#$&* course mth 158 200 7/15 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Is the number in front of the x the number that tells you how wide the parabola is stretched? So for -x^2, it would be stretched by 1 and for -2x^2 it is stretched by 2? ------------------------------------------------ Self-critique Rating: 2
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel like I am missing something? Do I need to have a graphing calculator to be able to graph these and see where they are plotted on the graph? ------------------------------------------------ Self-critique Rating: 1
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Given Solution: * * 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. ** What is the function after you then reflect the graph about the y axis? ** 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. ** What is the function after you then fhist the graph left 3 units? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To shift the function to the left the equation becomes y= sqrt(-x +3) + 2. confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not know how to begin this problem. I’m really confused on where to start and how you know what to do. I understand the concept of moving the graph by adding and subtract different things to the equation but I don’t see how this is completed. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with y = x^3 we replace x by x +2 to shift the graph 2 units left. We obtain y = (x + 2)^3. We then shift this graph -3 units vertically by adding -3 to the value of the function, obtaining y = (x + 2 )^3 - 3. The basic points of the y = x^3 graph are (-1, -1), (0, 0) and (1, 1). Shifted 2 units left and 3 units down these points become (-1 - 2, -1 -3 ) = (-3, -4), (0 - 2, 0 -3) = (-2, -3) and (1 - 2, 1 -3) = (-2, -2). 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not understanding where y=x^3 is coming from? How do you know that you must shift the graph to the left first? After a further look at the problem, I was confused at the numbers because in my book it is number 46 instead of 42. I did not see the equation in this document and was just looking at the problem number in my book. I’m still confused on how you knew to move it left and then vertically. ------------------------------------------------ Self-critique Rating: 1
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Given Solution: * * Note on the function y = 1 / x. Your text does not seem to have included an explanation of the graph of y = 1 / x in this section. Your author has good reasons for this (a very good discussion follows in the next chapter when it discusses rational functions), but I think the following explanation will be useful to many students. A short table for y = 1 / x helps explain the features of its graph: x y = 1 / x 1/10 10 1/2 2 1 1 2 1/2 10 1/10 Before reading further you should hand-sketch a graph corresponding to this table. You aren't likely to understand the rest of this explanation without taking a couple of minutes to do so. We see that as x increases from 1 to 2 to 10 the value of 1/x decreases from 1 to 1/2 to 1/10. As x continues to increase, y = 1/x will continue to decrease, but will still remain positive. Thus, as we move to the right the graph descends toward the x axis, never reaching it but becoming closer and closer, with no limit to how close it will eventually become. Now we move to the left of x = 1, moving through decreasing x values which approach 0. As x decreases from 1 to 1/2 to 1/10, the value of y = 1 / x increases from 1 to 2 to 10. As the value of x gets smaller and smaller, approaching 0 but never reaching 0, the value of 1 / x increases more and more rapidly. You should identify the points (1/10, 10), (1/2, 2), (1, 1), (2, 1/2) and (10, 1/10) on the graph below, and compare this graph to yours. The positive x axis is a horizontal asymptote for the graph: as we move to the right the graph approaches this line, with no limit to how closely it will eventually approach. The positive y axis is also an asymptote. The closer x approaches 0, the greater the value of the function. There is no limit to how closely x can approach 0, so there is no limit to how large the values of 1 / x can become. There is no limit to how closely the graph approaches the y axis, nor is there a limit to how large the y values can become. Now we consider the more abbreviated table below: x y = 1 / x -2 -1/2 -1 -1 -1/2 -2 1/2 2 1 1 2 1/2 This table includes three negative values of x. You should add the corresponding three points to your graph. If you do so it should become clear to you how the entire graph of y = 1 / x is constructed, and why. Having plotted these points the trend of the graph should be pretty clear to you. All negative values of x result in negative values of y = 1 / x. As we move to the left from x = -1, the graph moves closer to the x axis, and the negative x axis is a horizontal asymptote. As we move from x = -1 toward x = 0, the graph approaches the y axis as an asymptote. The graph is depicted below. You should locate the three new points (-2, -1/2), (-1, -1) and (-1/2, -2) on this graph. Be sure you understand how the horizontal asymptotes on the positive and negative x axes occur, as well as the vertical asymptotes on the positive and negative y axes. It is useful to also notice the symmetry behaviors of the graph. Now for an explanation of the problem: We start with the basic reciprocal function y = 1 / x, which has vertical asymptotes at the y axis and horizontal asymptotes 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 manner increases 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. ??? So when adding k units to the function to have it shift, you don’t just shift it up or down on the graph? You actually add 2 to the y unit? You added 2 to -4 in (-1, -4 + 2) = (-1, -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). STUDENT QUESTION: When you say multiplying a function by 4, are you saying