course Mth 163 I am definitely going to need further pracatice on these two assignments. Would it be possible for me to do the problems in the book and submit them to you and you could give me feedback as to what I am doing incorrectly? I have had quite the problem with this.
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14:28:01 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> Given f(x) = x^3 The value for f(-2) would be f(-2) = (-2)^3 f(-2) = -8 Te value for f(-a) would be f(-a) = (-a)^3 The value for f(x-4) would be f(x-4) = (x-4)^3 f(x-4) = (x-4)(x-4)(x-4) f(x-4) = x^2 -8x + 16 (x-4) f(x-4) = x^3 - 12x - 64 The value for f(x) - 4 would be f(x) - 4 = x^3 - 4
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14:30:40 ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. **
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RESPONSE --> I understood the more simple forms, I will do some more practice with regards to the more complicated forms. I should have also further simplified the -a question instead of leaving it to simply (-a)^2, I should have put a^2.
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14:32:52 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> For f(x) = 2^x For f(2) f(2) = 2^2 f(2) = 4 For f(-a) f(-a) = 2^(-a) For f(x+3) f(x+3) = 2^(x+3) For f(x) + 3 f(x) + 3 = 2^x + 3
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14:33:23 ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **
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RESPONSE --> I have a pretty firm understanding of these problems.
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14:33:29 13:06:29
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RESPONSE --> ok
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14:34:30 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> Some advantages of using meaningful names for functions would be to distinguish them verbally instead of being required to visually explain each item every time.
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14:34:39 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example...
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RESPONSE --> ok
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14:35:34 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.**
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RESPONSE --> I agree that the equation does take on more of a specialized meaning as we become more specific as to exactly what we are trying to find.
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14:36:10 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> I am not sure as I do not remember where these items are being drawn from.
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14:38:44 ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. **
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RESPONSE --> Ok this was on the f(x) notation problems. I turned these in separately by pasting out of a word document into the submission form.
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14:38:49 13:14:30
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RESPONSE --> ok
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14:39:29 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> Illumination(1) = 50 / 1^2 = 50 / 1 = 50 Illumination(2) = 50 / 2^2 = 50 / 4 = 12.5 Illumination(3) = 50 / 3^2 = 50 / 9 = 5.56 Illumination(distance) / illumination(2*distance) = (50 / distance^2) / (50 / 2*distance^2)
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14:40:13 ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. **
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RESPONSE --> I am going to make notes on this one and study on it further.
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14:41:57 query #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph.
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RESPONSE --> You would construct a graph by taking the x and y values given in the statement. (2, 80) (5,40) (10, 25) The y values are larger so I made my unit of value set to 10 so as not to make the graph unmanagable. The x axis unit of measure would be 1 so as not to make the graph too crowded.
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14:42:39 ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. **
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RESPONSE --> I also connected the points via lines, I was neglectful as to stating this fact.
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14:44:11 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> I would estimate the value to be 60 as I am not given an equation which I can use to give a true estimate. I am guessing since there is no equation that the value of f(x) would be 60.
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14:44:54 **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3**
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RESPONSE --> Oh I was not referring to the previous graph. It did not state that the question referred back to the previous question. I apologize for this.
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ѱӿTD_ assignment #004 lCٍ蛢w Precalculus I 02-05-2006
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14:47:33
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RESPONSE -->
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14:47:41 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> done
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14:47:46 ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. **
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RESPONSE --> done
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14:47:54 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> done
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14:48:03 ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **
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RESPONSE --> done
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14:48:08 13:06:29
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RESPONSE --> done
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14:48:13 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> done
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14:48:18 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example...
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RESPONSE --> done
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14:48:22 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.**
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RESPONSE --> done
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14:48:26 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> done
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14:48:32 ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. **
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RESPONSE --> done
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14:48:38 13:14:30
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RESPONSE --> done
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14:48:45 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> done
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14:48:52 ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. **
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RESPONSE --> done
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14:49:21 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> done
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14:49:26 **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3**
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RESPONSE --> done
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14:51:29 what is your estimate of the value f(7)?
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RESPONSE --> For the value of f(7) I would estimate the corresponding y coordinate to be 30.
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14:52:08 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. **
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RESPONSE --> I was close but did not use the slope to find the approximate value.
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14:53:06 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> The difference between f(7) and f(9) would be estimated at 2-3 units.
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14:53:48 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> I have decided to invest in graph paper as my estimates according to my graphs are truly not coming out correct.
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14:53:53 13:29:11
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RESPONSE --> ok
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14:55:00 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
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RESPONSE --> I would estimate the difference to be approximately 4 1/4 units.
