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3.1 #2 In exercises 1 – 4, you’re given an equation and the graph of a quadratic function. Do each of the following: The graph needed is in a separate pdf file sent. #3.1 #2 on page 303 a) Give the domain and range b) Give the coordinates of the vertex c) Give the equation of the axis d) Find the y-intercept e) Find the x-intercepts f x x 5 4 2 Ans: a) domain: (-∞,∞) Range: [-4, ∞) b) (5,-4) c) x = 5 d) (0, 21) e) (3, 0) and (7, 0) Note the arrows. Since you can take any value for x, the domain is all real numbers (-∞,∞). Your range is correct, except that since -4 is included use a bracket. The vertex is the minimum or maximum point at (5,-4). The equation of the axis is the vertical line containing the vertex. The y-intercept is the point on the graph with x=0. This isn’t shown on this graph, but it does exist: y = (0-5)2-4 = 25-4 = 21. For the x-intercepts, give the entire point. 3.1 #18 Graph each quadratic function. Give the vertex, axis, domain, and range. f x 3x 2 1 2 See Attached for Graph. Ans: Vertex: (2, 1) Axis: x = 2 Domain: (-∞, ∞) Range: (-∞, 1] By axis, I believe they mean the equation of the axis of symmetry. This is the line containing the vertex, x = 2. Since any number can be substituted for x, the domain is all real numbers (-∞,∞). You’re range was correct, but always write it from smaller to larger, and since 1 is included use a bracket on that side. 3.1 #24 Graph each quadratic function. Give the vertex, axis, domain, and range. f x 3x 2 24x 46 See Attached for Graph. Ans: Vertex: (4, 2) Axis: x = 4 Domain: (-∞,∞) Range: (-∞, 2] 3.1 #62 College Freshmen Pursuing Medical Degrees. Between 1996 and 2002, the percent of college freshmen who planned to get a professional degree in a medical field (that is, M.D., D.O., D.D.S., or D.V.M.) can be modeled by f x .07726x 2 .8004x 10.84 , where x = 0 represents 1996. (Source: Higher Education Research Institute, UCLA, 2002.) Based on this model, in what year did the percent of freshmen planning to get a medical degree reach its minimum? The minimum is the vertex at x = -b / (2a) = - (-0.8004) / [2 (0.07726)] = 5.2 This is in year 1996 + 5 = 2001 3.2 #2 Use synthetic division to perform each division. x 3 4x 2 5x 44 x6 Synthetic division is difficult to explain step by step here, but a good website explaining it is here: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg _tut37_syndiv.htm I will write the complete steps here: -6| 1 1 4 -6 -2 -5 12 7 44 -42 2 So the answer is x2 - 2x + 7 + 2/(x+6) 3.2 #14 Use synthetic division to perform each division. x 4 5x 3 6x 2 2x x 1 -1| 1 1 5 -1 4 -6 -4 -10 2 10 12 0 -12 -12 So the answer is x3 + 4x2 – 10x + 12 – 12/(x+1) 3.2 #20 Express f x in the form f x x k q x r for the given value of k. f x 2x 3 3x 2 16x 10 ; k = -4 -4| 2 2 3 -8 -5 -16 20 4 10 -16 -6 Since x – (-4) = x + 4, f(x) = (x + 4) (2x2 - 5x + 4) - 6 3.2 #28 For each polynomial function, use the remainder theorem and synthetic division to find f k . f x x 2 4x 5 ; k = 3 3| 1 1 -4 3 -1 5 -3 2 Since the remainder is 2, f(3) = 2. 3.2 #34 For each polynomial function, use the remainder theorem and synthetic division to find f k . f x x 4 6x 3 9x 2 3x 3 ; k = 4 4| 1 1 6 4 10 9 40 49 3 196 199 -3 796 793 Since the remainder is 793, f(4) = 793. 3.2 #44 Use synthetic division to decide whether the given number k is a zero of the given polynomial function. If it is not, give the value of f k . f x 2x 3 9x 2 16x 12 ; k = 1 1| 2 2 9 2 11 -16 11 -5 12 -5 7 So 1 is not a zero of f since f(1) is not 0. f(1) = 7 3.3 #14 Use the factor theorem and synthetic division to decide whether the second polynomial is a factor of the first. 2x 3 x 2 -1| 2 2 0 -2 -2 x 1 1 2 3 2 -3 -1 Since the remainder is not zero, x+1 is not a factor of 2x3 + x + 2. 