Assignment 5 qa

#$&*

course Mth 163

f 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

.

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 8 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: 

 y = (-3)^2 = 9

y = (-2)^2 = 4

y = (-1)^2 = 1

y = 0^2 = 0

y = 1^2 = 1

y = 2^2 = 4

y = 3^2 = 9

 

 

confidence rating #$&*:

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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: 

 y = 2^-3 = 1/8

y = 2^-2 = ¼

y = 2^-1 = ½

y = 2^0 = 1

y = 2^1 = 2

y = 2^2 = 4

y = 2^3 = 8

 

 

confidence rating #$&*:

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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:

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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: 

 y = (-3)^-2 = 1/9

 y = (-2)^-2 = 1/4

y = (-1)^-1 = 1

y = (0)^-2 = 0

y = 1^-2 = 1

y = 2^-2 = ¼

y = 3^-2 = 1/9

 

 

confidence rating #$&*:

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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):

 I don't understand. ???Were we trying to find the x values or the y values??? The question asks for the y values

 

 

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Self-critique rating:

@& Your table did report the y values correctly.

*@

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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: 

 y = (-3)^3 = -27

y = (-2)^3 = -8

y = (-1)^3 = -1

y = 0^3 = 0

y = 1^3 = 1

y = 2^3 = 8

y = 3^3 = 27

 

 

confidence rating #$&*:

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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:

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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: 

 y = x^2 is a parabola with the vertex at (0, 0)

y = 2^x graph is a curve. The slope increases as we go from left to right

y = x^-2 begins at (-3, 1/9), increases to (-1, 1) then decreases to (0, 0) then increases again to (1,1) then decreases again ending at (3, 1/9)

@& (0, 0) is not a point of this graph, and there is no point of this graph which is close to (0, 0).

0^-2 would be 1 / 0^2, or 1 / 0. Division by 0 is not defined.

As your other values show, the closer x gets to 0, the larger the value of x^-2. There is no limit to how close x can be to 0, so there is no limit to how large x^-2 can get. When x approaches 0, it should therefore be clear that x^-2 does not approach 0 but rather moves more and more quickly away from 0.*@

y = x^3 starts out with a steep slope from point (-3, -27) till it gets to point (-1, -1), then the slope still increases but at a much smaller rate until it gets to (1, 1)

 

 

confidence rating #$&*:

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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):

 My third graph does not resemble the given solution at all. I'm not sure what I did wrong here, but my answer to the third question was completely off as well

@& See my note above.*@

 

 

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Self-critique rating:

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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 table for y = x^2 + 3 are 3 more than the y values on the table for y = x^2

The graph is the same shape, just 3 points higher

 

 

confidence rating #$&*:

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2

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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 values on the new table for y are -64, -27, -8, -1, 0, 1, 8. On the table for y = x^3, they are -27, -8, -1, 0, 1, 8, 27.

The graph is parallell to the original graph, just moved 1 point to the right

 

 

confidence rating #$&*:

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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, 3/4, 3/2, 3, 6, 12 and 24. The y values for y = 2^x are 1/8, 1/4, 1/2, 1, 2, 4, 8. They old values are all multiplied by 3 to get the new values.

The shape is similar to the graph of y = 2^x, but each y point is 3 times as higher making the slope of the graph steeper

 

 

confidence rating #$&*:

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

2

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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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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:

#*&!

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