Precalculus Asst 22

#$&*

course Math 163

Submitted 11-08-10 @ 7:25pm.

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.

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.

022. `query 22

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Question: `qExplain why the function y = x^-p has a vertical asymptote at x = 0.

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Your solution:

To solve the function of y = x^-p, you must transform it into a fraction to eliminate the negative exponent, i.e. 1/x^p. Therefore, when you divide 1 by a smaller and smaller number you get larger and larger absolute values, resulting in a vertical asymptote at x = 0.

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Given Solution:

`a** x^-p = 1 / x^p.

As x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result

with larger and larger magnitude.

There is no limit to how close x can get to 0, so there is no limit to how many times x^p can divide into 1.

This results in y = x^p values that approach infinite distance from the x axis. The graph therefore approaches a vertical limit. **

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Self-critique (if necessary):

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Question: `qExplain why the function y = (x-h)^-p has a vertical asymptote at x = h.

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Your solution:

When x = h is the function of y = (x-h)^-p, or y = (h-h) ^-p, then the result is

y = (0)^-p

y = 1/0^p and as stated earlier, whenever you divide 1 by a smaller number the result is a larger absolute value which results in a vertical asymptote.

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Given Solution:

`a** (x-h)^-p = 1 / (x-h)^p.

As x gets closer to h, (x-h)^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude.

There is no limit to how close x can get to h, so there is no limit to how many times (x-h)^p can divide into 1.

This results in y = (x-h)^p values that approach infinite distance from the x axis as x approaches h. The graph therefore approaches a vertical limit.

This can also be seen as a horizontal shift of the y = x^-p function. Replacing x by x - h shifts the graph h units in the x direction, so the asymptote at x = 0 shifts to x = h. **

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Self-critique (if necessary):

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Question: `qExplain why the function y = (x-h)^-p is identical to that of x^-p except for the shift of h units in the x direction.

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Your solution:

When you replace x, in the original function of x^-p, by (x-h), you obtain the function y = (x-h)^-p, the division by 0 occurs when x = h, compared to x = 0 as in the original function. When comparing charts of these two functions you can see that a displacement of the y values occur in the positive x direction because of the added “h” value. This added “h” value will shift the graph “h” units to the right.

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Given Solution:

`aSTUDENT ANSWER: You end up with the exact same y values but at the different position of x changed by the h value.

INSTRUCTOR COMMENT: Good start. More specifically the x value at which a given y value occurs is shifted h

units, so that for example y = x^p is zero when x = 0, but y = (x - h)^p is zero when x = h.

To put this as a series of questions (you are welcome to insert answers to these questions, using #$&* before and after each insertion)::

Assume that p is positive.

For what value of x is x^p equal to zero?

For what value of x is (x - 5)^p equal to zero?

For what value of x is (x - 1)^p equal to zero?

For what value of x is (x - 12)^p equal to zero?

For what value of x is (x - h)^p equal to zero?

For example, the figure below depicts the p = 3 power functions x^3, (x-1)^3 and (x-5)^3.

Assume now that p is negative.

For what value of x does the graph of y = x^p have a vertical asymptote?

For what value of x does the graph of y = (x-1)^p have a vertical asymptote?

For what value of x does the graph of y = (x-5)^p have a vertical asymptote?

For what value of x does the graph of y = (x-12)^p have a vertical asymptote?

For what value of x does the graph of y = (x-h)^p have a vertical asymptote?

For example, the figure below depicts the p = -3 power functions x^-3 and (x-5)^-3.

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Self-critique (if necessary):

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Question: `qGive your table (increment .4) showing how the y = x^-3 function can be transformed first into y = (x - .4) ^ -3, then into y = -2 (x - .4) ^ -3, and finally into y = -2 (x - .4) ^ -3 + .6.

