course mth 163 憢sx蘺ۖȫ\`IǍBassignment #005
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11:33:00 query introduction to basic function families problem 1 on basic graphs Why is the graph of y = x a straight line?
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RESPONSE --> Because the y value is the same as the x value making the points lie in a straight line. confidence assessment: 1
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11:33:39 ** Since y = x the rise and run between any two points on the graph are equal, which makes the slope 1. A graph with constant slope is a straight line. **
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RESPONSE --> That says it much more elegantly. self critique assessment: 2
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11:36:43 why is y = x^2 symmetric about x = 0 (i.e., taking the same values on either side of x = 0)
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RESPONSE --> Because when you square a negative number the result is positive, therefore making the graph symmetrical confidence assessment: 2
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11:42:33 ** The graph of y = x^2 is symmetric about x = 0 because (-x)^2 = x^2. Thus for any point on the x axis the y values at that point and at the point on the opposite side of the origin are equal, so that the graph on one side of the y axis is a 'reflection' of the graph on the other side. **
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RESPONSE --> self critique assessment: 3
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11:45:19 why does y = 2^x keep increasing as x increases, and why does the graph approache the x axis for negative values of x
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RESPONSE --> the graph approaches the x axis for negative values of x becuase something raised to the negative power is shown as 1/x where x is raised to whatever power is listed. confidence assessment: 2
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11:45:59 ** GOOD STUDENT RESPONSE: y = 2^x will increase as x increases on the positive side because x is the value of the exponent. This will cause the y value to double from its last value when you move one unit in the positive x direction. On the negative side of the y axis y = 2^x will approach the x axis because a negative exponent causes the value to invert into a fractional value of itself--i.e., 2^(-x) = 1 / 2^x. As we move one unit at a time negatively the value will become one half of the previous value so it will never quite reach y = 0. **
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RESPONSE --> This is what I was trying to say, just not as well. self critique assessment: 2
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11:48:39 why is y = x^3 antisymmetric about x = 0 (i.e., taking the same values except for the - sign on opposite sides of x = 0)
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RESPONSE --> When multiplying an odd number of negative numbers the result will be a negative number. Therefore once you get (0, 0) the numbers will become positive again. confidence assessment: 2
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11:49:00 ** y = x^3 is antisymmetric because if you cube a negative number you get a negative, if you cube a positive number you get a positive, and the magnitude of the cubed number is the cube of the magnitude of the number. So for example (-3)^2 = -27 and 3^3 = 27; the points (-3, -27) and (3, 37) are antisymmetric, one being `down' while the other is `up'. GOOD STUDENT RESPONSE: y = x^3 is antisymmetric about x = 0 because the exponent is an odd number. This will cause negative x values to have a negative y result. The absolute value of the negative y result will be equivalent to its corresponding positive y value. **
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RESPONSE --> self critique assessment: 3
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11:51:53 why do y = x^-2 and y = x^-3 rise more and more steeply as x approaches 0, and why do their graphs approach the x axis as we move away from the y axis.
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RESPONSE --> With both of these, you will take the inverse of x. If it is a whole number then you will continue to get closer and closer to the x axis, but never fully touching it. It moves further away from the y axis because the smaller x is (especially if it is a fraction) the larger y becomes (since the fractions are inverted). confidence assessment: 2
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11:53:22 ** And as x approaches 0 the expressions x^-2 and x^-3, which mean 1 / x^2 and 1 / x^3, have smaller and smaller denominators. As the denominators approach zero their reciprocals grow beyond all bound. y = x^-2 and y = x^-3 rise more and more steeply as x approaches zero because they have negative exponents they become fractions of positive expressions x^2 and x^3 respectively which have less and less slope as they approach zero. As x^2 and x^3 approach zero and become fractional, x^-2 and x^-3 begin to increase more and more rapidly because thier functions are then a whole number; (1) being divided by a fraction in which the denominator is increasing at an increasing rate. As y = x^-2 and y = x^-3 move away from the y-axis they approach the x-axis because they have negative exponents. This makes them eqivalent to a fraction of 1 / x^2 or 1 / x^3. As x^2 and x^3 increase in absolute value, the values of y = x^-2 and y = x^-3 constantly close in on the x-axis by becoming a portion of the remaining distance closer, they will never reach x = zero though as this would be division by zero (since it is a fraction) **
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RESPONSE --> I didn't put that 0 raised to a negaive power can never occur, which is why the graph will never reach the x axis. self critique assessment: 2
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11:57:29 query problem 2. family y = x^2 + c Explain why the family has a series of identical parabolas, each 1 unit higher than the one below it.
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RESPONSE --> The variable c tells us how many units each equation will be above, in this case 1. confidence assessment: 2
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12:10:16 query problem 4. describe the graph of the exponential family y = A * 2^x for the values A = -3 to 3.
