course Mth 163 ̓hem]xassignment #006
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21:30:14 `q001. Note that this assignment has 10 questions Recall that the graph of y = x^2 + 3 was identical to the graph of y = x^2, except that it was raised 3 units. This function is of the form y = x^2 + c. In the case of this specific function, c = 3. What function would this form give us for c = -1? How would the graph of this function compare with the graph of y = x^2?
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RESPONSE --> y = x^2 - 1 would be the function for c = -1 The parabola would be very similar to the one for the function for y = x^2 except the parabola would be shifted to the left 1 unit. Also the vertex would be located at the (0, -1) coordinate. confidence assessment: 3
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22:11:16 `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> The final graph would have a vertex of coordinate (0, 3) meaning that they would all have the same shape except each graph for each function c = 3 down to c = -3 would fall one spot for each function. The first graph would have a vertex of (0, -3). confidence assessment: 3
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22:32:53 `q003. The function y = (x -1)^3 is of the form y = (x -k)^3 with k = 1. What function would this form give us for k = 3? How would the graph of this function compare with that of y = x^3?
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RESPONSE --> The function for k = 3 would be y = (x-3)^3. The graph of y = (x-3)^3 has the same ""s"" shape as the y = x^3 except that the graph for the y = x^3 slows and goes through the (0, 0) coordinate and then drastically increases upward. The other lies on the right side of this graph and passes through the x line at (3, 0) or around that coordinate. confidence assessment: 3
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22:38:23 `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> They will be very similar to the previous graph except that k = 2 would lie 2 units to the right of the graph for y = x^3 and k = 4 would lie 4 units to the right. confidence assessment: 3
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22:54:36 `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x?
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RESPONSE --> The function for A = 2 is y = 2 * 2^x. The graph for y = 2^x would look like a line that starts very close to x-axis at x-coordinate -3 then increases to 1 on the y-axis and then goes back down to the (0, 0) and then increases drastically. The other graph is very similar in that it keeps the same pattern and goes through the (0, 0) coordinate, but it is .25 less unit down on the negative side and it increases by double on the positive side of the y-axis. confidence assessment: 2
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22:57:27 `q006. Suppose we wish to graph the functions y = A * 2^x for values A ranging from 2 to 5. If we graph all such functions on the same set of coordinate axes, what will the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> The last graph increases by 5 times with each unit of x, while the others rise by 2, 3, and 4 times with each unit. confidence assessment: 3
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22:59:44 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> The slope of the line would be (12-8) / (9-3) which equals 4/6 or .667 confidence assessment: 3
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23:03:02 `q008. What are the coordinates of the t = 5 and t = 9 points on the graph of y = 2 t^2 + 3? What is the slope of the straight line between these points?
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RESPONSE --> The coordinate for t = 5 is (5, 53) and the thr coordinate for t = 9 is (9, 165). The slope of a line that passes through these points is (165-53) / (9-5) which is equal to 112/4 or 28 confidence assessment: 3
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23:06:09 The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3.
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RESPONSE --> Our slope of the line would represent our average rate of change except that it would be expressed with cm and time. Therefore the answer is 28cm/s self critique assessment: 3
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23:07:12 02-09-2008 23:07:12 `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval?
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NOTES ------->
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23:09:15 `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval?
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RESPONSE --> Because with every observation with a certain time we get a change in depth. Even if the depth is increasing or decreasing we can calculate the average depth change over a certain time period according to the slope. confidence assessment: 3
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23:09:41 end program
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RESPONSE --> OK self critique assessment: 3
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course Mth 163 ƌ[ᩴ`assignment #007
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23:58:37 ** 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 --> 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 self critique assessment: 3
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23:59:42 ** 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 --> A designates the x strectch factor while h affects a y shift & k affects an x shift self critique assessment: 3
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00:14:55 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 --> (20, 58) (40, -18) slope = (-18 - 58) / (40 - 20) = -76/20 or average rate of depth change is 3.8 depth/time confidence assessment: 3
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00:19:34 What is the average rate of depth change from t = 60 to t = 80?
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RESPONSE --> t = 60; (60, -78) t = 80; (80, -122) average rate of depth change is (-122 +78) / (80 - 60) which equals -44/20 or -2.2 depth/time confidence assessment: 3
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00:20:40 ** 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 --> 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 self critique assessment: 3
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00:21:46 ** 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 --> For each interval of `dt = 20 the rate changes by +.8. self critique assessment: 3
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00:23:05 ** 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 --> For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(50.5) - depth(49.5)]/(50.5 - 49.5) = -3 / 1 = -3 self critique assessment: 3
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00:23:27 ** 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 --> For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. self critique assessment: 3
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00:23:58 ** 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 match each other, and they also match the average rate for the interval 40 < t < 60, which is also centered at t = 50. self critique assessment: 3
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00:25:12 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 --> .46 degrees/minute self critique assessment: 3
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00:26:07 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: 3
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