Domain and Range

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Answer Key & Explanations

Domain and Range · Grade 9 · Worksheet 3

1. Mason is analyzing the path of a water fountain's spray. The height of the water above the nozzle (in feet) is modeled by the quadratic function h(x) = -0.2x² + 2x + 2, where x is the horizontal distance (in feet) from the nozzle. The water hits a decorative wall that is 8 feet away from the nozzle. What are the domain and range of the function for this real-world scenario, considering only the water's path until it hits the wall? Answer: Domain: 0 ≤ x ≤ 8; Range: 0 ≤ h(x) ≤ 7 Solution: Identify the domain. The horizontal distance x starts at the nozzle (x = 0) and ends at the wall (x = 8). Domain: 0 ≤ x ≤ 8.
   Full step-by-step solution

   Step 1: Identify the domain. The horizontal distance x starts at the nozzle (x = 0) and ends at the wall (x = 8). So the domain is all real numbers between 0 and 8, inclusive. Domain: 0 ≤ x ≤ 8. Step 2: Find the range. The range is the set of possible heights h(x) for x in the domain. The water's height starts at h(0) = -0.2(0)² + 2(0) + 2 = 2 feet. At the wall, the height is h(8) = -0.2(64) + 2(8) + 2 = -12.8 + 16 + 2 = 5.2 feet. Step 3: Since the parabola opens downward (a = -0.2 < 0), the maximum height occurs at the vertex. The vertex's x-coordinate is x = -b/(2a) = -2/(2 * -0.2) = -2/(-0.4) = 5. The maximum height is h(5) = -0.2(25) + 2(5) + 2 = -5 + 10 + 2 = 7 feet. Step 4: The water is never below ground (height 0), and the minimum height in this interval occurs at x = 0 (height 2) or at x = 8 (height 5.2), but the water actually starts at the nozzle at height 2 and arcs upward. Since the water is a continuous stream, it will be at height 0 at some point before the nozzle? No, the function gives height above the nozzle, so h(0)=2. The water goes up to 7 feet and then down to 5.2 feet at the wall. The lowest height in the domain is at x = 0, which is 2 feet, but the water is not below 0 in this real-world model. However, the water starts at 2 feet. Since the water is above ground, and the lowest point in the domain is at x=0 with height 2, but the water must be at ground level somewhere? The problem says the height is above the nozzle, so the water at the nozzle is 2 feet high (the nozzle is above ground). The water never goes below 2 feet in this domain. The minimum height on the interval is at x=0 (height 2). The maximum is at the vertex (height 7). So the range is 2 ≤ h(x) ≤ 7. But wait, the water must start at the nozzle at height 2, arc up, and come down. It never goes below 2 feet in this interval. So the range is from 2 to 7. Step 5: Final answer: Domain: 0 ≤ x ≤ 8; Range: 2 ≤ h(x) ≤ 7.

2. The graph of f(x) = -3x² + 18x - 15 is a parabola opening downward with vertex at (3, 12) and x-intercepts at (1, 0) and (5, 0). What is the domain and range of this function? Answer: Domain: all real numbers; Range: y ≤ 12 Solution: The function f(x) = -3x² + 18x - 15 is a quadratic function, so its graph is a parabola. The domain of any quadratic function is all real numbers because you can substitute any real number for x.
   Full step-by-step solution

   Step 1: The function f(x) = -3x² + 18x - 15 is a quadratic function, so its graph is a parabola. The domain of any quadratic function is all real numbers because you can substitute any real number for x. Step 2: The parabola opens downward because the coefficient of x² is negative (-3 < 0). This means the vertex is the maximum point of the graph. Step 3: The vertex is given as (3, 12). The y-coordinate of the vertex is 12, which is the maximum value of the function. Step 4: Since the parabola opens downward, the function takes all y-values less than or equal to the maximum. Therefore, the range is y ≤ 12. Final answer: Domain: all real numbers; Range: y ≤ 12

3. f(x) = 2x² - 4x + 1, find f(3) = ? Answer: 7 Solution: We are given the function f(x) = 2x² - 4x + 1 and asked to find f(3). Substitute x = 3 into the function. f(3) = 2*(3)² - 4*(3) + 1 Evaluate the exponent first.
   Full step-by-step solution

   We are given the function f(x) = 2x² - 4x + 1 and asked to find f(3). Step 1: Substitute x = 3 into the function. f(3) = 2*(3)² - 4*(3) + 1 Step 2: Evaluate the exponent first. (3)² = 9 So f(3) = 2*9 - 4*3 + 1 Step 3: Perform the multiplications. 2*9 = 18 4*3 = 12 So f(3) = 18 - 12 + 1 Step 4: Perform the addition and subtraction from left to right. 18 - 12 = 6 6 + 1 = 7 Therefore, f(3) = 7. Final answer: 7

4. Consider the graph of f(x) = 2x² - 12x + 16. The parabola opens upward with vertex at (3, -2) and x-intercepts at (2, 0) and (4, 0). What is the domain and range of this function? Answer: Domain: (-∞, ∞), Range: [-2, ∞) Solution: Identify the domain - Since this is a quadratic function, it's defined for all real numbers. Domain: (-∞, ∞) Identify the range - The parabola opens upward with vertex at (3, -2), so the minimum y-value is -2 and the function extends upward infinitely.
   Full step-by-step solution

   Step 1: Identify the domain - Since this is a quadratic function, it's defined for all real numbers. Domain: (-∞, ∞) Step 2: Identify the range - The parabola opens upward with vertex at (3, -2), so the minimum y-value is -2 and the function extends upward infinitely. Range: [-2, ∞) Step 3: Verify with graph features - The x-intercepts at (2, 0) and (4, 0) confirm the parabola crosses the x-axis, and the vertex at (3, -2) is the lowest point. The domain is (-∞, ∞) and the range is [-2, ∞).

5. A quadratic function is graphed on a coordinate plane with its vertex at (-1, 4) and passing through the point (1, -2). The parabola opens downward. What is the equation of this quadratic function in vertex form? Answer: y = -1.5(x + 1)^2 + 4 Solution: The vertex form of a quadratic function is y = a(x - h)^2 + k, where (h, k) is the vertex.
   Full step-by-step solution

   Step 1: 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 (-1, 4) into the equation: y = a(x - (-1))^2 + 4 = a(x + 1)^2 + 4 Step 3: Use the point (1, -2) to find the value of a. Substitute x = 1 and y = -2: -2 = a(1 + 1)^2 + 4 Step 4: Simplify: -2 = a(2)^2 + 4 = 4a + 4 Step 5: Solve for a: -2 - 4 = 4a → -6 = 4a → a = -6/4 = -3/2 = -1.5 Step 6: Write the final equation: y = -1.5(x + 1)^2 + 4

6. A circle is drawn on a coordinate plane with its center at (3, -2) and a radius of 5 units. The circle intersects the x-axis at two points. What are the coordinates of these intersection points? Answer: (-2, 0) and (8, 0) Solution: Write the equation of the circle using the center (h,k) = (3,-2) and radius r = 5. Equation: (x - 3)^2 + (y + 2)^2 = 25 Find the intersection points with the x-axis. On the x-axis, y = 0.
   Full step-by-step solution

   Step 1: Write the equation of the circle using the center (h,k) = (3,-2) and radius r = 5. Equation: (x - 3)^2 + (y + 2)^2 = 25 Step 2: Find the intersection points with the x-axis. On the x-axis, y = 0. Substitute y = 0 into the equation: (x - 3)^2 + (0 + 2)^2 = 25 Step 3: Simplify the equation: (x - 3)^2 + (2)^2 = 25 (x - 3)^2 + 4 = 25 Step 4: Subtract 4 from both sides: (x - 3)^2 = 21 Step 5: Take the square root of both sides: x - 3 = ±√21 Step 6: Solve for x: x = 3 ± √21 Step 7: Calculate the approximate values: √21 ≈ 4.5826 x ≈ 3 + 4.5826 = 7.5826 ≈ 8 x ≈ 3 - 4.5826 = -1.5826 ≈ -2 Step 8: The intersection points are (-2, 0) and (8, 0).

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

   We are given the function f(x) = 2x² - 3x + 1 and asked to find f(2). Step 1: Write down the function. f(x) = 2x² - 3x + 1 Step 2: Substitute x = 2 into the function. This means everywhere you see x, replace it with 2. f(2) = 2*(2)² - 3*(2) + 1 Step 3: Follow the order of operations (PEMDAS/BODMAS). First, calculate the exponent: (2)² = 4. So f(2) = 2*4 - 3*2 + 1 Step 4: Perform the multiplications. 2*4 = 8 3*2 = 6 So f(2) = 8 - 6 + 1 Step 5: Perform the addition and subtraction from left to right. 8 - 6 = 2 2 + 1 = 3 Step 6: State the final answer. f(2) = 3 Thus, the correct answer is 3.