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14:55:24 ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. **
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RESPONSE --> I did the drawing of lines, my graph just appears to be off.
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14:56:27 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> For some reason the rest of the question was not displayed on my ""Question"" side of this form. (query #4)
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14:57:38 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 **
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RESPONSE --> I am pretty sure I understand this concept but have pasted these into my notes.
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14:59:06 What equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30?
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RESPONSE --> T(150) T(t) = 80 - 30 T(50)
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14:59:28 ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] **
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RESPONSE --> ok
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14:59:55 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
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RESPONSE --> ????? I do not know where this question is referring?
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15:00:00 ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' **
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RESPONSE --> ok
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15:00:07 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> ?/????/?
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15:00:13 ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. **
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RESPONSE --> ok
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15:00:20 On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds?
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RESPONSE --> ?
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15:00:24 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. **
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RESPONSE --> ok
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15:00:31 On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds?
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RESPONSE --> ?
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15:00:36 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. **
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RESPONSE --> ok
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N{Ҋ` assignment #005 lCٍ蛢w Precalculus I 02-05-2006
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15:16:46 query introduction to basic function families problem 1 on basic graphs Why is the graph of y = x a straight line?
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RESPONSE --> For every value of x it will equal the value of y. This leads to a slope that allows the formation of a straight line.
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15:16:55 ** Since y = x the rise and run between any two points on the graph are equal, which makes the slope 1. A graph with constant slope is a straight line. **
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RESPONSE --> ok
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15:22:04 why is y = x^2 symmetric about x = 0 (i.e., taking the same values on either side of x = 0)
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RESPONSE --> 0 to any degree will still be equal to zero.
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15:23:45 ** The graph of y = x^2 is symmetric about x = 0 because (-x)^2 = x^2. Thus for any point on the x axis the y values at that point and at the point on the opposite side of the origin are equal, so that the graph on one side of the y axis is a 'reflection' of the graph on the other side. **
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RESPONSE --> Ok I thought I was just supposed to speak as to why the 0 always was 0 according to the preceeding equation. Any integer, negative or positive, that is squared becomes a positive and therefore causes a mirrot image from the negative and positive sides.
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15:25:26 why does y = 2^x keep increasing as x increases, and why does the graph approache the x axis for negative values of x
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RESPONSE --> y=2^x increases as x increases since we are taking the 2 to a higher degree. (2^2 will be smaller than 2^3) However if the exponent were to be negative the result would still be positive, just in a fractional form.
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15:25:36 ** GOOD STUDENT RESPONSE: y = 2^x will increase as x increases on the positive side because x is the value of the exponent. This will cause the y value to double from its last value when you move one unit in the positive x direction. On the negative side of the y axis y = 2^x will approach the x axis because a negative exponent causes the value to invert into a fractional value of itself--i.e., 2^(-x) = 1 / 2^x. As we move one unit at a time negatively the value will become one half of the previous value so it will never quite reach y = 0. **
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RESPONSE --> I understand.
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15:26:46 why is y = x^3 antisymmetric about x = 0 (i.e., taking the same values except for the - sign on opposite sides of x = 0)
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RESPONSE --> If using a negative number the result will be negative, while if using a positive number the result will be positive. I cannot therefore be symmetric about the x=0.
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15:26:52 ** y = x^3 is antisymmetric because if you cube a negative number you get a negative, if you cube a positive number you get a positive, and the magnitude of the cubed number is the cube of the magnitude of the number. So for example (-3)^2 = -27 and 3^3 = 27; the points (-3, -27) and (3, 37) are antisymmetric, one being `down' while the other is `up'. GOOD STUDENT RESPONSE: y = x^3 is antisymmetric about x = 0 because the exponent is an odd number. This will cause negative x values to have a negative y result. The absolute value of the negative y result will be equivalent to its corresponding positive y value. **
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RESPONSE --> ok
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15:28:07 why do y = x^-2 and y = x^-3 rise more and more steeply as x approaches 0, and why do their graphs approach the x axis as we move away from the y axis.
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RESPONSE --> Since the number is a negative, the smaller the number that we use in place of x, the closer we get to the y axis.
