course Mth 163 Vxiǖكassignment #003
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20:06:11 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 graph is a parabola with the vertex at (0,0). When the graph is stretched it increase to both the right and left. When stretching by 2 the next points would be (1,2) and (-1,2) when stretching by three (1,3) and (-1,3) for .5 (1,.5) and (-1,.2) confidence assessment: 1
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20:06:35 ** 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 self critique assessment: 3
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20:15:24 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 --> The vertex of y=x^2 +2x+1 would be (-1,0) x=-2/2(1) x=-2/2 x=-1 y=-1^2+2(-1)+1 y=1-2+1 y=0 The vertex of y=x^2+3x+1 would be (-1.5,-1.25) x=-3/2 x=-1.5 y=1(-1.5)^2+3(-1.5)+1 y=2.25-4.5+1 y=-1.25 confidence assessment: 2
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20:16:50 ** 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 I omitted the answer for the fundamental points. self critique assessment: 2
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20:29:22 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 --> It changed from (-.5,.75) to (-1,0) to (-1.5,-3.25) and finally to (-2,-3) X increased in .5 units as the parabola was stretched. confidence assessment: 1
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20:29:40 ** 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 --> ok self critique assessment: 3
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20:30:23 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
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RESPONSE --> It give the vertex which is centered and then the first point to the right and to the lef which allows a fairly accurate sketch of the graph. confidence assessment: 1
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20:31:08 ** 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 --> It also shows which way the parabola opens and how wide it will be. self critique assessment: 2
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20:33:11 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 crosses the x axis then the parabola will have zeros. If it never crosses the axis then there are no zeros. confidence assessment: 1
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20:34:28 ** 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 understand so if it has a zero then there will be one zero, if there is a negitive discriminant none, and if it is positive then 2 zeros. self critique assessment: 2
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20:36:14 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 --> Looking at the graphs the vertex lies halfway between the zeros. The points to the right and left of the vertex are one unit up and to the right or lef of the vertex/ confidence assessment: 1
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20:36:23 ** 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 --> self critique assessment: 3
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20:37:34 What was the shape of the curve connecting the vertices?
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RESPONSE --> It is in the shape of another parabola that would open to the right hand side. confidence assessment: 1
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20:37:54 ** 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 self critique assessment: 3
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20:39:00 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I saw that as the value of x increases for the equation then there is another parabola formed from the vertices. self critique assessment: 3
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