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RESPONSE --> For the graphs that are multiplied by the negative numbers, they will be below the x axis. All graphs will each be A times greater than the graph for 2^x. confidence assessment: 1
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12:11:21 ** This family includes the functions y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x, y = 0 * 2^x, y = 1 * 2^x, y = 2 * 2^2 and y = 3 * 2^x. Each function is obtained by vertically stretching the y = 2^x function. y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x all vertically stretch y = 2^x by a negative factor, so the graphs all lie below the x axis, asymptotic to the negative x axis and approaching negative infinity for positive x. They pass thru the y axis as the respective values y = -3, y = -2, y = -1. y = 1 * 2^x, y = 2 * 2^x, y = 3 * 2^x all vertically stretch y = 2^x by a positive factor, so the graphs all lie above the x axis, asymptotic to the negative x axis and approaching positive infinity for positive x. They pass thru the y axis as the respective values y = 1, y = 2, y = 3. y = 0 * 2^x is just y = 0, the x axis. Of course the functions for fractional values are also included (e.g., y = -2.374 * 2^x) but only the integer-valued functions need to be included in order to get a picture of the behavior of the family. ** STUDENT QUESTION: Ok, it was A = -3 to 3. I understand how to substitute these values into y = A * 2^x. I knew that is was an asymptote, but I'm a little confused as to how to graph the asymptote. INSTRUCTOR RESPONSE: For each value of A you have a different function. For A = -3, -2, -1, 0, 1, 2, 3you have seven different functions, so you will get 7 different graphs. Each graph will contain the points for all values of x. For example the A = -3 function is y = -3 * 2^x. This function has basic points (0, -3) and (1, -6). As x takes on the negative values -1, -2, -3, etc., the y values will be -1.5, -.75, -.375, etc.. As x continues through negative values the y values will approach zero. This makes the y axis a horizontal asymptote for the function. You should figure out the x = 0 and x = 1 values for every one of these seven functions, and you should be sure you understand why each function approaches the negative x axis as an asymptote. *&*&
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RESPONSE --> I said the same thing in a little less detail. self critique assessment: 2
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12:13:11 02-03-2007 12:13:11 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.
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NOTES -------> Each graph will move c units to the either the right or the left (depending on if c is positive or negative). These equations tell you the horizontal positioning in respect to the original equation 2^x
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12:13:13 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.
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RESPONSE --> confidence assessment: 2
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12:14:42 ** There are 7 graphs, including y = 2^x + 0 or just y = 2^x. The c = 1, 2, 3 functions are y = 2^x + 1, y = 2^x + 2 and y = 2^x + 3, which are shifted by 1, 2 and 3 units upward from the graph of y = 2^x. The c = -1, -2, -3 functions are y = 2^x - 1, y = 2^x - 2 and y = 2^x - 3, which are shifted by 1, 2 and 3 units downward from the graph of y = 2^x. **
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RESPONSE --> I said that the graphs will move horizontally, but this is not the case for this particular set. I understand that the graphs will move upward or downward. self critique assessment: 2
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12:36:54 query problem 5. power function families Describe the graph of the power function family y = A (x-h) ^ p + c for p = -3: A = 1, h = -3 to 3, c = 0.
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RESPONSE --> his family contains (after simplifying): y=(x+3)^-3 y=(x+2)^-3 y=(x+1)^-3 y=(x-0)^-3 y=(x-1)^-3 y=(x-2)^-3 y=(x-3)^-3 The graphs will move upward or downward depending on if h is positive or negative. confidence assessment: 1
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12:38:39 ** GOOD STUDENT RESPONSE: I sketched the graph of the power function family y = A (x-h)^p + c for p = -3: A = 1: h = -3 to 3, c = 0. Beginning on the left side of the graph the curve was infinitely close to its asmptote of y = 0. This was determined by the value of c. As we move from left to right the curves decreased at an increasing rate, approaching thier vertical asmptotes which was determined by thier individual values of h. The curves broke at x = c as this value was never possible due to division by zero. The curves resurfaced on the graph high on the right side of thier vertical asymptotes and from there they decreased at a decreasing rate, once again approaching thier horizontal asymptote of y = 0. INSTRUCTOR COMMENTS: Only the h value changes. p=-3, A=1 and c=0, so the functions are y = 1 * (x-h)^-3 or y = (x-h)^-3. For h = -3 to 3 the functions are y = (x - (-3))^-3, y = (x - (-2))^-3, y = (x - (-1))^-3, y = (x - 0)^-3, y = (x - 1)^-3, y = (x - 2)^-3, y = (x - 3)^-3. These graphs march from left to right, moving 1 unit each time. Be sure you see in terms of the tables why this happens. **
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RESPONSE --> In my response I wrote that the graphs moved up and down when I should have said left to right. self critique assessment: 2
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12:40:57 query problem 10 illumination. What function did you evaluate to get your results?
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RESPONSE --> the function ended up being f(x)=370*x^-1 confidence assessment: 2
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12:41:04 ** I determined the illumination y from a certain florescent bulb at the distances of 1, 2, 3, and 4 units using the generalized power function for p = -1 with A = 370, h = 0 and c = 0. This power function is y = A (x- h)^p + c = 370 (x - 0)^(-1) + 0, or just y = 370 x^-1. **
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RESPONSE --> self critique assessment: 3
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12:46:25 Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph.
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RESPONSE --> The distances are 370, 185, 123.33, 92.5 respectively confidence assessment: 3
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12:47:31 ** Student Solution: For x=1 we obtain y=370(-1-0) ^-1=370 For x=2 we obtain y=370(2-0)^-1=185 For x=3 we obtain y=370(3-0)^-1 =123.3 For x=4we obtain y=370(4-0)^-1=92.5**
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RESPONSE --> sometime the solutions show show to get the answers and sometimes they don't. I did use the same equations that the students solutions show to get my answer. self critique assessment: 2
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12:47:51 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 3
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12:47:58 ** STUDENT COMMENT: I have never worked with graphs in the power family, and very little in the exponential family. I am always amazed at the patterns that a function produces. It helps me understand the equation so much better than a list of numbers. I do feel that I need the data table with the graph to fully understand it. INSTRUCTOR RESPONSE: The data table is certainly helpful, especially when you see the reasons for the number patterns in the formula as well as you do. **
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RESPONSE --> self critique assessment: 23
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