course PHY 201 WOW this was a refresher I didnt even know I had forgotten. end program}ʅie|z
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21:59:33 `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 --> x y -3 -13 y=3(-3)-4 -9-4 -2 -10 y=3(-2)-4 -6-4 -1 -7 y=3(-1)-4 -3-4 1 -1 y=3(-1)-4 3-4 2 2 y=3(2)-4 6-4 3 5 y=3(3)-4 9-4 Creating a gently sloping line confidence assessment: 2
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22:01:01 `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 --> gotcha confidence assessment:
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22:01:22 The graph forms a straight line with no change in steepness.
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RESPONSE --> that is correct self critique assessment: 2
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22:06:45 `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 be honest I do not remember how to determine slope I know what it is and I know what it measures but this is what I have come up with. x/y = rise/run if x=0 then y=-4 so the slope is 0 if x=1 then y=-1 so the slope is -1 if x=2 then y =2 so the slope is 1 confidence assessment: 0
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22:09:15 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 --> Thats right thats the equation. the coefficient is the slope its M self critique assessment: 1
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22:49:53 `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 --> ok x y -3 9 y=(-3)^2 -2 4 y= (-2)^2 -1 1 y=(-1)^2 0 0 y-(0)^2 1 1 2 4 3 9 so we look like this 3 x 2 x 1 x 0 x -1 x -2 x -3 x 01 4 9 so if x=1 and y=1 if x=2 and y=4 if x=3 and y=9 if x=9 and y=81 so rise over run would be (1,1) (2,4) 1-2=1 1-4=3 3/1=3=slope this slope makes for an increasing graph with a slight steepness creating a right faceing ""V"". I would call this a increasing at a constant rate. confidence assessment: 0
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22:54:01 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 --> okay now I remember maybe I need to refresh my math memory before taking this class I am feeling a bit rusty self critique assessment: 1
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22:57:15 `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 --> this is the same question should i answer it again confidence assessment:
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22:57:38 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 --> ok self critique assessment:
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23:15:14 `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 --> x y y=srtx 0 0 1 1 2 1.41 3 1.73 4 2 5 2.23 6 2.45 7 2.65 8 2.83 9 3 increasing the graph is steep it is increasing at an increasing rate confidence assessment: 0
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23:20:22 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 --> o i see self critique assessment:
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23:34:06 `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 --> ok x y y=5(2^(x)) 0 0 1 -10 2 -20 3 -40 4 -80 The graph is decreasing the steepness is static Id say this graph is decreasing at an increasing rate because although the graph is digging downward it is 10 points 10 points 20 points.so thats not consistent although it is a pattern each number is doubling. confidence assessment: 0
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23:34:44 ** 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 missed that one huge self critique assessment:
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23:36:00 `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 --> increasing well if the car is traveling faster and faster it will be an increasing at an increasing rate graph confidence assessment: 2
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23:36:29 ** 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 --> Y E S ! ! ! self critique assessment: 2-3
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