course Mth158 This is much easier than transformations.
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16:57:12 4.1.42 (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
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RESPONSE --> This graph opens up because a=1 which is greater than 0. The axis of symmetry is = to x=-b/2a, which is equal to 1. This is also x int. To find y, I evaluate f(1)= -4, so the vertex is located at (1,-4) The domain of this function is the set of all x such that x is not equal to 3 or -1. The range is all real #.
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16:57:52 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 see where I messed up on x and y int. Otherwise, I think I am okay.
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17:00:37 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 --> I think this would be the equation resulting from (x+5)(x-3)= x^2+2x-15 a=1= x^2+2x+15 a=2= 2x^2+4x-30 a=-2= -2x^2-4x+30 a=5 = 5x^2+10x-75
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17:00:42 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
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17:02:59 Does the value of a affect the location of the vertex?
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RESPONSE --> I don't think so. vertex= x=-b/2a, if x int. not affected, the vertex remains the same
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17:03:42 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 --> A longer explanation I am still trying to grasp. I know what I understand, it is just hard to put into technical terms.
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17:04:40 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 --> Something was skipped here. Looking at this, I do see how a change in a does affect the y coordinates, and therefore, the vertices.
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17:07:00 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 --> 2000m of fence x feet along the highway. The other two sides would be 2000-x. The fourth side is not enclosed. The three remaining sides would be x, and 2000-x/ 2. I am not sure where to go. The area would normally be the total of the four sides (in this case, three sides).
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17:09:13 ** 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, I see how to do the area now. It is x(2000-x)/2= -x^2/2+1000x. For some reason, I was thinking there was more to it than that. Because this is a downward parabola, the maximum value is also the greatest area that can be enclosed= x=-b/2a= 1000
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17:14:22 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 --> Area of a rectangle is l*w= x(10-x)= -x^2+10x. This is a downward opening parabola because the coefficient of highest term is negative. a=-1 b=10 Vertex= x=-b/2a= -10/-2=5. Take 5 and plug into original area equation= -x^2+10x= (-5)^2+10(5)= 25+50= 75 units is the largest area that can be enclosed by the rectangle.
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17:16:17 ** 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 messed up the math somehow! I see how I did it- I took negative 5 and squared it when I should have squared five and THEN made that result negative and added it to 10*5. I see now.
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