course Mth 163 ???k???????L???assignment #023023. `query 23
......!!!!!!!!...................................
15:55:39 Query problem 2. Describe the sum of the two graphs.
......!!!!!!!!...................................
RESPONSE --> The ""black"" graph has values 8, 3, 0, -1, 0, 3, 8 at x = -3, -2, -1, 0, 1, 2, 3. The 'blue' graph has values 1.7, .8, .2, -.1, -.4, -.6, -.8 at the same x values as the black graph. The 'blue' graph takes a value of zero at x = -.4. The sum of the two graphs will correspond with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The sum will correspond with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:22:54 ** 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 like the graph questions. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:40:38 Where it is the sum graph higher than the 'black' graph, and where is it lower? Answer by giving specific intervals.
......!!!!!!!!...................................
RESPONSE --> The sum of the graph is higher than the 'black' graph where the ""blue"" graph is positive, and lower where the ""blue"" graph is negative. The ""blue"" graph is positive on the interval from x = -3 to x = -.4, approx.. This interval should be written as follows: [-3, -.4), or -3 <= x < -.4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:57:46 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:16:44 Where it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving specific intervals.
......!!!!!!!!...................................
RESPONSE --> The sum of the graphs is higher than the ""blue""graph where the ""black"" graph is positive, and lower where the ""black"" graph is negative. The ""black"" graph is positive on the interval from x = -1 to x = 1. This interval can be written (-1, 1) or -1 < x < 1. confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:35:51 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:49:45 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 --> The sum corresponds with the ""black"" graph where the ""blue"" graph is zero, which occurs at about x = -.4. The coordinates would be around (-.4, -.7), on the ""black"" graph. confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:00:48 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:14:45 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 corresponds with the ""blue"" graph where the ""black""graph is zero, which occurs at x = -1 as well as x = 1. The coordinates would be around (-1, .2) and (1, -.4), on the ""blue"" graph. confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:27:59 ** 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 --> I really enjoy this assignment! self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:49:28 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 --> confidence assessment:
.................................................
......!!!!!!!!...................................
19:07:33 ** 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 --> self critique assessment:
.................................................
......!!!!!!!!...................................
19:29:47 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 --> At x = 0, 1/2, 1 and 2 I got x^2 values 0, 1/4, 1 and 4, mean whilex^4 takes values 0, -1/16, -1 and -16, and -2 x^4 takes values 0, -1/8, -2 and -32. All graphs pass through the origin. The graphs of y = x^2 - x^4 and y = x^4 - 2 x^4 are both increasingly negative at the far right and far left of the graph. confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:48:27 ** 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 --> That is a whole lot of explaining to do. I can explain some of it, but eventually I just get confused with what I am trying to explain. This is obviously something I need to work on. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:13:06 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 --> 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 I had to increase the y values of y = x^2 -2 x^4 by greater and greater amounts. Therefore, it will take a longer amount of time for the negative values of -2 x^4 to over power the positive values of x^2 + x then to overpower the positive values of x^2 and the x intercept which will shift somewhat to the right. As I move away from x = 0 through negative values of x I found that the positive effect of y = x^2 is instantly overpowered by the negative values of y = x, therefore, obviously, there is no x intercept to the left of x = 0. The graph stays fairly close to the graph of y = x closest to(0, 0), but it gradually moves away from the graph as the values of x^2 and -2 x^4 become more important. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:47:35 ** 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 got this one. There is only so much one can say about a question like this. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:15:54 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> This assignment was harder for me because as you noticed on the last test, I am not always so good at explaining things. I often think I am explaining the right thing, but then it turns out I haven't said something the correct way, just like I did with the slope on Test 1. Also, when the explanation gets to be very very long, I often get confused and jumble up my explanation, so then it is almost unreadable. I think from now on as I am working the problem, I need to go ahead and write my answer is words as well to get a better hang of problems like this. self critique assessment: 3
.................................................