#$&* course MTH 163 October 1 @ 10:47 PM
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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). ** 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): ------------------------------------------------ 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - The vertex of the middle two quadratic equations were: for y = x^2 + 2x + 1 => (-1, 0) , fundamental points were (-2, 1) and (0, 1) and for y = x^2 + 3x +1 => (-1.5, -1.25) , fundamental points were (-1.75, -1.1875) and (-1.25, -1.1875) confidence rating #$&*: 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, -.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): Scaled my graph by .25 units, whereas the parameters indicated moving from the fundamental point by a = 1 units ------------------------------------------------ Self-critique Rating: ********************************************* 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: - They seemed to curve in the direction of quadrant III, as in a negative half-parabola (maximum vertex rather than minimum vertex) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - For me, personally, they provide a guideline in which to plug more values into the quadratic equation, to make a more even curve. For example, when finding what one designates R vertex and L vertex (being the points right and left of the vertex) one sees that the y value is equivalent to either value, thus starting a basic idea of the curve and where it starts. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery Zeros of a quadratic function: What was it that determined whether a function had zeros or not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - With the understanding in mind that one had labeled the five equations in the following order: 1) y = x^2 + x + 1 2) y = x^2 + 2x + 1 3) y = x^2 + 3x +1 4) y = x^2 + 4x + 1 5) y = -2x^2 + 4x + 3 - One finds that the first resulted in -1 +- `sqrt(-3) / (2) which of course is an imaginary number; which one had presumed to be a big red flag for NO ZEROES. - In the second equation, I had found that using -(2) +\- `sqrt(0) / (2) had resulted in an equivalent solution of [+] = -2 and [-] = -2. **Note that one designates [+] and [-] as the positive and negative values associated with the “opposite of b plus or minus…” portion of the quadratic formula.** However, according to the graphing and not the quadratic formula, it appears that only 3 and 4 should have zeroes… confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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 either side of the parabola (being separated by the aforementioned point) will be perfectly symmetrical. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The graph of each function is completely symmetric about a vertical line through the vertex. Thus if one of the zeros lies to the right of the vertex, the other must lie an equal distance to the left of the vertex. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat was the shape of the curve connecting the vertices? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - As previously mentioned, it carried a negative parabolic shape in quadrant III confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** 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. ""1) The main comment I had found pretty awesome was the parabolic shape the vertices had created when accurate graphing was finished. 2) Using the discriminant to see whether or not any zeroes exist in a function will be pretty handy, considering I could remember to actually use it. 3) Use of my trusty protractor from trigonometry coupled with a good (but now tattered and full) graphing paper composition notebook makes calculus a little less painstaking, as I’m a stickler for neatness both for satisfaction and understanding. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat was the shape of the curve connecting the vertices? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - As previously mentioned, it carried a negative parabolic shape in quadrant III confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** 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. ""1) The main comment I had found pretty awesome was the parabolic shape the vertices had created when accurate graphing was finished. 2) Using the discriminant to see whether or not any zeroes exist in a function will be pretty handy, considering I could remember to actually use it. 3) Use of my trusty protractor from trigonometry coupled with a good (but now tattered and full) graphing paper composition notebook makes calculus a little less painstaking, as I’m a stickler for neatness both for satisfaction and understanding. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course MTH 163 October 1 @ 10:47 PM
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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). ** 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): ------------------------------------------------ Self-critique Rating: