course Mth 163
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10:26:47 `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 --> This would still be a quadradic equation which by definition would give us a parabola. This would give us the same parabola except the vertex of y + x^2 - 1 would be at the coordinate (0,-1) whre as the vertex of the y = x^2 is at the coordinate (0,0). The parabola of y + x^2 -1 has been shifted down one unit.
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10:27:20 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Oh Yeah
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11:04:17 `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 --> First, the all the equations are quadratic, and by definition, they will all be parabolas. Starting with the graph y = x^2, the graph will be symetric of the y axis with the vertex on the coordinate (0,0). Adding to the equation -3, -2, -1, 0, 1, 2, 3 for the ""c"" in the equation y - x^2 + c, the graphs would be identical to the original except for the fact that the c = -3 would have its vertex at the coordinate (0,-3), y = -2 would have its vertex at (0,-2), c = -1 would have its vertex at (0,-1), c = 0 would be the same graph as y = x^2 with a vertex at (0,0), c = 1 would have its vertex at (0,1), c = 2 would have its vertex at (0,2), and c = 3 would have its vertex at (0,3)
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11:05:14 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 --> that's the same thing I answered except my reply was quite a bit more entailed and maybe could have been more simply stated.
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11:25:31 `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 --> This is still a form of an exponential function. The graphs would be the same except for a horizontal shift 3 points to the right.
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11:25:47 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 --> YES
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11:34:09 `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 --> All the graphs are the same except horizontal shifts to the right and left depending on if the ""k"" value is negative or positve. The 3 would be shifted 3 points to the right of the original graph of y = x^3, k + 2 is hifted 2 ponts to the right and so on...
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11:34:30 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 understand
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11:45:33 `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 --> This is an exponential function, which by definition would give a graph of an asymptote graph. The graph of y = 2 * 2^x compared to y = 2^x gives me the same graph but horizontal shift in the positive direction 2 points.
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11:45:47 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 --> got it
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11:51:42 `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 --> All the graphs are asymptote in shape raised from 2 points to 5 points from the original graph of y = 2^x
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11:51:54 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 --> OK
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12:09:25 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> In order to find the slope we first have to list two points on the line, and in this case they are given...((3,8) and (9,12)...plugging them into the linear equation of y = mx + b, we have 12 = 9m + b and from this equation we subtract 8 = 3m + b. This gives us 4 = 6m, divide both sides by 6 give us the slope of 4/6.
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12:10:32 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 --> I didn't figure it this way, I plugged it into the linear equation. I didn't reduce it or change it to a decimal equivilent.
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12:18:22 `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 coordinates of t = 5 and t = 9 on the graph of y = 2 t^2 + 3 are (5,53) and (9,165). The slope of this straight line is 28. Plugging in the x and y values into the linear equation for each set of values, then subtracting the two will give you this slope
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12:18:33 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 --> YES
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12:23:14 `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 --> the change should be 28 cm/s
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12:23:33 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 --> ok
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12:26:51 `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 --> Slope of a straight line represents the average rate of 'dy (change in depth)/ 'dt(change in time) of the quantity y with respect to the quantity t...by definition
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12:27:41 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 --> I found this in the notes and copied it...but this is by definition...comment is requested if one is needed
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