course Mth 163 It just hit me that the graphs and vertices you probably were referring to were the one's I got when I stretched to graph by 2,3,.5 and -3. I'm sorry, I worked this query in spurts and lost track of the problems. I will return to the query and reanswer the problems in my notes. My responses may not make a lot of sense to you. ?`?g????????F???assignment #003003. `query 3
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14:22:36 query graph of y = x^2 stretched vertically by different factors
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14:22:45 query graph of y = x^2 stretched vertically by different factors
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14:22:47 query graph of y = x^2 stretched vertically by different factors
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14:22:47 query graph of y = x^2 stretched vertically by different factors
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14:22:49 query graph of y = x^2 stretched vertically by different factors
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14:39:57 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 --> The y=x^2 graph is a perfect parabola. When I stretched it vertially by a factor of 2 is was still a parabola, but thinner or more steep. When I stretched it by a factor of 3 it was even steeper than than the second graph. When I stretched it by .5, it was was really wide, wider or less steep than the original y=x^2 graph and less steep than the other graphs as well. When I stretched it by a factor of -3, it was very steep, but upside down or in the negative values. It was a mirror image of the y=3x^2 graph.
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14:45:09 ** 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 --> Yes, my graphs had those points as well. All graphs were opening up except the factor -3 graph and it was opening down.
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14:46:43 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 --> ok
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15:08:19 ** 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 --> y=x^2+2x+1 vertex: x= -2/2(1)= -1 y= (-1)^2+2(-1)+1=0, so the vertex is (-1,0) and two points are (-2,1) and (0,1). for y=x^2+3x+1 the vertex is (-1.5, -1.25) and the two points are (-2.5,1.25) and (-.5,1.25). For some reason this program is giving me the answer before the queston. I worked the problems out on paper anyway, but it seems backward to me. I only found the prob.3 by reading the answer before it.
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15:13:57 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 --> I'm confused, I have not had four graphs or four problems. So far I've found the vertex and points for y=x^2+2x+1 and y=x^2+3x+1. I'm not sure what problem I'm being asked about. Did I miss a question? When I compare how the vertex of these two graphs moved though, the first vertex of (-1,0) lies above and to the left of the vertex of the second graph (-1.5,-1.25).
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15:14:35 ** 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 --> yes, i agree to this for the two that I did.
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15:21:13 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
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RESPONSE --> Because from the fundamental points, you continue to move one unit to the left and one unit up, one unit to the right and one unit up. After graphing the vertex (-1,0) and the two fundamental points (-2,1) and (0-1) for y=x^2+2x+1, I graphed the points (-3,2) and (1,2). If you continue plotting these points, you can continue the parabola by drawing lines from the vertex and up each side.
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15:23:09 ** 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 agree with this, but I'm getting student comments or answers that aren't mine in the ""Next Question/Answer box"" Last semester when I did the query exercises, I don't remember the program showing these.
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15:34:37 query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?
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RESPONSE --> If it crossed the x axis or not. If it does not have values of x it will not cross the x axis and will not have zeros? Maybe, I'm not sure.
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15:37:22 ** 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 --> ok, so if the graph does not cross the x axis, as in going into the negative digits, it will not have zeros, but if it lies on the x axis at one point (like the vertex of the parabola) it does have zero's?
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15:41:24 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 --> The vertex lies halfway between the fundamental points that I got from the vertex. When I obtained the vertex, I used it's points to obtain the two fundamental points by adding on point to each the x and y values. Are these two fundamental points the same as the zeros? If so, the vertex will always lie between those two points, right?
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15:42:19 ** 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 --> O.K. I understand. This is because you use the vertex to find those points and space them evenly on each side and above the vertex, correct?
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15:45:18 What was the shape of the curve connecting the vertices?
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RESPONSE --> The shape is U to V shaped depending on how steep the parabola is. The graph of y=x^2+2x+1 is broader (U shaped) as opposed to the one of y=2x^2+2x+1 for example, it was more steep (V shaped).
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15:49:37 ** 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 --> ok, again, I'm not sure which vertices you're referring to. I must have missed something because there are only two parabolas and no preceding problem except the y=x^2+2x+1 and y=x^2+3x+1 and those two are parabolas with one vertex lying below and to the left of the other.
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15:52:16 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I was surprised that the query program was a little confusing, but it has been a long time since last semester. No, seriously, it was good practice to review finding vertices and fundamental points and good practice to graph. I think I understood that the two fundamental points are the same as the zeros of a quadratic function? Or am I wrong?
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