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21:34:31 `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?......!!!!!!!!...................................
RESPONSE --> The function for the form of y = x^2 + c where c is equal to -1 would be y = x^2 -1. The graph of this function would be moved down by 1 unit as compared to the graph of y = x^2.
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21:35:28 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 --> ok
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21:42:58 `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 show the first graph of y = x^2 - 3 as being 3 units below the original with the origin at (0, 0). The second graph of y = x^2 - 2 would be 2 units underneath the original yet 1 above the previous equation. The third graph of y = x^2 - 1 would be 1 unit below the original equation, 1 unit above the second equation and 2 units above the first equation. The graph of the equation where c = 0 would be the same as the original equation of y = x^2. The graph where c = 1 would be 1 unit above the previous graph. The graph where c = 2 would be 2 units above the original graph of y = x^2. The graph where c = 3 would be 3 units above the original equation.
The entire graph area would effectively include a set of parabolas beginning at -3 on the x axis and raising one unit at a time until it reached 3 on the x axis..................................................
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21:43:17 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.......!!!!!!!!...................................
RESPONSE --> ok
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21:55:57 `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 --> For k = 3, we would get the function of y = (x - 3)^3.
The graph of y = x^3 is a wavy line so the graph of y = (x-3)^3 would be shifted up or down on the y axis, depending on the values given..................................................
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21:57:25 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 --> ok, I have made notes in my notebook. I originally had this answer and then proceeded to solve some equations using numbers to replace the variables.This must have confused me quite a bit as I did not get the correct answers.
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22:03:50 `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 --> Each function will move to the left on the y axis where the number of units it moves is determined by the value of k.
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22:04:20 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 --> I should have gone to the right instead of to the left.
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22:06:06 `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 this form would give us where A = 2 would be y = 2 * 2^x.
The y value would double due to the fact that the 2 is an extra multiplier there..................................................
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22:06:20 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 --> ok
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22:09:04 `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 final graph would be a series of vertically stretched lines. Each one being progressively stretched in proportion to the value of A.
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22:15:57 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 --> ok
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22:21:40 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> Where slope is equal to 'dy / 'dx
y2 - y1 / x2 - x1 = 12 - 8/ 9 - 3 = 4/6 or 2/3 So the slope of the line would be 2/3..................................................
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19:21:50 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..........!!!!!!!!...................................
RESPONSE --> ok
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19:27:13 `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 --> (5, ) and (9, )
y = 2(5)^2 + 3 = 10^2 + 3 = 103 y = 2(9)^2 + 3 = 18^2 + 3 = 324 + 3 = 327 So the points would be (5, 103) and (9, 327) 9 - 5 / 327 - 103 = 4 / 224 = 1 / 56 So the slope is 1 / 56..................................................
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19:28:43 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.......!!!!!!!!...................................
RESPONSE --> Oh I made the mistake of squaring the square I believe. I understand the concept of slope, I just made a foolish mistake and instead of multiplying by 2, I squared the numbers.
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19:33:11 `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 --> t = 5
y = 2 (5)^2 + 3 y = 2(25) + 3 y = 50 + 3 y = 53 t = 9 y = 2 (9)^2 + 3 y = 2(81) + 3 y = 162 + 3 y = 165 9 - 5 / 165 - 53 = 4 / 112 = 1 / 28 It would change at an average rate of 1 / 28 cm/second..................................................
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19:33:26 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.......!!!!!!!!...................................
RESPONSE --> ok
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19:39:23 `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 --> The average rate at which the depth changes is found by dividing the change in depth by the time it takes to make the change. We have to use 2 points to find those changes. As the rate at which the depth changes increases so will the slope of the graph as the change in the points on the graph will reflect that change. This also applies to changes as the depth change slows.
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19:39:32 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 --> ok
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Áiι€¥ì¢‰VŽé¶ÒÁ¥“Jö}åÜ ^Ñ•µµ Student Name: assignment #007
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