you would do this:1/x * 4/1 and if so, wouldn’t that give you 4/x then would you substitute the x = 1 and x = -1 to get the next set of points (-1, -4) and (1, 4)? INSTRUCTOR RESPONSE: That is so. However note that (1, 4) is 4 times as far from the x axis as the point (1, 1), and (-1, -4) is 4 times as far from the x axis as (-1, -1). Given a graph of y = 1 / x, we could have simply taken each point of the graph and moved it 4 times as far from the x axis, without ever calculating y = 4 / x. We might well want to calculate the values, but it's important to visualize, without the point-by-point calculations, how the graphs transform. For example the graph below depicts the two functions. You should identify the points (-1, -1) and (1, 1), and also the points (-1, -4) and (1, 4). Then look at the vertical blue line and convince yourself that the point where this line intersects the y = 4 / x graph is four times as far from the x axis as its intersection with the y = 1 / x graph. Then visualize several other vertical lines, and observe how the same statement applies no matter where you draw the vertical line (with the exception of the y axis, where neither function is defined). STUDENT QUESTION: So, by adding the 2, the original points (-1, -4) and (1, 4) were just used to show how 4/x would look, and then by adding the y = 4/x + 2 is the “real” graph we are trying to obtain? INSTRUCTOR RESPONSE: That's about right. When we add 2 to get the function y = 4 / x + 2, we are increasing the y values by 2. This 'raises' the y = 4 / x graph 2 units. The graph below depicts the previous graphs of y = 1 / x and y = 4 / x, as well as the 'new' graph y = 4 / x + 2. You can note that the point (1, 4) on the y = 4 / x graph transforms to the point (1, 6), and the point (-1, -4) to the point (-1, -2). The 'blue' vertical line intersects the 'new' graph at a point 2 units higher than the y = 4 / x graph, and the same will be true of any vertical line. STUDENT QUESTION: It says where y = 0, the y = 4/x + 2 graph will approach the line y = 0 + 2 = 2……throws me off. I don’t understand this part…..and the horizontal asymptotes to the right and left will consist of the line y = 2…I tried drawing the graph and looking at it, but it doesn’t appear that anything = 2. INSTRUCTOR RESPONSE: The x axis is the line y = 0. To see this, pick several points on the x axis. What is the y coordinate of each? Your answer will always be that the y coordinate is 0. If we raise each point of the line y = 0 by 2 units we will get the line y = 2. To see this, pick several points on the x axis and 'raise' each of them by 2 units in the y direction. Your new y coordinate will always be 2. Your points will all lie on the line y = 2. The y = 4 / x graph had an asymptote at the x axis. That is, it was asymptotic to the line y = 0. The y = 4 / x + 2 graph will therefore have an asymptote at y = 2. The graph below is the same as the previous graph, but the line y = 2 has been added. You should observe how the graph of y = 4 / x + 2 approaches this line as a horizontal asymptote. Here's a picture of the final graph of y = 4 / x + 2, along with the horizontal asymptote y = 2. Self-critique (if necessary) ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: * 3.5.58 / 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The beginning basic function you begin with is sqrt(x). It is then multiplied by -4 to stretch it four units making it lie in quadrant 4. The equation is then moved downward 1 by subtracting 1 from the equation giving us y= 4sqrt(x)-1. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** STUDENT QUESTION I see that you multiply the y values by -4 but why do you multiply the y values and not the x values? INSTRUCTOR RESPONSE Think about the graph of y = 2 x^2 compared to the graph of y = x^2. To make a brief table for either function you might use x values -1, 0 and 1. The second function gives you points (-1, 1), (0, 0) and (1, 1). The first function gives you points (-1, 2), (0, 0) and (1, 2). If you multiply the y values of the first function by 2, you get the y values of the second. The x values don't change. In general if the function is y = f(x), multiplying f(x) by a 2 doubles the y values, while adding, say, 5 would increase every y value by 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m confused on how you find the initial points. How do you know that this is true>>The points (0, 0), (1, 1) and (4, 2) lie on the graph of the original function y = sqrt(x). ?
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Why does the function for the first one move to the left instead of upward? I thought that when you add a number to X the function moves up.
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Given Solution: * * 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. ???? Why do you complete the square by add (4/2)^2?? 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). ** STUDENT QUESTION I don’t understand how the function went from f(x) = x^2 + 4x + 4 - 4 + 2 tof(x) = (x^2 + 4x + 4) - 2. How did you get the +4 when the previous step had a positive and a negative 4? Wouldn’t that cancel each other out? INSTRUCTOR RESPONSE The object was to get a perfect square out of x^2 + 4 x. That's why you squared 4/2, giving you 4. x^2 + 4 x + 4 is a perfect square. Of course we can't just add 4 to the original expression--that would change its value. However we can add and subtract 4, which doesn't change the value of the expression. This is how we got the expression x^2 + 4 x + 4 - 4 + 2. Grouping the perfect square our expression becomes (x^2 + 4 x + 4) - 4 + 2 To simplify this expression we need only find -4 + 2, which is -2. Our expression becomes (x^2 + 4 x + 4) - 2, which we can write as (x + 2)^2 - 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m still pretty confused on all of this. I’m going to try and go through more problems on MathXL. ------------------------------------------------ Self-critique Rating: 1