Polynomial Technology

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Polynomial Technology · Grade 12 · Worksheet 2

1. Liam is designing a roller coaster track that follows the path of the polynomial function f(x) = 2x³ - 9x² - 24x + 5. He needs to determine the exact coordinates of all local extrema points to ensure proper safety engineering. Find all critical points where the roller coaster reaches its highest and lowest points along this path. Answer: (-1, 18) and (4, -107) Solution: f(x) = 2x³ - 9x² - 24x + 5 f'(x) = d/dx [2x³] - d/dx [9x²] - d/dx [24x] + d/dx [5] f'(x) = 6x² - 18x - 24 f'(x) = 6x² - 18x - 24 Set derivative equal to zero to find critical points 6x² - 18x - 24 = 0 Divide through by 6: x² - 3x - 4 = 0 (x - 4)(x + 1) = 0 x = 4 or x = -1 These are the…
   Full step-by-step solution

   Let's find the local extrema step by step. --- **Step 1: Find the derivative of f(x)** We have: f(x) = 2x³ - 9x² - 24x + 5 Differentiate term by term: f'(x) = d/dx [2x³] - d/dx [9x²] - d/dx [24x] + d/dx [5] f'(x) = 6x² - 18x - 24 So: f'(x) = 6x² - 18x - 24 --- **Step 2: Set derivative equal to zero to find critical points** 6x² - 18x - 24 = 0 Divide through by 6: x² - 3x - 4 = 0 Factor: (x - 4)(x + 1) = 0 So: x = 4 or x = -1 These are the x-coordinates of the critical points. --- **Step 3: Find the y-coordinates by plugging into f(x)** For x = -1: f(-1) = 2(-1)³ - 9(-1)² - 24(-1) + 5 = 2(-1) - 9(1) + 24 + 5 = -2 - 9 + 24 + 5 = -11 + 24 + 5 = 13 + 5 = 18 So one point: (-1, 18) For x = 4: f(4) = 2(4)³ - 9(4)² - 24(4) + 5 = 2(64) - 9(16) - 96 + 5 = 128 - 144 - 96 + 5 = -16 - 96 + 5 = -112 + 5 = -107 So the other point: (4, -107) --- **Step 4: Determine which is local max and which is local min** We can use the second derivative test. f''(x) = derivative of f'(x) = 12x - 18 At x = -1: f''(-1) = 12(-1) - 18 = -12 - 18 = -30 Since f''(-1) < 0, the curve is concave down → local maximum at (-1, 18) At x = 4: f''(4) = 12(4) - 18 = 48 - 18 = 30 Since f''(4) > 0, the curve is concave up → local minimum at (4, -107) --- **Final Answer:** The local extrema are at (-1, 18) and (4, -107), where (-1, 18) is a local maximum and (4, -107) is a local minimum.

2. A biomedical engineer is modeling the concentration of a new drug in a patient's bloodstream using the polynomial function C(t) = -0.02t³ + 0.3t² + 0.5t, where C is the concentration in milligrams per liter and t is time in hours after administration. The drug is considered effective when the concentration is increasing and above 1 mg/L. Determine the time interval during which the drug remains effective. Answer: (2, 5) Solution: In pharmaceutical modeling, we often analyze when drugs reach therapeutic levels.
   Full step-by-step solution

   In pharmaceutical modeling, we often analyze when drugs reach therapeutic levels. The increasing concentration condition relates to the derivative being positive, while the minimum concentration requirement involves solving an inequality. This type of analysis helps determine optimal dosing schedules in medical applications.

3. An environmental engineer is modeling the population growth of an endangered species in a wildlife reserve using the polynomial function P(t) = -0.01t³ + 0.15t² + 0.6t + 20, where P(t) represents the population in hundreds and t is the time in years since conservation efforts began. The species is considered to be recovering when the population growth rate is positive and the population exceeds 2,500 individuals. During what time interval is the species considered to be recovering? Answer: 0 < t < 10 Solution: In population modeling, the derivative represents the instantaneous growth rate. A positive derivative indicates increasing population, while the function value itself represents the current population size. To determine recovery periods, we need to find where both conditions are met: growth…
   Full step-by-step solution

   In population modeling, the derivative represents the instantaneous growth rate. A positive derivative indicates increasing population, while the function value itself represents the current population size. To determine recovery periods, we need to find where both conditions are met: growth rate > 0 and population > threshold value.

