course Mth 163 ???~?????c??n?assignment #021021. `query 21
......!!!!!!!!...................................
12:25:33 What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 6?
......!!!!!!!!...................................
RESPONSE --> it could have 6 linear factors or 3 identical 3 distinct or 4 identical 2 distinct or any other combination of even real/postive numbers below six confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:26:01 ** You can have as many as 6 linear and 3 irreducible quadratic factors for a polynomial of degree 6. For a polynomial of degree 6: If you have no irreducible quadratic factors then to have degree 6 you will need 6 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 4 linear factors. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 2 linear factors. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you can have no linear factors. For a polynomial of degree 7: If you have no irreducible quadratic factors then to have degree 7 you will need 7 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 5 linear factors to give you degree 7. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 3 linear factors to give you degree 7. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you will need 1 linear factor to give you degree 7. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:27:00 For a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros.
......!!!!!!!!...................................
RESPONSE --> 2 zeros confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:27:59 ** there could be 1 root repeated 4 times, 2 roots with 1 repeated 3 times and the other distinct from it, 2 distinct roots each repeated twice, three distinct roots with one of them repeated twice, or four distinct roots. Explanation: You have one zero for every linear factor, so there will be four zeros. Since the degree is even the far-left and far-right behaviors will be the same, either both increasing very rapidly toward +infinity or both decreasing very rapidly toward -infinity. You can have 4 distinct zeros, which will result in a graph passing straight thru the x axis at each zero, passing one way (up or down) through one zero and the opposite way (down or up) through the next. You can have 2 repeated and 2 distinct zeros. At the repeated zero the graph will just touch the x axis as does a parabola at its vertex. The graph will pass straight through the x axis at the two distinct zeros. You can have 3 repeated and 1 distinct zero. At the 3 repeated zeros the graph will level off at the instant it passes thru the x axis, in the same way the y = x^3 graph levels off at x = 0. The graph will pass through the x axis at the one distinct zero. You can have two pairs of 2 repeated zeros. At each repeated zero the graph will just touch the x axis as does a parabola at its vertex. Since there are no single zeros (or any other zeros repeated an odd number of times) the graph will not pass through the x axis, so will remain either entirely above or below the x axis except at these two points. You can have four repeated zeros. At the repeated zero the graph will just touch the x axis, much as does a parabola at its vertex except that just as the y = x^4 function is somewhat flatter near its 'vertex' than the y = x^2 function, the graph will be flatter near this zero than would be a parabolic graph. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:28:13 Describe a typical graph for each of these possibilities. Describe by specifying the shape of the graph at each of its zeros, and describe the far-left and far-right belavior of the graph.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:29:18 ERRONEOUS STUDENT SOLUTION AND INSTRUCTOR CORRESION for 4 distincts, the graph curves up above the x axis then it crosses it 4 timeswithe the line retreating from the direction it came from for say 2 distincts and 2 repeats, line curves above x axis then it kisses the axis then it crosses it twice, retreating to the side it came from for 1 distinct and 3 it curves above x axis then it crosses once and kisses, finnaly heading off to the opposite side it came from INSTRUCTOR COMMENTS: {The graph can't go off in th opposite direction. Since it is a product of four linear factors and any number of quadratic factors its degree is even so its large-x behavior is the same for large positive as for large negative x. It doesn't kiss at a degree-3 root, it acts like the y = x^3 polynomial, leveling off just for an instant as it passes through the zero and on to the other side of the axis. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:30:26 problems 3-5
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:31:50 It doesn't matter if you don't have a graphing utility--you can answer these questions based on what you know about the shape of each power function. Why does a cubic polynomial, with is shape influenced by the y = x^3 power function, fit the first graph better than a quadratic or a linear polynomial? What can a cubic polynomial do with this data that a quadratic can't?
......!!!!!!!!...................................
