course mth 163 yآ}QvΉԇlw
......!!!!!!!!...................................
19:18:26 Explain why the function y = x^-p has a vertical asymptote at x = 0.
......!!!!!!!!...................................
RESPONSE --> Because a negative exponent means 1/x and anything over zero is undefined. This causes an asymptote at x = 0. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:22:40 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:29:49 Explain why the function y = (x-h)^-p has a vertical asymptote at x = h.
......!!!!!!!!...................................
RESPONSE --> Because when x = h, (x-h) = 0 and since it is raised to the -p, 0 can't be divided into any number. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:30:42 ** (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. **
......!!!!!!!!...................................
RESPONSE --> My answer was shorter, but meant the same thing. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:41:42 ** 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
......!!!!!!!!...................................
RESPONSE --> I've got to work on the fact that this thing sometimes doesn't let me respond. You get the same graph because the (x-h)^-p is based on the x^-p graph. You can see the shift when you begin to substitute numbers in. self critique assessment: 1
.................................................
......!!!!!!!!...................................
20:11:33 Explain how your table demonstrates this transformation and describe the graph that depicts the transformation.
......!!!!!!!!...................................
RESPONSE --> the arious columns of my table do show the differences mentioned in previous questions. The first column (x-.4)^-3 is .4 units higher than x. The second column -2(x-.4)^-3 is the previous graph with the points -2 times as far from the x coordinate. The final column -2(x-.4)^-3 +6 is the previous graph with a horizontal shift of 6 units to the right of the points. The graph shows these differences as well. confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:13:30 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.
......!!!!!!!!...................................
RESPONSE --> I forgot to say that the points were twice as far from the x axis and on opposite sides of the graph. And the changes in the asymptote. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:34:54 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.
......!!!!!!!!...................................
RESPONSE --> The first graph has the points (0, 0), (1, 1), (2, 1.4142), (3, 1.7321), (4, 2). The graph looks like a half a parabola with the vertex at (0, 0) where the line of symmetry is the x axis. The second graph has the points (0, 0), (1, 3), (2, 4.2426), (3, 5.1962), (4, 6). The points are three times as far from the x axis than the first graph. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:35:35 *&*& 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:42:21 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.
......!!!!!!!!...................................
RESPONSE --> In A f(x-h) + k, A is only multiplied by f(x-h), not k. In the other equation, A is multiplied the whole equation, k included. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:44:19 ** 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.
......!!!!!!!!...................................
RESPONSE --> my explaination was not as good as this one. I do understand the explaination and the differences in the graphs. self critique assessment: 1
.................................................
......!!!!!!!!...................................
......!!!!!!!!................................... 20:39:17 Query problem 2. Describe the sum of the two graphs.
......!!!!!!!!...................................
RESPONSE --> When you look at the y values for both the black and blue graphs you can add the ""numbers"" and get a parabola that (with the coordinates given) begins at (-3, 10), whose vertex is at (0, -1), and ends at around (3, 7) confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:41:07 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I didn't say it quite like that, but my answer agrees with the explaination.. I just didn't mention when the sum graph would coincide with each of hte two other graphs. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:44:11 Where it is the sum graph higher than the 'black' graph, and where is it lower? Answer by giving specific intervals.
......!!!!!!!!...................................
RESPONSE --> the sum graph will be above the black graph until the point near (0-1), after this point, the sum graph will be below the black graph confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:47:56 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I didn't say anything about when the graph was positive or negative, just where the point are. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:58:13 Where it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving specific intervals.
......!!!!!!!!...................................
RESPONSE --> The Sum graph is higher than the blue graph at the intervals of [-3, -.4) (as mentioned in the previous explaination) and [1, 3). The sum graph is lower at the interval [-1, 1) confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:58:57 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I think I was on the right track writing it out. self critique assessment: 1
.................................................
......!!!!!!!!...................................
22:00:41 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.
......!!!!!!!!...................................
RESPONSE --> (-1, 0) (0, -.4) confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:25:04 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I think I understand where the -.7 came from. I just had to look at it a couple of times. self critique assessment: 1
.................................................
......!!!!!!!!...................................
22:37:01 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.
......!!!!!!!!...................................
RESPONSE --> The sum graph coincides with the blue graph when the black graph has x = 0 which is at (-1, .2) and (1, -.4) confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:37:11 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
23:08:02 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.
......!!!!!!!!...................................
RESPONSE --> The quotient graph is further from the x axis when y on the blue graph is below 1. It is closer to the x axis if the blue graph's y is greater than 1. confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:11:15 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I didn't put all the information given, but had the jist of how the answer was found. self critique assessment: 1
.................................................
......!!!!!!!!...................................
23:22:14 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?
......!!!!!!!!...................................
RESPONSE --> The major difference is that the first ""final"" equation is x^2 - 2x^4, while the ""second"" equation given to sketch is x^2 - x^4. The x^4 in the first equation is vertically stretched by a factor of negative 2 whiile the other is not. confidence assessment: 2
.................................................
......!!!!!!!!...................................
23:23:30 ** 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. **
......!!!!!!!!...................................
RESPONSE --> You didn't have to look too hard at the graphs to analyze this question, that is why I didn't put as much info in my answer like the explainaion did. self critique assessment: 2
.................................................
......!!!!!!!!...................................
23:40:32 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?
......!!!!!!!!...................................
RESPONSE --> The zeros of this function are x = -.8, 0, .8, 1 The first ""peak"" of the first equation is raised from around (-.5, -.5) to about (-.8, .1) confidence assessment: 0
.................................................
......!!!!!!!!...................................
23:42:37 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I am going to have to gp back and reread the explainations just for extra reference. I read them and understand somewhat, but usually a second (or third) glance. self critique assessment: 2
.................................................