course Mth 163 ????????????assignment #005005. `query 5
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14:42:18 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 --> Every y point is equal to the x point, (1,1), (2,2), (3,3), and so on. confidence assessment: 1
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15:05:08 ** 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 --> The slope of the graph would be 1 since the rise = run when y = x. self critique assessment: 2
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15:10:48 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 x is raised to the power of 2, the y value will always be a positive number. Negative time negative equals a positve, so with y = x^2, -2, -3, -4 will have the same y values as 2, 3, 4. So the graph will be symmetric, the line on the right side of x = 0 will look the same as the line on the left side of x = 0. confidence assessment: 1
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15:13:16 ** 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 --> A reflection is a good way to describe the graph y = x^2. self critique assessment: 3
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15:20:55 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 y = 2^x graph keeps increasing as x increases because 2 is raised to the power of whatever the x value is. When 2^2 = 4, 2^3 = 8, 2^4 = 16, and so on. The second value is twice and big as the first, the third twice as big as the second, the fourth twice as big as the third. That's the pattern with y = 2^x. When the x value is negatvie, y = 2^x = 1/2^x; so 2^-2 would be 1 / 2^2 = 1 / 4. Since the values do increase when 2^x, 1 / 2^x will continue to get closer to the x-axis. confidence assessment: 2
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15:23:43 06-24-2008 15:23:43 ** 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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NOTES -------> For the negative side, each unit we move negatively, the value will be one half of the previous value.
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15:23:44 ** 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 --> self critique assessment: 2 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 x is a negative value, y will be a negative value as well. When x is positive, y will be positive as well. The graph is antisymmetric because it is partly in the quadrant where the values are positive and when it passes through 0, the rest of the graph is in the quadrant with only negative values. confidence assessment: 1
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14:52:48 ** 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 --> self critique assessment: 3
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14:58:08 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 --> When x is raised to a negative power it will become 1 / x ^ b. The only whole number will be 1, the rest of the values will be fractions: 1 / -3 ^ 2 = 1 / 9 and so on. Both graphs will approach the x axis, closer and closer but never touching because the values will always be fractions. confidence assessment: 1
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15:00:48 ** 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 --> The denominator is increasing at an increasing rate. They will never reach x = 0 because of the fractions. self critique assessment: 3
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15:08:13 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 --> When there is a variable added onto the equation, it raises the graph on the y axis. When x = 0 and c = 1, we get the y value as 1. When c = 2, we get y = 2 and so on. The vertex will have a x value of 0 and the y value will be what the c value is. So we would have (0,1), (0,2), (0,3), (0,4)........ confidence assessment: 2
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15:10:16 ** 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 --> The vertex of the graph will be shifted vertically by the amount of the c value, every time the c value increases, the vertex will raise by 1. self critique assessment: 2
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15:23:53 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 A values of -3, -2, and -1, the y values will be negative also. When the A values are positive, the y values will be positive. The -3 and 3 have the same values for y, just the ' - ' will be ' - ' and the positive, positive. When A = 0, y will be 0 and the graph will be a straight line. confidence assessment: 1
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15:27:58 ** 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 --> The negative A values will cause the graph y = 2^x to be vertically stretched, so they will lie below the x axis. Asymptotic to the negative x axis and approaching the negative infinity for positive x. The positive A values will vertically stretch the graph, so they lie above the x axis. Doing the same thing but just on the positive x axis side. self critique assessment: 3
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15:43:48 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 will begin close to the x axis, but not touching the x axis. Once passing through the y axis, the graph will begin to increase at an increasing rate. We have y = 2^x + -3, 2^x + -3, 2^x + -1, 2^x + 0, 2^x + 1, 2^x + 2, and 2^x + 3. With every 1 unit the c value increases, the graph points will rise 1 unit also. The graphs will look the same to each other, just raised up 1 unit from the one preceding. confidence assessment: 2
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15:44:55 ** 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 --> From the original y = 2^x, with the negative c values, the graphs will be shifted downward 1, 2, and 3 units. When the c values are positive, the graphs will be shifted 1, 2, and 3 upward from the graph. self critique assessment: 2
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16:05:26 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 --> We would get the functions: y = 1 (x - -3) ^ -3 + 0 y = 1 (x - -2) ^ -3 + 0 y = 1 (x - -1) ^ -3 + 0 y = 1 (x - 0) ^ -3 + 0 y = 1 ( x - 1) ^ -3 + 0 y = 1 ( x - 2) ^ -3 + 0 y = 1 ( x - 3) ^ -3 + 0 Whenever we have a value raised to a negative power we will get a fraction for the y value. The graph will be confidence assessment: 0
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16:08:13 07-01-2008 16:08:13 ** 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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NOTES -------> The graph would be close to its asmptote of y = 0. The curves do decrease at an increasing rate as we move from left to right, approaching their vertical asmptote. 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. 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.
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16:08:14 ** 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 --> self critique assessment: 2
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16:10:26 query problem 10 illumination. What function did you evaluate to get your results?
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RESPONSE --> I'm not sure what kind of function being asked for here for what results. I would assume the problem before which was the function of y = A (x-h) ^ p + c confidence assessment: 0
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16:12:47 ** 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 didn't look to see what the illumination problem was. I see what was done just by substituting the values into the function. self critique assessment: 3
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Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph.
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RESPONSE --> confidence assessment: 0
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08:51:10 ** 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 accidentally hit the enter response button before I answered the question. I see that we just substitute the 1, 2, 3, and 4 values in for the x variable and solve the function out for y. self critique assessment: 3
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08:52:34 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It's neat to see how we can take a simple function such as y = x^2, add something on to it, and see how it affects the graph. self critique assessment: 3
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08:53:28 ** 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 --> It does help me understand the equation better and how it affects the shape of the graph. self critique assessment: 3
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