RESPONSE --> a cubic function would be the best fit for this data it is obvious to see because of the curve and a quadratic function would not be able to stretch out to fit this data correctly confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:31:58 ** the concavity (i.e., the direction of curvature) of a cubic can change. Linear graphs don't curve, quadratic graphs can be concave either upward or downward but not both on the same graph. Cubics can change concavity from upward to downward. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:33:18 ** higher even degrees flatten out more near their 'vertices' **
......!!!!!!!!...................................
RESPONSE --> I meant to leave an answer but hit the button too soon i was actually thinking that a sixth degree would fit better not only because it would be flatter but would have more meetings with the line in that it has six curves self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:33:49 On problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph?
......!!!!!!!!...................................
RESPONSE --> I think that this one would have the same answer as the previous question confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:34:02 STUDENT RESPONSE: progressively more flexing, because more curves, and fit graph better the that of a lesser degree INSTRUCTOR COMMENT: On a degree-2 polynomial there is only one change of direction, which occurs at the vertex. For degrees 4 and 6, respectively, there can be as many as 3 and 5 changes of direction, respectively. For higher degrees the graph has more ability to 'wobble around' to follow the data points.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:39:09 What is the degree 2 Taylor approximation for f(t) = e^(2t), and what is your approximation to f(.5)? How close is your approximation to the actual value of e^(2t)?
......!!!!!!!!...................................
RESPONSE --> 2.7=e^2t and the approximation equation would be 1+X+X^2/2 and that gives you 1.625 confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:39:54 ** The degree 2 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! = 1 + 2 t + 4 t^2 / 2 = 1 + 2 t + 2 t^2. Therefore we have T(.5) = 1 + 2 * .5 + 2 * .5^2 = 1 + 1 + 2 * .25 = 1 + 1 + .5 = 2.5. The actual value of e^(2t) at t = .5 is f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx.. The approximation is .218 less than the actual function. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:41:34 By how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation?
......!!!!!!!!...................................
RESPONSE --> I got 2.6666 confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:41:56 ** The degree 3 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! + (2t)^3 / 3! = 1 + 2 t + 4 t^2 / 2 + 8 t^3 / 6 = 1 + 2 t + 2 t^2 + 4 t^3 / 3. Therefore we have T(.5) = 1 + 2 * .5 + 2 * .5^2 + 4 * .5^3 / 3 = 1 + 1 + 2 * .25 + 4 * .125 / 3 = 1 + 1 + .5 + .167 = 2.667. The actual value of e^(2t) at t = .5 is f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx.. The approximation is .051 less than the actual function. This is about 4 times closer than the approximation we obtained from the degree-2 polynomial. **
......!!!!!!!!...................................
RESPONSE --> ok this is making more since now self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:42:50 Describe your graph of the error vs. the degree of the approximation for degree 2, degree 3, degree 4 and degree 5 approximations to e^.5.
......!!!!!!!!...................................
RESPONSE --> The graphs are very similar with minor changes confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:42:59 ** The errors, rounded to the nearest thousandth, are: degree-2 error: -.218 degree-3 error: -.051 degree-4 error: -.010 degree-5 error: -.002 A graph of error vs. degree decreases rapidly toward the horizontal axis, showing that the error decreases rapidly toward zero as the degree increases. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:43:16 What are your degree four approximations for e^.2, e^.4, e^.6 e^.8 and e^1? Describe the graph of the approximation error vs. x.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:43:25 The following are the approximations and errors: 0.2 1.2214 1.221402758 2.75816E-06 0.4 1.4917 1.491824698 9.13643E-05 0.6 1.8214 1.8221188 0.0007188 0.8 2.2224 2.225540928 0.003140928 1 2.7083 2.718281828 0.009948495 The errors can be written as .0000027, .000091, .000071, .0031 and .0099. A graph of approximation error vs. x increases exponentially, with over a 10-fold increase with every increment os .2.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:44:39 ** The function is P2(x) = (x-1) - (x-1)^2/2. A table of values of ln(x), P2(x) and P2(x) - ln(x): x ln(x) P2(x) P2(x) - ln(x) .6 -0.5108256237 -0.48 0.03082562376 .8 -0.2231435513 -0.22 0.003143551314 1.2 0.1823215567 0.18 -0.002321556793 1.4 0.3364722366 0.32 -0.01647223662 At x = 1 we have ln(x) = ln(1) = 0, and P2(x) = P2(1) = (1-1) - (1-1)^2 / 2 = 0. There is no difference in values at x = 1. As we move away from x = 1 the approximation becomes less and less accurate. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:44:45 What is the error in the degree 2 approximation to ln(x) for x = .6, .8, 1.2 and 1.4? Why does the approximation get better as x approaches 1?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:44:51 ** The respective errors are .03, .00314, .00232, .016472. There is no error at x = 1, since both the function and the approximation give us 0. As we move away from 1 the approximation becomes less and less accurate. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:45:25 problem 12. What does the 1/x graph do than no quadratic function can do?
