Graph Key Features

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Answer Key & Explanations

Graph Key Features · Grade 9 · Worksheet 1

1. Nikau is studying the population P(t) = 290 * 2^(t/3) of bacteria in a lab experiment, where t is hours. From the graph of this function, identify the y-intercept and describe the interval(s) where the population is increasing or decreasing. Answer: 290 Solution: The function P(t) = 290 * 2^(t/3) is an exponential growth function. The y-intercept occurs when t = 0: P(0) = 290 * 2^(0) = 290 * 1 = 290. Since the base 2 > 1, the function is always increasing for all t ≥ 0.
   Full step-by-step solution

   Step 1: The function P(t) = 290 * 2^(t/3) is an exponential growth function. Step 2: The y-intercept occurs when t = 0: P(0) = 290 * 2^(0) = 290 * 1 = 290. Step 3: Since the base 2 > 1, the function is always increasing for all t ≥ 0. Step 4: There are no decreasing intervals; the population grows continuously. Answer: The y-intercept is 290 (initial population). The population is increasing on the interval (0, ∞).

2. The population of a colony of bacteria is modeled by the function P(t) = 500 * 2^(t/3), where t is the time in hours and P(t) is the population size. After how many hours will the bacteria population reach 32,000? Answer: 18 Solution: Set up the equation: 500 * 2^(t/3) = 32,000 Divide both sides by 500: 2^(t/3) = 32,000 ÷ 500 Calculate the division: 2^(t/3) = 64 Recognize that 64 is a power of 2: 64 = 2^6 Set the exponents equal: t/3 = 6 Multiply both sides by 3: t = 6 * 3 Calculate: t = 18 The answer is 18 hours.
   Full step-by-step solution

   Step 1: Set up the equation: 500 * 2^(t/3) = 32,000 Step 2: Divide both sides by 500: 2^(t/3) = 32,000 ÷ 500 Step 3: Calculate the division: 2^(t/3) = 64 Step 4: Recognize that 64 is a power of 2: 64 = 2^6 Step 5: Set the exponents equal: t/3 = 6 Step 6: Multiply both sides by 3: t = 6 * 3 Step 7: Calculate: t = 18 The answer is 18 hours.

3. Sophia graphs a quadratic function on a coordinate plane. The parabola opens downward and has its vertex at (7, 9). The graph also passes through the point (10, 0). Identify the x-intercepts of the function and state the interval(s) where the function is increasing. Answer: x-intercepts at x = 4 and x = 10; increasing on (-infinity, 7) Solution: Use vertex form: y = a(x - h)^2 + k, with vertex (h, k) = (7, 9). So y = a(x - 7)^2 + 9. Substitute the point (10, 0): 0 = a(10 - 7)^2 + 9 -> 0 = a(3)^2 + 9 -> 0 = 9a + 9 -> 9a = -9 -> a = -1.
   Full step-by-step solution

   Step 1: Use vertex form: y = a(x - h)^2 + k, with vertex (h, k) = (7, 9). So y = a(x - 7)^2 + 9. Step 2: Substitute the point (10, 0): 0 = a(10 - 7)^2 + 9 -> 0 = a(3)^2 + 9 -> 0 = 9a + 9 -> 9a = -9 -> a = -1. Step 3: The equation is y = -(x - 7)^2 + 9. Step 4: To find x-intercepts, set y = 0: 0 = -(x - 7)^2 + 9 -> (x - 7)^2 = 9 -> x - 7 = ±3 -> x = 7 + 3 = 10 or x = 7 - 3 = 4. Step 5: The x-intercepts are at x = 4 and x = 10. Step 6: Since the parabola opens downward (a = -1 < 0), it increases to the left of the vertex and decreases to the right. The vertex is at x = 7, so the function is increasing on the interval (-infinity, 7). Final answer: x-intercepts at x = 4 and x = 10; increasing on (-infinity, 7).

4. Emma is analyzing the trajectory of a soccer ball kicked from the ground. The ball's height above the ground is modeled by the function h(t) = -5t² + 20t, where h is the height in meters and t is the time in seconds after the kick. How many seconds after being kicked does the soccer ball reach its maximum height? Answer: 2 Solution: The function h(t) = -5t² + 20t is a quadratic function in the form at² + bt + c, where a = -5 and b = 20. For a quadratic function, the vertex (maximum or minimum point) occurs at t = -b/(2a).
   Full step-by-step solution

   Step 1: The function h(t) = -5t² + 20t is a quadratic function in the form at² + bt + c, where a = -5 and b = 20. Step 2: For a quadratic function, the vertex (maximum or minimum point) occurs at t = -b/(2a). Step 3: Substitute the values: t = -20/(2*(-5)) = -20/(-10) = 2. Step 4: The ball reaches its maximum height 2 seconds after being kicked. The answer is 2.

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

   Let's solve this step-by-step. --- **Step 1: Write the vertex form of a quadratic function.** Since the parabola opens downward and has vertex (h, k) = (2, 9), the vertex form is: y = a(x - 2)^2 + 9 where a < 0 because it opens downward. --- **Step 2: Use the given point (5, 0) to find a.** Substitute x = 5, y = 0 into the equation: 0 = a(5 - 2)^2 + 9 0 = a(3)^2 + 9 0 = 9a + 9 --- **Step 3: Solve for a.** 9a + 9 = 0 9a = -9 a = -1 --- **Step 4: Write the vertex form with a = -1.** y = -1(x - 2)^2 + 9 --- **Step 5: Expand to standard form y = ax^2 + bx + c.** First expand (x - 2)^2: (x - 2)^2 = x^2 - 4x + 4 Now 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 6: Verify with given conditions.** Vertex: x = -b/(2a) = -4/(2*(-1)) = -4/(-2) = 2 y-coordinate at x = 2: y = -(2)^2 + 4*2 + 5 = -4 + 8 + 5 = 9 → vertex (2, 9) ✓ Point (5, 0): y = -(25) + 20 + 5 = -25 + 25 = 0 ✓ Opens downward because a = -1 < 0 ✓ --- **Final answer:** y = -x^2 + 4x + 5

6. 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

7. 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)

8. A drone is flying over a field, and its height above ground is modeled by the quadratic function h(t) = -2t² + 12t + 5, where h is height in meters and t is time in seconds. After how many seconds does the drone reach its maximum height? Answer: 3 Solution: We are given the height function: h(t) = -2t^2 + 12t + 5. This is a quadratic function in the form h(t) = a t^2 + b t + c, where a = -2, b = 12, c = 5.
   Full step-by-step solution

   We are given the height function: h(t) = -2t^2 + 12t + 5. This is a quadratic function in the form h(t) = a t^2 + b t + c, where a = -2, b = 12, c = 5. Since a is negative (a = -2), the parabola opens downward, so the vertex of the parabola gives the maximum height. The vertex occurs at time t = -b / (2a). Step 1: Identify a and b. a = -2 b = 12 Step 2: Plug into the vertex formula. t = -b / (2a) = -12 / (2 * -2) Step 3: Simplify the denominator. 2 * a = 2 * (-2) = -4 Step 4: Simplify the fraction. t = -12 / (-4) = 12 / 4 = 3 So, the drone reaches its maximum height at t = 3 seconds. Final answer: 3