course Mth 163 Vxiǖكassignment #003
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20:06:11 query graph of y = x^2 stretched vertically by different factors. When you constructed a graph of y = x^2 and stretched it, describe the graphs you obtained when you stretched it vertically by a factor of 2, by a factor of 3, by a factor of .5 and by a factor of -.3.
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RESPONSE --> The graph is a parabola with the vertex at (0,0). When the graph is stretched it increase to both the right and left. When stretching by 2 the next points would be (1,2) and (-1,2) when stretching by three (1,3) and (-1,3) for .5 (1,.5) and (-1,.2) confidence assessment: 1
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20:06:35 ** STUDENT RESPONSE: The graphs of all of the plots were parabolic with all of the vertices at (0,0). Graphs of the plots swtretched by factors .5, 2 and 3 all had the vertex 'down' with factor 3 graph as the narrowest, then factor 2 graph and finallyfactor .5 as the broadest. Factor -.3 was the broadest of them all and its vertex was up. INSTRUCTOR COMMENT: Good descriptions. You should also mention that the factor 3 graph contained the points (-1, 3) and (1, 3), while the factor 2 graph had points (-1, 2) and (1,2) while the factor .5 graph had (-1, .5) and (1, .5). **
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RESPONSE --> ok self critique assessment: 3
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20:15:24 query prob 3: Location of the vertex and fundamental points. Determine the location of the vertex and fundamental points of each of the following quadratic functions: y = x^2 + 2x + 1 and y = x^2 + 3x + 1
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RESPONSE --> The vertex of y=x^2 +2x+1 would be (-1,0) x=-2/2(1) x=-2/2 x=-1 y=-1^2+2(-1)+1 y=1-2+1 y=0 The vertex of y=x^2+3x+1 would be (-1.5,-1.25) x=-3/2 x=-1.5 y=1(-1.5)^2+3(-1.5)+1 y=2.25-4.5+1 y=-1.25 confidence assessment: 2
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20:16:50 ** The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a). For y = x^2 + 2 x + 1 we have vertex at x = -2 / (2 * 1) = -1. At x = -1 we have y = (-1)^2 + 2 * -1 + 1 = 0. So one fundamental point is the vertex (-1,0). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-2,1) and (0,1). For y = x^2 + 3 x + 1 we have vertex at x = -3 / (2 * 1) = -3/2. At x = -3/2 we have y = (-3/2)^2 + 2 * (-3/2) + 1 = 1/4. So one fundamental point is the vertex (-3/2, -5/4). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-5/2, -1/4) and (-1/2,-1/4); or in decimal form at (-2.5, 1.25) and (-.5, 1.25). **
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RESPONSE --> ok I understand I omitted the answer for the fundamental points. self critique assessment: 2
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20:29:22 how did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem?
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RESPONSE --> It changed from (-.5,.75) to (-1,0) to (-1.5,-3.25) and finally to (-2,-3) X increased in .5 units as the parabola was stretched. confidence assessment: 1
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20:29:40 ** The vertices move downward and to the left, but not along a straight line. In fact the vertices lie along a different parabola of their own. **
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RESPONSE --> ok self critique assessment: 3
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20:30:23 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
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RESPONSE --> It give the vertex which is centered and then the first point to the right and to the lef which allows a fairly accurate sketch of the graph. confidence assessment: 1
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20:31:08 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: They provide the initial sweep & symetry around the vertex which defines the 'shape' and direction and allows you to extrapolate. INSTRUCTOR COMMENT: That's very nicely put. The vertex lies on a vertical line of symmetry and the points to right and left show you which way the parabola opens (upward or downward) and indicate the width of the parabola. **
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RESPONSE --> It also shows which way the parabola opens and how wide it will be. self critique assessment: 2
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20:33:11 query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?
