course Mth 163 assignment #005005. `query 5
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16:37:44 `questionNumber 50000 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 --> The rise and run will be the same making the line straight confidence assessment: 2
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16:37:54 `questionNumber 50000 ** 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 --> i see self critique assessment: 2
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16:38:25 `questionNumber 50000 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 --> B/c the negative and positive x's mirror themselves due to the fact that the y stays the same on both sides. confidence assessment: 2
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16:38:36 `questionNumber 50000 ** 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 --> i understand self critique assessment: 2
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16:39:22 `questionNumber 50000 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 --> It increases b/c it is doubled to the exponent (x). The negative value in exponential form causes fractions, thus keeping the line close to the axis. confidence assessment: 3
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16:39:34 `questionNumber 50000 ** 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 --> I understand self critique assessment: 3
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16:40:31 `questionNumber 50000 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 you take the x value to the third power you keep the neg. or pos. sign and its antisymmetric b/c as the pos. value goes higher the neg. value goes lower. confidence assessment: 2
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16:40:39 `questionNumber 50000 ** 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 --> i get it self critique assessment: 2
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16:41:16 `questionNumber 50000 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 --> They are negative in exponential power, thus resorting to fractions and because of the neg. exponential power it stays away from the y-axis. confidence assessment: 2
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16:41:38 `questionNumber 50000 ** 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 understand. self critique assessment: 2
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16:42:06 `questionNumber 50000 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 value of c increases by one when adding c which makes the parabola one higher. confidence assessment: 2
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16:42:14 `questionNumber 50000 ** GOOD STUDENT RESPONSE: The graph of y = x^2 + c, with c varying from -5 to 4 is a series of identical parabolas each 1 unit higher than the one below it. The c value in the quadratic equation has a direct impact on the vertical shift. The vertex of the graph will be shifted vertically by the amount of the c value, so every time c increases by 1 the graph is raised 1 unit. **
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RESPONSE --> i get it self critique assessment: 2
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16:42:38 `questionNumber 50000 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 --> There is a different graph for each equation confidence assessment: 2
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16:42:50 `questionNumber 50000 ** 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 see. self critique assessment: 2
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16:43:14 `questionNumber 50000 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.
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RESPONSE --> There is a graph for each equation. confidence assessment: 2
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16:43:22 `questionNumber 50000 ** 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 see. self critique assessment: 2
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16:43:35 `questionNumber 50000 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 --> The graphs go from left to right. confidence assessment: 2
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16:43:48 `questionNumber 50000 ** 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 --> i understand the solution self critique assessment: 2
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16:44:12 `questionNumber 50000 query problem 10 illumination. What function did you evaluate to get your results?
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RESPONSE --> I used power function confidence assessment: 2
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16:44:21 `questionNumber 50000 ** 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 --> I see self critique assessment: 2
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16:44:53 `questionNumber 50000 Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph.
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RESPONSE --> Plug into equation and get 1. 370 2. 185 3. 123 4. 92 confidence assessment: 2
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16:45:03 `questionNumber 50000 ** 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 --> i got it. self critique assessment: 2
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16:45:14 `questionNumber 50000 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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16:45:24 `questionNumber 50000 ** 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: 3
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