#$&* course Mth 163 1/24/13 around 7:00p.m. Question: `q001. Note that this assignment has 6 questions
.............................................
Given Solution: For the function y = - 0.45833 x^2 + 5.33333 x - 6.875 we have a = -0.45833, b = 5.33333 and c = -6.875. The quadratic formula therefore tells us that for our function we have y = 0 when x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638 and when x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006. STUDENT COMMENT This whole problem confuses me. I put in these numbers in my calculator but I get a different answer everytime. INSTRUCTOR RESPONSE It is possible your calculator doesn't follow the order of operations. Most do, but some do not. It is also possible that you are making an error with the order of operations. You do have some out-of-place parentheses (e.g., in the expression 4 * ( -0.4583) * ) -6.875)). You should evaluate the various quantities separately, then put them together. For example to calculate [ -5.33 + square rt (5.33^2 -4 * ( -0.4583) * ) -6.875) ] / (2 * (-0.45833) , begin by calculating 4 * ( -0.4583) * -6.875, then calculate 5.33^2, then subtract, then take the square root. Calculate 2 * (-0.45833) . Combine your results to calculate [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)). If you tell me what you enter into your calculator at each step, and what you get, I can tell you if you're making an error and, if so, how to avoid it in the future. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. Extend the smooth curve in your sketch to include both points at which y = 0. Estimate the x value at which y takes its maximum value. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My graph closely passes through both 1.47 and 10.16 in a parabola form. The graph comes to a point almost at the midway point between both values and at about x = 5.5 and about y = 8.6 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Your graph should clearly show how the parabola passes through the x axis at the points where x is approximately 1.5 (corresponding to the more accurate value 1.47638 found in the preceding problem) and where takes is a little more than 10 (corresponding to the more accurate value 10.16006 found in the preceding problem). The graph of the parabola will peak halfway between these x values, at approximately x = 6 (actually closer to x = 5.8), where the y value will be between 8 and 9. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q003. For the function of the preceding two questions, y will take its maximum value when x is halfway between the two values at which y = 0. Recall that these two values are approximately x = 1.48 and x = 10.16. At what x value will the function take its maximum value? What will be this value? What are the coordinates of the highest point on the graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Honestly I am not sure how to go about solving this problem??? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82. At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx.. Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After looking at your solution I understand completely and it makes sense. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a). At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we must figure out what x values the vertex is at so we use the given formula x = -b/2a. Plugging numbers into this we get x = -5.3333/ 2*-0.4583333. Where x = 5.81818 Then we go back to our original equation: y = - 0.458333 x^2 + 5.33333 x - 6.875 We plug in x=5.81818 and get the equation y = - 0.458333 *5.81818 ^2 + 5.33333 *5.81818 - 6.875. This gives us y = 8.63992. Therefore we get the vertex of (5.81818, 8.63992) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain x value at which function is maximized: x = -b / (2a) = - 5.33333 / (2 * -0.45833) = 5.81818. To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024. Thus the vertex of the parabola lies at (5.81818, 8.64024). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402). What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex? What is the value of y corresponding to each of these x values? By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For one unit to the right we just add 1 to the x value which gives us 6.81812. When we plug this into our equation we get y = 8.1795 For one unit to the left we just subtract 1 to the x value which gives us 4.81812. When we plug this into our equation we get y = 8.1802 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q006. In the preceding problem we saw an instance of the following rule: The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola. In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points. What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex? Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the coordinates of the vertex we must use the formula given before which was vertex = x = -b/2a. Plugging values into this we get x = -10/ 2* -1. X = 5 Then to get the y value we must go back to the quadratic model of -1x^2 + 10x +100 and plug in x = 5. When we do this we get y = 125. So our coordinates are (5, 125) For 1 unit to the right we get x = 6 by adding 1 and then put 6 into the quadratic model and get y = 124. For 1 unit to the left we get x = 4 by subtract1 and then put 4 into the quadratic model and get y = 124 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex. STUDENT COMMENT This problem was a little confusing. I wasn’t really sure the point of seeing if the parabola would touch the x axis? INSTRUCTOR RESPONSE On the x axis, the y value is zero. The points where the graph intersects the x axis are called the zeros of the function. The quadratic formula gives you the zeros of the function a x^2 + b x + c. These points are very important in applications, as you will see very soon. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 003. `query 3 ********************************************* Question: `qquery 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When graphing a sketch of y^2 we get a parabola. When we stretch it vertically by different factors like 2 units that would mean it would shift two units up from the line it has already made. When stretched vertically by a factor of 3 the graph has move 3 units up at each point from where the line originated. When stretched .5 units is again shifted from the original line up .5 units at each point. When stretched by a factor of -.3 the line moves downward from where it originated by .3 units. The units 2, 3 and .5 will all cause the parabola to become more narrow and closer to the y axis in the upward direction. The -.3 unit will make it become more narrow, but in a downward position. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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). ** The student question below is answered by a series of questions, which may be copied and submitted separately if you think it would be beneficial to do so. However you aren't required to submit these questions and answers: STUDENT QUESTION I’m ok on sketching the graphs, but I don’t know how to find the points in the different graphs. INSTRUCTOR RESPONSE This is explained more fully in the worksheets and on the DVDs. However here is a series of questions you should consider answering to understand how this works for the y = x^2 function: ********************************************* Question: What are the x = -1, x = 0 and x = 1 points of the y = x^2 function? (Answer: For x = -1, 0 and 1 the y = x^2 function takes values (-1)^2 = 1, 0^2 = 1 and 1^2 = 1. This gives us the points (-1, 1), (0, 0) and (1, 1).) ********************************************* Question: Plot these points on a graph, and construct a parabola through the points, using only these points. You should know pretty much what a parabola looks like, and these three points should be sufficient. For the parabola you constructed, what are the y coordinates corresponding to the following x coordinates? Don't cheat yourself and evaluate the y = x^2 function at this point, use the graph as a basis of your estimate. x = 1/2; x = 2; x = -1.5; x = -3. (answer: The further you are from x = 0, the greater will be your y value; so the y value for x = 1/2 will be the smallest and the y value for x = -3 will be the greatest) ********************************************* Question: What are the x values for the following y values? y = 3; y = 7; y = .5 (answer: note that each y value will correspond to two x values, one positive and one negative) ********************************************* Question: Now refine your graph by plotting the graph points for x = -3, -2, -1/2, 1/2, 2 and 3. Sketch a smooth curve and see how it differs from the graph you made based only on the three points. (answer: your second graph will likely be more accurate than the first; the two graphs will probably be pretty close between x = -1 and x = +1, and will probably be closer at the x = -2 and x = +2 points than at the x = -3 and x = +3 points. This exercise should help you better visualize the shape of a parabola. However chances are you did pretty well with your 3-point attempt, and will do better with your next.) ********************************************* Question: Start a new graph. Again plot the x = -1, x = 0 and x = 1 points of the y = x^2 graph, and sketch an approximate parabola based on these three points. On the same set of coordinate axes plot the x = -1, x = 0 and x = 1 points of the y = 3 x^2 graph, and sketch the approximate parabola that corresponds to these points. Describe your two parabolas, compare them and give the coordinates of the three points of each. (answer: The y = 3 x^2 parabola will be taller and skinnier than the y = x^2 parabola. The three points of the new parabola will be (-1, 3), (-1, 0) and (1, 3). Any vertical line will intersect both the original parabola and the new parabola. The intersection point for the new parabola will be 3 times as far from the x axis as the intersection point for the original parabola.) ********************************************* Question: Plot the points (4, 5), (5, 2) and (6, 5), and sketch the parabola defined by these three points. What is the vertex of the parabola? If you move 1 unit to the right of the vertex, how far up or down do you have to go to get to the parabola? If you move 1 unit to the left of the vertex, how far up or down do you have to go to get to the parabola? Is the shape of this parabola more like that of the y = x^2 parabola or the y = 3 x^2 parabola? (answer: The vertex is at the point (5, 2). If you move 1 unit to the right of the vertex you are at the point (6, 2), directly below the point (6, 5) of the parabola. You have to go up 3 units to get to the parabola. Similarly if you move 1 unit to the left of the vertex, you will be 3 units below the point (4, 5) of the parabola. Since movement 1 unit to the right or left corresponds to a vertical change of 3 units, and since the y = 3 x^2 parabola has the same characteristic, this parabola has a shape which is geometrically similar to that of y = 3 x^2.) ********************************************* Question: If you move 5 units to the right and two units up from the vertex of the y = 3 x^2 parabola, what are the coordinates of the point where you end up? Locate this point on your graph. If you move 5 units to the right and two units up from the x = 1 point of the y = 3 x^2 parabola, what are the coordinates of the point where you end up? Locate this point on your graph. If you move 5 units to the left and two units up from the x = 1 point of the y = 3 x^2 parabola, what are the coordinates of the point where you end up? Locate this point on your graph. What do you observe? (answer: The vertex of the y = 3 x^2 parabola is at (0, 0). 