course MTH 158 zsƥ͊͠Bassignment #025
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14:47:07 3.5.12 (was 3.4.6). Upward parabola vertex (0, 2). What equation matches this function?
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RESPONSE --> f(x) = x^2 + 2 confidence assessment: 2
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14:50:30 The correct equation is y = x^2 + 2. How can you tell it isn't y = 2 x^2 + 2?. How can you tell it isn't y = (x+2)^2.
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RESPONSE --> It isn't y = 2x^2 + 2 because the arc would be much wider due to the doubling of each x coordinate
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14:52:25 GOOD STUDENT ANSWERS: it isnt y = 2x^2 + 2 because that would make it a verical stretch and it isnt vertical it is just a parabola. it isnt y = (x+2)^2 because that would make it shift to the left two units and it did not. INSTRUCTOR NOTE: Good answers. Here is more detail: The basic y = x^2 parabola has its vertex at (0, 0) and also passes through (-1, 1) and (1, 1). y = 2 x^2 would stretch the basic y = x^2 parabola vertically by factor 2, which would move every point twice as far from the x axis. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). Then y = 2x^2 + 2 would shift this function vertically 2 units, so that the points (0, 0), (-1, 2) and (1, 2) would shift 2 units upward to (0, 2), (-1, 4) and (1, 4). The resulting graph coincides with the given graph at the vertex but because of the vertical stretch it is too steep to match the given graph. The function y = (x + 2)^2 shifts the basic y = x^2 parabola 2 units to the left so that the vertex would move to (-2, 0) and the points (-1, 1) and (1, 1) to (-3, 1) and (-1, 1). The resulting graph does not coincide with the given graph. y = x^2 + 2 would shift the basic y = x^2 parabola vertically 2 units, so that the points (0, 0), (-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1, 3). These points do coincide with points on the given graph, so it is the y = x^2 + 2 function that we are looking for. **
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RESPONSE --> Much wider arc/vertical stretch... very close, right? :) I see that it would make it shift to the left instead of up two units. self critique assessment: 2
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14:53:36 3.5.16 (was 3.4.10). Downward parabola.
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RESPONSE --> y = -2x^2 confidence assessment: 2
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14:56:11 The correct equation is y = -2 x^2. How can you tell it isn't y = 2 x^2?. How can you tell it isn't y = - x^2?
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RESPONSE --> It isn't y = 2x^2 because in that case it would be flipped so the arrows were pointing up I can't tell that it isn't y = -x^2.. for one thing, that wasn't one of the options in my book for the possible functions, but as I look at it I dont see why it couldn't be. self critique assessment: 2
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14:59:50 ** The basic y = x^2 parabola has its vertex at (0, 0) and also passes through (-1, 1) and (1, 1). y = -x^2 would stretch the basic y = x^2 parabola vertically by factor -1, which would move every point to another point which is the same distance from the x axis but on the other side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, -1) and (1, -1). y = -2 x^2 would stretch the basic y = x^2 parabola vertically by factor -2, which would move every point to another point which is twice the distance from the x axis but on the other side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, -2) and (1, -2). This graph coincides with the given graph. y = 2 x^2 would stretch the basic y = x^2 parabola vertically by factor 2, which would move every point twice the distance from the x axis and on the same side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). This graph is on the wrong side of the x axis from the given graph. **
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RESPONSE --> Alright... the difference in -x^2 and -2x^2 in the graph wasn't very obvious at all in the image in the book, and that's why I couldn't really tell. self critique assessment: 2
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15:00:41 3.5.18 (was 3.4.12). V with vertex at origin. What equation matches this function?
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RESPONSE --> y = 2|x| confidence assessment: 3
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15:41:01 The correct equation is y = 2 | x | . How can you tell it isn't y = | x | ?. How can you tell it isn't y = | x | + 2?
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RESPONSE --> If it were y = |x| then the graph would contain points such as (1,1), (2,2), (-1,1), and (-2, 2), and it doesn't. If it were y = |x| + 2, then the graph's origin point would go from (0,0) to (2,0), as well as have a smaller angle due to what I explained in the last one confidence assessment: 2
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15:41:55 ** The basic y = | x | graph is in the shape of a V with a 'vertex' at (0, 0) and also passes through (-1, 1) and (1, 1). The given graph has the point of the V at the origin but passes above (-1, 1) and (1, 1). y = | x | + 2 would shift the basic y = | x | graph vertically 2 units, so that the points (0, 0), (-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1, 3). The point of the V would lie at (0, 2). None of these points coincide with points on the given graph y = 2 | x | would stretch the basic y = | x | graph vertically by factor 2, which would move every point twice the distance from the x axis and on the same side. This would leave the point of the 'V' at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). This graph coincides with the given graph. **
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RESPONSE --> That was my answer, correct? self critique assessment: 2
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15:48:20 3.5.30 (was 3.4.24). Transformations on y = sqrt(x). What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function?
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RESPONSE --> f(x) = sqrt(3 - x) + 2 confidence assessment: 2
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15:49:15 What is the function after you shift the graph up 2 units?
