??????H?????Precalculus I 04-30-2006
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09:08:00 Explain why the function y = x^-p has a vertical asymptote at x = 0.
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RESPONSE --> Y = x^-p is according to definition the same as 1/x^p. As I plotted these points as x became larger the y axis became smaller. If this is graphed, it would show an asymptote along the x axis.
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09:08:19 ** 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 --> ok
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11:07:00 Explain why the function y = (x-h)^-p has a vertical asymptote at x = h.
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RESPONSE --> Y = (x-h)^-p by definition is the same as 1/(x-h)^p?fter thinking about this, I can't really understand? plotted some points but it still doesn't make a lot of sense.
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11:12:06 ** (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 --> I assume that this graph will show something like the previous graph with an asymptote along the x axis...If I've presumed incorrectly, please make not to me....thanks.
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12:44:47 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 --> It all depends on the ""h"" value?he point would be the same except for the factor of ""h"".
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12:44:57 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 --> ok
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13:35:02 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 --> Table is as follows: Y = x^-3 .12 -.00005787 .8 1.953 .4 15.625 0 not divisible -.4 -15.625 -.8 -1.953 -.12 -578.7037 y = (x - .4)^-3 .12 -45.55393586 .8 15.625 .4 not divisible 0 -15.625 -.4 -1..953125 -.8 -.5787037037 -.12 -7.111970869 y = -2(x - .4)^-3 .12 91.10787172 .8 31.25 .4 not divisible 0 -31.25 -.4 3.90625 -.8 1.157407407 -.12 -14.22394174 y = -2(x - .4)^3 + .6 .12 91.70787172 .8 31.85 .4 -.6 0 -30.65 -.4 4.50625 -.8 1.757407407 -.12 -13.62394174
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13:37:31 ** 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Great!
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13:52:39 Explain how your table demonstrates this transformation and describe the graph that depicts the transformation.
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RESPONSE --> The graph of y = x^-3 is a typical cubical graph going thru the center of the graph at 0,0 with a curve going toward positive connecting at .4,15.625 and .8,1.953 and is symetric to the lower left quadrant of the graph at the same negative positions. The graph of y = (x-.4)^-3 is the same graph except shifting all points to the right of the y axis by .4 points. The graph of -2(x - .4)^-3 is the same graph as the above except that it streches the graph by -2 points. The graph of y = -2(x - .4)^-3 + .6 shows that the graph shifts .6 units upward.
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14:06:58 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 --> Ok
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18:37:22 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 --> The graph of y = x^.5 has the basic coordinates of 0,0 and -1,-1 and 1,1 and -2,-1.414 and 2,1.414 which it itravels up from the negative coordinates thru the center of the graph at 0,0 at take toward the positive thru the positive coordinates. The graph of y = 3x^.5 has teh same shape as the previous but it is 3 times as far from the original graph going thru the points -2,-4.242...-1,-3...0,0...1,3...2,4.242. This graph is stretched by 3 points from y - x^.5
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18:37:38 *&*& 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 --> got it
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21:17:33 problem 6.
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RESPONSE --> I completed the graphs
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21:25:10 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 --> The first equation you would plot f(x), shift it on the x axis by -h units, stretch it by A units, then shift it on the y axis by k units. The second equation, you would plot f(x),shift it on the x axis by -h units, shift it on the y axis by k units and then stretch it by A units
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21:27:32 ** 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 see that I made a mistake in the order to complete the graph. I stated that it would be shifted on the x axis first, where you have stated it would be stretched first. I believe that the concept of the second equation was the same. The brackets were the key in helping me know what the order would be on the second equation.
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21:29:59 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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