3.3 #18 Factor f x into linear factors given that k is a zero of f x . f x 2x 3 3x 2 5x 6 ; k=1 Since 1 is a zero of the function, x – 1 is a factor. Divide it out. 1| 2 2 -3 2 -1 -5 -1 -6 6 -6 0 The quotient is 2x2 – x – 6. Factoring this, we get (x - 2)(2x + 3) So the complete factorization is (x - 1) (x - 2) (2x + 3) 3.3 #30 For each polynomial function, one zero is given. Find all others. f x x 3 4 x 2 5 ; 1 Divide out x – 1: 1| 1 1 4 1 5 0 5 5 -5 5 0 The quotient is x2 + 5x + 5. This has zeros 5 5 2 4(1)(5) 5 5 = -3.168, -1,382 x 2(1) 2 3.3 #34 For each polynomial function, one zero is given. Find all others. f x x 4 10x 3 27x 2 10x 26 ; i Divide out x – i i| 1 1 10 i 10+i 27 10 -1+10i -10+26i 26+10i 26i 26 -26 0 So we need to find the other zeros of x3 + (10+i)x2 + (26+10i)x + 26i The complex conjugate of a solution is often a solution, so let’s try –i. -i| 1 1 10+i -i 10 26+10i 26i -10i -26i 26 0 So –i is also a solution. We are left with x2 + 10x + 26. 10 10 2 4(1)( 26) 10 4 5 i 2(1) 2 So the solutions are ± i and -5 ± i x 3.3 #52 Find a polynomial function of degree 3 with real coefficients that satisfies the given conditions. Zeros of 2, -3, and 5: f 3 6 f(x) = a (x - 2) (x + 3) (x – 5) for some value of a. Plug in f(3) to find a: a (1) (6) (-2) = 6, -12 a = 6, a = -1/2. So f(x) = -1/2 (x-2) (x+3) (x-5) = -1/2 x3 + 2x2 + 11/2 x – 15 after multiplying it out. 3.3 #54 Find a polynomial function of degree 3 with real coefficients that satisfies the given conditions. Zero of 4 having multiplicity 2 and zero of 2 having multiplicity 1; f 1 18 Follow the same procedure. f(x) = a (x-4)(x-4)(x-2) a (-3) (-3) (-1) = -18, a = 2 f(x) = 2 (x-4)(x-4)(x-2) = 2x3 – 20x2 + 64x - 64 3.3 #56 Find a polynomial function of least degree having only real coefficients with zeros as given. 7 2i and 7 2i (x – 7 + 2i) (x – 7 – 2i) = x2 – 14x + 53 3.4 #12 Sketch the graph of each polynomial function. f x 5 5 x 4 See Attached for Graph. Looks good. 3.4 #18 Sketch the graph of each polynomial function. f x 1 x 34 3 See Attached for Graph. Looks good. 3.4 #32 Graph each polynomial function. Factor first if the expression is not in factored form. f x x 2 x 1x 1 See Attached for Graph. Looks good. 3.4 #44 Use the intermediate value theorem for polynomials to show that each polynomial function has a real zero between the numbers given. f x 3x 2 x 4 ; 1 and 2 f(1) = 3(1) – 1 – 4 = -2 f(2) = 3(4) – 2 – 4 = 6 Since f(x) is continuous and 0 lies between -2 and 6, there is a zero of f(x) between 1 and 2. 3.4 #54 Show that the real zeros of each polynomial function satisfy the given conditions. f x 2x 5 x 4 2x 3 2x 2 4x 4 ; no real zero greater than 1 For f(1) = 2 – 1 + 2 – 2 + 4 – 4 = 1 For x > 1, each positive term is greater than the next negative term, 2x5 > x4, 2x3 > 2x2, and 4x > 4, so the function f(x) > 0 for all x>1. Therefore the function has no real zeros greater than 1. 3.6 #14 Solve each variation problem. If m varies jointly as z and p, and m = 10 when z = 3 and p = 5, find m when z = 5 and p = 7. Since they vary jointly, m = k z p for some value of k. First, find k. 10 = k * 3 * 5, so k = 2/3. m = 2/3 z p = 2/3 (5) (7) = 70/3 = 23 1/3 3.6 #28 Current in a Circuit. The current in a simple electrical circuit varies inversely as the resistance. If the current is 50 amps when the resistance is 10 ohms, find the current if the resistance is 5 ohms. I = k/R for some value of k 50 = k/10, so k = 500 I = 500/R = 500/5 = 100 amps 3.6 #40 Long Distance Phone Calls. The number of long-distance phone calls between two cities in a certain time period varies directly as the populations p1 and p2 of the cities, and inversely as the distance between them. If 10,000 calls are made between two cities 500 mi apart, having populations of 50,000 and 125,000, find the number of calls between two cities 800 mi apart, having populations of 20,000 and 80,000. N = k p1 p2 / d 10000 = k (50000) (125000) / 500, k = 0.0008 N = 0.0008 p1 p2 / d = 0.0008 (20000) (80000) / 800 = 1600 3.6 #52 Concept Check. Work each problem. What happens to y if y is inversely proportional to x, and x is tripled? y=k/x If x is tripled, y = k / (3x) = 1/3 (k / x) so y is reduced to one-third of its original value.