course mth 163 brad morelock, btm2569@email.vcc.edu, 02-82-394, 11/11/09 at 4:40 qExplain why the function y = x^-p has a vertical asymptote at x = 0.
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:average ********************************************* Question: `qExplain why the function y = (x-h)^-p has a vertical asymptote at x = h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I assume that this is similair to the previuos question in that you can divide this into 1 as mant times as you want and the more times you do it the more vertical it can get confidence rating: average ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** (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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:average ********************************************* Question: `qExplain 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the y values would be the same but at different positions on the x axis confidence rating: average ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:average ********************************************* Question: `qGive 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `atable 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y=x^-3, y =-2(x-.4)^-3, y=-2(x-.4)^-3, y=-2(x-.4)^-3+.6 0.8, 1.953, 15.625, 31.25, confidence rating: below average ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The table is as follows (note that column headings might not line up correctly with the columns): x y=x^-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/0.6 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):i wasnt for sure were to use the incriment of .4 when i got to the next step, or even if i was supposed to after looking at the given solution i couldnt figure out the next step in the problem
| x | y = x^3 | y = (x-0.4)^(-3) | y = -2 (x-0.4)^(-3) | y = -2 (x-0.4)^(-3) + .6 |
| -0.8 | -1.953 | -0.579 | 1.16 | 1.76 |
| -0.4 | -15.625 | -1.953 | 3.90 | 4.50 |
| 0 | div by 0 | -15.625 | 31.25 | 32.85 |
| 0.4 | 15.625 | div by 0 | div by 0 | div by 0 |
| 0.8 | 1.953 | 15.625 | -31.25 | -30.65 |
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Given Solution: `ay = 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):see my solution ------------------------------------------------ Self-critique rating: ********************************************* Question: `qDescribe 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y=x^.5 cannot be a negative number all points are positve so it increases from the first point. when we add y=3x^.5 the graph stretches vertically. confidence rating: average ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a*&*& 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. problem 6. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why the graph of A f(x-h) + k is different than the graph of A [ f(x-h) + k ], and describe the difference. Your solution when you graph the second equation you have to work inside the brkets first which will you a different value than the first and thus this will lead to two different graohs completly confidence rating: average ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. Query Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:average