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15:28:48 ** And as x approaches 0 the expressions x^-2 and x^-3, which mean 1 / x^2 and 1 / x^3, have smaller and smaller denominators. As the denominators approach zero their reciprocals grow beyond all bound. y = x^-2 and y = x^-3 rise more and more steeply as x approaches zero because they have negative exponents they become fractions of positive expressions x^2 and x^3 respectively which have less and less slope as they approach zero. As x^2 and x^3 approach zero and become fractional, x^-2 and x^-3 begin to increase more and more rapidly because thier functions are then a whole number; (1) being divided by a fraction in which the denominator is increasing at an increasing rate. As y = x^-2 and y = x^-3 move away from the y-axis they approach the x-axis because they have negative exponents. This makes them eqivalent to a fraction of 1 / x^2 or 1 / x^3. As x^2 and x^3 increase in absolute value, the values of y = x^-2 and y = x^-3 constantly close in on the x-axis by becoming a portion of the remaining distance closer, they will never reach x = zero though as this would be division by zero (since it is a fraction) **
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RESPONSE --> I have saved this in my notes.
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15:29:55 query problem 2. family y = x^2 + c Explain why the family has a series of identical parabolas, each 1 unit higher than the one below it.
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RESPONSE --> The graph of x^2 is a parabola. Since you are adding ""c"" to the equation it moves the location of the parabola on the graph.
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15:30:00 ** GOOD STUDENT RESPONSE: The graph of y = x^2 + c, with c varying from -5 to 4 is a series of identical parabolas each 1 unit higher than the one below it. The c value in the quadratic equation has a direct impact on the vertical shift. The vertex of the graph will be shifted vertically by the amount of the c value, so every time c increases by 1 the graph is raised 1 unit. **
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RESPONSE --> ok
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15:32:36 query problem 4. describe the graph of the exponential family y = A * 2^x for the values A = -3 to 3.
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RESPONSE --> For each of the values of A, the location of the line on the graph will change location. The line will increase at an increasing rate.
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15:32:41 ** This family includes the functions y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x, y = 0 * 2^x, y = 1 * 2^x, y = 2 * 2^2 and y = 3 * 2^x. Each function is obtained by vertically stretching the y = 2^x function. y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x all vertically stretch y = 2^x by a negative factor, so the graphs all lie below the x axis, asymptotic to the negative x axis and approaching negative infinity for positive x. They pass thru the y axis as the respective values y = -3, y = -2, y = -1. y = 1 * 2^x, y = 2 * 2^x, y = 3 * 2^x all vertically stretch y = 2^x by a positive factor, so the graphs all lie above the x axis, asymptotic to the negative x axis and approaching positive infinity for positive x. They pass thru the y axis as the respective values y = 1, y = 2, y = 3. y = 0 * 2^x is just y = 0, the x axis. Of course the functions for fractional values are also included (e.g., y = -2.374 * 2^x) but only the integer-valued functions need to be included in order to get a picture of the behavior of the family. ** STUDENT QUESTION: Ok, it was A = -3 to 3. I understand how to substitute these values into y = A * 2^x. I knew that is was an asymptote, but I'm a little confused as to how to graph the asymptote. INSTRUCTOR RESPONSE: For each value of A you have a different function. For A = -3, -2, -1, 0, 1, 2, 3you have seven different functions, so you will get 7 different graphs. Each graph will contain the points for all values of x. For example the A = -3 function is y = -3 * 2^x. This function has basic points (0, -3) and (1, -6). As x takes on the negative values -1, -2, -3, etc., the y values will be -1.5, -.75, -.375, etc.. As x continues through negative values the y values will approach zero. This makes the y axis a horizontal asymptote for the function. You should figure out the x = 0 and x = 1 values for every one of these seven functions, and you should be sure you understand why each function approaches the negative x axis as an asymptote. *&*&
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RESPONSE --> ok
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15:33:10 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.
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RESPONSE --> The line will once again change locations based on the value of c.
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15:33:16 ** There are 7 graphs, including y = 2^x + 0 or just y = 2^x. The c = 1, 2, 3 functions are y = 2^x + 1, y = 2^x + 2 and y = 2^x + 3, which are shifted by 1, 2 and 3 units upward from the graph of y = 2^x. The c = -1, -2, -3 functions are y = 2^x - 1, y = 2^x - 2 and y = 2^x - 3, which are shifted by 1, 2 and 3 units downward from the graph of y = 2^x. **
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RESPONSE --> ok
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15:33:50 query problem 5. power function families Describe the graph of the power function family y = A (x-h) ^ p + c for p = -3: A = 1, h = -3 to 3, c = 0.
