course mth 163 }ل֞۫qassignment #003
......!!!!!!!!...................................
20:39:02 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.
......!!!!!!!!...................................
RESPONSE --> The graph stretched by a factor of 2 was two times taller than the original y = x^2 graph, and the graph stretched by 3 was even taller than that one. The one stretched by .5 was half as tall as the original one, and the -.3 one opened down and was about a third as tall as the original.
.................................................
......!!!!!!!!...................................
20:41:38 ** 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). **
......!!!!!!!!...................................
RESPONSE --> Okay, yes, I guess those points sort of define the characteristic of the graphs. Also, I neglected the most obvious part...that the graphs were all parabolas.
.................................................
......!!!!!!!!...................................
20:49:31 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
......!!!!!!!!...................................
RESPONSE --> y = x^2 + 2x + 1 the x coordinate of the vertex = -b/2a, so it would be -2/2 = -1. to get the y, I plug that back into the equation: y = (1)^2 + 2(1) + 1 = 1 - 2 + 1 = 0 so the vertex is (-1, 0) the fundamental points differ vertically from the vertex by a, which is 1, when the x is moved left or right one unit. This makes the fundamental points for this graph (-2, 1) and (0, 1) for y = x^2 + 3x + 1 the x coordinate of the vertex would be -3/2 then i plug that back into the equation: y = (-3/2)^2 + 3(-3/2) + 1 = (9/4) - (36/4) + (4/4) = -23/4 the fundamental points differ by a, which is 1 here as well, so they are (-1/2, -19/4) and (-5/2, -19/4)
.................................................
......!!!!!!!!...................................
20:57:18 ** 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). **
......!!!!!!!!...................................
RESPONSE --> Oh, on the second one, I messed up when I put -3/2 back into the equation. It should have been 9/4 - 18/4 + 4/4, which does give me -5/4. Then when I find the fundamental points, that makes them at (-5/2, -1/4) and (-1/2, -1/4).
.................................................
......!!!!!!!!...................................
21:00:29 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?
......!!!!!!!!...................................
RESPONSE --> The location of each one was progressively farther down and to the left
.................................................
......!!!!!!!!...................................
21:00:45 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Yes, I noticed that they seemed parabolic themselves, too.
.................................................
......!!!!!!!!...................................
21:01:33 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
......!!!!!!!!...................................
RESPONSE --> They show you how close to or far from the vertex those two points are, which gives you an idea of how vertically stretched or compressed the graph is.
.................................................
......!!!!!!!!...................................
21:02:08 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Also shows up and down, yes.
.................................................
......!!!!!!!!...................................
21:06:54 query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?
......!!!!!!!!...................................
RESPONSE --> Whether or not the quadratic formula actually gave you one or two answers determined if there were zeroes for a function. If what was under the square root symbol ended up being negative, then that told you that there were no zeroes, because there's no real number that can be solved for.
.................................................
......!!!!!!!!...................................
21:07:24 ** 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). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:13:20 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?
......!!!!!!!!...................................
RESPONSE --> The two fundamental points on either side of the vertex are equidistant from the vertex, and the two sides of the parabola mirror one another in shape as they open...for each y value that falls on the graph, there are two x values at equal distances from the vertex. So the zeroes should be no different.
.................................................
......!!!!!!!!...................................
21:13:30 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:14:22 What was the shape of the curve connecting the vertices?
......!!!!!!!!...................................
RESPONSE --> The shape of the curve connecting them seems to be a parabola.
.................................................
......!!!!!!!!...................................
21:15:29 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:21:31 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE -->
.................................................
䜤~y assignment #004 Ťexzw Precalculus I 09-17-2006
......!!!!!!!!...................................
23:25:14 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
......!!!!!!!!...................................
RESPONSE --> f(-2) = (-2)^3 = -8 f(-a) = (-a)^3 f(x-4) = (x-4)^3 = (x-4)(x-4)(x-4) = (x^2 - 8x + 16)(x-4) = x^3 - 4x^2 - 8x^2 + 32x + 16x - 64 = x^3 - 12x^2 + 48x -64 f(x) - 4 = x^3 - 4
.................................................
......!!!!!!!!...................................
23:27:14 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:29:00 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
......!!!!!!!!...................................
RESPONSE --> f(2) = 2^2 = 4 f(-a) = 2^(-a) f(x+3) = 2^(x+3) f(x) + 3 = 2^x + 3
.................................................
......!!!!!!!!...................................
23:29:13 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:30:55 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
......!!!!!!!!...................................
RESPONSE --> Using meaningful names helps you keep track of what it is you're actually finding. Using a one-letter name simplifies things when you're writing them out, but it gets hard to keep track of what exactly x and y and k and t and whatever else actually stand for, sometimes, if there are many of them.
.................................................
......!!!!!!!!...................................
23:31:34 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:32:01 '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.**
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:36:32 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
......!!!!!!!!...................................
RESPONSE --> value(0) = 1000(1.07)^0 value(0) = 1000(1) = 1000 value(2) = 1000(1.07^2) value(2) = 1144.9 value(t+3) = 1000*1.07^(t+3) value(t+3)/value(t) = [1000*1.07^(t+3)] / [1000*1.07^t) = [1.07^(t+3)] / 1.07^t
.................................................
......!!!!!!!!...................................
23:40:39 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Okay, I didn't simplify the last one enough. I forgot you could make 1.07^(t+3) into (1.07^t)(1.07^3).
.................................................
......!!!!!!!!...................................
23:44:00 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
......!!!!!!!!...................................
RESPONSE --> [50/distance^2] / [50/(2*distance^2) (50/distance^2) * [(2*distance^2)/50] the 50s cancel, which leaves (2*distance^2) / distance^2 so 4*distance^2/distance^2 = 4
.................................................
......!!!!!!!!...................................
23:46:12 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:52:24 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.
......!!!!!!!!...................................
RESPONSE --> since y = f(x), then for f(2) = 80, you know that 80 must be the y coordinate and 2 is the x coordinate. Similarly, for f(5) = 40, 40 is the y coordinate and 5 is the x, and so on. Then you simply plot the points (2, 80), (5, 40), and (10, 25) and draw a curve connecting them.
.................................................
......!!!!!!!!...................................
23:53:49 ** 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. **
......!!!!!!!!...................................
RESPONSE --> More or less same, except I marked them off in fives.
.................................................
......!!!!!!!!...................................
23:55:15 what is your estimate of value of x for which f(x) = 60?
......!!!!!!!!...................................
RESPONSE --> my estimate for x when f(x) = 60 is 4.
.................................................
......!!!!!!!!...................................
23:55:41 **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**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:55:55 what is your estimate of the value f(7)?
......!!!!!!!!...................................
RESPONSE --> I estimated it to be around 32.
.................................................
......!!!!!!!!...................................
23:56:30 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:57:29 what is your estimate of the difference between f(7) and f(9)?
......!!!!!!!!...................................
RESPONSE --> f(7) is about 32, and f(9) is about 26, so it'd be about 6.
.................................................
......!!!!!!!!...................................
23:57:39 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
00:00:04 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
......!!!!!!!!...................................
RESPONSE --> f(x) = 70 is at about 2.5 and f(x) = 30 is at about 7, so the difference would be about 4.5.
.................................................
......!!!!!!!!...................................
00:01:27 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
00:04:06 ** 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 **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:04:30 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?
......!!!!!!!!...................................
RESPONSE --> we would solve T(t) = 150. To find the length of time, we'd find T(80) and T(30), and then subtract them.
.................................................
......!!!!!!!!...................................
00:07:37 ** 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)] **
......!!!!!!!!...................................
RESPONSE --> Yes, subtract the smaller from the larger...I don't think I said that.
.................................................
......!!!!!!!!...................................
00:15:48 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
......!!!!!!!!...................................
놖ޫeLj٢Ch assignment #004 Ťexzw Precalculus I 09-18-2006 xאωմ|[܊ assignment #004 Ťexzw Precalculus I 09-18-2006
......!!!!!!!!...................................
04:01:27 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
......!!!!!!!!...................................
RESPONSE --> Solving 34 = f(t) gives you one time, and solving 47 = f(t) gives you a second time. So to find the length of time, I guess you'd subtract the second time from the first.
.................................................
......!!!!!!!!...................................
04:03:30 ** 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 ... ' **
......!!!!!!!!...................................
RESPONSE --> Okay, it's an absolute value of the difference between the two times.
.................................................
......!!!!!!!!...................................
04:09:26 By how much did the depth change between t = 23 seconds and t = 34 seconds?
......!!!!!!!!...................................
RESPONSE --> f(23) = .00226(23^2) - .907(23) + 89.7 = 70.03454 f(34) = .00226(34&2) - .907(34) + 89.7 = 61.47456 so the difference between them would be -8.5599.
.................................................
......!!!!!!!!...................................
04:09:40 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
04:12:21 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?
......!!!!!!!!...................................
RESPONSE --> There's a difference of 11 seconds between them, and a difference of -8.56 cm, so to get seconds per centimeter, I'd divide 11/-8.56 to get -1.29 sec/cm
.................................................
......!!!!!!!!...................................
04:17:16 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
04:17:55 On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds?
......!!!!!!!!...................................
RESPONSE --> To get centimeters per second, I'd divide the diffrence in depth (-8.56 cm)by the difference in time (11 seconds), which gives me -0.778 cm/sec.
.................................................
......!!!!!!!!...................................
04:23:04 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
04:28:33 query. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph.
......!!!!!!!!...................................
RESPONSE --> I used each square on the graph paper to represent 5 seconds on the x axis and 5 centimeters on the y. Then I plotted the points and drew the line as directed, through the middle of the points but not touching them.
.................................................
......!!!!!!!!...................................
04:28:53 ** 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). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
04:30:53 What 3 data point did you use as a basis for your model?
......!!!!!!!!...................................
RESPONSE --> I picked from my graph the points (15, 80) (35, 57) and (65, 39) for the model.
.................................................
......!!!!!!!!...................................
04:36:08 ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
04:46:36 What was your function model?
......!!!!!!!!...................................
RESPONSE --> My function model was f(t) = .01226a^2 - 1.8333b + 104.651.
.................................................
......!!!!!!!!...................................
04:46:52 ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089x^2 - 1.4992x + 98.8544. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
I͌Mxٞ assignment #004 Ťexzw Precalculus I 09-19-2006
......!!!!!!!!...................................
21:24:05 What is the average deviation for your model?
......!!!!!!!!...................................
RESPONSE --> My average deviation was 4.3875
.................................................
......!!!!!!!!...................................
21:24:31 ** 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.**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:26:26 How close is your model to the curve you sketched earlier?
......!!!!!!!!...................................
RESPONSE --> The graph of my model is pretty close to the curve I sketched in most places. Not perfect, but close.
.................................................
......!!!!!!!!...................................
21:26:35 ** 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.**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:27:28 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I was surprised how well the model I figured out matched the curve.
.................................................
......!!!!!!!!...................................
21:27:42 ** 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 **
......!!!!!!!!...................................
RESPONSE -->
.................................................
"