course Mth 158
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17:11:52 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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17:16:11 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 --> I got it correct. It isn't y = 2x^2+2 because its points are (0,2) and y = (x+2)^2 points are (-2.1,.02)
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17:16:12 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 -->
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17:16:35 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 --> correct.
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17:20:21 query 3.5.16 (was 3.4.10). Downward parabola.
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RESPONSE --> y = -2x^2
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17:22:27 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 cannot be it because it is an upward parabola. y= -x^2 is indeed a downward parabola but its points are different an it is wider.
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17:22:46 ** 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 --> correct
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17:24:40 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|x|
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17:26:28 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=|x| is not correct because the points are not the same and it is a wider parabola. y=2|x| is not correct because the intersect point is at (0,2)
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17:26:35 ** 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 --> correct
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17:39:31 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 --> 1) Shift up 2 units 2) Reflect about the y-axis 3) Shift left 3 units y = sqrt(x) 1) add 2: y= sqrt(x) + 2 2) replace by -x: y= -sqrt(x) + 2 3) add 3: y = -sqrt(x + 3) + 2
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17:40:45 What is the function after you shift the graph up 2 units?
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RESPONSE --> y = sqrt(x) + 2 You add the 2 on the outside of the original equation.
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17:41:01 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 --> correct
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17:41:48 What is the function after you then reflect the graph about the y axis?
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RESPONSE --> y = - sqrt(x) + 2 You add the negative sign on the outside of the original problem.
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17:41:57 ** 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 --> correct
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17:42:25 What is the function after you then fhist the graph left 3 units?
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RESPONSE --> y = sqrt(-x + 3) + 2
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17:42:31 ** 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 --> correct
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17:46:47 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+2)^3 - 3 Using the graph of y = x^3, shift left 2 units. Then shift down by -3
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17:47:18 ** 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 needed to shift the -3 to the left, not down.
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17:50:09 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 --> h(x) = (4) / (x) +2 I am not sure how to do this one because the 4 is getting divided by x. In the examples it does not show that. After reading your answers, I am sure I will understand.
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17:51:04 ** 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 needed to start with the reciprocal function y = 1/x I needed to move the point 4 times as far from the x axis.
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17:53:20 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 --> f (x) = -4 sqrt(x-1) Start with f(x) = sqrt(x) Then, shift left 1 unit. Then, move the point vertical 4 times. Then, reflect about the y-axis to make it negative.
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17:53:33 ** 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 --> correct
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13:35:19 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) Shift left two units H(x) = f(x+1) - 2 Shift left 1 unit and shift down 2 units g(x) - f(-x) Reflect about the y-axis
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13:35:55 ** 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
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13:49:27 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 --> x^2 + 4 = -2 divide by 2
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13:53:11 ** 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 completed the square wrong. But after reviewing your notes I understand how you did it.
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