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RESPONSE --> I will need further practice on this,
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15:34:13 ** GOOD STUDENT RESPONSE: I sketched the graph of the power function family y = A (x-h)^p + c for p = -3: A = 1: h = -3 to 3, c = 0. Beginning on the left side of the graph the curve was infinitely close to its asmptote of y = 0. This was determined by the value of c. As we move from left to right the curves decreased at an increasing rate, approaching thier vertical asmptotes which was determined by thier individual values of h. The curves broke at x = c as this value was never possible due to division by zero. The curves resurfaced on the graph high on the right side of thier vertical asymptotes and from there they decreased at a decreasing rate, once again approaching thier horizontal asymptote of y = 0. INSTRUCTOR COMMENTS: Only the h value changes. p=-3, A=1 and c=0, so the functions are y = 1 * (x-h)^-3 or y = (x-h)^-3. For h = -3 to 3 the functions are y = (x - (-3))^-3, y = (x - (-2))^-3, y = (x - (-1))^-3, y = (x - 0)^-3, y = (x - 1)^-3, y = (x - 2)^-3, y = (x - 3)^-3. These graphs march from left to right, moving 1 unit each time. Be sure you see in terms of the tables why this happens. **
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RESPONSE --> I have saved this into my notes.
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ؓ`߄`ᙵ Student Name: assignment #004
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12:01:49 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.
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RESPONSE --> When given the equation f(x) = x^2 + 4 Find the value of f(3) f(3) = (3^2) + 4 f(3) = 9 + 4 f(3) = 13 Find the value of f(7) f(7) = (7^2) + 4 f(7) = 49 + 4 f(7) = 53 Find the value of f(-5) f(-5) = (-5)^2 + 4 f(-5) = 25 + 4 f(-5) = 29 The graph that I sketched appears to be a parabola that would pass through the y axis at (0, 4). It begins at the point (-5, 29) and gradually rounds down to pass through its vertex at (0,4) and then starts back up touching on points(3,13) and (7, 53).
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12:03:30 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.
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RESPONSE --> I have a fair grasp of this concept but will definitely continue to practice as far as the graphing is concerned. It has never been my strong point but I am truly working on it.
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12:19:48 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.
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RESPONSE --> If f(x) = x^2 + 4 The symbolic expression for f(a) would be f(a) = (a^2) + 4 The symbolic expression of f(x+2) would be f(x+2) = (x+2)^2 +4 f(x+2) = [(x+2)(x+2)] + 4 f(x+2) = [x^2 + 2x + 2x + 4] + 4 f(x+2) = x^2 + 4x + 4 + 4 f(x+2) = x^2 + 4x + 8 The symbolic expression of f(x+h) would be f(x+h) = [(x+h)^2] + 4 f(x+h) = [(x+h)(x+h)] + 4 f(x+h) = [x^2 + xh + xh + h^2] + 4 f(x+h) = [x^2 + 2xh + h^2] + 4 The symbolic expression of f(x+h) - f(x) would be f(x+h) - f(x) = [(x+h)^2 + 4] - (x^2 + 4) f(x+h) - f(x) = [(x+h)(x+h) + 4] - (x^2 + 4) f(x+h) - f(x) = [x^2 + 2xh + h^2] - [x^2 + 4] f(x+h) - f(x) = x^2 + 2xh + h^2 - x^2 + 4 f(x+h) - f(x) = 2xh + h^2 + 4 f(x+h) - f(x) = h^2 + 2xh + 4 The symbolic expression of [f(x+h) - f(x)]/h [h^2 + 2xh + 4]/h (I got the amount in the parentheses from the previous problem.) [f(x+h)-f(x)]/h = h^2 + 2x + 4
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12:22:28 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
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RESPONSE --> The only place I can see that I did not understand this problem is in the last problem. I still had the 4 included in the expression. I failed to subtract the 4 from the first part of the equation.
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12:31:13 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.
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RESPONSE --> If f(x) = 5x + 7 then For f(x1): f(x1) = 5(x1) + 7 For f(x2): f(x2) = 5(x2) + 7 For [f(x2) - f(x1)] / (x2 - x1) [f(x2) - f(x1)] / (x2 - x1) [(5(x2) + 7) - (5(x1) + 7)] / (x2 - x1) [5(x2) - 5(x1) + 14] / (x2 - x1)
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12:32:33 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
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RESPONSE --> I thought I should have factored out the 5 but failed to do so. I do understand how that happened and I also failed once again to subtract the 7 as I had previously with the 4.
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12:35:38 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
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RESPONSE --> f(x) = 5x + 7 So in order for the value of x in f(x) to leave the equation equal to -3, x will have to be -3 = 5x + 7 -10 = 5x -2 = x So in order for the expression of f(x) = 5x + 7 the value of x would need to be -2. Verification f(-2) = 5(-2) + 7 f(-2) = -10 + 7 f(-2) = -3
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12:36:00 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.
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RESPONSE --> I understand this concept and was able to come up with the correct answer.
