Orientation- describing graphs

course Mth163

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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:

We make a table for y=3x-4as follows:

We make two columns, putting x in first column and y in next column.

Put numbers-3,-2,-1,0,1,2,3 in the x column.

when we substitute -3 in the expression for x we get -13 y = 3(-3)-4=-13, and so on for each substitution for the x .

The coordinates for y are -13,-10,-7,-4,-1,2,5

I graphed the line using the x and y coordinates from the table.

To get the y intercept, I substituted 0 for the x coordinate: y = 3(0)-4=-4

To get the x intercept, I substituted 0 for the y coordinate:

0=3x-4

4=3x I added 4 to both sides – what you do to one side you must do to the other

x=4/3 = 1 1/3 I divided both sides by 3, and changed to a mixed fraction

Confidence Assessment: mostly

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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 (if necessary):

I need to remember to write my intercepts as points.

Self-critique Rating:

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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:

No the steepness of the graph in the preceding exercise does not change, because the a straight line has the same slope in all places.

Confidence Assessment: positive

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Given Solution:

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

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Self-critique (if necessary):

correct

Self-critique Rating:

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Question: `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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Your solution:

rise / run = y/x this is the y coordinate over the x coordinate

y / x = 3 / 1 You go up 3 places on the y coordinate and over I place on the x coordinate

3 / 1 = 3 any # over 1 is a whole #

Confidence Assessment: mostly

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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.

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Self-critique (if necessary):

If I took y = 3x – 4, x=2

y = 3(2) – 4

y = 2

then x = 3

y = 3(3)-4

y = 5

rise is 5 – 2 = 3

run is 3 – 2 = 1

slope is rise / run = 3 / 1 = 3

Self-critique Rating:

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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 the graph is increasing.

Yes, the steepness of the graph changes because it is not a straight line anymore. The line is getting steeper as it goes up.

It is increasing at a constant rate.

Confidence Assessment: mostly

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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. *&*&.

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Self-critique (if necessary):

I understand you logic, but the y increased by a constant rate of 2.

Does a line that is increasing, mean that it has a positive slope?

Self-critique Rating:

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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:

The graph is decreasing because of the negative slope.

The steepness of the graph is changing because it is no longer a straight line.

I would say the slope is decreasing at a decreasing rate because it goes from 9, to 4, to 1, to 0.

Confidence Assessment: unsure

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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 is decreasing at a decreasing rate.

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Self-critique (if necessary):

I understand

Self-critique Rating:

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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 graph is increasing because it has a positive slope.

The steepness is changing because it is not a straight line.

The graph is increasing at a decreasing rate because the rate goes from 0, to 1, to 1.4, to 1.7, getting smaller each increase.

Confidence Assessment: possitive

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Given Solution:

`aIf 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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Self-critique (if necessary):

I got it.

Self-critique Rating:

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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:

Taking x and making it (-) changes it to a fraction with 1 over the ^ solution.

Then I changed the fraction to a mixed fraction in order to graph it properly.

The graph is decreasing because it goes from 5, to 4 ½, to 1 ½, to 5/8.

It is decreasing at a decreasing rate because it is getting smaller each time.

Confidence Assessment: positive

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Given Solution:

`a** 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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Self-critique (if necessary):

I understand

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:

I would say the graph would be increasing. The time would be getting higher, so the graph would be increasing at an increasing rate.

Confidence Assessment: mostly

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Given Solution:

`a** 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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Self-critique (if necessary):

In this problem to show I understood it better, I should have included that the speed of car increases and thus, so does the time.

Self-critique Rating:better

&#This looks good. Let me know if you have any questions. &#