course Mth 163 לЎkޓαӒf~zՆassignment #010
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20:13:14 `q001. Note that this assignment has 10 questions Sketch the function y = x and describe your graph. Describe how the graphs of y = .5 x and y = 2 x compare with the graph of y = x.
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RESPONSE --> The graph of y=x is a straight line that goes through (0,0) and has a slope of 1. It moves up the graph like (-1,-1) and (1,1). The graph of y=.5 lie closer to the x-axis. Some of the points are (0,0), (1, .5), and (-1, -.5) The graph y=2x are twice as far as the original graph. The points are (0,0), (1,2) and (-1,2). Depending on what is infront of the x is where the next point on the graph goes. All of these graphs start at (0,0). From there it just depends on what is with x as where the next point goes. confidence assessment: 2
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20:18:46 The graph of y = x consists of a straight line through the origin, having slope 1. This line has basic points (0,0) and (1,1). The points of the graph of y = .5 x all lie twice as close to the x-axis as the points of the graph of y = x. The point (0,0) of the y = x graph is already on the x-axis, so the corresponding point on the graph of y = .5 x is also (0,0). The point (1,1) of the y = x graph lies 1 unit above the x-axis, so the corresponding point on the graph of y = .5 x will lie twice as close, or .5 units above the x-axis, so that the corresponding point is (1, .5). The graph of y = .5 x Thus passes through the points (0,0) and (1,.5). Of course this result could have been found by simply plugging 0 and 1 into the function y = .5 x, but the point here is to see that we can get the same result if we think of moving all points twice as close. This order thinking will be useful when dealing with more complex functions. Thinking along similar lines we expect the points of the graph of y = 2 x to all lie twice as far from the x-axis as the points of the function y = x. Thus the two basic points (0,0) and (1,1) of the y = x graph will correspond to the points (0,0) and (1,2) on the graph of y = 2 x.
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RESPONSE --> I see what is being done. With all of these graphs you can visual what the graphs would look like even before you start. When you go to work them out and you put all of them on the graph and compare them you can see that they have the same origin but the points vary. It basically depends on what is infront of the variable x. When you have y=.5x then the grph is going to be closer to the origin. When you have y=2x then the graph is going to be twice as far from the origin. self critique assessment: 3
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20:26:44 `q002. If we were to sketch all the graphs of the form y = a x for which .5 < a < 2, what would our sketch look like?
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RESPONSE --> If you have y=.5x then the points in the graph will be (-1, -.5), (0,0), and (1, .5). If you have y=2x then the points will be (1,2), (0,0) and (-1, -2). With these figured out you would be able to figure out that graphs for .5 < a< 2 would like somewhere between y=.5x and y=2x. You would then have the slope of .5 and 2 so the graphs would lie between .5 and 2. confidence assessment: 2
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20:29:54 If a =.5 then our function is y = .5 x and the basic points will be (0,0) and (1,.5), as seen in the preceding problem. Similarly if a = 2 then our function is y = 2 x, with basic points (0,0) and (1,2). For .5 < a < 2, our functions will lie between the graphs of y = .5 x and y = 2 x. Since these two functions have slopes .5 and 2, the slopes of all the graphs will lie between .5 and 2. We could represent these functions by sketching dotted-line graphs of y = .5 x and y = 2 x (the dotted lines indicating that these graphs are not included in the family, because the < sign does not include equality). We could then sketch a series of several solid lines through the origin and lying between the two dotted-line graphs.
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RESPONSE --> I understand what is being done. If you need to find a graph that is between .5 < a< 2. You would need to figure out the points from the two main graphs. These would be y=.5x and y=2x. Once you have found these points then you can figure out that the graph would then lie somewhere between the graphs of .5 and 2. self critique assessment: 3
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20:44:07 `q003. Describe how the graphs of y = x - 2 and y = x + 3 compare with the graph of y = x. If we were to sketch all graphs of the form y = x + c for -2 < x < 3, what would our graph look like?
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RESPONSE --> The graph of y=x is a straight line. When you compare it to y=x-2 you see that it's points lie 2 places below thr origin. The graph y=x+3, the points are 3 units above the origin. To find y=x+c for -2 < x <3 it seems that you would taake the y=x-2 and y= x+3 and place those graph. You would see that the answer for y=x+c would lie somewhere between the other two. confidence assessment: 2
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20:51:45 The graph of y = x - 2 lies at every point 2 units below the corresponding point on the graph of y = x, so this graph is parallel to the graph of y = x and 2 units lower. Similarly the graph of y = x + 3 lies parallel to the graph of y = x and 3 units higher. To sketch the family y = x + c for -2 < x < 3, we first can draw dotted-line graphs of y = x - 2 and y = x + 3, then a series of several solid line graphs, all parallel to the graph of y = x, lying between the two dotted-line graphs. STUDENT COMMENT: I got a little confused with y = x + c part, but I understand the first part completely. ** The instructions said to sketch all graphs of the form y = x + c for -2 < x < 3. So for example c could be -1, 0, 1 or 2, giving us the functions y = x - 1, y = x, y = x + 1 andy x+ 2. c could also be -1.9, or .432, or 2.9, giving us functions y = x - 1.9, y = x + .432, y = x + 2.9. c can be any number between -2 and 3. These graphs are as described in the given solution. **
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RESPONSE --> I see how the other graps are done. I wasn't sure how to do the graph for y=x+c for -2
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21:01:26 `q004. Describe how the graph of y = 2 x compares with the graph of y = x. Describe how the graph of y = 2 x - 2 compares with the graph of y = 2 x.
