course Phy 201 I am still unable to find this on the file I have on my hard drive. I have been copying it and pasting it directly from the session after I finish it. What do you think that I have done incorrectly for these files not to be showing up on my hard drive? What do you suggest? I tried loading the software again and that did not work. I have windows Vista do you think this is a problem. Thanks for your help.
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22:01:35 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> The points go through the x axis at (0, 1.5) and the points go through the y axis at (0, -4). confidence assessment: 3
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22:02:37 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> Yes. This is the same answer that I got when working out this graph. self critique assessment: 2
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22:05:00 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> Yes. The graph starts out in at the point (-3, -13) and as it travels up into the quad II and ends at point (3,5) it does become slightly steeper as it rises. confidence assessment: 3
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22:07:29 The graph forms a straight line with no change in steepness.
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RESPONSE --> Yes I understand that the graph forms a straight line. I thought since it was rising that you would consider this part of becoming more steeper. I can clearly tell that the line doesn't change to go straight up but I misread what you were asking. But I do understand that you meant the overall slope of the line and not the direction in which is it going. self critique assessment: 2
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22:16:07 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> When plugging in two points into the formula (Y2-Y1)/(X2-X1) I used the points (-3,-13) and (3,5). After plugging them into the above formula I got : 5-(-13)/3-(-3)= 18/6 = 3 Slope = 3 confidence assessment: 3
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22:16:46 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> I got this correct and I understand how I got the answer. self critique assessment: 2
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22:20:43 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> After plugging in the numbers into the y = x^2 formula. The graph is increasing. The steepness of the graph does change. It becomes more and more steeper ending in quad II. The graph is increasing at an increasing rate. confidence assessment: 2
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22:21:43 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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RESPONSE --> I got this question correct and I understand how I got the answer. I got the same thing that you got. self critique assessment: 2
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22:26:13 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> After plugging in the numbers into the formula I got (0,0) (-1,-1) (-2,-4) (-3, -9) After graphing the number I saw that the graph was decreasing. The steepness of the graph does change becoming steeper as it decreases ending at the point (-3, -9) The graph is decreasing at a decreasing rate. confidence assessment: 2
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22:26:52 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> This is the same answer that I got and I understand and agree with it. self critique assessment: 2
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22:30:27 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> When keying in the numbers into the formula I got the following points: (0,0) (1,1) (2,1.41) (3, 1.73) After plotting these numbers I can see that the graph is increasing. The steepness of the graph is increasing at a pretty constant rate. It is a slight increase confidence assessment: 2
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22:32:26 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> I understand that you would say that the graph is increasing at a decreasing rate. It makes since now that I have read your answer. Since the graph is increasing by less and less each time it is increasing at a decreasing rate. self critique assessment: 2
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22:37:31 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> After substituting the numbers into the formula I got (0,5) (1, 2.5) (2, 1.25) (3, .625) The graph shows that it is decreasing. The steepness of the graph does change it is decreasing at a decreasing rate. confidence assessment: 2
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22:38:13 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> This is the exact same answer that I got and understand how I got it. self critique assessment: 2
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22:43:30 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph would be decreasing. As the car travels away the car gets farther and farther which is increasing and the time stays at a constant rate the car is getting farther and farther away and it is increasing at a decreasing rate. confidence assessment: 2
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22:45:15 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> I see that it would be a greater change in distance with each second traveling causing it to be a greater slope. I was just thinking about it a little different. I see that it would actually be increasing at an increasing rate. self critique assessment: 2
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