course Mth 163 003. `query 3
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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qquery 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 ********************************************* Your solution: y = x^2 + 2x + 1 A B C x = - b / (2 a) x= -2 / (2(1)) x= -1 (Vertex) y = x^2 + 2x + 1 y= (-1)^2 +2(-1) +1 y= 1-2+1=0 (-1, 0) y = x^2 + 3x + 1 A B C x = - b / (2 a). x= -3 / (2(1)) x= -1.5 y = x^2 + 3x + 1 y= -1.5^2 + 3(-1.5) +1 y= 2.25 -4.5 +1 y= 7.75 (-1.5, -1.25) Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to do the two fundamental points for each one, but I do understand it. It’s just one to the left and one to the right of each original set of points. Self-critique Rating: 2 ********************************************* Question: `qhow did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem? ********************************************* Your solution: The second was down farther and to the left father than the first graph. Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qHow do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola? ********************************************* Your solution: It gives you those extra coordinates you need to make the sketch. With only two pairs of coordinates it’s harder to visualize than if you have two added fundamental coordinates. It kind of lets you see the shape that it’s starting to make and then it can be finished easily. You can see with a line of symmetry right where the middle ( or peak in other words) of the parabola is. Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qquery Zeros of a quadratic function: What was it that determined whether a function had zeros or not? ********************************************* Your solution: I had to read the solution on this one. I will try to put it into my own words. I believe what it is saying is that the number that appears within the square root can only be solved if it is positive and if it turns out negative then it will in no way have zeros. If it is positive, then you will have two zeros and if the original number is zero, then you will have one zero. Confidence Assessment: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had to read the solution on this one because I wasn’t quite sure what the question meant or what the answer was. I’m having a hard time tying in this concept with everything else.
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qWhat was the shape of the curve connecting the vertices? ********************************************* Your solution: It is a parabola. Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** Query Add comments on any surprises or insights you experienced as a result of this assignment. The zero question kind of threw my for a loop at first, but I’m trying to understand it. " ""