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Your solution:

x y = x^-3 Y = (x – .4)^-3 Y=-2(x-.4)^-3 Y=-2(x-.4)^-3 + .6

-.8 -1.953 -.5787 1.157 1.757

-.4 -15.625 -1.953 3.906 4.506

0 0 -15.625 31.25 31.85

.4 15.625 0 0 .6

.8 1.953 15.625 -31.25 -30.65

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Given Solution:

`a** The table is as follows (note that column headings might not line up correctly with the columns):

x y = x^3 y = (x-0.4)^(-3) y = -2 (x-0.4)^(-3) y = -2 (x-0.4)^(-3) + .6

-0.8 -1.953 -0.579 1.16 1.76

-0.4 -15.625 -1.953 3.90 4.50

0 div by 0 -15.625 31.25 32.85

0.4 15.625 div by 0 div by 0 div by 0

0.8 1.953 15.625 -31.25 -30.65

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Self-critique (if necessary):

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Question: `qExplain how your table demonstrates this transformation and describe the graph that depicts the transformation.

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Your solution:

The transformation from y = x^-3 to y = (x-.4)^-3 is a horizontal shift of .4 units to the right. It has a vertical asymptote when x = .4.

The transformation of y = (x-.4)^-3 to y = -2 (x - .4)^-3 is a vertical stretch of -2, which moves the points 2 units more from the x axis. It has a vertical asymptote when x = .4.

The transformation of y = -2(x-.4)^-3 to y = -2 ( x - .4)^-3 + 6 is a vertical shift of .6 units the graph +6 units from the x-axis.

confidence rating #$&*:

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Given Solution:

`a

y = x^-3 transforms into y = (x - .4)^-3, shifting the basic points .4 unit to the right. The vertical asymptote at the y axis (x = 0) shifts to the vertical line x = .4. The x axis is a horizontal asymptote.

y = -2 (x - .4)^-3 vertically stretches the graph by factor -2, moving every point twice as far from the x axis and also to the opposite side of the x axis. This leaves the vertical line x = .4 as a vertical asymptote. The x axis remains a horizontal asymptote.

y = -2 ( x - .4)^-3 + 6 vertically shifts the graph +6 units. This has the effect of maintaining the shape of the graph but raising the horizontal asymptote to x = 6.

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Self-critique (if necessary):

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Question: `qDescribe your graphs of y = x ^ .5 and y = 3 x^.5. Describe how your graph depicts the ratios of y values between the two functions.

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Your solution:

The graph of y = x^.5 has the basic points of (0,0) (1,1) and (2,2.14).

The graph increases at a decreasing rate.

The graph of y = 3x^.5 has the basic points of (0,0) (1, 3) amd (2, 4.24). This function makes a vertical shift of the original graph 3 times as far from the x axis. Its increasing at a decreasing rate.

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Given Solution:

`a*&*& This is a power function y = x^p with p = .5.

The basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414).

• Attempting to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the conclusion that y = x^.5 is not defined for negative values of x.

• The graph therefore begins at the origin and increases at a decreasing rate.

• However since we can make x^.5 as large as we wish by making x sufficiently large, there is no horizontal asymptote.

y = 3 x^.5 vertically stretches the graph of y = x^.5 by factor 3, giving us basic points (0, 0), (.5, 2.12), (1, 3) and (2, 4.242).

• This graph is also increasing at a decreasing rate, staying 3 times as far from the x axis as the graph of the original y = x^.5.

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Self-critique (if necessary):

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Question: `qExplain why the graph of A f(x-h) + k is different than the graph of A [ f(x-h) + k ], and describe the difference.

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Your solution:

The difference is in the order of operations. For the graph of Af(x-h) + k, you would perform the function in parenthesis first then multiply the result by A and then add the value of k. This means that you would first vertically shift the graph A units, then horizontally shift it h units and vertically k units.

For the graph of A [ f(x-h) + k ], you still begin by performing the function inside the parentheses but since the brackets include the k value you would add it next and then multiply by A. So you horizontally shift h units and then vertically shift k units and then do the vertical stretch of A units.

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Given Solution:

`a** The first graph is obtained from y = f(x) by first vertically stretching by factor A, then horizontall shifting h units and finally vertically shifting k units.

The graph of A [f(x-h) + k] is obtained by first doing what is in brackets, horizontally shifting h units then vertically shifting k units before doing the vertical stretch by factor A. Thus the vertical stretch applies to the vertical shift in addition to the values of the function. This results in different y coordinates and a typically a very different graph.

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Self-critique (if necessary):

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Query Add comments on any surprises or insights you experienced as a result of this assignment.

This was a good refresher exercise.

Good work.