ԧW߳{ձK assignment #022 Kի Precalculus I 04-28-2006
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22:59:37 Explain why the function y = x^-p has a vertical asymptote at x = 0.
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RESPONSE --> because it is understood that the value of p is -1, therefore if you replace it with -p, you will then have y = x^1 and anytime you have a value which is raised to a power, the value stays the same. in this case it's 1 or x. therefore the result decreases toward 0 and doesn't level off. x will approach zero but never hit it.
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23:01:01 ** 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 --> i was almost correct. i believe i understand it more now.
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23:02:32 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 understand a little more
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23:11:37 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 --> to get from y = x^-3 to y = (x - .4)^-3 it is shifted .4 units. to get to y = -2(x - .4)^-3 it is shifted .4 units and -2 units. and to get to y = -2(x - .4)^-3 + .6, it is shifted .4 units and -2 units and raised to the -3 + .6 power. ?? i'm not sure if this is correct. i may be mistaking this method for something else...
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23:13:42 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 was wrong. what exactly are the 'colomns' that you're talking about here? are these the lines in the table?
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23:14:52 ** 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 --> oh, ok..nevermind...i got what you were meaning
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23:18:16 Explain how your table demonstrates this transformation and describe the graph that depicts the transformation.
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RESPONSE --> each transformation goes across the top with beginning x value in first column then each change to x to get the y values in resulting columns. It shows how the y values increase in magnitude as x approaches 0, forming the vertical asymptote at y = 0, and how the y values approach the y = 0 asymptote (the x axis) when x becomes large.
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23:19:45 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 --> i wasn't fully correct. i understand what is being done though.
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00:46:52 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 --> A y = 3 x^.5 point is 3 times as far from the x axis as the y = x^.5 point, and on the other side of the x axis.
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00:59:11 *&*& 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 understand the power function i believe.
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01:12:58 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 --> I'm not sure of the exact and correct answer to this question, but i do know that in order of operations, Expressions grouped together inside brackets [] are always evaluated first, therefore i know that would be one diffference from the 2nd equation and the first, since the first doesn't have any. it's kinda like saying a + b + c = d and then a [b + c] = d. in the first one, you solve by adding up the variables a, b, and c, but in the second one, a is multiplied by the product of b + c. am i on the right path?
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01:14:19 ** 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 --> i was halfway correct. i jsut didn't describe what was being done inside the brackets and the equation, with stretching and shifting.
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01:19:38 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I have become more aware and understanding of the material being covered in this query. i'm coming a long way in my comprehension of preCalculus. I learn the best and most when i see examples of problems being solved or dealing with equations when you can jsut plug values in. i think i'm slowly, but surely, getting the hang of this. would you agree?
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