4. A polynomial function f(x) is graphed on a coordinate plane. The graph shows the function crossing the x-axis at (-3, 0) and (2, 0), and it touches but does not cross the x-axis at (1, 0). The graph also passes through the point (0, -6). If the function has a positive leading coefficient, what is the equation of f(x) in factored form? Answer: f(x) = (x+3)(x-2)(x-1)^2 Solution: Identify the roots and their multiplicities from the graph. - The graph crosses the x-axis at \((-3, 0)\) and \((2, 0)\).
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Identify the roots and their multiplicities from the graph.** - The graph crosses the x-axis at \((-3, 0)\) and \((2, 0)\). Crossing means odd multiplicity, and for a polynomial without further info, the simplest odd multiplicity is 1. - The graph touches but does not cross at \((1, 0)\). Touching without crossing means even multiplicity, smallest even multiplicity is 2. So we have: Root \(x = -3\) with multiplicity 1 → factor \((x + 3)\) Root \(x = 2\) with multiplicity 1 → factor \((x - 2)\) Root \(x = 1\) with multiplicity 2 → factor \((x - 1)^2\) --- **Step 2: Write the general factored form.** \[ f(x) = a(x + 3)(x - 2)(x - 1)^2 \] where \(a\) is a constant (leading coefficient). --- **Step 3: Use the given point \((0, -6)\) to find \(a\).** Substitute \(x = 0\), \(f(0) = -6\): \[ f(0) = a(0 + 3)(0 - 2)(0 - 1)^2 \] \[ f(0) = a(3)(-2)(1) \] \[ f(0) = a(-6) \] \[ -6 = -6a \] \[ a = 1 \] --- **Step 4: Check the sign of the leading coefficient.** The problem says the leading coefficient is positive. With \(a = 1\), that is positive. Also, expanding symbolically: \((x+3)(x-2) = x^2 + x - 6\) Multiply by \((x-1)^2 = x^2 - 2x + 1\): The leading term is \(x^2 \cdot x^2 = x^4\), coefficient \(1\), positive. Good. --- **Step 5: Write the final factored form.** \[ f(x) = (x + 3)(x - 2)(x - 1)^2 \] --- **Final answer:** f(x) = (x+3)(x-2)(x-1)^2

5. Use Desmos to graph f(x)=x³-9x²+23x-15 and find the sum of all x-coordinates where the local maximum and minimum occur. Answer: 6 Solution: Find the derivative of f(x)=x³-9x²+23x-15 f'(x)=3x²-18x+23 Set the derivative equal to zero to find critical points 3x²-18x+23=0 x=[18±√(324-276)]/6 x=[18±√48]/6 x=[18±4√3]/6 x=3±(2√3)/3 The two critical points are x=3+2√3/3 and x=3-2√3/3 (3+2√3/3)+(3-2√3/3)=6 The answer is 6.
   Full step-by-step solution

   Step 1: Find the derivative of f(x)=x³-9x²+23x-15 f'(x)=3x²-18x+23 Step 2: Set the derivative equal to zero to find critical points 3x²-18x+23=0 Step 3: Use the quadratic formula to solve x=[18±√(324-276)]/6 x=[18±√48]/6 x=[18±4√3]/6 x=3±(2√3)/3 Step 4: The two critical points are x=3+2√3/3 and x=3-2√3/3 Step 5: Sum the x-coordinates of the local extrema (3+2√3/3)+(3-2√3/3)=6 The answer is 6.

6. Using Desmos or a graphing calculator, graph the polynomial function f(x) = -2x^4 + 11x^3 + 6x^2 - 61x - 56. Identify all x-intercepts (zeros), the y-intercept, and the coordinates of all local maxima and minima. Round all answers to the nearest hundredth. Answer: x-intercepts: (-1.75, 0), (-1, 0), (3.56, 0), (4.19, 0); y-intercept: (0, -56); local maximum: (2.68, 25.09); local minima: (-1.18, -0.52) and (4.19, 0) Solution: Graph f(x) = -2x^4 + 11x^3 + 6x^2 - 61x - 56 in Desmos or a graphing calculator. Find x-intercepts by using the zero/root feature. The graph crosses the x-axis at approximately x = -1.75, x = -1, x = 3.56, and x = 4.19.
   Full step-by-step solution

   Step 1: Graph f(x) = -2x^4 + 11x^3 + 6x^2 - 61x - 56 in Desmos or a graphing calculator. Step 2: Find x-intercepts by using the zero/root feature. The graph crosses the x-axis at approximately x = -1.75, x = -1, x = 3.56, and x = 4.19. So the zeros are (-1.75, 0), (-1, 0), (3.56, 0), and (4.19, 0). Step 3: Find the y-intercept by evaluating f(0) = -2(0)^4 + 11(0)^3 + 6(0)^2 - 61(0) - 56 = -56. So the y-intercept is (0, -56). Step 4: Find local maxima and minima. The graph shows a peak near x = 2.68. Using the maximum tool, the local maximum is approximately (2.68, 25.09). There are two valleys: one near x = -1.18, giving a local minimum of (-1.18, -0.52), and another at x = 4.19, giving a local minimum of (4.19, 0). Step 5: All values are rounded to the nearest hundredth as requested. The final answer: x-intercepts: (-1.75, 0), (-1, 0), (3.56, 0), (4.19, 0); y-intercept: (0, -56); local maximum: (2.68, 25.09); local minima: (-1.18, -0.52) and (4.19, 0).