course ÎVÌ”‹—¾®®¢ÒãßÍ깿E¼€¶š¤óassignment #028
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12:54:36 4.1.42 (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
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RESPONSE --> a = 1, b = -2, c = -3 Since a > 0, graph opens up -(-2/2(1)) = 1, 1^2 - 2(1) - 3 = -4, vertex is (1,-4) axis of symmetry is x = 1 y-intercept is -3 x-intercepts... (-2)^2 - 4(-3) = 16, 16 >0 so there will be 2 x-intercepts. x^2 - 2x - 3 = 0, (x-3)(x+1) = 0, x-ints are 3 and -1 domain is the set of all real numbers range is y | y >= -4 The function is increasing on (1, infinity) and decreasing on (-infinity, 1) confidence assessment: 2
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12:55:01 The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3. The graph of this quadratic function will open upwards, since a > 0. The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4). The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the quadratic formula. Here will will factor to get (x - 3) ( x + 1) = 0 so that x - 3 = 0 OR x + 1 = 0, giving us x = 3 OR x = -1. So the x intercepts are (-1, 0) and (3, 0). The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3).
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RESPONSE --> ok self critique assessment: 3
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22:02:56 4.1.67 (was 4.1.50). What are your quadratic functions whose x intercepts are -5 and 3, for the values a=1; a=2; a=-2; a=5?
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RESPONSE --> This is actually problem number 68 (a) f(x) = (1)(x+5)(x-3), x^2 + 2x - 15 f(x) = (2)(x+5)(x-3), 2x^2 + 4x - 30 f(x) = (-2)(x+5)(x-3), -2x - 4x + 30 f(x) = (5)(x+5)(x-3), 5x + 10x - 75 (b) y-intercepts are multiplied by a. x-intercepts remain the same. (c) The axis of symmetry is unaffected by the value of a. (d) y-coordinate of vertex is multiplied by the value of a (e) the x-coordinate of the vertx is the midpoint of the x-intercepts. confidence assessment: 2
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22:06:29 Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3). If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15. If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30. If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30. If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.
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RESPONSE --> ok self critique assessment: 3
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22:07:11 Does the value of a affect the location of the vertex?
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RESPONSE --> Can you just take the answers from what I already typed in the previous section? :) confidence assessment: 3
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22:07:29 In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1.
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RESPONSE --> ok self critique assessment: 3
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22:07:52 The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following: For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12. For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24. For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24. For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60. So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60).
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RESPONSE --> ok self critique assessment: 3
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22:30:01 4.1.78 (was 4.1.60). A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?
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RESPONSE --> A(x) = (2000 - 2x)x, 2000x - 2x^2, -2x^2 + 2000x x = - (2000 / 2(-2)) = - (2000 / -4) = 500 -2(500)^2 + 2000(500) = -50000+ 1000000 = 50,000 square meters. confidence assessment: 3
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22:30:15 If the farmer uses 2000 meters of fence, and fences x feet along the highway, then he have 2000 - x meters for the other two sides. So the dimensions of the rectangle are x meters by (2000 - x) / 2 meters. The area is therefore x * (2000 - x) / 2 = -x^2 / 2 + 1000 x. The graph of this function forms a downward-opening parabola with vertex at x = -b / (2 a) = -1000 / (2 * -1/2) = 1000. At x = 1000 the area is -x^2 / 2 + 1000 x = -1000^2 / 2 + 1000 * 1000 = 500,000, meaning 500,000 square meters. Since this is the 'highest' point of the area vs. dimension parabola, this is the maximum possible area. **
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RESPONSE --> ok self critique assessment: 3
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09:33:40 The dimensions of the rectangle are x and y = 10 - x. So the area is area = x ( 10 - x) = -x^2 + 10 x. The vertex of this rectangle is at x = -b / (2 a) = -10 / (2 * -1) = 5. Since the parabola opens downward this value of x results in a maximum area, which is -x^2 + 10 x = -5^2 + 10 * 5 = 25. **
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RESPONSE --> I accidentally clicked next question again... but I'll just ignore that side and type in my own answer... A(x) = x(10-x) = -x^2 + 10x a=-1, b =10, c =0 x = - (b/2a) = - (10/2(-1)) = 5 a(5) = -25 + 50 = 25 25 units^2 is the largest area that can be enclosed. self critique assessment: 2
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