......!!!!!!!!...................................
RESPONSE --> give us an asymptote confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:45:34 ** The y = 1/x graph has vertical asymptotes at the y axis and horizontal asymptotes at the x axis. The parabolas we get from quadratic functions do have neither vertical nor horizontal asymptotes. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:45:45 What are the errors in the quadratic approximation to 1/x at x = .6, .8, 1, 1.2, and 1.4? Describe a graph of the approximation error vs. x.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:45:50 ** The quadratic approximation to 1/x is the second-degree Taylor polynomial P2(x) = 1 - (x - 1) + (x - 1)^2. A table of values of 1/x, P2(x) and P2(x) - 1/x: x 1/x P2(x) P2(x) - 1/x .6 1.666666666 1.56 - 0.1066666666 .8 1.25 1.24 - 0.01 1.2 0.8333333333 0.84, 0.006666666666 1.4 0.7142857142 0.76 0.04571428571. A graph of appoximation error vs. x decreases at a decreasing rate to 0 at x = 1, then increases at an increasing rate for x > 1. This shows how the accuracy of the approximation decreases as we move away from x = 1. The graph of approximation error vs. x gets greater as we move away from x = 1.**
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
course Mth 163 {??????????????assignment #022
......!!!!!!!!...................................
13:01:45 Explain why the function y = x^-p has a vertical asymptote at x = 0.
......!!!!!!!!...................................
RESPONSE --> because 0^ to any negative number is not possible and the reason that this function would have a zero asymptote is because it doesnt have a vertical or horizontal shift confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:01:53 ** 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:02:21 Explain why the function y = (x-h)^-p has a vertical asymptote at x = h.
......!!!!!!!!...................................
RESPONSE --> the h shifts the graph horizontally confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:02:27 ** (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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:03:06 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.
......!!!!!!!!...................................
RESPONSE --> the h will shift if however much the value says to confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:03:16 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.
......!!!!!!!!...................................
RESPONSE --> oh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:03:33 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.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:03:45 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
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:03:54 ** 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 --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
13:04:56 Explain how your table demonstrates this transformation and describe the graph that depicts the transformation.
......!!!!!!!!...................................
RESPONSE --> the table that would be constructed for this would prove the theory to be true and the graph would look the same but have different asymptotes and different curves confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:05:00 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:07:04 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 two graphs are positively sloping and are completely positive they stop at the x axis before they get neg. The only difference in the two is that the 3x stretches the line and makes it increase more rapidly and more noticably confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:07:33 *&*& 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:07:39 problem 6.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:09:17 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 --> The different groupings change the order of operation which changes the x values confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:09:22 ** 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:11:06 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I actually already knew most of the stuff in this query the last question was a new concept but actually the query for asignment 21 was the one that taught me a few things it was kind of confusing and involved however I did learn quite a bit from that query self critique assessment: 3
.................................................