graphs_initprobs

course Phy121

I had some confusion understanding how the steepness of the graph for the square root of x was decreasing as opposed to increasing. I saw how it was increasing by less along the x-axis, but did not understand how that would mean it to be decreasing. The only comparison I could make was to deceleration perhaps? It's slowing down?

Your graph should look like the side view of a hill that appears to be gradually leveling off (though it turns out that the square root function never actually levels off).

Good work on these questions. See my notes.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

kaV񊴵 Student Name: assignment #002 002. Describing Graphs

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17:10:03 `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 --> x intercept: at 4/3 or 1 1/3 y intercept: at 4

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17:11:31 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 --> Sorry, I forgot the (-) sign in front of the 4.

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17:14:20 `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 --> No, it is a straight line, no change.

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17:15:25 The graph forms a straight line with no change in steepness.

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RESPONSE --> ok

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17:16:35 `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 --> The slope of the graph is 3. Slope is y=mx + b y = 3x -4, y + 4/x = 3

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17:17:19 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 --> ok, it is increasing from left to right and at a constant rate b/c the steepness does not change.

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17:21:29 `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 --> (0,0) (1,1) (2,4) (3,9)

The graph is increaseing from left to right.

The steepness does change, it becomes steeper as you increase.

I would say it is increasing at an increasing rate.

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17:21:41 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 --> Ok.

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17:23:20 `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 --> (0,0) (-1, 1) (-2, 4) (-3, 9)

It is decreasing from left to right.

Yes, it becomes less steep as you move from left to right.

This graph is decreasing at a decreasing rate.

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17:23:37 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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17:26:15 `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 --> (0,0) (1,1) (2, 1.4142) (3, 1.7321)

Increasing, very slightly.

It does, but barely. Slightly increases.

It increases at an increasing rate, but then starts to slow down and increases at a lesser rate.

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17:32:16 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 --> I found this answer a bit difficult to understand. I do see how it is increasing at a decreasing rate, but it is still increasing nonetheless. I did not understand how the steepness was decreasing. It still seems to be rising, but just by less and less. Like it's slowing down?

It's rising by less and less, meaning that while it's still increasing, it's getting less and less steep. So the rate of increase is decreasing.

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17:41:12 `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 --> (0,5) (1,2.5) (2, 1.25) (3, .625)

The graph is decreasing from left to right.

Yes, the steepness is decreasing.

Decreasing at a constant rate, by 1/2.

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17:43:56 ** 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 stated it was decreasing at a constant rate because each time it was decreasing by 50% of the y value, but obviously it is decreasing at less rate as we move right along the x-axis. (-1, 10) and then (1, 2.5). I had the word constant in my head, but what I had described as constant, really is consistency.

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17:46:51 `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 --> I would say y vs. t to be increasing since the car is traveling faster and faster. Its speed and distance continue to increase.

Increasing at an increasing rate since the car continues to speed up, thus increasing the distance.

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17:47:04 ** 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 --> Yeah!

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