course Mth 163 TÑÑÕ‹žR´ÅÄì‚Æ¡dÔÍù°W†‡ùîÎðtãassignment #003
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13:44:05 query graph of y = x^2 stretched vertically by different factors. When you constructed a graph of y = x^2 and stretched it, describe the graphs you obtained when you stretched it vertically by a factor of 2, by a factor of 3, by a factor of .5 and by a factor of -.3.
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RESPONSE --> Of all the graphs -.3 was the broadest of al of then with 3 being the narrowest. .5 and 2 was also broad. It is easy to see the difference when they are all placed on a graph. confidence assessment: 2
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13:46:59 ** STUDENT RESPONSE: The graphs of all of the plots were parabolic with all of the vertices at (0,0). Graphs of the plots swtretched by factors .5, 2 and 3 all had the vertex 'down' with factor 3 graph as the narrowest, then factor 2 graph and finallyfactor .5 as the broadest. Factor -.3 was the broadest of them all and its vertex was up. INSTRUCTOR COMMENT: Good descriptions. You should also mention that the factor 3 graph contained the points (-1, 3) and (1, 3), while the factor 2 graph had points (-1, 2) and (1,2) while the factor .5 graph had (-1, .5) and (1, .5). **
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RESPONSE --> I forgot to mention that everyone one had the same vertex of (0,0). I see how the points are. They are the same points that I have. self critique assessment: 3
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13:55:55 query prob 3: Location of the vertex and fundamental points. Determine the location of the vertex and fundamental points of each of the following quadratic functions: y = x^2 + 2x + 1 and y = x^2 + 3x + 1
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RESPONSE --> For the equation y=x^2+2x+1 the point is (-1,0). One place on each side of it would be (-2,1) and (0,1). For the equation y=x^2+3x+1 the point is (-3/2,-5/4). The points on each side are (-1/2, -1/4) (-5/2, -1/4) (-2.5,1.25) (-.5,1.25). confidence assessment: 2
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13:56:17 ** The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a). For y = x^2 + 2 x + 1 we have vertex at x = -2 / (2 * 1) = -1. At x = -1 we have y = (-1)^2 + 2 * -1 + 1 = 0. So one fundamental point is the vertex (-1,0). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-2,1) and (0,1). For y = x^2 + 3 x + 1 we have vertex at x = -3 / (2 * 1) = -3/2. At x = -3/2 we have y = (-3/2)^2 + 2 * (-3/2) + 1 = 1/4. So one fundamental point is the vertex (-3/2, -5/4). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-5/2, -1/4) and (-1/2,-1/4); or in decimal form at (-2.5, 1.25) and (-.5, 1.25). **
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RESPONSE --> I see how it is done and I understand how to find the solution. self critique assessment: 3
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13:58:28 how did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem?
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RESPONSE --> The vertices move downward and to the left, but not along a straight line. They also in a parabola of their own. confidence assessment: 2
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13:58:46 ** The vertices move downward and to the left, but not along a straight line. In fact the vertices lie along a different parabola of their own. **
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RESPONSE --> I see how it is done. I placed them on a graph and saw it. self critique assessment: 3
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14:01:34 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
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RESPONSE --> The vertex lines in a vertical line and the points on the left and right side show how the parabola opens up or down to show the width. confidence assessment: 2
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14:01:47 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: They provide the initial sweep & symetry around the vertex which defines the 'shape' and direction and allows you to extrapolate. INSTRUCTOR COMMENT: That's very nicely put. The vertex lies on a vertical line of symmetry and the points to right and left show you which way the parabola opens (upward or downward) and indicate the width of the parabola. **
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RESPONSE --> I see how it is done adn I understand how to see it. self critique assessment: 3
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14:05:48 query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?
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RESPONSE --> The vertex and the 2 points show whether ot not there are any zero's crossing. If there is any negative numbers you cannot take the square root of them so there cannot be any zeros but if you ake the square root of a positive number then there can be zeros. confidence assessment: 2
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14:06:32 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: Initially, I used the graph of my fundemental points as a visual aid as to whether zero crossings were present INSTRUCTOR COMMENT:There are two ways to answer this question. One is your answer--the vertex and two basic points clearly show you whether the function has zeros. The other way to answer is to note that if the discriminant b^2 - 4 a c, which appears inside the square root, is negative the function cannot have zeros (since you can't take the square root of a negative) and if the discriminant is positive you do have zeros (the quadratic formula will yield one zero if the discriminant is zero and will yield two zeros if the discriminant is positive). **
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RESPONSE --> I see how it is done and I completley understand how to go about finding out whether or not there will be any zeros present. self critique assessment: 3
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14:08:22 query #4. Questions about vertex between zeros and the shape of the curve connecting vertices: What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist?
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RESPONSE --> There actually isn't any way to be certainally sure that this is possible. You can take the formula and try to solve for y to get the answer. confidence assessment: 2
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14:08:53 ** The graph of each function is completely symmetric about a vertical line through the vertex. Thus if one of the zeros lies to the right of the vertex, the other must lie an equal distance to the left of the vertex. **
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RESPONSE --> I was some what confussed about this question and answer and tried to answer it to the best of my ability. self critique assessment: 3
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14:09:58 What was the shape of the curve connecting the vertices?
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RESPONSE --> It is in the shape of a ""u"" going upward. confidence assessment: 2
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14:10:25 ** In fact the vertices of y = x^2 + b x + 1 for various values of b (e.g., b = 2 and b = 3 for the parabolas of the preceding problem) themselves lie on a parabola. We're a ways from actually proving that, but it's an interesting idea. Don't actually do it but if you took any three of these vertices and fit them to the form y = a t^2 + b t + c you would get an equation for y vs. t, and all the other vertices would satisfy the equation. **
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RESPONSE --> I am not too sure about this one and how the answer was found. self critique assessment: 3
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14:11:19 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This assignment wasn't too bad and I was able to see how everything was done. There was one or two questions that I was a little confused on but if I study on it a little more I am sure I would be able to find the correct answer. self critique assessment: 3
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