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course Mth 173
002. Describing Graphs
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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 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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Your solution:
To graph the function y = 3x - 4, I constructed one column for the x values -3, -2, -1, 0, 1, 2, and 3; and another for the y values -13, -10, -7, -4, -1, 2, and 5 which I got after substituting the x values for the x variable in the function each at a time. Then when I graphed the function the points formed a straight line. To find the x-intercept I replaced the y with a zero in the function which is (4/3, 0). And to find the y-intercept I replaced the x with a zero in the function which (0, -4).
confidence rating #$&*:
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Self-Critique: OK
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Self-Critique Rating: OK
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Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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Your solution:
The steepness of the graph doesn“t change, it looks like a straight line going up from left to right.
confidence rating #$&*:
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Self-Critique: OK
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Self-Critique Rating: OK
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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)?
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Your solution:
To find the slope of the graph I chose two points P1(0, -4) and P2 (1,-1), then I calculated the slope formula (y2 - y1)/(x2-x1) which becomes (-1-(-4)/(1-0) which is 3. This indicates that the slope is 3, meaning that the graph rises 3 and runs 1.
confidence rating #$&*:
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Self-Critique: OK
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Self-Critique Rating: OK
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Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing?
Does the steepness of the graph change and if so, how?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution:
The table that I constructed of y vs. x had the x values 0, 1, 2, and 3; and the y values 0, 1, 4, and 9. It created the points (0, 0), (1, 1), (2, 4), and (3, 9). After I graphed the points and connected the line I noticed that the graph was increasing from left to right. The steepness of the graph is increasing becoming steeper each time I move one unit in the x-axis. I think the graph is increasing at an increasing rate because if we look at the y-values comparing to the x-values we can see that each time we move one unit right on the x-axis the y-axis almost doubles its values.
confidence rating #$&*:
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Self-Critique: OK
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Self-Critique Rating: OK
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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?
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Your solution:
My table of y vs. x had following x values -3, -2, -1, and 0; and the y values 9, 4, 1, and 0. When I graphed the points (-3, 9), (-2, 4), (-1, 1), and (0, 0) I noticed that they formed a slightly curved line which was decreasing from left to right. The graph is decreasing from left to right at a decreasing rate because by observing the x values and comparing them with the y values we see that as we move one unit right, the y axis values decrease several units. The steepness of the graph does not change because the graph is decreasing.
confidence rating #$&*:
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Self-Critique: OK
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Self-Critique Rating: OK
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Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing?
Does the steepness of the graph change and if so, how?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution:
The table I constructed had the x values of 0, 1, 2, and 3; and the y values of 0, 1, 1.414, and 1.732 which made the points (0, 0), (1, 1), (2, 1.414), and (3, 1.732). When I graphed these points and drew the line to connect them I noticed that this graph was increasing very slowly. For that reason the graph is increasing at a decreasing rate because if we look at the x values and compare them to the y values we see that for each unit we move in x axis, the y values almost stay the same even though they are still increasing. The steepness of the graph is increasing at a decreasing rate but it looks like it is going to become less and less steep.
confidence rating #$&*:
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Self-Critique: OK
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Self-Critique Rating: OK
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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?
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Your solution:
The table that I constructed had the x values of 0, 1, 2, and 3; and the y values of 5, 2.5, 1.25, 0.625 creating the points (0, 5), (1, 2.5), (2, 1.25), and (3, 0.625). The way I got these y values was by substituting the x values in the y function and solve for y. When I graphed the function I noticed that the line connecting the points looked like it was decreasing from left to right. For that reason the graph is decreasing at a decreasing rate because if we look at the x values and compare them to the y values we realize that for each unit we move in the x-axis, the values in the y axis decrease by half which means that the graph is slowly decreasing. The steepness of the graph is decreasing at a decreasing rate and it looks like the decrease is going to become less and less.
confidence rating #$&*:
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Self-Critique: In my solution I did not mention the fact that when there is number raised to a negative exponent the number becomes the inverse of the original number and the exponent becomes positive. For example, 2^ -2 is ½^2 which is ¼.
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Self-Critique Rating: 3
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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?
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Your solution:
A graph of y vs. t would be increasing because the car is moving faster and faster so the distance is getting greater and greater. On the other hand the time is moving at a constant rate so for each unit we move in the x axis (for time) the y axis is moving several units more (represents distance). Therefore the graph is increasing at an increasing rate because as the car moves faster and faster, its distance gets greater and greater but the time stays at a constant rate which leads to a bigger slope.
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Self-critique (if necessary):
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Self-critique rating:
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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?
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Your solution:
A graph of y vs. t would be increasing because the car is moving faster and faster so the distance is getting greater and greater. On the other hand the time is moving at a constant rate so for each unit we move in the x axis (for time) the y axis is moving several units more (represents distance). Therefore the graph is increasing at an increasing rate because as the car moves faster and faster, its distance gets greater and greater but the time stays at a constant rate which leads to a bigger slope.
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Self-critique (if necessary):
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Self-critique rating:
#*&!
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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?
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Your solution:
A graph of y vs. t would be increasing because the car is moving faster and faster so the distance is getting greater and greater. On the other hand the time is moving at a constant rate so for each unit we move in the x axis (for time) the y axis is moving several units more (represents distance). Therefore the graph is increasing at an increasing rate because as the car moves faster and faster, its distance gets greater and greater but the time stays at a constant rate which leads to a bigger slope.
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Self-critique (if necessary):
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Self-critique rating:
#*&!#*&!
&#Your work looks good. Let me know if you have any questions. &#