course Mth 163 assignment #006006. `query 6
......!!!!!!!!...................................
17:03:12 `questionNumber 60000 Query 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions?
......!!!!!!!!...................................
RESPONSE --> Quadratic=y=ax^2+bx+c Exponential=y=a*2^(kx)+c Power=a=(x-h)^p+c Linear=y=mx+b confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:03:23 `questionNumber 60000 ** 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. **
......!!!!!!!!...................................
RESPONSE --> got it self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:03:52 `questionNumber 60000 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)?
......!!!!!!!!...................................
RESPONSE --> K affects y and h affects x. a multiplies y away from x axis. confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:03:58 `questionNumber 60000 ** 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. **
......!!!!!!!!...................................
RESPONSE --> i see self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:05:33 `questionNumber 60000 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
......!!!!!!!!...................................
RESPONSE --> after plugging into equation depth(t) = .02t^2 - 5t + 150 the answers are 58 and -18. you find change in depth by -18-58=-76 Change in clock time=40-20=20 then divide -76/20= -3.8 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:05:41 `questionNumber 60000 ** 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 **
......!!!!!!!!...................................
RESPONSE --> i got it self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:07:18 `questionNumber 60000 What is the average rate of depth change from t = 60 to t = 80?
......!!!!!!!!...................................
RESPONSE --> plug in equations and get -122--78:-44(change in depth) gives you -44/20(change in clock time)=-2.2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:07:25 `questionNumber 60000 ** 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 **
......!!!!!!!!...................................
RESPONSE --> i see self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:07:45 `questionNumber 60000 describe your graph of y = .02t^2 - 5t + 150
......!!!!!!!!...................................
RESPONSE --> graph is parabola and opens up. confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:08:06 `questionNumber 60000 ** 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.**
......!!!!!!!!...................................
RESPONSE --> I didn't explain to that extent but see how answer was achieved. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:08:34 `questionNumber 60000 describe the pattern to the depth change rates
......!!!!!!!!...................................
RESPONSE --> Rates were solved in previous equations and the difference in them was -2.2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:09:32 `questionNumber 60000 ** 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. **
......!!!!!!!!...................................
RESPONSE --> i don't know what i was thinking but i understand how the answer was achieved. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:10:45 `questionNumber 60000 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?
......!!!!!!!!...................................
RESPONSE --> 1 sec. interval @ 50 is 49.5 and 50.5 so plug it in the depth(t)=.02^2-5t+150 then subtract, divide to get -3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:10:55 `questionNumber 60000 ** 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. **
......!!!!!!!!...................................
RESPONSE --> i got it. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:11:40 `questionNumber 60000 what is the average rate of change for the six-second time interval centered at the midpoint.
......!!!!!!!!...................................
RESPONSE --> Do same as last 50-3=47 and 50+3 is 53 Plug in and get -3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:11:48 `questionNumber 60000 ** 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. **
......!!!!!!!!...................................
RESPONSE --> i got it self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:11:59 `questionNumber 60000 What did you observe about your two results?
......!!!!!!!!...................................
RESPONSE --> They were the same confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:12:08 `questionNumber 60000 ** 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. **
......!!!!!!!!...................................
RESPONSE --> got it self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:12:24 `questionNumber 60000 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?
......!!!!!!!!...................................
RESPONSE --> approx .45 when plugged it. confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:12:34 `questionNumber 60000 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.**
......!!!!!!!!...................................
RESPONSE --> i was close. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:12:57 `questionNumber 60000 what is the average rate of change for the six-second time interval centered at the midpoint.
......!!!!!!!!...................................
RESPONSE --> approx.46 when plugged in confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:13:11 `questionNumber 60000 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. **
......!!!!!!!!...................................
RESPONSE --> i rounded also. self critique assessment: 2
.................................................