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RESPONSE --> If it crosses the x axis then the parabola will have zeros. If it never crosses the axis then there are no zeros. confidence assessment: 1
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20:34:28 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: Initially, I used the graph of my fundemental points as a visual aid as to whether zero crossings were present INSTRUCTOR COMMENT:There are two ways to answer this question. One is your answer--the vertex and two basic points clearly show you whether the function has zeros. The other way to answer is to note that if the discriminant b^2 - 4 a c, which appears inside the square root, is negative the function cannot have zeros (since you can't take the square root of a negative) and if the discriminant is positive you do have zeros (the quadratic formula will yield one zero if the discriminant is zero and will yield two zeros if the discriminant is positive). **
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RESPONSE --> I understand so if it has a zero then there will be one zero, if there is a negitive discriminant none, and if it is positive then 2 zeros. self critique assessment: 2
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20:36:14 query #4. Questions about vertex between zeros and the shape of the curve connecting vertices: What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist?
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RESPONSE --> Looking at the graphs the vertex lies halfway between the zeros. The points to the right and left of the vertex are one unit up and to the right or lef of the vertex/ confidence assessment: 1
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20:36:23 ** The graph of each function is completely symmetric about a vertical line through the vertex. Thus if one of the zeros lies to the right of the vertex, the other must lie an equal distance to the left of the vertex. **
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RESPONSE --> self critique assessment: 3
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20:37:34 What was the shape of the curve connecting the vertices?
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RESPONSE --> It is in the shape of another parabola that would open to the right hand side. confidence assessment: 1
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20:37:54 ** In fact the vertices of y = x^2 + b x + 1 for various values of b (e.g., b = 2 and b = 3 for the parabolas of the preceding problem) themselves lie on a parabola. We're a ways from actually proving that, but it's an interesting idea. Don't actually do it but if you took any three of these vertices and fit them to the form y = a t^2 + b t + c you would get an equation for y vs. t, and all the other vertices would satisfy the equation. **
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RESPONSE --> ok self critique assessment: 3
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20:39:00 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I saw that as the value of x increases for the equation then there is another parabola formed from the vertices. self critique assessment: 3
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ަ^ǗԜpyÄF assignment #004 004. `query 4 Precalculus I 02-10-2008 ۈ®pѓIß assignment #004 004. `query 4 Precalculus I 02-10-2008
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23:27:59 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> confidence assessment:
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23:30:43 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> f(-2)=(-2)^3 f(-a)=(-a)^3 f(x-4)=(x-4)^3 f(-4)=(.4)^3 confidence assessment: 2
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23:31:05 ** 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 --> ok self critique assessment: 3
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23:31:24 ** 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 --> self critique assessment:
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23:31:27 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> confidence assessment:
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23:34:44 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> f(2)=2^(2) f(-a)=2^(-a) f(x+3)=2^(x+3) f(x)+3=2^x+3 confidence assessment: 2
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23:35:00 ** 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 --> ok self critique assessment: 3
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23:35:19 ** 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 --> self critique assessment:
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23:37:22 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> Some of the advantages of using meaningful names for functions is that it is quicker when you are working with applying the function to a real world situation. The name of the value is already there. For example if you are calulating time then it is already labeled. confidence assessment: 1
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23:37:42 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... '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 --> ok self critique assessment: 3
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23:37:51 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> confidence assessment:
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23:37:54 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... '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 --> self critique assessment:
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23:37:57 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> confidence assessment:
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23:48:00 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> 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)/3] confidence assessment: 1
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23:51:08 ** 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 --> I got all of those correct except the last one and I do not understand where the the denominator came from self critique assessment: 1
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23:51:20 ** 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 --> self critique assessment:
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23:51:23 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> confidence assessment:
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23:57:39 ** 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 --> self critique assessment:
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23:59:06 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> 50/illumination(2distance)? I could not figure out what I was supposed to do here. confidence assessment: 0
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00:05:46 ** 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 --> Again on this I did not know how to determine what the denominator would be. To invert it I guess you just switch the numerator and denominator on the bottom of the demoninator? self critique assessment: 1
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00:06:05 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 --> confidence assessment:
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00:09:57 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 --> I used the coordinated pairs (2,80) (5,40) and (10,25) I drew the graph using grid paper and assigned each block to be approx 10 units. I graphed each point and they formed a curved line through the points. confidence assessment: 3
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00:10:10 ** 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 --> self critique assessment: 3
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00:11:27 ** 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 --> self critique assessment:
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00:11:31 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> confidence assessment:
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00:13:31 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> Since f(2)=80 looking at the graph then if f(x)=60 x would be about 3 since the line is curving downward. confidence assessment: 2
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00:13:49 **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 --> ok self critique assessment: 3
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00:16:01 what is your estimate of the value f(7)?
