#$&* course mth 163 7/11 2pm `query 22*********************************************
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Given Solution: `a** x^-p = 1 / x^p. As x gets closer to 0, x^p gets closer to 0. Dividing 1 by a number which gets closer and closer to 0 gives us a result with larger and larger magnitude. There is no limit to how close x can get to 0, so there is no limit to how many times x^p can divide into 1. This results in y = x^p values that approach infinite distance from the x axis. The graph therefore approaches a vertical limit. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why the function y = (x-h)^-p has a vertical asymptote at x = h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because y = (x-h) ^-p equals y = 1/ (x-h) ^p which means dividing 1 by whatever values get us closer and closer to the x axis. There is also a shift at work here shifting the function in the direction of x. So x = 0 is the same as x = h. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Because y = (x-h) ^-p is the same as x^-p if we ignore the h shift. If we don’t ignore the h shift the x value is changed by the value of h but the value of y still doesn’t change. 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 = -1.2, -.8, -.4, 0, .4, .8, 1.2 Y= x^-3: y= -.579, -1.953, -15.625, 0, 15.625, 1.953, .579 Y= (x-.4) ^-3: y= -.244, -.579, -1.953, -15.625, 0, 15.625, 1.953 Y= -2(x-4) ^-3: y= .488, 1.157, .023, 31.25, 0, -31.25, -3.906 Y= -2(x-4) ^-3+.6: y= 1.088, 1.757, 4.506, 31.85, 0, -30.65, -3.306 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The table is as follows (note that column headings might not line up correctly with the columns): x y = x^3 y = (x-0.4)^(-3) y = -2 (x-0.4)^(-3) y = -2 (x-0.4)^(-3) + .6 -0.8 -1.953 -0.579 1.16 1.76 -0.4 -15.625 -1.953 3.90 4.50 0 div by 0 -15.625 31.25 32.85 0.4 15.625 div by 0 div by 0 div by 0 0.8 1.953 15.625 -31.25 -30.65 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain how your table demonstrates this transformation and describe the graph that depicts the transformation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We start with y = x^-3 then we are going to shift it to the right by .4 with the function y = (x-.4) ^-3. Then we are going to stretch y = (x-.4) ^-3 by a factor of -2 using the function y = -2(x-.4) ^-3 which moves it 2 times as far from the x axis. Then we are going to shift the function vertically .6 by using the function y = -2(x-.4) ^-3+.6 confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: Both graphs are in the upper right section of the graph The basic points for x^.5 are (0, 0), (.5, .7), (1, 1), and (2, 1.4142) The basic points for 3x^.5 are (0, 0), (.5, 2.1), (1, 3), and (2, 4.2426) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a*&*& This is a power function y = x^p with p = .5. The basic points of y = x^.5 are (0, 0), (.5, .707), (1, 1), (2, 1.414). • Attempting to find a basic point at x = -1 we find that -1^-.5 is not a real number, leading us to the conclusion that y = x^.5 is not defined for negative values of x. • The graph therefore begins at the origin and increases at a decreasing rate. • However since we can make x^.5 as large as we wish by making x sufficiently large, there is no horizontal asymptote. y = 3 x^.5 vertically stretches the graph of y = x^.5 by factor 3, giving us basic points (0, 0), (.5, 2.12), (1, 3) and (2, 4.242). • This graph is also increasing at a decreasing rate, staying 3 times as far from the x axis as the graph of the original y = x^.5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why the graph of A f(x-h) + k is different than the graph of A [ f(x-h) + k ], and describe the difference. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The reason is A f(x-h) +k the A is only multiplying the x-h then add k. In the other we have x-h then add k then multiply that answer by A. These will be different graphs. confidence rating #$&*: 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: