Assignment 22

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course Mth 163

4/11/13 around 10a.m.

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:

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 a 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 will give us y = x^p values that approach infinite distance from the x axis. The graph thus approaches a vertical limit

confidence rating #$&*: 3

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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): ok

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Self-critique rating: 3

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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:

(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 gives us 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. If we substitute x by x - h it will shift the graph h units in the x direction, so the asymptote at x = 0 shifts to x = h.

confidence rating #$&*: 3

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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): ok

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Self-critique rating: 3

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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:

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.

confidence rating #$&*: 3

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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): ok

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Self-critique rating: 3

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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:

The table is as follows

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

confidence rating #$&*: 3

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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): ok

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Self-critique rating: 3

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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:

y = x^-3 changes 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.

confidence rating #$&*: 3

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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): ok

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Self-critique rating: 3

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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:

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)

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.

confidence rating #$&*: 3

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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): ok

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Self-critique rating: 3

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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:

We get the first graph from y = f(x) by first vertically stretching by factor then horizontal shifting h units and finally vertically shifting k units.

The graph of A [f(x-h) + k] is attained 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

confidence rating #$&*: 3

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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): ok

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Self-critique rating:3

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: Explain why the function y = (x - 3) ^ (-2) has a vertical asymptote at x = +3, while the function y = (x + 3)^(-2) has a vertical asymptote at x = -3.

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Your solution:The first function has a vertical asymptote at x = 3 because that is where the graph of the function intersect the vertical asymptote once, at (0,3). We have \

Both of these statements are true making this a vertical asymptote

The same is true for the next equation because the graph fo the function intersects the vertical asymptote at (0,-3).

confidence rating #$&*: 2

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Question: Use a table containing appropriate values to demonstrate the transformation of the y = x^(.5) function into the y = 2 (x - 3) ^ .5 - 1 function, as a series of transformations in which the original function is transformed first to y = 2 x^.5, then to y = 2 ( x - 3) ^ .5, then finally to the final form.

Briefly describe the original graph, and how the graph changes with each successive transformation.

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Your solution:

Our table is as follows:

Y = x^.5

X Y

4 2

3 1.73

2 1.414

1 1

0 0

Transforming this graph changes it to y = x with a shift by .5 units in the x direction

If we then transform it into y = 2 x^.5 we have multiplied it by 2 giving us the graph

Y = 2 x^.5

X Y

4 4

3 3.46

2 2.828

1 2

0 0

If we have transformed it into y = 2 ( x - 3) ^ .5 we have substituted x for x - 3 giving us

Y = 2 (x-3) ^.5

X Y

4 2

3 0

2 cannot be determined

1 cannot be determined

0 cannot be determined

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This transformation changes the x values, and would not result in x values 4, 3, 2, 1, 0.

It appears that you might have evaluated the function y = 2 ( x - 3 ) ^ .5 at the x values 4, 3, 2, 1, 0, rather than using the transformation to directly transform the graph. (Your transformed function would indeed be y = 2 ( x - 3 ) ^ .5 , but the method used here would not rely on evaluating that function. It would of course be useful to check yourself, after transforming the table, by evaluating the function. )

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Finally if we transform it into y = 2 (x-3) ^.5 - 1 we have changed x to x - 3 and then subtracted by 1 meaning we have an asymptote at -1.

The table we can construct from this gives us

Y = 2 (x-3) ^.5 - 1

X Y

4 .414

3 -1

2 cannot be determined

1 cannot be determined

0 cannot be determined

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Your x = 4 value does not follow from the previous table by this transformation (nor does it follow by evaluating the function, but as I've mention that's not the method appropriate to this question).

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confidence rating #$&*: 2

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Question: Starting with the function y = f(x), describe the series of graphical transformations required to transform its graph to the graph of y = A f(x - h) + k.

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Your solution:

If we start with y = f(x) we would first transform x into x -h giving us y = f(x-h). We would then add the factor of A multiplying the equation by it giving us y = A f(x-h). The finally add an asymptote where it crosses the axis giving us the final form of y = A f(x-h) + k

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This is good as far as it goes, but you haven't described the graphical transformations.

If you were given the graph of an unknown function f(x), how would you transform it into the function A f(x - h) + k?

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confidence rating #$&*: 3

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Self-critique rating: 3

"

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Good. You're doing a lot of things right.

However I do suggest you take a few minutes and revise the problems at the end.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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