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RESPONSE --> after shifting up 2 more units? Then it would be f(x) = sqrt(3 = x) + 4 confidence assessment: 2
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15:53:09 ERRONEOUS STUDENT RESPONSE: y = x^2 + 2 INSTRUCTOR CORRECTION: y = sqrt(x) is not the same as y = x^2. y = sqrt(x) is the square root of x, not the square of x. Shifting the graph of y = sqrt(x) up so units we would obtain the graph y = sqrt(x) + 2. **
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RESPONSE --> The question in my book was ""find the function that is finally graphed after the following transformations are applied to the graph of y = sqrt(x)."", then the ""transformations"" were 1 - shift up 2 units 2 - reflect about the y-axis 3 - shift left 3 units I take it that was not the one intended for the query.. did I get the answer right according to what I thought it was asking, though? self critique assessment: 1
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16:04:20 What is the function after you then reflect the graph about the y axis?
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RESPONSE --> With all three incorporated (shift up 2 units, reflect about the y axis, shift left 3 units), the answer I got was f(x) = sqrt (3 - x) + 2 confidence assessment: 2
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16:04:51 ** To reflect a graph about the y axis we replace x with -x. It is the y = sqrt(x) + 2 function that is being reflected so the function becomes y = sqrt(-x) + 2. **
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RESPONSE --> I'm still confused, what was it asking?? self critique assessment: 1
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16:05:22 What is the function after you then fhist the graph left 3 units?
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RESPONSE --> f(x) = sqrt(3 - x) + 2 confidence assessment: 2
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16:05:41 ** To shift a graph 3 units to the left we replace x with x + 3. It is the y = sqrt(-x) + 2 function that is being reflected so the function becomes y = sqrt( -(x+3) ) + 2. **
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RESPONSE --> Ok, so I got that right... self critique assessment: 3
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16:18:52 3.5.42 (was 3.4.36). f(x) = (x+2)^3 - 3. What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function? Describe your graph. Give three points on your graph and tell to which basic points on the graph of your basic function each corresponds.
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RESPONSE --> Starting with the basic function f(x) = x^3 Then shift the graph to the left 2 units, then vertically shift down 3 units. This would put the midpoint of the graph at (-2, -3), which corresponds with the original midpoint (0,0). Another point would be (-1, -2) which corresponds with the original point (1,1). The other point would be (-3, -4) which corresponds with the original point (-1, -1) confidence assessment: 2
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16:21:03 ** Starting with y = x^3 we replace x by x - 3 to shift the graph 3 units left. We obtain y = (x + 3)^3. We then shift this graph 2 units vertically by adding 2 to the value of the function, obtaining y = (x + 3 )^3 + 2. The basic points of the y = x^3 graph are (-1, -1), (0, 0) and (1, 1). Shifted 3 units left and 2 units up these points become (-1 - 3, -1 + 2) = (-4, 1), (0 - 3, 0 + 2) = (-3, 2) and (1 - 3, 1 + 2) = (-2, 3). The points you give might not be the same three points but should be obtained by a similar process, and you should specify both the original point and the point to which it is transformed. **
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RESPONSE --> I don't understand why you shift it 3 units left and 2 units up rather than 2 units left and 3 units down. That made sense to me. self critique assessment: 1
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19:26:29 3.5.58 (was 3.4.40). h(x) = 4 / x + 2. What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function? Describe your graph. Give three points on your graph and tell to which basic points on the graph of your basic function each corresponds.
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RESPONSE --> Using the graph of f(x) = 1/x , multiply by 4; vertical stretch of the graph by a factor of 4, then a vertical shift up 2 units. The original point (1,1) corresponds to the new point (1,6), original point (.5, 2) corresponds to (.5, 10), original point (2, .5) corresponds to (2, 4) confidence assessment: 2
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19:28:33 ** We start with the basic reciprocal function y = 1 / x, which has vertical asymptote at the y axis and horizontal asymptotes at the right and left along the x axis and passes through the points (-1, -1) and (1, 1). To get y = 4 / x we must multiply y = 1 / x by 4. Multiplying a function by 4 results in a vertical stretch by factor 4, moving every point 4 times as far from the x axis. This will not affect the location of the vertical or horizontal asymptotes but for example the points (-1, -1) and (1, 1) will be transformed to (-1, -4) and (1, 4). At every point the new graph will at this point be 4 times as far from the x axis. At this point we have the graph of y = 4 / x. The function we wish to obtain is y = 4 / x + 2. Adding 2 in this mannerincreases the y value of each point by 2. The point (-1, -4) will therefore become (-1, -4 + 2) = (-1, -2). The point (1, 4) will similarly be raised to (1, 6). Since every point is raised vertically by 2 units, with no horizontal shift, the y axis will remain an asymptote. As the y = 4 / x graph approaches the x axis, where y = 0, the y = 4 / x + 2 graph will approach the line y = 0 + 2 = 2, so the horizontal asymptotes to the right and left will consist of the line y = 2. Our final graph will have asymptotes at the y axis and at the line y = 2, with basic points (-1, -2) and (1, 6).
