course Mth 163 Gqɮեɯdassignment #022
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16:12:55 Explain why the function y = x^-p has a vertical asymptote at x = 0.
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RESPONSE --> x^-p = 1/x^p so if x=0, then it is equal to 1/0^p which reduces to 1/0 and you cannot divide by 0 so therefore there is an asymptote at x=0, the graph will just get closer and closer to 0.
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16:14:00 ** 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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16:17:09 Explain why the function y = (x-h)^-p has a vertical asymptote at x = h.
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RESPONSE --> y = (x-h)^-p = 1/(x-h)^p As the x value gets closer to h, the value of (x-h)^p gets closer to 0. X can get closer and closer to h so there is no limit on how many times (x-h)^p can be divided into one which makes an asymptote at x=h because, if x in fact was equal to h, then the equation (x-h)^p would equal to zero and there is no division by zero.
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16:18:38 ** (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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16:20:08 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 double clicked accidently, but I understand that the graph of y=x^p is shifted horizontally by h and does not stretch it in any way.
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16:29:12 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 --> x values: -1.2, -.8, -.4, 0, .4, .8, 1.2 y=x^(-3): -.5787, -1.9531, -15.625, no, 15.625, 1.9531, .5787 y=(x-.4)^(-3): -.2441, -.5787, -1.9531, -15.625, no, 15.625, 1.9531 y=-2(x-.4)^(-3): .4883, 1.1574, 3.9062, 32.25, no, -31.25, -3.906 y=-2(x-.4)^(-3)+.6: 1.0883, 1.7574, 4.5063, 31.85, no, -30.65, -3.3062
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16:29:26 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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16:30:03 ** 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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16:36:42 Explain how your table demonstrates this transformation and describe the graph that depicts the transformation.
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RESPONSE --> y+x^(-3) transforms into y=(x-.4)^(-3) by keeping the same shape but shifting horizontally to the right .4 units, this is clearly seen by the vertical asymptote moving from the y axis to x=.4 y=(x-.4)^(-3) transforms into y = -2(x-.4)^(-3) by being vertically stretched by a factor of -2 which moves every point 2 times away from the x axis as the original point is, the negative flips the graph.vertical and horizontal asymptotes of x=.4 and y = 0 still remain. y = -2(x-.4)^(-3) transforms into y = -2(x-.4)^(-3)+.6 by vertically shifting .6 units upwards. The graph's shape remains the same and the vertical asymptote but the horizontal asymptote is now y=.4 instead of y=0.
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16:39:54 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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16:42:37 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=3x^.5 is the graph of y=x^.5 vertically stretched by a factor of 3 so that all of the y values are 3 times the distance from the x axis as in the original gaph. The ratios were constantly 1/3 which shows that the y values of the graph y=x^.5 are 1/3 the amount of the graph y=3x^.5
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16:43:48 *&*& 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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16:43:52 problem 6.
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16:48:27 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 --> A f(x-h) + k is vertically stretched by a factor of A, horizontally shifted by h and vertically shifted by k. A [ f(x-h) + k ] is horizontally shifted by h, vertically shifted by k and then the vertical stretch of A is applied, which results in a different graph.
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16:48:45 ** 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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16:48:49 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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XtOzw~z assignment #023 ҶQ̫ Precalculus I 11-13-2006
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21:08:09 Query problem 2. Describe the sum of the two graphs.
