course Mth 163 022. `query 22 Precalculus I 11-07-2007
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12:07:33 Explain why the function y = x^-p has a vertical asymptote at x = 0.
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RESPONSE --> Because as x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number that gets constantly closer to 0 gave me a result with a much larger magnitude. There is no limit to how close x can get to 0, therefore, is no limit to how many times x^p can divide into 1. This results in y = x^p values that get close to infinite distance from the x -axis. The graph therefore, obviously, approaches a vertical limit. confidence assessment: 2
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12:25:50 ** x^-p = 1 / x^p. As x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to 0, so there is no limit to how many times x^p can divide into 1. This results in y = x^p values that approach infinite distance from the x axis. The graph therefore approaches a vertical limit. **
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RESPONSE --> When I reviewed the correct answer, it was a lot like my answer. I understand this question when I sit down and really think about the steps and when I take my time to make sure I do not mess up my answer in some way. self critique assessment: 3
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12:48:50 Explain why the function y = (x-h)^-p has a vertical asymptote at x = h.
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RESPONSE --> Because as x gets closer to h, (x-h)^p gets closer to 0. Dividing 1 by a number which gets very close to 0 which gave me a result with a much larger magnitude. There is no limit to how close x can get to h, therefore, there is no limit to how many times (x-h)^p can divide into 1. This results in y = (x-h)^p values that get close to an infinite distance from the x axis as x approaches h. The graph therefore approaches a vertical limit. confidence assessment: 3
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13:12:18 ** (x-h)^-p = 1 / (x-h)^p. As x gets closer to h, (x-h)^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to h, so there is no limit to how many times (x-h)^p can divide into 1. This results in y = (x-h)^p values that approach infinite distance from the x axis as x approaches h. The graph therefore approaches a vertical limit. This can also be seen as a horizontal shift of the y = x^-p function. Replacing x by x - h shifts the graph h units in the x direction, so the asymptote at x = 0 shifts to x = h. **
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RESPONSE --> This was the same process as the question before. I like having multiple problems that involve the same steps because if I get it wrong the first time, I can correct it the second time or if I know how to do the problem from the beginning I can use the alike questions to master the steps of that particular problem. self critique assessment: 3
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13:36:17 Explain why the function y = (x-h)^-p is identical to that of x^-p except for the shift of h units in the x direction.
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RESPONSE --> The x value at which a y value occurs is shifted h number of units. For example, y = x^p equals zero when x = 0, but y = (x - h)^p equals zero when x = h confidence assessment: 3
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13:50:29 STUDENT ANSWER: You end up with the exact same y values but at the different position of x changed by the h value. INSTRUCTOR COMMENT: Good start. More specifically the x value at which a given y value occurs is shifted h units, so that for example y = x^p is zero when x = 0, but y = (x - h)^p is zero when x = h.
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RESPONSE --> I thought this was an easy one. self critique assessment: 3
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13:59:28 Give your table (increment .4) showing how the y = x^-3 function can be transformed first into y = (x - .4) ^ -3, then into y = -2 (x - .4) ^ -3, and finally into y = -2 (x - .4) ^ -3 + .6.
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RESPONSE --> confidence assessment:
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14:16:02 table has each transformation across the top with beginning x value in first column then each change to x to get the y values in resulting columns
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RESPONSE --> I didn't understand the question, but I got it now. I was just supposed to explain how the graph would look. self critique assessment: 3
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14:22:48 ** The table is as follows (note that column headings might not line up correctly with the columns): x y=x^-3 y= (x-.4)^-3 y= -2(x-.4)^-3 y= -2(x-.4)^-3 +.6 0.8 1.953 15.625 31.25 31.85 0.4 15.625 div/0 0 0.6 0 div/0 -15.625 -31.25 -30.65 -0.4 -15.625 -1.953 3.906 4.506 -0.8 -1.953 -0.579 1.158 1.758
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RESPONSE --> I got it. After going back and reviewing the question and working out my own table, I see how the conclusion is reached. self critique assessment: 3
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14:39:11 Explain how your table demonstrates this transformation and describe the graph that depicts the transformation.
