Describing Graphs

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course PHY 201

Samantha RogersPHY 201

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:

x y

-3 -13

-2 -10

-1 -7

0 -4

1 -1

2 2

3 5

y = 3(-3) - 4 = -9 - 4 = -13

y = 3(-2) - 4 = -6 - 4 = -10

y = 3(-1) - 4 = -3 - 4 = -7

y = 3(0) - 4 = 0 - 4 -4

y = 3(1) - 4 = 3 - 4 = -1

y = 3(2) - 4 = 6 - 4 = 2

y = 3(3) - 4 = 9 - 4 = 5

y = 3(0)-4 = 0 - 4 = -4 y intercept is -4

0 = 3(x)-4 (add 4 to both sides)

4 = 3x (divide by 3 to both sides)

4/3= x x intercept is 4/3

confidence rating #$&*: 3

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Given Solution:

`aThe 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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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 graph of this function is linear, a straight line, therefore the steepness is consistent throughout.

confidence rating #$&*: 3

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Given Solution:

`aThe graph forms a straight line with no change in steepness.

STUDENT COMMENT

Ok, I may not understand what exactly it meant by steepness, I was thinking since it was increasing it would also be getting steeper?????

INSTRUCTOR RESPONSE

A graph can increase while getting steeper and steeper; or it can increase while getting less and less steep. Or it can increase with no change in steepness.

Analogies:

When you walk up a hill, typically as you approach the top the slope starts to level off--it gets less steep.
When you go up a ramp the steepness stays the same until you get to the end of the ramp.
When you start climbing a hill, typically it gets steeper for awhile, the stays at about a constant slope, then gets less steep toward the top.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

y intercept = -4

x intercept = 4/3

rise = 4

run = 4/3

rise/run = 4/(4/3) = 3

confidence rating #$&*:

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Given Solution:

`aBetween 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.

EXPANDED EXPLANATION

Any student who has completed Algebra I and Algebra II should be familiar with slope calculations. Most students are. However a number of students appear to be very fuzzy on the concept, and I suspect that not all prerequisite courses cover this concept adequately (though I am confident that it's done well at VHCC). Also a number of students haven't taken a math course in awhile, and might simply be a bit rusty with this idea. In any case the following expanded explanation might be helpful to some students:

Slope = rise / run.
The rise between two graph points is the change in the y coordinate. The run is the change in the x coordinate.
Our function is y = 3 x - 4. 
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. 
The graph therefore contains the points (2, 2) and (8, 20).
You should have made a graph showing these points. If not you should do so now.
As you go from point to point your y coordinate goes from 2 to 20. So the 'rise' between the points is 20 - 2 = 18.Your x coordinate goes from 2 to 8. So the 'run' between the points is 8 - 2 = 6.
The slope is rise / run = 18 / 6 = 3.
The numbers 2 and 8, which were used for the x values, were chosen arbitrarily. Any other two x values would have given you different coordinates, likely with different rise and run. However whatever two x values you use, you will get the same slope. The slope of this graph is constant, and is equal to 3.

STUDENT QUESTION

Am I not allowed to utilize my calculus tools, yet?

Couldn't I have just taken the derivative for the function, y = 3x -4 to obtain 3 as the slope?

However, I do know how to do both ways. Which is the more preferred method? 
INSTRUCTOR RESPONSE

This exercise develops a language for describing some aspects of graphs, and does not assume calculus tools.

Of course it's fine to use the calculus tools if you have them, as long as you understand the problem at the more basic level as well.

Unfortunately, not every student who has had a calculus course would know how to apply those tools to this situation (for example, I've had students from other institutions who have made A's in Applied Calculus courses from other (not particularly reputable) institutions, who didn't understand the concept of a slope).

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Self-critique (if necessary): OK

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Self-critique Rating: 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?

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Your solution:

y = x^2

x y

-3 -3^2 = -9

-2 -2^2 = -4

-1 -1^2 = -1

0 0^2 = 0

1 1^2 = 1

2 2^2 = 4

3 3^2 = 9

The graph is exponential and is increasing. Yes the steepness changes, it gets steeper as the x axis increases. The line is increasing at an increasing rate.

confidence rating #$&*: 2

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Given Solution:

`aGraph 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

STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate?

INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1?

In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.

STUDENT QUESTION: I am a little hazy on what the steepness is
INSTRUCTOR RESPONSE: The hill analogy I used above might be helpful. 
Formally, steepness could be defined as the magnitude of the slope, i.e., the absolute value of the slope. 
Two graphs with respective slopes 4 and -4 would be equally steep; both would have slope of magnitude 4. Both of these graphs would be steeper than, say a graph with slope 3 or -3.

NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope.

Since in this case the slope is positive, which implies that the function is increasing, the increasing slope therefore implies that the value of the function is increasing at an increasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave upward.

This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

x y

-3 -3^2 = -9

-2 -2^2 = -4

-1 -1^2 = -1

0 0^2 = 0

1 1^2 = 1

2 2^2 = 4

3 3^2 = 9

The graph is exponential and it is decreasing. The steepness of the graph is changing and it is less steep when going from left to right. The steepness is decreasing at a decreasing rate.

confidence rating #$&*: 3

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Given Solution:

`aFrom 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 on this interval is decreasing at a decreasing rate.

NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope.

Since in this case the slope is negative, which implies that the function is decreasing, the increasing slope therefore implies that the rate of decrease is decreasing. The value of the function is therefore decreasing at a decreasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as decreasing and concave upward.

This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up.

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Self-critique (if necessary): OK

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

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Self-critique Rating: 3

&#Your work looks good. See my notes. Let me know if you have any questions. &#