course Mth 163 ͱæÒ¨S³þXÇoÙRË΃÷N„ª¶÷ÍkÛÊêñúiÖÏassignment #003
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08:59:53 `q001. Note that this assignment has 6 questions The function y = a x^2 + b x + c takes the value y = 0 when x = [ -b + `sqrt(b^2 - 4 a c ] / (2 a) or when x = [ -b - `sqrt(b^2 - 4 a c ] / (2 a). For the function y = - 0.45833 x^2 + 5.33333 x - 6.875, which you obtained as a quadratic model of the points (1, -2), (3, 5) and (7, 8) in the preceding assignment, find the values of x for which y = 0. Compare to the estimates you made from the graph through (1,-2), (3, 5) and (7, 8) in Assignment 1.
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RESPONSE --> To find the values for x we substitute the values for a, b, and c into the function: y = a x ^ 2 + b x + c So for the first function: x = [ -5.33333 + 'sqrt[5.33333^2 - 4(-0.45833*-6.875)] / 2(-0.45833) = x = [ =5.33333 + 3.97987] / -0.9166 = x = -1.35346 / -0.9166 = 1.47661 The second version is: x = [-5.33333 - 'sqrt'[5.33333^2 - 4(-0.45833*-6.875)] / 2(-0.45833) = x = [-5.33333 - 3.97987] / 0.9166 = -9.3132 / 0.9166 = -10.16059 The values for x are 1.47661 and -10.16059 confidence assessment: 2
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09:02:15 For the function y = - 0.45833 x^2 + 5.33333 x - 6.875 we have a = -0.45833, b = 5.33333 and c = -6.875. The quadratic formula therefore tells us that for our function we have y = 0 when x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638 and when x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006.
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RESPONSE --> My decimal points were of just a few points and I did get -10.16059 for the second value. I see now that I kept the last part of the function positive, 2*(-0.45833). That would've made a big difference and I overlooked that. self critique assessment: 2
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09:14:34 `q002. Extend the smooth curve in your sketch to include both points at which y = 0. Estimate the x value at which y takes its maximum value.
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RESPONSE --> I'm not sure how to go about this, if I graphed this out correctly though I could estimate that the x value at which y takes its maximum value would be 7. The graph I have is a parabola that opens downward. confidence assessment: 0
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09:18:17 Your graph should clearly show how the parabola passes through the x axis at the points where x is approximately 1.5 (corresponding to the more accurate value 1.47638 found in the preceding problem) and where takes is a little more than 10 (corresponding to the more accurate value 10.16006 found in the preceding problem). The graph of the parabola will peak halfway between these x values, at approximately x = 6 (actually closer to x = 5.8), where the y value will be between 8 and 9.
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RESPONSE --> I thought there was more to the answer than that, but I was closer than I thought. My graph did pass through the 1.5 and 10.16 points on the x-axis. self critique assessment: 1
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09:21:37 `q003. For the function of the preceding two questions, y will take its maximum value when x is halfway between the two values at which y = 0. Recall that these two values are approximately x = 1.48 and x = 10.16. At what x value will the function take its maximum value? What will be this value? What are the coordinates of the highest point on the graph?
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RESPONSE --> Once again, according to the graph I have sketched out the function will take its maximum value when x is equal to 10.16. The highest point on this graph would be (7,8). confidence assessment: 1
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09:27:24 The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82. At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx.. Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64).
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RESPONSE --> I understand how to get to those answers. I was just trying to use the coordinates I aleady had for the graph. It makes sense to find the halfway point between 1,48 and 10.16 and then substitute that value into the function to find the y value. self critique assessment: 2
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09:41:41 `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a). At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value?
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RESPONSE --> The vertex is sometimes the highest point. In this graph of the parabola to find the vertex we use x = -b / (2a). If we substitute the values for a and b into this we find: x = -5.33333 / (2*-0.45833) = -5.33333/-0.91666 x = 5.81822 Now to find the corresponding y value we substitute the value for x into the function Y = -0.458333 x^2 + 5.33333 x - 6.875 y = -0.458333 * (5.81822^2) + 5.33333*5.81822 - 6.875 y = -0.45833 * 33.85168 + 31.03049 - 6.875 y = -15.51524 + 24.15549 y = 8.64025 confidence assessment: 1
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09:42:58 In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain x value at which function is maximized: x = -b / (2a) = 5.33333 / (2 * -0.45833) = 5.81818. To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024. Thus the vertex of the parabola lies at (5.81818, 8.64024).
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RESPONSE --> I rounded the values off to the 5th decimal points, not sure if that matters all that much. I arrived at the same answers for this, with the vertex being (5.81822, 8.64025). self critique assessment: 1
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20:37:54 `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402). What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex? What is the value of y corresponding to each of these x values? By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function?
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RESPONSE --> I'm not sure how to go about this problem but will give it a try. The value of x at a point of the parabola which lies 1 unit to the right of the vertex would be 6.8182. The value of x at a point on the parabola which lies one unit to the left of the vertex would be 4.8182. The corresponding y values for each of the x values would be: y = -0.45833(6.8182^2) + 5.33333*6.8182 - 6.875 y = -0.45833*46.48785 + 36.36371 - 6.875 y = -21.30678 + 29.48871 y = 8.18193 Y = -0.45833(4.8182^2) + 5.33333(4.8182)-6.875 y = -0.45833(23.21505) + 18.82205 y = -10.64015 + 18.82205 y = 8.1819 When x = 6.8182, y = 8.18193 When x = 4.8182, y = 8.1819 Both of the new y values have a differences of 0.4583 from the y value at the vertex. confidence assessment: 0
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20:39:49 03-28-2008 20:39:49 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
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NOTES -------> I really didn't think that's how the problem would be solved. Both my points for y came out to be around 8.1818, depending on what decimal point you round off at.
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20:39:50 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
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RESPONSE --> self critique assessment: 1
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20:47:03 `q006. In the preceding problem we saw an instance of the following rule: The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola. In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points. What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex? Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis?
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RESPONSE --> I believe that I am lost on this problem, I'm not sure if I must first solve the function with y = 0 and get the x value then. Or if I just choose numbers to plug into the function. If it was to first have y = 0, I'm still not sure if I would know how to solve the function. confidence assessment: 0
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20:49:48 03-28-2008 20:49:48 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
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NOTES -------> I forgot about the x = -b /2a to solve for the vertex. It was that first part that kept me confused. Once you find the x value for the vertex the rest of the problem is pretty easy, just plugging numbers in to find the other values.
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20:49:49 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
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RESPONSE --> self critique assessment: 1
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