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RESPONSE --> When you have graph y=2x every point on the graph will be twice as far from the origin. When this graph is compared to the graph of y=x you find that the points on the grpah y=2x will be twice as far from the origin than y=x/ When you have the graph of y=2x-2 every point on the graph will be 2 units below the graph y=2x. confidence assessment: 2
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21:07:04 The graph of y = 2 x lies at every point twice as far the x-axis as that of y = x. This graph passes through the points (0,0) and (1, 2), i.e., passing through the origin with slope 2. The graph of y = 2x - 2 will lie 2 units below the graph of y = 2 x. This graph will therefore have a slope of 2 and will pass-through the y axis at (0, -2).
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RESPONSE --> I understand how this is done. It is just a matter of understanding where the points will lie. For the graph y=2x-2 the coordinates will pass through (0,-2). It varies from the graph of y=2x because the first graph will be negative more while the other will be more positive. self critique assessment: 3
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21:14:55 `q005. Suppose we graph y = 2 x + c for all values of c for which -2 < c < 3. What with our graph look like?
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RESPONSE --> You would have to see what the points would be. One of the graphs would be y=x-2. The points on the graph would be two units below the origin. For the graph y=x + 3. The points on the graph would be three units above the origin. Therefore the graph for y= 2x +c for which -2 < c< 3 would lie somewhere between -2 and 3. confidence assessment: 2
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21:17:10 Each graph will lie c units vertically from the graph of y = 2 x, therefore having slope 2 the passing through the y-axis at the point (0, c). The family of functions defined by y = 2 x + c will therefore consist of a series of straight lines each with slope 2, passing through the y-axis between (0, -2) and (0, 3).
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RESPONSE --> When you place the points for the grpah the points would be (0,c), (0, -2), and (0,3). Therefore the points for the graph would basically lie somewhere between -2 and 3. self critique assessment: 3
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21:20:51 `q006. Sketch two points, not particularly close to one another, with one point in the second quadrant and the other in the first, with clearly different y values. Label the first point (x1, y1) and the second (x2, y2). Draw a straight line passing through both of these points and extending significantly beyond both. In terms of the symbols x1, x2, y1, and y2, what is slope of this straight line?
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RESPONSE --> To find the rise you would take y=y1 and y=y2 and the you would have a rise of y2-y1. The run is found the same way. The run would be x2-x1. The slope is then (y2-y1) / (x2-x1). confidence assessment: 2
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21:21:21 The rise of a line is from y = y1 to y = y2, a rise of y2-y1. The run is similarly found to be x2-x1. The slope is therefore slope = (y2-y1) / (x2-x1).
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RESPONSE --> I understand how it is done. It is easy to see and figure out how to find the rise, run, and slope. self critique assessment: 3
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21:24:07 `q007. On the sketch you made for the preceding problem, and add a point (x, y) on your straight line but not between the two points already labeled, and not too close to either. What is the slope from (x1, y1) to (x, y)?
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RESPONSE --> The slope looks to be from (x1, y1) to (x, y). The slope is then equal to the rise/run which is (y-y1)/ (x-x1). confidence assessment: 2
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21:24:43 The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1).
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RESPONSE --> I see how it is done. When you write down the points and see the graph I can understand how to find the slope, rise, and run. self critique assessment: 3
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21:25:34 `q008. Should the slope from (x1, y1) to (x, y) be greater than, equal to or less than the slope from (x1, y1) to (x2, y2)?
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RESPONSE --> When you figure out the slope between both lines you can see the they are a straight line. That would make the slopes equal. confidence assessment: 2
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21:26:00 The slope between any two points of a straight line must be the same. The two slopes must therefore be equal.
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RESPONSE --> I understand it. If the slopes are the same then thay are both a straight line and at the same time they are equal. self critique assessment: 3
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21:28:10 `q009. The slope from (x1, y1) to (x, y) is equal to the slope from (x1, y1) to (x2, y2). If you set the expressions you obtained earlier for the slopes equal to one another, what equation do you get?
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RESPONSE --> The slopes you would have are (y2-y1) / (x2-x1). Also there would be (y-y1) / (x-x1). Then if you set the two slopes equal to each other you would then have (y2-y1)/ (x2-x1) = (y2-y1) / (x2-x1). confidence assessment: 2
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21:28:56 The slopes are (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). Setting the two slopes equal we obtain the equation (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1).
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RESPONSE --> I see. If one slope is one equation and they are eqaul to each other then the two slopes are the same. self critique assessment: 3
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21:32:53 `q010. Setting the two slopes equal you should have obtained (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1). Solve this equation for y.
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RESPONSE --> You would take (y-y1) / (x-x1) = (y2-y1)/ (x2-x1). This would then equal (y2-y1)/(x2-x1). Then you have to multiply both sides by (x-x1) to get y by itself. When you do this you get (y-y1) = (y2-y1) / (x2-x1) * (x-x1). Then you add y1 to both sides. Now you have y= (y2-y1)/ (x2-x1) * (x-x1) + y1. confidence assessment: 2
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21:34:14 Starting with (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), we wish to isolate y on the left-hand side. We therefore begin by multiplying both sides by (x - x1) to obtain (y - y1) = (y2 - y1) / (x2 - x1) * (x - x1). We could then add y1 to both sides to obtain y = (y2 - y1) / (x2 - x1) * (x - x1) + y1.
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RESPONSE --> I see how this is done. In order to solve for y you have to have you equation out in front of you and then get y by itself on the left hand side. In order to do this you have to multiply both sides by (x-x1). After doing that you have to add y1 to both sides. when you do this you get the final equation with y= the answer. self critique assessment: 3
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