course Mth 163 I know I'm a little behind but I am going to take the Major Quiz this week and plan on getting to Test 1 next week. I will get caught back up because I've got more time now to work on this. Sorry! f????d?Z??????????assignment #006
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13:29:14 `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 --> We would have the function y = x^2 - 1. The graph is the same as the graph for y = x^2 except the points are 1 unit down on the graph y = x^2 - 1. The vertex is (0, -1) for y = x^2 - 1 and the vertex for the graph y = x^2 is (0, 0). confidence assessment: 2
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13:30:20 If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2.
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RESPONSE --> Every value on the table for y = x^2 - 1 will lie 1 unit lower than each point on the graph of y = x^2. self critique assessment: 3
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13:40:39 `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 graphs for y = x^2 - 3, y = x^2 -2, and y = x^2 -1 would all lie below the graph of y = x^2. Each would lie one more unit below than the graph before hand with the vertexes of (0, -3), (0, -2), and (0, -1). The graph for y = x^2 + 0 would be the same for the graph of y = x^2 with it's vertex at (0, 0). The graphs of y = x^2 + 1, y = x^2 + 2, and y = x^2 +3 would lie above the original graph. Each would lie one more unit above the graph before. The vertexes would be (0, 3), (0, 2), and (0,1). confidence assessment: 2
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13:43:17 The graph of the c= -3 function y = x^2 - 3 will lie 3 units lower than the graph of y = x^2. The graph of the c= -2 function y = x^2 - 2 will lie 22222 units lower than the graph of y = x^2. The progression should be obvious. The graph of the c= 3 function y = x^2 + 3 will lie 3 units higher than the graph of y = x^2. The final graph will therefore show a series of 7 functions, with the lowest three units below the parabolic graph of y = x^2 and the highest three units above the graph of this function. Each graph will lie one unit higher than its predecessor.
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RESPONSE --> The graphs will be the same shape just on a different unit. self critique assessment: 3
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13:56:39 `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 --> We would get y = (x - 3)^3 when k = 3. The graph of y = x^3 crosses at (0,0) and starts out in the negative quadrant, increases at a decreasing rate when it gets close to the y axis. Then increases at a increasing rate the further away from the y axis. The graph of y = (x-3)^3 also starts out in the negative quadrant, increasing at an increasing rate, then increasing at a decreasing rate the closer it gets to the y axis. The graph crosses the y axis at (3, 0) confidence assessment: 2
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13:57:27 Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3.
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RESPONSE --> The graph does lie 3 units to the right since the vertex lies at (3,0). The graph seems a little more stretched out than that of y = x^3. self critique assessment: 3
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14:03:44 `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 --> The graph for y = (x-2)^3 would lie 2 units to the right of the graph y = x^3. The graph for y = (x-3)^3 would lie 3 units to the right of the graph y =x^3. The graph for y = (x-4)^3 will lie 4 units to the right of y = x^3. The graphs are similar just to the right of each other. confidence assessment: 2
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14:05:38 The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right.
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RESPONSE --> The graphs will have the same shape self critique assessment: 3
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14:14:12 `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 --> We would get the function y = 2 * 2^x when A = 2. This graph is the same shape at the graph for y = 2^x. The graphs start close to the x axis and increase at an increasing rate. The difference is that the points for y = 2 * 2^x are twice as far from the x axis as those points of the function y = 2^x. confidence assessment: 2
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14:16:54 As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x.
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RESPONSE --> I did look back the the graph for y = 3 * 2^x and it helped me to see that the same would be true for the graph of y = 2 * 2^x. self critique assessment: 3
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14:19:00 `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 graph of y = 2 * 2^x will be twice as far from the x axis. The graph of y = 3 * 2^x will be 3 times as far from the x axis. The graph of y = 4 * 2^x will be 4 times as far from the x axis. The graph of y = 5 * 2^x will be 5 times as far from the x axis. confidence assessment: 2
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14:21:54 These graphs will range from 2 times as far to 5 times as far from the x-axis as the graph of y = 2^x, and will be from 2 to 5 times as steep. The y intercepts of these graphs will be (0,2), (0, 3), (0, 4), (0,5).
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RESPONSE --> The graphs will intercept at points (0,2), (0,3), (0,4), and (0,5). self critique assessment: 3
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14:25:07 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> We can find the slope by getting the rise over the run. 9 - 3 = 6 12 - 8 = 4 The rise over run is 6 / 4 or 3 / 2. The slope of the graph is 3 / 2, for every 3 units the line rises, it will move 2 units to the right. confidence assessment: 2
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14:26:51 07-08-2008 14:26:51 The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666....
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NOTES -------> I got the rise over run opposite. I see that the rise over run would be 2 / 3 instead of 3 / 2. So the rise of a graph is from the y points and the run for a graph is from the x points.
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14:26:52 The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666....
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RESPONSE --> self critique assessment: 2
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15:07:25 `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 --> When t = 5 we will find: y = 2 * 5^2 + 3 = 53 When t = 9 we wil find: y = 2 * 9^2 + 3 = 165 Our coordinates are (5, 53) and (9, 165) The slope of this line is: 165 - 53 = 112 9 - 5 = 4 112 / 4 or 28 The slope of the line with the coordinates (5, 53) and (9, 165) is 28 / 1 confidence assessment: 2
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15:08:02 The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28.
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RESPONSE --> I wasn't sure if it mattered to put 28 / 1, I did it just in case. self critique assessment: 3
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15:13:21 `q009. Suppose y = 2 t^2 + 3 represents the depth, in cm, of water in a container at clock time t, in seconds. At what average rate does the depth of water change between t = 5 seconds and t = 9 seconds?
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RESPONSE --> I believe to find the average rate of change in water depth we need to find the change in depth over the change in time. So, since we have the time of 5 seconds and 9 seconds we can find the water depth. In the previous problem we found the depths of 53 cm and 165 cm, respectively. The change in depth is 165 - 53 = 112 The change in time is 9 - 5 = 4 We have 112 / 4 or 28, which we found in the previous problem to be the slope of the graph. The average rate of the change in water depth is 28 cm per second. confidence assessment: 2
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15:14:47 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 --> With a graph the average rate of change with depth and time will be identical to the calculation of the slope of the graph. It is always the change in depth over the change in time; similar to the change in rise over the change in run. self critique assessment: 2
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15:18:18 `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 --> Similar to the slope, the average rate at which depth changes over the time interval is found by comparing the change in depth over the change in time. We have to have two numbers to find the average rate of change. We have to have two coordinates to find the slope of a graph as well. Both look at the difference between two points, how fast or slow did the change occur. confidence assessment: 1
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15:20:11 The rise of the graph represents the change in the depth y and the run represents the change clock time t. The slope, which is rise/run, therefore represents change in depth/change in clock time, which is the average rate at which the depth changes.
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RESPONSE --> self critique assessment: 3
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