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RESPONSE --> the blue graph has y values 1.75, .75, 0, -.1, -.25, -.5, -.75 the black graph has y values 8, 3, 0, -1, 0, 3, 8 So, the sum of these graphs has y values 9.75, 3.75, 0, -1.1, -.25, 2.5, 7.75
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21:08:43 ** The 'black' graph takes values 8, 3, 0, -1, 0, 3, 8 at x = -3, -2, -1, 0, 1, 2, 3. The 'blue' graph takes approximate values 1.7, .8, .2, -.1, -.4, -.6, -.8 at the same x values. The 'blue' graph takes value zero at approximately x = -.4. The sum of the two graphs will coincide with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The sum will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. **
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21:10:07 Where it is the sum graph higher than the 'black' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> the sum graph is higher than the black graph from x = -3 to -.4 and is lower from x=-.4 to 7.75
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21:10:57 ** The sum of the graphs is higher than the 'black' graph where the 'blue' graph is positive, lower where the 'blue' graph is negative. The 'blue' graph is positive on the interval from x = -3 to x = -.4, approx.. This interval can be written [-3, -.4), or -3 <= x < -.4. **
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21:13:03 Where it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> the sum graph is higher than the blue graph from x = -3 to -1 and from x = 1 to 3 and it is lower between x = -1 to 1
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21:13:37 ** The sum of the graphs is higher than the 'blue' graph where the 'black' graph is positive, lower where the 'black' graph is negative. The 'black' graph is positive on the interval from x = -1 to x = 1, not including the endpoints of the interval. This interval can be written (-1, 1) or -1 < x < 1. **
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21:15:39 Where does thus sum graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> The sum graph coincides with the black graph when the blue graph is zero. It does here because there is no value from the blue graph to add to the value of the black graph so the sum graph takes on the amount of the black graph. Occurs at x= -.4
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21:17:05 ** The sum coincides with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. The coordinates would be about (-.4, -.7), on the 'black' graph. **
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21:18:54 Where does thus sum graph coincide with the 'blue' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> The sum graph coincides with the blue graph when the black graph is zero. This is because there is no value of the black graph to be added to the blue graph so the sum graph takes on the value of the blue graph. Occurs at x=-1 and x=1
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21:19:02 ** The sum coincides with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The coordinates would be about (-1, .2) and (1, -.4), on the 'blue' graph. **
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21:46:26 Query problem 3 Describe the quotient graph obtained by dividing the 'black' graph by the 'blue' graph. You should answer the following questions: Where it is the quotient graph further from the x axis than the 'black' graph, and where is it closer? Answer by giving specific intervals, and explaining why you believe these to be the correct intervals. Where it is the quotient graph on the same side of the x axis as the 'black' graph, and where is it on the opposite side, and why? Answer by giving specific intervals. Where does thus quotient graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs. Where does the quotient graph have vertical asymptote(s), and why? Describe the graph at each vertical asymptote.
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RESPONSE --> The quotient graph will always be closer to the x axis when the blue graph gets within 1 unit of the x axis. This is because if you divide by a number less than one gives an answer larger then was put in. The quotient graph is on the same side of the x axis as the black graph between x=3 on and is on the opposite side of the x axis on x values less than three. This is because below x=3 a neg/neg = pos, so the quotient graph is the opp. sign of the black graph and also a pos/neg = neg, but the black is pos so the quotient graph is opposite. The quotient graph will coincide with the black graph where the black graph crosses the x axis (it's zeros) because 0/# = 0 so the quotient graph takes on the 0 value. There are vertical asymptotes where the blue graph crosses the x axis (it's zeros) because a #/0 = unable to do, no division by a zero, so, there is a vertical asymptote that results. This occurs at about x=5.5, here, the graph rises in a positive direction continuously while moving closer to the asymptote yet never touching it.
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22:03:12 ** The 'black' graph is periodic, passing through 0 at approximately x = -3.1, 0, 3.1, 6.3. This graph has peaks with y = 1.5, approx., at x = 1.6 and 7.8, approx., and valleys with y = -1.5 at x = -1.6 and x = 4.7 approx. The 'blue' graph appears to be parabolic, passing thru the y axis at x = -1 and reaching a minimum value around y = -1.1 somewhere near x = 1. This graph passes thru the x axis at x = 5.5, approx., and first exceeds y = 1 around x = 7.5. The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the origin, since division by a number whose magnitude is less than 1 gives a result whose magnitude is greater than the number being divided. This will occur to the left of x = 1, and between about x = 2 and x = 7.5. Between about x = 0 and x = 1 the 'blue' graph is more than 1 unit from the x axis and the quotient graph will be closer to the x axis than the 'black' graph. The same is true for x > 7.5, approx.. The 'black' graph is zero at or near x = -3.1, 0, 3.1, 6.3. At both of these points the 'blue' graph is nonzero so the quotient will be zero. The 'blue' graph is negative for x < 5.5, approx.. Since division by a negative number gives us the opposite sign as the number being divided, on this interval the quotient graph will be on the opposite side of the x axis from the 'black' graph. The 'blue' graph is positive for x > 5.5, approx.. Since division by a positive number