Math Quiz 2.1 Review Solution Key

2.1-2.5 Quiz Review Name: Hour: Honors Pre-Calculus Find the vertex, x-intercepts and y-intercept of the function. 1. f(x) = -3x² + 9x + 5 Vertex: (3/2, 47/4) X-intercepts: (9±√141)/6 Y-intercept: (0, 5) 2. f(x) = (x + 3/2)² + 1 Vertex: (-3/2, 1) X-intercepts: None Y-intercept: (0, 13/4) Convert the function to vertex form. 3. f(x) = 4x² - 4x + 21 f(x) = 4(x - 1/2)² + 20 4. f(x) = -2x² + 16x - 31 f(x) = -2(x - 4)² - 1 Write the standard form of the quadratic function that has the indicated vertex and whose graph passes through the given point. 5. Vertex: (-4, -1), Point: (-2,   4) f(x) = (5/4)(x + 4)² - 1 6. Vertex: (-1/4, 3/2), Point: (-2, 0) f(x) = (-24/49)(x + 1/4)² + 3/2 7. A textile manufacturer has daily production costs of C = 10000 - 110x + 0.45x², where C is the total cost and x is the number of units produced. Determine how many units should be produced each day to yield a minimum cost?   What is the minimum cost? 122 units produced $3277.80 Minimum Cost Problem 8: Find two positive numbers whose sum is 100 and the sum of whose squares is a minimum. Let x be one number. The other number is 100 - x. The sum of their squares is: x² + (100-x)² To minimize this sum, we can find the vertex of the parabola represented by this quadratic equation. Expanding the equation: x² + 10000 - 200x + x² = 2x² - 200x + 10000 The x-coordinate of the vertex is: x = -b / 2a = 200 / (2*2) = 50 So, the two numbers are 50 and 50. Problem 9: A football is kicked, and its height, f(x), in feet, can be modeled by f(x) = -0.01x² + 1.18x + 2, where x is the horizontal distance in feet from the point of impact. a) What distance from the kicker's foot does the ball reach its maximum height? The maximum height occurs at the vertex of the parabola. The x-coordinate of the vertex is: x = -b / 2a = -1.18 / (2 * -0.01) = 59 feet b) What is the maximum height reached by the ball? f(59) = -0.01(59)² + 1.18(59) + 2 = 36.81 feet c) How far must the nearest defensive player, who is 6 feet from the kicker, reach to block the punt? f(6) = -0.01(6)² + 1.18(6) + 2 = 8.72 feet d) If the ball is not blocked, how far down the field will it go before hitting the ground? We need to find the x-intercepts of the quadratic equation: -0.01x² + 1.18x + 2 = 0 Using the quadratic formula, we get: x ≈ 119.67 feet Problem 10: A farmer has 2400 feet of fencing to fence off a rectangular field bordering a river. a) Find a function that models the area of the field in terms of one of its sides. Let x be the side perpendicular to the river. The side parallel to the river will be (2400 - 2x) / 2 = 1200 - x. Area = x * (1200 - x) = -x² + 1200x b) What are the dimensions of the field that would produce the maximum area? The x-coordinate of the vertex of the parabola -x² + 1200x is: x = -b / 2a = 1200 / 2 = 600 So, the dimensions are 600 feet by 1200 - 600 = 600 feet. c) What is the maximum area? Maximum area = 600 * 600 = 360,000 square feet Find all the real zeros of the polynomial function (You do not need the Rational Root Theorem for this! This is from Section 2.2!) Determine the multiplicity of each zero. Use the leading coefficient test and end behavior to sketch the graph. 11. f(x) = x³   + 2x² - 9x - 18 x²(x+2) - 9(x+2) (x²+9)(x+2) (x+3)(x-3)(x+2) Zeros: -2, ±3 End behavior: x→-∞, f(x)→-∞ x→∞, f(x)→∞ 12. f(x) = -x²(x+3)³(2-x) Zeros: -3 (mult. 3), 0 (mult. 2), 2 End behavior: x→-∞, f(x)→∞ x→∞, f(x)→∞ 13. f(x) = x⁵ - 6x³ + 9x x(x⁴-6x²+9) x(x²-3)² x=0 x²-3=0 x²=3 x=±√3 (mult. 2) Zeros: ±√3 (mult. 2), 0 End behavior: x→-∞, f(x)→-∞ x→∞, f(x)→∞ 14. f(x) = 1/3(5x²+8x-4) y = 1/3(5x-2)(x+2) Zeros: 2/5, -2 End behavior: x→-∞, f(x)→∞ x→∞, f(x)→∞ For #15-18, find a polynomial function with real coefficients that satisfies the given information. Write your answer in expanded form (do not leave it in factored form).   The polynomial is cubic, tangent to the x-axis at -2, and contains the points (3,0) and (0,-2).   y = a(x+2)²(x-3) -2 = a(4)(-3) -2 = -12a a = 1/6 y = 1/6(x³+x²-8x-12) y = 1/6x³ + 1/6x² - 4/3x - 2 End behavior: x→-∞, f(x)→-∞ x→∞, f(x)→∞ 16. The polynomial has zeros: -4 with multiplicity 2, 1 + √2 with multiplicity 1, and falls to the right and left. (x+4)²(x-(1+√2))(x-(1-√2)) (x²+8x+16)[(x-1)-√2][(x-1)+√2] (x²+8x+16)[(x-1)²-2] (x²+8x+16)(x²-2x-1) x⁴+6x³-x²-40x-16 17. The polynomial has zeros -1 with multiplicity 2 and 2 + 5i. (x+1)²(x-(2+5i))(x-(2-5i)) (x²+2x+1)[(x-2)-5i][(x-2)+5i] (x²+2x+1)[(x-2)²-25i²] (x²+2x+1)(x²-4x+4+25) (x²+2x+1)(x²-4x+29) x⁴-2x³+22x²+54x+29 18. The polynomial has zeros -1, 2, and i and contains the point (1,8). (x+1)(x-2)(x-i)(x+i) y = a(x⁴-x³-x²-x-2) 8 = a(1-1-1-1-2) 8 = a(-4) a = -2 y = -2x⁴+2x³+2x²+2x+4 19. Given f(x) = 3x³ + 2x² - 19x + 6 a) Verify that (x + 3) is a factor of the function. Show your work below. | 3 | 2 -19 6 | |---|---|---|---| | | 9 33 -42 | | 3 | 11 -14 -36 | b) Find the remaining factors of f(x). 3x² - 7x + 2 (3x-1)(x-2) c) Use your results from part b to show the complete factorization of f(x). (x + 3)(3x - 1)(x - 2) d) List all real zeros of f(x). -3, 1/3, 2