#$&*
course Mth163
july 13,2013 02:02
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:
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The graphs of all functions approach the y axis as an asymptote, reflecting how division by smaller and smaller values gives absolute values which are larger and larger
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Why is it that division by smaller and smaller values results in larger and larger absolute values?
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Since the power is odd it will keep fractioning off but never really reach zero, because fractions cannot reach zero,even zero over zero is undef
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There is no assumption that the power is odd.
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Self-critique (if necessary):
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By saying limit does this mean that its asymptote is vertical as well???
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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:
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The displacement of y values in the positive x direction corresponds to shift of the graph h units to the right
The vertical asymptote at x=h corresponds to the division of zero which linked to asymptotes
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Good, but your teminology on this basic operation could lead to confusion. Correct terminology helps avoid confusion.
It's division by zero, not division of zero. Zero isn't being divided by anything. It's the attempt to divide by zero that alerts us to the likelihood of an asymptote.
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confidence rating #$&*: 2
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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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, h can be divided by one for how every many times because there is no limit on it
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h doesn't get divided by anything in this expression
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Your origin (0,0) has been shifted to (h,y) which causes the asymptote to move (because it was originally at (0,0) right???)
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Right. (I believe you meant (h, 0) rather than (h, y) ).
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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:
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Both negative power functions
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This would be true for any power, whether positive or negative. So this doesn't explain the identical graph shapes.
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confidence rating #$&*::
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Given Solution:
`aSTUDENT ANSWER: You end up with the exact same y values but at the different position of x changed by the h value.
INSTRUCTOR COMMENT: Good start. More specifically the x value at which a given y value occurs is shifted h
units, so that for example y = x^p is zero when x = 0, but y = (x - h)^p is zero when x = h.
To put this as a series of questions (you are welcome to insert answers to these questions, using #$&* before and after each insertion)::
Assume that p is positive.
For what value of x is x^p equal to zero?&&&0&&&
For what value of x is (x - 5)^p equal to zero?
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5=x
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For what value of x is (x - 1)^p equal to zero?
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X=1
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For what value of x is (x - 12)^p equal to zero?
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12=x
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For what value of x is (x - h)^p equal to zero?
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. h=x
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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?
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1/x
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x isn't equal to 1 / x, except for x values 1 and -1.
You were asked for a value of x.
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For what value of x does the graph of y = (x-1)^p have a vertical asymptote?
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1/1
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For what value of x does the graph of y = (x-5)^p have a vertical asymptote?
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1/5
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If x = 1/5, y = (x - 5)^ p has the value (-24/25) ^ p. For all x values near x = 1/5 the value of y will be close to (-24/25)^p.
There is no asymptote at x = 1/5.
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For what value of x does the graph of y = (x-12)^p have a vertical asymptote?
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1/12
If you plug in x = 1/12, you obtain a specific finite result for y. There is no vertical asymptote at x = 1/12.
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For what value of x does the graph of y = (x-h)^p have a vertical asymptote?
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1/h
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If you plug in x = 1/h, you obtain a specific expression for y, which will be finite for any nonzero value of h. There is no vertical asymptote at x = 1/h.
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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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your y value is the x shifted h units so when x=0 , o=h,x=h
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Self-critique rating:3
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Self-critique rating:
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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:
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X
.4
.8
1.2
1.6
2
. y=x^-3
15.6
2
.6
.2
.1
. y=(x-.4)^-3
Asymptote (undef)
15.6
2
.6
.2
. y= -2(x-.4)^-3
Asy.
-31.3
-3.9
-1.2
-0.5
. y= -2(x-.4)^-3+.6
Asy.
-30.7
-3.3
-0.6
0.1
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confidence rating #$&*:
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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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In my example I used y = x^-3 instead of y = x^3
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Self-critique rating:
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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:
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Going from linear x=y, you get a power function y = x^-3 with both vertical and horizontal zeros you get curves in the first and third quadrants
Now y = (x-0.4)^(-3) if I had not messed up before and put -3 instead of 3 at this point the quadrants in which your curves are in would be reversed and shifted to the right .4 units
y = -2 (x-0.4)^(-3) due to the negative 2 our graph is now vertically shifted by a factor of 2 and our curves are in quadrants two and four
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A vertical shift raises every point by the same amount.