5 units to the right of this vertex is the point (5, 0). 2 units up from this point is the point (5, 2) Similarly if you move 5 units to the right of the x = 1 point (1, 3) you are at (6, 3), and 2 units up from this is the point (6, 5). If you move 5 units to the right of the x = -1 point (-1, 3) you are at (4, 3), and 2 units up from this is the point (4, 5). The three new points are the three basic points you plotted for your new parabola. This demonstrates how the new parabola is horizontally shifted 5 units and vertically shifted 3 units from the y = 3 x^2 parabola.) ********************************************* Question: The three basic points of a parabola whose vertex is at the origina (0, 0) are the x = -1, x = 0 and x = 1 points. All the functions in the present Query question are of the form y = A x^2, for varying values of A. Any function of this form is the graph of a parabola with vertex (0, 0). The graph with factor 3 included the points (-1, 3) and (1, 3), which as we have seen corresponds to the function y = 3 x^2. This graph is of the form y = A x^2 with A = 3. The graph with factor 2 had points (-1, 2) and (1,2), and corresponds to the function y = 2 x^2. This graph is of the form y = A x^2 with A = 2. The graph with factor .5 had points (-1, .5) and (1, .5), and corresponds to the function y = .5 x^2. This graph is of the form y = A x^2 with A = .5. Explain your understanding of this (answer: ) ********************************************* Question: By contrast with these parabolas, the vertex of the last parabola you plotted in this series of questions was not the origin, but the point (5, 2). The function for this parabola is not of the form y = A x^2. The function for this last parabola is y = 3 ( x - 5)^2 + 2. You can't create this function from the form y = A x^2; pick any value of A you like, and you can't get the function y = 3 ( x - 5)^2 + 2. You can get the function y = 3 ( x - 5 )^2 + 2 from the form y = A (x - h)^2 + k, by choosing A = 3, h = 5 and k = 2. What function do you get if you let A = .5, h = 2 and k = -3? (answer: Substituting A = .5, h = 2 and k = -3 into the form y = A ( x - h )^2 + k you get the function y = .5 ( x - 2)^2 +(-3), which can be written more simply as y = .5 (x-2)^2 - 3. Later questions in this and subsequent Queries will further develop this idea.) #$&* &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand your answer, but I felt like the question asked more of what the graph would look like instead of where it lies. I did get the points (-1, 1) (0,0) & (1,1). I do fully understand all of the questions and responses from the student.
.............................................
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, -.25) and (-.5, -.25). ** STUDENT QUESTION The only discrepancy is the fundamental points on the second equation. I will have to look back at the video and see if there is a calculation for this. I can’t remember one. INSTRUCTOR RESPONSE For each of the graphs y = x^2 + 1, y = x^2 + x + 1, y = x^2 + 2 x + 1 and y = x^2 + 3 x + 1, you first find the vertex, which you did correctly above. Then for each vertex: • Starting at the vertex, you move 1 unit to the right, then a units in the vertical direction. (This means that if a is positive you move upward, if a is negative you move downward). Mark your point. This is what I refer to as 'the fundamental point 1 unit to the right of the vertex'. Then go back to the vertex, move 1 unit to the left, then a units in the vertical direction. Mark your point. This is what I refer to as 'the fundamental point 1 unit to the rightleft of the vertex'. In the case of the equation y = x^2 + 3 x + 1, the vertex is at (-3/2, -5/4). a = 1 so we move 1 unit right and 1 unit vertically, which puts us at the point (-3/2 + 1, -5/4 + 1) = (-1/2, -1/4). The we move 1 unit left and 1 unit vertically, which puts us at the point (-3/2 - 1, -5/4 + 1) = (-7/2, -1/4). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each vertex would move to the left and down. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** The vertices move downward and to the left, but not along a straight line. In fact the vertices lie along a different parabola of their own. You wouldn't have been expected know at this point that the vertices all lie along a parabola of their own, of course, though students occasionally conjecture that it is so. However that statement should make sense in terms of your picture, and it's a very interesting and unexpected connection.