#$&* course Mth 163 4/26 6:00 22 query*********************************************
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Given Solution:`a** x^-p = 1 / x^p. As x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to 0, so there is no limit to how many times x^p can divide into 1. This results in y = x^p values that approach infinite distance from the x axis. The graph therefore approaches a vertical limit. ** ********************************************* Question: `qExplain why the function y = (x-h)^-p has a vertical asymptote at x = h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x-h)^-p = 1 / (x-h)^p. As x gets closer to h, (x-h)^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to h, so there is no limit to how many times (x-h)^p can divide into 1. This results in y = (x-h)^p values that approach infinite distance from the x axis as x approaches h. The graph therefore approaches a vertical limit. This can also be seen as a horizontal shift of the y = x^-p function. Replacing x by x - h shifts the graph h units in the x direction, so the asymptote at x = 0 shifts to x = h confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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. ** ********************************************* 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: You end up with the exact same y values but at the different position of x changed by the h value 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. ********************************************* 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-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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The table is as follows (note that column headings might not line up correctly with the columns): x y = x^3 y = (x-0.4)^(-3) y = -2 (x-0.4)^(-3) y = -2 (x-0.4)^(-3) + .6 -0.8 -1.953 -0.579 1.16 1.76 -0.4 -15.625 -1.953 3.90 4.50 0 div by 0 -15.625 31.25 32.85 0.4 15.625 div by 0 div by 0 div by 0 0.8 1.953 15.625 -31.25 -30.65 ********************************************* Question: `qExplain how your table demonstrates this transformation and describe the graph that depicts the transformation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When transferred it shifts 4 points to the right. The axis is then horizontal to the asymptote, it vertically covers by -2 moving twice as far from the x axis and the opposite confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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. ********************************************* 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: basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414). Trying to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the assumption that y = x^.5 is not defined for negative values of x. The graph thus begins at the origin and increases at a decreasing rate. Yet since we can make x^.5 as large as we wish by making x suitably 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a*&*& This is a power function y = x^p with p = .5. The basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414). Attempting to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the conclusion that y = x^.5 is not defined for negative values of x. The graph therefore begins at the origin and increases at a decreasing rate. However since we can make x^.5 as large as we wish by making x sufficiently large, there is no horizontal asymptote. y = 3 x^.5 vertically stretches the graph of y = x^.5 by factor 3, giving us basic points (0, 0), (.5, 2.12), (1, 3) and (2, 4.242). This graph is also increasing at a decreasing rate, staying 3 times as far from the x axis as the graph of the original y = x^.5. ********************************************* Question: `qExplain why the graph of A f(x-h) + k is different than the graph of A [ f(x-h) + k ], and describe the difference. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first graph is found from y = f(x) by first vertically elongating by factor A, then horizontally shifting h units and finally vertically shifting k units. The graph of A [f(x-h) + k] is gained by first doing what is in brackets, horizontally shifting h units then vertically shifting k units before doing the vertical stretch by factor A. Therefore the vertical stretch applies to the vertical shift in addition to the values of the function. This results in unlike y coordinates and typically a very different graph. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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:
#$&* course Mth 163 4/26 6:00 22 query*********************************************
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Given Solution:`a** x^-p = 1 / x^p. As x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to 0, so there is no limit to how many times x^p can divide into 1. This results in y = x^p values that approach infinite distance from the x axis. The graph therefore approaches a vertical limit. ** ********************************************* Question: `qExplain why the function y = (x-h)^-p has a vertical asymptote at x = h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x-h)^-p = 1 / (x-h)^p. As x gets closer to h, (x-h)^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to h, so there is no limit to how many times (x-h)^p can divide into 1. This results in y = (x-h)^p values that approach infinite distance from the x axis as x approaches h. The graph therefore approaches a vertical limit. This can also be seen as a horizontal shift of the y = x^-p function. Replacing x by x - h shifts the graph h units in the x direction, so the asymptote at x = 0 shifts to x = h confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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. ** ********************************************* 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: You end up with the exact same y values but at the different position of x changed by the h value 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. ********************************************* 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-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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The table is as follows (note that column headings might not line up correctly with the columns): x y = x^3 y = (x-0.4)^(-3) y = -2 (x-0.4)^(-3) y = -2 (x-0.4)^(-3) + .6 -0.8 -1.953 -0.579 1.16 1.76 -0.4 -15.625 -1.953 3.90 4.50 0 div by 0 -15.625 31.25 32.85 0.4 15.625 div by 0 div by 0 div by 0 0.8 1.953 15.625 -31.25 -30.65 ********************************************* Question: `qExplain how your table demonstrates this transformation and describe the graph that depicts the transformation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When transferred it shifts 4 points to the right. The axis is then horizontal to the asymptote, it vertically covers by -2 moving twice as far from the x axis and the opposite confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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. ********************************************* 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: basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414). Trying to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the assumption that y = x^.5 is not defined for negative values of x. The graph thus begins at the origin and increases at a decreasing rate. Yet since we can make x^.5 as large as we wish by making x suitably 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a*&*& This is a power function y = x^p with p = .5. The basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414). Attempting to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the conclusion that y = x^.5 is not defined for negative values of x. The graph therefore begins at the origin and increases at a decreasing rate. However since we can make x^.5 as large as we wish by making x sufficiently large, there is no horizontal asymptote. y = 3 x^.5 vertically stretches the graph of y = x^.5 by factor 3, giving us basic points (0, 0), (.5, 2.12), (1, 3) and (2, 4.242). This graph is also increasing at a decreasing rate, staying 3 times as far from the x axis as the graph of the original y = x^.5. ********************************************* Question: `qExplain why the graph of A f(x-h) + k is different than the graph of A [ f(x-h) + k ], and describe the difference. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first graph is found from y = f(x) by first vertically elongating by factor A, then horizontally shifting h units and finally vertically shifting k units. The graph of A [f(x-h) + k] is gained by first doing what is in brackets, horizontally shifting h units then vertically shifting k units before doing the vertical stretch by factor A. Therefore the vertical stretch applies to the vertical shift in addition to the values of the function. This results in unlike y coordinates and typically a very different graph. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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: #*&!