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RESPONSE --> since f(10)=2 since it is decreasing if f(7) would be approx 27.5since the point would be slightly higher than25 confidence assessment: 2
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00:16:51 ** 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 see where 34 would have been a better answer. I was not looking at the rates of change between the points. self critique assessment: 2
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00:18:18 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) was approx 6 confidence assessment: 2
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00:18:33 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> ok self critique assessment: 3
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00:21:44 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 used an estimation of approx11 and 2.2 would be be a difference of about 8.8 confidence assessment: 2
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00:22:04 ** 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 --> ok self critique assessment: 3
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00:22:28 **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 --> self critique assessment:
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00:22:33 what is your estimate of the value f(7)?
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RESPONSE --> confidence assessment:
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00:22:40 ** 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 --> self critique assessment:
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00:22:44 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> confidence assessment:
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00:22:47 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> self critique assessment:
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00:22:52 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 --> confidence assessment:
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00:23:16 ** 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 --> self critique assessment:
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00:23:18 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> confidence assessment:
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㽵zzaR assignment #004 004. `query 4 Precalculus I 02-12-2008
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23:13:39 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> confidence assessment: 0
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{ƝC|ݾw assignment #004 004. `query 4 Precalculus I 02-12-2008
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23:18:01 ** 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 was not sure what the question was asking. I understand the order of the expression is very important when determing change. To get the average of two numbers they are added togeter and then divided by 2. self critique assessment: 1
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23:26:33 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 --> The equation T(t)=150 would have to be solved first and then T(t)=80 and T(t)=30 then we could subtract to get the length of time. confidence assessment: 2
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23:27:00 ** 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 self critique assessment: 3
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23:30:29 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 --> You could solve f(x)=34 to get x1 and then solve f(x)=47 to get x2 and then to get the depth between the two subrtact x1 from x2 or x2 from x1 depending on which one was the larger number. confidence assessment: 1
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23:30:46 ** 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 self critique assessment: 3
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23:43:26 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> If f(x)=23 then f(x)=.00226(23)^2-4.3(23)+92 =1.19554-98.9+92 =-5.70446 If f(x)=34 then f(x)=.00226(34)^2-4.3(34)+92 =2.61256-146.2+92 =-51.5874 f(34)-f(23)=-45.88298 The depth changed by -45.88298 confidence assessment: 2
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23:44:00 ** 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 --> The math looks similar but I used the model from the notes y=.00226t^2-4.3t+92 self critique assessment: 3
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23:49:42 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 --> 34-23=11secs this would be the change in time or t since we determined that the change in depth was 45.88298 we can divide the change in time by the change in depth to get the answer 11/45.88298=.23974 sec per cm confidence assessment: 1
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23:49:58 ** 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 --> It looks pretty close. self critique assessment: 3
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23:51:55 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 --> This would be the opposite change in depth divided by change in secs. 45.88298/11=4.17118 cm per sec confidence assessment: 2
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23:52:15 ** 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 self critique assessment: 3
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00:00:56 query. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph.
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RESPONSE --> I sketched all the given data points on a graph paper assigning each line 10, I drew a smooth curve through the middle of data points without touching any of them but coming as close as possible as indicated. confidence assessment: 3
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00:01:41 ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). **
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RESPONSE --> I choose the points (60,40) (32,60) and (15,85) self critique assessment: 2
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00:05:44 What 3 data point did you use as a basis for your model?
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RESPONSE --> confidence assessment:
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00:14:08 ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.**
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RESPONSE --> self critique assessment: 0
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00:14:15 How close is your model to the curve you sketched earlier?
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RESPONSE --> confidence assessment: 0
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00:14:21 ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.**
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RESPONSE --> self critique assessment: 0
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00:14:27 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 0
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00:14:32 ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation **
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RESPONSE --> self critique assessment: 0
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