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course Mth163
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.
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Your solution:
I will be using the same X values as in the example,-3,-2,-1,2,3(I took out one of the neg ones)
1. Y=3(-3)+4
2. Y=-9+4
3. Y=-5
1. Y=3(-2)+4
2. -6+4
3. Y=-2
1. Y=3(-1)+4
2. Y=-3+4
3. Y=1
1. Y=3(2)+4
2. Y=6+4
3. Y=10
1. Y=3(3)+4
2. Y=9+4
3. Y=13
Now we have our x&y coordinates; (-3,5),(-2,-2),(-1,1),(2,10),(3,13)
Graph seems to be linear, except for the (-1,1) outlier. It is positive staring from the bottom left (third quadrant) increasing to the right (first quadrant). This graph also has a slope of 4 rise/run=12/3=4.
Confidence Assessment: 3
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Self-critique (if necessary): first of all I’m glad we are reviewing, alright I see what I did wrong I was supposed to set y=0 to find x, and set x=0 to find y so in my self-critique here I will do that.
1. y = 3x - 4
2. set y=0
3. 0=3x-4
4. Add 4 to each side, 0+4=3x-4+4,4’s cancel out
5. Then divide by 3, 4/3=3x/3, threes cancel out
6. So you are left with 4/3=x
7. (4/3,0)
Now I am going to set x=0
1. X=0
2. Y=3(0)-4
3. Y=-4
4. (0,-4)
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Self-critique 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.
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Your solution:
The steepness of the graph seems to be decreasing due to the fact that it starts in the top left of the graph (2nd quadrant) and lowers into the bottom right hand side of the graph (4th quadrant).
confidence rating #$&*: 3
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Self-critique (if necessary):
So by steepness are you talking about slope??? The example about the ramp helped at first I thought you were referring to the direction of the line, but slope being constant makes more sense. I think I understand. The rise would be 3 and run 1, it should stay this way throughout the whole graph, right???
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Right. Good.
However steepness is not quite the same as slope. A positive slope can be steep, as can a negative slope.
We can say this:
One slope is steeper than another if and only if the magnitude of its slope (i.e., the absolute value of the slope) is greater than that of the other.
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Self-critique Rating: 2
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Question: `q003. What is the slope of the graph of the preceding two exercises (the function is y = 3x - 4;slope is rise / run between two points of the graph)?
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Your solution:
3/1 (as explained in the previous self-critique) the rise would be 3 and run 1, it should stay this way throughout the whole graph. To find this I simply drew rise on my graph and connected it with my run at the next point. (using the original points, (0,-4),(4/3,0))
confidence rating #$&*: 3
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Self-critique (if necessary):
Even though I got a slope of 3 I would like to know where you got these numbers, When x = 2, we substitute 2 for x to get y = 3 * 2 - 4, which is equal to 2.
When x = 8, we substitute 8 for x to get y = 3 * 8 - 4, which is equal to 20., where these just random number choices or did you use these ones for a particular reason???
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Good question, but you didn't include the given solution, so I can't reference it directly (do remember to never remove anything from a document).
But I believe the wording used there was 'for example'. The choice of those numbers was completely arbitrry. Any two numbers you choose will give you the same result.
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Self-critique Rating: 2.5
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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:
I would say that the graph is increasing, at first you would think the slope resembled the identity function but then it curves up on you looking like a half quadratic or parabola and this most diffinitanely indicates that the slope changes but I can prove this by just citing my graphs points start off by (0,0) origin then (1,1) so at this point my slope is one, then as it goes from (1,1) to (2,4) the slope becomes 3, then as it went from the previous point to (3,9) the slope became 5, so our graph is not only increasing it is curving as well I believe our graph is increasing at an increasing rate.
Self-critique: I did not include all the points in the beginning which are x values are 0,1,2,3 and corresponding y values are 0,1,4,9
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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 x values are -3,-2,-1,0 my corresponding y values are 9,4,1,0 so my coordinates are (-3,9),(-2,4),(-1,1),(0,0) once you graph these points you see that you have gotten the other half or your parabola (from the previous question). This half of the graph is decreasing, since the graph is doing so this will only effect the signs of the changing slope this time it goes form -1,-3,-5. Since this is the other half of the last question I’m going to say that this graph is decreasing at a decreasing rate.
confidence rating #$&*: 3
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Self-critique (if necessary):
Can you explain the steepness a little more in terms of slope???
I’m not entirely sure if I got the steepness right.
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Self-critique Rating: 3
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Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing?
Does the steepness of the graph change and if so, how?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution:
My coordinates are (0,0),(1,1),(2,1.414),(3,1.732) When I graphed my coordinates I got a perfect looking squared root function. The steepness of the graph decreases using y=mx+b (plugging 0 in for b) I got these results on slope/steepness starts off at 0 then increases to 1 even though you would think the graph is increasing it is actually decreasing (according to slope) your next slope is .707 then decreases again (ever so slightly) to .5773 (I’m not sure how to show the 3 is repeating on the key board, would you rather just have me round it as many of my past teachers prefer???) I believe the graph is decreasing at an increased rate (even though it’s pretty close to constant, does close count in this case???)
confidence rating #$&*: 2.8
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Self-critique (if necessary): Its decrease is happing at an increased rate, that makes much more sense (wasn’t quite sure how to word that)
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Self-critique Rating: 3
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The numbers are increasing, but by less and less each time. So they are increasing at a decreasing rate.
It's important for you to leave the given solutions, and everything else, in these documents for possible future reference.
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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:
My coordinates are (0,5),(1,2.5),(2,1.25),(3,.625) my graph is decreasing, using y=mx+b the steepness of the graph is decreasing at an increasing rate I can prove with the pattern of my slopes which are (in order starting with the first (x,y) coordinates) 0,-2.5,-1.875, -4.375
confidence rating #$&*: 3
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Self-critique (if necessary): due to the fact that the numbers are getting smaller and smaller due to the halving the numbers are decreasing at a much slower rate each time, it takes longer for whatever that process is to happen??? If something happened in a less time then wouldn’t it be faster???, I think I’m just confusing myself right now, can you apply it to a real world scenario???
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Self-critique Rating: 2
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The numbers 5, 2.5, 1.25, .625 are decreasing but by less and less each time x changes by 1 unit.
So the graph decreases at a decreasing rate.
This situation could, for example, describe the decay of a radioactive substance. The less of it you have, the fewer decays and the less is lost.
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Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster.
If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing?
Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution:
The graph would be increasing at an increasing rate, the car would be moving away from faster and faster making the distance father and father between you and the car and the time shorter and shorter.
confidence rating #$&*: 3
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Self-critique Rating: OK
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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:
My coordinates are (-3,8),(-2,6),(-1,4),(0,1),(1,0),(2,1),(3,4) from the coordinates (-3,8) to (1,0) the graph is decreasing, however from points (1,0) to (2,4) the graph is increasing. The steepness changes several times on this graph on both sides you will notice that the slope go from 1 to 2, but it stops there for the increasing side of the graph on the decreasing side it seems to have stabilize (maybe temporarily) to a slope of 2 (from points (-1, 4) to (-2, 6), (still staying 2) from points (-2, 6) to (-3, 8) the decreasing half of the graph was decreasing at an increased rate, while the increasing half was increasing at an increasing rate. I believe this graph is an example of a quadratic or parabolic function.
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Good.
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confidence rating #$&*::3
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Self-critique Rating: I cannot give a self-critique rating because there is nothing on the WS for me to check my answer by.
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Good thinking, and excellent questions.
Do be sure you include the complete document with subsequent assignments.
Check my notes in answer to many of your questions.
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