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Ǫo Student Name: assignment #005
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12:44:15 `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 --> For y = x^2 evaluate for x values of For x value of -3 y = (-3)^2 y = 9 For x value of -2 y = (-2)^2 y = 4 For the x value of -1 y = (-1)^2 y = 1 For the x value of 0 y = (0)^2 y = 0 For the x value of 1 y = (1)^2 y = 1 For the x value of 2 y = (2)^2 y = 4 For the x value of 3 y = (3)^2 y = 9
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12:44:33 You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.
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RESPONSE --> I was able to obtain these values from the previous exercise.
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12:51:40 `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 --> For the expression y = 2^x For the x value of -3 y = (2^-3) y = 1 / 2^3 y = 1 / 8 For the x value of -2 y = (2^-2) y = 1 / 2^2 y = 1 / 4 For the x value of -1 y = (2^-1) y = 1 / 2^1 y = 1 / 2 For the x value of 0 y = (2^0) y = 1 For the x value of 1 y = 2^1 y = 2 For the x value of 2 y = 2^2 y = 4 For the value of 3 y = 2^3 y = 8
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12:52:17 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 was able to get the correct responses for this exercise.
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12:58:34 `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 --> Evaluate expression y = x^-2 For the x value of -3 y = -3^-2 y = 1/ (-3^2) y = 1 / 9 For the x value of -2 y = -2^-2 y = 1 / -2^2 y = 1 / 4 For the x value of -1 y = -1^-2 y = 1 / -1^2 y = 1 / 1 y = 1 For the x value of 0 y = 0^-2 y = 1 / 0^2 y = 0 For the x value of 1 y = 1^-2 y = 1 / 1^2 y = 1 For the x value of 2 y = 2^-2 y = 1 / 2^2 y = 1 / 4 For the x value of 3 y = 3^-2 y = 1 / 3^2 y = 1 / 9
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12:59:47 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 was able to get the correct values for the previous equation. However, I did include the o instead of stating it as being an undefined value.
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13:02:40 `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 --> y = x^3 For x value -3 y = -3^3 y = -27 For x value -2 y = -2^3 y = -8 For x value -1 y = -1^3 y = -1 For x value 0 y = 0^3 y = 0 For x value 1 y = 1^3 y = 1 For x value 2 y = 2^3 y = 8 For x value 3 y = 3^3 y = 27
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13:04:13 The y values should be -27, -8, -1, 0, 1, 4, 9.
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RESPONSE --> I do not understand how we got the last 2 numbers ( both 4 and 9). I thought we were cubing the x values which should therefore give us an answer of 8 for the x value of 2 and for the x value of 3 why would the answer not be 27?
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13:13:32 `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 --> The graph that would correspond to the y = x^2 equation would be a parabola which would cross the x and y axis at (0,0) and be symmetrical. The graph for y = 2^x would be the graph of a line increasing at an increasing rate. The graph for y = x^-2 would be a line that begins at (-3, 1/9) and rises to a point at (-1, 1) and then comes down to touch the line at (0,0) then goes back up to a point at (1,1) then comes back down to point with coordinates (3,1/9). The graph for y = x^3 would be a parabola.
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13:15:47 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 --> I made a mistake in stating that the final graph for y = x^3 was a parabola. I should have stated that it was a line that starts at (-3,-27) and rises up fairly quickly to pass through the x axis at (0,0) then rises again on the positive side of the x axis.
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13:19:22 `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 --> For the expression y = x^2 + 3 x y -3 12 -2 7 -1 4 0 3 1 4 2 7 3 12 As compared to values on the table for y = x^2, the values are all increased by a value of 3. There was a shift to the right of 3 spaces.
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13:20:54 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 --> I stated there was a shift to the right of 3, my graph is not of the utmost accurate and therefore it did shift on my graph but I do also note a shift upward on the graph also of 3 points.
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13:25:34 `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 --> For y = (x-1)^3 x y -3 -64 -2 -27 -1 -8 0 -1 1 0 2 1 3 8 Where the value for -3 for the equation y = x^3 here for the equation y = (x-1)^3 the value for -3 is -64. So the -3 value in the last expression would coordinate to the value of -4 in the equation y = x^3. There is a shift in the y = (x-1)^3 equation.
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13:26:31 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 --> I understood this question and was able to come up with the correct answers though I should have been more descriptive in regards to the relationship between the equations and the shift.
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13:31:41 `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 --> y = 3 * 2^x x y -3 3/8 -2 3/4 -1 3/2 0 3 1 6 2 12 3 24 The y values in the above equation are all 3 times the amount in the original equation of y = 2^x. There is a shift in the graph on the y axis.
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13:31:58 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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RESPONSE --> I understand this concept.
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