course Mth 163 áÏö‹¨Pß°‰›¤¦|¸Lµ›ŸÝ¼^ÉÌCØè»Ó
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10:54:37 `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 --> y = x is a straight line going through the origin, having a slope 1. This line has 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 ---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, therefore, the corresponding point on the graph of y = .5 x will lie .5 units above the x-axis. The corresponding point is (1, .5). The graph of y = .5 x passes through the points (0,0) and (1,.5) confidence assessment: 2
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11:55:24 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 tried to explain the graphs as best I could and my answers were a lot like the actual answer, so that's good. self critique assessment: 3
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11:57:35 `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 a =.5 then our function is y = .5 x with points (0,0) and (1,.5) If a = 2 then our function is y = 2 x, with points (0,0) and (1,2) For .5 < a < 2-- the slopes of all the graphs will lie between .5 and 2 confidence assessment: 3
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11:57:42 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 --> Correct self critique assessment: 3
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11:58:36 `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 --> confidence assessment: 3
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11:59:20 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 --> Sorry I somehow skipped this question. I understand how to do it though. self critique assessment: 3
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12:01:31 `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 --> The graph of y = 2 x lies at every point twice as far the x-axis as the y = x graph The graph passes through the points (0,0) Slope-2 The graph of y = 2x - 2 will lie 2 units below the graph of y = 2 x Slope- 2 Pass through the y-axis at (0, -2) confidence assessment: 2
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12:01:38 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 got it. self critique assessment: 3
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12:02:59 `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 --> The graphs will lie c units upright from the graph of y = 2 x, tthe slope will be 2, passing through the y-axis at the point (0, c) confidence assessment: 2
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12:16:49 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 --> I sort of got ir right, but I understand it all now. self critique assessment: 2
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12:17:53 `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 --> The rise from y = y1 to y = y2 is a rise of y2-y1 Run--- x2-x1 Slope -- (y2-y1) / (x2-x1) confidence assessment: 3
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12:18:00 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 --> CORRECT. self critique assessment: 3
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12:18:56 `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 --> From (x1, y1) to (x, y) the slope will be: Slope(rise/run) = (y - y1) / (x - x1) confidence assessment: 3
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12:19:04 The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1).
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RESPONSE --> This one was simple. self critique assessment: 3
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12:19:42 `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 --> The slope between any two straight lines should be equal. confidence assessment: 3
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12:20:01 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 --> A little different wording, but I knew what I meant. I got it right. self critique assessment: 3
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12:20:50 `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 --> slopes --- (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). When the two equasions are set equal to each other, we get: (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1) confidence assessment: 3
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12:21:03 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 answered this correctly. self critique assessment: 3
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12:22:14 `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 --> We should multiply both sides by (x - x1) to get (y - y1) = (y2 - y1) / (x2 - x1) * (x - x1) We could then add y 1 to both sides to get y = (y2 - y1) / (x2 - x1) * (x - x1) + y1 confidence assessment: 3
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12:22:23 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 --> Correct~! self critique assessment: 3
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