course Mth 158 x՞TK[߃assignment #028
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06:17:30 4.1.42 (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
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RESPONSE --> The graph opens upward, vertex is (3,-2), axis of symmetry is 3, x-intercepts (1,0) (5,0) Domain x>= -2 confidence assessment: 2
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06:18:36 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 --> I tried to use just the graph which was wrong..I think I understand. self critique assessment: 3
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06:29:04 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 --> Book problem was intercepts -3 and 1 f(x)=1(x+3)(x-1)=x^2+2x-3 f(x)=2(x+3)(x-1)=2x^2-4x+6 f(x)=-2(x+3)(x-1)=-2x^2-4x+6 f(x)=5(x+3) (x-1)=5x^2+10x-15 confidence assessment: 3
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06:29:20 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 --> It looks like the same form. self critique assessment: 3
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06:30:39 Does the value of a affect the location of the vertex?
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RESPONSE --> no the vertex will remain in the same place halfway between the x-intercepts confidence assessment: 1
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06:32:50 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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06:34:12 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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06:44:39 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 --> two sides of the rectangle would =x and the other two sides would equal 2000-x x(2000-x)/2=2000x-x^2/a=-x^2/2+1000x vertex at x=-1000/2(-1/2) If x=1000 -x^2/2+1000x=-1000^2/2+1000(1000)=500,000 So 500,000 would be the most he could fence without fencing the side closest to the highway. confidence assessment: 2
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06:44:54 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 --> Looks correct. self critique assessment: 3
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06:53:31 4.1.101 (was 4.1.80). A rectangle has one vertex on the line y=10-x, x>0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area that can be enclosed by the rectangle.
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RESPONSE --> x(10-x)=10x-x^2 or -x^2+10x vertex =-10/2 (-1)=5 -(5^2)+10 (5)=-25+50=25 25 is the maximum A of the rectangle. confidence assessment: 2
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06:53:40 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 --> ok self critique assessment: 3
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