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RESPONSE --> After constructing my table, I get y = x^-3 transforms into y = (x - .4)^-3, which shifts the basic point .4 units to the right. The vertical asymptote at the y axis (x = 0) shifts to the vertical line x = .4. The x axis is a horizontal asymptote. y = -2 (x - .4)^-3 vertically stretches the graph by a factor of ( -2), moving every point double as far from the x axis as well as moving it to the opposite side of the x axis. This leaves the vertical line x = .4 as a vertical asymptote. The x axis remains a horizontal asymptote. y = -2 ( x - .4)^-3 + 6 vertically shifts the graph up 6 units. This raises the horizontal asymptote to x = 6. confidence assessment: 2
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14:47:42 y = x^-3 transforms into y = (x - .4)^-3, shifting the basic points .4 unit to the right. The vertical asymptote at the y axis (x = 0) shifts to the vertical line x = .4. The x axis is a horizontal asymptote. y = -2 (x - .4)^-3 vertically stretches the graph by factor -2, moving every point twice as far from the x axis and also to the opposite side of the x axis. This leaves the vertical line x = .4 as a vertical asymptote. The x axis remains a horizontal asymptote. y = -2 ( x - .4)^-3 + 6 vertically shifts the graph +6 units. This has the effect of maintaining the shape of the graph but raising the horizontal asymptote to x = 6.
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RESPONSE --> This problem was simple after constructing the table to look the appropriate way. self critique assessment: 3
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14:58:24 Describe your graphs of y = x ^ .5 and y = 3 x^.5. Describe how your graph depicts the ratios of y values between the two functions.
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RESPONSE --> This is a power function y = x^p with p = .5. The basic points of y = x^.5 would be (0, 0), (.5, .707), (1, 1), (2, 1.414). y = 3 x^.5 vertically stretches the graph of y = x^.5 by a factor of 3, giving me the basic points of (0, 0), (.5, 2.12), (1, 3) and (2, 4.242). This graph is also increasing at a decreasing rate, staying 3 times as far from the x axis as the graph of the original y = x^.5. confidence assessment: 3
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15:11:22 *&*& This is a power function y = x^p with p = .5. The basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414). Attempting to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the conclusion that y = x^.5 is not defined for negative values of x. The graph therefore begins at the origin and increases at a decreasing rate. However since we can make x^.5 as large as we wish by making x sufficiently large, there is no horizontal asymptote. y = 3 x^.5 vertically stretches the graph of y = x^.5 by factor 3, giving us basic ponits (0, 0), (.5, 2.12), (1, 3) and (2, 4.242). This graph is also increasing at a decreasing rate, staying 3 times as far from the x axis as the graph of the original y = x^.5.
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RESPONSE --> I left out a little bit, but i really like this question because this is something like what we learned in Biology the other day, so now I know exactly how to do it and I can use it here as well as in Biology. self critique assessment: 3
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15:24:30 problem 6.
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RESPONSE --> self critique assessment:
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15:40:12 Explain why the graph of A f(x-h) + k is different than the graph of A [ f(x-h) + k ], and describe the difference.
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RESPONSE --> First, the first graph is obtained from y = f(x) by vertically stretching factor A, then horizontally shifting h units and then eventually vertically shifting k units. The graph of A [f(x-h) + k] is found by first solving what is located inside of the brackets, horizontally shifting h units then vertically shifting k units before doing the vertical stretch by factor A. Therefore, the vertical stretch applies to the vertical shift in addition to the values of the function. This results in different y coordinates as well as a graph that often looks different. confidence assessment: 3
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15:55:21 ** The first graph is obtained from y = f(x) by first vertically stretching by factor A, then horizontall shifting h units and finally vertically shifting k units. The graph of A [f(x-h) + k] is obtained by first doing what is in brackets, horizontally shifting h units then vertically shifting k units before doing the vertical stretch by factor A. Thus the vertical stretch applies to the vertical shift in addition to the values of the function. This results in different y coordinates and a typically a very different graph.
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RESPONSE --> Thumbs up! self critique assessment: 3
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16:15:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I really feel like I am catching on to everything very well. I hope I do as well on Test 2 as I did on Test 1. Hopefully, this time I can make sure I don't make silly careless mistakes and maybe get an high A. Thank you for everything Mr. Smith. self critique assessment: 3