into the function: f(4) = 2(4)³ - 11(4)² + 12(4) + 9 Step 3: Calculate the exponents first: (4)³ = 64 and (4)² = 16 Step 4: Multiply: 2 × 64 = 128, -11 × 16 = -176, and 12 × 4 = 48 Step 5: Rewrite the expression: f(4) = 128 - 176 + 48 + 9 Step 6: Perform the operations from left to right: 128 - 176 = -48, then -48 + 48 = 0, then 0 + 9 = 9 The answer is 9.

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

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

6. Aisha is designing a rectangular garden with a fixed perimeter of 60 meters. She wants to model the area of the garden as a function of its width. If the width is represented by x meters, what is the maximum possible area the garden can have? Answer: 225 Solution: Let the width be x meters and the length be L meters. The perimeter is 60 meters, so 2x + 2L = 60. Simplify to get x + L = 30, so L = 30 - x.
   Full step-by-step solution

   Step 1: Let the width be x meters and the length be L meters. Step 2: The perimeter is 60 meters, so 2x + 2L = 60. Step 3: Simplify to get x + L = 30, so L = 30 - x. Step 4: The area A = length × width = (30 - x)(x) = 30x - x². Step 5: This is a quadratic function A(x) = -x² + 30x, which opens downward. Step 6: The maximum occurs at the vertex x = -b/(2a) = -30/(2×(-1)) = 15. Step 7: Substitute x = 15 into the area function: A = 30(15) - (15)² = 450 - 225 = 225. Step 8: The maximum possible area is 225 square meters.

7. Maya is designing a skateboard ramp with a parabolic shape. The height of the ramp above the ground is modeled by the function h(x) = -0.5x² + 4x, where h is the height in feet and x is the horizontal distance in feet from the starting point. What is the maximum height the ramp reaches, and at what horizontal distance does this occur? Answer: 8 feet at 4 feet Solution: The function is h(x) = -0.5x² + 4x. Since the coefficient of x² is negative (-0.5), the parabola opens downward, so the vertex gives the maximum height.
   Full step-by-step solution

   Step 1: The function is h(x) = -0.5x² + 4x. Since the coefficient of x² is negative (-0.5), the parabola opens downward, so the vertex gives the maximum height. Step 2: To find the x-coordinate of the vertex, use the formula x = -b/(2a), where a = -0.5 and b = 4. Step 3: Calculate x = -4/(2 × -0.5) = -4/(-1) = 4 feet. Step 4: Substitute x = 4 into the function to find the maximum height: h(4) = -0.5(4)² + 4(4) = -0.5(16) + 16 = -8 + 16 = 8 feet. Step 5: Therefore, the maximum height is 8 feet, and it occurs at a horizontal distance of 4 feet from the starting point.