Assugnment 25

You did well on most of these questions. You still have a little way to go with the transformations, but you're well on the way. See my notes.

Let me know if you have specific questions.

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14:21:21 query 3.5.12 (was 3.4.6). Upward parabola vertex (0, 2). What equation matches this function?

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RESPONSE --> y=x^2+2

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14:22:06 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 --> The graph is not stretched and it is not shifted left.

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14:22:16 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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14:23:00 query 3.5.16 (was 3.4.10). Downward parabola.

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RESPONSE --> The equation for this is y= -2x^2

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14:23:31 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 faces downward so it is not 2x^2, and it is stretched, so it is not the second equation

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14:23:36 ** 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 --> ok

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14:24:14 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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14:24:34 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 --> It is stretched and not shifted, so it cannot be these two equations

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14:24:41 ** 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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14:26:46 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 --> I start with y= sqrt x which has points at (0,0), (1,1), (4,2), and (9,3). I then shift up the y axis 2 units, and then for every positive x coordinate, I graph its negative with the same y coordinates as before. Finally, I shift the graph left along the x axis 3 units.

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14:27:16 What is the function after you shift the graph up 2 units?

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RESPONSE --> It would be y= 2 *sqrt x

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14:27:54 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 --> I did the wrong operation, but now I see why.

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14:28:42 What is the function after you then reflect the graph about the y axis?

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RESPONSE --> y= sqrt -x

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14:29:01 ** 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 --> okay, so I do carry over the +2, wasn't sure, but I see now

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14:31:14 What is the function after you then fhist the graph left 3 units?

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RESPONSE --> I think this is y= [sqrt -x+2 ]+3

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14:31:30 ** 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 --> I had this somewhat out of order

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14:35:12 query 3.5.42 (was 3.4.36). f(x) = (x - 3)^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 --> I think this is a cube function, so I would start with the x^3. The point would be (0,0), (1,1), (-1,-1). Then I would shift the graph vertically three units, which would be (0,3), (1,4), (-1,3). I would then shift the grap right along the x axis three units.

This is what you would do if the function is f(x) = (x - 3)^3 + 3.

I'm not sure what the correct function is, but the function originally given by the Query did not match the given solution.

I've changed the function to f(x) = (x - 3)^3 + 3 and modified the given solution to match this function. Be sure you understand the complete explanation.

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14:35:53

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RESPONSE --> I think I switched the units involved with each shift. I will have to look over this some more.

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14:38:06 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 --> This is a basic reciprocal function so I would start with f(x)=1/x. The points are (1,1) and (-1,-1).I believe the graph should shift vertically 2 units, but I am not sure what to do with the 4 in the numerator.

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14:38:46 ** 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 see now. There was a stretch involved that I was not expecting.

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15:41:32 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 --> I would start with the sqrt of x which has points at (0,0), (1,1), and (4,2). Then I would stretch it four units vertically in the negative direction. I am not sure what to do after this

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15:42:17 ** 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 --> okay, I have drawn this out and I see now what to do.

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15:43:01 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 --> I have no idea really how to do this

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15:48:34 ** 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 will continue working on this, as I find it very complicated. I have graphed these coordinates and will continue to study it until I can figure this out.

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15:50:46 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 --> I am having trouble with this too. I know how to complete the square, but I am not sure where to go

You should included the step on completing the square so I could help you address any errors in that process.

After you've completed the square, the rest isn't too bad.

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15:53:18 ** 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 --> In looking at this problem again, I am more confident about doing it. I now see how to write this equation out, which was the area where I was confused.

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