Describing Graphs

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course MTH 272

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:

x y = 3x - 4

-3 3(-3) - 4 = -13

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

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

0 3(0) - 4 = -4

1 3(1) - 4 = -1

2 3(2) - 4 = 2

3 3(3) - 4 = 5

13

-3 3

-13

X-intercept is 0 = 3x - 4

4 = 3x x = 4/3

= (4/3, 0)

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

= (0, -4)

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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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 forms a continuous straight line. There is no change in the steepness of the graph throughout the line.

confidence rating #$&*:: 3

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Given Solution:

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

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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:

The slope of the graph is 3.

confidence rating #$&*: 3

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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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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:

X Y = x^2

0 0

1 1

2 4

3 9

This graph is increasing. The graph becomes steeper because it is increasing at an increasing rate.

confidence rating #$&*: 3

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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

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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 = x^2

-3 9

-2 4

-1 1

0 0

The graph is decreasing. It is becoming less steep which means it 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.

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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:

X Y = sqrt(x)

0 0

1 1

2 1.414

3 1.732

The graph is increasing and is becoming less steep. This means the graph is increasing at a decreasing rate.

confidence rating #$&*: 3

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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 represents 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 described

take another look at your plot and make a note in your response indicating any difficulties.

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Self-critique Rating: 3

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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:

X Y = 5 * 2^(-x)

0 5

1 2.5

2 1.25

3 0.625

This graph is decreasing and becoming less steep. The graph is decreasing at a decreasing rate.

confidence rating #$&*: 3

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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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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:

If y represents the distance from you to the car and t represents the time in seconds since the car started out, the graph of y vs. t would be

increasing. It would also be increasing at an increasing rate since the car is accelerating.

confidence rating #$&*: 3

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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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