course Mth 163 ɩҩWx״assignment #005 005. `query 5 Precalculus I 02-06-2008
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18:10:14 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 --> It is a straight line because it is a linear equation with points (-3,3) (-2,2) (-1,1) (0,0) (1,1) (2,2) and (3,3). confidence assessment: 2
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18:11:11 ** 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 --> A straight line graph has a slope of 1 between two points and therefore it makes a straight line. self critique assessment: 3
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18:12:06 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 --> The reason y=x^2 is because it is a quadratic formula and it is a parabola and has symmetric. confidence assessment: 2
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18:12:57 ** 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 --> On both sides of a graph the points are equal so the graph is in the form of a parabola and looks like it is reflecting itself. self critique assessment: 3
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18:14:07 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 keeps increasing because it is an exponential function and it is increasing or decreasing and is a horizontal asymptote. confidence assessment: 2
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18:16:42 ** 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 --> If x increases the formula will increase with it. This will then cause y to double from the previous point. As y approaches x it becomes negative because of the exponent. Over time it will never quite reach 0. self critique assessment: 3
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18:38:06 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 --> y=x^3 is antisymmetric because it is a power function and is either odd or even points. confidence assessment: 2
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18:39:30 ** 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 --> If you are to cube a negitive then you get a negitive and if you cube a positive then you get a positive. They are antisymmetric with one being up or and the other being down. self critique assessment: 3
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18:40:45 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 --> y=x^-2 and y=x^-3 and becomes steeper because it is a power function and has either a positive or negative or a horizontal and verticle asymptotes. confidence assessment: 2
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18:42:58 ** 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 --> As it approaches 0 x will either be 1/x^2 or 1/x^3 and the denominators will get smaller and smaller in a fractor. The graph becomes more and more steep and they will either be positive or negative. self critique assessment: 3
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18:44:13 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 family of y-x^2+c has identical parabolas because it has a constant m family which gives the graph the same slope. confidence assessment: 2
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18:46:05 ** 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 --> The graph will vary by 1 unit and it will either be higher or lower than the previous one. The c value will have a vertical shift. So the graph increases by 1 and then raised by 1 unit. self critique assessment: 3
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18:49:46 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 -3 the graph slopes in a negative way starting at the point (-4, -.1875). As it slopes the x axes increases and the y value increases at a negaitive rate. For 3, the graph slpes in a positive way starting at the point (4, .1875). As it slopes the x axes increases and the y value increases at a positive rate. confidence assessment: 2
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18:53:28 ** 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 --> A= -3 to 3 so I substitute these values into y=A*2^x. It is an asymptote and so when the values of -3, -2, -1, 0, 1, 2, and 3 are put into the formula you get a total of 7 graphs. You have to place them on the graph to inderstand them and to be able to see how the formula works. self critique assessment: 3
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18:56:49 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.
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RESPONSE --> With the formula y=2^x+c you need to take the values c= -3 to 3. You place this numbers in the formula with all the possible choices. You have -3, -2, -1, 0, 1, 2, and 3. You place these numbers into the formula tog et 7 different answers and graphs. confidence assessment: 2
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18:57:48 ** 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 --> The seven graphs all include the orginial formula and in thos you place the number you want to use in place of c to get you answer. Some of the graphs will be positive and some will be negaitive. self critique assessment: 3
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19:05:41 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 --> For the first equation of y=1(x-0)^-2-3 the graph starts out negative and then peaks in the positive and then slopes back down to be negaitive. The equatioin y=1(x-0)^-2+3 the whole graph is positive throughout the whole thing. The next equation y=1(x+3)^-3+0 part of the graph is negaitive and the other part is positive. This graph never meets like the other ones does. confidence assessment: 2
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19:07:10 ** 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 --> The h value is the only thing that changes in the problems. Everything else is almost constant. The graphs change from right to left and change one unit at a time. self critique assessment: 3
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19:09:38 query problem 10 illumination. What function did you evaluate to get your results?
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RESPONSE --> The illuminations that I got for the function y=A(x-h)^p+c and then I subsituted the numbers to get y=370(x-0)^-1+0. The ordered pairs were (1,370) (2,185) (3,123.33) and (4,92.5). The whole graph is positive. confidence assessment: 2
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19:13:16 ** I determined the illumination y from a certain florescent bulb at the distances of 1, 2, 3, and 4 units using the generalized power function for p = -1 with A = 370, h = 0 and c = 0. This power function is y = A (x- h)^p + c = 370 (x - 0)^(-1) + 0, or just y = 370 x^-1. **
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RESPONSE --> The illuminations are determined of a florescent buld bu the distances of 1, 2, 3, and 4 units. You subsitute the numbers of p=-1, A=370, h=0, and c=0. so the fomula turns out to be y=370x^-1. self critique assessment: 3
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19:15:20 Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph.
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RESPONSE --> The units are (1,370) (2,185) (3,123.33) and (4,92.5). The graph stars off positive and decreases over time. confidence assessment: 2
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19:16:25 ** Student Solution: For x=1 we obtain y=370(-1-0) ^-1=370 For x=2 we obtain y=370(2-0)^-1=185 For x=3 we obtain y=370(3-0)^-1 =123.3 For x=4we obtain y=370(4-0)^-1=92.5**
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RESPONSE --> For each point you solve them using the formula. For x=1 you get 370. For x=2 you get 185. for x=3 you get 123.3. For x=4 you get 92.5. self critique assessment: 3
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19:17:40 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I was glad to be able to understand more about the direction of a graph and how the family functions help with the formulas. There were some problems that were a little hard and took longer to do but after working them and getting the right answer it was good to know that I was on the right track. self critique assessment: 3
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19:18:50 ** STUDENT COMMENT: I have never worked with graphs in the power family, and very little in the exponential family. I am always amazed at the patterns that a function produces. It helps me understand the equation so much better than a list of numbers. I do feel that I need the data table with the graph to fully understand it. INSTRUCTOR RESPONSE: The data table is certainly helpful, especially when you see the reasons for the number patterns in the formula as well as you do. **
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RESPONSE --> I also have never done graphs in a power family and the practice was good. It is interesting to see the pattern that are created throughout them. The data table is so helpful and when you are working it you cn visualize what it will look like. self critique assessment: 3
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