course Mth jиլ˞㒌R͂assignment #006
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12:01:07 Query 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions?
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RESPONSE --> The four basic function families are y=x, y=x^2, y=2^x, and y=x^p. confidence assessment: 2
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12:03:22 ** STUDENT RESPONSE: Linear is y=mx+b Quadratic is y=ax^2 + bx +c Exponential is y= A*2^ (kx)+c Power = A (x-h)^p+c INSTRUCTOR COMMENTS: These are the generalized forms. The basic functions are y = x, y = x^2, y = 2^x and y = x^p. **
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RESPONSE --> The linear is y=mx+b Quadratic is y=ax^+bx+x Exponential is y=A*2^(kx)+c Power= A(x-h)^p+c I did not read the question all the way through and s I only answered part of the question. I now realize whatt the functions are. self critique assessment: 3
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12:05:54 For a function f(x), what is the significance of the function A f(x-h) + k and how does its graph compare to that of f(x)?
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RESPONSE --> k is the y shift h is the x shift A is the multiplier A f(x-h)+k is used by x which is he vertical stretch by A and the horizontal shift h and vertical shift k. confidence assessment: 2
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12:10:12 ** STUDENT RESPONSE: A designates the x strectch factor while h affects a y shift & k affects an x shift INSTRUCTOR COMMENTS: k is the y shift for a simple enough reason -- if you add k to the y value it raises the graph by k units. h is the x shift. The reason isn't quite as simple, but not that hard to understand. It's because when x is replaced by x - h the y values on the table shift 'forward' by h units. A is a multiplier. When all y values are multiplied by A that moves them all A times as far from the x axis, which is what causes the stretch. Thus A f(x-h) + k is obtained from f(x) by vertical stretch A, horizontal shift h and vertical shift k. The two aren't the same, but of course they're closely related. **
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RESPONSE --> k is the y shift and when you add k to the y value you are able to raise the graph by k number of units. h is the x shift and when x is replaced by x-h the y values on the table seem to shift forward by h units. A is the multiplie when the y values are multiplied by A that moves A number of times as far as the x axis will stretch. This makes A f(x-h)+k is found from f(x) and is vertically stretched by A, a horizontal shift h and vertical shift by k. self critique assessment: 3
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13:19:56 query introduction to rates and slopes, problem 1 ave vel for function depth(t) = .02t^2 - 5t + 150 give the average rate of depth change from t = 20 to t = 40
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RESPONSE --> The depth(20) when pluged into the formula equals 58. The depth(40) when pluged into the formula equals -18. The change in depth when you subtract -18 from 58 is -76. The change in clock time when you subtract 40 from 20 is 20. THe average rate is found by taking -76 and dividing it by 20 which gives you -3.8. confidence assessment: 2
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13:22:32 ** depth(20) = .02(20^2) - 5(20) + 150 = 58 depth(40) = .02(40^2) - 5(40) + 150 = -18 change in depth = depth(40) - depth(20) = -18 - 58 = -76 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -76 / 20 = -3.8 **
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RESPONSE --> The depth(20)= .02(20^2)-5(20)+150=58 The depth(40)= .02(40^2)-5(40+150=-18 Then you take the change in time and you have 40-40=20 and then you take the change in depth which is -18-58=-74. You then take the two and divide it which is -76/20= -3.8. self critique assessment: 3
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13:26:39 What is the average rate of depth change from t = 60 to t = 80?
