course Mth 151 It has been around 7 years since I have taken a math course. Sorry for being a little rusty. I found a little extra time to work on the q_a_ graphing Monday night after I had submitted my typewriter notation. I have copied the part of the graphing that I did on 5/29/06 and then copied the part from 5/30/06 after I finished tonight. Is this okay or would you prefer I just send the entire two files (5/29/06 and 5/30/06).
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20:27:15 `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. 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 --> y=3x-4 substitute -3 into the expression and you get -13. y = 3(-3) -4 or y = -9 -4 or y = -9+ -4 or y = -13 X Y _____________________ -3 -13 -2 -10 -1 -7 0 -4 1 -1 2 2 3 5 The Y intercept is 0,-4 They X intercept is 4/3 , 0
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20:30:07 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 was unclear about what I should do. I used the numbers from the example to plug into the equation also. I had forgotten how to find the x and y intercepts so I looked in the book to figure those out.
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20:50:17 `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 --> Don't unerstand how to do this. Looked up the definition for slope and found the equation. My best guess is m = y2 - y1 0- -4 4 ---------- --------- ---------- x2 - x1 4/3 - 0 4/3
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20:50:49 The graph forms a straight line with no change in steepness.
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RESPONSE --> Didn't understand how to do this.
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20:51:55 `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 --> I attempted this in the previous problem. I don't know if I am doing this correctly.
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21:25:23 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 --> If choosing the points X Y ---------------------------- 2 2 3 5 5-2 3 ___ ______ 3-2 1 or 3
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21:32:51 `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 --> Increasing Increasing at a constant rate because the steepness of a straight line doesn't change.
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21:38:27 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.
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RESPONSE --> I worked out the problem and got the values for y 0,1,4,9. I did this part correctly but in the previous problem you stated that the steepness of a straight line doesn't increase. That is why I said it is increasing at a constant rate. Can you explain this?
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22:05:38 `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 --> X Y _________________ -3 9 -2 4 -1 1 0 0 Slope y2-y1 9-4 = 5 __________________ x2-x1 3- -2 = 1 or 5 the slope would be 5, 3, 1 I would say the graph is decreasing and decreasing at a decreasing rate. Since y is going down quickly.
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22:09:49 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
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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 _______________ 0 0 1 1 2 1.41 3 1.73 Slope is 1, .4/1, .33/1 The graph is increasing at a decreasing rate because when moving from left to right the y values are increasing 0, 1, 1.4, 1.73 but at a slow rate so it would be increasing at a decreasing rate.
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21:51:30 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 --> ok
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22:10:27 `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 --> X Y _____________ 0 5 1 -10 2 -20 3 -40 The steepness of the graph is increasing. When moving from left to right the graph is decreasing but doing so very quickly y values 5, -10, -20, -40. So the graph is decreasing at an increasing rate.
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22:13:41 ** 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 didn't recall the formula a^(-b)= 1/(a^b). Therefore I didn't get the correct values for y and did not do my graph correctly.
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22:27:18 `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 --> As the car moves away from you, it is traveling faster and faster. The distance the car moves away from you gets greater and greater. However, time is moving at a constant rate. The graph is increasing. The graph is increasing at an increasing rate.
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22:27:43 ** 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 --> ok
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