course Mth 163 This is assignment six the last assignment covered by the major quiz. You told me that if I could get through the major quiz that an incomplete was possiible so that I may finish the work. I am taking the quiz first thing Friday morning and will email you when I have completed it. Is this still sufficient for an incomplete at this time? ۮ䟪Iassignment #006
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01:34:20 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 --> Linear Functions y=f(x)=x generalized function y=f(x)=mx+b Quadratic Functions y=f(x)=x^2 generalized function y=f(x)=ax^2+bx+c Exponential functions y=f(x)=2^x generalized function y=f(x)=Ab^x+c or y=f(x)=A(2^kx)+c Power Functions y=f(x)=x^p generaized function y=f(x)=A(x-h)^p+c confidence assessment: 3
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01:35:02 ** 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 --> ok self critique assessment: 3
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01:41:33 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 --> the A value impacts the vertical shift while the value of h impacts the horizontal shift of the graph confidence assessment: 2
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01:46:06 ** 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 --> got it. still learning to speak math :) self critique assessment: 3
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01:58:11 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 --> average depth change is 38 confidence assessment: 2
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02:01:35 ** 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 --> got it I created more data than I needed. I understand my mistakes. self critique assessment: 3
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02:09:38 What is the average rate of depth change from t = 60 to t = 80?
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RESPONSE --> using the previous function information depth(60)=.02(60)^2-5(60)+150=-78 depth(80)=.02(80)^2-5(80)+150=-122 `ddepth=depth(80)-depth(60)=-44 `dctime=80-60=20 avg rate of `ddepth= `ddepth/`dctime=-44/20=-2.2 confidence assessment: 3
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02:10:01 ** 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 --> yes! self critique assessment: 3
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02:29:00 describe your graph of y = .02t^2 - 5t + 150
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RESPONSE --> Using the given information in the previous questions points on the graph were (0,150) (20,58) (40,-18) (60,-78) (80,-122) Vertex (125, -162.5) graphing the given points my sketched looks like it could almost be a straight line , but there is a slight curve that crosses the x axis around x=35 my graph appears to be half of a parabola confidence assessment: 2
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02:30:23 ** 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 --> ok self critique assessment: 2
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02:37:28 describe the pattern to the depth change rates
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RESPONSE --> as time increases the depth change decreases confidence assessment: 2
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02:40:03 ** 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 --> got it. self critique assessment: 2
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02:50:39 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 --> average rate of depth change for 1 second interval at teh midpoint of 50 = -3 confidence assessment: 2
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02:52:22 ** 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 --> I went about getting the answer differently, but my answer was the same. self critique assessment: 2
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02:59:32 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> 6 second interval = 47 to 53 avg rate of change= depth(53)-depth(47)/53-47 =-58.82--40.82/53-47 =-18/6 =-3 confidence assessment: 3
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02:59:49 ** 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 --> yea! self critique assessment: 2
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03:01:41 What did you observe about your two results?
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RESPONSE --> they were equal to each other which tells me that the average rate change per second for the six seconds included in both problems was constant confidence assessment: 2
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03:02:06 ** 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 --> self critique assessment: 2
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03:26:40 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 --> 1 second interval with a midpoint of 50 = 49.5 to 50.5 average rate of change = `dtemp/`dtime temp(50.5)-temp(49.5)/50.5-49.5 =38.0305-38.49025/1 = -.45975 confidence assessment: 3
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03:28:02 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 --> do you try to subject the data points to avoid negative answers? my result was negative self critique assessment: 1
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03:28:56 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> -.4603375 this is slightly different from the 1 second interval result confidence assessment: 2
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03:29:31 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 --> ok self critique assessment: 2
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