course Mth 152 After working on all these problems ... I believe that I did them in error. There was a no beside this section on Mth 152. I did not complete the entire section ... please let me know if that needs to be finished. Next work submitted will be from Asst with yes beside it.Thank you.
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16:02:06 `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 --> confidence assessment:
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Ā˰咚 assignment #002 002. Describing Graphs qa initial problems 01-12-2008
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16:15:58 `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 --> We make a table for y=3x-4 as follows: We construct two columns, the first column is 'x' and the second column is 'y'. We put the numbers -3, -2, -1, 0, 1, 2, 3 in the 'x' column. When we substitue -3 into the expression, we get y = 3(-3) - 4= -13. When we substitute -2 into the expression y = 3(-2) - 4 = -10. Substituting the remaining numbers for 'x' we get the 'y' values -7, -4, -1, 2 and 5. 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. These points lie on a straight line and we then contruct a line through the points. confidence assessment: 2
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16:19:30 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 --> My graph does confirm that the y-intercept is at (0,4) when graphing the values for y=3x-4 self critique assessment: 3
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16:21:09 `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 of the graph in the preceding exercise (of the function y=3x-4) does not change. As the x value goes up by 1, the y value always goes up by 3. confidence assessment: 3
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16:21:35 The graph forms a straight line with no change in steepness.
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RESPONSE --> I was correct ... the graph forms a straight line with no change in steepness. self critique assessment: 3
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16:29:43 `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 --> To calculate the slope of a line we use the formula (slope = rise/run = change in y/change in x = m). Using the two points (2,2) and (3,5) we would fill in the formula as m = 3-2 / 5-2 = 1/3. Therefore the slope of function ia y + 3x - 4 is 1/3. confidence assessment: 3
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16:32:56 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 missed that one ... had the y values as the denominator when it should have been the numerator. My answer was 1/3 and it should have been 3. self critique assessment: 1
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16:55:58 `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 --> y = x^2 We first make a table with two columns ... the left column being the 'x' value the right column being the 'y' value. When using 0, 1, 2, 3 for the value of x then the value of y is 0, 1, 4, 9. x values are place in x column and y values are placed beside subsequent x values in y column. The graph is definitely increasing. The steepness of the graph is getting higher. As the x value continues to go up only one at a time, the y value increases since the increasing value of x is squared. The graph is increasing at an increasing rate. confidence assessment: 2
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16:56:43 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 --> My answer was correct. self critique assessment: 2
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17:01:02 `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 --> Table for y = x^2 We make a table with 'x' values in left column and 'y' values in right column. With x values being -3, -2, -1, 0 the y values will be 9, 4, 1, 0. The graph is decreasing. The steepness of the graph is also decreasing. As the x value gets closer to 0, the steepness gets less. The graph is decreasing at a decreasing rate. confidence assessment: 3
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17:01:33 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 --> My answer was correct. self critique assessment: 2
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17:09:46 `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 --> Table for y= 'sqrt(x) When x values in the table are 0, 1, 2, 3 the subsequent y values are 0, 1, 1.45, 1.73. The graph is increasing. The steepness of the graph however is decreasing. As the x values go up and move to the right, the y values are also going up but at a decreasing rate. Therefore the graph is increasing at a decreasing rate. confidence assessment: 2
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17:10:08 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 --> self critique assessment: 2
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17:17:12 `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 --> Table for y = 5* 2^(-x) When the x values are 0, 1, 2, 3 the subsequential y values are 0, -10, -20 ,-40. The graph is decreasing and the steepness of the graph is increasing. As the x value increases, the y value decreases by a greater amount each time. The graph is decreasing at an increasing rate. confidence assessment: 2
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17:18:22 ** 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 --> It's been a while since I had basic algebra ... sure messed that one up. self critique assessment: 1
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KGV҂meܜ assignment #002 002. Describing Graphs qa initial problems 01-12-2008 ت[¿Rϑ assignment #002 002. Describing Graphs qa initial problems 01-12-2008