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course MTH 158
Alvin LomansMth 158
20 January 2015
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Question: `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 submit your work as instructed.
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.
Your solution: The y intercepts for -3, -2, -1, 1, 2, 3 are -13, -10, -7, -1, 2, 5. The line is a line like this / that moves left to right. Crosses y intercept at (0,-4) and the x intercept at (4/3,0).
confidence rating #$&*: 3
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Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
Your solution: The steepness continues from what I can see it doesn’t seem like it will level out no matter what number you put in it continues to climb. At some point it might steady off but it doesn’t seem like it with the numbers that I inserted.
confidence rating #$&*: 2
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Self Critique: I should have mentioned the top out point which was mentioned by the instructor in his answer but other than that I described what happens with the graph as it continues to climb.
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Question: `q003. What is the slope of the graph of the preceding two exercises (the function is y = 3x - 4; slope is rise / run between two points of the graph)?
Your solution: I chose x=2 and x=6 which gives you 2 for x=2 and 14 for x=6. Next step you would minus those two which equals 14-2 which is 12 that is my y intercept. The x intercept would be 6-2 which is 4. Then you would divide those two which would be 12/4 which is 3.
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Question: `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?
Your solution: The graph is increasing. I put in all the way up to 10 and it continued to climb it never got to a tipping point. The graph is increasing at a constant rate.
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Question: `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?
Your solution: The line is decreasing. The steepness of the graph changes its actually goes from decreasing to increasing as you reach (0,0) making a U. I believe its none because it is a parabola.
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Question: `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?
Your solution: The graph is increasing. The steepness of the graph doesn’t change it continues at a decreasing rate. The steepness of the graph does change at one point down the line it becomes more level.
Self Critique: I could have done a better job with explaining the steepness of the problem.
Self Critique Rating: 2
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Question: `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?
Your solution: The graph is decreasing. The steepness of the graph does not change. I believe it is decreasing at a decreasing rate.
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Question: `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?
Your solution: It would be increasing because the car would be moving away from you at a certain speed that is increasing. It would be increasing at an increasing rate.
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Question: `q009. As you saw above, on the interval from x = -3 to x = 3 the graph of y = x^2 is decreasing at a decreasing rate up to x = 0 and increasing at an increasing rate beyond x = 0.
How would you describe the behavior of the graph of y = (x - 1)^2 between x = -3 and x = 3?
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Your solution: The graph makes an parabola which is sort of like a U. It starts out decreasing and then it gets to the point (1,0) then starts to increase.
confidence rating #$&*: 3
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Self Critique Rating: 3
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