assignment #025 025. `query 25 College Algebra 08-31-2008
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18:27:45 query 3.5.12 (was 3.4.6). Upward parabola vertex (0, 2). What equation matches this function?
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RESPONSE --> a, y = x^2 + 2
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18:27:52 query 3.5.12 (was 3.4.6). Upward parabola vertex (0, 2). What equation matches this function?
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RESPONSE -->
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18:44:54 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 --> y = x^2 +2 has the points (1,3) and ( 2,6) and a y intercept of (0,2); y= 2x^2 +2 has the same y intercept, but contains the points (1, 11) and (2, 11). The graph of y = x +2 would be a parabala shifted 2 units right giving it an intercept of (-2, 0)
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18:47:22 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 --> ok
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18:49:03 query 3.5.16 (was 3.4.10). Downward parabola.
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RESPONSE --> y = -2x^2
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18:55:11 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 --> y = 2x^2 would have all y values being positive values resulting in a graph above the x axis. If a foacor of -2 existed, the y values would become negative since -x in this case would require the graph to be reflected about the y axis.
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19:00:40 ** 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 --> saved as notes
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19:01:56 query 3.5.18 (was 3.4.12). V with vertex at origin. What equation matches this function?
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RESPONSE --> y = 2 * absolute value of x
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19:13:23 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 --> y = l x l has the point (1,1 ) ; y =2 l xl has a vertical stretch of 2 which would give it the same x, but a y value multiplied by a factor of 2: (1,2).The intercepts of both these graphs is (0,0) and (0,0). Y = lxl +2 would start at ( 0,2) since all y values are shifted up 2 units. The fraph would move closer to the y axis.
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19:13:46 ** 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 --> ok
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19:17:29 query 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 --> y = sqrtx y = sqrt x +2 y = sqrt -x +2 y = (sqrt -x+ 2) +3
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19:18:46 What is the function after you shift the graph up 2 units?
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RESPONSE --> y = 2 sqrt x
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19:22:48 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) + 2 up so units we would obtain the graph y = sqrt(x) + 2. **
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RESPONSE --> This was the response i gave to begin with. I changed the response here to y = 2sqrt(x) because I thought I must haved missed the answer when I read the following question, I have y = sqrt (x) +2 on my homework paper.
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19:23:27 What is the function after you then reflect the graph about the y axis?
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RESPONSE --> y = sqrt(-x) +2
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19:23:36 ** 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 --> ok
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19:24:17 What is the function after you then fhist the graph left 3 units?
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RESPONSE --> y = sqrt (-x) +2 +3
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19:25:10 ** 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 --> noted
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19:27:30 query 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 --> f(x) = x^3 I shifted the graph 2 units left and then down by a factor of -3
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19:30:37 ** 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 --> noted on homework page
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20:08:35 query 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 --> y = 1/x 1. vertical stretch of 4 2.move 2 units left (.4, 2.35 ), (1,1), (2, .47) are on the basic graph. My final graph has the corresponding points of (.4, 11), (1,6), and (2, 4)
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20:09:23 ** 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 --> saved on my notes
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20:17:06 query 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 --> 1.vertical stretch 4 2. move 1 unit right my graph: (1,0), (2,4), 5,8) basic graph (0,). (1,1) (4,2)
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20:21:30 ** 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 x axis and to the opposite side of the x 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 --> I did the shifts in the correct order orginially, but changed my response here and did the stretch first. I am confused about which one to do first. I also miscopied the -4 as 4, which obviously gives no chance of a correct solution.
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20:41:02 query 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) moves the points 2 units left. ( -6,2), (-4,2), (-2,0), (2,2) H(x) = f(x +1) -2moves the graph 1 unit left and then down 2 units ( -5, -4), (1,0), (-1, -2), (3, 0) g(x) = f( -x) : reflect about y axis ( 4,2) , ( -2,2), (0,0), ( -4,2)
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20:45:30 ** 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 --> I don't understand. I will ask my tutor on Tuesday
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23:20:21 query 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 + 4x +2 = ( x^2 +4x + 4) +2 -4 = ( x+2)^2 -2 start with y = x^2: shift left 2 units ; then down 2 units. The graph of x^ is a parabala with the points ((2,4) , (1,1) , (0,0) (-1,1), (-2, 4). My graph has corresponding points of (0,2) , (-1,1), (-2,-2),(-3, -1), (-4,2)
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23:22:03 ** 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 --> I think my answer is correct?
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