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RESPONSE --> The depth(60) is -78 The depth(80) is -122 THe change in depth is -44 THe change in time is 20 The average rate is -2.2. confidence assessment: 2
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13:28:32 ** depth(60) = .02(60^2) - 5(60) + 150 = -78 depth(80) = .02(80^2) - 5(80) + 150 = -122 change in depth = depth(80) - depth(60) = -122 - (-78) = -44 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -44 / 20 = -2.2 **
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RESPONSE --> Depth(60)= .02(60^2)-5(60)+150=-78 Depth(80)- .03(80^2)-5(80)+150= -122 The change in depth is -122-(-78) = -44 The change in clock time is 20. The average rate of depth change is -44/20= -2.2. self critique assessment: 3
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13:32:27 describe your graph of y = .02t^2 - 5t + 150
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RESPONSE --> y=.02t^2-5t+150. You have x=-b/(2a)=-(-5)/2*.02=125. The graph opens upward and corsses abpit t=35 and t=215. Til the graph gets to t=125 the graph is decreasing at a decreasing rate. confidence assessment: 2
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13:37:16 ** The graph is a parabola. y = .02t^2 - 5t + 150 has vertex at x = -b / (2a) = -(-5) / (2 * .02) = 125, at which point y = .02 (125^2) - 5(125) + 150 = -162.5. The graph opens upward, intercepting the x axis at about t = 35 and t = 215. Up to t = 125 the graph is decreasing at a decreasing rate. That is, it's decreasing but the slopes, which are negative, are increasing toward 0.**
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RESPONSE --> self critique assessment:
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13:39:36 describe the pattern to the depth change rates
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RESPONSE --> The change is rate is -3.8, -3, and -2.2. (20,40), (40,60) and (60,80). Each has an interval of `dt=20 with a rate change of .8. confidence assessment: 2
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13:41:02 ** Rates are -3.8, -3 and -2.2 for the three intervals (20,40), (40,60) and (60,80). For each interval of `dt = 20 the rate changes by +.8. **
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RESPONSE --> The raes are -3.8, -1, and -2.2 for the three intervals at (20,40), (40,60), and (60,80). For each of the intervals at `dt=20 and the rate of change would be .8 positive. self critique assessment: 3
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13:44:27 query problem 2. ave rates at midpoint times what is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint?
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RESPONSE --> The average rate of depth at the chage for a 1 second time interval and centered at the 50 sec midpoint is -3. confidence assessment: 2
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13:46:46 ** The 1-sec interval centered at t = 50 is 49.5 < t < 50.5. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(50.5) - depth(49.5)]/(50.5 - 49.5) = -3 / 1 = -3. **
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RESPONSE --> When you take the 1-second time interval and it is centered at t=50, 49.5
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13:48:39 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> The average rate of change for a 6 second interva; that is centered at the midpoint is -3. confidence assessment: 2
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13:50:14 ** The 6-sec interval centered at t = 50 is 47 < t < 53. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. **
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RESPONSE --> The 6 second interval is centered at t=50 is 47
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13:52:01 What did you observe about your two results?
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RESPONSE --> The two rates seem to match each other at the average rate for 40
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13:53:34 ** The two rates match each other, and they also match the average rate for the interval 40 < t < 60, which is also centered at t = 50. For a quadratic function, and only for a quadratic function, the rate is the same for all intervals having the same midpoint. This is a property unique to quadratic functions. **
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RESPONSE --> The two rates will match each other when they are matched with the average for each interval at 40
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13:54:24 query problem 3. ave rates at midpt times for temperature function Temperature(t) = 75(2^(-.05t)) + 25. What is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint?
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RESPONSE --> The average is .46degrees/minute confidence assessment: 2
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13:55:05 STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: More precisely .4595 deg/min, and this does not agree exactly with the result for the 6-second interval. Remember that you need to look at enough significant figures to see if there is a difference between apparently identical results.**
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RESPONSE --> Without rounding it, it is .4595degrees/minutes and this is at 6 second intervals. self critique assessment: 3
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13:58:19 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> The average rate of change for the 6 second time intercal at the midpoint is .46degrees/minute. confidence assessment: 2
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13:59:45 STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: The 1- and 6-second results might possibly be the same to two significant figures, but they aren't the same. Be sure to recalculate these according to my notes above and verify this for yourself. The average rate for the 6-second interval is .4603 deg/min. It differs from the average rate .4595 deg/min, calculated over the 1-second interval, by almost .001 deg/min. This differs from the behavior of the quadratic. For a quadratic that the results for all intervals centered at the same point will all agree. This is not the case for the present function, which is exponential. Exact agreement is a characteristic of quadratic functions, and of no other type. **
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RESPONSE --> The 1 and 6 seconds could be the same as the 2 figures but they are not. The average rate at the 6second interval is .4603degrees/minute. It is different from .4595degrees/minute with the 1 second interval by about .001 degrees/minute. self critique assessment: 3
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