#$&* course MTH 163 October 13th, 2012 @ 3:47PMSee my comments within the self-inquiries, especially the last regarding comments and suprises in the assignment.
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Given Solution: ** Since y = x the rise and run between any two points on the graph are equal, which makes the slope 1. A graph with constant slope is a straight line. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhy is y = x^2 symmetric about x = 0 (i.e., taking the same values on either side of x = 0) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Because of any negative number being raised to an even exponent, the result will carry the opposite value. In other words, (-3)^2 = 9 instead of -9. I feel like I need to note that, to any other student that sees something wrong with this when using a TI-84 Plus Silver Edition calculator like mine, one needs to put the negative number in parentheses in order for this to work. Otherwise, -3^2 will be -9. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: See my note about using calculators to check negative coefficients with positive, even exponents.
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhy is y = x^3 antisymmetric about x = 0 (i.e., taking the same values except for the - sign on opposite sides of x = 0) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - For the similar reason a quadratic graph occurs, apart from the type of exponent being even or odd. When an odd exponent is tacked onto a negative coefficient, that number increases in a negative fashion just like the positive side of x. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** y = x^3 is antisymmetric because if you cube a negative number you get a negative, if you cube a positive number you get a positive, and the magnitude of the cubed number is the cube of the magnitude of the number. So for example (-3)^2 = -27 and 3^3 = 27; the points (-3, -27) and (3, 37) are antisymmetric, one being `down' while the other is `up'. GOOD STUDENT RESPONSE: y = x^3 is antisymmetric about x = 0 because the exponent is an odd number. This will cause negative x values to have a negative y result. The absolute value of the negative y result will be equivalent to its corresponding positive y value. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhy do y = x^-2 and y = x^-3 rise more and more steeply as x approaches 0, and why do their graphs approach the x axis as we move away from the y axis. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Didn’t quite understand this one; more to follow. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ended up completely lost by this particular problem; could you paraphrase the last stanza? ------------------------------------------------ Self-critique Rating:
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 4. describe the graph of the exponential family y = A * 2^x for the values A = -3 to 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - It seems the variable, ‘A’, constitutes where the y axis will be crossed on either the slope or rise of the graph itself; as the value for ‘A’ increases so will the point where the y axis will be crossed. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. *&*& STUDENT QUESTION: I again am a little confused on this one. I know that vertical shift is being affected and will approach the x axis as the A values approach 0, and will move away from the x axis as the positive A values increase. INSTRUCTOR RESPONSE A vertical shift occurs when the points of one graph are all raised by the same amount. However that is not the case here. The basic points of the y = 2^x function are (-1, 1/2), (0, 1) and (1, 2). Now consider the A = 3 function y = 3 * 2^x. The basic points are (-1, 3/2), (0, 3) and (1, 6). The x = -1 point (-1, 3/2) of y = 3 * 2^x is 1 unit higher than the x = -1 point (-1, 1/2) of the y = 2^x function. The x = 0 point ( 0, 3/2) of y = 3 * 2^x is 2 units higher than the x = 0 point ( 0, 1) of the y = 2^x function. The x = 1 point ( 1, 3/2) of y = 3 * 2^x is 4 units higher than the x = 2 point ( 1, 2) of the y = 2^x function. So this is not a vertical shift. However each point of the y = 3 * 2^x function is 3 times further from the x axis than the corresponding point of the y = 2^x function: (-1,3/2) is 3/2 of a unit from the x axis, 3 times as far as the point (-1, 1/2). (0, 3) is 3 units from the x axis, 3 times as far as the point (0, 1). (1, 6) is 6 units from the x axis, 3 times as far as the point (1, 2). A vertical shift would occur, for example, for the function y = 2^x + 3. The basic points of the y = 2^x + 3 function would be (-1, 7/2), (0, 4) and (1, 6), each point being 3 units higher than the corresponding basic point of the y = 2^x function. As with most descriptions, in order to best understand the explanation you need to sketch it out. In this case you should sketch out these points on a graph and see how they are related. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How would I graph these equations? In particular, y = A* 2 ^ (kx) + c I want to be able to graph them without the use of a calculator to check or give me an idea of what they should look like. ------------------------------------------------ Self-critique Rating:
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
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Given Solution: ** 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. ** STUDENT COMMENT/QUESTION Lost..... Its not clicking for me. I understand its going to change by 1, but I need to see more information. The book is not helping me and I did a search on the Internet and can only come up with a DNA Biology math reference. Where can I look for more explanation? Thank You INSTRUCTOR RESPONSE: This is easier than you think, once you see it. Your are told that p = -3. So y = A ( x - h)^p + c becomes y = A ( x - h)^(-3) + c. Then you're told that A = 1, so y = 1 ( x - h)^(-3) + c, or just y = (x - h)^(-3) + c. Let's skip the h part for a minute and notice that c = 0. So now we have y = (x - h)^(-3) + 0 or just y = (x - h)^(-3). Now to deal with h, which is said to vary from -3 to 3. There are an infinite number of values between -3 and 3 and of course you're not expected to write a separate function for each of them. You could get the general idea using h values -3, 0 and 3. This gives you the functions y = (x - (-3) ) ^ (-3) y = (x - 0) ^ (-3) y = (x - 3) ^ (-3). Simplifying these you have y = (x + 3) ^ (-3) y = x ^ (-3) y = (x - 3) ^ (-3). The graph of x^(-3) has a vertical asymptote at the y axis (note that the y axis is at x = 0). To the right of the asymptote it decreases at a decreasing rate toward a horizontal asymptote with the positive x axis. As x approaches 0 through the negative numbers, the graph decreases at an increasing rate and forms its asymptote with the negative y axis. The graph of y = (x + 3)^(-3) has its vertical axis at x = -3; i.e., it's shifted 3 units to the left of the graph of y = x^(-3). The graph of y = (x - 3)^(-3) has its vertical axis at x = 3; i.e., it's shifted 3 units to the right of the graph of y = x^(-3). So the graphs 'march' across the x-y plane, from left to right, with vertical asymptotes varying from x = -3 to x = +3. We could fill in the h = -2, -1, 1 and 2 graphs, and as many graphs between these as we might wish, but the pattern should be clear from the three graphs discussed here. STUDENT QUESTION For this equation, I had trouble graphing and seeing how the changes were made. I didn’t understand how to substitute these values and get a shift. For example: y = A(x-h)^p + c then substituting the values you have y= 1 (x - (-3))^-3 + 0 I can see there is no vertical stretch on the y axis and then inside the parenthesis you would have (x + 3) raised to a power of -3 and finally there is no shift on the c value. Does this mean the graph would shift left or right according to the given values? But how do you solve (x + 3)^-3
#$&* course MTH 163 October 13th, 2012 @ 3:47PMSee my comments within the self-inquiries, especially the last regarding comments and suprises in the assignment.
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: