Query 5

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course Mth 163

1-28-13 1:50

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.

005.

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Question: `q001. Note that this assignment has 10 questions

Evaluate the function y = x^2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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Your solution:

X^2 y

-3 9

-2 4

-1 1

0 0

1 1

2 4

3 9

confidence rating #$&*: 3

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Given Solution:

You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.

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Self-critique (if necessary):

ok

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Self-critique rating:ok

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Question: `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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Your solution:

X Y

-3 .125

-2 .25

-1 .5

0 1 (anything raised 0 equals 1)

1 2

2 4

3 8

confidence rating #$&*: 3

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Given Solution:

By the laws of exponents, b^-x = 1 / b^x. So for example 2^-2 = 1 / 2^2 = 1/4.

Your y values will be 1/8, 1/4, 1/2, 1, 2, 4 and 8. Note that we have used the fact that for any b, b^0 = 1.

It is a common error to say that 2^0 is 0. Note that this error would interfere with the pattern or progression of the y values.

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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Your solution:

X y

-3 .111111111 (1/9)

-2 .25

-1 1

0 ------ undefined

1 1

2 .25

3 .111111111 (1/9)

confidence rating #$&*: 2

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Given Solution:

By the laws of exponents, x^-p = 1 / x^p. So x^-2 = 1 / x^2, and your x values should be 1/9, 1/4, and 1. Since 1 / 0^2 = 1 / 0 and division by zero is not defined, the x = 0 value is undefined. The last three values will be 1, 1/4, and 1/9.

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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Your solution:

X y

-3 -27

-2 -8

-1 -1

0 0

1 1

2 8

3 27

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

The y values should be -27, -8, -1, 0, 1, 8, 27

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `q005. Sketch graphs for y = x^2, y = 2^x, y = x^-2 and y = x^3, using the values you obtained in the preceding four problems. Describe the graph of each function.

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Your solution:

Graph of y=x^2 with the coordinates (-3,9) (-2,4) (-1,1) (0,0) (1,1) (2,4) (3,9) is a parabola with a vertex at (0,0). The equation of the parabola is y=1x^2.

The graph of y=2^x has coordinates at (-3,1/8) (-2,1/4) (-1,1/2) (0,1) (1,2) (2,4) and (3,8) and it is an exponential function. The horizontal asymptote is at y=0.

The graph of y=x^-2 has the coordinates (-3,1/9) (-2,1/4) (-1,-1) (zero is a vertical asymptote) (1,1) (2,1/4) and (3,1/9). The graph is of a logarithmic function with a vertical asymptote at x=0 and a horizontal asymptote at y=0.

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The graph of a logarithmic function does have a vertical asymptote at the y axis, but it passes on both sides of the x axis, and has a different progression of slopes than the x^-2 function.

The graph of the logarithmic function is inverse to the graph of the exponential function (e.g., the base-2 logarithm is inverse to the y = 2^x function of the preceding exercise).

The x^-2 function is a power function, of the form y = x^p with p = -2.

These aren't distinctions you are expected to make at this point, but since you're apparently familiar with logarithmic functions I decided to include this note. You'll see this very clearly in upcoming assignments.

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The graph x^3 is a cube function with the coordinates (-3, -27) (-2,-8) (-1,-1) (0,0) (1,1) (2,8) and (3,27). I like to call this graph an “s function” because it looks like an s with the domain and range being all real numbers.

confidence rating #$&*: 3

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Given Solution:

The graph of y = x^2 is a parabola with its vertex at the origin. It is worth noting that the graph is symmetric with respect to the y-axis. That is, the graph to the left of the y-axis is a mirror image of the graph to the right of the y-axis.

The graph of y = 2^x begins at x = -3 with value 1/8, which is relatively close to zero. The graph therefore starts to the left, close to the x-axis. With each succeeding unit of x, with x moving to the right, the y value doubles. This causes the graph to rise more and more quickly as we move from left to right. The graph intercepts the y-axis at y = 1.

The graph of y = x^-2 rises more and more rapidly as we approach the y-axis from the left. It might not be clear from the values obtained here that this progression continues, with the y values increasing beyond bound, but this is the case. This behavior is mirrored on the other side of the y-axis, so that the graph rises as we approach the y-axis from either side. In fact the graph rises without bound as we approach the y-axis from either side. The y-axis is therefore a vertical asymptote for this graph.

The graph of y = x ^ 3 has negative y values whenever x is negative and positive y values whenever x is positive. As we approach x = 0 from the left, through negative x values, the y values increase toward zero, but the rate of increase slows so that the graph actually levels off for an instant at the point (0,0) before beginning to increase again. To the right of x = 0 the graph increases faster and faster.