gives us the same sign as the number being divided, on this interval the quotient graph will be on the same side of the x axis as the 'black' graph. The quotient graph will therefore start at the left with positive y values, about 3 times as far from the x axis as the 'black' graph (this since the value of the 'blue' graph is about -1/3, and division by -1/3 reverses the sign and gives us a result with 3 times the magnitude of the divisor). The quotient graph will have y value about 2.5 at x = -1.6, where the 'black' graph 'peaks', but the quotient graph will 'peak' slightly to the left of this point due to the increasing magnitude of the 'blue' graph. The quotient graph will then reach y = 0 / (-1) = 0 at x = 0 and, since the 'black' graph then becomes positive while the 'blue' graph remains negative, the quotient graph will become negative. Between x = 0 and x = 2 the magnitude of the 'blue' graph is a little greater than 1, so the quotient graph will be a little closer to the x axis than the 'black' graph (while remaining on the other side of the x axis). At x = 3.1 approx. the 'black graph is again zero, so the quotient graph will meet the x axis at this point. Past x = 3.1 the quotient graph will become positive, since the signs of both graphs are negative. As we approach x = 5.5, where the value of the 'blue' graph is zero, the quotient will increase more and more rapidly in magnitude (this since the result of dividing a negative number by a negative number near zero is a large positive number, larger the closer the divisor is to zero). The result will be a vertical asymptote at x = 5.5, with the y value approaching +infinity as x approaches 5.5 from the left. Just past x = 5.5 the 'blue' values become positive. Dividing a negative number by a positive number near zero results in a very large negative value, so that on this side of x = 5.5 the asymptote will rise up from -infinity. The quotient graph passes through the x axis near x = 6.3, where the 'black' graph is again zero. To the right of this point both graphs have positive values and the quotient graph will be positive. Around x = 7.5, where the 'blue' value is 1, the graph will coincide with the 'black' graph, giving us a point near (7.5, 1.3). Past this point the 'blue' value is greater than 1 so that the quotient graph will become nearer the x axis than the 'black' graph, increasingly so as x (and hence the 'blue' value) increases. This will result in a 'peak' of the quotient graph somewhere around x = 7.5, a bit to the left of the peak of the 'black' graph. **
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22:10:15 Query problems 7-8 Sketch the graph of y = x^2 - 2 x^4 by first sketching the graphs of y = x^2 and y = -2 x^4. How does the result compare to the graph of y = x^2 - x^4, and how do you explain the difference?
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RESPONSE --> Both graphs go off to an infinite neg. number to the left and to the right of the y axis, both pass through origin it appears However, the sum graph rises above x axis, turns back down through origin, goes again above x axis then decreases at an increasing rate where the other graph is more of a parabola. This difference is due to the two different equations added together to make the sum graph, one dominates over the other at different times causing the new graph to be a mesh of the two.
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22:11:23 ** At x = 0, 1/2, 1 and 2 we have x^2 values 0, 1/4, 1 and 4, while -x^4 takes values 0, -1/16, -1 and -16, and -2 x^4 takes values 0, -1/8, -2 and -32. All graphs clearly pass through the origin. The graphs of y = x^2 - x^4 and y = x^4 - 2 x^4 are both increasingly negative at far right and far left. Graphical addition will show that y = x^2 - x^4 takes value 0 and hence passes thru the x axis when the graphs have equal but opposite y values, which occurs at x = 1 and x = -1. To the left of x = -1 and to the right of x = 1 the negative values of -x^4 overwhelm the positive values of x^2 and the sum graph will be increasingly negative, with values dominated by -x^4. Near x = 0 the graph of y = -x^4 is 'flatter' than that of y = x^2 and the x^2 values win out, making the sum graph positive. y = x^2 - 2 x^4 will take value 0 where the graphs are equal and opposite in value; this occurs somewhere between x = .8 and x = .9, and also between x = -.9 and x = -.8, which places the zeros closer to the y axis than those of the graph of y = x^2 - x^4. The graph of y = -2 x^4 is still flatter near x = 0 than the graph of y = x^2, but not as flat as the graph of y = -x^4, so while the sum graph will be positive between the zeros the values won't be as great. Outside the zeros the sum graph will be increasingly negative, with values dominated by -2x^4. **
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22:16:53 How does the shape of the graph change when you add x to get y = -2 x^4 + x^2 + x, and how do you explain this change?
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RESPONSE --> Adding pos. values will increase y values. adding neg values will increase neg y values adding a pos/neg will depend on which value is greater on if the answer will turn out pos/neg.
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22:19:06 ** At x = 0 there is no change in the y value, so the graph still passes through (0, 0). As x increases through positive numbers we will have to increase the y values of y = x^2 -2 x^4 by greater and greater amounts. So it will take a little longer for the negative values of -2 x^4 to 'overwhelm' the positive values of x^2 + x than to overcome the positive values of x^2 and the x intercept will shift a bit to the right. As we move away from x = 0 through negative values of x we will find that the positive effect of y = x^2 is immediately overcome by the negative values of y = x, so there is no x intercept to the left of x = 0. The graph in fact stays fairly close to the graph of y = x near (0, 0), gradually moving away from that graph as the values of x^2 and -2 x^4 become more and more significant. **
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RESPONSE --> my answer was quite generalized and not pertaining to this specific graph.
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22:19:09 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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22:19:11 none
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