course Mth163
I submitted this assignment on Monday 8th. I did not receive it back so I am resubmitting this assignment again.
Form Confirmation
Thank you for submitting the following information:
identifyingInfo: Submitting Assignment
yourCourse: Mth163
Contact_FirstName: Tina
Contact_LastName: Graham
firstPendLength: Assignment - 3
email: tmg2239@email.vccs.edu
S1
response
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm . Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 003. Question: `q001. Note that this assignment has 6 questions The function y = a x^2 + b x + c takes the value y = 0 when x = [ -b + `sqrt(b^2 - 4 a c ] / (2 a) or when x = [ -b - `sqrt(b^2 - 4 a c ] / (2 a). For the function y = - 0.45833 x^2 + 5.33333 x - 6.875, which you obtained as a quadratic model of the points (1, -2), (3, 5) and (7, 8) in the preceding assignment, find the values of x for which y = 0. Compare to the estimates you made from the graph through (1,-2), (3, 5) and (7, 8) in Assignment 1. your solution: ->->->->->->->->->->->-> In your quadratic formula y = 0.45833x^2 +5.33333x -6.875 , if the function y = a x^2 + b x + c, then a = 0.45833, b = 5.33333, and c = -6.875 when the value of y = 0 Then x = [-b + ‘sqrt (b^2 – 4 a c) / 2a = -1.4763797 X = [-5.33333 - ‘sqrt(5.33333^2 – 4 * 0.45833 * -6.875) / (2 * 0.45833) = -10.1600611 Confidence Assessment: mostly 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. Self-critique (if necessary): O.k. Self-critique Rating: 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. your solution: ->->->->->->->->->->->-> Using the two given points found in the last problem approx. 1.5 and 10.2, and the coordinates from the first problem, we sketch the parabola. Due to a parabola being symmetrical, you need to find the mid-point between the two given points above, which aprox. 5.85, this being the x value at which y takes its maximum value. Confidence Assessment: 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: 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? your solution: ->->->->->->->->->->->-> the function will get its max value when x = 5.85 In order to find the y value we add the x value into the original equation. y = -0.45833 * 5.85^2 + 5.33333 * 5.85 – 6.875 y = -15.685 + 31.199805 – 6.875 y = 8.639805 or approx. y = 8.64 The coordinates of the highest point of the graph is ( 5.85, 8.64) Confidence Assessment: mostly 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): I feel like our x values is slightly different because I used the whole number instead of rounding off when calculating for that number. Self-critique Rating: 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? your solution: ->->->->->->->->->->->-> I will take the a & b value of this equation which is a = -0.458333, b = 5.33333 and put it into the equation to find the vertex x = -5.33333 / 2 * -0.458333 x = -5.33333 / -0.916666 x = 5.81818 To find the y value we plug the x value into the original equation. y = -0.45833 * 5.81818^2 + 5.33333 * 5.81818 – 6.875 y = -15.51503 + 31.03027 – 6.875 y = 8.64024 Confidence Assessment: 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, need to give complete (x,y) coordinates. Self-critique Rating: 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? your solution: ->->->->->->->->->->->-> The value of the x that lies one unit to the right of the vertex (5.8182, 8.6402) will be x = 6.8182 and I unit to the left will be x = 4.8182. In order to find the y correspondence to the x values we must plug these values in to the original equation. y = -0.45833 * 6.8182^2 + 5.33333 * 6.8182 – 6.875 y = -21.30677+ 36.36371 – 6.875 y = 8.18193 and y = -0.45833 * 4.8182^2 + 5.33333 * 4.8182 – 6.875 y = -10.64015 + 25.69705 – 6.875 y = 8.1819 The two y coordinates is y = 8.18193 and y = 8.1819 These y values differ from the original y value of approx. between 0.45831 and 0.45834 Confidence Assessment: mostly 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): I recalculated my equations and still came up with the two y values which slightly different than yours. I don’t understand how you derived at those numbers and are you rounding off differently than I am? Sometimes you seem to round up and sometimes you seem to not round at all even if we are supposed to in a mathematical since, this may be our differences. If