** STUDENT QUESTION I don’t think I understand the concept of this. How would they lie along a parabola of their own? INSTRUCTOR RESPONSE Sketch a new graph and plot just those vertices. They don't lie along a straight line, they lie along a curve. That curve happens to be a parabola. Don't take the time to do it, but if you picked three of those points and correctly solved the simultaneous equations to get a quadratic model, the fourth point would also lie on the resulting parabola. It's a little beyond the scope of this course to actually prove that the vertices of all such parabolas in fact form a parabola of their own. You could do it, but it would take way more work than it's worth. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As long as you have the vertex and some points that let you know what direction you are going you can be able to infer that it would be the sketch of a parabola. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT ANSWER AND INSTRUCTOR COMMENT: They provide the initial sweep & symmetry 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. ** ANOTHER VERY GOOD STUDENT STATEMENT: Vertex provides the anchor point and central line of symmetry, and the other two points give the direction and shape of the curve. STUDENT COMMENT: With my drawing skills, 3 points doesn't seem to be enough. However, the three fundamental points do provide the vertex, or in other terms, the line of symmetry or the parabola, as well as an additional point to each side of the vertex to determine the approximate shape of the curve, which would indicate whether the parabola opens upwards or downwards at the very least. INSTRUCTOR RESPONSE: You're not going to get a completely accurate picture, but you can see the direction in which the parabola opens and get a pretty good idea how wide or narrow it's going to be. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery Zeros of a quadratic function: What was it that determined whether a function had zeros or not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You can just look at your vertices and see if they have 0s in them. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT ANSWER AND INSTRUCTOR COMMENT: Initially, I used the graph of my fundemental points as a visual aid as to whether zero crossings were present INSTRUCTOR COMMENT:There are two ways to answer this question. One is your answer--the vertex and two basic points clearly show you whether the function has zeros. The other way to answer is to note that if the discriminant b^2 - 4 a c, which appears inside the square root, is negative the function cannot have zeros (since you can't take the square root of a negative) and if the discriminant is positive you do have zeros (the quadratic formula will yield one zero if the discriminant is zero and will yield two zeros if the discriminant is positive). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery #4. Questions about vertex between zeros and the shape of the curve connecting vertices: What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The fact that the graphs are symmetric gives indication that the parabola lies on a line halfway between its two zeros. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat was the shape of the curve connecting the vertices? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A U shape confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** In fact the vertices of y = x^2 + b x + 1 for various values of b (e.g., b = 2 and b = 3 for the parabolas of the preceding problem) themselves lie on a parabola. We're a ways from actually proving that, but it's an interesting idea. Don't actually do it but if you took any three of these vertices and fit them to the form y = a t^2 + b t + c you would get an equation for y vs. t, and all the other vertices would satisfy the equation. ** The figure below shows the graphs of the above functions, which have the form y = x^2 + b x + 1 for b = 2 and 3 (i.e., when b = 2 the form y = x^2 + b x + 1 becomes y = x^2 + 2 x + 1; when b = 3 the form gives us x^2 + 3 x + 1), as well as all the functions we get for b = -5, -4, -3, ..., 3, 4, 5.. You should be able to identify the parabolas you graphed among these. The figure below depicts the same graphs, and includes as well the parabola that passes through the vertices of all the other parabolas. The equation of that new parabola was obtained by using the vertices (0, 1), (-1, 0) and (-1.5, -1.25) of the b = 0, b = 2 and b = 3 parabolas. Using those vertices as our three selected points, we could easily write down then (tediously) solve the equations to get the parabolic model that fits the points. Query Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat was the shape of the curve connecting the vertices? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A U shape confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** The figure below shows the graphs of the above functions, which have the form y = x^2 + b x + 1 for b = 2 and 3 (i.e., when b = 2 the form y = x^2 + b x + 1 becomes y = x^2 + 2 x + 1; when b = 3 the form gives us x^2 + 3 x + 1), as well as all the functions we get for b = -5, -4, -3, ..., 3, 4, 5.. You should be able to identify the parabolas you graphed among these. The figure below depicts the same graphs, and includes as well the parabola that passes through the vertices of all the other parabolas. The equation of that new parabola was obtained by using the vertices (0, 1), (-1, 0) and (-1.5, -1.25) of the b = 0, b = 2 and b = 3 parabolas. Using those vertices as our three selected points, we could easily write down then (tediously) solve the equations to get the parabolic model that fits the points. Query Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!