A vertical stretch moves every point further from the x axis by some constant factor, multiplying the distance of each point from the x axis by that factor.
Which occurs here, and why?
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y = -2 (x-0.4)^(-3) + .6 going from a neg power function with horizontal and vertical asy. The graph here appears to have gone linear perhaps the .6 evened the equation out???
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The values in your table certainly don't indicate linearity. If you graph those values, what does your graph look like?
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confidence rating #$&*: 2
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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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Self-critique rating:
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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:
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. x
0
1
2
3
4
. y=x^.5
0
1
1.4
1.7
2
. y=x^.5
0
3
4.2
5.2
6
So the differences between my functions are 2,2.8,3.5,4 (I look the abs value because I was not sure whether to subtract the first column from the second or vice versa)
The graph is moved my a factor or 3, vertically stretches all of your other points
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confidence rating #$&*:
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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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Next time find basic points first,
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Self-critique rating:
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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:
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A f(x-h) + k here you are just vertically stretching the y by a factor of A
A [ f(x-h) + k ] here you are actually saying Af(x-h) + Ak, you are vertically stretching your whole graph by a factor of A
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You haven't mentioned the vertical shifts of these graphs.
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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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Order , what’s inside done first then stretched by factor A where ever it appears in the equation
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Self-critique rating:
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:
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What your zero is determines where your asymptote lies
In the first equation the zero of (x-3)=3=x
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(x-3)=3=x
is not a true statement. x - 3 is not equal to 3, and it's not equal to x.
I know what you mean, but you work on saying what you mean clearly, using correct terminology to connect correctly formed mathematical expressions.
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In the second equation (x + 3)=-3=x
Both are double roots, possibility 8 degree???
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confidence rating #$&*: 3
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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:
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(when I put decimals in front of letters it is to keep them from capitalizing)
. x
1
2.5
3
. y=x^.5
1
1.6
1.7
. y=2x^.5
2
3.3
3.5
. y=2(x-3)^.5
. e^(1.6i)*2.8
. e^(1.6i)*1.4
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You would not use the same x values for this function. You would use the x values that give you the same y values as the preceding.
For y = 2 x ^ .5 a correct table would be
0 0
1 2
2 2.8
3 3.5
4 4
This is pretty much like your table but x values 0 and 4 have been added, and your y value for x = 2 was not correct.
For the new function y = 2 ( x - 3 ) ^ .5 the x values must be 3 units larger in order to give you the same results. For example x = 4 would give you x - 3 = 1, so for x = 4 you would be taking the .5 power of 1.
The table, with all x values increased by 3, would then be
3 0
4 2
5 2.8
6 3.5
7 4
This table shows why replacing x by x - 3 shifts the graph 3 units to the right.
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0
.y=2(x-3)^.5-1
. e^(1.9i)*3
. e^(2.2i)*1.7
-1
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The appropriate table is based on the previous table
3 0
4 2
5 2.8
6 3.5
7 4
The y value would be decreased by 1 to give you
3 -1
4 1
5 1.8
6 2.5
7 3
This would demonstrate clearly why the graph shifts 1 unit downward.
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From y = x^.5 to y = 2 x^.5 the entire graph vertically stretches by a factor of 2
Then at 2 ( x - 3) ^ .5 the graph shifts over 3 units to the right,( then V stretch by a factor of two then compressed by .5)
Finally at y=2(x-3)^.5-1 the graph moves down one unit as a whole from the last step
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confidence rating #$&*: 3
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Question: Starting with the function y = f(x), describe the series of graphical transformations required to transform its graph to the graph of y = A f(x - h) + k.
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Your solution:
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1. Horizontally shift h units
2. Then V stretch by a factor of A
3. Finally add the rest to k
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That last statement could be more specific:
You would add k to each of the resulting y coordinates.
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confidence rating #$&*: 3
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Self-critique rating:
Query Add comments on any surprises or insights you experienced as a result of this assignment.
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Zero raised to a negative even number equals infinity (at least on my cal) and if its neg odd its undefined (which would be an asymptote right???)
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"
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Your calculator is not a correct mathematical instrument, though it does most mathematics correctly.
Zero raised to a negative power is undefined. It doesn't matter what the power.
When you raise numbers close to zero to negative powers, you get results of large magnitude, and the closer those numbers are to zero, the larger that magnitude. The magnitudes increase without bound as the denominator continues to approach zero, which is the reason for the vertical asymptotes of the graph.
The reasons for all this cannot be understood by using a calculator. Then need to be understood in terms of your understanding of division and what division means, as well as what powers mean, and the calculator should play no part in developing this understanding.
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Self-critique (if necessary):
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Self-critique rating:
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You've got a general picture of these functions, but you don't appear to really understand what vertical asymptotes mean and how they arise.
You're not doing badly with the ideas of shifts and stretches, but you don't yet completely understand the specifics of why they happen. Be sure to see my notes on how to use tables to demonstrate them.
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I'm going to ask you to submit your answers to the following questions, which should clarify vertical asymptotes and add to your understanding of the meaning of division:
If you understand this it won't take you long. If you don't, it's worth the time you spend on it:
Sketch a line segment at least a couple of inches long.
Sketch a line segment 1/3 the length of the original.
How many of the shorter segment are required to equal the length of the longer?
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what is 1 divided by 1/3 and how do your line segments reveal the meaning of this division?
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Now sketch a line segment 1/10 the length of the original.
How many of the shorter segment are required to equal the length of the longer?
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what is 1 divided by 1/10 and how do your line segments reveal the meaning of this division?
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How can you make sense of the division of 1 by 1/n using this geometrical reasoning?
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On a y vs. x graph, scale the x axis so that the interval from 0 to 1 on that axis matches the length of the original line segment you
sketched above.
Mark the points 1/3 and 1/10 on the x axis.
What is the value of y = 1 / x for x = 1, 1/3 and 1/10, and how is this related to the sketches you made above?
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Scale your y axis so it accommodates these y values.
Plot the points on a graph of y = 1 / x corresponding to x values 1, 1/3 and 1/10. Describe your graph.
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What are the values of x = 1 / n for n = 1, 2, 3, 5 and 10?
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Which of these x values were used in sketching your graph?
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As n gets larger, what happens to the point you would mark on the x axis to represent x = 1 / n?
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Mark a new point on your x axis which is as close as possible to the origin, but still distinct from the origin.
How many times closer to the origin is this point than the point you marked for x = 1/10?
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If this point represents x = 1 / n, what do you think is the value of n?
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If you were to sketch a very short line segment whose length is equal to the distance from the origin to your new point, how many such
segments would be required to equal the length of your very first line segment, the one you drew for the very first question?
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If x is the value represented by your new point, what then is the value of 1 / x?
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Where would the corresponding point on the graph of y = 1 / x be? Would you be able to draw it on your graph?
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Imagine you had a powerful magnifying glass that could magnify the small interval between the origin and your new point until it looked as
long as your very first line segment. Imagine you have precise instruments that could mark a microscopic point on this interval as close to
the origin as possible, but still distinct from the origin, as well as the microscopic line segment from the origin to this point.
How many times closer to the origin would this point be than the new point you marked previously?
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How many times further from the x axis would the corresponding graph point be, compared to the preceding graph point?
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How far from the x axis would be the corresponding point on the graph of y = 1 / x?
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If we continue to repeat this process, every time magnifying the interval to mark an additional points closer to the origin and plotting the
corresponding point on the graph of y = 1 / x, what happens to the graph?
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What does all this have to do with vertical asymptotes?
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How does this help you understand the graph of y = x^(-p)?
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