we move one unit to the right then the y value will always differ from the y value by the coefficient of x ^2 Self-critique Rating: 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? your solution: ->->->->->->->->->->->-> To find the vertex of the quadratic model y = -1 x^2 + 10x + 100, we must use the vertex formula of x = -b/2a: x = -10 /2 * -1 x = -10 / -2 x = 5 we plug the x value into our original model, y = -1x^2 + 10 x +100 y = -1*5^2 + 10 * 5 +100 y = -25 +50 + 100 y = 125 The vertex is (5,125) Because a = -1 the points 1 unit to the right an left of (5,125) will be (4,124) and (6, 124) Because the parabola is decreasing as we move across, the line will eventually cross the x axis. Confidence Assessment: mostly 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. Self-critique (if necessary): ok Self-critique Rating: 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. Your solution: To stretch a graph with a factor of 2 you would have an equation of y = 2x^2, factor of 3 would be y = 3x^2, .5 being y = .5x^2. They all have a down vertex with x = 1, -1, 2, -2, 3, -3. 3 was the narrowest graph with 2 being a little wider and .5 the widest. To stretch a graph with a factor of -3, you would have an equation of y = -3x^2, but this graph would have an up vertex with x = 1,-1,2,-2,3,-3. This graph is just as narrow as the 3 but flipped in the other direction. Confidence Assessment: positive 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): ok 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: ->->->->->->->->->->->-> Using the original quadratic formula of y = ax^2 + bx + c to find a,b,c in the above equations, then using them in the vertex formula of x = -b / 2a y = x^2 + 2x + 1 x = -2 / 2 * 1 x = -1 then plugging that back into the equation to find y y = -1^2 + 2 * -1 + 1 y = 0 The vertex is ( -1,0) Lets say we went 1 unit to the left and the right, the other fundamental points would be (0,1) and ( -2,1) y = x^2 + 3x + 1 using the vertex formula x = -b / 2a x = -3/ 2 * 1 x = -3 / 2 then we plug it back into the original equation y = -3/2 ^2 + 3 * -3/2 + 1 y = 9/4 – 12/4 + 4/4 y = ¼ The vertex is (-3/2, ¼) in decimal form ( -1.5,0.25), to move 1 unit to the left and right you have the points ( -2.5, 1.25) and ( -0.5, 1.25) Confidence Assessment: positive 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): ok 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? your solution: ->->->->->->->->->->->-> It moves but not in a straight line. Confidence Assessment: mostly 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): The vertex moves downward and to left. Self-critique Rating: Question: `qHow do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola? your solution: ->->->->->->->->->->->-> The vertex is always in the line of symmetry and the other points show you if the graph will go up or down it also will give you an idea if your graph will be narrow or wide. Confidence Assessment: mostly 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): ok Self-critique Rating: Question: `qquery Zeros of a quadratic function: What was it that determined whether a function had zeros or not? your solution: ->->->->->->->->->->->-> Graphing the points given helped me see if o was in the function Confidence Assessment: unsure 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: 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? your solution: ->->->->->->->->->->->-> the vertex of a parabola is a line of symmetry that always crosses the x axis at 0 Confidence Assessment: unsure 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): A symmetrical line must have equal distance of points on both sides. Self-critique Rating: Question: `qWhat was the shape of the curve connecting the vertices? your solution: ->->->->->->->->->->->-> The graphs in these exercises were all parabolas Confidence Assessment: mostly 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.
Return to the form."
Your work looks good. However I can't find the original submission when searching for your username.
If you want this posted in its original format, you're welcome to resubmit a copy of your original document. You might find the format of the confirmation message difficult to read (I've inserted ->->->-> markings to help locate your answers, so it's no problem for me).