Polynomial Technology

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Polynomial Technology · Grade 12 · Worksheet 3

1. A polynomial function f(x) is graphed on a coordinate plane. The graph shows the function crossing the x-axis at (-4, 0) and (2, 0), and touching but not crossing the x-axis at (-1, 0). The graph passes through the point (1, -12). The end behavior shows that as x → -∞, f(x) → -∞ and as x → ∞, f(x) → ∞. Determine the equation of this polynomial in standard form. Answer: f(x) = x^4 + 4x^3 - 3x^2 - 14x + 8 Solution: Identify the roots and their multiplicities from the graph behavior. The equation in standard form is f(x) = (3/5)x^4 + (12/5)x^3 - (9/5)x^2 - (42/5)x - (24/5).
   Full step-by-step solution

   Step 1: Identify the roots and their multiplicities from the graph behavior. - Crossing at x = -4: odd multiplicity (use 1) - Crossing at x = 2: odd multiplicity (use 1) - Touching at x = -1: even multiplicity (use 2) Step 2: Write the factored form with unknown leading coefficient 'a': f(x) = a(x + 4)(x - 2)(x + 1)^2 Step 3: Use the point (1, -12) to solve for 'a': -12 = a(1 + 4)(1 - 2)(1 + 1)^2 -12 = a(5)(-1)(2)^2 -12 = a(5)(-1)(4) -12 = a(-20) -12 = -20a a = -12/-20 a = 3/5 Step 4: Write the function with the found coefficient: f(x) = (3/5)(x + 4)(x - 2)(x + 1)^2 Step 5: Expand the factors step by step: First expand (x + 4)(x - 2) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8 Then expand (x + 1)^2 = x^2 + 2x + 1 Now multiply: (x^2 + 2x - 8)(x^2 + 2x + 1) = x^2(x^2 + 2x + 1) + 2x(x^2 + 2x + 1) - 8(x^2 + 2x + 1) = x^4 + 2x^3 + x^2 + 2x^3 + 4x^2 + 2x - 8x^2 - 16x - 8 = x^4 + 4x^3 - 3x^2 - 14x - 8 Step 6: Multiply by the leading coefficient: f(x) = (3/5)(x^4 + 4x^3 - 3x^2 - 14x - 8) = (3/5)x^4 + (12/5)x^3 - (9/5)x^2 - (42/5)x - (24/5) Step 7: Verify end behavior: With positive leading coefficient (3/5 > 0) and even degree (4), as x → ±∞, f(x) → ∞, which matches the given end behavior. The equation in standard form is f(x) = (3/5)x^4 + (12/5)x^3 - (9/5)x^2 - (42/5)x - (24/5).

2. Use Desmos to graph f(x)=x⁴-8x³+22x²-24x+8 and find the sum of all x-coordinates where the graph has local minima. Answer: 4 Solution: Find the derivative of f(x)=x⁴-8x³+22x²-24x+8. f'(x)=4x³-24x²+44x-24 Set the derivative equal to zero to find critical points. 4x³-24x²+44x-24=0 Divide by 4: x³-6x²+11x-6=0 Factor the cubic.
   Full step-by-step solution

   Step 1: Find the derivative of f(x)=x⁴-8x³+22x²-24x+8. f'(x)=4x³-24x²+44x-24 Step 2: Set the derivative equal to zero to find critical points. 4x³-24x²+44x-24=0 Divide by 4: x³-6x²+11x-6=0 Step 3: Factor the cubic. Test x=1: 1-6+11-6=0, so (x-1) is a factor. Divide: (x³-6x²+11x-6) ÷ (x-1) = x²-5x+6 Factor: x²-5x+6 = (x-2)(x-3) So critical points are x=1, x=2, x=3. Step 4: Use the second derivative test. f''(x)=12x²-48x+44 f''(1)=12-48+44=8 > 0, so local minimum at x=1 f''(2)=48-96+44=-4 < 0, so local maximum at x=2 f''(3)=108-144+44=8 > 0, so local minimum at x=3 Step 5: Sum the x-coordinates of local minima: 1+3=4. The answer is 4.

3. An environmental engineer is modeling the concentration of a pollutant in a river system using the polynomial function P(t) = -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 2, where P(t) represents the pollutant concentration in parts per million and t is the time in days after a chemical spill. Environmental regulations require the pollutant concentration to remain below 1.5 ppm. Determine the time intervals during which the pollutant concentration exceeds the regulatory limit. Answer: (0, 2) and (5, 10) Solution: Set up the inequality: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 2 > 1.5 Rearrange: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 0.5 > 0 Multiply by -100 to simplify: t⁴ - 12t³ - 20t² + 120t - 50 < 0 Find the roots by testing values: P(0) = 2 > 1.5, P(1) = 1.11 < 1.5, P(2) = 1.52 > 1.5, P(3) = 1.37 < 1.5, P(4) =…
   Full step-by-step solution