Be sure to note whether your graph had all these characteristics, and whether your description included these characteristics. Note also any characteristics included in your description that were not included here.

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `q006. Make a table for y = x^2 + 3, using x values -3, -2, -1, 0, 1, 2, 3. How do the y values on the table compare to the y values on the table for y = x^2? How does the graph of y = x^2 + 3 compare to the graph of y = x^2?

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Your solution:

The y values on the y=x^2 graph are all 3 less than their mirrored values on the y=x^2+3 graph.

The x^2+3 graph has a vertex at (0,3) and the x^2 has a vertex at (0,0).

confidence rating #$&*: 3

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Given Solution:

A list of the y values will include, in order, y = 12, 7, 4, 3, 4, 7, 12.

A list for y = x^2 would include, in order, y = 9, 4, 1, 0, 1, 4, 9.

The values for y = x^2 + 3 are each 3 units greater than those for the function y = x^2.

The graph of y = x^2 + 3 therefore lies 3 units higher at each point than the graph of y = x^2.

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `q007. Make a table for y = (x -1)^3, using x values -3, -2, -1, 0, 1, 2, 3. How did the values on the table compare to the values on the table for y = x^3? Describe the relationship between the graph of y = (x -1)^3 and y = x^3.

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Your solution:

The y values are: -64, -27, -8, -1, 0, 1, 8 and the y values for x^3 are: -27, -8, -1, 0, 1, 8, and 27.

The points are moved one unit to the right. The graph is also shifted one to the right from its parent graph.

confidence rating #$&*: 3

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Given Solution:

The values you obtained should have been -64, -27, -8, -1, 0, 1, 8.

The values for y = x^3 are -27, -8, -1, 0, 1, 8, 27.

The values of y = (x-1)^3 are shifted 1 position to the right relative to the values of y = x^3. The graph of y = (x-1)^3 is similarly shifted 1 unit to the right of the graph of y = x^3.

STUDENT QUESTION

I assumed the graph was shifted 1 unit down since the graph passes through (0, -1) instead of origin. Then again, it passes through (1, 0), so could it be said that the graph is shifted 1 unit down OR 1 unit to the right?

INSTRUCTOR RESPONSE

Based on those two points that would be correct. Nowever, for example, (-2, -8) shifts to (-1, -8), a shift to the right, but not to (-2, -9), as would be the case if this was a downward shift.

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `q008. Make a table for y = 3 * 2^x, using x values -3, -2, -1, 0, 1, 2, 3. How do the values on the table compare to the values on the table for y = 2^x? Describe the relationship between the graph of y = 3 * 2^x and y = 2^x.

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Your solution:

The y values for 3*2^x are: 3/8, ¾, 3/2, 6, 12, and 24

The values are each 3 times larger than the values for 2^x.

The graph is of an exponential function but the values are moved three times as much from the x axis.

confidence rating #$&*: 3

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Given Solution:

You should have obtained y values 3/8, 3/4, 3/2, 3, 6, 12 and 24.

Comparing these with the values 1/8, 1/4, 1/2, 1, 2, 4, 8 of the function y = 2^x we see that the values are each 3 times as great.

The graph of y = 3 * 2^x has an overall shape similar to that of y = 2^x, but each point lies 3 times as far from the x-axis. It is also worth noting that at every point the graph of y = 3 * 2^x is three times as 'high' as the corresponding point of y = 2^x.

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Self-critique (if necessary):

ok

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Self-critique rating: ok

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Question: `q009. How do the values on a table for y = (x + 2)^2 compare to those for y = x^2? Use x values -3, -2, -1, 0, 1, 2, 3 to construct each table. What is the axis of symmetry for this function?

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Your solution:

The y values for (x+2)^2 are: 1,0, 1, 4, 9 ,16, and 25

The y values for x^2 are: 9, 4, 1, 0, 1, 4, and 9

The axis of symmetry for (x+2)^2 is (-2, 0).

confidence rating #$&*:

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Question: `q010. Explain in terms of the values of y = x^2 for the numbers x = -2, -1, 0, 1, 2 why we expect the graph of y = x^2 to be symmetric about the y axis.

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Your solution:

A graph is symmetric with respect to the y-axis if, for every point (x, y) on the graph, the point (-x, y) is also on the graph. We consider the parent function an even function.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Self-critique rating:

ok

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Self-critique (if necessary):

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Self-critique rating:

ok

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Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

Excellent.

I inserted one note that really goes beyond what we're doing right now. You seem to have the background to understand that note, but if not don't worry about it; we'll see it in much more detail later.

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