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RESPONSE --> I'm sorry I forgot to put points of the symmetrical graph, but I guess my answer was still correct, right? self critique assessment: 2
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19:55:04 3.5.60 (was 3.4.54). f(x) = -4 sqrt(x-1). What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function? Describe your graph. Give three points on your graph and tell to which basic points on the graph of your basic function each corresponds.
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RESPONSE --> Using the graph of f(x) = sqrt(x), move the graph 1 unit to the left, reflect about y axis, then multiply times 4 for a vertical stretch. Original point (0,0) corresponds with new point (1,0), point (1,1) corresponds with new point (0,4) and point (4,2) corresponds with new point (-3, 8) confidence assessment: 2
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19:56:52 ** Starting with the basic function y = sqrt(x) we replace x by x - 1, which shifts the graph right 1 unit, then we stretch the graph by factor -4, which moves every point 4 times further from the y axis and to the opposite side of the y axis. The points (0, 0), (1, 1) and (4, 2) lie on the graph of the original function y = sqrt(x). Shifting each point 1 unit to the right we have the points (1, 0), (2, 1) and (5, 2). Then multiplying each y value by -4 we get the points (1, 0), (2, -4) and (5, -8). Note that each of these points is 4 times further from the x axis than the point from which it came, and on the opposite side of the x axis. **
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RESPONSE --> why is it moved to the opposite side of the x axis and not the y axis? and why does it shift right 1 unit and not left one unit as the book seems to portray in its earlier explanations? self critique assessment: 1
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20:18:25 3.5.66 (was 3.4.60). Piecewise linear (-4, -2) to (-2, -2) to (2, 2) to (4, -2). Describe your graphs of G(x) = f(x+2), H(x) = f(x+1) - 2 and g(x) = f(-x). Give the four points on each of these graphs that correspond to the four points labeled on the original graph.
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RESPONSE --> G(x) = f(x+2), I believe that you shift the whole graph 2 units to the left so that (-4, -2) becomes (-6, -2), (-2, -2) becomes (-4, -2), (2,2) becomes (0,2) and (4, -2) becomes (2, -2) h(x) = f(x+1) - 2, I believe that you shift the whole graph 1 unit to the left, as well as 2 units down vertically, so that (-4, -2) becomes (-5, -4), (-2, -2) becomes (-3, -4), (2,2) becomes (1, 0) and (4,-2) becomes (3, -4) g(x) = f(-x), becomes a mirror image of itself due to the multiplication of all x's by -1... as (-4, -2) becomes (4, -2), (-2, -2) becomes (2,-2), (2,2) becomes (-2, 2) and (4, -2) becomes (-4, -2) confidence assessment: 2
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20:19:08 ** G(x) = f(x+2) shifts the points of the f(x) function 2 units to the left, so instead of going from (-4, -2) to (-2, -2) to (2, 2) to (4, -2) the G(x) function goes from (-4-2, -2) to (-2-2, -2) to (2-2, 2) to (4-2, -2), i.e., from (-6, -2) to (-4, -2) to (0, 2) to (2, -2). H(x) = f(x+2) - 2 shifts the points of the f(x) function 1 unit to the left and 2 units down, so instead of going from (-4, -2) to (-2, -2) to (2, 2) to (4, -2) the H(x) function goes from (-4-1, -2-2) to (-2-1, -2-2) to (2-1, 2-2) to (4-1, -2-2), i.e., from (-5, -4) to (-3, 0) to (1, 0) to (3, -4). g(x) = f(-x) replaces x with -x, which shifts the graph about the y axis. Instead of going from (-4, -2) to (-2, -2) to (2, 2) to (4, -2) the g(x) function goes from (4, -2) to (2, -2) to (-2, 2) to (-4, -2) You should carefully sketch all these graphs so you can see how the transformations affect the graphs. **
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RESPONSE --> OK self critique assessment: 3
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21:18:39 3.5.78 (was 3.4.72). Complete square and graph f(x) = x^2 + 4 x + 2. Give the function in the designated form. Describe your graph this function.
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RESPONSE --> f(x) = x^2 + 4 x + 2 (x^2 + 4x + 4) - 2 (x+2)^2 - 2 f(x) = (x+2)^2 - 2 Using the graph of f(x) = x^2, shift left 2 units then shift vertically down 2 units. Points now include (-2, -2), (-1, -1) and (-3, -1) confidence assessment: 2
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21:19:02 ** To complete the square on f(x) = x^2 + 4x + 2 we first look at x^2 + 4x and note that to complete the square on this expression we must add (4/2)^2 = 4. Going back to our original expression we write f(x) = x^2 + 4x + 2 as f(x) = x^2 + 4x + 4 - 4 + 2 the group and simplify to get f(x) = (x^2 + 4x + 4) - 2. Since the term in parentheses is a perfect square we write this as f(x) = (x+2)^2 - 2. This shifts the graph of the basic y = x^2 function 2 units to the left and 2 units down, so the basic points (-1, 1), (0, 0) and (1, 1) of the y = x^2 graph shift to (-3, -1), (-2, -2) and (-1, -1). **
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RESPONSE --> OK self critique assessment: 3
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