Graph Key Features

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Answer Key & Explanations

Graph Key Features · Grade 9 · Worksheet 3

1. Mere's function graph shows a parabola with vertex at (4, 16), x-intercepts at (0, 0) and (8, 0), and y-intercept at (0, 0). Identify the intervals where the function is increasing and decreasing. Answer: Increasing: (-∞, 4); Decreasing: (4, ∞) Solution: The vertex is at (4, 16), which is the highest point on the graph since the y-value is positive and the parabola opens downward.
   Full step-by-step solution

   Step 1: The vertex is at (4, 16), which is the highest point on the graph since the y-value is positive and the parabola opens downward. Step 2: Since the parabola opens downward, the function increases as it approaches the vertex from the left and decreases after passing the vertex. Step 3: The increasing interval is from negative infinity to the x-coordinate of the vertex: (-∞, 4) Step 4: The decreasing interval is from the x-coordinate of the vertex to positive infinity: (4, ∞) The function is increasing on (-∞, 4) and decreasing on (4, ∞).

2. Emma is analyzing the graph of a quadratic function that opens downward. The graph shows x-intercepts at (-5, 0) and (5, 0), and a y-intercept at (0, -25). The vertex is at (0, -25). Over what interval is the function increasing? Answer: (-∞, 0) Solution: Identify that the parabola opens downward since the vertex is at (0, -25) and the y-intercept is also at (0, -25), indicating this is the maximum point.
   Full step-by-step solution

   Step 1: Identify that the parabola opens downward since the vertex is at (0, -25) and the y-intercept is also at (0, -25), indicating this is the maximum point. Step 2: For a downward-opening parabola, the function increases on the interval from negative infinity to the x-coordinate of the vertex. Step 3: The vertex is at (0, -25), so the x-coordinate is 0. Step 4: Therefore, the function is increasing on the interval (-∞, 0). The answer is (-∞, 0).

3. The function f(x) = -2x² + 12x - 10 models the height (in meters) of a projectile above ground level, where x is the time in seconds after launch. At what time does the projectile reach its maximum height? Answer: 3 Solution: f(x) = -2x² + 12x - 10 This function models the height of a projectile over time x (in seconds).
   Full step-by-step solution

   We are given the quadratic function: f(x) = -2x² + 12x - 10 This function models the height of a projectile over time x (in seconds). Since the coefficient of x² is negative (-2), the parabola opens downward, meaning the vertex of the parabola is the maximum point. For a quadratic in the form f(x) = ax² + bx + c, the x-coordinate of the vertex is given by: x = -b / (2a) Here: a = -2 b = 12 Step 1: Apply the vertex formula. x = -b / (2a) x = -12 / (2 * -2) Step 2: Simplify the denominator. 2 * -2 = -4 Step 3: Simplify the fraction. x = -12 / (-4) x = 12 / 4 x = 3 So the projectile reaches its maximum height at x = 3 seconds. Final answer: 3

4. Emma is analyzing the graph of f(x) = 3x² - 15x + 7. Identify the vertex coordinates. Answer: (2.5, -11.75) Solution: Identify coefficients from f(x) = 3x² - 15x + 7: a = 3, b = -15, c = 7 Calculate x-coordinate using formula x = -b/(2a): x = -(-15)/(2×3) = 15/6 = 2.5 Substitute x = 2.5 into the function: f(2.5) = 3(2.5)² - 15(2.5) + 7 Calculate (2.5)² = 6.25 Multiply: 3 × 6.25 = 18.75 and 15 × 2.5 = 37.5…
   Full step-by-step solution

   Step 1: Identify coefficients from f(x) = 3x² - 15x + 7: a = 3, b = -15, c = 7 Step 2: Calculate x-coordinate using formula x = -b/(2a): x = -(-15)/(2×3) = 15/6 = 2.5 Step 3: Substitute x = 2.5 into the function: f(2.5) = 3(2.5)² - 15(2.5) + 7 Step 4: Calculate (2.5)² = 6.25 Step 5: Multiply: 3 × 6.25 = 18.75 and 15 × 2.5 = 37.5 Step 6: Rewrite: f(2.5) = 18.75 - 37.5 + 7 Step 7: Perform operations: 18.75 - 37.5 = -18.75, then -18.75 + 7 = -11.75 Step 8: The vertex coordinates are (2.5, -11.75)

5. A quadratic function is graphed on a coordinate plane. The parabola opens upward and has its vertex at (-3, -4). The parabola passes through the point (1, 12). What is the equation of this quadratic function in standard form? Answer: x² + 6x + 5 Solution: Start with the vertex form of a quadratic equation: y = a(x - h)² + k, where (h, k) is the vertex. Substitute the vertex (-3, -4): y = a(x - (-3))² + (-4) = a(x + 3)² - 4.
   Full step-by-step solution

   Step 1: Start with the vertex form of a quadratic equation: y = a(x - h)² + k, where (h, k) is the vertex. Step 2: Substitute the vertex (-3, -4): y = a(x - (-3))² + (-4) = a(x + 3)² - 4. Step 3: Substitute the point (1, 12) into the equation: 12 = a(1 + 3)² - 4. Step 4: Simplify: 12 = a(4)² - 4 → 12 = 16a - 4. Step 5: Solve for a: 12 + 4 = 16a → 16 = 16a → a = 1. Step 6: Write the equation with a = 1: y = (x + 3)² - 4. Step 7: Expand to standard form: y = (x² + 6x + 9) - 4 = x² + 6x + 5. The equation in standard form is x² + 6x + 5.

6. Aroha is analyzing the graph of f(x) = 2x³ - 15x² + 24x + 10. Over which interval is the function decreasing? Answer: (1, 4) Solution: Find the derivative of f(x) = 2x³ - 15x² + 24x + 10 f'(x) = 6x² - 30x + 24 Set the derivative equal to zero to find critical points 6x² - 30x + 24 = 0 Divide the equation by 6 x² - 5x + 4 = 0 (x - 1)(x - 4) = 0 x = 1 or x = 4 For x < 1 (test x = 0): f'(0) = 6(0)² - 30(0) + 24 = 24 > 0…
   Full step-by-step solution

   Step 1: Find the derivative of f(x) = 2x³ - 15x² + 24x + 10 f'(x) = 6x² - 30x + 24 Step 2: Set the derivative equal to zero to find critical points 6x² - 30x + 24 = 0 Step 3: Divide the equation by 6 x² - 5x + 4 = 0 Step 4: Factor the quadratic equation (x - 1)(x - 4) = 0 Step 5: Solve for x x = 1 or x = 4 Step 6: Test intervals around the critical points For x < 1 (test x = 0): f'(0) = 6(0)² - 30(0) + 24 = 24 > 0 (increasing) For 1 < x < 4 (test x = 2): f'(2) = 6(4) - 30(2) + 24 = 24 - 60 + 24 = -12 < 0 (decreasing) For x > 4 (test x = 5): f'(5) = 6(25) - 30(5) + 24 = 150 - 150 + 24 = 24 > 0 (increasing) The function is decreasing on the interval (1, 4).

7. Sophia is analyzing the graph of f(x) = -x² + 8x - 7. The graph has x-intercepts at (1,0) and (7,0), a y-intercept at (0,-7), and a vertex at (4,9). Over which interval is the function increasing? Answer: (-∞, 4) Solution: Identify the direction of the parabola. Since the coefficient of x² is -1 (negative), the parabola opens downward.
   Full step-by-step solution

   Step 1: Identify the direction of the parabola. Since the coefficient of x² is -1 (negative), the parabola opens downward. Step 2: For a downward-opening parabola, the function increases on the left side of the vertex and decreases on the right side. Step 3: The vertex is at (4,9), so the function increases as x approaches 4 from the left. Step 4: Therefore, the function is increasing on the interval (-∞, 4).

8. Isabella is analyzing the graph of f(x) = -2x² + 12x - 7. Identify: (a) the y-intercept, (b) the vertex coordinates, (c) the interval where the function is increasing. Answer: (0,-7);(3,11);(-∞,3) Solution: Find the y-intercept by evaluating f(0): f(0) = -2(0)² + 12(0) - 7 = -7. Find the vertex coordinates using x = -b/(2a). Here a = -2, b = 12, so x = -12/(2×-2) = -12/-4 = 3.
   Full step-by-step solution

   Step 1: Find the y-intercept by evaluating f(0): f(0) = -2(0)² + 12(0) - 7 = -7. So the y-intercept is (0, -7). Step 2: Find the vertex coordinates using x = -b/(2a). Here a = -2, b = 12, so x = -12/(2×-2) = -12/-4 = 3. Step 3: Find the y-coordinate of the vertex by evaluating f(3): f(3) = -2(3)² + 12(3) - 7 = -2(9) + 36 - 7 = -18 + 36 - 7 = 11. So the vertex is (3, 11). Step 4: Determine intervals of increase/decrease. Since a = -2 (negative), the parabola opens downward, so the function increases on the interval (-∞, 3) and decreases on (3, ∞). Final answers: y-intercept: (0, -7); vertex: (3, 11); increasing interval: (-∞, 3)

9. A quadratic function is graphed on a coordinate plane. The parabola opens downward and has its vertex at the point (2, 9). The parabola also passes through the point (5, 0). What is the equation of this quadratic function in standard form? Answer: y = -x² + 4x + 5 Solution: - Vertex at (2, 9) - Passes through (5, 0) y = a(x - h)^2 + k where (h, k) is the vertex.
   Full step-by-step solution

   We are told the quadratic function has: - Vertex at (2, 9) - Passes through (5, 0) - Parabola opens downward --- **Step 1: Use vertex form of a quadratic** The vertex form is: y = a(x - h)^2 + k where (h, k) is the vertex. Here, h = 2, k = 9, so: y = a(x - 2)^2 + 9 --- **Step 2: Use the point (5, 0) to find a** Substitute x = 5, y = 0: 0 = a(5 - 2)^2 + 9 0 = a(3)^2 + 9 0 = 9a + 9 Solve for a: 9a = -9 a = -1 --- **Step 3: Write the equation in vertex form** y = -1(x - 2)^2 + 9 --- **Step 4: Expand to standard form y = ax^2 + bx + c** First expand (x - 2)^2: (x - 2)^2 = x^2 - 4x + 4 Multiply by -1: -1(x^2 - 4x + 4) = -x^2 + 4x - 4 Now add 9: y = -x^2 + 4x - 4 + 9 y = -x^2 + 4x + 5 --- **Step 5: Check with given conditions** Vertex: x = -b/(2a) = -4/(2*(-1)) = -4/(-2) = 2, correct. y-coordinate at x=2: -(4) + 4*2 + 5 = -4 + 8 + 5 = 9, correct. Point (5, 0): -(25) + 4*5 + 5 = -25 + 20 + 5 = 0, correct. Opens downward (a = -1), correct. --- **Final answer:** y = -x^2 + 4x + 5