#$&* course Math 163 Submitted 11-08-10 @ 7:25pm. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** x^-p = 1 / x^p. As x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to 0, so there is no limit to how many times x^p can divide into 1. This results in y = x^p values that approach infinite distance from the x axis. The graph therefore approaches a vertical limit. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qExplain why the function y = (x-h)^-p has a vertical asymptote at x = h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When x = h is the function of y = (x-h)^-p, or y = (h-h) ^-p, then the result is y = (0)^-p y = 1/0^p and as stated earlier, whenever you divide 1 by a smaller number the result is a larger absolute value which results in a vertical asymptote. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** (x-h)^-p = 1 / (x-h)^p. As x gets closer to h, (x-h)^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to h, so there is no limit to how many times (x-h)^p can divide into 1. This results in y = (x-h)^p values that approach infinite distance from the x axis as x approaches h. The graph therefore approaches a vertical limit. This can also be seen as a horizontal shift of the y = x^-p function. Replacing x by x - h shifts the graph h units in the x direction, so the asymptote at x = 0 shifts to x = h. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qExplain why the function y = (x-h)^-p is identical to that of x^-p except for the shift of h units in the x direction. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When you replace x, in the original function of x^-p, by (x-h), you obtain the function y = (x-h)^-p, the division by 0 occurs when x = h, compared to x = 0 as in the original function. When comparing charts of these two functions you can see that a displacement of the y values occur in the positive x direction because of the added “h” value. This added “h” value will shift the graph “h” units to the right. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aSTUDENT ANSWER: You end up with the exact same y values but at the different position of x changed by the h value. INSTRUCTOR COMMENT: Good start. More specifically the x value at which a given y value occurs is shifted h units, so that for example y = x^p is zero when x = 0, but y = (x - h)^p is zero when x = h. To put this as a series of questions (you are welcome to insert answers to these questions, using #$&* before and after each insertion):: Assume that p is positive. For what value of x is x^p equal to zero? For what value of x is (x - 5)^p equal to zero? For what value of x is (x - 1)^p equal to zero? For what value of x is (x - 12)^p equal to zero? For what value of x is (x - h)^p equal to zero? For example, the figure below depicts the p = 3 power functions x^3, (x-1)^3 and (x-5)^3. Assume now that p is negative. For what value of x does the graph of y = x^p have a vertical asymptote? For what value of x does the graph of y = (x-1)^p have a vertical asymptote? For what value of x does the graph of y = (x-5)^p have a vertical asymptote? For what value of x does the graph of y = (x-12)^p have a vertical asymptote? For what value of x does the graph of y = (x-h)^p have a vertical asymptote? For example, the figure below depicts the p = -3 power functions x^-3 and (x-5)^-3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qGive your table (increment .4) showing how the y = x^-3 function can be transformed first into y = (x - .4) ^ -3, then into y = -2 (x - .4) ^ -3, and finally into y = -2 (x - .4) ^ -3 + .6. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x y = x^-3 Y = (x – .4)^-3 Y=-2(x-.4)^-3 Y=-2(x-.4)^-3 + .6 -.8 -1.953 -.5787 1.157 1.757 -.4 -15.625 -1.953 3.906 4.506 0 0 -15.625 31.25 31.85 .4 15.625 0 0 .6 .8 1.953 15.625 -31.25 -30.65 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The table is as follows (note that column headings might not line up correctly with the columns): x y = x^3 y = (x-0.4)^(-3) y = -2 (x-0.4)^(-3) y = -2 (x-0.4)^(-3) + .6 -0.8 -1.953 -0.579 1.16 1.76 -0.4 -15.625 -1.953 3.90 4.50 0 div by 0 -15.625 31.25 32.85 0.4 15.625 div by 0 div by 0 div by 0 0.8 1.953 15.625 -31.25 -30.65 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qExplain how your table demonstrates this transformation and describe the graph that depicts the transformation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The transformation from y = x^-3 to y = (x-.4)^-3 is a horizontal shift of .4 units to the right. It has a vertical asymptote when x = .4. The transformation of y = (x-.4)^-3 to y = -2 (x - .4)^-3 is a vertical stretch of -2, which moves the points 2 units more from the x axis. It has a vertical asymptote when x = .4. The transformation of y = -2(x-.4)^-3 to y = -2 ( x - .4)^-3 + 6 is a vertical shift of .6 units the graph +6 units from the x-axis. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a y = x^-3 transforms into y = (x - .4)^-3, shifting the basic points .4 unit to the right. The vertical asymptote at the y axis (x = 0) shifts to the vertical line x = .4. The x axis is a horizontal asymptote. y = -2 (x - .4)^-3 vertically stretches the graph by factor -2, moving every point twice as far from the x axis and also to the opposite side of the x axis. This leaves the vertical line x = .4 as a vertical asymptote. The x axis remains a horizontal asymptote. y = -2 ( x - .4)^-3 + 6 vertically shifts the graph +6 units. This has the effect of maintaining the shape of the graph but raising the horizontal asymptote to x = 6. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qDescribe your graphs of y = x ^ .5 and y = 3 x^.5. Describe how your graph depicts the ratios of y values between the two functions. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of y = x^.5 has the basic points of (0,0) (1,1) and (2,2.14). The graph increases at a decreasing rate. The graph of y = 3x^.5 has the basic points of (0,0) (1, 3) amd (2, 4.24). This function makes a vertical shift of the original graph 3 times as far from the x axis. Its increasing at a decreasing rate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a*&*& This is a power function y = x^p with p = .5. The basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414). • Attempting to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the conclusion that y = x^.5 is not defined for negative values of x. • The graph therefore begins at the origin and increases at a decreasing rate. • However since we can make x^.5 as large as we wish by making x sufficiently large, there is no horizontal asymptote. y = 3 x^.5 vertically stretches the graph of y = x^.5 by factor 3, giving us basic points (0, 0), (.5, 2.12), (1, 3) and (2, 4.242). • This graph is also increasing at a decreasing rate, staying 3 times as far from the x axis as the graph of the original y = x^.5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qExplain why the graph of A f(x-h) + k is different than the graph of A [ f(x-h) + k ], and describe the difference. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The difference is in the order of operations. For the graph of Af(x-h) + k, you would perform the function in parenthesis first then multiply the result by A and then add the value of k. This means that you would first vertically shift the graph A units, then horizontally shift it h units and vertically k units. For the graph of A [ f(x-h) + k ], you still begin by performing the function inside the parentheses but since the brackets include the k value you would add it next and then multiply by A. So you horizontally shift h units and then vertically shift k units and then do the vertical stretch of A units. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The first graph is obtained from y = f(x) by first vertically stretching by factor A, then horizontall shifting h units and finally vertically shifting k units. The graph of A [f(x-h) + k] is obtained by first doing what is in brackets, horizontally shifting h units then vertically shifting k units before doing the vertical stretch by factor A. Thus the vertical stretch applies to the vertical shift in addition to the values of the function. This results in different y coordinates and a typically a very different graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: Query Add comments on any surprises or insights you experienced as a result of this assignment. This was a good refresher exercise.