assignment 15

course mth 163

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Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

assignment #005

005.

Precalculus I

02-21-2007

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16:47:11

`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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RESPONSE -->

y values are 0,1,4,9

confidence assessment: 3

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16:47:34

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

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RESPONSE -->

yes but i didn't write them all out- sorry

self critique assessment: 3

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16:52:16

`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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RESPONSE -->

.125,.25,.5,2,4,8

confidence assessment: 3

Decimal answers on questions of this nature always cause me to worry about calculator misuse, and trigger the following note. From your subsequent answers is doesn't appear that you have been misusing the calculator, but this note should clarify what I consider appropriate and inappropriate use.

If you got .125, .25 and .5 from your knowledge that 1/8 is .125, 1/4 is .25 and .5 is 1/2, then these are appropriate answers.

If they resulted from plugging numbers into a calculator rather than doing the calculations yourself, then you should go back and be very sure you understand the calculations, which are essential to understanding the nature of this function.

If would be appropriate to use the calculator to find, say, 2^.75, or 2^15. It's not appropriate to use it to calculate, say, 2^-3; it's important to understand the application of the laws of exponents, in which case it's quicker and easier to just write it down than to punch the buttons.

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16:53:05

By velocity 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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RESPONSE -->

I made the ""common error""

self critique assessment: 3

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16:55:55

`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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RESPONSE -->

-1/9,-1/4,-1,1,1/4,1/9

confidence assessment: 3

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16:57:42

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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RESPONSE -->

I don't know the difference between the last problem with raising x^0=1, and 0^-2 is undefined. Is anything raised to the zero power one?

Anything raised to the 0 power is one.

There's a big difference between raising something to the 0 power and raising the number 0 to a power.

0 raised to any positive power is 0. However by the laws of exponents, 0 raised to a negative power would be the reciprocal of 0 raised to a positive power, which would give you 0 in the denominator.

self critique assessment: 3

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16:58:57

`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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RESPONSE -->

-27,-8,-1,0,1,8,27

confidence assessment: 3

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16:59:05

The y values should be -27, -8, -1, 0, 1, 4, 9.

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RESPONSE -->

ok

self critique assessment: 3

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17:01:12

`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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RESPONSE -->

I don't remember all of the previous values, however, I believe each would be a straight line

confidence assessment: 3

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17:02:06

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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RESPONSE -->

way off!!!!! Is there a way to go back and look at questions?

unfortunately there is not, except that you can always make a copy of the existing SEND file in another folder and open it (never open the original).

It is expected that you are jotting down notes and calculations, and also that the calculations required for this questionhave been done using something other than a calculator (i.e., mental calculations or simple paper calculations--save the calculator for things you can't easily figure out otherwise). On a problem like this, even if you haven't jotted anything down you should be able to quickly and easily reconstruct your tables. If you use the calculator for these simple calculations, you won't develop an internal understanding of the functions.

self critique assessment: 3

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17:09:09

`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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RESPONSE -->

After making a table the values of the two equations vary by 3. The graph y=x^2+3 shifts up on the y axis

confidence assessment: 3

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17:09:34

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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RESPONSE -->

ok I just didn't list all the values

self critique assessment: 3

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17:20:33

`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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RESPONSE -->

table for y=(x-1)^3 is -64,-27,-8,-1,0,1,8

table for y=x^3 is -27,-8,-1,0,1,8,27

y=(x-1)^3 shifted a unit to the right compared to y=x^3. Just the opposite of what one might think

confidence assessment: 3

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17:20:52

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.

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RESPONSE -->

ok

self critique assessment: 3

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17:36:44

`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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RESPONSE -->

for y=3*2^x 3/8,3/4,3/2,3,6,12,24

for y=2^x 1/8,1/4,1/2,1,2,4,8

the graph will increase up by a factor of 3

confidence assessment: 2

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17:37:37

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 far from the x axis as the graph of y = 2^x.

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RESPONSE -->

your last sentence makes no sense to me?

The original sentence was probably composed using dictation software, which misinterpreted my words. I've corrected them above.

self critique assessment: 3

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You had a few glitches early, but you did very well on the tougher questions in the second half of the assignment and you appear to be in very good shape here. Nice work.