Assignment 5

course MTh 163

02/07 8:35pm

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.

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=x^2 x y

Y=-3^2 -3 9

Y=9 -2 4

-1 1

Y=-2^2 0 0

Y=4 1 1

2 4

Y=-1^2 3 9

Y=1

Y=0^2

Y=0

Y=1^2

Y=1

Y=2^2

Y=4

Y=3^2

Y=9

confidence rating #$&* 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):

I understand this and it makes sense to me!!

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=2^x

Y=2^-3 x y

Y=-8 -3 -8

-2 4

Y=2^-2 -1 -2

Y=4 0 0

1 1

Y=2^-1 2 4

Y=-2 3 8

Y=2^0

Y=0

Y=2^1

Y=2

Y=2^2

Y=4

Y=2^3

Y=8

confidence rating #$&* ?

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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):

Where did the b come from? I thought I knew what I was doing and now I have no idea what I’m suppose to do.

I just Googled 'laws of exponents'. The second site that came up was

http://oakroadsystems.com/math/expolaws.htm

I can recommend this for a review of the laws of exponents, which are an Algebra I topic but with which students in Mth 163 are often pretty rusty.

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Self-critique rating #$&* Not OK

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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?

Your solution

Y=x^-2

Y=-3^-2

Y=9

Y=-2^-2

Y=4

Y=-1^-2

Y=1

Y=0^-2

Y=0

Y=1^-2

Y=1

Y=2^-2

Y=4

Y=3^-2

Y=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 the formula, I tried looking up laws of exponents and didn’t find anything helpful. I really did think I this figured out.

The site I quoted above includes the following (see the site for better formatting):

What Is an Exponent, Anyway?

There’s nothing mysterious! An exponent is simply shorthand for multiplying that number of identical factors. So 4³ is the same as (4)(4)(4), three identical factors of 4. And x³ is just three factors of x, (x)(x)(x).

One warning: Remember the order of operations. Exponents are the first operation (in the absence of grouping symbols like parentheses), so the exponent applies only to what it’s directly attached to. 3x³ is 3(x)(x)(x), not (3x)(3x)(3x). If we wanted (3x)(3x)(3x), we’d need to use grouping: (3x)³.

Negative Exponents

A negative exponent means to divide by that number of factors instead of multiplying. So 4-3 is the same as 1/(43), and x-3 = 1/x3.

As you know, you can’t divide by zero. So there’s a restriction that x-n = 1/xn only when x is not zero. When x = 0, x-n is undefined.

A little later, we’ll look at negative exponents in the bottom of a fraction.

Fractional Exponents

A fractional exponent — specifically, an exponent of the form 1/n — means to take the nth root instead of multiplying or dividing. For example, 4(1/3) is the 3rd root (cube root) of 4.

Here’s All You Need to Memorize

[[ what appears here is a picture, which can't be included here but which is easy to read on the website ]]

And that’s it for memory work. Period. If you memorize these three definitions, you can work everything else out by combining them and by counting:

Granted, there’s a little bit of hand waving in my statement that you can work everything else out. Let me make good on that promise, by showing you how all the other laws of exponents come from just the three definitions above. The idea is that you won’t need to memorize the other laws — or if you do choose to memorize them, you’ll know why they work and you’ll find them easier to memorize accurately.

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Self-critique rating #$&*Not OK

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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=x^3

Y=-3^3

Y=-27

Y=-2^3

Y=-8

Y=-1^3

Y=-1

Y=0^3

Y=0

Y=1^3

Y=1

Y=2^3

Y=8

Y=3^3

Y=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):I get this it makes sense to me.

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Self-critique rating #$&* OK

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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: The graph makes a perfect parabola from questions 1 and 4. The points fit together with one steeper than the other. I didn’t get the answers from the other two because I don’t understand the formula

confidence rating #$&* 1/2

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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): I have a hard time describing the graphs, when I read the solution it sounds really good. I got half of the graph to work but the half that had the negative exponents I don’t understand.

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Self-critique rating #$&*Confused

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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:

Y=x^2+3

Y=-3^2+3

Y=9+3

Y=12

Y=-2^2+3

Y=4+3

Y=7

Y=-1^2+3

Y=1+3

Y=4

Y=0^2+3

Y=0+3

Y=3

Y=1^2+3

Y=1+3

Y=4

Y=2^2+3

Y=4+3

Y=7

Y=3^2+3

Y=9+3

Y=12

The graph would be larger by three spaces

confidence rating #$&* 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):Again this makes sense to me

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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:

Y=(x-1)^3 x y

Y=(-3-1)^3 -3 -64

Y=-4^3 -2 -27

Y=-64 -1 -8

0 -1

Y=(-2-1)^3 1 0

Y=-3^3 2 1

Y=-27 3 8

Y=(-1-1)^3

Y=-2^3

Y=-8

Y=(0-1)^3

Y=-1^3

Y=-1

Y=(1-1)^3

Y=0^3

Y=0

Y=(2-1)^3

Y=1^3

Y=1

Y=(3-1)^3

Y=2^3

Y=8

confidence rating #$&* 3

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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):I thought I had the right answer but apparently I didn’t go far enough.

I feel very uncertain about these, I think I understand but, I am not very confident.

I expect you're OK with this part.

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Self-critique rating #$&*

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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:

Y=3*2^x

Y=3*2^-1

Y=3*-2

Y=-6

Y=3*2^-2

Y=3*4

Y=12

Y=3*2^-3

Y=3*-8

Y=-24

Y=3*2^0

Y=3*2

Y=6

Y=3*2^1

Y=3*2

Y=6

Y=3*2^2

Y=3*4

Y=12

Y=3*2^3

Y=3*8

Y=24

confidence rating #$&* -1

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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 the past that of y = 2^x.

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Self-critique (if necessary):I am having trouble with the negative exponents formula.

Please advise

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Self-critique rating #$&*Not ok

I'm not going to explain negative exponents, simply because it's an Algebra I topic and online references do a better job than I could here. I've given you at least one reference. I'll be glad to answer questions about what you see at the site to which I referred you.

It's not a difficult concept, once you get used to it, and you have seen it before (very fundamental in algebra courses). You just haven't used it in awhile. So I think you'll get it. Once you have, go back and revisit those problems.

Otherwise you appear to be doing very well.