course MTH 163 I've been a little behind on my assignments. An unexpected trip out of town kind of threw me off. I'll be spending the next few days working my way back up to speed. ^ְE_Yassignment #005
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17:49:28 `q001. Note that this assignment has 8 questions Evaluate the function y = x^2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> y = -3^2 = 9 y = -2^2 = 4 y = -1^2 = 1 y = 0^2 = 0 y = 1^2 = 1 y = 2^2 = 4 y = 3^2 = 9 confidence assessment: 2
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17:49:43 You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.
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RESPONSE --> I understand. self critique assessment: 2
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17:49:58 `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> y = 2^-3 = 0.125 y = 2^-2 = 0.25 y = 2^-1 = 0.5 y = 2^0 = 1 y = 2^1 = 2 y = 2^2 = 4 y = 2^3 = 8 confidence assessment: 2
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17:50:08 By velocity exponents, b^-x = 1 / b^x. So for example 2^-2 = 1 / 2^2 = 1/4. Your y values will be 1/8, 1/4, 1/2, 1, 2, 4 and 8. Note that we have used the fact that for any b, b^0 = 1. It is a common error to say that 2^0 is 0. Note that this error would interfere with the pattern or progression of the y values.
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RESPONSE --> I understand. self critique assessment: 2
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17:50:27 `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> y = -3^-2 = 0.111 y = -2^-2 = 0.25 y = -1^-2 = 1 y = 0^-2 = 0 y = 1^-2 = 1 y = 2^-2 = 0.25 y = 3^-2 = 0.111 confidence assessment: 2
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17:50:38 By the laws of exponents, x^-p = 1 / x^p. So x^-2 = 1 / x^2, and your x values should be 1/9, 1/4, and 1. Since 1 / 0^2 = 1 / 0 and division by zero is not defined, the x = 0 value is undefined. The last three values will be 1, 1/4, and 1/9.
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RESPONSE --> I had forgotten that 1/0 yields an undefined answer. Otherwise I understand. self critique assessment: 1
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17:50:58 `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> y = -3^3 = -27 y = -2^3 = -8 y = -1^3 = -1 y = 0^3 = 0 y = 1^3 = 1 y = 2^3 = 8 y = 3^3 = 27 confidence assessment: 2
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17:51:11 The y values should be -27, -8, -1, 0, 1, 4, 9.
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RESPONSE --> I don't understand the last two y values. I don't see how 2^3 = 4 and how 3^3 = 9. self critique assessment: 1
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17:51:29 `q005. Sketch graphs for y = x^2, y = 2^x, y = x^-2 and y = x^3, using the values you obtained in the preceding four problems. Describe the graph of each function.
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RESPONSE --> y = x^2 gives a parabola that opens upwards. y = 2^x gives an asymtope that flows upwards to the right.
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17:51:45 The graph of y = x^2 is a parabola with its vertex at the origin. It is worth noting that the graph is symmetric with respect to the y-axis. That is, the graph to the left of the y-axis is a mirror image of the graph to the right of the y-axis. The graph of y = 2^x begins at x = -3 with value 1/8, which is relatively close to zero. The graph therefore starts to the left, close to the x-axis. With each succeeding unit of x, with x moving to the right, the y value doubles. This causes the graph to rise more and more quickly as we move from left to right. The graph intercepts the y-axis at y = 1. The graph of y = x^-2 rises more and more rapidly as we approach the y-axis from the left. It might not be clear from the values obtained here that this progression continues, with the y values increasing beyond bound, but this is the case. This behavior is mirrored on the other side of the y-axis, so that the graph rises as we approach the y-axis from either side. In fact the graph rises without bound as we approach the y-axis from either side. The y-axis is therefore a vertical asymptote for this graph. The graph of y = x ^ 3 has negative y values whenever x is negative and positive y values whenever x is positive. As we approach x = 0 from the left, through negative x values, the y values increase toward zero, but the rate of increase slows so that the graph actually levels off for an instant at the point (0,0) before beginning to increase again. To the right of x = 0 the graph increases faster and faster. Be sure to note whether your graph had all these characteristics, and whether your description included these characteristics. Note also any characteristics included in your description that were not included here.
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RESPONSE --> My graphs did contain all these characteristics, but my description wasn't quite as detailed as explained here self critique assessment: 1
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17:51:58 `q006. Make a table for y = x^2 + 3, using x values -3, -2, -1, 0, 1, 2, 3. How do the y values on the table compare to the y values on the table for y = x^2? How does the graph of y = x^2 + 3 compare to the graph of y = x^2?
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RESPONSE --> The table for y = x^2 + 3 contains the y values 12, 7, 4, 3, 4, 7, and 12. The table for y = x^2 contains the y values 9, 4, 1, 0, 1, 4, and 9. In other words, all of the y values in the first table are three points higher than the y values in the second table. The shape of the graph of the first table appears identical to the shape of the graph of the second table. It has just shifted three points up the y-axis. confidence assessment: 2
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17:52:09 A list of the y values will include, in order, y = 12, 7, 4, 3, 4, 7, 12. A list for y = x^2 would include, in order, y = 9, 4, 1, 0, 1, 4, 9. The values for y = x^2 + 3 are each 3 units greater than those for the function y = x^2. The graph of y = x^2 + 3 therefore lies 3 units higher at each point than the graph of y = x^2.
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RESPONSE --> I understand. self critique assessment: 2
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17:52:23 `q007. Make a table for y = (x -1)^3, using x values -3, -2, -1, 0, 1, 2, 3. How did the values on the table compare to the values on the table for y = x^3? Describe the relationship between the graph of y = (x -1)^3 and y = x^3.
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RESPONSE --> The y valaes for the table of y = (x-1)^3 are -64, -27, -8, -1, 0, 1, and 8. The y values for the table y = x^3 are -27, -8, -1, 0, 1, 8, and 27. The y values on the first table are identical to those on the second table, except one point lower to the corresponding x values on the second table. The shape of the two graphs are identical. The first one is just shifted one point to the left on the x-axis. confidence assessment: 2
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17:52:34 The values you obtained should have been -64, -27, -8, -1, 0, 1, 8. The values for y = x^3 are -27, -8, -1, 0, 1, 8, 27. The values of y = (x-1)^3 are shifted 1 position to the right relative to the values of y = x^3. The graph of y = (x-1)^3 is similarly shifted 1 unit to the right of the graph of y = x^3.
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RESPONSE --> I had accidentally stated 'left' in my answer when I really meant 'right.' I understand. self critique assessment: 1
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17:52:46 `q008. Make a table for y = 3 * 2^x, using x values -3, -2, -1, 0, 1, 2, 3. How do the values on the table compare to the values on the table for y = 2^x? Describe the relationship between the graph of y = 3 * 2^x and y = 2^x.
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RESPONSE --> The y values for the table of y = 3*2^x are 0.375, 0.75, 1.5, 3, 6, 12, and 24. The y values for the table of y = 2^x are 0.125, 0.25, 0.5, 1, 2, 4, and 8. The y values of the first table are three times larger than the values of the second table. The shape of the two graphs are identical. The position of the second graph is just shifted slightly to the right. confidence assessment: 2
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17:52:58 You should have obtained y values 3/8, 3/4, 3/2, 3, 6, 12 and 24. Comparing these with the values 1/8, 1/4, 1/2, 1, 2, 4, 8 of the function y = 2^x we see that the values are each 3 times as great. The graph of y = 3 * 2^x has an overall shape similar to that of y = 2^x, but each point lies 3 times as far from the x-axis. It is also worth noting that at every point the graph of y = 3 * 2^x is three times as the past that of y = 2^x.
