course Mth 163 assignment #003003. `query 3
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15:11:29 `questionNumber 30000 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 --> The pics are all parabolas meeting at the center (0,0). All the graphs y values remained corresponding and they all pointed down except the negative. confidence assessment: 2
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15:11:46 `questionNumber 30000 ** 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 --> ok i understand self critique assessment: 3
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15:17:12 `questionNumber 30000 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 --> x=(2)/2(1)= -1 x= -1 y = -1^2 + 2(-1) + 1=0 First fundamental point= (-1,0) Find second and third by adding one and subtracting one for right and left. 2nd and 3rd points= (0,1) and (-2,1) y = x^2 + 3x + 1 x=(-3)/2(1)= -3/2 y = -3^2 + 2(-3/2) + 1=-1/4 first fund. point =(-3/2, -1/4) 2nd and 3rd= (-5/2, -1/4) and (-1/2, -1/4) confidence assessment: 2
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15:17:32 `questionNumber 30000 ** 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 --> ok i understand self critique assessment: 3
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15:18:13 `questionNumber 30000 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 --> vertices move towards the bottom or below and they go to the left but not in any order. confidence assessment: 3
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15:18:34 `questionNumber 30000 ** 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 understand self critique assessment: 3
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15:19:14 `questionNumber 30000 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
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RESPONSE --> If you have the 3 fundamental points then they give you the basic outline so you know the shape and which way they are going. confidence assessment: 3
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15:19:39 `questionNumber 30000 ** 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 --> ok i understand self critique assessment: 3
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15:20:17 `questionNumber 30000 query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?
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RESPONSE --> All you have to do is look at the graph and see if there are any crossings at the zeros. confidence assessment: 2
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15:20:28 `questionNumber 30000 ** 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 I get it. self critique assessment:
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15:21:08 `questionNumber 30000 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 --> b/c the graph is equal on both sides so if one zero is on one side, there must be one the same distance on the other side. confidence assessment: 2
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15:21:19 `questionNumber 30000 ** 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 see self critique assessment: 3
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15:21:38 `questionNumber 30000 What was the shape of the curve connecting the vertices?
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RESPONSE --> The shape of the curve to me looks to be a parabola. confidence assessment: 2
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15:21:51 `questionNumber 30000 ** 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 i understand. self critique assessment: 3
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15:22:11 `questionNumber 30000 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment:
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