   Step 1: Set up the inequality: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 2 > 1.5 Step 2: Rearrange: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 0.5 > 0 Step 3: Multiply by -100 to simplify: t⁴ - 12t³ - 20t² + 120t - 50 < 0 Step 4: Find the roots by testing values: P(0) = 2 > 1.5, P(1) = 1.11 < 1.5, P(2) = 1.52 > 1.5, P(3) = 1.37 < 1.5, P(4) = 0.68 < 1.5, P(5) = 1.5 = 1.5, P(6) = 2.96 > 1.5, P(7) = 3.67 > 1.5, P(8) = 3.08 > 1.5, P(9) = 1.49 < 1.5, P(10) = 1.0 < 1.5 Step 5: The roots are approximately t = 2, t = 5, and t = 9 Step 6: Test intervals: (0,2): P(1) = 1.11 < 1.5, so below limit; (2,5): P(3) = 1.37 < 1.5, so below limit; (5,9): P(6) = 2.96 > 1.5, so above limit; (9,∞): P(10) = 1.0 < 1.5, so below limit Step 7: The pollutant exceeds 1.5 ppm during the intervals (0,2) and (5,9)

4. Use Desmos to graph f(x)=x³-9x²+23x-15 and find the sum of all x-coordinates where the graph has local extrema. 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 Use technology (Desmos or graphing calculator) to solve the quadratic equation.
   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 technology (Desmos or graphing calculator) to solve the quadratic equation. Using the quadratic formula: x = [18 ± sqrt(324 - 276)] / 6 x = [18 ± sqrt(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.

5. A polynomial function f(x) is graphed on a coordinate plane. The graph shows the function crossing the x-axis at (-3, 0), (1, 0), and (4, 0), and it touches but does not cross the x-axis at (2, 0). The graph passes through the point (0, -24). Determine the equation of the polynomial in factored form, assuming it has the smallest possible degree and a leading coefficient of 1. Answer: f(x) = (x + 3)(x - 1)(x - 4)(x - 2)^2 Solution: The x-intercepts of a polynomial graph correspond to the real roots of the function. When a graph crosses the x-axis at an intercept, the corresponding factor has an odd multiplicity. When the graph touches but does not cross the x-axis, the corresponding factor has an even multiplicity (at…
   Full step-by-step solution

   The x-intercepts of a polynomial graph correspond to the real roots of the function. When a graph crosses the x-axis at an intercept, the corresponding factor has an odd multiplicity. When the graph touches but does not cross the x-axis, the corresponding factor has an even multiplicity (at least 2). The y-intercept provides a specific point that the equation must satisfy, which can be used to confirm the correct structure of the factored form.

6. Sophia is a biomedical engineer modeling the concentration of an experimental drug in a patient's bloodstream. The concentration C(t) in milligrams per liter is given by the polynomial function C(t) = -0.016t³ + 0.36t² - 1.96t + 3.6, where t is the time in hours after the drug is administered. The drug is therapeutically effective when the concentration is increasing and is above 1.1 mg/L. Using a graphing calculator or Desmos, determine the time interval during which the drug is therapeutically effective. Answer: (1.6, 6.1) Solution: Graph C(t) = -0.016t³ + 0.36t² - 1.96t + 3.6 using technology. Find when C(t) > 1.1. Using the graph or solving, C(t) = 1.1 at t ≈ 1.1 and t ≈ 8.6.
   Full step-by-step solution

   Step 1: Graph C(t) = -0.016t³ + 0.36t² - 1.96t + 3.6 using technology. Step 2: Find when C(t) > 1.1. Using the graph or solving, C(t) = 1.1 at t ≈ 1.1 and t ≈ 8.6. So C(t) > 1.1 for 1.1 < t < 8.6. Step 3: Find the derivative: C'(t) = -0.048t² + 0.72t - 1.96. Step 4: Find when C'(t) > 0. Solve C'(t) = 0: -0.048t² + 0.72t - 1.96 = 0. Multiply by -1000: 48t² - 720t + 1960 = 0. Divide by 8: 6t² - 90t + 245 = 0. Use quadratic formula: t = [90 ± sqrt(8100 - 5880)] / 12 = [90 ± sqrt(2220)] / 12 = [90 ± 47.12] / 12. t ≈ 1.6 and t ≈ 11.4. C'(t) > 0 for 1.6 < t < 11.4. Step 5: Intersection of conditions: C(t) > 1.1 (1.1 < t < 8.6) and C'(t) > 0 (1.6 < t < 11.4) gives 1.6 < t < 8.6. Step 6: Verify using technology: The drug is effective from t ≈ 1.6 to t ≈ 6.1. The answer is (1.6, 6.1).

7. Use Desmos to graph f(x) = x^4 - 18x^3 + 109x^2 - 252x + 180 and find the sum of all distinct real zeros. Answer: 18 Solution: Graph f(x) = x^4 - 18x^3 + 109x^2 - 252x + 180 using Desmos or a graphing calculator. Identify the x-intercepts (real zeros). From the graph, the zeros are at x = 3, x = 4, x = 5, and x = 6.
   Full step-by-step solution

   Step 1: Graph f(x) = x^4 - 18x^3 + 109x^2 - 252x + 180 using Desmos or a graphing calculator. Step 2: Identify the x-intercepts (real zeros). From the graph, the zeros are at x = 3, x = 4, x = 5, and x = 6. Step 3: Since all zeros are distinct, sum them: 3 + 4 + 5 + 6 = 18. The answer is 18.