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RESPONSE --> I didn't construct my graph large enoguh to see that each point of the first table lied three times as far from the x-axis than the second. Going back and fiddling with it I see that now. self critique assessment: 1
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qLxϹLNÇ assignment #005 005. `query 5 Precalculus I 09-28-2007
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17:55:22 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 --> When y and x have the same values, the rise and the run will the same, therefore the slope will be prefectly square, giving way to a perfectly straight diagonal line. confidence assessment: 2
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17:55:41 ** 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 understand. self critique assessment: 2
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17:58:55 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 --> When a negative number is squared it yields a positive result (a negative times a negative is a positive.) The resulting positive x-values match the positive x-values given for the squared positive y-values. This gives a graph symmetrical on both sides of x=0. confidence assessment: 2
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17:59:09 ** 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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18:02:56 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 --> Beginning at the x-axis, all positive y-values double for each x-value. This causes the graph to increase rapidly. All the negative y-values decrease by half for each x-value. This causes the graph to decrease slowly. confidence assessment: 2
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18:04:15 ** 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: 2
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18:07:25 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 --> The graph on the positive side of the x-axis increases due to positive results. However, the graph on the negative side of the x-axis decreases due to negative results. When a negative number is cubed it yields a negative result, unlike a square which yields a positive result. confidence assessment: 2
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18:07:48 ** 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 understand. self critique assessment: 2
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18:13:43 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 --> On the negative side of x=0 the value decreases by a fraction of it's corresponding x-value. That is if the x-value is a, the corresponding y-value will be 1/a. On the positive side of x=0. You could look at this as the same reason the graph rises so steeply as x approaches 0. confidence assessment: 2
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18:14:35 ** 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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18:18:01 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 heading of the exaplins that there is a corresponding graph for each value of c=-5 to c=4. In each graph the factors A and h remain constant. The only difference is the factor c, which determines vertical shift. This therefore positions each graph one apart vertically on the axis. confidence assessment: 2
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18:18:15 ** 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 understand. self critique assessment: 2
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19:48:15 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 corresponding to the A-values -3 to -1 the graph slowly increases as it approaches the x-axis and rapidly decreases as it approaches the y-axis. For the graphs corresponding to the A-values 1 to 3 the graphs slowly decrease as they approach the x-axis and rapidly increase as they approach they y-axis. The two sets are perfectly symmetrical of each other at the x-axis. The graph corresponding to the A-value 0 lies directly on the x-axis. confidence assessment: 2
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19:49:15 ** 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 understand. self critique assessment: 2
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20:25:55 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.
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RESPONSE --> All the graphs have an identical shape in that they stretch in a straight line from negative infinity begin rapidly curving upwards. The graph corresponding to the c-value -3 stretches in from y-value -3 and crosses the y-axis at y-value 0-2. The graph corresponding to the c-value -2 stretches in from y-value -2 and crosses the y-axis at y-value -1. The graph corresponding to the c-value -1 stretches in from y-value -1 and crosses the y-axis at y-value 0. The graph corresponding to the c-value 0 stretches in from y-value 0 and crosses the y-axis at y-value 1. The graph corresponding to the c-value 1 stretches in from y-value 1 and crosses the y-axis at y-value 2. The graph corresponding to the c-value 2 stretches in from y-value 2 and crosses the y-axis at y-value 3. The graph corresponding to the c-value 3 stretches in from y-value 3 and crosses the y-axis at y-value 4. confidence assessment: 2
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20:26:21 ** 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 understand. self critique assessment: 2
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20:36:36 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 in the family are: y = (x+3) ^ -3 y = (x+2) ^ -3 y = (x+1) ^ -3 y = x ^ -3 y = (x-1) ^ -3 y = (x-2) ^ -3 y = (x-3) ^ -3 All the graphs have the same shape in that they stretch in from infinity negative x, decrease slowly and then rapidly to infinity negative y, shoot straight up along the y-axis, increase slowy and then rapidly to infinity positive y, and stretch on to infinity positive x. Being as that the only difference in each is the h (the horizontal shift), each graph lies one value apart from each other on the x-axis from -3 to 3. confidence assessment: 2
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20:37:02 ** 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. self critique assessment: 2
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20:40:43 query problem 10 illumination. What function did you evaluate to get your results?
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RESPONSE --> I used the function f(x) = A (x-h)^p + c to get my results, obtaining the formula f(x) = 370 (x^-1) For the distance of 1: f(x) = 370 (1^-1) = 370 (1) = 370 For the distance of 2: f(x) = 370 (2^-1) = 370 (1/2) = 185 For the distance of 3: f(x) = 370 (3^-1) = 370 (1/3) = 123.333 For the distance of 4: f(x) = 370 (4^-1) = 370 (1/4) = 92.5 confidence assessment: 2
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20:41:44 ** 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 went ahead and gave my results, giving a ""little too much answer"", but I understand. self critique assessment: 2
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20:42:49 Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph.
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RESPONSE --> My results again: For the distance of 1: f(x) = 370 (1^-1) = 370 (1) = 370 For the distance of 2: f(x) = 370 (2^-1) = 370 (1/2) = 185 For the distance of 3: f(x) = 370 (3^-1) = 370 (1/3) = 123.333 For the distance of 4: f(x) = 370 (4^-1) = 370 (1/4) = 92.5 confidence assessment: 2
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20:43:00 ** 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 understand. self critique assessment: 2
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20:44:16 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I'm slowly starting to understand and comprehend how complex equations can look on a graph. self critique assessment: 2
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20:44:43 ** 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 --> I agree with the student's comment totally. self critique assessment: 2
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