course Mth163 B˳K՜Fassignment #002
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13:19:57 `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 --> Wtih the numbers -3, -2, -1, -, 1, 2, 3 in the x column we get the following for the y column: 3(-3) - 4 = -13 3(-2) - 4 = -10 3(-1) - 4 = -7 3(0) - 4 = -4 3(1) - 4 = -1 3(2) - 4 = 2 3(3) - 4 = 5 The coordinates for the graph will then be (-3 , -13), (-2, -10), (-1, -7), (0, -4), (1, -1), (2, 2), and (3, 5). The graph that corresponds to these coordinates is a straight line that intersects the y-axis at (-4) and intersects the x-axis between the 1 and 2 marks. confidence assessment: 2
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13:23:48 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 --> I didn't describe it using quite those terms but I do believe that I was around the same area. The y-intercept was when y = (-4) and x = 0. I got the x intercept between two coordinates: (1, -1) and (2, 2). It is close to (4/3, 0). self critique assessment: 2
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13:26:27 `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 --> The steepness in the graph for the function y = 3x - 4 does not change. It is constant, for every 1 step the x coordinate increases, the y coordinate increases by 3: x = -3, -2, -1, 0, 1, 2, 3 y = -13, -10, -7, -4, -1, 2, 5 You can see that between each of the x numbers there is a difference on 1 and the difference between the y is 3. confidence assessment: 3
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13:26:56 The graph forms a straight line with no change in steepness.
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RESPONSE --> I came to the same conclusion with the coordinates used. self critique assessment: 3
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13:28:30 `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 --> The slope of the graph using the function y = 3x - 4 would be (1 / 3). For every one step the x axis goes up, the y axis coordinate goes to the right by 3. confidence assessment: 3
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13:34:01 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 did the actual rise over run between each individual point which is 1 / 3. I didn't use two sets of coordinates to find the rise over run but I see how it works. I do see that the graph is a constant rate and the it is increasing because it's going up the x axis as the goes to the right on the y axis. self critique assessment: 2
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13:44:17 `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 --> First is to find the coordinates for the graph using the function y = x^2, when x = 0, 1, 2, 3. x = 0, y = 0^2 = 0 x = 1, y = 1^2 = 1 x = 2, y = 2^2 = 4 x = 3, y = 3^2 = 9 So the coordinates for the graph are (0, 0), (1, 1), (2, 4), and (3, 9). The graph is increasing. The steepness of the graph does change, there is a great difference between the points (1,1) and (2, 4). The rise is still 1 between each point on the x axis. The points on the y axis, however, have a different number of spaces between each of them. The first two points have a difference of 1, the second two, a difference of 3, and then a difference of 5. The graph is increasing at an increasing rate because there is not the same rise/run for each set of coordinates. confidence assessment: 2
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13:46:36 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 understand how that graph works and how it is increasing at an increasing rate. There is a difference between each y coordinate and if we were to graph the points when x = 4, 5, 6 we could understand it even more. self critique assessment: 2
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13:55:20 `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 --> First we find the coordinates for the graph with the function y = x^2: x = -3, y = -3 ^ 2 = 9 x = -2, y = -2 ^ 2 = 4 x = -1, y = -1 ^ 2 = 1 x = 0, y = 0 ^ 2 = 0 The set of coordinates are : (-3, 9), (-2, 4), (-1, 1), and (0, 0). The graph is decreasing, it goes downward as the moves from the left to right. The steepness of the graph does change. For every one point over the coordinates move along the x - axis, there is a different amount of points between the coordinates on the y axis. The graph is decreasing at an increasing rate. confidence assessment: 2
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13:56:42 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 --> It would be decreasing at a decreasing rate, I was looking at it backwards and thought it would be decreasing at an increasing rate. self critique assessment: 2
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14:08:37 `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 --> The coordinates are as follows for the function y = sqrt(x): x = 0, y = sqrt(0) = 0 x = 1, y = sqrt(1) = 1 x = 2, y = sqrt(2) = 1.4 x = 3, y = sqrt(3) = 1.7 (0, 0), (1, 1), (2, 1.4), (3, 1.7). The graph is increasing, even though between the last two points there is only a 0.3 increase, it's still an increase. The steepness of the graph does change, the steepness decreases between the points on the x - axis. I would say the graph is increasing at a decreasing rate because the steepness of the graph when you move from the left to right doesn't get any greater. The difference between the last two points is just 0.3, compared to the difference of 1 between the first two points. confidence assessment: 2
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14:12:04 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 agree the graph is increasing at a decreasing rate, it is exactly like a hill that starts out steep but is easier closer to the top. Even though the rate of increase isn't as great, there is still an increase. I thought the steepness would still be increasing since the y value still changes for each point, still increasing from the first. I see that each value does change less than the other, so I understand how the steepness would be decreasing. self critique assessment: 2
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14:22:16 `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 --> First to find the coordinates for the function y = 5 * 2^(-x) x = 0, y = 5 * 2^(0) = 5 * 0 = 0 x = 1, y = 5 * 2^(-1) = 5 * -2 = -10 x = 2, y = 5* 2^(-2) = 5 * -4 = -20 x = 3, y = 5 * 2 ^(-3) = 5* -6 = -30 (0, 0), (1, -10), (2, -20), (3, -30) The graph is decreasing from left to right. The steepness of the graph is constant, the difference is 10 between each point. Therefore, the graph would be decreasing at a constant rate since the slope is the same between each point. confidence assessment: 2
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14:25:03 ** 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 --> I did not remember the a^(-b) = 1 / (a^b) so my numbers were all thrown off. The graph was still decreasing but it was a constant decrease with my numbers. I see how it would be decreasing at a decreasing rate with the correct numbers. self critique assessment: 2
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14:28:03 `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 increasing because the car is getting farther away(y) the more time(t) that passes since it left. If the car travels faster, the farther is goes it would make the graph increasing at an increasing rate. The distance would get greater with each passing second from the previous. confidence assessment: 2
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14:29:11 ** 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 --> The key is that the car increases speed for every second it travels, if it didn't increase it's speed, the graph would be constant. self critique assessment: 1
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